MECHANICAL VIBRATIONS WORK BOOK CUM LECTURE NOTES

(FOR SIXTH SEMESTER MECHANICAL STUDENTS)

(FOR PRIVATE CIRCULATION ONLY)

JAGADEESHA T Associate Professor

Mechanical Engineering Department

Name USN

Section

ST JOSEPH ENGVAMANJOOR M

INEERING COLLEGE ANGALORE ndash 575 028

2

M

3

CHAPTER 1- INTRODUCTION TO VIBRATION LEARNING OBJECTIVES

] Introduction to vibration

] Terminologies used in Vibration

] Simple Harmonic Motion

] Addition of Harmonics Principle of super position applied to SHM

] Introduction to Fourier analysis Beats

] Problems related to SHM and Fourier analysis

Vibration is defined as a motion which repeats after equal interval of time and is also a periodic motion The swinging of a pendulum is a simple example of vibration Vibration occurs in all bodies which are having mass and elasticity They are caused due to several reasons such as presence of unbalanced force in rotating machines elastic nature of the system external application of force or wind loads and earthquakes Vibrations are undesirable in most engineering systems and desirable in few cases A body is said to vibrate if it has periodic motion Mechanical vibration is the study of oscillatory motions of a dynamic system An oscillatory motion is a repeated motion with equal interval of time Example for useful vibration

o General industries ndash crushers jackhammer concrete compactor etc o Medical and health ndash electric massage high frequency vibration probe for

heart disease treatment o Music ndash string instruments ie guitar etc

Example for unwanted vibration

o Poor ride comfort in vehicle due to road irregularities o Sea sickness when traveling on ships boats etc o Earthquakes o Fatigue failures in machine and structures

Cycle The movement of vibrating body from the mean to its extreme position in one direction then to mean then to another extreme position and back to mean is called as cycle of vibration Mean position

Time period It is the time taken to complete one cycle It is equal to the time for the vector to rotate through 2π radians ` Frequency It is the number of cycles per unit time Amplitude It is the maximum displacement of a vibrating body from the mean position Phase difference it is angle between the two rotating vectors executing simple harmonic motion of same frequency The first vector is x1= X sin ( wt ) The second vector is x2= X sin ( wt+φ) Where φ is the phase difference between x1 and x Resonance it is the frequency of the externalfrequency of the system a condition known aresonance the system undergoes dangerously larg Damping It is the resistance offered to the motion Periodic motion If the motion is repeated after equal intervals of timsimplest type of periodic motion is harmonic motio Aperiodic motion If the motion does not repeat after equal interval of

rads ωωωω

Reference

rads ωωωω

2

force coincides with the s resonance occurs Dure oscillations

of a vibrating body

e it is called periodic motin

time it is called aperiodic

3

natural ing the

on The

motion

rads ωωωω

DESCRIBING MOTIONS OF VIBRATING SYSTEMS Periodic motions Described as sine or cosine functions [sin (ωt) and cos(ωt)] ω = radian frequency (radsec) ω = 2πf where f is frequency (Hz) Period = time between two adjacent peaks or valleys P = 1f

Simple harmonic motion ( ) ( )φω += tAty sin Displacement ( ) ( )φωω += tAty cosamp Velocity ( ) ( )φωω +minus= tAty sin2amp Acceleration Classification of vibrations One method of classifying mechanical vibrations is based on degreesThe number of degrees of freedom for a system is the number of independent variables necessary to completely describe the motion of ein the system Based on degrees of freedom we can classify mechanicas follows 1Single Degree of freedom Systems 2Two Degrees of freedom Systems 3Multidegree of freedom Systems 4Continuous Systems or systems with infinite degrees of freedom Another broad classification of vibrations is 1 Free and forced vibrations 2 Damped and undamped vibrations

Leads Lags

4

of freedom kinematically very particle al vibrations

5

Sometime vibration problems are classified as 1 Linear vibrations 2 Non-linear vibrations 3 Random vibrations 4 Transient vibrations 5 Longitudinal vibrations 6 Transverse vibrations 7 Torsional vibrations Free vibration If a system after initial disturbance is left to vibrate on its own the resulting vibration is known as free vibrations Free vibration takes when a system vibrates under the action of forces inherent in the system and when the external forces are absent The frequency of free vibration of a system is called natural frequency Natural frequency is a property of a dynamical system Forced vibration Vibration that takes place under the excitation of external forces is called forced vibration the forced vibration takes place at different forced frequencies or external frequencies Damped vibration If any energy is lost or dissipated during oscillations then the vibration is known as damped vibration Undamped vibration if no energy is lost or dissipated during oscillations such vibrations are known as undamped vibration Linear vibration If all the basic component of a vibrating system behave linearly the resulting vibration is known as linear vibration The differential equations govern linear vibratory system are linear If the vibration is linear the principle of superposition holds and mathematical techniques of analysis are well developed Non linear vibration If any of the basic components of a vibrating system behave non linearly the resulting vibration is known as non linear vibration The differential equations that govern non linear vibratory system are non-linear If the vibration is non linear the principle of superposition does not hold good and techniques of analysis is well known Deterministic vibration If the magnitude of excitation on a vibrating system is known at any given time the resulting vibration is known as deterministic vibration Random vibration If the magnitude of excitation acting on a vibratory system at a given time cannot be predicted the resulting vibration is known as non deterministic or random vibration

6

Longitudinal vibration Consider a body of mass m carried on one end of a slender shaft and other end being fixed If the mass vibrates parallel to the spindle axis it is said to be execute longitudinal vibration Transverse vibration If the mass vibrates perpendicular to the spindle axis it is said to execute the transverse vibration Torsional vibration If the shaft gets alternatively twisted and un twisted on account of an alternate torque on the disk it is said to execute the torsional vibration SIMPLE HARMONIC MOTION A Vibration with acceleration proportional to the displacement and directed toward the mean position is known as SHM ( Simple Harmonic Motion) Consider a spring mass system as shown in the figure along with the displacement time diagram The equation of motion of the mass can be written as follows From the right angled triangle OAB we have Where x= displacement at any instant of time X = amplitude of vibration W= angular velocity or frequency in radsecond

m1

7

The velocity of the mass m at an instant of time t is given by V = The acceleration of mass m is given by Hence we can conclude that in SHM the acceleration is proportional to displacement and is directed towards mean position observing the equations 2 and 3 the velocity and acceleration are harmonic with the same frequency but lead a displacement vector by π2 and π radians respectively

t

t

t X

x

X= A sin ωt

x-Displacement X-amplitude T-Periodic Time f-Frequency f=1T ω=Frequency in radians per second t= time

8

Rotor with inertia J1

kt1 kt2

Rotor with Inertia J2

Degrees of freedom The minimum number of independent coordinates required to determine completely the positions of all the parts of a system at any instant of time is called degrees of freedom One degree of freedom Two degree of freedom Three degree of freedom Infinite degree of freedom

m1

m2

m1

m2

m3

Cantilever Beam

Continuous system ( consider the mass of the beam )

m1

m1

K1

m2

K

x1

x2

F1

9

G

K1 K2

mJ

a b

Exercise Specify the no of degree of freedom for the following 1 2 3

4

5

4m 2m m 3k k k

Cantilever Beam

Continuous system ( Neglect the mass of the beam )

k1 k2 m1 m2

y y

m2

θ1

θ2

L1

L2

M1

M2

10

O

A

B

Addition of SIMPLE HARMONIC MOTION ( SHM) The addition of two simple harmonic motion having frequency yields a resultant which is simple harmonic having the same frequency Consider tow simple harmonic motions of 1x and 2x having the same frequency and phase difference φ as given below

)(sin11 tXx ωωωω====

)(sin22 φω += tXx

Adding x = x1 +x2 Hence the resultant displacement is also SHM of amplitude X and phase angle θ

θθθθ

ωωωωt

φφφφ

11

-6

-4

-2

0

2

4

6

0 005 01 015 02 025

Tutorial problems on Simple Harmonic Motion 1 Add the following harmonic motion analytically and verify the solution graphically 1) X1= 3 sin (((( ))))30 t ++++ωωωω X2= 4 cos (((( ))))10 t ++++ωωωω ( VTU Jan 2005) 2) X1= 2 cos (((( ))))05 t ++++ωωωω X2= 5 sin (((( ))))1 t ++++ωωωω ( VTU July 2006)

3) X1= 10 cos (((( ))))4 t ππππωωωω ++++ X2= 8 sin (((( ))))6 t ππππωωωω ++++ ( VTU Dec 2007)

2 A body is subjected to two harmonic motions X1= 8 cos (((( ))))6 t ππππωωωω ++++ X2= 15 sin (((( ))))6 t ππππωωωω ++++ what harmonic motion

should be given to the body to bring it to equilibrium (VTU July 2005) 3 Split the harmonic motion x = 10 sin (((( ))))6 t ππππωωωω ++++ into two harmonic motions

having the phase of zero and the other of 45o 4 Show that resultant motion of harmonic motion given below is zero X1= X sin (((( ))))t ωωωω X2= X sin (((( ))))3

2 t ππππωωωω ++++ X3= X sin (((( ))))34 t ππππωωωω ++++

5 The displacement of a vibrating body is given by x = 5 sin (3141t + 4

ππππ )

Draw the variation of displacement for one cycle of vibration Also determine the displacement of body after 011 second ( repeat the problem for velocity and acceleration and draw graph using Excel and compare ) Time Displacement

0 0025 5 005

0075 0 01

0125 015 -353

0175 0 02

Calculate the remaining values 6 The Motion of a particle is represented by x = 4 ( ) t sin ω sketch the variation of the displacement velocity and acceleration and determine the max value of these quantities Assume ( )5 =ω ( Try to use MATLABExcel) sketch all on same graph

12

BEATS When two harmonic motions with frequencies close to one another are added The resulting motion exhibits a phenomenon known as Beats A Beat Frequency is the result of two closely spaced frequencies going into and out of synchronization with one another Let us consider tow harmonic motion of same amplitude and slightly different frequencies X1 = X Cos ( ) tωωωω X2 = X Cos ( )( ) tδδδδωωωω + Where δ is a small quantity The addition of above two harmonics can be written as X = X1 + X2

X=

+

ttX2

cos2

cos2δδδδωωωω

δδδδ

The above equation shown graphically in Figure The resulting motion represents cosine wave with frequency ( ) 2

ππππωωωω + and with a varying amplitude 2 cos( ( ) 2δδδδ t

whenever the amplitude reaches a maximum it is called the beat The frequency δ at which amplitude builds up and dies down between o and 2 X is known as beat frequency Beat phenomenon is found in machines structures and electric power houses In machines the beat phenomenon occurs when the forcing frequency is close to the natural frequency of the system

+2X

-2X

ωωωωππππ2

ELEMENTS OF VIBRATION The elements of constitute vibrating systems are

1 Mass or Inertia element - m 2 Spring - k 3 Damper - c 4 Excitation F(t)

Mass or Inertia element The mass or inertia element is assumed to be rigid body During vibration velomass changes hence kinetic energy can be gained or loosed The force appthe mass from newton second law of motion can be written as F= ma Work dthe mass is stored in the form of kinetic energy given by KE= frac12 M V2

Combination of masses In practical cases for simple analysis we replace several masses by equivalent mass Case1 Translational masses connected by rigid bar Let the masses M1 M2 and M3 are attached to a rigid bar at locates 1 respectively as shown in the figure The equivalent mass meq be assumelocated at 1 is as shown in figure (b) Let the displacement of masses M1 M2 and M3 be x1 x2 x3 and simivelocities of respective masses be x1 x2 and x3 We can express the velomasses m2 and m3 in terms of m1

Elements of Vibration

Passive element

m C K

Active element

F(t)

Conservative element

( Mass and Spring)

Non conservative element

( Damper)

C k

m F(t)

Voig

13

city of lied on one on

a single

2 and 3 d to be

larly the cities of

t Model

Translationa Let a mass mmass momearrangement These two mamass Meq or

2

1

33

2

1

221

+

+=

ll

Mll

MMMeq

14

is the required answer

l and rotational masses coupled together

having a translational velocity x be coupled to another mass having nt of inertia Io with a rotational velocity θ as in rack and pinion

shown in the figure

sses can be combined to obtain either a single equivalent translational a single equivalent mass moment of inertia Jeq

15

Equivalent translational masses

Kinetic energy of the equivalent mass =

2

21

eqeq XM amp

Kinetic energy of the two masses =

+

20

2

21

21

θθθθampamp JXM

Meq= m

+ 2RJ

m o is the required answer

Also determine the equivalent rotational mass Jeq

Jeq= m [ ]oJmR +2 is the required answer Spring element Whenever there is a relative motion between the two ends of the spring a force is developed called spring force or restoring force The spring force is proportional to the amount of deformation x and then F α x or F = kx Where k is stiffness of the spring or spring constant or spring gradient The spring stiffness is equal to spring force per unit deromation

The spring stiffness k = mNxF

Workdone in deforming a spring is equal to the strain energy or potential energy Strain energy = potential energy = area of the triangle OAB Stiffness of beams Cantilever beam consider a cantilever beam with an end mass shown in the figure The mass of the beam is assumed to be negligible The static deflection of beam at free end is given by

Similarly derive the expression for Simply supported beam and fixed support beam

x

k

m

x

m

x

m

x

mNlIE

st 192

3

=δδδδ

mNEI

Wlst

48

3

=δδδδ

F

K=stiffness

F

16

17

Stiffness of slender bar subjected to longitudinal vibrations For a system executing the longitudinal vibrations as shown in the figure let us derive the expression for stiffness

Torsional Stiffness of bar It is the amount of torque required to cause a unit angular deformation in the element

Torsional stiffness = Kt = θθθθT

Combination of stiffness Determination of equivalent spring stiffness when the springs are arranged in series Consider two springs of stiffness K1 and K2 acted upon by the force F

The deflection of spring k1 is x1 = 1KF

The deflection of spring k1 is x2 = Let these two springs be replaced by an equivalent stiffness Keq upon which a force F acts and due to which its deflection is given by

x = eqKF

x= x1+x2

m

E A

l

60 cm 70 cm

Determination of equivalent spring stiffness when the springs are arranged in parallel Force acting on K1 spring = F1=k1x Force acting on K2 spring = F2 Force required for an equivalent spring keq to defined by x given by F= Keq x But F = F1 +F2 Tutorial problems on Equivalent stiffness of springs 1Determine the equivalent stiffness for the system shown in figure 2 Determine the equivalent stiffness for the system shown in figure

3 Determine the equivale

M

3

M

Nm 102 6x

18

nt stiffness for the system shown in figure

38 kg

Nm 101 6x Nm 102 6x

Nm 106x

2k

k

k

2k

3k

k

2k

k

19

60 cm 80 cm 50 cm

50 cm 60 cm

4Determine the equivalent stiffness for the system shown in figure

5Replace the following torsional stiffness by a single shaft having radius 4cm and find the length required for the equivalent shaft Assume the material of given system and equivalent system is same R1= 3cm R2= 5cm DAMPING Every vibration energy is gradually converted into heat or sound Hence the displacement during vibration gradually reduces The mechanism by which vibration energy is gradually converted into heat or sound is known as damping A damper is assumed to have either mass or elasticity Hence damping is modeled as one or more of the following types Viscous damping Coulomb or dry friction damping materials or solid or hysteric damping Viscous damping Viscous damping is most commonly used damping mechanism in vibration analysis When the mechanical system vibrates in a fluid medium such as air gas water or oil the resistance offered by the fluid to the moving body causes energy to be dissipated In this case the amount of dissipated energy depends on many factors such as size or shape of the vibrating body the viscosity of the fluid the frequency of vibration and velocity of fluid Resistance due to viscous damping is directly proportional to the velocity of vibration

dF V αααα

dF = xC amp

Where C= damping coefficient Fd = damping force Examples of Viscous damping

1) Fluid film between sliding surface 2) Fluid flow around a piston in a cylinder 3) Fluid flow through orifice 4) Fluid flow around a journal in a bearing

Rreq =4cm

leqn

stress

Strain

Coulomb damping or dry friction damping Here a damping force is constant in magnitude but opposite in direction to that of the motion of vibrating body It is caused by the friction between the surfaces that are dry or have insufficient lubrication Material or solid or hysteric damping

When the materials are deformed energy is absorbed and dissipated by the material The effect is due to friction between the internal planes which slip or slide as the deformation takes place When a body having the material damping is subjected to vibration the stress strain diagram shows the hysteresis loop as shown in the figure The area of the loop denotes the energy lost per unit volume of the body per cycle

20

21

ττττ ττττ2 ττττ3

ττττ

X(t)

FOURIER SERIES The simplest of periodic motion happens to be SHM It is simple to handle but the motion of many vibrating system is not harmonic (but periodic) Few examples are shown below

Forces acting on machines are generally periodic but this may not be harmonic for example the excitation force in a punching machine is periodic and it can be represented as shown in figure 13 Vibration analysis of system subjected to periodic but nonharmonic forces can be done with the help of Fourier series The problem becomes a multifrequency excitation problem The principle of linear superposition is applied and the total response is the sum of the response due to each of the individual frequency term

Any periodic motion can be expressed as an infinite sum of sines and cosines terms If x(t) is a periodic function with period t its Fourier representation is given by

X(t) = (((( )))) (((( )))) (((( ))))t sint cost cos2 111 ωωωωωωωωωωωω baaao ++++++++++++

= (((( )))) (((( ))))t sint cos2 1

ωωωωωωωω nnn

o bnaa

++++++++sumsumsumsuminfininfininfininfin

====

tππππωωωω

2==== = Fundamental frequency ndash (1)

where ao an bn are constants

ττττ ττττ2 ττττ3

ττττ

X(t)

22

t

X(t)

035

025

Determination of constants To find ao Integrate both sides of equation(1) over any interval ττττ All intergrals on the RHS of the equation are zero except the one containing ao

(((( ))))dttxao

o 2

2

intintintint====ωωωωππππ

ππππωωωω

= (((( ))))dttxo

2intintintintττττ

ττττ

To find an multiply equation 1 by cos (((( ))))tnωωωω and Integrate over any interval ττττ All intergrals

(((( )))) (((( )))) dttxao

n tncos 2

2

ωωωωππππωωωω ωωωω

ππππ

intintintint==== (((( )))) (((( )))) dttxo

tncos 2

ωωωωττττ

ττττ

intintintint====

To find bn multiply equation 1 by sin (((( ))))tnωωωω and Integrate over any interval ττττ All intergrals

(((( )))) (((( )))) dttxao

n tnsin 2

2

ωωωωππππωωωω ωωωω

ππππ

intintintint==== (((( )))) (((( )))) dttxo

tnsin 2

ωωωωττττ

ττττ

intintintint====

Find the Fourier series for the curve shown below

ττττ ττττ2 ττττ3

t

X(t)

E

ττττ ττττ2 ττττ3

t

X(t)

ππππ2

t

X(t) ππππ

1 Represent for the periodic motion shown in the figure

23

CHAPTER 2 UNDAMPED FREE VIBRATION LEARNING OBJECTIVES

] Introduction to undamped free vibration

] Terminologies used in undamped free vibration

] Three methods to solve the undamped free vibration problems

] Problems related to undamped free vibration problems

Free vibrations are oscillations about a systems equilibrium position that occur in the absence of an external excitation force If during vibrations there is no loss of energy it is known as undamped vibration The first step in solving a vibration problem is setting up the differential equation of motion Consider spring mass system which is assumed to move only along the vertical direction as shown below Let m be the mass of the block and k be the stiffness of the spring When block of mass m is attached to spring the deflection of spring will be ∆ known as static deflection In the static equilibrium position the free body diagram of forces acting on the mass is shown in Figure(b) Hence mg= kA Once the system is disturbed the system executes vibrations

Let at any instant of time t the mass is displaced from the equilibrium position x the different forces acting on the system are shown in figure (d) From Newtonrsquos second law of motion

sum = maF

Inertia force ( disturbing force) = restoring force

mgxkxm ++∆minus= )(ampamp

0)( =+ xkxm ampamp

or 0)( =+ xmk

xampamp

equation 2 is the differential equation of motion for spring mfigure Comparing equation (2) with the equation of SHM +xampampsince the vibrations of the above system are free( without the forces) we can write

sec radmk

n =ω

24

ass system shown in 0)(2 =xω

resistance of external

time period km

2 f1

n

πτ ==

from the equation(1) mg = k∆

∆=

q

mk

Difference between the translation ( rectilinear) and rotational system of vibration

Translatory Rotational

In the analysis thFORCES are co Linear stiffness K Problems 1A mass of 10kFind the natural 2 A spring massnatural frequencnatural frequenc

kt

m

sec radmk

n =ω

k

25

e disturbing and restoring nsidered

In the analysis In the analysis MASS Moment of Inertia (J) is considered

in Nm is considered

g when suspended from a spring causes a static deflection of 1cm frequency of system

system has a spring stiffness K Nm and a mass of m Kg It has a y of vibration 12 Hz An extra 2kg mass coupled to it then the y reduces by 2 Hz find K and m

secrad n =ω

26

3 A steel wire of 2mm diameter and 30m long It is fixed at the upper end and carries a mass of m kg at its free end Find m so that the frequency of longitudinal vibration is 4 Hz 4 A spring mass system has a natural period of 02 seconds What will be the new period if the spring constant is 1) increased by 50 2) decreased by 50 5 A spring mass system has a natural frequency of 10 Hz when the spring constant is reduced by 800 Nm the frequency is altered by 45 Find the mass and spring constant of the original system 6 Determine the natural frequency of system shown in fig is which shaft is supported in SHORT bearings 7 Determine the natural frequency of system shown in fig is which shaft is supported in LONG bearings where l is the length of bearing and E ndash youngrsquos modulus and I is moment of Inertia 7 Determine the natural frequency of system shown in fig is which shaft is supported in LONG bearings 8 A light cantilever beam of rectangular section( 5 cm deep x 25cm wide) has a mass fixed at its free end Find the ratio of frequency of free lateral vibration in vertical plane to that in horizontal 9 Determine the natural frequency of simple pendulum 10 Homogeneous square plate of size l and mass m is suspended from the mid point of one of its sides as shown in figure Find the frequency of vibration

l

l

27

11 A compound pendulum which is rigid body of mass m and it is pivoted at O The point of pivot is at distance d from the centre of gravity It is free to rotate about its axis Find the frequency of oscillation of such pendulum 12 A connecting rod shown in fig is supported at the wrist pin end It is displaced and allowed to oscillate The mass of rod is 5kg and centre of gravity is 20 cm from the pivot point O If the frequency of oscillation is 40 cyclesminute calculate the mass moment of inertia about its CG

13 A semi circular homogenous disc of radius r and mass m is pivoted freely about its centre as shown in figure Determine the natural frequency of oscillation 14A simply supported beam of square cross section 5mmx5mm and length 1m carrying a mass of 0575 kg at the middle is found to have natural frequency of 30 radsec Determine youngrsquos modulus of elasticity of beam 15 A spring mass system k1 and m have a natural frequency f1 Determine the value of k2 of another spring in terms of k1 which when placed in series with k1 lowers the natural frequency to 23 f1

COMPLETE SOLUTION OF SYSTEM EXECUTING SHM

The equation of motion of system executing SHM can be represented by

0)( =+ xkxm ampamp ----------------------------------------(1)

022

2=+

dtdx

dt

dx ω

The general solution of equation (1) can be expressed as

------------------------(2)

Where A and B are arbitrary constant which can be determined from the initial conditions of the system Two initial conditions are to be specified to evaluate these

constants x=x0 at t=0 and xamp = Vo at t=0 substituting in the equation (2) Energy method In a conservative system the total energy is constant The dwell as natural frequency can be determined by the princienergy For free vibration of undamped system at any instant and partly potential The kinetic energy T is stored in the masswhere as the potential energy U is stored in the form of sdeformation or work done in a force field such as gravity The total energy being constant T+U = constant Its rate of cha

t)(sin Bt)cos( AX ω+ω=

t)(sin V

t)cos( xx o0 ω

ω+ω=

Is the required complete

solution

28

ifferential equation as ple of conservation of of time is partly kinetic by virtue of its velocity train energy in elastic

nge

Is given by [ ] 0UTdtd

=+

From this we get a differential equation of motion as well as natural frequency of the system Determine the natural frequency of spring mass system using energy method

Determine the natural frequency of system shown in figure Determine the natural frequency of the system shown in figure Is there any limitation on the value of K Discuss Determine the natural frequency of s

m

l k

m

a

θ

m

a

θ

l

k

29

k

k

ystem shown below Neglect the mass of ball

l

m

30

A string shown in figure is under tension T which can be assumed to remain constant for small displacements Find the natural frequency of vertical vibrations of spring An acrobat 120kg walks on a tight rope as shown in figure The frequency of vibration in the given position is vertical direction is 30 rads Find the tension in the rope A manometer has a uniform bore of cross section area A If the column of liquid of length L and Density ρ is set into motion as shown in figure Find the frequency of the resulting oscillation

m

l

a

T T

36m 8m

31

k

R

k

m

r

m

Find the expression for natural frequency of system shown in the figure Neglect the mass of the cantilever beam Study the special case i) k=Infinity ii) I = infinity Determine the expression for the natural frequency of a system shown in figure The two discs are keyed to a common shaft and have a combined mass moment of inertia about centre of oscillation O It is assumed that the card attached to mass m does not stretch and is always under tension Determine the natural frequency of a system shown

m

l

32

Determine the expression for the natural frequency of the system shown in figure Assume that the wires connecting the masses do not stretch and are always in tension

Determine the natural frequency of spring mass system taking the MASS OF SPRING (ms ) into account

M3 M1

M2

k2 k1

m

k

33

RAYLEIGHrsquoS METHOD

This is the extension of energy method Here we can determine the natural frequency of a conservative system without finding the equation of motion The natural frequency is a function of the rate of change of Kinetic energy and potential energy of the system From the principle of conservation of energy we have T+U = Constant T= KE and U =PE We can write the above equation as T1+U1 = --------------------------(1) Where 1 and 2 represents tow instances of time Let 1 be the time when the mass is passing through static equilibrium position Let 2 be the time corresponding to the mass displacement of the mass At this instant the velocity f the mass is zero and hence Substituting in equation (1) we get ( )Max = ( )Max Above equation leads directly to natural frequency Determine the natural frequency of spring mass system using Rayleighrsquos method m

34

r

B

K K

Home work

Determine the natural frequency of pendulum using Rayleighrsquos method

A Cylinder of mass m and moment of inertia Jo is to role without slipping but is sustained by spring K as shown in figure Determine the natural frequency of oscillation

Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping

a

Determine the natural frequency of system shown where in cylindrical disc rolls over the inextensible spring A cylinder of radius r rolls without slipping on cylindrical the equation of motion and hence determine the naoscillations about the lowest point shown Repeat the above problem ndash Instead of cylinder assume swithout slipping on concave surface as shown above

r

K

35

portion of radius R Derive tural frequency of small

phere of radius with r rolls

The torsional pendulum has a natural frequency of 5 Hz What length of steel wire of diameter 2 mm should be used for pendulum The inertia of mass fixed at the free end is 00098 kgm2 Take g= 083x1011 Nm2 Determine the torsional natural frequency of the system shown in figure Neglect the mass moment of inertia of the shaft ( Figure not proportionate) Determine the natural frequency of simple pendulum considering the mass of the rod into account

Using the energy method find tThe chord may be assumed ine

Kt1

Kt2

20cm

06m 08m 04m

20cm 20cm 20cm

m

M

he natural frxtensible in

k

M

36

equency of the system shown in figure the spring mass pulley system

A rod of length l and depth H having a mass m rests on a semicylidrical surface os radius r It is tipped slightly and let go Find the frequency of oscillation

A thin bar of l having a mass m rests on a semicylidrical surface radius r It is tipped slightly and let go Find the frequency of oscillation

l

h

R

l

37

38

CHAPTER 3

DAMPED FREE VIBRATION OF SINGLE DEGREE FREEDOM SYSTEM LEARNING OBJECTIVES

] Introduction to damped free vibration

] Terminologies used in damped free vibration

] Three different types of damping

] Concept of critical damping and its importance

] Study of response of viscous damped system for cases of under damping

critical and over damping

] Concept of logarithmic decrement

] Problems related to damped free vibration problems

Viscously damped free vibration of single degree freedom sytem Consider a single degree freedom system with viscous damping as shown in figure For a viscous damped system the damping force is

where c is a constant of proportionality It is called damping coefficient or coefficient of viscous damping

Symbolically it is designated by a dashpot as shown in Fig 3

x xampx

39

At any instant of time let the mass be displaced by x from the equilibrium position Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

Okxxcxm =minusminus= ampampamp

Okxxcxm =++ ampampamp ----------------------------------------------------------(1)

or

0kxdtdx

Cdt

xdM

2

2

=++

Assuming the solution of equation as

The first term in the above equation m2Ct

eminus

is an exponentially decaying function of time The behavior of the terms in the parenthesis however depends on whether numerical value within the radicals is positive zero or negative

CASE 1 if 0mk

m2c

2

=minus

40

CASE 2 if 0mk

m2c

2

=gt

CASE 3 if 0mk

m2c

2

=lt

41

Critically damped coefficient Critically damped coefficient is defined as the value of the damping coeffient for which radical in equation

mk

m2c

m2c

S2

21 minus

plusmn

= becomes zero

Damping Ratio ( Damping factor) ( ξ ) It is defined as the ratio of damping coefficient to the critical damping coefficient Note The damping coefficient c is defined as the actual amount of damping while the damping coefficient is a constant depending upon the mass and stiffness of the system The roots of characteristics equation can be now written as Note if c=cc or ξ =1 then system is critically damped if cgt cc or ξ gt1 then system is over damped I if clt cc or ξ lt1 then system is under damped

42

Take home quiz Derive General solution of different damped system

1 Critically damped system 2 Under damped system 3 Over damped system

Logarithmic Decrement It is used to determine the amount of damping present in system This measure The rate of decay of free vibration The larger the damping the greater will be rate of decay Logarithmic decrement is defined as the natural logarithm of ratio of any two successive amplitudes

=δ

2

1

xx

ln

Expression for logarithmic decrement (δ ) Consider a damped free vibration which is expressed by

( ) ( ) φ+ξminusω= ξω t 1sinXex 2n

tn

x is the displacement at any instant of time X and φ are arbitrary constants

21

2

ξminus

πξ=δ

43

Tutorial problems 1 Derive an expression for damping ratio in terms of logarithmic decrement

2 Show that logarithmic decrement

=δ

n

o

xx

lnn1

where xo is the initial amplitude xn

is amplitude after n cycles 3 A mass of 1 kg is to be supported on a spring having stiffness K= 9800 Nm The damping constant c= 59 Nsm Determine the natural frequency of system Find also logarithmic decrement and amplitude after 3 cycles if initial displacement is 03 cm 4 A vibrating system is defined by the following parameters M = 3kg k= 100 Nm damping coefficient = 3 Nsm determine Damping factor natural frequency damped vibration logarithmic decrement ratio of consecutive amplitude no of cycles after which the original amplitude is reduced to 20 5A single degree damped vibrating system a suspended mass of 18kg makes 15 oscillations in 003 seconds The amplitude decreases to 025 of the initial value after 5 oscillations Determine the stiffness of spring Logarithmic decrement damping factor and damping coefficient 6 A damped vibration record of a spring mass system shows the following data Amplitude at the end of 2nd cycle is 9mm amplitude at eh end of 3rd cycle is 6mm amplitude at the end of 4th cycle is 4mm stiffness of spring =8000 Nm weight =20N Find the logarithmic decrement damping force at unit velocity periodic time of vibration 7 For the system shown in figure the characteristics of dashpot is such that the constant force of 49 N is applied to the piston it velocity is found to be constant at 012 ms Determine the value of C would you expect the complete system to be periodic or aperiodic 8 For a torsional system shown in the figuref1 is natural frequency in air and f2 is the natural frequency when immersed in oil Determine the damping coefficient

Viscous fluid

Disc of mass moment of Inertia J

2a

C

m M

b

a a

k

9 A thin plate of area A and weigh w is fixed to the end of the spring and allow to oscillate in a viscous fluid as shown in figure It f1 is the frequency of oscillation in air and f2 in liquid Show that

22

21 ff

gA2

minusπω

=micro where the damping force on the plate

AV2Fd micro= V being the velocity of vibration 10 A torsional pendulum has a natural frequency of 175 cyclessecond vibrating in vacuum The mass moment of inertia is 2 kgm2 It is immersed in oil and it is observed that the natural frequency is 142 cyclessecond Find the damping torque If the disc is replaced by 34 deg When in oil find the displacement at the end of the first complete cycle 11 An automobile can be modeled as a mass placed on 4 shock absorbers each consisting of a spring and a damper Such that each spring is equally loaded Determine the stiffness damping constant of each shock absorber So that natural frequency is 2Hz and system is critically damped The mass of vehicle is 200kg 12 Write the differential equation of motion for the system shown in the figure below Determine the natural frequency of the damped oscillation and critical damping coefficient

Determine the differential equation of motion and find the critical cdamped natural frequency for system shown in figure

a

C

m M

M =200kg

44

200kg kgkg

oefficient and

45

Derive the equation of motion and find the critical damping coefficient and damped natural frequency for the system shown below Write the differential equation of motion and determine the expression for critical damping coefficient and damped natural frequency of system shown in figure

Set up a differential equation of motion for system shown in figure and determine the following Undamped natural frequencies critical damping coefficient damping ratio damped natural frequency

b

C

kM

a

m

46

A mass of 300N is resting on two springs of stiffness 3000 Nm each and a dashpot of damping coefficient 150 Nsm as shown in figure If the initial velocity of 10cms is given to mass at its equilibrium position what will be the displacement from the equilibrium position at the end of first second Derive the formula used A gun barrel of mass 600kg has a recoil spring of stiffness 294 knm if the barrel recoils 13m on firing determine i) The initial recoil velocity of barrel ii) The critical damping coefficient of dashpot which is engaged at the end of recoil stroke Iii) Time required for the barrel to return at the end of recoil stroke Iv) time required for the barrel to return to a position 5cm from the initial position

W= 3000 N

FIRING

CHAPTER 4

FORCED VIBRATION LEARNING OBJECTIVES

] Introduction to forced free vibration

] Terminologies used in forced vibration

] Concept of magnification factor

] Study of reciprocating and rotating unbalance

] Concept of base excitation and support excitation

] Concept of Transmissibility ratio

] Problems related to forced vibration

The vibration that takes place under the excitation of external forces are called

forced vibration When a system is subjected to harmonic excitation it is forced to vibrate at the same frequency as that of excitation Most of the mechanical systems are prone to steady state forced vibration which cause fatigue failure The vibration problem also becomes very important when any of excitation frequencies coincide with one of natural frequencies of a system Under this condition resonance occurs during which large amplitude of vibrations are observed To avoid resonance external frequency (operating speed) may be changed or properties of the system may be changed to alter the natural frequency In some cases sufficient amount of damping may be provided to avoid large amplitude during resonance Thus the problem of force vibration is very important in mechanical design

FORCED VIBRATIONS OF SINGLE DOF SYSTEMS UNDER HARMONIC EXCITATION

` [ ] tsin FF o ω=

x

sin FF o=

Fo

xampamp

[ ] tω

xamp

47

48

Consider a spring mass system having a viscous damping excited by a harmonic force [ ] tsin FF o ω= as shown in figure The various forces acting on the system is shown in figure Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

O)tsin(Fkxxcxm o =ω+minusminus= ampampamp

)tsin(Fkxxcxm o ω=++ ampampamp ----------------------------------------------------------(1)

or

)tsin(Fkxdtdx

Cdt

xdM o2

2

ω=++

Equation (1) represents the differential equation of motion of the system The solution to this equation consists of two parts namely complementary function and a particular integral

The first of solution is also termed as transient solution of homogeneous solution and can be obtained from the homogeneous differential equation 0kxxcxm =++ ampampamp

The second part of solution particular integral is also termed as steady state solution ad is of form )tsin(Xx φminusω= Where X= steady state amplitude φ is phase difference between external force and displacement

)tsin(Xx φminusω=

The vector representation of the shown

49

The force polygon for the above vector diagram can be written as follows

( )φminusωt O

KX

xm 2ω

Fo

xCω

tω

φ

50

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

is the governing equation

Zero frequency deflection (Xst) It is deflection observed at zero frequency ( when the external frequency 0=ω

( ) kF

001

kF

x o

22

o

st =

+minus=

Magnification factor (M) It is defined as the ratio of steady state amplitude to zero frequency deflection xst

M = arX

X=

A plot of magnification factor versus frequency ratio

damping factor ( )ξ is called frequency plot of a harmon All the frequency terms starts from magnification factor of damping ratio ( )ξ The amplitude of vibration at any on the value of damping The amplitude will be higher for higher damping The amplitude of vibration at reso

amplitude tends to zero as

ωω

n tends to infinity

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

Freq

ζ =

ζ = 1

Magnification Factor M = XXst

kFo

51

ωω

nfor varying values of

ically excite vibrating system

M=1 what ever may be value particular frequency depends

for lesser damping and lesser nance will be very high The

uency Ratio r = (ωωn)

ζ = 0 0

ζ = 025 ζ = 0375 ζ = 0 5

ζ = 0707

ζ = 2

52

Plot of phase angle φ versus frequency ratio

ωω

n is shown in the

figure The phase angle φ is zero for all the values of damping ratio

at

ωω

n=0 The phase

angle φ is 90 for all the values of damping ratio

at

ωω

n=1 ie at

resonance The phase angle between o to 90 deg for all the values of

ωω

n= 0 to 1 and

between 90 deg to 180 deg for all the values of

ωω

ngt1

Class work Derive an expression for frequency at peak amplitude

2np 21 ξminusω=ω=ω is the required equation

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

53

Derive expression amplitude at resonance ( In dimensional form) In dimensional form We know that steady state amplitude

( ) ( )

ω+ωminus=

222

o

Cmk

Fx Equation in dimensional form

We know that at resonance mk

n =ω=ω

( )

ω+

minus

=2

n

2

oR

Cmk

mk

Fx

( )

ω=

2n

oR

C

Fx

( )

ω=

n

oR C

Fx

Class work Derive an expression for amplitude at resonance ( in non dimensional form)

Tutorial Problems on Forced Vibration 1A weight of 60N is suspended by a spring stiffness 12 kNm is forced to vibrate by a harmonic force of 10N assuming viscous damping of 0086 kN-sm Find i) resonance frequency ii) amplitude at resonance iii) phase angle at resonance iv) frequency corresponding to peak amplitude 2 A periodic torque having a maximum value of 05888Nm at a frequency of 4 rads is impressed upon a flywheel suspended from a wire The wheel has mass moment of inertia 012 kgm2 and the wire has stiffness 1176 Nmrad A viscous dashpot applies damping couple of 0784 Nm an angular velocity of 2 rads calculate i)maximum angular displacement from rest position ii) maximum couple applied to dashpot Iii) angle by which the angular displacement lags the torque A mass of 1019 kg is suspended from a spring stiffness 20 Nmm viscous damping causes the amplitude to decrease to 110th of initial value in 4 complete cycle If a periodic force of 20 cos(50 t) is applied to the mass Find the amplitude of forced vibration What would be the amplitude of oscillation if the applied force coincides with the natural period

m=1019 kg

[ ] t50cos 30F =

54

55

ROTATING UNBALANCE Unbalance in rotating machinery is a common source of vibration excitation Consider a machine of mass m having mounted on a foundation of stiffness k and damping coefficient C as shown in figure The unbalance is represented by an eccentric rotating mass m with eccentricity (e) and having a angular velocity (ω) let x be the displacement of non rotating mass from the static equilibrium position The displacement of rotating mass m from the static equilibrium is ( ) tsin( ex ω+ Applying Newtonrsquos law of motion we have

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

56

RECIPROCATING UNBALANCE Unbalance in reciprocating machinery is a common source of vibration excitation Consider a machine of mass M having mounted on a foundation of stiffness K and damping coefficient C as shown in figure The unbalance is represented by a reciprocating mass (m) having crank radus (e) and connecting rod of length (l) Let ωωωω be angular velocity of the crank Let x be the displacement of non reciprocating mass from the static equilibrium position at any instant of time The displacement of reciprocating mass m from the static equilibrium position is

)2sin(sin +++ tle

tex ωωωωωωωω

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

57

Note The equations obtained in reciprocating unbalance is same as the rotating unbalance and having the following differences Rotating Unbalance Reciprocating Unbalance Here mass of the rotating element is considered

Here mass of reciprocating element is considered

Here the eccentricity (e) which is the distance between the centre of rotation and CG

Here the eccentricity e is the crank radius and crank radius = stroke length2

The frequency plot of non dimensional factor( ( )me

MX versus frequency ratio

ωω

n for various values of ( )ξξξξ is shown in figure This frequency plot is same for

the rotating and reciprocating unbalance with the following points are noted from the diagram

1 Damping ratio plays very important role during resonance in controlling the amplitude

2 When

ωω

n=0 then ( )me

MX =0 for all the values of damping ratio ( )ξξξξ

3 When

ωω

nis higher than ( )me

MX tends to unity

4 At very high speeds ie at very high frequency ratio

ωω

n there is no effect

of damping The frequency plot of phase angle versus the frequency ratio is given as shown in figure The explanation is same as in forced harmonic vibration

58

VIBRATION ISOLATION

Vibratory forces and motions generated by machine and other caused are unavoidable Howeever their effect on the other dynamical systems can be minimized by proper isolator design An isolation system is either to protect o delicate machine from excessive vibration transmitted to it from its supporting structure or to prevent vibratory forces and motions generated by machines from being transmitted to its surroundings The effectiveness of isolation may be measured in terms of forces and motion transmitted The first type is known as force isolation and the second type is known as motion isolation

Force isolation (Force Transmissibility) When vibratory forces generated by machine are isolated the effectiveness of forced isolation can be made by defining the term transmissibility The term transmissibility is defined as the ratio of force transmitted to the foundation to that of the external impressed force

Force Transmissibility = o

tr

FF

force Impressed Externalfoundation toted transmitForce

=

Consider viscous damped spring mass system as shown in figure Upon this system an external harmonic force ) t (Sin Fo ω is applied The difference forces acting on system are inertia forces damping force spring force and external force All these forces are represented on force polygon shown below Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these two forces

( ) ( )22 xckx ω+=

59

This is the expression for force transmissibility in dimensional form To represent same in the non dimensional form divide numerator and denominator by k

2

n

22

n

2

n

o

tr

21

21

FF

TR

ωω

ξ+

ωω

minus

ωω

ξ+

==

PHASE DIFFERNECE BETWEEN EXTERNAL FORCE AND DISPLACEMENT Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these

ωω

minus

ωω

ξ=

ωminusω

=φ minusminus2

n

n1

2

1

1

2tan

mKC

tan

The phase difference between force transmitted to the foundation Ftr to the displacement

kc

KxxC

tanω

=ω

=α or

ωω

ξ=

ω=α minusminus

n

11 2tanKx

xCtan

The phase difference between external force Fo and force transmitted to foundationis β

60

β= αminusφ =

ωω

minus

ωω

ξminus

2

n

n1

1

2tan -

ωω

ξminus

n

1 2tan

Displacement Transmissibility It is the ratio of amplitude of steady state vibration of machine to that of the base motion Forced vibration due to support motion Consider a spring mass damper system subjected to harmonic excitation )tsin(Yy ω= at the base as shown in figure Let x be the absolute displacement of the mass at any instant of time t The different forces acting on system are shown in free body diagram Applying Newtonrsquos Law of motion Ma = sumF

)yx(k)yx(cxm minusminusminusminus= ampampampamp

0)yx(k)yx(cxm =minus+minus+ ampampampamp

61

Frequency plot of transmissibility versus frequency ratio

62

The plot of transmissibility versus frequency ratio is as shown in figure The following points can be noted from it

i) The value of transmissibility is unity at frequency ratio

ωω

n=0 and 2

ii) For an undamped system ( )ξ =0 transmissibility tends to infin at resonance iii) The value of transmissibility is less than unity for the value of frequency ratio

greater than 2

iv) For frequency ratio

ωω

nlt 2 smaller damping ratio lead to larger values of

transmissibility on the other hand for frequency ratio

ωω

ngt 2 smaller damping

ratio leads to smaller value of transmissibility

The frequency plot of transmissibility (TR) versus frequency ratio

ωω

n can be

broadly divided into two zones

1

ωω

n= 0 to 2

2

ωω

n= 2 to infin

CASE 1

When

ωω

n= 0 to 2

In this zone the value of transmissibility is always greater than 1 for all values of damping ratio In this zone the transmissibility decreases with increase in damping Hence we can conclude that in this zone damping is useful in isolating vibration forces being transmitted to the support CASE 2

When

ωω

n= 2 to infin

In this zone the value of transmissibility is less than 1 for all values of damping ratio In this zone the transmissibility increases with increase in damping Hence we can conclude that in this zone damping will adversely affect in isolating vibratory forces

63

Tutorial problems on Forced Vibration 1 Machine of mass 75kg is mounted on spring of stiffness 12000 Nm with an assumed damping factor ξ =02 piston within the machine of mass 2 kg has a reciprocating motion with a stroke of 75 cm and speed of 15 cyclessecond Assuming the motion of piston to be harmonic determine amplitude of machine ( X) Transmissibility and force transmitted to foundation phase angle of transmitted force wrt to exciting force 2 Motor weighs 600N is mounted on a simple beam that has a spring rate of 400 Ncm The motor armature base 160 N and has an eccentricity of 0005cm What will be the amplitude of motion when it runs at 1760 RPM Neglect damping the deflection of armature shaft and weight of the beam 3 A Heavy machine of mass 300 kg is supported on resilient foundation the static deflection of foundation due to weight of machine is found to be 75mm It is observed that the machine vibrates with an amplitude of 10mm when the base of the foundation is subjected to harmonic oscillation at the undampeds natural frequency of system with an amplitude of 25mm Find the damping constant of foundation Ii) dynamic force amplitude on the base iii) amplitude of the displacement of machine relative to base 4 An air compressor of mass 450 kg operates at a constant speed of 1750 RPM The reciprocating masses are 10 kg the crank radius is 100mm If the damper for mounting introduces a damping factor 015 i) specify the springs of mounting such that only 20 of the unbalanced force is transmitted to the foundation ii) determine the magnitude of the transmitted force bull Repeat the above problem by assuming that there is no damping in system

5 A body of a car having mass 1500 kg and is mounted on 4 equal springs which deflect through 225 cm under the weight of the body A total damping forces the 4 shock absorbers is 46 N at a velocity of 1cmsec The car is placed with all the four wheels on a test platform which is move up and down at resonant speed With an amplitude of 25cm Find the amplitude of the car body on its springs assuming its CG to be in the centre of wheel base 6A 75 kg machine which is mounted on springs of stiffness 116x105 Nm with the damping of 20=ξ A 2kg piston within the machine as reciprocating motion with a stroke of 008m and speed of 3000 RPM Assuming the motion of piston to be harmonic determine the amplitude of vibration of machine

7 The spring of an automobile are compressed 10cm under its own weight Find the critical speed when a trailer is traveling over road with a people approximaterly by a sine wave of amplitude 75 cm and wave length 15m What will be the amplitude of vibration at 64kmhour 8 An eccentric mass excites us used to determine vibratory characteristics of a structure of mass 25kg as given in figure At a speed of 1000 RPM the eccentric mass to beat top position at the instant the structure was moved upwards through its static equilibrium position and corresponding amplitude was 20mm The eccentric mass is 05kg at an eccentricity of 10mm Determine i) Undamped natural frequency of the system ii) Damping ratio iii) Steady state amplitude and phase angle at 1500 RPM iv) Force transmitted to the ground at 1200 RPM

M=25kg

C

K

Eccentric mass

64

65

CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Support motion Consider a spring mass damper system as shown in figure Let it excited by motion

of support let y be the harmonic displacement of support From Newtonrsquos second law of motion

)yx(k)yx(cxm minusminusminusminus= ampampampamp

)1(0)yx(k)yx(cxm minusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminus=minus+minus+ ampampampamp

Let z be the relative displacement of the mass with respect to support then

Z=x-y

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

2

M

3

CHAPTER 1- INTRODUCTION TO VIBRATION LEARNING OBJECTIVES

] Introduction to vibration

] Terminologies used in Vibration

] Simple Harmonic Motion

] Addition of Harmonics Principle of super position applied to SHM

] Introduction to Fourier analysis Beats

] Problems related to SHM and Fourier analysis

Vibration is defined as a motion which repeats after equal interval of time and is also a periodic motion The swinging of a pendulum is a simple example of vibration Vibration occurs in all bodies which are having mass and elasticity They are caused due to several reasons such as presence of unbalanced force in rotating machines elastic nature of the system external application of force or wind loads and earthquakes Vibrations are undesirable in most engineering systems and desirable in few cases A body is said to vibrate if it has periodic motion Mechanical vibration is the study of oscillatory motions of a dynamic system An oscillatory motion is a repeated motion with equal interval of time Example for useful vibration

o General industries ndash crushers jackhammer concrete compactor etc o Medical and health ndash electric massage high frequency vibration probe for

heart disease treatment o Music ndash string instruments ie guitar etc

Example for unwanted vibration

o Poor ride comfort in vehicle due to road irregularities o Sea sickness when traveling on ships boats etc o Earthquakes o Fatigue failures in machine and structures

Cycle The movement of vibrating body from the mean to its extreme position in one direction then to mean then to another extreme position and back to mean is called as cycle of vibration Mean position

Time period It is the time taken to complete one cycle It is equal to the time for the vector to rotate through 2π radians ` Frequency It is the number of cycles per unit time Amplitude It is the maximum displacement of a vibrating body from the mean position Phase difference it is angle between the two rotating vectors executing simple harmonic motion of same frequency The first vector is x1= X sin ( wt ) The second vector is x2= X sin ( wt+φ) Where φ is the phase difference between x1 and x Resonance it is the frequency of the externalfrequency of the system a condition known aresonance the system undergoes dangerously larg Damping It is the resistance offered to the motion Periodic motion If the motion is repeated after equal intervals of timsimplest type of periodic motion is harmonic motio Aperiodic motion If the motion does not repeat after equal interval of

rads ωωωω

Reference

rads ωωωω

2

force coincides with the s resonance occurs Dure oscillations

of a vibrating body

e it is called periodic motin

time it is called aperiodic

3

natural ing the

on The

motion

rads ωωωω

DESCRIBING MOTIONS OF VIBRATING SYSTEMS Periodic motions Described as sine or cosine functions [sin (ωt) and cos(ωt)] ω = radian frequency (radsec) ω = 2πf where f is frequency (Hz) Period = time between two adjacent peaks or valleys P = 1f

Simple harmonic motion ( ) ( )φω += tAty sin Displacement ( ) ( )φωω += tAty cosamp Velocity ( ) ( )φωω +minus= tAty sin2amp Acceleration Classification of vibrations One method of classifying mechanical vibrations is based on degreesThe number of degrees of freedom for a system is the number of independent variables necessary to completely describe the motion of ein the system Based on degrees of freedom we can classify mechanicas follows 1Single Degree of freedom Systems 2Two Degrees of freedom Systems 3Multidegree of freedom Systems 4Continuous Systems or systems with infinite degrees of freedom Another broad classification of vibrations is 1 Free and forced vibrations 2 Damped and undamped vibrations

Leads Lags

4

of freedom kinematically very particle al vibrations

5

Sometime vibration problems are classified as 1 Linear vibrations 2 Non-linear vibrations 3 Random vibrations 4 Transient vibrations 5 Longitudinal vibrations 6 Transverse vibrations 7 Torsional vibrations Free vibration If a system after initial disturbance is left to vibrate on its own the resulting vibration is known as free vibrations Free vibration takes when a system vibrates under the action of forces inherent in the system and when the external forces are absent The frequency of free vibration of a system is called natural frequency Natural frequency is a property of a dynamical system Forced vibration Vibration that takes place under the excitation of external forces is called forced vibration the forced vibration takes place at different forced frequencies or external frequencies Damped vibration If any energy is lost or dissipated during oscillations then the vibration is known as damped vibration Undamped vibration if no energy is lost or dissipated during oscillations such vibrations are known as undamped vibration Linear vibration If all the basic component of a vibrating system behave linearly the resulting vibration is known as linear vibration The differential equations govern linear vibratory system are linear If the vibration is linear the principle of superposition holds and mathematical techniques of analysis are well developed Non linear vibration If any of the basic components of a vibrating system behave non linearly the resulting vibration is known as non linear vibration The differential equations that govern non linear vibratory system are non-linear If the vibration is non linear the principle of superposition does not hold good and techniques of analysis is well known Deterministic vibration If the magnitude of excitation on a vibrating system is known at any given time the resulting vibration is known as deterministic vibration Random vibration If the magnitude of excitation acting on a vibratory system at a given time cannot be predicted the resulting vibration is known as non deterministic or random vibration

6

Longitudinal vibration Consider a body of mass m carried on one end of a slender shaft and other end being fixed If the mass vibrates parallel to the spindle axis it is said to be execute longitudinal vibration Transverse vibration If the mass vibrates perpendicular to the spindle axis it is said to execute the transverse vibration Torsional vibration If the shaft gets alternatively twisted and un twisted on account of an alternate torque on the disk it is said to execute the torsional vibration SIMPLE HARMONIC MOTION A Vibration with acceleration proportional to the displacement and directed toward the mean position is known as SHM ( Simple Harmonic Motion) Consider a spring mass system as shown in the figure along with the displacement time diagram The equation of motion of the mass can be written as follows From the right angled triangle OAB we have Where x= displacement at any instant of time X = amplitude of vibration W= angular velocity or frequency in radsecond

m1

7

The velocity of the mass m at an instant of time t is given by V = The acceleration of mass m is given by Hence we can conclude that in SHM the acceleration is proportional to displacement and is directed towards mean position observing the equations 2 and 3 the velocity and acceleration are harmonic with the same frequency but lead a displacement vector by π2 and π radians respectively

t

t

t X

x

X= A sin ωt

x-Displacement X-amplitude T-Periodic Time f-Frequency f=1T ω=Frequency in radians per second t= time

8

Rotor with inertia J1

kt1 kt2

Rotor with Inertia J2

Degrees of freedom The minimum number of independent coordinates required to determine completely the positions of all the parts of a system at any instant of time is called degrees of freedom One degree of freedom Two degree of freedom Three degree of freedom Infinite degree of freedom

m1

m2

m1

m2

m3

Cantilever Beam

Continuous system ( consider the mass of the beam )

m1

m1

K1

m2

K

x1

x2

F1

9

G

K1 K2

mJ

a b

Exercise Specify the no of degree of freedom for the following 1 2 3

4

5

4m 2m m 3k k k

Cantilever Beam

Continuous system ( Neglect the mass of the beam )

k1 k2 m1 m2

y y

m2

θ1

θ2

L1

L2

M1

M2

10

O

A

B

Addition of SIMPLE HARMONIC MOTION ( SHM) The addition of two simple harmonic motion having frequency yields a resultant which is simple harmonic having the same frequency Consider tow simple harmonic motions of 1x and 2x having the same frequency and phase difference φ as given below

)(sin11 tXx ωωωω====

)(sin22 φω += tXx

Adding x = x1 +x2 Hence the resultant displacement is also SHM of amplitude X and phase angle θ

θθθθ

ωωωωt

φφφφ

11

-6

-4

-2

0

2

4

6

0 005 01 015 02 025

Tutorial problems on Simple Harmonic Motion 1 Add the following harmonic motion analytically and verify the solution graphically 1) X1= 3 sin (((( ))))30 t ++++ωωωω X2= 4 cos (((( ))))10 t ++++ωωωω ( VTU Jan 2005) 2) X1= 2 cos (((( ))))05 t ++++ωωωω X2= 5 sin (((( ))))1 t ++++ωωωω ( VTU July 2006)

3) X1= 10 cos (((( ))))4 t ππππωωωω ++++ X2= 8 sin (((( ))))6 t ππππωωωω ++++ ( VTU Dec 2007)

2 A body is subjected to two harmonic motions X1= 8 cos (((( ))))6 t ππππωωωω ++++ X2= 15 sin (((( ))))6 t ππππωωωω ++++ what harmonic motion

should be given to the body to bring it to equilibrium (VTU July 2005) 3 Split the harmonic motion x = 10 sin (((( ))))6 t ππππωωωω ++++ into two harmonic motions

having the phase of zero and the other of 45o 4 Show that resultant motion of harmonic motion given below is zero X1= X sin (((( ))))t ωωωω X2= X sin (((( ))))3

2 t ππππωωωω ++++ X3= X sin (((( ))))34 t ππππωωωω ++++

5 The displacement of a vibrating body is given by x = 5 sin (3141t + 4

ππππ )

Draw the variation of displacement for one cycle of vibration Also determine the displacement of body after 011 second ( repeat the problem for velocity and acceleration and draw graph using Excel and compare ) Time Displacement

0 0025 5 005

0075 0 01

0125 015 -353

0175 0 02

Calculate the remaining values 6 The Motion of a particle is represented by x = 4 ( ) t sin ω sketch the variation of the displacement velocity and acceleration and determine the max value of these quantities Assume ( )5 =ω ( Try to use MATLABExcel) sketch all on same graph

12

BEATS When two harmonic motions with frequencies close to one another are added The resulting motion exhibits a phenomenon known as Beats A Beat Frequency is the result of two closely spaced frequencies going into and out of synchronization with one another Let us consider tow harmonic motion of same amplitude and slightly different frequencies X1 = X Cos ( ) tωωωω X2 = X Cos ( )( ) tδδδδωωωω + Where δ is a small quantity The addition of above two harmonics can be written as X = X1 + X2

X=

+

ttX2

cos2

cos2δδδδωωωω

δδδδ

The above equation shown graphically in Figure The resulting motion represents cosine wave with frequency ( ) 2

ππππωωωω + and with a varying amplitude 2 cos( ( ) 2δδδδ t

whenever the amplitude reaches a maximum it is called the beat The frequency δ at which amplitude builds up and dies down between o and 2 X is known as beat frequency Beat phenomenon is found in machines structures and electric power houses In machines the beat phenomenon occurs when the forcing frequency is close to the natural frequency of the system

+2X

-2X

ωωωωππππ2

ELEMENTS OF VIBRATION The elements of constitute vibrating systems are

1 Mass or Inertia element - m 2 Spring - k 3 Damper - c 4 Excitation F(t)

Mass or Inertia element The mass or inertia element is assumed to be rigid body During vibration velomass changes hence kinetic energy can be gained or loosed The force appthe mass from newton second law of motion can be written as F= ma Work dthe mass is stored in the form of kinetic energy given by KE= frac12 M V2

Combination of masses In practical cases for simple analysis we replace several masses by equivalent mass Case1 Translational masses connected by rigid bar Let the masses M1 M2 and M3 are attached to a rigid bar at locates 1 respectively as shown in the figure The equivalent mass meq be assumelocated at 1 is as shown in figure (b) Let the displacement of masses M1 M2 and M3 be x1 x2 x3 and simivelocities of respective masses be x1 x2 and x3 We can express the velomasses m2 and m3 in terms of m1

Elements of Vibration

Passive element

m C K

Active element

F(t)

Conservative element

( Mass and Spring)

Non conservative element

( Damper)

C k

m F(t)

Voig

13

city of lied on one on

a single

2 and 3 d to be

larly the cities of

t Model

Translationa Let a mass mmass momearrangement These two mamass Meq or

2

1

33

2

1

221

+

+=

ll

Mll

MMMeq

14

is the required answer

l and rotational masses coupled together

having a translational velocity x be coupled to another mass having nt of inertia Io with a rotational velocity θ as in rack and pinion

shown in the figure

sses can be combined to obtain either a single equivalent translational a single equivalent mass moment of inertia Jeq

15

Equivalent translational masses

Kinetic energy of the equivalent mass =

2

21

eqeq XM amp

Kinetic energy of the two masses =

+

20

2

21

21

θθθθampamp JXM

Meq= m

+ 2RJ

m o is the required answer

Also determine the equivalent rotational mass Jeq

Jeq= m [ ]oJmR +2 is the required answer Spring element Whenever there is a relative motion between the two ends of the spring a force is developed called spring force or restoring force The spring force is proportional to the amount of deformation x and then F α x or F = kx Where k is stiffness of the spring or spring constant or spring gradient The spring stiffness is equal to spring force per unit deromation

The spring stiffness k = mNxF

Workdone in deforming a spring is equal to the strain energy or potential energy Strain energy = potential energy = area of the triangle OAB Stiffness of beams Cantilever beam consider a cantilever beam with an end mass shown in the figure The mass of the beam is assumed to be negligible The static deflection of beam at free end is given by

Similarly derive the expression for Simply supported beam and fixed support beam

x

k

m

x

m

x

m

x

mNlIE

st 192

3

=δδδδ

mNEI

Wlst

48

3

=δδδδ

F

K=stiffness

F

16

17

Stiffness of slender bar subjected to longitudinal vibrations For a system executing the longitudinal vibrations as shown in the figure let us derive the expression for stiffness

Torsional Stiffness of bar It is the amount of torque required to cause a unit angular deformation in the element

Torsional stiffness = Kt = θθθθT

Combination of stiffness Determination of equivalent spring stiffness when the springs are arranged in series Consider two springs of stiffness K1 and K2 acted upon by the force F

The deflection of spring k1 is x1 = 1KF

The deflection of spring k1 is x2 = Let these two springs be replaced by an equivalent stiffness Keq upon which a force F acts and due to which its deflection is given by

x = eqKF

x= x1+x2

m

E A

l

60 cm 70 cm

Determination of equivalent spring stiffness when the springs are arranged in parallel Force acting on K1 spring = F1=k1x Force acting on K2 spring = F2 Force required for an equivalent spring keq to defined by x given by F= Keq x But F = F1 +F2 Tutorial problems on Equivalent stiffness of springs 1Determine the equivalent stiffness for the system shown in figure 2 Determine the equivalent stiffness for the system shown in figure

3 Determine the equivale

M

3

M

Nm 102 6x

18

nt stiffness for the system shown in figure

38 kg

Nm 101 6x Nm 102 6x

Nm 106x

2k

k

k

2k

3k

k

2k

k

19

60 cm 80 cm 50 cm

50 cm 60 cm

4Determine the equivalent stiffness for the system shown in figure

5Replace the following torsional stiffness by a single shaft having radius 4cm and find the length required for the equivalent shaft Assume the material of given system and equivalent system is same R1= 3cm R2= 5cm DAMPING Every vibration energy is gradually converted into heat or sound Hence the displacement during vibration gradually reduces The mechanism by which vibration energy is gradually converted into heat or sound is known as damping A damper is assumed to have either mass or elasticity Hence damping is modeled as one or more of the following types Viscous damping Coulomb or dry friction damping materials or solid or hysteric damping Viscous damping Viscous damping is most commonly used damping mechanism in vibration analysis When the mechanical system vibrates in a fluid medium such as air gas water or oil the resistance offered by the fluid to the moving body causes energy to be dissipated In this case the amount of dissipated energy depends on many factors such as size or shape of the vibrating body the viscosity of the fluid the frequency of vibration and velocity of fluid Resistance due to viscous damping is directly proportional to the velocity of vibration

dF V αααα

dF = xC amp

Where C= damping coefficient Fd = damping force Examples of Viscous damping

1) Fluid film between sliding surface 2) Fluid flow around a piston in a cylinder 3) Fluid flow through orifice 4) Fluid flow around a journal in a bearing

Rreq =4cm

leqn

stress

Strain

Coulomb damping or dry friction damping Here a damping force is constant in magnitude but opposite in direction to that of the motion of vibrating body It is caused by the friction between the surfaces that are dry or have insufficient lubrication Material or solid or hysteric damping

When the materials are deformed energy is absorbed and dissipated by the material The effect is due to friction between the internal planes which slip or slide as the deformation takes place When a body having the material damping is subjected to vibration the stress strain diagram shows the hysteresis loop as shown in the figure The area of the loop denotes the energy lost per unit volume of the body per cycle

20

21

ττττ ττττ2 ττττ3

ττττ

X(t)

FOURIER SERIES The simplest of periodic motion happens to be SHM It is simple to handle but the motion of many vibrating system is not harmonic (but periodic) Few examples are shown below

Forces acting on machines are generally periodic but this may not be harmonic for example the excitation force in a punching machine is periodic and it can be represented as shown in figure 13 Vibration analysis of system subjected to periodic but nonharmonic forces can be done with the help of Fourier series The problem becomes a multifrequency excitation problem The principle of linear superposition is applied and the total response is the sum of the response due to each of the individual frequency term

Any periodic motion can be expressed as an infinite sum of sines and cosines terms If x(t) is a periodic function with period t its Fourier representation is given by

X(t) = (((( )))) (((( )))) (((( ))))t sint cost cos2 111 ωωωωωωωωωωωω baaao ++++++++++++

= (((( )))) (((( ))))t sint cos2 1

ωωωωωωωω nnn

o bnaa

++++++++sumsumsumsuminfininfininfininfin

====

tππππωωωω

2==== = Fundamental frequency ndash (1)

where ao an bn are constants

ττττ ττττ2 ττττ3

ττττ

X(t)

22

t

X(t)

035

025

Determination of constants To find ao Integrate both sides of equation(1) over any interval ττττ All intergrals on the RHS of the equation are zero except the one containing ao

(((( ))))dttxao

o 2

2

intintintint====ωωωωππππ

ππππωωωω

= (((( ))))dttxo

2intintintintττττ

ττττ

To find an multiply equation 1 by cos (((( ))))tnωωωω and Integrate over any interval ττττ All intergrals

(((( )))) (((( )))) dttxao

n tncos 2

2

ωωωωππππωωωω ωωωω

ππππ

intintintint==== (((( )))) (((( )))) dttxo

tncos 2

ωωωωττττ

ττττ

intintintint====

To find bn multiply equation 1 by sin (((( ))))tnωωωω and Integrate over any interval ττττ All intergrals

(((( )))) (((( )))) dttxao

n tnsin 2

2

ωωωωππππωωωω ωωωω

ππππ

intintintint==== (((( )))) (((( )))) dttxo

tnsin 2

ωωωωττττ

ττττ

intintintint====

Find the Fourier series for the curve shown below

ττττ ττττ2 ττττ3

t

X(t)

E

ττττ ττττ2 ττττ3

t

X(t)

ππππ2

t

X(t) ππππ

1 Represent for the periodic motion shown in the figure

23

CHAPTER 2 UNDAMPED FREE VIBRATION LEARNING OBJECTIVES

] Introduction to undamped free vibration

] Terminologies used in undamped free vibration

] Three methods to solve the undamped free vibration problems

] Problems related to undamped free vibration problems

Free vibrations are oscillations about a systems equilibrium position that occur in the absence of an external excitation force If during vibrations there is no loss of energy it is known as undamped vibration The first step in solving a vibration problem is setting up the differential equation of motion Consider spring mass system which is assumed to move only along the vertical direction as shown below Let m be the mass of the block and k be the stiffness of the spring When block of mass m is attached to spring the deflection of spring will be ∆ known as static deflection In the static equilibrium position the free body diagram of forces acting on the mass is shown in Figure(b) Hence mg= kA Once the system is disturbed the system executes vibrations

Let at any instant of time t the mass is displaced from the equilibrium position x the different forces acting on the system are shown in figure (d) From Newtonrsquos second law of motion

sum = maF

Inertia force ( disturbing force) = restoring force

mgxkxm ++∆minus= )(ampamp

0)( =+ xkxm ampamp

or 0)( =+ xmk

xampamp

equation 2 is the differential equation of motion for spring mfigure Comparing equation (2) with the equation of SHM +xampampsince the vibrations of the above system are free( without the forces) we can write

sec radmk

n =ω

24

ass system shown in 0)(2 =xω

resistance of external

time period km

2 f1

n

πτ ==

from the equation(1) mg = k∆

∆=

q

mk

Difference between the translation ( rectilinear) and rotational system of vibration

Translatory Rotational

In the analysis thFORCES are co Linear stiffness K Problems 1A mass of 10kFind the natural 2 A spring massnatural frequencnatural frequenc

kt

m

sec radmk

n =ω

k

25

e disturbing and restoring nsidered

In the analysis In the analysis MASS Moment of Inertia (J) is considered

in Nm is considered

g when suspended from a spring causes a static deflection of 1cm frequency of system

system has a spring stiffness K Nm and a mass of m Kg It has a y of vibration 12 Hz An extra 2kg mass coupled to it then the y reduces by 2 Hz find K and m

secrad n =ω

26

3 A steel wire of 2mm diameter and 30m long It is fixed at the upper end and carries a mass of m kg at its free end Find m so that the frequency of longitudinal vibration is 4 Hz 4 A spring mass system has a natural period of 02 seconds What will be the new period if the spring constant is 1) increased by 50 2) decreased by 50 5 A spring mass system has a natural frequency of 10 Hz when the spring constant is reduced by 800 Nm the frequency is altered by 45 Find the mass and spring constant of the original system 6 Determine the natural frequency of system shown in fig is which shaft is supported in SHORT bearings 7 Determine the natural frequency of system shown in fig is which shaft is supported in LONG bearings where l is the length of bearing and E ndash youngrsquos modulus and I is moment of Inertia 7 Determine the natural frequency of system shown in fig is which shaft is supported in LONG bearings 8 A light cantilever beam of rectangular section( 5 cm deep x 25cm wide) has a mass fixed at its free end Find the ratio of frequency of free lateral vibration in vertical plane to that in horizontal 9 Determine the natural frequency of simple pendulum 10 Homogeneous square plate of size l and mass m is suspended from the mid point of one of its sides as shown in figure Find the frequency of vibration

l

l

27

11 A compound pendulum which is rigid body of mass m and it is pivoted at O The point of pivot is at distance d from the centre of gravity It is free to rotate about its axis Find the frequency of oscillation of such pendulum 12 A connecting rod shown in fig is supported at the wrist pin end It is displaced and allowed to oscillate The mass of rod is 5kg and centre of gravity is 20 cm from the pivot point O If the frequency of oscillation is 40 cyclesminute calculate the mass moment of inertia about its CG

13 A semi circular homogenous disc of radius r and mass m is pivoted freely about its centre as shown in figure Determine the natural frequency of oscillation 14A simply supported beam of square cross section 5mmx5mm and length 1m carrying a mass of 0575 kg at the middle is found to have natural frequency of 30 radsec Determine youngrsquos modulus of elasticity of beam 15 A spring mass system k1 and m have a natural frequency f1 Determine the value of k2 of another spring in terms of k1 which when placed in series with k1 lowers the natural frequency to 23 f1

COMPLETE SOLUTION OF SYSTEM EXECUTING SHM

The equation of motion of system executing SHM can be represented by

0)( =+ xkxm ampamp ----------------------------------------(1)

022

2=+

dtdx

dt

dx ω

The general solution of equation (1) can be expressed as

------------------------(2)

Where A and B are arbitrary constant which can be determined from the initial conditions of the system Two initial conditions are to be specified to evaluate these

constants x=x0 at t=0 and xamp = Vo at t=0 substituting in the equation (2) Energy method In a conservative system the total energy is constant The dwell as natural frequency can be determined by the princienergy For free vibration of undamped system at any instant and partly potential The kinetic energy T is stored in the masswhere as the potential energy U is stored in the form of sdeformation or work done in a force field such as gravity The total energy being constant T+U = constant Its rate of cha

t)(sin Bt)cos( AX ω+ω=

t)(sin V

t)cos( xx o0 ω

ω+ω=

Is the required complete

solution

28

ifferential equation as ple of conservation of of time is partly kinetic by virtue of its velocity train energy in elastic

nge

Is given by [ ] 0UTdtd

=+

From this we get a differential equation of motion as well as natural frequency of the system Determine the natural frequency of spring mass system using energy method

Determine the natural frequency of system shown in figure Determine the natural frequency of the system shown in figure Is there any limitation on the value of K Discuss Determine the natural frequency of s

m

l k

m

a

θ

m

a

θ

l

k

29

k

k

ystem shown below Neglect the mass of ball

l

m

30

A string shown in figure is under tension T which can be assumed to remain constant for small displacements Find the natural frequency of vertical vibrations of spring An acrobat 120kg walks on a tight rope as shown in figure The frequency of vibration in the given position is vertical direction is 30 rads Find the tension in the rope A manometer has a uniform bore of cross section area A If the column of liquid of length L and Density ρ is set into motion as shown in figure Find the frequency of the resulting oscillation

m

l

a

T T

36m 8m

31

k

R

k

m

r

m

Find the expression for natural frequency of system shown in the figure Neglect the mass of the cantilever beam Study the special case i) k=Infinity ii) I = infinity Determine the expression for the natural frequency of a system shown in figure The two discs are keyed to a common shaft and have a combined mass moment of inertia about centre of oscillation O It is assumed that the card attached to mass m does not stretch and is always under tension Determine the natural frequency of a system shown

m

l

32

Determine the expression for the natural frequency of the system shown in figure Assume that the wires connecting the masses do not stretch and are always in tension

Determine the natural frequency of spring mass system taking the MASS OF SPRING (ms ) into account

M3 M1

M2

k2 k1

m

k

33

RAYLEIGHrsquoS METHOD

This is the extension of energy method Here we can determine the natural frequency of a conservative system without finding the equation of motion The natural frequency is a function of the rate of change of Kinetic energy and potential energy of the system From the principle of conservation of energy we have T+U = Constant T= KE and U =PE We can write the above equation as T1+U1 = --------------------------(1) Where 1 and 2 represents tow instances of time Let 1 be the time when the mass is passing through static equilibrium position Let 2 be the time corresponding to the mass displacement of the mass At this instant the velocity f the mass is zero and hence Substituting in equation (1) we get ( )Max = ( )Max Above equation leads directly to natural frequency Determine the natural frequency of spring mass system using Rayleighrsquos method m

34

r

B

K K

Home work

Determine the natural frequency of pendulum using Rayleighrsquos method

A Cylinder of mass m and moment of inertia Jo is to role without slipping but is sustained by spring K as shown in figure Determine the natural frequency of oscillation

Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping

a

Determine the natural frequency of system shown where in cylindrical disc rolls over the inextensible spring A cylinder of radius r rolls without slipping on cylindrical the equation of motion and hence determine the naoscillations about the lowest point shown Repeat the above problem ndash Instead of cylinder assume swithout slipping on concave surface as shown above

r

K

35

portion of radius R Derive tural frequency of small

phere of radius with r rolls

The torsional pendulum has a natural frequency of 5 Hz What length of steel wire of diameter 2 mm should be used for pendulum The inertia of mass fixed at the free end is 00098 kgm2 Take g= 083x1011 Nm2 Determine the torsional natural frequency of the system shown in figure Neglect the mass moment of inertia of the shaft ( Figure not proportionate) Determine the natural frequency of simple pendulum considering the mass of the rod into account

Using the energy method find tThe chord may be assumed ine

Kt1

Kt2

20cm

06m 08m 04m

20cm 20cm 20cm

m

M

he natural frxtensible in

k

M

36

equency of the system shown in figure the spring mass pulley system

A rod of length l and depth H having a mass m rests on a semicylidrical surface os radius r It is tipped slightly and let go Find the frequency of oscillation

A thin bar of l having a mass m rests on a semicylidrical surface radius r It is tipped slightly and let go Find the frequency of oscillation

l

h

R

l

37

38

CHAPTER 3

DAMPED FREE VIBRATION OF SINGLE DEGREE FREEDOM SYSTEM LEARNING OBJECTIVES

] Introduction to damped free vibration

] Terminologies used in damped free vibration

] Three different types of damping

] Concept of critical damping and its importance

] Study of response of viscous damped system for cases of under damping

critical and over damping

] Concept of logarithmic decrement

] Problems related to damped free vibration problems

Viscously damped free vibration of single degree freedom sytem Consider a single degree freedom system with viscous damping as shown in figure For a viscous damped system the damping force is

where c is a constant of proportionality It is called damping coefficient or coefficient of viscous damping

Symbolically it is designated by a dashpot as shown in Fig 3

x xampx

39

At any instant of time let the mass be displaced by x from the equilibrium position Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

Okxxcxm =minusminus= ampampamp

Okxxcxm =++ ampampamp ----------------------------------------------------------(1)

or

0kxdtdx

Cdt

xdM

2

2

=++

Assuming the solution of equation as

The first term in the above equation m2Ct

eminus

is an exponentially decaying function of time The behavior of the terms in the parenthesis however depends on whether numerical value within the radicals is positive zero or negative

CASE 1 if 0mk

m2c

2

=minus

40

CASE 2 if 0mk

m2c

2

=gt

CASE 3 if 0mk

m2c

2

=lt

41

Critically damped coefficient Critically damped coefficient is defined as the value of the damping coeffient for which radical in equation

mk

m2c

m2c

S2

21 minus

plusmn

= becomes zero

Damping Ratio ( Damping factor) ( ξ ) It is defined as the ratio of damping coefficient to the critical damping coefficient Note The damping coefficient c is defined as the actual amount of damping while the damping coefficient is a constant depending upon the mass and stiffness of the system The roots of characteristics equation can be now written as Note if c=cc or ξ =1 then system is critically damped if cgt cc or ξ gt1 then system is over damped I if clt cc or ξ lt1 then system is under damped

42

Take home quiz Derive General solution of different damped system

1 Critically damped system 2 Under damped system 3 Over damped system

Logarithmic Decrement It is used to determine the amount of damping present in system This measure The rate of decay of free vibration The larger the damping the greater will be rate of decay Logarithmic decrement is defined as the natural logarithm of ratio of any two successive amplitudes

=δ

2

1

xx

ln

Expression for logarithmic decrement (δ ) Consider a damped free vibration which is expressed by

( ) ( ) φ+ξminusω= ξω t 1sinXex 2n

tn

x is the displacement at any instant of time X and φ are arbitrary constants

21

2

ξminus

πξ=δ

43

Tutorial problems 1 Derive an expression for damping ratio in terms of logarithmic decrement

2 Show that logarithmic decrement

=δ

n

o

xx

lnn1

where xo is the initial amplitude xn

is amplitude after n cycles 3 A mass of 1 kg is to be supported on a spring having stiffness K= 9800 Nm The damping constant c= 59 Nsm Determine the natural frequency of system Find also logarithmic decrement and amplitude after 3 cycles if initial displacement is 03 cm 4 A vibrating system is defined by the following parameters M = 3kg k= 100 Nm damping coefficient = 3 Nsm determine Damping factor natural frequency damped vibration logarithmic decrement ratio of consecutive amplitude no of cycles after which the original amplitude is reduced to 20 5A single degree damped vibrating system a suspended mass of 18kg makes 15 oscillations in 003 seconds The amplitude decreases to 025 of the initial value after 5 oscillations Determine the stiffness of spring Logarithmic decrement damping factor and damping coefficient 6 A damped vibration record of a spring mass system shows the following data Amplitude at the end of 2nd cycle is 9mm amplitude at eh end of 3rd cycle is 6mm amplitude at the end of 4th cycle is 4mm stiffness of spring =8000 Nm weight =20N Find the logarithmic decrement damping force at unit velocity periodic time of vibration 7 For the system shown in figure the characteristics of dashpot is such that the constant force of 49 N is applied to the piston it velocity is found to be constant at 012 ms Determine the value of C would you expect the complete system to be periodic or aperiodic 8 For a torsional system shown in the figuref1 is natural frequency in air and f2 is the natural frequency when immersed in oil Determine the damping coefficient

Viscous fluid

Disc of mass moment of Inertia J

2a

C

m M

b

a a

k

9 A thin plate of area A and weigh w is fixed to the end of the spring and allow to oscillate in a viscous fluid as shown in figure It f1 is the frequency of oscillation in air and f2 in liquid Show that

22

21 ff

gA2

minusπω

=micro where the damping force on the plate

AV2Fd micro= V being the velocity of vibration 10 A torsional pendulum has a natural frequency of 175 cyclessecond vibrating in vacuum The mass moment of inertia is 2 kgm2 It is immersed in oil and it is observed that the natural frequency is 142 cyclessecond Find the damping torque If the disc is replaced by 34 deg When in oil find the displacement at the end of the first complete cycle 11 An automobile can be modeled as a mass placed on 4 shock absorbers each consisting of a spring and a damper Such that each spring is equally loaded Determine the stiffness damping constant of each shock absorber So that natural frequency is 2Hz and system is critically damped The mass of vehicle is 200kg 12 Write the differential equation of motion for the system shown in the figure below Determine the natural frequency of the damped oscillation and critical damping coefficient

Determine the differential equation of motion and find the critical cdamped natural frequency for system shown in figure

a

C

m M

M =200kg

44

200kg kgkg

oefficient and

45

Derive the equation of motion and find the critical damping coefficient and damped natural frequency for the system shown below Write the differential equation of motion and determine the expression for critical damping coefficient and damped natural frequency of system shown in figure

Set up a differential equation of motion for system shown in figure and determine the following Undamped natural frequencies critical damping coefficient damping ratio damped natural frequency

b

C

kM

a

m

46

A mass of 300N is resting on two springs of stiffness 3000 Nm each and a dashpot of damping coefficient 150 Nsm as shown in figure If the initial velocity of 10cms is given to mass at its equilibrium position what will be the displacement from the equilibrium position at the end of first second Derive the formula used A gun barrel of mass 600kg has a recoil spring of stiffness 294 knm if the barrel recoils 13m on firing determine i) The initial recoil velocity of barrel ii) The critical damping coefficient of dashpot which is engaged at the end of recoil stroke Iii) Time required for the barrel to return at the end of recoil stroke Iv) time required for the barrel to return to a position 5cm from the initial position

W= 3000 N

FIRING

CHAPTER 4

FORCED VIBRATION LEARNING OBJECTIVES

] Introduction to forced free vibration

] Terminologies used in forced vibration

] Concept of magnification factor

] Study of reciprocating and rotating unbalance

] Concept of base excitation and support excitation

] Concept of Transmissibility ratio

] Problems related to forced vibration

The vibration that takes place under the excitation of external forces are called

forced vibration When a system is subjected to harmonic excitation it is forced to vibrate at the same frequency as that of excitation Most of the mechanical systems are prone to steady state forced vibration which cause fatigue failure The vibration problem also becomes very important when any of excitation frequencies coincide with one of natural frequencies of a system Under this condition resonance occurs during which large amplitude of vibrations are observed To avoid resonance external frequency (operating speed) may be changed or properties of the system may be changed to alter the natural frequency In some cases sufficient amount of damping may be provided to avoid large amplitude during resonance Thus the problem of force vibration is very important in mechanical design

FORCED VIBRATIONS OF SINGLE DOF SYSTEMS UNDER HARMONIC EXCITATION

` [ ] tsin FF o ω=

x

sin FF o=

Fo

xampamp

[ ] tω

xamp

47

48

Consider a spring mass system having a viscous damping excited by a harmonic force [ ] tsin FF o ω= as shown in figure The various forces acting on the system is shown in figure Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

O)tsin(Fkxxcxm o =ω+minusminus= ampampamp

)tsin(Fkxxcxm o ω=++ ampampamp ----------------------------------------------------------(1)

or

)tsin(Fkxdtdx

Cdt

xdM o2

2

ω=++

Equation (1) represents the differential equation of motion of the system The solution to this equation consists of two parts namely complementary function and a particular integral

The first of solution is also termed as transient solution of homogeneous solution and can be obtained from the homogeneous differential equation 0kxxcxm =++ ampampamp

The second part of solution particular integral is also termed as steady state solution ad is of form )tsin(Xx φminusω= Where X= steady state amplitude φ is phase difference between external force and displacement

)tsin(Xx φminusω=

The vector representation of the shown

49

The force polygon for the above vector diagram can be written as follows

( )φminusωt O

KX

xm 2ω

Fo

xCω

tω

φ

50

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

is the governing equation

Zero frequency deflection (Xst) It is deflection observed at zero frequency ( when the external frequency 0=ω

( ) kF

001

kF

x o

22

o

st =

+minus=

Magnification factor (M) It is defined as the ratio of steady state amplitude to zero frequency deflection xst

M = arX

X=

A plot of magnification factor versus frequency ratio

damping factor ( )ξ is called frequency plot of a harmon All the frequency terms starts from magnification factor of damping ratio ( )ξ The amplitude of vibration at any on the value of damping The amplitude will be higher for higher damping The amplitude of vibration at reso

amplitude tends to zero as

ωω

n tends to infinity

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

Freq

ζ =

ζ = 1

Magnification Factor M = XXst

kFo

51

ωω

nfor varying values of

ically excite vibrating system

M=1 what ever may be value particular frequency depends

for lesser damping and lesser nance will be very high The

uency Ratio r = (ωωn)

ζ = 0 0

ζ = 025 ζ = 0375 ζ = 0 5

ζ = 0707

ζ = 2

52

Plot of phase angle φ versus frequency ratio

ωω

n is shown in the

figure The phase angle φ is zero for all the values of damping ratio

at

ωω

n=0 The phase

angle φ is 90 for all the values of damping ratio

at

ωω

n=1 ie at

resonance The phase angle between o to 90 deg for all the values of

ωω

n= 0 to 1 and

between 90 deg to 180 deg for all the values of

ωω

ngt1

Class work Derive an expression for frequency at peak amplitude

2np 21 ξminusω=ω=ω is the required equation

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

53

Derive expression amplitude at resonance ( In dimensional form) In dimensional form We know that steady state amplitude

( ) ( )

ω+ωminus=

222

o

Cmk

Fx Equation in dimensional form

We know that at resonance mk

n =ω=ω

( )

ω+

minus

=2

n

2

oR

Cmk

mk

Fx

( )

ω=

2n

oR

C

Fx

( )

ω=

n

oR C

Fx

Class work Derive an expression for amplitude at resonance ( in non dimensional form)

Tutorial Problems on Forced Vibration 1A weight of 60N is suspended by a spring stiffness 12 kNm is forced to vibrate by a harmonic force of 10N assuming viscous damping of 0086 kN-sm Find i) resonance frequency ii) amplitude at resonance iii) phase angle at resonance iv) frequency corresponding to peak amplitude 2 A periodic torque having a maximum value of 05888Nm at a frequency of 4 rads is impressed upon a flywheel suspended from a wire The wheel has mass moment of inertia 012 kgm2 and the wire has stiffness 1176 Nmrad A viscous dashpot applies damping couple of 0784 Nm an angular velocity of 2 rads calculate i)maximum angular displacement from rest position ii) maximum couple applied to dashpot Iii) angle by which the angular displacement lags the torque A mass of 1019 kg is suspended from a spring stiffness 20 Nmm viscous damping causes the amplitude to decrease to 110th of initial value in 4 complete cycle If a periodic force of 20 cos(50 t) is applied to the mass Find the amplitude of forced vibration What would be the amplitude of oscillation if the applied force coincides with the natural period

m=1019 kg

[ ] t50cos 30F =

54

55

ROTATING UNBALANCE Unbalance in rotating machinery is a common source of vibration excitation Consider a machine of mass m having mounted on a foundation of stiffness k and damping coefficient C as shown in figure The unbalance is represented by an eccentric rotating mass m with eccentricity (e) and having a angular velocity (ω) let x be the displacement of non rotating mass from the static equilibrium position The displacement of rotating mass m from the static equilibrium is ( ) tsin( ex ω+ Applying Newtonrsquos law of motion we have

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

56

RECIPROCATING UNBALANCE Unbalance in reciprocating machinery is a common source of vibration excitation Consider a machine of mass M having mounted on a foundation of stiffness K and damping coefficient C as shown in figure The unbalance is represented by a reciprocating mass (m) having crank radus (e) and connecting rod of length (l) Let ωωωω be angular velocity of the crank Let x be the displacement of non reciprocating mass from the static equilibrium position at any instant of time The displacement of reciprocating mass m from the static equilibrium position is

)2sin(sin +++ tle

tex ωωωωωωωω

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

57

Note The equations obtained in reciprocating unbalance is same as the rotating unbalance and having the following differences Rotating Unbalance Reciprocating Unbalance Here mass of the rotating element is considered

Here mass of reciprocating element is considered

Here the eccentricity (e) which is the distance between the centre of rotation and CG

Here the eccentricity e is the crank radius and crank radius = stroke length2

The frequency plot of non dimensional factor( ( )me

MX versus frequency ratio

ωω

n for various values of ( )ξξξξ is shown in figure This frequency plot is same for

the rotating and reciprocating unbalance with the following points are noted from the diagram

1 Damping ratio plays very important role during resonance in controlling the amplitude

2 When

ωω

n=0 then ( )me

MX =0 for all the values of damping ratio ( )ξξξξ

3 When

ωω

nis higher than ( )me

MX tends to unity

4 At very high speeds ie at very high frequency ratio

ωω

n there is no effect

of damping The frequency plot of phase angle versus the frequency ratio is given as shown in figure The explanation is same as in forced harmonic vibration

58

VIBRATION ISOLATION

Vibratory forces and motions generated by machine and other caused are unavoidable Howeever their effect on the other dynamical systems can be minimized by proper isolator design An isolation system is either to protect o delicate machine from excessive vibration transmitted to it from its supporting structure or to prevent vibratory forces and motions generated by machines from being transmitted to its surroundings The effectiveness of isolation may be measured in terms of forces and motion transmitted The first type is known as force isolation and the second type is known as motion isolation

Force isolation (Force Transmissibility) When vibratory forces generated by machine are isolated the effectiveness of forced isolation can be made by defining the term transmissibility The term transmissibility is defined as the ratio of force transmitted to the foundation to that of the external impressed force

Force Transmissibility = o

tr

FF

force Impressed Externalfoundation toted transmitForce

=

Consider viscous damped spring mass system as shown in figure Upon this system an external harmonic force ) t (Sin Fo ω is applied The difference forces acting on system are inertia forces damping force spring force and external force All these forces are represented on force polygon shown below Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these two forces

( ) ( )22 xckx ω+=

59

This is the expression for force transmissibility in dimensional form To represent same in the non dimensional form divide numerator and denominator by k

2

n

22

n

2

n

o

tr

21

21

FF

TR

ωω

ξ+

ωω

minus

ωω

ξ+

==

PHASE DIFFERNECE BETWEEN EXTERNAL FORCE AND DISPLACEMENT Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these

ωω

minus

ωω

ξ=

ωminusω

=φ minusminus2

n

n1

2

1

1

2tan

mKC

tan

The phase difference between force transmitted to the foundation Ftr to the displacement

kc

KxxC

tanω

=ω

=α or

ωω

ξ=

ω=α minusminus

n

11 2tanKx

xCtan

The phase difference between external force Fo and force transmitted to foundationis β

60

β= αminusφ =

ωω

minus

ωω

ξminus

2

n

n1

1

2tan -

ωω

ξminus

n

1 2tan

Displacement Transmissibility It is the ratio of amplitude of steady state vibration of machine to that of the base motion Forced vibration due to support motion Consider a spring mass damper system subjected to harmonic excitation )tsin(Yy ω= at the base as shown in figure Let x be the absolute displacement of the mass at any instant of time t The different forces acting on system are shown in free body diagram Applying Newtonrsquos Law of motion Ma = sumF

)yx(k)yx(cxm minusminusminusminus= ampampampamp

0)yx(k)yx(cxm =minus+minus+ ampampampamp

61

Frequency plot of transmissibility versus frequency ratio

62

The plot of transmissibility versus frequency ratio is as shown in figure The following points can be noted from it

i) The value of transmissibility is unity at frequency ratio

ωω

n=0 and 2

ii) For an undamped system ( )ξ =0 transmissibility tends to infin at resonance iii) The value of transmissibility is less than unity for the value of frequency ratio

greater than 2

iv) For frequency ratio

ωω

nlt 2 smaller damping ratio lead to larger values of

transmissibility on the other hand for frequency ratio

ωω

ngt 2 smaller damping

ratio leads to smaller value of transmissibility

The frequency plot of transmissibility (TR) versus frequency ratio

ωω

n can be

broadly divided into two zones

1

ωω

n= 0 to 2

2

ωω

n= 2 to infin

CASE 1

When

ωω

n= 0 to 2

In this zone the value of transmissibility is always greater than 1 for all values of damping ratio In this zone the transmissibility decreases with increase in damping Hence we can conclude that in this zone damping is useful in isolating vibration forces being transmitted to the support CASE 2

When

ωω

n= 2 to infin

In this zone the value of transmissibility is less than 1 for all values of damping ratio In this zone the transmissibility increases with increase in damping Hence we can conclude that in this zone damping will adversely affect in isolating vibratory forces

63

Tutorial problems on Forced Vibration 1 Machine of mass 75kg is mounted on spring of stiffness 12000 Nm with an assumed damping factor ξ =02 piston within the machine of mass 2 kg has a reciprocating motion with a stroke of 75 cm and speed of 15 cyclessecond Assuming the motion of piston to be harmonic determine amplitude of machine ( X) Transmissibility and force transmitted to foundation phase angle of transmitted force wrt to exciting force 2 Motor weighs 600N is mounted on a simple beam that has a spring rate of 400 Ncm The motor armature base 160 N and has an eccentricity of 0005cm What will be the amplitude of motion when it runs at 1760 RPM Neglect damping the deflection of armature shaft and weight of the beam 3 A Heavy machine of mass 300 kg is supported on resilient foundation the static deflection of foundation due to weight of machine is found to be 75mm It is observed that the machine vibrates with an amplitude of 10mm when the base of the foundation is subjected to harmonic oscillation at the undampeds natural frequency of system with an amplitude of 25mm Find the damping constant of foundation Ii) dynamic force amplitude on the base iii) amplitude of the displacement of machine relative to base 4 An air compressor of mass 450 kg operates at a constant speed of 1750 RPM The reciprocating masses are 10 kg the crank radius is 100mm If the damper for mounting introduces a damping factor 015 i) specify the springs of mounting such that only 20 of the unbalanced force is transmitted to the foundation ii) determine the magnitude of the transmitted force bull Repeat the above problem by assuming that there is no damping in system

5 A body of a car having mass 1500 kg and is mounted on 4 equal springs which deflect through 225 cm under the weight of the body A total damping forces the 4 shock absorbers is 46 N at a velocity of 1cmsec The car is placed with all the four wheels on a test platform which is move up and down at resonant speed With an amplitude of 25cm Find the amplitude of the car body on its springs assuming its CG to be in the centre of wheel base 6A 75 kg machine which is mounted on springs of stiffness 116x105 Nm with the damping of 20=ξ A 2kg piston within the machine as reciprocating motion with a stroke of 008m and speed of 3000 RPM Assuming the motion of piston to be harmonic determine the amplitude of vibration of machine

7 The spring of an automobile are compressed 10cm under its own weight Find the critical speed when a trailer is traveling over road with a people approximaterly by a sine wave of amplitude 75 cm and wave length 15m What will be the amplitude of vibration at 64kmhour 8 An eccentric mass excites us used to determine vibratory characteristics of a structure of mass 25kg as given in figure At a speed of 1000 RPM the eccentric mass to beat top position at the instant the structure was moved upwards through its static equilibrium position and corresponding amplitude was 20mm The eccentric mass is 05kg at an eccentricity of 10mm Determine i) Undamped natural frequency of the system ii) Damping ratio iii) Steady state amplitude and phase angle at 1500 RPM iv) Force transmitted to the ground at 1200 RPM

M=25kg

C

K

Eccentric mass

64

65

CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Support motion Consider a spring mass damper system as shown in figure Let it excited by motion

of support let y be the harmonic displacement of support From Newtonrsquos second law of motion

)yx(k)yx(cxm minusminusminusminus= ampampampamp

)1(0)yx(k)yx(cxm minusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminus=minus+minus+ ampampampamp

Let z be the relative displacement of the mass with respect to support then

Z=x-y

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

Time period It is the time taken to complete one cycle It is equal to the time for the vector to rotate through 2π radians ` Frequency It is the number of cycles per unit time Amplitude It is the maximum displacement of a vibrating body from the mean position Phase difference it is angle between the two rotating vectors executing simple harmonic motion of same frequency The first vector is x1= X sin ( wt ) The second vector is x2= X sin ( wt+φ) Where φ is the phase difference between x1 and x Resonance it is the frequency of the externalfrequency of the system a condition known aresonance the system undergoes dangerously larg Damping It is the resistance offered to the motion Periodic motion If the motion is repeated after equal intervals of timsimplest type of periodic motion is harmonic motio Aperiodic motion If the motion does not repeat after equal interval of

rads ωωωω

Reference

rads ωωωω

2

force coincides with the s resonance occurs Dure oscillations

of a vibrating body

e it is called periodic motin

time it is called aperiodic

3

natural ing the

on The

motion

rads ωωωω

DESCRIBING MOTIONS OF VIBRATING SYSTEMS Periodic motions Described as sine or cosine functions [sin (ωt) and cos(ωt)] ω = radian frequency (radsec) ω = 2πf where f is frequency (Hz) Period = time between two adjacent peaks or valleys P = 1f

Simple harmonic motion ( ) ( )φω += tAty sin Displacement ( ) ( )φωω += tAty cosamp Velocity ( ) ( )φωω +minus= tAty sin2amp Acceleration Classification of vibrations One method of classifying mechanical vibrations is based on degreesThe number of degrees of freedom for a system is the number of independent variables necessary to completely describe the motion of ein the system Based on degrees of freedom we can classify mechanicas follows 1Single Degree of freedom Systems 2Two Degrees of freedom Systems 3Multidegree of freedom Systems 4Continuous Systems or systems with infinite degrees of freedom Another broad classification of vibrations is 1 Free and forced vibrations 2 Damped and undamped vibrations

Leads Lags

4

of freedom kinematically very particle al vibrations

5

Sometime vibration problems are classified as 1 Linear vibrations 2 Non-linear vibrations 3 Random vibrations 4 Transient vibrations 5 Longitudinal vibrations 6 Transverse vibrations 7 Torsional vibrations Free vibration If a system after initial disturbance is left to vibrate on its own the resulting vibration is known as free vibrations Free vibration takes when a system vibrates under the action of forces inherent in the system and when the external forces are absent The frequency of free vibration of a system is called natural frequency Natural frequency is a property of a dynamical system Forced vibration Vibration that takes place under the excitation of external forces is called forced vibration the forced vibration takes place at different forced frequencies or external frequencies Damped vibration If any energy is lost or dissipated during oscillations then the vibration is known as damped vibration Undamped vibration if no energy is lost or dissipated during oscillations such vibrations are known as undamped vibration Linear vibration If all the basic component of a vibrating system behave linearly the resulting vibration is known as linear vibration The differential equations govern linear vibratory system are linear If the vibration is linear the principle of superposition holds and mathematical techniques of analysis are well developed Non linear vibration If any of the basic components of a vibrating system behave non linearly the resulting vibration is known as non linear vibration The differential equations that govern non linear vibratory system are non-linear If the vibration is non linear the principle of superposition does not hold good and techniques of analysis is well known Deterministic vibration If the magnitude of excitation on a vibrating system is known at any given time the resulting vibration is known as deterministic vibration Random vibration If the magnitude of excitation acting on a vibratory system at a given time cannot be predicted the resulting vibration is known as non deterministic or random vibration

6

Longitudinal vibration Consider a body of mass m carried on one end of a slender shaft and other end being fixed If the mass vibrates parallel to the spindle axis it is said to be execute longitudinal vibration Transverse vibration If the mass vibrates perpendicular to the spindle axis it is said to execute the transverse vibration Torsional vibration If the shaft gets alternatively twisted and un twisted on account of an alternate torque on the disk it is said to execute the torsional vibration SIMPLE HARMONIC MOTION A Vibration with acceleration proportional to the displacement and directed toward the mean position is known as SHM ( Simple Harmonic Motion) Consider a spring mass system as shown in the figure along with the displacement time diagram The equation of motion of the mass can be written as follows From the right angled triangle OAB we have Where x= displacement at any instant of time X = amplitude of vibration W= angular velocity or frequency in radsecond

m1

7

The velocity of the mass m at an instant of time t is given by V = The acceleration of mass m is given by Hence we can conclude that in SHM the acceleration is proportional to displacement and is directed towards mean position observing the equations 2 and 3 the velocity and acceleration are harmonic with the same frequency but lead a displacement vector by π2 and π radians respectively

t

t

t X

x

X= A sin ωt

x-Displacement X-amplitude T-Periodic Time f-Frequency f=1T ω=Frequency in radians per second t= time

8

Rotor with inertia J1

kt1 kt2

Rotor with Inertia J2

Degrees of freedom The minimum number of independent coordinates required to determine completely the positions of all the parts of a system at any instant of time is called degrees of freedom One degree of freedom Two degree of freedom Three degree of freedom Infinite degree of freedom

m1

m2

m1

m2

m3

Cantilever Beam

Continuous system ( consider the mass of the beam )

m1

m1

K1

m2

K

x1

x2

F1

9

G

K1 K2

mJ

a b

Exercise Specify the no of degree of freedom for the following 1 2 3

4

5

4m 2m m 3k k k

Cantilever Beam

Continuous system ( Neglect the mass of the beam )

k1 k2 m1 m2

y y

m2

θ1

θ2

L1

L2

M1

M2

10

O

A

B

Addition of SIMPLE HARMONIC MOTION ( SHM) The addition of two simple harmonic motion having frequency yields a resultant which is simple harmonic having the same frequency Consider tow simple harmonic motions of 1x and 2x having the same frequency and phase difference φ as given below

)(sin11 tXx ωωωω====

)(sin22 φω += tXx

Adding x = x1 +x2 Hence the resultant displacement is also SHM of amplitude X and phase angle θ

θθθθ

ωωωωt

φφφφ

11

-6

-4

-2

0

2

4

6

0 005 01 015 02 025

Tutorial problems on Simple Harmonic Motion 1 Add the following harmonic motion analytically and verify the solution graphically 1) X1= 3 sin (((( ))))30 t ++++ωωωω X2= 4 cos (((( ))))10 t ++++ωωωω ( VTU Jan 2005) 2) X1= 2 cos (((( ))))05 t ++++ωωωω X2= 5 sin (((( ))))1 t ++++ωωωω ( VTU July 2006)

3) X1= 10 cos (((( ))))4 t ππππωωωω ++++ X2= 8 sin (((( ))))6 t ππππωωωω ++++ ( VTU Dec 2007)

2 A body is subjected to two harmonic motions X1= 8 cos (((( ))))6 t ππππωωωω ++++ X2= 15 sin (((( ))))6 t ππππωωωω ++++ what harmonic motion

should be given to the body to bring it to equilibrium (VTU July 2005) 3 Split the harmonic motion x = 10 sin (((( ))))6 t ππππωωωω ++++ into two harmonic motions

having the phase of zero and the other of 45o 4 Show that resultant motion of harmonic motion given below is zero X1= X sin (((( ))))t ωωωω X2= X sin (((( ))))3

2 t ππππωωωω ++++ X3= X sin (((( ))))34 t ππππωωωω ++++

5 The displacement of a vibrating body is given by x = 5 sin (3141t + 4

ππππ )

Draw the variation of displacement for one cycle of vibration Also determine the displacement of body after 011 second ( repeat the problem for velocity and acceleration and draw graph using Excel and compare ) Time Displacement

0 0025 5 005

0075 0 01

0125 015 -353

0175 0 02

Calculate the remaining values 6 The Motion of a particle is represented by x = 4 ( ) t sin ω sketch the variation of the displacement velocity and acceleration and determine the max value of these quantities Assume ( )5 =ω ( Try to use MATLABExcel) sketch all on same graph

12

BEATS When two harmonic motions with frequencies close to one another are added The resulting motion exhibits a phenomenon known as Beats A Beat Frequency is the result of two closely spaced frequencies going into and out of synchronization with one another Let us consider tow harmonic motion of same amplitude and slightly different frequencies X1 = X Cos ( ) tωωωω X2 = X Cos ( )( ) tδδδδωωωω + Where δ is a small quantity The addition of above two harmonics can be written as X = X1 + X2

X=

+

ttX2

cos2

cos2δδδδωωωω

δδδδ

The above equation shown graphically in Figure The resulting motion represents cosine wave with frequency ( ) 2

ππππωωωω + and with a varying amplitude 2 cos( ( ) 2δδδδ t

whenever the amplitude reaches a maximum it is called the beat The frequency δ at which amplitude builds up and dies down between o and 2 X is known as beat frequency Beat phenomenon is found in machines structures and electric power houses In machines the beat phenomenon occurs when the forcing frequency is close to the natural frequency of the system

+2X

-2X

ωωωωππππ2

ELEMENTS OF VIBRATION The elements of constitute vibrating systems are

1 Mass or Inertia element - m 2 Spring - k 3 Damper - c 4 Excitation F(t)

Mass or Inertia element The mass or inertia element is assumed to be rigid body During vibration velomass changes hence kinetic energy can be gained or loosed The force appthe mass from newton second law of motion can be written as F= ma Work dthe mass is stored in the form of kinetic energy given by KE= frac12 M V2

Combination of masses In practical cases for simple analysis we replace several masses by equivalent mass Case1 Translational masses connected by rigid bar Let the masses M1 M2 and M3 are attached to a rigid bar at locates 1 respectively as shown in the figure The equivalent mass meq be assumelocated at 1 is as shown in figure (b) Let the displacement of masses M1 M2 and M3 be x1 x2 x3 and simivelocities of respective masses be x1 x2 and x3 We can express the velomasses m2 and m3 in terms of m1

Elements of Vibration

Passive element

m C K

Active element

F(t)

Conservative element

( Mass and Spring)

Non conservative element

( Damper)

C k

m F(t)

Voig

13

city of lied on one on

a single

2 and 3 d to be

larly the cities of

t Model

Translationa Let a mass mmass momearrangement These two mamass Meq or

2

1

33

2

1

221

+

+=

ll

Mll

MMMeq

14

is the required answer

l and rotational masses coupled together

having a translational velocity x be coupled to another mass having nt of inertia Io with a rotational velocity θ as in rack and pinion

shown in the figure

sses can be combined to obtain either a single equivalent translational a single equivalent mass moment of inertia Jeq

15

Equivalent translational masses

Kinetic energy of the equivalent mass =

2

21

eqeq XM amp

Kinetic energy of the two masses =

+

20

2

21

21

θθθθampamp JXM

Meq= m

+ 2RJ

m o is the required answer

Also determine the equivalent rotational mass Jeq

Jeq= m [ ]oJmR +2 is the required answer Spring element Whenever there is a relative motion between the two ends of the spring a force is developed called spring force or restoring force The spring force is proportional to the amount of deformation x and then F α x or F = kx Where k is stiffness of the spring or spring constant or spring gradient The spring stiffness is equal to spring force per unit deromation

The spring stiffness k = mNxF

Workdone in deforming a spring is equal to the strain energy or potential energy Strain energy = potential energy = area of the triangle OAB Stiffness of beams Cantilever beam consider a cantilever beam with an end mass shown in the figure The mass of the beam is assumed to be negligible The static deflection of beam at free end is given by

Similarly derive the expression for Simply supported beam and fixed support beam

x

k

m

x

m

x

m

x

mNlIE

st 192

3

=δδδδ

mNEI

Wlst

48

3

=δδδδ

F

K=stiffness

F

16

17

Stiffness of slender bar subjected to longitudinal vibrations For a system executing the longitudinal vibrations as shown in the figure let us derive the expression for stiffness

Torsional Stiffness of bar It is the amount of torque required to cause a unit angular deformation in the element

Torsional stiffness = Kt = θθθθT

Combination of stiffness Determination of equivalent spring stiffness when the springs are arranged in series Consider two springs of stiffness K1 and K2 acted upon by the force F

The deflection of spring k1 is x1 = 1KF

The deflection of spring k1 is x2 = Let these two springs be replaced by an equivalent stiffness Keq upon which a force F acts and due to which its deflection is given by

x = eqKF

x= x1+x2

m

E A

l

60 cm 70 cm

Determination of equivalent spring stiffness when the springs are arranged in parallel Force acting on K1 spring = F1=k1x Force acting on K2 spring = F2 Force required for an equivalent spring keq to defined by x given by F= Keq x But F = F1 +F2 Tutorial problems on Equivalent stiffness of springs 1Determine the equivalent stiffness for the system shown in figure 2 Determine the equivalent stiffness for the system shown in figure

3 Determine the equivale

M

3

M

Nm 102 6x

18

nt stiffness for the system shown in figure

38 kg

Nm 101 6x Nm 102 6x

Nm 106x

2k

k

k

2k

3k

k

2k

k

19

60 cm 80 cm 50 cm

50 cm 60 cm

4Determine the equivalent stiffness for the system shown in figure

5Replace the following torsional stiffness by a single shaft having radius 4cm and find the length required for the equivalent shaft Assume the material of given system and equivalent system is same R1= 3cm R2= 5cm DAMPING Every vibration energy is gradually converted into heat or sound Hence the displacement during vibration gradually reduces The mechanism by which vibration energy is gradually converted into heat or sound is known as damping A damper is assumed to have either mass or elasticity Hence damping is modeled as one or more of the following types Viscous damping Coulomb or dry friction damping materials or solid or hysteric damping Viscous damping Viscous damping is most commonly used damping mechanism in vibration analysis When the mechanical system vibrates in a fluid medium such as air gas water or oil the resistance offered by the fluid to the moving body causes energy to be dissipated In this case the amount of dissipated energy depends on many factors such as size or shape of the vibrating body the viscosity of the fluid the frequency of vibration and velocity of fluid Resistance due to viscous damping is directly proportional to the velocity of vibration

dF V αααα

dF = xC amp

Where C= damping coefficient Fd = damping force Examples of Viscous damping

1) Fluid film between sliding surface 2) Fluid flow around a piston in a cylinder 3) Fluid flow through orifice 4) Fluid flow around a journal in a bearing

Rreq =4cm

leqn

stress

Strain

Coulomb damping or dry friction damping Here a damping force is constant in magnitude but opposite in direction to that of the motion of vibrating body It is caused by the friction between the surfaces that are dry or have insufficient lubrication Material or solid or hysteric damping

When the materials are deformed energy is absorbed and dissipated by the material The effect is due to friction between the internal planes which slip or slide as the deformation takes place When a body having the material damping is subjected to vibration the stress strain diagram shows the hysteresis loop as shown in the figure The area of the loop denotes the energy lost per unit volume of the body per cycle

20

21

ττττ ττττ2 ττττ3

ττττ

X(t)

FOURIER SERIES The simplest of periodic motion happens to be SHM It is simple to handle but the motion of many vibrating system is not harmonic (but periodic) Few examples are shown below

Forces acting on machines are generally periodic but this may not be harmonic for example the excitation force in a punching machine is periodic and it can be represented as shown in figure 13 Vibration analysis of system subjected to periodic but nonharmonic forces can be done with the help of Fourier series The problem becomes a multifrequency excitation problem The principle of linear superposition is applied and the total response is the sum of the response due to each of the individual frequency term

Any periodic motion can be expressed as an infinite sum of sines and cosines terms If x(t) is a periodic function with period t its Fourier representation is given by

X(t) = (((( )))) (((( )))) (((( ))))t sint cost cos2 111 ωωωωωωωωωωωω baaao ++++++++++++

= (((( )))) (((( ))))t sint cos2 1

ωωωωωωωω nnn

o bnaa

++++++++sumsumsumsuminfininfininfininfin

====

tππππωωωω

2==== = Fundamental frequency ndash (1)

where ao an bn are constants

ττττ ττττ2 ττττ3

ττττ

X(t)

22

t

X(t)

035

025

Determination of constants To find ao Integrate both sides of equation(1) over any interval ττττ All intergrals on the RHS of the equation are zero except the one containing ao

(((( ))))dttxao

o 2

2

intintintint====ωωωωππππ

ππππωωωω

= (((( ))))dttxo

2intintintintττττ

ττττ

To find an multiply equation 1 by cos (((( ))))tnωωωω and Integrate over any interval ττττ All intergrals

(((( )))) (((( )))) dttxao

n tncos 2

2

ωωωωππππωωωω ωωωω

ππππ

intintintint==== (((( )))) (((( )))) dttxo

tncos 2

ωωωωττττ

ττττ

intintintint====

To find bn multiply equation 1 by sin (((( ))))tnωωωω and Integrate over any interval ττττ All intergrals

(((( )))) (((( )))) dttxao

n tnsin 2

2

ωωωωππππωωωω ωωωω

ππππ

intintintint==== (((( )))) (((( )))) dttxo

tnsin 2

ωωωωττττ

ττττ

intintintint====

Find the Fourier series for the curve shown below

ττττ ττττ2 ττττ3

t

X(t)

E

ττττ ττττ2 ττττ3

t

X(t)

ππππ2

t

X(t) ππππ

1 Represent for the periodic motion shown in the figure

23

CHAPTER 2 UNDAMPED FREE VIBRATION LEARNING OBJECTIVES

] Introduction to undamped free vibration

] Terminologies used in undamped free vibration

] Three methods to solve the undamped free vibration problems

] Problems related to undamped free vibration problems

Free vibrations are oscillations about a systems equilibrium position that occur in the absence of an external excitation force If during vibrations there is no loss of energy it is known as undamped vibration The first step in solving a vibration problem is setting up the differential equation of motion Consider spring mass system which is assumed to move only along the vertical direction as shown below Let m be the mass of the block and k be the stiffness of the spring When block of mass m is attached to spring the deflection of spring will be ∆ known as static deflection In the static equilibrium position the free body diagram of forces acting on the mass is shown in Figure(b) Hence mg= kA Once the system is disturbed the system executes vibrations

Let at any instant of time t the mass is displaced from the equilibrium position x the different forces acting on the system are shown in figure (d) From Newtonrsquos second law of motion

sum = maF

Inertia force ( disturbing force) = restoring force

mgxkxm ++∆minus= )(ampamp

0)( =+ xkxm ampamp

or 0)( =+ xmk

xampamp

equation 2 is the differential equation of motion for spring mfigure Comparing equation (2) with the equation of SHM +xampampsince the vibrations of the above system are free( without the forces) we can write

sec radmk

n =ω

24

ass system shown in 0)(2 =xω

resistance of external

time period km

2 f1

n

πτ ==

from the equation(1) mg = k∆

∆=

q

mk

Difference between the translation ( rectilinear) and rotational system of vibration

Translatory Rotational

In the analysis thFORCES are co Linear stiffness K Problems 1A mass of 10kFind the natural 2 A spring massnatural frequencnatural frequenc

kt

m

sec radmk

n =ω

k

25

e disturbing and restoring nsidered

In the analysis In the analysis MASS Moment of Inertia (J) is considered

in Nm is considered

g when suspended from a spring causes a static deflection of 1cm frequency of system

system has a spring stiffness K Nm and a mass of m Kg It has a y of vibration 12 Hz An extra 2kg mass coupled to it then the y reduces by 2 Hz find K and m

secrad n =ω

26

3 A steel wire of 2mm diameter and 30m long It is fixed at the upper end and carries a mass of m kg at its free end Find m so that the frequency of longitudinal vibration is 4 Hz 4 A spring mass system has a natural period of 02 seconds What will be the new period if the spring constant is 1) increased by 50 2) decreased by 50 5 A spring mass system has a natural frequency of 10 Hz when the spring constant is reduced by 800 Nm the frequency is altered by 45 Find the mass and spring constant of the original system 6 Determine the natural frequency of system shown in fig is which shaft is supported in SHORT bearings 7 Determine the natural frequency of system shown in fig is which shaft is supported in LONG bearings where l is the length of bearing and E ndash youngrsquos modulus and I is moment of Inertia 7 Determine the natural frequency of system shown in fig is which shaft is supported in LONG bearings 8 A light cantilever beam of rectangular section( 5 cm deep x 25cm wide) has a mass fixed at its free end Find the ratio of frequency of free lateral vibration in vertical plane to that in horizontal 9 Determine the natural frequency of simple pendulum 10 Homogeneous square plate of size l and mass m is suspended from the mid point of one of its sides as shown in figure Find the frequency of vibration

l

l

27

11 A compound pendulum which is rigid body of mass m and it is pivoted at O The point of pivot is at distance d from the centre of gravity It is free to rotate about its axis Find the frequency of oscillation of such pendulum 12 A connecting rod shown in fig is supported at the wrist pin end It is displaced and allowed to oscillate The mass of rod is 5kg and centre of gravity is 20 cm from the pivot point O If the frequency of oscillation is 40 cyclesminute calculate the mass moment of inertia about its CG

13 A semi circular homogenous disc of radius r and mass m is pivoted freely about its centre as shown in figure Determine the natural frequency of oscillation 14A simply supported beam of square cross section 5mmx5mm and length 1m carrying a mass of 0575 kg at the middle is found to have natural frequency of 30 radsec Determine youngrsquos modulus of elasticity of beam 15 A spring mass system k1 and m have a natural frequency f1 Determine the value of k2 of another spring in terms of k1 which when placed in series with k1 lowers the natural frequency to 23 f1

COMPLETE SOLUTION OF SYSTEM EXECUTING SHM

The equation of motion of system executing SHM can be represented by

0)( =+ xkxm ampamp ----------------------------------------(1)

022

2=+

dtdx

dt

dx ω

The general solution of equation (1) can be expressed as

------------------------(2)

Where A and B are arbitrary constant which can be determined from the initial conditions of the system Two initial conditions are to be specified to evaluate these

constants x=x0 at t=0 and xamp = Vo at t=0 substituting in the equation (2) Energy method In a conservative system the total energy is constant The dwell as natural frequency can be determined by the princienergy For free vibration of undamped system at any instant and partly potential The kinetic energy T is stored in the masswhere as the potential energy U is stored in the form of sdeformation or work done in a force field such as gravity The total energy being constant T+U = constant Its rate of cha

t)(sin Bt)cos( AX ω+ω=

t)(sin V

t)cos( xx o0 ω

ω+ω=

Is the required complete

solution

28

ifferential equation as ple of conservation of of time is partly kinetic by virtue of its velocity train energy in elastic

nge

Is given by [ ] 0UTdtd

=+

From this we get a differential equation of motion as well as natural frequency of the system Determine the natural frequency of spring mass system using energy method

Determine the natural frequency of system shown in figure Determine the natural frequency of the system shown in figure Is there any limitation on the value of K Discuss Determine the natural frequency of s

m

l k

m

a

θ

m

a

θ

l

k

29

k

k

ystem shown below Neglect the mass of ball

l

m

30

A string shown in figure is under tension T which can be assumed to remain constant for small displacements Find the natural frequency of vertical vibrations of spring An acrobat 120kg walks on a tight rope as shown in figure The frequency of vibration in the given position is vertical direction is 30 rads Find the tension in the rope A manometer has a uniform bore of cross section area A If the column of liquid of length L and Density ρ is set into motion as shown in figure Find the frequency of the resulting oscillation

m

l

a

T T

36m 8m

31

k

R

k

m

r

m

Find the expression for natural frequency of system shown in the figure Neglect the mass of the cantilever beam Study the special case i) k=Infinity ii) I = infinity Determine the expression for the natural frequency of a system shown in figure The two discs are keyed to a common shaft and have a combined mass moment of inertia about centre of oscillation O It is assumed that the card attached to mass m does not stretch and is always under tension Determine the natural frequency of a system shown

m

l

32

Determine the expression for the natural frequency of the system shown in figure Assume that the wires connecting the masses do not stretch and are always in tension

Determine the natural frequency of spring mass system taking the MASS OF SPRING (ms ) into account

M3 M1

M2

k2 k1

m

k

33

RAYLEIGHrsquoS METHOD

This is the extension of energy method Here we can determine the natural frequency of a conservative system without finding the equation of motion The natural frequency is a function of the rate of change of Kinetic energy and potential energy of the system From the principle of conservation of energy we have T+U = Constant T= KE and U =PE We can write the above equation as T1+U1 = --------------------------(1) Where 1 and 2 represents tow instances of time Let 1 be the time when the mass is passing through static equilibrium position Let 2 be the time corresponding to the mass displacement of the mass At this instant the velocity f the mass is zero and hence Substituting in equation (1) we get ( )Max = ( )Max Above equation leads directly to natural frequency Determine the natural frequency of spring mass system using Rayleighrsquos method m

34

r

B

K K

Home work

Determine the natural frequency of pendulum using Rayleighrsquos method

A Cylinder of mass m and moment of inertia Jo is to role without slipping but is sustained by spring K as shown in figure Determine the natural frequency of oscillation

Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping

a

Determine the natural frequency of system shown where in cylindrical disc rolls over the inextensible spring A cylinder of radius r rolls without slipping on cylindrical the equation of motion and hence determine the naoscillations about the lowest point shown Repeat the above problem ndash Instead of cylinder assume swithout slipping on concave surface as shown above

r

K

35

portion of radius R Derive tural frequency of small

phere of radius with r rolls

The torsional pendulum has a natural frequency of 5 Hz What length of steel wire of diameter 2 mm should be used for pendulum The inertia of mass fixed at the free end is 00098 kgm2 Take g= 083x1011 Nm2 Determine the torsional natural frequency of the system shown in figure Neglect the mass moment of inertia of the shaft ( Figure not proportionate) Determine the natural frequency of simple pendulum considering the mass of the rod into account

Using the energy method find tThe chord may be assumed ine

Kt1

Kt2

20cm

06m 08m 04m

20cm 20cm 20cm

m

M

he natural frxtensible in

k

M

36

equency of the system shown in figure the spring mass pulley system

A rod of length l and depth H having a mass m rests on a semicylidrical surface os radius r It is tipped slightly and let go Find the frequency of oscillation

A thin bar of l having a mass m rests on a semicylidrical surface radius r It is tipped slightly and let go Find the frequency of oscillation

l

h

R

l

37

38

CHAPTER 3

DAMPED FREE VIBRATION OF SINGLE DEGREE FREEDOM SYSTEM LEARNING OBJECTIVES

] Introduction to damped free vibration

] Terminologies used in damped free vibration

] Three different types of damping

] Concept of critical damping and its importance

] Study of response of viscous damped system for cases of under damping

critical and over damping

] Concept of logarithmic decrement

] Problems related to damped free vibration problems

Viscously damped free vibration of single degree freedom sytem Consider a single degree freedom system with viscous damping as shown in figure For a viscous damped system the damping force is

where c is a constant of proportionality It is called damping coefficient or coefficient of viscous damping

Symbolically it is designated by a dashpot as shown in Fig 3

x xampx

39

At any instant of time let the mass be displaced by x from the equilibrium position Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

Okxxcxm =minusminus= ampampamp

Okxxcxm =++ ampampamp ----------------------------------------------------------(1)

or

0kxdtdx

Cdt

xdM

2

2

=++

Assuming the solution of equation as

The first term in the above equation m2Ct

eminus

is an exponentially decaying function of time The behavior of the terms in the parenthesis however depends on whether numerical value within the radicals is positive zero or negative

CASE 1 if 0mk

m2c

2

=minus

40

CASE 2 if 0mk

m2c

2

=gt

CASE 3 if 0mk

m2c

2

=lt

41

Critically damped coefficient Critically damped coefficient is defined as the value of the damping coeffient for which radical in equation

mk

m2c

m2c

S2

21 minus

plusmn

= becomes zero

Damping Ratio ( Damping factor) ( ξ ) It is defined as the ratio of damping coefficient to the critical damping coefficient Note The damping coefficient c is defined as the actual amount of damping while the damping coefficient is a constant depending upon the mass and stiffness of the system The roots of characteristics equation can be now written as Note if c=cc or ξ =1 then system is critically damped if cgt cc or ξ gt1 then system is over damped I if clt cc or ξ lt1 then system is under damped

42

Take home quiz Derive General solution of different damped system

1 Critically damped system 2 Under damped system 3 Over damped system

Logarithmic Decrement It is used to determine the amount of damping present in system This measure The rate of decay of free vibration The larger the damping the greater will be rate of decay Logarithmic decrement is defined as the natural logarithm of ratio of any two successive amplitudes

=δ

2

1

xx

ln

Expression for logarithmic decrement (δ ) Consider a damped free vibration which is expressed by

( ) ( ) φ+ξminusω= ξω t 1sinXex 2n

tn

x is the displacement at any instant of time X and φ are arbitrary constants

21

2

ξminus

πξ=δ

43

Tutorial problems 1 Derive an expression for damping ratio in terms of logarithmic decrement

2 Show that logarithmic decrement

=δ

n

o

xx

lnn1

where xo is the initial amplitude xn

is amplitude after n cycles 3 A mass of 1 kg is to be supported on a spring having stiffness K= 9800 Nm The damping constant c= 59 Nsm Determine the natural frequency of system Find also logarithmic decrement and amplitude after 3 cycles if initial displacement is 03 cm 4 A vibrating system is defined by the following parameters M = 3kg k= 100 Nm damping coefficient = 3 Nsm determine Damping factor natural frequency damped vibration logarithmic decrement ratio of consecutive amplitude no of cycles after which the original amplitude is reduced to 20 5A single degree damped vibrating system a suspended mass of 18kg makes 15 oscillations in 003 seconds The amplitude decreases to 025 of the initial value after 5 oscillations Determine the stiffness of spring Logarithmic decrement damping factor and damping coefficient 6 A damped vibration record of a spring mass system shows the following data Amplitude at the end of 2nd cycle is 9mm amplitude at eh end of 3rd cycle is 6mm amplitude at the end of 4th cycle is 4mm stiffness of spring =8000 Nm weight =20N Find the logarithmic decrement damping force at unit velocity periodic time of vibration 7 For the system shown in figure the characteristics of dashpot is such that the constant force of 49 N is applied to the piston it velocity is found to be constant at 012 ms Determine the value of C would you expect the complete system to be periodic or aperiodic 8 For a torsional system shown in the figuref1 is natural frequency in air and f2 is the natural frequency when immersed in oil Determine the damping coefficient

Viscous fluid

Disc of mass moment of Inertia J

2a

C

m M

b

a a

k

9 A thin plate of area A and weigh w is fixed to the end of the spring and allow to oscillate in a viscous fluid as shown in figure It f1 is the frequency of oscillation in air and f2 in liquid Show that

22

21 ff

gA2

minusπω

=micro where the damping force on the plate

AV2Fd micro= V being the velocity of vibration 10 A torsional pendulum has a natural frequency of 175 cyclessecond vibrating in vacuum The mass moment of inertia is 2 kgm2 It is immersed in oil and it is observed that the natural frequency is 142 cyclessecond Find the damping torque If the disc is replaced by 34 deg When in oil find the displacement at the end of the first complete cycle 11 An automobile can be modeled as a mass placed on 4 shock absorbers each consisting of a spring and a damper Such that each spring is equally loaded Determine the stiffness damping constant of each shock absorber So that natural frequency is 2Hz and system is critically damped The mass of vehicle is 200kg 12 Write the differential equation of motion for the system shown in the figure below Determine the natural frequency of the damped oscillation and critical damping coefficient

Determine the differential equation of motion and find the critical cdamped natural frequency for system shown in figure

a

C

m M

M =200kg

44

200kg kgkg

oefficient and

45

Derive the equation of motion and find the critical damping coefficient and damped natural frequency for the system shown below Write the differential equation of motion and determine the expression for critical damping coefficient and damped natural frequency of system shown in figure

Set up a differential equation of motion for system shown in figure and determine the following Undamped natural frequencies critical damping coefficient damping ratio damped natural frequency

b

C

kM

a

m

46

A mass of 300N is resting on two springs of stiffness 3000 Nm each and a dashpot of damping coefficient 150 Nsm as shown in figure If the initial velocity of 10cms is given to mass at its equilibrium position what will be the displacement from the equilibrium position at the end of first second Derive the formula used A gun barrel of mass 600kg has a recoil spring of stiffness 294 knm if the barrel recoils 13m on firing determine i) The initial recoil velocity of barrel ii) The critical damping coefficient of dashpot which is engaged at the end of recoil stroke Iii) Time required for the barrel to return at the end of recoil stroke Iv) time required for the barrel to return to a position 5cm from the initial position

W= 3000 N

FIRING

CHAPTER 4

FORCED VIBRATION LEARNING OBJECTIVES

] Introduction to forced free vibration

] Terminologies used in forced vibration

] Concept of magnification factor

] Study of reciprocating and rotating unbalance

] Concept of base excitation and support excitation

] Concept of Transmissibility ratio

] Problems related to forced vibration

The vibration that takes place under the excitation of external forces are called

forced vibration When a system is subjected to harmonic excitation it is forced to vibrate at the same frequency as that of excitation Most of the mechanical systems are prone to steady state forced vibration which cause fatigue failure The vibration problem also becomes very important when any of excitation frequencies coincide with one of natural frequencies of a system Under this condition resonance occurs during which large amplitude of vibrations are observed To avoid resonance external frequency (operating speed) may be changed or properties of the system may be changed to alter the natural frequency In some cases sufficient amount of damping may be provided to avoid large amplitude during resonance Thus the problem of force vibration is very important in mechanical design

FORCED VIBRATIONS OF SINGLE DOF SYSTEMS UNDER HARMONIC EXCITATION

` [ ] tsin FF o ω=

x

sin FF o=

Fo

xampamp

[ ] tω

xamp

47

48

Consider a spring mass system having a viscous damping excited by a harmonic force [ ] tsin FF o ω= as shown in figure The various forces acting on the system is shown in figure Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

O)tsin(Fkxxcxm o =ω+minusminus= ampampamp

)tsin(Fkxxcxm o ω=++ ampampamp ----------------------------------------------------------(1)

or

)tsin(Fkxdtdx

Cdt

xdM o2

2

ω=++

Equation (1) represents the differential equation of motion of the system The solution to this equation consists of two parts namely complementary function and a particular integral

The first of solution is also termed as transient solution of homogeneous solution and can be obtained from the homogeneous differential equation 0kxxcxm =++ ampampamp

The second part of solution particular integral is also termed as steady state solution ad is of form )tsin(Xx φminusω= Where X= steady state amplitude φ is phase difference between external force and displacement

)tsin(Xx φminusω=

The vector representation of the shown

49

The force polygon for the above vector diagram can be written as follows

( )φminusωt O

KX

xm 2ω

Fo

xCω

tω

φ

50

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

is the governing equation

Zero frequency deflection (Xst) It is deflection observed at zero frequency ( when the external frequency 0=ω

( ) kF

001

kF

x o

22

o

st =

+minus=

Magnification factor (M) It is defined as the ratio of steady state amplitude to zero frequency deflection xst

M = arX

X=

A plot of magnification factor versus frequency ratio

damping factor ( )ξ is called frequency plot of a harmon All the frequency terms starts from magnification factor of damping ratio ( )ξ The amplitude of vibration at any on the value of damping The amplitude will be higher for higher damping The amplitude of vibration at reso

amplitude tends to zero as

ωω

n tends to infinity

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

Freq

ζ =

ζ = 1

Magnification Factor M = XXst

kFo

51

ωω

nfor varying values of

ically excite vibrating system

M=1 what ever may be value particular frequency depends

for lesser damping and lesser nance will be very high The

uency Ratio r = (ωωn)

ζ = 0 0

ζ = 025 ζ = 0375 ζ = 0 5

ζ = 0707

ζ = 2

52

Plot of phase angle φ versus frequency ratio

ωω

n is shown in the

figure The phase angle φ is zero for all the values of damping ratio

at

ωω

n=0 The phase

angle φ is 90 for all the values of damping ratio

at

ωω

n=1 ie at

resonance The phase angle between o to 90 deg for all the values of

ωω

n= 0 to 1 and

between 90 deg to 180 deg for all the values of

ωω

ngt1

Class work Derive an expression for frequency at peak amplitude

2np 21 ξminusω=ω=ω is the required equation

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

53

Derive expression amplitude at resonance ( In dimensional form) In dimensional form We know that steady state amplitude

( ) ( )

ω+ωminus=

222

o

Cmk

Fx Equation in dimensional form

We know that at resonance mk

n =ω=ω

( )

ω+

minus

=2

n

2

oR

Cmk

mk

Fx

( )

ω=

2n

oR

C

Fx

( )

ω=

n

oR C

Fx

Class work Derive an expression for amplitude at resonance ( in non dimensional form)

Tutorial Problems on Forced Vibration 1A weight of 60N is suspended by a spring stiffness 12 kNm is forced to vibrate by a harmonic force of 10N assuming viscous damping of 0086 kN-sm Find i) resonance frequency ii) amplitude at resonance iii) phase angle at resonance iv) frequency corresponding to peak amplitude 2 A periodic torque having a maximum value of 05888Nm at a frequency of 4 rads is impressed upon a flywheel suspended from a wire The wheel has mass moment of inertia 012 kgm2 and the wire has stiffness 1176 Nmrad A viscous dashpot applies damping couple of 0784 Nm an angular velocity of 2 rads calculate i)maximum angular displacement from rest position ii) maximum couple applied to dashpot Iii) angle by which the angular displacement lags the torque A mass of 1019 kg is suspended from a spring stiffness 20 Nmm viscous damping causes the amplitude to decrease to 110th of initial value in 4 complete cycle If a periodic force of 20 cos(50 t) is applied to the mass Find the amplitude of forced vibration What would be the amplitude of oscillation if the applied force coincides with the natural period

m=1019 kg

[ ] t50cos 30F =

54

55

ROTATING UNBALANCE Unbalance in rotating machinery is a common source of vibration excitation Consider a machine of mass m having mounted on a foundation of stiffness k and damping coefficient C as shown in figure The unbalance is represented by an eccentric rotating mass m with eccentricity (e) and having a angular velocity (ω) let x be the displacement of non rotating mass from the static equilibrium position The displacement of rotating mass m from the static equilibrium is ( ) tsin( ex ω+ Applying Newtonrsquos law of motion we have

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

56

RECIPROCATING UNBALANCE Unbalance in reciprocating machinery is a common source of vibration excitation Consider a machine of mass M having mounted on a foundation of stiffness K and damping coefficient C as shown in figure The unbalance is represented by a reciprocating mass (m) having crank radus (e) and connecting rod of length (l) Let ωωωω be angular velocity of the crank Let x be the displacement of non reciprocating mass from the static equilibrium position at any instant of time The displacement of reciprocating mass m from the static equilibrium position is

)2sin(sin +++ tle

tex ωωωωωωωω

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

57

Note The equations obtained in reciprocating unbalance is same as the rotating unbalance and having the following differences Rotating Unbalance Reciprocating Unbalance Here mass of the rotating element is considered

Here mass of reciprocating element is considered

Here the eccentricity (e) which is the distance between the centre of rotation and CG

Here the eccentricity e is the crank radius and crank radius = stroke length2

The frequency plot of non dimensional factor( ( )me

MX versus frequency ratio

ωω

n for various values of ( )ξξξξ is shown in figure This frequency plot is same for

the rotating and reciprocating unbalance with the following points are noted from the diagram

1 Damping ratio plays very important role during resonance in controlling the amplitude

2 When

ωω

n=0 then ( )me

MX =0 for all the values of damping ratio ( )ξξξξ

3 When

ωω

nis higher than ( )me

MX tends to unity

4 At very high speeds ie at very high frequency ratio

ωω

n there is no effect

of damping The frequency plot of phase angle versus the frequency ratio is given as shown in figure The explanation is same as in forced harmonic vibration

58

VIBRATION ISOLATION

Vibratory forces and motions generated by machine and other caused are unavoidable Howeever their effect on the other dynamical systems can be minimized by proper isolator design An isolation system is either to protect o delicate machine from excessive vibration transmitted to it from its supporting structure or to prevent vibratory forces and motions generated by machines from being transmitted to its surroundings The effectiveness of isolation may be measured in terms of forces and motion transmitted The first type is known as force isolation and the second type is known as motion isolation

Force isolation (Force Transmissibility) When vibratory forces generated by machine are isolated the effectiveness of forced isolation can be made by defining the term transmissibility The term transmissibility is defined as the ratio of force transmitted to the foundation to that of the external impressed force

Force Transmissibility = o

tr

FF

force Impressed Externalfoundation toted transmitForce

=

Consider viscous damped spring mass system as shown in figure Upon this system an external harmonic force ) t (Sin Fo ω is applied The difference forces acting on system are inertia forces damping force spring force and external force All these forces are represented on force polygon shown below Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these two forces

( ) ( )22 xckx ω+=

59

This is the expression for force transmissibility in dimensional form To represent same in the non dimensional form divide numerator and denominator by k

2

n

22

n

2

n

o

tr

21

21

FF

TR

ωω

ξ+

ωω

minus

ωω

ξ+

==

PHASE DIFFERNECE BETWEEN EXTERNAL FORCE AND DISPLACEMENT Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these

ωω

minus

ωω

ξ=

ωminusω

=φ minusminus2

n

n1

2

1

1

2tan

mKC

tan

The phase difference between force transmitted to the foundation Ftr to the displacement

kc

KxxC

tanω

=ω

=α or

ωω

ξ=

ω=α minusminus

n

11 2tanKx

xCtan

The phase difference between external force Fo and force transmitted to foundationis β

60

β= αminusφ =

ωω

minus

ωω

ξminus

2

n

n1

1

2tan -

ωω

ξminus

n

1 2tan

Displacement Transmissibility It is the ratio of amplitude of steady state vibration of machine to that of the base motion Forced vibration due to support motion Consider a spring mass damper system subjected to harmonic excitation )tsin(Yy ω= at the base as shown in figure Let x be the absolute displacement of the mass at any instant of time t The different forces acting on system are shown in free body diagram Applying Newtonrsquos Law of motion Ma = sumF

)yx(k)yx(cxm minusminusminusminus= ampampampamp

0)yx(k)yx(cxm =minus+minus+ ampampampamp

61

Frequency plot of transmissibility versus frequency ratio

62

The plot of transmissibility versus frequency ratio is as shown in figure The following points can be noted from it

i) The value of transmissibility is unity at frequency ratio

ωω

n=0 and 2

ii) For an undamped system ( )ξ =0 transmissibility tends to infin at resonance iii) The value of transmissibility is less than unity for the value of frequency ratio

greater than 2

iv) For frequency ratio

ωω

nlt 2 smaller damping ratio lead to larger values of

transmissibility on the other hand for frequency ratio

ωω

ngt 2 smaller damping

ratio leads to smaller value of transmissibility

The frequency plot of transmissibility (TR) versus frequency ratio

ωω

n can be

broadly divided into two zones

1

ωω

n= 0 to 2

2

ωω

n= 2 to infin

CASE 1

When

ωω

n= 0 to 2

In this zone the value of transmissibility is always greater than 1 for all values of damping ratio In this zone the transmissibility decreases with increase in damping Hence we can conclude that in this zone damping is useful in isolating vibration forces being transmitted to the support CASE 2

When

ωω

n= 2 to infin

In this zone the value of transmissibility is less than 1 for all values of damping ratio In this zone the transmissibility increases with increase in damping Hence we can conclude that in this zone damping will adversely affect in isolating vibratory forces

63

Tutorial problems on Forced Vibration 1 Machine of mass 75kg is mounted on spring of stiffness 12000 Nm with an assumed damping factor ξ =02 piston within the machine of mass 2 kg has a reciprocating motion with a stroke of 75 cm and speed of 15 cyclessecond Assuming the motion of piston to be harmonic determine amplitude of machine ( X) Transmissibility and force transmitted to foundation phase angle of transmitted force wrt to exciting force 2 Motor weighs 600N is mounted on a simple beam that has a spring rate of 400 Ncm The motor armature base 160 N and has an eccentricity of 0005cm What will be the amplitude of motion when it runs at 1760 RPM Neglect damping the deflection of armature shaft and weight of the beam 3 A Heavy machine of mass 300 kg is supported on resilient foundation the static deflection of foundation due to weight of machine is found to be 75mm It is observed that the machine vibrates with an amplitude of 10mm when the base of the foundation is subjected to harmonic oscillation at the undampeds natural frequency of system with an amplitude of 25mm Find the damping constant of foundation Ii) dynamic force amplitude on the base iii) amplitude of the displacement of machine relative to base 4 An air compressor of mass 450 kg operates at a constant speed of 1750 RPM The reciprocating masses are 10 kg the crank radius is 100mm If the damper for mounting introduces a damping factor 015 i) specify the springs of mounting such that only 20 of the unbalanced force is transmitted to the foundation ii) determine the magnitude of the transmitted force bull Repeat the above problem by assuming that there is no damping in system

5 A body of a car having mass 1500 kg and is mounted on 4 equal springs which deflect through 225 cm under the weight of the body A total damping forces the 4 shock absorbers is 46 N at a velocity of 1cmsec The car is placed with all the four wheels on a test platform which is move up and down at resonant speed With an amplitude of 25cm Find the amplitude of the car body on its springs assuming its CG to be in the centre of wheel base 6A 75 kg machine which is mounted on springs of stiffness 116x105 Nm with the damping of 20=ξ A 2kg piston within the machine as reciprocating motion with a stroke of 008m and speed of 3000 RPM Assuming the motion of piston to be harmonic determine the amplitude of vibration of machine

7 The spring of an automobile are compressed 10cm under its own weight Find the critical speed when a trailer is traveling over road with a people approximaterly by a sine wave of amplitude 75 cm and wave length 15m What will be the amplitude of vibration at 64kmhour 8 An eccentric mass excites us used to determine vibratory characteristics of a structure of mass 25kg as given in figure At a speed of 1000 RPM the eccentric mass to beat top position at the instant the structure was moved upwards through its static equilibrium position and corresponding amplitude was 20mm The eccentric mass is 05kg at an eccentricity of 10mm Determine i) Undamped natural frequency of the system ii) Damping ratio iii) Steady state amplitude and phase angle at 1500 RPM iv) Force transmitted to the ground at 1200 RPM

M=25kg

C

K

Eccentric mass

64

65

CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Support motion Consider a spring mass damper system as shown in figure Let it excited by motion

of support let y be the harmonic displacement of support From Newtonrsquos second law of motion

)yx(k)yx(cxm minusminusminusminus= ampampampamp

)1(0)yx(k)yx(cxm minusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminus=minus+minus+ ampampampamp

Let z be the relative displacement of the mass with respect to support then

Z=x-y

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

DESCRIBING MOTIONS OF VIBRATING SYSTEMS Periodic motions Described as sine or cosine functions [sin (ωt) and cos(ωt)] ω = radian frequency (radsec) ω = 2πf where f is frequency (Hz) Period = time between two adjacent peaks or valleys P = 1f

Simple harmonic motion ( ) ( )φω += tAty sin Displacement ( ) ( )φωω += tAty cosamp Velocity ( ) ( )φωω +minus= tAty sin2amp Acceleration Classification of vibrations One method of classifying mechanical vibrations is based on degreesThe number of degrees of freedom for a system is the number of independent variables necessary to completely describe the motion of ein the system Based on degrees of freedom we can classify mechanicas follows 1Single Degree of freedom Systems 2Two Degrees of freedom Systems 3Multidegree of freedom Systems 4Continuous Systems or systems with infinite degrees of freedom Another broad classification of vibrations is 1 Free and forced vibrations 2 Damped and undamped vibrations

Leads Lags

4

of freedom kinematically very particle al vibrations

5

Sometime vibration problems are classified as 1 Linear vibrations 2 Non-linear vibrations 3 Random vibrations 4 Transient vibrations 5 Longitudinal vibrations 6 Transverse vibrations 7 Torsional vibrations Free vibration If a system after initial disturbance is left to vibrate on its own the resulting vibration is known as free vibrations Free vibration takes when a system vibrates under the action of forces inherent in the system and when the external forces are absent The frequency of free vibration of a system is called natural frequency Natural frequency is a property of a dynamical system Forced vibration Vibration that takes place under the excitation of external forces is called forced vibration the forced vibration takes place at different forced frequencies or external frequencies Damped vibration If any energy is lost or dissipated during oscillations then the vibration is known as damped vibration Undamped vibration if no energy is lost or dissipated during oscillations such vibrations are known as undamped vibration Linear vibration If all the basic component of a vibrating system behave linearly the resulting vibration is known as linear vibration The differential equations govern linear vibratory system are linear If the vibration is linear the principle of superposition holds and mathematical techniques of analysis are well developed Non linear vibration If any of the basic components of a vibrating system behave non linearly the resulting vibration is known as non linear vibration The differential equations that govern non linear vibratory system are non-linear If the vibration is non linear the principle of superposition does not hold good and techniques of analysis is well known Deterministic vibration If the magnitude of excitation on a vibrating system is known at any given time the resulting vibration is known as deterministic vibration Random vibration If the magnitude of excitation acting on a vibratory system at a given time cannot be predicted the resulting vibration is known as non deterministic or random vibration

6

Longitudinal vibration Consider a body of mass m carried on one end of a slender shaft and other end being fixed If the mass vibrates parallel to the spindle axis it is said to be execute longitudinal vibration Transverse vibration If the mass vibrates perpendicular to the spindle axis it is said to execute the transverse vibration Torsional vibration If the shaft gets alternatively twisted and un twisted on account of an alternate torque on the disk it is said to execute the torsional vibration SIMPLE HARMONIC MOTION A Vibration with acceleration proportional to the displacement and directed toward the mean position is known as SHM ( Simple Harmonic Motion) Consider a spring mass system as shown in the figure along with the displacement time diagram The equation of motion of the mass can be written as follows From the right angled triangle OAB we have Where x= displacement at any instant of time X = amplitude of vibration W= angular velocity or frequency in radsecond

m1

7

The velocity of the mass m at an instant of time t is given by V = The acceleration of mass m is given by Hence we can conclude that in SHM the acceleration is proportional to displacement and is directed towards mean position observing the equations 2 and 3 the velocity and acceleration are harmonic with the same frequency but lead a displacement vector by π2 and π radians respectively

t

t

t X

x

X= A sin ωt

x-Displacement X-amplitude T-Periodic Time f-Frequency f=1T ω=Frequency in radians per second t= time

8

Rotor with inertia J1

kt1 kt2

Rotor with Inertia J2

Degrees of freedom The minimum number of independent coordinates required to determine completely the positions of all the parts of a system at any instant of time is called degrees of freedom One degree of freedom Two degree of freedom Three degree of freedom Infinite degree of freedom

m1

m2

m1

m2

m3

Cantilever Beam

Continuous system ( consider the mass of the beam )

m1

m1

K1

m2

K

x1

x2

F1

9

G

K1 K2

mJ

a b

Exercise Specify the no of degree of freedom for the following 1 2 3

4

5

4m 2m m 3k k k

Cantilever Beam

Continuous system ( Neglect the mass of the beam )

k1 k2 m1 m2

y y

m2

θ1

θ2

L1

L2

M1

M2

10

O

A

B

Addition of SIMPLE HARMONIC MOTION ( SHM) The addition of two simple harmonic motion having frequency yields a resultant which is simple harmonic having the same frequency Consider tow simple harmonic motions of 1x and 2x having the same frequency and phase difference φ as given below

)(sin11 tXx ωωωω====

)(sin22 φω += tXx

Adding x = x1 +x2 Hence the resultant displacement is also SHM of amplitude X and phase angle θ

θθθθ

ωωωωt

φφφφ

11

-6

-4

-2

0

2

4

6

0 005 01 015 02 025

Tutorial problems on Simple Harmonic Motion 1 Add the following harmonic motion analytically and verify the solution graphically 1) X1= 3 sin (((( ))))30 t ++++ωωωω X2= 4 cos (((( ))))10 t ++++ωωωω ( VTU Jan 2005) 2) X1= 2 cos (((( ))))05 t ++++ωωωω X2= 5 sin (((( ))))1 t ++++ωωωω ( VTU July 2006)

3) X1= 10 cos (((( ))))4 t ππππωωωω ++++ X2= 8 sin (((( ))))6 t ππππωωωω ++++ ( VTU Dec 2007)

2 A body is subjected to two harmonic motions X1= 8 cos (((( ))))6 t ππππωωωω ++++ X2= 15 sin (((( ))))6 t ππππωωωω ++++ what harmonic motion

should be given to the body to bring it to equilibrium (VTU July 2005) 3 Split the harmonic motion x = 10 sin (((( ))))6 t ππππωωωω ++++ into two harmonic motions

having the phase of zero and the other of 45o 4 Show that resultant motion of harmonic motion given below is zero X1= X sin (((( ))))t ωωωω X2= X sin (((( ))))3

2 t ππππωωωω ++++ X3= X sin (((( ))))34 t ππππωωωω ++++

5 The displacement of a vibrating body is given by x = 5 sin (3141t + 4

ππππ )

Draw the variation of displacement for one cycle of vibration Also determine the displacement of body after 011 second ( repeat the problem for velocity and acceleration and draw graph using Excel and compare ) Time Displacement

0 0025 5 005

0075 0 01

0125 015 -353

0175 0 02

Calculate the remaining values 6 The Motion of a particle is represented by x = 4 ( ) t sin ω sketch the variation of the displacement velocity and acceleration and determine the max value of these quantities Assume ( )5 =ω ( Try to use MATLABExcel) sketch all on same graph

12

BEATS When two harmonic motions with frequencies close to one another are added The resulting motion exhibits a phenomenon known as Beats A Beat Frequency is the result of two closely spaced frequencies going into and out of synchronization with one another Let us consider tow harmonic motion of same amplitude and slightly different frequencies X1 = X Cos ( ) tωωωω X2 = X Cos ( )( ) tδδδδωωωω + Where δ is a small quantity The addition of above two harmonics can be written as X = X1 + X2

X=

+

ttX2

cos2

cos2δδδδωωωω

δδδδ

The above equation shown graphically in Figure The resulting motion represents cosine wave with frequency ( ) 2

ππππωωωω + and with a varying amplitude 2 cos( ( ) 2δδδδ t

whenever the amplitude reaches a maximum it is called the beat The frequency δ at which amplitude builds up and dies down between o and 2 X is known as beat frequency Beat phenomenon is found in machines structures and electric power houses In machines the beat phenomenon occurs when the forcing frequency is close to the natural frequency of the system

+2X

-2X

ωωωωππππ2

ELEMENTS OF VIBRATION The elements of constitute vibrating systems are

1 Mass or Inertia element - m 2 Spring - k 3 Damper - c 4 Excitation F(t)

Mass or Inertia element The mass or inertia element is assumed to be rigid body During vibration velomass changes hence kinetic energy can be gained or loosed The force appthe mass from newton second law of motion can be written as F= ma Work dthe mass is stored in the form of kinetic energy given by KE= frac12 M V2

Combination of masses In practical cases for simple analysis we replace several masses by equivalent mass Case1 Translational masses connected by rigid bar Let the masses M1 M2 and M3 are attached to a rigid bar at locates 1 respectively as shown in the figure The equivalent mass meq be assumelocated at 1 is as shown in figure (b) Let the displacement of masses M1 M2 and M3 be x1 x2 x3 and simivelocities of respective masses be x1 x2 and x3 We can express the velomasses m2 and m3 in terms of m1

Elements of Vibration

Passive element

m C K

Active element

F(t)

Conservative element

( Mass and Spring)

Non conservative element

( Damper)

C k

m F(t)

Voig

13

city of lied on one on

a single

2 and 3 d to be

larly the cities of

t Model

Translationa Let a mass mmass momearrangement These two mamass Meq or

2

1

33

2

1

221

+

+=

ll

Mll

MMMeq

14

is the required answer

l and rotational masses coupled together

having a translational velocity x be coupled to another mass having nt of inertia Io with a rotational velocity θ as in rack and pinion

shown in the figure

sses can be combined to obtain either a single equivalent translational a single equivalent mass moment of inertia Jeq

15

Equivalent translational masses

Kinetic energy of the equivalent mass =

2

21

eqeq XM amp

Kinetic energy of the two masses =

+

20

2

21

21

θθθθampamp JXM

Meq= m

+ 2RJ

m o is the required answer

Also determine the equivalent rotational mass Jeq

Jeq= m [ ]oJmR +2 is the required answer Spring element Whenever there is a relative motion between the two ends of the spring a force is developed called spring force or restoring force The spring force is proportional to the amount of deformation x and then F α x or F = kx Where k is stiffness of the spring or spring constant or spring gradient The spring stiffness is equal to spring force per unit deromation

The spring stiffness k = mNxF

Workdone in deforming a spring is equal to the strain energy or potential energy Strain energy = potential energy = area of the triangle OAB Stiffness of beams Cantilever beam consider a cantilever beam with an end mass shown in the figure The mass of the beam is assumed to be negligible The static deflection of beam at free end is given by

Similarly derive the expression for Simply supported beam and fixed support beam

x

k

m

x

m

x

m

x

mNlIE

st 192

3

=δδδδ

mNEI

Wlst

48

3

=δδδδ

F

K=stiffness

F

16

17

Stiffness of slender bar subjected to longitudinal vibrations For a system executing the longitudinal vibrations as shown in the figure let us derive the expression for stiffness

Torsional Stiffness of bar It is the amount of torque required to cause a unit angular deformation in the element

Torsional stiffness = Kt = θθθθT

Combination of stiffness Determination of equivalent spring stiffness when the springs are arranged in series Consider two springs of stiffness K1 and K2 acted upon by the force F

The deflection of spring k1 is x1 = 1KF

The deflection of spring k1 is x2 = Let these two springs be replaced by an equivalent stiffness Keq upon which a force F acts and due to which its deflection is given by

x = eqKF

x= x1+x2

m

E A

l

60 cm 70 cm

Determination of equivalent spring stiffness when the springs are arranged in parallel Force acting on K1 spring = F1=k1x Force acting on K2 spring = F2 Force required for an equivalent spring keq to defined by x given by F= Keq x But F = F1 +F2 Tutorial problems on Equivalent stiffness of springs 1Determine the equivalent stiffness for the system shown in figure 2 Determine the equivalent stiffness for the system shown in figure

3 Determine the equivale

M

3

M

Nm 102 6x

18

nt stiffness for the system shown in figure

38 kg

Nm 101 6x Nm 102 6x

Nm 106x

2k

k

k

2k

3k

k

2k

k

19

60 cm 80 cm 50 cm

50 cm 60 cm

4Determine the equivalent stiffness for the system shown in figure

5Replace the following torsional stiffness by a single shaft having radius 4cm and find the length required for the equivalent shaft Assume the material of given system and equivalent system is same R1= 3cm R2= 5cm DAMPING Every vibration energy is gradually converted into heat or sound Hence the displacement during vibration gradually reduces The mechanism by which vibration energy is gradually converted into heat or sound is known as damping A damper is assumed to have either mass or elasticity Hence damping is modeled as one or more of the following types Viscous damping Coulomb or dry friction damping materials or solid or hysteric damping Viscous damping Viscous damping is most commonly used damping mechanism in vibration analysis When the mechanical system vibrates in a fluid medium such as air gas water or oil the resistance offered by the fluid to the moving body causes energy to be dissipated In this case the amount of dissipated energy depends on many factors such as size or shape of the vibrating body the viscosity of the fluid the frequency of vibration and velocity of fluid Resistance due to viscous damping is directly proportional to the velocity of vibration

dF V αααα

dF = xC amp

Where C= damping coefficient Fd = damping force Examples of Viscous damping

1) Fluid film between sliding surface 2) Fluid flow around a piston in a cylinder 3) Fluid flow through orifice 4) Fluid flow around a journal in a bearing

Rreq =4cm

leqn

stress

Strain

Coulomb damping or dry friction damping Here a damping force is constant in magnitude but opposite in direction to that of the motion of vibrating body It is caused by the friction between the surfaces that are dry or have insufficient lubrication Material or solid or hysteric damping

When the materials are deformed energy is absorbed and dissipated by the material The effect is due to friction between the internal planes which slip or slide as the deformation takes place When a body having the material damping is subjected to vibration the stress strain diagram shows the hysteresis loop as shown in the figure The area of the loop denotes the energy lost per unit volume of the body per cycle

20

21

ττττ ττττ2 ττττ3

ττττ

X(t)

FOURIER SERIES The simplest of periodic motion happens to be SHM It is simple to handle but the motion of many vibrating system is not harmonic (but periodic) Few examples are shown below

Forces acting on machines are generally periodic but this may not be harmonic for example the excitation force in a punching machine is periodic and it can be represented as shown in figure 13 Vibration analysis of system subjected to periodic but nonharmonic forces can be done with the help of Fourier series The problem becomes a multifrequency excitation problem The principle of linear superposition is applied and the total response is the sum of the response due to each of the individual frequency term

Any periodic motion can be expressed as an infinite sum of sines and cosines terms If x(t) is a periodic function with period t its Fourier representation is given by

X(t) = (((( )))) (((( )))) (((( ))))t sint cost cos2 111 ωωωωωωωωωωωω baaao ++++++++++++

= (((( )))) (((( ))))t sint cos2 1

ωωωωωωωω nnn

o bnaa

++++++++sumsumsumsuminfininfininfininfin

====

tππππωωωω

2==== = Fundamental frequency ndash (1)

where ao an bn are constants

ττττ ττττ2 ττττ3

ττττ

X(t)

22

t

X(t)

035

025

Determination of constants To find ao Integrate both sides of equation(1) over any interval ττττ All intergrals on the RHS of the equation are zero except the one containing ao

(((( ))))dttxao

o 2

2

intintintint====ωωωωππππ

ππππωωωω

= (((( ))))dttxo

2intintintintττττ

ττττ

To find an multiply equation 1 by cos (((( ))))tnωωωω and Integrate over any interval ττττ All intergrals

(((( )))) (((( )))) dttxao

n tncos 2

2

ωωωωππππωωωω ωωωω

ππππ

intintintint==== (((( )))) (((( )))) dttxo

tncos 2

ωωωωττττ

ττττ

intintintint====

To find bn multiply equation 1 by sin (((( ))))tnωωωω and Integrate over any interval ττττ All intergrals

(((( )))) (((( )))) dttxao

n tnsin 2

2

ωωωωππππωωωω ωωωω

ππππ

intintintint==== (((( )))) (((( )))) dttxo

tnsin 2

ωωωωττττ

ττττ

intintintint====

Find the Fourier series for the curve shown below

ττττ ττττ2 ττττ3

t

X(t)

E

ττττ ττττ2 ττττ3

t

X(t)

ππππ2

t

X(t) ππππ

1 Represent for the periodic motion shown in the figure

23

CHAPTER 2 UNDAMPED FREE VIBRATION LEARNING OBJECTIVES

] Introduction to undamped free vibration

] Terminologies used in undamped free vibration

] Three methods to solve the undamped free vibration problems

] Problems related to undamped free vibration problems

Free vibrations are oscillations about a systems equilibrium position that occur in the absence of an external excitation force If during vibrations there is no loss of energy it is known as undamped vibration The first step in solving a vibration problem is setting up the differential equation of motion Consider spring mass system which is assumed to move only along the vertical direction as shown below Let m be the mass of the block and k be the stiffness of the spring When block of mass m is attached to spring the deflection of spring will be ∆ known as static deflection In the static equilibrium position the free body diagram of forces acting on the mass is shown in Figure(b) Hence mg= kA Once the system is disturbed the system executes vibrations

Let at any instant of time t the mass is displaced from the equilibrium position x the different forces acting on the system are shown in figure (d) From Newtonrsquos second law of motion

sum = maF

Inertia force ( disturbing force) = restoring force

mgxkxm ++∆minus= )(ampamp

0)( =+ xkxm ampamp

or 0)( =+ xmk

xampamp

equation 2 is the differential equation of motion for spring mfigure Comparing equation (2) with the equation of SHM +xampampsince the vibrations of the above system are free( without the forces) we can write

sec radmk

n =ω

24

ass system shown in 0)(2 =xω

resistance of external

time period km

2 f1

n

πτ ==

from the equation(1) mg = k∆

∆=

q

mk

Difference between the translation ( rectilinear) and rotational system of vibration

Translatory Rotational

In the analysis thFORCES are co Linear stiffness K Problems 1A mass of 10kFind the natural 2 A spring massnatural frequencnatural frequenc

kt

m

sec radmk

n =ω

k

25

e disturbing and restoring nsidered

In the analysis In the analysis MASS Moment of Inertia (J) is considered

in Nm is considered

g when suspended from a spring causes a static deflection of 1cm frequency of system

system has a spring stiffness K Nm and a mass of m Kg It has a y of vibration 12 Hz An extra 2kg mass coupled to it then the y reduces by 2 Hz find K and m

secrad n =ω

26

3 A steel wire of 2mm diameter and 30m long It is fixed at the upper end and carries a mass of m kg at its free end Find m so that the frequency of longitudinal vibration is 4 Hz 4 A spring mass system has a natural period of 02 seconds What will be the new period if the spring constant is 1) increased by 50 2) decreased by 50 5 A spring mass system has a natural frequency of 10 Hz when the spring constant is reduced by 800 Nm the frequency is altered by 45 Find the mass and spring constant of the original system 6 Determine the natural frequency of system shown in fig is which shaft is supported in SHORT bearings 7 Determine the natural frequency of system shown in fig is which shaft is supported in LONG bearings where l is the length of bearing and E ndash youngrsquos modulus and I is moment of Inertia 7 Determine the natural frequency of system shown in fig is which shaft is supported in LONG bearings 8 A light cantilever beam of rectangular section( 5 cm deep x 25cm wide) has a mass fixed at its free end Find the ratio of frequency of free lateral vibration in vertical plane to that in horizontal 9 Determine the natural frequency of simple pendulum 10 Homogeneous square plate of size l and mass m is suspended from the mid point of one of its sides as shown in figure Find the frequency of vibration

l

l

27

11 A compound pendulum which is rigid body of mass m and it is pivoted at O The point of pivot is at distance d from the centre of gravity It is free to rotate about its axis Find the frequency of oscillation of such pendulum 12 A connecting rod shown in fig is supported at the wrist pin end It is displaced and allowed to oscillate The mass of rod is 5kg and centre of gravity is 20 cm from the pivot point O If the frequency of oscillation is 40 cyclesminute calculate the mass moment of inertia about its CG

13 A semi circular homogenous disc of radius r and mass m is pivoted freely about its centre as shown in figure Determine the natural frequency of oscillation 14A simply supported beam of square cross section 5mmx5mm and length 1m carrying a mass of 0575 kg at the middle is found to have natural frequency of 30 radsec Determine youngrsquos modulus of elasticity of beam 15 A spring mass system k1 and m have a natural frequency f1 Determine the value of k2 of another spring in terms of k1 which when placed in series with k1 lowers the natural frequency to 23 f1

COMPLETE SOLUTION OF SYSTEM EXECUTING SHM

The equation of motion of system executing SHM can be represented by

0)( =+ xkxm ampamp ----------------------------------------(1)

022

2=+

dtdx

dt

dx ω

The general solution of equation (1) can be expressed as

------------------------(2)

Where A and B are arbitrary constant which can be determined from the initial conditions of the system Two initial conditions are to be specified to evaluate these

constants x=x0 at t=0 and xamp = Vo at t=0 substituting in the equation (2) Energy method In a conservative system the total energy is constant The dwell as natural frequency can be determined by the princienergy For free vibration of undamped system at any instant and partly potential The kinetic energy T is stored in the masswhere as the potential energy U is stored in the form of sdeformation or work done in a force field such as gravity The total energy being constant T+U = constant Its rate of cha

t)(sin Bt)cos( AX ω+ω=

t)(sin V

t)cos( xx o0 ω

ω+ω=

Is the required complete

solution

28

ifferential equation as ple of conservation of of time is partly kinetic by virtue of its velocity train energy in elastic

nge

Is given by [ ] 0UTdtd

=+

From this we get a differential equation of motion as well as natural frequency of the system Determine the natural frequency of spring mass system using energy method

Determine the natural frequency of system shown in figure Determine the natural frequency of the system shown in figure Is there any limitation on the value of K Discuss Determine the natural frequency of s

m

l k

m

a

θ

m

a

θ

l

k

29

k

k

ystem shown below Neglect the mass of ball

l

m

30

A string shown in figure is under tension T which can be assumed to remain constant for small displacements Find the natural frequency of vertical vibrations of spring An acrobat 120kg walks on a tight rope as shown in figure The frequency of vibration in the given position is vertical direction is 30 rads Find the tension in the rope A manometer has a uniform bore of cross section area A If the column of liquid of length L and Density ρ is set into motion as shown in figure Find the frequency of the resulting oscillation

m

l

a

T T

36m 8m

31

k

R

k

m

r

m

Find the expression for natural frequency of system shown in the figure Neglect the mass of the cantilever beam Study the special case i) k=Infinity ii) I = infinity Determine the expression for the natural frequency of a system shown in figure The two discs are keyed to a common shaft and have a combined mass moment of inertia about centre of oscillation O It is assumed that the card attached to mass m does not stretch and is always under tension Determine the natural frequency of a system shown

m

l

32

Determine the expression for the natural frequency of the system shown in figure Assume that the wires connecting the masses do not stretch and are always in tension

Determine the natural frequency of spring mass system taking the MASS OF SPRING (ms ) into account

M3 M1

M2

k2 k1

m

k

33

RAYLEIGHrsquoS METHOD

This is the extension of energy method Here we can determine the natural frequency of a conservative system without finding the equation of motion The natural frequency is a function of the rate of change of Kinetic energy and potential energy of the system From the principle of conservation of energy we have T+U = Constant T= KE and U =PE We can write the above equation as T1+U1 = --------------------------(1) Where 1 and 2 represents tow instances of time Let 1 be the time when the mass is passing through static equilibrium position Let 2 be the time corresponding to the mass displacement of the mass At this instant the velocity f the mass is zero and hence Substituting in equation (1) we get ( )Max = ( )Max Above equation leads directly to natural frequency Determine the natural frequency of spring mass system using Rayleighrsquos method m

34

r

B

K K

Home work

Determine the natural frequency of pendulum using Rayleighrsquos method

A Cylinder of mass m and moment of inertia Jo is to role without slipping but is sustained by spring K as shown in figure Determine the natural frequency of oscillation

Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping

a

Determine the natural frequency of system shown where in cylindrical disc rolls over the inextensible spring A cylinder of radius r rolls without slipping on cylindrical the equation of motion and hence determine the naoscillations about the lowest point shown Repeat the above problem ndash Instead of cylinder assume swithout slipping on concave surface as shown above

r

K

35

portion of radius R Derive tural frequency of small

phere of radius with r rolls

The torsional pendulum has a natural frequency of 5 Hz What length of steel wire of diameter 2 mm should be used for pendulum The inertia of mass fixed at the free end is 00098 kgm2 Take g= 083x1011 Nm2 Determine the torsional natural frequency of the system shown in figure Neglect the mass moment of inertia of the shaft ( Figure not proportionate) Determine the natural frequency of simple pendulum considering the mass of the rod into account

Using the energy method find tThe chord may be assumed ine

Kt1

Kt2

20cm

06m 08m 04m

20cm 20cm 20cm

m

M

he natural frxtensible in

k

M

36

equency of the system shown in figure the spring mass pulley system

A rod of length l and depth H having a mass m rests on a semicylidrical surface os radius r It is tipped slightly and let go Find the frequency of oscillation

A thin bar of l having a mass m rests on a semicylidrical surface radius r It is tipped slightly and let go Find the frequency of oscillation

l

h

R

l

37

38

CHAPTER 3

DAMPED FREE VIBRATION OF SINGLE DEGREE FREEDOM SYSTEM LEARNING OBJECTIVES

] Introduction to damped free vibration

] Terminologies used in damped free vibration

] Three different types of damping

] Concept of critical damping and its importance

] Study of response of viscous damped system for cases of under damping

critical and over damping

] Concept of logarithmic decrement

] Problems related to damped free vibration problems

Viscously damped free vibration of single degree freedom sytem Consider a single degree freedom system with viscous damping as shown in figure For a viscous damped system the damping force is

where c is a constant of proportionality It is called damping coefficient or coefficient of viscous damping

Symbolically it is designated by a dashpot as shown in Fig 3

x xampx

39

At any instant of time let the mass be displaced by x from the equilibrium position Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

Okxxcxm =minusminus= ampampamp

Okxxcxm =++ ampampamp ----------------------------------------------------------(1)

or

0kxdtdx

Cdt

xdM

2

2

=++

Assuming the solution of equation as

The first term in the above equation m2Ct

eminus

is an exponentially decaying function of time The behavior of the terms in the parenthesis however depends on whether numerical value within the radicals is positive zero or negative

CASE 1 if 0mk

m2c

2

=minus

40

CASE 2 if 0mk

m2c

2

=gt

CASE 3 if 0mk

m2c

2

=lt

41

Critically damped coefficient Critically damped coefficient is defined as the value of the damping coeffient for which radical in equation

mk

m2c

m2c

S2

21 minus

plusmn

= becomes zero

Damping Ratio ( Damping factor) ( ξ ) It is defined as the ratio of damping coefficient to the critical damping coefficient Note The damping coefficient c is defined as the actual amount of damping while the damping coefficient is a constant depending upon the mass and stiffness of the system The roots of characteristics equation can be now written as Note if c=cc or ξ =1 then system is critically damped if cgt cc or ξ gt1 then system is over damped I if clt cc or ξ lt1 then system is under damped

42

Take home quiz Derive General solution of different damped system

1 Critically damped system 2 Under damped system 3 Over damped system

Logarithmic Decrement It is used to determine the amount of damping present in system This measure The rate of decay of free vibration The larger the damping the greater will be rate of decay Logarithmic decrement is defined as the natural logarithm of ratio of any two successive amplitudes

=δ

2

1

xx

ln

Expression for logarithmic decrement (δ ) Consider a damped free vibration which is expressed by

( ) ( ) φ+ξminusω= ξω t 1sinXex 2n

tn

x is the displacement at any instant of time X and φ are arbitrary constants

21

2

ξminus

πξ=δ

43

Tutorial problems 1 Derive an expression for damping ratio in terms of logarithmic decrement

2 Show that logarithmic decrement

=δ

n

o

xx

lnn1

where xo is the initial amplitude xn

is amplitude after n cycles 3 A mass of 1 kg is to be supported on a spring having stiffness K= 9800 Nm The damping constant c= 59 Nsm Determine the natural frequency of system Find also logarithmic decrement and amplitude after 3 cycles if initial displacement is 03 cm 4 A vibrating system is defined by the following parameters M = 3kg k= 100 Nm damping coefficient = 3 Nsm determine Damping factor natural frequency damped vibration logarithmic decrement ratio of consecutive amplitude no of cycles after which the original amplitude is reduced to 20 5A single degree damped vibrating system a suspended mass of 18kg makes 15 oscillations in 003 seconds The amplitude decreases to 025 of the initial value after 5 oscillations Determine the stiffness of spring Logarithmic decrement damping factor and damping coefficient 6 A damped vibration record of a spring mass system shows the following data Amplitude at the end of 2nd cycle is 9mm amplitude at eh end of 3rd cycle is 6mm amplitude at the end of 4th cycle is 4mm stiffness of spring =8000 Nm weight =20N Find the logarithmic decrement damping force at unit velocity periodic time of vibration 7 For the system shown in figure the characteristics of dashpot is such that the constant force of 49 N is applied to the piston it velocity is found to be constant at 012 ms Determine the value of C would you expect the complete system to be periodic or aperiodic 8 For a torsional system shown in the figuref1 is natural frequency in air and f2 is the natural frequency when immersed in oil Determine the damping coefficient

Viscous fluid

Disc of mass moment of Inertia J

2a

C

m M

b

a a

k

9 A thin plate of area A and weigh w is fixed to the end of the spring and allow to oscillate in a viscous fluid as shown in figure It f1 is the frequency of oscillation in air and f2 in liquid Show that

22

21 ff

gA2

minusπω

=micro where the damping force on the plate

AV2Fd micro= V being the velocity of vibration 10 A torsional pendulum has a natural frequency of 175 cyclessecond vibrating in vacuum The mass moment of inertia is 2 kgm2 It is immersed in oil and it is observed that the natural frequency is 142 cyclessecond Find the damping torque If the disc is replaced by 34 deg When in oil find the displacement at the end of the first complete cycle 11 An automobile can be modeled as a mass placed on 4 shock absorbers each consisting of a spring and a damper Such that each spring is equally loaded Determine the stiffness damping constant of each shock absorber So that natural frequency is 2Hz and system is critically damped The mass of vehicle is 200kg 12 Write the differential equation of motion for the system shown in the figure below Determine the natural frequency of the damped oscillation and critical damping coefficient

Determine the differential equation of motion and find the critical cdamped natural frequency for system shown in figure

a

C

m M

M =200kg

44

200kg kgkg

oefficient and

45

Derive the equation of motion and find the critical damping coefficient and damped natural frequency for the system shown below Write the differential equation of motion and determine the expression for critical damping coefficient and damped natural frequency of system shown in figure

Set up a differential equation of motion for system shown in figure and determine the following Undamped natural frequencies critical damping coefficient damping ratio damped natural frequency

b

C

kM

a

m

46

A mass of 300N is resting on two springs of stiffness 3000 Nm each and a dashpot of damping coefficient 150 Nsm as shown in figure If the initial velocity of 10cms is given to mass at its equilibrium position what will be the displacement from the equilibrium position at the end of first second Derive the formula used A gun barrel of mass 600kg has a recoil spring of stiffness 294 knm if the barrel recoils 13m on firing determine i) The initial recoil velocity of barrel ii) The critical damping coefficient of dashpot which is engaged at the end of recoil stroke Iii) Time required for the barrel to return at the end of recoil stroke Iv) time required for the barrel to return to a position 5cm from the initial position

W= 3000 N

FIRING

CHAPTER 4

FORCED VIBRATION LEARNING OBJECTIVES

] Introduction to forced free vibration

] Terminologies used in forced vibration

] Concept of magnification factor

] Study of reciprocating and rotating unbalance

] Concept of base excitation and support excitation

] Concept of Transmissibility ratio

] Problems related to forced vibration

The vibration that takes place under the excitation of external forces are called

forced vibration When a system is subjected to harmonic excitation it is forced to vibrate at the same frequency as that of excitation Most of the mechanical systems are prone to steady state forced vibration which cause fatigue failure The vibration problem also becomes very important when any of excitation frequencies coincide with one of natural frequencies of a system Under this condition resonance occurs during which large amplitude of vibrations are observed To avoid resonance external frequency (operating speed) may be changed or properties of the system may be changed to alter the natural frequency In some cases sufficient amount of damping may be provided to avoid large amplitude during resonance Thus the problem of force vibration is very important in mechanical design

FORCED VIBRATIONS OF SINGLE DOF SYSTEMS UNDER HARMONIC EXCITATION

` [ ] tsin FF o ω=

x

sin FF o=

Fo

xampamp

[ ] tω

xamp

47

48

Consider a spring mass system having a viscous damping excited by a harmonic force [ ] tsin FF o ω= as shown in figure The various forces acting on the system is shown in figure Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

O)tsin(Fkxxcxm o =ω+minusminus= ampampamp

)tsin(Fkxxcxm o ω=++ ampampamp ----------------------------------------------------------(1)

or

)tsin(Fkxdtdx

Cdt

xdM o2

2

ω=++

Equation (1) represents the differential equation of motion of the system The solution to this equation consists of two parts namely complementary function and a particular integral

The first of solution is also termed as transient solution of homogeneous solution and can be obtained from the homogeneous differential equation 0kxxcxm =++ ampampamp

The second part of solution particular integral is also termed as steady state solution ad is of form )tsin(Xx φminusω= Where X= steady state amplitude φ is phase difference between external force and displacement

)tsin(Xx φminusω=

The vector representation of the shown

49

The force polygon for the above vector diagram can be written as follows

( )φminusωt O

KX

xm 2ω

Fo

xCω

tω

φ

50

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

is the governing equation

Zero frequency deflection (Xst) It is deflection observed at zero frequency ( when the external frequency 0=ω

( ) kF

001

kF

x o

22

o

st =

+minus=

Magnification factor (M) It is defined as the ratio of steady state amplitude to zero frequency deflection xst

M = arX

X=

A plot of magnification factor versus frequency ratio

damping factor ( )ξ is called frequency plot of a harmon All the frequency terms starts from magnification factor of damping ratio ( )ξ The amplitude of vibration at any on the value of damping The amplitude will be higher for higher damping The amplitude of vibration at reso

amplitude tends to zero as

ωω

n tends to infinity

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

Freq

ζ =

ζ = 1

Magnification Factor M = XXst

kFo

51

ωω

nfor varying values of

ically excite vibrating system

M=1 what ever may be value particular frequency depends

for lesser damping and lesser nance will be very high The

uency Ratio r = (ωωn)

ζ = 0 0

ζ = 025 ζ = 0375 ζ = 0 5

ζ = 0707

ζ = 2

52

Plot of phase angle φ versus frequency ratio

ωω

n is shown in the

figure The phase angle φ is zero for all the values of damping ratio

at

ωω

n=0 The phase

angle φ is 90 for all the values of damping ratio

at

ωω

n=1 ie at

resonance The phase angle between o to 90 deg for all the values of

ωω

n= 0 to 1 and

between 90 deg to 180 deg for all the values of

ωω

ngt1

Class work Derive an expression for frequency at peak amplitude

2np 21 ξminusω=ω=ω is the required equation

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

53

Derive expression amplitude at resonance ( In dimensional form) In dimensional form We know that steady state amplitude

( ) ( )

ω+ωminus=

222

o

Cmk

Fx Equation in dimensional form

We know that at resonance mk

n =ω=ω

( )

ω+

minus

=2

n

2

oR

Cmk

mk

Fx

( )

ω=

2n

oR

C

Fx

( )

ω=

n

oR C

Fx

Class work Derive an expression for amplitude at resonance ( in non dimensional form)

Tutorial Problems on Forced Vibration 1A weight of 60N is suspended by a spring stiffness 12 kNm is forced to vibrate by a harmonic force of 10N assuming viscous damping of 0086 kN-sm Find i) resonance frequency ii) amplitude at resonance iii) phase angle at resonance iv) frequency corresponding to peak amplitude 2 A periodic torque having a maximum value of 05888Nm at a frequency of 4 rads is impressed upon a flywheel suspended from a wire The wheel has mass moment of inertia 012 kgm2 and the wire has stiffness 1176 Nmrad A viscous dashpot applies damping couple of 0784 Nm an angular velocity of 2 rads calculate i)maximum angular displacement from rest position ii) maximum couple applied to dashpot Iii) angle by which the angular displacement lags the torque A mass of 1019 kg is suspended from a spring stiffness 20 Nmm viscous damping causes the amplitude to decrease to 110th of initial value in 4 complete cycle If a periodic force of 20 cos(50 t) is applied to the mass Find the amplitude of forced vibration What would be the amplitude of oscillation if the applied force coincides with the natural period

m=1019 kg

[ ] t50cos 30F =

54

55

ROTATING UNBALANCE Unbalance in rotating machinery is a common source of vibration excitation Consider a machine of mass m having mounted on a foundation of stiffness k and damping coefficient C as shown in figure The unbalance is represented by an eccentric rotating mass m with eccentricity (e) and having a angular velocity (ω) let x be the displacement of non rotating mass from the static equilibrium position The displacement of rotating mass m from the static equilibrium is ( ) tsin( ex ω+ Applying Newtonrsquos law of motion we have

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

56

RECIPROCATING UNBALANCE Unbalance in reciprocating machinery is a common source of vibration excitation Consider a machine of mass M having mounted on a foundation of stiffness K and damping coefficient C as shown in figure The unbalance is represented by a reciprocating mass (m) having crank radus (e) and connecting rod of length (l) Let ωωωω be angular velocity of the crank Let x be the displacement of non reciprocating mass from the static equilibrium position at any instant of time The displacement of reciprocating mass m from the static equilibrium position is

)2sin(sin +++ tle

tex ωωωωωωωω

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

57

Note The equations obtained in reciprocating unbalance is same as the rotating unbalance and having the following differences Rotating Unbalance Reciprocating Unbalance Here mass of the rotating element is considered

Here mass of reciprocating element is considered

Here the eccentricity (e) which is the distance between the centre of rotation and CG

Here the eccentricity e is the crank radius and crank radius = stroke length2

The frequency plot of non dimensional factor( ( )me

MX versus frequency ratio

ωω

n for various values of ( )ξξξξ is shown in figure This frequency plot is same for

the rotating and reciprocating unbalance with the following points are noted from the diagram

1 Damping ratio plays very important role during resonance in controlling the amplitude

2 When

ωω

n=0 then ( )me

MX =0 for all the values of damping ratio ( )ξξξξ

3 When

ωω

nis higher than ( )me

MX tends to unity

4 At very high speeds ie at very high frequency ratio

ωω

n there is no effect

of damping The frequency plot of phase angle versus the frequency ratio is given as shown in figure The explanation is same as in forced harmonic vibration

58

VIBRATION ISOLATION

Vibratory forces and motions generated by machine and other caused are unavoidable Howeever their effect on the other dynamical systems can be minimized by proper isolator design An isolation system is either to protect o delicate machine from excessive vibration transmitted to it from its supporting structure or to prevent vibratory forces and motions generated by machines from being transmitted to its surroundings The effectiveness of isolation may be measured in terms of forces and motion transmitted The first type is known as force isolation and the second type is known as motion isolation

Force isolation (Force Transmissibility) When vibratory forces generated by machine are isolated the effectiveness of forced isolation can be made by defining the term transmissibility The term transmissibility is defined as the ratio of force transmitted to the foundation to that of the external impressed force

Force Transmissibility = o

tr

FF

force Impressed Externalfoundation toted transmitForce

=

Consider viscous damped spring mass system as shown in figure Upon this system an external harmonic force ) t (Sin Fo ω is applied The difference forces acting on system are inertia forces damping force spring force and external force All these forces are represented on force polygon shown below Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these two forces

( ) ( )22 xckx ω+=

59

This is the expression for force transmissibility in dimensional form To represent same in the non dimensional form divide numerator and denominator by k

2

n

22

n

2

n

o

tr

21

21

FF

TR

ωω

ξ+

ωω

minus

ωω

ξ+

==

PHASE DIFFERNECE BETWEEN EXTERNAL FORCE AND DISPLACEMENT Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these

ωω

minus

ωω

ξ=

ωminusω

=φ minusminus2

n

n1

2

1

1

2tan

mKC

tan

The phase difference between force transmitted to the foundation Ftr to the displacement

kc

KxxC

tanω

=ω

=α or

ωω

ξ=

ω=α minusminus

n

11 2tanKx

xCtan

The phase difference between external force Fo and force transmitted to foundationis β

60

β= αminusφ =

ωω

minus

ωω

ξminus

2

n

n1

1

2tan -

ωω

ξminus

n

1 2tan

Displacement Transmissibility It is the ratio of amplitude of steady state vibration of machine to that of the base motion Forced vibration due to support motion Consider a spring mass damper system subjected to harmonic excitation )tsin(Yy ω= at the base as shown in figure Let x be the absolute displacement of the mass at any instant of time t The different forces acting on system are shown in free body diagram Applying Newtonrsquos Law of motion Ma = sumF

)yx(k)yx(cxm minusminusminusminus= ampampampamp

0)yx(k)yx(cxm =minus+minus+ ampampampamp

61

Frequency plot of transmissibility versus frequency ratio

62

The plot of transmissibility versus frequency ratio is as shown in figure The following points can be noted from it

i) The value of transmissibility is unity at frequency ratio

ωω

n=0 and 2

ii) For an undamped system ( )ξ =0 transmissibility tends to infin at resonance iii) The value of transmissibility is less than unity for the value of frequency ratio

greater than 2

iv) For frequency ratio

ωω

nlt 2 smaller damping ratio lead to larger values of

transmissibility on the other hand for frequency ratio

ωω

ngt 2 smaller damping

ratio leads to smaller value of transmissibility

The frequency plot of transmissibility (TR) versus frequency ratio

ωω

n can be

broadly divided into two zones

1

ωω

n= 0 to 2

2

ωω

n= 2 to infin

CASE 1

When

ωω

n= 0 to 2

In this zone the value of transmissibility is always greater than 1 for all values of damping ratio In this zone the transmissibility decreases with increase in damping Hence we can conclude that in this zone damping is useful in isolating vibration forces being transmitted to the support CASE 2

When

ωω

n= 2 to infin

In this zone the value of transmissibility is less than 1 for all values of damping ratio In this zone the transmissibility increases with increase in damping Hence we can conclude that in this zone damping will adversely affect in isolating vibratory forces

63

Tutorial problems on Forced Vibration 1 Machine of mass 75kg is mounted on spring of stiffness 12000 Nm with an assumed damping factor ξ =02 piston within the machine of mass 2 kg has a reciprocating motion with a stroke of 75 cm and speed of 15 cyclessecond Assuming the motion of piston to be harmonic determine amplitude of machine ( X) Transmissibility and force transmitted to foundation phase angle of transmitted force wrt to exciting force 2 Motor weighs 600N is mounted on a simple beam that has a spring rate of 400 Ncm The motor armature base 160 N and has an eccentricity of 0005cm What will be the amplitude of motion when it runs at 1760 RPM Neglect damping the deflection of armature shaft and weight of the beam 3 A Heavy machine of mass 300 kg is supported on resilient foundation the static deflection of foundation due to weight of machine is found to be 75mm It is observed that the machine vibrates with an amplitude of 10mm when the base of the foundation is subjected to harmonic oscillation at the undampeds natural frequency of system with an amplitude of 25mm Find the damping constant of foundation Ii) dynamic force amplitude on the base iii) amplitude of the displacement of machine relative to base 4 An air compressor of mass 450 kg operates at a constant speed of 1750 RPM The reciprocating masses are 10 kg the crank radius is 100mm If the damper for mounting introduces a damping factor 015 i) specify the springs of mounting such that only 20 of the unbalanced force is transmitted to the foundation ii) determine the magnitude of the transmitted force bull Repeat the above problem by assuming that there is no damping in system

5 A body of a car having mass 1500 kg and is mounted on 4 equal springs which deflect through 225 cm under the weight of the body A total damping forces the 4 shock absorbers is 46 N at a velocity of 1cmsec The car is placed with all the four wheels on a test platform which is move up and down at resonant speed With an amplitude of 25cm Find the amplitude of the car body on its springs assuming its CG to be in the centre of wheel base 6A 75 kg machine which is mounted on springs of stiffness 116x105 Nm with the damping of 20=ξ A 2kg piston within the machine as reciprocating motion with a stroke of 008m and speed of 3000 RPM Assuming the motion of piston to be harmonic determine the amplitude of vibration of machine

7 The spring of an automobile are compressed 10cm under its own weight Find the critical speed when a trailer is traveling over road with a people approximaterly by a sine wave of amplitude 75 cm and wave length 15m What will be the amplitude of vibration at 64kmhour 8 An eccentric mass excites us used to determine vibratory characteristics of a structure of mass 25kg as given in figure At a speed of 1000 RPM the eccentric mass to beat top position at the instant the structure was moved upwards through its static equilibrium position and corresponding amplitude was 20mm The eccentric mass is 05kg at an eccentricity of 10mm Determine i) Undamped natural frequency of the system ii) Damping ratio iii) Steady state amplitude and phase angle at 1500 RPM iv) Force transmitted to the ground at 1200 RPM

M=25kg

C

K

Eccentric mass

64

65

CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Support motion Consider a spring mass damper system as shown in figure Let it excited by motion

of support let y be the harmonic displacement of support From Newtonrsquos second law of motion

)yx(k)yx(cxm minusminusminusminus= ampampampamp

)1(0)yx(k)yx(cxm minusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminus=minus+minus+ ampampampamp

Let z be the relative displacement of the mass with respect to support then

Z=x-y

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

5

Sometime vibration problems are classified as 1 Linear vibrations 2 Non-linear vibrations 3 Random vibrations 4 Transient vibrations 5 Longitudinal vibrations 6 Transverse vibrations 7 Torsional vibrations Free vibration If a system after initial disturbance is left to vibrate on its own the resulting vibration is known as free vibrations Free vibration takes when a system vibrates under the action of forces inherent in the system and when the external forces are absent The frequency of free vibration of a system is called natural frequency Natural frequency is a property of a dynamical system Forced vibration Vibration that takes place under the excitation of external forces is called forced vibration the forced vibration takes place at different forced frequencies or external frequencies Damped vibration If any energy is lost or dissipated during oscillations then the vibration is known as damped vibration Undamped vibration if no energy is lost or dissipated during oscillations such vibrations are known as undamped vibration Linear vibration If all the basic component of a vibrating system behave linearly the resulting vibration is known as linear vibration The differential equations govern linear vibratory system are linear If the vibration is linear the principle of superposition holds and mathematical techniques of analysis are well developed Non linear vibration If any of the basic components of a vibrating system behave non linearly the resulting vibration is known as non linear vibration The differential equations that govern non linear vibratory system are non-linear If the vibration is non linear the principle of superposition does not hold good and techniques of analysis is well known Deterministic vibration If the magnitude of excitation on a vibrating system is known at any given time the resulting vibration is known as deterministic vibration Random vibration If the magnitude of excitation acting on a vibratory system at a given time cannot be predicted the resulting vibration is known as non deterministic or random vibration

6

Longitudinal vibration Consider a body of mass m carried on one end of a slender shaft and other end being fixed If the mass vibrates parallel to the spindle axis it is said to be execute longitudinal vibration Transverse vibration If the mass vibrates perpendicular to the spindle axis it is said to execute the transverse vibration Torsional vibration If the shaft gets alternatively twisted and un twisted on account of an alternate torque on the disk it is said to execute the torsional vibration SIMPLE HARMONIC MOTION A Vibration with acceleration proportional to the displacement and directed toward the mean position is known as SHM ( Simple Harmonic Motion) Consider a spring mass system as shown in the figure along with the displacement time diagram The equation of motion of the mass can be written as follows From the right angled triangle OAB we have Where x= displacement at any instant of time X = amplitude of vibration W= angular velocity or frequency in radsecond

m1

7

The velocity of the mass m at an instant of time t is given by V = The acceleration of mass m is given by Hence we can conclude that in SHM the acceleration is proportional to displacement and is directed towards mean position observing the equations 2 and 3 the velocity and acceleration are harmonic with the same frequency but lead a displacement vector by π2 and π radians respectively

t

t

t X

x

X= A sin ωt

x-Displacement X-amplitude T-Periodic Time f-Frequency f=1T ω=Frequency in radians per second t= time

8

Rotor with inertia J1

kt1 kt2

Rotor with Inertia J2

Degrees of freedom The minimum number of independent coordinates required to determine completely the positions of all the parts of a system at any instant of time is called degrees of freedom One degree of freedom Two degree of freedom Three degree of freedom Infinite degree of freedom

m1

m2

m1

m2

m3

Cantilever Beam

Continuous system ( consider the mass of the beam )

m1

m1

K1

m2

K

x1

x2

F1

9

G

K1 K2

mJ

a b

Exercise Specify the no of degree of freedom for the following 1 2 3

4

5

4m 2m m 3k k k

Cantilever Beam

Continuous system ( Neglect the mass of the beam )

k1 k2 m1 m2

y y

m2

θ1

θ2

L1

L2

M1

M2

10

O

A

B

Addition of SIMPLE HARMONIC MOTION ( SHM) The addition of two simple harmonic motion having frequency yields a resultant which is simple harmonic having the same frequency Consider tow simple harmonic motions of 1x and 2x having the same frequency and phase difference φ as given below

)(sin11 tXx ωωωω====

)(sin22 φω += tXx

Adding x = x1 +x2 Hence the resultant displacement is also SHM of amplitude X and phase angle θ

θθθθ

ωωωωt

φφφφ

11

-6

-4

-2

0

2

4

6

0 005 01 015 02 025

Tutorial problems on Simple Harmonic Motion 1 Add the following harmonic motion analytically and verify the solution graphically 1) X1= 3 sin (((( ))))30 t ++++ωωωω X2= 4 cos (((( ))))10 t ++++ωωωω ( VTU Jan 2005) 2) X1= 2 cos (((( ))))05 t ++++ωωωω X2= 5 sin (((( ))))1 t ++++ωωωω ( VTU July 2006)

3) X1= 10 cos (((( ))))4 t ππππωωωω ++++ X2= 8 sin (((( ))))6 t ππππωωωω ++++ ( VTU Dec 2007)

2 A body is subjected to two harmonic motions X1= 8 cos (((( ))))6 t ππππωωωω ++++ X2= 15 sin (((( ))))6 t ππππωωωω ++++ what harmonic motion

should be given to the body to bring it to equilibrium (VTU July 2005) 3 Split the harmonic motion x = 10 sin (((( ))))6 t ππππωωωω ++++ into two harmonic motions

having the phase of zero and the other of 45o 4 Show that resultant motion of harmonic motion given below is zero X1= X sin (((( ))))t ωωωω X2= X sin (((( ))))3

2 t ππππωωωω ++++ X3= X sin (((( ))))34 t ππππωωωω ++++

5 The displacement of a vibrating body is given by x = 5 sin (3141t + 4

ππππ )

Draw the variation of displacement for one cycle of vibration Also determine the displacement of body after 011 second ( repeat the problem for velocity and acceleration and draw graph using Excel and compare ) Time Displacement

0 0025 5 005

0075 0 01

0125 015 -353

0175 0 02

Calculate the remaining values 6 The Motion of a particle is represented by x = 4 ( ) t sin ω sketch the variation of the displacement velocity and acceleration and determine the max value of these quantities Assume ( )5 =ω ( Try to use MATLABExcel) sketch all on same graph

12

BEATS When two harmonic motions with frequencies close to one another are added The resulting motion exhibits a phenomenon known as Beats A Beat Frequency is the result of two closely spaced frequencies going into and out of synchronization with one another Let us consider tow harmonic motion of same amplitude and slightly different frequencies X1 = X Cos ( ) tωωωω X2 = X Cos ( )( ) tδδδδωωωω + Where δ is a small quantity The addition of above two harmonics can be written as X = X1 + X2

X=

+

ttX2

cos2

cos2δδδδωωωω

δδδδ

The above equation shown graphically in Figure The resulting motion represents cosine wave with frequency ( ) 2

ππππωωωω + and with a varying amplitude 2 cos( ( ) 2δδδδ t

whenever the amplitude reaches a maximum it is called the beat The frequency δ at which amplitude builds up and dies down between o and 2 X is known as beat frequency Beat phenomenon is found in machines structures and electric power houses In machines the beat phenomenon occurs when the forcing frequency is close to the natural frequency of the system

+2X

-2X

ωωωωππππ2

ELEMENTS OF VIBRATION The elements of constitute vibrating systems are

1 Mass or Inertia element - m 2 Spring - k 3 Damper - c 4 Excitation F(t)

Mass or Inertia element The mass or inertia element is assumed to be rigid body During vibration velomass changes hence kinetic energy can be gained or loosed The force appthe mass from newton second law of motion can be written as F= ma Work dthe mass is stored in the form of kinetic energy given by KE= frac12 M V2

Combination of masses In practical cases for simple analysis we replace several masses by equivalent mass Case1 Translational masses connected by rigid bar Let the masses M1 M2 and M3 are attached to a rigid bar at locates 1 respectively as shown in the figure The equivalent mass meq be assumelocated at 1 is as shown in figure (b) Let the displacement of masses M1 M2 and M3 be x1 x2 x3 and simivelocities of respective masses be x1 x2 and x3 We can express the velomasses m2 and m3 in terms of m1

Elements of Vibration

Passive element

m C K

Active element

F(t)

Conservative element

( Mass and Spring)

Non conservative element

( Damper)

C k

m F(t)

Voig

13

city of lied on one on

a single

2 and 3 d to be

larly the cities of

t Model

Translationa Let a mass mmass momearrangement These two mamass Meq or

2

1

33

2

1

221

+

+=

ll

Mll

MMMeq

14

is the required answer

l and rotational masses coupled together

having a translational velocity x be coupled to another mass having nt of inertia Io with a rotational velocity θ as in rack and pinion

shown in the figure

sses can be combined to obtain either a single equivalent translational a single equivalent mass moment of inertia Jeq

15

Equivalent translational masses

Kinetic energy of the equivalent mass =

2

21

eqeq XM amp

Kinetic energy of the two masses =

+

20

2

21

21

θθθθampamp JXM

Meq= m

+ 2RJ

m o is the required answer

Also determine the equivalent rotational mass Jeq

Jeq= m [ ]oJmR +2 is the required answer Spring element Whenever there is a relative motion between the two ends of the spring a force is developed called spring force or restoring force The spring force is proportional to the amount of deformation x and then F α x or F = kx Where k is stiffness of the spring or spring constant or spring gradient The spring stiffness is equal to spring force per unit deromation

The spring stiffness k = mNxF

Workdone in deforming a spring is equal to the strain energy or potential energy Strain energy = potential energy = area of the triangle OAB Stiffness of beams Cantilever beam consider a cantilever beam with an end mass shown in the figure The mass of the beam is assumed to be negligible The static deflection of beam at free end is given by

Similarly derive the expression for Simply supported beam and fixed support beam

x

k

m

x

m

x

m

x

mNlIE

st 192

3

=δδδδ

mNEI

Wlst

48

3

=δδδδ

F

K=stiffness

F

16

17

Stiffness of slender bar subjected to longitudinal vibrations For a system executing the longitudinal vibrations as shown in the figure let us derive the expression for stiffness

Torsional Stiffness of bar It is the amount of torque required to cause a unit angular deformation in the element

Torsional stiffness = Kt = θθθθT

Combination of stiffness Determination of equivalent spring stiffness when the springs are arranged in series Consider two springs of stiffness K1 and K2 acted upon by the force F

The deflection of spring k1 is x1 = 1KF

The deflection of spring k1 is x2 = Let these two springs be replaced by an equivalent stiffness Keq upon which a force F acts and due to which its deflection is given by

x = eqKF

x= x1+x2

m

E A

l

60 cm 70 cm

Determination of equivalent spring stiffness when the springs are arranged in parallel Force acting on K1 spring = F1=k1x Force acting on K2 spring = F2 Force required for an equivalent spring keq to defined by x given by F= Keq x But F = F1 +F2 Tutorial problems on Equivalent stiffness of springs 1Determine the equivalent stiffness for the system shown in figure 2 Determine the equivalent stiffness for the system shown in figure

3 Determine the equivale

M

3

M

Nm 102 6x

18

nt stiffness for the system shown in figure

38 kg

Nm 101 6x Nm 102 6x

Nm 106x

2k

k

k

2k

3k

k

2k

k

19

60 cm 80 cm 50 cm

50 cm 60 cm

4Determine the equivalent stiffness for the system shown in figure

5Replace the following torsional stiffness by a single shaft having radius 4cm and find the length required for the equivalent shaft Assume the material of given system and equivalent system is same R1= 3cm R2= 5cm DAMPING Every vibration energy is gradually converted into heat or sound Hence the displacement during vibration gradually reduces The mechanism by which vibration energy is gradually converted into heat or sound is known as damping A damper is assumed to have either mass or elasticity Hence damping is modeled as one or more of the following types Viscous damping Coulomb or dry friction damping materials or solid or hysteric damping Viscous damping Viscous damping is most commonly used damping mechanism in vibration analysis When the mechanical system vibrates in a fluid medium such as air gas water or oil the resistance offered by the fluid to the moving body causes energy to be dissipated In this case the amount of dissipated energy depends on many factors such as size or shape of the vibrating body the viscosity of the fluid the frequency of vibration and velocity of fluid Resistance due to viscous damping is directly proportional to the velocity of vibration

dF V αααα

dF = xC amp

Where C= damping coefficient Fd = damping force Examples of Viscous damping

1) Fluid film between sliding surface 2) Fluid flow around a piston in a cylinder 3) Fluid flow through orifice 4) Fluid flow around a journal in a bearing

Rreq =4cm

leqn

stress

Strain

Coulomb damping or dry friction damping Here a damping force is constant in magnitude but opposite in direction to that of the motion of vibrating body It is caused by the friction between the surfaces that are dry or have insufficient lubrication Material or solid or hysteric damping

When the materials are deformed energy is absorbed and dissipated by the material The effect is due to friction between the internal planes which slip or slide as the deformation takes place When a body having the material damping is subjected to vibration the stress strain diagram shows the hysteresis loop as shown in the figure The area of the loop denotes the energy lost per unit volume of the body per cycle

20

21

ττττ ττττ2 ττττ3

ττττ

X(t)

FOURIER SERIES The simplest of periodic motion happens to be SHM It is simple to handle but the motion of many vibrating system is not harmonic (but periodic) Few examples are shown below

Forces acting on machines are generally periodic but this may not be harmonic for example the excitation force in a punching machine is periodic and it can be represented as shown in figure 13 Vibration analysis of system subjected to periodic but nonharmonic forces can be done with the help of Fourier series The problem becomes a multifrequency excitation problem The principle of linear superposition is applied and the total response is the sum of the response due to each of the individual frequency term

Any periodic motion can be expressed as an infinite sum of sines and cosines terms If x(t) is a periodic function with period t its Fourier representation is given by

X(t) = (((( )))) (((( )))) (((( ))))t sint cost cos2 111 ωωωωωωωωωωωω baaao ++++++++++++

= (((( )))) (((( ))))t sint cos2 1

ωωωωωωωω nnn

o bnaa

++++++++sumsumsumsuminfininfininfininfin

====

tππππωωωω

2==== = Fundamental frequency ndash (1)

where ao an bn are constants

ττττ ττττ2 ττττ3

ττττ

X(t)

22

t

X(t)

035

025

Determination of constants To find ao Integrate both sides of equation(1) over any interval ττττ All intergrals on the RHS of the equation are zero except the one containing ao

(((( ))))dttxao

o 2

2

intintintint====ωωωωππππ

ππππωωωω

= (((( ))))dttxo

2intintintintττττ

ττττ

To find an multiply equation 1 by cos (((( ))))tnωωωω and Integrate over any interval ττττ All intergrals

(((( )))) (((( )))) dttxao

n tncos 2

2

ωωωωππππωωωω ωωωω

ππππ

intintintint==== (((( )))) (((( )))) dttxo

tncos 2

ωωωωττττ

ττττ

intintintint====

To find bn multiply equation 1 by sin (((( ))))tnωωωω and Integrate over any interval ττττ All intergrals

(((( )))) (((( )))) dttxao

n tnsin 2

2

ωωωωππππωωωω ωωωω

ππππ

intintintint==== (((( )))) (((( )))) dttxo

tnsin 2

ωωωωττττ

ττττ

intintintint====

Find the Fourier series for the curve shown below

ττττ ττττ2 ττττ3

t

X(t)

E

ττττ ττττ2 ττττ3

t

X(t)

ππππ2

t

X(t) ππππ

1 Represent for the periodic motion shown in the figure

23

CHAPTER 2 UNDAMPED FREE VIBRATION LEARNING OBJECTIVES

] Introduction to undamped free vibration

] Terminologies used in undamped free vibration

] Three methods to solve the undamped free vibration problems

] Problems related to undamped free vibration problems

Free vibrations are oscillations about a systems equilibrium position that occur in the absence of an external excitation force If during vibrations there is no loss of energy it is known as undamped vibration The first step in solving a vibration problem is setting up the differential equation of motion Consider spring mass system which is assumed to move only along the vertical direction as shown below Let m be the mass of the block and k be the stiffness of the spring When block of mass m is attached to spring the deflection of spring will be ∆ known as static deflection In the static equilibrium position the free body diagram of forces acting on the mass is shown in Figure(b) Hence mg= kA Once the system is disturbed the system executes vibrations

Let at any instant of time t the mass is displaced from the equilibrium position x the different forces acting on the system are shown in figure (d) From Newtonrsquos second law of motion

sum = maF

Inertia force ( disturbing force) = restoring force

mgxkxm ++∆minus= )(ampamp

0)( =+ xkxm ampamp

or 0)( =+ xmk

xampamp

equation 2 is the differential equation of motion for spring mfigure Comparing equation (2) with the equation of SHM +xampampsince the vibrations of the above system are free( without the forces) we can write

sec radmk

n =ω

24

ass system shown in 0)(2 =xω

resistance of external

time period km

2 f1

n

πτ ==

from the equation(1) mg = k∆

∆=

q

mk

Difference between the translation ( rectilinear) and rotational system of vibration

Translatory Rotational

In the analysis thFORCES are co Linear stiffness K Problems 1A mass of 10kFind the natural 2 A spring massnatural frequencnatural frequenc

kt

m

sec radmk

n =ω

k

25

e disturbing and restoring nsidered

In the analysis In the analysis MASS Moment of Inertia (J) is considered

in Nm is considered

g when suspended from a spring causes a static deflection of 1cm frequency of system

system has a spring stiffness K Nm and a mass of m Kg It has a y of vibration 12 Hz An extra 2kg mass coupled to it then the y reduces by 2 Hz find K and m

secrad n =ω

26

3 A steel wire of 2mm diameter and 30m long It is fixed at the upper end and carries a mass of m kg at its free end Find m so that the frequency of longitudinal vibration is 4 Hz 4 A spring mass system has a natural period of 02 seconds What will be the new period if the spring constant is 1) increased by 50 2) decreased by 50 5 A spring mass system has a natural frequency of 10 Hz when the spring constant is reduced by 800 Nm the frequency is altered by 45 Find the mass and spring constant of the original system 6 Determine the natural frequency of system shown in fig is which shaft is supported in SHORT bearings 7 Determine the natural frequency of system shown in fig is which shaft is supported in LONG bearings where l is the length of bearing and E ndash youngrsquos modulus and I is moment of Inertia 7 Determine the natural frequency of system shown in fig is which shaft is supported in LONG bearings 8 A light cantilever beam of rectangular section( 5 cm deep x 25cm wide) has a mass fixed at its free end Find the ratio of frequency of free lateral vibration in vertical plane to that in horizontal 9 Determine the natural frequency of simple pendulum 10 Homogeneous square plate of size l and mass m is suspended from the mid point of one of its sides as shown in figure Find the frequency of vibration

l

l

27

11 A compound pendulum which is rigid body of mass m and it is pivoted at O The point of pivot is at distance d from the centre of gravity It is free to rotate about its axis Find the frequency of oscillation of such pendulum 12 A connecting rod shown in fig is supported at the wrist pin end It is displaced and allowed to oscillate The mass of rod is 5kg and centre of gravity is 20 cm from the pivot point O If the frequency of oscillation is 40 cyclesminute calculate the mass moment of inertia about its CG

13 A semi circular homogenous disc of radius r and mass m is pivoted freely about its centre as shown in figure Determine the natural frequency of oscillation 14A simply supported beam of square cross section 5mmx5mm and length 1m carrying a mass of 0575 kg at the middle is found to have natural frequency of 30 radsec Determine youngrsquos modulus of elasticity of beam 15 A spring mass system k1 and m have a natural frequency f1 Determine the value of k2 of another spring in terms of k1 which when placed in series with k1 lowers the natural frequency to 23 f1

COMPLETE SOLUTION OF SYSTEM EXECUTING SHM

The equation of motion of system executing SHM can be represented by

0)( =+ xkxm ampamp ----------------------------------------(1)

022

2=+

dtdx

dt

dx ω

The general solution of equation (1) can be expressed as

------------------------(2)

Where A and B are arbitrary constant which can be determined from the initial conditions of the system Two initial conditions are to be specified to evaluate these

constants x=x0 at t=0 and xamp = Vo at t=0 substituting in the equation (2) Energy method In a conservative system the total energy is constant The dwell as natural frequency can be determined by the princienergy For free vibration of undamped system at any instant and partly potential The kinetic energy T is stored in the masswhere as the potential energy U is stored in the form of sdeformation or work done in a force field such as gravity The total energy being constant T+U = constant Its rate of cha

t)(sin Bt)cos( AX ω+ω=

t)(sin V

t)cos( xx o0 ω

ω+ω=

Is the required complete

solution

28

ifferential equation as ple of conservation of of time is partly kinetic by virtue of its velocity train energy in elastic

nge

Is given by [ ] 0UTdtd

=+

From this we get a differential equation of motion as well as natural frequency of the system Determine the natural frequency of spring mass system using energy method

Determine the natural frequency of system shown in figure Determine the natural frequency of the system shown in figure Is there any limitation on the value of K Discuss Determine the natural frequency of s

m

l k

m

a

θ

m

a

θ

l

k

29

k

k

ystem shown below Neglect the mass of ball

l

m

30

A string shown in figure is under tension T which can be assumed to remain constant for small displacements Find the natural frequency of vertical vibrations of spring An acrobat 120kg walks on a tight rope as shown in figure The frequency of vibration in the given position is vertical direction is 30 rads Find the tension in the rope A manometer has a uniform bore of cross section area A If the column of liquid of length L and Density ρ is set into motion as shown in figure Find the frequency of the resulting oscillation

m

l

a

T T

36m 8m

31

k

R

k

m

r

m

Find the expression for natural frequency of system shown in the figure Neglect the mass of the cantilever beam Study the special case i) k=Infinity ii) I = infinity Determine the expression for the natural frequency of a system shown in figure The two discs are keyed to a common shaft and have a combined mass moment of inertia about centre of oscillation O It is assumed that the card attached to mass m does not stretch and is always under tension Determine the natural frequency of a system shown

m

l

32

Determine the expression for the natural frequency of the system shown in figure Assume that the wires connecting the masses do not stretch and are always in tension

Determine the natural frequency of spring mass system taking the MASS OF SPRING (ms ) into account

M3 M1

M2

k2 k1

m

k

33

RAYLEIGHrsquoS METHOD

This is the extension of energy method Here we can determine the natural frequency of a conservative system without finding the equation of motion The natural frequency is a function of the rate of change of Kinetic energy and potential energy of the system From the principle of conservation of energy we have T+U = Constant T= KE and U =PE We can write the above equation as T1+U1 = --------------------------(1) Where 1 and 2 represents tow instances of time Let 1 be the time when the mass is passing through static equilibrium position Let 2 be the time corresponding to the mass displacement of the mass At this instant the velocity f the mass is zero and hence Substituting in equation (1) we get ( )Max = ( )Max Above equation leads directly to natural frequency Determine the natural frequency of spring mass system using Rayleighrsquos method m

34

r

B

K K

Home work

Determine the natural frequency of pendulum using Rayleighrsquos method

A Cylinder of mass m and moment of inertia Jo is to role without slipping but is sustained by spring K as shown in figure Determine the natural frequency of oscillation

Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping

a

Determine the natural frequency of system shown where in cylindrical disc rolls over the inextensible spring A cylinder of radius r rolls without slipping on cylindrical the equation of motion and hence determine the naoscillations about the lowest point shown Repeat the above problem ndash Instead of cylinder assume swithout slipping on concave surface as shown above

r

K

35

portion of radius R Derive tural frequency of small

phere of radius with r rolls

The torsional pendulum has a natural frequency of 5 Hz What length of steel wire of diameter 2 mm should be used for pendulum The inertia of mass fixed at the free end is 00098 kgm2 Take g= 083x1011 Nm2 Determine the torsional natural frequency of the system shown in figure Neglect the mass moment of inertia of the shaft ( Figure not proportionate) Determine the natural frequency of simple pendulum considering the mass of the rod into account

Using the energy method find tThe chord may be assumed ine

Kt1

Kt2

20cm

06m 08m 04m

20cm 20cm 20cm

m

M

he natural frxtensible in

k

M

36

equency of the system shown in figure the spring mass pulley system

A rod of length l and depth H having a mass m rests on a semicylidrical surface os radius r It is tipped slightly and let go Find the frequency of oscillation

A thin bar of l having a mass m rests on a semicylidrical surface radius r It is tipped slightly and let go Find the frequency of oscillation

l

h

R

l

37

38

CHAPTER 3

DAMPED FREE VIBRATION OF SINGLE DEGREE FREEDOM SYSTEM LEARNING OBJECTIVES

] Introduction to damped free vibration

] Terminologies used in damped free vibration

] Three different types of damping

] Concept of critical damping and its importance

] Study of response of viscous damped system for cases of under damping

critical and over damping

] Concept of logarithmic decrement

] Problems related to damped free vibration problems

Viscously damped free vibration of single degree freedom sytem Consider a single degree freedom system with viscous damping as shown in figure For a viscous damped system the damping force is

where c is a constant of proportionality It is called damping coefficient or coefficient of viscous damping

Symbolically it is designated by a dashpot as shown in Fig 3

x xampx

39

At any instant of time let the mass be displaced by x from the equilibrium position Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

Okxxcxm =minusminus= ampampamp

Okxxcxm =++ ampampamp ----------------------------------------------------------(1)

or

0kxdtdx

Cdt

xdM

2

2

=++

Assuming the solution of equation as

The first term in the above equation m2Ct

eminus

is an exponentially decaying function of time The behavior of the terms in the parenthesis however depends on whether numerical value within the radicals is positive zero or negative

CASE 1 if 0mk

m2c

2

=minus

40

CASE 2 if 0mk

m2c

2

=gt

CASE 3 if 0mk

m2c

2

=lt

41

Critically damped coefficient Critically damped coefficient is defined as the value of the damping coeffient for which radical in equation

mk

m2c

m2c

S2

21 minus

plusmn

= becomes zero

Damping Ratio ( Damping factor) ( ξ ) It is defined as the ratio of damping coefficient to the critical damping coefficient Note The damping coefficient c is defined as the actual amount of damping while the damping coefficient is a constant depending upon the mass and stiffness of the system The roots of characteristics equation can be now written as Note if c=cc or ξ =1 then system is critically damped if cgt cc or ξ gt1 then system is over damped I if clt cc or ξ lt1 then system is under damped

42

Take home quiz Derive General solution of different damped system

1 Critically damped system 2 Under damped system 3 Over damped system

Logarithmic Decrement It is used to determine the amount of damping present in system This measure The rate of decay of free vibration The larger the damping the greater will be rate of decay Logarithmic decrement is defined as the natural logarithm of ratio of any two successive amplitudes

=δ

2

1

xx

ln

Expression for logarithmic decrement (δ ) Consider a damped free vibration which is expressed by

( ) ( ) φ+ξminusω= ξω t 1sinXex 2n

tn

x is the displacement at any instant of time X and φ are arbitrary constants

21

2

ξminus

πξ=δ

43

Tutorial problems 1 Derive an expression for damping ratio in terms of logarithmic decrement

2 Show that logarithmic decrement

=δ

n

o

xx

lnn1

where xo is the initial amplitude xn

is amplitude after n cycles 3 A mass of 1 kg is to be supported on a spring having stiffness K= 9800 Nm The damping constant c= 59 Nsm Determine the natural frequency of system Find also logarithmic decrement and amplitude after 3 cycles if initial displacement is 03 cm 4 A vibrating system is defined by the following parameters M = 3kg k= 100 Nm damping coefficient = 3 Nsm determine Damping factor natural frequency damped vibration logarithmic decrement ratio of consecutive amplitude no of cycles after which the original amplitude is reduced to 20 5A single degree damped vibrating system a suspended mass of 18kg makes 15 oscillations in 003 seconds The amplitude decreases to 025 of the initial value after 5 oscillations Determine the stiffness of spring Logarithmic decrement damping factor and damping coefficient 6 A damped vibration record of a spring mass system shows the following data Amplitude at the end of 2nd cycle is 9mm amplitude at eh end of 3rd cycle is 6mm amplitude at the end of 4th cycle is 4mm stiffness of spring =8000 Nm weight =20N Find the logarithmic decrement damping force at unit velocity periodic time of vibration 7 For the system shown in figure the characteristics of dashpot is such that the constant force of 49 N is applied to the piston it velocity is found to be constant at 012 ms Determine the value of C would you expect the complete system to be periodic or aperiodic 8 For a torsional system shown in the figuref1 is natural frequency in air and f2 is the natural frequency when immersed in oil Determine the damping coefficient

Viscous fluid

Disc of mass moment of Inertia J

2a

C

m M

b

a a

k

9 A thin plate of area A and weigh w is fixed to the end of the spring and allow to oscillate in a viscous fluid as shown in figure It f1 is the frequency of oscillation in air and f2 in liquid Show that

22

21 ff

gA2

minusπω

=micro where the damping force on the plate

AV2Fd micro= V being the velocity of vibration 10 A torsional pendulum has a natural frequency of 175 cyclessecond vibrating in vacuum The mass moment of inertia is 2 kgm2 It is immersed in oil and it is observed that the natural frequency is 142 cyclessecond Find the damping torque If the disc is replaced by 34 deg When in oil find the displacement at the end of the first complete cycle 11 An automobile can be modeled as a mass placed on 4 shock absorbers each consisting of a spring and a damper Such that each spring is equally loaded Determine the stiffness damping constant of each shock absorber So that natural frequency is 2Hz and system is critically damped The mass of vehicle is 200kg 12 Write the differential equation of motion for the system shown in the figure below Determine the natural frequency of the damped oscillation and critical damping coefficient

Determine the differential equation of motion and find the critical cdamped natural frequency for system shown in figure

a

C

m M

M =200kg

44

200kg kgkg

oefficient and

45

Derive the equation of motion and find the critical damping coefficient and damped natural frequency for the system shown below Write the differential equation of motion and determine the expression for critical damping coefficient and damped natural frequency of system shown in figure

Set up a differential equation of motion for system shown in figure and determine the following Undamped natural frequencies critical damping coefficient damping ratio damped natural frequency

b

C

kM

a

m

46

A mass of 300N is resting on two springs of stiffness 3000 Nm each and a dashpot of damping coefficient 150 Nsm as shown in figure If the initial velocity of 10cms is given to mass at its equilibrium position what will be the displacement from the equilibrium position at the end of first second Derive the formula used A gun barrel of mass 600kg has a recoil spring of stiffness 294 knm if the barrel recoils 13m on firing determine i) The initial recoil velocity of barrel ii) The critical damping coefficient of dashpot which is engaged at the end of recoil stroke Iii) Time required for the barrel to return at the end of recoil stroke Iv) time required for the barrel to return to a position 5cm from the initial position

W= 3000 N

FIRING

CHAPTER 4

FORCED VIBRATION LEARNING OBJECTIVES

] Introduction to forced free vibration

] Terminologies used in forced vibration

] Concept of magnification factor

] Study of reciprocating and rotating unbalance

] Concept of base excitation and support excitation

] Concept of Transmissibility ratio

] Problems related to forced vibration

The vibration that takes place under the excitation of external forces are called

forced vibration When a system is subjected to harmonic excitation it is forced to vibrate at the same frequency as that of excitation Most of the mechanical systems are prone to steady state forced vibration which cause fatigue failure The vibration problem also becomes very important when any of excitation frequencies coincide with one of natural frequencies of a system Under this condition resonance occurs during which large amplitude of vibrations are observed To avoid resonance external frequency (operating speed) may be changed or properties of the system may be changed to alter the natural frequency In some cases sufficient amount of damping may be provided to avoid large amplitude during resonance Thus the problem of force vibration is very important in mechanical design

FORCED VIBRATIONS OF SINGLE DOF SYSTEMS UNDER HARMONIC EXCITATION

` [ ] tsin FF o ω=

x

sin FF o=

Fo

xampamp

[ ] tω

xamp

47

48

Consider a spring mass system having a viscous damping excited by a harmonic force [ ] tsin FF o ω= as shown in figure The various forces acting on the system is shown in figure Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

O)tsin(Fkxxcxm o =ω+minusminus= ampampamp

)tsin(Fkxxcxm o ω=++ ampampamp ----------------------------------------------------------(1)

or

)tsin(Fkxdtdx

Cdt

xdM o2

2

ω=++

Equation (1) represents the differential equation of motion of the system The solution to this equation consists of two parts namely complementary function and a particular integral

The first of solution is also termed as transient solution of homogeneous solution and can be obtained from the homogeneous differential equation 0kxxcxm =++ ampampamp

The second part of solution particular integral is also termed as steady state solution ad is of form )tsin(Xx φminusω= Where X= steady state amplitude φ is phase difference between external force and displacement

)tsin(Xx φminusω=

The vector representation of the shown

49

The force polygon for the above vector diagram can be written as follows

( )φminusωt O

KX

xm 2ω

Fo

xCω

tω

φ

50

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

is the governing equation

Zero frequency deflection (Xst) It is deflection observed at zero frequency ( when the external frequency 0=ω

( ) kF

001

kF

x o

22

o

st =

+minus=

Magnification factor (M) It is defined as the ratio of steady state amplitude to zero frequency deflection xst

M = arX

X=

A plot of magnification factor versus frequency ratio

damping factor ( )ξ is called frequency plot of a harmon All the frequency terms starts from magnification factor of damping ratio ( )ξ The amplitude of vibration at any on the value of damping The amplitude will be higher for higher damping The amplitude of vibration at reso

amplitude tends to zero as

ωω

n tends to infinity

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

Freq

ζ =

ζ = 1

Magnification Factor M = XXst

kFo

51

ωω

nfor varying values of

ically excite vibrating system

M=1 what ever may be value particular frequency depends

for lesser damping and lesser nance will be very high The

uency Ratio r = (ωωn)

ζ = 0 0

ζ = 025 ζ = 0375 ζ = 0 5

ζ = 0707

ζ = 2

52

Plot of phase angle φ versus frequency ratio

ωω

n is shown in the

figure The phase angle φ is zero for all the values of damping ratio

at

ωω

n=0 The phase

angle φ is 90 for all the values of damping ratio

at

ωω

n=1 ie at

resonance The phase angle between o to 90 deg for all the values of

ωω

n= 0 to 1 and

between 90 deg to 180 deg for all the values of

ωω

ngt1

Class work Derive an expression for frequency at peak amplitude

2np 21 ξminusω=ω=ω is the required equation

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

53

Derive expression amplitude at resonance ( In dimensional form) In dimensional form We know that steady state amplitude

( ) ( )

ω+ωminus=

222

o

Cmk

Fx Equation in dimensional form

We know that at resonance mk

n =ω=ω

( )

ω+

minus

=2

n

2

oR

Cmk

mk

Fx

( )

ω=

2n

oR

C

Fx

( )

ω=

n

oR C

Fx

Class work Derive an expression for amplitude at resonance ( in non dimensional form)

Tutorial Problems on Forced Vibration 1A weight of 60N is suspended by a spring stiffness 12 kNm is forced to vibrate by a harmonic force of 10N assuming viscous damping of 0086 kN-sm Find i) resonance frequency ii) amplitude at resonance iii) phase angle at resonance iv) frequency corresponding to peak amplitude 2 A periodic torque having a maximum value of 05888Nm at a frequency of 4 rads is impressed upon a flywheel suspended from a wire The wheel has mass moment of inertia 012 kgm2 and the wire has stiffness 1176 Nmrad A viscous dashpot applies damping couple of 0784 Nm an angular velocity of 2 rads calculate i)maximum angular displacement from rest position ii) maximum couple applied to dashpot Iii) angle by which the angular displacement lags the torque A mass of 1019 kg is suspended from a spring stiffness 20 Nmm viscous damping causes the amplitude to decrease to 110th of initial value in 4 complete cycle If a periodic force of 20 cos(50 t) is applied to the mass Find the amplitude of forced vibration What would be the amplitude of oscillation if the applied force coincides with the natural period

m=1019 kg

[ ] t50cos 30F =

54

55

ROTATING UNBALANCE Unbalance in rotating machinery is a common source of vibration excitation Consider a machine of mass m having mounted on a foundation of stiffness k and damping coefficient C as shown in figure The unbalance is represented by an eccentric rotating mass m with eccentricity (e) and having a angular velocity (ω) let x be the displacement of non rotating mass from the static equilibrium position The displacement of rotating mass m from the static equilibrium is ( ) tsin( ex ω+ Applying Newtonrsquos law of motion we have

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

56

RECIPROCATING UNBALANCE Unbalance in reciprocating machinery is a common source of vibration excitation Consider a machine of mass M having mounted on a foundation of stiffness K and damping coefficient C as shown in figure The unbalance is represented by a reciprocating mass (m) having crank radus (e) and connecting rod of length (l) Let ωωωω be angular velocity of the crank Let x be the displacement of non reciprocating mass from the static equilibrium position at any instant of time The displacement of reciprocating mass m from the static equilibrium position is

)2sin(sin +++ tle

tex ωωωωωωωω

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

57

Note The equations obtained in reciprocating unbalance is same as the rotating unbalance and having the following differences Rotating Unbalance Reciprocating Unbalance Here mass of the rotating element is considered

Here mass of reciprocating element is considered

Here the eccentricity (e) which is the distance between the centre of rotation and CG

Here the eccentricity e is the crank radius and crank radius = stroke length2

The frequency plot of non dimensional factor( ( )me

MX versus frequency ratio

ωω

n for various values of ( )ξξξξ is shown in figure This frequency plot is same for

the rotating and reciprocating unbalance with the following points are noted from the diagram

1 Damping ratio plays very important role during resonance in controlling the amplitude

2 When

ωω

n=0 then ( )me

MX =0 for all the values of damping ratio ( )ξξξξ

3 When

ωω

nis higher than ( )me

MX tends to unity

4 At very high speeds ie at very high frequency ratio

ωω

n there is no effect

of damping The frequency plot of phase angle versus the frequency ratio is given as shown in figure The explanation is same as in forced harmonic vibration

58

VIBRATION ISOLATION

Vibratory forces and motions generated by machine and other caused are unavoidable Howeever their effect on the other dynamical systems can be minimized by proper isolator design An isolation system is either to protect o delicate machine from excessive vibration transmitted to it from its supporting structure or to prevent vibratory forces and motions generated by machines from being transmitted to its surroundings The effectiveness of isolation may be measured in terms of forces and motion transmitted The first type is known as force isolation and the second type is known as motion isolation

Force isolation (Force Transmissibility) When vibratory forces generated by machine are isolated the effectiveness of forced isolation can be made by defining the term transmissibility The term transmissibility is defined as the ratio of force transmitted to the foundation to that of the external impressed force

Force Transmissibility = o

tr

FF

force Impressed Externalfoundation toted transmitForce

=

Consider viscous damped spring mass system as shown in figure Upon this system an external harmonic force ) t (Sin Fo ω is applied The difference forces acting on system are inertia forces damping force spring force and external force All these forces are represented on force polygon shown below Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these two forces

( ) ( )22 xckx ω+=

59

This is the expression for force transmissibility in dimensional form To represent same in the non dimensional form divide numerator and denominator by k

2

n

22

n

2

n

o

tr

21

21

FF

TR

ωω

ξ+

ωω

minus

ωω

ξ+

==

PHASE DIFFERNECE BETWEEN EXTERNAL FORCE AND DISPLACEMENT Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these

ωω

minus

ωω

ξ=

ωminusω

=φ minusminus2

n

n1

2

1

1

2tan

mKC

tan

The phase difference between force transmitted to the foundation Ftr to the displacement

kc

KxxC

tanω

=ω

=α or

ωω

ξ=

ω=α minusminus

n

11 2tanKx

xCtan

The phase difference between external force Fo and force transmitted to foundationis β

60

β= αminusφ =

ωω

minus

ωω

ξminus

2

n

n1

1

2tan -

ωω

ξminus

n

1 2tan

Displacement Transmissibility It is the ratio of amplitude of steady state vibration of machine to that of the base motion Forced vibration due to support motion Consider a spring mass damper system subjected to harmonic excitation )tsin(Yy ω= at the base as shown in figure Let x be the absolute displacement of the mass at any instant of time t The different forces acting on system are shown in free body diagram Applying Newtonrsquos Law of motion Ma = sumF

)yx(k)yx(cxm minusminusminusminus= ampampampamp

0)yx(k)yx(cxm =minus+minus+ ampampampamp

61

Frequency plot of transmissibility versus frequency ratio

62

The plot of transmissibility versus frequency ratio is as shown in figure The following points can be noted from it

i) The value of transmissibility is unity at frequency ratio

ωω

n=0 and 2

ii) For an undamped system ( )ξ =0 transmissibility tends to infin at resonance iii) The value of transmissibility is less than unity for the value of frequency ratio

greater than 2

iv) For frequency ratio

ωω

nlt 2 smaller damping ratio lead to larger values of

transmissibility on the other hand for frequency ratio

ωω

ngt 2 smaller damping

ratio leads to smaller value of transmissibility

The frequency plot of transmissibility (TR) versus frequency ratio

ωω

n can be

broadly divided into two zones

1

ωω

n= 0 to 2

2

ωω

n= 2 to infin

CASE 1

When

ωω

n= 0 to 2

In this zone the value of transmissibility is always greater than 1 for all values of damping ratio In this zone the transmissibility decreases with increase in damping Hence we can conclude that in this zone damping is useful in isolating vibration forces being transmitted to the support CASE 2

When

ωω

n= 2 to infin

In this zone the value of transmissibility is less than 1 for all values of damping ratio In this zone the transmissibility increases with increase in damping Hence we can conclude that in this zone damping will adversely affect in isolating vibratory forces

63

Tutorial problems on Forced Vibration 1 Machine of mass 75kg is mounted on spring of stiffness 12000 Nm with an assumed damping factor ξ =02 piston within the machine of mass 2 kg has a reciprocating motion with a stroke of 75 cm and speed of 15 cyclessecond Assuming the motion of piston to be harmonic determine amplitude of machine ( X) Transmissibility and force transmitted to foundation phase angle of transmitted force wrt to exciting force 2 Motor weighs 600N is mounted on a simple beam that has a spring rate of 400 Ncm The motor armature base 160 N and has an eccentricity of 0005cm What will be the amplitude of motion when it runs at 1760 RPM Neglect damping the deflection of armature shaft and weight of the beam 3 A Heavy machine of mass 300 kg is supported on resilient foundation the static deflection of foundation due to weight of machine is found to be 75mm It is observed that the machine vibrates with an amplitude of 10mm when the base of the foundation is subjected to harmonic oscillation at the undampeds natural frequency of system with an amplitude of 25mm Find the damping constant of foundation Ii) dynamic force amplitude on the base iii) amplitude of the displacement of machine relative to base 4 An air compressor of mass 450 kg operates at a constant speed of 1750 RPM The reciprocating masses are 10 kg the crank radius is 100mm If the damper for mounting introduces a damping factor 015 i) specify the springs of mounting such that only 20 of the unbalanced force is transmitted to the foundation ii) determine the magnitude of the transmitted force bull Repeat the above problem by assuming that there is no damping in system

5 A body of a car having mass 1500 kg and is mounted on 4 equal springs which deflect through 225 cm under the weight of the body A total damping forces the 4 shock absorbers is 46 N at a velocity of 1cmsec The car is placed with all the four wheels on a test platform which is move up and down at resonant speed With an amplitude of 25cm Find the amplitude of the car body on its springs assuming its CG to be in the centre of wheel base 6A 75 kg machine which is mounted on springs of stiffness 116x105 Nm with the damping of 20=ξ A 2kg piston within the machine as reciprocating motion with a stroke of 008m and speed of 3000 RPM Assuming the motion of piston to be harmonic determine the amplitude of vibration of machine

7 The spring of an automobile are compressed 10cm under its own weight Find the critical speed when a trailer is traveling over road with a people approximaterly by a sine wave of amplitude 75 cm and wave length 15m What will be the amplitude of vibration at 64kmhour 8 An eccentric mass excites us used to determine vibratory characteristics of a structure of mass 25kg as given in figure At a speed of 1000 RPM the eccentric mass to beat top position at the instant the structure was moved upwards through its static equilibrium position and corresponding amplitude was 20mm The eccentric mass is 05kg at an eccentricity of 10mm Determine i) Undamped natural frequency of the system ii) Damping ratio iii) Steady state amplitude and phase angle at 1500 RPM iv) Force transmitted to the ground at 1200 RPM

M=25kg

C

K

Eccentric mass

64

65

CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Support motion Consider a spring mass damper system as shown in figure Let it excited by motion

of support let y be the harmonic displacement of support From Newtonrsquos second law of motion

)yx(k)yx(cxm minusminusminusminus= ampampampamp

)1(0)yx(k)yx(cxm minusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminus=minus+minus+ ampampampamp

Let z be the relative displacement of the mass with respect to support then

Z=x-y

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

6

Longitudinal vibration Consider a body of mass m carried on one end of a slender shaft and other end being fixed If the mass vibrates parallel to the spindle axis it is said to be execute longitudinal vibration Transverse vibration If the mass vibrates perpendicular to the spindle axis it is said to execute the transverse vibration Torsional vibration If the shaft gets alternatively twisted and un twisted on account of an alternate torque on the disk it is said to execute the torsional vibration SIMPLE HARMONIC MOTION A Vibration with acceleration proportional to the displacement and directed toward the mean position is known as SHM ( Simple Harmonic Motion) Consider a spring mass system as shown in the figure along with the displacement time diagram The equation of motion of the mass can be written as follows From the right angled triangle OAB we have Where x= displacement at any instant of time X = amplitude of vibration W= angular velocity or frequency in radsecond

m1

7

The velocity of the mass m at an instant of time t is given by V = The acceleration of mass m is given by Hence we can conclude that in SHM the acceleration is proportional to displacement and is directed towards mean position observing the equations 2 and 3 the velocity and acceleration are harmonic with the same frequency but lead a displacement vector by π2 and π radians respectively

t

t

t X

x

X= A sin ωt

x-Displacement X-amplitude T-Periodic Time f-Frequency f=1T ω=Frequency in radians per second t= time

8

Rotor with inertia J1

kt1 kt2

Rotor with Inertia J2

Degrees of freedom The minimum number of independent coordinates required to determine completely the positions of all the parts of a system at any instant of time is called degrees of freedom One degree of freedom Two degree of freedom Three degree of freedom Infinite degree of freedom

m1

m2

m1

m2

m3

Cantilever Beam

Continuous system ( consider the mass of the beam )

m1

m1

K1

m2

K

x1

x2

F1

9

G

K1 K2

mJ

a b

Exercise Specify the no of degree of freedom for the following 1 2 3

4

5

4m 2m m 3k k k

Cantilever Beam

Continuous system ( Neglect the mass of the beam )

k1 k2 m1 m2

y y

m2

θ1

θ2

L1

L2

M1

M2

10

O

A

B

Addition of SIMPLE HARMONIC MOTION ( SHM) The addition of two simple harmonic motion having frequency yields a resultant which is simple harmonic having the same frequency Consider tow simple harmonic motions of 1x and 2x having the same frequency and phase difference φ as given below

)(sin11 tXx ωωωω====

)(sin22 φω += tXx

Adding x = x1 +x2 Hence the resultant displacement is also SHM of amplitude X and phase angle θ

θθθθ

ωωωωt

φφφφ

11

-6

-4

-2

0

2

4

6

0 005 01 015 02 025

Tutorial problems on Simple Harmonic Motion 1 Add the following harmonic motion analytically and verify the solution graphically 1) X1= 3 sin (((( ))))30 t ++++ωωωω X2= 4 cos (((( ))))10 t ++++ωωωω ( VTU Jan 2005) 2) X1= 2 cos (((( ))))05 t ++++ωωωω X2= 5 sin (((( ))))1 t ++++ωωωω ( VTU July 2006)

3) X1= 10 cos (((( ))))4 t ππππωωωω ++++ X2= 8 sin (((( ))))6 t ππππωωωω ++++ ( VTU Dec 2007)

2 A body is subjected to two harmonic motions X1= 8 cos (((( ))))6 t ππππωωωω ++++ X2= 15 sin (((( ))))6 t ππππωωωω ++++ what harmonic motion

should be given to the body to bring it to equilibrium (VTU July 2005) 3 Split the harmonic motion x = 10 sin (((( ))))6 t ππππωωωω ++++ into two harmonic motions

having the phase of zero and the other of 45o 4 Show that resultant motion of harmonic motion given below is zero X1= X sin (((( ))))t ωωωω X2= X sin (((( ))))3

2 t ππππωωωω ++++ X3= X sin (((( ))))34 t ππππωωωω ++++

5 The displacement of a vibrating body is given by x = 5 sin (3141t + 4

ππππ )

Draw the variation of displacement for one cycle of vibration Also determine the displacement of body after 011 second ( repeat the problem for velocity and acceleration and draw graph using Excel and compare ) Time Displacement

0 0025 5 005

0075 0 01

0125 015 -353

0175 0 02

Calculate the remaining values 6 The Motion of a particle is represented by x = 4 ( ) t sin ω sketch the variation of the displacement velocity and acceleration and determine the max value of these quantities Assume ( )5 =ω ( Try to use MATLABExcel) sketch all on same graph

12

BEATS When two harmonic motions with frequencies close to one another are added The resulting motion exhibits a phenomenon known as Beats A Beat Frequency is the result of two closely spaced frequencies going into and out of synchronization with one another Let us consider tow harmonic motion of same amplitude and slightly different frequencies X1 = X Cos ( ) tωωωω X2 = X Cos ( )( ) tδδδδωωωω + Where δ is a small quantity The addition of above two harmonics can be written as X = X1 + X2

X=

+

ttX2

cos2

cos2δδδδωωωω

δδδδ

The above equation shown graphically in Figure The resulting motion represents cosine wave with frequency ( ) 2

ππππωωωω + and with a varying amplitude 2 cos( ( ) 2δδδδ t

whenever the amplitude reaches a maximum it is called the beat The frequency δ at which amplitude builds up and dies down between o and 2 X is known as beat frequency Beat phenomenon is found in machines structures and electric power houses In machines the beat phenomenon occurs when the forcing frequency is close to the natural frequency of the system

+2X

-2X

ωωωωππππ2

ELEMENTS OF VIBRATION The elements of constitute vibrating systems are

1 Mass or Inertia element - m 2 Spring - k 3 Damper - c 4 Excitation F(t)

Mass or Inertia element The mass or inertia element is assumed to be rigid body During vibration velomass changes hence kinetic energy can be gained or loosed The force appthe mass from newton second law of motion can be written as F= ma Work dthe mass is stored in the form of kinetic energy given by KE= frac12 M V2

Combination of masses In practical cases for simple analysis we replace several masses by equivalent mass Case1 Translational masses connected by rigid bar Let the masses M1 M2 and M3 are attached to a rigid bar at locates 1 respectively as shown in the figure The equivalent mass meq be assumelocated at 1 is as shown in figure (b) Let the displacement of masses M1 M2 and M3 be x1 x2 x3 and simivelocities of respective masses be x1 x2 and x3 We can express the velomasses m2 and m3 in terms of m1

Elements of Vibration

Passive element

m C K

Active element

F(t)

Conservative element

( Mass and Spring)

Non conservative element

( Damper)

C k

m F(t)

Voig

13

city of lied on one on

a single

2 and 3 d to be

larly the cities of

t Model

Translationa Let a mass mmass momearrangement These two mamass Meq or

2

1

33

2

1

221

+

+=

ll

Mll

MMMeq

14

is the required answer

l and rotational masses coupled together

having a translational velocity x be coupled to another mass having nt of inertia Io with a rotational velocity θ as in rack and pinion

shown in the figure

sses can be combined to obtain either a single equivalent translational a single equivalent mass moment of inertia Jeq

15

Equivalent translational masses

Kinetic energy of the equivalent mass =

2

21

eqeq XM amp

Kinetic energy of the two masses =

+

20

2

21

21

θθθθampamp JXM

Meq= m

+ 2RJ

m o is the required answer

Also determine the equivalent rotational mass Jeq

Jeq= m [ ]oJmR +2 is the required answer Spring element Whenever there is a relative motion between the two ends of the spring a force is developed called spring force or restoring force The spring force is proportional to the amount of deformation x and then F α x or F = kx Where k is stiffness of the spring or spring constant or spring gradient The spring stiffness is equal to spring force per unit deromation

The spring stiffness k = mNxF

Workdone in deforming a spring is equal to the strain energy or potential energy Strain energy = potential energy = area of the triangle OAB Stiffness of beams Cantilever beam consider a cantilever beam with an end mass shown in the figure The mass of the beam is assumed to be negligible The static deflection of beam at free end is given by

Similarly derive the expression for Simply supported beam and fixed support beam

x

k

m

x

m

x

m

x

mNlIE

st 192

3

=δδδδ

mNEI

Wlst

48

3

=δδδδ

F

K=stiffness

F

16

17

Stiffness of slender bar subjected to longitudinal vibrations For a system executing the longitudinal vibrations as shown in the figure let us derive the expression for stiffness

Torsional Stiffness of bar It is the amount of torque required to cause a unit angular deformation in the element

Torsional stiffness = Kt = θθθθT

Combination of stiffness Determination of equivalent spring stiffness when the springs are arranged in series Consider two springs of stiffness K1 and K2 acted upon by the force F

The deflection of spring k1 is x1 = 1KF

The deflection of spring k1 is x2 = Let these two springs be replaced by an equivalent stiffness Keq upon which a force F acts and due to which its deflection is given by

x = eqKF

x= x1+x2

m

E A

l

60 cm 70 cm

Determination of equivalent spring stiffness when the springs are arranged in parallel Force acting on K1 spring = F1=k1x Force acting on K2 spring = F2 Force required for an equivalent spring keq to defined by x given by F= Keq x But F = F1 +F2 Tutorial problems on Equivalent stiffness of springs 1Determine the equivalent stiffness for the system shown in figure 2 Determine the equivalent stiffness for the system shown in figure

3 Determine the equivale

M

3

M

Nm 102 6x

18

nt stiffness for the system shown in figure

38 kg

Nm 101 6x Nm 102 6x

Nm 106x

2k

k

k

2k

3k

k

2k

k

19

60 cm 80 cm 50 cm

50 cm 60 cm

4Determine the equivalent stiffness for the system shown in figure

5Replace the following torsional stiffness by a single shaft having radius 4cm and find the length required for the equivalent shaft Assume the material of given system and equivalent system is same R1= 3cm R2= 5cm DAMPING Every vibration energy is gradually converted into heat or sound Hence the displacement during vibration gradually reduces The mechanism by which vibration energy is gradually converted into heat or sound is known as damping A damper is assumed to have either mass or elasticity Hence damping is modeled as one or more of the following types Viscous damping Coulomb or dry friction damping materials or solid or hysteric damping Viscous damping Viscous damping is most commonly used damping mechanism in vibration analysis When the mechanical system vibrates in a fluid medium such as air gas water or oil the resistance offered by the fluid to the moving body causes energy to be dissipated In this case the amount of dissipated energy depends on many factors such as size or shape of the vibrating body the viscosity of the fluid the frequency of vibration and velocity of fluid Resistance due to viscous damping is directly proportional to the velocity of vibration

dF V αααα

dF = xC amp

Where C= damping coefficient Fd = damping force Examples of Viscous damping

1) Fluid film between sliding surface 2) Fluid flow around a piston in a cylinder 3) Fluid flow through orifice 4) Fluid flow around a journal in a bearing

Rreq =4cm

leqn

stress

Strain

Coulomb damping or dry friction damping Here a damping force is constant in magnitude but opposite in direction to that of the motion of vibrating body It is caused by the friction between the surfaces that are dry or have insufficient lubrication Material or solid or hysteric damping

When the materials are deformed energy is absorbed and dissipated by the material The effect is due to friction between the internal planes which slip or slide as the deformation takes place When a body having the material damping is subjected to vibration the stress strain diagram shows the hysteresis loop as shown in the figure The area of the loop denotes the energy lost per unit volume of the body per cycle

20

21

ττττ ττττ2 ττττ3

ττττ

X(t)

FOURIER SERIES The simplest of periodic motion happens to be SHM It is simple to handle but the motion of many vibrating system is not harmonic (but periodic) Few examples are shown below

Forces acting on machines are generally periodic but this may not be harmonic for example the excitation force in a punching machine is periodic and it can be represented as shown in figure 13 Vibration analysis of system subjected to periodic but nonharmonic forces can be done with the help of Fourier series The problem becomes a multifrequency excitation problem The principle of linear superposition is applied and the total response is the sum of the response due to each of the individual frequency term

Any periodic motion can be expressed as an infinite sum of sines and cosines terms If x(t) is a periodic function with period t its Fourier representation is given by

X(t) = (((( )))) (((( )))) (((( ))))t sint cost cos2 111 ωωωωωωωωωωωω baaao ++++++++++++

= (((( )))) (((( ))))t sint cos2 1

ωωωωωωωω nnn

o bnaa

++++++++sumsumsumsuminfininfininfininfin

====

tππππωωωω

2==== = Fundamental frequency ndash (1)

where ao an bn are constants

ττττ ττττ2 ττττ3

ττττ

X(t)

22

t

X(t)

035

025

Determination of constants To find ao Integrate both sides of equation(1) over any interval ττττ All intergrals on the RHS of the equation are zero except the one containing ao

(((( ))))dttxao

o 2

2

intintintint====ωωωωππππ

ππππωωωω

= (((( ))))dttxo

2intintintintττττ

ττττ

To find an multiply equation 1 by cos (((( ))))tnωωωω and Integrate over any interval ττττ All intergrals

(((( )))) (((( )))) dttxao

n tncos 2

2

ωωωωππππωωωω ωωωω

ππππ

intintintint==== (((( )))) (((( )))) dttxo

tncos 2

ωωωωττττ

ττττ

intintintint====

To find bn multiply equation 1 by sin (((( ))))tnωωωω and Integrate over any interval ττττ All intergrals

(((( )))) (((( )))) dttxao

n tnsin 2

2

ωωωωππππωωωω ωωωω

ππππ

intintintint==== (((( )))) (((( )))) dttxo

tnsin 2

ωωωωττττ

ττττ

intintintint====

Find the Fourier series for the curve shown below

ττττ ττττ2 ττττ3

t

X(t)

E

ττττ ττττ2 ττττ3

t

X(t)

ππππ2

t

X(t) ππππ

1 Represent for the periodic motion shown in the figure

23

CHAPTER 2 UNDAMPED FREE VIBRATION LEARNING OBJECTIVES

] Introduction to undamped free vibration

] Terminologies used in undamped free vibration

] Three methods to solve the undamped free vibration problems

] Problems related to undamped free vibration problems

Free vibrations are oscillations about a systems equilibrium position that occur in the absence of an external excitation force If during vibrations there is no loss of energy it is known as undamped vibration The first step in solving a vibration problem is setting up the differential equation of motion Consider spring mass system which is assumed to move only along the vertical direction as shown below Let m be the mass of the block and k be the stiffness of the spring When block of mass m is attached to spring the deflection of spring will be ∆ known as static deflection In the static equilibrium position the free body diagram of forces acting on the mass is shown in Figure(b) Hence mg= kA Once the system is disturbed the system executes vibrations

Let at any instant of time t the mass is displaced from the equilibrium position x the different forces acting on the system are shown in figure (d) From Newtonrsquos second law of motion

sum = maF

Inertia force ( disturbing force) = restoring force

mgxkxm ++∆minus= )(ampamp

0)( =+ xkxm ampamp

or 0)( =+ xmk

xampamp

equation 2 is the differential equation of motion for spring mfigure Comparing equation (2) with the equation of SHM +xampampsince the vibrations of the above system are free( without the forces) we can write

sec radmk

n =ω

24

ass system shown in 0)(2 =xω

resistance of external

time period km

2 f1

n

πτ ==

from the equation(1) mg = k∆

∆=

q

mk

Difference between the translation ( rectilinear) and rotational system of vibration

Translatory Rotational

In the analysis thFORCES are co Linear stiffness K Problems 1A mass of 10kFind the natural 2 A spring massnatural frequencnatural frequenc

kt

m

sec radmk

n =ω

k

25

e disturbing and restoring nsidered

In the analysis In the analysis MASS Moment of Inertia (J) is considered

in Nm is considered

g when suspended from a spring causes a static deflection of 1cm frequency of system

system has a spring stiffness K Nm and a mass of m Kg It has a y of vibration 12 Hz An extra 2kg mass coupled to it then the y reduces by 2 Hz find K and m

secrad n =ω

26

3 A steel wire of 2mm diameter and 30m long It is fixed at the upper end and carries a mass of m kg at its free end Find m so that the frequency of longitudinal vibration is 4 Hz 4 A spring mass system has a natural period of 02 seconds What will be the new period if the spring constant is 1) increased by 50 2) decreased by 50 5 A spring mass system has a natural frequency of 10 Hz when the spring constant is reduced by 800 Nm the frequency is altered by 45 Find the mass and spring constant of the original system 6 Determine the natural frequency of system shown in fig is which shaft is supported in SHORT bearings 7 Determine the natural frequency of system shown in fig is which shaft is supported in LONG bearings where l is the length of bearing and E ndash youngrsquos modulus and I is moment of Inertia 7 Determine the natural frequency of system shown in fig is which shaft is supported in LONG bearings 8 A light cantilever beam of rectangular section( 5 cm deep x 25cm wide) has a mass fixed at its free end Find the ratio of frequency of free lateral vibration in vertical plane to that in horizontal 9 Determine the natural frequency of simple pendulum 10 Homogeneous square plate of size l and mass m is suspended from the mid point of one of its sides as shown in figure Find the frequency of vibration

l

l

27

11 A compound pendulum which is rigid body of mass m and it is pivoted at O The point of pivot is at distance d from the centre of gravity It is free to rotate about its axis Find the frequency of oscillation of such pendulum 12 A connecting rod shown in fig is supported at the wrist pin end It is displaced and allowed to oscillate The mass of rod is 5kg and centre of gravity is 20 cm from the pivot point O If the frequency of oscillation is 40 cyclesminute calculate the mass moment of inertia about its CG

13 A semi circular homogenous disc of radius r and mass m is pivoted freely about its centre as shown in figure Determine the natural frequency of oscillation 14A simply supported beam of square cross section 5mmx5mm and length 1m carrying a mass of 0575 kg at the middle is found to have natural frequency of 30 radsec Determine youngrsquos modulus of elasticity of beam 15 A spring mass system k1 and m have a natural frequency f1 Determine the value of k2 of another spring in terms of k1 which when placed in series with k1 lowers the natural frequency to 23 f1

COMPLETE SOLUTION OF SYSTEM EXECUTING SHM

The equation of motion of system executing SHM can be represented by

0)( =+ xkxm ampamp ----------------------------------------(1)

022

2=+

dtdx

dt

dx ω

The general solution of equation (1) can be expressed as

------------------------(2)

Where A and B are arbitrary constant which can be determined from the initial conditions of the system Two initial conditions are to be specified to evaluate these

constants x=x0 at t=0 and xamp = Vo at t=0 substituting in the equation (2) Energy method In a conservative system the total energy is constant The dwell as natural frequency can be determined by the princienergy For free vibration of undamped system at any instant and partly potential The kinetic energy T is stored in the masswhere as the potential energy U is stored in the form of sdeformation or work done in a force field such as gravity The total energy being constant T+U = constant Its rate of cha

t)(sin Bt)cos( AX ω+ω=

t)(sin V

t)cos( xx o0 ω

ω+ω=

Is the required complete

solution

28

ifferential equation as ple of conservation of of time is partly kinetic by virtue of its velocity train energy in elastic

nge

Is given by [ ] 0UTdtd

=+

From this we get a differential equation of motion as well as natural frequency of the system Determine the natural frequency of spring mass system using energy method

Determine the natural frequency of system shown in figure Determine the natural frequency of the system shown in figure Is there any limitation on the value of K Discuss Determine the natural frequency of s

m

l k

m

a

θ

m

a

θ

l

k

29

k

k

ystem shown below Neglect the mass of ball

l

m

30

A string shown in figure is under tension T which can be assumed to remain constant for small displacements Find the natural frequency of vertical vibrations of spring An acrobat 120kg walks on a tight rope as shown in figure The frequency of vibration in the given position is vertical direction is 30 rads Find the tension in the rope A manometer has a uniform bore of cross section area A If the column of liquid of length L and Density ρ is set into motion as shown in figure Find the frequency of the resulting oscillation

m

l

a

T T

36m 8m

31

k

R

k

m

r

m

Find the expression for natural frequency of system shown in the figure Neglect the mass of the cantilever beam Study the special case i) k=Infinity ii) I = infinity Determine the expression for the natural frequency of a system shown in figure The two discs are keyed to a common shaft and have a combined mass moment of inertia about centre of oscillation O It is assumed that the card attached to mass m does not stretch and is always under tension Determine the natural frequency of a system shown

m

l

32

Determine the expression for the natural frequency of the system shown in figure Assume that the wires connecting the masses do not stretch and are always in tension

Determine the natural frequency of spring mass system taking the MASS OF SPRING (ms ) into account

M3 M1

M2

k2 k1

m

k

33

RAYLEIGHrsquoS METHOD

This is the extension of energy method Here we can determine the natural frequency of a conservative system without finding the equation of motion The natural frequency is a function of the rate of change of Kinetic energy and potential energy of the system From the principle of conservation of energy we have T+U = Constant T= KE and U =PE We can write the above equation as T1+U1 = --------------------------(1) Where 1 and 2 represents tow instances of time Let 1 be the time when the mass is passing through static equilibrium position Let 2 be the time corresponding to the mass displacement of the mass At this instant the velocity f the mass is zero and hence Substituting in equation (1) we get ( )Max = ( )Max Above equation leads directly to natural frequency Determine the natural frequency of spring mass system using Rayleighrsquos method m

34

r

B

K K

Home work

Determine the natural frequency of pendulum using Rayleighrsquos method

A Cylinder of mass m and moment of inertia Jo is to role without slipping but is sustained by spring K as shown in figure Determine the natural frequency of oscillation

Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping

a

Determine the natural frequency of system shown where in cylindrical disc rolls over the inextensible spring A cylinder of radius r rolls without slipping on cylindrical the equation of motion and hence determine the naoscillations about the lowest point shown Repeat the above problem ndash Instead of cylinder assume swithout slipping on concave surface as shown above

r

K

35

portion of radius R Derive tural frequency of small

phere of radius with r rolls

The torsional pendulum has a natural frequency of 5 Hz What length of steel wire of diameter 2 mm should be used for pendulum The inertia of mass fixed at the free end is 00098 kgm2 Take g= 083x1011 Nm2 Determine the torsional natural frequency of the system shown in figure Neglect the mass moment of inertia of the shaft ( Figure not proportionate) Determine the natural frequency of simple pendulum considering the mass of the rod into account

Using the energy method find tThe chord may be assumed ine

Kt1

Kt2

20cm

06m 08m 04m

20cm 20cm 20cm

m

M

he natural frxtensible in

k

M

36

equency of the system shown in figure the spring mass pulley system

A rod of length l and depth H having a mass m rests on a semicylidrical surface os radius r It is tipped slightly and let go Find the frequency of oscillation

A thin bar of l having a mass m rests on a semicylidrical surface radius r It is tipped slightly and let go Find the frequency of oscillation

l

h

R

l

37

38

CHAPTER 3

DAMPED FREE VIBRATION OF SINGLE DEGREE FREEDOM SYSTEM LEARNING OBJECTIVES

] Introduction to damped free vibration

] Terminologies used in damped free vibration

] Three different types of damping

] Concept of critical damping and its importance

] Study of response of viscous damped system for cases of under damping

critical and over damping

] Concept of logarithmic decrement

] Problems related to damped free vibration problems

Viscously damped free vibration of single degree freedom sytem Consider a single degree freedom system with viscous damping as shown in figure For a viscous damped system the damping force is

where c is a constant of proportionality It is called damping coefficient or coefficient of viscous damping

Symbolically it is designated by a dashpot as shown in Fig 3

x xampx

39

At any instant of time let the mass be displaced by x from the equilibrium position Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

Okxxcxm =minusminus= ampampamp

Okxxcxm =++ ampampamp ----------------------------------------------------------(1)

or

0kxdtdx

Cdt

xdM

2

2

=++

Assuming the solution of equation as

The first term in the above equation m2Ct

eminus

is an exponentially decaying function of time The behavior of the terms in the parenthesis however depends on whether numerical value within the radicals is positive zero or negative

CASE 1 if 0mk

m2c

2

=minus

40

CASE 2 if 0mk

m2c

2

=gt

CASE 3 if 0mk

m2c

2

=lt

41

Critically damped coefficient Critically damped coefficient is defined as the value of the damping coeffient for which radical in equation

mk

m2c

m2c

S2

21 minus

plusmn

= becomes zero

Damping Ratio ( Damping factor) ( ξ ) It is defined as the ratio of damping coefficient to the critical damping coefficient Note The damping coefficient c is defined as the actual amount of damping while the damping coefficient is a constant depending upon the mass and stiffness of the system The roots of characteristics equation can be now written as Note if c=cc or ξ =1 then system is critically damped if cgt cc or ξ gt1 then system is over damped I if clt cc or ξ lt1 then system is under damped

42

Take home quiz Derive General solution of different damped system

1 Critically damped system 2 Under damped system 3 Over damped system

Logarithmic Decrement It is used to determine the amount of damping present in system This measure The rate of decay of free vibration The larger the damping the greater will be rate of decay Logarithmic decrement is defined as the natural logarithm of ratio of any two successive amplitudes

=δ

2

1

xx

ln

Expression for logarithmic decrement (δ ) Consider a damped free vibration which is expressed by

( ) ( ) φ+ξminusω= ξω t 1sinXex 2n

tn

x is the displacement at any instant of time X and φ are arbitrary constants

21

2

ξminus

πξ=δ

43

Tutorial problems 1 Derive an expression for damping ratio in terms of logarithmic decrement

2 Show that logarithmic decrement

=δ

n

o

xx

lnn1

where xo is the initial amplitude xn

is amplitude after n cycles 3 A mass of 1 kg is to be supported on a spring having stiffness K= 9800 Nm The damping constant c= 59 Nsm Determine the natural frequency of system Find also logarithmic decrement and amplitude after 3 cycles if initial displacement is 03 cm 4 A vibrating system is defined by the following parameters M = 3kg k= 100 Nm damping coefficient = 3 Nsm determine Damping factor natural frequency damped vibration logarithmic decrement ratio of consecutive amplitude no of cycles after which the original amplitude is reduced to 20 5A single degree damped vibrating system a suspended mass of 18kg makes 15 oscillations in 003 seconds The amplitude decreases to 025 of the initial value after 5 oscillations Determine the stiffness of spring Logarithmic decrement damping factor and damping coefficient 6 A damped vibration record of a spring mass system shows the following data Amplitude at the end of 2nd cycle is 9mm amplitude at eh end of 3rd cycle is 6mm amplitude at the end of 4th cycle is 4mm stiffness of spring =8000 Nm weight =20N Find the logarithmic decrement damping force at unit velocity periodic time of vibration 7 For the system shown in figure the characteristics of dashpot is such that the constant force of 49 N is applied to the piston it velocity is found to be constant at 012 ms Determine the value of C would you expect the complete system to be periodic or aperiodic 8 For a torsional system shown in the figuref1 is natural frequency in air and f2 is the natural frequency when immersed in oil Determine the damping coefficient

Viscous fluid

Disc of mass moment of Inertia J

2a

C

m M

b

a a

k

9 A thin plate of area A and weigh w is fixed to the end of the spring and allow to oscillate in a viscous fluid as shown in figure It f1 is the frequency of oscillation in air and f2 in liquid Show that

22

21 ff

gA2

minusπω

=micro where the damping force on the plate

AV2Fd micro= V being the velocity of vibration 10 A torsional pendulum has a natural frequency of 175 cyclessecond vibrating in vacuum The mass moment of inertia is 2 kgm2 It is immersed in oil and it is observed that the natural frequency is 142 cyclessecond Find the damping torque If the disc is replaced by 34 deg When in oil find the displacement at the end of the first complete cycle 11 An automobile can be modeled as a mass placed on 4 shock absorbers each consisting of a spring and a damper Such that each spring is equally loaded Determine the stiffness damping constant of each shock absorber So that natural frequency is 2Hz and system is critically damped The mass of vehicle is 200kg 12 Write the differential equation of motion for the system shown in the figure below Determine the natural frequency of the damped oscillation and critical damping coefficient

Determine the differential equation of motion and find the critical cdamped natural frequency for system shown in figure

a

C

m M

M =200kg

44

200kg kgkg

oefficient and

45

Derive the equation of motion and find the critical damping coefficient and damped natural frequency for the system shown below Write the differential equation of motion and determine the expression for critical damping coefficient and damped natural frequency of system shown in figure

Set up a differential equation of motion for system shown in figure and determine the following Undamped natural frequencies critical damping coefficient damping ratio damped natural frequency

b

C

kM

a

m

46

A mass of 300N is resting on two springs of stiffness 3000 Nm each and a dashpot of damping coefficient 150 Nsm as shown in figure If the initial velocity of 10cms is given to mass at its equilibrium position what will be the displacement from the equilibrium position at the end of first second Derive the formula used A gun barrel of mass 600kg has a recoil spring of stiffness 294 knm if the barrel recoils 13m on firing determine i) The initial recoil velocity of barrel ii) The critical damping coefficient of dashpot which is engaged at the end of recoil stroke Iii) Time required for the barrel to return at the end of recoil stroke Iv) time required for the barrel to return to a position 5cm from the initial position

W= 3000 N

FIRING

CHAPTER 4

FORCED VIBRATION LEARNING OBJECTIVES

] Introduction to forced free vibration

] Terminologies used in forced vibration

] Concept of magnification factor

] Study of reciprocating and rotating unbalance

] Concept of base excitation and support excitation

] Concept of Transmissibility ratio

] Problems related to forced vibration

The vibration that takes place under the excitation of external forces are called

forced vibration When a system is subjected to harmonic excitation it is forced to vibrate at the same frequency as that of excitation Most of the mechanical systems are prone to steady state forced vibration which cause fatigue failure The vibration problem also becomes very important when any of excitation frequencies coincide with one of natural frequencies of a system Under this condition resonance occurs during which large amplitude of vibrations are observed To avoid resonance external frequency (operating speed) may be changed or properties of the system may be changed to alter the natural frequency In some cases sufficient amount of damping may be provided to avoid large amplitude during resonance Thus the problem of force vibration is very important in mechanical design

FORCED VIBRATIONS OF SINGLE DOF SYSTEMS UNDER HARMONIC EXCITATION

` [ ] tsin FF o ω=

x

sin FF o=

Fo

xampamp

[ ] tω

xamp

47

48

Consider a spring mass system having a viscous damping excited by a harmonic force [ ] tsin FF o ω= as shown in figure The various forces acting on the system is shown in figure Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

O)tsin(Fkxxcxm o =ω+minusminus= ampampamp

)tsin(Fkxxcxm o ω=++ ampampamp ----------------------------------------------------------(1)

or

)tsin(Fkxdtdx

Cdt

xdM o2

2

ω=++

Equation (1) represents the differential equation of motion of the system The solution to this equation consists of two parts namely complementary function and a particular integral

The first of solution is also termed as transient solution of homogeneous solution and can be obtained from the homogeneous differential equation 0kxxcxm =++ ampampamp

The second part of solution particular integral is also termed as steady state solution ad is of form )tsin(Xx φminusω= Where X= steady state amplitude φ is phase difference between external force and displacement

)tsin(Xx φminusω=

The vector representation of the shown

49

The force polygon for the above vector diagram can be written as follows

( )φminusωt O

KX

xm 2ω

Fo

xCω

tω

φ

50

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

is the governing equation

Zero frequency deflection (Xst) It is deflection observed at zero frequency ( when the external frequency 0=ω

( ) kF

001

kF

x o

22

o

st =

+minus=

Magnification factor (M) It is defined as the ratio of steady state amplitude to zero frequency deflection xst

M = arX

X=

A plot of magnification factor versus frequency ratio

damping factor ( )ξ is called frequency plot of a harmon All the frequency terms starts from magnification factor of damping ratio ( )ξ The amplitude of vibration at any on the value of damping The amplitude will be higher for higher damping The amplitude of vibration at reso

amplitude tends to zero as

ωω

n tends to infinity

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

Freq

ζ =

ζ = 1

Magnification Factor M = XXst

kFo

51

ωω

nfor varying values of

ically excite vibrating system

M=1 what ever may be value particular frequency depends

for lesser damping and lesser nance will be very high The

uency Ratio r = (ωωn)

ζ = 0 0

ζ = 025 ζ = 0375 ζ = 0 5

ζ = 0707

ζ = 2

52

Plot of phase angle φ versus frequency ratio

ωω

n is shown in the

figure The phase angle φ is zero for all the values of damping ratio

at

ωω

n=0 The phase

angle φ is 90 for all the values of damping ratio

at

ωω

n=1 ie at

resonance The phase angle between o to 90 deg for all the values of

ωω

n= 0 to 1 and

between 90 deg to 180 deg for all the values of

ωω

ngt1

Class work Derive an expression for frequency at peak amplitude

2np 21 ξminusω=ω=ω is the required equation

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

53

Derive expression amplitude at resonance ( In dimensional form) In dimensional form We know that steady state amplitude

( ) ( )

ω+ωminus=

222

o

Cmk

Fx Equation in dimensional form

We know that at resonance mk

n =ω=ω

( )

ω+

minus

=2

n

2

oR

Cmk

mk

Fx

( )

ω=

2n

oR

C

Fx

( )

ω=

n

oR C

Fx

Class work Derive an expression for amplitude at resonance ( in non dimensional form)

Tutorial Problems on Forced Vibration 1A weight of 60N is suspended by a spring stiffness 12 kNm is forced to vibrate by a harmonic force of 10N assuming viscous damping of 0086 kN-sm Find i) resonance frequency ii) amplitude at resonance iii) phase angle at resonance iv) frequency corresponding to peak amplitude 2 A periodic torque having a maximum value of 05888Nm at a frequency of 4 rads is impressed upon a flywheel suspended from a wire The wheel has mass moment of inertia 012 kgm2 and the wire has stiffness 1176 Nmrad A viscous dashpot applies damping couple of 0784 Nm an angular velocity of 2 rads calculate i)maximum angular displacement from rest position ii) maximum couple applied to dashpot Iii) angle by which the angular displacement lags the torque A mass of 1019 kg is suspended from a spring stiffness 20 Nmm viscous damping causes the amplitude to decrease to 110th of initial value in 4 complete cycle If a periodic force of 20 cos(50 t) is applied to the mass Find the amplitude of forced vibration What would be the amplitude of oscillation if the applied force coincides with the natural period

m=1019 kg

[ ] t50cos 30F =

54

55

ROTATING UNBALANCE Unbalance in rotating machinery is a common source of vibration excitation Consider a machine of mass m having mounted on a foundation of stiffness k and damping coefficient C as shown in figure The unbalance is represented by an eccentric rotating mass m with eccentricity (e) and having a angular velocity (ω) let x be the displacement of non rotating mass from the static equilibrium position The displacement of rotating mass m from the static equilibrium is ( ) tsin( ex ω+ Applying Newtonrsquos law of motion we have

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

56

RECIPROCATING UNBALANCE Unbalance in reciprocating machinery is a common source of vibration excitation Consider a machine of mass M having mounted on a foundation of stiffness K and damping coefficient C as shown in figure The unbalance is represented by a reciprocating mass (m) having crank radus (e) and connecting rod of length (l) Let ωωωω be angular velocity of the crank Let x be the displacement of non reciprocating mass from the static equilibrium position at any instant of time The displacement of reciprocating mass m from the static equilibrium position is

)2sin(sin +++ tle

tex ωωωωωωωω

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

57

Note The equations obtained in reciprocating unbalance is same as the rotating unbalance and having the following differences Rotating Unbalance Reciprocating Unbalance Here mass of the rotating element is considered

Here mass of reciprocating element is considered

Here the eccentricity (e) which is the distance between the centre of rotation and CG

Here the eccentricity e is the crank radius and crank radius = stroke length2

The frequency plot of non dimensional factor( ( )me

MX versus frequency ratio

ωω

n for various values of ( )ξξξξ is shown in figure This frequency plot is same for

the rotating and reciprocating unbalance with the following points are noted from the diagram

1 Damping ratio plays very important role during resonance in controlling the amplitude

2 When

ωω

n=0 then ( )me

MX =0 for all the values of damping ratio ( )ξξξξ

3 When

ωω

nis higher than ( )me

MX tends to unity

4 At very high speeds ie at very high frequency ratio

ωω

n there is no effect

of damping The frequency plot of phase angle versus the frequency ratio is given as shown in figure The explanation is same as in forced harmonic vibration

58

VIBRATION ISOLATION

Vibratory forces and motions generated by machine and other caused are unavoidable Howeever their effect on the other dynamical systems can be minimized by proper isolator design An isolation system is either to protect o delicate machine from excessive vibration transmitted to it from its supporting structure or to prevent vibratory forces and motions generated by machines from being transmitted to its surroundings The effectiveness of isolation may be measured in terms of forces and motion transmitted The first type is known as force isolation and the second type is known as motion isolation

Force isolation (Force Transmissibility) When vibratory forces generated by machine are isolated the effectiveness of forced isolation can be made by defining the term transmissibility The term transmissibility is defined as the ratio of force transmitted to the foundation to that of the external impressed force

Force Transmissibility = o

tr

FF

force Impressed Externalfoundation toted transmitForce

=

Consider viscous damped spring mass system as shown in figure Upon this system an external harmonic force ) t (Sin Fo ω is applied The difference forces acting on system are inertia forces damping force spring force and external force All these forces are represented on force polygon shown below Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these two forces

( ) ( )22 xckx ω+=

59

This is the expression for force transmissibility in dimensional form To represent same in the non dimensional form divide numerator and denominator by k

2

n

22

n

2

n

o

tr

21

21

FF

TR

ωω

ξ+

ωω

minus

ωω

ξ+

==

PHASE DIFFERNECE BETWEEN EXTERNAL FORCE AND DISPLACEMENT Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these

ωω

minus

ωω

ξ=

ωminusω

=φ minusminus2

n

n1

2

1

1

2tan

mKC

tan

The phase difference between force transmitted to the foundation Ftr to the displacement

kc

KxxC

tanω

=ω

=α or

ωω

ξ=

ω=α minusminus

n

11 2tanKx

xCtan

The phase difference between external force Fo and force transmitted to foundationis β

60

β= αminusφ =

ωω

minus

ωω

ξminus

2

n

n1

1

2tan -

ωω

ξminus

n

1 2tan

Displacement Transmissibility It is the ratio of amplitude of steady state vibration of machine to that of the base motion Forced vibration due to support motion Consider a spring mass damper system subjected to harmonic excitation )tsin(Yy ω= at the base as shown in figure Let x be the absolute displacement of the mass at any instant of time t The different forces acting on system are shown in free body diagram Applying Newtonrsquos Law of motion Ma = sumF

)yx(k)yx(cxm minusminusminusminus= ampampampamp

0)yx(k)yx(cxm =minus+minus+ ampampampamp

61

Frequency plot of transmissibility versus frequency ratio

62

The plot of transmissibility versus frequency ratio is as shown in figure The following points can be noted from it

i) The value of transmissibility is unity at frequency ratio

ωω

n=0 and 2

ii) For an undamped system ( )ξ =0 transmissibility tends to infin at resonance iii) The value of transmissibility is less than unity for the value of frequency ratio

greater than 2

iv) For frequency ratio

ωω

nlt 2 smaller damping ratio lead to larger values of

transmissibility on the other hand for frequency ratio

ωω

ngt 2 smaller damping

ratio leads to smaller value of transmissibility

The frequency plot of transmissibility (TR) versus frequency ratio

ωω

n can be

broadly divided into two zones

1

ωω

n= 0 to 2

2

ωω

n= 2 to infin

CASE 1

When

ωω

n= 0 to 2

In this zone the value of transmissibility is always greater than 1 for all values of damping ratio In this zone the transmissibility decreases with increase in damping Hence we can conclude that in this zone damping is useful in isolating vibration forces being transmitted to the support CASE 2

When

ωω

n= 2 to infin

In this zone the value of transmissibility is less than 1 for all values of damping ratio In this zone the transmissibility increases with increase in damping Hence we can conclude that in this zone damping will adversely affect in isolating vibratory forces

63

Tutorial problems on Forced Vibration 1 Machine of mass 75kg is mounted on spring of stiffness 12000 Nm with an assumed damping factor ξ =02 piston within the machine of mass 2 kg has a reciprocating motion with a stroke of 75 cm and speed of 15 cyclessecond Assuming the motion of piston to be harmonic determine amplitude of machine ( X) Transmissibility and force transmitted to foundation phase angle of transmitted force wrt to exciting force 2 Motor weighs 600N is mounted on a simple beam that has a spring rate of 400 Ncm The motor armature base 160 N and has an eccentricity of 0005cm What will be the amplitude of motion when it runs at 1760 RPM Neglect damping the deflection of armature shaft and weight of the beam 3 A Heavy machine of mass 300 kg is supported on resilient foundation the static deflection of foundation due to weight of machine is found to be 75mm It is observed that the machine vibrates with an amplitude of 10mm when the base of the foundation is subjected to harmonic oscillation at the undampeds natural frequency of system with an amplitude of 25mm Find the damping constant of foundation Ii) dynamic force amplitude on the base iii) amplitude of the displacement of machine relative to base 4 An air compressor of mass 450 kg operates at a constant speed of 1750 RPM The reciprocating masses are 10 kg the crank radius is 100mm If the damper for mounting introduces a damping factor 015 i) specify the springs of mounting such that only 20 of the unbalanced force is transmitted to the foundation ii) determine the magnitude of the transmitted force bull Repeat the above problem by assuming that there is no damping in system

5 A body of a car having mass 1500 kg and is mounted on 4 equal springs which deflect through 225 cm under the weight of the body A total damping forces the 4 shock absorbers is 46 N at a velocity of 1cmsec The car is placed with all the four wheels on a test platform which is move up and down at resonant speed With an amplitude of 25cm Find the amplitude of the car body on its springs assuming its CG to be in the centre of wheel base 6A 75 kg machine which is mounted on springs of stiffness 116x105 Nm with the damping of 20=ξ A 2kg piston within the machine as reciprocating motion with a stroke of 008m and speed of 3000 RPM Assuming the motion of piston to be harmonic determine the amplitude of vibration of machine

7 The spring of an automobile are compressed 10cm under its own weight Find the critical speed when a trailer is traveling over road with a people approximaterly by a sine wave of amplitude 75 cm and wave length 15m What will be the amplitude of vibration at 64kmhour 8 An eccentric mass excites us used to determine vibratory characteristics of a structure of mass 25kg as given in figure At a speed of 1000 RPM the eccentric mass to beat top position at the instant the structure was moved upwards through its static equilibrium position and corresponding amplitude was 20mm The eccentric mass is 05kg at an eccentricity of 10mm Determine i) Undamped natural frequency of the system ii) Damping ratio iii) Steady state amplitude and phase angle at 1500 RPM iv) Force transmitted to the ground at 1200 RPM

M=25kg

C

K

Eccentric mass

64

65

CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Support motion Consider a spring mass damper system as shown in figure Let it excited by motion

of support let y be the harmonic displacement of support From Newtonrsquos second law of motion

)yx(k)yx(cxm minusminusminusminus= ampampampamp

)1(0)yx(k)yx(cxm minusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminus=minus+minus+ ampampampamp

Let z be the relative displacement of the mass with respect to support then

Z=x-y

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

7

The velocity of the mass m at an instant of time t is given by V = The acceleration of mass m is given by Hence we can conclude that in SHM the acceleration is proportional to displacement and is directed towards mean position observing the equations 2 and 3 the velocity and acceleration are harmonic with the same frequency but lead a displacement vector by π2 and π radians respectively

t

t

t X

x

X= A sin ωt

x-Displacement X-amplitude T-Periodic Time f-Frequency f=1T ω=Frequency in radians per second t= time

8

Rotor with inertia J1

kt1 kt2

Rotor with Inertia J2

Degrees of freedom The minimum number of independent coordinates required to determine completely the positions of all the parts of a system at any instant of time is called degrees of freedom One degree of freedom Two degree of freedom Three degree of freedom Infinite degree of freedom

m1

m2

m1

m2

m3

Cantilever Beam

Continuous system ( consider the mass of the beam )

m1

m1

K1

m2

K

x1

x2

F1

9

G

K1 K2

mJ

a b

Exercise Specify the no of degree of freedom for the following 1 2 3

4

5

4m 2m m 3k k k

Cantilever Beam

Continuous system ( Neglect the mass of the beam )

k1 k2 m1 m2

y y

m2

θ1

θ2

L1

L2

M1

M2

10

O

A

B

Addition of SIMPLE HARMONIC MOTION ( SHM) The addition of two simple harmonic motion having frequency yields a resultant which is simple harmonic having the same frequency Consider tow simple harmonic motions of 1x and 2x having the same frequency and phase difference φ as given below

)(sin11 tXx ωωωω====

)(sin22 φω += tXx

Adding x = x1 +x2 Hence the resultant displacement is also SHM of amplitude X and phase angle θ

θθθθ

ωωωωt

φφφφ

11

-6

-4

-2

0

2

4

6

0 005 01 015 02 025

Tutorial problems on Simple Harmonic Motion 1 Add the following harmonic motion analytically and verify the solution graphically 1) X1= 3 sin (((( ))))30 t ++++ωωωω X2= 4 cos (((( ))))10 t ++++ωωωω ( VTU Jan 2005) 2) X1= 2 cos (((( ))))05 t ++++ωωωω X2= 5 sin (((( ))))1 t ++++ωωωω ( VTU July 2006)

3) X1= 10 cos (((( ))))4 t ππππωωωω ++++ X2= 8 sin (((( ))))6 t ππππωωωω ++++ ( VTU Dec 2007)

2 A body is subjected to two harmonic motions X1= 8 cos (((( ))))6 t ππππωωωω ++++ X2= 15 sin (((( ))))6 t ππππωωωω ++++ what harmonic motion

should be given to the body to bring it to equilibrium (VTU July 2005) 3 Split the harmonic motion x = 10 sin (((( ))))6 t ππππωωωω ++++ into two harmonic motions

having the phase of zero and the other of 45o 4 Show that resultant motion of harmonic motion given below is zero X1= X sin (((( ))))t ωωωω X2= X sin (((( ))))3

2 t ππππωωωω ++++ X3= X sin (((( ))))34 t ππππωωωω ++++

5 The displacement of a vibrating body is given by x = 5 sin (3141t + 4

ππππ )

Draw the variation of displacement for one cycle of vibration Also determine the displacement of body after 011 second ( repeat the problem for velocity and acceleration and draw graph using Excel and compare ) Time Displacement

0 0025 5 005

0075 0 01

0125 015 -353

0175 0 02

Calculate the remaining values 6 The Motion of a particle is represented by x = 4 ( ) t sin ω sketch the variation of the displacement velocity and acceleration and determine the max value of these quantities Assume ( )5 =ω ( Try to use MATLABExcel) sketch all on same graph

12

BEATS When two harmonic motions with frequencies close to one another are added The resulting motion exhibits a phenomenon known as Beats A Beat Frequency is the result of two closely spaced frequencies going into and out of synchronization with one another Let us consider tow harmonic motion of same amplitude and slightly different frequencies X1 = X Cos ( ) tωωωω X2 = X Cos ( )( ) tδδδδωωωω + Where δ is a small quantity The addition of above two harmonics can be written as X = X1 + X2

X=

+

ttX2

cos2

cos2δδδδωωωω

δδδδ

The above equation shown graphically in Figure The resulting motion represents cosine wave with frequency ( ) 2

ππππωωωω + and with a varying amplitude 2 cos( ( ) 2δδδδ t

whenever the amplitude reaches a maximum it is called the beat The frequency δ at which amplitude builds up and dies down between o and 2 X is known as beat frequency Beat phenomenon is found in machines structures and electric power houses In machines the beat phenomenon occurs when the forcing frequency is close to the natural frequency of the system

+2X

-2X

ωωωωππππ2

ELEMENTS OF VIBRATION The elements of constitute vibrating systems are

1 Mass or Inertia element - m 2 Spring - k 3 Damper - c 4 Excitation F(t)

Mass or Inertia element The mass or inertia element is assumed to be rigid body During vibration velomass changes hence kinetic energy can be gained or loosed The force appthe mass from newton second law of motion can be written as F= ma Work dthe mass is stored in the form of kinetic energy given by KE= frac12 M V2

Combination of masses In practical cases for simple analysis we replace several masses by equivalent mass Case1 Translational masses connected by rigid bar Let the masses M1 M2 and M3 are attached to a rigid bar at locates 1 respectively as shown in the figure The equivalent mass meq be assumelocated at 1 is as shown in figure (b) Let the displacement of masses M1 M2 and M3 be x1 x2 x3 and simivelocities of respective masses be x1 x2 and x3 We can express the velomasses m2 and m3 in terms of m1

Elements of Vibration

Passive element

m C K

Active element

F(t)

Conservative element

( Mass and Spring)

Non conservative element

( Damper)

C k

m F(t)

Voig

13

city of lied on one on

a single

2 and 3 d to be

larly the cities of

t Model

Translationa Let a mass mmass momearrangement These two mamass Meq or

2

1

33

2

1

221

+

+=

ll

Mll

MMMeq

14

is the required answer

l and rotational masses coupled together

having a translational velocity x be coupled to another mass having nt of inertia Io with a rotational velocity θ as in rack and pinion

shown in the figure

sses can be combined to obtain either a single equivalent translational a single equivalent mass moment of inertia Jeq

15

Equivalent translational masses

Kinetic energy of the equivalent mass =

2

21

eqeq XM amp

Kinetic energy of the two masses =

+

20

2

21

21

θθθθampamp JXM

Meq= m

+ 2RJ

m o is the required answer

Also determine the equivalent rotational mass Jeq

Jeq= m [ ]oJmR +2 is the required answer Spring element Whenever there is a relative motion between the two ends of the spring a force is developed called spring force or restoring force The spring force is proportional to the amount of deformation x and then F α x or F = kx Where k is stiffness of the spring or spring constant or spring gradient The spring stiffness is equal to spring force per unit deromation

The spring stiffness k = mNxF

Workdone in deforming a spring is equal to the strain energy or potential energy Strain energy = potential energy = area of the triangle OAB Stiffness of beams Cantilever beam consider a cantilever beam with an end mass shown in the figure The mass of the beam is assumed to be negligible The static deflection of beam at free end is given by

Similarly derive the expression for Simply supported beam and fixed support beam

x

k

m

x

m

x

m

x

mNlIE

st 192

3

=δδδδ

mNEI

Wlst

48

3

=δδδδ

F

K=stiffness

F

16

17

Stiffness of slender bar subjected to longitudinal vibrations For a system executing the longitudinal vibrations as shown in the figure let us derive the expression for stiffness

Torsional Stiffness of bar It is the amount of torque required to cause a unit angular deformation in the element

Torsional stiffness = Kt = θθθθT

Combination of stiffness Determination of equivalent spring stiffness when the springs are arranged in series Consider two springs of stiffness K1 and K2 acted upon by the force F

The deflection of spring k1 is x1 = 1KF

The deflection of spring k1 is x2 = Let these two springs be replaced by an equivalent stiffness Keq upon which a force F acts and due to which its deflection is given by

x = eqKF

x= x1+x2

m

E A

l

60 cm 70 cm

Determination of equivalent spring stiffness when the springs are arranged in parallel Force acting on K1 spring = F1=k1x Force acting on K2 spring = F2 Force required for an equivalent spring keq to defined by x given by F= Keq x But F = F1 +F2 Tutorial problems on Equivalent stiffness of springs 1Determine the equivalent stiffness for the system shown in figure 2 Determine the equivalent stiffness for the system shown in figure

3 Determine the equivale

M

3

M

Nm 102 6x

18

nt stiffness for the system shown in figure

38 kg

Nm 101 6x Nm 102 6x

Nm 106x

2k

k

k

2k

3k

k

2k

k

19

60 cm 80 cm 50 cm

50 cm 60 cm

4Determine the equivalent stiffness for the system shown in figure

5Replace the following torsional stiffness by a single shaft having radius 4cm and find the length required for the equivalent shaft Assume the material of given system and equivalent system is same R1= 3cm R2= 5cm DAMPING Every vibration energy is gradually converted into heat or sound Hence the displacement during vibration gradually reduces The mechanism by which vibration energy is gradually converted into heat or sound is known as damping A damper is assumed to have either mass or elasticity Hence damping is modeled as one or more of the following types Viscous damping Coulomb or dry friction damping materials or solid or hysteric damping Viscous damping Viscous damping is most commonly used damping mechanism in vibration analysis When the mechanical system vibrates in a fluid medium such as air gas water or oil the resistance offered by the fluid to the moving body causes energy to be dissipated In this case the amount of dissipated energy depends on many factors such as size or shape of the vibrating body the viscosity of the fluid the frequency of vibration and velocity of fluid Resistance due to viscous damping is directly proportional to the velocity of vibration

dF V αααα

dF = xC amp

Where C= damping coefficient Fd = damping force Examples of Viscous damping

1) Fluid film between sliding surface 2) Fluid flow around a piston in a cylinder 3) Fluid flow through orifice 4) Fluid flow around a journal in a bearing

Rreq =4cm

leqn

stress

Strain

Coulomb damping or dry friction damping Here a damping force is constant in magnitude but opposite in direction to that of the motion of vibrating body It is caused by the friction between the surfaces that are dry or have insufficient lubrication Material or solid or hysteric damping

When the materials are deformed energy is absorbed and dissipated by the material The effect is due to friction between the internal planes which slip or slide as the deformation takes place When a body having the material damping is subjected to vibration the stress strain diagram shows the hysteresis loop as shown in the figure The area of the loop denotes the energy lost per unit volume of the body per cycle

20

21

ττττ ττττ2 ττττ3

ττττ

X(t)

FOURIER SERIES The simplest of periodic motion happens to be SHM It is simple to handle but the motion of many vibrating system is not harmonic (but periodic) Few examples are shown below

Forces acting on machines are generally periodic but this may not be harmonic for example the excitation force in a punching machine is periodic and it can be represented as shown in figure 13 Vibration analysis of system subjected to periodic but nonharmonic forces can be done with the help of Fourier series The problem becomes a multifrequency excitation problem The principle of linear superposition is applied and the total response is the sum of the response due to each of the individual frequency term

Any periodic motion can be expressed as an infinite sum of sines and cosines terms If x(t) is a periodic function with period t its Fourier representation is given by

X(t) = (((( )))) (((( )))) (((( ))))t sint cost cos2 111 ωωωωωωωωωωωω baaao ++++++++++++

= (((( )))) (((( ))))t sint cos2 1

ωωωωωωωω nnn

o bnaa

++++++++sumsumsumsuminfininfininfininfin

====

tππππωωωω

2==== = Fundamental frequency ndash (1)

where ao an bn are constants

ττττ ττττ2 ττττ3

ττττ

X(t)

22

t

X(t)

035

025

Determination of constants To find ao Integrate both sides of equation(1) over any interval ττττ All intergrals on the RHS of the equation are zero except the one containing ao

(((( ))))dttxao

o 2

2

intintintint====ωωωωππππ

ππππωωωω

= (((( ))))dttxo

2intintintintττττ

ττττ

To find an multiply equation 1 by cos (((( ))))tnωωωω and Integrate over any interval ττττ All intergrals

(((( )))) (((( )))) dttxao

n tncos 2

2

ωωωωππππωωωω ωωωω

ππππ

intintintint==== (((( )))) (((( )))) dttxo

tncos 2

ωωωωττττ

ττττ

intintintint====

To find bn multiply equation 1 by sin (((( ))))tnωωωω and Integrate over any interval ττττ All intergrals

(((( )))) (((( )))) dttxao

n tnsin 2

2

ωωωωππππωωωω ωωωω

ππππ

intintintint==== (((( )))) (((( )))) dttxo

tnsin 2

ωωωωττττ

ττττ

intintintint====

Find the Fourier series for the curve shown below

ττττ ττττ2 ττττ3

t

X(t)

E

ττττ ττττ2 ττττ3

t

X(t)

ππππ2

t

X(t) ππππ

1 Represent for the periodic motion shown in the figure

23

CHAPTER 2 UNDAMPED FREE VIBRATION LEARNING OBJECTIVES

] Introduction to undamped free vibration

] Terminologies used in undamped free vibration

] Three methods to solve the undamped free vibration problems

] Problems related to undamped free vibration problems

Free vibrations are oscillations about a systems equilibrium position that occur in the absence of an external excitation force If during vibrations there is no loss of energy it is known as undamped vibration The first step in solving a vibration problem is setting up the differential equation of motion Consider spring mass system which is assumed to move only along the vertical direction as shown below Let m be the mass of the block and k be the stiffness of the spring When block of mass m is attached to spring the deflection of spring will be ∆ known as static deflection In the static equilibrium position the free body diagram of forces acting on the mass is shown in Figure(b) Hence mg= kA Once the system is disturbed the system executes vibrations

Let at any instant of time t the mass is displaced from the equilibrium position x the different forces acting on the system are shown in figure (d) From Newtonrsquos second law of motion

sum = maF

Inertia force ( disturbing force) = restoring force

mgxkxm ++∆minus= )(ampamp

0)( =+ xkxm ampamp

or 0)( =+ xmk

xampamp

equation 2 is the differential equation of motion for spring mfigure Comparing equation (2) with the equation of SHM +xampampsince the vibrations of the above system are free( without the forces) we can write

sec radmk

n =ω

24

ass system shown in 0)(2 =xω

resistance of external

time period km

2 f1

n

πτ ==

from the equation(1) mg = k∆

∆=

q

mk

Difference between the translation ( rectilinear) and rotational system of vibration

Translatory Rotational

In the analysis thFORCES are co Linear stiffness K Problems 1A mass of 10kFind the natural 2 A spring massnatural frequencnatural frequenc

kt

m

sec radmk

n =ω

k

25

e disturbing and restoring nsidered

In the analysis In the analysis MASS Moment of Inertia (J) is considered

in Nm is considered

g when suspended from a spring causes a static deflection of 1cm frequency of system

system has a spring stiffness K Nm and a mass of m Kg It has a y of vibration 12 Hz An extra 2kg mass coupled to it then the y reduces by 2 Hz find K and m

secrad n =ω

26

3 A steel wire of 2mm diameter and 30m long It is fixed at the upper end and carries a mass of m kg at its free end Find m so that the frequency of longitudinal vibration is 4 Hz 4 A spring mass system has a natural period of 02 seconds What will be the new period if the spring constant is 1) increased by 50 2) decreased by 50 5 A spring mass system has a natural frequency of 10 Hz when the spring constant is reduced by 800 Nm the frequency is altered by 45 Find the mass and spring constant of the original system 6 Determine the natural frequency of system shown in fig is which shaft is supported in SHORT bearings 7 Determine the natural frequency of system shown in fig is which shaft is supported in LONG bearings where l is the length of bearing and E ndash youngrsquos modulus and I is moment of Inertia 7 Determine the natural frequency of system shown in fig is which shaft is supported in LONG bearings 8 A light cantilever beam of rectangular section( 5 cm deep x 25cm wide) has a mass fixed at its free end Find the ratio of frequency of free lateral vibration in vertical plane to that in horizontal 9 Determine the natural frequency of simple pendulum 10 Homogeneous square plate of size l and mass m is suspended from the mid point of one of its sides as shown in figure Find the frequency of vibration

l

l

27

11 A compound pendulum which is rigid body of mass m and it is pivoted at O The point of pivot is at distance d from the centre of gravity It is free to rotate about its axis Find the frequency of oscillation of such pendulum 12 A connecting rod shown in fig is supported at the wrist pin end It is displaced and allowed to oscillate The mass of rod is 5kg and centre of gravity is 20 cm from the pivot point O If the frequency of oscillation is 40 cyclesminute calculate the mass moment of inertia about its CG

13 A semi circular homogenous disc of radius r and mass m is pivoted freely about its centre as shown in figure Determine the natural frequency of oscillation 14A simply supported beam of square cross section 5mmx5mm and length 1m carrying a mass of 0575 kg at the middle is found to have natural frequency of 30 radsec Determine youngrsquos modulus of elasticity of beam 15 A spring mass system k1 and m have a natural frequency f1 Determine the value of k2 of another spring in terms of k1 which when placed in series with k1 lowers the natural frequency to 23 f1

COMPLETE SOLUTION OF SYSTEM EXECUTING SHM

The equation of motion of system executing SHM can be represented by

0)( =+ xkxm ampamp ----------------------------------------(1)

022

2=+

dtdx

dt

dx ω

The general solution of equation (1) can be expressed as

------------------------(2)

Where A and B are arbitrary constant which can be determined from the initial conditions of the system Two initial conditions are to be specified to evaluate these

constants x=x0 at t=0 and xamp = Vo at t=0 substituting in the equation (2) Energy method In a conservative system the total energy is constant The dwell as natural frequency can be determined by the princienergy For free vibration of undamped system at any instant and partly potential The kinetic energy T is stored in the masswhere as the potential energy U is stored in the form of sdeformation or work done in a force field such as gravity The total energy being constant T+U = constant Its rate of cha

t)(sin Bt)cos( AX ω+ω=

t)(sin V

t)cos( xx o0 ω

ω+ω=

Is the required complete

solution

28

ifferential equation as ple of conservation of of time is partly kinetic by virtue of its velocity train energy in elastic

nge

Is given by [ ] 0UTdtd

=+

From this we get a differential equation of motion as well as natural frequency of the system Determine the natural frequency of spring mass system using energy method

Determine the natural frequency of system shown in figure Determine the natural frequency of the system shown in figure Is there any limitation on the value of K Discuss Determine the natural frequency of s

m

l k

m

a

θ

m

a

θ

l

k

29

k

k

ystem shown below Neglect the mass of ball

l

m

30

A string shown in figure is under tension T which can be assumed to remain constant for small displacements Find the natural frequency of vertical vibrations of spring An acrobat 120kg walks on a tight rope as shown in figure The frequency of vibration in the given position is vertical direction is 30 rads Find the tension in the rope A manometer has a uniform bore of cross section area A If the column of liquid of length L and Density ρ is set into motion as shown in figure Find the frequency of the resulting oscillation

m

l

a

T T

36m 8m

31

k

R

k

m

r

m

Find the expression for natural frequency of system shown in the figure Neglect the mass of the cantilever beam Study the special case i) k=Infinity ii) I = infinity Determine the expression for the natural frequency of a system shown in figure The two discs are keyed to a common shaft and have a combined mass moment of inertia about centre of oscillation O It is assumed that the card attached to mass m does not stretch and is always under tension Determine the natural frequency of a system shown

m

l

32

Determine the expression for the natural frequency of the system shown in figure Assume that the wires connecting the masses do not stretch and are always in tension

Determine the natural frequency of spring mass system taking the MASS OF SPRING (ms ) into account

M3 M1

M2

k2 k1

m

k

33

RAYLEIGHrsquoS METHOD

This is the extension of energy method Here we can determine the natural frequency of a conservative system without finding the equation of motion The natural frequency is a function of the rate of change of Kinetic energy and potential energy of the system From the principle of conservation of energy we have T+U = Constant T= KE and U =PE We can write the above equation as T1+U1 = --------------------------(1) Where 1 and 2 represents tow instances of time Let 1 be the time when the mass is passing through static equilibrium position Let 2 be the time corresponding to the mass displacement of the mass At this instant the velocity f the mass is zero and hence Substituting in equation (1) we get ( )Max = ( )Max Above equation leads directly to natural frequency Determine the natural frequency of spring mass system using Rayleighrsquos method m

34

r

B

K K

Home work

Determine the natural frequency of pendulum using Rayleighrsquos method

A Cylinder of mass m and moment of inertia Jo is to role without slipping but is sustained by spring K as shown in figure Determine the natural frequency of oscillation

Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping

a

Determine the natural frequency of system shown where in cylindrical disc rolls over the inextensible spring A cylinder of radius r rolls without slipping on cylindrical the equation of motion and hence determine the naoscillations about the lowest point shown Repeat the above problem ndash Instead of cylinder assume swithout slipping on concave surface as shown above

r

K

35

portion of radius R Derive tural frequency of small

phere of radius with r rolls

The torsional pendulum has a natural frequency of 5 Hz What length of steel wire of diameter 2 mm should be used for pendulum The inertia of mass fixed at the free end is 00098 kgm2 Take g= 083x1011 Nm2 Determine the torsional natural frequency of the system shown in figure Neglect the mass moment of inertia of the shaft ( Figure not proportionate) Determine the natural frequency of simple pendulum considering the mass of the rod into account

Using the energy method find tThe chord may be assumed ine

Kt1

Kt2

20cm

06m 08m 04m

20cm 20cm 20cm

m

M

he natural frxtensible in

k

M

36

equency of the system shown in figure the spring mass pulley system

A rod of length l and depth H having a mass m rests on a semicylidrical surface os radius r It is tipped slightly and let go Find the frequency of oscillation

A thin bar of l having a mass m rests on a semicylidrical surface radius r It is tipped slightly and let go Find the frequency of oscillation

l

h

R

l

37

38

CHAPTER 3

DAMPED FREE VIBRATION OF SINGLE DEGREE FREEDOM SYSTEM LEARNING OBJECTIVES

] Introduction to damped free vibration

] Terminologies used in damped free vibration

] Three different types of damping

] Concept of critical damping and its importance

] Study of response of viscous damped system for cases of under damping

critical and over damping

] Concept of logarithmic decrement

] Problems related to damped free vibration problems

Viscously damped free vibration of single degree freedom sytem Consider a single degree freedom system with viscous damping as shown in figure For a viscous damped system the damping force is

where c is a constant of proportionality It is called damping coefficient or coefficient of viscous damping

Symbolically it is designated by a dashpot as shown in Fig 3

x xampx

39

At any instant of time let the mass be displaced by x from the equilibrium position Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

Okxxcxm =minusminus= ampampamp

Okxxcxm =++ ampampamp ----------------------------------------------------------(1)

or

0kxdtdx

Cdt

xdM

2

2

=++

Assuming the solution of equation as

The first term in the above equation m2Ct

eminus

is an exponentially decaying function of time The behavior of the terms in the parenthesis however depends on whether numerical value within the radicals is positive zero or negative

CASE 1 if 0mk

m2c

2

=minus

40

CASE 2 if 0mk

m2c

2

=gt

CASE 3 if 0mk

m2c

2

=lt

41

Critically damped coefficient Critically damped coefficient is defined as the value of the damping coeffient for which radical in equation

mk

m2c

m2c

S2

21 minus

plusmn

= becomes zero

Damping Ratio ( Damping factor) ( ξ ) It is defined as the ratio of damping coefficient to the critical damping coefficient Note The damping coefficient c is defined as the actual amount of damping while the damping coefficient is a constant depending upon the mass and stiffness of the system The roots of characteristics equation can be now written as Note if c=cc or ξ =1 then system is critically damped if cgt cc or ξ gt1 then system is over damped I if clt cc or ξ lt1 then system is under damped

42

Take home quiz Derive General solution of different damped system

1 Critically damped system 2 Under damped system 3 Over damped system

Logarithmic Decrement It is used to determine the amount of damping present in system This measure The rate of decay of free vibration The larger the damping the greater will be rate of decay Logarithmic decrement is defined as the natural logarithm of ratio of any two successive amplitudes

=δ

2

1

xx

ln

Expression for logarithmic decrement (δ ) Consider a damped free vibration which is expressed by

( ) ( ) φ+ξminusω= ξω t 1sinXex 2n

tn

x is the displacement at any instant of time X and φ are arbitrary constants

21

2

ξminus

πξ=δ

43

Tutorial problems 1 Derive an expression for damping ratio in terms of logarithmic decrement

2 Show that logarithmic decrement

=δ

n

o

xx

lnn1

where xo is the initial amplitude xn

is amplitude after n cycles 3 A mass of 1 kg is to be supported on a spring having stiffness K= 9800 Nm The damping constant c= 59 Nsm Determine the natural frequency of system Find also logarithmic decrement and amplitude after 3 cycles if initial displacement is 03 cm 4 A vibrating system is defined by the following parameters M = 3kg k= 100 Nm damping coefficient = 3 Nsm determine Damping factor natural frequency damped vibration logarithmic decrement ratio of consecutive amplitude no of cycles after which the original amplitude is reduced to 20 5A single degree damped vibrating system a suspended mass of 18kg makes 15 oscillations in 003 seconds The amplitude decreases to 025 of the initial value after 5 oscillations Determine the stiffness of spring Logarithmic decrement damping factor and damping coefficient 6 A damped vibration record of a spring mass system shows the following data Amplitude at the end of 2nd cycle is 9mm amplitude at eh end of 3rd cycle is 6mm amplitude at the end of 4th cycle is 4mm stiffness of spring =8000 Nm weight =20N Find the logarithmic decrement damping force at unit velocity periodic time of vibration 7 For the system shown in figure the characteristics of dashpot is such that the constant force of 49 N is applied to the piston it velocity is found to be constant at 012 ms Determine the value of C would you expect the complete system to be periodic or aperiodic 8 For a torsional system shown in the figuref1 is natural frequency in air and f2 is the natural frequency when immersed in oil Determine the damping coefficient

Viscous fluid

Disc of mass moment of Inertia J

2a

C

m M

b

a a

k

9 A thin plate of area A and weigh w is fixed to the end of the spring and allow to oscillate in a viscous fluid as shown in figure It f1 is the frequency of oscillation in air and f2 in liquid Show that

22

21 ff

gA2

minusπω

=micro where the damping force on the plate

AV2Fd micro= V being the velocity of vibration 10 A torsional pendulum has a natural frequency of 175 cyclessecond vibrating in vacuum The mass moment of inertia is 2 kgm2 It is immersed in oil and it is observed that the natural frequency is 142 cyclessecond Find the damping torque If the disc is replaced by 34 deg When in oil find the displacement at the end of the first complete cycle 11 An automobile can be modeled as a mass placed on 4 shock absorbers each consisting of a spring and a damper Such that each spring is equally loaded Determine the stiffness damping constant of each shock absorber So that natural frequency is 2Hz and system is critically damped The mass of vehicle is 200kg 12 Write the differential equation of motion for the system shown in the figure below Determine the natural frequency of the damped oscillation and critical damping coefficient

Determine the differential equation of motion and find the critical cdamped natural frequency for system shown in figure

a

C

m M

M =200kg

44

200kg kgkg

oefficient and

45

Derive the equation of motion and find the critical damping coefficient and damped natural frequency for the system shown below Write the differential equation of motion and determine the expression for critical damping coefficient and damped natural frequency of system shown in figure

Set up a differential equation of motion for system shown in figure and determine the following Undamped natural frequencies critical damping coefficient damping ratio damped natural frequency

b

C

kM

a

m

46

A mass of 300N is resting on two springs of stiffness 3000 Nm each and a dashpot of damping coefficient 150 Nsm as shown in figure If the initial velocity of 10cms is given to mass at its equilibrium position what will be the displacement from the equilibrium position at the end of first second Derive the formula used A gun barrel of mass 600kg has a recoil spring of stiffness 294 knm if the barrel recoils 13m on firing determine i) The initial recoil velocity of barrel ii) The critical damping coefficient of dashpot which is engaged at the end of recoil stroke Iii) Time required for the barrel to return at the end of recoil stroke Iv) time required for the barrel to return to a position 5cm from the initial position

W= 3000 N

FIRING

CHAPTER 4

FORCED VIBRATION LEARNING OBJECTIVES

] Introduction to forced free vibration

] Terminologies used in forced vibration

] Concept of magnification factor

] Study of reciprocating and rotating unbalance

] Concept of base excitation and support excitation

] Concept of Transmissibility ratio

] Problems related to forced vibration

The vibration that takes place under the excitation of external forces are called

forced vibration When a system is subjected to harmonic excitation it is forced to vibrate at the same frequency as that of excitation Most of the mechanical systems are prone to steady state forced vibration which cause fatigue failure The vibration problem also becomes very important when any of excitation frequencies coincide with one of natural frequencies of a system Under this condition resonance occurs during which large amplitude of vibrations are observed To avoid resonance external frequency (operating speed) may be changed or properties of the system may be changed to alter the natural frequency In some cases sufficient amount of damping may be provided to avoid large amplitude during resonance Thus the problem of force vibration is very important in mechanical design

FORCED VIBRATIONS OF SINGLE DOF SYSTEMS UNDER HARMONIC EXCITATION

` [ ] tsin FF o ω=

x

sin FF o=

Fo

xampamp

[ ] tω

xamp

47

48

Consider a spring mass system having a viscous damping excited by a harmonic force [ ] tsin FF o ω= as shown in figure The various forces acting on the system is shown in figure Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

O)tsin(Fkxxcxm o =ω+minusminus= ampampamp

)tsin(Fkxxcxm o ω=++ ampampamp ----------------------------------------------------------(1)

or

)tsin(Fkxdtdx

Cdt

xdM o2

2

ω=++

Equation (1) represents the differential equation of motion of the system The solution to this equation consists of two parts namely complementary function and a particular integral

The first of solution is also termed as transient solution of homogeneous solution and can be obtained from the homogeneous differential equation 0kxxcxm =++ ampampamp

The second part of solution particular integral is also termed as steady state solution ad is of form )tsin(Xx φminusω= Where X= steady state amplitude φ is phase difference between external force and displacement

)tsin(Xx φminusω=

The vector representation of the shown

49

The force polygon for the above vector diagram can be written as follows

( )φminusωt O

KX

xm 2ω

Fo

xCω

tω

φ

50

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

is the governing equation

Zero frequency deflection (Xst) It is deflection observed at zero frequency ( when the external frequency 0=ω

( ) kF

001

kF

x o

22

o

st =

+minus=

Magnification factor (M) It is defined as the ratio of steady state amplitude to zero frequency deflection xst

M = arX

X=

A plot of magnification factor versus frequency ratio

damping factor ( )ξ is called frequency plot of a harmon All the frequency terms starts from magnification factor of damping ratio ( )ξ The amplitude of vibration at any on the value of damping The amplitude will be higher for higher damping The amplitude of vibration at reso

amplitude tends to zero as

ωω

n tends to infinity

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

Freq

ζ =

ζ = 1

Magnification Factor M = XXst

kFo

51

ωω

nfor varying values of

ically excite vibrating system

M=1 what ever may be value particular frequency depends

for lesser damping and lesser nance will be very high The

uency Ratio r = (ωωn)

ζ = 0 0

ζ = 025 ζ = 0375 ζ = 0 5

ζ = 0707

ζ = 2

52

Plot of phase angle φ versus frequency ratio

ωω

n is shown in the

figure The phase angle φ is zero for all the values of damping ratio

at

ωω

n=0 The phase

angle φ is 90 for all the values of damping ratio

at

ωω

n=1 ie at

resonance The phase angle between o to 90 deg for all the values of

ωω

n= 0 to 1 and

between 90 deg to 180 deg for all the values of

ωω

ngt1

Class work Derive an expression for frequency at peak amplitude

2np 21 ξminusω=ω=ω is the required equation

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

53

Derive expression amplitude at resonance ( In dimensional form) In dimensional form We know that steady state amplitude

( ) ( )

ω+ωminus=

222

o

Cmk

Fx Equation in dimensional form

We know that at resonance mk

n =ω=ω

( )

ω+

minus

=2

n

2

oR

Cmk

mk

Fx

( )

ω=

2n

oR

C

Fx

( )

ω=

n

oR C

Fx

Class work Derive an expression for amplitude at resonance ( in non dimensional form)

Tutorial Problems on Forced Vibration 1A weight of 60N is suspended by a spring stiffness 12 kNm is forced to vibrate by a harmonic force of 10N assuming viscous damping of 0086 kN-sm Find i) resonance frequency ii) amplitude at resonance iii) phase angle at resonance iv) frequency corresponding to peak amplitude 2 A periodic torque having a maximum value of 05888Nm at a frequency of 4 rads is impressed upon a flywheel suspended from a wire The wheel has mass moment of inertia 012 kgm2 and the wire has stiffness 1176 Nmrad A viscous dashpot applies damping couple of 0784 Nm an angular velocity of 2 rads calculate i)maximum angular displacement from rest position ii) maximum couple applied to dashpot Iii) angle by which the angular displacement lags the torque A mass of 1019 kg is suspended from a spring stiffness 20 Nmm viscous damping causes the amplitude to decrease to 110th of initial value in 4 complete cycle If a periodic force of 20 cos(50 t) is applied to the mass Find the amplitude of forced vibration What would be the amplitude of oscillation if the applied force coincides with the natural period

m=1019 kg

[ ] t50cos 30F =

54

55

ROTATING UNBALANCE Unbalance in rotating machinery is a common source of vibration excitation Consider a machine of mass m having mounted on a foundation of stiffness k and damping coefficient C as shown in figure The unbalance is represented by an eccentric rotating mass m with eccentricity (e) and having a angular velocity (ω) let x be the displacement of non rotating mass from the static equilibrium position The displacement of rotating mass m from the static equilibrium is ( ) tsin( ex ω+ Applying Newtonrsquos law of motion we have

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

56

RECIPROCATING UNBALANCE Unbalance in reciprocating machinery is a common source of vibration excitation Consider a machine of mass M having mounted on a foundation of stiffness K and damping coefficient C as shown in figure The unbalance is represented by a reciprocating mass (m) having crank radus (e) and connecting rod of length (l) Let ωωωω be angular velocity of the crank Let x be the displacement of non reciprocating mass from the static equilibrium position at any instant of time The displacement of reciprocating mass m from the static equilibrium position is

)2sin(sin +++ tle

tex ωωωωωωωω

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

57

Note The equations obtained in reciprocating unbalance is same as the rotating unbalance and having the following differences Rotating Unbalance Reciprocating Unbalance Here mass of the rotating element is considered

Here mass of reciprocating element is considered

Here the eccentricity (e) which is the distance between the centre of rotation and CG

Here the eccentricity e is the crank radius and crank radius = stroke length2

The frequency plot of non dimensional factor( ( )me

MX versus frequency ratio

ωω

n for various values of ( )ξξξξ is shown in figure This frequency plot is same for

the rotating and reciprocating unbalance with the following points are noted from the diagram

1 Damping ratio plays very important role during resonance in controlling the amplitude

2 When

ωω

n=0 then ( )me

MX =0 for all the values of damping ratio ( )ξξξξ

3 When

ωω

nis higher than ( )me

MX tends to unity

4 At very high speeds ie at very high frequency ratio

ωω

n there is no effect

of damping The frequency plot of phase angle versus the frequency ratio is given as shown in figure The explanation is same as in forced harmonic vibration

58

VIBRATION ISOLATION

Vibratory forces and motions generated by machine and other caused are unavoidable Howeever their effect on the other dynamical systems can be minimized by proper isolator design An isolation system is either to protect o delicate machine from excessive vibration transmitted to it from its supporting structure or to prevent vibratory forces and motions generated by machines from being transmitted to its surroundings The effectiveness of isolation may be measured in terms of forces and motion transmitted The first type is known as force isolation and the second type is known as motion isolation

Force isolation (Force Transmissibility) When vibratory forces generated by machine are isolated the effectiveness of forced isolation can be made by defining the term transmissibility The term transmissibility is defined as the ratio of force transmitted to the foundation to that of the external impressed force

Force Transmissibility = o

tr

FF

force Impressed Externalfoundation toted transmitForce

=

Consider viscous damped spring mass system as shown in figure Upon this system an external harmonic force ) t (Sin Fo ω is applied The difference forces acting on system are inertia forces damping force spring force and external force All these forces are represented on force polygon shown below Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these two forces

( ) ( )22 xckx ω+=

59

This is the expression for force transmissibility in dimensional form To represent same in the non dimensional form divide numerator and denominator by k

2

n

22

n

2

n

o

tr

21

21

FF

TR

ωω

ξ+

ωω

minus

ωω

ξ+

==

PHASE DIFFERNECE BETWEEN EXTERNAL FORCE AND DISPLACEMENT Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these

ωω

minus

ωω

ξ=

ωminusω

=φ minusminus2

n

n1

2

1

1

2tan

mKC

tan

The phase difference between force transmitted to the foundation Ftr to the displacement

kc

KxxC

tanω

=ω

=α or

ωω

ξ=

ω=α minusminus

n

11 2tanKx

xCtan

The phase difference between external force Fo and force transmitted to foundationis β

60

β= αminusφ =

ωω

minus

ωω

ξminus

2

n

n1

1

2tan -

ωω

ξminus

n

1 2tan

Displacement Transmissibility It is the ratio of amplitude of steady state vibration of machine to that of the base motion Forced vibration due to support motion Consider a spring mass damper system subjected to harmonic excitation )tsin(Yy ω= at the base as shown in figure Let x be the absolute displacement of the mass at any instant of time t The different forces acting on system are shown in free body diagram Applying Newtonrsquos Law of motion Ma = sumF

)yx(k)yx(cxm minusminusminusminus= ampampampamp

0)yx(k)yx(cxm =minus+minus+ ampampampamp

61

Frequency plot of transmissibility versus frequency ratio

62

The plot of transmissibility versus frequency ratio is as shown in figure The following points can be noted from it

i) The value of transmissibility is unity at frequency ratio

ωω

n=0 and 2

ii) For an undamped system ( )ξ =0 transmissibility tends to infin at resonance iii) The value of transmissibility is less than unity for the value of frequency ratio

greater than 2

iv) For frequency ratio

ωω

nlt 2 smaller damping ratio lead to larger values of

transmissibility on the other hand for frequency ratio

ωω

ngt 2 smaller damping

ratio leads to smaller value of transmissibility

The frequency plot of transmissibility (TR) versus frequency ratio

ωω

n can be

broadly divided into two zones

1

ωω

n= 0 to 2

2

ωω

n= 2 to infin

CASE 1

When

ωω

n= 0 to 2

In this zone the value of transmissibility is always greater than 1 for all values of damping ratio In this zone the transmissibility decreases with increase in damping Hence we can conclude that in this zone damping is useful in isolating vibration forces being transmitted to the support CASE 2

When

ωω

n= 2 to infin

In this zone the value of transmissibility is less than 1 for all values of damping ratio In this zone the transmissibility increases with increase in damping Hence we can conclude that in this zone damping will adversely affect in isolating vibratory forces

63

Tutorial problems on Forced Vibration 1 Machine of mass 75kg is mounted on spring of stiffness 12000 Nm with an assumed damping factor ξ =02 piston within the machine of mass 2 kg has a reciprocating motion with a stroke of 75 cm and speed of 15 cyclessecond Assuming the motion of piston to be harmonic determine amplitude of machine ( X) Transmissibility and force transmitted to foundation phase angle of transmitted force wrt to exciting force 2 Motor weighs 600N is mounted on a simple beam that has a spring rate of 400 Ncm The motor armature base 160 N and has an eccentricity of 0005cm What will be the amplitude of motion when it runs at 1760 RPM Neglect damping the deflection of armature shaft and weight of the beam 3 A Heavy machine of mass 300 kg is supported on resilient foundation the static deflection of foundation due to weight of machine is found to be 75mm It is observed that the machine vibrates with an amplitude of 10mm when the base of the foundation is subjected to harmonic oscillation at the undampeds natural frequency of system with an amplitude of 25mm Find the damping constant of foundation Ii) dynamic force amplitude on the base iii) amplitude of the displacement of machine relative to base 4 An air compressor of mass 450 kg operates at a constant speed of 1750 RPM The reciprocating masses are 10 kg the crank radius is 100mm If the damper for mounting introduces a damping factor 015 i) specify the springs of mounting such that only 20 of the unbalanced force is transmitted to the foundation ii) determine the magnitude of the transmitted force bull Repeat the above problem by assuming that there is no damping in system

5 A body of a car having mass 1500 kg and is mounted on 4 equal springs which deflect through 225 cm under the weight of the body A total damping forces the 4 shock absorbers is 46 N at a velocity of 1cmsec The car is placed with all the four wheels on a test platform which is move up and down at resonant speed With an amplitude of 25cm Find the amplitude of the car body on its springs assuming its CG to be in the centre of wheel base 6A 75 kg machine which is mounted on springs of stiffness 116x105 Nm with the damping of 20=ξ A 2kg piston within the machine as reciprocating motion with a stroke of 008m and speed of 3000 RPM Assuming the motion of piston to be harmonic determine the amplitude of vibration of machine

7 The spring of an automobile are compressed 10cm under its own weight Find the critical speed when a trailer is traveling over road with a people approximaterly by a sine wave of amplitude 75 cm and wave length 15m What will be the amplitude of vibration at 64kmhour 8 An eccentric mass excites us used to determine vibratory characteristics of a structure of mass 25kg as given in figure At a speed of 1000 RPM the eccentric mass to beat top position at the instant the structure was moved upwards through its static equilibrium position and corresponding amplitude was 20mm The eccentric mass is 05kg at an eccentricity of 10mm Determine i) Undamped natural frequency of the system ii) Damping ratio iii) Steady state amplitude and phase angle at 1500 RPM iv) Force transmitted to the ground at 1200 RPM

M=25kg

C

K

Eccentric mass

64

65

CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Support motion Consider a spring mass damper system as shown in figure Let it excited by motion

of support let y be the harmonic displacement of support From Newtonrsquos second law of motion

)yx(k)yx(cxm minusminusminusminus= ampampampamp

)1(0)yx(k)yx(cxm minusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminus=minus+minus+ ampampampamp

Let z be the relative displacement of the mass with respect to support then

Z=x-y

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

8

Rotor with inertia J1

kt1 kt2

Rotor with Inertia J2

Degrees of freedom The minimum number of independent coordinates required to determine completely the positions of all the parts of a system at any instant of time is called degrees of freedom One degree of freedom Two degree of freedom Three degree of freedom Infinite degree of freedom

m1

m2

m1

m2

m3

Cantilever Beam

Continuous system ( consider the mass of the beam )

m1

m1

K1

m2

K

x1

x2

F1

9

G

K1 K2

mJ

a b

Exercise Specify the no of degree of freedom for the following 1 2 3

4

5

4m 2m m 3k k k

Cantilever Beam

Continuous system ( Neglect the mass of the beam )

k1 k2 m1 m2

y y

m2

θ1

θ2

L1

L2

M1

M2

10

O

A

B

Addition of SIMPLE HARMONIC MOTION ( SHM) The addition of two simple harmonic motion having frequency yields a resultant which is simple harmonic having the same frequency Consider tow simple harmonic motions of 1x and 2x having the same frequency and phase difference φ as given below

)(sin11 tXx ωωωω====

)(sin22 φω += tXx

Adding x = x1 +x2 Hence the resultant displacement is also SHM of amplitude X and phase angle θ

θθθθ

ωωωωt

φφφφ

11

-6

-4

-2

0

2

4

6

0 005 01 015 02 025

Tutorial problems on Simple Harmonic Motion 1 Add the following harmonic motion analytically and verify the solution graphically 1) X1= 3 sin (((( ))))30 t ++++ωωωω X2= 4 cos (((( ))))10 t ++++ωωωω ( VTU Jan 2005) 2) X1= 2 cos (((( ))))05 t ++++ωωωω X2= 5 sin (((( ))))1 t ++++ωωωω ( VTU July 2006)

3) X1= 10 cos (((( ))))4 t ππππωωωω ++++ X2= 8 sin (((( ))))6 t ππππωωωω ++++ ( VTU Dec 2007)

2 A body is subjected to two harmonic motions X1= 8 cos (((( ))))6 t ππππωωωω ++++ X2= 15 sin (((( ))))6 t ππππωωωω ++++ what harmonic motion

should be given to the body to bring it to equilibrium (VTU July 2005) 3 Split the harmonic motion x = 10 sin (((( ))))6 t ππππωωωω ++++ into two harmonic motions

having the phase of zero and the other of 45o 4 Show that resultant motion of harmonic motion given below is zero X1= X sin (((( ))))t ωωωω X2= X sin (((( ))))3

2 t ππππωωωω ++++ X3= X sin (((( ))))34 t ππππωωωω ++++

5 The displacement of a vibrating body is given by x = 5 sin (3141t + 4

ππππ )

Draw the variation of displacement for one cycle of vibration Also determine the displacement of body after 011 second ( repeat the problem for velocity and acceleration and draw graph using Excel and compare ) Time Displacement

0 0025 5 005

0075 0 01

0125 015 -353

0175 0 02

Calculate the remaining values 6 The Motion of a particle is represented by x = 4 ( ) t sin ω sketch the variation of the displacement velocity and acceleration and determine the max value of these quantities Assume ( )5 =ω ( Try to use MATLABExcel) sketch all on same graph

12

BEATS When two harmonic motions with frequencies close to one another are added The resulting motion exhibits a phenomenon known as Beats A Beat Frequency is the result of two closely spaced frequencies going into and out of synchronization with one another Let us consider tow harmonic motion of same amplitude and slightly different frequencies X1 = X Cos ( ) tωωωω X2 = X Cos ( )( ) tδδδδωωωω + Where δ is a small quantity The addition of above two harmonics can be written as X = X1 + X2

X=

+

ttX2

cos2

cos2δδδδωωωω

δδδδ

The above equation shown graphically in Figure The resulting motion represents cosine wave with frequency ( ) 2

ππππωωωω + and with a varying amplitude 2 cos( ( ) 2δδδδ t

whenever the amplitude reaches a maximum it is called the beat The frequency δ at which amplitude builds up and dies down between o and 2 X is known as beat frequency Beat phenomenon is found in machines structures and electric power houses In machines the beat phenomenon occurs when the forcing frequency is close to the natural frequency of the system

+2X

-2X

ωωωωππππ2

ELEMENTS OF VIBRATION The elements of constitute vibrating systems are

1 Mass or Inertia element - m 2 Spring - k 3 Damper - c 4 Excitation F(t)

Mass or Inertia element The mass or inertia element is assumed to be rigid body During vibration velomass changes hence kinetic energy can be gained or loosed The force appthe mass from newton second law of motion can be written as F= ma Work dthe mass is stored in the form of kinetic energy given by KE= frac12 M V2

Combination of masses In practical cases for simple analysis we replace several masses by equivalent mass Case1 Translational masses connected by rigid bar Let the masses M1 M2 and M3 are attached to a rigid bar at locates 1 respectively as shown in the figure The equivalent mass meq be assumelocated at 1 is as shown in figure (b) Let the displacement of masses M1 M2 and M3 be x1 x2 x3 and simivelocities of respective masses be x1 x2 and x3 We can express the velomasses m2 and m3 in terms of m1

Elements of Vibration

Passive element

m C K

Active element

F(t)

Conservative element

( Mass and Spring)

Non conservative element

( Damper)

C k

m F(t)

Voig

13

city of lied on one on

a single

2 and 3 d to be

larly the cities of

t Model

Translationa Let a mass mmass momearrangement These two mamass Meq or

2

1

33

2

1

221

+

+=

ll

Mll

MMMeq

14

is the required answer

l and rotational masses coupled together

having a translational velocity x be coupled to another mass having nt of inertia Io with a rotational velocity θ as in rack and pinion

shown in the figure

sses can be combined to obtain either a single equivalent translational a single equivalent mass moment of inertia Jeq

15

Equivalent translational masses

Kinetic energy of the equivalent mass =

2

21

eqeq XM amp

Kinetic energy of the two masses =

+

20

2

21

21

θθθθampamp JXM

Meq= m

+ 2RJ

m o is the required answer

Also determine the equivalent rotational mass Jeq

Jeq= m [ ]oJmR +2 is the required answer Spring element Whenever there is a relative motion between the two ends of the spring a force is developed called spring force or restoring force The spring force is proportional to the amount of deformation x and then F α x or F = kx Where k is stiffness of the spring or spring constant or spring gradient The spring stiffness is equal to spring force per unit deromation

The spring stiffness k = mNxF

Workdone in deforming a spring is equal to the strain energy or potential energy Strain energy = potential energy = area of the triangle OAB Stiffness of beams Cantilever beam consider a cantilever beam with an end mass shown in the figure The mass of the beam is assumed to be negligible The static deflection of beam at free end is given by

Similarly derive the expression for Simply supported beam and fixed support beam

x

k

m

x

m

x

m

x

mNlIE

st 192

3

=δδδδ

mNEI

Wlst

48

3

=δδδδ

F

K=stiffness

F

16

17

Stiffness of slender bar subjected to longitudinal vibrations For a system executing the longitudinal vibrations as shown in the figure let us derive the expression for stiffness

Torsional Stiffness of bar It is the amount of torque required to cause a unit angular deformation in the element

Torsional stiffness = Kt = θθθθT

Combination of stiffness Determination of equivalent spring stiffness when the springs are arranged in series Consider two springs of stiffness K1 and K2 acted upon by the force F

The deflection of spring k1 is x1 = 1KF

The deflection of spring k1 is x2 = Let these two springs be replaced by an equivalent stiffness Keq upon which a force F acts and due to which its deflection is given by

x = eqKF

x= x1+x2

m

E A

l

60 cm 70 cm

Determination of equivalent spring stiffness when the springs are arranged in parallel Force acting on K1 spring = F1=k1x Force acting on K2 spring = F2 Force required for an equivalent spring keq to defined by x given by F= Keq x But F = F1 +F2 Tutorial problems on Equivalent stiffness of springs 1Determine the equivalent stiffness for the system shown in figure 2 Determine the equivalent stiffness for the system shown in figure

3 Determine the equivale

M

3

M

Nm 102 6x

18

nt stiffness for the system shown in figure

38 kg

Nm 101 6x Nm 102 6x

Nm 106x

2k

k

k

2k

3k

k

2k

k

19

60 cm 80 cm 50 cm

50 cm 60 cm

4Determine the equivalent stiffness for the system shown in figure

5Replace the following torsional stiffness by a single shaft having radius 4cm and find the length required for the equivalent shaft Assume the material of given system and equivalent system is same R1= 3cm R2= 5cm DAMPING Every vibration energy is gradually converted into heat or sound Hence the displacement during vibration gradually reduces The mechanism by which vibration energy is gradually converted into heat or sound is known as damping A damper is assumed to have either mass or elasticity Hence damping is modeled as one or more of the following types Viscous damping Coulomb or dry friction damping materials or solid or hysteric damping Viscous damping Viscous damping is most commonly used damping mechanism in vibration analysis When the mechanical system vibrates in a fluid medium such as air gas water or oil the resistance offered by the fluid to the moving body causes energy to be dissipated In this case the amount of dissipated energy depends on many factors such as size or shape of the vibrating body the viscosity of the fluid the frequency of vibration and velocity of fluid Resistance due to viscous damping is directly proportional to the velocity of vibration

dF V αααα

dF = xC amp

Where C= damping coefficient Fd = damping force Examples of Viscous damping

1) Fluid film between sliding surface 2) Fluid flow around a piston in a cylinder 3) Fluid flow through orifice 4) Fluid flow around a journal in a bearing

Rreq =4cm

leqn

stress

Strain

Coulomb damping or dry friction damping Here a damping force is constant in magnitude but opposite in direction to that of the motion of vibrating body It is caused by the friction between the surfaces that are dry or have insufficient lubrication Material or solid or hysteric damping

When the materials are deformed energy is absorbed and dissipated by the material The effect is due to friction between the internal planes which slip or slide as the deformation takes place When a body having the material damping is subjected to vibration the stress strain diagram shows the hysteresis loop as shown in the figure The area of the loop denotes the energy lost per unit volume of the body per cycle

20

21

ττττ ττττ2 ττττ3

ττττ

X(t)

FOURIER SERIES The simplest of periodic motion happens to be SHM It is simple to handle but the motion of many vibrating system is not harmonic (but periodic) Few examples are shown below

Forces acting on machines are generally periodic but this may not be harmonic for example the excitation force in a punching machine is periodic and it can be represented as shown in figure 13 Vibration analysis of system subjected to periodic but nonharmonic forces can be done with the help of Fourier series The problem becomes a multifrequency excitation problem The principle of linear superposition is applied and the total response is the sum of the response due to each of the individual frequency term

Any periodic motion can be expressed as an infinite sum of sines and cosines terms If x(t) is a periodic function with period t its Fourier representation is given by

X(t) = (((( )))) (((( )))) (((( ))))t sint cost cos2 111 ωωωωωωωωωωωω baaao ++++++++++++

= (((( )))) (((( ))))t sint cos2 1

ωωωωωωωω nnn

o bnaa

++++++++sumsumsumsuminfininfininfininfin

====

tππππωωωω

2==== = Fundamental frequency ndash (1)

where ao an bn are constants

ττττ ττττ2 ττττ3

ττττ

X(t)

22

t

X(t)

035

025

Determination of constants To find ao Integrate both sides of equation(1) over any interval ττττ All intergrals on the RHS of the equation are zero except the one containing ao

(((( ))))dttxao

o 2

2

intintintint====ωωωωππππ

ππππωωωω

= (((( ))))dttxo

2intintintintττττ

ττττ

To find an multiply equation 1 by cos (((( ))))tnωωωω and Integrate over any interval ττττ All intergrals

(((( )))) (((( )))) dttxao

n tncos 2

2

ωωωωππππωωωω ωωωω

ππππ

intintintint==== (((( )))) (((( )))) dttxo

tncos 2

ωωωωττττ

ττττ

intintintint====

To find bn multiply equation 1 by sin (((( ))))tnωωωω and Integrate over any interval ττττ All intergrals

(((( )))) (((( )))) dttxao

n tnsin 2

2

ωωωωππππωωωω ωωωω

ππππ

intintintint==== (((( )))) (((( )))) dttxo

tnsin 2

ωωωωττττ

ττττ

intintintint====

Find the Fourier series for the curve shown below

ττττ ττττ2 ττττ3

t

X(t)

E

ττττ ττττ2 ττττ3

t

X(t)

ππππ2

t

X(t) ππππ

1 Represent for the periodic motion shown in the figure

23

CHAPTER 2 UNDAMPED FREE VIBRATION LEARNING OBJECTIVES

] Introduction to undamped free vibration

] Terminologies used in undamped free vibration

] Three methods to solve the undamped free vibration problems

] Problems related to undamped free vibration problems

Free vibrations are oscillations about a systems equilibrium position that occur in the absence of an external excitation force If during vibrations there is no loss of energy it is known as undamped vibration The first step in solving a vibration problem is setting up the differential equation of motion Consider spring mass system which is assumed to move only along the vertical direction as shown below Let m be the mass of the block and k be the stiffness of the spring When block of mass m is attached to spring the deflection of spring will be ∆ known as static deflection In the static equilibrium position the free body diagram of forces acting on the mass is shown in Figure(b) Hence mg= kA Once the system is disturbed the system executes vibrations

Let at any instant of time t the mass is displaced from the equilibrium position x the different forces acting on the system are shown in figure (d) From Newtonrsquos second law of motion

sum = maF

Inertia force ( disturbing force) = restoring force

mgxkxm ++∆minus= )(ampamp

0)( =+ xkxm ampamp

or 0)( =+ xmk

xampamp

equation 2 is the differential equation of motion for spring mfigure Comparing equation (2) with the equation of SHM +xampampsince the vibrations of the above system are free( without the forces) we can write

sec radmk

n =ω

24

ass system shown in 0)(2 =xω

resistance of external

time period km

2 f1

n

πτ ==

from the equation(1) mg = k∆

∆=

q

mk

Difference between the translation ( rectilinear) and rotational system of vibration

Translatory Rotational

In the analysis thFORCES are co Linear stiffness K Problems 1A mass of 10kFind the natural 2 A spring massnatural frequencnatural frequenc

kt

m

sec radmk

n =ω

k

25

e disturbing and restoring nsidered

In the analysis In the analysis MASS Moment of Inertia (J) is considered

in Nm is considered

g when suspended from a spring causes a static deflection of 1cm frequency of system

system has a spring stiffness K Nm and a mass of m Kg It has a y of vibration 12 Hz An extra 2kg mass coupled to it then the y reduces by 2 Hz find K and m

secrad n =ω

26

3 A steel wire of 2mm diameter and 30m long It is fixed at the upper end and carries a mass of m kg at its free end Find m so that the frequency of longitudinal vibration is 4 Hz 4 A spring mass system has a natural period of 02 seconds What will be the new period if the spring constant is 1) increased by 50 2) decreased by 50 5 A spring mass system has a natural frequency of 10 Hz when the spring constant is reduced by 800 Nm the frequency is altered by 45 Find the mass and spring constant of the original system 6 Determine the natural frequency of system shown in fig is which shaft is supported in SHORT bearings 7 Determine the natural frequency of system shown in fig is which shaft is supported in LONG bearings where l is the length of bearing and E ndash youngrsquos modulus and I is moment of Inertia 7 Determine the natural frequency of system shown in fig is which shaft is supported in LONG bearings 8 A light cantilever beam of rectangular section( 5 cm deep x 25cm wide) has a mass fixed at its free end Find the ratio of frequency of free lateral vibration in vertical plane to that in horizontal 9 Determine the natural frequency of simple pendulum 10 Homogeneous square plate of size l and mass m is suspended from the mid point of one of its sides as shown in figure Find the frequency of vibration

l

l

27

11 A compound pendulum which is rigid body of mass m and it is pivoted at O The point of pivot is at distance d from the centre of gravity It is free to rotate about its axis Find the frequency of oscillation of such pendulum 12 A connecting rod shown in fig is supported at the wrist pin end It is displaced and allowed to oscillate The mass of rod is 5kg and centre of gravity is 20 cm from the pivot point O If the frequency of oscillation is 40 cyclesminute calculate the mass moment of inertia about its CG

13 A semi circular homogenous disc of radius r and mass m is pivoted freely about its centre as shown in figure Determine the natural frequency of oscillation 14A simply supported beam of square cross section 5mmx5mm and length 1m carrying a mass of 0575 kg at the middle is found to have natural frequency of 30 radsec Determine youngrsquos modulus of elasticity of beam 15 A spring mass system k1 and m have a natural frequency f1 Determine the value of k2 of another spring in terms of k1 which when placed in series with k1 lowers the natural frequency to 23 f1

COMPLETE SOLUTION OF SYSTEM EXECUTING SHM

The equation of motion of system executing SHM can be represented by

0)( =+ xkxm ampamp ----------------------------------------(1)

022

2=+

dtdx

dt

dx ω

The general solution of equation (1) can be expressed as

------------------------(2)

Where A and B are arbitrary constant which can be determined from the initial conditions of the system Two initial conditions are to be specified to evaluate these

constants x=x0 at t=0 and xamp = Vo at t=0 substituting in the equation (2) Energy method In a conservative system the total energy is constant The dwell as natural frequency can be determined by the princienergy For free vibration of undamped system at any instant and partly potential The kinetic energy T is stored in the masswhere as the potential energy U is stored in the form of sdeformation or work done in a force field such as gravity The total energy being constant T+U = constant Its rate of cha

t)(sin Bt)cos( AX ω+ω=

t)(sin V

t)cos( xx o0 ω

ω+ω=

Is the required complete

solution

28

ifferential equation as ple of conservation of of time is partly kinetic by virtue of its velocity train energy in elastic

nge

Is given by [ ] 0UTdtd

=+

From this we get a differential equation of motion as well as natural frequency of the system Determine the natural frequency of spring mass system using energy method

Determine the natural frequency of system shown in figure Determine the natural frequency of the system shown in figure Is there any limitation on the value of K Discuss Determine the natural frequency of s

m

l k

m

a

θ

m

a

θ

l

k

29

k

k

ystem shown below Neglect the mass of ball

l

m

30

A string shown in figure is under tension T which can be assumed to remain constant for small displacements Find the natural frequency of vertical vibrations of spring An acrobat 120kg walks on a tight rope as shown in figure The frequency of vibration in the given position is vertical direction is 30 rads Find the tension in the rope A manometer has a uniform bore of cross section area A If the column of liquid of length L and Density ρ is set into motion as shown in figure Find the frequency of the resulting oscillation

m

l

a

T T

36m 8m

31

k

R

k

m

r

m

Find the expression for natural frequency of system shown in the figure Neglect the mass of the cantilever beam Study the special case i) k=Infinity ii) I = infinity Determine the expression for the natural frequency of a system shown in figure The two discs are keyed to a common shaft and have a combined mass moment of inertia about centre of oscillation O It is assumed that the card attached to mass m does not stretch and is always under tension Determine the natural frequency of a system shown

m

l

32

Determine the expression for the natural frequency of the system shown in figure Assume that the wires connecting the masses do not stretch and are always in tension

Determine the natural frequency of spring mass system taking the MASS OF SPRING (ms ) into account

M3 M1

M2

k2 k1

m

k

33

RAYLEIGHrsquoS METHOD

This is the extension of energy method Here we can determine the natural frequency of a conservative system without finding the equation of motion The natural frequency is a function of the rate of change of Kinetic energy and potential energy of the system From the principle of conservation of energy we have T+U = Constant T= KE and U =PE We can write the above equation as T1+U1 = --------------------------(1) Where 1 and 2 represents tow instances of time Let 1 be the time when the mass is passing through static equilibrium position Let 2 be the time corresponding to the mass displacement of the mass At this instant the velocity f the mass is zero and hence Substituting in equation (1) we get ( )Max = ( )Max Above equation leads directly to natural frequency Determine the natural frequency of spring mass system using Rayleighrsquos method m

34

r

B

K K

Home work

Determine the natural frequency of pendulum using Rayleighrsquos method

A Cylinder of mass m and moment of inertia Jo is to role without slipping but is sustained by spring K as shown in figure Determine the natural frequency of oscillation

Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping

a

Determine the natural frequency of system shown where in cylindrical disc rolls over the inextensible spring A cylinder of radius r rolls without slipping on cylindrical the equation of motion and hence determine the naoscillations about the lowest point shown Repeat the above problem ndash Instead of cylinder assume swithout slipping on concave surface as shown above

r

K

35

portion of radius R Derive tural frequency of small

phere of radius with r rolls

The torsional pendulum has a natural frequency of 5 Hz What length of steel wire of diameter 2 mm should be used for pendulum The inertia of mass fixed at the free end is 00098 kgm2 Take g= 083x1011 Nm2 Determine the torsional natural frequency of the system shown in figure Neglect the mass moment of inertia of the shaft ( Figure not proportionate) Determine the natural frequency of simple pendulum considering the mass of the rod into account

Using the energy method find tThe chord may be assumed ine

Kt1

Kt2

20cm

06m 08m 04m

20cm 20cm 20cm

m

M

he natural frxtensible in

k

M

36

equency of the system shown in figure the spring mass pulley system

A rod of length l and depth H having a mass m rests on a semicylidrical surface os radius r It is tipped slightly and let go Find the frequency of oscillation

A thin bar of l having a mass m rests on a semicylidrical surface radius r It is tipped slightly and let go Find the frequency of oscillation

l

h

R

l

37

38

CHAPTER 3

DAMPED FREE VIBRATION OF SINGLE DEGREE FREEDOM SYSTEM LEARNING OBJECTIVES

] Introduction to damped free vibration

] Terminologies used in damped free vibration

] Three different types of damping

] Concept of critical damping and its importance

] Study of response of viscous damped system for cases of under damping

critical and over damping

] Concept of logarithmic decrement

] Problems related to damped free vibration problems

Viscously damped free vibration of single degree freedom sytem Consider a single degree freedom system with viscous damping as shown in figure For a viscous damped system the damping force is

where c is a constant of proportionality It is called damping coefficient or coefficient of viscous damping

Symbolically it is designated by a dashpot as shown in Fig 3

x xampx

39

At any instant of time let the mass be displaced by x from the equilibrium position Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

Okxxcxm =minusminus= ampampamp

Okxxcxm =++ ampampamp ----------------------------------------------------------(1)

or

0kxdtdx

Cdt

xdM

2

2

=++

Assuming the solution of equation as

The first term in the above equation m2Ct

eminus

is an exponentially decaying function of time The behavior of the terms in the parenthesis however depends on whether numerical value within the radicals is positive zero or negative

CASE 1 if 0mk

m2c

2

=minus

40

CASE 2 if 0mk

m2c

2

=gt

CASE 3 if 0mk

m2c

2

=lt

41

Critically damped coefficient Critically damped coefficient is defined as the value of the damping coeffient for which radical in equation

mk

m2c

m2c

S2

21 minus

plusmn

= becomes zero

Damping Ratio ( Damping factor) ( ξ ) It is defined as the ratio of damping coefficient to the critical damping coefficient Note The damping coefficient c is defined as the actual amount of damping while the damping coefficient is a constant depending upon the mass and stiffness of the system The roots of characteristics equation can be now written as Note if c=cc or ξ =1 then system is critically damped if cgt cc or ξ gt1 then system is over damped I if clt cc or ξ lt1 then system is under damped

42

Take home quiz Derive General solution of different damped system

1 Critically damped system 2 Under damped system 3 Over damped system

Logarithmic Decrement It is used to determine the amount of damping present in system This measure The rate of decay of free vibration The larger the damping the greater will be rate of decay Logarithmic decrement is defined as the natural logarithm of ratio of any two successive amplitudes

=δ

2

1

xx

ln

Expression for logarithmic decrement (δ ) Consider a damped free vibration which is expressed by

( ) ( ) φ+ξminusω= ξω t 1sinXex 2n

tn

x is the displacement at any instant of time X and φ are arbitrary constants

21

2

ξminus

πξ=δ

43

Tutorial problems 1 Derive an expression for damping ratio in terms of logarithmic decrement

2 Show that logarithmic decrement

=δ

n

o

xx

lnn1

where xo is the initial amplitude xn

is amplitude after n cycles 3 A mass of 1 kg is to be supported on a spring having stiffness K= 9800 Nm The damping constant c= 59 Nsm Determine the natural frequency of system Find also logarithmic decrement and amplitude after 3 cycles if initial displacement is 03 cm 4 A vibrating system is defined by the following parameters M = 3kg k= 100 Nm damping coefficient = 3 Nsm determine Damping factor natural frequency damped vibration logarithmic decrement ratio of consecutive amplitude no of cycles after which the original amplitude is reduced to 20 5A single degree damped vibrating system a suspended mass of 18kg makes 15 oscillations in 003 seconds The amplitude decreases to 025 of the initial value after 5 oscillations Determine the stiffness of spring Logarithmic decrement damping factor and damping coefficient 6 A damped vibration record of a spring mass system shows the following data Amplitude at the end of 2nd cycle is 9mm amplitude at eh end of 3rd cycle is 6mm amplitude at the end of 4th cycle is 4mm stiffness of spring =8000 Nm weight =20N Find the logarithmic decrement damping force at unit velocity periodic time of vibration 7 For the system shown in figure the characteristics of dashpot is such that the constant force of 49 N is applied to the piston it velocity is found to be constant at 012 ms Determine the value of C would you expect the complete system to be periodic or aperiodic 8 For a torsional system shown in the figuref1 is natural frequency in air and f2 is the natural frequency when immersed in oil Determine the damping coefficient

Viscous fluid

Disc of mass moment of Inertia J

2a

C

m M

b

a a

k

9 A thin plate of area A and weigh w is fixed to the end of the spring and allow to oscillate in a viscous fluid as shown in figure It f1 is the frequency of oscillation in air and f2 in liquid Show that

22

21 ff

gA2

minusπω

=micro where the damping force on the plate

AV2Fd micro= V being the velocity of vibration 10 A torsional pendulum has a natural frequency of 175 cyclessecond vibrating in vacuum The mass moment of inertia is 2 kgm2 It is immersed in oil and it is observed that the natural frequency is 142 cyclessecond Find the damping torque If the disc is replaced by 34 deg When in oil find the displacement at the end of the first complete cycle 11 An automobile can be modeled as a mass placed on 4 shock absorbers each consisting of a spring and a damper Such that each spring is equally loaded Determine the stiffness damping constant of each shock absorber So that natural frequency is 2Hz and system is critically damped The mass of vehicle is 200kg 12 Write the differential equation of motion for the system shown in the figure below Determine the natural frequency of the damped oscillation and critical damping coefficient

Determine the differential equation of motion and find the critical cdamped natural frequency for system shown in figure

a

C

m M

M =200kg

44

200kg kgkg

oefficient and

45

Derive the equation of motion and find the critical damping coefficient and damped natural frequency for the system shown below Write the differential equation of motion and determine the expression for critical damping coefficient and damped natural frequency of system shown in figure

Set up a differential equation of motion for system shown in figure and determine the following Undamped natural frequencies critical damping coefficient damping ratio damped natural frequency

b

C

kM

a

m

46

A mass of 300N is resting on two springs of stiffness 3000 Nm each and a dashpot of damping coefficient 150 Nsm as shown in figure If the initial velocity of 10cms is given to mass at its equilibrium position what will be the displacement from the equilibrium position at the end of first second Derive the formula used A gun barrel of mass 600kg has a recoil spring of stiffness 294 knm if the barrel recoils 13m on firing determine i) The initial recoil velocity of barrel ii) The critical damping coefficient of dashpot which is engaged at the end of recoil stroke Iii) Time required for the barrel to return at the end of recoil stroke Iv) time required for the barrel to return to a position 5cm from the initial position

W= 3000 N

FIRING

CHAPTER 4

FORCED VIBRATION LEARNING OBJECTIVES

] Introduction to forced free vibration

] Terminologies used in forced vibration

] Concept of magnification factor

] Study of reciprocating and rotating unbalance

] Concept of base excitation and support excitation

] Concept of Transmissibility ratio

] Problems related to forced vibration

The vibration that takes place under the excitation of external forces are called

forced vibration When a system is subjected to harmonic excitation it is forced to vibrate at the same frequency as that of excitation Most of the mechanical systems are prone to steady state forced vibration which cause fatigue failure The vibration problem also becomes very important when any of excitation frequencies coincide with one of natural frequencies of a system Under this condition resonance occurs during which large amplitude of vibrations are observed To avoid resonance external frequency (operating speed) may be changed or properties of the system may be changed to alter the natural frequency In some cases sufficient amount of damping may be provided to avoid large amplitude during resonance Thus the problem of force vibration is very important in mechanical design

FORCED VIBRATIONS OF SINGLE DOF SYSTEMS UNDER HARMONIC EXCITATION

` [ ] tsin FF o ω=

x

sin FF o=

Fo

xampamp

[ ] tω

xamp

47

48

Consider a spring mass system having a viscous damping excited by a harmonic force [ ] tsin FF o ω= as shown in figure The various forces acting on the system is shown in figure Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

O)tsin(Fkxxcxm o =ω+minusminus= ampampamp

)tsin(Fkxxcxm o ω=++ ampampamp ----------------------------------------------------------(1)

or

)tsin(Fkxdtdx

Cdt

xdM o2

2

ω=++

Equation (1) represents the differential equation of motion of the system The solution to this equation consists of two parts namely complementary function and a particular integral

The first of solution is also termed as transient solution of homogeneous solution and can be obtained from the homogeneous differential equation 0kxxcxm =++ ampampamp

The second part of solution particular integral is also termed as steady state solution ad is of form )tsin(Xx φminusω= Where X= steady state amplitude φ is phase difference between external force and displacement

)tsin(Xx φminusω=

The vector representation of the shown

49

The force polygon for the above vector diagram can be written as follows

( )φminusωt O

KX

xm 2ω

Fo

xCω

tω

φ

50

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

is the governing equation

Zero frequency deflection (Xst) It is deflection observed at zero frequency ( when the external frequency 0=ω

( ) kF

001

kF

x o

22

o

st =

+minus=

Magnification factor (M) It is defined as the ratio of steady state amplitude to zero frequency deflection xst

M = arX

X=

A plot of magnification factor versus frequency ratio

damping factor ( )ξ is called frequency plot of a harmon All the frequency terms starts from magnification factor of damping ratio ( )ξ The amplitude of vibration at any on the value of damping The amplitude will be higher for higher damping The amplitude of vibration at reso

amplitude tends to zero as

ωω

n tends to infinity

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

Freq

ζ =

ζ = 1

Magnification Factor M = XXst

kFo

51

ωω

nfor varying values of

ically excite vibrating system

M=1 what ever may be value particular frequency depends

for lesser damping and lesser nance will be very high The

uency Ratio r = (ωωn)

ζ = 0 0

ζ = 025 ζ = 0375 ζ = 0 5

ζ = 0707

ζ = 2

52

Plot of phase angle φ versus frequency ratio

ωω

n is shown in the

figure The phase angle φ is zero for all the values of damping ratio

at

ωω

n=0 The phase

angle φ is 90 for all the values of damping ratio

at

ωω

n=1 ie at

resonance The phase angle between o to 90 deg for all the values of

ωω

n= 0 to 1 and

between 90 deg to 180 deg for all the values of

ωω

ngt1

Class work Derive an expression for frequency at peak amplitude

2np 21 ξminusω=ω=ω is the required equation

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

53

Derive expression amplitude at resonance ( In dimensional form) In dimensional form We know that steady state amplitude

( ) ( )

ω+ωminus=

222

o

Cmk

Fx Equation in dimensional form

We know that at resonance mk

n =ω=ω

( )

ω+

minus

=2

n

2

oR

Cmk

mk

Fx

( )

ω=

2n

oR

C

Fx

( )

ω=

n

oR C

Fx

Class work Derive an expression for amplitude at resonance ( in non dimensional form)

Tutorial Problems on Forced Vibration 1A weight of 60N is suspended by a spring stiffness 12 kNm is forced to vibrate by a harmonic force of 10N assuming viscous damping of 0086 kN-sm Find i) resonance frequency ii) amplitude at resonance iii) phase angle at resonance iv) frequency corresponding to peak amplitude 2 A periodic torque having a maximum value of 05888Nm at a frequency of 4 rads is impressed upon a flywheel suspended from a wire The wheel has mass moment of inertia 012 kgm2 and the wire has stiffness 1176 Nmrad A viscous dashpot applies damping couple of 0784 Nm an angular velocity of 2 rads calculate i)maximum angular displacement from rest position ii) maximum couple applied to dashpot Iii) angle by which the angular displacement lags the torque A mass of 1019 kg is suspended from a spring stiffness 20 Nmm viscous damping causes the amplitude to decrease to 110th of initial value in 4 complete cycle If a periodic force of 20 cos(50 t) is applied to the mass Find the amplitude of forced vibration What would be the amplitude of oscillation if the applied force coincides with the natural period

m=1019 kg

[ ] t50cos 30F =

54

55

ROTATING UNBALANCE Unbalance in rotating machinery is a common source of vibration excitation Consider a machine of mass m having mounted on a foundation of stiffness k and damping coefficient C as shown in figure The unbalance is represented by an eccentric rotating mass m with eccentricity (e) and having a angular velocity (ω) let x be the displacement of non rotating mass from the static equilibrium position The displacement of rotating mass m from the static equilibrium is ( ) tsin( ex ω+ Applying Newtonrsquos law of motion we have

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

56

RECIPROCATING UNBALANCE Unbalance in reciprocating machinery is a common source of vibration excitation Consider a machine of mass M having mounted on a foundation of stiffness K and damping coefficient C as shown in figure The unbalance is represented by a reciprocating mass (m) having crank radus (e) and connecting rod of length (l) Let ωωωω be angular velocity of the crank Let x be the displacement of non reciprocating mass from the static equilibrium position at any instant of time The displacement of reciprocating mass m from the static equilibrium position is

)2sin(sin +++ tle

tex ωωωωωωωω

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

57

Note The equations obtained in reciprocating unbalance is same as the rotating unbalance and having the following differences Rotating Unbalance Reciprocating Unbalance Here mass of the rotating element is considered

Here mass of reciprocating element is considered

Here the eccentricity (e) which is the distance between the centre of rotation and CG

Here the eccentricity e is the crank radius and crank radius = stroke length2

The frequency plot of non dimensional factor( ( )me

MX versus frequency ratio

ωω

n for various values of ( )ξξξξ is shown in figure This frequency plot is same for

the rotating and reciprocating unbalance with the following points are noted from the diagram

1 Damping ratio plays very important role during resonance in controlling the amplitude

2 When

ωω

n=0 then ( )me

MX =0 for all the values of damping ratio ( )ξξξξ

3 When

ωω

nis higher than ( )me

MX tends to unity

4 At very high speeds ie at very high frequency ratio

ωω

n there is no effect

of damping The frequency plot of phase angle versus the frequency ratio is given as shown in figure The explanation is same as in forced harmonic vibration

58

VIBRATION ISOLATION

Vibratory forces and motions generated by machine and other caused are unavoidable Howeever their effect on the other dynamical systems can be minimized by proper isolator design An isolation system is either to protect o delicate machine from excessive vibration transmitted to it from its supporting structure or to prevent vibratory forces and motions generated by machines from being transmitted to its surroundings The effectiveness of isolation may be measured in terms of forces and motion transmitted The first type is known as force isolation and the second type is known as motion isolation

Force isolation (Force Transmissibility) When vibratory forces generated by machine are isolated the effectiveness of forced isolation can be made by defining the term transmissibility The term transmissibility is defined as the ratio of force transmitted to the foundation to that of the external impressed force

Force Transmissibility = o

tr

FF

force Impressed Externalfoundation toted transmitForce

=

Consider viscous damped spring mass system as shown in figure Upon this system an external harmonic force ) t (Sin Fo ω is applied The difference forces acting on system are inertia forces damping force spring force and external force All these forces are represented on force polygon shown below Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these two forces

( ) ( )22 xckx ω+=

59

This is the expression for force transmissibility in dimensional form To represent same in the non dimensional form divide numerator and denominator by k

2

n

22

n

2

n

o

tr

21

21

FF

TR

ωω

ξ+

ωω

minus

ωω

ξ+

==

PHASE DIFFERNECE BETWEEN EXTERNAL FORCE AND DISPLACEMENT Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these

ωω

minus

ωω

ξ=

ωminusω

=φ minusminus2

n

n1

2

1

1

2tan

mKC

tan

The phase difference between force transmitted to the foundation Ftr to the displacement

kc

KxxC

tanω

=ω

=α or

ωω

ξ=

ω=α minusminus

n

11 2tanKx

xCtan

The phase difference between external force Fo and force transmitted to foundationis β

60

β= αminusφ =

ωω

minus

ωω

ξminus

2

n

n1

1

2tan -

ωω

ξminus

n

1 2tan

Displacement Transmissibility It is the ratio of amplitude of steady state vibration of machine to that of the base motion Forced vibration due to support motion Consider a spring mass damper system subjected to harmonic excitation )tsin(Yy ω= at the base as shown in figure Let x be the absolute displacement of the mass at any instant of time t The different forces acting on system are shown in free body diagram Applying Newtonrsquos Law of motion Ma = sumF

)yx(k)yx(cxm minusminusminusminus= ampampampamp

0)yx(k)yx(cxm =minus+minus+ ampampampamp

61

Frequency plot of transmissibility versus frequency ratio

62

The plot of transmissibility versus frequency ratio is as shown in figure The following points can be noted from it

i) The value of transmissibility is unity at frequency ratio

ωω

n=0 and 2

ii) For an undamped system ( )ξ =0 transmissibility tends to infin at resonance iii) The value of transmissibility is less than unity for the value of frequency ratio

greater than 2

iv) For frequency ratio

ωω

nlt 2 smaller damping ratio lead to larger values of

transmissibility on the other hand for frequency ratio

ωω

ngt 2 smaller damping

ratio leads to smaller value of transmissibility

The frequency plot of transmissibility (TR) versus frequency ratio

ωω

n can be

broadly divided into two zones

1

ωω

n= 0 to 2

2

ωω

n= 2 to infin

CASE 1

When

ωω

n= 0 to 2

In this zone the value of transmissibility is always greater than 1 for all values of damping ratio In this zone the transmissibility decreases with increase in damping Hence we can conclude that in this zone damping is useful in isolating vibration forces being transmitted to the support CASE 2

When

ωω

n= 2 to infin

In this zone the value of transmissibility is less than 1 for all values of damping ratio In this zone the transmissibility increases with increase in damping Hence we can conclude that in this zone damping will adversely affect in isolating vibratory forces

63

Tutorial problems on Forced Vibration 1 Machine of mass 75kg is mounted on spring of stiffness 12000 Nm with an assumed damping factor ξ =02 piston within the machine of mass 2 kg has a reciprocating motion with a stroke of 75 cm and speed of 15 cyclessecond Assuming the motion of piston to be harmonic determine amplitude of machine ( X) Transmissibility and force transmitted to foundation phase angle of transmitted force wrt to exciting force 2 Motor weighs 600N is mounted on a simple beam that has a spring rate of 400 Ncm The motor armature base 160 N and has an eccentricity of 0005cm What will be the amplitude of motion when it runs at 1760 RPM Neglect damping the deflection of armature shaft and weight of the beam 3 A Heavy machine of mass 300 kg is supported on resilient foundation the static deflection of foundation due to weight of machine is found to be 75mm It is observed that the machine vibrates with an amplitude of 10mm when the base of the foundation is subjected to harmonic oscillation at the undampeds natural frequency of system with an amplitude of 25mm Find the damping constant of foundation Ii) dynamic force amplitude on the base iii) amplitude of the displacement of machine relative to base 4 An air compressor of mass 450 kg operates at a constant speed of 1750 RPM The reciprocating masses are 10 kg the crank radius is 100mm If the damper for mounting introduces a damping factor 015 i) specify the springs of mounting such that only 20 of the unbalanced force is transmitted to the foundation ii) determine the magnitude of the transmitted force bull Repeat the above problem by assuming that there is no damping in system

5 A body of a car having mass 1500 kg and is mounted on 4 equal springs which deflect through 225 cm under the weight of the body A total damping forces the 4 shock absorbers is 46 N at a velocity of 1cmsec The car is placed with all the four wheels on a test platform which is move up and down at resonant speed With an amplitude of 25cm Find the amplitude of the car body on its springs assuming its CG to be in the centre of wheel base 6A 75 kg machine which is mounted on springs of stiffness 116x105 Nm with the damping of 20=ξ A 2kg piston within the machine as reciprocating motion with a stroke of 008m and speed of 3000 RPM Assuming the motion of piston to be harmonic determine the amplitude of vibration of machine

7 The spring of an automobile are compressed 10cm under its own weight Find the critical speed when a trailer is traveling over road with a people approximaterly by a sine wave of amplitude 75 cm and wave length 15m What will be the amplitude of vibration at 64kmhour 8 An eccentric mass excites us used to determine vibratory characteristics of a structure of mass 25kg as given in figure At a speed of 1000 RPM the eccentric mass to beat top position at the instant the structure was moved upwards through its static equilibrium position and corresponding amplitude was 20mm The eccentric mass is 05kg at an eccentricity of 10mm Determine i) Undamped natural frequency of the system ii) Damping ratio iii) Steady state amplitude and phase angle at 1500 RPM iv) Force transmitted to the ground at 1200 RPM

M=25kg

C

K

Eccentric mass

64

65

CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Support motion Consider a spring mass damper system as shown in figure Let it excited by motion

of support let y be the harmonic displacement of support From Newtonrsquos second law of motion

)yx(k)yx(cxm minusminusminusminus= ampampampamp

)1(0)yx(k)yx(cxm minusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminus=minus+minus+ ampampampamp

Let z be the relative displacement of the mass with respect to support then

Z=x-y

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

9

G

K1 K2

mJ

a b

Exercise Specify the no of degree of freedom for the following 1 2 3

4

5

4m 2m m 3k k k

Cantilever Beam

Continuous system ( Neglect the mass of the beam )

k1 k2 m1 m2

y y

m2

θ1

θ2

L1

L2

M1

M2

10

O

A

B

Addition of SIMPLE HARMONIC MOTION ( SHM) The addition of two simple harmonic motion having frequency yields a resultant which is simple harmonic having the same frequency Consider tow simple harmonic motions of 1x and 2x having the same frequency and phase difference φ as given below

)(sin11 tXx ωωωω====

)(sin22 φω += tXx

Adding x = x1 +x2 Hence the resultant displacement is also SHM of amplitude X and phase angle θ

θθθθ

ωωωωt

φφφφ

11

-6

-4

-2

0

2

4

6

0 005 01 015 02 025

Tutorial problems on Simple Harmonic Motion 1 Add the following harmonic motion analytically and verify the solution graphically 1) X1= 3 sin (((( ))))30 t ++++ωωωω X2= 4 cos (((( ))))10 t ++++ωωωω ( VTU Jan 2005) 2) X1= 2 cos (((( ))))05 t ++++ωωωω X2= 5 sin (((( ))))1 t ++++ωωωω ( VTU July 2006)

3) X1= 10 cos (((( ))))4 t ππππωωωω ++++ X2= 8 sin (((( ))))6 t ππππωωωω ++++ ( VTU Dec 2007)

2 A body is subjected to two harmonic motions X1= 8 cos (((( ))))6 t ππππωωωω ++++ X2= 15 sin (((( ))))6 t ππππωωωω ++++ what harmonic motion

should be given to the body to bring it to equilibrium (VTU July 2005) 3 Split the harmonic motion x = 10 sin (((( ))))6 t ππππωωωω ++++ into two harmonic motions

having the phase of zero and the other of 45o 4 Show that resultant motion of harmonic motion given below is zero X1= X sin (((( ))))t ωωωω X2= X sin (((( ))))3

2 t ππππωωωω ++++ X3= X sin (((( ))))34 t ππππωωωω ++++

5 The displacement of a vibrating body is given by x = 5 sin (3141t + 4

ππππ )

Draw the variation of displacement for one cycle of vibration Also determine the displacement of body after 011 second ( repeat the problem for velocity and acceleration and draw graph using Excel and compare ) Time Displacement

0 0025 5 005

0075 0 01

0125 015 -353

0175 0 02

Calculate the remaining values 6 The Motion of a particle is represented by x = 4 ( ) t sin ω sketch the variation of the displacement velocity and acceleration and determine the max value of these quantities Assume ( )5 =ω ( Try to use MATLABExcel) sketch all on same graph

12

BEATS When two harmonic motions with frequencies close to one another are added The resulting motion exhibits a phenomenon known as Beats A Beat Frequency is the result of two closely spaced frequencies going into and out of synchronization with one another Let us consider tow harmonic motion of same amplitude and slightly different frequencies X1 = X Cos ( ) tωωωω X2 = X Cos ( )( ) tδδδδωωωω + Where δ is a small quantity The addition of above two harmonics can be written as X = X1 + X2

X=

+

ttX2

cos2

cos2δδδδωωωω

δδδδ

The above equation shown graphically in Figure The resulting motion represents cosine wave with frequency ( ) 2

ππππωωωω + and with a varying amplitude 2 cos( ( ) 2δδδδ t

whenever the amplitude reaches a maximum it is called the beat The frequency δ at which amplitude builds up and dies down between o and 2 X is known as beat frequency Beat phenomenon is found in machines structures and electric power houses In machines the beat phenomenon occurs when the forcing frequency is close to the natural frequency of the system

+2X

-2X

ωωωωππππ2

ELEMENTS OF VIBRATION The elements of constitute vibrating systems are

1 Mass or Inertia element - m 2 Spring - k 3 Damper - c 4 Excitation F(t)

Mass or Inertia element The mass or inertia element is assumed to be rigid body During vibration velomass changes hence kinetic energy can be gained or loosed The force appthe mass from newton second law of motion can be written as F= ma Work dthe mass is stored in the form of kinetic energy given by KE= frac12 M V2

Combination of masses In practical cases for simple analysis we replace several masses by equivalent mass Case1 Translational masses connected by rigid bar Let the masses M1 M2 and M3 are attached to a rigid bar at locates 1 respectively as shown in the figure The equivalent mass meq be assumelocated at 1 is as shown in figure (b) Let the displacement of masses M1 M2 and M3 be x1 x2 x3 and simivelocities of respective masses be x1 x2 and x3 We can express the velomasses m2 and m3 in terms of m1

Elements of Vibration

Passive element

m C K

Active element

F(t)

Conservative element

( Mass and Spring)

Non conservative element

( Damper)

C k

m F(t)

Voig

13

city of lied on one on

a single

2 and 3 d to be

larly the cities of

t Model

Translationa Let a mass mmass momearrangement These two mamass Meq or

2

1

33

2

1

221

+

+=

ll

Mll

MMMeq

14

is the required answer

l and rotational masses coupled together

having a translational velocity x be coupled to another mass having nt of inertia Io with a rotational velocity θ as in rack and pinion

shown in the figure

sses can be combined to obtain either a single equivalent translational a single equivalent mass moment of inertia Jeq

15

Equivalent translational masses

Kinetic energy of the equivalent mass =

2

21

eqeq XM amp

Kinetic energy of the two masses =

+

20

2

21

21

θθθθampamp JXM

Meq= m

+ 2RJ

m o is the required answer

Also determine the equivalent rotational mass Jeq

Jeq= m [ ]oJmR +2 is the required answer Spring element Whenever there is a relative motion between the two ends of the spring a force is developed called spring force or restoring force The spring force is proportional to the amount of deformation x and then F α x or F = kx Where k is stiffness of the spring or spring constant or spring gradient The spring stiffness is equal to spring force per unit deromation

The spring stiffness k = mNxF

Workdone in deforming a spring is equal to the strain energy or potential energy Strain energy = potential energy = area of the triangle OAB Stiffness of beams Cantilever beam consider a cantilever beam with an end mass shown in the figure The mass of the beam is assumed to be negligible The static deflection of beam at free end is given by

Similarly derive the expression for Simply supported beam and fixed support beam

x

k

m

x

m

x

m

x

mNlIE

st 192

3

=δδδδ

mNEI

Wlst

48

3

=δδδδ

F

K=stiffness

F

16

17

Stiffness of slender bar subjected to longitudinal vibrations For a system executing the longitudinal vibrations as shown in the figure let us derive the expression for stiffness

Torsional Stiffness of bar It is the amount of torque required to cause a unit angular deformation in the element

Torsional stiffness = Kt = θθθθT

Combination of stiffness Determination of equivalent spring stiffness when the springs are arranged in series Consider two springs of stiffness K1 and K2 acted upon by the force F

The deflection of spring k1 is x1 = 1KF

The deflection of spring k1 is x2 = Let these two springs be replaced by an equivalent stiffness Keq upon which a force F acts and due to which its deflection is given by

x = eqKF

x= x1+x2

m

E A

l

60 cm 70 cm

Determination of equivalent spring stiffness when the springs are arranged in parallel Force acting on K1 spring = F1=k1x Force acting on K2 spring = F2 Force required for an equivalent spring keq to defined by x given by F= Keq x But F = F1 +F2 Tutorial problems on Equivalent stiffness of springs 1Determine the equivalent stiffness for the system shown in figure 2 Determine the equivalent stiffness for the system shown in figure

3 Determine the equivale

M

3

M

Nm 102 6x

18

nt stiffness for the system shown in figure

38 kg

Nm 101 6x Nm 102 6x

Nm 106x

2k

k

k

2k

3k

k

2k

k

19

60 cm 80 cm 50 cm

50 cm 60 cm

4Determine the equivalent stiffness for the system shown in figure

5Replace the following torsional stiffness by a single shaft having radius 4cm and find the length required for the equivalent shaft Assume the material of given system and equivalent system is same R1= 3cm R2= 5cm DAMPING Every vibration energy is gradually converted into heat or sound Hence the displacement during vibration gradually reduces The mechanism by which vibration energy is gradually converted into heat or sound is known as damping A damper is assumed to have either mass or elasticity Hence damping is modeled as one or more of the following types Viscous damping Coulomb or dry friction damping materials or solid or hysteric damping Viscous damping Viscous damping is most commonly used damping mechanism in vibration analysis When the mechanical system vibrates in a fluid medium such as air gas water or oil the resistance offered by the fluid to the moving body causes energy to be dissipated In this case the amount of dissipated energy depends on many factors such as size or shape of the vibrating body the viscosity of the fluid the frequency of vibration and velocity of fluid Resistance due to viscous damping is directly proportional to the velocity of vibration

dF V αααα

dF = xC amp

Where C= damping coefficient Fd = damping force Examples of Viscous damping

1) Fluid film between sliding surface 2) Fluid flow around a piston in a cylinder 3) Fluid flow through orifice 4) Fluid flow around a journal in a bearing

Rreq =4cm

leqn

stress

Strain

Coulomb damping or dry friction damping Here a damping force is constant in magnitude but opposite in direction to that of the motion of vibrating body It is caused by the friction between the surfaces that are dry or have insufficient lubrication Material or solid or hysteric damping

When the materials are deformed energy is absorbed and dissipated by the material The effect is due to friction between the internal planes which slip or slide as the deformation takes place When a body having the material damping is subjected to vibration the stress strain diagram shows the hysteresis loop as shown in the figure The area of the loop denotes the energy lost per unit volume of the body per cycle

20

21

ττττ ττττ2 ττττ3

ττττ

X(t)

FOURIER SERIES The simplest of periodic motion happens to be SHM It is simple to handle but the motion of many vibrating system is not harmonic (but periodic) Few examples are shown below

Forces acting on machines are generally periodic but this may not be harmonic for example the excitation force in a punching machine is periodic and it can be represented as shown in figure 13 Vibration analysis of system subjected to periodic but nonharmonic forces can be done with the help of Fourier series The problem becomes a multifrequency excitation problem The principle of linear superposition is applied and the total response is the sum of the response due to each of the individual frequency term

Any periodic motion can be expressed as an infinite sum of sines and cosines terms If x(t) is a periodic function with period t its Fourier representation is given by

X(t) = (((( )))) (((( )))) (((( ))))t sint cost cos2 111 ωωωωωωωωωωωω baaao ++++++++++++

= (((( )))) (((( ))))t sint cos2 1

ωωωωωωωω nnn

o bnaa

++++++++sumsumsumsuminfininfininfininfin

====

tππππωωωω

2==== = Fundamental frequency ndash (1)

where ao an bn are constants

ττττ ττττ2 ττττ3

ττττ

X(t)

22

t

X(t)

035

025

Determination of constants To find ao Integrate both sides of equation(1) over any interval ττττ All intergrals on the RHS of the equation are zero except the one containing ao

(((( ))))dttxao

o 2

2

intintintint====ωωωωππππ

ππππωωωω

= (((( ))))dttxo

2intintintintττττ

ττττ

To find an multiply equation 1 by cos (((( ))))tnωωωω and Integrate over any interval ττττ All intergrals

(((( )))) (((( )))) dttxao

n tncos 2

2

ωωωωππππωωωω ωωωω

ππππ

intintintint==== (((( )))) (((( )))) dttxo

tncos 2

ωωωωττττ

ττττ

intintintint====

To find bn multiply equation 1 by sin (((( ))))tnωωωω and Integrate over any interval ττττ All intergrals

(((( )))) (((( )))) dttxao

n tnsin 2

2

ωωωωππππωωωω ωωωω

ππππ

intintintint==== (((( )))) (((( )))) dttxo

tnsin 2

ωωωωττττ

ττττ

intintintint====

Find the Fourier series for the curve shown below

ττττ ττττ2 ττττ3

t

X(t)

E

ττττ ττττ2 ττττ3

t

X(t)

ππππ2

t

X(t) ππππ

1 Represent for the periodic motion shown in the figure

23

CHAPTER 2 UNDAMPED FREE VIBRATION LEARNING OBJECTIVES

] Introduction to undamped free vibration

] Terminologies used in undamped free vibration

] Three methods to solve the undamped free vibration problems

] Problems related to undamped free vibration problems

Free vibrations are oscillations about a systems equilibrium position that occur in the absence of an external excitation force If during vibrations there is no loss of energy it is known as undamped vibration The first step in solving a vibration problem is setting up the differential equation of motion Consider spring mass system which is assumed to move only along the vertical direction as shown below Let m be the mass of the block and k be the stiffness of the spring When block of mass m is attached to spring the deflection of spring will be ∆ known as static deflection In the static equilibrium position the free body diagram of forces acting on the mass is shown in Figure(b) Hence mg= kA Once the system is disturbed the system executes vibrations

Let at any instant of time t the mass is displaced from the equilibrium position x the different forces acting on the system are shown in figure (d) From Newtonrsquos second law of motion

sum = maF

Inertia force ( disturbing force) = restoring force

mgxkxm ++∆minus= )(ampamp

0)( =+ xkxm ampamp

or 0)( =+ xmk

xampamp

equation 2 is the differential equation of motion for spring mfigure Comparing equation (2) with the equation of SHM +xampampsince the vibrations of the above system are free( without the forces) we can write

sec radmk

n =ω

24

ass system shown in 0)(2 =xω

resistance of external

time period km

2 f1

n

πτ ==

from the equation(1) mg = k∆

∆=

q

mk

Difference between the translation ( rectilinear) and rotational system of vibration

Translatory Rotational

In the analysis thFORCES are co Linear stiffness K Problems 1A mass of 10kFind the natural 2 A spring massnatural frequencnatural frequenc

kt

m

sec radmk

n =ω

k

25

e disturbing and restoring nsidered

In the analysis In the analysis MASS Moment of Inertia (J) is considered

in Nm is considered

g when suspended from a spring causes a static deflection of 1cm frequency of system

system has a spring stiffness K Nm and a mass of m Kg It has a y of vibration 12 Hz An extra 2kg mass coupled to it then the y reduces by 2 Hz find K and m

secrad n =ω

26

3 A steel wire of 2mm diameter and 30m long It is fixed at the upper end and carries a mass of m kg at its free end Find m so that the frequency of longitudinal vibration is 4 Hz 4 A spring mass system has a natural period of 02 seconds What will be the new period if the spring constant is 1) increased by 50 2) decreased by 50 5 A spring mass system has a natural frequency of 10 Hz when the spring constant is reduced by 800 Nm the frequency is altered by 45 Find the mass and spring constant of the original system 6 Determine the natural frequency of system shown in fig is which shaft is supported in SHORT bearings 7 Determine the natural frequency of system shown in fig is which shaft is supported in LONG bearings where l is the length of bearing and E ndash youngrsquos modulus and I is moment of Inertia 7 Determine the natural frequency of system shown in fig is which shaft is supported in LONG bearings 8 A light cantilever beam of rectangular section( 5 cm deep x 25cm wide) has a mass fixed at its free end Find the ratio of frequency of free lateral vibration in vertical plane to that in horizontal 9 Determine the natural frequency of simple pendulum 10 Homogeneous square plate of size l and mass m is suspended from the mid point of one of its sides as shown in figure Find the frequency of vibration

l

l

27

11 A compound pendulum which is rigid body of mass m and it is pivoted at O The point of pivot is at distance d from the centre of gravity It is free to rotate about its axis Find the frequency of oscillation of such pendulum 12 A connecting rod shown in fig is supported at the wrist pin end It is displaced and allowed to oscillate The mass of rod is 5kg and centre of gravity is 20 cm from the pivot point O If the frequency of oscillation is 40 cyclesminute calculate the mass moment of inertia about its CG

13 A semi circular homogenous disc of radius r and mass m is pivoted freely about its centre as shown in figure Determine the natural frequency of oscillation 14A simply supported beam of square cross section 5mmx5mm and length 1m carrying a mass of 0575 kg at the middle is found to have natural frequency of 30 radsec Determine youngrsquos modulus of elasticity of beam 15 A spring mass system k1 and m have a natural frequency f1 Determine the value of k2 of another spring in terms of k1 which when placed in series with k1 lowers the natural frequency to 23 f1

COMPLETE SOLUTION OF SYSTEM EXECUTING SHM

The equation of motion of system executing SHM can be represented by

0)( =+ xkxm ampamp ----------------------------------------(1)

022

2=+

dtdx

dt

dx ω

The general solution of equation (1) can be expressed as

------------------------(2)

Where A and B are arbitrary constant which can be determined from the initial conditions of the system Two initial conditions are to be specified to evaluate these

constants x=x0 at t=0 and xamp = Vo at t=0 substituting in the equation (2) Energy method In a conservative system the total energy is constant The dwell as natural frequency can be determined by the princienergy For free vibration of undamped system at any instant and partly potential The kinetic energy T is stored in the masswhere as the potential energy U is stored in the form of sdeformation or work done in a force field such as gravity The total energy being constant T+U = constant Its rate of cha

t)(sin Bt)cos( AX ω+ω=

t)(sin V

t)cos( xx o0 ω

ω+ω=

Is the required complete

solution

28

ifferential equation as ple of conservation of of time is partly kinetic by virtue of its velocity train energy in elastic

nge

Is given by [ ] 0UTdtd

=+

From this we get a differential equation of motion as well as natural frequency of the system Determine the natural frequency of spring mass system using energy method

Determine the natural frequency of system shown in figure Determine the natural frequency of the system shown in figure Is there any limitation on the value of K Discuss Determine the natural frequency of s

m

l k

m

a

θ

m

a

θ

l

k

29

k

k

ystem shown below Neglect the mass of ball

l

m

30

A string shown in figure is under tension T which can be assumed to remain constant for small displacements Find the natural frequency of vertical vibrations of spring An acrobat 120kg walks on a tight rope as shown in figure The frequency of vibration in the given position is vertical direction is 30 rads Find the tension in the rope A manometer has a uniform bore of cross section area A If the column of liquid of length L and Density ρ is set into motion as shown in figure Find the frequency of the resulting oscillation

m

l

a

T T

36m 8m

31

k

R

k

m

r

m

Find the expression for natural frequency of system shown in the figure Neglect the mass of the cantilever beam Study the special case i) k=Infinity ii) I = infinity Determine the expression for the natural frequency of a system shown in figure The two discs are keyed to a common shaft and have a combined mass moment of inertia about centre of oscillation O It is assumed that the card attached to mass m does not stretch and is always under tension Determine the natural frequency of a system shown

m

l

32

Determine the expression for the natural frequency of the system shown in figure Assume that the wires connecting the masses do not stretch and are always in tension

Determine the natural frequency of spring mass system taking the MASS OF SPRING (ms ) into account

M3 M1

M2

k2 k1

m

k

33

RAYLEIGHrsquoS METHOD

This is the extension of energy method Here we can determine the natural frequency of a conservative system without finding the equation of motion The natural frequency is a function of the rate of change of Kinetic energy and potential energy of the system From the principle of conservation of energy we have T+U = Constant T= KE and U =PE We can write the above equation as T1+U1 = --------------------------(1) Where 1 and 2 represents tow instances of time Let 1 be the time when the mass is passing through static equilibrium position Let 2 be the time corresponding to the mass displacement of the mass At this instant the velocity f the mass is zero and hence Substituting in equation (1) we get ( )Max = ( )Max Above equation leads directly to natural frequency Determine the natural frequency of spring mass system using Rayleighrsquos method m

34

r

B

K K

Home work

Determine the natural frequency of pendulum using Rayleighrsquos method

A Cylinder of mass m and moment of inertia Jo is to role without slipping but is sustained by spring K as shown in figure Determine the natural frequency of oscillation

Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping

a

Determine the natural frequency of system shown where in cylindrical disc rolls over the inextensible spring A cylinder of radius r rolls without slipping on cylindrical the equation of motion and hence determine the naoscillations about the lowest point shown Repeat the above problem ndash Instead of cylinder assume swithout slipping on concave surface as shown above

r

K

35

portion of radius R Derive tural frequency of small

phere of radius with r rolls

The torsional pendulum has a natural frequency of 5 Hz What length of steel wire of diameter 2 mm should be used for pendulum The inertia of mass fixed at the free end is 00098 kgm2 Take g= 083x1011 Nm2 Determine the torsional natural frequency of the system shown in figure Neglect the mass moment of inertia of the shaft ( Figure not proportionate) Determine the natural frequency of simple pendulum considering the mass of the rod into account

Using the energy method find tThe chord may be assumed ine

Kt1

Kt2

20cm

06m 08m 04m

20cm 20cm 20cm

m

M

he natural frxtensible in

k

M

36

equency of the system shown in figure the spring mass pulley system

A rod of length l and depth H having a mass m rests on a semicylidrical surface os radius r It is tipped slightly and let go Find the frequency of oscillation

A thin bar of l having a mass m rests on a semicylidrical surface radius r It is tipped slightly and let go Find the frequency of oscillation

l

h

R

l

37

38

CHAPTER 3

DAMPED FREE VIBRATION OF SINGLE DEGREE FREEDOM SYSTEM LEARNING OBJECTIVES

] Introduction to damped free vibration

] Terminologies used in damped free vibration

] Three different types of damping

] Concept of critical damping and its importance

] Study of response of viscous damped system for cases of under damping

critical and over damping

] Concept of logarithmic decrement

] Problems related to damped free vibration problems

Viscously damped free vibration of single degree freedom sytem Consider a single degree freedom system with viscous damping as shown in figure For a viscous damped system the damping force is

where c is a constant of proportionality It is called damping coefficient or coefficient of viscous damping

Symbolically it is designated by a dashpot as shown in Fig 3

x xampx

39

At any instant of time let the mass be displaced by x from the equilibrium position Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

Okxxcxm =minusminus= ampampamp

Okxxcxm =++ ampampamp ----------------------------------------------------------(1)

or

0kxdtdx

Cdt

xdM

2

2

=++

Assuming the solution of equation as

The first term in the above equation m2Ct

eminus

is an exponentially decaying function of time The behavior of the terms in the parenthesis however depends on whether numerical value within the radicals is positive zero or negative

CASE 1 if 0mk

m2c

2

=minus

40

CASE 2 if 0mk

m2c

2

=gt

CASE 3 if 0mk

m2c

2

=lt

41

Critically damped coefficient Critically damped coefficient is defined as the value of the damping coeffient for which radical in equation

mk

m2c

m2c

S2

21 minus

plusmn

= becomes zero

Damping Ratio ( Damping factor) ( ξ ) It is defined as the ratio of damping coefficient to the critical damping coefficient Note The damping coefficient c is defined as the actual amount of damping while the damping coefficient is a constant depending upon the mass and stiffness of the system The roots of characteristics equation can be now written as Note if c=cc or ξ =1 then system is critically damped if cgt cc or ξ gt1 then system is over damped I if clt cc or ξ lt1 then system is under damped

42

Take home quiz Derive General solution of different damped system

1 Critically damped system 2 Under damped system 3 Over damped system

Logarithmic Decrement It is used to determine the amount of damping present in system This measure The rate of decay of free vibration The larger the damping the greater will be rate of decay Logarithmic decrement is defined as the natural logarithm of ratio of any two successive amplitudes

=δ

2

1

xx

ln

Expression for logarithmic decrement (δ ) Consider a damped free vibration which is expressed by

( ) ( ) φ+ξminusω= ξω t 1sinXex 2n

tn

x is the displacement at any instant of time X and φ are arbitrary constants

21

2

ξminus

πξ=δ

43

Tutorial problems 1 Derive an expression for damping ratio in terms of logarithmic decrement

2 Show that logarithmic decrement

=δ

n

o

xx

lnn1

where xo is the initial amplitude xn

is amplitude after n cycles 3 A mass of 1 kg is to be supported on a spring having stiffness K= 9800 Nm The damping constant c= 59 Nsm Determine the natural frequency of system Find also logarithmic decrement and amplitude after 3 cycles if initial displacement is 03 cm 4 A vibrating system is defined by the following parameters M = 3kg k= 100 Nm damping coefficient = 3 Nsm determine Damping factor natural frequency damped vibration logarithmic decrement ratio of consecutive amplitude no of cycles after which the original amplitude is reduced to 20 5A single degree damped vibrating system a suspended mass of 18kg makes 15 oscillations in 003 seconds The amplitude decreases to 025 of the initial value after 5 oscillations Determine the stiffness of spring Logarithmic decrement damping factor and damping coefficient 6 A damped vibration record of a spring mass system shows the following data Amplitude at the end of 2nd cycle is 9mm amplitude at eh end of 3rd cycle is 6mm amplitude at the end of 4th cycle is 4mm stiffness of spring =8000 Nm weight =20N Find the logarithmic decrement damping force at unit velocity periodic time of vibration 7 For the system shown in figure the characteristics of dashpot is such that the constant force of 49 N is applied to the piston it velocity is found to be constant at 012 ms Determine the value of C would you expect the complete system to be periodic or aperiodic 8 For a torsional system shown in the figuref1 is natural frequency in air and f2 is the natural frequency when immersed in oil Determine the damping coefficient

Viscous fluid

Disc of mass moment of Inertia J

2a

C

m M

b

a a

k

9 A thin plate of area A and weigh w is fixed to the end of the spring and allow to oscillate in a viscous fluid as shown in figure It f1 is the frequency of oscillation in air and f2 in liquid Show that

22

21 ff

gA2

minusπω

=micro where the damping force on the plate

AV2Fd micro= V being the velocity of vibration 10 A torsional pendulum has a natural frequency of 175 cyclessecond vibrating in vacuum The mass moment of inertia is 2 kgm2 It is immersed in oil and it is observed that the natural frequency is 142 cyclessecond Find the damping torque If the disc is replaced by 34 deg When in oil find the displacement at the end of the first complete cycle 11 An automobile can be modeled as a mass placed on 4 shock absorbers each consisting of a spring and a damper Such that each spring is equally loaded Determine the stiffness damping constant of each shock absorber So that natural frequency is 2Hz and system is critically damped The mass of vehicle is 200kg 12 Write the differential equation of motion for the system shown in the figure below Determine the natural frequency of the damped oscillation and critical damping coefficient

Determine the differential equation of motion and find the critical cdamped natural frequency for system shown in figure

a

C

m M

M =200kg

44

200kg kgkg

oefficient and

45

Derive the equation of motion and find the critical damping coefficient and damped natural frequency for the system shown below Write the differential equation of motion and determine the expression for critical damping coefficient and damped natural frequency of system shown in figure

Set up a differential equation of motion for system shown in figure and determine the following Undamped natural frequencies critical damping coefficient damping ratio damped natural frequency

b

C

kM

a

m

46

A mass of 300N is resting on two springs of stiffness 3000 Nm each and a dashpot of damping coefficient 150 Nsm as shown in figure If the initial velocity of 10cms is given to mass at its equilibrium position what will be the displacement from the equilibrium position at the end of first second Derive the formula used A gun barrel of mass 600kg has a recoil spring of stiffness 294 knm if the barrel recoils 13m on firing determine i) The initial recoil velocity of barrel ii) The critical damping coefficient of dashpot which is engaged at the end of recoil stroke Iii) Time required for the barrel to return at the end of recoil stroke Iv) time required for the barrel to return to a position 5cm from the initial position

W= 3000 N

FIRING

CHAPTER 4

FORCED VIBRATION LEARNING OBJECTIVES

] Introduction to forced free vibration

] Terminologies used in forced vibration

] Concept of magnification factor

] Study of reciprocating and rotating unbalance

] Concept of base excitation and support excitation

] Concept of Transmissibility ratio

] Problems related to forced vibration

The vibration that takes place under the excitation of external forces are called

forced vibration When a system is subjected to harmonic excitation it is forced to vibrate at the same frequency as that of excitation Most of the mechanical systems are prone to steady state forced vibration which cause fatigue failure The vibration problem also becomes very important when any of excitation frequencies coincide with one of natural frequencies of a system Under this condition resonance occurs during which large amplitude of vibrations are observed To avoid resonance external frequency (operating speed) may be changed or properties of the system may be changed to alter the natural frequency In some cases sufficient amount of damping may be provided to avoid large amplitude during resonance Thus the problem of force vibration is very important in mechanical design

FORCED VIBRATIONS OF SINGLE DOF SYSTEMS UNDER HARMONIC EXCITATION

` [ ] tsin FF o ω=

x

sin FF o=

Fo

xampamp

[ ] tω

xamp

47

48

Consider a spring mass system having a viscous damping excited by a harmonic force [ ] tsin FF o ω= as shown in figure The various forces acting on the system is shown in figure Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

O)tsin(Fkxxcxm o =ω+minusminus= ampampamp

)tsin(Fkxxcxm o ω=++ ampampamp ----------------------------------------------------------(1)

or

)tsin(Fkxdtdx

Cdt

xdM o2

2

ω=++

Equation (1) represents the differential equation of motion of the system The solution to this equation consists of two parts namely complementary function and a particular integral

The first of solution is also termed as transient solution of homogeneous solution and can be obtained from the homogeneous differential equation 0kxxcxm =++ ampampamp

The second part of solution particular integral is also termed as steady state solution ad is of form )tsin(Xx φminusω= Where X= steady state amplitude φ is phase difference between external force and displacement

)tsin(Xx φminusω=

The vector representation of the shown

49

The force polygon for the above vector diagram can be written as follows

( )φminusωt O

KX

xm 2ω

Fo

xCω

tω

φ

50

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

is the governing equation

Zero frequency deflection (Xst) It is deflection observed at zero frequency ( when the external frequency 0=ω

( ) kF

001

kF

x o

22

o

st =

+minus=

Magnification factor (M) It is defined as the ratio of steady state amplitude to zero frequency deflection xst

M = arX

X=

A plot of magnification factor versus frequency ratio

damping factor ( )ξ is called frequency plot of a harmon All the frequency terms starts from magnification factor of damping ratio ( )ξ The amplitude of vibration at any on the value of damping The amplitude will be higher for higher damping The amplitude of vibration at reso

amplitude tends to zero as

ωω

n tends to infinity

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

Freq

ζ =

ζ = 1

Magnification Factor M = XXst

kFo

51

ωω

nfor varying values of

ically excite vibrating system

M=1 what ever may be value particular frequency depends

for lesser damping and lesser nance will be very high The

uency Ratio r = (ωωn)

ζ = 0 0

ζ = 025 ζ = 0375 ζ = 0 5

ζ = 0707

ζ = 2

52

Plot of phase angle φ versus frequency ratio

ωω

n is shown in the

figure The phase angle φ is zero for all the values of damping ratio

at

ωω

n=0 The phase

angle φ is 90 for all the values of damping ratio

at

ωω

n=1 ie at

resonance The phase angle between o to 90 deg for all the values of

ωω

n= 0 to 1 and

between 90 deg to 180 deg for all the values of

ωω

ngt1

Class work Derive an expression for frequency at peak amplitude

2np 21 ξminusω=ω=ω is the required equation

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

53

Derive expression amplitude at resonance ( In dimensional form) In dimensional form We know that steady state amplitude

( ) ( )

ω+ωminus=

222

o

Cmk

Fx Equation in dimensional form

We know that at resonance mk

n =ω=ω

( )

ω+

minus

=2

n

2

oR

Cmk

mk

Fx

( )

ω=

2n

oR

C

Fx

( )

ω=

n

oR C

Fx

Class work Derive an expression for amplitude at resonance ( in non dimensional form)

Tutorial Problems on Forced Vibration 1A weight of 60N is suspended by a spring stiffness 12 kNm is forced to vibrate by a harmonic force of 10N assuming viscous damping of 0086 kN-sm Find i) resonance frequency ii) amplitude at resonance iii) phase angle at resonance iv) frequency corresponding to peak amplitude 2 A periodic torque having a maximum value of 05888Nm at a frequency of 4 rads is impressed upon a flywheel suspended from a wire The wheel has mass moment of inertia 012 kgm2 and the wire has stiffness 1176 Nmrad A viscous dashpot applies damping couple of 0784 Nm an angular velocity of 2 rads calculate i)maximum angular displacement from rest position ii) maximum couple applied to dashpot Iii) angle by which the angular displacement lags the torque A mass of 1019 kg is suspended from a spring stiffness 20 Nmm viscous damping causes the amplitude to decrease to 110th of initial value in 4 complete cycle If a periodic force of 20 cos(50 t) is applied to the mass Find the amplitude of forced vibration What would be the amplitude of oscillation if the applied force coincides with the natural period

m=1019 kg

[ ] t50cos 30F =

54

55

ROTATING UNBALANCE Unbalance in rotating machinery is a common source of vibration excitation Consider a machine of mass m having mounted on a foundation of stiffness k and damping coefficient C as shown in figure The unbalance is represented by an eccentric rotating mass m with eccentricity (e) and having a angular velocity (ω) let x be the displacement of non rotating mass from the static equilibrium position The displacement of rotating mass m from the static equilibrium is ( ) tsin( ex ω+ Applying Newtonrsquos law of motion we have

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

56

RECIPROCATING UNBALANCE Unbalance in reciprocating machinery is a common source of vibration excitation Consider a machine of mass M having mounted on a foundation of stiffness K and damping coefficient C as shown in figure The unbalance is represented by a reciprocating mass (m) having crank radus (e) and connecting rod of length (l) Let ωωωω be angular velocity of the crank Let x be the displacement of non reciprocating mass from the static equilibrium position at any instant of time The displacement of reciprocating mass m from the static equilibrium position is

)2sin(sin +++ tle

tex ωωωωωωωω

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

57

Note The equations obtained in reciprocating unbalance is same as the rotating unbalance and having the following differences Rotating Unbalance Reciprocating Unbalance Here mass of the rotating element is considered

Here mass of reciprocating element is considered

Here the eccentricity (e) which is the distance between the centre of rotation and CG

Here the eccentricity e is the crank radius and crank radius = stroke length2

The frequency plot of non dimensional factor( ( )me

MX versus frequency ratio

ωω

n for various values of ( )ξξξξ is shown in figure This frequency plot is same for

the rotating and reciprocating unbalance with the following points are noted from the diagram

1 Damping ratio plays very important role during resonance in controlling the amplitude

2 When

ωω

n=0 then ( )me

MX =0 for all the values of damping ratio ( )ξξξξ

3 When

ωω

nis higher than ( )me

MX tends to unity

4 At very high speeds ie at very high frequency ratio

ωω

n there is no effect

of damping The frequency plot of phase angle versus the frequency ratio is given as shown in figure The explanation is same as in forced harmonic vibration

58

VIBRATION ISOLATION

Vibratory forces and motions generated by machine and other caused are unavoidable Howeever their effect on the other dynamical systems can be minimized by proper isolator design An isolation system is either to protect o delicate machine from excessive vibration transmitted to it from its supporting structure or to prevent vibratory forces and motions generated by machines from being transmitted to its surroundings The effectiveness of isolation may be measured in terms of forces and motion transmitted The first type is known as force isolation and the second type is known as motion isolation

Force isolation (Force Transmissibility) When vibratory forces generated by machine are isolated the effectiveness of forced isolation can be made by defining the term transmissibility The term transmissibility is defined as the ratio of force transmitted to the foundation to that of the external impressed force

Force Transmissibility = o

tr

FF

force Impressed Externalfoundation toted transmitForce

=

Consider viscous damped spring mass system as shown in figure Upon this system an external harmonic force ) t (Sin Fo ω is applied The difference forces acting on system are inertia forces damping force spring force and external force All these forces are represented on force polygon shown below Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these two forces

( ) ( )22 xckx ω+=

59

This is the expression for force transmissibility in dimensional form To represent same in the non dimensional form divide numerator and denominator by k

2

n

22

n

2

n

o

tr

21

21

FF

TR

ωω

ξ+

ωω

minus

ωω

ξ+

==

PHASE DIFFERNECE BETWEEN EXTERNAL FORCE AND DISPLACEMENT Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these

ωω

minus

ωω

ξ=

ωminusω

=φ minusminus2

n

n1

2

1

1

2tan

mKC

tan

The phase difference between force transmitted to the foundation Ftr to the displacement

kc

KxxC

tanω

=ω

=α or

ωω

ξ=

ω=α minusminus

n

11 2tanKx

xCtan

The phase difference between external force Fo and force transmitted to foundationis β

60

β= αminusφ =

ωω

minus

ωω

ξminus

2

n

n1

1

2tan -

ωω

ξminus

n

1 2tan

Displacement Transmissibility It is the ratio of amplitude of steady state vibration of machine to that of the base motion Forced vibration due to support motion Consider a spring mass damper system subjected to harmonic excitation )tsin(Yy ω= at the base as shown in figure Let x be the absolute displacement of the mass at any instant of time t The different forces acting on system are shown in free body diagram Applying Newtonrsquos Law of motion Ma = sumF

)yx(k)yx(cxm minusminusminusminus= ampampampamp

0)yx(k)yx(cxm =minus+minus+ ampampampamp

61

Frequency plot of transmissibility versus frequency ratio

62

The plot of transmissibility versus frequency ratio is as shown in figure The following points can be noted from it

i) The value of transmissibility is unity at frequency ratio

ωω

n=0 and 2

ii) For an undamped system ( )ξ =0 transmissibility tends to infin at resonance iii) The value of transmissibility is less than unity for the value of frequency ratio

greater than 2

iv) For frequency ratio

ωω

nlt 2 smaller damping ratio lead to larger values of

transmissibility on the other hand for frequency ratio

ωω

ngt 2 smaller damping

ratio leads to smaller value of transmissibility

The frequency plot of transmissibility (TR) versus frequency ratio

ωω

n can be

broadly divided into two zones

1

ωω

n= 0 to 2

2

ωω

n= 2 to infin

CASE 1

When

ωω

n= 0 to 2

In this zone the value of transmissibility is always greater than 1 for all values of damping ratio In this zone the transmissibility decreases with increase in damping Hence we can conclude that in this zone damping is useful in isolating vibration forces being transmitted to the support CASE 2

When

ωω

n= 2 to infin

In this zone the value of transmissibility is less than 1 for all values of damping ratio In this zone the transmissibility increases with increase in damping Hence we can conclude that in this zone damping will adversely affect in isolating vibratory forces

63

Tutorial problems on Forced Vibration 1 Machine of mass 75kg is mounted on spring of stiffness 12000 Nm with an assumed damping factor ξ =02 piston within the machine of mass 2 kg has a reciprocating motion with a stroke of 75 cm and speed of 15 cyclessecond Assuming the motion of piston to be harmonic determine amplitude of machine ( X) Transmissibility and force transmitted to foundation phase angle of transmitted force wrt to exciting force 2 Motor weighs 600N is mounted on a simple beam that has a spring rate of 400 Ncm The motor armature base 160 N and has an eccentricity of 0005cm What will be the amplitude of motion when it runs at 1760 RPM Neglect damping the deflection of armature shaft and weight of the beam 3 A Heavy machine of mass 300 kg is supported on resilient foundation the static deflection of foundation due to weight of machine is found to be 75mm It is observed that the machine vibrates with an amplitude of 10mm when the base of the foundation is subjected to harmonic oscillation at the undampeds natural frequency of system with an amplitude of 25mm Find the damping constant of foundation Ii) dynamic force amplitude on the base iii) amplitude of the displacement of machine relative to base 4 An air compressor of mass 450 kg operates at a constant speed of 1750 RPM The reciprocating masses are 10 kg the crank radius is 100mm If the damper for mounting introduces a damping factor 015 i) specify the springs of mounting such that only 20 of the unbalanced force is transmitted to the foundation ii) determine the magnitude of the transmitted force bull Repeat the above problem by assuming that there is no damping in system

5 A body of a car having mass 1500 kg and is mounted on 4 equal springs which deflect through 225 cm under the weight of the body A total damping forces the 4 shock absorbers is 46 N at a velocity of 1cmsec The car is placed with all the four wheels on a test platform which is move up and down at resonant speed With an amplitude of 25cm Find the amplitude of the car body on its springs assuming its CG to be in the centre of wheel base 6A 75 kg machine which is mounted on springs of stiffness 116x105 Nm with the damping of 20=ξ A 2kg piston within the machine as reciprocating motion with a stroke of 008m and speed of 3000 RPM Assuming the motion of piston to be harmonic determine the amplitude of vibration of machine

7 The spring of an automobile are compressed 10cm under its own weight Find the critical speed when a trailer is traveling over road with a people approximaterly by a sine wave of amplitude 75 cm and wave length 15m What will be the amplitude of vibration at 64kmhour 8 An eccentric mass excites us used to determine vibratory characteristics of a structure of mass 25kg as given in figure At a speed of 1000 RPM the eccentric mass to beat top position at the instant the structure was moved upwards through its static equilibrium position and corresponding amplitude was 20mm The eccentric mass is 05kg at an eccentricity of 10mm Determine i) Undamped natural frequency of the system ii) Damping ratio iii) Steady state amplitude and phase angle at 1500 RPM iv) Force transmitted to the ground at 1200 RPM

M=25kg

C

K

Eccentric mass

64

65

CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Support motion Consider a spring mass damper system as shown in figure Let it excited by motion

of support let y be the harmonic displacement of support From Newtonrsquos second law of motion

)yx(k)yx(cxm minusminusminusminus= ampampampamp

)1(0)yx(k)yx(cxm minusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminus=minus+minus+ ampampampamp

Let z be the relative displacement of the mass with respect to support then

Z=x-y

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

10

O

A

B

Addition of SIMPLE HARMONIC MOTION ( SHM) The addition of two simple harmonic motion having frequency yields a resultant which is simple harmonic having the same frequency Consider tow simple harmonic motions of 1x and 2x having the same frequency and phase difference φ as given below

)(sin11 tXx ωωωω====

)(sin22 φω += tXx

Adding x = x1 +x2 Hence the resultant displacement is also SHM of amplitude X and phase angle θ

θθθθ

ωωωωt

φφφφ

11

-6

-4

-2

0

2

4

6

0 005 01 015 02 025

Tutorial problems on Simple Harmonic Motion 1 Add the following harmonic motion analytically and verify the solution graphically 1) X1= 3 sin (((( ))))30 t ++++ωωωω X2= 4 cos (((( ))))10 t ++++ωωωω ( VTU Jan 2005) 2) X1= 2 cos (((( ))))05 t ++++ωωωω X2= 5 sin (((( ))))1 t ++++ωωωω ( VTU July 2006)

3) X1= 10 cos (((( ))))4 t ππππωωωω ++++ X2= 8 sin (((( ))))6 t ππππωωωω ++++ ( VTU Dec 2007)

2 A body is subjected to two harmonic motions X1= 8 cos (((( ))))6 t ππππωωωω ++++ X2= 15 sin (((( ))))6 t ππππωωωω ++++ what harmonic motion

should be given to the body to bring it to equilibrium (VTU July 2005) 3 Split the harmonic motion x = 10 sin (((( ))))6 t ππππωωωω ++++ into two harmonic motions

having the phase of zero and the other of 45o 4 Show that resultant motion of harmonic motion given below is zero X1= X sin (((( ))))t ωωωω X2= X sin (((( ))))3

2 t ππππωωωω ++++ X3= X sin (((( ))))34 t ππππωωωω ++++

5 The displacement of a vibrating body is given by x = 5 sin (3141t + 4

ππππ )

Draw the variation of displacement for one cycle of vibration Also determine the displacement of body after 011 second ( repeat the problem for velocity and acceleration and draw graph using Excel and compare ) Time Displacement

0 0025 5 005

0075 0 01

0125 015 -353

0175 0 02

Calculate the remaining values 6 The Motion of a particle is represented by x = 4 ( ) t sin ω sketch the variation of the displacement velocity and acceleration and determine the max value of these quantities Assume ( )5 =ω ( Try to use MATLABExcel) sketch all on same graph

12

BEATS When two harmonic motions with frequencies close to one another are added The resulting motion exhibits a phenomenon known as Beats A Beat Frequency is the result of two closely spaced frequencies going into and out of synchronization with one another Let us consider tow harmonic motion of same amplitude and slightly different frequencies X1 = X Cos ( ) tωωωω X2 = X Cos ( )( ) tδδδδωωωω + Where δ is a small quantity The addition of above two harmonics can be written as X = X1 + X2

X=

+

ttX2

cos2

cos2δδδδωωωω

δδδδ

The above equation shown graphically in Figure The resulting motion represents cosine wave with frequency ( ) 2

ππππωωωω + and with a varying amplitude 2 cos( ( ) 2δδδδ t

whenever the amplitude reaches a maximum it is called the beat The frequency δ at which amplitude builds up and dies down between o and 2 X is known as beat frequency Beat phenomenon is found in machines structures and electric power houses In machines the beat phenomenon occurs when the forcing frequency is close to the natural frequency of the system

+2X

-2X

ωωωωππππ2

ELEMENTS OF VIBRATION The elements of constitute vibrating systems are

1 Mass or Inertia element - m 2 Spring - k 3 Damper - c 4 Excitation F(t)

Mass or Inertia element The mass or inertia element is assumed to be rigid body During vibration velomass changes hence kinetic energy can be gained or loosed The force appthe mass from newton second law of motion can be written as F= ma Work dthe mass is stored in the form of kinetic energy given by KE= frac12 M V2

Combination of masses In practical cases for simple analysis we replace several masses by equivalent mass Case1 Translational masses connected by rigid bar Let the masses M1 M2 and M3 are attached to a rigid bar at locates 1 respectively as shown in the figure The equivalent mass meq be assumelocated at 1 is as shown in figure (b) Let the displacement of masses M1 M2 and M3 be x1 x2 x3 and simivelocities of respective masses be x1 x2 and x3 We can express the velomasses m2 and m3 in terms of m1

Elements of Vibration

Passive element

m C K

Active element

F(t)

Conservative element

( Mass and Spring)

Non conservative element

( Damper)

C k

m F(t)

Voig

13

city of lied on one on

a single

2 and 3 d to be

larly the cities of

t Model

Translationa Let a mass mmass momearrangement These two mamass Meq or

2

1

33

2

1

221

+

+=

ll

Mll

MMMeq

14

is the required answer

l and rotational masses coupled together

having a translational velocity x be coupled to another mass having nt of inertia Io with a rotational velocity θ as in rack and pinion

shown in the figure

sses can be combined to obtain either a single equivalent translational a single equivalent mass moment of inertia Jeq

15

Equivalent translational masses

Kinetic energy of the equivalent mass =

2

21

eqeq XM amp

Kinetic energy of the two masses =

+

20

2

21

21

θθθθampamp JXM

Meq= m

+ 2RJ

m o is the required answer

Also determine the equivalent rotational mass Jeq

Jeq= m [ ]oJmR +2 is the required answer Spring element Whenever there is a relative motion between the two ends of the spring a force is developed called spring force or restoring force The spring force is proportional to the amount of deformation x and then F α x or F = kx Where k is stiffness of the spring or spring constant or spring gradient The spring stiffness is equal to spring force per unit deromation

The spring stiffness k = mNxF

Workdone in deforming a spring is equal to the strain energy or potential energy Strain energy = potential energy = area of the triangle OAB Stiffness of beams Cantilever beam consider a cantilever beam with an end mass shown in the figure The mass of the beam is assumed to be negligible The static deflection of beam at free end is given by

Similarly derive the expression for Simply supported beam and fixed support beam

x

k

m

x

m

x

m

x

mNlIE

st 192

3

=δδδδ

mNEI

Wlst

48

3

=δδδδ

F

K=stiffness

F

16

17

Stiffness of slender bar subjected to longitudinal vibrations For a system executing the longitudinal vibrations as shown in the figure let us derive the expression for stiffness

Torsional Stiffness of bar It is the amount of torque required to cause a unit angular deformation in the element

Torsional stiffness = Kt = θθθθT

Combination of stiffness Determination of equivalent spring stiffness when the springs are arranged in series Consider two springs of stiffness K1 and K2 acted upon by the force F

The deflection of spring k1 is x1 = 1KF

The deflection of spring k1 is x2 = Let these two springs be replaced by an equivalent stiffness Keq upon which a force F acts and due to which its deflection is given by

x = eqKF

x= x1+x2

m

E A

l

60 cm 70 cm

Determination of equivalent spring stiffness when the springs are arranged in parallel Force acting on K1 spring = F1=k1x Force acting on K2 spring = F2 Force required for an equivalent spring keq to defined by x given by F= Keq x But F = F1 +F2 Tutorial problems on Equivalent stiffness of springs 1Determine the equivalent stiffness for the system shown in figure 2 Determine the equivalent stiffness for the system shown in figure

3 Determine the equivale

M

3

M

Nm 102 6x

18

nt stiffness for the system shown in figure

38 kg

Nm 101 6x Nm 102 6x

Nm 106x

2k

k

k

2k

3k

k

2k

k

19

60 cm 80 cm 50 cm

50 cm 60 cm

4Determine the equivalent stiffness for the system shown in figure

5Replace the following torsional stiffness by a single shaft having radius 4cm and find the length required for the equivalent shaft Assume the material of given system and equivalent system is same R1= 3cm R2= 5cm DAMPING Every vibration energy is gradually converted into heat or sound Hence the displacement during vibration gradually reduces The mechanism by which vibration energy is gradually converted into heat or sound is known as damping A damper is assumed to have either mass or elasticity Hence damping is modeled as one or more of the following types Viscous damping Coulomb or dry friction damping materials or solid or hysteric damping Viscous damping Viscous damping is most commonly used damping mechanism in vibration analysis When the mechanical system vibrates in a fluid medium such as air gas water or oil the resistance offered by the fluid to the moving body causes energy to be dissipated In this case the amount of dissipated energy depends on many factors such as size or shape of the vibrating body the viscosity of the fluid the frequency of vibration and velocity of fluid Resistance due to viscous damping is directly proportional to the velocity of vibration

dF V αααα

dF = xC amp

Where C= damping coefficient Fd = damping force Examples of Viscous damping

1) Fluid film between sliding surface 2) Fluid flow around a piston in a cylinder 3) Fluid flow through orifice 4) Fluid flow around a journal in a bearing

Rreq =4cm

leqn

stress

Strain

Coulomb damping or dry friction damping Here a damping force is constant in magnitude but opposite in direction to that of the motion of vibrating body It is caused by the friction between the surfaces that are dry or have insufficient lubrication Material or solid or hysteric damping

When the materials are deformed energy is absorbed and dissipated by the material The effect is due to friction between the internal planes which slip or slide as the deformation takes place When a body having the material damping is subjected to vibration the stress strain diagram shows the hysteresis loop as shown in the figure The area of the loop denotes the energy lost per unit volume of the body per cycle

20

21

ττττ ττττ2 ττττ3

ττττ

X(t)

FOURIER SERIES The simplest of periodic motion happens to be SHM It is simple to handle but the motion of many vibrating system is not harmonic (but periodic) Few examples are shown below

Forces acting on machines are generally periodic but this may not be harmonic for example the excitation force in a punching machine is periodic and it can be represented as shown in figure 13 Vibration analysis of system subjected to periodic but nonharmonic forces can be done with the help of Fourier series The problem becomes a multifrequency excitation problem The principle of linear superposition is applied and the total response is the sum of the response due to each of the individual frequency term

Any periodic motion can be expressed as an infinite sum of sines and cosines terms If x(t) is a periodic function with period t its Fourier representation is given by

X(t) = (((( )))) (((( )))) (((( ))))t sint cost cos2 111 ωωωωωωωωωωωω baaao ++++++++++++

= (((( )))) (((( ))))t sint cos2 1

ωωωωωωωω nnn

o bnaa

++++++++sumsumsumsuminfininfininfininfin

====

tππππωωωω

2==== = Fundamental frequency ndash (1)

where ao an bn are constants

ττττ ττττ2 ττττ3

ττττ

X(t)

22

t

X(t)

035

025

Determination of constants To find ao Integrate both sides of equation(1) over any interval ττττ All intergrals on the RHS of the equation are zero except the one containing ao

(((( ))))dttxao

o 2

2

intintintint====ωωωωππππ

ππππωωωω

= (((( ))))dttxo

2intintintintττττ

ττττ

To find an multiply equation 1 by cos (((( ))))tnωωωω and Integrate over any interval ττττ All intergrals

(((( )))) (((( )))) dttxao

n tncos 2

2

ωωωωππππωωωω ωωωω

ππππ

intintintint==== (((( )))) (((( )))) dttxo

tncos 2

ωωωωττττ

ττττ

intintintint====

To find bn multiply equation 1 by sin (((( ))))tnωωωω and Integrate over any interval ττττ All intergrals

(((( )))) (((( )))) dttxao

n tnsin 2

2

ωωωωππππωωωω ωωωω

ππππ

intintintint==== (((( )))) (((( )))) dttxo

tnsin 2

ωωωωττττ

ττττ

intintintint====

Find the Fourier series for the curve shown below

ττττ ττττ2 ττττ3

t

X(t)

E

ττττ ττττ2 ττττ3

t

X(t)

ππππ2

t

X(t) ππππ

1 Represent for the periodic motion shown in the figure

23

CHAPTER 2 UNDAMPED FREE VIBRATION LEARNING OBJECTIVES

] Introduction to undamped free vibration

] Terminologies used in undamped free vibration

] Three methods to solve the undamped free vibration problems

] Problems related to undamped free vibration problems

Free vibrations are oscillations about a systems equilibrium position that occur in the absence of an external excitation force If during vibrations there is no loss of energy it is known as undamped vibration The first step in solving a vibration problem is setting up the differential equation of motion Consider spring mass system which is assumed to move only along the vertical direction as shown below Let m be the mass of the block and k be the stiffness of the spring When block of mass m is attached to spring the deflection of spring will be ∆ known as static deflection In the static equilibrium position the free body diagram of forces acting on the mass is shown in Figure(b) Hence mg= kA Once the system is disturbed the system executes vibrations

Let at any instant of time t the mass is displaced from the equilibrium position x the different forces acting on the system are shown in figure (d) From Newtonrsquos second law of motion

sum = maF

Inertia force ( disturbing force) = restoring force

mgxkxm ++∆minus= )(ampamp

0)( =+ xkxm ampamp

or 0)( =+ xmk

xampamp

equation 2 is the differential equation of motion for spring mfigure Comparing equation (2) with the equation of SHM +xampampsince the vibrations of the above system are free( without the forces) we can write

sec radmk

n =ω

24

ass system shown in 0)(2 =xω

resistance of external

time period km

2 f1

n

πτ ==

from the equation(1) mg = k∆

∆=

q

mk

Difference between the translation ( rectilinear) and rotational system of vibration

Translatory Rotational

In the analysis thFORCES are co Linear stiffness K Problems 1A mass of 10kFind the natural 2 A spring massnatural frequencnatural frequenc

kt

m

sec radmk

n =ω

k

25

e disturbing and restoring nsidered

In the analysis In the analysis MASS Moment of Inertia (J) is considered

in Nm is considered

g when suspended from a spring causes a static deflection of 1cm frequency of system

system has a spring stiffness K Nm and a mass of m Kg It has a y of vibration 12 Hz An extra 2kg mass coupled to it then the y reduces by 2 Hz find K and m

secrad n =ω

26

3 A steel wire of 2mm diameter and 30m long It is fixed at the upper end and carries a mass of m kg at its free end Find m so that the frequency of longitudinal vibration is 4 Hz 4 A spring mass system has a natural period of 02 seconds What will be the new period if the spring constant is 1) increased by 50 2) decreased by 50 5 A spring mass system has a natural frequency of 10 Hz when the spring constant is reduced by 800 Nm the frequency is altered by 45 Find the mass and spring constant of the original system 6 Determine the natural frequency of system shown in fig is which shaft is supported in SHORT bearings 7 Determine the natural frequency of system shown in fig is which shaft is supported in LONG bearings where l is the length of bearing and E ndash youngrsquos modulus and I is moment of Inertia 7 Determine the natural frequency of system shown in fig is which shaft is supported in LONG bearings 8 A light cantilever beam of rectangular section( 5 cm deep x 25cm wide) has a mass fixed at its free end Find the ratio of frequency of free lateral vibration in vertical plane to that in horizontal 9 Determine the natural frequency of simple pendulum 10 Homogeneous square plate of size l and mass m is suspended from the mid point of one of its sides as shown in figure Find the frequency of vibration

l

l

27

11 A compound pendulum which is rigid body of mass m and it is pivoted at O The point of pivot is at distance d from the centre of gravity It is free to rotate about its axis Find the frequency of oscillation of such pendulum 12 A connecting rod shown in fig is supported at the wrist pin end It is displaced and allowed to oscillate The mass of rod is 5kg and centre of gravity is 20 cm from the pivot point O If the frequency of oscillation is 40 cyclesminute calculate the mass moment of inertia about its CG

13 A semi circular homogenous disc of radius r and mass m is pivoted freely about its centre as shown in figure Determine the natural frequency of oscillation 14A simply supported beam of square cross section 5mmx5mm and length 1m carrying a mass of 0575 kg at the middle is found to have natural frequency of 30 radsec Determine youngrsquos modulus of elasticity of beam 15 A spring mass system k1 and m have a natural frequency f1 Determine the value of k2 of another spring in terms of k1 which when placed in series with k1 lowers the natural frequency to 23 f1

COMPLETE SOLUTION OF SYSTEM EXECUTING SHM

The equation of motion of system executing SHM can be represented by

0)( =+ xkxm ampamp ----------------------------------------(1)

022

2=+

dtdx

dt

dx ω

The general solution of equation (1) can be expressed as

------------------------(2)

Where A and B are arbitrary constant which can be determined from the initial conditions of the system Two initial conditions are to be specified to evaluate these

constants x=x0 at t=0 and xamp = Vo at t=0 substituting in the equation (2) Energy method In a conservative system the total energy is constant The dwell as natural frequency can be determined by the princienergy For free vibration of undamped system at any instant and partly potential The kinetic energy T is stored in the masswhere as the potential energy U is stored in the form of sdeformation or work done in a force field such as gravity The total energy being constant T+U = constant Its rate of cha

t)(sin Bt)cos( AX ω+ω=

t)(sin V

t)cos( xx o0 ω

ω+ω=

Is the required complete

solution

28

ifferential equation as ple of conservation of of time is partly kinetic by virtue of its velocity train energy in elastic

nge

Is given by [ ] 0UTdtd

=+

From this we get a differential equation of motion as well as natural frequency of the system Determine the natural frequency of spring mass system using energy method

Determine the natural frequency of system shown in figure Determine the natural frequency of the system shown in figure Is there any limitation on the value of K Discuss Determine the natural frequency of s

m

l k

m

a

θ

m

a

θ

l

k

29

k

k

ystem shown below Neglect the mass of ball

l

m

30

A string shown in figure is under tension T which can be assumed to remain constant for small displacements Find the natural frequency of vertical vibrations of spring An acrobat 120kg walks on a tight rope as shown in figure The frequency of vibration in the given position is vertical direction is 30 rads Find the tension in the rope A manometer has a uniform bore of cross section area A If the column of liquid of length L and Density ρ is set into motion as shown in figure Find the frequency of the resulting oscillation

m

l

a

T T

36m 8m

31

k

R

k

m

r

m

Find the expression for natural frequency of system shown in the figure Neglect the mass of the cantilever beam Study the special case i) k=Infinity ii) I = infinity Determine the expression for the natural frequency of a system shown in figure The two discs are keyed to a common shaft and have a combined mass moment of inertia about centre of oscillation O It is assumed that the card attached to mass m does not stretch and is always under tension Determine the natural frequency of a system shown

m

l

32

Determine the expression for the natural frequency of the system shown in figure Assume that the wires connecting the masses do not stretch and are always in tension

Determine the natural frequency of spring mass system taking the MASS OF SPRING (ms ) into account

M3 M1

M2

k2 k1

m

k

33

RAYLEIGHrsquoS METHOD

This is the extension of energy method Here we can determine the natural frequency of a conservative system without finding the equation of motion The natural frequency is a function of the rate of change of Kinetic energy and potential energy of the system From the principle of conservation of energy we have T+U = Constant T= KE and U =PE We can write the above equation as T1+U1 = --------------------------(1) Where 1 and 2 represents tow instances of time Let 1 be the time when the mass is passing through static equilibrium position Let 2 be the time corresponding to the mass displacement of the mass At this instant the velocity f the mass is zero and hence Substituting in equation (1) we get ( )Max = ( )Max Above equation leads directly to natural frequency Determine the natural frequency of spring mass system using Rayleighrsquos method m

34

r

B

K K

Home work

Determine the natural frequency of pendulum using Rayleighrsquos method

A Cylinder of mass m and moment of inertia Jo is to role without slipping but is sustained by spring K as shown in figure Determine the natural frequency of oscillation

Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping

a

Determine the natural frequency of system shown where in cylindrical disc rolls over the inextensible spring A cylinder of radius r rolls without slipping on cylindrical the equation of motion and hence determine the naoscillations about the lowest point shown Repeat the above problem ndash Instead of cylinder assume swithout slipping on concave surface as shown above

r

K

35

portion of radius R Derive tural frequency of small

phere of radius with r rolls

The torsional pendulum has a natural frequency of 5 Hz What length of steel wire of diameter 2 mm should be used for pendulum The inertia of mass fixed at the free end is 00098 kgm2 Take g= 083x1011 Nm2 Determine the torsional natural frequency of the system shown in figure Neglect the mass moment of inertia of the shaft ( Figure not proportionate) Determine the natural frequency of simple pendulum considering the mass of the rod into account

Using the energy method find tThe chord may be assumed ine

Kt1

Kt2

20cm

06m 08m 04m

20cm 20cm 20cm

m

M

he natural frxtensible in

k

M

36

equency of the system shown in figure the spring mass pulley system

A rod of length l and depth H having a mass m rests on a semicylidrical surface os radius r It is tipped slightly and let go Find the frequency of oscillation

A thin bar of l having a mass m rests on a semicylidrical surface radius r It is tipped slightly and let go Find the frequency of oscillation

l

h

R

l

37

38

CHAPTER 3

DAMPED FREE VIBRATION OF SINGLE DEGREE FREEDOM SYSTEM LEARNING OBJECTIVES

] Introduction to damped free vibration

] Terminologies used in damped free vibration

] Three different types of damping

] Concept of critical damping and its importance

] Study of response of viscous damped system for cases of under damping

critical and over damping

] Concept of logarithmic decrement

] Problems related to damped free vibration problems

Viscously damped free vibration of single degree freedom sytem Consider a single degree freedom system with viscous damping as shown in figure For a viscous damped system the damping force is

where c is a constant of proportionality It is called damping coefficient or coefficient of viscous damping

Symbolically it is designated by a dashpot as shown in Fig 3

x xampx

39

At any instant of time let the mass be displaced by x from the equilibrium position Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

Okxxcxm =minusminus= ampampamp

Okxxcxm =++ ampampamp ----------------------------------------------------------(1)

or

0kxdtdx

Cdt

xdM

2

2

=++

Assuming the solution of equation as

The first term in the above equation m2Ct

eminus

is an exponentially decaying function of time The behavior of the terms in the parenthesis however depends on whether numerical value within the radicals is positive zero or negative

CASE 1 if 0mk

m2c

2

=minus

40

CASE 2 if 0mk

m2c

2

=gt

CASE 3 if 0mk

m2c

2

=lt

41

Critically damped coefficient Critically damped coefficient is defined as the value of the damping coeffient for which radical in equation

mk

m2c

m2c

S2

21 minus

plusmn

= becomes zero

Damping Ratio ( Damping factor) ( ξ ) It is defined as the ratio of damping coefficient to the critical damping coefficient Note The damping coefficient c is defined as the actual amount of damping while the damping coefficient is a constant depending upon the mass and stiffness of the system The roots of characteristics equation can be now written as Note if c=cc or ξ =1 then system is critically damped if cgt cc or ξ gt1 then system is over damped I if clt cc or ξ lt1 then system is under damped

42

Take home quiz Derive General solution of different damped system

1 Critically damped system 2 Under damped system 3 Over damped system

Logarithmic Decrement It is used to determine the amount of damping present in system This measure The rate of decay of free vibration The larger the damping the greater will be rate of decay Logarithmic decrement is defined as the natural logarithm of ratio of any two successive amplitudes

=δ

2

1

xx

ln

Expression for logarithmic decrement (δ ) Consider a damped free vibration which is expressed by

( ) ( ) φ+ξminusω= ξω t 1sinXex 2n

tn

x is the displacement at any instant of time X and φ are arbitrary constants

21

2

ξminus

πξ=δ

43

Tutorial problems 1 Derive an expression for damping ratio in terms of logarithmic decrement

2 Show that logarithmic decrement

=δ

n

o

xx

lnn1

where xo is the initial amplitude xn

is amplitude after n cycles 3 A mass of 1 kg is to be supported on a spring having stiffness K= 9800 Nm The damping constant c= 59 Nsm Determine the natural frequency of system Find also logarithmic decrement and amplitude after 3 cycles if initial displacement is 03 cm 4 A vibrating system is defined by the following parameters M = 3kg k= 100 Nm damping coefficient = 3 Nsm determine Damping factor natural frequency damped vibration logarithmic decrement ratio of consecutive amplitude no of cycles after which the original amplitude is reduced to 20 5A single degree damped vibrating system a suspended mass of 18kg makes 15 oscillations in 003 seconds The amplitude decreases to 025 of the initial value after 5 oscillations Determine the stiffness of spring Logarithmic decrement damping factor and damping coefficient 6 A damped vibration record of a spring mass system shows the following data Amplitude at the end of 2nd cycle is 9mm amplitude at eh end of 3rd cycle is 6mm amplitude at the end of 4th cycle is 4mm stiffness of spring =8000 Nm weight =20N Find the logarithmic decrement damping force at unit velocity periodic time of vibration 7 For the system shown in figure the characteristics of dashpot is such that the constant force of 49 N is applied to the piston it velocity is found to be constant at 012 ms Determine the value of C would you expect the complete system to be periodic or aperiodic 8 For a torsional system shown in the figuref1 is natural frequency in air and f2 is the natural frequency when immersed in oil Determine the damping coefficient

Viscous fluid

Disc of mass moment of Inertia J

2a

C

m M

b

a a

k

9 A thin plate of area A and weigh w is fixed to the end of the spring and allow to oscillate in a viscous fluid as shown in figure It f1 is the frequency of oscillation in air and f2 in liquid Show that

22

21 ff

gA2

minusπω

=micro where the damping force on the plate

AV2Fd micro= V being the velocity of vibration 10 A torsional pendulum has a natural frequency of 175 cyclessecond vibrating in vacuum The mass moment of inertia is 2 kgm2 It is immersed in oil and it is observed that the natural frequency is 142 cyclessecond Find the damping torque If the disc is replaced by 34 deg When in oil find the displacement at the end of the first complete cycle 11 An automobile can be modeled as a mass placed on 4 shock absorbers each consisting of a spring and a damper Such that each spring is equally loaded Determine the stiffness damping constant of each shock absorber So that natural frequency is 2Hz and system is critically damped The mass of vehicle is 200kg 12 Write the differential equation of motion for the system shown in the figure below Determine the natural frequency of the damped oscillation and critical damping coefficient

Determine the differential equation of motion and find the critical cdamped natural frequency for system shown in figure

a

C

m M

M =200kg

44

200kg kgkg

oefficient and

45

Derive the equation of motion and find the critical damping coefficient and damped natural frequency for the system shown below Write the differential equation of motion and determine the expression for critical damping coefficient and damped natural frequency of system shown in figure

Set up a differential equation of motion for system shown in figure and determine the following Undamped natural frequencies critical damping coefficient damping ratio damped natural frequency

b

C

kM

a

m

46

A mass of 300N is resting on two springs of stiffness 3000 Nm each and a dashpot of damping coefficient 150 Nsm as shown in figure If the initial velocity of 10cms is given to mass at its equilibrium position what will be the displacement from the equilibrium position at the end of first second Derive the formula used A gun barrel of mass 600kg has a recoil spring of stiffness 294 knm if the barrel recoils 13m on firing determine i) The initial recoil velocity of barrel ii) The critical damping coefficient of dashpot which is engaged at the end of recoil stroke Iii) Time required for the barrel to return at the end of recoil stroke Iv) time required for the barrel to return to a position 5cm from the initial position

W= 3000 N

FIRING

CHAPTER 4

FORCED VIBRATION LEARNING OBJECTIVES

] Introduction to forced free vibration

] Terminologies used in forced vibration

] Concept of magnification factor

] Study of reciprocating and rotating unbalance

] Concept of base excitation and support excitation

] Concept of Transmissibility ratio

] Problems related to forced vibration

The vibration that takes place under the excitation of external forces are called

forced vibration When a system is subjected to harmonic excitation it is forced to vibrate at the same frequency as that of excitation Most of the mechanical systems are prone to steady state forced vibration which cause fatigue failure The vibration problem also becomes very important when any of excitation frequencies coincide with one of natural frequencies of a system Under this condition resonance occurs during which large amplitude of vibrations are observed To avoid resonance external frequency (operating speed) may be changed or properties of the system may be changed to alter the natural frequency In some cases sufficient amount of damping may be provided to avoid large amplitude during resonance Thus the problem of force vibration is very important in mechanical design

FORCED VIBRATIONS OF SINGLE DOF SYSTEMS UNDER HARMONIC EXCITATION

` [ ] tsin FF o ω=

x

sin FF o=

Fo

xampamp

[ ] tω

xamp

47

48

Consider a spring mass system having a viscous damping excited by a harmonic force [ ] tsin FF o ω= as shown in figure The various forces acting on the system is shown in figure Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

O)tsin(Fkxxcxm o =ω+minusminus= ampampamp

)tsin(Fkxxcxm o ω=++ ampampamp ----------------------------------------------------------(1)

or

)tsin(Fkxdtdx

Cdt

xdM o2

2

ω=++

Equation (1) represents the differential equation of motion of the system The solution to this equation consists of two parts namely complementary function and a particular integral

The first of solution is also termed as transient solution of homogeneous solution and can be obtained from the homogeneous differential equation 0kxxcxm =++ ampampamp

The second part of solution particular integral is also termed as steady state solution ad is of form )tsin(Xx φminusω= Where X= steady state amplitude φ is phase difference between external force and displacement

)tsin(Xx φminusω=

The vector representation of the shown

49

The force polygon for the above vector diagram can be written as follows

( )φminusωt O

KX

xm 2ω

Fo

xCω

tω

φ

50

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

is the governing equation

Zero frequency deflection (Xst) It is deflection observed at zero frequency ( when the external frequency 0=ω

( ) kF

001

kF

x o

22

o

st =

+minus=

Magnification factor (M) It is defined as the ratio of steady state amplitude to zero frequency deflection xst

M = arX

X=

A plot of magnification factor versus frequency ratio

damping factor ( )ξ is called frequency plot of a harmon All the frequency terms starts from magnification factor of damping ratio ( )ξ The amplitude of vibration at any on the value of damping The amplitude will be higher for higher damping The amplitude of vibration at reso

amplitude tends to zero as

ωω

n tends to infinity

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

Freq

ζ =

ζ = 1

Magnification Factor M = XXst

kFo

51

ωω

nfor varying values of

ically excite vibrating system

M=1 what ever may be value particular frequency depends

for lesser damping and lesser nance will be very high The

uency Ratio r = (ωωn)

ζ = 0 0

ζ = 025 ζ = 0375 ζ = 0 5

ζ = 0707

ζ = 2

52

Plot of phase angle φ versus frequency ratio

ωω

n is shown in the

figure The phase angle φ is zero for all the values of damping ratio

at

ωω

n=0 The phase

angle φ is 90 for all the values of damping ratio

at

ωω

n=1 ie at

resonance The phase angle between o to 90 deg for all the values of

ωω

n= 0 to 1 and

between 90 deg to 180 deg for all the values of

ωω

ngt1

Class work Derive an expression for frequency at peak amplitude

2np 21 ξminusω=ω=ω is the required equation

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

53

Derive expression amplitude at resonance ( In dimensional form) In dimensional form We know that steady state amplitude

( ) ( )

ω+ωminus=

222

o

Cmk

Fx Equation in dimensional form

We know that at resonance mk

n =ω=ω

( )

ω+

minus

=2

n

2

oR

Cmk

mk

Fx

( )

ω=

2n

oR

C

Fx

( )

ω=

n

oR C

Fx

Class work Derive an expression for amplitude at resonance ( in non dimensional form)

Tutorial Problems on Forced Vibration 1A weight of 60N is suspended by a spring stiffness 12 kNm is forced to vibrate by a harmonic force of 10N assuming viscous damping of 0086 kN-sm Find i) resonance frequency ii) amplitude at resonance iii) phase angle at resonance iv) frequency corresponding to peak amplitude 2 A periodic torque having a maximum value of 05888Nm at a frequency of 4 rads is impressed upon a flywheel suspended from a wire The wheel has mass moment of inertia 012 kgm2 and the wire has stiffness 1176 Nmrad A viscous dashpot applies damping couple of 0784 Nm an angular velocity of 2 rads calculate i)maximum angular displacement from rest position ii) maximum couple applied to dashpot Iii) angle by which the angular displacement lags the torque A mass of 1019 kg is suspended from a spring stiffness 20 Nmm viscous damping causes the amplitude to decrease to 110th of initial value in 4 complete cycle If a periodic force of 20 cos(50 t) is applied to the mass Find the amplitude of forced vibration What would be the amplitude of oscillation if the applied force coincides with the natural period

m=1019 kg

[ ] t50cos 30F =

54

55

ROTATING UNBALANCE Unbalance in rotating machinery is a common source of vibration excitation Consider a machine of mass m having mounted on a foundation of stiffness k and damping coefficient C as shown in figure The unbalance is represented by an eccentric rotating mass m with eccentricity (e) and having a angular velocity (ω) let x be the displacement of non rotating mass from the static equilibrium position The displacement of rotating mass m from the static equilibrium is ( ) tsin( ex ω+ Applying Newtonrsquos law of motion we have

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

56

RECIPROCATING UNBALANCE Unbalance in reciprocating machinery is a common source of vibration excitation Consider a machine of mass M having mounted on a foundation of stiffness K and damping coefficient C as shown in figure The unbalance is represented by a reciprocating mass (m) having crank radus (e) and connecting rod of length (l) Let ωωωω be angular velocity of the crank Let x be the displacement of non reciprocating mass from the static equilibrium position at any instant of time The displacement of reciprocating mass m from the static equilibrium position is

)2sin(sin +++ tle

tex ωωωωωωωω

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

57

Note The equations obtained in reciprocating unbalance is same as the rotating unbalance and having the following differences Rotating Unbalance Reciprocating Unbalance Here mass of the rotating element is considered

Here mass of reciprocating element is considered

Here the eccentricity (e) which is the distance between the centre of rotation and CG

Here the eccentricity e is the crank radius and crank radius = stroke length2

The frequency plot of non dimensional factor( ( )me

MX versus frequency ratio

ωω

n for various values of ( )ξξξξ is shown in figure This frequency plot is same for

the rotating and reciprocating unbalance with the following points are noted from the diagram

1 Damping ratio plays very important role during resonance in controlling the amplitude

2 When

ωω

n=0 then ( )me

MX =0 for all the values of damping ratio ( )ξξξξ

3 When

ωω

nis higher than ( )me

MX tends to unity

4 At very high speeds ie at very high frequency ratio

ωω

n there is no effect

of damping The frequency plot of phase angle versus the frequency ratio is given as shown in figure The explanation is same as in forced harmonic vibration

58

VIBRATION ISOLATION

Vibratory forces and motions generated by machine and other caused are unavoidable Howeever their effect on the other dynamical systems can be minimized by proper isolator design An isolation system is either to protect o delicate machine from excessive vibration transmitted to it from its supporting structure or to prevent vibratory forces and motions generated by machines from being transmitted to its surroundings The effectiveness of isolation may be measured in terms of forces and motion transmitted The first type is known as force isolation and the second type is known as motion isolation

Force isolation (Force Transmissibility) When vibratory forces generated by machine are isolated the effectiveness of forced isolation can be made by defining the term transmissibility The term transmissibility is defined as the ratio of force transmitted to the foundation to that of the external impressed force

Force Transmissibility = o

tr

FF

force Impressed Externalfoundation toted transmitForce

=

Consider viscous damped spring mass system as shown in figure Upon this system an external harmonic force ) t (Sin Fo ω is applied The difference forces acting on system are inertia forces damping force spring force and external force All these forces are represented on force polygon shown below Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these two forces

( ) ( )22 xckx ω+=

59

This is the expression for force transmissibility in dimensional form To represent same in the non dimensional form divide numerator and denominator by k

2

n

22

n

2

n

o

tr

21

21

FF

TR

ωω

ξ+

ωω

minus

ωω

ξ+

==

PHASE DIFFERNECE BETWEEN EXTERNAL FORCE AND DISPLACEMENT Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these

ωω

minus

ωω

ξ=

ωminusω

=φ minusminus2

n

n1

2

1

1

2tan

mKC

tan

The phase difference between force transmitted to the foundation Ftr to the displacement

kc

KxxC

tanω

=ω

=α or

ωω

ξ=

ω=α minusminus

n

11 2tanKx

xCtan

The phase difference between external force Fo and force transmitted to foundationis β

60

β= αminusφ =

ωω

minus

ωω

ξminus

2

n

n1

1

2tan -

ωω

ξminus

n

1 2tan

Displacement Transmissibility It is the ratio of amplitude of steady state vibration of machine to that of the base motion Forced vibration due to support motion Consider a spring mass damper system subjected to harmonic excitation )tsin(Yy ω= at the base as shown in figure Let x be the absolute displacement of the mass at any instant of time t The different forces acting on system are shown in free body diagram Applying Newtonrsquos Law of motion Ma = sumF

)yx(k)yx(cxm minusminusminusminus= ampampampamp

0)yx(k)yx(cxm =minus+minus+ ampampampamp

61

Frequency plot of transmissibility versus frequency ratio

62

The plot of transmissibility versus frequency ratio is as shown in figure The following points can be noted from it

i) The value of transmissibility is unity at frequency ratio

ωω

n=0 and 2

ii) For an undamped system ( )ξ =0 transmissibility tends to infin at resonance iii) The value of transmissibility is less than unity for the value of frequency ratio

greater than 2

iv) For frequency ratio

ωω

nlt 2 smaller damping ratio lead to larger values of

transmissibility on the other hand for frequency ratio

ωω

ngt 2 smaller damping

ratio leads to smaller value of transmissibility

The frequency plot of transmissibility (TR) versus frequency ratio

ωω

n can be

broadly divided into two zones

1

ωω

n= 0 to 2

2

ωω

n= 2 to infin

CASE 1

When

ωω

n= 0 to 2

In this zone the value of transmissibility is always greater than 1 for all values of damping ratio In this zone the transmissibility decreases with increase in damping Hence we can conclude that in this zone damping is useful in isolating vibration forces being transmitted to the support CASE 2

When

ωω

n= 2 to infin

In this zone the value of transmissibility is less than 1 for all values of damping ratio In this zone the transmissibility increases with increase in damping Hence we can conclude that in this zone damping will adversely affect in isolating vibratory forces

63

Tutorial problems on Forced Vibration 1 Machine of mass 75kg is mounted on spring of stiffness 12000 Nm with an assumed damping factor ξ =02 piston within the machine of mass 2 kg has a reciprocating motion with a stroke of 75 cm and speed of 15 cyclessecond Assuming the motion of piston to be harmonic determine amplitude of machine ( X) Transmissibility and force transmitted to foundation phase angle of transmitted force wrt to exciting force 2 Motor weighs 600N is mounted on a simple beam that has a spring rate of 400 Ncm The motor armature base 160 N and has an eccentricity of 0005cm What will be the amplitude of motion when it runs at 1760 RPM Neglect damping the deflection of armature shaft and weight of the beam 3 A Heavy machine of mass 300 kg is supported on resilient foundation the static deflection of foundation due to weight of machine is found to be 75mm It is observed that the machine vibrates with an amplitude of 10mm when the base of the foundation is subjected to harmonic oscillation at the undampeds natural frequency of system with an amplitude of 25mm Find the damping constant of foundation Ii) dynamic force amplitude on the base iii) amplitude of the displacement of machine relative to base 4 An air compressor of mass 450 kg operates at a constant speed of 1750 RPM The reciprocating masses are 10 kg the crank radius is 100mm If the damper for mounting introduces a damping factor 015 i) specify the springs of mounting such that only 20 of the unbalanced force is transmitted to the foundation ii) determine the magnitude of the transmitted force bull Repeat the above problem by assuming that there is no damping in system

5 A body of a car having mass 1500 kg and is mounted on 4 equal springs which deflect through 225 cm under the weight of the body A total damping forces the 4 shock absorbers is 46 N at a velocity of 1cmsec The car is placed with all the four wheels on a test platform which is move up and down at resonant speed With an amplitude of 25cm Find the amplitude of the car body on its springs assuming its CG to be in the centre of wheel base 6A 75 kg machine which is mounted on springs of stiffness 116x105 Nm with the damping of 20=ξ A 2kg piston within the machine as reciprocating motion with a stroke of 008m and speed of 3000 RPM Assuming the motion of piston to be harmonic determine the amplitude of vibration of machine

7 The spring of an automobile are compressed 10cm under its own weight Find the critical speed when a trailer is traveling over road with a people approximaterly by a sine wave of amplitude 75 cm and wave length 15m What will be the amplitude of vibration at 64kmhour 8 An eccentric mass excites us used to determine vibratory characteristics of a structure of mass 25kg as given in figure At a speed of 1000 RPM the eccentric mass to beat top position at the instant the structure was moved upwards through its static equilibrium position and corresponding amplitude was 20mm The eccentric mass is 05kg at an eccentricity of 10mm Determine i) Undamped natural frequency of the system ii) Damping ratio iii) Steady state amplitude and phase angle at 1500 RPM iv) Force transmitted to the ground at 1200 RPM

M=25kg

C

K

Eccentric mass

64

65

CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Support motion Consider a spring mass damper system as shown in figure Let it excited by motion

of support let y be the harmonic displacement of support From Newtonrsquos second law of motion

)yx(k)yx(cxm minusminusminusminus= ampampampamp

)1(0)yx(k)yx(cxm minusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminus=minus+minus+ ampampampamp

Let z be the relative displacement of the mass with respect to support then

Z=x-y

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

11

-6

-4

-2

0

2

4

6

0 005 01 015 02 025

Tutorial problems on Simple Harmonic Motion 1 Add the following harmonic motion analytically and verify the solution graphically 1) X1= 3 sin (((( ))))30 t ++++ωωωω X2= 4 cos (((( ))))10 t ++++ωωωω ( VTU Jan 2005) 2) X1= 2 cos (((( ))))05 t ++++ωωωω X2= 5 sin (((( ))))1 t ++++ωωωω ( VTU July 2006)

3) X1= 10 cos (((( ))))4 t ππππωωωω ++++ X2= 8 sin (((( ))))6 t ππππωωωω ++++ ( VTU Dec 2007)

2 A body is subjected to two harmonic motions X1= 8 cos (((( ))))6 t ππππωωωω ++++ X2= 15 sin (((( ))))6 t ππππωωωω ++++ what harmonic motion

should be given to the body to bring it to equilibrium (VTU July 2005) 3 Split the harmonic motion x = 10 sin (((( ))))6 t ππππωωωω ++++ into two harmonic motions

having the phase of zero and the other of 45o 4 Show that resultant motion of harmonic motion given below is zero X1= X sin (((( ))))t ωωωω X2= X sin (((( ))))3

2 t ππππωωωω ++++ X3= X sin (((( ))))34 t ππππωωωω ++++

5 The displacement of a vibrating body is given by x = 5 sin (3141t + 4

ππππ )

Draw the variation of displacement for one cycle of vibration Also determine the displacement of body after 011 second ( repeat the problem for velocity and acceleration and draw graph using Excel and compare ) Time Displacement

0 0025 5 005

0075 0 01

0125 015 -353

0175 0 02

Calculate the remaining values 6 The Motion of a particle is represented by x = 4 ( ) t sin ω sketch the variation of the displacement velocity and acceleration and determine the max value of these quantities Assume ( )5 =ω ( Try to use MATLABExcel) sketch all on same graph

12

BEATS When two harmonic motions with frequencies close to one another are added The resulting motion exhibits a phenomenon known as Beats A Beat Frequency is the result of two closely spaced frequencies going into and out of synchronization with one another Let us consider tow harmonic motion of same amplitude and slightly different frequencies X1 = X Cos ( ) tωωωω X2 = X Cos ( )( ) tδδδδωωωω + Where δ is a small quantity The addition of above two harmonics can be written as X = X1 + X2

X=

+

ttX2

cos2

cos2δδδδωωωω

δδδδ

The above equation shown graphically in Figure The resulting motion represents cosine wave with frequency ( ) 2

ππππωωωω + and with a varying amplitude 2 cos( ( ) 2δδδδ t

whenever the amplitude reaches a maximum it is called the beat The frequency δ at which amplitude builds up and dies down between o and 2 X is known as beat frequency Beat phenomenon is found in machines structures and electric power houses In machines the beat phenomenon occurs when the forcing frequency is close to the natural frequency of the system

+2X

-2X

ωωωωππππ2

ELEMENTS OF VIBRATION The elements of constitute vibrating systems are

1 Mass or Inertia element - m 2 Spring - k 3 Damper - c 4 Excitation F(t)

Mass or Inertia element The mass or inertia element is assumed to be rigid body During vibration velomass changes hence kinetic energy can be gained or loosed The force appthe mass from newton second law of motion can be written as F= ma Work dthe mass is stored in the form of kinetic energy given by KE= frac12 M V2

Combination of masses In practical cases for simple analysis we replace several masses by equivalent mass Case1 Translational masses connected by rigid bar Let the masses M1 M2 and M3 are attached to a rigid bar at locates 1 respectively as shown in the figure The equivalent mass meq be assumelocated at 1 is as shown in figure (b) Let the displacement of masses M1 M2 and M3 be x1 x2 x3 and simivelocities of respective masses be x1 x2 and x3 We can express the velomasses m2 and m3 in terms of m1

Elements of Vibration

Passive element

m C K

Active element

F(t)

Conservative element

( Mass and Spring)

Non conservative element

( Damper)

C k

m F(t)

Voig

13

city of lied on one on

a single

2 and 3 d to be

larly the cities of

t Model

Translationa Let a mass mmass momearrangement These two mamass Meq or

2

1

33

2

1

221

+

+=

ll

Mll

MMMeq

14

is the required answer

l and rotational masses coupled together

having a translational velocity x be coupled to another mass having nt of inertia Io with a rotational velocity θ as in rack and pinion

shown in the figure

sses can be combined to obtain either a single equivalent translational a single equivalent mass moment of inertia Jeq

15

Equivalent translational masses

Kinetic energy of the equivalent mass =

2

21

eqeq XM amp

Kinetic energy of the two masses =

+

20

2

21

21

θθθθampamp JXM

Meq= m

+ 2RJ

m o is the required answer

Also determine the equivalent rotational mass Jeq

Jeq= m [ ]oJmR +2 is the required answer Spring element Whenever there is a relative motion between the two ends of the spring a force is developed called spring force or restoring force The spring force is proportional to the amount of deformation x and then F α x or F = kx Where k is stiffness of the spring or spring constant or spring gradient The spring stiffness is equal to spring force per unit deromation

The spring stiffness k = mNxF

Workdone in deforming a spring is equal to the strain energy or potential energy Strain energy = potential energy = area of the triangle OAB Stiffness of beams Cantilever beam consider a cantilever beam with an end mass shown in the figure The mass of the beam is assumed to be negligible The static deflection of beam at free end is given by

Similarly derive the expression for Simply supported beam and fixed support beam

x

k

m

x

m

x

m

x

mNlIE

st 192

3

=δδδδ

mNEI

Wlst

48

3

=δδδδ

F

K=stiffness

F

16

17

Stiffness of slender bar subjected to longitudinal vibrations For a system executing the longitudinal vibrations as shown in the figure let us derive the expression for stiffness

Torsional Stiffness of bar It is the amount of torque required to cause a unit angular deformation in the element

Torsional stiffness = Kt = θθθθT

Combination of stiffness Determination of equivalent spring stiffness when the springs are arranged in series Consider two springs of stiffness K1 and K2 acted upon by the force F

The deflection of spring k1 is x1 = 1KF

The deflection of spring k1 is x2 = Let these two springs be replaced by an equivalent stiffness Keq upon which a force F acts and due to which its deflection is given by

x = eqKF

x= x1+x2

m

E A

l

60 cm 70 cm

Determination of equivalent spring stiffness when the springs are arranged in parallel Force acting on K1 spring = F1=k1x Force acting on K2 spring = F2 Force required for an equivalent spring keq to defined by x given by F= Keq x But F = F1 +F2 Tutorial problems on Equivalent stiffness of springs 1Determine the equivalent stiffness for the system shown in figure 2 Determine the equivalent stiffness for the system shown in figure

3 Determine the equivale

M

3

M

Nm 102 6x

18

nt stiffness for the system shown in figure

38 kg

Nm 101 6x Nm 102 6x

Nm 106x

2k

k

k

2k

3k

k

2k

k

19

60 cm 80 cm 50 cm

50 cm 60 cm

4Determine the equivalent stiffness for the system shown in figure

5Replace the following torsional stiffness by a single shaft having radius 4cm and find the length required for the equivalent shaft Assume the material of given system and equivalent system is same R1= 3cm R2= 5cm DAMPING Every vibration energy is gradually converted into heat or sound Hence the displacement during vibration gradually reduces The mechanism by which vibration energy is gradually converted into heat or sound is known as damping A damper is assumed to have either mass or elasticity Hence damping is modeled as one or more of the following types Viscous damping Coulomb or dry friction damping materials or solid or hysteric damping Viscous damping Viscous damping is most commonly used damping mechanism in vibration analysis When the mechanical system vibrates in a fluid medium such as air gas water or oil the resistance offered by the fluid to the moving body causes energy to be dissipated In this case the amount of dissipated energy depends on many factors such as size or shape of the vibrating body the viscosity of the fluid the frequency of vibration and velocity of fluid Resistance due to viscous damping is directly proportional to the velocity of vibration

dF V αααα

dF = xC amp

Where C= damping coefficient Fd = damping force Examples of Viscous damping

1) Fluid film between sliding surface 2) Fluid flow around a piston in a cylinder 3) Fluid flow through orifice 4) Fluid flow around a journal in a bearing

Rreq =4cm

leqn

stress

Strain

Coulomb damping or dry friction damping Here a damping force is constant in magnitude but opposite in direction to that of the motion of vibrating body It is caused by the friction between the surfaces that are dry or have insufficient lubrication Material or solid or hysteric damping

When the materials are deformed energy is absorbed and dissipated by the material The effect is due to friction between the internal planes which slip or slide as the deformation takes place When a body having the material damping is subjected to vibration the stress strain diagram shows the hysteresis loop as shown in the figure The area of the loop denotes the energy lost per unit volume of the body per cycle

20

21

ττττ ττττ2 ττττ3

ττττ

X(t)

FOURIER SERIES The simplest of periodic motion happens to be SHM It is simple to handle but the motion of many vibrating system is not harmonic (but periodic) Few examples are shown below

Forces acting on machines are generally periodic but this may not be harmonic for example the excitation force in a punching machine is periodic and it can be represented as shown in figure 13 Vibration analysis of system subjected to periodic but nonharmonic forces can be done with the help of Fourier series The problem becomes a multifrequency excitation problem The principle of linear superposition is applied and the total response is the sum of the response due to each of the individual frequency term

Any periodic motion can be expressed as an infinite sum of sines and cosines terms If x(t) is a periodic function with period t its Fourier representation is given by

X(t) = (((( )))) (((( )))) (((( ))))t sint cost cos2 111 ωωωωωωωωωωωω baaao ++++++++++++

= (((( )))) (((( ))))t sint cos2 1

ωωωωωωωω nnn

o bnaa

++++++++sumsumsumsuminfininfininfininfin

====

tππππωωωω

2==== = Fundamental frequency ndash (1)

where ao an bn are constants

ττττ ττττ2 ττττ3

ττττ

X(t)

22

t

X(t)

035

025

Determination of constants To find ao Integrate both sides of equation(1) over any interval ττττ All intergrals on the RHS of the equation are zero except the one containing ao

(((( ))))dttxao

o 2

2

intintintint====ωωωωππππ

ππππωωωω

= (((( ))))dttxo

2intintintintττττ

ττττ

To find an multiply equation 1 by cos (((( ))))tnωωωω and Integrate over any interval ττττ All intergrals

(((( )))) (((( )))) dttxao

n tncos 2

2

ωωωωππππωωωω ωωωω

ππππ

intintintint==== (((( )))) (((( )))) dttxo

tncos 2

ωωωωττττ

ττττ

intintintint====

To find bn multiply equation 1 by sin (((( ))))tnωωωω and Integrate over any interval ττττ All intergrals

(((( )))) (((( )))) dttxao

n tnsin 2

2

ωωωωππππωωωω ωωωω

ππππ

intintintint==== (((( )))) (((( )))) dttxo

tnsin 2

ωωωωττττ

ττττ

intintintint====

Find the Fourier series for the curve shown below

ττττ ττττ2 ττττ3

t

X(t)

E

ττττ ττττ2 ττττ3

t

X(t)

ππππ2

t

X(t) ππππ

1 Represent for the periodic motion shown in the figure

23

CHAPTER 2 UNDAMPED FREE VIBRATION LEARNING OBJECTIVES

] Introduction to undamped free vibration

] Terminologies used in undamped free vibration

] Three methods to solve the undamped free vibration problems

] Problems related to undamped free vibration problems

Free vibrations are oscillations about a systems equilibrium position that occur in the absence of an external excitation force If during vibrations there is no loss of energy it is known as undamped vibration The first step in solving a vibration problem is setting up the differential equation of motion Consider spring mass system which is assumed to move only along the vertical direction as shown below Let m be the mass of the block and k be the stiffness of the spring When block of mass m is attached to spring the deflection of spring will be ∆ known as static deflection In the static equilibrium position the free body diagram of forces acting on the mass is shown in Figure(b) Hence mg= kA Once the system is disturbed the system executes vibrations

Let at any instant of time t the mass is displaced from the equilibrium position x the different forces acting on the system are shown in figure (d) From Newtonrsquos second law of motion

sum = maF

Inertia force ( disturbing force) = restoring force

mgxkxm ++∆minus= )(ampamp

0)( =+ xkxm ampamp

or 0)( =+ xmk

xampamp

equation 2 is the differential equation of motion for spring mfigure Comparing equation (2) with the equation of SHM +xampampsince the vibrations of the above system are free( without the forces) we can write

sec radmk

n =ω

24

ass system shown in 0)(2 =xω

resistance of external

time period km

2 f1

n

πτ ==

from the equation(1) mg = k∆

∆=

q

mk

Difference between the translation ( rectilinear) and rotational system of vibration

Translatory Rotational

In the analysis thFORCES are co Linear stiffness K Problems 1A mass of 10kFind the natural 2 A spring massnatural frequencnatural frequenc

kt

m

sec radmk

n =ω

k

25

e disturbing and restoring nsidered

In the analysis In the analysis MASS Moment of Inertia (J) is considered

in Nm is considered

g when suspended from a spring causes a static deflection of 1cm frequency of system

system has a spring stiffness K Nm and a mass of m Kg It has a y of vibration 12 Hz An extra 2kg mass coupled to it then the y reduces by 2 Hz find K and m

secrad n =ω

26

3 A steel wire of 2mm diameter and 30m long It is fixed at the upper end and carries a mass of m kg at its free end Find m so that the frequency of longitudinal vibration is 4 Hz 4 A spring mass system has a natural period of 02 seconds What will be the new period if the spring constant is 1) increased by 50 2) decreased by 50 5 A spring mass system has a natural frequency of 10 Hz when the spring constant is reduced by 800 Nm the frequency is altered by 45 Find the mass and spring constant of the original system 6 Determine the natural frequency of system shown in fig is which shaft is supported in SHORT bearings 7 Determine the natural frequency of system shown in fig is which shaft is supported in LONG bearings where l is the length of bearing and E ndash youngrsquos modulus and I is moment of Inertia 7 Determine the natural frequency of system shown in fig is which shaft is supported in LONG bearings 8 A light cantilever beam of rectangular section( 5 cm deep x 25cm wide) has a mass fixed at its free end Find the ratio of frequency of free lateral vibration in vertical plane to that in horizontal 9 Determine the natural frequency of simple pendulum 10 Homogeneous square plate of size l and mass m is suspended from the mid point of one of its sides as shown in figure Find the frequency of vibration

l

l

27

11 A compound pendulum which is rigid body of mass m and it is pivoted at O The point of pivot is at distance d from the centre of gravity It is free to rotate about its axis Find the frequency of oscillation of such pendulum 12 A connecting rod shown in fig is supported at the wrist pin end It is displaced and allowed to oscillate The mass of rod is 5kg and centre of gravity is 20 cm from the pivot point O If the frequency of oscillation is 40 cyclesminute calculate the mass moment of inertia about its CG

13 A semi circular homogenous disc of radius r and mass m is pivoted freely about its centre as shown in figure Determine the natural frequency of oscillation 14A simply supported beam of square cross section 5mmx5mm and length 1m carrying a mass of 0575 kg at the middle is found to have natural frequency of 30 radsec Determine youngrsquos modulus of elasticity of beam 15 A spring mass system k1 and m have a natural frequency f1 Determine the value of k2 of another spring in terms of k1 which when placed in series with k1 lowers the natural frequency to 23 f1

COMPLETE SOLUTION OF SYSTEM EXECUTING SHM

The equation of motion of system executing SHM can be represented by

0)( =+ xkxm ampamp ----------------------------------------(1)

022

2=+

dtdx

dt

dx ω

The general solution of equation (1) can be expressed as

------------------------(2)

Where A and B are arbitrary constant which can be determined from the initial conditions of the system Two initial conditions are to be specified to evaluate these

constants x=x0 at t=0 and xamp = Vo at t=0 substituting in the equation (2) Energy method In a conservative system the total energy is constant The dwell as natural frequency can be determined by the princienergy For free vibration of undamped system at any instant and partly potential The kinetic energy T is stored in the masswhere as the potential energy U is stored in the form of sdeformation or work done in a force field such as gravity The total energy being constant T+U = constant Its rate of cha

t)(sin Bt)cos( AX ω+ω=

t)(sin V

t)cos( xx o0 ω

ω+ω=

Is the required complete

solution

28

ifferential equation as ple of conservation of of time is partly kinetic by virtue of its velocity train energy in elastic

nge

Is given by [ ] 0UTdtd

=+

From this we get a differential equation of motion as well as natural frequency of the system Determine the natural frequency of spring mass system using energy method

Determine the natural frequency of system shown in figure Determine the natural frequency of the system shown in figure Is there any limitation on the value of K Discuss Determine the natural frequency of s

m

l k

m

a

θ

m

a

θ

l

k

29

k

k

ystem shown below Neglect the mass of ball

l

m

30

A string shown in figure is under tension T which can be assumed to remain constant for small displacements Find the natural frequency of vertical vibrations of spring An acrobat 120kg walks on a tight rope as shown in figure The frequency of vibration in the given position is vertical direction is 30 rads Find the tension in the rope A manometer has a uniform bore of cross section area A If the column of liquid of length L and Density ρ is set into motion as shown in figure Find the frequency of the resulting oscillation

m

l

a

T T

36m 8m

31

k

R

k

m

r

m

Find the expression for natural frequency of system shown in the figure Neglect the mass of the cantilever beam Study the special case i) k=Infinity ii) I = infinity Determine the expression for the natural frequency of a system shown in figure The two discs are keyed to a common shaft and have a combined mass moment of inertia about centre of oscillation O It is assumed that the card attached to mass m does not stretch and is always under tension Determine the natural frequency of a system shown

m

l

32

Determine the expression for the natural frequency of the system shown in figure Assume that the wires connecting the masses do not stretch and are always in tension

Determine the natural frequency of spring mass system taking the MASS OF SPRING (ms ) into account

M3 M1

M2

k2 k1

m

k

33

RAYLEIGHrsquoS METHOD

This is the extension of energy method Here we can determine the natural frequency of a conservative system without finding the equation of motion The natural frequency is a function of the rate of change of Kinetic energy and potential energy of the system From the principle of conservation of energy we have T+U = Constant T= KE and U =PE We can write the above equation as T1+U1 = --------------------------(1) Where 1 and 2 represents tow instances of time Let 1 be the time when the mass is passing through static equilibrium position Let 2 be the time corresponding to the mass displacement of the mass At this instant the velocity f the mass is zero and hence Substituting in equation (1) we get ( )Max = ( )Max Above equation leads directly to natural frequency Determine the natural frequency of spring mass system using Rayleighrsquos method m

34

r

B

K K

Home work

Determine the natural frequency of pendulum using Rayleighrsquos method

A Cylinder of mass m and moment of inertia Jo is to role without slipping but is sustained by spring K as shown in figure Determine the natural frequency of oscillation

Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping

a

Determine the natural frequency of system shown where in cylindrical disc rolls over the inextensible spring A cylinder of radius r rolls without slipping on cylindrical the equation of motion and hence determine the naoscillations about the lowest point shown Repeat the above problem ndash Instead of cylinder assume swithout slipping on concave surface as shown above

r

K

35

portion of radius R Derive tural frequency of small

phere of radius with r rolls

The torsional pendulum has a natural frequency of 5 Hz What length of steel wire of diameter 2 mm should be used for pendulum The inertia of mass fixed at the free end is 00098 kgm2 Take g= 083x1011 Nm2 Determine the torsional natural frequency of the system shown in figure Neglect the mass moment of inertia of the shaft ( Figure not proportionate) Determine the natural frequency of simple pendulum considering the mass of the rod into account

Using the energy method find tThe chord may be assumed ine

Kt1

Kt2

20cm

06m 08m 04m

20cm 20cm 20cm

m

M

he natural frxtensible in

k

M

36

equency of the system shown in figure the spring mass pulley system

A rod of length l and depth H having a mass m rests on a semicylidrical surface os radius r It is tipped slightly and let go Find the frequency of oscillation

A thin bar of l having a mass m rests on a semicylidrical surface radius r It is tipped slightly and let go Find the frequency of oscillation

l

h

R

l

37

38

CHAPTER 3

DAMPED FREE VIBRATION OF SINGLE DEGREE FREEDOM SYSTEM LEARNING OBJECTIVES

] Introduction to damped free vibration

] Terminologies used in damped free vibration

] Three different types of damping

] Concept of critical damping and its importance

] Study of response of viscous damped system for cases of under damping

critical and over damping

] Concept of logarithmic decrement

] Problems related to damped free vibration problems

Viscously damped free vibration of single degree freedom sytem Consider a single degree freedom system with viscous damping as shown in figure For a viscous damped system the damping force is

where c is a constant of proportionality It is called damping coefficient or coefficient of viscous damping

Symbolically it is designated by a dashpot as shown in Fig 3

x xampx

39

At any instant of time let the mass be displaced by x from the equilibrium position Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

Okxxcxm =minusminus= ampampamp

Okxxcxm =++ ampampamp ----------------------------------------------------------(1)

or

0kxdtdx

Cdt

xdM

2

2

=++

Assuming the solution of equation as

The first term in the above equation m2Ct

eminus

is an exponentially decaying function of time The behavior of the terms in the parenthesis however depends on whether numerical value within the radicals is positive zero or negative

CASE 1 if 0mk

m2c

2

=minus

40

CASE 2 if 0mk

m2c

2

=gt

CASE 3 if 0mk

m2c

2

=lt

41

Critically damped coefficient Critically damped coefficient is defined as the value of the damping coeffient for which radical in equation

mk

m2c

m2c

S2

21 minus

plusmn

= becomes zero

Damping Ratio ( Damping factor) ( ξ ) It is defined as the ratio of damping coefficient to the critical damping coefficient Note The damping coefficient c is defined as the actual amount of damping while the damping coefficient is a constant depending upon the mass and stiffness of the system The roots of characteristics equation can be now written as Note if c=cc or ξ =1 then system is critically damped if cgt cc or ξ gt1 then system is over damped I if clt cc or ξ lt1 then system is under damped

42

Take home quiz Derive General solution of different damped system

1 Critically damped system 2 Under damped system 3 Over damped system

Logarithmic Decrement It is used to determine the amount of damping present in system This measure The rate of decay of free vibration The larger the damping the greater will be rate of decay Logarithmic decrement is defined as the natural logarithm of ratio of any two successive amplitudes

=δ

2

1

xx

ln

Expression for logarithmic decrement (δ ) Consider a damped free vibration which is expressed by

( ) ( ) φ+ξminusω= ξω t 1sinXex 2n

tn

x is the displacement at any instant of time X and φ are arbitrary constants

21

2

ξminus

πξ=δ

43

Tutorial problems 1 Derive an expression for damping ratio in terms of logarithmic decrement

2 Show that logarithmic decrement

=δ

n

o

xx

lnn1

where xo is the initial amplitude xn

is amplitude after n cycles 3 A mass of 1 kg is to be supported on a spring having stiffness K= 9800 Nm The damping constant c= 59 Nsm Determine the natural frequency of system Find also logarithmic decrement and amplitude after 3 cycles if initial displacement is 03 cm 4 A vibrating system is defined by the following parameters M = 3kg k= 100 Nm damping coefficient = 3 Nsm determine Damping factor natural frequency damped vibration logarithmic decrement ratio of consecutive amplitude no of cycles after which the original amplitude is reduced to 20 5A single degree damped vibrating system a suspended mass of 18kg makes 15 oscillations in 003 seconds The amplitude decreases to 025 of the initial value after 5 oscillations Determine the stiffness of spring Logarithmic decrement damping factor and damping coefficient 6 A damped vibration record of a spring mass system shows the following data Amplitude at the end of 2nd cycle is 9mm amplitude at eh end of 3rd cycle is 6mm amplitude at the end of 4th cycle is 4mm stiffness of spring =8000 Nm weight =20N Find the logarithmic decrement damping force at unit velocity periodic time of vibration 7 For the system shown in figure the characteristics of dashpot is such that the constant force of 49 N is applied to the piston it velocity is found to be constant at 012 ms Determine the value of C would you expect the complete system to be periodic or aperiodic 8 For a torsional system shown in the figuref1 is natural frequency in air and f2 is the natural frequency when immersed in oil Determine the damping coefficient

Viscous fluid

Disc of mass moment of Inertia J

2a

C

m M

b

a a

k

9 A thin plate of area A and weigh w is fixed to the end of the spring and allow to oscillate in a viscous fluid as shown in figure It f1 is the frequency of oscillation in air and f2 in liquid Show that

22

21 ff

gA2

minusπω

=micro where the damping force on the plate

AV2Fd micro= V being the velocity of vibration 10 A torsional pendulum has a natural frequency of 175 cyclessecond vibrating in vacuum The mass moment of inertia is 2 kgm2 It is immersed in oil and it is observed that the natural frequency is 142 cyclessecond Find the damping torque If the disc is replaced by 34 deg When in oil find the displacement at the end of the first complete cycle 11 An automobile can be modeled as a mass placed on 4 shock absorbers each consisting of a spring and a damper Such that each spring is equally loaded Determine the stiffness damping constant of each shock absorber So that natural frequency is 2Hz and system is critically damped The mass of vehicle is 200kg 12 Write the differential equation of motion for the system shown in the figure below Determine the natural frequency of the damped oscillation and critical damping coefficient

Determine the differential equation of motion and find the critical cdamped natural frequency for system shown in figure

a

C

m M

M =200kg

44

200kg kgkg

oefficient and

45

Derive the equation of motion and find the critical damping coefficient and damped natural frequency for the system shown below Write the differential equation of motion and determine the expression for critical damping coefficient and damped natural frequency of system shown in figure

Set up a differential equation of motion for system shown in figure and determine the following Undamped natural frequencies critical damping coefficient damping ratio damped natural frequency

b

C

kM

a

m

46

A mass of 300N is resting on two springs of stiffness 3000 Nm each and a dashpot of damping coefficient 150 Nsm as shown in figure If the initial velocity of 10cms is given to mass at its equilibrium position what will be the displacement from the equilibrium position at the end of first second Derive the formula used A gun barrel of mass 600kg has a recoil spring of stiffness 294 knm if the barrel recoils 13m on firing determine i) The initial recoil velocity of barrel ii) The critical damping coefficient of dashpot which is engaged at the end of recoil stroke Iii) Time required for the barrel to return at the end of recoil stroke Iv) time required for the barrel to return to a position 5cm from the initial position

W= 3000 N

FIRING

CHAPTER 4

FORCED VIBRATION LEARNING OBJECTIVES

] Introduction to forced free vibration

] Terminologies used in forced vibration

] Concept of magnification factor

] Study of reciprocating and rotating unbalance

] Concept of base excitation and support excitation

] Concept of Transmissibility ratio

] Problems related to forced vibration

The vibration that takes place under the excitation of external forces are called

forced vibration When a system is subjected to harmonic excitation it is forced to vibrate at the same frequency as that of excitation Most of the mechanical systems are prone to steady state forced vibration which cause fatigue failure The vibration problem also becomes very important when any of excitation frequencies coincide with one of natural frequencies of a system Under this condition resonance occurs during which large amplitude of vibrations are observed To avoid resonance external frequency (operating speed) may be changed or properties of the system may be changed to alter the natural frequency In some cases sufficient amount of damping may be provided to avoid large amplitude during resonance Thus the problem of force vibration is very important in mechanical design

FORCED VIBRATIONS OF SINGLE DOF SYSTEMS UNDER HARMONIC EXCITATION

` [ ] tsin FF o ω=

x

sin FF o=

Fo

xampamp

[ ] tω

xamp

47

48

Consider a spring mass system having a viscous damping excited by a harmonic force [ ] tsin FF o ω= as shown in figure The various forces acting on the system is shown in figure Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

O)tsin(Fkxxcxm o =ω+minusminus= ampampamp

)tsin(Fkxxcxm o ω=++ ampampamp ----------------------------------------------------------(1)

or

)tsin(Fkxdtdx

Cdt

xdM o2

2

ω=++

Equation (1) represents the differential equation of motion of the system The solution to this equation consists of two parts namely complementary function and a particular integral

The first of solution is also termed as transient solution of homogeneous solution and can be obtained from the homogeneous differential equation 0kxxcxm =++ ampampamp

The second part of solution particular integral is also termed as steady state solution ad is of form )tsin(Xx φminusω= Where X= steady state amplitude φ is phase difference between external force and displacement

)tsin(Xx φminusω=

The vector representation of the shown

49

The force polygon for the above vector diagram can be written as follows

( )φminusωt O

KX

xm 2ω

Fo

xCω

tω

φ

50

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

is the governing equation

Zero frequency deflection (Xst) It is deflection observed at zero frequency ( when the external frequency 0=ω

( ) kF

001

kF

x o

22

o

st =

+minus=

Magnification factor (M) It is defined as the ratio of steady state amplitude to zero frequency deflection xst

M = arX

X=

A plot of magnification factor versus frequency ratio

damping factor ( )ξ is called frequency plot of a harmon All the frequency terms starts from magnification factor of damping ratio ( )ξ The amplitude of vibration at any on the value of damping The amplitude will be higher for higher damping The amplitude of vibration at reso

amplitude tends to zero as

ωω

n tends to infinity

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

Freq

ζ =

ζ = 1

Magnification Factor M = XXst

kFo

51

ωω

nfor varying values of

ically excite vibrating system

M=1 what ever may be value particular frequency depends

for lesser damping and lesser nance will be very high The

uency Ratio r = (ωωn)

ζ = 0 0

ζ = 025 ζ = 0375 ζ = 0 5

ζ = 0707

ζ = 2

52

Plot of phase angle φ versus frequency ratio

ωω

n is shown in the

figure The phase angle φ is zero for all the values of damping ratio

at

ωω

n=0 The phase

angle φ is 90 for all the values of damping ratio

at

ωω

n=1 ie at

resonance The phase angle between o to 90 deg for all the values of

ωω

n= 0 to 1 and

between 90 deg to 180 deg for all the values of

ωω

ngt1

Class work Derive an expression for frequency at peak amplitude

2np 21 ξminusω=ω=ω is the required equation

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

53

Derive expression amplitude at resonance ( In dimensional form) In dimensional form We know that steady state amplitude

( ) ( )

ω+ωminus=

222

o

Cmk

Fx Equation in dimensional form

We know that at resonance mk

n =ω=ω

( )

ω+

minus

=2

n

2

oR

Cmk

mk

Fx

( )

ω=

2n

oR

C

Fx

( )

ω=

n

oR C

Fx

Class work Derive an expression for amplitude at resonance ( in non dimensional form)

Tutorial Problems on Forced Vibration 1A weight of 60N is suspended by a spring stiffness 12 kNm is forced to vibrate by a harmonic force of 10N assuming viscous damping of 0086 kN-sm Find i) resonance frequency ii) amplitude at resonance iii) phase angle at resonance iv) frequency corresponding to peak amplitude 2 A periodic torque having a maximum value of 05888Nm at a frequency of 4 rads is impressed upon a flywheel suspended from a wire The wheel has mass moment of inertia 012 kgm2 and the wire has stiffness 1176 Nmrad A viscous dashpot applies damping couple of 0784 Nm an angular velocity of 2 rads calculate i)maximum angular displacement from rest position ii) maximum couple applied to dashpot Iii) angle by which the angular displacement lags the torque A mass of 1019 kg is suspended from a spring stiffness 20 Nmm viscous damping causes the amplitude to decrease to 110th of initial value in 4 complete cycle If a periodic force of 20 cos(50 t) is applied to the mass Find the amplitude of forced vibration What would be the amplitude of oscillation if the applied force coincides with the natural period

m=1019 kg

[ ] t50cos 30F =

54

55

ROTATING UNBALANCE Unbalance in rotating machinery is a common source of vibration excitation Consider a machine of mass m having mounted on a foundation of stiffness k and damping coefficient C as shown in figure The unbalance is represented by an eccentric rotating mass m with eccentricity (e) and having a angular velocity (ω) let x be the displacement of non rotating mass from the static equilibrium position The displacement of rotating mass m from the static equilibrium is ( ) tsin( ex ω+ Applying Newtonrsquos law of motion we have

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

56

RECIPROCATING UNBALANCE Unbalance in reciprocating machinery is a common source of vibration excitation Consider a machine of mass M having mounted on a foundation of stiffness K and damping coefficient C as shown in figure The unbalance is represented by a reciprocating mass (m) having crank radus (e) and connecting rod of length (l) Let ωωωω be angular velocity of the crank Let x be the displacement of non reciprocating mass from the static equilibrium position at any instant of time The displacement of reciprocating mass m from the static equilibrium position is

)2sin(sin +++ tle

tex ωωωωωωωω

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

57

Note The equations obtained in reciprocating unbalance is same as the rotating unbalance and having the following differences Rotating Unbalance Reciprocating Unbalance Here mass of the rotating element is considered

Here mass of reciprocating element is considered

Here the eccentricity (e) which is the distance between the centre of rotation and CG

Here the eccentricity e is the crank radius and crank radius = stroke length2

The frequency plot of non dimensional factor( ( )me

MX versus frequency ratio

ωω

n for various values of ( )ξξξξ is shown in figure This frequency plot is same for

the rotating and reciprocating unbalance with the following points are noted from the diagram

1 Damping ratio plays very important role during resonance in controlling the amplitude

2 When

ωω

n=0 then ( )me

MX =0 for all the values of damping ratio ( )ξξξξ

3 When

ωω

nis higher than ( )me

MX tends to unity

4 At very high speeds ie at very high frequency ratio

ωω

n there is no effect

of damping The frequency plot of phase angle versus the frequency ratio is given as shown in figure The explanation is same as in forced harmonic vibration

58

VIBRATION ISOLATION

Vibratory forces and motions generated by machine and other caused are unavoidable Howeever their effect on the other dynamical systems can be minimized by proper isolator design An isolation system is either to protect o delicate machine from excessive vibration transmitted to it from its supporting structure or to prevent vibratory forces and motions generated by machines from being transmitted to its surroundings The effectiveness of isolation may be measured in terms of forces and motion transmitted The first type is known as force isolation and the second type is known as motion isolation

Force isolation (Force Transmissibility) When vibratory forces generated by machine are isolated the effectiveness of forced isolation can be made by defining the term transmissibility The term transmissibility is defined as the ratio of force transmitted to the foundation to that of the external impressed force

Force Transmissibility = o

tr

FF

force Impressed Externalfoundation toted transmitForce

=

Consider viscous damped spring mass system as shown in figure Upon this system an external harmonic force ) t (Sin Fo ω is applied The difference forces acting on system are inertia forces damping force spring force and external force All these forces are represented on force polygon shown below Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these two forces

( ) ( )22 xckx ω+=

59

This is the expression for force transmissibility in dimensional form To represent same in the non dimensional form divide numerator and denominator by k

2

n

22

n

2

n

o

tr

21

21

FF

TR

ωω

ξ+

ωω

minus

ωω

ξ+

==

PHASE DIFFERNECE BETWEEN EXTERNAL FORCE AND DISPLACEMENT Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these

ωω

minus

ωω

ξ=

ωminusω

=φ minusminus2

n

n1

2

1

1

2tan

mKC

tan

The phase difference between force transmitted to the foundation Ftr to the displacement

kc

KxxC

tanω

=ω

=α or

ωω

ξ=

ω=α minusminus

n

11 2tanKx

xCtan

The phase difference between external force Fo and force transmitted to foundationis β

60

β= αminusφ =

ωω

minus

ωω

ξminus

2

n

n1

1

2tan -

ωω

ξminus

n

1 2tan

Displacement Transmissibility It is the ratio of amplitude of steady state vibration of machine to that of the base motion Forced vibration due to support motion Consider a spring mass damper system subjected to harmonic excitation )tsin(Yy ω= at the base as shown in figure Let x be the absolute displacement of the mass at any instant of time t The different forces acting on system are shown in free body diagram Applying Newtonrsquos Law of motion Ma = sumF

)yx(k)yx(cxm minusminusminusminus= ampampampamp

0)yx(k)yx(cxm =minus+minus+ ampampampamp

61

Frequency plot of transmissibility versus frequency ratio

62

The plot of transmissibility versus frequency ratio is as shown in figure The following points can be noted from it

i) The value of transmissibility is unity at frequency ratio

ωω

n=0 and 2

ii) For an undamped system ( )ξ =0 transmissibility tends to infin at resonance iii) The value of transmissibility is less than unity for the value of frequency ratio

greater than 2

iv) For frequency ratio

ωω

nlt 2 smaller damping ratio lead to larger values of

transmissibility on the other hand for frequency ratio

ωω

ngt 2 smaller damping

ratio leads to smaller value of transmissibility

The frequency plot of transmissibility (TR) versus frequency ratio

ωω

n can be

broadly divided into two zones

1

ωω

n= 0 to 2

2

ωω

n= 2 to infin

CASE 1

When

ωω

n= 0 to 2

In this zone the value of transmissibility is always greater than 1 for all values of damping ratio In this zone the transmissibility decreases with increase in damping Hence we can conclude that in this zone damping is useful in isolating vibration forces being transmitted to the support CASE 2

When

ωω

n= 2 to infin

In this zone the value of transmissibility is less than 1 for all values of damping ratio In this zone the transmissibility increases with increase in damping Hence we can conclude that in this zone damping will adversely affect in isolating vibratory forces

63

Tutorial problems on Forced Vibration 1 Machine of mass 75kg is mounted on spring of stiffness 12000 Nm with an assumed damping factor ξ =02 piston within the machine of mass 2 kg has a reciprocating motion with a stroke of 75 cm and speed of 15 cyclessecond Assuming the motion of piston to be harmonic determine amplitude of machine ( X) Transmissibility and force transmitted to foundation phase angle of transmitted force wrt to exciting force 2 Motor weighs 600N is mounted on a simple beam that has a spring rate of 400 Ncm The motor armature base 160 N and has an eccentricity of 0005cm What will be the amplitude of motion when it runs at 1760 RPM Neglect damping the deflection of armature shaft and weight of the beam 3 A Heavy machine of mass 300 kg is supported on resilient foundation the static deflection of foundation due to weight of machine is found to be 75mm It is observed that the machine vibrates with an amplitude of 10mm when the base of the foundation is subjected to harmonic oscillation at the undampeds natural frequency of system with an amplitude of 25mm Find the damping constant of foundation Ii) dynamic force amplitude on the base iii) amplitude of the displacement of machine relative to base 4 An air compressor of mass 450 kg operates at a constant speed of 1750 RPM The reciprocating masses are 10 kg the crank radius is 100mm If the damper for mounting introduces a damping factor 015 i) specify the springs of mounting such that only 20 of the unbalanced force is transmitted to the foundation ii) determine the magnitude of the transmitted force bull Repeat the above problem by assuming that there is no damping in system

5 A body of a car having mass 1500 kg and is mounted on 4 equal springs which deflect through 225 cm under the weight of the body A total damping forces the 4 shock absorbers is 46 N at a velocity of 1cmsec The car is placed with all the four wheels on a test platform which is move up and down at resonant speed With an amplitude of 25cm Find the amplitude of the car body on its springs assuming its CG to be in the centre of wheel base 6A 75 kg machine which is mounted on springs of stiffness 116x105 Nm with the damping of 20=ξ A 2kg piston within the machine as reciprocating motion with a stroke of 008m and speed of 3000 RPM Assuming the motion of piston to be harmonic determine the amplitude of vibration of machine

7 The spring of an automobile are compressed 10cm under its own weight Find the critical speed when a trailer is traveling over road with a people approximaterly by a sine wave of amplitude 75 cm and wave length 15m What will be the amplitude of vibration at 64kmhour 8 An eccentric mass excites us used to determine vibratory characteristics of a structure of mass 25kg as given in figure At a speed of 1000 RPM the eccentric mass to beat top position at the instant the structure was moved upwards through its static equilibrium position and corresponding amplitude was 20mm The eccentric mass is 05kg at an eccentricity of 10mm Determine i) Undamped natural frequency of the system ii) Damping ratio iii) Steady state amplitude and phase angle at 1500 RPM iv) Force transmitted to the ground at 1200 RPM

M=25kg

C

K

Eccentric mass

64

65

CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Support motion Consider a spring mass damper system as shown in figure Let it excited by motion

of support let y be the harmonic displacement of support From Newtonrsquos second law of motion

)yx(k)yx(cxm minusminusminusminus= ampampampamp

)1(0)yx(k)yx(cxm minusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminus=minus+minus+ ampampampamp

Let z be the relative displacement of the mass with respect to support then

Z=x-y

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

12

BEATS When two harmonic motions with frequencies close to one another are added The resulting motion exhibits a phenomenon known as Beats A Beat Frequency is the result of two closely spaced frequencies going into and out of synchronization with one another Let us consider tow harmonic motion of same amplitude and slightly different frequencies X1 = X Cos ( ) tωωωω X2 = X Cos ( )( ) tδδδδωωωω + Where δ is a small quantity The addition of above two harmonics can be written as X = X1 + X2

X=

+

ttX2

cos2

cos2δδδδωωωω

δδδδ

The above equation shown graphically in Figure The resulting motion represents cosine wave with frequency ( ) 2

ππππωωωω + and with a varying amplitude 2 cos( ( ) 2δδδδ t

whenever the amplitude reaches a maximum it is called the beat The frequency δ at which amplitude builds up and dies down between o and 2 X is known as beat frequency Beat phenomenon is found in machines structures and electric power houses In machines the beat phenomenon occurs when the forcing frequency is close to the natural frequency of the system

+2X

-2X

ωωωωππππ2

ELEMENTS OF VIBRATION The elements of constitute vibrating systems are

1 Mass or Inertia element - m 2 Spring - k 3 Damper - c 4 Excitation F(t)

Mass or Inertia element The mass or inertia element is assumed to be rigid body During vibration velomass changes hence kinetic energy can be gained or loosed The force appthe mass from newton second law of motion can be written as F= ma Work dthe mass is stored in the form of kinetic energy given by KE= frac12 M V2

Combination of masses In practical cases for simple analysis we replace several masses by equivalent mass Case1 Translational masses connected by rigid bar Let the masses M1 M2 and M3 are attached to a rigid bar at locates 1 respectively as shown in the figure The equivalent mass meq be assumelocated at 1 is as shown in figure (b) Let the displacement of masses M1 M2 and M3 be x1 x2 x3 and simivelocities of respective masses be x1 x2 and x3 We can express the velomasses m2 and m3 in terms of m1

Elements of Vibration

Passive element

m C K

Active element

F(t)

Conservative element

( Mass and Spring)

Non conservative element

( Damper)

C k

m F(t)

Voig

13

city of lied on one on

a single

2 and 3 d to be

larly the cities of

t Model

Translationa Let a mass mmass momearrangement These two mamass Meq or

2

1

33

2

1

221

+

+=

ll

Mll

MMMeq

14

is the required answer

l and rotational masses coupled together

having a translational velocity x be coupled to another mass having nt of inertia Io with a rotational velocity θ as in rack and pinion

shown in the figure

sses can be combined to obtain either a single equivalent translational a single equivalent mass moment of inertia Jeq

15

Equivalent translational masses

Kinetic energy of the equivalent mass =

2

21

eqeq XM amp

Kinetic energy of the two masses =

+

20

2

21

21

θθθθampamp JXM

Meq= m

+ 2RJ

m o is the required answer

Also determine the equivalent rotational mass Jeq

Jeq= m [ ]oJmR +2 is the required answer Spring element Whenever there is a relative motion between the two ends of the spring a force is developed called spring force or restoring force The spring force is proportional to the amount of deformation x and then F α x or F = kx Where k is stiffness of the spring or spring constant or spring gradient The spring stiffness is equal to spring force per unit deromation

The spring stiffness k = mNxF

Workdone in deforming a spring is equal to the strain energy or potential energy Strain energy = potential energy = area of the triangle OAB Stiffness of beams Cantilever beam consider a cantilever beam with an end mass shown in the figure The mass of the beam is assumed to be negligible The static deflection of beam at free end is given by

Similarly derive the expression for Simply supported beam and fixed support beam

x

k

m

x

m

x

m

x

mNlIE

st 192

3

=δδδδ

mNEI

Wlst

48

3

=δδδδ

F

K=stiffness

F

16

17

Stiffness of slender bar subjected to longitudinal vibrations For a system executing the longitudinal vibrations as shown in the figure let us derive the expression for stiffness

Torsional Stiffness of bar It is the amount of torque required to cause a unit angular deformation in the element

Torsional stiffness = Kt = θθθθT

Combination of stiffness Determination of equivalent spring stiffness when the springs are arranged in series Consider two springs of stiffness K1 and K2 acted upon by the force F

The deflection of spring k1 is x1 = 1KF

The deflection of spring k1 is x2 = Let these two springs be replaced by an equivalent stiffness Keq upon which a force F acts and due to which its deflection is given by

x = eqKF

x= x1+x2

m

E A

l

60 cm 70 cm

Determination of equivalent spring stiffness when the springs are arranged in parallel Force acting on K1 spring = F1=k1x Force acting on K2 spring = F2 Force required for an equivalent spring keq to defined by x given by F= Keq x But F = F1 +F2 Tutorial problems on Equivalent stiffness of springs 1Determine the equivalent stiffness for the system shown in figure 2 Determine the equivalent stiffness for the system shown in figure

3 Determine the equivale

M

3

M

Nm 102 6x

18

nt stiffness for the system shown in figure

38 kg

Nm 101 6x Nm 102 6x

Nm 106x

2k

k

k

2k

3k

k

2k

k

19

60 cm 80 cm 50 cm

50 cm 60 cm

4Determine the equivalent stiffness for the system shown in figure

5Replace the following torsional stiffness by a single shaft having radius 4cm and find the length required for the equivalent shaft Assume the material of given system and equivalent system is same R1= 3cm R2= 5cm DAMPING Every vibration energy is gradually converted into heat or sound Hence the displacement during vibration gradually reduces The mechanism by which vibration energy is gradually converted into heat or sound is known as damping A damper is assumed to have either mass or elasticity Hence damping is modeled as one or more of the following types Viscous damping Coulomb or dry friction damping materials or solid or hysteric damping Viscous damping Viscous damping is most commonly used damping mechanism in vibration analysis When the mechanical system vibrates in a fluid medium such as air gas water or oil the resistance offered by the fluid to the moving body causes energy to be dissipated In this case the amount of dissipated energy depends on many factors such as size or shape of the vibrating body the viscosity of the fluid the frequency of vibration and velocity of fluid Resistance due to viscous damping is directly proportional to the velocity of vibration

dF V αααα

dF = xC amp

Where C= damping coefficient Fd = damping force Examples of Viscous damping

1) Fluid film between sliding surface 2) Fluid flow around a piston in a cylinder 3) Fluid flow through orifice 4) Fluid flow around a journal in a bearing

Rreq =4cm

leqn

stress

Strain

Coulomb damping or dry friction damping Here a damping force is constant in magnitude but opposite in direction to that of the motion of vibrating body It is caused by the friction between the surfaces that are dry or have insufficient lubrication Material or solid or hysteric damping

When the materials are deformed energy is absorbed and dissipated by the material The effect is due to friction between the internal planes which slip or slide as the deformation takes place When a body having the material damping is subjected to vibration the stress strain diagram shows the hysteresis loop as shown in the figure The area of the loop denotes the energy lost per unit volume of the body per cycle

20

21

ττττ ττττ2 ττττ3

ττττ

X(t)

FOURIER SERIES The simplest of periodic motion happens to be SHM It is simple to handle but the motion of many vibrating system is not harmonic (but periodic) Few examples are shown below

Forces acting on machines are generally periodic but this may not be harmonic for example the excitation force in a punching machine is periodic and it can be represented as shown in figure 13 Vibration analysis of system subjected to periodic but nonharmonic forces can be done with the help of Fourier series The problem becomes a multifrequency excitation problem The principle of linear superposition is applied and the total response is the sum of the response due to each of the individual frequency term

Any periodic motion can be expressed as an infinite sum of sines and cosines terms If x(t) is a periodic function with period t its Fourier representation is given by

X(t) = (((( )))) (((( )))) (((( ))))t sint cost cos2 111 ωωωωωωωωωωωω baaao ++++++++++++

= (((( )))) (((( ))))t sint cos2 1

ωωωωωωωω nnn

o bnaa

++++++++sumsumsumsuminfininfininfininfin

====

tππππωωωω

2==== = Fundamental frequency ndash (1)

where ao an bn are constants

ττττ ττττ2 ττττ3

ττττ

X(t)

22

t

X(t)

035

025

Determination of constants To find ao Integrate both sides of equation(1) over any interval ττττ All intergrals on the RHS of the equation are zero except the one containing ao

(((( ))))dttxao

o 2

2

intintintint====ωωωωππππ

ππππωωωω

= (((( ))))dttxo

2intintintintττττ

ττττ

To find an multiply equation 1 by cos (((( ))))tnωωωω and Integrate over any interval ττττ All intergrals

(((( )))) (((( )))) dttxao

n tncos 2

2

ωωωωππππωωωω ωωωω

ππππ

intintintint==== (((( )))) (((( )))) dttxo

tncos 2

ωωωωττττ

ττττ

intintintint====

To find bn multiply equation 1 by sin (((( ))))tnωωωω and Integrate over any interval ττττ All intergrals

(((( )))) (((( )))) dttxao

n tnsin 2

2

ωωωωππππωωωω ωωωω

ππππ

intintintint==== (((( )))) (((( )))) dttxo

tnsin 2

ωωωωττττ

ττττ

intintintint====

Find the Fourier series for the curve shown below

ττττ ττττ2 ττττ3

t

X(t)

E

ττττ ττττ2 ττττ3

t

X(t)

ππππ2

t

X(t) ππππ

1 Represent for the periodic motion shown in the figure

23

CHAPTER 2 UNDAMPED FREE VIBRATION LEARNING OBJECTIVES

] Introduction to undamped free vibration

] Terminologies used in undamped free vibration

] Three methods to solve the undamped free vibration problems

] Problems related to undamped free vibration problems

Free vibrations are oscillations about a systems equilibrium position that occur in the absence of an external excitation force If during vibrations there is no loss of energy it is known as undamped vibration The first step in solving a vibration problem is setting up the differential equation of motion Consider spring mass system which is assumed to move only along the vertical direction as shown below Let m be the mass of the block and k be the stiffness of the spring When block of mass m is attached to spring the deflection of spring will be ∆ known as static deflection In the static equilibrium position the free body diagram of forces acting on the mass is shown in Figure(b) Hence mg= kA Once the system is disturbed the system executes vibrations

Let at any instant of time t the mass is displaced from the equilibrium position x the different forces acting on the system are shown in figure (d) From Newtonrsquos second law of motion

sum = maF

Inertia force ( disturbing force) = restoring force

mgxkxm ++∆minus= )(ampamp

0)( =+ xkxm ampamp

or 0)( =+ xmk

xampamp

equation 2 is the differential equation of motion for spring mfigure Comparing equation (2) with the equation of SHM +xampampsince the vibrations of the above system are free( without the forces) we can write

sec radmk

n =ω

24

ass system shown in 0)(2 =xω

resistance of external

time period km

2 f1

n

πτ ==

from the equation(1) mg = k∆

∆=

q

mk

Difference between the translation ( rectilinear) and rotational system of vibration

Translatory Rotational

In the analysis thFORCES are co Linear stiffness K Problems 1A mass of 10kFind the natural 2 A spring massnatural frequencnatural frequenc

kt

m

sec radmk

n =ω

k

25

e disturbing and restoring nsidered

In the analysis In the analysis MASS Moment of Inertia (J) is considered

in Nm is considered

g when suspended from a spring causes a static deflection of 1cm frequency of system

system has a spring stiffness K Nm and a mass of m Kg It has a y of vibration 12 Hz An extra 2kg mass coupled to it then the y reduces by 2 Hz find K and m

secrad n =ω

26

3 A steel wire of 2mm diameter and 30m long It is fixed at the upper end and carries a mass of m kg at its free end Find m so that the frequency of longitudinal vibration is 4 Hz 4 A spring mass system has a natural period of 02 seconds What will be the new period if the spring constant is 1) increased by 50 2) decreased by 50 5 A spring mass system has a natural frequency of 10 Hz when the spring constant is reduced by 800 Nm the frequency is altered by 45 Find the mass and spring constant of the original system 6 Determine the natural frequency of system shown in fig is which shaft is supported in SHORT bearings 7 Determine the natural frequency of system shown in fig is which shaft is supported in LONG bearings where l is the length of bearing and E ndash youngrsquos modulus and I is moment of Inertia 7 Determine the natural frequency of system shown in fig is which shaft is supported in LONG bearings 8 A light cantilever beam of rectangular section( 5 cm deep x 25cm wide) has a mass fixed at its free end Find the ratio of frequency of free lateral vibration in vertical plane to that in horizontal 9 Determine the natural frequency of simple pendulum 10 Homogeneous square plate of size l and mass m is suspended from the mid point of one of its sides as shown in figure Find the frequency of vibration

l

l

27

11 A compound pendulum which is rigid body of mass m and it is pivoted at O The point of pivot is at distance d from the centre of gravity It is free to rotate about its axis Find the frequency of oscillation of such pendulum 12 A connecting rod shown in fig is supported at the wrist pin end It is displaced and allowed to oscillate The mass of rod is 5kg and centre of gravity is 20 cm from the pivot point O If the frequency of oscillation is 40 cyclesminute calculate the mass moment of inertia about its CG

13 A semi circular homogenous disc of radius r and mass m is pivoted freely about its centre as shown in figure Determine the natural frequency of oscillation 14A simply supported beam of square cross section 5mmx5mm and length 1m carrying a mass of 0575 kg at the middle is found to have natural frequency of 30 radsec Determine youngrsquos modulus of elasticity of beam 15 A spring mass system k1 and m have a natural frequency f1 Determine the value of k2 of another spring in terms of k1 which when placed in series with k1 lowers the natural frequency to 23 f1

COMPLETE SOLUTION OF SYSTEM EXECUTING SHM

The equation of motion of system executing SHM can be represented by

0)( =+ xkxm ampamp ----------------------------------------(1)

022

2=+

dtdx

dt

dx ω

The general solution of equation (1) can be expressed as

------------------------(2)

Where A and B are arbitrary constant which can be determined from the initial conditions of the system Two initial conditions are to be specified to evaluate these

constants x=x0 at t=0 and xamp = Vo at t=0 substituting in the equation (2) Energy method In a conservative system the total energy is constant The dwell as natural frequency can be determined by the princienergy For free vibration of undamped system at any instant and partly potential The kinetic energy T is stored in the masswhere as the potential energy U is stored in the form of sdeformation or work done in a force field such as gravity The total energy being constant T+U = constant Its rate of cha

t)(sin Bt)cos( AX ω+ω=

t)(sin V

t)cos( xx o0 ω

ω+ω=

Is the required complete

solution

28

ifferential equation as ple of conservation of of time is partly kinetic by virtue of its velocity train energy in elastic

nge

Is given by [ ] 0UTdtd

=+

From this we get a differential equation of motion as well as natural frequency of the system Determine the natural frequency of spring mass system using energy method

Determine the natural frequency of system shown in figure Determine the natural frequency of the system shown in figure Is there any limitation on the value of K Discuss Determine the natural frequency of s

m

l k

m

a

θ

m

a

θ

l

k

29

k

k

ystem shown below Neglect the mass of ball

l

m

30

A string shown in figure is under tension T which can be assumed to remain constant for small displacements Find the natural frequency of vertical vibrations of spring An acrobat 120kg walks on a tight rope as shown in figure The frequency of vibration in the given position is vertical direction is 30 rads Find the tension in the rope A manometer has a uniform bore of cross section area A If the column of liquid of length L and Density ρ is set into motion as shown in figure Find the frequency of the resulting oscillation

m

l

a

T T

36m 8m

31

k

R

k

m

r

m

Find the expression for natural frequency of system shown in the figure Neglect the mass of the cantilever beam Study the special case i) k=Infinity ii) I = infinity Determine the expression for the natural frequency of a system shown in figure The two discs are keyed to a common shaft and have a combined mass moment of inertia about centre of oscillation O It is assumed that the card attached to mass m does not stretch and is always under tension Determine the natural frequency of a system shown

m

l

32

Determine the expression for the natural frequency of the system shown in figure Assume that the wires connecting the masses do not stretch and are always in tension

Determine the natural frequency of spring mass system taking the MASS OF SPRING (ms ) into account

M3 M1

M2

k2 k1

m

k

33

RAYLEIGHrsquoS METHOD

This is the extension of energy method Here we can determine the natural frequency of a conservative system without finding the equation of motion The natural frequency is a function of the rate of change of Kinetic energy and potential energy of the system From the principle of conservation of energy we have T+U = Constant T= KE and U =PE We can write the above equation as T1+U1 = --------------------------(1) Where 1 and 2 represents tow instances of time Let 1 be the time when the mass is passing through static equilibrium position Let 2 be the time corresponding to the mass displacement of the mass At this instant the velocity f the mass is zero and hence Substituting in equation (1) we get ( )Max = ( )Max Above equation leads directly to natural frequency Determine the natural frequency of spring mass system using Rayleighrsquos method m

34

r

B

K K

Home work

Determine the natural frequency of pendulum using Rayleighrsquos method

A Cylinder of mass m and moment of inertia Jo is to role without slipping but is sustained by spring K as shown in figure Determine the natural frequency of oscillation

Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping

a

Determine the natural frequency of system shown where in cylindrical disc rolls over the inextensible spring A cylinder of radius r rolls without slipping on cylindrical the equation of motion and hence determine the naoscillations about the lowest point shown Repeat the above problem ndash Instead of cylinder assume swithout slipping on concave surface as shown above

r

K

35

portion of radius R Derive tural frequency of small

phere of radius with r rolls

The torsional pendulum has a natural frequency of 5 Hz What length of steel wire of diameter 2 mm should be used for pendulum The inertia of mass fixed at the free end is 00098 kgm2 Take g= 083x1011 Nm2 Determine the torsional natural frequency of the system shown in figure Neglect the mass moment of inertia of the shaft ( Figure not proportionate) Determine the natural frequency of simple pendulum considering the mass of the rod into account

Using the energy method find tThe chord may be assumed ine

Kt1

Kt2

20cm

06m 08m 04m

20cm 20cm 20cm

m

M

he natural frxtensible in

k

M

36

equency of the system shown in figure the spring mass pulley system

A rod of length l and depth H having a mass m rests on a semicylidrical surface os radius r It is tipped slightly and let go Find the frequency of oscillation

A thin bar of l having a mass m rests on a semicylidrical surface radius r It is tipped slightly and let go Find the frequency of oscillation

l

h

R

l

37

38

CHAPTER 3

DAMPED FREE VIBRATION OF SINGLE DEGREE FREEDOM SYSTEM LEARNING OBJECTIVES

] Introduction to damped free vibration

] Terminologies used in damped free vibration

] Three different types of damping

] Concept of critical damping and its importance

] Study of response of viscous damped system for cases of under damping

critical and over damping

] Concept of logarithmic decrement

] Problems related to damped free vibration problems

Viscously damped free vibration of single degree freedom sytem Consider a single degree freedom system with viscous damping as shown in figure For a viscous damped system the damping force is

where c is a constant of proportionality It is called damping coefficient or coefficient of viscous damping

Symbolically it is designated by a dashpot as shown in Fig 3

x xampx

39

At any instant of time let the mass be displaced by x from the equilibrium position Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

Okxxcxm =minusminus= ampampamp

Okxxcxm =++ ampampamp ----------------------------------------------------------(1)

or

0kxdtdx

Cdt

xdM

2

2

=++

Assuming the solution of equation as

The first term in the above equation m2Ct

eminus

is an exponentially decaying function of time The behavior of the terms in the parenthesis however depends on whether numerical value within the radicals is positive zero or negative

CASE 1 if 0mk

m2c

2

=minus

40

CASE 2 if 0mk

m2c

2

=gt

CASE 3 if 0mk

m2c

2

=lt

41

Critically damped coefficient Critically damped coefficient is defined as the value of the damping coeffient for which radical in equation

mk

m2c

m2c

S2

21 minus

plusmn

= becomes zero

Damping Ratio ( Damping factor) ( ξ ) It is defined as the ratio of damping coefficient to the critical damping coefficient Note The damping coefficient c is defined as the actual amount of damping while the damping coefficient is a constant depending upon the mass and stiffness of the system The roots of characteristics equation can be now written as Note if c=cc or ξ =1 then system is critically damped if cgt cc or ξ gt1 then system is over damped I if clt cc or ξ lt1 then system is under damped

42

Take home quiz Derive General solution of different damped system

1 Critically damped system 2 Under damped system 3 Over damped system

Logarithmic Decrement It is used to determine the amount of damping present in system This measure The rate of decay of free vibration The larger the damping the greater will be rate of decay Logarithmic decrement is defined as the natural logarithm of ratio of any two successive amplitudes

=δ

2

1

xx

ln

Expression for logarithmic decrement (δ ) Consider a damped free vibration which is expressed by

( ) ( ) φ+ξminusω= ξω t 1sinXex 2n

tn

x is the displacement at any instant of time X and φ are arbitrary constants

21

2

ξminus

πξ=δ

43

Tutorial problems 1 Derive an expression for damping ratio in terms of logarithmic decrement

2 Show that logarithmic decrement

=δ

n

o

xx

lnn1

where xo is the initial amplitude xn

is amplitude after n cycles 3 A mass of 1 kg is to be supported on a spring having stiffness K= 9800 Nm The damping constant c= 59 Nsm Determine the natural frequency of system Find also logarithmic decrement and amplitude after 3 cycles if initial displacement is 03 cm 4 A vibrating system is defined by the following parameters M = 3kg k= 100 Nm damping coefficient = 3 Nsm determine Damping factor natural frequency damped vibration logarithmic decrement ratio of consecutive amplitude no of cycles after which the original amplitude is reduced to 20 5A single degree damped vibrating system a suspended mass of 18kg makes 15 oscillations in 003 seconds The amplitude decreases to 025 of the initial value after 5 oscillations Determine the stiffness of spring Logarithmic decrement damping factor and damping coefficient 6 A damped vibration record of a spring mass system shows the following data Amplitude at the end of 2nd cycle is 9mm amplitude at eh end of 3rd cycle is 6mm amplitude at the end of 4th cycle is 4mm stiffness of spring =8000 Nm weight =20N Find the logarithmic decrement damping force at unit velocity periodic time of vibration 7 For the system shown in figure the characteristics of dashpot is such that the constant force of 49 N is applied to the piston it velocity is found to be constant at 012 ms Determine the value of C would you expect the complete system to be periodic or aperiodic 8 For a torsional system shown in the figuref1 is natural frequency in air and f2 is the natural frequency when immersed in oil Determine the damping coefficient

Viscous fluid

Disc of mass moment of Inertia J

2a

C

m M

b

a a

k

9 A thin plate of area A and weigh w is fixed to the end of the spring and allow to oscillate in a viscous fluid as shown in figure It f1 is the frequency of oscillation in air and f2 in liquid Show that

22

21 ff

gA2

minusπω

=micro where the damping force on the plate

AV2Fd micro= V being the velocity of vibration 10 A torsional pendulum has a natural frequency of 175 cyclessecond vibrating in vacuum The mass moment of inertia is 2 kgm2 It is immersed in oil and it is observed that the natural frequency is 142 cyclessecond Find the damping torque If the disc is replaced by 34 deg When in oil find the displacement at the end of the first complete cycle 11 An automobile can be modeled as a mass placed on 4 shock absorbers each consisting of a spring and a damper Such that each spring is equally loaded Determine the stiffness damping constant of each shock absorber So that natural frequency is 2Hz and system is critically damped The mass of vehicle is 200kg 12 Write the differential equation of motion for the system shown in the figure below Determine the natural frequency of the damped oscillation and critical damping coefficient

Determine the differential equation of motion and find the critical cdamped natural frequency for system shown in figure

a

C

m M

M =200kg

44

200kg kgkg

oefficient and

45

Derive the equation of motion and find the critical damping coefficient and damped natural frequency for the system shown below Write the differential equation of motion and determine the expression for critical damping coefficient and damped natural frequency of system shown in figure

Set up a differential equation of motion for system shown in figure and determine the following Undamped natural frequencies critical damping coefficient damping ratio damped natural frequency

b

C

kM

a

m

46

A mass of 300N is resting on two springs of stiffness 3000 Nm each and a dashpot of damping coefficient 150 Nsm as shown in figure If the initial velocity of 10cms is given to mass at its equilibrium position what will be the displacement from the equilibrium position at the end of first second Derive the formula used A gun barrel of mass 600kg has a recoil spring of stiffness 294 knm if the barrel recoils 13m on firing determine i) The initial recoil velocity of barrel ii) The critical damping coefficient of dashpot which is engaged at the end of recoil stroke Iii) Time required for the barrel to return at the end of recoil stroke Iv) time required for the barrel to return to a position 5cm from the initial position

W= 3000 N

FIRING

CHAPTER 4

FORCED VIBRATION LEARNING OBJECTIVES

] Introduction to forced free vibration

] Terminologies used in forced vibration

] Concept of magnification factor

] Study of reciprocating and rotating unbalance

] Concept of base excitation and support excitation

] Concept of Transmissibility ratio

] Problems related to forced vibration

The vibration that takes place under the excitation of external forces are called

forced vibration When a system is subjected to harmonic excitation it is forced to vibrate at the same frequency as that of excitation Most of the mechanical systems are prone to steady state forced vibration which cause fatigue failure The vibration problem also becomes very important when any of excitation frequencies coincide with one of natural frequencies of a system Under this condition resonance occurs during which large amplitude of vibrations are observed To avoid resonance external frequency (operating speed) may be changed or properties of the system may be changed to alter the natural frequency In some cases sufficient amount of damping may be provided to avoid large amplitude during resonance Thus the problem of force vibration is very important in mechanical design

FORCED VIBRATIONS OF SINGLE DOF SYSTEMS UNDER HARMONIC EXCITATION

` [ ] tsin FF o ω=

x

sin FF o=

Fo

xampamp

[ ] tω

xamp

47

48

Consider a spring mass system having a viscous damping excited by a harmonic force [ ] tsin FF o ω= as shown in figure The various forces acting on the system is shown in figure Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

O)tsin(Fkxxcxm o =ω+minusminus= ampampamp

)tsin(Fkxxcxm o ω=++ ampampamp ----------------------------------------------------------(1)

or

)tsin(Fkxdtdx

Cdt

xdM o2

2

ω=++

Equation (1) represents the differential equation of motion of the system The solution to this equation consists of two parts namely complementary function and a particular integral

The first of solution is also termed as transient solution of homogeneous solution and can be obtained from the homogeneous differential equation 0kxxcxm =++ ampampamp

The second part of solution particular integral is also termed as steady state solution ad is of form )tsin(Xx φminusω= Where X= steady state amplitude φ is phase difference between external force and displacement

)tsin(Xx φminusω=

The vector representation of the shown

49

The force polygon for the above vector diagram can be written as follows

( )φminusωt O

KX

xm 2ω

Fo

xCω

tω

φ

50

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

is the governing equation

Zero frequency deflection (Xst) It is deflection observed at zero frequency ( when the external frequency 0=ω

( ) kF

001

kF

x o

22

o

st =

+minus=

Magnification factor (M) It is defined as the ratio of steady state amplitude to zero frequency deflection xst

M = arX

X=

A plot of magnification factor versus frequency ratio

damping factor ( )ξ is called frequency plot of a harmon All the frequency terms starts from magnification factor of damping ratio ( )ξ The amplitude of vibration at any on the value of damping The amplitude will be higher for higher damping The amplitude of vibration at reso

amplitude tends to zero as

ωω

n tends to infinity

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

Freq

ζ =

ζ = 1

Magnification Factor M = XXst

kFo

51

ωω

nfor varying values of

ically excite vibrating system

M=1 what ever may be value particular frequency depends

for lesser damping and lesser nance will be very high The

uency Ratio r = (ωωn)

ζ = 0 0

ζ = 025 ζ = 0375 ζ = 0 5

ζ = 0707

ζ = 2

52

Plot of phase angle φ versus frequency ratio

ωω

n is shown in the

figure The phase angle φ is zero for all the values of damping ratio

at

ωω

n=0 The phase

angle φ is 90 for all the values of damping ratio

at

ωω

n=1 ie at

resonance The phase angle between o to 90 deg for all the values of

ωω

n= 0 to 1 and

between 90 deg to 180 deg for all the values of

ωω

ngt1

Class work Derive an expression for frequency at peak amplitude

2np 21 ξminusω=ω=ω is the required equation

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

53

Derive expression amplitude at resonance ( In dimensional form) In dimensional form We know that steady state amplitude

( ) ( )

ω+ωminus=

222

o

Cmk

Fx Equation in dimensional form

We know that at resonance mk

n =ω=ω

( )

ω+

minus

=2

n

2

oR

Cmk

mk

Fx

( )

ω=

2n

oR

C

Fx

( )

ω=

n

oR C

Fx

Class work Derive an expression for amplitude at resonance ( in non dimensional form)

Tutorial Problems on Forced Vibration 1A weight of 60N is suspended by a spring stiffness 12 kNm is forced to vibrate by a harmonic force of 10N assuming viscous damping of 0086 kN-sm Find i) resonance frequency ii) amplitude at resonance iii) phase angle at resonance iv) frequency corresponding to peak amplitude 2 A periodic torque having a maximum value of 05888Nm at a frequency of 4 rads is impressed upon a flywheel suspended from a wire The wheel has mass moment of inertia 012 kgm2 and the wire has stiffness 1176 Nmrad A viscous dashpot applies damping couple of 0784 Nm an angular velocity of 2 rads calculate i)maximum angular displacement from rest position ii) maximum couple applied to dashpot Iii) angle by which the angular displacement lags the torque A mass of 1019 kg is suspended from a spring stiffness 20 Nmm viscous damping causes the amplitude to decrease to 110th of initial value in 4 complete cycle If a periodic force of 20 cos(50 t) is applied to the mass Find the amplitude of forced vibration What would be the amplitude of oscillation if the applied force coincides with the natural period

m=1019 kg

[ ] t50cos 30F =

54

55

ROTATING UNBALANCE Unbalance in rotating machinery is a common source of vibration excitation Consider a machine of mass m having mounted on a foundation of stiffness k and damping coefficient C as shown in figure The unbalance is represented by an eccentric rotating mass m with eccentricity (e) and having a angular velocity (ω) let x be the displacement of non rotating mass from the static equilibrium position The displacement of rotating mass m from the static equilibrium is ( ) tsin( ex ω+ Applying Newtonrsquos law of motion we have

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

56

RECIPROCATING UNBALANCE Unbalance in reciprocating machinery is a common source of vibration excitation Consider a machine of mass M having mounted on a foundation of stiffness K and damping coefficient C as shown in figure The unbalance is represented by a reciprocating mass (m) having crank radus (e) and connecting rod of length (l) Let ωωωω be angular velocity of the crank Let x be the displacement of non reciprocating mass from the static equilibrium position at any instant of time The displacement of reciprocating mass m from the static equilibrium position is

)2sin(sin +++ tle

tex ωωωωωωωω

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

57

Note The equations obtained in reciprocating unbalance is same as the rotating unbalance and having the following differences Rotating Unbalance Reciprocating Unbalance Here mass of the rotating element is considered

Here mass of reciprocating element is considered

Here the eccentricity (e) which is the distance between the centre of rotation and CG

Here the eccentricity e is the crank radius and crank radius = stroke length2

The frequency plot of non dimensional factor( ( )me

MX versus frequency ratio

ωω

n for various values of ( )ξξξξ is shown in figure This frequency plot is same for

the rotating and reciprocating unbalance with the following points are noted from the diagram

1 Damping ratio plays very important role during resonance in controlling the amplitude

2 When

ωω

n=0 then ( )me

MX =0 for all the values of damping ratio ( )ξξξξ

3 When

ωω

nis higher than ( )me

MX tends to unity

4 At very high speeds ie at very high frequency ratio

ωω

n there is no effect

of damping The frequency plot of phase angle versus the frequency ratio is given as shown in figure The explanation is same as in forced harmonic vibration

58

VIBRATION ISOLATION

Vibratory forces and motions generated by machine and other caused are unavoidable Howeever their effect on the other dynamical systems can be minimized by proper isolator design An isolation system is either to protect o delicate machine from excessive vibration transmitted to it from its supporting structure or to prevent vibratory forces and motions generated by machines from being transmitted to its surroundings The effectiveness of isolation may be measured in terms of forces and motion transmitted The first type is known as force isolation and the second type is known as motion isolation

Force isolation (Force Transmissibility) When vibratory forces generated by machine are isolated the effectiveness of forced isolation can be made by defining the term transmissibility The term transmissibility is defined as the ratio of force transmitted to the foundation to that of the external impressed force

Force Transmissibility = o

tr

FF

force Impressed Externalfoundation toted transmitForce

=

Consider viscous damped spring mass system as shown in figure Upon this system an external harmonic force ) t (Sin Fo ω is applied The difference forces acting on system are inertia forces damping force spring force and external force All these forces are represented on force polygon shown below Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these two forces

( ) ( )22 xckx ω+=

59

This is the expression for force transmissibility in dimensional form To represent same in the non dimensional form divide numerator and denominator by k

2

n

22

n

2

n

o

tr

21

21

FF

TR

ωω

ξ+

ωω

minus

ωω

ξ+

==

PHASE DIFFERNECE BETWEEN EXTERNAL FORCE AND DISPLACEMENT Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these

ωω

minus

ωω

ξ=

ωminusω

=φ minusminus2

n

n1

2

1

1

2tan

mKC

tan

The phase difference between force transmitted to the foundation Ftr to the displacement

kc

KxxC

tanω

=ω

=α or

ωω

ξ=

ω=α minusminus

n

11 2tanKx

xCtan

The phase difference between external force Fo and force transmitted to foundationis β

60

β= αminusφ =

ωω

minus

ωω

ξminus

2

n

n1

1

2tan -

ωω

ξminus

n

1 2tan

Displacement Transmissibility It is the ratio of amplitude of steady state vibration of machine to that of the base motion Forced vibration due to support motion Consider a spring mass damper system subjected to harmonic excitation )tsin(Yy ω= at the base as shown in figure Let x be the absolute displacement of the mass at any instant of time t The different forces acting on system are shown in free body diagram Applying Newtonrsquos Law of motion Ma = sumF

)yx(k)yx(cxm minusminusminusminus= ampampampamp

0)yx(k)yx(cxm =minus+minus+ ampampampamp

61

Frequency plot of transmissibility versus frequency ratio

62

The plot of transmissibility versus frequency ratio is as shown in figure The following points can be noted from it

i) The value of transmissibility is unity at frequency ratio

ωω

n=0 and 2

ii) For an undamped system ( )ξ =0 transmissibility tends to infin at resonance iii) The value of transmissibility is less than unity for the value of frequency ratio

greater than 2

iv) For frequency ratio

ωω

nlt 2 smaller damping ratio lead to larger values of

transmissibility on the other hand for frequency ratio

ωω

ngt 2 smaller damping

ratio leads to smaller value of transmissibility

The frequency plot of transmissibility (TR) versus frequency ratio

ωω

n can be

broadly divided into two zones

1

ωω

n= 0 to 2

2

ωω

n= 2 to infin

CASE 1

When

ωω

n= 0 to 2

In this zone the value of transmissibility is always greater than 1 for all values of damping ratio In this zone the transmissibility decreases with increase in damping Hence we can conclude that in this zone damping is useful in isolating vibration forces being transmitted to the support CASE 2

When

ωω

n= 2 to infin

In this zone the value of transmissibility is less than 1 for all values of damping ratio In this zone the transmissibility increases with increase in damping Hence we can conclude that in this zone damping will adversely affect in isolating vibratory forces

63

Tutorial problems on Forced Vibration 1 Machine of mass 75kg is mounted on spring of stiffness 12000 Nm with an assumed damping factor ξ =02 piston within the machine of mass 2 kg has a reciprocating motion with a stroke of 75 cm and speed of 15 cyclessecond Assuming the motion of piston to be harmonic determine amplitude of machine ( X) Transmissibility and force transmitted to foundation phase angle of transmitted force wrt to exciting force 2 Motor weighs 600N is mounted on a simple beam that has a spring rate of 400 Ncm The motor armature base 160 N and has an eccentricity of 0005cm What will be the amplitude of motion when it runs at 1760 RPM Neglect damping the deflection of armature shaft and weight of the beam 3 A Heavy machine of mass 300 kg is supported on resilient foundation the static deflection of foundation due to weight of machine is found to be 75mm It is observed that the machine vibrates with an amplitude of 10mm when the base of the foundation is subjected to harmonic oscillation at the undampeds natural frequency of system with an amplitude of 25mm Find the damping constant of foundation Ii) dynamic force amplitude on the base iii) amplitude of the displacement of machine relative to base 4 An air compressor of mass 450 kg operates at a constant speed of 1750 RPM The reciprocating masses are 10 kg the crank radius is 100mm If the damper for mounting introduces a damping factor 015 i) specify the springs of mounting such that only 20 of the unbalanced force is transmitted to the foundation ii) determine the magnitude of the transmitted force bull Repeat the above problem by assuming that there is no damping in system

5 A body of a car having mass 1500 kg and is mounted on 4 equal springs which deflect through 225 cm under the weight of the body A total damping forces the 4 shock absorbers is 46 N at a velocity of 1cmsec The car is placed with all the four wheels on a test platform which is move up and down at resonant speed With an amplitude of 25cm Find the amplitude of the car body on its springs assuming its CG to be in the centre of wheel base 6A 75 kg machine which is mounted on springs of stiffness 116x105 Nm with the damping of 20=ξ A 2kg piston within the machine as reciprocating motion with a stroke of 008m and speed of 3000 RPM Assuming the motion of piston to be harmonic determine the amplitude of vibration of machine

7 The spring of an automobile are compressed 10cm under its own weight Find the critical speed when a trailer is traveling over road with a people approximaterly by a sine wave of amplitude 75 cm and wave length 15m What will be the amplitude of vibration at 64kmhour 8 An eccentric mass excites us used to determine vibratory characteristics of a structure of mass 25kg as given in figure At a speed of 1000 RPM the eccentric mass to beat top position at the instant the structure was moved upwards through its static equilibrium position and corresponding amplitude was 20mm The eccentric mass is 05kg at an eccentricity of 10mm Determine i) Undamped natural frequency of the system ii) Damping ratio iii) Steady state amplitude and phase angle at 1500 RPM iv) Force transmitted to the ground at 1200 RPM

M=25kg

C

K

Eccentric mass

64

65

CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Support motion Consider a spring mass damper system as shown in figure Let it excited by motion

of support let y be the harmonic displacement of support From Newtonrsquos second law of motion

)yx(k)yx(cxm minusminusminusminus= ampampampamp

)1(0)yx(k)yx(cxm minusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminus=minus+minus+ ampampampamp

Let z be the relative displacement of the mass with respect to support then

Z=x-y

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

ELEMENTS OF VIBRATION The elements of constitute vibrating systems are

1 Mass or Inertia element - m 2 Spring - k 3 Damper - c 4 Excitation F(t)

Mass or Inertia element The mass or inertia element is assumed to be rigid body During vibration velomass changes hence kinetic energy can be gained or loosed The force appthe mass from newton second law of motion can be written as F= ma Work dthe mass is stored in the form of kinetic energy given by KE= frac12 M V2

Combination of masses In practical cases for simple analysis we replace several masses by equivalent mass Case1 Translational masses connected by rigid bar Let the masses M1 M2 and M3 are attached to a rigid bar at locates 1 respectively as shown in the figure The equivalent mass meq be assumelocated at 1 is as shown in figure (b) Let the displacement of masses M1 M2 and M3 be x1 x2 x3 and simivelocities of respective masses be x1 x2 and x3 We can express the velomasses m2 and m3 in terms of m1

Elements of Vibration

Passive element

m C K

Active element

F(t)

Conservative element

( Mass and Spring)

Non conservative element

( Damper)

C k

m F(t)

Voig

13

city of lied on one on

a single

2 and 3 d to be

larly the cities of

t Model

Translationa Let a mass mmass momearrangement These two mamass Meq or

2

1

33

2

1

221

+

+=

ll

Mll

MMMeq

14

is the required answer

l and rotational masses coupled together

having a translational velocity x be coupled to another mass having nt of inertia Io with a rotational velocity θ as in rack and pinion

shown in the figure

sses can be combined to obtain either a single equivalent translational a single equivalent mass moment of inertia Jeq

15

Equivalent translational masses

Kinetic energy of the equivalent mass =

2

21

eqeq XM amp

Kinetic energy of the two masses =

+

20

2

21

21

θθθθampamp JXM

Meq= m

+ 2RJ

m o is the required answer

Also determine the equivalent rotational mass Jeq

Jeq= m [ ]oJmR +2 is the required answer Spring element Whenever there is a relative motion between the two ends of the spring a force is developed called spring force or restoring force The spring force is proportional to the amount of deformation x and then F α x or F = kx Where k is stiffness of the spring or spring constant or spring gradient The spring stiffness is equal to spring force per unit deromation

The spring stiffness k = mNxF

Workdone in deforming a spring is equal to the strain energy or potential energy Strain energy = potential energy = area of the triangle OAB Stiffness of beams Cantilever beam consider a cantilever beam with an end mass shown in the figure The mass of the beam is assumed to be negligible The static deflection of beam at free end is given by

Similarly derive the expression for Simply supported beam and fixed support beam

x

k

m

x

m

x

m

x

mNlIE

st 192

3

=δδδδ

mNEI

Wlst

48

3

=δδδδ

F

K=stiffness

F

16

17

Stiffness of slender bar subjected to longitudinal vibrations For a system executing the longitudinal vibrations as shown in the figure let us derive the expression for stiffness

Torsional Stiffness of bar It is the amount of torque required to cause a unit angular deformation in the element

Torsional stiffness = Kt = θθθθT

Combination of stiffness Determination of equivalent spring stiffness when the springs are arranged in series Consider two springs of stiffness K1 and K2 acted upon by the force F

The deflection of spring k1 is x1 = 1KF

The deflection of spring k1 is x2 = Let these two springs be replaced by an equivalent stiffness Keq upon which a force F acts and due to which its deflection is given by

x = eqKF

x= x1+x2

m

E A

l

60 cm 70 cm

Determination of equivalent spring stiffness when the springs are arranged in parallel Force acting on K1 spring = F1=k1x Force acting on K2 spring = F2 Force required for an equivalent spring keq to defined by x given by F= Keq x But F = F1 +F2 Tutorial problems on Equivalent stiffness of springs 1Determine the equivalent stiffness for the system shown in figure 2 Determine the equivalent stiffness for the system shown in figure

3 Determine the equivale

M

3

M

Nm 102 6x

18

nt stiffness for the system shown in figure

38 kg

Nm 101 6x Nm 102 6x

Nm 106x

2k

k

k

2k

3k

k

2k

k

19

60 cm 80 cm 50 cm

50 cm 60 cm

4Determine the equivalent stiffness for the system shown in figure

5Replace the following torsional stiffness by a single shaft having radius 4cm and find the length required for the equivalent shaft Assume the material of given system and equivalent system is same R1= 3cm R2= 5cm DAMPING Every vibration energy is gradually converted into heat or sound Hence the displacement during vibration gradually reduces The mechanism by which vibration energy is gradually converted into heat or sound is known as damping A damper is assumed to have either mass or elasticity Hence damping is modeled as one or more of the following types Viscous damping Coulomb or dry friction damping materials or solid or hysteric damping Viscous damping Viscous damping is most commonly used damping mechanism in vibration analysis When the mechanical system vibrates in a fluid medium such as air gas water or oil the resistance offered by the fluid to the moving body causes energy to be dissipated In this case the amount of dissipated energy depends on many factors such as size or shape of the vibrating body the viscosity of the fluid the frequency of vibration and velocity of fluid Resistance due to viscous damping is directly proportional to the velocity of vibration

dF V αααα

dF = xC amp

Where C= damping coefficient Fd = damping force Examples of Viscous damping

1) Fluid film between sliding surface 2) Fluid flow around a piston in a cylinder 3) Fluid flow through orifice 4) Fluid flow around a journal in a bearing

Rreq =4cm

leqn

stress

Strain

Coulomb damping or dry friction damping Here a damping force is constant in magnitude but opposite in direction to that of the motion of vibrating body It is caused by the friction between the surfaces that are dry or have insufficient lubrication Material or solid or hysteric damping

When the materials are deformed energy is absorbed and dissipated by the material The effect is due to friction between the internal planes which slip or slide as the deformation takes place When a body having the material damping is subjected to vibration the stress strain diagram shows the hysteresis loop as shown in the figure The area of the loop denotes the energy lost per unit volume of the body per cycle

20

21

ττττ ττττ2 ττττ3

ττττ

X(t)

FOURIER SERIES The simplest of periodic motion happens to be SHM It is simple to handle but the motion of many vibrating system is not harmonic (but periodic) Few examples are shown below

Forces acting on machines are generally periodic but this may not be harmonic for example the excitation force in a punching machine is periodic and it can be represented as shown in figure 13 Vibration analysis of system subjected to periodic but nonharmonic forces can be done with the help of Fourier series The problem becomes a multifrequency excitation problem The principle of linear superposition is applied and the total response is the sum of the response due to each of the individual frequency term

Any periodic motion can be expressed as an infinite sum of sines and cosines terms If x(t) is a periodic function with period t its Fourier representation is given by

X(t) = (((( )))) (((( )))) (((( ))))t sint cost cos2 111 ωωωωωωωωωωωω baaao ++++++++++++

= (((( )))) (((( ))))t sint cos2 1

ωωωωωωωω nnn

o bnaa

++++++++sumsumsumsuminfininfininfininfin

====

tππππωωωω

2==== = Fundamental frequency ndash (1)

where ao an bn are constants

ττττ ττττ2 ττττ3

ττττ

X(t)

22

t

X(t)

035

025

Determination of constants To find ao Integrate both sides of equation(1) over any interval ττττ All intergrals on the RHS of the equation are zero except the one containing ao

(((( ))))dttxao

o 2

2

intintintint====ωωωωππππ

ππππωωωω

= (((( ))))dttxo

2intintintintττττ

ττττ

To find an multiply equation 1 by cos (((( ))))tnωωωω and Integrate over any interval ττττ All intergrals

(((( )))) (((( )))) dttxao

n tncos 2

2

ωωωωππππωωωω ωωωω

ππππ

intintintint==== (((( )))) (((( )))) dttxo

tncos 2

ωωωωττττ

ττττ

intintintint====

To find bn multiply equation 1 by sin (((( ))))tnωωωω and Integrate over any interval ττττ All intergrals

(((( )))) (((( )))) dttxao

n tnsin 2

2

ωωωωππππωωωω ωωωω

ππππ

intintintint==== (((( )))) (((( )))) dttxo

tnsin 2

ωωωωττττ

ττττ

intintintint====

Find the Fourier series for the curve shown below

ττττ ττττ2 ττττ3

t

X(t)

E

ττττ ττττ2 ττττ3

t

X(t)

ππππ2

t

X(t) ππππ

1 Represent for the periodic motion shown in the figure

23

CHAPTER 2 UNDAMPED FREE VIBRATION LEARNING OBJECTIVES

] Introduction to undamped free vibration

] Terminologies used in undamped free vibration

] Three methods to solve the undamped free vibration problems

] Problems related to undamped free vibration problems

Free vibrations are oscillations about a systems equilibrium position that occur in the absence of an external excitation force If during vibrations there is no loss of energy it is known as undamped vibration The first step in solving a vibration problem is setting up the differential equation of motion Consider spring mass system which is assumed to move only along the vertical direction as shown below Let m be the mass of the block and k be the stiffness of the spring When block of mass m is attached to spring the deflection of spring will be ∆ known as static deflection In the static equilibrium position the free body diagram of forces acting on the mass is shown in Figure(b) Hence mg= kA Once the system is disturbed the system executes vibrations

Let at any instant of time t the mass is displaced from the equilibrium position x the different forces acting on the system are shown in figure (d) From Newtonrsquos second law of motion

sum = maF

Inertia force ( disturbing force) = restoring force

mgxkxm ++∆minus= )(ampamp

0)( =+ xkxm ampamp

or 0)( =+ xmk

xampamp

equation 2 is the differential equation of motion for spring mfigure Comparing equation (2) with the equation of SHM +xampampsince the vibrations of the above system are free( without the forces) we can write

sec radmk

n =ω

24

ass system shown in 0)(2 =xω

resistance of external

time period km

2 f1

n

πτ ==

from the equation(1) mg = k∆

∆=

q

mk

Difference between the translation ( rectilinear) and rotational system of vibration

Translatory Rotational

In the analysis thFORCES are co Linear stiffness K Problems 1A mass of 10kFind the natural 2 A spring massnatural frequencnatural frequenc

kt

m

sec radmk

n =ω

k

25

e disturbing and restoring nsidered

In the analysis In the analysis MASS Moment of Inertia (J) is considered

in Nm is considered

g when suspended from a spring causes a static deflection of 1cm frequency of system

system has a spring stiffness K Nm and a mass of m Kg It has a y of vibration 12 Hz An extra 2kg mass coupled to it then the y reduces by 2 Hz find K and m

secrad n =ω

26

3 A steel wire of 2mm diameter and 30m long It is fixed at the upper end and carries a mass of m kg at its free end Find m so that the frequency of longitudinal vibration is 4 Hz 4 A spring mass system has a natural period of 02 seconds What will be the new period if the spring constant is 1) increased by 50 2) decreased by 50 5 A spring mass system has a natural frequency of 10 Hz when the spring constant is reduced by 800 Nm the frequency is altered by 45 Find the mass and spring constant of the original system 6 Determine the natural frequency of system shown in fig is which shaft is supported in SHORT bearings 7 Determine the natural frequency of system shown in fig is which shaft is supported in LONG bearings where l is the length of bearing and E ndash youngrsquos modulus and I is moment of Inertia 7 Determine the natural frequency of system shown in fig is which shaft is supported in LONG bearings 8 A light cantilever beam of rectangular section( 5 cm deep x 25cm wide) has a mass fixed at its free end Find the ratio of frequency of free lateral vibration in vertical plane to that in horizontal 9 Determine the natural frequency of simple pendulum 10 Homogeneous square plate of size l and mass m is suspended from the mid point of one of its sides as shown in figure Find the frequency of vibration

l

l

27

11 A compound pendulum which is rigid body of mass m and it is pivoted at O The point of pivot is at distance d from the centre of gravity It is free to rotate about its axis Find the frequency of oscillation of such pendulum 12 A connecting rod shown in fig is supported at the wrist pin end It is displaced and allowed to oscillate The mass of rod is 5kg and centre of gravity is 20 cm from the pivot point O If the frequency of oscillation is 40 cyclesminute calculate the mass moment of inertia about its CG

13 A semi circular homogenous disc of radius r and mass m is pivoted freely about its centre as shown in figure Determine the natural frequency of oscillation 14A simply supported beam of square cross section 5mmx5mm and length 1m carrying a mass of 0575 kg at the middle is found to have natural frequency of 30 radsec Determine youngrsquos modulus of elasticity of beam 15 A spring mass system k1 and m have a natural frequency f1 Determine the value of k2 of another spring in terms of k1 which when placed in series with k1 lowers the natural frequency to 23 f1

COMPLETE SOLUTION OF SYSTEM EXECUTING SHM

The equation of motion of system executing SHM can be represented by

0)( =+ xkxm ampamp ----------------------------------------(1)

022

2=+

dtdx

dt

dx ω

The general solution of equation (1) can be expressed as

------------------------(2)

Where A and B are arbitrary constant which can be determined from the initial conditions of the system Two initial conditions are to be specified to evaluate these

constants x=x0 at t=0 and xamp = Vo at t=0 substituting in the equation (2) Energy method In a conservative system the total energy is constant The dwell as natural frequency can be determined by the princienergy For free vibration of undamped system at any instant and partly potential The kinetic energy T is stored in the masswhere as the potential energy U is stored in the form of sdeformation or work done in a force field such as gravity The total energy being constant T+U = constant Its rate of cha

t)(sin Bt)cos( AX ω+ω=

t)(sin V

t)cos( xx o0 ω

ω+ω=

Is the required complete

solution

28

ifferential equation as ple of conservation of of time is partly kinetic by virtue of its velocity train energy in elastic

nge

Is given by [ ] 0UTdtd

=+

From this we get a differential equation of motion as well as natural frequency of the system Determine the natural frequency of spring mass system using energy method

Determine the natural frequency of system shown in figure Determine the natural frequency of the system shown in figure Is there any limitation on the value of K Discuss Determine the natural frequency of s

m

l k

m

a

θ

m

a

θ

l

k

29

k

k

ystem shown below Neglect the mass of ball

l

m

30

A string shown in figure is under tension T which can be assumed to remain constant for small displacements Find the natural frequency of vertical vibrations of spring An acrobat 120kg walks on a tight rope as shown in figure The frequency of vibration in the given position is vertical direction is 30 rads Find the tension in the rope A manometer has a uniform bore of cross section area A If the column of liquid of length L and Density ρ is set into motion as shown in figure Find the frequency of the resulting oscillation

m

l

a

T T

36m 8m

31

k

R

k

m

r

m

Find the expression for natural frequency of system shown in the figure Neglect the mass of the cantilever beam Study the special case i) k=Infinity ii) I = infinity Determine the expression for the natural frequency of a system shown in figure The two discs are keyed to a common shaft and have a combined mass moment of inertia about centre of oscillation O It is assumed that the card attached to mass m does not stretch and is always under tension Determine the natural frequency of a system shown

m

l

32

Determine the expression for the natural frequency of the system shown in figure Assume that the wires connecting the masses do not stretch and are always in tension

Determine the natural frequency of spring mass system taking the MASS OF SPRING (ms ) into account

M3 M1

M2

k2 k1

m

k

33

RAYLEIGHrsquoS METHOD

This is the extension of energy method Here we can determine the natural frequency of a conservative system without finding the equation of motion The natural frequency is a function of the rate of change of Kinetic energy and potential energy of the system From the principle of conservation of energy we have T+U = Constant T= KE and U =PE We can write the above equation as T1+U1 = --------------------------(1) Where 1 and 2 represents tow instances of time Let 1 be the time when the mass is passing through static equilibrium position Let 2 be the time corresponding to the mass displacement of the mass At this instant the velocity f the mass is zero and hence Substituting in equation (1) we get ( )Max = ( )Max Above equation leads directly to natural frequency Determine the natural frequency of spring mass system using Rayleighrsquos method m

34

r

B

K K

Home work

Determine the natural frequency of pendulum using Rayleighrsquos method

A Cylinder of mass m and moment of inertia Jo is to role without slipping but is sustained by spring K as shown in figure Determine the natural frequency of oscillation

Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping

a

Determine the natural frequency of system shown where in cylindrical disc rolls over the inextensible spring A cylinder of radius r rolls without slipping on cylindrical the equation of motion and hence determine the naoscillations about the lowest point shown Repeat the above problem ndash Instead of cylinder assume swithout slipping on concave surface as shown above

r

K

35

portion of radius R Derive tural frequency of small

phere of radius with r rolls

The torsional pendulum has a natural frequency of 5 Hz What length of steel wire of diameter 2 mm should be used for pendulum The inertia of mass fixed at the free end is 00098 kgm2 Take g= 083x1011 Nm2 Determine the torsional natural frequency of the system shown in figure Neglect the mass moment of inertia of the shaft ( Figure not proportionate) Determine the natural frequency of simple pendulum considering the mass of the rod into account

Using the energy method find tThe chord may be assumed ine

Kt1

Kt2

20cm

06m 08m 04m

20cm 20cm 20cm

m

M

he natural frxtensible in

k

M

36

equency of the system shown in figure the spring mass pulley system

A rod of length l and depth H having a mass m rests on a semicylidrical surface os radius r It is tipped slightly and let go Find the frequency of oscillation

A thin bar of l having a mass m rests on a semicylidrical surface radius r It is tipped slightly and let go Find the frequency of oscillation

l

h

R

l

37

38

CHAPTER 3

DAMPED FREE VIBRATION OF SINGLE DEGREE FREEDOM SYSTEM LEARNING OBJECTIVES

] Introduction to damped free vibration

] Terminologies used in damped free vibration

] Three different types of damping

] Concept of critical damping and its importance

] Study of response of viscous damped system for cases of under damping

critical and over damping

] Concept of logarithmic decrement

] Problems related to damped free vibration problems

Viscously damped free vibration of single degree freedom sytem Consider a single degree freedom system with viscous damping as shown in figure For a viscous damped system the damping force is

where c is a constant of proportionality It is called damping coefficient or coefficient of viscous damping

Symbolically it is designated by a dashpot as shown in Fig 3

x xampx

39

At any instant of time let the mass be displaced by x from the equilibrium position Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

Okxxcxm =minusminus= ampampamp

Okxxcxm =++ ampampamp ----------------------------------------------------------(1)

or

0kxdtdx

Cdt

xdM

2

2

=++

Assuming the solution of equation as

The first term in the above equation m2Ct

eminus

is an exponentially decaying function of time The behavior of the terms in the parenthesis however depends on whether numerical value within the radicals is positive zero or negative

CASE 1 if 0mk

m2c

2

=minus

40

CASE 2 if 0mk

m2c

2

=gt

CASE 3 if 0mk

m2c

2

=lt

41

Critically damped coefficient Critically damped coefficient is defined as the value of the damping coeffient for which radical in equation

mk

m2c

m2c

S2

21 minus

plusmn

= becomes zero

Damping Ratio ( Damping factor) ( ξ ) It is defined as the ratio of damping coefficient to the critical damping coefficient Note The damping coefficient c is defined as the actual amount of damping while the damping coefficient is a constant depending upon the mass and stiffness of the system The roots of characteristics equation can be now written as Note if c=cc or ξ =1 then system is critically damped if cgt cc or ξ gt1 then system is over damped I if clt cc or ξ lt1 then system is under damped

42

Take home quiz Derive General solution of different damped system

1 Critically damped system 2 Under damped system 3 Over damped system

Logarithmic Decrement It is used to determine the amount of damping present in system This measure The rate of decay of free vibration The larger the damping the greater will be rate of decay Logarithmic decrement is defined as the natural logarithm of ratio of any two successive amplitudes

=δ

2

1

xx

ln

Expression for logarithmic decrement (δ ) Consider a damped free vibration which is expressed by

( ) ( ) φ+ξminusω= ξω t 1sinXex 2n

tn

x is the displacement at any instant of time X and φ are arbitrary constants

21

2

ξminus

πξ=δ

43

Tutorial problems 1 Derive an expression for damping ratio in terms of logarithmic decrement

2 Show that logarithmic decrement

=δ

n

o

xx

lnn1

where xo is the initial amplitude xn

is amplitude after n cycles 3 A mass of 1 kg is to be supported on a spring having stiffness K= 9800 Nm The damping constant c= 59 Nsm Determine the natural frequency of system Find also logarithmic decrement and amplitude after 3 cycles if initial displacement is 03 cm 4 A vibrating system is defined by the following parameters M = 3kg k= 100 Nm damping coefficient = 3 Nsm determine Damping factor natural frequency damped vibration logarithmic decrement ratio of consecutive amplitude no of cycles after which the original amplitude is reduced to 20 5A single degree damped vibrating system a suspended mass of 18kg makes 15 oscillations in 003 seconds The amplitude decreases to 025 of the initial value after 5 oscillations Determine the stiffness of spring Logarithmic decrement damping factor and damping coefficient 6 A damped vibration record of a spring mass system shows the following data Amplitude at the end of 2nd cycle is 9mm amplitude at eh end of 3rd cycle is 6mm amplitude at the end of 4th cycle is 4mm stiffness of spring =8000 Nm weight =20N Find the logarithmic decrement damping force at unit velocity periodic time of vibration 7 For the system shown in figure the characteristics of dashpot is such that the constant force of 49 N is applied to the piston it velocity is found to be constant at 012 ms Determine the value of C would you expect the complete system to be periodic or aperiodic 8 For a torsional system shown in the figuref1 is natural frequency in air and f2 is the natural frequency when immersed in oil Determine the damping coefficient

Viscous fluid

Disc of mass moment of Inertia J

2a

C

m M

b

a a

k

9 A thin plate of area A and weigh w is fixed to the end of the spring and allow to oscillate in a viscous fluid as shown in figure It f1 is the frequency of oscillation in air and f2 in liquid Show that

22

21 ff

gA2

minusπω

=micro where the damping force on the plate

AV2Fd micro= V being the velocity of vibration 10 A torsional pendulum has a natural frequency of 175 cyclessecond vibrating in vacuum The mass moment of inertia is 2 kgm2 It is immersed in oil and it is observed that the natural frequency is 142 cyclessecond Find the damping torque If the disc is replaced by 34 deg When in oil find the displacement at the end of the first complete cycle 11 An automobile can be modeled as a mass placed on 4 shock absorbers each consisting of a spring and a damper Such that each spring is equally loaded Determine the stiffness damping constant of each shock absorber So that natural frequency is 2Hz and system is critically damped The mass of vehicle is 200kg 12 Write the differential equation of motion for the system shown in the figure below Determine the natural frequency of the damped oscillation and critical damping coefficient

Determine the differential equation of motion and find the critical cdamped natural frequency for system shown in figure

a

C

m M

M =200kg

44

200kg kgkg

oefficient and

45

Derive the equation of motion and find the critical damping coefficient and damped natural frequency for the system shown below Write the differential equation of motion and determine the expression for critical damping coefficient and damped natural frequency of system shown in figure

Set up a differential equation of motion for system shown in figure and determine the following Undamped natural frequencies critical damping coefficient damping ratio damped natural frequency

b

C

kM

a

m

46

A mass of 300N is resting on two springs of stiffness 3000 Nm each and a dashpot of damping coefficient 150 Nsm as shown in figure If the initial velocity of 10cms is given to mass at its equilibrium position what will be the displacement from the equilibrium position at the end of first second Derive the formula used A gun barrel of mass 600kg has a recoil spring of stiffness 294 knm if the barrel recoils 13m on firing determine i) The initial recoil velocity of barrel ii) The critical damping coefficient of dashpot which is engaged at the end of recoil stroke Iii) Time required for the barrel to return at the end of recoil stroke Iv) time required for the barrel to return to a position 5cm from the initial position

W= 3000 N

FIRING

CHAPTER 4

FORCED VIBRATION LEARNING OBJECTIVES

] Introduction to forced free vibration

] Terminologies used in forced vibration

] Concept of magnification factor

] Study of reciprocating and rotating unbalance

] Concept of base excitation and support excitation

] Concept of Transmissibility ratio

] Problems related to forced vibration

The vibration that takes place under the excitation of external forces are called

forced vibration When a system is subjected to harmonic excitation it is forced to vibrate at the same frequency as that of excitation Most of the mechanical systems are prone to steady state forced vibration which cause fatigue failure The vibration problem also becomes very important when any of excitation frequencies coincide with one of natural frequencies of a system Under this condition resonance occurs during which large amplitude of vibrations are observed To avoid resonance external frequency (operating speed) may be changed or properties of the system may be changed to alter the natural frequency In some cases sufficient amount of damping may be provided to avoid large amplitude during resonance Thus the problem of force vibration is very important in mechanical design

FORCED VIBRATIONS OF SINGLE DOF SYSTEMS UNDER HARMONIC EXCITATION

` [ ] tsin FF o ω=

x

sin FF o=

Fo

xampamp

[ ] tω

xamp

47

48

Consider a spring mass system having a viscous damping excited by a harmonic force [ ] tsin FF o ω= as shown in figure The various forces acting on the system is shown in figure Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

O)tsin(Fkxxcxm o =ω+minusminus= ampampamp

)tsin(Fkxxcxm o ω=++ ampampamp ----------------------------------------------------------(1)

or

)tsin(Fkxdtdx

Cdt

xdM o2

2

ω=++

Equation (1) represents the differential equation of motion of the system The solution to this equation consists of two parts namely complementary function and a particular integral

The first of solution is also termed as transient solution of homogeneous solution and can be obtained from the homogeneous differential equation 0kxxcxm =++ ampampamp

The second part of solution particular integral is also termed as steady state solution ad is of form )tsin(Xx φminusω= Where X= steady state amplitude φ is phase difference between external force and displacement

)tsin(Xx φminusω=

The vector representation of the shown

49

The force polygon for the above vector diagram can be written as follows

( )φminusωt O

KX

xm 2ω

Fo

xCω

tω

φ

50

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

is the governing equation

Zero frequency deflection (Xst) It is deflection observed at zero frequency ( when the external frequency 0=ω

( ) kF

001

kF

x o

22

o

st =

+minus=

Magnification factor (M) It is defined as the ratio of steady state amplitude to zero frequency deflection xst

M = arX

X=

A plot of magnification factor versus frequency ratio

damping factor ( )ξ is called frequency plot of a harmon All the frequency terms starts from magnification factor of damping ratio ( )ξ The amplitude of vibration at any on the value of damping The amplitude will be higher for higher damping The amplitude of vibration at reso

amplitude tends to zero as

ωω

n tends to infinity

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

Freq

ζ =

ζ = 1

Magnification Factor M = XXst

kFo

51

ωω

nfor varying values of

ically excite vibrating system

M=1 what ever may be value particular frequency depends

for lesser damping and lesser nance will be very high The

uency Ratio r = (ωωn)

ζ = 0 0

ζ = 025 ζ = 0375 ζ = 0 5

ζ = 0707

ζ = 2

52

Plot of phase angle φ versus frequency ratio

ωω

n is shown in the

figure The phase angle φ is zero for all the values of damping ratio

at

ωω

n=0 The phase

angle φ is 90 for all the values of damping ratio

at

ωω

n=1 ie at

resonance The phase angle between o to 90 deg for all the values of

ωω

n= 0 to 1 and

between 90 deg to 180 deg for all the values of

ωω

ngt1

Class work Derive an expression for frequency at peak amplitude

2np 21 ξminusω=ω=ω is the required equation

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

53

Derive expression amplitude at resonance ( In dimensional form) In dimensional form We know that steady state amplitude

( ) ( )

ω+ωminus=

222

o

Cmk

Fx Equation in dimensional form

We know that at resonance mk

n =ω=ω

( )

ω+

minus

=2

n

2

oR

Cmk

mk

Fx

( )

ω=

2n

oR

C

Fx

( )

ω=

n

oR C

Fx

Class work Derive an expression for amplitude at resonance ( in non dimensional form)

Tutorial Problems on Forced Vibration 1A weight of 60N is suspended by a spring stiffness 12 kNm is forced to vibrate by a harmonic force of 10N assuming viscous damping of 0086 kN-sm Find i) resonance frequency ii) amplitude at resonance iii) phase angle at resonance iv) frequency corresponding to peak amplitude 2 A periodic torque having a maximum value of 05888Nm at a frequency of 4 rads is impressed upon a flywheel suspended from a wire The wheel has mass moment of inertia 012 kgm2 and the wire has stiffness 1176 Nmrad A viscous dashpot applies damping couple of 0784 Nm an angular velocity of 2 rads calculate i)maximum angular displacement from rest position ii) maximum couple applied to dashpot Iii) angle by which the angular displacement lags the torque A mass of 1019 kg is suspended from a spring stiffness 20 Nmm viscous damping causes the amplitude to decrease to 110th of initial value in 4 complete cycle If a periodic force of 20 cos(50 t) is applied to the mass Find the amplitude of forced vibration What would be the amplitude of oscillation if the applied force coincides with the natural period

m=1019 kg

[ ] t50cos 30F =

54

55

ROTATING UNBALANCE Unbalance in rotating machinery is a common source of vibration excitation Consider a machine of mass m having mounted on a foundation of stiffness k and damping coefficient C as shown in figure The unbalance is represented by an eccentric rotating mass m with eccentricity (e) and having a angular velocity (ω) let x be the displacement of non rotating mass from the static equilibrium position The displacement of rotating mass m from the static equilibrium is ( ) tsin( ex ω+ Applying Newtonrsquos law of motion we have

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

56

RECIPROCATING UNBALANCE Unbalance in reciprocating machinery is a common source of vibration excitation Consider a machine of mass M having mounted on a foundation of stiffness K and damping coefficient C as shown in figure The unbalance is represented by a reciprocating mass (m) having crank radus (e) and connecting rod of length (l) Let ωωωω be angular velocity of the crank Let x be the displacement of non reciprocating mass from the static equilibrium position at any instant of time The displacement of reciprocating mass m from the static equilibrium position is

)2sin(sin +++ tle

tex ωωωωωωωω

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

57

Note The equations obtained in reciprocating unbalance is same as the rotating unbalance and having the following differences Rotating Unbalance Reciprocating Unbalance Here mass of the rotating element is considered

Here mass of reciprocating element is considered

Here the eccentricity (e) which is the distance between the centre of rotation and CG

Here the eccentricity e is the crank radius and crank radius = stroke length2

The frequency plot of non dimensional factor( ( )me

MX versus frequency ratio

ωω

n for various values of ( )ξξξξ is shown in figure This frequency plot is same for

the rotating and reciprocating unbalance with the following points are noted from the diagram

1 Damping ratio plays very important role during resonance in controlling the amplitude

2 When

ωω

n=0 then ( )me

MX =0 for all the values of damping ratio ( )ξξξξ

3 When

ωω

nis higher than ( )me

MX tends to unity

4 At very high speeds ie at very high frequency ratio

ωω

n there is no effect

of damping The frequency plot of phase angle versus the frequency ratio is given as shown in figure The explanation is same as in forced harmonic vibration

58

VIBRATION ISOLATION

Vibratory forces and motions generated by machine and other caused are unavoidable Howeever their effect on the other dynamical systems can be minimized by proper isolator design An isolation system is either to protect o delicate machine from excessive vibration transmitted to it from its supporting structure or to prevent vibratory forces and motions generated by machines from being transmitted to its surroundings The effectiveness of isolation may be measured in terms of forces and motion transmitted The first type is known as force isolation and the second type is known as motion isolation

Force isolation (Force Transmissibility) When vibratory forces generated by machine are isolated the effectiveness of forced isolation can be made by defining the term transmissibility The term transmissibility is defined as the ratio of force transmitted to the foundation to that of the external impressed force

Force Transmissibility = o

tr

FF

force Impressed Externalfoundation toted transmitForce

=

Consider viscous damped spring mass system as shown in figure Upon this system an external harmonic force ) t (Sin Fo ω is applied The difference forces acting on system are inertia forces damping force spring force and external force All these forces are represented on force polygon shown below Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these two forces

( ) ( )22 xckx ω+=

59

This is the expression for force transmissibility in dimensional form To represent same in the non dimensional form divide numerator and denominator by k

2

n

22

n

2

n

o

tr

21

21

FF

TR

ωω

ξ+

ωω

minus

ωω

ξ+

==

PHASE DIFFERNECE BETWEEN EXTERNAL FORCE AND DISPLACEMENT Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these

ωω

minus

ωω

ξ=

ωminusω

=φ minusminus2

n

n1

2

1

1

2tan

mKC

tan

The phase difference between force transmitted to the foundation Ftr to the displacement

kc

KxxC

tanω

=ω

=α or

ωω

ξ=

ω=α minusminus

n

11 2tanKx

xCtan

The phase difference between external force Fo and force transmitted to foundationis β

60

β= αminusφ =

ωω

minus

ωω

ξminus

2

n

n1

1

2tan -

ωω

ξminus

n

1 2tan

Displacement Transmissibility It is the ratio of amplitude of steady state vibration of machine to that of the base motion Forced vibration due to support motion Consider a spring mass damper system subjected to harmonic excitation )tsin(Yy ω= at the base as shown in figure Let x be the absolute displacement of the mass at any instant of time t The different forces acting on system are shown in free body diagram Applying Newtonrsquos Law of motion Ma = sumF

)yx(k)yx(cxm minusminusminusminus= ampampampamp

0)yx(k)yx(cxm =minus+minus+ ampampampamp

61

Frequency plot of transmissibility versus frequency ratio

62

The plot of transmissibility versus frequency ratio is as shown in figure The following points can be noted from it

i) The value of transmissibility is unity at frequency ratio

ωω

n=0 and 2

ii) For an undamped system ( )ξ =0 transmissibility tends to infin at resonance iii) The value of transmissibility is less than unity for the value of frequency ratio

greater than 2

iv) For frequency ratio

ωω

nlt 2 smaller damping ratio lead to larger values of

transmissibility on the other hand for frequency ratio

ωω

ngt 2 smaller damping

ratio leads to smaller value of transmissibility

The frequency plot of transmissibility (TR) versus frequency ratio

ωω

n can be

broadly divided into two zones

1

ωω

n= 0 to 2

2

ωω

n= 2 to infin

CASE 1

When

ωω

n= 0 to 2

In this zone the value of transmissibility is always greater than 1 for all values of damping ratio In this zone the transmissibility decreases with increase in damping Hence we can conclude that in this zone damping is useful in isolating vibration forces being transmitted to the support CASE 2

When

ωω

n= 2 to infin

In this zone the value of transmissibility is less than 1 for all values of damping ratio In this zone the transmissibility increases with increase in damping Hence we can conclude that in this zone damping will adversely affect in isolating vibratory forces

63

Tutorial problems on Forced Vibration 1 Machine of mass 75kg is mounted on spring of stiffness 12000 Nm with an assumed damping factor ξ =02 piston within the machine of mass 2 kg has a reciprocating motion with a stroke of 75 cm and speed of 15 cyclessecond Assuming the motion of piston to be harmonic determine amplitude of machine ( X) Transmissibility and force transmitted to foundation phase angle of transmitted force wrt to exciting force 2 Motor weighs 600N is mounted on a simple beam that has a spring rate of 400 Ncm The motor armature base 160 N and has an eccentricity of 0005cm What will be the amplitude of motion when it runs at 1760 RPM Neglect damping the deflection of armature shaft and weight of the beam 3 A Heavy machine of mass 300 kg is supported on resilient foundation the static deflection of foundation due to weight of machine is found to be 75mm It is observed that the machine vibrates with an amplitude of 10mm when the base of the foundation is subjected to harmonic oscillation at the undampeds natural frequency of system with an amplitude of 25mm Find the damping constant of foundation Ii) dynamic force amplitude on the base iii) amplitude of the displacement of machine relative to base 4 An air compressor of mass 450 kg operates at a constant speed of 1750 RPM The reciprocating masses are 10 kg the crank radius is 100mm If the damper for mounting introduces a damping factor 015 i) specify the springs of mounting such that only 20 of the unbalanced force is transmitted to the foundation ii) determine the magnitude of the transmitted force bull Repeat the above problem by assuming that there is no damping in system

5 A body of a car having mass 1500 kg and is mounted on 4 equal springs which deflect through 225 cm under the weight of the body A total damping forces the 4 shock absorbers is 46 N at a velocity of 1cmsec The car is placed with all the four wheels on a test platform which is move up and down at resonant speed With an amplitude of 25cm Find the amplitude of the car body on its springs assuming its CG to be in the centre of wheel base 6A 75 kg machine which is mounted on springs of stiffness 116x105 Nm with the damping of 20=ξ A 2kg piston within the machine as reciprocating motion with a stroke of 008m and speed of 3000 RPM Assuming the motion of piston to be harmonic determine the amplitude of vibration of machine

7 The spring of an automobile are compressed 10cm under its own weight Find the critical speed when a trailer is traveling over road with a people approximaterly by a sine wave of amplitude 75 cm and wave length 15m What will be the amplitude of vibration at 64kmhour 8 An eccentric mass excites us used to determine vibratory characteristics of a structure of mass 25kg as given in figure At a speed of 1000 RPM the eccentric mass to beat top position at the instant the structure was moved upwards through its static equilibrium position and corresponding amplitude was 20mm The eccentric mass is 05kg at an eccentricity of 10mm Determine i) Undamped natural frequency of the system ii) Damping ratio iii) Steady state amplitude and phase angle at 1500 RPM iv) Force transmitted to the ground at 1200 RPM

M=25kg

C

K

Eccentric mass

64

65

CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Support motion Consider a spring mass damper system as shown in figure Let it excited by motion

of support let y be the harmonic displacement of support From Newtonrsquos second law of motion

)yx(k)yx(cxm minusminusminusminus= ampampampamp

)1(0)yx(k)yx(cxm minusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminus=minus+minus+ ampampampamp

Let z be the relative displacement of the mass with respect to support then

Z=x-y

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

Translationa Let a mass mmass momearrangement These two mamass Meq or

2

1

33

2

1

221

+

+=

ll

Mll

MMMeq

14

is the required answer

l and rotational masses coupled together

having a translational velocity x be coupled to another mass having nt of inertia Io with a rotational velocity θ as in rack and pinion

shown in the figure

sses can be combined to obtain either a single equivalent translational a single equivalent mass moment of inertia Jeq

15

Equivalent translational masses

Kinetic energy of the equivalent mass =

2

21

eqeq XM amp

Kinetic energy of the two masses =

+

20

2

21

21

θθθθampamp JXM

Meq= m

+ 2RJ

m o is the required answer

Also determine the equivalent rotational mass Jeq

Jeq= m [ ]oJmR +2 is the required answer Spring element Whenever there is a relative motion between the two ends of the spring a force is developed called spring force or restoring force The spring force is proportional to the amount of deformation x and then F α x or F = kx Where k is stiffness of the spring or spring constant or spring gradient The spring stiffness is equal to spring force per unit deromation

The spring stiffness k = mNxF

Workdone in deforming a spring is equal to the strain energy or potential energy Strain energy = potential energy = area of the triangle OAB Stiffness of beams Cantilever beam consider a cantilever beam with an end mass shown in the figure The mass of the beam is assumed to be negligible The static deflection of beam at free end is given by

Similarly derive the expression for Simply supported beam and fixed support beam

x

k

m

x

m

x

m

x

mNlIE

st 192

3

=δδδδ

mNEI

Wlst

48

3

=δδδδ

F

K=stiffness

F

16

17

Stiffness of slender bar subjected to longitudinal vibrations For a system executing the longitudinal vibrations as shown in the figure let us derive the expression for stiffness

Torsional Stiffness of bar It is the amount of torque required to cause a unit angular deformation in the element

Torsional stiffness = Kt = θθθθT

Combination of stiffness Determination of equivalent spring stiffness when the springs are arranged in series Consider two springs of stiffness K1 and K2 acted upon by the force F

The deflection of spring k1 is x1 = 1KF

The deflection of spring k1 is x2 = Let these two springs be replaced by an equivalent stiffness Keq upon which a force F acts and due to which its deflection is given by

x = eqKF

x= x1+x2

m

E A

l

60 cm 70 cm

Determination of equivalent spring stiffness when the springs are arranged in parallel Force acting on K1 spring = F1=k1x Force acting on K2 spring = F2 Force required for an equivalent spring keq to defined by x given by F= Keq x But F = F1 +F2 Tutorial problems on Equivalent stiffness of springs 1Determine the equivalent stiffness for the system shown in figure 2 Determine the equivalent stiffness for the system shown in figure

3 Determine the equivale

M

3

M

Nm 102 6x

18

nt stiffness for the system shown in figure

38 kg

Nm 101 6x Nm 102 6x

Nm 106x

2k

k

k

2k

3k

k

2k

k

19

60 cm 80 cm 50 cm

50 cm 60 cm

4Determine the equivalent stiffness for the system shown in figure

5Replace the following torsional stiffness by a single shaft having radius 4cm and find the length required for the equivalent shaft Assume the material of given system and equivalent system is same R1= 3cm R2= 5cm DAMPING Every vibration energy is gradually converted into heat or sound Hence the displacement during vibration gradually reduces The mechanism by which vibration energy is gradually converted into heat or sound is known as damping A damper is assumed to have either mass or elasticity Hence damping is modeled as one or more of the following types Viscous damping Coulomb or dry friction damping materials or solid or hysteric damping Viscous damping Viscous damping is most commonly used damping mechanism in vibration analysis When the mechanical system vibrates in a fluid medium such as air gas water or oil the resistance offered by the fluid to the moving body causes energy to be dissipated In this case the amount of dissipated energy depends on many factors such as size or shape of the vibrating body the viscosity of the fluid the frequency of vibration and velocity of fluid Resistance due to viscous damping is directly proportional to the velocity of vibration

dF V αααα

dF = xC amp

Where C= damping coefficient Fd = damping force Examples of Viscous damping

1) Fluid film between sliding surface 2) Fluid flow around a piston in a cylinder 3) Fluid flow through orifice 4) Fluid flow around a journal in a bearing

Rreq =4cm

leqn

stress

Strain

Coulomb damping or dry friction damping Here a damping force is constant in magnitude but opposite in direction to that of the motion of vibrating body It is caused by the friction between the surfaces that are dry or have insufficient lubrication Material or solid or hysteric damping

When the materials are deformed energy is absorbed and dissipated by the material The effect is due to friction between the internal planes which slip or slide as the deformation takes place When a body having the material damping is subjected to vibration the stress strain diagram shows the hysteresis loop as shown in the figure The area of the loop denotes the energy lost per unit volume of the body per cycle

20

21

ττττ ττττ2 ττττ3

ττττ

X(t)

FOURIER SERIES The simplest of periodic motion happens to be SHM It is simple to handle but the motion of many vibrating system is not harmonic (but periodic) Few examples are shown below

Forces acting on machines are generally periodic but this may not be harmonic for example the excitation force in a punching machine is periodic and it can be represented as shown in figure 13 Vibration analysis of system subjected to periodic but nonharmonic forces can be done with the help of Fourier series The problem becomes a multifrequency excitation problem The principle of linear superposition is applied and the total response is the sum of the response due to each of the individual frequency term

Any periodic motion can be expressed as an infinite sum of sines and cosines terms If x(t) is a periodic function with period t its Fourier representation is given by

X(t) = (((( )))) (((( )))) (((( ))))t sint cost cos2 111 ωωωωωωωωωωωω baaao ++++++++++++

= (((( )))) (((( ))))t sint cos2 1

ωωωωωωωω nnn

o bnaa

++++++++sumsumsumsuminfininfininfininfin

====

tππππωωωω

2==== = Fundamental frequency ndash (1)

where ao an bn are constants

ττττ ττττ2 ττττ3

ττττ

X(t)

22

t

X(t)

035

025

Determination of constants To find ao Integrate both sides of equation(1) over any interval ττττ All intergrals on the RHS of the equation are zero except the one containing ao

(((( ))))dttxao

o 2

2

intintintint====ωωωωππππ

ππππωωωω

= (((( ))))dttxo

2intintintintττττ

ττττ

To find an multiply equation 1 by cos (((( ))))tnωωωω and Integrate over any interval ττττ All intergrals

(((( )))) (((( )))) dttxao

n tncos 2

2

ωωωωππππωωωω ωωωω

ππππ

intintintint==== (((( )))) (((( )))) dttxo

tncos 2

ωωωωττττ

ττττ

intintintint====

To find bn multiply equation 1 by sin (((( ))))tnωωωω and Integrate over any interval ττττ All intergrals

(((( )))) (((( )))) dttxao

n tnsin 2

2

ωωωωππππωωωω ωωωω

ππππ

intintintint==== (((( )))) (((( )))) dttxo

tnsin 2

ωωωωττττ

ττττ

intintintint====

Find the Fourier series for the curve shown below

ττττ ττττ2 ττττ3

t

X(t)

E

ττττ ττττ2 ττττ3

t

X(t)

ππππ2

t

X(t) ππππ

1 Represent for the periodic motion shown in the figure

23

CHAPTER 2 UNDAMPED FREE VIBRATION LEARNING OBJECTIVES

] Introduction to undamped free vibration

] Terminologies used in undamped free vibration

] Three methods to solve the undamped free vibration problems

] Problems related to undamped free vibration problems

Free vibrations are oscillations about a systems equilibrium position that occur in the absence of an external excitation force If during vibrations there is no loss of energy it is known as undamped vibration The first step in solving a vibration problem is setting up the differential equation of motion Consider spring mass system which is assumed to move only along the vertical direction as shown below Let m be the mass of the block and k be the stiffness of the spring When block of mass m is attached to spring the deflection of spring will be ∆ known as static deflection In the static equilibrium position the free body diagram of forces acting on the mass is shown in Figure(b) Hence mg= kA Once the system is disturbed the system executes vibrations

Let at any instant of time t the mass is displaced from the equilibrium position x the different forces acting on the system are shown in figure (d) From Newtonrsquos second law of motion

sum = maF

Inertia force ( disturbing force) = restoring force

mgxkxm ++∆minus= )(ampamp

0)( =+ xkxm ampamp

or 0)( =+ xmk

xampamp

equation 2 is the differential equation of motion for spring mfigure Comparing equation (2) with the equation of SHM +xampampsince the vibrations of the above system are free( without the forces) we can write

sec radmk

n =ω

24

ass system shown in 0)(2 =xω

resistance of external

time period km

2 f1

n

πτ ==

from the equation(1) mg = k∆

∆=

q

mk

Difference between the translation ( rectilinear) and rotational system of vibration

Translatory Rotational

In the analysis thFORCES are co Linear stiffness K Problems 1A mass of 10kFind the natural 2 A spring massnatural frequencnatural frequenc

kt

m

sec radmk

n =ω

k

25

e disturbing and restoring nsidered

In the analysis In the analysis MASS Moment of Inertia (J) is considered

in Nm is considered

g when suspended from a spring causes a static deflection of 1cm frequency of system

system has a spring stiffness K Nm and a mass of m Kg It has a y of vibration 12 Hz An extra 2kg mass coupled to it then the y reduces by 2 Hz find K and m

secrad n =ω

26

3 A steel wire of 2mm diameter and 30m long It is fixed at the upper end and carries a mass of m kg at its free end Find m so that the frequency of longitudinal vibration is 4 Hz 4 A spring mass system has a natural period of 02 seconds What will be the new period if the spring constant is 1) increased by 50 2) decreased by 50 5 A spring mass system has a natural frequency of 10 Hz when the spring constant is reduced by 800 Nm the frequency is altered by 45 Find the mass and spring constant of the original system 6 Determine the natural frequency of system shown in fig is which shaft is supported in SHORT bearings 7 Determine the natural frequency of system shown in fig is which shaft is supported in LONG bearings where l is the length of bearing and E ndash youngrsquos modulus and I is moment of Inertia 7 Determine the natural frequency of system shown in fig is which shaft is supported in LONG bearings 8 A light cantilever beam of rectangular section( 5 cm deep x 25cm wide) has a mass fixed at its free end Find the ratio of frequency of free lateral vibration in vertical plane to that in horizontal 9 Determine the natural frequency of simple pendulum 10 Homogeneous square plate of size l and mass m is suspended from the mid point of one of its sides as shown in figure Find the frequency of vibration

l

l

27

11 A compound pendulum which is rigid body of mass m and it is pivoted at O The point of pivot is at distance d from the centre of gravity It is free to rotate about its axis Find the frequency of oscillation of such pendulum 12 A connecting rod shown in fig is supported at the wrist pin end It is displaced and allowed to oscillate The mass of rod is 5kg and centre of gravity is 20 cm from the pivot point O If the frequency of oscillation is 40 cyclesminute calculate the mass moment of inertia about its CG

13 A semi circular homogenous disc of radius r and mass m is pivoted freely about its centre as shown in figure Determine the natural frequency of oscillation 14A simply supported beam of square cross section 5mmx5mm and length 1m carrying a mass of 0575 kg at the middle is found to have natural frequency of 30 radsec Determine youngrsquos modulus of elasticity of beam 15 A spring mass system k1 and m have a natural frequency f1 Determine the value of k2 of another spring in terms of k1 which when placed in series with k1 lowers the natural frequency to 23 f1

COMPLETE SOLUTION OF SYSTEM EXECUTING SHM

The equation of motion of system executing SHM can be represented by

0)( =+ xkxm ampamp ----------------------------------------(1)

022

2=+

dtdx

dt

dx ω

The general solution of equation (1) can be expressed as

------------------------(2)

Where A and B are arbitrary constant which can be determined from the initial conditions of the system Two initial conditions are to be specified to evaluate these

constants x=x0 at t=0 and xamp = Vo at t=0 substituting in the equation (2) Energy method In a conservative system the total energy is constant The dwell as natural frequency can be determined by the princienergy For free vibration of undamped system at any instant and partly potential The kinetic energy T is stored in the masswhere as the potential energy U is stored in the form of sdeformation or work done in a force field such as gravity The total energy being constant T+U = constant Its rate of cha

t)(sin Bt)cos( AX ω+ω=

t)(sin V

t)cos( xx o0 ω

ω+ω=

Is the required complete

solution

28

ifferential equation as ple of conservation of of time is partly kinetic by virtue of its velocity train energy in elastic

nge

Is given by [ ] 0UTdtd

=+

From this we get a differential equation of motion as well as natural frequency of the system Determine the natural frequency of spring mass system using energy method

Determine the natural frequency of system shown in figure Determine the natural frequency of the system shown in figure Is there any limitation on the value of K Discuss Determine the natural frequency of s

m

l k

m

a

θ

m

a

θ

l

k

29

k

k

ystem shown below Neglect the mass of ball

l

m

30

A string shown in figure is under tension T which can be assumed to remain constant for small displacements Find the natural frequency of vertical vibrations of spring An acrobat 120kg walks on a tight rope as shown in figure The frequency of vibration in the given position is vertical direction is 30 rads Find the tension in the rope A manometer has a uniform bore of cross section area A If the column of liquid of length L and Density ρ is set into motion as shown in figure Find the frequency of the resulting oscillation

m

l

a

T T

36m 8m

31

k

R

k

m

r

m

Find the expression for natural frequency of system shown in the figure Neglect the mass of the cantilever beam Study the special case i) k=Infinity ii) I = infinity Determine the expression for the natural frequency of a system shown in figure The two discs are keyed to a common shaft and have a combined mass moment of inertia about centre of oscillation O It is assumed that the card attached to mass m does not stretch and is always under tension Determine the natural frequency of a system shown

m

l

32

Determine the expression for the natural frequency of the system shown in figure Assume that the wires connecting the masses do not stretch and are always in tension

Determine the natural frequency of spring mass system taking the MASS OF SPRING (ms ) into account

M3 M1

M2

k2 k1

m

k

33

RAYLEIGHrsquoS METHOD

This is the extension of energy method Here we can determine the natural frequency of a conservative system without finding the equation of motion The natural frequency is a function of the rate of change of Kinetic energy and potential energy of the system From the principle of conservation of energy we have T+U = Constant T= KE and U =PE We can write the above equation as T1+U1 = --------------------------(1) Where 1 and 2 represents tow instances of time Let 1 be the time when the mass is passing through static equilibrium position Let 2 be the time corresponding to the mass displacement of the mass At this instant the velocity f the mass is zero and hence Substituting in equation (1) we get ( )Max = ( )Max Above equation leads directly to natural frequency Determine the natural frequency of spring mass system using Rayleighrsquos method m

34

r

B

K K

Home work

Determine the natural frequency of pendulum using Rayleighrsquos method

A Cylinder of mass m and moment of inertia Jo is to role without slipping but is sustained by spring K as shown in figure Determine the natural frequency of oscillation

Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping

a

Determine the natural frequency of system shown where in cylindrical disc rolls over the inextensible spring A cylinder of radius r rolls without slipping on cylindrical the equation of motion and hence determine the naoscillations about the lowest point shown Repeat the above problem ndash Instead of cylinder assume swithout slipping on concave surface as shown above

r

K

35

portion of radius R Derive tural frequency of small

phere of radius with r rolls

The torsional pendulum has a natural frequency of 5 Hz What length of steel wire of diameter 2 mm should be used for pendulum The inertia of mass fixed at the free end is 00098 kgm2 Take g= 083x1011 Nm2 Determine the torsional natural frequency of the system shown in figure Neglect the mass moment of inertia of the shaft ( Figure not proportionate) Determine the natural frequency of simple pendulum considering the mass of the rod into account

Using the energy method find tThe chord may be assumed ine

Kt1

Kt2

20cm

06m 08m 04m

20cm 20cm 20cm

m

M

he natural frxtensible in

k

M

36

equency of the system shown in figure the spring mass pulley system

A rod of length l and depth H having a mass m rests on a semicylidrical surface os radius r It is tipped slightly and let go Find the frequency of oscillation

A thin bar of l having a mass m rests on a semicylidrical surface radius r It is tipped slightly and let go Find the frequency of oscillation

l

h

R

l

37

38

CHAPTER 3

DAMPED FREE VIBRATION OF SINGLE DEGREE FREEDOM SYSTEM LEARNING OBJECTIVES

] Introduction to damped free vibration

] Terminologies used in damped free vibration

] Three different types of damping

] Concept of critical damping and its importance

] Study of response of viscous damped system for cases of under damping

critical and over damping

] Concept of logarithmic decrement

] Problems related to damped free vibration problems

Viscously damped free vibration of single degree freedom sytem Consider a single degree freedom system with viscous damping as shown in figure For a viscous damped system the damping force is

where c is a constant of proportionality It is called damping coefficient or coefficient of viscous damping

Symbolically it is designated by a dashpot as shown in Fig 3

x xampx

39

At any instant of time let the mass be displaced by x from the equilibrium position Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

Okxxcxm =minusminus= ampampamp

Okxxcxm =++ ampampamp ----------------------------------------------------------(1)

or

0kxdtdx

Cdt

xdM

2

2

=++

Assuming the solution of equation as

The first term in the above equation m2Ct

eminus

is an exponentially decaying function of time The behavior of the terms in the parenthesis however depends on whether numerical value within the radicals is positive zero or negative

CASE 1 if 0mk

m2c

2

=minus

40

CASE 2 if 0mk

m2c

2

=gt

CASE 3 if 0mk

m2c

2

=lt

41

Critically damped coefficient Critically damped coefficient is defined as the value of the damping coeffient for which radical in equation

mk

m2c

m2c

S2

21 minus

plusmn

= becomes zero

Damping Ratio ( Damping factor) ( ξ ) It is defined as the ratio of damping coefficient to the critical damping coefficient Note The damping coefficient c is defined as the actual amount of damping while the damping coefficient is a constant depending upon the mass and stiffness of the system The roots of characteristics equation can be now written as Note if c=cc or ξ =1 then system is critically damped if cgt cc or ξ gt1 then system is over damped I if clt cc or ξ lt1 then system is under damped

42

Take home quiz Derive General solution of different damped system

1 Critically damped system 2 Under damped system 3 Over damped system

Logarithmic Decrement It is used to determine the amount of damping present in system This measure The rate of decay of free vibration The larger the damping the greater will be rate of decay Logarithmic decrement is defined as the natural logarithm of ratio of any two successive amplitudes

=δ

2

1

xx

ln

Expression for logarithmic decrement (δ ) Consider a damped free vibration which is expressed by

( ) ( ) φ+ξminusω= ξω t 1sinXex 2n

tn

x is the displacement at any instant of time X and φ are arbitrary constants

21

2

ξminus

πξ=δ

43

Tutorial problems 1 Derive an expression for damping ratio in terms of logarithmic decrement

2 Show that logarithmic decrement

=δ

n

o

xx

lnn1

where xo is the initial amplitude xn

is amplitude after n cycles 3 A mass of 1 kg is to be supported on a spring having stiffness K= 9800 Nm The damping constant c= 59 Nsm Determine the natural frequency of system Find also logarithmic decrement and amplitude after 3 cycles if initial displacement is 03 cm 4 A vibrating system is defined by the following parameters M = 3kg k= 100 Nm damping coefficient = 3 Nsm determine Damping factor natural frequency damped vibration logarithmic decrement ratio of consecutive amplitude no of cycles after which the original amplitude is reduced to 20 5A single degree damped vibrating system a suspended mass of 18kg makes 15 oscillations in 003 seconds The amplitude decreases to 025 of the initial value after 5 oscillations Determine the stiffness of spring Logarithmic decrement damping factor and damping coefficient 6 A damped vibration record of a spring mass system shows the following data Amplitude at the end of 2nd cycle is 9mm amplitude at eh end of 3rd cycle is 6mm amplitude at the end of 4th cycle is 4mm stiffness of spring =8000 Nm weight =20N Find the logarithmic decrement damping force at unit velocity periodic time of vibration 7 For the system shown in figure the characteristics of dashpot is such that the constant force of 49 N is applied to the piston it velocity is found to be constant at 012 ms Determine the value of C would you expect the complete system to be periodic or aperiodic 8 For a torsional system shown in the figuref1 is natural frequency in air and f2 is the natural frequency when immersed in oil Determine the damping coefficient

Viscous fluid

Disc of mass moment of Inertia J

2a

C

m M

b

a a

k

9 A thin plate of area A and weigh w is fixed to the end of the spring and allow to oscillate in a viscous fluid as shown in figure It f1 is the frequency of oscillation in air and f2 in liquid Show that

22

21 ff

gA2

minusπω

=micro where the damping force on the plate

AV2Fd micro= V being the velocity of vibration 10 A torsional pendulum has a natural frequency of 175 cyclessecond vibrating in vacuum The mass moment of inertia is 2 kgm2 It is immersed in oil and it is observed that the natural frequency is 142 cyclessecond Find the damping torque If the disc is replaced by 34 deg When in oil find the displacement at the end of the first complete cycle 11 An automobile can be modeled as a mass placed on 4 shock absorbers each consisting of a spring and a damper Such that each spring is equally loaded Determine the stiffness damping constant of each shock absorber So that natural frequency is 2Hz and system is critically damped The mass of vehicle is 200kg 12 Write the differential equation of motion for the system shown in the figure below Determine the natural frequency of the damped oscillation and critical damping coefficient

Determine the differential equation of motion and find the critical cdamped natural frequency for system shown in figure

a

C

m M

M =200kg

44

200kg kgkg

oefficient and

45

Derive the equation of motion and find the critical damping coefficient and damped natural frequency for the system shown below Write the differential equation of motion and determine the expression for critical damping coefficient and damped natural frequency of system shown in figure

Set up a differential equation of motion for system shown in figure and determine the following Undamped natural frequencies critical damping coefficient damping ratio damped natural frequency

b

C

kM

a

m

46

A mass of 300N is resting on two springs of stiffness 3000 Nm each and a dashpot of damping coefficient 150 Nsm as shown in figure If the initial velocity of 10cms is given to mass at its equilibrium position what will be the displacement from the equilibrium position at the end of first second Derive the formula used A gun barrel of mass 600kg has a recoil spring of stiffness 294 knm if the barrel recoils 13m on firing determine i) The initial recoil velocity of barrel ii) The critical damping coefficient of dashpot which is engaged at the end of recoil stroke Iii) Time required for the barrel to return at the end of recoil stroke Iv) time required for the barrel to return to a position 5cm from the initial position

W= 3000 N

FIRING

CHAPTER 4

FORCED VIBRATION LEARNING OBJECTIVES

] Introduction to forced free vibration

] Terminologies used in forced vibration

] Concept of magnification factor

] Study of reciprocating and rotating unbalance

] Concept of base excitation and support excitation

] Concept of Transmissibility ratio

] Problems related to forced vibration

The vibration that takes place under the excitation of external forces are called

forced vibration When a system is subjected to harmonic excitation it is forced to vibrate at the same frequency as that of excitation Most of the mechanical systems are prone to steady state forced vibration which cause fatigue failure The vibration problem also becomes very important when any of excitation frequencies coincide with one of natural frequencies of a system Under this condition resonance occurs during which large amplitude of vibrations are observed To avoid resonance external frequency (operating speed) may be changed or properties of the system may be changed to alter the natural frequency In some cases sufficient amount of damping may be provided to avoid large amplitude during resonance Thus the problem of force vibration is very important in mechanical design

FORCED VIBRATIONS OF SINGLE DOF SYSTEMS UNDER HARMONIC EXCITATION

` [ ] tsin FF o ω=

x

sin FF o=

Fo

xampamp

[ ] tω

xamp

47

48

Consider a spring mass system having a viscous damping excited by a harmonic force [ ] tsin FF o ω= as shown in figure The various forces acting on the system is shown in figure Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

O)tsin(Fkxxcxm o =ω+minusminus= ampampamp

)tsin(Fkxxcxm o ω=++ ampampamp ----------------------------------------------------------(1)

or

)tsin(Fkxdtdx

Cdt

xdM o2

2

ω=++

Equation (1) represents the differential equation of motion of the system The solution to this equation consists of two parts namely complementary function and a particular integral

The first of solution is also termed as transient solution of homogeneous solution and can be obtained from the homogeneous differential equation 0kxxcxm =++ ampampamp

The second part of solution particular integral is also termed as steady state solution ad is of form )tsin(Xx φminusω= Where X= steady state amplitude φ is phase difference between external force and displacement

)tsin(Xx φminusω=

The vector representation of the shown

49

The force polygon for the above vector diagram can be written as follows

( )φminusωt O

KX

xm 2ω

Fo

xCω

tω

φ

50

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

is the governing equation

Zero frequency deflection (Xst) It is deflection observed at zero frequency ( when the external frequency 0=ω

( ) kF

001

kF

x o

22

o

st =

+minus=

Magnification factor (M) It is defined as the ratio of steady state amplitude to zero frequency deflection xst

M = arX

X=

A plot of magnification factor versus frequency ratio

damping factor ( )ξ is called frequency plot of a harmon All the frequency terms starts from magnification factor of damping ratio ( )ξ The amplitude of vibration at any on the value of damping The amplitude will be higher for higher damping The amplitude of vibration at reso

amplitude tends to zero as

ωω

n tends to infinity

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

Freq

ζ =

ζ = 1

Magnification Factor M = XXst

kFo

51

ωω

nfor varying values of

ically excite vibrating system

M=1 what ever may be value particular frequency depends

for lesser damping and lesser nance will be very high The

uency Ratio r = (ωωn)

ζ = 0 0

ζ = 025 ζ = 0375 ζ = 0 5

ζ = 0707

ζ = 2

52

Plot of phase angle φ versus frequency ratio

ωω

n is shown in the

figure The phase angle φ is zero for all the values of damping ratio

at

ωω

n=0 The phase

angle φ is 90 for all the values of damping ratio

at

ωω

n=1 ie at

resonance The phase angle between o to 90 deg for all the values of

ωω

n= 0 to 1 and

between 90 deg to 180 deg for all the values of

ωω

ngt1

Class work Derive an expression for frequency at peak amplitude

2np 21 ξminusω=ω=ω is the required equation

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

53

Derive expression amplitude at resonance ( In dimensional form) In dimensional form We know that steady state amplitude

( ) ( )

ω+ωminus=

222

o

Cmk

Fx Equation in dimensional form

We know that at resonance mk

n =ω=ω

( )

ω+

minus

=2

n

2

oR

Cmk

mk

Fx

( )

ω=

2n

oR

C

Fx

( )

ω=

n

oR C

Fx

Class work Derive an expression for amplitude at resonance ( in non dimensional form)

Tutorial Problems on Forced Vibration 1A weight of 60N is suspended by a spring stiffness 12 kNm is forced to vibrate by a harmonic force of 10N assuming viscous damping of 0086 kN-sm Find i) resonance frequency ii) amplitude at resonance iii) phase angle at resonance iv) frequency corresponding to peak amplitude 2 A periodic torque having a maximum value of 05888Nm at a frequency of 4 rads is impressed upon a flywheel suspended from a wire The wheel has mass moment of inertia 012 kgm2 and the wire has stiffness 1176 Nmrad A viscous dashpot applies damping couple of 0784 Nm an angular velocity of 2 rads calculate i)maximum angular displacement from rest position ii) maximum couple applied to dashpot Iii) angle by which the angular displacement lags the torque A mass of 1019 kg is suspended from a spring stiffness 20 Nmm viscous damping causes the amplitude to decrease to 110th of initial value in 4 complete cycle If a periodic force of 20 cos(50 t) is applied to the mass Find the amplitude of forced vibration What would be the amplitude of oscillation if the applied force coincides with the natural period

m=1019 kg

[ ] t50cos 30F =

54

55

ROTATING UNBALANCE Unbalance in rotating machinery is a common source of vibration excitation Consider a machine of mass m having mounted on a foundation of stiffness k and damping coefficient C as shown in figure The unbalance is represented by an eccentric rotating mass m with eccentricity (e) and having a angular velocity (ω) let x be the displacement of non rotating mass from the static equilibrium position The displacement of rotating mass m from the static equilibrium is ( ) tsin( ex ω+ Applying Newtonrsquos law of motion we have

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

56

RECIPROCATING UNBALANCE Unbalance in reciprocating machinery is a common source of vibration excitation Consider a machine of mass M having mounted on a foundation of stiffness K and damping coefficient C as shown in figure The unbalance is represented by a reciprocating mass (m) having crank radus (e) and connecting rod of length (l) Let ωωωω be angular velocity of the crank Let x be the displacement of non reciprocating mass from the static equilibrium position at any instant of time The displacement of reciprocating mass m from the static equilibrium position is

)2sin(sin +++ tle

tex ωωωωωωωω

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

57

Note The equations obtained in reciprocating unbalance is same as the rotating unbalance and having the following differences Rotating Unbalance Reciprocating Unbalance Here mass of the rotating element is considered

Here mass of reciprocating element is considered

Here the eccentricity (e) which is the distance between the centre of rotation and CG

Here the eccentricity e is the crank radius and crank radius = stroke length2

The frequency plot of non dimensional factor( ( )me

MX versus frequency ratio

ωω

n for various values of ( )ξξξξ is shown in figure This frequency plot is same for

the rotating and reciprocating unbalance with the following points are noted from the diagram

1 Damping ratio plays very important role during resonance in controlling the amplitude

2 When

ωω

n=0 then ( )me

MX =0 for all the values of damping ratio ( )ξξξξ

3 When

ωω

nis higher than ( )me

MX tends to unity

4 At very high speeds ie at very high frequency ratio

ωω

n there is no effect

of damping The frequency plot of phase angle versus the frequency ratio is given as shown in figure The explanation is same as in forced harmonic vibration

58

VIBRATION ISOLATION

Vibratory forces and motions generated by machine and other caused are unavoidable Howeever their effect on the other dynamical systems can be minimized by proper isolator design An isolation system is either to protect o delicate machine from excessive vibration transmitted to it from its supporting structure or to prevent vibratory forces and motions generated by machines from being transmitted to its surroundings The effectiveness of isolation may be measured in terms of forces and motion transmitted The first type is known as force isolation and the second type is known as motion isolation

Force isolation (Force Transmissibility) When vibratory forces generated by machine are isolated the effectiveness of forced isolation can be made by defining the term transmissibility The term transmissibility is defined as the ratio of force transmitted to the foundation to that of the external impressed force

Force Transmissibility = o

tr

FF

force Impressed Externalfoundation toted transmitForce

=

Consider viscous damped spring mass system as shown in figure Upon this system an external harmonic force ) t (Sin Fo ω is applied The difference forces acting on system are inertia forces damping force spring force and external force All these forces are represented on force polygon shown below Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these two forces

( ) ( )22 xckx ω+=

59

This is the expression for force transmissibility in dimensional form To represent same in the non dimensional form divide numerator and denominator by k

2

n

22

n

2

n

o

tr

21

21

FF

TR

ωω

ξ+

ωω

minus

ωω

ξ+

==

PHASE DIFFERNECE BETWEEN EXTERNAL FORCE AND DISPLACEMENT Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these

ωω

minus

ωω

ξ=

ωminusω

=φ minusminus2

n

n1

2

1

1

2tan

mKC

tan

The phase difference between force transmitted to the foundation Ftr to the displacement

kc

KxxC

tanω

=ω

=α or

ωω

ξ=

ω=α minusminus

n

11 2tanKx

xCtan

The phase difference between external force Fo and force transmitted to foundationis β

60

β= αminusφ =

ωω

minus

ωω

ξminus

2

n

n1

1

2tan -

ωω

ξminus

n

1 2tan

Displacement Transmissibility It is the ratio of amplitude of steady state vibration of machine to that of the base motion Forced vibration due to support motion Consider a spring mass damper system subjected to harmonic excitation )tsin(Yy ω= at the base as shown in figure Let x be the absolute displacement of the mass at any instant of time t The different forces acting on system are shown in free body diagram Applying Newtonrsquos Law of motion Ma = sumF

)yx(k)yx(cxm minusminusminusminus= ampampampamp

0)yx(k)yx(cxm =minus+minus+ ampampampamp

61

Frequency plot of transmissibility versus frequency ratio

62

The plot of transmissibility versus frequency ratio is as shown in figure The following points can be noted from it

i) The value of transmissibility is unity at frequency ratio

ωω

n=0 and 2

ii) For an undamped system ( )ξ =0 transmissibility tends to infin at resonance iii) The value of transmissibility is less than unity for the value of frequency ratio

greater than 2

iv) For frequency ratio

ωω

nlt 2 smaller damping ratio lead to larger values of

transmissibility on the other hand for frequency ratio

ωω

ngt 2 smaller damping

ratio leads to smaller value of transmissibility

The frequency plot of transmissibility (TR) versus frequency ratio

ωω

n can be

broadly divided into two zones

1

ωω

n= 0 to 2

2

ωω

n= 2 to infin

CASE 1

When

ωω

n= 0 to 2

In this zone the value of transmissibility is always greater than 1 for all values of damping ratio In this zone the transmissibility decreases with increase in damping Hence we can conclude that in this zone damping is useful in isolating vibration forces being transmitted to the support CASE 2

When

ωω

n= 2 to infin

In this zone the value of transmissibility is less than 1 for all values of damping ratio In this zone the transmissibility increases with increase in damping Hence we can conclude that in this zone damping will adversely affect in isolating vibratory forces

63

Tutorial problems on Forced Vibration 1 Machine of mass 75kg is mounted on spring of stiffness 12000 Nm with an assumed damping factor ξ =02 piston within the machine of mass 2 kg has a reciprocating motion with a stroke of 75 cm and speed of 15 cyclessecond Assuming the motion of piston to be harmonic determine amplitude of machine ( X) Transmissibility and force transmitted to foundation phase angle of transmitted force wrt to exciting force 2 Motor weighs 600N is mounted on a simple beam that has a spring rate of 400 Ncm The motor armature base 160 N and has an eccentricity of 0005cm What will be the amplitude of motion when it runs at 1760 RPM Neglect damping the deflection of armature shaft and weight of the beam 3 A Heavy machine of mass 300 kg is supported on resilient foundation the static deflection of foundation due to weight of machine is found to be 75mm It is observed that the machine vibrates with an amplitude of 10mm when the base of the foundation is subjected to harmonic oscillation at the undampeds natural frequency of system with an amplitude of 25mm Find the damping constant of foundation Ii) dynamic force amplitude on the base iii) amplitude of the displacement of machine relative to base 4 An air compressor of mass 450 kg operates at a constant speed of 1750 RPM The reciprocating masses are 10 kg the crank radius is 100mm If the damper for mounting introduces a damping factor 015 i) specify the springs of mounting such that only 20 of the unbalanced force is transmitted to the foundation ii) determine the magnitude of the transmitted force bull Repeat the above problem by assuming that there is no damping in system

5 A body of a car having mass 1500 kg and is mounted on 4 equal springs which deflect through 225 cm under the weight of the body A total damping forces the 4 shock absorbers is 46 N at a velocity of 1cmsec The car is placed with all the four wheels on a test platform which is move up and down at resonant speed With an amplitude of 25cm Find the amplitude of the car body on its springs assuming its CG to be in the centre of wheel base 6A 75 kg machine which is mounted on springs of stiffness 116x105 Nm with the damping of 20=ξ A 2kg piston within the machine as reciprocating motion with a stroke of 008m and speed of 3000 RPM Assuming the motion of piston to be harmonic determine the amplitude of vibration of machine

7 The spring of an automobile are compressed 10cm under its own weight Find the critical speed when a trailer is traveling over road with a people approximaterly by a sine wave of amplitude 75 cm and wave length 15m What will be the amplitude of vibration at 64kmhour 8 An eccentric mass excites us used to determine vibratory characteristics of a structure of mass 25kg as given in figure At a speed of 1000 RPM the eccentric mass to beat top position at the instant the structure was moved upwards through its static equilibrium position and corresponding amplitude was 20mm The eccentric mass is 05kg at an eccentricity of 10mm Determine i) Undamped natural frequency of the system ii) Damping ratio iii) Steady state amplitude and phase angle at 1500 RPM iv) Force transmitted to the ground at 1200 RPM

M=25kg

C

K

Eccentric mass

64

65

CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Support motion Consider a spring mass damper system as shown in figure Let it excited by motion

of support let y be the harmonic displacement of support From Newtonrsquos second law of motion

)yx(k)yx(cxm minusminusminusminus= ampampampamp

)1(0)yx(k)yx(cxm minusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminus=minus+minus+ ampampampamp

Let z be the relative displacement of the mass with respect to support then

Z=x-y

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

15

Equivalent translational masses

Kinetic energy of the equivalent mass =

2

21

eqeq XM amp

Kinetic energy of the two masses =

+

20

2

21

21

θθθθampamp JXM

Meq= m

+ 2RJ

m o is the required answer

Also determine the equivalent rotational mass Jeq

Jeq= m [ ]oJmR +2 is the required answer Spring element Whenever there is a relative motion between the two ends of the spring a force is developed called spring force or restoring force The spring force is proportional to the amount of deformation x and then F α x or F = kx Where k is stiffness of the spring or spring constant or spring gradient The spring stiffness is equal to spring force per unit deromation

The spring stiffness k = mNxF

Workdone in deforming a spring is equal to the strain energy or potential energy Strain energy = potential energy = area of the triangle OAB Stiffness of beams Cantilever beam consider a cantilever beam with an end mass shown in the figure The mass of the beam is assumed to be negligible The static deflection of beam at free end is given by

Similarly derive the expression for Simply supported beam and fixed support beam

x

k

m

x

m

x

m

x

mNlIE

st 192

3

=δδδδ

mNEI

Wlst

48

3

=δδδδ

F

K=stiffness

F

16

17

Stiffness of slender bar subjected to longitudinal vibrations For a system executing the longitudinal vibrations as shown in the figure let us derive the expression for stiffness

Torsional Stiffness of bar It is the amount of torque required to cause a unit angular deformation in the element

Torsional stiffness = Kt = θθθθT

Combination of stiffness Determination of equivalent spring stiffness when the springs are arranged in series Consider two springs of stiffness K1 and K2 acted upon by the force F

The deflection of spring k1 is x1 = 1KF

The deflection of spring k1 is x2 = Let these two springs be replaced by an equivalent stiffness Keq upon which a force F acts and due to which its deflection is given by

x = eqKF

x= x1+x2

m

E A

l

60 cm 70 cm

Determination of equivalent spring stiffness when the springs are arranged in parallel Force acting on K1 spring = F1=k1x Force acting on K2 spring = F2 Force required for an equivalent spring keq to defined by x given by F= Keq x But F = F1 +F2 Tutorial problems on Equivalent stiffness of springs 1Determine the equivalent stiffness for the system shown in figure 2 Determine the equivalent stiffness for the system shown in figure

3 Determine the equivale

M

3

M

Nm 102 6x

18

nt stiffness for the system shown in figure

38 kg

Nm 101 6x Nm 102 6x

Nm 106x

2k

k

k

2k

3k

k

2k

k

19

60 cm 80 cm 50 cm

50 cm 60 cm

4Determine the equivalent stiffness for the system shown in figure

5Replace the following torsional stiffness by a single shaft having radius 4cm and find the length required for the equivalent shaft Assume the material of given system and equivalent system is same R1= 3cm R2= 5cm DAMPING Every vibration energy is gradually converted into heat or sound Hence the displacement during vibration gradually reduces The mechanism by which vibration energy is gradually converted into heat or sound is known as damping A damper is assumed to have either mass or elasticity Hence damping is modeled as one or more of the following types Viscous damping Coulomb or dry friction damping materials or solid or hysteric damping Viscous damping Viscous damping is most commonly used damping mechanism in vibration analysis When the mechanical system vibrates in a fluid medium such as air gas water or oil the resistance offered by the fluid to the moving body causes energy to be dissipated In this case the amount of dissipated energy depends on many factors such as size or shape of the vibrating body the viscosity of the fluid the frequency of vibration and velocity of fluid Resistance due to viscous damping is directly proportional to the velocity of vibration

dF V αααα

dF = xC amp

Where C= damping coefficient Fd = damping force Examples of Viscous damping

1) Fluid film between sliding surface 2) Fluid flow around a piston in a cylinder 3) Fluid flow through orifice 4) Fluid flow around a journal in a bearing

Rreq =4cm

leqn

stress

Strain

Coulomb damping or dry friction damping Here a damping force is constant in magnitude but opposite in direction to that of the motion of vibrating body It is caused by the friction between the surfaces that are dry or have insufficient lubrication Material or solid or hysteric damping

When the materials are deformed energy is absorbed and dissipated by the material The effect is due to friction between the internal planes which slip or slide as the deformation takes place When a body having the material damping is subjected to vibration the stress strain diagram shows the hysteresis loop as shown in the figure The area of the loop denotes the energy lost per unit volume of the body per cycle

20

21

ττττ ττττ2 ττττ3

ττττ

X(t)

FOURIER SERIES The simplest of periodic motion happens to be SHM It is simple to handle but the motion of many vibrating system is not harmonic (but periodic) Few examples are shown below

Forces acting on machines are generally periodic but this may not be harmonic for example the excitation force in a punching machine is periodic and it can be represented as shown in figure 13 Vibration analysis of system subjected to periodic but nonharmonic forces can be done with the help of Fourier series The problem becomes a multifrequency excitation problem The principle of linear superposition is applied and the total response is the sum of the response due to each of the individual frequency term

Any periodic motion can be expressed as an infinite sum of sines and cosines terms If x(t) is a periodic function with period t its Fourier representation is given by

X(t) = (((( )))) (((( )))) (((( ))))t sint cost cos2 111 ωωωωωωωωωωωω baaao ++++++++++++

= (((( )))) (((( ))))t sint cos2 1

ωωωωωωωω nnn

o bnaa

++++++++sumsumsumsuminfininfininfininfin

====

tππππωωωω

2==== = Fundamental frequency ndash (1)

where ao an bn are constants

ττττ ττττ2 ττττ3

ττττ

X(t)

22

t

X(t)

035

025

Determination of constants To find ao Integrate both sides of equation(1) over any interval ττττ All intergrals on the RHS of the equation are zero except the one containing ao

(((( ))))dttxao

o 2

2

intintintint====ωωωωππππ

ππππωωωω

= (((( ))))dttxo

2intintintintττττ

ττττ

To find an multiply equation 1 by cos (((( ))))tnωωωω and Integrate over any interval ττττ All intergrals

(((( )))) (((( )))) dttxao

n tncos 2

2

ωωωωππππωωωω ωωωω

ππππ

intintintint==== (((( )))) (((( )))) dttxo

tncos 2

ωωωωττττ

ττττ

intintintint====

To find bn multiply equation 1 by sin (((( ))))tnωωωω and Integrate over any interval ττττ All intergrals

(((( )))) (((( )))) dttxao

n tnsin 2

2

ωωωωππππωωωω ωωωω

ππππ

intintintint==== (((( )))) (((( )))) dttxo

tnsin 2

ωωωωττττ

ττττ

intintintint====

Find the Fourier series for the curve shown below

ττττ ττττ2 ττττ3

t

X(t)

E

ττττ ττττ2 ττττ3

t

X(t)

ππππ2

t

X(t) ππππ

1 Represent for the periodic motion shown in the figure

23

CHAPTER 2 UNDAMPED FREE VIBRATION LEARNING OBJECTIVES

] Introduction to undamped free vibration

] Terminologies used in undamped free vibration

] Three methods to solve the undamped free vibration problems

] Problems related to undamped free vibration problems

Free vibrations are oscillations about a systems equilibrium position that occur in the absence of an external excitation force If during vibrations there is no loss of energy it is known as undamped vibration The first step in solving a vibration problem is setting up the differential equation of motion Consider spring mass system which is assumed to move only along the vertical direction as shown below Let m be the mass of the block and k be the stiffness of the spring When block of mass m is attached to spring the deflection of spring will be ∆ known as static deflection In the static equilibrium position the free body diagram of forces acting on the mass is shown in Figure(b) Hence mg= kA Once the system is disturbed the system executes vibrations

Let at any instant of time t the mass is displaced from the equilibrium position x the different forces acting on the system are shown in figure (d) From Newtonrsquos second law of motion

sum = maF

Inertia force ( disturbing force) = restoring force

mgxkxm ++∆minus= )(ampamp

0)( =+ xkxm ampamp

or 0)( =+ xmk

xampamp

equation 2 is the differential equation of motion for spring mfigure Comparing equation (2) with the equation of SHM +xampampsince the vibrations of the above system are free( without the forces) we can write

sec radmk

n =ω

24

ass system shown in 0)(2 =xω

resistance of external

time period km

2 f1

n

πτ ==

from the equation(1) mg = k∆

∆=

q

mk

Difference between the translation ( rectilinear) and rotational system of vibration

Translatory Rotational

In the analysis thFORCES are co Linear stiffness K Problems 1A mass of 10kFind the natural 2 A spring massnatural frequencnatural frequenc

kt

m

sec radmk

n =ω

k

25

e disturbing and restoring nsidered

In the analysis In the analysis MASS Moment of Inertia (J) is considered

in Nm is considered

g when suspended from a spring causes a static deflection of 1cm frequency of system

system has a spring stiffness K Nm and a mass of m Kg It has a y of vibration 12 Hz An extra 2kg mass coupled to it then the y reduces by 2 Hz find K and m

secrad n =ω

26

3 A steel wire of 2mm diameter and 30m long It is fixed at the upper end and carries a mass of m kg at its free end Find m so that the frequency of longitudinal vibration is 4 Hz 4 A spring mass system has a natural period of 02 seconds What will be the new period if the spring constant is 1) increased by 50 2) decreased by 50 5 A spring mass system has a natural frequency of 10 Hz when the spring constant is reduced by 800 Nm the frequency is altered by 45 Find the mass and spring constant of the original system 6 Determine the natural frequency of system shown in fig is which shaft is supported in SHORT bearings 7 Determine the natural frequency of system shown in fig is which shaft is supported in LONG bearings where l is the length of bearing and E ndash youngrsquos modulus and I is moment of Inertia 7 Determine the natural frequency of system shown in fig is which shaft is supported in LONG bearings 8 A light cantilever beam of rectangular section( 5 cm deep x 25cm wide) has a mass fixed at its free end Find the ratio of frequency of free lateral vibration in vertical plane to that in horizontal 9 Determine the natural frequency of simple pendulum 10 Homogeneous square plate of size l and mass m is suspended from the mid point of one of its sides as shown in figure Find the frequency of vibration

l

l

27

11 A compound pendulum which is rigid body of mass m and it is pivoted at O The point of pivot is at distance d from the centre of gravity It is free to rotate about its axis Find the frequency of oscillation of such pendulum 12 A connecting rod shown in fig is supported at the wrist pin end It is displaced and allowed to oscillate The mass of rod is 5kg and centre of gravity is 20 cm from the pivot point O If the frequency of oscillation is 40 cyclesminute calculate the mass moment of inertia about its CG

13 A semi circular homogenous disc of radius r and mass m is pivoted freely about its centre as shown in figure Determine the natural frequency of oscillation 14A simply supported beam of square cross section 5mmx5mm and length 1m carrying a mass of 0575 kg at the middle is found to have natural frequency of 30 radsec Determine youngrsquos modulus of elasticity of beam 15 A spring mass system k1 and m have a natural frequency f1 Determine the value of k2 of another spring in terms of k1 which when placed in series with k1 lowers the natural frequency to 23 f1

COMPLETE SOLUTION OF SYSTEM EXECUTING SHM

The equation of motion of system executing SHM can be represented by

0)( =+ xkxm ampamp ----------------------------------------(1)

022

2=+

dtdx

dt

dx ω

The general solution of equation (1) can be expressed as

------------------------(2)

Where A and B are arbitrary constant which can be determined from the initial conditions of the system Two initial conditions are to be specified to evaluate these

constants x=x0 at t=0 and xamp = Vo at t=0 substituting in the equation (2) Energy method In a conservative system the total energy is constant The dwell as natural frequency can be determined by the princienergy For free vibration of undamped system at any instant and partly potential The kinetic energy T is stored in the masswhere as the potential energy U is stored in the form of sdeformation or work done in a force field such as gravity The total energy being constant T+U = constant Its rate of cha

t)(sin Bt)cos( AX ω+ω=

t)(sin V

t)cos( xx o0 ω

ω+ω=

Is the required complete

solution

28

ifferential equation as ple of conservation of of time is partly kinetic by virtue of its velocity train energy in elastic

nge

Is given by [ ] 0UTdtd

=+

From this we get a differential equation of motion as well as natural frequency of the system Determine the natural frequency of spring mass system using energy method

Determine the natural frequency of system shown in figure Determine the natural frequency of the system shown in figure Is there any limitation on the value of K Discuss Determine the natural frequency of s

m

l k

m

a

θ

m

a

θ

l

k

29

k

k

ystem shown below Neglect the mass of ball

l

m

30

A string shown in figure is under tension T which can be assumed to remain constant for small displacements Find the natural frequency of vertical vibrations of spring An acrobat 120kg walks on a tight rope as shown in figure The frequency of vibration in the given position is vertical direction is 30 rads Find the tension in the rope A manometer has a uniform bore of cross section area A If the column of liquid of length L and Density ρ is set into motion as shown in figure Find the frequency of the resulting oscillation

m

l

a

T T

36m 8m

31

k

R

k

m

r

m

Find the expression for natural frequency of system shown in the figure Neglect the mass of the cantilever beam Study the special case i) k=Infinity ii) I = infinity Determine the expression for the natural frequency of a system shown in figure The two discs are keyed to a common shaft and have a combined mass moment of inertia about centre of oscillation O It is assumed that the card attached to mass m does not stretch and is always under tension Determine the natural frequency of a system shown

m

l

32

Determine the expression for the natural frequency of the system shown in figure Assume that the wires connecting the masses do not stretch and are always in tension

Determine the natural frequency of spring mass system taking the MASS OF SPRING (ms ) into account

M3 M1

M2

k2 k1

m

k

33

RAYLEIGHrsquoS METHOD

This is the extension of energy method Here we can determine the natural frequency of a conservative system without finding the equation of motion The natural frequency is a function of the rate of change of Kinetic energy and potential energy of the system From the principle of conservation of energy we have T+U = Constant T= KE and U =PE We can write the above equation as T1+U1 = --------------------------(1) Where 1 and 2 represents tow instances of time Let 1 be the time when the mass is passing through static equilibrium position Let 2 be the time corresponding to the mass displacement of the mass At this instant the velocity f the mass is zero and hence Substituting in equation (1) we get ( )Max = ( )Max Above equation leads directly to natural frequency Determine the natural frequency of spring mass system using Rayleighrsquos method m

34

r

B

K K

Home work

Determine the natural frequency of pendulum using Rayleighrsquos method

A Cylinder of mass m and moment of inertia Jo is to role without slipping but is sustained by spring K as shown in figure Determine the natural frequency of oscillation

Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping

a

Determine the natural frequency of system shown where in cylindrical disc rolls over the inextensible spring A cylinder of radius r rolls without slipping on cylindrical the equation of motion and hence determine the naoscillations about the lowest point shown Repeat the above problem ndash Instead of cylinder assume swithout slipping on concave surface as shown above

r

K

35

portion of radius R Derive tural frequency of small

phere of radius with r rolls

The torsional pendulum has a natural frequency of 5 Hz What length of steel wire of diameter 2 mm should be used for pendulum The inertia of mass fixed at the free end is 00098 kgm2 Take g= 083x1011 Nm2 Determine the torsional natural frequency of the system shown in figure Neglect the mass moment of inertia of the shaft ( Figure not proportionate) Determine the natural frequency of simple pendulum considering the mass of the rod into account

Using the energy method find tThe chord may be assumed ine

Kt1

Kt2

20cm

06m 08m 04m

20cm 20cm 20cm

m

M

he natural frxtensible in

k

M

36

equency of the system shown in figure the spring mass pulley system

A rod of length l and depth H having a mass m rests on a semicylidrical surface os radius r It is tipped slightly and let go Find the frequency of oscillation

A thin bar of l having a mass m rests on a semicylidrical surface radius r It is tipped slightly and let go Find the frequency of oscillation

l

h

R

l

37

38

CHAPTER 3

DAMPED FREE VIBRATION OF SINGLE DEGREE FREEDOM SYSTEM LEARNING OBJECTIVES

] Introduction to damped free vibration

] Terminologies used in damped free vibration

] Three different types of damping

] Concept of critical damping and its importance

] Study of response of viscous damped system for cases of under damping

critical and over damping

] Concept of logarithmic decrement

] Problems related to damped free vibration problems

Viscously damped free vibration of single degree freedom sytem Consider a single degree freedom system with viscous damping as shown in figure For a viscous damped system the damping force is

where c is a constant of proportionality It is called damping coefficient or coefficient of viscous damping

Symbolically it is designated by a dashpot as shown in Fig 3

x xampx

39

At any instant of time let the mass be displaced by x from the equilibrium position Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

Okxxcxm =minusminus= ampampamp

Okxxcxm =++ ampampamp ----------------------------------------------------------(1)

or

0kxdtdx

Cdt

xdM

2

2

=++

Assuming the solution of equation as

The first term in the above equation m2Ct

eminus

is an exponentially decaying function of time The behavior of the terms in the parenthesis however depends on whether numerical value within the radicals is positive zero or negative

CASE 1 if 0mk

m2c

2

=minus

40

CASE 2 if 0mk

m2c

2

=gt

CASE 3 if 0mk

m2c

2

=lt

41

Critically damped coefficient Critically damped coefficient is defined as the value of the damping coeffient for which radical in equation

mk

m2c

m2c

S2

21 minus

plusmn

= becomes zero

Damping Ratio ( Damping factor) ( ξ ) It is defined as the ratio of damping coefficient to the critical damping coefficient Note The damping coefficient c is defined as the actual amount of damping while the damping coefficient is a constant depending upon the mass and stiffness of the system The roots of characteristics equation can be now written as Note if c=cc or ξ =1 then system is critically damped if cgt cc or ξ gt1 then system is over damped I if clt cc or ξ lt1 then system is under damped

42

Take home quiz Derive General solution of different damped system

1 Critically damped system 2 Under damped system 3 Over damped system

Logarithmic Decrement It is used to determine the amount of damping present in system This measure The rate of decay of free vibration The larger the damping the greater will be rate of decay Logarithmic decrement is defined as the natural logarithm of ratio of any two successive amplitudes

=δ

2

1

xx

ln

Expression for logarithmic decrement (δ ) Consider a damped free vibration which is expressed by

( ) ( ) φ+ξminusω= ξω t 1sinXex 2n

tn

x is the displacement at any instant of time X and φ are arbitrary constants

21

2

ξminus

πξ=δ

43

Tutorial problems 1 Derive an expression for damping ratio in terms of logarithmic decrement

2 Show that logarithmic decrement

=δ

n

o

xx

lnn1

where xo is the initial amplitude xn

is amplitude after n cycles 3 A mass of 1 kg is to be supported on a spring having stiffness K= 9800 Nm The damping constant c= 59 Nsm Determine the natural frequency of system Find also logarithmic decrement and amplitude after 3 cycles if initial displacement is 03 cm 4 A vibrating system is defined by the following parameters M = 3kg k= 100 Nm damping coefficient = 3 Nsm determine Damping factor natural frequency damped vibration logarithmic decrement ratio of consecutive amplitude no of cycles after which the original amplitude is reduced to 20 5A single degree damped vibrating system a suspended mass of 18kg makes 15 oscillations in 003 seconds The amplitude decreases to 025 of the initial value after 5 oscillations Determine the stiffness of spring Logarithmic decrement damping factor and damping coefficient 6 A damped vibration record of a spring mass system shows the following data Amplitude at the end of 2nd cycle is 9mm amplitude at eh end of 3rd cycle is 6mm amplitude at the end of 4th cycle is 4mm stiffness of spring =8000 Nm weight =20N Find the logarithmic decrement damping force at unit velocity periodic time of vibration 7 For the system shown in figure the characteristics of dashpot is such that the constant force of 49 N is applied to the piston it velocity is found to be constant at 012 ms Determine the value of C would you expect the complete system to be periodic or aperiodic 8 For a torsional system shown in the figuref1 is natural frequency in air and f2 is the natural frequency when immersed in oil Determine the damping coefficient

Viscous fluid

Disc of mass moment of Inertia J

2a

C

m M

b

a a

k

9 A thin plate of area A and weigh w is fixed to the end of the spring and allow to oscillate in a viscous fluid as shown in figure It f1 is the frequency of oscillation in air and f2 in liquid Show that

22

21 ff

gA2

minusπω

=micro where the damping force on the plate

AV2Fd micro= V being the velocity of vibration 10 A torsional pendulum has a natural frequency of 175 cyclessecond vibrating in vacuum The mass moment of inertia is 2 kgm2 It is immersed in oil and it is observed that the natural frequency is 142 cyclessecond Find the damping torque If the disc is replaced by 34 deg When in oil find the displacement at the end of the first complete cycle 11 An automobile can be modeled as a mass placed on 4 shock absorbers each consisting of a spring and a damper Such that each spring is equally loaded Determine the stiffness damping constant of each shock absorber So that natural frequency is 2Hz and system is critically damped The mass of vehicle is 200kg 12 Write the differential equation of motion for the system shown in the figure below Determine the natural frequency of the damped oscillation and critical damping coefficient

Determine the differential equation of motion and find the critical cdamped natural frequency for system shown in figure

a

C

m M

M =200kg

44

200kg kgkg

oefficient and

45

Derive the equation of motion and find the critical damping coefficient and damped natural frequency for the system shown below Write the differential equation of motion and determine the expression for critical damping coefficient and damped natural frequency of system shown in figure

Set up a differential equation of motion for system shown in figure and determine the following Undamped natural frequencies critical damping coefficient damping ratio damped natural frequency

b

C

kM

a

m

46

A mass of 300N is resting on two springs of stiffness 3000 Nm each and a dashpot of damping coefficient 150 Nsm as shown in figure If the initial velocity of 10cms is given to mass at its equilibrium position what will be the displacement from the equilibrium position at the end of first second Derive the formula used A gun barrel of mass 600kg has a recoil spring of stiffness 294 knm if the barrel recoils 13m on firing determine i) The initial recoil velocity of barrel ii) The critical damping coefficient of dashpot which is engaged at the end of recoil stroke Iii) Time required for the barrel to return at the end of recoil stroke Iv) time required for the barrel to return to a position 5cm from the initial position

W= 3000 N

FIRING

CHAPTER 4

FORCED VIBRATION LEARNING OBJECTIVES

] Introduction to forced free vibration

] Terminologies used in forced vibration

] Concept of magnification factor

] Study of reciprocating and rotating unbalance

] Concept of base excitation and support excitation

] Concept of Transmissibility ratio

] Problems related to forced vibration

The vibration that takes place under the excitation of external forces are called

forced vibration When a system is subjected to harmonic excitation it is forced to vibrate at the same frequency as that of excitation Most of the mechanical systems are prone to steady state forced vibration which cause fatigue failure The vibration problem also becomes very important when any of excitation frequencies coincide with one of natural frequencies of a system Under this condition resonance occurs during which large amplitude of vibrations are observed To avoid resonance external frequency (operating speed) may be changed or properties of the system may be changed to alter the natural frequency In some cases sufficient amount of damping may be provided to avoid large amplitude during resonance Thus the problem of force vibration is very important in mechanical design

FORCED VIBRATIONS OF SINGLE DOF SYSTEMS UNDER HARMONIC EXCITATION

` [ ] tsin FF o ω=

x

sin FF o=

Fo

xampamp

[ ] tω

xamp

47

48

Consider a spring mass system having a viscous damping excited by a harmonic force [ ] tsin FF o ω= as shown in figure The various forces acting on the system is shown in figure Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

O)tsin(Fkxxcxm o =ω+minusminus= ampampamp

)tsin(Fkxxcxm o ω=++ ampampamp ----------------------------------------------------------(1)

or

)tsin(Fkxdtdx

Cdt

xdM o2

2

ω=++

Equation (1) represents the differential equation of motion of the system The solution to this equation consists of two parts namely complementary function and a particular integral

The first of solution is also termed as transient solution of homogeneous solution and can be obtained from the homogeneous differential equation 0kxxcxm =++ ampampamp

The second part of solution particular integral is also termed as steady state solution ad is of form )tsin(Xx φminusω= Where X= steady state amplitude φ is phase difference between external force and displacement

)tsin(Xx φminusω=

The vector representation of the shown

49

The force polygon for the above vector diagram can be written as follows

( )φminusωt O

KX

xm 2ω

Fo

xCω

tω

φ

50

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

is the governing equation

Zero frequency deflection (Xst) It is deflection observed at zero frequency ( when the external frequency 0=ω

( ) kF

001

kF

x o

22

o

st =

+minus=

Magnification factor (M) It is defined as the ratio of steady state amplitude to zero frequency deflection xst

M = arX

X=

A plot of magnification factor versus frequency ratio

damping factor ( )ξ is called frequency plot of a harmon All the frequency terms starts from magnification factor of damping ratio ( )ξ The amplitude of vibration at any on the value of damping The amplitude will be higher for higher damping The amplitude of vibration at reso

amplitude tends to zero as

ωω

n tends to infinity

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

Freq

ζ =

ζ = 1

Magnification Factor M = XXst

kFo

51

ωω

nfor varying values of

ically excite vibrating system

M=1 what ever may be value particular frequency depends

for lesser damping and lesser nance will be very high The

uency Ratio r = (ωωn)

ζ = 0 0

ζ = 025 ζ = 0375 ζ = 0 5

ζ = 0707

ζ = 2

52

Plot of phase angle φ versus frequency ratio

ωω

n is shown in the

figure The phase angle φ is zero for all the values of damping ratio

at

ωω

n=0 The phase

angle φ is 90 for all the values of damping ratio

at

ωω

n=1 ie at

resonance The phase angle between o to 90 deg for all the values of

ωω

n= 0 to 1 and

between 90 deg to 180 deg for all the values of

ωω

ngt1

Class work Derive an expression for frequency at peak amplitude

2np 21 ξminusω=ω=ω is the required equation

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

53

Derive expression amplitude at resonance ( In dimensional form) In dimensional form We know that steady state amplitude

( ) ( )

ω+ωminus=

222

o

Cmk

Fx Equation in dimensional form

We know that at resonance mk

n =ω=ω

( )

ω+

minus

=2

n

2

oR

Cmk

mk

Fx

( )

ω=

2n

oR

C

Fx

( )

ω=

n

oR C

Fx

Class work Derive an expression for amplitude at resonance ( in non dimensional form)

Tutorial Problems on Forced Vibration 1A weight of 60N is suspended by a spring stiffness 12 kNm is forced to vibrate by a harmonic force of 10N assuming viscous damping of 0086 kN-sm Find i) resonance frequency ii) amplitude at resonance iii) phase angle at resonance iv) frequency corresponding to peak amplitude 2 A periodic torque having a maximum value of 05888Nm at a frequency of 4 rads is impressed upon a flywheel suspended from a wire The wheel has mass moment of inertia 012 kgm2 and the wire has stiffness 1176 Nmrad A viscous dashpot applies damping couple of 0784 Nm an angular velocity of 2 rads calculate i)maximum angular displacement from rest position ii) maximum couple applied to dashpot Iii) angle by which the angular displacement lags the torque A mass of 1019 kg is suspended from a spring stiffness 20 Nmm viscous damping causes the amplitude to decrease to 110th of initial value in 4 complete cycle If a periodic force of 20 cos(50 t) is applied to the mass Find the amplitude of forced vibration What would be the amplitude of oscillation if the applied force coincides with the natural period

m=1019 kg

[ ] t50cos 30F =

54

55

ROTATING UNBALANCE Unbalance in rotating machinery is a common source of vibration excitation Consider a machine of mass m having mounted on a foundation of stiffness k and damping coefficient C as shown in figure The unbalance is represented by an eccentric rotating mass m with eccentricity (e) and having a angular velocity (ω) let x be the displacement of non rotating mass from the static equilibrium position The displacement of rotating mass m from the static equilibrium is ( ) tsin( ex ω+ Applying Newtonrsquos law of motion we have

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

56

RECIPROCATING UNBALANCE Unbalance in reciprocating machinery is a common source of vibration excitation Consider a machine of mass M having mounted on a foundation of stiffness K and damping coefficient C as shown in figure The unbalance is represented by a reciprocating mass (m) having crank radus (e) and connecting rod of length (l) Let ωωωω be angular velocity of the crank Let x be the displacement of non reciprocating mass from the static equilibrium position at any instant of time The displacement of reciprocating mass m from the static equilibrium position is

)2sin(sin +++ tle

tex ωωωωωωωω

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

57

Note The equations obtained in reciprocating unbalance is same as the rotating unbalance and having the following differences Rotating Unbalance Reciprocating Unbalance Here mass of the rotating element is considered

Here mass of reciprocating element is considered

Here the eccentricity (e) which is the distance between the centre of rotation and CG

Here the eccentricity e is the crank radius and crank radius = stroke length2

The frequency plot of non dimensional factor( ( )me

MX versus frequency ratio

ωω

n for various values of ( )ξξξξ is shown in figure This frequency plot is same for

the rotating and reciprocating unbalance with the following points are noted from the diagram

1 Damping ratio plays very important role during resonance in controlling the amplitude

2 When

ωω

n=0 then ( )me

MX =0 for all the values of damping ratio ( )ξξξξ

3 When

ωω

nis higher than ( )me

MX tends to unity

4 At very high speeds ie at very high frequency ratio

ωω

n there is no effect

of damping The frequency plot of phase angle versus the frequency ratio is given as shown in figure The explanation is same as in forced harmonic vibration

58

VIBRATION ISOLATION

Vibratory forces and motions generated by machine and other caused are unavoidable Howeever their effect on the other dynamical systems can be minimized by proper isolator design An isolation system is either to protect o delicate machine from excessive vibration transmitted to it from its supporting structure or to prevent vibratory forces and motions generated by machines from being transmitted to its surroundings The effectiveness of isolation may be measured in terms of forces and motion transmitted The first type is known as force isolation and the second type is known as motion isolation

Force isolation (Force Transmissibility) When vibratory forces generated by machine are isolated the effectiveness of forced isolation can be made by defining the term transmissibility The term transmissibility is defined as the ratio of force transmitted to the foundation to that of the external impressed force

Force Transmissibility = o

tr

FF

force Impressed Externalfoundation toted transmitForce

=

Consider viscous damped spring mass system as shown in figure Upon this system an external harmonic force ) t (Sin Fo ω is applied The difference forces acting on system are inertia forces damping force spring force and external force All these forces are represented on force polygon shown below Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these two forces

( ) ( )22 xckx ω+=

59

This is the expression for force transmissibility in dimensional form To represent same in the non dimensional form divide numerator and denominator by k

2

n

22

n

2

n

o

tr

21

21

FF

TR

ωω

ξ+

ωω

minus

ωω

ξ+

==

PHASE DIFFERNECE BETWEEN EXTERNAL FORCE AND DISPLACEMENT Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these

ωω

minus

ωω

ξ=

ωminusω

=φ minusminus2

n

n1

2

1

1

2tan

mKC

tan

The phase difference between force transmitted to the foundation Ftr to the displacement

kc

KxxC

tanω

=ω

=α or

ωω

ξ=

ω=α minusminus

n

11 2tanKx

xCtan

The phase difference between external force Fo and force transmitted to foundationis β

60

β= αminusφ =

ωω

minus

ωω

ξminus

2

n

n1

1

2tan -

ωω

ξminus

n

1 2tan

Displacement Transmissibility It is the ratio of amplitude of steady state vibration of machine to that of the base motion Forced vibration due to support motion Consider a spring mass damper system subjected to harmonic excitation )tsin(Yy ω= at the base as shown in figure Let x be the absolute displacement of the mass at any instant of time t The different forces acting on system are shown in free body diagram Applying Newtonrsquos Law of motion Ma = sumF

)yx(k)yx(cxm minusminusminusminus= ampampampamp

0)yx(k)yx(cxm =minus+minus+ ampampampamp

61

Frequency plot of transmissibility versus frequency ratio

62

The plot of transmissibility versus frequency ratio is as shown in figure The following points can be noted from it

i) The value of transmissibility is unity at frequency ratio

ωω

n=0 and 2

ii) For an undamped system ( )ξ =0 transmissibility tends to infin at resonance iii) The value of transmissibility is less than unity for the value of frequency ratio

greater than 2

iv) For frequency ratio

ωω

nlt 2 smaller damping ratio lead to larger values of

transmissibility on the other hand for frequency ratio

ωω

ngt 2 smaller damping

ratio leads to smaller value of transmissibility

The frequency plot of transmissibility (TR) versus frequency ratio

ωω

n can be

broadly divided into two zones

1

ωω

n= 0 to 2

2

ωω

n= 2 to infin

CASE 1

When

ωω

n= 0 to 2

In this zone the value of transmissibility is always greater than 1 for all values of damping ratio In this zone the transmissibility decreases with increase in damping Hence we can conclude that in this zone damping is useful in isolating vibration forces being transmitted to the support CASE 2

When

ωω

n= 2 to infin

In this zone the value of transmissibility is less than 1 for all values of damping ratio In this zone the transmissibility increases with increase in damping Hence we can conclude that in this zone damping will adversely affect in isolating vibratory forces

63

Tutorial problems on Forced Vibration 1 Machine of mass 75kg is mounted on spring of stiffness 12000 Nm with an assumed damping factor ξ =02 piston within the machine of mass 2 kg has a reciprocating motion with a stroke of 75 cm and speed of 15 cyclessecond Assuming the motion of piston to be harmonic determine amplitude of machine ( X) Transmissibility and force transmitted to foundation phase angle of transmitted force wrt to exciting force 2 Motor weighs 600N is mounted on a simple beam that has a spring rate of 400 Ncm The motor armature base 160 N and has an eccentricity of 0005cm What will be the amplitude of motion when it runs at 1760 RPM Neglect damping the deflection of armature shaft and weight of the beam 3 A Heavy machine of mass 300 kg is supported on resilient foundation the static deflection of foundation due to weight of machine is found to be 75mm It is observed that the machine vibrates with an amplitude of 10mm when the base of the foundation is subjected to harmonic oscillation at the undampeds natural frequency of system with an amplitude of 25mm Find the damping constant of foundation Ii) dynamic force amplitude on the base iii) amplitude of the displacement of machine relative to base 4 An air compressor of mass 450 kg operates at a constant speed of 1750 RPM The reciprocating masses are 10 kg the crank radius is 100mm If the damper for mounting introduces a damping factor 015 i) specify the springs of mounting such that only 20 of the unbalanced force is transmitted to the foundation ii) determine the magnitude of the transmitted force bull Repeat the above problem by assuming that there is no damping in system

5 A body of a car having mass 1500 kg and is mounted on 4 equal springs which deflect through 225 cm under the weight of the body A total damping forces the 4 shock absorbers is 46 N at a velocity of 1cmsec The car is placed with all the four wheels on a test platform which is move up and down at resonant speed With an amplitude of 25cm Find the amplitude of the car body on its springs assuming its CG to be in the centre of wheel base 6A 75 kg machine which is mounted on springs of stiffness 116x105 Nm with the damping of 20=ξ A 2kg piston within the machine as reciprocating motion with a stroke of 008m and speed of 3000 RPM Assuming the motion of piston to be harmonic determine the amplitude of vibration of machine

7 The spring of an automobile are compressed 10cm under its own weight Find the critical speed when a trailer is traveling over road with a people approximaterly by a sine wave of amplitude 75 cm and wave length 15m What will be the amplitude of vibration at 64kmhour 8 An eccentric mass excites us used to determine vibratory characteristics of a structure of mass 25kg as given in figure At a speed of 1000 RPM the eccentric mass to beat top position at the instant the structure was moved upwards through its static equilibrium position and corresponding amplitude was 20mm The eccentric mass is 05kg at an eccentricity of 10mm Determine i) Undamped natural frequency of the system ii) Damping ratio iii) Steady state amplitude and phase angle at 1500 RPM iv) Force transmitted to the ground at 1200 RPM

M=25kg

C

K

Eccentric mass

64

65

CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Support motion Consider a spring mass damper system as shown in figure Let it excited by motion

of support let y be the harmonic displacement of support From Newtonrsquos second law of motion

)yx(k)yx(cxm minusminusminusminus= ampampampamp

)1(0)yx(k)yx(cxm minusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminus=minus+minus+ ampampampamp

Let z be the relative displacement of the mass with respect to support then

Z=x-y

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

Workdone in deforming a spring is equal to the strain energy or potential energy Strain energy = potential energy = area of the triangle OAB Stiffness of beams Cantilever beam consider a cantilever beam with an end mass shown in the figure The mass of the beam is assumed to be negligible The static deflection of beam at free end is given by

Similarly derive the expression for Simply supported beam and fixed support beam

x

k

m

x

m

x

m

x

mNlIE

st 192

3

=δδδδ

mNEI

Wlst

48

3

=δδδδ

F

K=stiffness

F

16

17

Stiffness of slender bar subjected to longitudinal vibrations For a system executing the longitudinal vibrations as shown in the figure let us derive the expression for stiffness

Torsional Stiffness of bar It is the amount of torque required to cause a unit angular deformation in the element

Torsional stiffness = Kt = θθθθT

Combination of stiffness Determination of equivalent spring stiffness when the springs are arranged in series Consider two springs of stiffness K1 and K2 acted upon by the force F

The deflection of spring k1 is x1 = 1KF

The deflection of spring k1 is x2 = Let these two springs be replaced by an equivalent stiffness Keq upon which a force F acts and due to which its deflection is given by

x = eqKF

x= x1+x2

m

E A

l

60 cm 70 cm

Determination of equivalent spring stiffness when the springs are arranged in parallel Force acting on K1 spring = F1=k1x Force acting on K2 spring = F2 Force required for an equivalent spring keq to defined by x given by F= Keq x But F = F1 +F2 Tutorial problems on Equivalent stiffness of springs 1Determine the equivalent stiffness for the system shown in figure 2 Determine the equivalent stiffness for the system shown in figure

3 Determine the equivale

M

3

M

Nm 102 6x

18

nt stiffness for the system shown in figure

38 kg

Nm 101 6x Nm 102 6x

Nm 106x

2k

k

k

2k

3k

k

2k

k

19

60 cm 80 cm 50 cm

50 cm 60 cm

4Determine the equivalent stiffness for the system shown in figure

5Replace the following torsional stiffness by a single shaft having radius 4cm and find the length required for the equivalent shaft Assume the material of given system and equivalent system is same R1= 3cm R2= 5cm DAMPING Every vibration energy is gradually converted into heat or sound Hence the displacement during vibration gradually reduces The mechanism by which vibration energy is gradually converted into heat or sound is known as damping A damper is assumed to have either mass or elasticity Hence damping is modeled as one or more of the following types Viscous damping Coulomb or dry friction damping materials or solid or hysteric damping Viscous damping Viscous damping is most commonly used damping mechanism in vibration analysis When the mechanical system vibrates in a fluid medium such as air gas water or oil the resistance offered by the fluid to the moving body causes energy to be dissipated In this case the amount of dissipated energy depends on many factors such as size or shape of the vibrating body the viscosity of the fluid the frequency of vibration and velocity of fluid Resistance due to viscous damping is directly proportional to the velocity of vibration

dF V αααα

dF = xC amp

Where C= damping coefficient Fd = damping force Examples of Viscous damping

1) Fluid film between sliding surface 2) Fluid flow around a piston in a cylinder 3) Fluid flow through orifice 4) Fluid flow around a journal in a bearing

Rreq =4cm

leqn

stress

Strain

Coulomb damping or dry friction damping Here a damping force is constant in magnitude but opposite in direction to that of the motion of vibrating body It is caused by the friction between the surfaces that are dry or have insufficient lubrication Material or solid or hysteric damping

When the materials are deformed energy is absorbed and dissipated by the material The effect is due to friction between the internal planes which slip or slide as the deformation takes place When a body having the material damping is subjected to vibration the stress strain diagram shows the hysteresis loop as shown in the figure The area of the loop denotes the energy lost per unit volume of the body per cycle

20

21

ττττ ττττ2 ττττ3

ττττ

X(t)

FOURIER SERIES The simplest of periodic motion happens to be SHM It is simple to handle but the motion of many vibrating system is not harmonic (but periodic) Few examples are shown below

Forces acting on machines are generally periodic but this may not be harmonic for example the excitation force in a punching machine is periodic and it can be represented as shown in figure 13 Vibration analysis of system subjected to periodic but nonharmonic forces can be done with the help of Fourier series The problem becomes a multifrequency excitation problem The principle of linear superposition is applied and the total response is the sum of the response due to each of the individual frequency term

Any periodic motion can be expressed as an infinite sum of sines and cosines terms If x(t) is a periodic function with period t its Fourier representation is given by

X(t) = (((( )))) (((( )))) (((( ))))t sint cost cos2 111 ωωωωωωωωωωωω baaao ++++++++++++

= (((( )))) (((( ))))t sint cos2 1

ωωωωωωωω nnn

o bnaa

++++++++sumsumsumsuminfininfininfininfin

====

tππππωωωω

2==== = Fundamental frequency ndash (1)

where ao an bn are constants

ττττ ττττ2 ττττ3

ττττ

X(t)

22

t

X(t)

035

025

Determination of constants To find ao Integrate both sides of equation(1) over any interval ττττ All intergrals on the RHS of the equation are zero except the one containing ao

(((( ))))dttxao

o 2

2

intintintint====ωωωωππππ

ππππωωωω

= (((( ))))dttxo

2intintintintττττ

ττττ

To find an multiply equation 1 by cos (((( ))))tnωωωω and Integrate over any interval ττττ All intergrals

(((( )))) (((( )))) dttxao

n tncos 2

2

ωωωωππππωωωω ωωωω

ππππ

intintintint==== (((( )))) (((( )))) dttxo

tncos 2

ωωωωττττ

ττττ

intintintint====

To find bn multiply equation 1 by sin (((( ))))tnωωωω and Integrate over any interval ττττ All intergrals

(((( )))) (((( )))) dttxao

n tnsin 2

2

ωωωωππππωωωω ωωωω

ππππ

intintintint==== (((( )))) (((( )))) dttxo

tnsin 2

ωωωωττττ

ττττ

intintintint====

Find the Fourier series for the curve shown below

ττττ ττττ2 ττττ3

t

X(t)

E

ττττ ττττ2 ττττ3

t

X(t)

ππππ2

t

X(t) ππππ

1 Represent for the periodic motion shown in the figure

23

CHAPTER 2 UNDAMPED FREE VIBRATION LEARNING OBJECTIVES

] Introduction to undamped free vibration

] Terminologies used in undamped free vibration

] Three methods to solve the undamped free vibration problems

] Problems related to undamped free vibration problems

Free vibrations are oscillations about a systems equilibrium position that occur in the absence of an external excitation force If during vibrations there is no loss of energy it is known as undamped vibration The first step in solving a vibration problem is setting up the differential equation of motion Consider spring mass system which is assumed to move only along the vertical direction as shown below Let m be the mass of the block and k be the stiffness of the spring When block of mass m is attached to spring the deflection of spring will be ∆ known as static deflection In the static equilibrium position the free body diagram of forces acting on the mass is shown in Figure(b) Hence mg= kA Once the system is disturbed the system executes vibrations

Let at any instant of time t the mass is displaced from the equilibrium position x the different forces acting on the system are shown in figure (d) From Newtonrsquos second law of motion

sum = maF

Inertia force ( disturbing force) = restoring force

mgxkxm ++∆minus= )(ampamp

0)( =+ xkxm ampamp

or 0)( =+ xmk

xampamp

equation 2 is the differential equation of motion for spring mfigure Comparing equation (2) with the equation of SHM +xampampsince the vibrations of the above system are free( without the forces) we can write

sec radmk

n =ω

24

ass system shown in 0)(2 =xω

resistance of external

time period km

2 f1

n

πτ ==

from the equation(1) mg = k∆

∆=

q

mk

Difference between the translation ( rectilinear) and rotational system of vibration

Translatory Rotational

In the analysis thFORCES are co Linear stiffness K Problems 1A mass of 10kFind the natural 2 A spring massnatural frequencnatural frequenc

kt

m

sec radmk

n =ω

k

25

e disturbing and restoring nsidered

In the analysis In the analysis MASS Moment of Inertia (J) is considered

in Nm is considered

g when suspended from a spring causes a static deflection of 1cm frequency of system

system has a spring stiffness K Nm and a mass of m Kg It has a y of vibration 12 Hz An extra 2kg mass coupled to it then the y reduces by 2 Hz find K and m

secrad n =ω

26

3 A steel wire of 2mm diameter and 30m long It is fixed at the upper end and carries a mass of m kg at its free end Find m so that the frequency of longitudinal vibration is 4 Hz 4 A spring mass system has a natural period of 02 seconds What will be the new period if the spring constant is 1) increased by 50 2) decreased by 50 5 A spring mass system has a natural frequency of 10 Hz when the spring constant is reduced by 800 Nm the frequency is altered by 45 Find the mass and spring constant of the original system 6 Determine the natural frequency of system shown in fig is which shaft is supported in SHORT bearings 7 Determine the natural frequency of system shown in fig is which shaft is supported in LONG bearings where l is the length of bearing and E ndash youngrsquos modulus and I is moment of Inertia 7 Determine the natural frequency of system shown in fig is which shaft is supported in LONG bearings 8 A light cantilever beam of rectangular section( 5 cm deep x 25cm wide) has a mass fixed at its free end Find the ratio of frequency of free lateral vibration in vertical plane to that in horizontal 9 Determine the natural frequency of simple pendulum 10 Homogeneous square plate of size l and mass m is suspended from the mid point of one of its sides as shown in figure Find the frequency of vibration

l

l

27

11 A compound pendulum which is rigid body of mass m and it is pivoted at O The point of pivot is at distance d from the centre of gravity It is free to rotate about its axis Find the frequency of oscillation of such pendulum 12 A connecting rod shown in fig is supported at the wrist pin end It is displaced and allowed to oscillate The mass of rod is 5kg and centre of gravity is 20 cm from the pivot point O If the frequency of oscillation is 40 cyclesminute calculate the mass moment of inertia about its CG

13 A semi circular homogenous disc of radius r and mass m is pivoted freely about its centre as shown in figure Determine the natural frequency of oscillation 14A simply supported beam of square cross section 5mmx5mm and length 1m carrying a mass of 0575 kg at the middle is found to have natural frequency of 30 radsec Determine youngrsquos modulus of elasticity of beam 15 A spring mass system k1 and m have a natural frequency f1 Determine the value of k2 of another spring in terms of k1 which when placed in series with k1 lowers the natural frequency to 23 f1

COMPLETE SOLUTION OF SYSTEM EXECUTING SHM

The equation of motion of system executing SHM can be represented by

0)( =+ xkxm ampamp ----------------------------------------(1)

022

2=+

dtdx

dt

dx ω

The general solution of equation (1) can be expressed as

------------------------(2)

Where A and B are arbitrary constant which can be determined from the initial conditions of the system Two initial conditions are to be specified to evaluate these

constants x=x0 at t=0 and xamp = Vo at t=0 substituting in the equation (2) Energy method In a conservative system the total energy is constant The dwell as natural frequency can be determined by the princienergy For free vibration of undamped system at any instant and partly potential The kinetic energy T is stored in the masswhere as the potential energy U is stored in the form of sdeformation or work done in a force field such as gravity The total energy being constant T+U = constant Its rate of cha

t)(sin Bt)cos( AX ω+ω=

t)(sin V

t)cos( xx o0 ω

ω+ω=

Is the required complete

solution

28

ifferential equation as ple of conservation of of time is partly kinetic by virtue of its velocity train energy in elastic

nge

Is given by [ ] 0UTdtd

=+

From this we get a differential equation of motion as well as natural frequency of the system Determine the natural frequency of spring mass system using energy method

Determine the natural frequency of system shown in figure Determine the natural frequency of the system shown in figure Is there any limitation on the value of K Discuss Determine the natural frequency of s

m

l k

m

a

θ

m

a

θ

l

k

29

k

k

ystem shown below Neglect the mass of ball

l

m

30

A string shown in figure is under tension T which can be assumed to remain constant for small displacements Find the natural frequency of vertical vibrations of spring An acrobat 120kg walks on a tight rope as shown in figure The frequency of vibration in the given position is vertical direction is 30 rads Find the tension in the rope A manometer has a uniform bore of cross section area A If the column of liquid of length L and Density ρ is set into motion as shown in figure Find the frequency of the resulting oscillation

m

l

a

T T

36m 8m

31

k

R

k

m

r

m

Find the expression for natural frequency of system shown in the figure Neglect the mass of the cantilever beam Study the special case i) k=Infinity ii) I = infinity Determine the expression for the natural frequency of a system shown in figure The two discs are keyed to a common shaft and have a combined mass moment of inertia about centre of oscillation O It is assumed that the card attached to mass m does not stretch and is always under tension Determine the natural frequency of a system shown

m

l

32

Determine the expression for the natural frequency of the system shown in figure Assume that the wires connecting the masses do not stretch and are always in tension

Determine the natural frequency of spring mass system taking the MASS OF SPRING (ms ) into account

M3 M1

M2

k2 k1

m

k

33

RAYLEIGHrsquoS METHOD

This is the extension of energy method Here we can determine the natural frequency of a conservative system without finding the equation of motion The natural frequency is a function of the rate of change of Kinetic energy and potential energy of the system From the principle of conservation of energy we have T+U = Constant T= KE and U =PE We can write the above equation as T1+U1 = --------------------------(1) Where 1 and 2 represents tow instances of time Let 1 be the time when the mass is passing through static equilibrium position Let 2 be the time corresponding to the mass displacement of the mass At this instant the velocity f the mass is zero and hence Substituting in equation (1) we get ( )Max = ( )Max Above equation leads directly to natural frequency Determine the natural frequency of spring mass system using Rayleighrsquos method m

34

r

B

K K

Home work

Determine the natural frequency of pendulum using Rayleighrsquos method

A Cylinder of mass m and moment of inertia Jo is to role without slipping but is sustained by spring K as shown in figure Determine the natural frequency of oscillation

Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping

a

Determine the natural frequency of system shown where in cylindrical disc rolls over the inextensible spring A cylinder of radius r rolls without slipping on cylindrical the equation of motion and hence determine the naoscillations about the lowest point shown Repeat the above problem ndash Instead of cylinder assume swithout slipping on concave surface as shown above

r

K

35

portion of radius R Derive tural frequency of small

phere of radius with r rolls

The torsional pendulum has a natural frequency of 5 Hz What length of steel wire of diameter 2 mm should be used for pendulum The inertia of mass fixed at the free end is 00098 kgm2 Take g= 083x1011 Nm2 Determine the torsional natural frequency of the system shown in figure Neglect the mass moment of inertia of the shaft ( Figure not proportionate) Determine the natural frequency of simple pendulum considering the mass of the rod into account

Using the energy method find tThe chord may be assumed ine

Kt1

Kt2

20cm

06m 08m 04m

20cm 20cm 20cm

m

M

he natural frxtensible in

k

M

36

equency of the system shown in figure the spring mass pulley system

A rod of length l and depth H having a mass m rests on a semicylidrical surface os radius r It is tipped slightly and let go Find the frequency of oscillation

A thin bar of l having a mass m rests on a semicylidrical surface radius r It is tipped slightly and let go Find the frequency of oscillation

l

h

R

l

37

38

CHAPTER 3

DAMPED FREE VIBRATION OF SINGLE DEGREE FREEDOM SYSTEM LEARNING OBJECTIVES

] Introduction to damped free vibration

] Terminologies used in damped free vibration

] Three different types of damping

] Concept of critical damping and its importance

] Study of response of viscous damped system for cases of under damping

critical and over damping

] Concept of logarithmic decrement

] Problems related to damped free vibration problems

Viscously damped free vibration of single degree freedom sytem Consider a single degree freedom system with viscous damping as shown in figure For a viscous damped system the damping force is

where c is a constant of proportionality It is called damping coefficient or coefficient of viscous damping

Symbolically it is designated by a dashpot as shown in Fig 3

x xampx

39

At any instant of time let the mass be displaced by x from the equilibrium position Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

Okxxcxm =minusminus= ampampamp

Okxxcxm =++ ampampamp ----------------------------------------------------------(1)

or

0kxdtdx

Cdt

xdM

2

2

=++

Assuming the solution of equation as

The first term in the above equation m2Ct

eminus

is an exponentially decaying function of time The behavior of the terms in the parenthesis however depends on whether numerical value within the radicals is positive zero or negative

CASE 1 if 0mk

m2c

2

=minus

40

CASE 2 if 0mk

m2c

2

=gt

CASE 3 if 0mk

m2c

2

=lt

41

Critically damped coefficient Critically damped coefficient is defined as the value of the damping coeffient for which radical in equation

mk

m2c

m2c

S2

21 minus

plusmn

= becomes zero

Damping Ratio ( Damping factor) ( ξ ) It is defined as the ratio of damping coefficient to the critical damping coefficient Note The damping coefficient c is defined as the actual amount of damping while the damping coefficient is a constant depending upon the mass and stiffness of the system The roots of characteristics equation can be now written as Note if c=cc or ξ =1 then system is critically damped if cgt cc or ξ gt1 then system is over damped I if clt cc or ξ lt1 then system is under damped

42

Take home quiz Derive General solution of different damped system

1 Critically damped system 2 Under damped system 3 Over damped system

Logarithmic Decrement It is used to determine the amount of damping present in system This measure The rate of decay of free vibration The larger the damping the greater will be rate of decay Logarithmic decrement is defined as the natural logarithm of ratio of any two successive amplitudes

=δ

2

1

xx

ln

Expression for logarithmic decrement (δ ) Consider a damped free vibration which is expressed by

( ) ( ) φ+ξminusω= ξω t 1sinXex 2n

tn

x is the displacement at any instant of time X and φ are arbitrary constants

21

2

ξminus

πξ=δ

43

Tutorial problems 1 Derive an expression for damping ratio in terms of logarithmic decrement

2 Show that logarithmic decrement

=δ

n

o

xx

lnn1

where xo is the initial amplitude xn

is amplitude after n cycles 3 A mass of 1 kg is to be supported on a spring having stiffness K= 9800 Nm The damping constant c= 59 Nsm Determine the natural frequency of system Find also logarithmic decrement and amplitude after 3 cycles if initial displacement is 03 cm 4 A vibrating system is defined by the following parameters M = 3kg k= 100 Nm damping coefficient = 3 Nsm determine Damping factor natural frequency damped vibration logarithmic decrement ratio of consecutive amplitude no of cycles after which the original amplitude is reduced to 20 5A single degree damped vibrating system a suspended mass of 18kg makes 15 oscillations in 003 seconds The amplitude decreases to 025 of the initial value after 5 oscillations Determine the stiffness of spring Logarithmic decrement damping factor and damping coefficient 6 A damped vibration record of a spring mass system shows the following data Amplitude at the end of 2nd cycle is 9mm amplitude at eh end of 3rd cycle is 6mm amplitude at the end of 4th cycle is 4mm stiffness of spring =8000 Nm weight =20N Find the logarithmic decrement damping force at unit velocity periodic time of vibration 7 For the system shown in figure the characteristics of dashpot is such that the constant force of 49 N is applied to the piston it velocity is found to be constant at 012 ms Determine the value of C would you expect the complete system to be periodic or aperiodic 8 For a torsional system shown in the figuref1 is natural frequency in air and f2 is the natural frequency when immersed in oil Determine the damping coefficient

Viscous fluid

Disc of mass moment of Inertia J

2a

C

m M

b

a a

k

9 A thin plate of area A and weigh w is fixed to the end of the spring and allow to oscillate in a viscous fluid as shown in figure It f1 is the frequency of oscillation in air and f2 in liquid Show that

22

21 ff

gA2

minusπω

=micro where the damping force on the plate

AV2Fd micro= V being the velocity of vibration 10 A torsional pendulum has a natural frequency of 175 cyclessecond vibrating in vacuum The mass moment of inertia is 2 kgm2 It is immersed in oil and it is observed that the natural frequency is 142 cyclessecond Find the damping torque If the disc is replaced by 34 deg When in oil find the displacement at the end of the first complete cycle 11 An automobile can be modeled as a mass placed on 4 shock absorbers each consisting of a spring and a damper Such that each spring is equally loaded Determine the stiffness damping constant of each shock absorber So that natural frequency is 2Hz and system is critically damped The mass of vehicle is 200kg 12 Write the differential equation of motion for the system shown in the figure below Determine the natural frequency of the damped oscillation and critical damping coefficient

Determine the differential equation of motion and find the critical cdamped natural frequency for system shown in figure

a

C

m M

M =200kg

44

200kg kgkg

oefficient and

45

Derive the equation of motion and find the critical damping coefficient and damped natural frequency for the system shown below Write the differential equation of motion and determine the expression for critical damping coefficient and damped natural frequency of system shown in figure

Set up a differential equation of motion for system shown in figure and determine the following Undamped natural frequencies critical damping coefficient damping ratio damped natural frequency

b

C

kM

a

m

46

A mass of 300N is resting on two springs of stiffness 3000 Nm each and a dashpot of damping coefficient 150 Nsm as shown in figure If the initial velocity of 10cms is given to mass at its equilibrium position what will be the displacement from the equilibrium position at the end of first second Derive the formula used A gun barrel of mass 600kg has a recoil spring of stiffness 294 knm if the barrel recoils 13m on firing determine i) The initial recoil velocity of barrel ii) The critical damping coefficient of dashpot which is engaged at the end of recoil stroke Iii) Time required for the barrel to return at the end of recoil stroke Iv) time required for the barrel to return to a position 5cm from the initial position

W= 3000 N

FIRING

CHAPTER 4

FORCED VIBRATION LEARNING OBJECTIVES

] Introduction to forced free vibration

] Terminologies used in forced vibration

] Concept of magnification factor

] Study of reciprocating and rotating unbalance

] Concept of base excitation and support excitation

] Concept of Transmissibility ratio

] Problems related to forced vibration

The vibration that takes place under the excitation of external forces are called

forced vibration When a system is subjected to harmonic excitation it is forced to vibrate at the same frequency as that of excitation Most of the mechanical systems are prone to steady state forced vibration which cause fatigue failure The vibration problem also becomes very important when any of excitation frequencies coincide with one of natural frequencies of a system Under this condition resonance occurs during which large amplitude of vibrations are observed To avoid resonance external frequency (operating speed) may be changed or properties of the system may be changed to alter the natural frequency In some cases sufficient amount of damping may be provided to avoid large amplitude during resonance Thus the problem of force vibration is very important in mechanical design

FORCED VIBRATIONS OF SINGLE DOF SYSTEMS UNDER HARMONIC EXCITATION

` [ ] tsin FF o ω=

x

sin FF o=

Fo

xampamp

[ ] tω

xamp

47

48

Consider a spring mass system having a viscous damping excited by a harmonic force [ ] tsin FF o ω= as shown in figure The various forces acting on the system is shown in figure Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

O)tsin(Fkxxcxm o =ω+minusminus= ampampamp

)tsin(Fkxxcxm o ω=++ ampampamp ----------------------------------------------------------(1)

or

)tsin(Fkxdtdx

Cdt

xdM o2

2

ω=++

Equation (1) represents the differential equation of motion of the system The solution to this equation consists of two parts namely complementary function and a particular integral

The first of solution is also termed as transient solution of homogeneous solution and can be obtained from the homogeneous differential equation 0kxxcxm =++ ampampamp

The second part of solution particular integral is also termed as steady state solution ad is of form )tsin(Xx φminusω= Where X= steady state amplitude φ is phase difference between external force and displacement

)tsin(Xx φminusω=

The vector representation of the shown

49

The force polygon for the above vector diagram can be written as follows

( )φminusωt O

KX

xm 2ω

Fo

xCω

tω

φ

50

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

is the governing equation

Zero frequency deflection (Xst) It is deflection observed at zero frequency ( when the external frequency 0=ω

( ) kF

001

kF

x o

22

o

st =

+minus=

Magnification factor (M) It is defined as the ratio of steady state amplitude to zero frequency deflection xst

M = arX

X=

A plot of magnification factor versus frequency ratio

damping factor ( )ξ is called frequency plot of a harmon All the frequency terms starts from magnification factor of damping ratio ( )ξ The amplitude of vibration at any on the value of damping The amplitude will be higher for higher damping The amplitude of vibration at reso

amplitude tends to zero as

ωω

n tends to infinity

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

Freq

ζ =

ζ = 1

Magnification Factor M = XXst

kFo

51

ωω

nfor varying values of

ically excite vibrating system

M=1 what ever may be value particular frequency depends

for lesser damping and lesser nance will be very high The

uency Ratio r = (ωωn)

ζ = 0 0

ζ = 025 ζ = 0375 ζ = 0 5

ζ = 0707

ζ = 2

52

Plot of phase angle φ versus frequency ratio

ωω

n is shown in the

figure The phase angle φ is zero for all the values of damping ratio

at

ωω

n=0 The phase

angle φ is 90 for all the values of damping ratio

at

ωω

n=1 ie at

resonance The phase angle between o to 90 deg for all the values of

ωω

n= 0 to 1 and

between 90 deg to 180 deg for all the values of

ωω

ngt1

Class work Derive an expression for frequency at peak amplitude

2np 21 ξminusω=ω=ω is the required equation

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

53

Derive expression amplitude at resonance ( In dimensional form) In dimensional form We know that steady state amplitude

( ) ( )

ω+ωminus=

222

o

Cmk

Fx Equation in dimensional form

We know that at resonance mk

n =ω=ω

( )

ω+

minus

=2

n

2

oR

Cmk

mk

Fx

( )

ω=

2n

oR

C

Fx

( )

ω=

n

oR C

Fx

Class work Derive an expression for amplitude at resonance ( in non dimensional form)

Tutorial Problems on Forced Vibration 1A weight of 60N is suspended by a spring stiffness 12 kNm is forced to vibrate by a harmonic force of 10N assuming viscous damping of 0086 kN-sm Find i) resonance frequency ii) amplitude at resonance iii) phase angle at resonance iv) frequency corresponding to peak amplitude 2 A periodic torque having a maximum value of 05888Nm at a frequency of 4 rads is impressed upon a flywheel suspended from a wire The wheel has mass moment of inertia 012 kgm2 and the wire has stiffness 1176 Nmrad A viscous dashpot applies damping couple of 0784 Nm an angular velocity of 2 rads calculate i)maximum angular displacement from rest position ii) maximum couple applied to dashpot Iii) angle by which the angular displacement lags the torque A mass of 1019 kg is suspended from a spring stiffness 20 Nmm viscous damping causes the amplitude to decrease to 110th of initial value in 4 complete cycle If a periodic force of 20 cos(50 t) is applied to the mass Find the amplitude of forced vibration What would be the amplitude of oscillation if the applied force coincides with the natural period

m=1019 kg

[ ] t50cos 30F =

54

55

ROTATING UNBALANCE Unbalance in rotating machinery is a common source of vibration excitation Consider a machine of mass m having mounted on a foundation of stiffness k and damping coefficient C as shown in figure The unbalance is represented by an eccentric rotating mass m with eccentricity (e) and having a angular velocity (ω) let x be the displacement of non rotating mass from the static equilibrium position The displacement of rotating mass m from the static equilibrium is ( ) tsin( ex ω+ Applying Newtonrsquos law of motion we have

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

56

RECIPROCATING UNBALANCE Unbalance in reciprocating machinery is a common source of vibration excitation Consider a machine of mass M having mounted on a foundation of stiffness K and damping coefficient C as shown in figure The unbalance is represented by a reciprocating mass (m) having crank radus (e) and connecting rod of length (l) Let ωωωω be angular velocity of the crank Let x be the displacement of non reciprocating mass from the static equilibrium position at any instant of time The displacement of reciprocating mass m from the static equilibrium position is

)2sin(sin +++ tle

tex ωωωωωωωω

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

57

Note The equations obtained in reciprocating unbalance is same as the rotating unbalance and having the following differences Rotating Unbalance Reciprocating Unbalance Here mass of the rotating element is considered

Here mass of reciprocating element is considered

Here the eccentricity (e) which is the distance between the centre of rotation and CG

Here the eccentricity e is the crank radius and crank radius = stroke length2

The frequency plot of non dimensional factor( ( )me

MX versus frequency ratio

ωω

n for various values of ( )ξξξξ is shown in figure This frequency plot is same for

the rotating and reciprocating unbalance with the following points are noted from the diagram

1 Damping ratio plays very important role during resonance in controlling the amplitude

2 When

ωω

n=0 then ( )me

MX =0 for all the values of damping ratio ( )ξξξξ

3 When

ωω

nis higher than ( )me

MX tends to unity

4 At very high speeds ie at very high frequency ratio

ωω

n there is no effect

of damping The frequency plot of phase angle versus the frequency ratio is given as shown in figure The explanation is same as in forced harmonic vibration

58

VIBRATION ISOLATION

Vibratory forces and motions generated by machine and other caused are unavoidable Howeever their effect on the other dynamical systems can be minimized by proper isolator design An isolation system is either to protect o delicate machine from excessive vibration transmitted to it from its supporting structure or to prevent vibratory forces and motions generated by machines from being transmitted to its surroundings The effectiveness of isolation may be measured in terms of forces and motion transmitted The first type is known as force isolation and the second type is known as motion isolation

Force isolation (Force Transmissibility) When vibratory forces generated by machine are isolated the effectiveness of forced isolation can be made by defining the term transmissibility The term transmissibility is defined as the ratio of force transmitted to the foundation to that of the external impressed force

Force Transmissibility = o

tr

FF

force Impressed Externalfoundation toted transmitForce

=

Consider viscous damped spring mass system as shown in figure Upon this system an external harmonic force ) t (Sin Fo ω is applied The difference forces acting on system are inertia forces damping force spring force and external force All these forces are represented on force polygon shown below Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these two forces

( ) ( )22 xckx ω+=

59

This is the expression for force transmissibility in dimensional form To represent same in the non dimensional form divide numerator and denominator by k

2

n

22

n

2

n

o

tr

21

21

FF

TR

ωω

ξ+

ωω

minus

ωω

ξ+

==

PHASE DIFFERNECE BETWEEN EXTERNAL FORCE AND DISPLACEMENT Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these

ωω

minus

ωω

ξ=

ωminusω

=φ minusminus2

n

n1

2

1

1

2tan

mKC

tan

The phase difference between force transmitted to the foundation Ftr to the displacement

kc

KxxC

tanω

=ω

=α or

ωω

ξ=

ω=α minusminus

n

11 2tanKx

xCtan

The phase difference between external force Fo and force transmitted to foundationis β

60

β= αminusφ =

ωω

minus

ωω

ξminus

2

n

n1

1

2tan -

ωω

ξminus

n

1 2tan

Displacement Transmissibility It is the ratio of amplitude of steady state vibration of machine to that of the base motion Forced vibration due to support motion Consider a spring mass damper system subjected to harmonic excitation )tsin(Yy ω= at the base as shown in figure Let x be the absolute displacement of the mass at any instant of time t The different forces acting on system are shown in free body diagram Applying Newtonrsquos Law of motion Ma = sumF

)yx(k)yx(cxm minusminusminusminus= ampampampamp

0)yx(k)yx(cxm =minus+minus+ ampampampamp

61

Frequency plot of transmissibility versus frequency ratio

62

The plot of transmissibility versus frequency ratio is as shown in figure The following points can be noted from it

i) The value of transmissibility is unity at frequency ratio

ωω

n=0 and 2

ii) For an undamped system ( )ξ =0 transmissibility tends to infin at resonance iii) The value of transmissibility is less than unity for the value of frequency ratio

greater than 2

iv) For frequency ratio

ωω

nlt 2 smaller damping ratio lead to larger values of

transmissibility on the other hand for frequency ratio

ωω

ngt 2 smaller damping

ratio leads to smaller value of transmissibility

The frequency plot of transmissibility (TR) versus frequency ratio

ωω

n can be

broadly divided into two zones

1

ωω

n= 0 to 2

2

ωω

n= 2 to infin

CASE 1

When

ωω

n= 0 to 2

In this zone the value of transmissibility is always greater than 1 for all values of damping ratio In this zone the transmissibility decreases with increase in damping Hence we can conclude that in this zone damping is useful in isolating vibration forces being transmitted to the support CASE 2

When

ωω

n= 2 to infin

In this zone the value of transmissibility is less than 1 for all values of damping ratio In this zone the transmissibility increases with increase in damping Hence we can conclude that in this zone damping will adversely affect in isolating vibratory forces

63

Tutorial problems on Forced Vibration 1 Machine of mass 75kg is mounted on spring of stiffness 12000 Nm with an assumed damping factor ξ =02 piston within the machine of mass 2 kg has a reciprocating motion with a stroke of 75 cm and speed of 15 cyclessecond Assuming the motion of piston to be harmonic determine amplitude of machine ( X) Transmissibility and force transmitted to foundation phase angle of transmitted force wrt to exciting force 2 Motor weighs 600N is mounted on a simple beam that has a spring rate of 400 Ncm The motor armature base 160 N and has an eccentricity of 0005cm What will be the amplitude of motion when it runs at 1760 RPM Neglect damping the deflection of armature shaft and weight of the beam 3 A Heavy machine of mass 300 kg is supported on resilient foundation the static deflection of foundation due to weight of machine is found to be 75mm It is observed that the machine vibrates with an amplitude of 10mm when the base of the foundation is subjected to harmonic oscillation at the undampeds natural frequency of system with an amplitude of 25mm Find the damping constant of foundation Ii) dynamic force amplitude on the base iii) amplitude of the displacement of machine relative to base 4 An air compressor of mass 450 kg operates at a constant speed of 1750 RPM The reciprocating masses are 10 kg the crank radius is 100mm If the damper for mounting introduces a damping factor 015 i) specify the springs of mounting such that only 20 of the unbalanced force is transmitted to the foundation ii) determine the magnitude of the transmitted force bull Repeat the above problem by assuming that there is no damping in system

5 A body of a car having mass 1500 kg and is mounted on 4 equal springs which deflect through 225 cm under the weight of the body A total damping forces the 4 shock absorbers is 46 N at a velocity of 1cmsec The car is placed with all the four wheels on a test platform which is move up and down at resonant speed With an amplitude of 25cm Find the amplitude of the car body on its springs assuming its CG to be in the centre of wheel base 6A 75 kg machine which is mounted on springs of stiffness 116x105 Nm with the damping of 20=ξ A 2kg piston within the machine as reciprocating motion with a stroke of 008m and speed of 3000 RPM Assuming the motion of piston to be harmonic determine the amplitude of vibration of machine

7 The spring of an automobile are compressed 10cm under its own weight Find the critical speed when a trailer is traveling over road with a people approximaterly by a sine wave of amplitude 75 cm and wave length 15m What will be the amplitude of vibration at 64kmhour 8 An eccentric mass excites us used to determine vibratory characteristics of a structure of mass 25kg as given in figure At a speed of 1000 RPM the eccentric mass to beat top position at the instant the structure was moved upwards through its static equilibrium position and corresponding amplitude was 20mm The eccentric mass is 05kg at an eccentricity of 10mm Determine i) Undamped natural frequency of the system ii) Damping ratio iii) Steady state amplitude and phase angle at 1500 RPM iv) Force transmitted to the ground at 1200 RPM

M=25kg

C

K

Eccentric mass

64

65

CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Support motion Consider a spring mass damper system as shown in figure Let it excited by motion

of support let y be the harmonic displacement of support From Newtonrsquos second law of motion

)yx(k)yx(cxm minusminusminusminus= ampampampamp

)1(0)yx(k)yx(cxm minusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminus=minus+minus+ ampampampamp

Let z be the relative displacement of the mass with respect to support then

Z=x-y

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

17

Stiffness of slender bar subjected to longitudinal vibrations For a system executing the longitudinal vibrations as shown in the figure let us derive the expression for stiffness

Torsional Stiffness of bar It is the amount of torque required to cause a unit angular deformation in the element

Torsional stiffness = Kt = θθθθT

Combination of stiffness Determination of equivalent spring stiffness when the springs are arranged in series Consider two springs of stiffness K1 and K2 acted upon by the force F

The deflection of spring k1 is x1 = 1KF

The deflection of spring k1 is x2 = Let these two springs be replaced by an equivalent stiffness Keq upon which a force F acts and due to which its deflection is given by

x = eqKF

x= x1+x2

m

E A

l

60 cm 70 cm

Determination of equivalent spring stiffness when the springs are arranged in parallel Force acting on K1 spring = F1=k1x Force acting on K2 spring = F2 Force required for an equivalent spring keq to defined by x given by F= Keq x But F = F1 +F2 Tutorial problems on Equivalent stiffness of springs 1Determine the equivalent stiffness for the system shown in figure 2 Determine the equivalent stiffness for the system shown in figure

3 Determine the equivale

M

3

M

Nm 102 6x

18

nt stiffness for the system shown in figure

38 kg

Nm 101 6x Nm 102 6x

Nm 106x

2k

k

k

2k

3k

k

2k

k

19

60 cm 80 cm 50 cm

50 cm 60 cm

4Determine the equivalent stiffness for the system shown in figure

5Replace the following torsional stiffness by a single shaft having radius 4cm and find the length required for the equivalent shaft Assume the material of given system and equivalent system is same R1= 3cm R2= 5cm DAMPING Every vibration energy is gradually converted into heat or sound Hence the displacement during vibration gradually reduces The mechanism by which vibration energy is gradually converted into heat or sound is known as damping A damper is assumed to have either mass or elasticity Hence damping is modeled as one or more of the following types Viscous damping Coulomb or dry friction damping materials or solid or hysteric damping Viscous damping Viscous damping is most commonly used damping mechanism in vibration analysis When the mechanical system vibrates in a fluid medium such as air gas water or oil the resistance offered by the fluid to the moving body causes energy to be dissipated In this case the amount of dissipated energy depends on many factors such as size or shape of the vibrating body the viscosity of the fluid the frequency of vibration and velocity of fluid Resistance due to viscous damping is directly proportional to the velocity of vibration

dF V αααα

dF = xC amp

Where C= damping coefficient Fd = damping force Examples of Viscous damping

1) Fluid film between sliding surface 2) Fluid flow around a piston in a cylinder 3) Fluid flow through orifice 4) Fluid flow around a journal in a bearing

Rreq =4cm

leqn

stress

Strain

Coulomb damping or dry friction damping Here a damping force is constant in magnitude but opposite in direction to that of the motion of vibrating body It is caused by the friction between the surfaces that are dry or have insufficient lubrication Material or solid or hysteric damping

When the materials are deformed energy is absorbed and dissipated by the material The effect is due to friction between the internal planes which slip or slide as the deformation takes place When a body having the material damping is subjected to vibration the stress strain diagram shows the hysteresis loop as shown in the figure The area of the loop denotes the energy lost per unit volume of the body per cycle

20

21

ττττ ττττ2 ττττ3

ττττ

X(t)

FOURIER SERIES The simplest of periodic motion happens to be SHM It is simple to handle but the motion of many vibrating system is not harmonic (but periodic) Few examples are shown below

Forces acting on machines are generally periodic but this may not be harmonic for example the excitation force in a punching machine is periodic and it can be represented as shown in figure 13 Vibration analysis of system subjected to periodic but nonharmonic forces can be done with the help of Fourier series The problem becomes a multifrequency excitation problem The principle of linear superposition is applied and the total response is the sum of the response due to each of the individual frequency term

Any periodic motion can be expressed as an infinite sum of sines and cosines terms If x(t) is a periodic function with period t its Fourier representation is given by

X(t) = (((( )))) (((( )))) (((( ))))t sint cost cos2 111 ωωωωωωωωωωωω baaao ++++++++++++

= (((( )))) (((( ))))t sint cos2 1

ωωωωωωωω nnn

o bnaa

++++++++sumsumsumsuminfininfininfininfin

====

tππππωωωω

2==== = Fundamental frequency ndash (1)

where ao an bn are constants

ττττ ττττ2 ττττ3

ττττ

X(t)

22

t

X(t)

035

025

Determination of constants To find ao Integrate both sides of equation(1) over any interval ττττ All intergrals on the RHS of the equation are zero except the one containing ao

(((( ))))dttxao

o 2

2

intintintint====ωωωωππππ

ππππωωωω

= (((( ))))dttxo

2intintintintττττ

ττττ

To find an multiply equation 1 by cos (((( ))))tnωωωω and Integrate over any interval ττττ All intergrals

(((( )))) (((( )))) dttxao

n tncos 2

2

ωωωωππππωωωω ωωωω

ππππ

intintintint==== (((( )))) (((( )))) dttxo

tncos 2

ωωωωττττ

ττττ

intintintint====

To find bn multiply equation 1 by sin (((( ))))tnωωωω and Integrate over any interval ττττ All intergrals

(((( )))) (((( )))) dttxao

n tnsin 2

2

ωωωωππππωωωω ωωωω

ππππ

intintintint==== (((( )))) (((( )))) dttxo

tnsin 2

ωωωωττττ

ττττ

intintintint====

Find the Fourier series for the curve shown below

ττττ ττττ2 ττττ3

t

X(t)

E

ττττ ττττ2 ττττ3

t

X(t)

ππππ2

t

X(t) ππππ

1 Represent for the periodic motion shown in the figure

23

CHAPTER 2 UNDAMPED FREE VIBRATION LEARNING OBJECTIVES

] Introduction to undamped free vibration

] Terminologies used in undamped free vibration

] Three methods to solve the undamped free vibration problems

] Problems related to undamped free vibration problems

Free vibrations are oscillations about a systems equilibrium position that occur in the absence of an external excitation force If during vibrations there is no loss of energy it is known as undamped vibration The first step in solving a vibration problem is setting up the differential equation of motion Consider spring mass system which is assumed to move only along the vertical direction as shown below Let m be the mass of the block and k be the stiffness of the spring When block of mass m is attached to spring the deflection of spring will be ∆ known as static deflection In the static equilibrium position the free body diagram of forces acting on the mass is shown in Figure(b) Hence mg= kA Once the system is disturbed the system executes vibrations

Let at any instant of time t the mass is displaced from the equilibrium position x the different forces acting on the system are shown in figure (d) From Newtonrsquos second law of motion

sum = maF

Inertia force ( disturbing force) = restoring force

mgxkxm ++∆minus= )(ampamp

0)( =+ xkxm ampamp

or 0)( =+ xmk

xampamp

equation 2 is the differential equation of motion for spring mfigure Comparing equation (2) with the equation of SHM +xampampsince the vibrations of the above system are free( without the forces) we can write

sec radmk

n =ω

24

ass system shown in 0)(2 =xω

resistance of external

time period km

2 f1

n

πτ ==

from the equation(1) mg = k∆

∆=

q

mk

Difference between the translation ( rectilinear) and rotational system of vibration

Translatory Rotational

In the analysis thFORCES are co Linear stiffness K Problems 1A mass of 10kFind the natural 2 A spring massnatural frequencnatural frequenc

kt

m

sec radmk

n =ω

k

25

e disturbing and restoring nsidered

In the analysis In the analysis MASS Moment of Inertia (J) is considered

in Nm is considered

g when suspended from a spring causes a static deflection of 1cm frequency of system

system has a spring stiffness K Nm and a mass of m Kg It has a y of vibration 12 Hz An extra 2kg mass coupled to it then the y reduces by 2 Hz find K and m

secrad n =ω

26

3 A steel wire of 2mm diameter and 30m long It is fixed at the upper end and carries a mass of m kg at its free end Find m so that the frequency of longitudinal vibration is 4 Hz 4 A spring mass system has a natural period of 02 seconds What will be the new period if the spring constant is 1) increased by 50 2) decreased by 50 5 A spring mass system has a natural frequency of 10 Hz when the spring constant is reduced by 800 Nm the frequency is altered by 45 Find the mass and spring constant of the original system 6 Determine the natural frequency of system shown in fig is which shaft is supported in SHORT bearings 7 Determine the natural frequency of system shown in fig is which shaft is supported in LONG bearings where l is the length of bearing and E ndash youngrsquos modulus and I is moment of Inertia 7 Determine the natural frequency of system shown in fig is which shaft is supported in LONG bearings 8 A light cantilever beam of rectangular section( 5 cm deep x 25cm wide) has a mass fixed at its free end Find the ratio of frequency of free lateral vibration in vertical plane to that in horizontal 9 Determine the natural frequency of simple pendulum 10 Homogeneous square plate of size l and mass m is suspended from the mid point of one of its sides as shown in figure Find the frequency of vibration

l

l

27

11 A compound pendulum which is rigid body of mass m and it is pivoted at O The point of pivot is at distance d from the centre of gravity It is free to rotate about its axis Find the frequency of oscillation of such pendulum 12 A connecting rod shown in fig is supported at the wrist pin end It is displaced and allowed to oscillate The mass of rod is 5kg and centre of gravity is 20 cm from the pivot point O If the frequency of oscillation is 40 cyclesminute calculate the mass moment of inertia about its CG

13 A semi circular homogenous disc of radius r and mass m is pivoted freely about its centre as shown in figure Determine the natural frequency of oscillation 14A simply supported beam of square cross section 5mmx5mm and length 1m carrying a mass of 0575 kg at the middle is found to have natural frequency of 30 radsec Determine youngrsquos modulus of elasticity of beam 15 A spring mass system k1 and m have a natural frequency f1 Determine the value of k2 of another spring in terms of k1 which when placed in series with k1 lowers the natural frequency to 23 f1

COMPLETE SOLUTION OF SYSTEM EXECUTING SHM

The equation of motion of system executing SHM can be represented by

0)( =+ xkxm ampamp ----------------------------------------(1)

022

2=+

dtdx

dt

dx ω

The general solution of equation (1) can be expressed as

------------------------(2)

Where A and B are arbitrary constant which can be determined from the initial conditions of the system Two initial conditions are to be specified to evaluate these

constants x=x0 at t=0 and xamp = Vo at t=0 substituting in the equation (2) Energy method In a conservative system the total energy is constant The dwell as natural frequency can be determined by the princienergy For free vibration of undamped system at any instant and partly potential The kinetic energy T is stored in the masswhere as the potential energy U is stored in the form of sdeformation or work done in a force field such as gravity The total energy being constant T+U = constant Its rate of cha

t)(sin Bt)cos( AX ω+ω=

t)(sin V

t)cos( xx o0 ω

ω+ω=

Is the required complete

solution

28

ifferential equation as ple of conservation of of time is partly kinetic by virtue of its velocity train energy in elastic

nge

Is given by [ ] 0UTdtd

=+

From this we get a differential equation of motion as well as natural frequency of the system Determine the natural frequency of spring mass system using energy method

Determine the natural frequency of system shown in figure Determine the natural frequency of the system shown in figure Is there any limitation on the value of K Discuss Determine the natural frequency of s

m

l k

m

a

θ

m

a

θ

l

k

29

k

k

ystem shown below Neglect the mass of ball

l

m

30

A string shown in figure is under tension T which can be assumed to remain constant for small displacements Find the natural frequency of vertical vibrations of spring An acrobat 120kg walks on a tight rope as shown in figure The frequency of vibration in the given position is vertical direction is 30 rads Find the tension in the rope A manometer has a uniform bore of cross section area A If the column of liquid of length L and Density ρ is set into motion as shown in figure Find the frequency of the resulting oscillation

m

l

a

T T

36m 8m

31

k

R

k

m

r

m

Find the expression for natural frequency of system shown in the figure Neglect the mass of the cantilever beam Study the special case i) k=Infinity ii) I = infinity Determine the expression for the natural frequency of a system shown in figure The two discs are keyed to a common shaft and have a combined mass moment of inertia about centre of oscillation O It is assumed that the card attached to mass m does not stretch and is always under tension Determine the natural frequency of a system shown

m

l

32

Determine the expression for the natural frequency of the system shown in figure Assume that the wires connecting the masses do not stretch and are always in tension

Determine the natural frequency of spring mass system taking the MASS OF SPRING (ms ) into account

M3 M1

M2

k2 k1

m

k

33

RAYLEIGHrsquoS METHOD

This is the extension of energy method Here we can determine the natural frequency of a conservative system without finding the equation of motion The natural frequency is a function of the rate of change of Kinetic energy and potential energy of the system From the principle of conservation of energy we have T+U = Constant T= KE and U =PE We can write the above equation as T1+U1 = --------------------------(1) Where 1 and 2 represents tow instances of time Let 1 be the time when the mass is passing through static equilibrium position Let 2 be the time corresponding to the mass displacement of the mass At this instant the velocity f the mass is zero and hence Substituting in equation (1) we get ( )Max = ( )Max Above equation leads directly to natural frequency Determine the natural frequency of spring mass system using Rayleighrsquos method m

34

r

B

K K

Home work

Determine the natural frequency of pendulum using Rayleighrsquos method

A Cylinder of mass m and moment of inertia Jo is to role without slipping but is sustained by spring K as shown in figure Determine the natural frequency of oscillation

Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping

a

Determine the natural frequency of system shown where in cylindrical disc rolls over the inextensible spring A cylinder of radius r rolls without slipping on cylindrical the equation of motion and hence determine the naoscillations about the lowest point shown Repeat the above problem ndash Instead of cylinder assume swithout slipping on concave surface as shown above

r

K

35

portion of radius R Derive tural frequency of small

phere of radius with r rolls

The torsional pendulum has a natural frequency of 5 Hz What length of steel wire of diameter 2 mm should be used for pendulum The inertia of mass fixed at the free end is 00098 kgm2 Take g= 083x1011 Nm2 Determine the torsional natural frequency of the system shown in figure Neglect the mass moment of inertia of the shaft ( Figure not proportionate) Determine the natural frequency of simple pendulum considering the mass of the rod into account

Using the energy method find tThe chord may be assumed ine

Kt1

Kt2

20cm

06m 08m 04m

20cm 20cm 20cm

m

M

he natural frxtensible in

k

M

36

equency of the system shown in figure the spring mass pulley system

A rod of length l and depth H having a mass m rests on a semicylidrical surface os radius r It is tipped slightly and let go Find the frequency of oscillation

A thin bar of l having a mass m rests on a semicylidrical surface radius r It is tipped slightly and let go Find the frequency of oscillation

l

h

R

l

37

38

CHAPTER 3

DAMPED FREE VIBRATION OF SINGLE DEGREE FREEDOM SYSTEM LEARNING OBJECTIVES

] Introduction to damped free vibration

] Terminologies used in damped free vibration

] Three different types of damping

] Concept of critical damping and its importance

] Study of response of viscous damped system for cases of under damping

critical and over damping

] Concept of logarithmic decrement

] Problems related to damped free vibration problems

Viscously damped free vibration of single degree freedom sytem Consider a single degree freedom system with viscous damping as shown in figure For a viscous damped system the damping force is

where c is a constant of proportionality It is called damping coefficient or coefficient of viscous damping

Symbolically it is designated by a dashpot as shown in Fig 3

x xampx

39

At any instant of time let the mass be displaced by x from the equilibrium position Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

Okxxcxm =minusminus= ampampamp

Okxxcxm =++ ampampamp ----------------------------------------------------------(1)

or

0kxdtdx

Cdt

xdM

2

2

=++

Assuming the solution of equation as

The first term in the above equation m2Ct

eminus

is an exponentially decaying function of time The behavior of the terms in the parenthesis however depends on whether numerical value within the radicals is positive zero or negative

CASE 1 if 0mk

m2c

2

=minus

40

CASE 2 if 0mk

m2c

2

=gt

CASE 3 if 0mk

m2c

2

=lt

41

Critically damped coefficient Critically damped coefficient is defined as the value of the damping coeffient for which radical in equation

mk

m2c

m2c

S2

21 minus

plusmn

= becomes zero

Damping Ratio ( Damping factor) ( ξ ) It is defined as the ratio of damping coefficient to the critical damping coefficient Note The damping coefficient c is defined as the actual amount of damping while the damping coefficient is a constant depending upon the mass and stiffness of the system The roots of characteristics equation can be now written as Note if c=cc or ξ =1 then system is critically damped if cgt cc or ξ gt1 then system is over damped I if clt cc or ξ lt1 then system is under damped

42

Take home quiz Derive General solution of different damped system

1 Critically damped system 2 Under damped system 3 Over damped system

Logarithmic Decrement It is used to determine the amount of damping present in system This measure The rate of decay of free vibration The larger the damping the greater will be rate of decay Logarithmic decrement is defined as the natural logarithm of ratio of any two successive amplitudes

=δ

2

1

xx

ln

Expression for logarithmic decrement (δ ) Consider a damped free vibration which is expressed by

( ) ( ) φ+ξminusω= ξω t 1sinXex 2n

tn

x is the displacement at any instant of time X and φ are arbitrary constants

21

2

ξminus

πξ=δ

43

Tutorial problems 1 Derive an expression for damping ratio in terms of logarithmic decrement

2 Show that logarithmic decrement

=δ

n

o

xx

lnn1

where xo is the initial amplitude xn

is amplitude after n cycles 3 A mass of 1 kg is to be supported on a spring having stiffness K= 9800 Nm The damping constant c= 59 Nsm Determine the natural frequency of system Find also logarithmic decrement and amplitude after 3 cycles if initial displacement is 03 cm 4 A vibrating system is defined by the following parameters M = 3kg k= 100 Nm damping coefficient = 3 Nsm determine Damping factor natural frequency damped vibration logarithmic decrement ratio of consecutive amplitude no of cycles after which the original amplitude is reduced to 20 5A single degree damped vibrating system a suspended mass of 18kg makes 15 oscillations in 003 seconds The amplitude decreases to 025 of the initial value after 5 oscillations Determine the stiffness of spring Logarithmic decrement damping factor and damping coefficient 6 A damped vibration record of a spring mass system shows the following data Amplitude at the end of 2nd cycle is 9mm amplitude at eh end of 3rd cycle is 6mm amplitude at the end of 4th cycle is 4mm stiffness of spring =8000 Nm weight =20N Find the logarithmic decrement damping force at unit velocity periodic time of vibration 7 For the system shown in figure the characteristics of dashpot is such that the constant force of 49 N is applied to the piston it velocity is found to be constant at 012 ms Determine the value of C would you expect the complete system to be periodic or aperiodic 8 For a torsional system shown in the figuref1 is natural frequency in air and f2 is the natural frequency when immersed in oil Determine the damping coefficient

Viscous fluid

Disc of mass moment of Inertia J

2a

C

m M

b

a a

k

9 A thin plate of area A and weigh w is fixed to the end of the spring and allow to oscillate in a viscous fluid as shown in figure It f1 is the frequency of oscillation in air and f2 in liquid Show that

22

21 ff

gA2

minusπω

=micro where the damping force on the plate

AV2Fd micro= V being the velocity of vibration 10 A torsional pendulum has a natural frequency of 175 cyclessecond vibrating in vacuum The mass moment of inertia is 2 kgm2 It is immersed in oil and it is observed that the natural frequency is 142 cyclessecond Find the damping torque If the disc is replaced by 34 deg When in oil find the displacement at the end of the first complete cycle 11 An automobile can be modeled as a mass placed on 4 shock absorbers each consisting of a spring and a damper Such that each spring is equally loaded Determine the stiffness damping constant of each shock absorber So that natural frequency is 2Hz and system is critically damped The mass of vehicle is 200kg 12 Write the differential equation of motion for the system shown in the figure below Determine the natural frequency of the damped oscillation and critical damping coefficient

Determine the differential equation of motion and find the critical cdamped natural frequency for system shown in figure

a

C

m M

M =200kg

44

200kg kgkg

oefficient and

45

Derive the equation of motion and find the critical damping coefficient and damped natural frequency for the system shown below Write the differential equation of motion and determine the expression for critical damping coefficient and damped natural frequency of system shown in figure

Set up a differential equation of motion for system shown in figure and determine the following Undamped natural frequencies critical damping coefficient damping ratio damped natural frequency

b

C

kM

a

m

46

A mass of 300N is resting on two springs of stiffness 3000 Nm each and a dashpot of damping coefficient 150 Nsm as shown in figure If the initial velocity of 10cms is given to mass at its equilibrium position what will be the displacement from the equilibrium position at the end of first second Derive the formula used A gun barrel of mass 600kg has a recoil spring of stiffness 294 knm if the barrel recoils 13m on firing determine i) The initial recoil velocity of barrel ii) The critical damping coefficient of dashpot which is engaged at the end of recoil stroke Iii) Time required for the barrel to return at the end of recoil stroke Iv) time required for the barrel to return to a position 5cm from the initial position

W= 3000 N

FIRING

CHAPTER 4

FORCED VIBRATION LEARNING OBJECTIVES

] Introduction to forced free vibration

] Terminologies used in forced vibration

] Concept of magnification factor

] Study of reciprocating and rotating unbalance

] Concept of base excitation and support excitation

] Concept of Transmissibility ratio

] Problems related to forced vibration

The vibration that takes place under the excitation of external forces are called

forced vibration When a system is subjected to harmonic excitation it is forced to vibrate at the same frequency as that of excitation Most of the mechanical systems are prone to steady state forced vibration which cause fatigue failure The vibration problem also becomes very important when any of excitation frequencies coincide with one of natural frequencies of a system Under this condition resonance occurs during which large amplitude of vibrations are observed To avoid resonance external frequency (operating speed) may be changed or properties of the system may be changed to alter the natural frequency In some cases sufficient amount of damping may be provided to avoid large amplitude during resonance Thus the problem of force vibration is very important in mechanical design

FORCED VIBRATIONS OF SINGLE DOF SYSTEMS UNDER HARMONIC EXCITATION

` [ ] tsin FF o ω=

x

sin FF o=

Fo

xampamp

[ ] tω

xamp

47

48

Consider a spring mass system having a viscous damping excited by a harmonic force [ ] tsin FF o ω= as shown in figure The various forces acting on the system is shown in figure Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

O)tsin(Fkxxcxm o =ω+minusminus= ampampamp

)tsin(Fkxxcxm o ω=++ ampampamp ----------------------------------------------------------(1)

or

)tsin(Fkxdtdx

Cdt

xdM o2

2

ω=++

Equation (1) represents the differential equation of motion of the system The solution to this equation consists of two parts namely complementary function and a particular integral

The first of solution is also termed as transient solution of homogeneous solution and can be obtained from the homogeneous differential equation 0kxxcxm =++ ampampamp

The second part of solution particular integral is also termed as steady state solution ad is of form )tsin(Xx φminusω= Where X= steady state amplitude φ is phase difference between external force and displacement

)tsin(Xx φminusω=

The vector representation of the shown

49

The force polygon for the above vector diagram can be written as follows

( )φminusωt O

KX

xm 2ω

Fo

xCω

tω

φ

50

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

is the governing equation

Zero frequency deflection (Xst) It is deflection observed at zero frequency ( when the external frequency 0=ω

( ) kF

001

kF

x o

22

o

st =

+minus=

Magnification factor (M) It is defined as the ratio of steady state amplitude to zero frequency deflection xst

M = arX

X=

A plot of magnification factor versus frequency ratio

damping factor ( )ξ is called frequency plot of a harmon All the frequency terms starts from magnification factor of damping ratio ( )ξ The amplitude of vibration at any on the value of damping The amplitude will be higher for higher damping The amplitude of vibration at reso

amplitude tends to zero as

ωω

n tends to infinity

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

Freq

ζ =

ζ = 1

Magnification Factor M = XXst

kFo

51

ωω

nfor varying values of

ically excite vibrating system

M=1 what ever may be value particular frequency depends

for lesser damping and lesser nance will be very high The

uency Ratio r = (ωωn)

ζ = 0 0

ζ = 025 ζ = 0375 ζ = 0 5

ζ = 0707

ζ = 2

52

Plot of phase angle φ versus frequency ratio

ωω

n is shown in the

figure The phase angle φ is zero for all the values of damping ratio

at

ωω

n=0 The phase

angle φ is 90 for all the values of damping ratio

at

ωω

n=1 ie at

resonance The phase angle between o to 90 deg for all the values of

ωω

n= 0 to 1 and

between 90 deg to 180 deg for all the values of

ωω

ngt1

Class work Derive an expression for frequency at peak amplitude

2np 21 ξminusω=ω=ω is the required equation

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

53

Derive expression amplitude at resonance ( In dimensional form) In dimensional form We know that steady state amplitude

( ) ( )

ω+ωminus=

222

o

Cmk

Fx Equation in dimensional form

We know that at resonance mk

n =ω=ω

( )

ω+

minus

=2

n

2

oR

Cmk

mk

Fx

( )

ω=

2n

oR

C

Fx

( )

ω=

n

oR C

Fx

Class work Derive an expression for amplitude at resonance ( in non dimensional form)

Tutorial Problems on Forced Vibration 1A weight of 60N is suspended by a spring stiffness 12 kNm is forced to vibrate by a harmonic force of 10N assuming viscous damping of 0086 kN-sm Find i) resonance frequency ii) amplitude at resonance iii) phase angle at resonance iv) frequency corresponding to peak amplitude 2 A periodic torque having a maximum value of 05888Nm at a frequency of 4 rads is impressed upon a flywheel suspended from a wire The wheel has mass moment of inertia 012 kgm2 and the wire has stiffness 1176 Nmrad A viscous dashpot applies damping couple of 0784 Nm an angular velocity of 2 rads calculate i)maximum angular displacement from rest position ii) maximum couple applied to dashpot Iii) angle by which the angular displacement lags the torque A mass of 1019 kg is suspended from a spring stiffness 20 Nmm viscous damping causes the amplitude to decrease to 110th of initial value in 4 complete cycle If a periodic force of 20 cos(50 t) is applied to the mass Find the amplitude of forced vibration What would be the amplitude of oscillation if the applied force coincides with the natural period

m=1019 kg

[ ] t50cos 30F =

54

55

ROTATING UNBALANCE Unbalance in rotating machinery is a common source of vibration excitation Consider a machine of mass m having mounted on a foundation of stiffness k and damping coefficient C as shown in figure The unbalance is represented by an eccentric rotating mass m with eccentricity (e) and having a angular velocity (ω) let x be the displacement of non rotating mass from the static equilibrium position The displacement of rotating mass m from the static equilibrium is ( ) tsin( ex ω+ Applying Newtonrsquos law of motion we have

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

56

RECIPROCATING UNBALANCE Unbalance in reciprocating machinery is a common source of vibration excitation Consider a machine of mass M having mounted on a foundation of stiffness K and damping coefficient C as shown in figure The unbalance is represented by a reciprocating mass (m) having crank radus (e) and connecting rod of length (l) Let ωωωω be angular velocity of the crank Let x be the displacement of non reciprocating mass from the static equilibrium position at any instant of time The displacement of reciprocating mass m from the static equilibrium position is

)2sin(sin +++ tle

tex ωωωωωωωω

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

57

Note The equations obtained in reciprocating unbalance is same as the rotating unbalance and having the following differences Rotating Unbalance Reciprocating Unbalance Here mass of the rotating element is considered

Here mass of reciprocating element is considered

Here the eccentricity (e) which is the distance between the centre of rotation and CG

Here the eccentricity e is the crank radius and crank radius = stroke length2

The frequency plot of non dimensional factor( ( )me

MX versus frequency ratio

ωω

n for various values of ( )ξξξξ is shown in figure This frequency plot is same for

the rotating and reciprocating unbalance with the following points are noted from the diagram

1 Damping ratio plays very important role during resonance in controlling the amplitude

2 When

ωω

n=0 then ( )me

MX =0 for all the values of damping ratio ( )ξξξξ

3 When

ωω

nis higher than ( )me

MX tends to unity

4 At very high speeds ie at very high frequency ratio

ωω

n there is no effect

of damping The frequency plot of phase angle versus the frequency ratio is given as shown in figure The explanation is same as in forced harmonic vibration

58

VIBRATION ISOLATION

Vibratory forces and motions generated by machine and other caused are unavoidable Howeever their effect on the other dynamical systems can be minimized by proper isolator design An isolation system is either to protect o delicate machine from excessive vibration transmitted to it from its supporting structure or to prevent vibratory forces and motions generated by machines from being transmitted to its surroundings The effectiveness of isolation may be measured in terms of forces and motion transmitted The first type is known as force isolation and the second type is known as motion isolation

Force isolation (Force Transmissibility) When vibratory forces generated by machine are isolated the effectiveness of forced isolation can be made by defining the term transmissibility The term transmissibility is defined as the ratio of force transmitted to the foundation to that of the external impressed force

Force Transmissibility = o

tr

FF

force Impressed Externalfoundation toted transmitForce

=

Consider viscous damped spring mass system as shown in figure Upon this system an external harmonic force ) t (Sin Fo ω is applied The difference forces acting on system are inertia forces damping force spring force and external force All these forces are represented on force polygon shown below Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these two forces

( ) ( )22 xckx ω+=

59

This is the expression for force transmissibility in dimensional form To represent same in the non dimensional form divide numerator and denominator by k

2

n

22

n

2

n

o

tr

21

21

FF

TR

ωω

ξ+

ωω

minus

ωω

ξ+

==

PHASE DIFFERNECE BETWEEN EXTERNAL FORCE AND DISPLACEMENT Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these

ωω

minus

ωω

ξ=

ωminusω

=φ minusminus2

n

n1

2

1

1

2tan

mKC

tan

The phase difference between force transmitted to the foundation Ftr to the displacement

kc

KxxC

tanω

=ω

=α or

ωω

ξ=

ω=α minusminus

n

11 2tanKx

xCtan

The phase difference between external force Fo and force transmitted to foundationis β

60

β= αminusφ =

ωω

minus

ωω

ξminus

2

n

n1

1

2tan -

ωω

ξminus

n

1 2tan

Displacement Transmissibility It is the ratio of amplitude of steady state vibration of machine to that of the base motion Forced vibration due to support motion Consider a spring mass damper system subjected to harmonic excitation )tsin(Yy ω= at the base as shown in figure Let x be the absolute displacement of the mass at any instant of time t The different forces acting on system are shown in free body diagram Applying Newtonrsquos Law of motion Ma = sumF

)yx(k)yx(cxm minusminusminusminus= ampampampamp

0)yx(k)yx(cxm =minus+minus+ ampampampamp

61

Frequency plot of transmissibility versus frequency ratio

62

The plot of transmissibility versus frequency ratio is as shown in figure The following points can be noted from it

i) The value of transmissibility is unity at frequency ratio

ωω

n=0 and 2

ii) For an undamped system ( )ξ =0 transmissibility tends to infin at resonance iii) The value of transmissibility is less than unity for the value of frequency ratio

greater than 2

iv) For frequency ratio

ωω

nlt 2 smaller damping ratio lead to larger values of

transmissibility on the other hand for frequency ratio

ωω

ngt 2 smaller damping

ratio leads to smaller value of transmissibility

The frequency plot of transmissibility (TR) versus frequency ratio

ωω

n can be

broadly divided into two zones

1

ωω

n= 0 to 2

2

ωω

n= 2 to infin

CASE 1

When

ωω

n= 0 to 2

In this zone the value of transmissibility is always greater than 1 for all values of damping ratio In this zone the transmissibility decreases with increase in damping Hence we can conclude that in this zone damping is useful in isolating vibration forces being transmitted to the support CASE 2

When

ωω

n= 2 to infin

In this zone the value of transmissibility is less than 1 for all values of damping ratio In this zone the transmissibility increases with increase in damping Hence we can conclude that in this zone damping will adversely affect in isolating vibratory forces

63

Tutorial problems on Forced Vibration 1 Machine of mass 75kg is mounted on spring of stiffness 12000 Nm with an assumed damping factor ξ =02 piston within the machine of mass 2 kg has a reciprocating motion with a stroke of 75 cm and speed of 15 cyclessecond Assuming the motion of piston to be harmonic determine amplitude of machine ( X) Transmissibility and force transmitted to foundation phase angle of transmitted force wrt to exciting force 2 Motor weighs 600N is mounted on a simple beam that has a spring rate of 400 Ncm The motor armature base 160 N and has an eccentricity of 0005cm What will be the amplitude of motion when it runs at 1760 RPM Neglect damping the deflection of armature shaft and weight of the beam 3 A Heavy machine of mass 300 kg is supported on resilient foundation the static deflection of foundation due to weight of machine is found to be 75mm It is observed that the machine vibrates with an amplitude of 10mm when the base of the foundation is subjected to harmonic oscillation at the undampeds natural frequency of system with an amplitude of 25mm Find the damping constant of foundation Ii) dynamic force amplitude on the base iii) amplitude of the displacement of machine relative to base 4 An air compressor of mass 450 kg operates at a constant speed of 1750 RPM The reciprocating masses are 10 kg the crank radius is 100mm If the damper for mounting introduces a damping factor 015 i) specify the springs of mounting such that only 20 of the unbalanced force is transmitted to the foundation ii) determine the magnitude of the transmitted force bull Repeat the above problem by assuming that there is no damping in system

5 A body of a car having mass 1500 kg and is mounted on 4 equal springs which deflect through 225 cm under the weight of the body A total damping forces the 4 shock absorbers is 46 N at a velocity of 1cmsec The car is placed with all the four wheels on a test platform which is move up and down at resonant speed With an amplitude of 25cm Find the amplitude of the car body on its springs assuming its CG to be in the centre of wheel base 6A 75 kg machine which is mounted on springs of stiffness 116x105 Nm with the damping of 20=ξ A 2kg piston within the machine as reciprocating motion with a stroke of 008m and speed of 3000 RPM Assuming the motion of piston to be harmonic determine the amplitude of vibration of machine

7 The spring of an automobile are compressed 10cm under its own weight Find the critical speed when a trailer is traveling over road with a people approximaterly by a sine wave of amplitude 75 cm and wave length 15m What will be the amplitude of vibration at 64kmhour 8 An eccentric mass excites us used to determine vibratory characteristics of a structure of mass 25kg as given in figure At a speed of 1000 RPM the eccentric mass to beat top position at the instant the structure was moved upwards through its static equilibrium position and corresponding amplitude was 20mm The eccentric mass is 05kg at an eccentricity of 10mm Determine i) Undamped natural frequency of the system ii) Damping ratio iii) Steady state amplitude and phase angle at 1500 RPM iv) Force transmitted to the ground at 1200 RPM

M=25kg

C

K

Eccentric mass

64

65

CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Support motion Consider a spring mass damper system as shown in figure Let it excited by motion

of support let y be the harmonic displacement of support From Newtonrsquos second law of motion

)yx(k)yx(cxm minusminusminusminus= ampampampamp

)1(0)yx(k)yx(cxm minusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminus=minus+minus+ ampampampamp

Let z be the relative displacement of the mass with respect to support then

Z=x-y

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

60 cm 70 cm

Determination of equivalent spring stiffness when the springs are arranged in parallel Force acting on K1 spring = F1=k1x Force acting on K2 spring = F2 Force required for an equivalent spring keq to defined by x given by F= Keq x But F = F1 +F2 Tutorial problems on Equivalent stiffness of springs 1Determine the equivalent stiffness for the system shown in figure 2 Determine the equivalent stiffness for the system shown in figure

3 Determine the equivale

M

3

M

Nm 102 6x

18

nt stiffness for the system shown in figure

38 kg

Nm 101 6x Nm 102 6x

Nm 106x

2k

k

k

2k

3k

k

2k

k

19

60 cm 80 cm 50 cm

50 cm 60 cm

4Determine the equivalent stiffness for the system shown in figure

5Replace the following torsional stiffness by a single shaft having radius 4cm and find the length required for the equivalent shaft Assume the material of given system and equivalent system is same R1= 3cm R2= 5cm DAMPING Every vibration energy is gradually converted into heat or sound Hence the displacement during vibration gradually reduces The mechanism by which vibration energy is gradually converted into heat or sound is known as damping A damper is assumed to have either mass or elasticity Hence damping is modeled as one or more of the following types Viscous damping Coulomb or dry friction damping materials or solid or hysteric damping Viscous damping Viscous damping is most commonly used damping mechanism in vibration analysis When the mechanical system vibrates in a fluid medium such as air gas water or oil the resistance offered by the fluid to the moving body causes energy to be dissipated In this case the amount of dissipated energy depends on many factors such as size or shape of the vibrating body the viscosity of the fluid the frequency of vibration and velocity of fluid Resistance due to viscous damping is directly proportional to the velocity of vibration

dF V αααα

dF = xC amp

Where C= damping coefficient Fd = damping force Examples of Viscous damping

1) Fluid film between sliding surface 2) Fluid flow around a piston in a cylinder 3) Fluid flow through orifice 4) Fluid flow around a journal in a bearing

Rreq =4cm

leqn

stress

Strain

Coulomb damping or dry friction damping Here a damping force is constant in magnitude but opposite in direction to that of the motion of vibrating body It is caused by the friction between the surfaces that are dry or have insufficient lubrication Material or solid or hysteric damping

When the materials are deformed energy is absorbed and dissipated by the material The effect is due to friction between the internal planes which slip or slide as the deformation takes place When a body having the material damping is subjected to vibration the stress strain diagram shows the hysteresis loop as shown in the figure The area of the loop denotes the energy lost per unit volume of the body per cycle

20

21

ττττ ττττ2 ττττ3

ττττ

X(t)

FOURIER SERIES The simplest of periodic motion happens to be SHM It is simple to handle but the motion of many vibrating system is not harmonic (but periodic) Few examples are shown below

Forces acting on machines are generally periodic but this may not be harmonic for example the excitation force in a punching machine is periodic and it can be represented as shown in figure 13 Vibration analysis of system subjected to periodic but nonharmonic forces can be done with the help of Fourier series The problem becomes a multifrequency excitation problem The principle of linear superposition is applied and the total response is the sum of the response due to each of the individual frequency term

Any periodic motion can be expressed as an infinite sum of sines and cosines terms If x(t) is a periodic function with period t its Fourier representation is given by

X(t) = (((( )))) (((( )))) (((( ))))t sint cost cos2 111 ωωωωωωωωωωωω baaao ++++++++++++

= (((( )))) (((( ))))t sint cos2 1

ωωωωωωωω nnn

o bnaa

++++++++sumsumsumsuminfininfininfininfin

====

tππππωωωω

2==== = Fundamental frequency ndash (1)

where ao an bn are constants

ττττ ττττ2 ττττ3

ττττ

X(t)

22

t

X(t)

035

025

Determination of constants To find ao Integrate both sides of equation(1) over any interval ττττ All intergrals on the RHS of the equation are zero except the one containing ao

(((( ))))dttxao

o 2

2

intintintint====ωωωωππππ

ππππωωωω

= (((( ))))dttxo

2intintintintττττ

ττττ

To find an multiply equation 1 by cos (((( ))))tnωωωω and Integrate over any interval ττττ All intergrals

(((( )))) (((( )))) dttxao

n tncos 2

2

ωωωωππππωωωω ωωωω

ππππ

intintintint==== (((( )))) (((( )))) dttxo

tncos 2

ωωωωττττ

ττττ

intintintint====

To find bn multiply equation 1 by sin (((( ))))tnωωωω and Integrate over any interval ττττ All intergrals

(((( )))) (((( )))) dttxao

n tnsin 2

2

ωωωωππππωωωω ωωωω

ππππ

intintintint==== (((( )))) (((( )))) dttxo

tnsin 2

ωωωωττττ

ττττ

intintintint====

Find the Fourier series for the curve shown below

ττττ ττττ2 ττττ3

t

X(t)

E

ττττ ττττ2 ττττ3

t

X(t)

ππππ2

t

X(t) ππππ

1 Represent for the periodic motion shown in the figure

23

CHAPTER 2 UNDAMPED FREE VIBRATION LEARNING OBJECTIVES

] Introduction to undamped free vibration

] Terminologies used in undamped free vibration

] Three methods to solve the undamped free vibration problems

] Problems related to undamped free vibration problems

Free vibrations are oscillations about a systems equilibrium position that occur in the absence of an external excitation force If during vibrations there is no loss of energy it is known as undamped vibration The first step in solving a vibration problem is setting up the differential equation of motion Consider spring mass system which is assumed to move only along the vertical direction as shown below Let m be the mass of the block and k be the stiffness of the spring When block of mass m is attached to spring the deflection of spring will be ∆ known as static deflection In the static equilibrium position the free body diagram of forces acting on the mass is shown in Figure(b) Hence mg= kA Once the system is disturbed the system executes vibrations

Let at any instant of time t the mass is displaced from the equilibrium position x the different forces acting on the system are shown in figure (d) From Newtonrsquos second law of motion

sum = maF

Inertia force ( disturbing force) = restoring force

mgxkxm ++∆minus= )(ampamp

0)( =+ xkxm ampamp

or 0)( =+ xmk

xampamp

equation 2 is the differential equation of motion for spring mfigure Comparing equation (2) with the equation of SHM +xampampsince the vibrations of the above system are free( without the forces) we can write

sec radmk

n =ω

24

ass system shown in 0)(2 =xω

resistance of external

time period km

2 f1

n

πτ ==

from the equation(1) mg = k∆

∆=

q

mk

Difference between the translation ( rectilinear) and rotational system of vibration

Translatory Rotational

In the analysis thFORCES are co Linear stiffness K Problems 1A mass of 10kFind the natural 2 A spring massnatural frequencnatural frequenc

kt

m

sec radmk

n =ω

k

25

e disturbing and restoring nsidered

In the analysis In the analysis MASS Moment of Inertia (J) is considered

in Nm is considered

g when suspended from a spring causes a static deflection of 1cm frequency of system

system has a spring stiffness K Nm and a mass of m Kg It has a y of vibration 12 Hz An extra 2kg mass coupled to it then the y reduces by 2 Hz find K and m

secrad n =ω

26

3 A steel wire of 2mm diameter and 30m long It is fixed at the upper end and carries a mass of m kg at its free end Find m so that the frequency of longitudinal vibration is 4 Hz 4 A spring mass system has a natural period of 02 seconds What will be the new period if the spring constant is 1) increased by 50 2) decreased by 50 5 A spring mass system has a natural frequency of 10 Hz when the spring constant is reduced by 800 Nm the frequency is altered by 45 Find the mass and spring constant of the original system 6 Determine the natural frequency of system shown in fig is which shaft is supported in SHORT bearings 7 Determine the natural frequency of system shown in fig is which shaft is supported in LONG bearings where l is the length of bearing and E ndash youngrsquos modulus and I is moment of Inertia 7 Determine the natural frequency of system shown in fig is which shaft is supported in LONG bearings 8 A light cantilever beam of rectangular section( 5 cm deep x 25cm wide) has a mass fixed at its free end Find the ratio of frequency of free lateral vibration in vertical plane to that in horizontal 9 Determine the natural frequency of simple pendulum 10 Homogeneous square plate of size l and mass m is suspended from the mid point of one of its sides as shown in figure Find the frequency of vibration

l

l

27

11 A compound pendulum which is rigid body of mass m and it is pivoted at O The point of pivot is at distance d from the centre of gravity It is free to rotate about its axis Find the frequency of oscillation of such pendulum 12 A connecting rod shown in fig is supported at the wrist pin end It is displaced and allowed to oscillate The mass of rod is 5kg and centre of gravity is 20 cm from the pivot point O If the frequency of oscillation is 40 cyclesminute calculate the mass moment of inertia about its CG

13 A semi circular homogenous disc of radius r and mass m is pivoted freely about its centre as shown in figure Determine the natural frequency of oscillation 14A simply supported beam of square cross section 5mmx5mm and length 1m carrying a mass of 0575 kg at the middle is found to have natural frequency of 30 radsec Determine youngrsquos modulus of elasticity of beam 15 A spring mass system k1 and m have a natural frequency f1 Determine the value of k2 of another spring in terms of k1 which when placed in series with k1 lowers the natural frequency to 23 f1

COMPLETE SOLUTION OF SYSTEM EXECUTING SHM

The equation of motion of system executing SHM can be represented by

0)( =+ xkxm ampamp ----------------------------------------(1)

022

2=+

dtdx

dt

dx ω

The general solution of equation (1) can be expressed as

------------------------(2)

Where A and B are arbitrary constant which can be determined from the initial conditions of the system Two initial conditions are to be specified to evaluate these

constants x=x0 at t=0 and xamp = Vo at t=0 substituting in the equation (2) Energy method In a conservative system the total energy is constant The dwell as natural frequency can be determined by the princienergy For free vibration of undamped system at any instant and partly potential The kinetic energy T is stored in the masswhere as the potential energy U is stored in the form of sdeformation or work done in a force field such as gravity The total energy being constant T+U = constant Its rate of cha

t)(sin Bt)cos( AX ω+ω=

t)(sin V

t)cos( xx o0 ω

ω+ω=

Is the required complete

solution

28

ifferential equation as ple of conservation of of time is partly kinetic by virtue of its velocity train energy in elastic

nge

Is given by [ ] 0UTdtd

=+

From this we get a differential equation of motion as well as natural frequency of the system Determine the natural frequency of spring mass system using energy method

Determine the natural frequency of system shown in figure Determine the natural frequency of the system shown in figure Is there any limitation on the value of K Discuss Determine the natural frequency of s

m

l k

m

a

θ

m

a

θ

l

k

29

k

k

ystem shown below Neglect the mass of ball

l

m

30

A string shown in figure is under tension T which can be assumed to remain constant for small displacements Find the natural frequency of vertical vibrations of spring An acrobat 120kg walks on a tight rope as shown in figure The frequency of vibration in the given position is vertical direction is 30 rads Find the tension in the rope A manometer has a uniform bore of cross section area A If the column of liquid of length L and Density ρ is set into motion as shown in figure Find the frequency of the resulting oscillation

m

l

a

T T

36m 8m

31

k

R

k

m

r

m

Find the expression for natural frequency of system shown in the figure Neglect the mass of the cantilever beam Study the special case i) k=Infinity ii) I = infinity Determine the expression for the natural frequency of a system shown in figure The two discs are keyed to a common shaft and have a combined mass moment of inertia about centre of oscillation O It is assumed that the card attached to mass m does not stretch and is always under tension Determine the natural frequency of a system shown

m

l

32

Determine the expression for the natural frequency of the system shown in figure Assume that the wires connecting the masses do not stretch and are always in tension

Determine the natural frequency of spring mass system taking the MASS OF SPRING (ms ) into account

M3 M1

M2

k2 k1

m

k

33

RAYLEIGHrsquoS METHOD

This is the extension of energy method Here we can determine the natural frequency of a conservative system without finding the equation of motion The natural frequency is a function of the rate of change of Kinetic energy and potential energy of the system From the principle of conservation of energy we have T+U = Constant T= KE and U =PE We can write the above equation as T1+U1 = --------------------------(1) Where 1 and 2 represents tow instances of time Let 1 be the time when the mass is passing through static equilibrium position Let 2 be the time corresponding to the mass displacement of the mass At this instant the velocity f the mass is zero and hence Substituting in equation (1) we get ( )Max = ( )Max Above equation leads directly to natural frequency Determine the natural frequency of spring mass system using Rayleighrsquos method m

34

r

B

K K

Home work

Determine the natural frequency of pendulum using Rayleighrsquos method

A Cylinder of mass m and moment of inertia Jo is to role without slipping but is sustained by spring K as shown in figure Determine the natural frequency of oscillation

Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping

a

Determine the natural frequency of system shown where in cylindrical disc rolls over the inextensible spring A cylinder of radius r rolls without slipping on cylindrical the equation of motion and hence determine the naoscillations about the lowest point shown Repeat the above problem ndash Instead of cylinder assume swithout slipping on concave surface as shown above

r

K

35

portion of radius R Derive tural frequency of small

phere of radius with r rolls

The torsional pendulum has a natural frequency of 5 Hz What length of steel wire of diameter 2 mm should be used for pendulum The inertia of mass fixed at the free end is 00098 kgm2 Take g= 083x1011 Nm2 Determine the torsional natural frequency of the system shown in figure Neglect the mass moment of inertia of the shaft ( Figure not proportionate) Determine the natural frequency of simple pendulum considering the mass of the rod into account

Using the energy method find tThe chord may be assumed ine

Kt1

Kt2

20cm

06m 08m 04m

20cm 20cm 20cm

m

M

he natural frxtensible in

k

M

36

equency of the system shown in figure the spring mass pulley system

A rod of length l and depth H having a mass m rests on a semicylidrical surface os radius r It is tipped slightly and let go Find the frequency of oscillation

A thin bar of l having a mass m rests on a semicylidrical surface radius r It is tipped slightly and let go Find the frequency of oscillation

l

h

R

l

37

38

CHAPTER 3

DAMPED FREE VIBRATION OF SINGLE DEGREE FREEDOM SYSTEM LEARNING OBJECTIVES

] Introduction to damped free vibration

] Terminologies used in damped free vibration

] Three different types of damping

] Concept of critical damping and its importance

] Study of response of viscous damped system for cases of under damping

critical and over damping

] Concept of logarithmic decrement

] Problems related to damped free vibration problems

Viscously damped free vibration of single degree freedom sytem Consider a single degree freedom system with viscous damping as shown in figure For a viscous damped system the damping force is

where c is a constant of proportionality It is called damping coefficient or coefficient of viscous damping

Symbolically it is designated by a dashpot as shown in Fig 3

x xampx

39

At any instant of time let the mass be displaced by x from the equilibrium position Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

Okxxcxm =minusminus= ampampamp

Okxxcxm =++ ampampamp ----------------------------------------------------------(1)

or

0kxdtdx

Cdt

xdM

2

2

=++

Assuming the solution of equation as

The first term in the above equation m2Ct

eminus

is an exponentially decaying function of time The behavior of the terms in the parenthesis however depends on whether numerical value within the radicals is positive zero or negative

CASE 1 if 0mk

m2c

2

=minus

40

CASE 2 if 0mk

m2c

2

=gt

CASE 3 if 0mk

m2c

2

=lt

41

Critically damped coefficient Critically damped coefficient is defined as the value of the damping coeffient for which radical in equation

mk

m2c

m2c

S2

21 minus

plusmn

= becomes zero

Damping Ratio ( Damping factor) ( ξ ) It is defined as the ratio of damping coefficient to the critical damping coefficient Note The damping coefficient c is defined as the actual amount of damping while the damping coefficient is a constant depending upon the mass and stiffness of the system The roots of characteristics equation can be now written as Note if c=cc or ξ =1 then system is critically damped if cgt cc or ξ gt1 then system is over damped I if clt cc or ξ lt1 then system is under damped

42

Take home quiz Derive General solution of different damped system

1 Critically damped system 2 Under damped system 3 Over damped system

Logarithmic Decrement It is used to determine the amount of damping present in system This measure The rate of decay of free vibration The larger the damping the greater will be rate of decay Logarithmic decrement is defined as the natural logarithm of ratio of any two successive amplitudes

=δ

2

1

xx

ln

Expression for logarithmic decrement (δ ) Consider a damped free vibration which is expressed by

( ) ( ) φ+ξminusω= ξω t 1sinXex 2n

tn

x is the displacement at any instant of time X and φ are arbitrary constants

21

2

ξminus

πξ=δ

43

Tutorial problems 1 Derive an expression for damping ratio in terms of logarithmic decrement

2 Show that logarithmic decrement

=δ

n

o

xx

lnn1

where xo is the initial amplitude xn

is amplitude after n cycles 3 A mass of 1 kg is to be supported on a spring having stiffness K= 9800 Nm The damping constant c= 59 Nsm Determine the natural frequency of system Find also logarithmic decrement and amplitude after 3 cycles if initial displacement is 03 cm 4 A vibrating system is defined by the following parameters M = 3kg k= 100 Nm damping coefficient = 3 Nsm determine Damping factor natural frequency damped vibration logarithmic decrement ratio of consecutive amplitude no of cycles after which the original amplitude is reduced to 20 5A single degree damped vibrating system a suspended mass of 18kg makes 15 oscillations in 003 seconds The amplitude decreases to 025 of the initial value after 5 oscillations Determine the stiffness of spring Logarithmic decrement damping factor and damping coefficient 6 A damped vibration record of a spring mass system shows the following data Amplitude at the end of 2nd cycle is 9mm amplitude at eh end of 3rd cycle is 6mm amplitude at the end of 4th cycle is 4mm stiffness of spring =8000 Nm weight =20N Find the logarithmic decrement damping force at unit velocity periodic time of vibration 7 For the system shown in figure the characteristics of dashpot is such that the constant force of 49 N is applied to the piston it velocity is found to be constant at 012 ms Determine the value of C would you expect the complete system to be periodic or aperiodic 8 For a torsional system shown in the figuref1 is natural frequency in air and f2 is the natural frequency when immersed in oil Determine the damping coefficient

Viscous fluid

Disc of mass moment of Inertia J

2a

C

m M

b

a a

k

9 A thin plate of area A and weigh w is fixed to the end of the spring and allow to oscillate in a viscous fluid as shown in figure It f1 is the frequency of oscillation in air and f2 in liquid Show that

22

21 ff

gA2

minusπω

=micro where the damping force on the plate

AV2Fd micro= V being the velocity of vibration 10 A torsional pendulum has a natural frequency of 175 cyclessecond vibrating in vacuum The mass moment of inertia is 2 kgm2 It is immersed in oil and it is observed that the natural frequency is 142 cyclessecond Find the damping torque If the disc is replaced by 34 deg When in oil find the displacement at the end of the first complete cycle 11 An automobile can be modeled as a mass placed on 4 shock absorbers each consisting of a spring and a damper Such that each spring is equally loaded Determine the stiffness damping constant of each shock absorber So that natural frequency is 2Hz and system is critically damped The mass of vehicle is 200kg 12 Write the differential equation of motion for the system shown in the figure below Determine the natural frequency of the damped oscillation and critical damping coefficient

Determine the differential equation of motion and find the critical cdamped natural frequency for system shown in figure

a

C

m M

M =200kg

44

200kg kgkg

oefficient and

45

Derive the equation of motion and find the critical damping coefficient and damped natural frequency for the system shown below Write the differential equation of motion and determine the expression for critical damping coefficient and damped natural frequency of system shown in figure

Set up a differential equation of motion for system shown in figure and determine the following Undamped natural frequencies critical damping coefficient damping ratio damped natural frequency

b

C

kM

a

m

46

A mass of 300N is resting on two springs of stiffness 3000 Nm each and a dashpot of damping coefficient 150 Nsm as shown in figure If the initial velocity of 10cms is given to mass at its equilibrium position what will be the displacement from the equilibrium position at the end of first second Derive the formula used A gun barrel of mass 600kg has a recoil spring of stiffness 294 knm if the barrel recoils 13m on firing determine i) The initial recoil velocity of barrel ii) The critical damping coefficient of dashpot which is engaged at the end of recoil stroke Iii) Time required for the barrel to return at the end of recoil stroke Iv) time required for the barrel to return to a position 5cm from the initial position

W= 3000 N

FIRING

CHAPTER 4

FORCED VIBRATION LEARNING OBJECTIVES

] Introduction to forced free vibration

] Terminologies used in forced vibration

] Concept of magnification factor

] Study of reciprocating and rotating unbalance

] Concept of base excitation and support excitation

] Concept of Transmissibility ratio

] Problems related to forced vibration

The vibration that takes place under the excitation of external forces are called

forced vibration When a system is subjected to harmonic excitation it is forced to vibrate at the same frequency as that of excitation Most of the mechanical systems are prone to steady state forced vibration which cause fatigue failure The vibration problem also becomes very important when any of excitation frequencies coincide with one of natural frequencies of a system Under this condition resonance occurs during which large amplitude of vibrations are observed To avoid resonance external frequency (operating speed) may be changed or properties of the system may be changed to alter the natural frequency In some cases sufficient amount of damping may be provided to avoid large amplitude during resonance Thus the problem of force vibration is very important in mechanical design

FORCED VIBRATIONS OF SINGLE DOF SYSTEMS UNDER HARMONIC EXCITATION

` [ ] tsin FF o ω=

x

sin FF o=

Fo

xampamp

[ ] tω

xamp

47

48

Consider a spring mass system having a viscous damping excited by a harmonic force [ ] tsin FF o ω= as shown in figure The various forces acting on the system is shown in figure Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

O)tsin(Fkxxcxm o =ω+minusminus= ampampamp

)tsin(Fkxxcxm o ω=++ ampampamp ----------------------------------------------------------(1)

or

)tsin(Fkxdtdx

Cdt

xdM o2

2

ω=++

Equation (1) represents the differential equation of motion of the system The solution to this equation consists of two parts namely complementary function and a particular integral

The first of solution is also termed as transient solution of homogeneous solution and can be obtained from the homogeneous differential equation 0kxxcxm =++ ampampamp

The second part of solution particular integral is also termed as steady state solution ad is of form )tsin(Xx φminusω= Where X= steady state amplitude φ is phase difference between external force and displacement

)tsin(Xx φminusω=

The vector representation of the shown

49

The force polygon for the above vector diagram can be written as follows

( )φminusωt O

KX

xm 2ω

Fo

xCω

tω

φ

50

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

is the governing equation

Zero frequency deflection (Xst) It is deflection observed at zero frequency ( when the external frequency 0=ω

( ) kF

001

kF

x o

22

o

st =

+minus=

Magnification factor (M) It is defined as the ratio of steady state amplitude to zero frequency deflection xst

M = arX

X=

A plot of magnification factor versus frequency ratio

damping factor ( )ξ is called frequency plot of a harmon All the frequency terms starts from magnification factor of damping ratio ( )ξ The amplitude of vibration at any on the value of damping The amplitude will be higher for higher damping The amplitude of vibration at reso

amplitude tends to zero as

ωω

n tends to infinity

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

Freq

ζ =

ζ = 1

Magnification Factor M = XXst

kFo

51

ωω

nfor varying values of

ically excite vibrating system

M=1 what ever may be value particular frequency depends

for lesser damping and lesser nance will be very high The

uency Ratio r = (ωωn)

ζ = 0 0

ζ = 025 ζ = 0375 ζ = 0 5

ζ = 0707

ζ = 2

52

Plot of phase angle φ versus frequency ratio

ωω

n is shown in the

figure The phase angle φ is zero for all the values of damping ratio

at

ωω

n=0 The phase

angle φ is 90 for all the values of damping ratio

at

ωω

n=1 ie at

resonance The phase angle between o to 90 deg for all the values of

ωω

n= 0 to 1 and

between 90 deg to 180 deg for all the values of

ωω

ngt1

Class work Derive an expression for frequency at peak amplitude

2np 21 ξminusω=ω=ω is the required equation

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

53

Derive expression amplitude at resonance ( In dimensional form) In dimensional form We know that steady state amplitude

( ) ( )

ω+ωminus=

222

o

Cmk

Fx Equation in dimensional form

We know that at resonance mk

n =ω=ω

( )

ω+

minus

=2

n

2

oR

Cmk

mk

Fx

( )

ω=

2n

oR

C

Fx

( )

ω=

n

oR C

Fx

Class work Derive an expression for amplitude at resonance ( in non dimensional form)

Tutorial Problems on Forced Vibration 1A weight of 60N is suspended by a spring stiffness 12 kNm is forced to vibrate by a harmonic force of 10N assuming viscous damping of 0086 kN-sm Find i) resonance frequency ii) amplitude at resonance iii) phase angle at resonance iv) frequency corresponding to peak amplitude 2 A periodic torque having a maximum value of 05888Nm at a frequency of 4 rads is impressed upon a flywheel suspended from a wire The wheel has mass moment of inertia 012 kgm2 and the wire has stiffness 1176 Nmrad A viscous dashpot applies damping couple of 0784 Nm an angular velocity of 2 rads calculate i)maximum angular displacement from rest position ii) maximum couple applied to dashpot Iii) angle by which the angular displacement lags the torque A mass of 1019 kg is suspended from a spring stiffness 20 Nmm viscous damping causes the amplitude to decrease to 110th of initial value in 4 complete cycle If a periodic force of 20 cos(50 t) is applied to the mass Find the amplitude of forced vibration What would be the amplitude of oscillation if the applied force coincides with the natural period

m=1019 kg

[ ] t50cos 30F =

54

55

ROTATING UNBALANCE Unbalance in rotating machinery is a common source of vibration excitation Consider a machine of mass m having mounted on a foundation of stiffness k and damping coefficient C as shown in figure The unbalance is represented by an eccentric rotating mass m with eccentricity (e) and having a angular velocity (ω) let x be the displacement of non rotating mass from the static equilibrium position The displacement of rotating mass m from the static equilibrium is ( ) tsin( ex ω+ Applying Newtonrsquos law of motion we have

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

56

RECIPROCATING UNBALANCE Unbalance in reciprocating machinery is a common source of vibration excitation Consider a machine of mass M having mounted on a foundation of stiffness K and damping coefficient C as shown in figure The unbalance is represented by a reciprocating mass (m) having crank radus (e) and connecting rod of length (l) Let ωωωω be angular velocity of the crank Let x be the displacement of non reciprocating mass from the static equilibrium position at any instant of time The displacement of reciprocating mass m from the static equilibrium position is

)2sin(sin +++ tle

tex ωωωωωωωω

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

57

Note The equations obtained in reciprocating unbalance is same as the rotating unbalance and having the following differences Rotating Unbalance Reciprocating Unbalance Here mass of the rotating element is considered

Here mass of reciprocating element is considered

Here the eccentricity (e) which is the distance between the centre of rotation and CG

Here the eccentricity e is the crank radius and crank radius = stroke length2

The frequency plot of non dimensional factor( ( )me

MX versus frequency ratio

ωω

n for various values of ( )ξξξξ is shown in figure This frequency plot is same for

the rotating and reciprocating unbalance with the following points are noted from the diagram

1 Damping ratio plays very important role during resonance in controlling the amplitude

2 When

ωω

n=0 then ( )me

MX =0 for all the values of damping ratio ( )ξξξξ

3 When

ωω

nis higher than ( )me

MX tends to unity

4 At very high speeds ie at very high frequency ratio

ωω

n there is no effect

of damping The frequency plot of phase angle versus the frequency ratio is given as shown in figure The explanation is same as in forced harmonic vibration

58

VIBRATION ISOLATION

Vibratory forces and motions generated by machine and other caused are unavoidable Howeever their effect on the other dynamical systems can be minimized by proper isolator design An isolation system is either to protect o delicate machine from excessive vibration transmitted to it from its supporting structure or to prevent vibratory forces and motions generated by machines from being transmitted to its surroundings The effectiveness of isolation may be measured in terms of forces and motion transmitted The first type is known as force isolation and the second type is known as motion isolation

Force isolation (Force Transmissibility) When vibratory forces generated by machine are isolated the effectiveness of forced isolation can be made by defining the term transmissibility The term transmissibility is defined as the ratio of force transmitted to the foundation to that of the external impressed force

Force Transmissibility = o

tr

FF

force Impressed Externalfoundation toted transmitForce

=

Consider viscous damped spring mass system as shown in figure Upon this system an external harmonic force ) t (Sin Fo ω is applied The difference forces acting on system are inertia forces damping force spring force and external force All these forces are represented on force polygon shown below Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these two forces

( ) ( )22 xckx ω+=

59

This is the expression for force transmissibility in dimensional form To represent same in the non dimensional form divide numerator and denominator by k

2

n

22

n

2

n

o

tr

21

21

FF

TR

ωω

ξ+

ωω

minus

ωω

ξ+

==

PHASE DIFFERNECE BETWEEN EXTERNAL FORCE AND DISPLACEMENT Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these

ωω

minus

ωω

ξ=

ωminusω

=φ minusminus2

n

n1

2

1

1

2tan

mKC

tan

The phase difference between force transmitted to the foundation Ftr to the displacement

kc

KxxC

tanω

=ω

=α or

ωω

ξ=

ω=α minusminus

n

11 2tanKx

xCtan

The phase difference between external force Fo and force transmitted to foundationis β

60

β= αminusφ =

ωω

minus

ωω

ξminus

2

n

n1

1

2tan -

ωω

ξminus

n

1 2tan

Displacement Transmissibility It is the ratio of amplitude of steady state vibration of machine to that of the base motion Forced vibration due to support motion Consider a spring mass damper system subjected to harmonic excitation )tsin(Yy ω= at the base as shown in figure Let x be the absolute displacement of the mass at any instant of time t The different forces acting on system are shown in free body diagram Applying Newtonrsquos Law of motion Ma = sumF

)yx(k)yx(cxm minusminusminusminus= ampampampamp

0)yx(k)yx(cxm =minus+minus+ ampampampamp

61

Frequency plot of transmissibility versus frequency ratio

62

The plot of transmissibility versus frequency ratio is as shown in figure The following points can be noted from it

i) The value of transmissibility is unity at frequency ratio

ωω

n=0 and 2

ii) For an undamped system ( )ξ =0 transmissibility tends to infin at resonance iii) The value of transmissibility is less than unity for the value of frequency ratio

greater than 2

iv) For frequency ratio

ωω

nlt 2 smaller damping ratio lead to larger values of

transmissibility on the other hand for frequency ratio

ωω

ngt 2 smaller damping

ratio leads to smaller value of transmissibility

The frequency plot of transmissibility (TR) versus frequency ratio

ωω

n can be

broadly divided into two zones

1

ωω

n= 0 to 2

2

ωω

n= 2 to infin

CASE 1

When

ωω

n= 0 to 2

In this zone the value of transmissibility is always greater than 1 for all values of damping ratio In this zone the transmissibility decreases with increase in damping Hence we can conclude that in this zone damping is useful in isolating vibration forces being transmitted to the support CASE 2

When

ωω

n= 2 to infin

In this zone the value of transmissibility is less than 1 for all values of damping ratio In this zone the transmissibility increases with increase in damping Hence we can conclude that in this zone damping will adversely affect in isolating vibratory forces

63

Tutorial problems on Forced Vibration 1 Machine of mass 75kg is mounted on spring of stiffness 12000 Nm with an assumed damping factor ξ =02 piston within the machine of mass 2 kg has a reciprocating motion with a stroke of 75 cm and speed of 15 cyclessecond Assuming the motion of piston to be harmonic determine amplitude of machine ( X) Transmissibility and force transmitted to foundation phase angle of transmitted force wrt to exciting force 2 Motor weighs 600N is mounted on a simple beam that has a spring rate of 400 Ncm The motor armature base 160 N and has an eccentricity of 0005cm What will be the amplitude of motion when it runs at 1760 RPM Neglect damping the deflection of armature shaft and weight of the beam 3 A Heavy machine of mass 300 kg is supported on resilient foundation the static deflection of foundation due to weight of machine is found to be 75mm It is observed that the machine vibrates with an amplitude of 10mm when the base of the foundation is subjected to harmonic oscillation at the undampeds natural frequency of system with an amplitude of 25mm Find the damping constant of foundation Ii) dynamic force amplitude on the base iii) amplitude of the displacement of machine relative to base 4 An air compressor of mass 450 kg operates at a constant speed of 1750 RPM The reciprocating masses are 10 kg the crank radius is 100mm If the damper for mounting introduces a damping factor 015 i) specify the springs of mounting such that only 20 of the unbalanced force is transmitted to the foundation ii) determine the magnitude of the transmitted force bull Repeat the above problem by assuming that there is no damping in system

5 A body of a car having mass 1500 kg and is mounted on 4 equal springs which deflect through 225 cm under the weight of the body A total damping forces the 4 shock absorbers is 46 N at a velocity of 1cmsec The car is placed with all the four wheels on a test platform which is move up and down at resonant speed With an amplitude of 25cm Find the amplitude of the car body on its springs assuming its CG to be in the centre of wheel base 6A 75 kg machine which is mounted on springs of stiffness 116x105 Nm with the damping of 20=ξ A 2kg piston within the machine as reciprocating motion with a stroke of 008m and speed of 3000 RPM Assuming the motion of piston to be harmonic determine the amplitude of vibration of machine

7 The spring of an automobile are compressed 10cm under its own weight Find the critical speed when a trailer is traveling over road with a people approximaterly by a sine wave of amplitude 75 cm and wave length 15m What will be the amplitude of vibration at 64kmhour 8 An eccentric mass excites us used to determine vibratory characteristics of a structure of mass 25kg as given in figure At a speed of 1000 RPM the eccentric mass to beat top position at the instant the structure was moved upwards through its static equilibrium position and corresponding amplitude was 20mm The eccentric mass is 05kg at an eccentricity of 10mm Determine i) Undamped natural frequency of the system ii) Damping ratio iii) Steady state amplitude and phase angle at 1500 RPM iv) Force transmitted to the ground at 1200 RPM

M=25kg

C

K

Eccentric mass

64

65

CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Support motion Consider a spring mass damper system as shown in figure Let it excited by motion

of support let y be the harmonic displacement of support From Newtonrsquos second law of motion

)yx(k)yx(cxm minusminusminusminus= ampampampamp

)1(0)yx(k)yx(cxm minusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminus=minus+minus+ ampampampamp

Let z be the relative displacement of the mass with respect to support then

Z=x-y

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

19

60 cm 80 cm 50 cm

50 cm 60 cm

4Determine the equivalent stiffness for the system shown in figure

5Replace the following torsional stiffness by a single shaft having radius 4cm and find the length required for the equivalent shaft Assume the material of given system and equivalent system is same R1= 3cm R2= 5cm DAMPING Every vibration energy is gradually converted into heat or sound Hence the displacement during vibration gradually reduces The mechanism by which vibration energy is gradually converted into heat or sound is known as damping A damper is assumed to have either mass or elasticity Hence damping is modeled as one or more of the following types Viscous damping Coulomb or dry friction damping materials or solid or hysteric damping Viscous damping Viscous damping is most commonly used damping mechanism in vibration analysis When the mechanical system vibrates in a fluid medium such as air gas water or oil the resistance offered by the fluid to the moving body causes energy to be dissipated In this case the amount of dissipated energy depends on many factors such as size or shape of the vibrating body the viscosity of the fluid the frequency of vibration and velocity of fluid Resistance due to viscous damping is directly proportional to the velocity of vibration

dF V αααα

dF = xC amp

Where C= damping coefficient Fd = damping force Examples of Viscous damping

1) Fluid film between sliding surface 2) Fluid flow around a piston in a cylinder 3) Fluid flow through orifice 4) Fluid flow around a journal in a bearing

Rreq =4cm

leqn

stress

Strain

Coulomb damping or dry friction damping Here a damping force is constant in magnitude but opposite in direction to that of the motion of vibrating body It is caused by the friction between the surfaces that are dry or have insufficient lubrication Material or solid or hysteric damping

When the materials are deformed energy is absorbed and dissipated by the material The effect is due to friction between the internal planes which slip or slide as the deformation takes place When a body having the material damping is subjected to vibration the stress strain diagram shows the hysteresis loop as shown in the figure The area of the loop denotes the energy lost per unit volume of the body per cycle

20

21

ττττ ττττ2 ττττ3

ττττ

X(t)

FOURIER SERIES The simplest of periodic motion happens to be SHM It is simple to handle but the motion of many vibrating system is not harmonic (but periodic) Few examples are shown below

Forces acting on machines are generally periodic but this may not be harmonic for example the excitation force in a punching machine is periodic and it can be represented as shown in figure 13 Vibration analysis of system subjected to periodic but nonharmonic forces can be done with the help of Fourier series The problem becomes a multifrequency excitation problem The principle of linear superposition is applied and the total response is the sum of the response due to each of the individual frequency term

Any periodic motion can be expressed as an infinite sum of sines and cosines terms If x(t) is a periodic function with period t its Fourier representation is given by

X(t) = (((( )))) (((( )))) (((( ))))t sint cost cos2 111 ωωωωωωωωωωωω baaao ++++++++++++

= (((( )))) (((( ))))t sint cos2 1

ωωωωωωωω nnn

o bnaa

++++++++sumsumsumsuminfininfininfininfin

====

tππππωωωω

2==== = Fundamental frequency ndash (1)

where ao an bn are constants

ττττ ττττ2 ττττ3

ττττ

X(t)

22

t

X(t)

035

025

Determination of constants To find ao Integrate both sides of equation(1) over any interval ττττ All intergrals on the RHS of the equation are zero except the one containing ao

(((( ))))dttxao

o 2

2

intintintint====ωωωωππππ

ππππωωωω

= (((( ))))dttxo

2intintintintττττ

ττττ

To find an multiply equation 1 by cos (((( ))))tnωωωω and Integrate over any interval ττττ All intergrals

(((( )))) (((( )))) dttxao

n tncos 2

2

ωωωωππππωωωω ωωωω

ππππ

intintintint==== (((( )))) (((( )))) dttxo

tncos 2

ωωωωττττ

ττττ

intintintint====

To find bn multiply equation 1 by sin (((( ))))tnωωωω and Integrate over any interval ττττ All intergrals

(((( )))) (((( )))) dttxao

n tnsin 2

2

ωωωωππππωωωω ωωωω

ππππ

intintintint==== (((( )))) (((( )))) dttxo

tnsin 2

ωωωωττττ

ττττ

intintintint====

Find the Fourier series for the curve shown below

ττττ ττττ2 ττττ3

t

X(t)

E

ττττ ττττ2 ττττ3

t

X(t)

ππππ2

t

X(t) ππππ

1 Represent for the periodic motion shown in the figure

23

CHAPTER 2 UNDAMPED FREE VIBRATION LEARNING OBJECTIVES

] Introduction to undamped free vibration

] Terminologies used in undamped free vibration

] Three methods to solve the undamped free vibration problems

] Problems related to undamped free vibration problems

Free vibrations are oscillations about a systems equilibrium position that occur in the absence of an external excitation force If during vibrations there is no loss of energy it is known as undamped vibration The first step in solving a vibration problem is setting up the differential equation of motion Consider spring mass system which is assumed to move only along the vertical direction as shown below Let m be the mass of the block and k be the stiffness of the spring When block of mass m is attached to spring the deflection of spring will be ∆ known as static deflection In the static equilibrium position the free body diagram of forces acting on the mass is shown in Figure(b) Hence mg= kA Once the system is disturbed the system executes vibrations

Let at any instant of time t the mass is displaced from the equilibrium position x the different forces acting on the system are shown in figure (d) From Newtonrsquos second law of motion

sum = maF

Inertia force ( disturbing force) = restoring force

mgxkxm ++∆minus= )(ampamp

0)( =+ xkxm ampamp

or 0)( =+ xmk

xampamp

equation 2 is the differential equation of motion for spring mfigure Comparing equation (2) with the equation of SHM +xampampsince the vibrations of the above system are free( without the forces) we can write

sec radmk

n =ω

24

ass system shown in 0)(2 =xω

resistance of external

time period km

2 f1

n

πτ ==

from the equation(1) mg = k∆

∆=

q

mk

Difference between the translation ( rectilinear) and rotational system of vibration

Translatory Rotational

In the analysis thFORCES are co Linear stiffness K Problems 1A mass of 10kFind the natural 2 A spring massnatural frequencnatural frequenc

kt

m

sec radmk

n =ω

k

25

e disturbing and restoring nsidered

In the analysis In the analysis MASS Moment of Inertia (J) is considered

in Nm is considered

g when suspended from a spring causes a static deflection of 1cm frequency of system

system has a spring stiffness K Nm and a mass of m Kg It has a y of vibration 12 Hz An extra 2kg mass coupled to it then the y reduces by 2 Hz find K and m

secrad n =ω

26

3 A steel wire of 2mm diameter and 30m long It is fixed at the upper end and carries a mass of m kg at its free end Find m so that the frequency of longitudinal vibration is 4 Hz 4 A spring mass system has a natural period of 02 seconds What will be the new period if the spring constant is 1) increased by 50 2) decreased by 50 5 A spring mass system has a natural frequency of 10 Hz when the spring constant is reduced by 800 Nm the frequency is altered by 45 Find the mass and spring constant of the original system 6 Determine the natural frequency of system shown in fig is which shaft is supported in SHORT bearings 7 Determine the natural frequency of system shown in fig is which shaft is supported in LONG bearings where l is the length of bearing and E ndash youngrsquos modulus and I is moment of Inertia 7 Determine the natural frequency of system shown in fig is which shaft is supported in LONG bearings 8 A light cantilever beam of rectangular section( 5 cm deep x 25cm wide) has a mass fixed at its free end Find the ratio of frequency of free lateral vibration in vertical plane to that in horizontal 9 Determine the natural frequency of simple pendulum 10 Homogeneous square plate of size l and mass m is suspended from the mid point of one of its sides as shown in figure Find the frequency of vibration

l

l

27

11 A compound pendulum which is rigid body of mass m and it is pivoted at O The point of pivot is at distance d from the centre of gravity It is free to rotate about its axis Find the frequency of oscillation of such pendulum 12 A connecting rod shown in fig is supported at the wrist pin end It is displaced and allowed to oscillate The mass of rod is 5kg and centre of gravity is 20 cm from the pivot point O If the frequency of oscillation is 40 cyclesminute calculate the mass moment of inertia about its CG

13 A semi circular homogenous disc of radius r and mass m is pivoted freely about its centre as shown in figure Determine the natural frequency of oscillation 14A simply supported beam of square cross section 5mmx5mm and length 1m carrying a mass of 0575 kg at the middle is found to have natural frequency of 30 radsec Determine youngrsquos modulus of elasticity of beam 15 A spring mass system k1 and m have a natural frequency f1 Determine the value of k2 of another spring in terms of k1 which when placed in series with k1 lowers the natural frequency to 23 f1

COMPLETE SOLUTION OF SYSTEM EXECUTING SHM

The equation of motion of system executing SHM can be represented by

0)( =+ xkxm ampamp ----------------------------------------(1)

022

2=+

dtdx

dt

dx ω

The general solution of equation (1) can be expressed as

------------------------(2)

Where A and B are arbitrary constant which can be determined from the initial conditions of the system Two initial conditions are to be specified to evaluate these

constants x=x0 at t=0 and xamp = Vo at t=0 substituting in the equation (2) Energy method In a conservative system the total energy is constant The dwell as natural frequency can be determined by the princienergy For free vibration of undamped system at any instant and partly potential The kinetic energy T is stored in the masswhere as the potential energy U is stored in the form of sdeformation or work done in a force field such as gravity The total energy being constant T+U = constant Its rate of cha

t)(sin Bt)cos( AX ω+ω=

t)(sin V

t)cos( xx o0 ω

ω+ω=

Is the required complete

solution

28

ifferential equation as ple of conservation of of time is partly kinetic by virtue of its velocity train energy in elastic

nge

Is given by [ ] 0UTdtd

=+

From this we get a differential equation of motion as well as natural frequency of the system Determine the natural frequency of spring mass system using energy method

Determine the natural frequency of system shown in figure Determine the natural frequency of the system shown in figure Is there any limitation on the value of K Discuss Determine the natural frequency of s

m

l k

m

a

θ

m

a

θ

l

k

29

k

k

ystem shown below Neglect the mass of ball

l

m

30

A string shown in figure is under tension T which can be assumed to remain constant for small displacements Find the natural frequency of vertical vibrations of spring An acrobat 120kg walks on a tight rope as shown in figure The frequency of vibration in the given position is vertical direction is 30 rads Find the tension in the rope A manometer has a uniform bore of cross section area A If the column of liquid of length L and Density ρ is set into motion as shown in figure Find the frequency of the resulting oscillation

m

l

a

T T

36m 8m

31

k

R

k

m

r

m

Find the expression for natural frequency of system shown in the figure Neglect the mass of the cantilever beam Study the special case i) k=Infinity ii) I = infinity Determine the expression for the natural frequency of a system shown in figure The two discs are keyed to a common shaft and have a combined mass moment of inertia about centre of oscillation O It is assumed that the card attached to mass m does not stretch and is always under tension Determine the natural frequency of a system shown

m

l

32

Determine the expression for the natural frequency of the system shown in figure Assume that the wires connecting the masses do not stretch and are always in tension

Determine the natural frequency of spring mass system taking the MASS OF SPRING (ms ) into account

M3 M1

M2

k2 k1

m

k

33

RAYLEIGHrsquoS METHOD

This is the extension of energy method Here we can determine the natural frequency of a conservative system without finding the equation of motion The natural frequency is a function of the rate of change of Kinetic energy and potential energy of the system From the principle of conservation of energy we have T+U = Constant T= KE and U =PE We can write the above equation as T1+U1 = --------------------------(1) Where 1 and 2 represents tow instances of time Let 1 be the time when the mass is passing through static equilibrium position Let 2 be the time corresponding to the mass displacement of the mass At this instant the velocity f the mass is zero and hence Substituting in equation (1) we get ( )Max = ( )Max Above equation leads directly to natural frequency Determine the natural frequency of spring mass system using Rayleighrsquos method m

34

r

B

K K

Home work

Determine the natural frequency of pendulum using Rayleighrsquos method

A Cylinder of mass m and moment of inertia Jo is to role without slipping but is sustained by spring K as shown in figure Determine the natural frequency of oscillation

Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping

a

Determine the natural frequency of system shown where in cylindrical disc rolls over the inextensible spring A cylinder of radius r rolls without slipping on cylindrical the equation of motion and hence determine the naoscillations about the lowest point shown Repeat the above problem ndash Instead of cylinder assume swithout slipping on concave surface as shown above

r

K

35

portion of radius R Derive tural frequency of small

phere of radius with r rolls

The torsional pendulum has a natural frequency of 5 Hz What length of steel wire of diameter 2 mm should be used for pendulum The inertia of mass fixed at the free end is 00098 kgm2 Take g= 083x1011 Nm2 Determine the torsional natural frequency of the system shown in figure Neglect the mass moment of inertia of the shaft ( Figure not proportionate) Determine the natural frequency of simple pendulum considering the mass of the rod into account

Using the energy method find tThe chord may be assumed ine

Kt1

Kt2

20cm

06m 08m 04m

20cm 20cm 20cm

m

M

he natural frxtensible in

k

M

36

equency of the system shown in figure the spring mass pulley system

A rod of length l and depth H having a mass m rests on a semicylidrical surface os radius r It is tipped slightly and let go Find the frequency of oscillation

A thin bar of l having a mass m rests on a semicylidrical surface radius r It is tipped slightly and let go Find the frequency of oscillation

l

h

R

l

37

38

CHAPTER 3

DAMPED FREE VIBRATION OF SINGLE DEGREE FREEDOM SYSTEM LEARNING OBJECTIVES

] Introduction to damped free vibration

] Terminologies used in damped free vibration

] Three different types of damping

] Concept of critical damping and its importance

] Study of response of viscous damped system for cases of under damping

critical and over damping

] Concept of logarithmic decrement

] Problems related to damped free vibration problems

Viscously damped free vibration of single degree freedom sytem Consider a single degree freedom system with viscous damping as shown in figure For a viscous damped system the damping force is

where c is a constant of proportionality It is called damping coefficient or coefficient of viscous damping

Symbolically it is designated by a dashpot as shown in Fig 3

x xampx

39

At any instant of time let the mass be displaced by x from the equilibrium position Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

Okxxcxm =minusminus= ampampamp

Okxxcxm =++ ampampamp ----------------------------------------------------------(1)

or

0kxdtdx

Cdt

xdM

2

2

=++

Assuming the solution of equation as

The first term in the above equation m2Ct

eminus

is an exponentially decaying function of time The behavior of the terms in the parenthesis however depends on whether numerical value within the radicals is positive zero or negative

CASE 1 if 0mk

m2c

2

=minus

40

CASE 2 if 0mk

m2c

2

=gt

CASE 3 if 0mk

m2c

2

=lt

41

Critically damped coefficient Critically damped coefficient is defined as the value of the damping coeffient for which radical in equation

mk

m2c

m2c

S2

21 minus

plusmn

= becomes zero

Damping Ratio ( Damping factor) ( ξ ) It is defined as the ratio of damping coefficient to the critical damping coefficient Note The damping coefficient c is defined as the actual amount of damping while the damping coefficient is a constant depending upon the mass and stiffness of the system The roots of characteristics equation can be now written as Note if c=cc or ξ =1 then system is critically damped if cgt cc or ξ gt1 then system is over damped I if clt cc or ξ lt1 then system is under damped

42

Take home quiz Derive General solution of different damped system

1 Critically damped system 2 Under damped system 3 Over damped system

Logarithmic Decrement It is used to determine the amount of damping present in system This measure The rate of decay of free vibration The larger the damping the greater will be rate of decay Logarithmic decrement is defined as the natural logarithm of ratio of any two successive amplitudes

=δ

2

1

xx

ln

Expression for logarithmic decrement (δ ) Consider a damped free vibration which is expressed by

( ) ( ) φ+ξminusω= ξω t 1sinXex 2n

tn

x is the displacement at any instant of time X and φ are arbitrary constants

21

2

ξminus

πξ=δ

43

Tutorial problems 1 Derive an expression for damping ratio in terms of logarithmic decrement

2 Show that logarithmic decrement

=δ

n

o

xx

lnn1

where xo is the initial amplitude xn

is amplitude after n cycles 3 A mass of 1 kg is to be supported on a spring having stiffness K= 9800 Nm The damping constant c= 59 Nsm Determine the natural frequency of system Find also logarithmic decrement and amplitude after 3 cycles if initial displacement is 03 cm 4 A vibrating system is defined by the following parameters M = 3kg k= 100 Nm damping coefficient = 3 Nsm determine Damping factor natural frequency damped vibration logarithmic decrement ratio of consecutive amplitude no of cycles after which the original amplitude is reduced to 20 5A single degree damped vibrating system a suspended mass of 18kg makes 15 oscillations in 003 seconds The amplitude decreases to 025 of the initial value after 5 oscillations Determine the stiffness of spring Logarithmic decrement damping factor and damping coefficient 6 A damped vibration record of a spring mass system shows the following data Amplitude at the end of 2nd cycle is 9mm amplitude at eh end of 3rd cycle is 6mm amplitude at the end of 4th cycle is 4mm stiffness of spring =8000 Nm weight =20N Find the logarithmic decrement damping force at unit velocity periodic time of vibration 7 For the system shown in figure the characteristics of dashpot is such that the constant force of 49 N is applied to the piston it velocity is found to be constant at 012 ms Determine the value of C would you expect the complete system to be periodic or aperiodic 8 For a torsional system shown in the figuref1 is natural frequency in air and f2 is the natural frequency when immersed in oil Determine the damping coefficient

Viscous fluid

Disc of mass moment of Inertia J

2a

C

m M

b

a a

k

9 A thin plate of area A and weigh w is fixed to the end of the spring and allow to oscillate in a viscous fluid as shown in figure It f1 is the frequency of oscillation in air and f2 in liquid Show that

22

21 ff

gA2

minusπω

=micro where the damping force on the plate

AV2Fd micro= V being the velocity of vibration 10 A torsional pendulum has a natural frequency of 175 cyclessecond vibrating in vacuum The mass moment of inertia is 2 kgm2 It is immersed in oil and it is observed that the natural frequency is 142 cyclessecond Find the damping torque If the disc is replaced by 34 deg When in oil find the displacement at the end of the first complete cycle 11 An automobile can be modeled as a mass placed on 4 shock absorbers each consisting of a spring and a damper Such that each spring is equally loaded Determine the stiffness damping constant of each shock absorber So that natural frequency is 2Hz and system is critically damped The mass of vehicle is 200kg 12 Write the differential equation of motion for the system shown in the figure below Determine the natural frequency of the damped oscillation and critical damping coefficient

Determine the differential equation of motion and find the critical cdamped natural frequency for system shown in figure

a

C

m M

M =200kg

44

200kg kgkg

oefficient and

45

Derive the equation of motion and find the critical damping coefficient and damped natural frequency for the system shown below Write the differential equation of motion and determine the expression for critical damping coefficient and damped natural frequency of system shown in figure

Set up a differential equation of motion for system shown in figure and determine the following Undamped natural frequencies critical damping coefficient damping ratio damped natural frequency

b

C

kM

a

m

46

A mass of 300N is resting on two springs of stiffness 3000 Nm each and a dashpot of damping coefficient 150 Nsm as shown in figure If the initial velocity of 10cms is given to mass at its equilibrium position what will be the displacement from the equilibrium position at the end of first second Derive the formula used A gun barrel of mass 600kg has a recoil spring of stiffness 294 knm if the barrel recoils 13m on firing determine i) The initial recoil velocity of barrel ii) The critical damping coefficient of dashpot which is engaged at the end of recoil stroke Iii) Time required for the barrel to return at the end of recoil stroke Iv) time required for the barrel to return to a position 5cm from the initial position

W= 3000 N

FIRING

CHAPTER 4

FORCED VIBRATION LEARNING OBJECTIVES

] Introduction to forced free vibration

] Terminologies used in forced vibration

] Concept of magnification factor

] Study of reciprocating and rotating unbalance

] Concept of base excitation and support excitation

] Concept of Transmissibility ratio

] Problems related to forced vibration

The vibration that takes place under the excitation of external forces are called

forced vibration When a system is subjected to harmonic excitation it is forced to vibrate at the same frequency as that of excitation Most of the mechanical systems are prone to steady state forced vibration which cause fatigue failure The vibration problem also becomes very important when any of excitation frequencies coincide with one of natural frequencies of a system Under this condition resonance occurs during which large amplitude of vibrations are observed To avoid resonance external frequency (operating speed) may be changed or properties of the system may be changed to alter the natural frequency In some cases sufficient amount of damping may be provided to avoid large amplitude during resonance Thus the problem of force vibration is very important in mechanical design

FORCED VIBRATIONS OF SINGLE DOF SYSTEMS UNDER HARMONIC EXCITATION

` [ ] tsin FF o ω=

x

sin FF o=

Fo

xampamp

[ ] tω

xamp

47

48

Consider a spring mass system having a viscous damping excited by a harmonic force [ ] tsin FF o ω= as shown in figure The various forces acting on the system is shown in figure Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

O)tsin(Fkxxcxm o =ω+minusminus= ampampamp

)tsin(Fkxxcxm o ω=++ ampampamp ----------------------------------------------------------(1)

or

)tsin(Fkxdtdx

Cdt

xdM o2

2

ω=++

Equation (1) represents the differential equation of motion of the system The solution to this equation consists of two parts namely complementary function and a particular integral

The first of solution is also termed as transient solution of homogeneous solution and can be obtained from the homogeneous differential equation 0kxxcxm =++ ampampamp

The second part of solution particular integral is also termed as steady state solution ad is of form )tsin(Xx φminusω= Where X= steady state amplitude φ is phase difference between external force and displacement

)tsin(Xx φminusω=

The vector representation of the shown

49

The force polygon for the above vector diagram can be written as follows

( )φminusωt O

KX

xm 2ω

Fo

xCω

tω

φ

50

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

is the governing equation

Zero frequency deflection (Xst) It is deflection observed at zero frequency ( when the external frequency 0=ω

( ) kF

001

kF

x o

22

o

st =

+minus=

Magnification factor (M) It is defined as the ratio of steady state amplitude to zero frequency deflection xst

M = arX

X=

A plot of magnification factor versus frequency ratio

damping factor ( )ξ is called frequency plot of a harmon All the frequency terms starts from magnification factor of damping ratio ( )ξ The amplitude of vibration at any on the value of damping The amplitude will be higher for higher damping The amplitude of vibration at reso

amplitude tends to zero as

ωω

n tends to infinity

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

Freq

ζ =

ζ = 1

Magnification Factor M = XXst

kFo

51

ωω

nfor varying values of

ically excite vibrating system

M=1 what ever may be value particular frequency depends

for lesser damping and lesser nance will be very high The

uency Ratio r = (ωωn)

ζ = 0 0

ζ = 025 ζ = 0375 ζ = 0 5

ζ = 0707

ζ = 2

52

Plot of phase angle φ versus frequency ratio

ωω

n is shown in the

figure The phase angle φ is zero for all the values of damping ratio

at

ωω

n=0 The phase

angle φ is 90 for all the values of damping ratio

at

ωω

n=1 ie at

resonance The phase angle between o to 90 deg for all the values of

ωω

n= 0 to 1 and

between 90 deg to 180 deg for all the values of

ωω

ngt1

Class work Derive an expression for frequency at peak amplitude

2np 21 ξminusω=ω=ω is the required equation

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

53

Derive expression amplitude at resonance ( In dimensional form) In dimensional form We know that steady state amplitude

( ) ( )

ω+ωminus=

222

o

Cmk

Fx Equation in dimensional form

We know that at resonance mk

n =ω=ω

( )

ω+

minus

=2

n

2

oR

Cmk

mk

Fx

( )

ω=

2n

oR

C

Fx

( )

ω=

n

oR C

Fx

Class work Derive an expression for amplitude at resonance ( in non dimensional form)

Tutorial Problems on Forced Vibration 1A weight of 60N is suspended by a spring stiffness 12 kNm is forced to vibrate by a harmonic force of 10N assuming viscous damping of 0086 kN-sm Find i) resonance frequency ii) amplitude at resonance iii) phase angle at resonance iv) frequency corresponding to peak amplitude 2 A periodic torque having a maximum value of 05888Nm at a frequency of 4 rads is impressed upon a flywheel suspended from a wire The wheel has mass moment of inertia 012 kgm2 and the wire has stiffness 1176 Nmrad A viscous dashpot applies damping couple of 0784 Nm an angular velocity of 2 rads calculate i)maximum angular displacement from rest position ii) maximum couple applied to dashpot Iii) angle by which the angular displacement lags the torque A mass of 1019 kg is suspended from a spring stiffness 20 Nmm viscous damping causes the amplitude to decrease to 110th of initial value in 4 complete cycle If a periodic force of 20 cos(50 t) is applied to the mass Find the amplitude of forced vibration What would be the amplitude of oscillation if the applied force coincides with the natural period

m=1019 kg

[ ] t50cos 30F =

54

55

ROTATING UNBALANCE Unbalance in rotating machinery is a common source of vibration excitation Consider a machine of mass m having mounted on a foundation of stiffness k and damping coefficient C as shown in figure The unbalance is represented by an eccentric rotating mass m with eccentricity (e) and having a angular velocity (ω) let x be the displacement of non rotating mass from the static equilibrium position The displacement of rotating mass m from the static equilibrium is ( ) tsin( ex ω+ Applying Newtonrsquos law of motion we have

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

56

RECIPROCATING UNBALANCE Unbalance in reciprocating machinery is a common source of vibration excitation Consider a machine of mass M having mounted on a foundation of stiffness K and damping coefficient C as shown in figure The unbalance is represented by a reciprocating mass (m) having crank radus (e) and connecting rod of length (l) Let ωωωω be angular velocity of the crank Let x be the displacement of non reciprocating mass from the static equilibrium position at any instant of time The displacement of reciprocating mass m from the static equilibrium position is

)2sin(sin +++ tle

tex ωωωωωωωω

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

57

Note The equations obtained in reciprocating unbalance is same as the rotating unbalance and having the following differences Rotating Unbalance Reciprocating Unbalance Here mass of the rotating element is considered

Here mass of reciprocating element is considered

Here the eccentricity (e) which is the distance between the centre of rotation and CG

Here the eccentricity e is the crank radius and crank radius = stroke length2

The frequency plot of non dimensional factor( ( )me

MX versus frequency ratio

ωω

n for various values of ( )ξξξξ is shown in figure This frequency plot is same for

the rotating and reciprocating unbalance with the following points are noted from the diagram

1 Damping ratio plays very important role during resonance in controlling the amplitude

2 When

ωω

n=0 then ( )me

MX =0 for all the values of damping ratio ( )ξξξξ

3 When

ωω

nis higher than ( )me

MX tends to unity

4 At very high speeds ie at very high frequency ratio

ωω

n there is no effect

of damping The frequency plot of phase angle versus the frequency ratio is given as shown in figure The explanation is same as in forced harmonic vibration

58

VIBRATION ISOLATION

Vibratory forces and motions generated by machine and other caused are unavoidable Howeever their effect on the other dynamical systems can be minimized by proper isolator design An isolation system is either to protect o delicate machine from excessive vibration transmitted to it from its supporting structure or to prevent vibratory forces and motions generated by machines from being transmitted to its surroundings The effectiveness of isolation may be measured in terms of forces and motion transmitted The first type is known as force isolation and the second type is known as motion isolation

Force isolation (Force Transmissibility) When vibratory forces generated by machine are isolated the effectiveness of forced isolation can be made by defining the term transmissibility The term transmissibility is defined as the ratio of force transmitted to the foundation to that of the external impressed force

Force Transmissibility = o

tr

FF

force Impressed Externalfoundation toted transmitForce

=

Consider viscous damped spring mass system as shown in figure Upon this system an external harmonic force ) t (Sin Fo ω is applied The difference forces acting on system are inertia forces damping force spring force and external force All these forces are represented on force polygon shown below Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these two forces

( ) ( )22 xckx ω+=

59

This is the expression for force transmissibility in dimensional form To represent same in the non dimensional form divide numerator and denominator by k

2

n

22

n

2

n

o

tr

21

21

FF

TR

ωω

ξ+

ωω

minus

ωω

ξ+

==

PHASE DIFFERNECE BETWEEN EXTERNAL FORCE AND DISPLACEMENT Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these

ωω

minus

ωω

ξ=

ωminusω

=φ minusminus2

n

n1

2

1

1

2tan

mKC

tan

The phase difference between force transmitted to the foundation Ftr to the displacement

kc

KxxC

tanω

=ω

=α or

ωω

ξ=

ω=α minusminus

n

11 2tanKx

xCtan

The phase difference between external force Fo and force transmitted to foundationis β

60

β= αminusφ =

ωω

minus

ωω

ξminus

2

n

n1

1

2tan -

ωω

ξminus

n

1 2tan

Displacement Transmissibility It is the ratio of amplitude of steady state vibration of machine to that of the base motion Forced vibration due to support motion Consider a spring mass damper system subjected to harmonic excitation )tsin(Yy ω= at the base as shown in figure Let x be the absolute displacement of the mass at any instant of time t The different forces acting on system are shown in free body diagram Applying Newtonrsquos Law of motion Ma = sumF

)yx(k)yx(cxm minusminusminusminus= ampampampamp

0)yx(k)yx(cxm =minus+minus+ ampampampamp

61

Frequency plot of transmissibility versus frequency ratio

62

The plot of transmissibility versus frequency ratio is as shown in figure The following points can be noted from it

i) The value of transmissibility is unity at frequency ratio

ωω

n=0 and 2

ii) For an undamped system ( )ξ =0 transmissibility tends to infin at resonance iii) The value of transmissibility is less than unity for the value of frequency ratio

greater than 2

iv) For frequency ratio

ωω

nlt 2 smaller damping ratio lead to larger values of

transmissibility on the other hand for frequency ratio

ωω

ngt 2 smaller damping

ratio leads to smaller value of transmissibility

The frequency plot of transmissibility (TR) versus frequency ratio

ωω

n can be

broadly divided into two zones

1

ωω

n= 0 to 2

2

ωω

n= 2 to infin

CASE 1

When

ωω

n= 0 to 2

In this zone the value of transmissibility is always greater than 1 for all values of damping ratio In this zone the transmissibility decreases with increase in damping Hence we can conclude that in this zone damping is useful in isolating vibration forces being transmitted to the support CASE 2

When

ωω

n= 2 to infin

In this zone the value of transmissibility is less than 1 for all values of damping ratio In this zone the transmissibility increases with increase in damping Hence we can conclude that in this zone damping will adversely affect in isolating vibratory forces

63

Tutorial problems on Forced Vibration 1 Machine of mass 75kg is mounted on spring of stiffness 12000 Nm with an assumed damping factor ξ =02 piston within the machine of mass 2 kg has a reciprocating motion with a stroke of 75 cm and speed of 15 cyclessecond Assuming the motion of piston to be harmonic determine amplitude of machine ( X) Transmissibility and force transmitted to foundation phase angle of transmitted force wrt to exciting force 2 Motor weighs 600N is mounted on a simple beam that has a spring rate of 400 Ncm The motor armature base 160 N and has an eccentricity of 0005cm What will be the amplitude of motion when it runs at 1760 RPM Neglect damping the deflection of armature shaft and weight of the beam 3 A Heavy machine of mass 300 kg is supported on resilient foundation the static deflection of foundation due to weight of machine is found to be 75mm It is observed that the machine vibrates with an amplitude of 10mm when the base of the foundation is subjected to harmonic oscillation at the undampeds natural frequency of system with an amplitude of 25mm Find the damping constant of foundation Ii) dynamic force amplitude on the base iii) amplitude of the displacement of machine relative to base 4 An air compressor of mass 450 kg operates at a constant speed of 1750 RPM The reciprocating masses are 10 kg the crank radius is 100mm If the damper for mounting introduces a damping factor 015 i) specify the springs of mounting such that only 20 of the unbalanced force is transmitted to the foundation ii) determine the magnitude of the transmitted force bull Repeat the above problem by assuming that there is no damping in system

5 A body of a car having mass 1500 kg and is mounted on 4 equal springs which deflect through 225 cm under the weight of the body A total damping forces the 4 shock absorbers is 46 N at a velocity of 1cmsec The car is placed with all the four wheels on a test platform which is move up and down at resonant speed With an amplitude of 25cm Find the amplitude of the car body on its springs assuming its CG to be in the centre of wheel base 6A 75 kg machine which is mounted on springs of stiffness 116x105 Nm with the damping of 20=ξ A 2kg piston within the machine as reciprocating motion with a stroke of 008m and speed of 3000 RPM Assuming the motion of piston to be harmonic determine the amplitude of vibration of machine

7 The spring of an automobile are compressed 10cm under its own weight Find the critical speed when a trailer is traveling over road with a people approximaterly by a sine wave of amplitude 75 cm and wave length 15m What will be the amplitude of vibration at 64kmhour 8 An eccentric mass excites us used to determine vibratory characteristics of a structure of mass 25kg as given in figure At a speed of 1000 RPM the eccentric mass to beat top position at the instant the structure was moved upwards through its static equilibrium position and corresponding amplitude was 20mm The eccentric mass is 05kg at an eccentricity of 10mm Determine i) Undamped natural frequency of the system ii) Damping ratio iii) Steady state amplitude and phase angle at 1500 RPM iv) Force transmitted to the ground at 1200 RPM

M=25kg

C

K

Eccentric mass

64

65

CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Support motion Consider a spring mass damper system as shown in figure Let it excited by motion

of support let y be the harmonic displacement of support From Newtonrsquos second law of motion

)yx(k)yx(cxm minusminusminusminus= ampampampamp

)1(0)yx(k)yx(cxm minusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminus=minus+minus+ ampampampamp

Let z be the relative displacement of the mass with respect to support then

Z=x-y

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

stress

Strain

Coulomb damping or dry friction damping Here a damping force is constant in magnitude but opposite in direction to that of the motion of vibrating body It is caused by the friction between the surfaces that are dry or have insufficient lubrication Material or solid or hysteric damping

When the materials are deformed energy is absorbed and dissipated by the material The effect is due to friction between the internal planes which slip or slide as the deformation takes place When a body having the material damping is subjected to vibration the stress strain diagram shows the hysteresis loop as shown in the figure The area of the loop denotes the energy lost per unit volume of the body per cycle

20

21

ττττ ττττ2 ττττ3

ττττ

X(t)

FOURIER SERIES The simplest of periodic motion happens to be SHM It is simple to handle but the motion of many vibrating system is not harmonic (but periodic) Few examples are shown below

Forces acting on machines are generally periodic but this may not be harmonic for example the excitation force in a punching machine is periodic and it can be represented as shown in figure 13 Vibration analysis of system subjected to periodic but nonharmonic forces can be done with the help of Fourier series The problem becomes a multifrequency excitation problem The principle of linear superposition is applied and the total response is the sum of the response due to each of the individual frequency term

Any periodic motion can be expressed as an infinite sum of sines and cosines terms If x(t) is a periodic function with period t its Fourier representation is given by

X(t) = (((( )))) (((( )))) (((( ))))t sint cost cos2 111 ωωωωωωωωωωωω baaao ++++++++++++

= (((( )))) (((( ))))t sint cos2 1

ωωωωωωωω nnn

o bnaa

++++++++sumsumsumsuminfininfininfininfin

====

tππππωωωω

2==== = Fundamental frequency ndash (1)

where ao an bn are constants

ττττ ττττ2 ττττ3

ττττ

X(t)

22

t

X(t)

035

025

Determination of constants To find ao Integrate both sides of equation(1) over any interval ττττ All intergrals on the RHS of the equation are zero except the one containing ao

(((( ))))dttxao

o 2

2

intintintint====ωωωωππππ

ππππωωωω

= (((( ))))dttxo

2intintintintττττ

ττττ

To find an multiply equation 1 by cos (((( ))))tnωωωω and Integrate over any interval ττττ All intergrals

(((( )))) (((( )))) dttxao

n tncos 2

2

ωωωωππππωωωω ωωωω

ππππ

intintintint==== (((( )))) (((( )))) dttxo

tncos 2

ωωωωττττ

ττττ

intintintint====

To find bn multiply equation 1 by sin (((( ))))tnωωωω and Integrate over any interval ττττ All intergrals

(((( )))) (((( )))) dttxao

n tnsin 2

2

ωωωωππππωωωω ωωωω

ππππ

intintintint==== (((( )))) (((( )))) dttxo

tnsin 2

ωωωωττττ

ττττ

intintintint====

Find the Fourier series for the curve shown below

ττττ ττττ2 ττττ3

t

X(t)

E

ττττ ττττ2 ττττ3

t

X(t)

ππππ2

t

X(t) ππππ

1 Represent for the periodic motion shown in the figure

23

CHAPTER 2 UNDAMPED FREE VIBRATION LEARNING OBJECTIVES

] Introduction to undamped free vibration

] Terminologies used in undamped free vibration

] Three methods to solve the undamped free vibration problems

] Problems related to undamped free vibration problems

Free vibrations are oscillations about a systems equilibrium position that occur in the absence of an external excitation force If during vibrations there is no loss of energy it is known as undamped vibration The first step in solving a vibration problem is setting up the differential equation of motion Consider spring mass system which is assumed to move only along the vertical direction as shown below Let m be the mass of the block and k be the stiffness of the spring When block of mass m is attached to spring the deflection of spring will be ∆ known as static deflection In the static equilibrium position the free body diagram of forces acting on the mass is shown in Figure(b) Hence mg= kA Once the system is disturbed the system executes vibrations

Let at any instant of time t the mass is displaced from the equilibrium position x the different forces acting on the system are shown in figure (d) From Newtonrsquos second law of motion

sum = maF

Inertia force ( disturbing force) = restoring force

mgxkxm ++∆minus= )(ampamp

0)( =+ xkxm ampamp

or 0)( =+ xmk

xampamp

equation 2 is the differential equation of motion for spring mfigure Comparing equation (2) with the equation of SHM +xampampsince the vibrations of the above system are free( without the forces) we can write

sec radmk

n =ω

24

ass system shown in 0)(2 =xω

resistance of external

time period km

2 f1

n

πτ ==

from the equation(1) mg = k∆

∆=

q

mk

Difference between the translation ( rectilinear) and rotational system of vibration

Translatory Rotational

In the analysis thFORCES are co Linear stiffness K Problems 1A mass of 10kFind the natural 2 A spring massnatural frequencnatural frequenc

kt

m

sec radmk

n =ω

k

25

e disturbing and restoring nsidered

In the analysis In the analysis MASS Moment of Inertia (J) is considered

in Nm is considered

g when suspended from a spring causes a static deflection of 1cm frequency of system

system has a spring stiffness K Nm and a mass of m Kg It has a y of vibration 12 Hz An extra 2kg mass coupled to it then the y reduces by 2 Hz find K and m

secrad n =ω

26

3 A steel wire of 2mm diameter and 30m long It is fixed at the upper end and carries a mass of m kg at its free end Find m so that the frequency of longitudinal vibration is 4 Hz 4 A spring mass system has a natural period of 02 seconds What will be the new period if the spring constant is 1) increased by 50 2) decreased by 50 5 A spring mass system has a natural frequency of 10 Hz when the spring constant is reduced by 800 Nm the frequency is altered by 45 Find the mass and spring constant of the original system 6 Determine the natural frequency of system shown in fig is which shaft is supported in SHORT bearings 7 Determine the natural frequency of system shown in fig is which shaft is supported in LONG bearings where l is the length of bearing and E ndash youngrsquos modulus and I is moment of Inertia 7 Determine the natural frequency of system shown in fig is which shaft is supported in LONG bearings 8 A light cantilever beam of rectangular section( 5 cm deep x 25cm wide) has a mass fixed at its free end Find the ratio of frequency of free lateral vibration in vertical plane to that in horizontal 9 Determine the natural frequency of simple pendulum 10 Homogeneous square plate of size l and mass m is suspended from the mid point of one of its sides as shown in figure Find the frequency of vibration

l

l

27

11 A compound pendulum which is rigid body of mass m and it is pivoted at O The point of pivot is at distance d from the centre of gravity It is free to rotate about its axis Find the frequency of oscillation of such pendulum 12 A connecting rod shown in fig is supported at the wrist pin end It is displaced and allowed to oscillate The mass of rod is 5kg and centre of gravity is 20 cm from the pivot point O If the frequency of oscillation is 40 cyclesminute calculate the mass moment of inertia about its CG

13 A semi circular homogenous disc of radius r and mass m is pivoted freely about its centre as shown in figure Determine the natural frequency of oscillation 14A simply supported beam of square cross section 5mmx5mm and length 1m carrying a mass of 0575 kg at the middle is found to have natural frequency of 30 radsec Determine youngrsquos modulus of elasticity of beam 15 A spring mass system k1 and m have a natural frequency f1 Determine the value of k2 of another spring in terms of k1 which when placed in series with k1 lowers the natural frequency to 23 f1

COMPLETE SOLUTION OF SYSTEM EXECUTING SHM

The equation of motion of system executing SHM can be represented by

0)( =+ xkxm ampamp ----------------------------------------(1)

022

2=+

dtdx

dt

dx ω

The general solution of equation (1) can be expressed as

------------------------(2)

Where A and B are arbitrary constant which can be determined from the initial conditions of the system Two initial conditions are to be specified to evaluate these

constants x=x0 at t=0 and xamp = Vo at t=0 substituting in the equation (2) Energy method In a conservative system the total energy is constant The dwell as natural frequency can be determined by the princienergy For free vibration of undamped system at any instant and partly potential The kinetic energy T is stored in the masswhere as the potential energy U is stored in the form of sdeformation or work done in a force field such as gravity The total energy being constant T+U = constant Its rate of cha

t)(sin Bt)cos( AX ω+ω=

t)(sin V

t)cos( xx o0 ω

ω+ω=

Is the required complete

solution

28

ifferential equation as ple of conservation of of time is partly kinetic by virtue of its velocity train energy in elastic

nge

Is given by [ ] 0UTdtd

=+

From this we get a differential equation of motion as well as natural frequency of the system Determine the natural frequency of spring mass system using energy method

Determine the natural frequency of system shown in figure Determine the natural frequency of the system shown in figure Is there any limitation on the value of K Discuss Determine the natural frequency of s

m

l k

m

a

θ

m

a

θ

l

k

29

k

k

ystem shown below Neglect the mass of ball

l

m

30

A string shown in figure is under tension T which can be assumed to remain constant for small displacements Find the natural frequency of vertical vibrations of spring An acrobat 120kg walks on a tight rope as shown in figure The frequency of vibration in the given position is vertical direction is 30 rads Find the tension in the rope A manometer has a uniform bore of cross section area A If the column of liquid of length L and Density ρ is set into motion as shown in figure Find the frequency of the resulting oscillation

m

l

a

T T

36m 8m

31

k

R

k

m

r

m

Find the expression for natural frequency of system shown in the figure Neglect the mass of the cantilever beam Study the special case i) k=Infinity ii) I = infinity Determine the expression for the natural frequency of a system shown in figure The two discs are keyed to a common shaft and have a combined mass moment of inertia about centre of oscillation O It is assumed that the card attached to mass m does not stretch and is always under tension Determine the natural frequency of a system shown

m

l

32

Determine the expression for the natural frequency of the system shown in figure Assume that the wires connecting the masses do not stretch and are always in tension

Determine the natural frequency of spring mass system taking the MASS OF SPRING (ms ) into account

M3 M1

M2

k2 k1

m

k

33

RAYLEIGHrsquoS METHOD

This is the extension of energy method Here we can determine the natural frequency of a conservative system without finding the equation of motion The natural frequency is a function of the rate of change of Kinetic energy and potential energy of the system From the principle of conservation of energy we have T+U = Constant T= KE and U =PE We can write the above equation as T1+U1 = --------------------------(1) Where 1 and 2 represents tow instances of time Let 1 be the time when the mass is passing through static equilibrium position Let 2 be the time corresponding to the mass displacement of the mass At this instant the velocity f the mass is zero and hence Substituting in equation (1) we get ( )Max = ( )Max Above equation leads directly to natural frequency Determine the natural frequency of spring mass system using Rayleighrsquos method m

34

r

B

K K

Home work

Determine the natural frequency of pendulum using Rayleighrsquos method

A Cylinder of mass m and moment of inertia Jo is to role without slipping but is sustained by spring K as shown in figure Determine the natural frequency of oscillation

Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping

a

Determine the natural frequency of system shown where in cylindrical disc rolls over the inextensible spring A cylinder of radius r rolls without slipping on cylindrical the equation of motion and hence determine the naoscillations about the lowest point shown Repeat the above problem ndash Instead of cylinder assume swithout slipping on concave surface as shown above

r

K

35

portion of radius R Derive tural frequency of small

phere of radius with r rolls

The torsional pendulum has a natural frequency of 5 Hz What length of steel wire of diameter 2 mm should be used for pendulum The inertia of mass fixed at the free end is 00098 kgm2 Take g= 083x1011 Nm2 Determine the torsional natural frequency of the system shown in figure Neglect the mass moment of inertia of the shaft ( Figure not proportionate) Determine the natural frequency of simple pendulum considering the mass of the rod into account

Using the energy method find tThe chord may be assumed ine

Kt1

Kt2

20cm

06m 08m 04m

20cm 20cm 20cm

m

M

he natural frxtensible in

k

M

36

equency of the system shown in figure the spring mass pulley system

A rod of length l and depth H having a mass m rests on a semicylidrical surface os radius r It is tipped slightly and let go Find the frequency of oscillation

A thin bar of l having a mass m rests on a semicylidrical surface radius r It is tipped slightly and let go Find the frequency of oscillation

l

h

R

l

37

38

CHAPTER 3

DAMPED FREE VIBRATION OF SINGLE DEGREE FREEDOM SYSTEM LEARNING OBJECTIVES

] Introduction to damped free vibration

] Terminologies used in damped free vibration

] Three different types of damping

] Concept of critical damping and its importance

] Study of response of viscous damped system for cases of under damping

critical and over damping

] Concept of logarithmic decrement

] Problems related to damped free vibration problems

Viscously damped free vibration of single degree freedom sytem Consider a single degree freedom system with viscous damping as shown in figure For a viscous damped system the damping force is

where c is a constant of proportionality It is called damping coefficient or coefficient of viscous damping

Symbolically it is designated by a dashpot as shown in Fig 3

x xampx

39

At any instant of time let the mass be displaced by x from the equilibrium position Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

Okxxcxm =minusminus= ampampamp

Okxxcxm =++ ampampamp ----------------------------------------------------------(1)

or

0kxdtdx

Cdt

xdM

2

2

=++

Assuming the solution of equation as

The first term in the above equation m2Ct

eminus

is an exponentially decaying function of time The behavior of the terms in the parenthesis however depends on whether numerical value within the radicals is positive zero or negative

CASE 1 if 0mk

m2c

2

=minus

40

CASE 2 if 0mk

m2c

2

=gt

CASE 3 if 0mk

m2c

2

=lt

41

Critically damped coefficient Critically damped coefficient is defined as the value of the damping coeffient for which radical in equation

mk

m2c

m2c

S2

21 minus

plusmn

= becomes zero

Damping Ratio ( Damping factor) ( ξ ) It is defined as the ratio of damping coefficient to the critical damping coefficient Note The damping coefficient c is defined as the actual amount of damping while the damping coefficient is a constant depending upon the mass and stiffness of the system The roots of characteristics equation can be now written as Note if c=cc or ξ =1 then system is critically damped if cgt cc or ξ gt1 then system is over damped I if clt cc or ξ lt1 then system is under damped

42

Take home quiz Derive General solution of different damped system

1 Critically damped system 2 Under damped system 3 Over damped system

Logarithmic Decrement It is used to determine the amount of damping present in system This measure The rate of decay of free vibration The larger the damping the greater will be rate of decay Logarithmic decrement is defined as the natural logarithm of ratio of any two successive amplitudes

=δ

2

1

xx

ln

Expression for logarithmic decrement (δ ) Consider a damped free vibration which is expressed by

( ) ( ) φ+ξminusω= ξω t 1sinXex 2n

tn

x is the displacement at any instant of time X and φ are arbitrary constants

21

2

ξminus

πξ=δ

43

Tutorial problems 1 Derive an expression for damping ratio in terms of logarithmic decrement

2 Show that logarithmic decrement

=δ

n

o

xx

lnn1

where xo is the initial amplitude xn

is amplitude after n cycles 3 A mass of 1 kg is to be supported on a spring having stiffness K= 9800 Nm The damping constant c= 59 Nsm Determine the natural frequency of system Find also logarithmic decrement and amplitude after 3 cycles if initial displacement is 03 cm 4 A vibrating system is defined by the following parameters M = 3kg k= 100 Nm damping coefficient = 3 Nsm determine Damping factor natural frequency damped vibration logarithmic decrement ratio of consecutive amplitude no of cycles after which the original amplitude is reduced to 20 5A single degree damped vibrating system a suspended mass of 18kg makes 15 oscillations in 003 seconds The amplitude decreases to 025 of the initial value after 5 oscillations Determine the stiffness of spring Logarithmic decrement damping factor and damping coefficient 6 A damped vibration record of a spring mass system shows the following data Amplitude at the end of 2nd cycle is 9mm amplitude at eh end of 3rd cycle is 6mm amplitude at the end of 4th cycle is 4mm stiffness of spring =8000 Nm weight =20N Find the logarithmic decrement damping force at unit velocity periodic time of vibration 7 For the system shown in figure the characteristics of dashpot is such that the constant force of 49 N is applied to the piston it velocity is found to be constant at 012 ms Determine the value of C would you expect the complete system to be periodic or aperiodic 8 For a torsional system shown in the figuref1 is natural frequency in air and f2 is the natural frequency when immersed in oil Determine the damping coefficient

Viscous fluid

Disc of mass moment of Inertia J

2a

C

m M

b

a a

k

9 A thin plate of area A and weigh w is fixed to the end of the spring and allow to oscillate in a viscous fluid as shown in figure It f1 is the frequency of oscillation in air and f2 in liquid Show that

22

21 ff

gA2

minusπω

=micro where the damping force on the plate

AV2Fd micro= V being the velocity of vibration 10 A torsional pendulum has a natural frequency of 175 cyclessecond vibrating in vacuum The mass moment of inertia is 2 kgm2 It is immersed in oil and it is observed that the natural frequency is 142 cyclessecond Find the damping torque If the disc is replaced by 34 deg When in oil find the displacement at the end of the first complete cycle 11 An automobile can be modeled as a mass placed on 4 shock absorbers each consisting of a spring and a damper Such that each spring is equally loaded Determine the stiffness damping constant of each shock absorber So that natural frequency is 2Hz and system is critically damped The mass of vehicle is 200kg 12 Write the differential equation of motion for the system shown in the figure below Determine the natural frequency of the damped oscillation and critical damping coefficient

Determine the differential equation of motion and find the critical cdamped natural frequency for system shown in figure

a

C

m M

M =200kg

44

200kg kgkg

oefficient and

45

Derive the equation of motion and find the critical damping coefficient and damped natural frequency for the system shown below Write the differential equation of motion and determine the expression for critical damping coefficient and damped natural frequency of system shown in figure

Set up a differential equation of motion for system shown in figure and determine the following Undamped natural frequencies critical damping coefficient damping ratio damped natural frequency

b

C

kM

a

m

46

A mass of 300N is resting on two springs of stiffness 3000 Nm each and a dashpot of damping coefficient 150 Nsm as shown in figure If the initial velocity of 10cms is given to mass at its equilibrium position what will be the displacement from the equilibrium position at the end of first second Derive the formula used A gun barrel of mass 600kg has a recoil spring of stiffness 294 knm if the barrel recoils 13m on firing determine i) The initial recoil velocity of barrel ii) The critical damping coefficient of dashpot which is engaged at the end of recoil stroke Iii) Time required for the barrel to return at the end of recoil stroke Iv) time required for the barrel to return to a position 5cm from the initial position

W= 3000 N

FIRING

CHAPTER 4

FORCED VIBRATION LEARNING OBJECTIVES

] Introduction to forced free vibration

] Terminologies used in forced vibration

] Concept of magnification factor

] Study of reciprocating and rotating unbalance

] Concept of base excitation and support excitation

] Concept of Transmissibility ratio

] Problems related to forced vibration

The vibration that takes place under the excitation of external forces are called

forced vibration When a system is subjected to harmonic excitation it is forced to vibrate at the same frequency as that of excitation Most of the mechanical systems are prone to steady state forced vibration which cause fatigue failure The vibration problem also becomes very important when any of excitation frequencies coincide with one of natural frequencies of a system Under this condition resonance occurs during which large amplitude of vibrations are observed To avoid resonance external frequency (operating speed) may be changed or properties of the system may be changed to alter the natural frequency In some cases sufficient amount of damping may be provided to avoid large amplitude during resonance Thus the problem of force vibration is very important in mechanical design

FORCED VIBRATIONS OF SINGLE DOF SYSTEMS UNDER HARMONIC EXCITATION

` [ ] tsin FF o ω=

x

sin FF o=

Fo

xampamp

[ ] tω

xamp

47

48

Consider a spring mass system having a viscous damping excited by a harmonic force [ ] tsin FF o ω= as shown in figure The various forces acting on the system is shown in figure Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

O)tsin(Fkxxcxm o =ω+minusminus= ampampamp

)tsin(Fkxxcxm o ω=++ ampampamp ----------------------------------------------------------(1)

or

)tsin(Fkxdtdx

Cdt

xdM o2

2

ω=++

Equation (1) represents the differential equation of motion of the system The solution to this equation consists of two parts namely complementary function and a particular integral

The first of solution is also termed as transient solution of homogeneous solution and can be obtained from the homogeneous differential equation 0kxxcxm =++ ampampamp

The second part of solution particular integral is also termed as steady state solution ad is of form )tsin(Xx φminusω= Where X= steady state amplitude φ is phase difference between external force and displacement

)tsin(Xx φminusω=

The vector representation of the shown

49

The force polygon for the above vector diagram can be written as follows

( )φminusωt O

KX

xm 2ω

Fo

xCω

tω

φ

50

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

is the governing equation

Zero frequency deflection (Xst) It is deflection observed at zero frequency ( when the external frequency 0=ω

( ) kF

001

kF

x o

22

o

st =

+minus=

Magnification factor (M) It is defined as the ratio of steady state amplitude to zero frequency deflection xst

M = arX

X=

A plot of magnification factor versus frequency ratio

damping factor ( )ξ is called frequency plot of a harmon All the frequency terms starts from magnification factor of damping ratio ( )ξ The amplitude of vibration at any on the value of damping The amplitude will be higher for higher damping The amplitude of vibration at reso

amplitude tends to zero as

ωω

n tends to infinity

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

Freq

ζ =

ζ = 1

Magnification Factor M = XXst

kFo

51

ωω

nfor varying values of

ically excite vibrating system

M=1 what ever may be value particular frequency depends

for lesser damping and lesser nance will be very high The

uency Ratio r = (ωωn)

ζ = 0 0

ζ = 025 ζ = 0375 ζ = 0 5

ζ = 0707

ζ = 2

52

Plot of phase angle φ versus frequency ratio

ωω

n is shown in the

figure The phase angle φ is zero for all the values of damping ratio

at

ωω

n=0 The phase

angle φ is 90 for all the values of damping ratio

at

ωω

n=1 ie at

resonance The phase angle between o to 90 deg for all the values of

ωω

n= 0 to 1 and

between 90 deg to 180 deg for all the values of

ωω

ngt1

Class work Derive an expression for frequency at peak amplitude

2np 21 ξminusω=ω=ω is the required equation

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

53

Derive expression amplitude at resonance ( In dimensional form) In dimensional form We know that steady state amplitude

( ) ( )

ω+ωminus=

222

o

Cmk

Fx Equation in dimensional form

We know that at resonance mk

n =ω=ω

( )

ω+

minus

=2

n

2

oR

Cmk

mk

Fx

( )

ω=

2n

oR

C

Fx

( )

ω=

n

oR C

Fx

Class work Derive an expression for amplitude at resonance ( in non dimensional form)

Tutorial Problems on Forced Vibration 1A weight of 60N is suspended by a spring stiffness 12 kNm is forced to vibrate by a harmonic force of 10N assuming viscous damping of 0086 kN-sm Find i) resonance frequency ii) amplitude at resonance iii) phase angle at resonance iv) frequency corresponding to peak amplitude 2 A periodic torque having a maximum value of 05888Nm at a frequency of 4 rads is impressed upon a flywheel suspended from a wire The wheel has mass moment of inertia 012 kgm2 and the wire has stiffness 1176 Nmrad A viscous dashpot applies damping couple of 0784 Nm an angular velocity of 2 rads calculate i)maximum angular displacement from rest position ii) maximum couple applied to dashpot Iii) angle by which the angular displacement lags the torque A mass of 1019 kg is suspended from a spring stiffness 20 Nmm viscous damping causes the amplitude to decrease to 110th of initial value in 4 complete cycle If a periodic force of 20 cos(50 t) is applied to the mass Find the amplitude of forced vibration What would be the amplitude of oscillation if the applied force coincides with the natural period

m=1019 kg

[ ] t50cos 30F =

54

55

ROTATING UNBALANCE Unbalance in rotating machinery is a common source of vibration excitation Consider a machine of mass m having mounted on a foundation of stiffness k and damping coefficient C as shown in figure The unbalance is represented by an eccentric rotating mass m with eccentricity (e) and having a angular velocity (ω) let x be the displacement of non rotating mass from the static equilibrium position The displacement of rotating mass m from the static equilibrium is ( ) tsin( ex ω+ Applying Newtonrsquos law of motion we have

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

56

RECIPROCATING UNBALANCE Unbalance in reciprocating machinery is a common source of vibration excitation Consider a machine of mass M having mounted on a foundation of stiffness K and damping coefficient C as shown in figure The unbalance is represented by a reciprocating mass (m) having crank radus (e) and connecting rod of length (l) Let ωωωω be angular velocity of the crank Let x be the displacement of non reciprocating mass from the static equilibrium position at any instant of time The displacement of reciprocating mass m from the static equilibrium position is

)2sin(sin +++ tle

tex ωωωωωωωω

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

57

Note The equations obtained in reciprocating unbalance is same as the rotating unbalance and having the following differences Rotating Unbalance Reciprocating Unbalance Here mass of the rotating element is considered

Here mass of reciprocating element is considered

Here the eccentricity (e) which is the distance between the centre of rotation and CG

Here the eccentricity e is the crank radius and crank radius = stroke length2

The frequency plot of non dimensional factor( ( )me

MX versus frequency ratio

ωω

n for various values of ( )ξξξξ is shown in figure This frequency plot is same for

the rotating and reciprocating unbalance with the following points are noted from the diagram

1 Damping ratio plays very important role during resonance in controlling the amplitude

2 When

ωω

n=0 then ( )me

MX =0 for all the values of damping ratio ( )ξξξξ

3 When

ωω

nis higher than ( )me

MX tends to unity

4 At very high speeds ie at very high frequency ratio

ωω

n there is no effect

of damping The frequency plot of phase angle versus the frequency ratio is given as shown in figure The explanation is same as in forced harmonic vibration

58

VIBRATION ISOLATION

Vibratory forces and motions generated by machine and other caused are unavoidable Howeever their effect on the other dynamical systems can be minimized by proper isolator design An isolation system is either to protect o delicate machine from excessive vibration transmitted to it from its supporting structure or to prevent vibratory forces and motions generated by machines from being transmitted to its surroundings The effectiveness of isolation may be measured in terms of forces and motion transmitted The first type is known as force isolation and the second type is known as motion isolation

Force isolation (Force Transmissibility) When vibratory forces generated by machine are isolated the effectiveness of forced isolation can be made by defining the term transmissibility The term transmissibility is defined as the ratio of force transmitted to the foundation to that of the external impressed force

Force Transmissibility = o

tr

FF

force Impressed Externalfoundation toted transmitForce

=

Consider viscous damped spring mass system as shown in figure Upon this system an external harmonic force ) t (Sin Fo ω is applied The difference forces acting on system are inertia forces damping force spring force and external force All these forces are represented on force polygon shown below Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these two forces

( ) ( )22 xckx ω+=

59

This is the expression for force transmissibility in dimensional form To represent same in the non dimensional form divide numerator and denominator by k

2

n

22

n

2

n

o

tr

21

21

FF

TR

ωω

ξ+

ωω

minus

ωω

ξ+

==

PHASE DIFFERNECE BETWEEN EXTERNAL FORCE AND DISPLACEMENT Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these

ωω

minus

ωω

ξ=

ωminusω

=φ minusminus2

n

n1

2

1

1

2tan

mKC

tan

The phase difference between force transmitted to the foundation Ftr to the displacement

kc

KxxC

tanω

=ω

=α or

ωω

ξ=

ω=α minusminus

n

11 2tanKx

xCtan

The phase difference between external force Fo and force transmitted to foundationis β

60

β= αminusφ =

ωω

minus

ωω

ξminus

2

n

n1

1

2tan -

ωω

ξminus

n

1 2tan

Displacement Transmissibility It is the ratio of amplitude of steady state vibration of machine to that of the base motion Forced vibration due to support motion Consider a spring mass damper system subjected to harmonic excitation )tsin(Yy ω= at the base as shown in figure Let x be the absolute displacement of the mass at any instant of time t The different forces acting on system are shown in free body diagram Applying Newtonrsquos Law of motion Ma = sumF

)yx(k)yx(cxm minusminusminusminus= ampampampamp

0)yx(k)yx(cxm =minus+minus+ ampampampamp

61

Frequency plot of transmissibility versus frequency ratio

62

The plot of transmissibility versus frequency ratio is as shown in figure The following points can be noted from it

i) The value of transmissibility is unity at frequency ratio

ωω

n=0 and 2

ii) For an undamped system ( )ξ =0 transmissibility tends to infin at resonance iii) The value of transmissibility is less than unity for the value of frequency ratio

greater than 2

iv) For frequency ratio

ωω

nlt 2 smaller damping ratio lead to larger values of

transmissibility on the other hand for frequency ratio

ωω

ngt 2 smaller damping

ratio leads to smaller value of transmissibility

The frequency plot of transmissibility (TR) versus frequency ratio

ωω

n can be

broadly divided into two zones

1

ωω

n= 0 to 2

2

ωω

n= 2 to infin

CASE 1

When

ωω

n= 0 to 2

In this zone the value of transmissibility is always greater than 1 for all values of damping ratio In this zone the transmissibility decreases with increase in damping Hence we can conclude that in this zone damping is useful in isolating vibration forces being transmitted to the support CASE 2

When

ωω

n= 2 to infin

In this zone the value of transmissibility is less than 1 for all values of damping ratio In this zone the transmissibility increases with increase in damping Hence we can conclude that in this zone damping will adversely affect in isolating vibratory forces

63

Tutorial problems on Forced Vibration 1 Machine of mass 75kg is mounted on spring of stiffness 12000 Nm with an assumed damping factor ξ =02 piston within the machine of mass 2 kg has a reciprocating motion with a stroke of 75 cm and speed of 15 cyclessecond Assuming the motion of piston to be harmonic determine amplitude of machine ( X) Transmissibility and force transmitted to foundation phase angle of transmitted force wrt to exciting force 2 Motor weighs 600N is mounted on a simple beam that has a spring rate of 400 Ncm The motor armature base 160 N and has an eccentricity of 0005cm What will be the amplitude of motion when it runs at 1760 RPM Neglect damping the deflection of armature shaft and weight of the beam 3 A Heavy machine of mass 300 kg is supported on resilient foundation the static deflection of foundation due to weight of machine is found to be 75mm It is observed that the machine vibrates with an amplitude of 10mm when the base of the foundation is subjected to harmonic oscillation at the undampeds natural frequency of system with an amplitude of 25mm Find the damping constant of foundation Ii) dynamic force amplitude on the base iii) amplitude of the displacement of machine relative to base 4 An air compressor of mass 450 kg operates at a constant speed of 1750 RPM The reciprocating masses are 10 kg the crank radius is 100mm If the damper for mounting introduces a damping factor 015 i) specify the springs of mounting such that only 20 of the unbalanced force is transmitted to the foundation ii) determine the magnitude of the transmitted force bull Repeat the above problem by assuming that there is no damping in system

5 A body of a car having mass 1500 kg and is mounted on 4 equal springs which deflect through 225 cm under the weight of the body A total damping forces the 4 shock absorbers is 46 N at a velocity of 1cmsec The car is placed with all the four wheels on a test platform which is move up and down at resonant speed With an amplitude of 25cm Find the amplitude of the car body on its springs assuming its CG to be in the centre of wheel base 6A 75 kg machine which is mounted on springs of stiffness 116x105 Nm with the damping of 20=ξ A 2kg piston within the machine as reciprocating motion with a stroke of 008m and speed of 3000 RPM Assuming the motion of piston to be harmonic determine the amplitude of vibration of machine

7 The spring of an automobile are compressed 10cm under its own weight Find the critical speed when a trailer is traveling over road with a people approximaterly by a sine wave of amplitude 75 cm and wave length 15m What will be the amplitude of vibration at 64kmhour 8 An eccentric mass excites us used to determine vibratory characteristics of a structure of mass 25kg as given in figure At a speed of 1000 RPM the eccentric mass to beat top position at the instant the structure was moved upwards through its static equilibrium position and corresponding amplitude was 20mm The eccentric mass is 05kg at an eccentricity of 10mm Determine i) Undamped natural frequency of the system ii) Damping ratio iii) Steady state amplitude and phase angle at 1500 RPM iv) Force transmitted to the ground at 1200 RPM

M=25kg

C

K

Eccentric mass

64

65

CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Support motion Consider a spring mass damper system as shown in figure Let it excited by motion

of support let y be the harmonic displacement of support From Newtonrsquos second law of motion

)yx(k)yx(cxm minusminusminusminus= ampampampamp

)1(0)yx(k)yx(cxm minusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminus=minus+minus+ ampampampamp

Let z be the relative displacement of the mass with respect to support then

Z=x-y

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

21

ττττ ττττ2 ττττ3

ττττ

X(t)

FOURIER SERIES The simplest of periodic motion happens to be SHM It is simple to handle but the motion of many vibrating system is not harmonic (but periodic) Few examples are shown below

Forces acting on machines are generally periodic but this may not be harmonic for example the excitation force in a punching machine is periodic and it can be represented as shown in figure 13 Vibration analysis of system subjected to periodic but nonharmonic forces can be done with the help of Fourier series The problem becomes a multifrequency excitation problem The principle of linear superposition is applied and the total response is the sum of the response due to each of the individual frequency term

Any periodic motion can be expressed as an infinite sum of sines and cosines terms If x(t) is a periodic function with period t its Fourier representation is given by

X(t) = (((( )))) (((( )))) (((( ))))t sint cost cos2 111 ωωωωωωωωωωωω baaao ++++++++++++

= (((( )))) (((( ))))t sint cos2 1

ωωωωωωωω nnn

o bnaa

++++++++sumsumsumsuminfininfininfininfin

====

tππππωωωω

2==== = Fundamental frequency ndash (1)

where ao an bn are constants

ττττ ττττ2 ττττ3

ττττ

X(t)

22

t

X(t)

035

025

Determination of constants To find ao Integrate both sides of equation(1) over any interval ττττ All intergrals on the RHS of the equation are zero except the one containing ao

(((( ))))dttxao

o 2

2

intintintint====ωωωωππππ

ππππωωωω

= (((( ))))dttxo

2intintintintττττ

ττττ

To find an multiply equation 1 by cos (((( ))))tnωωωω and Integrate over any interval ττττ All intergrals

(((( )))) (((( )))) dttxao

n tncos 2

2

ωωωωππππωωωω ωωωω

ππππ

intintintint==== (((( )))) (((( )))) dttxo

tncos 2

ωωωωττττ

ττττ

intintintint====

To find bn multiply equation 1 by sin (((( ))))tnωωωω and Integrate over any interval ττττ All intergrals

(((( )))) (((( )))) dttxao

n tnsin 2

2

ωωωωππππωωωω ωωωω

ππππ

intintintint==== (((( )))) (((( )))) dttxo

tnsin 2

ωωωωττττ

ττττ

intintintint====

Find the Fourier series for the curve shown below

ττττ ττττ2 ττττ3

t

X(t)

E

ττττ ττττ2 ττττ3

t

X(t)

ππππ2

t

X(t) ππππ

1 Represent for the periodic motion shown in the figure

23

CHAPTER 2 UNDAMPED FREE VIBRATION LEARNING OBJECTIVES

] Introduction to undamped free vibration

] Terminologies used in undamped free vibration

] Three methods to solve the undamped free vibration problems

] Problems related to undamped free vibration problems

Free vibrations are oscillations about a systems equilibrium position that occur in the absence of an external excitation force If during vibrations there is no loss of energy it is known as undamped vibration The first step in solving a vibration problem is setting up the differential equation of motion Consider spring mass system which is assumed to move only along the vertical direction as shown below Let m be the mass of the block and k be the stiffness of the spring When block of mass m is attached to spring the deflection of spring will be ∆ known as static deflection In the static equilibrium position the free body diagram of forces acting on the mass is shown in Figure(b) Hence mg= kA Once the system is disturbed the system executes vibrations

Let at any instant of time t the mass is displaced from the equilibrium position x the different forces acting on the system are shown in figure (d) From Newtonrsquos second law of motion

sum = maF

Inertia force ( disturbing force) = restoring force

mgxkxm ++∆minus= )(ampamp

0)( =+ xkxm ampamp

or 0)( =+ xmk

xampamp

equation 2 is the differential equation of motion for spring mfigure Comparing equation (2) with the equation of SHM +xampampsince the vibrations of the above system are free( without the forces) we can write

sec radmk

n =ω

24

ass system shown in 0)(2 =xω

resistance of external

time period km

2 f1

n

πτ ==

from the equation(1) mg = k∆

∆=

q

mk

Difference between the translation ( rectilinear) and rotational system of vibration

Translatory Rotational

In the analysis thFORCES are co Linear stiffness K Problems 1A mass of 10kFind the natural 2 A spring massnatural frequencnatural frequenc

kt

m

sec radmk

n =ω

k

25

e disturbing and restoring nsidered

In the analysis In the analysis MASS Moment of Inertia (J) is considered

in Nm is considered

g when suspended from a spring causes a static deflection of 1cm frequency of system

system has a spring stiffness K Nm and a mass of m Kg It has a y of vibration 12 Hz An extra 2kg mass coupled to it then the y reduces by 2 Hz find K and m

secrad n =ω

26

3 A steel wire of 2mm diameter and 30m long It is fixed at the upper end and carries a mass of m kg at its free end Find m so that the frequency of longitudinal vibration is 4 Hz 4 A spring mass system has a natural period of 02 seconds What will be the new period if the spring constant is 1) increased by 50 2) decreased by 50 5 A spring mass system has a natural frequency of 10 Hz when the spring constant is reduced by 800 Nm the frequency is altered by 45 Find the mass and spring constant of the original system 6 Determine the natural frequency of system shown in fig is which shaft is supported in SHORT bearings 7 Determine the natural frequency of system shown in fig is which shaft is supported in LONG bearings where l is the length of bearing and E ndash youngrsquos modulus and I is moment of Inertia 7 Determine the natural frequency of system shown in fig is which shaft is supported in LONG bearings 8 A light cantilever beam of rectangular section( 5 cm deep x 25cm wide) has a mass fixed at its free end Find the ratio of frequency of free lateral vibration in vertical plane to that in horizontal 9 Determine the natural frequency of simple pendulum 10 Homogeneous square plate of size l and mass m is suspended from the mid point of one of its sides as shown in figure Find the frequency of vibration

l

l

27

11 A compound pendulum which is rigid body of mass m and it is pivoted at O The point of pivot is at distance d from the centre of gravity It is free to rotate about its axis Find the frequency of oscillation of such pendulum 12 A connecting rod shown in fig is supported at the wrist pin end It is displaced and allowed to oscillate The mass of rod is 5kg and centre of gravity is 20 cm from the pivot point O If the frequency of oscillation is 40 cyclesminute calculate the mass moment of inertia about its CG

13 A semi circular homogenous disc of radius r and mass m is pivoted freely about its centre as shown in figure Determine the natural frequency of oscillation 14A simply supported beam of square cross section 5mmx5mm and length 1m carrying a mass of 0575 kg at the middle is found to have natural frequency of 30 radsec Determine youngrsquos modulus of elasticity of beam 15 A spring mass system k1 and m have a natural frequency f1 Determine the value of k2 of another spring in terms of k1 which when placed in series with k1 lowers the natural frequency to 23 f1

COMPLETE SOLUTION OF SYSTEM EXECUTING SHM

The equation of motion of system executing SHM can be represented by

0)( =+ xkxm ampamp ----------------------------------------(1)

022

2=+

dtdx

dt

dx ω

The general solution of equation (1) can be expressed as

------------------------(2)

Where A and B are arbitrary constant which can be determined from the initial conditions of the system Two initial conditions are to be specified to evaluate these

constants x=x0 at t=0 and xamp = Vo at t=0 substituting in the equation (2) Energy method In a conservative system the total energy is constant The dwell as natural frequency can be determined by the princienergy For free vibration of undamped system at any instant and partly potential The kinetic energy T is stored in the masswhere as the potential energy U is stored in the form of sdeformation or work done in a force field such as gravity The total energy being constant T+U = constant Its rate of cha

t)(sin Bt)cos( AX ω+ω=

t)(sin V

t)cos( xx o0 ω

ω+ω=

Is the required complete

solution

28

ifferential equation as ple of conservation of of time is partly kinetic by virtue of its velocity train energy in elastic

nge

Is given by [ ] 0UTdtd

=+

From this we get a differential equation of motion as well as natural frequency of the system Determine the natural frequency of spring mass system using energy method

Determine the natural frequency of system shown in figure Determine the natural frequency of the system shown in figure Is there any limitation on the value of K Discuss Determine the natural frequency of s

m

l k

m

a

θ

m

a

θ

l

k

29

k

k

ystem shown below Neglect the mass of ball

l

m

30

A string shown in figure is under tension T which can be assumed to remain constant for small displacements Find the natural frequency of vertical vibrations of spring An acrobat 120kg walks on a tight rope as shown in figure The frequency of vibration in the given position is vertical direction is 30 rads Find the tension in the rope A manometer has a uniform bore of cross section area A If the column of liquid of length L and Density ρ is set into motion as shown in figure Find the frequency of the resulting oscillation

m

l

a

T T

36m 8m

31

k

R

k

m

r

m

Find the expression for natural frequency of system shown in the figure Neglect the mass of the cantilever beam Study the special case i) k=Infinity ii) I = infinity Determine the expression for the natural frequency of a system shown in figure The two discs are keyed to a common shaft and have a combined mass moment of inertia about centre of oscillation O It is assumed that the card attached to mass m does not stretch and is always under tension Determine the natural frequency of a system shown

m

l

32

Determine the expression for the natural frequency of the system shown in figure Assume that the wires connecting the masses do not stretch and are always in tension

Determine the natural frequency of spring mass system taking the MASS OF SPRING (ms ) into account

M3 M1

M2

k2 k1

m

k

33

RAYLEIGHrsquoS METHOD

This is the extension of energy method Here we can determine the natural frequency of a conservative system without finding the equation of motion The natural frequency is a function of the rate of change of Kinetic energy and potential energy of the system From the principle of conservation of energy we have T+U = Constant T= KE and U =PE We can write the above equation as T1+U1 = --------------------------(1) Where 1 and 2 represents tow instances of time Let 1 be the time when the mass is passing through static equilibrium position Let 2 be the time corresponding to the mass displacement of the mass At this instant the velocity f the mass is zero and hence Substituting in equation (1) we get ( )Max = ( )Max Above equation leads directly to natural frequency Determine the natural frequency of spring mass system using Rayleighrsquos method m

34

r

B

K K

Home work

Determine the natural frequency of pendulum using Rayleighrsquos method

A Cylinder of mass m and moment of inertia Jo is to role without slipping but is sustained by spring K as shown in figure Determine the natural frequency of oscillation

Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping

a

Determine the natural frequency of system shown where in cylindrical disc rolls over the inextensible spring A cylinder of radius r rolls without slipping on cylindrical the equation of motion and hence determine the naoscillations about the lowest point shown Repeat the above problem ndash Instead of cylinder assume swithout slipping on concave surface as shown above

r

K

35

portion of radius R Derive tural frequency of small

phere of radius with r rolls

The torsional pendulum has a natural frequency of 5 Hz What length of steel wire of diameter 2 mm should be used for pendulum The inertia of mass fixed at the free end is 00098 kgm2 Take g= 083x1011 Nm2 Determine the torsional natural frequency of the system shown in figure Neglect the mass moment of inertia of the shaft ( Figure not proportionate) Determine the natural frequency of simple pendulum considering the mass of the rod into account

Using the energy method find tThe chord may be assumed ine

Kt1

Kt2

20cm

06m 08m 04m

20cm 20cm 20cm

m

M

he natural frxtensible in

k

M

36

equency of the system shown in figure the spring mass pulley system

A rod of length l and depth H having a mass m rests on a semicylidrical surface os radius r It is tipped slightly and let go Find the frequency of oscillation

A thin bar of l having a mass m rests on a semicylidrical surface radius r It is tipped slightly and let go Find the frequency of oscillation

l

h

R

l

37

38

CHAPTER 3

DAMPED FREE VIBRATION OF SINGLE DEGREE FREEDOM SYSTEM LEARNING OBJECTIVES

] Introduction to damped free vibration

] Terminologies used in damped free vibration

] Three different types of damping

] Concept of critical damping and its importance

] Study of response of viscous damped system for cases of under damping

critical and over damping

] Concept of logarithmic decrement

] Problems related to damped free vibration problems

Viscously damped free vibration of single degree freedom sytem Consider a single degree freedom system with viscous damping as shown in figure For a viscous damped system the damping force is

where c is a constant of proportionality It is called damping coefficient or coefficient of viscous damping

Symbolically it is designated by a dashpot as shown in Fig 3

x xampx

39

At any instant of time let the mass be displaced by x from the equilibrium position Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

Okxxcxm =minusminus= ampampamp

Okxxcxm =++ ampampamp ----------------------------------------------------------(1)

or

0kxdtdx

Cdt

xdM

2

2

=++

Assuming the solution of equation as

The first term in the above equation m2Ct

eminus

is an exponentially decaying function of time The behavior of the terms in the parenthesis however depends on whether numerical value within the radicals is positive zero or negative

CASE 1 if 0mk

m2c

2

=minus

40

CASE 2 if 0mk

m2c

2

=gt

CASE 3 if 0mk

m2c

2

=lt

41

Critically damped coefficient Critically damped coefficient is defined as the value of the damping coeffient for which radical in equation

mk

m2c

m2c

S2

21 minus

plusmn

= becomes zero

Damping Ratio ( Damping factor) ( ξ ) It is defined as the ratio of damping coefficient to the critical damping coefficient Note The damping coefficient c is defined as the actual amount of damping while the damping coefficient is a constant depending upon the mass and stiffness of the system The roots of characteristics equation can be now written as Note if c=cc or ξ =1 then system is critically damped if cgt cc or ξ gt1 then system is over damped I if clt cc or ξ lt1 then system is under damped

42

Take home quiz Derive General solution of different damped system

1 Critically damped system 2 Under damped system 3 Over damped system

Logarithmic Decrement It is used to determine the amount of damping present in system This measure The rate of decay of free vibration The larger the damping the greater will be rate of decay Logarithmic decrement is defined as the natural logarithm of ratio of any two successive amplitudes

=δ

2

1

xx

ln

Expression for logarithmic decrement (δ ) Consider a damped free vibration which is expressed by

( ) ( ) φ+ξminusω= ξω t 1sinXex 2n

tn

x is the displacement at any instant of time X and φ are arbitrary constants

21

2

ξminus

πξ=δ

43

Tutorial problems 1 Derive an expression for damping ratio in terms of logarithmic decrement

2 Show that logarithmic decrement

=δ

n

o

xx

lnn1

where xo is the initial amplitude xn

is amplitude after n cycles 3 A mass of 1 kg is to be supported on a spring having stiffness K= 9800 Nm The damping constant c= 59 Nsm Determine the natural frequency of system Find also logarithmic decrement and amplitude after 3 cycles if initial displacement is 03 cm 4 A vibrating system is defined by the following parameters M = 3kg k= 100 Nm damping coefficient = 3 Nsm determine Damping factor natural frequency damped vibration logarithmic decrement ratio of consecutive amplitude no of cycles after which the original amplitude is reduced to 20 5A single degree damped vibrating system a suspended mass of 18kg makes 15 oscillations in 003 seconds The amplitude decreases to 025 of the initial value after 5 oscillations Determine the stiffness of spring Logarithmic decrement damping factor and damping coefficient 6 A damped vibration record of a spring mass system shows the following data Amplitude at the end of 2nd cycle is 9mm amplitude at eh end of 3rd cycle is 6mm amplitude at the end of 4th cycle is 4mm stiffness of spring =8000 Nm weight =20N Find the logarithmic decrement damping force at unit velocity periodic time of vibration 7 For the system shown in figure the characteristics of dashpot is such that the constant force of 49 N is applied to the piston it velocity is found to be constant at 012 ms Determine the value of C would you expect the complete system to be periodic or aperiodic 8 For a torsional system shown in the figuref1 is natural frequency in air and f2 is the natural frequency when immersed in oil Determine the damping coefficient

Viscous fluid

Disc of mass moment of Inertia J

2a

C

m M

b

a a

k

9 A thin plate of area A and weigh w is fixed to the end of the spring and allow to oscillate in a viscous fluid as shown in figure It f1 is the frequency of oscillation in air and f2 in liquid Show that

22

21 ff

gA2

minusπω

=micro where the damping force on the plate

AV2Fd micro= V being the velocity of vibration 10 A torsional pendulum has a natural frequency of 175 cyclessecond vibrating in vacuum The mass moment of inertia is 2 kgm2 It is immersed in oil and it is observed that the natural frequency is 142 cyclessecond Find the damping torque If the disc is replaced by 34 deg When in oil find the displacement at the end of the first complete cycle 11 An automobile can be modeled as a mass placed on 4 shock absorbers each consisting of a spring and a damper Such that each spring is equally loaded Determine the stiffness damping constant of each shock absorber So that natural frequency is 2Hz and system is critically damped The mass of vehicle is 200kg 12 Write the differential equation of motion for the system shown in the figure below Determine the natural frequency of the damped oscillation and critical damping coefficient

Determine the differential equation of motion and find the critical cdamped natural frequency for system shown in figure

a

C

m M

M =200kg

44

200kg kgkg

oefficient and

45

Derive the equation of motion and find the critical damping coefficient and damped natural frequency for the system shown below Write the differential equation of motion and determine the expression for critical damping coefficient and damped natural frequency of system shown in figure

Set up a differential equation of motion for system shown in figure and determine the following Undamped natural frequencies critical damping coefficient damping ratio damped natural frequency

b

C

kM

a

m

46

A mass of 300N is resting on two springs of stiffness 3000 Nm each and a dashpot of damping coefficient 150 Nsm as shown in figure If the initial velocity of 10cms is given to mass at its equilibrium position what will be the displacement from the equilibrium position at the end of first second Derive the formula used A gun barrel of mass 600kg has a recoil spring of stiffness 294 knm if the barrel recoils 13m on firing determine i) The initial recoil velocity of barrel ii) The critical damping coefficient of dashpot which is engaged at the end of recoil stroke Iii) Time required for the barrel to return at the end of recoil stroke Iv) time required for the barrel to return to a position 5cm from the initial position

W= 3000 N

FIRING

CHAPTER 4

FORCED VIBRATION LEARNING OBJECTIVES

] Introduction to forced free vibration

] Terminologies used in forced vibration

] Concept of magnification factor

] Study of reciprocating and rotating unbalance

] Concept of base excitation and support excitation

] Concept of Transmissibility ratio

] Problems related to forced vibration

The vibration that takes place under the excitation of external forces are called

forced vibration When a system is subjected to harmonic excitation it is forced to vibrate at the same frequency as that of excitation Most of the mechanical systems are prone to steady state forced vibration which cause fatigue failure The vibration problem also becomes very important when any of excitation frequencies coincide with one of natural frequencies of a system Under this condition resonance occurs during which large amplitude of vibrations are observed To avoid resonance external frequency (operating speed) may be changed or properties of the system may be changed to alter the natural frequency In some cases sufficient amount of damping may be provided to avoid large amplitude during resonance Thus the problem of force vibration is very important in mechanical design

FORCED VIBRATIONS OF SINGLE DOF SYSTEMS UNDER HARMONIC EXCITATION

` [ ] tsin FF o ω=

x

sin FF o=

Fo

xampamp

[ ] tω

xamp

47

48

Consider a spring mass system having a viscous damping excited by a harmonic force [ ] tsin FF o ω= as shown in figure The various forces acting on the system is shown in figure Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

O)tsin(Fkxxcxm o =ω+minusminus= ampampamp

)tsin(Fkxxcxm o ω=++ ampampamp ----------------------------------------------------------(1)

or

)tsin(Fkxdtdx

Cdt

xdM o2

2

ω=++

Equation (1) represents the differential equation of motion of the system The solution to this equation consists of two parts namely complementary function and a particular integral

The first of solution is also termed as transient solution of homogeneous solution and can be obtained from the homogeneous differential equation 0kxxcxm =++ ampampamp

The second part of solution particular integral is also termed as steady state solution ad is of form )tsin(Xx φminusω= Where X= steady state amplitude φ is phase difference between external force and displacement

)tsin(Xx φminusω=

The vector representation of the shown

49

The force polygon for the above vector diagram can be written as follows

( )φminusωt O

KX

xm 2ω

Fo

xCω

tω

φ

50

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

is the governing equation

Zero frequency deflection (Xst) It is deflection observed at zero frequency ( when the external frequency 0=ω

( ) kF

001

kF

x o

22

o

st =

+minus=

Magnification factor (M) It is defined as the ratio of steady state amplitude to zero frequency deflection xst

M = arX

X=

A plot of magnification factor versus frequency ratio

damping factor ( )ξ is called frequency plot of a harmon All the frequency terms starts from magnification factor of damping ratio ( )ξ The amplitude of vibration at any on the value of damping The amplitude will be higher for higher damping The amplitude of vibration at reso

amplitude tends to zero as

ωω

n tends to infinity

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

Freq

ζ =

ζ = 1

Magnification Factor M = XXst

kFo

51

ωω

nfor varying values of

ically excite vibrating system

M=1 what ever may be value particular frequency depends

for lesser damping and lesser nance will be very high The

uency Ratio r = (ωωn)

ζ = 0 0

ζ = 025 ζ = 0375 ζ = 0 5

ζ = 0707

ζ = 2

52

Plot of phase angle φ versus frequency ratio

ωω

n is shown in the

figure The phase angle φ is zero for all the values of damping ratio

at

ωω

n=0 The phase

angle φ is 90 for all the values of damping ratio

at

ωω

n=1 ie at

resonance The phase angle between o to 90 deg for all the values of

ωω

n= 0 to 1 and

between 90 deg to 180 deg for all the values of

ωω

ngt1

Class work Derive an expression for frequency at peak amplitude

2np 21 ξminusω=ω=ω is the required equation

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

53

Derive expression amplitude at resonance ( In dimensional form) In dimensional form We know that steady state amplitude

( ) ( )

ω+ωminus=

222

o

Cmk

Fx Equation in dimensional form

We know that at resonance mk

n =ω=ω

( )

ω+

minus

=2

n

2

oR

Cmk

mk

Fx

( )

ω=

2n

oR

C

Fx

( )

ω=

n

oR C

Fx

Class work Derive an expression for amplitude at resonance ( in non dimensional form)

Tutorial Problems on Forced Vibration 1A weight of 60N is suspended by a spring stiffness 12 kNm is forced to vibrate by a harmonic force of 10N assuming viscous damping of 0086 kN-sm Find i) resonance frequency ii) amplitude at resonance iii) phase angle at resonance iv) frequency corresponding to peak amplitude 2 A periodic torque having a maximum value of 05888Nm at a frequency of 4 rads is impressed upon a flywheel suspended from a wire The wheel has mass moment of inertia 012 kgm2 and the wire has stiffness 1176 Nmrad A viscous dashpot applies damping couple of 0784 Nm an angular velocity of 2 rads calculate i)maximum angular displacement from rest position ii) maximum couple applied to dashpot Iii) angle by which the angular displacement lags the torque A mass of 1019 kg is suspended from a spring stiffness 20 Nmm viscous damping causes the amplitude to decrease to 110th of initial value in 4 complete cycle If a periodic force of 20 cos(50 t) is applied to the mass Find the amplitude of forced vibration What would be the amplitude of oscillation if the applied force coincides with the natural period

m=1019 kg

[ ] t50cos 30F =

54

55

ROTATING UNBALANCE Unbalance in rotating machinery is a common source of vibration excitation Consider a machine of mass m having mounted on a foundation of stiffness k and damping coefficient C as shown in figure The unbalance is represented by an eccentric rotating mass m with eccentricity (e) and having a angular velocity (ω) let x be the displacement of non rotating mass from the static equilibrium position The displacement of rotating mass m from the static equilibrium is ( ) tsin( ex ω+ Applying Newtonrsquos law of motion we have

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

56

RECIPROCATING UNBALANCE Unbalance in reciprocating machinery is a common source of vibration excitation Consider a machine of mass M having mounted on a foundation of stiffness K and damping coefficient C as shown in figure The unbalance is represented by a reciprocating mass (m) having crank radus (e) and connecting rod of length (l) Let ωωωω be angular velocity of the crank Let x be the displacement of non reciprocating mass from the static equilibrium position at any instant of time The displacement of reciprocating mass m from the static equilibrium position is

)2sin(sin +++ tle

tex ωωωωωωωω

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

57

Note The equations obtained in reciprocating unbalance is same as the rotating unbalance and having the following differences Rotating Unbalance Reciprocating Unbalance Here mass of the rotating element is considered

Here mass of reciprocating element is considered

Here the eccentricity (e) which is the distance between the centre of rotation and CG

Here the eccentricity e is the crank radius and crank radius = stroke length2

The frequency plot of non dimensional factor( ( )me

MX versus frequency ratio

ωω

n for various values of ( )ξξξξ is shown in figure This frequency plot is same for

the rotating and reciprocating unbalance with the following points are noted from the diagram

1 Damping ratio plays very important role during resonance in controlling the amplitude

2 When

ωω

n=0 then ( )me

MX =0 for all the values of damping ratio ( )ξξξξ

3 When

ωω

nis higher than ( )me

MX tends to unity

4 At very high speeds ie at very high frequency ratio

ωω

n there is no effect

of damping The frequency plot of phase angle versus the frequency ratio is given as shown in figure The explanation is same as in forced harmonic vibration

58

VIBRATION ISOLATION

Vibratory forces and motions generated by machine and other caused are unavoidable Howeever their effect on the other dynamical systems can be minimized by proper isolator design An isolation system is either to protect o delicate machine from excessive vibration transmitted to it from its supporting structure or to prevent vibratory forces and motions generated by machines from being transmitted to its surroundings The effectiveness of isolation may be measured in terms of forces and motion transmitted The first type is known as force isolation and the second type is known as motion isolation

Force isolation (Force Transmissibility) When vibratory forces generated by machine are isolated the effectiveness of forced isolation can be made by defining the term transmissibility The term transmissibility is defined as the ratio of force transmitted to the foundation to that of the external impressed force

Force Transmissibility = o

tr

FF

force Impressed Externalfoundation toted transmitForce

=

Consider viscous damped spring mass system as shown in figure Upon this system an external harmonic force ) t (Sin Fo ω is applied The difference forces acting on system are inertia forces damping force spring force and external force All these forces are represented on force polygon shown below Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these two forces

( ) ( )22 xckx ω+=

59

This is the expression for force transmissibility in dimensional form To represent same in the non dimensional form divide numerator and denominator by k

2

n

22

n

2

n

o

tr

21

21

FF

TR

ωω

ξ+

ωω

minus

ωω

ξ+

==

PHASE DIFFERNECE BETWEEN EXTERNAL FORCE AND DISPLACEMENT Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these

ωω

minus

ωω

ξ=

ωminusω

=φ minusminus2

n

n1

2

1

1

2tan

mKC

tan

The phase difference between force transmitted to the foundation Ftr to the displacement

kc

KxxC

tanω

=ω

=α or

ωω

ξ=

ω=α minusminus

n

11 2tanKx

xCtan

The phase difference between external force Fo and force transmitted to foundationis β

60

β= αminusφ =

ωω

minus

ωω

ξminus

2

n

n1

1

2tan -

ωω

ξminus

n

1 2tan

Displacement Transmissibility It is the ratio of amplitude of steady state vibration of machine to that of the base motion Forced vibration due to support motion Consider a spring mass damper system subjected to harmonic excitation )tsin(Yy ω= at the base as shown in figure Let x be the absolute displacement of the mass at any instant of time t The different forces acting on system are shown in free body diagram Applying Newtonrsquos Law of motion Ma = sumF

)yx(k)yx(cxm minusminusminusminus= ampampampamp

0)yx(k)yx(cxm =minus+minus+ ampampampamp

61

Frequency plot of transmissibility versus frequency ratio

62

The plot of transmissibility versus frequency ratio is as shown in figure The following points can be noted from it

i) The value of transmissibility is unity at frequency ratio

ωω

n=0 and 2

ii) For an undamped system ( )ξ =0 transmissibility tends to infin at resonance iii) The value of transmissibility is less than unity for the value of frequency ratio

greater than 2

iv) For frequency ratio

ωω

nlt 2 smaller damping ratio lead to larger values of

transmissibility on the other hand for frequency ratio

ωω

ngt 2 smaller damping

ratio leads to smaller value of transmissibility

The frequency plot of transmissibility (TR) versus frequency ratio

ωω

n can be

broadly divided into two zones

1

ωω

n= 0 to 2

2

ωω

n= 2 to infin

CASE 1

When

ωω

n= 0 to 2

In this zone the value of transmissibility is always greater than 1 for all values of damping ratio In this zone the transmissibility decreases with increase in damping Hence we can conclude that in this zone damping is useful in isolating vibration forces being transmitted to the support CASE 2

When

ωω

n= 2 to infin

In this zone the value of transmissibility is less than 1 for all values of damping ratio In this zone the transmissibility increases with increase in damping Hence we can conclude that in this zone damping will adversely affect in isolating vibratory forces

63

Tutorial problems on Forced Vibration 1 Machine of mass 75kg is mounted on spring of stiffness 12000 Nm with an assumed damping factor ξ =02 piston within the machine of mass 2 kg has a reciprocating motion with a stroke of 75 cm and speed of 15 cyclessecond Assuming the motion of piston to be harmonic determine amplitude of machine ( X) Transmissibility and force transmitted to foundation phase angle of transmitted force wrt to exciting force 2 Motor weighs 600N is mounted on a simple beam that has a spring rate of 400 Ncm The motor armature base 160 N and has an eccentricity of 0005cm What will be the amplitude of motion when it runs at 1760 RPM Neglect damping the deflection of armature shaft and weight of the beam 3 A Heavy machine of mass 300 kg is supported on resilient foundation the static deflection of foundation due to weight of machine is found to be 75mm It is observed that the machine vibrates with an amplitude of 10mm when the base of the foundation is subjected to harmonic oscillation at the undampeds natural frequency of system with an amplitude of 25mm Find the damping constant of foundation Ii) dynamic force amplitude on the base iii) amplitude of the displacement of machine relative to base 4 An air compressor of mass 450 kg operates at a constant speed of 1750 RPM The reciprocating masses are 10 kg the crank radius is 100mm If the damper for mounting introduces a damping factor 015 i) specify the springs of mounting such that only 20 of the unbalanced force is transmitted to the foundation ii) determine the magnitude of the transmitted force bull Repeat the above problem by assuming that there is no damping in system

5 A body of a car having mass 1500 kg and is mounted on 4 equal springs which deflect through 225 cm under the weight of the body A total damping forces the 4 shock absorbers is 46 N at a velocity of 1cmsec The car is placed with all the four wheels on a test platform which is move up and down at resonant speed With an amplitude of 25cm Find the amplitude of the car body on its springs assuming its CG to be in the centre of wheel base 6A 75 kg machine which is mounted on springs of stiffness 116x105 Nm with the damping of 20=ξ A 2kg piston within the machine as reciprocating motion with a stroke of 008m and speed of 3000 RPM Assuming the motion of piston to be harmonic determine the amplitude of vibration of machine

7 The spring of an automobile are compressed 10cm under its own weight Find the critical speed when a trailer is traveling over road with a people approximaterly by a sine wave of amplitude 75 cm and wave length 15m What will be the amplitude of vibration at 64kmhour 8 An eccentric mass excites us used to determine vibratory characteristics of a structure of mass 25kg as given in figure At a speed of 1000 RPM the eccentric mass to beat top position at the instant the structure was moved upwards through its static equilibrium position and corresponding amplitude was 20mm The eccentric mass is 05kg at an eccentricity of 10mm Determine i) Undamped natural frequency of the system ii) Damping ratio iii) Steady state amplitude and phase angle at 1500 RPM iv) Force transmitted to the ground at 1200 RPM

M=25kg

C

K

Eccentric mass

64

65

CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Support motion Consider a spring mass damper system as shown in figure Let it excited by motion

of support let y be the harmonic displacement of support From Newtonrsquos second law of motion

)yx(k)yx(cxm minusminusminusminus= ampampampamp

)1(0)yx(k)yx(cxm minusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminus=minus+minus+ ampampampamp

Let z be the relative displacement of the mass with respect to support then

Z=x-y

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

22

t

X(t)

035

025

Determination of constants To find ao Integrate both sides of equation(1) over any interval ττττ All intergrals on the RHS of the equation are zero except the one containing ao

(((( ))))dttxao

o 2

2

intintintint====ωωωωππππ

ππππωωωω

= (((( ))))dttxo

2intintintintττττ

ττττ

To find an multiply equation 1 by cos (((( ))))tnωωωω and Integrate over any interval ττττ All intergrals

(((( )))) (((( )))) dttxao

n tncos 2

2

ωωωωππππωωωω ωωωω

ππππ

intintintint==== (((( )))) (((( )))) dttxo

tncos 2

ωωωωττττ

ττττ

intintintint====

To find bn multiply equation 1 by sin (((( ))))tnωωωω and Integrate over any interval ττττ All intergrals

(((( )))) (((( )))) dttxao

n tnsin 2

2

ωωωωππππωωωω ωωωω

ππππ

intintintint==== (((( )))) (((( )))) dttxo

tnsin 2

ωωωωττττ

ττττ

intintintint====

Find the Fourier series for the curve shown below

ττττ ττττ2 ττττ3

t

X(t)

E

ττττ ττττ2 ττττ3

t

X(t)

ππππ2

t

X(t) ππππ

1 Represent for the periodic motion shown in the figure

23

CHAPTER 2 UNDAMPED FREE VIBRATION LEARNING OBJECTIVES

] Introduction to undamped free vibration

] Terminologies used in undamped free vibration

] Three methods to solve the undamped free vibration problems

] Problems related to undamped free vibration problems

Free vibrations are oscillations about a systems equilibrium position that occur in the absence of an external excitation force If during vibrations there is no loss of energy it is known as undamped vibration The first step in solving a vibration problem is setting up the differential equation of motion Consider spring mass system which is assumed to move only along the vertical direction as shown below Let m be the mass of the block and k be the stiffness of the spring When block of mass m is attached to spring the deflection of spring will be ∆ known as static deflection In the static equilibrium position the free body diagram of forces acting on the mass is shown in Figure(b) Hence mg= kA Once the system is disturbed the system executes vibrations

Let at any instant of time t the mass is displaced from the equilibrium position x the different forces acting on the system are shown in figure (d) From Newtonrsquos second law of motion

sum = maF

Inertia force ( disturbing force) = restoring force

mgxkxm ++∆minus= )(ampamp

0)( =+ xkxm ampamp

or 0)( =+ xmk

xampamp

equation 2 is the differential equation of motion for spring mfigure Comparing equation (2) with the equation of SHM +xampampsince the vibrations of the above system are free( without the forces) we can write

sec radmk

n =ω

24

ass system shown in 0)(2 =xω

resistance of external

time period km

2 f1

n

πτ ==

from the equation(1) mg = k∆

∆=

q

mk

Difference between the translation ( rectilinear) and rotational system of vibration

Translatory Rotational

In the analysis thFORCES are co Linear stiffness K Problems 1A mass of 10kFind the natural 2 A spring massnatural frequencnatural frequenc

kt

m

sec radmk

n =ω

k

25

e disturbing and restoring nsidered

In the analysis In the analysis MASS Moment of Inertia (J) is considered

in Nm is considered

g when suspended from a spring causes a static deflection of 1cm frequency of system

system has a spring stiffness K Nm and a mass of m Kg It has a y of vibration 12 Hz An extra 2kg mass coupled to it then the y reduces by 2 Hz find K and m

secrad n =ω

26

3 A steel wire of 2mm diameter and 30m long It is fixed at the upper end and carries a mass of m kg at its free end Find m so that the frequency of longitudinal vibration is 4 Hz 4 A spring mass system has a natural period of 02 seconds What will be the new period if the spring constant is 1) increased by 50 2) decreased by 50 5 A spring mass system has a natural frequency of 10 Hz when the spring constant is reduced by 800 Nm the frequency is altered by 45 Find the mass and spring constant of the original system 6 Determine the natural frequency of system shown in fig is which shaft is supported in SHORT bearings 7 Determine the natural frequency of system shown in fig is which shaft is supported in LONG bearings where l is the length of bearing and E ndash youngrsquos modulus and I is moment of Inertia 7 Determine the natural frequency of system shown in fig is which shaft is supported in LONG bearings 8 A light cantilever beam of rectangular section( 5 cm deep x 25cm wide) has a mass fixed at its free end Find the ratio of frequency of free lateral vibration in vertical plane to that in horizontal 9 Determine the natural frequency of simple pendulum 10 Homogeneous square plate of size l and mass m is suspended from the mid point of one of its sides as shown in figure Find the frequency of vibration

l

l

27

11 A compound pendulum which is rigid body of mass m and it is pivoted at O The point of pivot is at distance d from the centre of gravity It is free to rotate about its axis Find the frequency of oscillation of such pendulum 12 A connecting rod shown in fig is supported at the wrist pin end It is displaced and allowed to oscillate The mass of rod is 5kg and centre of gravity is 20 cm from the pivot point O If the frequency of oscillation is 40 cyclesminute calculate the mass moment of inertia about its CG

13 A semi circular homogenous disc of radius r and mass m is pivoted freely about its centre as shown in figure Determine the natural frequency of oscillation 14A simply supported beam of square cross section 5mmx5mm and length 1m carrying a mass of 0575 kg at the middle is found to have natural frequency of 30 radsec Determine youngrsquos modulus of elasticity of beam 15 A spring mass system k1 and m have a natural frequency f1 Determine the value of k2 of another spring in terms of k1 which when placed in series with k1 lowers the natural frequency to 23 f1

COMPLETE SOLUTION OF SYSTEM EXECUTING SHM

The equation of motion of system executing SHM can be represented by

0)( =+ xkxm ampamp ----------------------------------------(1)

022

2=+

dtdx

dt

dx ω

The general solution of equation (1) can be expressed as

------------------------(2)

Where A and B are arbitrary constant which can be determined from the initial conditions of the system Two initial conditions are to be specified to evaluate these

constants x=x0 at t=0 and xamp = Vo at t=0 substituting in the equation (2) Energy method In a conservative system the total energy is constant The dwell as natural frequency can be determined by the princienergy For free vibration of undamped system at any instant and partly potential The kinetic energy T is stored in the masswhere as the potential energy U is stored in the form of sdeformation or work done in a force field such as gravity The total energy being constant T+U = constant Its rate of cha

t)(sin Bt)cos( AX ω+ω=

t)(sin V

t)cos( xx o0 ω

ω+ω=

Is the required complete

solution

28

ifferential equation as ple of conservation of of time is partly kinetic by virtue of its velocity train energy in elastic

nge

Is given by [ ] 0UTdtd

=+

From this we get a differential equation of motion as well as natural frequency of the system Determine the natural frequency of spring mass system using energy method

Determine the natural frequency of system shown in figure Determine the natural frequency of the system shown in figure Is there any limitation on the value of K Discuss Determine the natural frequency of s

m

l k

m

a

θ

m

a

θ

l

k

29

k

k

ystem shown below Neglect the mass of ball

l

m

30

A string shown in figure is under tension T which can be assumed to remain constant for small displacements Find the natural frequency of vertical vibrations of spring An acrobat 120kg walks on a tight rope as shown in figure The frequency of vibration in the given position is vertical direction is 30 rads Find the tension in the rope A manometer has a uniform bore of cross section area A If the column of liquid of length L and Density ρ is set into motion as shown in figure Find the frequency of the resulting oscillation

m

l

a

T T

36m 8m

31

k

R

k

m

r

m

Find the expression for natural frequency of system shown in the figure Neglect the mass of the cantilever beam Study the special case i) k=Infinity ii) I = infinity Determine the expression for the natural frequency of a system shown in figure The two discs are keyed to a common shaft and have a combined mass moment of inertia about centre of oscillation O It is assumed that the card attached to mass m does not stretch and is always under tension Determine the natural frequency of a system shown

m

l

32

Determine the expression for the natural frequency of the system shown in figure Assume that the wires connecting the masses do not stretch and are always in tension

Determine the natural frequency of spring mass system taking the MASS OF SPRING (ms ) into account

M3 M1

M2

k2 k1

m

k

33

RAYLEIGHrsquoS METHOD

This is the extension of energy method Here we can determine the natural frequency of a conservative system without finding the equation of motion The natural frequency is a function of the rate of change of Kinetic energy and potential energy of the system From the principle of conservation of energy we have T+U = Constant T= KE and U =PE We can write the above equation as T1+U1 = --------------------------(1) Where 1 and 2 represents tow instances of time Let 1 be the time when the mass is passing through static equilibrium position Let 2 be the time corresponding to the mass displacement of the mass At this instant the velocity f the mass is zero and hence Substituting in equation (1) we get ( )Max = ( )Max Above equation leads directly to natural frequency Determine the natural frequency of spring mass system using Rayleighrsquos method m

34

r

B

K K

Home work

Determine the natural frequency of pendulum using Rayleighrsquos method

A Cylinder of mass m and moment of inertia Jo is to role without slipping but is sustained by spring K as shown in figure Determine the natural frequency of oscillation

Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping

a

Determine the natural frequency of system shown where in cylindrical disc rolls over the inextensible spring A cylinder of radius r rolls without slipping on cylindrical the equation of motion and hence determine the naoscillations about the lowest point shown Repeat the above problem ndash Instead of cylinder assume swithout slipping on concave surface as shown above

r

K

35

portion of radius R Derive tural frequency of small

phere of radius with r rolls

The torsional pendulum has a natural frequency of 5 Hz What length of steel wire of diameter 2 mm should be used for pendulum The inertia of mass fixed at the free end is 00098 kgm2 Take g= 083x1011 Nm2 Determine the torsional natural frequency of the system shown in figure Neglect the mass moment of inertia of the shaft ( Figure not proportionate) Determine the natural frequency of simple pendulum considering the mass of the rod into account

Using the energy method find tThe chord may be assumed ine

Kt1

Kt2

20cm

06m 08m 04m

20cm 20cm 20cm

m

M

he natural frxtensible in

k

M

36

equency of the system shown in figure the spring mass pulley system

A rod of length l and depth H having a mass m rests on a semicylidrical surface os radius r It is tipped slightly and let go Find the frequency of oscillation

A thin bar of l having a mass m rests on a semicylidrical surface radius r It is tipped slightly and let go Find the frequency of oscillation

l

h

R

l

37

38

CHAPTER 3

DAMPED FREE VIBRATION OF SINGLE DEGREE FREEDOM SYSTEM LEARNING OBJECTIVES

] Introduction to damped free vibration

] Terminologies used in damped free vibration

] Three different types of damping

] Concept of critical damping and its importance

] Study of response of viscous damped system for cases of under damping

critical and over damping

] Concept of logarithmic decrement

] Problems related to damped free vibration problems

Viscously damped free vibration of single degree freedom sytem Consider a single degree freedom system with viscous damping as shown in figure For a viscous damped system the damping force is

where c is a constant of proportionality It is called damping coefficient or coefficient of viscous damping

Symbolically it is designated by a dashpot as shown in Fig 3

x xampx

39

At any instant of time let the mass be displaced by x from the equilibrium position Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

Okxxcxm =minusminus= ampampamp

Okxxcxm =++ ampampamp ----------------------------------------------------------(1)

or

0kxdtdx

Cdt

xdM

2

2

=++

Assuming the solution of equation as

The first term in the above equation m2Ct

eminus

is an exponentially decaying function of time The behavior of the terms in the parenthesis however depends on whether numerical value within the radicals is positive zero or negative

CASE 1 if 0mk

m2c

2

=minus

40

CASE 2 if 0mk

m2c

2

=gt

CASE 3 if 0mk

m2c

2

=lt

41

Critically damped coefficient Critically damped coefficient is defined as the value of the damping coeffient for which radical in equation

mk

m2c

m2c

S2

21 minus

plusmn

= becomes zero

Damping Ratio ( Damping factor) ( ξ ) It is defined as the ratio of damping coefficient to the critical damping coefficient Note The damping coefficient c is defined as the actual amount of damping while the damping coefficient is a constant depending upon the mass and stiffness of the system The roots of characteristics equation can be now written as Note if c=cc or ξ =1 then system is critically damped if cgt cc or ξ gt1 then system is over damped I if clt cc or ξ lt1 then system is under damped

42

Take home quiz Derive General solution of different damped system

1 Critically damped system 2 Under damped system 3 Over damped system

Logarithmic Decrement It is used to determine the amount of damping present in system This measure The rate of decay of free vibration The larger the damping the greater will be rate of decay Logarithmic decrement is defined as the natural logarithm of ratio of any two successive amplitudes

=δ

2

1

xx

ln

Expression for logarithmic decrement (δ ) Consider a damped free vibration which is expressed by

( ) ( ) φ+ξminusω= ξω t 1sinXex 2n

tn

x is the displacement at any instant of time X and φ are arbitrary constants

21

2

ξminus

πξ=δ

43

Tutorial problems 1 Derive an expression for damping ratio in terms of logarithmic decrement

2 Show that logarithmic decrement

=δ

n

o

xx

lnn1

where xo is the initial amplitude xn

is amplitude after n cycles 3 A mass of 1 kg is to be supported on a spring having stiffness K= 9800 Nm The damping constant c= 59 Nsm Determine the natural frequency of system Find also logarithmic decrement and amplitude after 3 cycles if initial displacement is 03 cm 4 A vibrating system is defined by the following parameters M = 3kg k= 100 Nm damping coefficient = 3 Nsm determine Damping factor natural frequency damped vibration logarithmic decrement ratio of consecutive amplitude no of cycles after which the original amplitude is reduced to 20 5A single degree damped vibrating system a suspended mass of 18kg makes 15 oscillations in 003 seconds The amplitude decreases to 025 of the initial value after 5 oscillations Determine the stiffness of spring Logarithmic decrement damping factor and damping coefficient 6 A damped vibration record of a spring mass system shows the following data Amplitude at the end of 2nd cycle is 9mm amplitude at eh end of 3rd cycle is 6mm amplitude at the end of 4th cycle is 4mm stiffness of spring =8000 Nm weight =20N Find the logarithmic decrement damping force at unit velocity periodic time of vibration 7 For the system shown in figure the characteristics of dashpot is such that the constant force of 49 N is applied to the piston it velocity is found to be constant at 012 ms Determine the value of C would you expect the complete system to be periodic or aperiodic 8 For a torsional system shown in the figuref1 is natural frequency in air and f2 is the natural frequency when immersed in oil Determine the damping coefficient

Viscous fluid

Disc of mass moment of Inertia J

2a

C

m M

b

a a

k

9 A thin plate of area A and weigh w is fixed to the end of the spring and allow to oscillate in a viscous fluid as shown in figure It f1 is the frequency of oscillation in air and f2 in liquid Show that

22

21 ff

gA2

minusπω

=micro where the damping force on the plate

AV2Fd micro= V being the velocity of vibration 10 A torsional pendulum has a natural frequency of 175 cyclessecond vibrating in vacuum The mass moment of inertia is 2 kgm2 It is immersed in oil and it is observed that the natural frequency is 142 cyclessecond Find the damping torque If the disc is replaced by 34 deg When in oil find the displacement at the end of the first complete cycle 11 An automobile can be modeled as a mass placed on 4 shock absorbers each consisting of a spring and a damper Such that each spring is equally loaded Determine the stiffness damping constant of each shock absorber So that natural frequency is 2Hz and system is critically damped The mass of vehicle is 200kg 12 Write the differential equation of motion for the system shown in the figure below Determine the natural frequency of the damped oscillation and critical damping coefficient

Determine the differential equation of motion and find the critical cdamped natural frequency for system shown in figure

a

C

m M

M =200kg

44

200kg kgkg

oefficient and

45

Derive the equation of motion and find the critical damping coefficient and damped natural frequency for the system shown below Write the differential equation of motion and determine the expression for critical damping coefficient and damped natural frequency of system shown in figure

Set up a differential equation of motion for system shown in figure and determine the following Undamped natural frequencies critical damping coefficient damping ratio damped natural frequency

b

C

kM

a

m

46

A mass of 300N is resting on two springs of stiffness 3000 Nm each and a dashpot of damping coefficient 150 Nsm as shown in figure If the initial velocity of 10cms is given to mass at its equilibrium position what will be the displacement from the equilibrium position at the end of first second Derive the formula used A gun barrel of mass 600kg has a recoil spring of stiffness 294 knm if the barrel recoils 13m on firing determine i) The initial recoil velocity of barrel ii) The critical damping coefficient of dashpot which is engaged at the end of recoil stroke Iii) Time required for the barrel to return at the end of recoil stroke Iv) time required for the barrel to return to a position 5cm from the initial position

W= 3000 N

FIRING

CHAPTER 4

FORCED VIBRATION LEARNING OBJECTIVES

] Introduction to forced free vibration

] Terminologies used in forced vibration

] Concept of magnification factor

] Study of reciprocating and rotating unbalance

] Concept of base excitation and support excitation

] Concept of Transmissibility ratio

] Problems related to forced vibration

The vibration that takes place under the excitation of external forces are called

forced vibration When a system is subjected to harmonic excitation it is forced to vibrate at the same frequency as that of excitation Most of the mechanical systems are prone to steady state forced vibration which cause fatigue failure The vibration problem also becomes very important when any of excitation frequencies coincide with one of natural frequencies of a system Under this condition resonance occurs during which large amplitude of vibrations are observed To avoid resonance external frequency (operating speed) may be changed or properties of the system may be changed to alter the natural frequency In some cases sufficient amount of damping may be provided to avoid large amplitude during resonance Thus the problem of force vibration is very important in mechanical design

FORCED VIBRATIONS OF SINGLE DOF SYSTEMS UNDER HARMONIC EXCITATION

` [ ] tsin FF o ω=

x

sin FF o=

Fo

xampamp

[ ] tω

xamp

47

48

Consider a spring mass system having a viscous damping excited by a harmonic force [ ] tsin FF o ω= as shown in figure The various forces acting on the system is shown in figure Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

O)tsin(Fkxxcxm o =ω+minusminus= ampampamp

)tsin(Fkxxcxm o ω=++ ampampamp ----------------------------------------------------------(1)

or

)tsin(Fkxdtdx

Cdt

xdM o2

2

ω=++

Equation (1) represents the differential equation of motion of the system The solution to this equation consists of two parts namely complementary function and a particular integral

The first of solution is also termed as transient solution of homogeneous solution and can be obtained from the homogeneous differential equation 0kxxcxm =++ ampampamp

The second part of solution particular integral is also termed as steady state solution ad is of form )tsin(Xx φminusω= Where X= steady state amplitude φ is phase difference between external force and displacement

)tsin(Xx φminusω=

The vector representation of the shown

49

The force polygon for the above vector diagram can be written as follows

( )φminusωt O

KX

xm 2ω

Fo

xCω

tω

φ

50

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

is the governing equation

Zero frequency deflection (Xst) It is deflection observed at zero frequency ( when the external frequency 0=ω

( ) kF

001

kF

x o

22

o

st =

+minus=

Magnification factor (M) It is defined as the ratio of steady state amplitude to zero frequency deflection xst

M = arX

X=

A plot of magnification factor versus frequency ratio

damping factor ( )ξ is called frequency plot of a harmon All the frequency terms starts from magnification factor of damping ratio ( )ξ The amplitude of vibration at any on the value of damping The amplitude will be higher for higher damping The amplitude of vibration at reso

amplitude tends to zero as

ωω

n tends to infinity

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

Freq

ζ =

ζ = 1

Magnification Factor M = XXst

kFo

51

ωω

nfor varying values of

ically excite vibrating system

M=1 what ever may be value particular frequency depends

for lesser damping and lesser nance will be very high The

uency Ratio r = (ωωn)

ζ = 0 0

ζ = 025 ζ = 0375 ζ = 0 5

ζ = 0707

ζ = 2

52

Plot of phase angle φ versus frequency ratio

ωω

n is shown in the

figure The phase angle φ is zero for all the values of damping ratio

at

ωω

n=0 The phase

angle φ is 90 for all the values of damping ratio

at

ωω

n=1 ie at

resonance The phase angle between o to 90 deg for all the values of

ωω

n= 0 to 1 and

between 90 deg to 180 deg for all the values of

ωω

ngt1

Class work Derive an expression for frequency at peak amplitude

2np 21 ξminusω=ω=ω is the required equation

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

53

Derive expression amplitude at resonance ( In dimensional form) In dimensional form We know that steady state amplitude

( ) ( )

ω+ωminus=

222

o

Cmk

Fx Equation in dimensional form

We know that at resonance mk

n =ω=ω

( )

ω+

minus

=2

n

2

oR

Cmk

mk

Fx

( )

ω=

2n

oR

C

Fx

( )

ω=

n

oR C

Fx

Class work Derive an expression for amplitude at resonance ( in non dimensional form)

Tutorial Problems on Forced Vibration 1A weight of 60N is suspended by a spring stiffness 12 kNm is forced to vibrate by a harmonic force of 10N assuming viscous damping of 0086 kN-sm Find i) resonance frequency ii) amplitude at resonance iii) phase angle at resonance iv) frequency corresponding to peak amplitude 2 A periodic torque having a maximum value of 05888Nm at a frequency of 4 rads is impressed upon a flywheel suspended from a wire The wheel has mass moment of inertia 012 kgm2 and the wire has stiffness 1176 Nmrad A viscous dashpot applies damping couple of 0784 Nm an angular velocity of 2 rads calculate i)maximum angular displacement from rest position ii) maximum couple applied to dashpot Iii) angle by which the angular displacement lags the torque A mass of 1019 kg is suspended from a spring stiffness 20 Nmm viscous damping causes the amplitude to decrease to 110th of initial value in 4 complete cycle If a periodic force of 20 cos(50 t) is applied to the mass Find the amplitude of forced vibration What would be the amplitude of oscillation if the applied force coincides with the natural period

m=1019 kg

[ ] t50cos 30F =

54

55

ROTATING UNBALANCE Unbalance in rotating machinery is a common source of vibration excitation Consider a machine of mass m having mounted on a foundation of stiffness k and damping coefficient C as shown in figure The unbalance is represented by an eccentric rotating mass m with eccentricity (e) and having a angular velocity (ω) let x be the displacement of non rotating mass from the static equilibrium position The displacement of rotating mass m from the static equilibrium is ( ) tsin( ex ω+ Applying Newtonrsquos law of motion we have

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

56

RECIPROCATING UNBALANCE Unbalance in reciprocating machinery is a common source of vibration excitation Consider a machine of mass M having mounted on a foundation of stiffness K and damping coefficient C as shown in figure The unbalance is represented by a reciprocating mass (m) having crank radus (e) and connecting rod of length (l) Let ωωωω be angular velocity of the crank Let x be the displacement of non reciprocating mass from the static equilibrium position at any instant of time The displacement of reciprocating mass m from the static equilibrium position is

)2sin(sin +++ tle

tex ωωωωωωωω

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

57

Note The equations obtained in reciprocating unbalance is same as the rotating unbalance and having the following differences Rotating Unbalance Reciprocating Unbalance Here mass of the rotating element is considered

Here mass of reciprocating element is considered

Here the eccentricity (e) which is the distance between the centre of rotation and CG

Here the eccentricity e is the crank radius and crank radius = stroke length2

The frequency plot of non dimensional factor( ( )me

MX versus frequency ratio

ωω

n for various values of ( )ξξξξ is shown in figure This frequency plot is same for

the rotating and reciprocating unbalance with the following points are noted from the diagram

1 Damping ratio plays very important role during resonance in controlling the amplitude

2 When

ωω

n=0 then ( )me

MX =0 for all the values of damping ratio ( )ξξξξ

3 When

ωω

nis higher than ( )me

MX tends to unity

4 At very high speeds ie at very high frequency ratio

ωω

n there is no effect

of damping The frequency plot of phase angle versus the frequency ratio is given as shown in figure The explanation is same as in forced harmonic vibration

58

VIBRATION ISOLATION

Vibratory forces and motions generated by machine and other caused are unavoidable Howeever their effect on the other dynamical systems can be minimized by proper isolator design An isolation system is either to protect o delicate machine from excessive vibration transmitted to it from its supporting structure or to prevent vibratory forces and motions generated by machines from being transmitted to its surroundings The effectiveness of isolation may be measured in terms of forces and motion transmitted The first type is known as force isolation and the second type is known as motion isolation

Force isolation (Force Transmissibility) When vibratory forces generated by machine are isolated the effectiveness of forced isolation can be made by defining the term transmissibility The term transmissibility is defined as the ratio of force transmitted to the foundation to that of the external impressed force

Force Transmissibility = o

tr

FF

force Impressed Externalfoundation toted transmitForce

=

Consider viscous damped spring mass system as shown in figure Upon this system an external harmonic force ) t (Sin Fo ω is applied The difference forces acting on system are inertia forces damping force spring force and external force All these forces are represented on force polygon shown below Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these two forces

( ) ( )22 xckx ω+=

59

This is the expression for force transmissibility in dimensional form To represent same in the non dimensional form divide numerator and denominator by k

2

n

22

n

2

n

o

tr

21

21

FF

TR

ωω

ξ+

ωω

minus

ωω

ξ+

==

PHASE DIFFERNECE BETWEEN EXTERNAL FORCE AND DISPLACEMENT Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these

ωω

minus

ωω

ξ=

ωminusω

=φ minusminus2

n

n1

2

1

1

2tan

mKC

tan

The phase difference between force transmitted to the foundation Ftr to the displacement

kc

KxxC

tanω

=ω

=α or

ωω

ξ=

ω=α minusminus

n

11 2tanKx

xCtan

The phase difference between external force Fo and force transmitted to foundationis β

60

β= αminusφ =

ωω

minus

ωω

ξminus

2

n

n1

1

2tan -

ωω

ξminus

n

1 2tan

Displacement Transmissibility It is the ratio of amplitude of steady state vibration of machine to that of the base motion Forced vibration due to support motion Consider a spring mass damper system subjected to harmonic excitation )tsin(Yy ω= at the base as shown in figure Let x be the absolute displacement of the mass at any instant of time t The different forces acting on system are shown in free body diagram Applying Newtonrsquos Law of motion Ma = sumF

)yx(k)yx(cxm minusminusminusminus= ampampampamp

0)yx(k)yx(cxm =minus+minus+ ampampampamp

61

Frequency plot of transmissibility versus frequency ratio

62

The plot of transmissibility versus frequency ratio is as shown in figure The following points can be noted from it

i) The value of transmissibility is unity at frequency ratio

ωω

n=0 and 2

ii) For an undamped system ( )ξ =0 transmissibility tends to infin at resonance iii) The value of transmissibility is less than unity for the value of frequency ratio

greater than 2

iv) For frequency ratio

ωω

nlt 2 smaller damping ratio lead to larger values of

transmissibility on the other hand for frequency ratio

ωω

ngt 2 smaller damping

ratio leads to smaller value of transmissibility

The frequency plot of transmissibility (TR) versus frequency ratio

ωω

n can be

broadly divided into two zones

1

ωω

n= 0 to 2

2

ωω

n= 2 to infin

CASE 1

When

ωω

n= 0 to 2

In this zone the value of transmissibility is always greater than 1 for all values of damping ratio In this zone the transmissibility decreases with increase in damping Hence we can conclude that in this zone damping is useful in isolating vibration forces being transmitted to the support CASE 2

When

ωω

n= 2 to infin

In this zone the value of transmissibility is less than 1 for all values of damping ratio In this zone the transmissibility increases with increase in damping Hence we can conclude that in this zone damping will adversely affect in isolating vibratory forces

63

Tutorial problems on Forced Vibration 1 Machine of mass 75kg is mounted on spring of stiffness 12000 Nm with an assumed damping factor ξ =02 piston within the machine of mass 2 kg has a reciprocating motion with a stroke of 75 cm and speed of 15 cyclessecond Assuming the motion of piston to be harmonic determine amplitude of machine ( X) Transmissibility and force transmitted to foundation phase angle of transmitted force wrt to exciting force 2 Motor weighs 600N is mounted on a simple beam that has a spring rate of 400 Ncm The motor armature base 160 N and has an eccentricity of 0005cm What will be the amplitude of motion when it runs at 1760 RPM Neglect damping the deflection of armature shaft and weight of the beam 3 A Heavy machine of mass 300 kg is supported on resilient foundation the static deflection of foundation due to weight of machine is found to be 75mm It is observed that the machine vibrates with an amplitude of 10mm when the base of the foundation is subjected to harmonic oscillation at the undampeds natural frequency of system with an amplitude of 25mm Find the damping constant of foundation Ii) dynamic force amplitude on the base iii) amplitude of the displacement of machine relative to base 4 An air compressor of mass 450 kg operates at a constant speed of 1750 RPM The reciprocating masses are 10 kg the crank radius is 100mm If the damper for mounting introduces a damping factor 015 i) specify the springs of mounting such that only 20 of the unbalanced force is transmitted to the foundation ii) determine the magnitude of the transmitted force bull Repeat the above problem by assuming that there is no damping in system

5 A body of a car having mass 1500 kg and is mounted on 4 equal springs which deflect through 225 cm under the weight of the body A total damping forces the 4 shock absorbers is 46 N at a velocity of 1cmsec The car is placed with all the four wheels on a test platform which is move up and down at resonant speed With an amplitude of 25cm Find the amplitude of the car body on its springs assuming its CG to be in the centre of wheel base 6A 75 kg machine which is mounted on springs of stiffness 116x105 Nm with the damping of 20=ξ A 2kg piston within the machine as reciprocating motion with a stroke of 008m and speed of 3000 RPM Assuming the motion of piston to be harmonic determine the amplitude of vibration of machine

7 The spring of an automobile are compressed 10cm under its own weight Find the critical speed when a trailer is traveling over road with a people approximaterly by a sine wave of amplitude 75 cm and wave length 15m What will be the amplitude of vibration at 64kmhour 8 An eccentric mass excites us used to determine vibratory characteristics of a structure of mass 25kg as given in figure At a speed of 1000 RPM the eccentric mass to beat top position at the instant the structure was moved upwards through its static equilibrium position and corresponding amplitude was 20mm The eccentric mass is 05kg at an eccentricity of 10mm Determine i) Undamped natural frequency of the system ii) Damping ratio iii) Steady state amplitude and phase angle at 1500 RPM iv) Force transmitted to the ground at 1200 RPM

M=25kg

C

K

Eccentric mass

64

65

CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Support motion Consider a spring mass damper system as shown in figure Let it excited by motion

of support let y be the harmonic displacement of support From Newtonrsquos second law of motion

)yx(k)yx(cxm minusminusminusminus= ampampampamp

)1(0)yx(k)yx(cxm minusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminus=minus+minus+ ampampampamp

Let z be the relative displacement of the mass with respect to support then

Z=x-y

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

23

CHAPTER 2 UNDAMPED FREE VIBRATION LEARNING OBJECTIVES

] Introduction to undamped free vibration

] Terminologies used in undamped free vibration

] Three methods to solve the undamped free vibration problems

] Problems related to undamped free vibration problems

Free vibrations are oscillations about a systems equilibrium position that occur in the absence of an external excitation force If during vibrations there is no loss of energy it is known as undamped vibration The first step in solving a vibration problem is setting up the differential equation of motion Consider spring mass system which is assumed to move only along the vertical direction as shown below Let m be the mass of the block and k be the stiffness of the spring When block of mass m is attached to spring the deflection of spring will be ∆ known as static deflection In the static equilibrium position the free body diagram of forces acting on the mass is shown in Figure(b) Hence mg= kA Once the system is disturbed the system executes vibrations

Let at any instant of time t the mass is displaced from the equilibrium position x the different forces acting on the system are shown in figure (d) From Newtonrsquos second law of motion

sum = maF

Inertia force ( disturbing force) = restoring force

mgxkxm ++∆minus= )(ampamp

0)( =+ xkxm ampamp

or 0)( =+ xmk

xampamp

equation 2 is the differential equation of motion for spring mfigure Comparing equation (2) with the equation of SHM +xampampsince the vibrations of the above system are free( without the forces) we can write

sec radmk

n =ω

24

ass system shown in 0)(2 =xω

resistance of external

time period km

2 f1

n

πτ ==

from the equation(1) mg = k∆

∆=

q

mk

Difference between the translation ( rectilinear) and rotational system of vibration

Translatory Rotational

In the analysis thFORCES are co Linear stiffness K Problems 1A mass of 10kFind the natural 2 A spring massnatural frequencnatural frequenc

kt

m

sec radmk

n =ω

k

25

e disturbing and restoring nsidered

In the analysis In the analysis MASS Moment of Inertia (J) is considered

in Nm is considered

g when suspended from a spring causes a static deflection of 1cm frequency of system

system has a spring stiffness K Nm and a mass of m Kg It has a y of vibration 12 Hz An extra 2kg mass coupled to it then the y reduces by 2 Hz find K and m

secrad n =ω

26

3 A steel wire of 2mm diameter and 30m long It is fixed at the upper end and carries a mass of m kg at its free end Find m so that the frequency of longitudinal vibration is 4 Hz 4 A spring mass system has a natural period of 02 seconds What will be the new period if the spring constant is 1) increased by 50 2) decreased by 50 5 A spring mass system has a natural frequency of 10 Hz when the spring constant is reduced by 800 Nm the frequency is altered by 45 Find the mass and spring constant of the original system 6 Determine the natural frequency of system shown in fig is which shaft is supported in SHORT bearings 7 Determine the natural frequency of system shown in fig is which shaft is supported in LONG bearings where l is the length of bearing and E ndash youngrsquos modulus and I is moment of Inertia 7 Determine the natural frequency of system shown in fig is which shaft is supported in LONG bearings 8 A light cantilever beam of rectangular section( 5 cm deep x 25cm wide) has a mass fixed at its free end Find the ratio of frequency of free lateral vibration in vertical plane to that in horizontal 9 Determine the natural frequency of simple pendulum 10 Homogeneous square plate of size l and mass m is suspended from the mid point of one of its sides as shown in figure Find the frequency of vibration

l

l

27

11 A compound pendulum which is rigid body of mass m and it is pivoted at O The point of pivot is at distance d from the centre of gravity It is free to rotate about its axis Find the frequency of oscillation of such pendulum 12 A connecting rod shown in fig is supported at the wrist pin end It is displaced and allowed to oscillate The mass of rod is 5kg and centre of gravity is 20 cm from the pivot point O If the frequency of oscillation is 40 cyclesminute calculate the mass moment of inertia about its CG

13 A semi circular homogenous disc of radius r and mass m is pivoted freely about its centre as shown in figure Determine the natural frequency of oscillation 14A simply supported beam of square cross section 5mmx5mm and length 1m carrying a mass of 0575 kg at the middle is found to have natural frequency of 30 radsec Determine youngrsquos modulus of elasticity of beam 15 A spring mass system k1 and m have a natural frequency f1 Determine the value of k2 of another spring in terms of k1 which when placed in series with k1 lowers the natural frequency to 23 f1

COMPLETE SOLUTION OF SYSTEM EXECUTING SHM

The equation of motion of system executing SHM can be represented by

0)( =+ xkxm ampamp ----------------------------------------(1)

022

2=+

dtdx

dt

dx ω

The general solution of equation (1) can be expressed as

------------------------(2)

Where A and B are arbitrary constant which can be determined from the initial conditions of the system Two initial conditions are to be specified to evaluate these

constants x=x0 at t=0 and xamp = Vo at t=0 substituting in the equation (2) Energy method In a conservative system the total energy is constant The dwell as natural frequency can be determined by the princienergy For free vibration of undamped system at any instant and partly potential The kinetic energy T is stored in the masswhere as the potential energy U is stored in the form of sdeformation or work done in a force field such as gravity The total energy being constant T+U = constant Its rate of cha

t)(sin Bt)cos( AX ω+ω=

t)(sin V

t)cos( xx o0 ω

ω+ω=

Is the required complete

solution

28

ifferential equation as ple of conservation of of time is partly kinetic by virtue of its velocity train energy in elastic

nge

Is given by [ ] 0UTdtd

=+

From this we get a differential equation of motion as well as natural frequency of the system Determine the natural frequency of spring mass system using energy method

Determine the natural frequency of system shown in figure Determine the natural frequency of the system shown in figure Is there any limitation on the value of K Discuss Determine the natural frequency of s

m

l k

m

a

θ

m

a

θ

l

k

29

k

k

ystem shown below Neglect the mass of ball

l

m

30

A string shown in figure is under tension T which can be assumed to remain constant for small displacements Find the natural frequency of vertical vibrations of spring An acrobat 120kg walks on a tight rope as shown in figure The frequency of vibration in the given position is vertical direction is 30 rads Find the tension in the rope A manometer has a uniform bore of cross section area A If the column of liquid of length L and Density ρ is set into motion as shown in figure Find the frequency of the resulting oscillation

m

l

a

T T

36m 8m

31

k

R

k

m

r

m

Find the expression for natural frequency of system shown in the figure Neglect the mass of the cantilever beam Study the special case i) k=Infinity ii) I = infinity Determine the expression for the natural frequency of a system shown in figure The two discs are keyed to a common shaft and have a combined mass moment of inertia about centre of oscillation O It is assumed that the card attached to mass m does not stretch and is always under tension Determine the natural frequency of a system shown

m

l

32

Determine the expression for the natural frequency of the system shown in figure Assume that the wires connecting the masses do not stretch and are always in tension

Determine the natural frequency of spring mass system taking the MASS OF SPRING (ms ) into account

M3 M1

M2

k2 k1

m

k

33

RAYLEIGHrsquoS METHOD

This is the extension of energy method Here we can determine the natural frequency of a conservative system without finding the equation of motion The natural frequency is a function of the rate of change of Kinetic energy and potential energy of the system From the principle of conservation of energy we have T+U = Constant T= KE and U =PE We can write the above equation as T1+U1 = --------------------------(1) Where 1 and 2 represents tow instances of time Let 1 be the time when the mass is passing through static equilibrium position Let 2 be the time corresponding to the mass displacement of the mass At this instant the velocity f the mass is zero and hence Substituting in equation (1) we get ( )Max = ( )Max Above equation leads directly to natural frequency Determine the natural frequency of spring mass system using Rayleighrsquos method m

34

r

B

K K

Home work

Determine the natural frequency of pendulum using Rayleighrsquos method

A Cylinder of mass m and moment of inertia Jo is to role without slipping but is sustained by spring K as shown in figure Determine the natural frequency of oscillation

Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping

a

Determine the natural frequency of system shown where in cylindrical disc rolls over the inextensible spring A cylinder of radius r rolls without slipping on cylindrical the equation of motion and hence determine the naoscillations about the lowest point shown Repeat the above problem ndash Instead of cylinder assume swithout slipping on concave surface as shown above

r

K

35

portion of radius R Derive tural frequency of small

phere of radius with r rolls

The torsional pendulum has a natural frequency of 5 Hz What length of steel wire of diameter 2 mm should be used for pendulum The inertia of mass fixed at the free end is 00098 kgm2 Take g= 083x1011 Nm2 Determine the torsional natural frequency of the system shown in figure Neglect the mass moment of inertia of the shaft ( Figure not proportionate) Determine the natural frequency of simple pendulum considering the mass of the rod into account

Using the energy method find tThe chord may be assumed ine

Kt1

Kt2

20cm

06m 08m 04m

20cm 20cm 20cm

m

M

he natural frxtensible in

k

M

36

equency of the system shown in figure the spring mass pulley system

A rod of length l and depth H having a mass m rests on a semicylidrical surface os radius r It is tipped slightly and let go Find the frequency of oscillation

A thin bar of l having a mass m rests on a semicylidrical surface radius r It is tipped slightly and let go Find the frequency of oscillation

l

h

R

l

37

38

CHAPTER 3

DAMPED FREE VIBRATION OF SINGLE DEGREE FREEDOM SYSTEM LEARNING OBJECTIVES

] Introduction to damped free vibration

] Terminologies used in damped free vibration

] Three different types of damping

] Concept of critical damping and its importance

] Study of response of viscous damped system for cases of under damping

critical and over damping

] Concept of logarithmic decrement

] Problems related to damped free vibration problems

Viscously damped free vibration of single degree freedom sytem Consider a single degree freedom system with viscous damping as shown in figure For a viscous damped system the damping force is

where c is a constant of proportionality It is called damping coefficient or coefficient of viscous damping

Symbolically it is designated by a dashpot as shown in Fig 3

x xampx

39

At any instant of time let the mass be displaced by x from the equilibrium position Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

Okxxcxm =minusminus= ampampamp

Okxxcxm =++ ampampamp ----------------------------------------------------------(1)

or

0kxdtdx

Cdt

xdM

2

2

=++

Assuming the solution of equation as

The first term in the above equation m2Ct

eminus

is an exponentially decaying function of time The behavior of the terms in the parenthesis however depends on whether numerical value within the radicals is positive zero or negative

CASE 1 if 0mk

m2c

2

=minus

40

CASE 2 if 0mk

m2c

2

=gt

CASE 3 if 0mk

m2c

2

=lt

41

Critically damped coefficient Critically damped coefficient is defined as the value of the damping coeffient for which radical in equation

mk

m2c

m2c

S2

21 minus

plusmn

= becomes zero

Damping Ratio ( Damping factor) ( ξ ) It is defined as the ratio of damping coefficient to the critical damping coefficient Note The damping coefficient c is defined as the actual amount of damping while the damping coefficient is a constant depending upon the mass and stiffness of the system The roots of characteristics equation can be now written as Note if c=cc or ξ =1 then system is critically damped if cgt cc or ξ gt1 then system is over damped I if clt cc or ξ lt1 then system is under damped

42

Take home quiz Derive General solution of different damped system

1 Critically damped system 2 Under damped system 3 Over damped system

Logarithmic Decrement It is used to determine the amount of damping present in system This measure The rate of decay of free vibration The larger the damping the greater will be rate of decay Logarithmic decrement is defined as the natural logarithm of ratio of any two successive amplitudes

=δ

2

1

xx

ln

Expression for logarithmic decrement (δ ) Consider a damped free vibration which is expressed by

( ) ( ) φ+ξminusω= ξω t 1sinXex 2n

tn

x is the displacement at any instant of time X and φ are arbitrary constants

21

2

ξminus

πξ=δ

43

Tutorial problems 1 Derive an expression for damping ratio in terms of logarithmic decrement

2 Show that logarithmic decrement

=δ

n

o

xx

lnn1

where xo is the initial amplitude xn

is amplitude after n cycles 3 A mass of 1 kg is to be supported on a spring having stiffness K= 9800 Nm The damping constant c= 59 Nsm Determine the natural frequency of system Find also logarithmic decrement and amplitude after 3 cycles if initial displacement is 03 cm 4 A vibrating system is defined by the following parameters M = 3kg k= 100 Nm damping coefficient = 3 Nsm determine Damping factor natural frequency damped vibration logarithmic decrement ratio of consecutive amplitude no of cycles after which the original amplitude is reduced to 20 5A single degree damped vibrating system a suspended mass of 18kg makes 15 oscillations in 003 seconds The amplitude decreases to 025 of the initial value after 5 oscillations Determine the stiffness of spring Logarithmic decrement damping factor and damping coefficient 6 A damped vibration record of a spring mass system shows the following data Amplitude at the end of 2nd cycle is 9mm amplitude at eh end of 3rd cycle is 6mm amplitude at the end of 4th cycle is 4mm stiffness of spring =8000 Nm weight =20N Find the logarithmic decrement damping force at unit velocity periodic time of vibration 7 For the system shown in figure the characteristics of dashpot is such that the constant force of 49 N is applied to the piston it velocity is found to be constant at 012 ms Determine the value of C would you expect the complete system to be periodic or aperiodic 8 For a torsional system shown in the figuref1 is natural frequency in air and f2 is the natural frequency when immersed in oil Determine the damping coefficient

Viscous fluid

Disc of mass moment of Inertia J

2a

C

m M

b

a a

k

9 A thin plate of area A and weigh w is fixed to the end of the spring and allow to oscillate in a viscous fluid as shown in figure It f1 is the frequency of oscillation in air and f2 in liquid Show that

22

21 ff

gA2

minusπω

=micro where the damping force on the plate

AV2Fd micro= V being the velocity of vibration 10 A torsional pendulum has a natural frequency of 175 cyclessecond vibrating in vacuum The mass moment of inertia is 2 kgm2 It is immersed in oil and it is observed that the natural frequency is 142 cyclessecond Find the damping torque If the disc is replaced by 34 deg When in oil find the displacement at the end of the first complete cycle 11 An automobile can be modeled as a mass placed on 4 shock absorbers each consisting of a spring and a damper Such that each spring is equally loaded Determine the stiffness damping constant of each shock absorber So that natural frequency is 2Hz and system is critically damped The mass of vehicle is 200kg 12 Write the differential equation of motion for the system shown in the figure below Determine the natural frequency of the damped oscillation and critical damping coefficient

Determine the differential equation of motion and find the critical cdamped natural frequency for system shown in figure

a

C

m M

M =200kg

44

200kg kgkg

oefficient and

45

Derive the equation of motion and find the critical damping coefficient and damped natural frequency for the system shown below Write the differential equation of motion and determine the expression for critical damping coefficient and damped natural frequency of system shown in figure

Set up a differential equation of motion for system shown in figure and determine the following Undamped natural frequencies critical damping coefficient damping ratio damped natural frequency

b

C

kM

a

m

46

A mass of 300N is resting on two springs of stiffness 3000 Nm each and a dashpot of damping coefficient 150 Nsm as shown in figure If the initial velocity of 10cms is given to mass at its equilibrium position what will be the displacement from the equilibrium position at the end of first second Derive the formula used A gun barrel of mass 600kg has a recoil spring of stiffness 294 knm if the barrel recoils 13m on firing determine i) The initial recoil velocity of barrel ii) The critical damping coefficient of dashpot which is engaged at the end of recoil stroke Iii) Time required for the barrel to return at the end of recoil stroke Iv) time required for the barrel to return to a position 5cm from the initial position

W= 3000 N

FIRING

CHAPTER 4

FORCED VIBRATION LEARNING OBJECTIVES

] Introduction to forced free vibration

] Terminologies used in forced vibration

] Concept of magnification factor

] Study of reciprocating and rotating unbalance

] Concept of base excitation and support excitation

] Concept of Transmissibility ratio

] Problems related to forced vibration

The vibration that takes place under the excitation of external forces are called

forced vibration When a system is subjected to harmonic excitation it is forced to vibrate at the same frequency as that of excitation Most of the mechanical systems are prone to steady state forced vibration which cause fatigue failure The vibration problem also becomes very important when any of excitation frequencies coincide with one of natural frequencies of a system Under this condition resonance occurs during which large amplitude of vibrations are observed To avoid resonance external frequency (operating speed) may be changed or properties of the system may be changed to alter the natural frequency In some cases sufficient amount of damping may be provided to avoid large amplitude during resonance Thus the problem of force vibration is very important in mechanical design

FORCED VIBRATIONS OF SINGLE DOF SYSTEMS UNDER HARMONIC EXCITATION

` [ ] tsin FF o ω=

x

sin FF o=

Fo

xampamp

[ ] tω

xamp

47

48

Consider a spring mass system having a viscous damping excited by a harmonic force [ ] tsin FF o ω= as shown in figure The various forces acting on the system is shown in figure Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

O)tsin(Fkxxcxm o =ω+minusminus= ampampamp

)tsin(Fkxxcxm o ω=++ ampampamp ----------------------------------------------------------(1)

or

)tsin(Fkxdtdx

Cdt

xdM o2

2

ω=++

Equation (1) represents the differential equation of motion of the system The solution to this equation consists of two parts namely complementary function and a particular integral

The first of solution is also termed as transient solution of homogeneous solution and can be obtained from the homogeneous differential equation 0kxxcxm =++ ampampamp

The second part of solution particular integral is also termed as steady state solution ad is of form )tsin(Xx φminusω= Where X= steady state amplitude φ is phase difference between external force and displacement

)tsin(Xx φminusω=

The vector representation of the shown

49

The force polygon for the above vector diagram can be written as follows

( )φminusωt O

KX

xm 2ω

Fo

xCω

tω

φ

50

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

is the governing equation

Zero frequency deflection (Xst) It is deflection observed at zero frequency ( when the external frequency 0=ω

( ) kF

001

kF

x o

22

o

st =

+minus=

Magnification factor (M) It is defined as the ratio of steady state amplitude to zero frequency deflection xst

M = arX

X=

A plot of magnification factor versus frequency ratio

damping factor ( )ξ is called frequency plot of a harmon All the frequency terms starts from magnification factor of damping ratio ( )ξ The amplitude of vibration at any on the value of damping The amplitude will be higher for higher damping The amplitude of vibration at reso

amplitude tends to zero as

ωω

n tends to infinity

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

Freq

ζ =

ζ = 1

Magnification Factor M = XXst

kFo

51

ωω

nfor varying values of

ically excite vibrating system

M=1 what ever may be value particular frequency depends

for lesser damping and lesser nance will be very high The

uency Ratio r = (ωωn)

ζ = 0 0

ζ = 025 ζ = 0375 ζ = 0 5

ζ = 0707

ζ = 2

52

Plot of phase angle φ versus frequency ratio

ωω

n is shown in the

figure The phase angle φ is zero for all the values of damping ratio

at

ωω

n=0 The phase

angle φ is 90 for all the values of damping ratio

at

ωω

n=1 ie at

resonance The phase angle between o to 90 deg for all the values of

ωω

n= 0 to 1 and

between 90 deg to 180 deg for all the values of

ωω

ngt1

Class work Derive an expression for frequency at peak amplitude

2np 21 ξminusω=ω=ω is the required equation

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

53

Derive expression amplitude at resonance ( In dimensional form) In dimensional form We know that steady state amplitude

( ) ( )

ω+ωminus=

222

o

Cmk

Fx Equation in dimensional form

We know that at resonance mk

n =ω=ω

( )

ω+

minus

=2

n

2

oR

Cmk

mk

Fx

( )

ω=

2n

oR

C

Fx

( )

ω=

n

oR C

Fx

Class work Derive an expression for amplitude at resonance ( in non dimensional form)

Tutorial Problems on Forced Vibration 1A weight of 60N is suspended by a spring stiffness 12 kNm is forced to vibrate by a harmonic force of 10N assuming viscous damping of 0086 kN-sm Find i) resonance frequency ii) amplitude at resonance iii) phase angle at resonance iv) frequency corresponding to peak amplitude 2 A periodic torque having a maximum value of 05888Nm at a frequency of 4 rads is impressed upon a flywheel suspended from a wire The wheel has mass moment of inertia 012 kgm2 and the wire has stiffness 1176 Nmrad A viscous dashpot applies damping couple of 0784 Nm an angular velocity of 2 rads calculate i)maximum angular displacement from rest position ii) maximum couple applied to dashpot Iii) angle by which the angular displacement lags the torque A mass of 1019 kg is suspended from a spring stiffness 20 Nmm viscous damping causes the amplitude to decrease to 110th of initial value in 4 complete cycle If a periodic force of 20 cos(50 t) is applied to the mass Find the amplitude of forced vibration What would be the amplitude of oscillation if the applied force coincides with the natural period

m=1019 kg

[ ] t50cos 30F =

54

55

ROTATING UNBALANCE Unbalance in rotating machinery is a common source of vibration excitation Consider a machine of mass m having mounted on a foundation of stiffness k and damping coefficient C as shown in figure The unbalance is represented by an eccentric rotating mass m with eccentricity (e) and having a angular velocity (ω) let x be the displacement of non rotating mass from the static equilibrium position The displacement of rotating mass m from the static equilibrium is ( ) tsin( ex ω+ Applying Newtonrsquos law of motion we have

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

56

RECIPROCATING UNBALANCE Unbalance in reciprocating machinery is a common source of vibration excitation Consider a machine of mass M having mounted on a foundation of stiffness K and damping coefficient C as shown in figure The unbalance is represented by a reciprocating mass (m) having crank radus (e) and connecting rod of length (l) Let ωωωω be angular velocity of the crank Let x be the displacement of non reciprocating mass from the static equilibrium position at any instant of time The displacement of reciprocating mass m from the static equilibrium position is

)2sin(sin +++ tle

tex ωωωωωωωω

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

57

Note The equations obtained in reciprocating unbalance is same as the rotating unbalance and having the following differences Rotating Unbalance Reciprocating Unbalance Here mass of the rotating element is considered

Here mass of reciprocating element is considered

Here the eccentricity (e) which is the distance between the centre of rotation and CG

Here the eccentricity e is the crank radius and crank radius = stroke length2

The frequency plot of non dimensional factor( ( )me

MX versus frequency ratio

ωω

n for various values of ( )ξξξξ is shown in figure This frequency plot is same for

the rotating and reciprocating unbalance with the following points are noted from the diagram

1 Damping ratio plays very important role during resonance in controlling the amplitude

2 When

ωω

n=0 then ( )me

MX =0 for all the values of damping ratio ( )ξξξξ

3 When

ωω

nis higher than ( )me

MX tends to unity

4 At very high speeds ie at very high frequency ratio

ωω

n there is no effect

of damping The frequency plot of phase angle versus the frequency ratio is given as shown in figure The explanation is same as in forced harmonic vibration

58

VIBRATION ISOLATION

Vibratory forces and motions generated by machine and other caused are unavoidable Howeever their effect on the other dynamical systems can be minimized by proper isolator design An isolation system is either to protect o delicate machine from excessive vibration transmitted to it from its supporting structure or to prevent vibratory forces and motions generated by machines from being transmitted to its surroundings The effectiveness of isolation may be measured in terms of forces and motion transmitted The first type is known as force isolation and the second type is known as motion isolation

Force isolation (Force Transmissibility) When vibratory forces generated by machine are isolated the effectiveness of forced isolation can be made by defining the term transmissibility The term transmissibility is defined as the ratio of force transmitted to the foundation to that of the external impressed force

Force Transmissibility = o

tr

FF

force Impressed Externalfoundation toted transmitForce

=

Consider viscous damped spring mass system as shown in figure Upon this system an external harmonic force ) t (Sin Fo ω is applied The difference forces acting on system are inertia forces damping force spring force and external force All these forces are represented on force polygon shown below Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these two forces

( ) ( )22 xckx ω+=

59

This is the expression for force transmissibility in dimensional form To represent same in the non dimensional form divide numerator and denominator by k

2

n

22

n

2

n

o

tr

21

21

FF

TR

ωω

ξ+

ωω

minus

ωω

ξ+

==

PHASE DIFFERNECE BETWEEN EXTERNAL FORCE AND DISPLACEMENT Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these

ωω

minus

ωω

ξ=

ωminusω

=φ minusminus2

n

n1

2

1

1

2tan

mKC

tan

The phase difference between force transmitted to the foundation Ftr to the displacement

kc

KxxC

tanω

=ω

=α or

ωω

ξ=

ω=α minusminus

n

11 2tanKx

xCtan

The phase difference between external force Fo and force transmitted to foundationis β

60

β= αminusφ =

ωω

minus

ωω

ξminus

2

n

n1

1

2tan -

ωω

ξminus

n

1 2tan

Displacement Transmissibility It is the ratio of amplitude of steady state vibration of machine to that of the base motion Forced vibration due to support motion Consider a spring mass damper system subjected to harmonic excitation )tsin(Yy ω= at the base as shown in figure Let x be the absolute displacement of the mass at any instant of time t The different forces acting on system are shown in free body diagram Applying Newtonrsquos Law of motion Ma = sumF

)yx(k)yx(cxm minusminusminusminus= ampampampamp

0)yx(k)yx(cxm =minus+minus+ ampampampamp

61

Frequency plot of transmissibility versus frequency ratio

62

The plot of transmissibility versus frequency ratio is as shown in figure The following points can be noted from it

i) The value of transmissibility is unity at frequency ratio

ωω

n=0 and 2

ii) For an undamped system ( )ξ =0 transmissibility tends to infin at resonance iii) The value of transmissibility is less than unity for the value of frequency ratio

greater than 2

iv) For frequency ratio

ωω

nlt 2 smaller damping ratio lead to larger values of

transmissibility on the other hand for frequency ratio

ωω

ngt 2 smaller damping

ratio leads to smaller value of transmissibility

The frequency plot of transmissibility (TR) versus frequency ratio

ωω

n can be

broadly divided into two zones

1

ωω

n= 0 to 2

2

ωω

n= 2 to infin

CASE 1

When

ωω

n= 0 to 2

In this zone the value of transmissibility is always greater than 1 for all values of damping ratio In this zone the transmissibility decreases with increase in damping Hence we can conclude that in this zone damping is useful in isolating vibration forces being transmitted to the support CASE 2

When

ωω

n= 2 to infin

In this zone the value of transmissibility is less than 1 for all values of damping ratio In this zone the transmissibility increases with increase in damping Hence we can conclude that in this zone damping will adversely affect in isolating vibratory forces

63

Tutorial problems on Forced Vibration 1 Machine of mass 75kg is mounted on spring of stiffness 12000 Nm with an assumed damping factor ξ =02 piston within the machine of mass 2 kg has a reciprocating motion with a stroke of 75 cm and speed of 15 cyclessecond Assuming the motion of piston to be harmonic determine amplitude of machine ( X) Transmissibility and force transmitted to foundation phase angle of transmitted force wrt to exciting force 2 Motor weighs 600N is mounted on a simple beam that has a spring rate of 400 Ncm The motor armature base 160 N and has an eccentricity of 0005cm What will be the amplitude of motion when it runs at 1760 RPM Neglect damping the deflection of armature shaft and weight of the beam 3 A Heavy machine of mass 300 kg is supported on resilient foundation the static deflection of foundation due to weight of machine is found to be 75mm It is observed that the machine vibrates with an amplitude of 10mm when the base of the foundation is subjected to harmonic oscillation at the undampeds natural frequency of system with an amplitude of 25mm Find the damping constant of foundation Ii) dynamic force amplitude on the base iii) amplitude of the displacement of machine relative to base 4 An air compressor of mass 450 kg operates at a constant speed of 1750 RPM The reciprocating masses are 10 kg the crank radius is 100mm If the damper for mounting introduces a damping factor 015 i) specify the springs of mounting such that only 20 of the unbalanced force is transmitted to the foundation ii) determine the magnitude of the transmitted force bull Repeat the above problem by assuming that there is no damping in system

5 A body of a car having mass 1500 kg and is mounted on 4 equal springs which deflect through 225 cm under the weight of the body A total damping forces the 4 shock absorbers is 46 N at a velocity of 1cmsec The car is placed with all the four wheels on a test platform which is move up and down at resonant speed With an amplitude of 25cm Find the amplitude of the car body on its springs assuming its CG to be in the centre of wheel base 6A 75 kg machine which is mounted on springs of stiffness 116x105 Nm with the damping of 20=ξ A 2kg piston within the machine as reciprocating motion with a stroke of 008m and speed of 3000 RPM Assuming the motion of piston to be harmonic determine the amplitude of vibration of machine

7 The spring of an automobile are compressed 10cm under its own weight Find the critical speed when a trailer is traveling over road with a people approximaterly by a sine wave of amplitude 75 cm and wave length 15m What will be the amplitude of vibration at 64kmhour 8 An eccentric mass excites us used to determine vibratory characteristics of a structure of mass 25kg as given in figure At a speed of 1000 RPM the eccentric mass to beat top position at the instant the structure was moved upwards through its static equilibrium position and corresponding amplitude was 20mm The eccentric mass is 05kg at an eccentricity of 10mm Determine i) Undamped natural frequency of the system ii) Damping ratio iii) Steady state amplitude and phase angle at 1500 RPM iv) Force transmitted to the ground at 1200 RPM

M=25kg

C

K

Eccentric mass

64

65

CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Support motion Consider a spring mass damper system as shown in figure Let it excited by motion

of support let y be the harmonic displacement of support From Newtonrsquos second law of motion

)yx(k)yx(cxm minusminusminusminus= ampampampamp

)1(0)yx(k)yx(cxm minusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminus=minus+minus+ ampampampamp

Let z be the relative displacement of the mass with respect to support then

Z=x-y

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

Let at any instant of time t the mass is displaced from the equilibrium position x the different forces acting on the system are shown in figure (d) From Newtonrsquos second law of motion

sum = maF

Inertia force ( disturbing force) = restoring force

mgxkxm ++∆minus= )(ampamp

0)( =+ xkxm ampamp

or 0)( =+ xmk

xampamp

equation 2 is the differential equation of motion for spring mfigure Comparing equation (2) with the equation of SHM +xampampsince the vibrations of the above system are free( without the forces) we can write

sec radmk

n =ω

24

ass system shown in 0)(2 =xω

resistance of external

time period km

2 f1

n

πτ ==

from the equation(1) mg = k∆

∆=

q

mk

Difference between the translation ( rectilinear) and rotational system of vibration

Translatory Rotational

In the analysis thFORCES are co Linear stiffness K Problems 1A mass of 10kFind the natural 2 A spring massnatural frequencnatural frequenc

kt

m

sec radmk

n =ω

k

25

e disturbing and restoring nsidered

In the analysis In the analysis MASS Moment of Inertia (J) is considered

in Nm is considered

g when suspended from a spring causes a static deflection of 1cm frequency of system

system has a spring stiffness K Nm and a mass of m Kg It has a y of vibration 12 Hz An extra 2kg mass coupled to it then the y reduces by 2 Hz find K and m

secrad n =ω

26

3 A steel wire of 2mm diameter and 30m long It is fixed at the upper end and carries a mass of m kg at its free end Find m so that the frequency of longitudinal vibration is 4 Hz 4 A spring mass system has a natural period of 02 seconds What will be the new period if the spring constant is 1) increased by 50 2) decreased by 50 5 A spring mass system has a natural frequency of 10 Hz when the spring constant is reduced by 800 Nm the frequency is altered by 45 Find the mass and spring constant of the original system 6 Determine the natural frequency of system shown in fig is which shaft is supported in SHORT bearings 7 Determine the natural frequency of system shown in fig is which shaft is supported in LONG bearings where l is the length of bearing and E ndash youngrsquos modulus and I is moment of Inertia 7 Determine the natural frequency of system shown in fig is which shaft is supported in LONG bearings 8 A light cantilever beam of rectangular section( 5 cm deep x 25cm wide) has a mass fixed at its free end Find the ratio of frequency of free lateral vibration in vertical plane to that in horizontal 9 Determine the natural frequency of simple pendulum 10 Homogeneous square plate of size l and mass m is suspended from the mid point of one of its sides as shown in figure Find the frequency of vibration

l

l

27

11 A compound pendulum which is rigid body of mass m and it is pivoted at O The point of pivot is at distance d from the centre of gravity It is free to rotate about its axis Find the frequency of oscillation of such pendulum 12 A connecting rod shown in fig is supported at the wrist pin end It is displaced and allowed to oscillate The mass of rod is 5kg and centre of gravity is 20 cm from the pivot point O If the frequency of oscillation is 40 cyclesminute calculate the mass moment of inertia about its CG

13 A semi circular homogenous disc of radius r and mass m is pivoted freely about its centre as shown in figure Determine the natural frequency of oscillation 14A simply supported beam of square cross section 5mmx5mm and length 1m carrying a mass of 0575 kg at the middle is found to have natural frequency of 30 radsec Determine youngrsquos modulus of elasticity of beam 15 A spring mass system k1 and m have a natural frequency f1 Determine the value of k2 of another spring in terms of k1 which when placed in series with k1 lowers the natural frequency to 23 f1

COMPLETE SOLUTION OF SYSTEM EXECUTING SHM

The equation of motion of system executing SHM can be represented by

0)( =+ xkxm ampamp ----------------------------------------(1)

022

2=+

dtdx

dt

dx ω

The general solution of equation (1) can be expressed as

------------------------(2)

Where A and B are arbitrary constant which can be determined from the initial conditions of the system Two initial conditions are to be specified to evaluate these

constants x=x0 at t=0 and xamp = Vo at t=0 substituting in the equation (2) Energy method In a conservative system the total energy is constant The dwell as natural frequency can be determined by the princienergy For free vibration of undamped system at any instant and partly potential The kinetic energy T is stored in the masswhere as the potential energy U is stored in the form of sdeformation or work done in a force field such as gravity The total energy being constant T+U = constant Its rate of cha

t)(sin Bt)cos( AX ω+ω=

t)(sin V

t)cos( xx o0 ω

ω+ω=

Is the required complete

solution

28

ifferential equation as ple of conservation of of time is partly kinetic by virtue of its velocity train energy in elastic

nge

Is given by [ ] 0UTdtd

=+

From this we get a differential equation of motion as well as natural frequency of the system Determine the natural frequency of spring mass system using energy method

Determine the natural frequency of system shown in figure Determine the natural frequency of the system shown in figure Is there any limitation on the value of K Discuss Determine the natural frequency of s

m

l k

m

a

θ

m

a

θ

l

k

29

k

k

ystem shown below Neglect the mass of ball

l

m

30

A string shown in figure is under tension T which can be assumed to remain constant for small displacements Find the natural frequency of vertical vibrations of spring An acrobat 120kg walks on a tight rope as shown in figure The frequency of vibration in the given position is vertical direction is 30 rads Find the tension in the rope A manometer has a uniform bore of cross section area A If the column of liquid of length L and Density ρ is set into motion as shown in figure Find the frequency of the resulting oscillation

m

l

a

T T

36m 8m

31

k

R

k

m

r

m

Find the expression for natural frequency of system shown in the figure Neglect the mass of the cantilever beam Study the special case i) k=Infinity ii) I = infinity Determine the expression for the natural frequency of a system shown in figure The two discs are keyed to a common shaft and have a combined mass moment of inertia about centre of oscillation O It is assumed that the card attached to mass m does not stretch and is always under tension Determine the natural frequency of a system shown

m

l

32

Determine the expression for the natural frequency of the system shown in figure Assume that the wires connecting the masses do not stretch and are always in tension

Determine the natural frequency of spring mass system taking the MASS OF SPRING (ms ) into account

M3 M1

M2

k2 k1

m

k

33

RAYLEIGHrsquoS METHOD

This is the extension of energy method Here we can determine the natural frequency of a conservative system without finding the equation of motion The natural frequency is a function of the rate of change of Kinetic energy and potential energy of the system From the principle of conservation of energy we have T+U = Constant T= KE and U =PE We can write the above equation as T1+U1 = --------------------------(1) Where 1 and 2 represents tow instances of time Let 1 be the time when the mass is passing through static equilibrium position Let 2 be the time corresponding to the mass displacement of the mass At this instant the velocity f the mass is zero and hence Substituting in equation (1) we get ( )Max = ( )Max Above equation leads directly to natural frequency Determine the natural frequency of spring mass system using Rayleighrsquos method m

34

r

B

K K

Home work

Determine the natural frequency of pendulum using Rayleighrsquos method

A Cylinder of mass m and moment of inertia Jo is to role without slipping but is sustained by spring K as shown in figure Determine the natural frequency of oscillation

Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping

a

Determine the natural frequency of system shown where in cylindrical disc rolls over the inextensible spring A cylinder of radius r rolls without slipping on cylindrical the equation of motion and hence determine the naoscillations about the lowest point shown Repeat the above problem ndash Instead of cylinder assume swithout slipping on concave surface as shown above

r

K

35

portion of radius R Derive tural frequency of small

phere of radius with r rolls

The torsional pendulum has a natural frequency of 5 Hz What length of steel wire of diameter 2 mm should be used for pendulum The inertia of mass fixed at the free end is 00098 kgm2 Take g= 083x1011 Nm2 Determine the torsional natural frequency of the system shown in figure Neglect the mass moment of inertia of the shaft ( Figure not proportionate) Determine the natural frequency of simple pendulum considering the mass of the rod into account

Using the energy method find tThe chord may be assumed ine

Kt1

Kt2

20cm

06m 08m 04m

20cm 20cm 20cm

m

M

he natural frxtensible in

k

M

36

equency of the system shown in figure the spring mass pulley system

A rod of length l and depth H having a mass m rests on a semicylidrical surface os radius r It is tipped slightly and let go Find the frequency of oscillation

A thin bar of l having a mass m rests on a semicylidrical surface radius r It is tipped slightly and let go Find the frequency of oscillation

l

h

R

l

37

38

CHAPTER 3

DAMPED FREE VIBRATION OF SINGLE DEGREE FREEDOM SYSTEM LEARNING OBJECTIVES

] Introduction to damped free vibration

] Terminologies used in damped free vibration

] Three different types of damping

] Concept of critical damping and its importance

] Study of response of viscous damped system for cases of under damping

critical and over damping

] Concept of logarithmic decrement

] Problems related to damped free vibration problems

Viscously damped free vibration of single degree freedom sytem Consider a single degree freedom system with viscous damping as shown in figure For a viscous damped system the damping force is

where c is a constant of proportionality It is called damping coefficient or coefficient of viscous damping

Symbolically it is designated by a dashpot as shown in Fig 3

x xampx

39

At any instant of time let the mass be displaced by x from the equilibrium position Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

Okxxcxm =minusminus= ampampamp

Okxxcxm =++ ampampamp ----------------------------------------------------------(1)

or

0kxdtdx

Cdt

xdM

2

2

=++

Assuming the solution of equation as

The first term in the above equation m2Ct

eminus

is an exponentially decaying function of time The behavior of the terms in the parenthesis however depends on whether numerical value within the radicals is positive zero or negative

CASE 1 if 0mk

m2c

2

=minus

40

CASE 2 if 0mk

m2c

2

=gt

CASE 3 if 0mk

m2c

2

=lt

41

Critically damped coefficient Critically damped coefficient is defined as the value of the damping coeffient for which radical in equation

mk

m2c

m2c

S2

21 minus

plusmn

= becomes zero

Damping Ratio ( Damping factor) ( ξ ) It is defined as the ratio of damping coefficient to the critical damping coefficient Note The damping coefficient c is defined as the actual amount of damping while the damping coefficient is a constant depending upon the mass and stiffness of the system The roots of characteristics equation can be now written as Note if c=cc or ξ =1 then system is critically damped if cgt cc or ξ gt1 then system is over damped I if clt cc or ξ lt1 then system is under damped

42

Take home quiz Derive General solution of different damped system

1 Critically damped system 2 Under damped system 3 Over damped system

Logarithmic Decrement It is used to determine the amount of damping present in system This measure The rate of decay of free vibration The larger the damping the greater will be rate of decay Logarithmic decrement is defined as the natural logarithm of ratio of any two successive amplitudes

=δ

2

1

xx

ln

Expression for logarithmic decrement (δ ) Consider a damped free vibration which is expressed by

( ) ( ) φ+ξminusω= ξω t 1sinXex 2n

tn

x is the displacement at any instant of time X and φ are arbitrary constants

21

2

ξminus

πξ=δ

43

Tutorial problems 1 Derive an expression for damping ratio in terms of logarithmic decrement

2 Show that logarithmic decrement

=δ

n

o

xx

lnn1

where xo is the initial amplitude xn

is amplitude after n cycles 3 A mass of 1 kg is to be supported on a spring having stiffness K= 9800 Nm The damping constant c= 59 Nsm Determine the natural frequency of system Find also logarithmic decrement and amplitude after 3 cycles if initial displacement is 03 cm 4 A vibrating system is defined by the following parameters M = 3kg k= 100 Nm damping coefficient = 3 Nsm determine Damping factor natural frequency damped vibration logarithmic decrement ratio of consecutive amplitude no of cycles after which the original amplitude is reduced to 20 5A single degree damped vibrating system a suspended mass of 18kg makes 15 oscillations in 003 seconds The amplitude decreases to 025 of the initial value after 5 oscillations Determine the stiffness of spring Logarithmic decrement damping factor and damping coefficient 6 A damped vibration record of a spring mass system shows the following data Amplitude at the end of 2nd cycle is 9mm amplitude at eh end of 3rd cycle is 6mm amplitude at the end of 4th cycle is 4mm stiffness of spring =8000 Nm weight =20N Find the logarithmic decrement damping force at unit velocity periodic time of vibration 7 For the system shown in figure the characteristics of dashpot is such that the constant force of 49 N is applied to the piston it velocity is found to be constant at 012 ms Determine the value of C would you expect the complete system to be periodic or aperiodic 8 For a torsional system shown in the figuref1 is natural frequency in air and f2 is the natural frequency when immersed in oil Determine the damping coefficient

Viscous fluid

Disc of mass moment of Inertia J

2a

C

m M

b

a a

k

9 A thin plate of area A and weigh w is fixed to the end of the spring and allow to oscillate in a viscous fluid as shown in figure It f1 is the frequency of oscillation in air and f2 in liquid Show that

22

21 ff

gA2

minusπω

=micro where the damping force on the plate

AV2Fd micro= V being the velocity of vibration 10 A torsional pendulum has a natural frequency of 175 cyclessecond vibrating in vacuum The mass moment of inertia is 2 kgm2 It is immersed in oil and it is observed that the natural frequency is 142 cyclessecond Find the damping torque If the disc is replaced by 34 deg When in oil find the displacement at the end of the first complete cycle 11 An automobile can be modeled as a mass placed on 4 shock absorbers each consisting of a spring and a damper Such that each spring is equally loaded Determine the stiffness damping constant of each shock absorber So that natural frequency is 2Hz and system is critically damped The mass of vehicle is 200kg 12 Write the differential equation of motion for the system shown in the figure below Determine the natural frequency of the damped oscillation and critical damping coefficient

Determine the differential equation of motion and find the critical cdamped natural frequency for system shown in figure

a

C

m M

M =200kg

44

200kg kgkg

oefficient and

45

Derive the equation of motion and find the critical damping coefficient and damped natural frequency for the system shown below Write the differential equation of motion and determine the expression for critical damping coefficient and damped natural frequency of system shown in figure

Set up a differential equation of motion for system shown in figure and determine the following Undamped natural frequencies critical damping coefficient damping ratio damped natural frequency

b

C

kM

a

m

46

A mass of 300N is resting on two springs of stiffness 3000 Nm each and a dashpot of damping coefficient 150 Nsm as shown in figure If the initial velocity of 10cms is given to mass at its equilibrium position what will be the displacement from the equilibrium position at the end of first second Derive the formula used A gun barrel of mass 600kg has a recoil spring of stiffness 294 knm if the barrel recoils 13m on firing determine i) The initial recoil velocity of barrel ii) The critical damping coefficient of dashpot which is engaged at the end of recoil stroke Iii) Time required for the barrel to return at the end of recoil stroke Iv) time required for the barrel to return to a position 5cm from the initial position

W= 3000 N

FIRING

CHAPTER 4

FORCED VIBRATION LEARNING OBJECTIVES

] Introduction to forced free vibration

] Terminologies used in forced vibration

] Concept of magnification factor

] Study of reciprocating and rotating unbalance

] Concept of base excitation and support excitation

] Concept of Transmissibility ratio

] Problems related to forced vibration

The vibration that takes place under the excitation of external forces are called

forced vibration When a system is subjected to harmonic excitation it is forced to vibrate at the same frequency as that of excitation Most of the mechanical systems are prone to steady state forced vibration which cause fatigue failure The vibration problem also becomes very important when any of excitation frequencies coincide with one of natural frequencies of a system Under this condition resonance occurs during which large amplitude of vibrations are observed To avoid resonance external frequency (operating speed) may be changed or properties of the system may be changed to alter the natural frequency In some cases sufficient amount of damping may be provided to avoid large amplitude during resonance Thus the problem of force vibration is very important in mechanical design

FORCED VIBRATIONS OF SINGLE DOF SYSTEMS UNDER HARMONIC EXCITATION

` [ ] tsin FF o ω=

x

sin FF o=

Fo

xampamp

[ ] tω

xamp

47

48

Consider a spring mass system having a viscous damping excited by a harmonic force [ ] tsin FF o ω= as shown in figure The various forces acting on the system is shown in figure Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

O)tsin(Fkxxcxm o =ω+minusminus= ampampamp

)tsin(Fkxxcxm o ω=++ ampampamp ----------------------------------------------------------(1)

or

)tsin(Fkxdtdx

Cdt

xdM o2

2

ω=++

Equation (1) represents the differential equation of motion of the system The solution to this equation consists of two parts namely complementary function and a particular integral

The first of solution is also termed as transient solution of homogeneous solution and can be obtained from the homogeneous differential equation 0kxxcxm =++ ampampamp

The second part of solution particular integral is also termed as steady state solution ad is of form )tsin(Xx φminusω= Where X= steady state amplitude φ is phase difference between external force and displacement

)tsin(Xx φminusω=

The vector representation of the shown

49

The force polygon for the above vector diagram can be written as follows

( )φminusωt O

KX

xm 2ω

Fo

xCω

tω

φ

50

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

is the governing equation

Zero frequency deflection (Xst) It is deflection observed at zero frequency ( when the external frequency 0=ω

( ) kF

001

kF

x o

22

o

st =

+minus=

Magnification factor (M) It is defined as the ratio of steady state amplitude to zero frequency deflection xst

M = arX

X=

A plot of magnification factor versus frequency ratio

damping factor ( )ξ is called frequency plot of a harmon All the frequency terms starts from magnification factor of damping ratio ( )ξ The amplitude of vibration at any on the value of damping The amplitude will be higher for higher damping The amplitude of vibration at reso

amplitude tends to zero as

ωω

n tends to infinity

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

Freq

ζ =

ζ = 1

Magnification Factor M = XXst

kFo

51

ωω

nfor varying values of

ically excite vibrating system

M=1 what ever may be value particular frequency depends

for lesser damping and lesser nance will be very high The

uency Ratio r = (ωωn)

ζ = 0 0

ζ = 025 ζ = 0375 ζ = 0 5

ζ = 0707

ζ = 2

52

Plot of phase angle φ versus frequency ratio

ωω

n is shown in the

figure The phase angle φ is zero for all the values of damping ratio

at

ωω

n=0 The phase

angle φ is 90 for all the values of damping ratio

at

ωω

n=1 ie at

resonance The phase angle between o to 90 deg for all the values of

ωω

n= 0 to 1 and

between 90 deg to 180 deg for all the values of

ωω

ngt1

Class work Derive an expression for frequency at peak amplitude

2np 21 ξminusω=ω=ω is the required equation

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

53

Derive expression amplitude at resonance ( In dimensional form) In dimensional form We know that steady state amplitude

( ) ( )

ω+ωminus=

222

o

Cmk

Fx Equation in dimensional form

We know that at resonance mk

n =ω=ω

( )

ω+

minus

=2

n

2

oR

Cmk

mk

Fx

( )

ω=

2n

oR

C

Fx

( )

ω=

n

oR C

Fx

Class work Derive an expression for amplitude at resonance ( in non dimensional form)

Tutorial Problems on Forced Vibration 1A weight of 60N is suspended by a spring stiffness 12 kNm is forced to vibrate by a harmonic force of 10N assuming viscous damping of 0086 kN-sm Find i) resonance frequency ii) amplitude at resonance iii) phase angle at resonance iv) frequency corresponding to peak amplitude 2 A periodic torque having a maximum value of 05888Nm at a frequency of 4 rads is impressed upon a flywheel suspended from a wire The wheel has mass moment of inertia 012 kgm2 and the wire has stiffness 1176 Nmrad A viscous dashpot applies damping couple of 0784 Nm an angular velocity of 2 rads calculate i)maximum angular displacement from rest position ii) maximum couple applied to dashpot Iii) angle by which the angular displacement lags the torque A mass of 1019 kg is suspended from a spring stiffness 20 Nmm viscous damping causes the amplitude to decrease to 110th of initial value in 4 complete cycle If a periodic force of 20 cos(50 t) is applied to the mass Find the amplitude of forced vibration What would be the amplitude of oscillation if the applied force coincides with the natural period

m=1019 kg

[ ] t50cos 30F =

54

55

ROTATING UNBALANCE Unbalance in rotating machinery is a common source of vibration excitation Consider a machine of mass m having mounted on a foundation of stiffness k and damping coefficient C as shown in figure The unbalance is represented by an eccentric rotating mass m with eccentricity (e) and having a angular velocity (ω) let x be the displacement of non rotating mass from the static equilibrium position The displacement of rotating mass m from the static equilibrium is ( ) tsin( ex ω+ Applying Newtonrsquos law of motion we have

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

56

RECIPROCATING UNBALANCE Unbalance in reciprocating machinery is a common source of vibration excitation Consider a machine of mass M having mounted on a foundation of stiffness K and damping coefficient C as shown in figure The unbalance is represented by a reciprocating mass (m) having crank radus (e) and connecting rod of length (l) Let ωωωω be angular velocity of the crank Let x be the displacement of non reciprocating mass from the static equilibrium position at any instant of time The displacement of reciprocating mass m from the static equilibrium position is

)2sin(sin +++ tle

tex ωωωωωωωω

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

57

Note The equations obtained in reciprocating unbalance is same as the rotating unbalance and having the following differences Rotating Unbalance Reciprocating Unbalance Here mass of the rotating element is considered

Here mass of reciprocating element is considered

Here the eccentricity (e) which is the distance between the centre of rotation and CG

Here the eccentricity e is the crank radius and crank radius = stroke length2

The frequency plot of non dimensional factor( ( )me

MX versus frequency ratio

ωω

n for various values of ( )ξξξξ is shown in figure This frequency plot is same for

the rotating and reciprocating unbalance with the following points are noted from the diagram

1 Damping ratio plays very important role during resonance in controlling the amplitude

2 When

ωω

n=0 then ( )me

MX =0 for all the values of damping ratio ( )ξξξξ

3 When

ωω

nis higher than ( )me

MX tends to unity

4 At very high speeds ie at very high frequency ratio

ωω

n there is no effect

of damping The frequency plot of phase angle versus the frequency ratio is given as shown in figure The explanation is same as in forced harmonic vibration

58

VIBRATION ISOLATION

Vibratory forces and motions generated by machine and other caused are unavoidable Howeever their effect on the other dynamical systems can be minimized by proper isolator design An isolation system is either to protect o delicate machine from excessive vibration transmitted to it from its supporting structure or to prevent vibratory forces and motions generated by machines from being transmitted to its surroundings The effectiveness of isolation may be measured in terms of forces and motion transmitted The first type is known as force isolation and the second type is known as motion isolation

Force isolation (Force Transmissibility) When vibratory forces generated by machine are isolated the effectiveness of forced isolation can be made by defining the term transmissibility The term transmissibility is defined as the ratio of force transmitted to the foundation to that of the external impressed force

Force Transmissibility = o

tr

FF

force Impressed Externalfoundation toted transmitForce

=

Consider viscous damped spring mass system as shown in figure Upon this system an external harmonic force ) t (Sin Fo ω is applied The difference forces acting on system are inertia forces damping force spring force and external force All these forces are represented on force polygon shown below Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these two forces

( ) ( )22 xckx ω+=

59

This is the expression for force transmissibility in dimensional form To represent same in the non dimensional form divide numerator and denominator by k

2

n

22

n

2

n

o

tr

21

21

FF

TR

ωω

ξ+

ωω

minus

ωω

ξ+

==

PHASE DIFFERNECE BETWEEN EXTERNAL FORCE AND DISPLACEMENT Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these

ωω

minus

ωω

ξ=

ωminusω

=φ minusminus2

n

n1

2

1

1

2tan

mKC

tan

The phase difference between force transmitted to the foundation Ftr to the displacement

kc

KxxC

tanω

=ω

=α or

ωω

ξ=

ω=α minusminus

n

11 2tanKx

xCtan

The phase difference between external force Fo and force transmitted to foundationis β

60

β= αminusφ =

ωω

minus

ωω

ξminus

2

n

n1

1

2tan -

ωω

ξminus

n

1 2tan

Displacement Transmissibility It is the ratio of amplitude of steady state vibration of machine to that of the base motion Forced vibration due to support motion Consider a spring mass damper system subjected to harmonic excitation )tsin(Yy ω= at the base as shown in figure Let x be the absolute displacement of the mass at any instant of time t The different forces acting on system are shown in free body diagram Applying Newtonrsquos Law of motion Ma = sumF

)yx(k)yx(cxm minusminusminusminus= ampampampamp

0)yx(k)yx(cxm =minus+minus+ ampampampamp

61

Frequency plot of transmissibility versus frequency ratio

62

The plot of transmissibility versus frequency ratio is as shown in figure The following points can be noted from it

i) The value of transmissibility is unity at frequency ratio

ωω

n=0 and 2

ii) For an undamped system ( )ξ =0 transmissibility tends to infin at resonance iii) The value of transmissibility is less than unity for the value of frequency ratio

greater than 2

iv) For frequency ratio

ωω

nlt 2 smaller damping ratio lead to larger values of

transmissibility on the other hand for frequency ratio

ωω

ngt 2 smaller damping

ratio leads to smaller value of transmissibility

The frequency plot of transmissibility (TR) versus frequency ratio

ωω

n can be

broadly divided into two zones

1

ωω

n= 0 to 2

2

ωω

n= 2 to infin

CASE 1

When

ωω

n= 0 to 2

In this zone the value of transmissibility is always greater than 1 for all values of damping ratio In this zone the transmissibility decreases with increase in damping Hence we can conclude that in this zone damping is useful in isolating vibration forces being transmitted to the support CASE 2

When

ωω

n= 2 to infin

In this zone the value of transmissibility is less than 1 for all values of damping ratio In this zone the transmissibility increases with increase in damping Hence we can conclude that in this zone damping will adversely affect in isolating vibratory forces

63

Tutorial problems on Forced Vibration 1 Machine of mass 75kg is mounted on spring of stiffness 12000 Nm with an assumed damping factor ξ =02 piston within the machine of mass 2 kg has a reciprocating motion with a stroke of 75 cm and speed of 15 cyclessecond Assuming the motion of piston to be harmonic determine amplitude of machine ( X) Transmissibility and force transmitted to foundation phase angle of transmitted force wrt to exciting force 2 Motor weighs 600N is mounted on a simple beam that has a spring rate of 400 Ncm The motor armature base 160 N and has an eccentricity of 0005cm What will be the amplitude of motion when it runs at 1760 RPM Neglect damping the deflection of armature shaft and weight of the beam 3 A Heavy machine of mass 300 kg is supported on resilient foundation the static deflection of foundation due to weight of machine is found to be 75mm It is observed that the machine vibrates with an amplitude of 10mm when the base of the foundation is subjected to harmonic oscillation at the undampeds natural frequency of system with an amplitude of 25mm Find the damping constant of foundation Ii) dynamic force amplitude on the base iii) amplitude of the displacement of machine relative to base 4 An air compressor of mass 450 kg operates at a constant speed of 1750 RPM The reciprocating masses are 10 kg the crank radius is 100mm If the damper for mounting introduces a damping factor 015 i) specify the springs of mounting such that only 20 of the unbalanced force is transmitted to the foundation ii) determine the magnitude of the transmitted force bull Repeat the above problem by assuming that there is no damping in system

5 A body of a car having mass 1500 kg and is mounted on 4 equal springs which deflect through 225 cm under the weight of the body A total damping forces the 4 shock absorbers is 46 N at a velocity of 1cmsec The car is placed with all the four wheels on a test platform which is move up and down at resonant speed With an amplitude of 25cm Find the amplitude of the car body on its springs assuming its CG to be in the centre of wheel base 6A 75 kg machine which is mounted on springs of stiffness 116x105 Nm with the damping of 20=ξ A 2kg piston within the machine as reciprocating motion with a stroke of 008m and speed of 3000 RPM Assuming the motion of piston to be harmonic determine the amplitude of vibration of machine

7 The spring of an automobile are compressed 10cm under its own weight Find the critical speed when a trailer is traveling over road with a people approximaterly by a sine wave of amplitude 75 cm and wave length 15m What will be the amplitude of vibration at 64kmhour 8 An eccentric mass excites us used to determine vibratory characteristics of a structure of mass 25kg as given in figure At a speed of 1000 RPM the eccentric mass to beat top position at the instant the structure was moved upwards through its static equilibrium position and corresponding amplitude was 20mm The eccentric mass is 05kg at an eccentricity of 10mm Determine i) Undamped natural frequency of the system ii) Damping ratio iii) Steady state amplitude and phase angle at 1500 RPM iv) Force transmitted to the ground at 1200 RPM

M=25kg

C

K

Eccentric mass

64

65

CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Support motion Consider a spring mass damper system as shown in figure Let it excited by motion

of support let y be the harmonic displacement of support From Newtonrsquos second law of motion

)yx(k)yx(cxm minusminusminusminus= ampampampamp

)1(0)yx(k)yx(cxm minusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminus=minus+minus+ ampampampamp

Let z be the relative displacement of the mass with respect to support then

Z=x-y

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

time period km

2 f1

n

πτ ==

from the equation(1) mg = k∆

∆=

q

mk

Difference between the translation ( rectilinear) and rotational system of vibration

Translatory Rotational

In the analysis thFORCES are co Linear stiffness K Problems 1A mass of 10kFind the natural 2 A spring massnatural frequencnatural frequenc

kt

m

sec radmk

n =ω

k

25

e disturbing and restoring nsidered

In the analysis In the analysis MASS Moment of Inertia (J) is considered

in Nm is considered

g when suspended from a spring causes a static deflection of 1cm frequency of system

system has a spring stiffness K Nm and a mass of m Kg It has a y of vibration 12 Hz An extra 2kg mass coupled to it then the y reduces by 2 Hz find K and m

secrad n =ω

26

3 A steel wire of 2mm diameter and 30m long It is fixed at the upper end and carries a mass of m kg at its free end Find m so that the frequency of longitudinal vibration is 4 Hz 4 A spring mass system has a natural period of 02 seconds What will be the new period if the spring constant is 1) increased by 50 2) decreased by 50 5 A spring mass system has a natural frequency of 10 Hz when the spring constant is reduced by 800 Nm the frequency is altered by 45 Find the mass and spring constant of the original system 6 Determine the natural frequency of system shown in fig is which shaft is supported in SHORT bearings 7 Determine the natural frequency of system shown in fig is which shaft is supported in LONG bearings where l is the length of bearing and E ndash youngrsquos modulus and I is moment of Inertia 7 Determine the natural frequency of system shown in fig is which shaft is supported in LONG bearings 8 A light cantilever beam of rectangular section( 5 cm deep x 25cm wide) has a mass fixed at its free end Find the ratio of frequency of free lateral vibration in vertical plane to that in horizontal 9 Determine the natural frequency of simple pendulum 10 Homogeneous square plate of size l and mass m is suspended from the mid point of one of its sides as shown in figure Find the frequency of vibration

l

l

27

11 A compound pendulum which is rigid body of mass m and it is pivoted at O The point of pivot is at distance d from the centre of gravity It is free to rotate about its axis Find the frequency of oscillation of such pendulum 12 A connecting rod shown in fig is supported at the wrist pin end It is displaced and allowed to oscillate The mass of rod is 5kg and centre of gravity is 20 cm from the pivot point O If the frequency of oscillation is 40 cyclesminute calculate the mass moment of inertia about its CG

13 A semi circular homogenous disc of radius r and mass m is pivoted freely about its centre as shown in figure Determine the natural frequency of oscillation 14A simply supported beam of square cross section 5mmx5mm and length 1m carrying a mass of 0575 kg at the middle is found to have natural frequency of 30 radsec Determine youngrsquos modulus of elasticity of beam 15 A spring mass system k1 and m have a natural frequency f1 Determine the value of k2 of another spring in terms of k1 which when placed in series with k1 lowers the natural frequency to 23 f1

COMPLETE SOLUTION OF SYSTEM EXECUTING SHM

The equation of motion of system executing SHM can be represented by

0)( =+ xkxm ampamp ----------------------------------------(1)

022

2=+

dtdx

dt

dx ω

The general solution of equation (1) can be expressed as

------------------------(2)

Where A and B are arbitrary constant which can be determined from the initial conditions of the system Two initial conditions are to be specified to evaluate these

constants x=x0 at t=0 and xamp = Vo at t=0 substituting in the equation (2) Energy method In a conservative system the total energy is constant The dwell as natural frequency can be determined by the princienergy For free vibration of undamped system at any instant and partly potential The kinetic energy T is stored in the masswhere as the potential energy U is stored in the form of sdeformation or work done in a force field such as gravity The total energy being constant T+U = constant Its rate of cha

t)(sin Bt)cos( AX ω+ω=

t)(sin V

t)cos( xx o0 ω

ω+ω=

Is the required complete

solution

28

ifferential equation as ple of conservation of of time is partly kinetic by virtue of its velocity train energy in elastic

nge

Is given by [ ] 0UTdtd

=+

From this we get a differential equation of motion as well as natural frequency of the system Determine the natural frequency of spring mass system using energy method

Determine the natural frequency of system shown in figure Determine the natural frequency of the system shown in figure Is there any limitation on the value of K Discuss Determine the natural frequency of s

m

l k

m

a

θ

m

a

θ

l

k

29

k

k

ystem shown below Neglect the mass of ball

l

m

30

A string shown in figure is under tension T which can be assumed to remain constant for small displacements Find the natural frequency of vertical vibrations of spring An acrobat 120kg walks on a tight rope as shown in figure The frequency of vibration in the given position is vertical direction is 30 rads Find the tension in the rope A manometer has a uniform bore of cross section area A If the column of liquid of length L and Density ρ is set into motion as shown in figure Find the frequency of the resulting oscillation

m

l

a

T T

36m 8m

31

k

R

k

m

r

m

Find the expression for natural frequency of system shown in the figure Neglect the mass of the cantilever beam Study the special case i) k=Infinity ii) I = infinity Determine the expression for the natural frequency of a system shown in figure The two discs are keyed to a common shaft and have a combined mass moment of inertia about centre of oscillation O It is assumed that the card attached to mass m does not stretch and is always under tension Determine the natural frequency of a system shown

m

l

32

Determine the expression for the natural frequency of the system shown in figure Assume that the wires connecting the masses do not stretch and are always in tension

Determine the natural frequency of spring mass system taking the MASS OF SPRING (ms ) into account

M3 M1

M2

k2 k1

m

k

33

RAYLEIGHrsquoS METHOD

This is the extension of energy method Here we can determine the natural frequency of a conservative system without finding the equation of motion The natural frequency is a function of the rate of change of Kinetic energy and potential energy of the system From the principle of conservation of energy we have T+U = Constant T= KE and U =PE We can write the above equation as T1+U1 = --------------------------(1) Where 1 and 2 represents tow instances of time Let 1 be the time when the mass is passing through static equilibrium position Let 2 be the time corresponding to the mass displacement of the mass At this instant the velocity f the mass is zero and hence Substituting in equation (1) we get ( )Max = ( )Max Above equation leads directly to natural frequency Determine the natural frequency of spring mass system using Rayleighrsquos method m

34

r

B

K K

Home work

Determine the natural frequency of pendulum using Rayleighrsquos method

A Cylinder of mass m and moment of inertia Jo is to role without slipping but is sustained by spring K as shown in figure Determine the natural frequency of oscillation

Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping

a

Determine the natural frequency of system shown where in cylindrical disc rolls over the inextensible spring A cylinder of radius r rolls without slipping on cylindrical the equation of motion and hence determine the naoscillations about the lowest point shown Repeat the above problem ndash Instead of cylinder assume swithout slipping on concave surface as shown above

r

K

35

portion of radius R Derive tural frequency of small

phere of radius with r rolls

The torsional pendulum has a natural frequency of 5 Hz What length of steel wire of diameter 2 mm should be used for pendulum The inertia of mass fixed at the free end is 00098 kgm2 Take g= 083x1011 Nm2 Determine the torsional natural frequency of the system shown in figure Neglect the mass moment of inertia of the shaft ( Figure not proportionate) Determine the natural frequency of simple pendulum considering the mass of the rod into account

Using the energy method find tThe chord may be assumed ine

Kt1

Kt2

20cm

06m 08m 04m

20cm 20cm 20cm

m

M

he natural frxtensible in

k

M

36

equency of the system shown in figure the spring mass pulley system

A rod of length l and depth H having a mass m rests on a semicylidrical surface os radius r It is tipped slightly and let go Find the frequency of oscillation

A thin bar of l having a mass m rests on a semicylidrical surface radius r It is tipped slightly and let go Find the frequency of oscillation

l

h

R

l

37

38

CHAPTER 3

DAMPED FREE VIBRATION OF SINGLE DEGREE FREEDOM SYSTEM LEARNING OBJECTIVES

] Introduction to damped free vibration

] Terminologies used in damped free vibration

] Three different types of damping

] Concept of critical damping and its importance

] Study of response of viscous damped system for cases of under damping

critical and over damping

] Concept of logarithmic decrement

] Problems related to damped free vibration problems

Viscously damped free vibration of single degree freedom sytem Consider a single degree freedom system with viscous damping as shown in figure For a viscous damped system the damping force is

where c is a constant of proportionality It is called damping coefficient or coefficient of viscous damping

Symbolically it is designated by a dashpot as shown in Fig 3

x xampx

39

At any instant of time let the mass be displaced by x from the equilibrium position Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

Okxxcxm =minusminus= ampampamp

Okxxcxm =++ ampampamp ----------------------------------------------------------(1)

or

0kxdtdx

Cdt

xdM

2

2

=++

Assuming the solution of equation as

The first term in the above equation m2Ct

eminus

is an exponentially decaying function of time The behavior of the terms in the parenthesis however depends on whether numerical value within the radicals is positive zero or negative

CASE 1 if 0mk

m2c

2

=minus

40

CASE 2 if 0mk

m2c

2

=gt

CASE 3 if 0mk

m2c

2

=lt

41

Critically damped coefficient Critically damped coefficient is defined as the value of the damping coeffient for which radical in equation

mk

m2c

m2c

S2

21 minus

plusmn

= becomes zero

Damping Ratio ( Damping factor) ( ξ ) It is defined as the ratio of damping coefficient to the critical damping coefficient Note The damping coefficient c is defined as the actual amount of damping while the damping coefficient is a constant depending upon the mass and stiffness of the system The roots of characteristics equation can be now written as Note if c=cc or ξ =1 then system is critically damped if cgt cc or ξ gt1 then system is over damped I if clt cc or ξ lt1 then system is under damped

42

Take home quiz Derive General solution of different damped system

1 Critically damped system 2 Under damped system 3 Over damped system

Logarithmic Decrement It is used to determine the amount of damping present in system This measure The rate of decay of free vibration The larger the damping the greater will be rate of decay Logarithmic decrement is defined as the natural logarithm of ratio of any two successive amplitudes

=δ

2

1

xx

ln

Expression for logarithmic decrement (δ ) Consider a damped free vibration which is expressed by

( ) ( ) φ+ξminusω= ξω t 1sinXex 2n

tn

x is the displacement at any instant of time X and φ are arbitrary constants

21

2

ξminus

πξ=δ

43

Tutorial problems 1 Derive an expression for damping ratio in terms of logarithmic decrement

2 Show that logarithmic decrement

=δ

n

o

xx

lnn1

where xo is the initial amplitude xn

is amplitude after n cycles 3 A mass of 1 kg is to be supported on a spring having stiffness K= 9800 Nm The damping constant c= 59 Nsm Determine the natural frequency of system Find also logarithmic decrement and amplitude after 3 cycles if initial displacement is 03 cm 4 A vibrating system is defined by the following parameters M = 3kg k= 100 Nm damping coefficient = 3 Nsm determine Damping factor natural frequency damped vibration logarithmic decrement ratio of consecutive amplitude no of cycles after which the original amplitude is reduced to 20 5A single degree damped vibrating system a suspended mass of 18kg makes 15 oscillations in 003 seconds The amplitude decreases to 025 of the initial value after 5 oscillations Determine the stiffness of spring Logarithmic decrement damping factor and damping coefficient 6 A damped vibration record of a spring mass system shows the following data Amplitude at the end of 2nd cycle is 9mm amplitude at eh end of 3rd cycle is 6mm amplitude at the end of 4th cycle is 4mm stiffness of spring =8000 Nm weight =20N Find the logarithmic decrement damping force at unit velocity periodic time of vibration 7 For the system shown in figure the characteristics of dashpot is such that the constant force of 49 N is applied to the piston it velocity is found to be constant at 012 ms Determine the value of C would you expect the complete system to be periodic or aperiodic 8 For a torsional system shown in the figuref1 is natural frequency in air and f2 is the natural frequency when immersed in oil Determine the damping coefficient

Viscous fluid

Disc of mass moment of Inertia J

2a

C

m M

b

a a

k

9 A thin plate of area A and weigh w is fixed to the end of the spring and allow to oscillate in a viscous fluid as shown in figure It f1 is the frequency of oscillation in air and f2 in liquid Show that

22

21 ff

gA2

minusπω

=micro where the damping force on the plate

AV2Fd micro= V being the velocity of vibration 10 A torsional pendulum has a natural frequency of 175 cyclessecond vibrating in vacuum The mass moment of inertia is 2 kgm2 It is immersed in oil and it is observed that the natural frequency is 142 cyclessecond Find the damping torque If the disc is replaced by 34 deg When in oil find the displacement at the end of the first complete cycle 11 An automobile can be modeled as a mass placed on 4 shock absorbers each consisting of a spring and a damper Such that each spring is equally loaded Determine the stiffness damping constant of each shock absorber So that natural frequency is 2Hz and system is critically damped The mass of vehicle is 200kg 12 Write the differential equation of motion for the system shown in the figure below Determine the natural frequency of the damped oscillation and critical damping coefficient

Determine the differential equation of motion and find the critical cdamped natural frequency for system shown in figure

a

C

m M

M =200kg

44

200kg kgkg

oefficient and

45

Derive the equation of motion and find the critical damping coefficient and damped natural frequency for the system shown below Write the differential equation of motion and determine the expression for critical damping coefficient and damped natural frequency of system shown in figure

Set up a differential equation of motion for system shown in figure and determine the following Undamped natural frequencies critical damping coefficient damping ratio damped natural frequency

b

C

kM

a

m

46

A mass of 300N is resting on two springs of stiffness 3000 Nm each and a dashpot of damping coefficient 150 Nsm as shown in figure If the initial velocity of 10cms is given to mass at its equilibrium position what will be the displacement from the equilibrium position at the end of first second Derive the formula used A gun barrel of mass 600kg has a recoil spring of stiffness 294 knm if the barrel recoils 13m on firing determine i) The initial recoil velocity of barrel ii) The critical damping coefficient of dashpot which is engaged at the end of recoil stroke Iii) Time required for the barrel to return at the end of recoil stroke Iv) time required for the barrel to return to a position 5cm from the initial position

W= 3000 N

FIRING

CHAPTER 4

FORCED VIBRATION LEARNING OBJECTIVES

] Introduction to forced free vibration

] Terminologies used in forced vibration

] Concept of magnification factor

] Study of reciprocating and rotating unbalance

] Concept of base excitation and support excitation

] Concept of Transmissibility ratio

] Problems related to forced vibration

The vibration that takes place under the excitation of external forces are called

forced vibration When a system is subjected to harmonic excitation it is forced to vibrate at the same frequency as that of excitation Most of the mechanical systems are prone to steady state forced vibration which cause fatigue failure The vibration problem also becomes very important when any of excitation frequencies coincide with one of natural frequencies of a system Under this condition resonance occurs during which large amplitude of vibrations are observed To avoid resonance external frequency (operating speed) may be changed or properties of the system may be changed to alter the natural frequency In some cases sufficient amount of damping may be provided to avoid large amplitude during resonance Thus the problem of force vibration is very important in mechanical design

FORCED VIBRATIONS OF SINGLE DOF SYSTEMS UNDER HARMONIC EXCITATION

` [ ] tsin FF o ω=

x

sin FF o=

Fo

xampamp

[ ] tω

xamp

47

48

Consider a spring mass system having a viscous damping excited by a harmonic force [ ] tsin FF o ω= as shown in figure The various forces acting on the system is shown in figure Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

O)tsin(Fkxxcxm o =ω+minusminus= ampampamp

)tsin(Fkxxcxm o ω=++ ampampamp ----------------------------------------------------------(1)

or

)tsin(Fkxdtdx

Cdt

xdM o2

2

ω=++

Equation (1) represents the differential equation of motion of the system The solution to this equation consists of two parts namely complementary function and a particular integral

The first of solution is also termed as transient solution of homogeneous solution and can be obtained from the homogeneous differential equation 0kxxcxm =++ ampampamp

The second part of solution particular integral is also termed as steady state solution ad is of form )tsin(Xx φminusω= Where X= steady state amplitude φ is phase difference between external force and displacement

)tsin(Xx φminusω=

The vector representation of the shown

49

The force polygon for the above vector diagram can be written as follows

( )φminusωt O

KX

xm 2ω

Fo

xCω

tω

φ

50

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

is the governing equation

Zero frequency deflection (Xst) It is deflection observed at zero frequency ( when the external frequency 0=ω

( ) kF

001

kF

x o

22

o

st =

+minus=

Magnification factor (M) It is defined as the ratio of steady state amplitude to zero frequency deflection xst

M = arX

X=

A plot of magnification factor versus frequency ratio

damping factor ( )ξ is called frequency plot of a harmon All the frequency terms starts from magnification factor of damping ratio ( )ξ The amplitude of vibration at any on the value of damping The amplitude will be higher for higher damping The amplitude of vibration at reso

amplitude tends to zero as

ωω

n tends to infinity

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

Freq

ζ =

ζ = 1

Magnification Factor M = XXst

kFo

51

ωω

nfor varying values of

ically excite vibrating system

M=1 what ever may be value particular frequency depends

for lesser damping and lesser nance will be very high The

uency Ratio r = (ωωn)

ζ = 0 0

ζ = 025 ζ = 0375 ζ = 0 5

ζ = 0707

ζ = 2

52

Plot of phase angle φ versus frequency ratio

ωω

n is shown in the

figure The phase angle φ is zero for all the values of damping ratio

at

ωω

n=0 The phase

angle φ is 90 for all the values of damping ratio

at

ωω

n=1 ie at

resonance The phase angle between o to 90 deg for all the values of

ωω

n= 0 to 1 and

between 90 deg to 180 deg for all the values of

ωω

ngt1

Class work Derive an expression for frequency at peak amplitude

2np 21 ξminusω=ω=ω is the required equation

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

53

Derive expression amplitude at resonance ( In dimensional form) In dimensional form We know that steady state amplitude

( ) ( )

ω+ωminus=

222

o

Cmk

Fx Equation in dimensional form

We know that at resonance mk

n =ω=ω

( )

ω+

minus

=2

n

2

oR

Cmk

mk

Fx

( )

ω=

2n

oR

C

Fx

( )

ω=

n

oR C

Fx

Class work Derive an expression for amplitude at resonance ( in non dimensional form)

Tutorial Problems on Forced Vibration 1A weight of 60N is suspended by a spring stiffness 12 kNm is forced to vibrate by a harmonic force of 10N assuming viscous damping of 0086 kN-sm Find i) resonance frequency ii) amplitude at resonance iii) phase angle at resonance iv) frequency corresponding to peak amplitude 2 A periodic torque having a maximum value of 05888Nm at a frequency of 4 rads is impressed upon a flywheel suspended from a wire The wheel has mass moment of inertia 012 kgm2 and the wire has stiffness 1176 Nmrad A viscous dashpot applies damping couple of 0784 Nm an angular velocity of 2 rads calculate i)maximum angular displacement from rest position ii) maximum couple applied to dashpot Iii) angle by which the angular displacement lags the torque A mass of 1019 kg is suspended from a spring stiffness 20 Nmm viscous damping causes the amplitude to decrease to 110th of initial value in 4 complete cycle If a periodic force of 20 cos(50 t) is applied to the mass Find the amplitude of forced vibration What would be the amplitude of oscillation if the applied force coincides with the natural period

m=1019 kg

[ ] t50cos 30F =

54

55

ROTATING UNBALANCE Unbalance in rotating machinery is a common source of vibration excitation Consider a machine of mass m having mounted on a foundation of stiffness k and damping coefficient C as shown in figure The unbalance is represented by an eccentric rotating mass m with eccentricity (e) and having a angular velocity (ω) let x be the displacement of non rotating mass from the static equilibrium position The displacement of rotating mass m from the static equilibrium is ( ) tsin( ex ω+ Applying Newtonrsquos law of motion we have

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

56

RECIPROCATING UNBALANCE Unbalance in reciprocating machinery is a common source of vibration excitation Consider a machine of mass M having mounted on a foundation of stiffness K and damping coefficient C as shown in figure The unbalance is represented by a reciprocating mass (m) having crank radus (e) and connecting rod of length (l) Let ωωωω be angular velocity of the crank Let x be the displacement of non reciprocating mass from the static equilibrium position at any instant of time The displacement of reciprocating mass m from the static equilibrium position is

)2sin(sin +++ tle

tex ωωωωωωωω

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

57

Note The equations obtained in reciprocating unbalance is same as the rotating unbalance and having the following differences Rotating Unbalance Reciprocating Unbalance Here mass of the rotating element is considered

Here mass of reciprocating element is considered

Here the eccentricity (e) which is the distance between the centre of rotation and CG

Here the eccentricity e is the crank radius and crank radius = stroke length2

The frequency plot of non dimensional factor( ( )me

MX versus frequency ratio

ωω

n for various values of ( )ξξξξ is shown in figure This frequency plot is same for

the rotating and reciprocating unbalance with the following points are noted from the diagram

1 Damping ratio plays very important role during resonance in controlling the amplitude

2 When

ωω

n=0 then ( )me

MX =0 for all the values of damping ratio ( )ξξξξ

3 When

ωω

nis higher than ( )me

MX tends to unity

4 At very high speeds ie at very high frequency ratio

ωω

n there is no effect

of damping The frequency plot of phase angle versus the frequency ratio is given as shown in figure The explanation is same as in forced harmonic vibration

58

VIBRATION ISOLATION

Vibratory forces and motions generated by machine and other caused are unavoidable Howeever their effect on the other dynamical systems can be minimized by proper isolator design An isolation system is either to protect o delicate machine from excessive vibration transmitted to it from its supporting structure or to prevent vibratory forces and motions generated by machines from being transmitted to its surroundings The effectiveness of isolation may be measured in terms of forces and motion transmitted The first type is known as force isolation and the second type is known as motion isolation

Force isolation (Force Transmissibility) When vibratory forces generated by machine are isolated the effectiveness of forced isolation can be made by defining the term transmissibility The term transmissibility is defined as the ratio of force transmitted to the foundation to that of the external impressed force

Force Transmissibility = o

tr

FF

force Impressed Externalfoundation toted transmitForce

=

Consider viscous damped spring mass system as shown in figure Upon this system an external harmonic force ) t (Sin Fo ω is applied The difference forces acting on system are inertia forces damping force spring force and external force All these forces are represented on force polygon shown below Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these two forces

( ) ( )22 xckx ω+=

59

This is the expression for force transmissibility in dimensional form To represent same in the non dimensional form divide numerator and denominator by k

2

n

22

n

2

n

o

tr

21

21

FF

TR

ωω

ξ+

ωω

minus

ωω

ξ+

==

PHASE DIFFERNECE BETWEEN EXTERNAL FORCE AND DISPLACEMENT Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these

ωω

minus

ωω

ξ=

ωminusω

=φ minusminus2

n

n1

2

1

1

2tan

mKC

tan

The phase difference between force transmitted to the foundation Ftr to the displacement

kc

KxxC

tanω

=ω

=α or

ωω

ξ=

ω=α minusminus

n

11 2tanKx

xCtan

The phase difference between external force Fo and force transmitted to foundationis β

60

β= αminusφ =

ωω

minus

ωω

ξminus

2

n

n1

1

2tan -

ωω

ξminus

n

1 2tan

Displacement Transmissibility It is the ratio of amplitude of steady state vibration of machine to that of the base motion Forced vibration due to support motion Consider a spring mass damper system subjected to harmonic excitation )tsin(Yy ω= at the base as shown in figure Let x be the absolute displacement of the mass at any instant of time t The different forces acting on system are shown in free body diagram Applying Newtonrsquos Law of motion Ma = sumF

)yx(k)yx(cxm minusminusminusminus= ampampampamp

0)yx(k)yx(cxm =minus+minus+ ampampampamp

61

Frequency plot of transmissibility versus frequency ratio

62

The plot of transmissibility versus frequency ratio is as shown in figure The following points can be noted from it

i) The value of transmissibility is unity at frequency ratio

ωω

n=0 and 2

ii) For an undamped system ( )ξ =0 transmissibility tends to infin at resonance iii) The value of transmissibility is less than unity for the value of frequency ratio

greater than 2

iv) For frequency ratio

ωω

nlt 2 smaller damping ratio lead to larger values of

transmissibility on the other hand for frequency ratio

ωω

ngt 2 smaller damping

ratio leads to smaller value of transmissibility

The frequency plot of transmissibility (TR) versus frequency ratio

ωω

n can be

broadly divided into two zones

1

ωω

n= 0 to 2

2

ωω

n= 2 to infin

CASE 1

When

ωω

n= 0 to 2

In this zone the value of transmissibility is always greater than 1 for all values of damping ratio In this zone the transmissibility decreases with increase in damping Hence we can conclude that in this zone damping is useful in isolating vibration forces being transmitted to the support CASE 2

When

ωω

n= 2 to infin

In this zone the value of transmissibility is less than 1 for all values of damping ratio In this zone the transmissibility increases with increase in damping Hence we can conclude that in this zone damping will adversely affect in isolating vibratory forces

63

Tutorial problems on Forced Vibration 1 Machine of mass 75kg is mounted on spring of stiffness 12000 Nm with an assumed damping factor ξ =02 piston within the machine of mass 2 kg has a reciprocating motion with a stroke of 75 cm and speed of 15 cyclessecond Assuming the motion of piston to be harmonic determine amplitude of machine ( X) Transmissibility and force transmitted to foundation phase angle of transmitted force wrt to exciting force 2 Motor weighs 600N is mounted on a simple beam that has a spring rate of 400 Ncm The motor armature base 160 N and has an eccentricity of 0005cm What will be the amplitude of motion when it runs at 1760 RPM Neglect damping the deflection of armature shaft and weight of the beam 3 A Heavy machine of mass 300 kg is supported on resilient foundation the static deflection of foundation due to weight of machine is found to be 75mm It is observed that the machine vibrates with an amplitude of 10mm when the base of the foundation is subjected to harmonic oscillation at the undampeds natural frequency of system with an amplitude of 25mm Find the damping constant of foundation Ii) dynamic force amplitude on the base iii) amplitude of the displacement of machine relative to base 4 An air compressor of mass 450 kg operates at a constant speed of 1750 RPM The reciprocating masses are 10 kg the crank radius is 100mm If the damper for mounting introduces a damping factor 015 i) specify the springs of mounting such that only 20 of the unbalanced force is transmitted to the foundation ii) determine the magnitude of the transmitted force bull Repeat the above problem by assuming that there is no damping in system

5 A body of a car having mass 1500 kg and is mounted on 4 equal springs which deflect through 225 cm under the weight of the body A total damping forces the 4 shock absorbers is 46 N at a velocity of 1cmsec The car is placed with all the four wheels on a test platform which is move up and down at resonant speed With an amplitude of 25cm Find the amplitude of the car body on its springs assuming its CG to be in the centre of wheel base 6A 75 kg machine which is mounted on springs of stiffness 116x105 Nm with the damping of 20=ξ A 2kg piston within the machine as reciprocating motion with a stroke of 008m and speed of 3000 RPM Assuming the motion of piston to be harmonic determine the amplitude of vibration of machine

7 The spring of an automobile are compressed 10cm under its own weight Find the critical speed when a trailer is traveling over road with a people approximaterly by a sine wave of amplitude 75 cm and wave length 15m What will be the amplitude of vibration at 64kmhour 8 An eccentric mass excites us used to determine vibratory characteristics of a structure of mass 25kg as given in figure At a speed of 1000 RPM the eccentric mass to beat top position at the instant the structure was moved upwards through its static equilibrium position and corresponding amplitude was 20mm The eccentric mass is 05kg at an eccentricity of 10mm Determine i) Undamped natural frequency of the system ii) Damping ratio iii) Steady state amplitude and phase angle at 1500 RPM iv) Force transmitted to the ground at 1200 RPM

M=25kg

C

K

Eccentric mass

64

65

CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Support motion Consider a spring mass damper system as shown in figure Let it excited by motion

of support let y be the harmonic displacement of support From Newtonrsquos second law of motion

)yx(k)yx(cxm minusminusminusminus= ampampampamp

)1(0)yx(k)yx(cxm minusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminus=minus+minus+ ampampampamp

Let z be the relative displacement of the mass with respect to support then

Z=x-y

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

26

3 A steel wire of 2mm diameter and 30m long It is fixed at the upper end and carries a mass of m kg at its free end Find m so that the frequency of longitudinal vibration is 4 Hz 4 A spring mass system has a natural period of 02 seconds What will be the new period if the spring constant is 1) increased by 50 2) decreased by 50 5 A spring mass system has a natural frequency of 10 Hz when the spring constant is reduced by 800 Nm the frequency is altered by 45 Find the mass and spring constant of the original system 6 Determine the natural frequency of system shown in fig is which shaft is supported in SHORT bearings 7 Determine the natural frequency of system shown in fig is which shaft is supported in LONG bearings where l is the length of bearing and E ndash youngrsquos modulus and I is moment of Inertia 7 Determine the natural frequency of system shown in fig is which shaft is supported in LONG bearings 8 A light cantilever beam of rectangular section( 5 cm deep x 25cm wide) has a mass fixed at its free end Find the ratio of frequency of free lateral vibration in vertical plane to that in horizontal 9 Determine the natural frequency of simple pendulum 10 Homogeneous square plate of size l and mass m is suspended from the mid point of one of its sides as shown in figure Find the frequency of vibration

l

l

27

11 A compound pendulum which is rigid body of mass m and it is pivoted at O The point of pivot is at distance d from the centre of gravity It is free to rotate about its axis Find the frequency of oscillation of such pendulum 12 A connecting rod shown in fig is supported at the wrist pin end It is displaced and allowed to oscillate The mass of rod is 5kg and centre of gravity is 20 cm from the pivot point O If the frequency of oscillation is 40 cyclesminute calculate the mass moment of inertia about its CG

13 A semi circular homogenous disc of radius r and mass m is pivoted freely about its centre as shown in figure Determine the natural frequency of oscillation 14A simply supported beam of square cross section 5mmx5mm and length 1m carrying a mass of 0575 kg at the middle is found to have natural frequency of 30 radsec Determine youngrsquos modulus of elasticity of beam 15 A spring mass system k1 and m have a natural frequency f1 Determine the value of k2 of another spring in terms of k1 which when placed in series with k1 lowers the natural frequency to 23 f1

COMPLETE SOLUTION OF SYSTEM EXECUTING SHM

The equation of motion of system executing SHM can be represented by

0)( =+ xkxm ampamp ----------------------------------------(1)

022

2=+

dtdx

dt

dx ω

The general solution of equation (1) can be expressed as

------------------------(2)

Where A and B are arbitrary constant which can be determined from the initial conditions of the system Two initial conditions are to be specified to evaluate these

constants x=x0 at t=0 and xamp = Vo at t=0 substituting in the equation (2) Energy method In a conservative system the total energy is constant The dwell as natural frequency can be determined by the princienergy For free vibration of undamped system at any instant and partly potential The kinetic energy T is stored in the masswhere as the potential energy U is stored in the form of sdeformation or work done in a force field such as gravity The total energy being constant T+U = constant Its rate of cha

t)(sin Bt)cos( AX ω+ω=

t)(sin V

t)cos( xx o0 ω

ω+ω=

Is the required complete

solution

28

ifferential equation as ple of conservation of of time is partly kinetic by virtue of its velocity train energy in elastic

nge

Is given by [ ] 0UTdtd

=+

From this we get a differential equation of motion as well as natural frequency of the system Determine the natural frequency of spring mass system using energy method

Determine the natural frequency of system shown in figure Determine the natural frequency of the system shown in figure Is there any limitation on the value of K Discuss Determine the natural frequency of s

m

l k

m

a

θ

m

a

θ

l

k

29

k

k

ystem shown below Neglect the mass of ball

l

m

30

A string shown in figure is under tension T which can be assumed to remain constant for small displacements Find the natural frequency of vertical vibrations of spring An acrobat 120kg walks on a tight rope as shown in figure The frequency of vibration in the given position is vertical direction is 30 rads Find the tension in the rope A manometer has a uniform bore of cross section area A If the column of liquid of length L and Density ρ is set into motion as shown in figure Find the frequency of the resulting oscillation

m

l

a

T T

36m 8m

31

k

R

k

m

r

m

Find the expression for natural frequency of system shown in the figure Neglect the mass of the cantilever beam Study the special case i) k=Infinity ii) I = infinity Determine the expression for the natural frequency of a system shown in figure The two discs are keyed to a common shaft and have a combined mass moment of inertia about centre of oscillation O It is assumed that the card attached to mass m does not stretch and is always under tension Determine the natural frequency of a system shown

m

l

32

Determine the expression for the natural frequency of the system shown in figure Assume that the wires connecting the masses do not stretch and are always in tension

Determine the natural frequency of spring mass system taking the MASS OF SPRING (ms ) into account

M3 M1

M2

k2 k1

m

k

33

RAYLEIGHrsquoS METHOD

This is the extension of energy method Here we can determine the natural frequency of a conservative system without finding the equation of motion The natural frequency is a function of the rate of change of Kinetic energy and potential energy of the system From the principle of conservation of energy we have T+U = Constant T= KE and U =PE We can write the above equation as T1+U1 = --------------------------(1) Where 1 and 2 represents tow instances of time Let 1 be the time when the mass is passing through static equilibrium position Let 2 be the time corresponding to the mass displacement of the mass At this instant the velocity f the mass is zero and hence Substituting in equation (1) we get ( )Max = ( )Max Above equation leads directly to natural frequency Determine the natural frequency of spring mass system using Rayleighrsquos method m

34

r

B

K K

Home work

Determine the natural frequency of pendulum using Rayleighrsquos method

A Cylinder of mass m and moment of inertia Jo is to role without slipping but is sustained by spring K as shown in figure Determine the natural frequency of oscillation

Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping

a

Determine the natural frequency of system shown where in cylindrical disc rolls over the inextensible spring A cylinder of radius r rolls without slipping on cylindrical the equation of motion and hence determine the naoscillations about the lowest point shown Repeat the above problem ndash Instead of cylinder assume swithout slipping on concave surface as shown above

r

K

35

portion of radius R Derive tural frequency of small

phere of radius with r rolls

The torsional pendulum has a natural frequency of 5 Hz What length of steel wire of diameter 2 mm should be used for pendulum The inertia of mass fixed at the free end is 00098 kgm2 Take g= 083x1011 Nm2 Determine the torsional natural frequency of the system shown in figure Neglect the mass moment of inertia of the shaft ( Figure not proportionate) Determine the natural frequency of simple pendulum considering the mass of the rod into account

Using the energy method find tThe chord may be assumed ine

Kt1

Kt2

20cm

06m 08m 04m

20cm 20cm 20cm

m

M

he natural frxtensible in

k

M

36

equency of the system shown in figure the spring mass pulley system

A rod of length l and depth H having a mass m rests on a semicylidrical surface os radius r It is tipped slightly and let go Find the frequency of oscillation

A thin bar of l having a mass m rests on a semicylidrical surface radius r It is tipped slightly and let go Find the frequency of oscillation

l

h

R

l

37

38

CHAPTER 3

DAMPED FREE VIBRATION OF SINGLE DEGREE FREEDOM SYSTEM LEARNING OBJECTIVES

] Introduction to damped free vibration

] Terminologies used in damped free vibration

] Three different types of damping

] Concept of critical damping and its importance

] Study of response of viscous damped system for cases of under damping

critical and over damping

] Concept of logarithmic decrement

] Problems related to damped free vibration problems

Viscously damped free vibration of single degree freedom sytem Consider a single degree freedom system with viscous damping as shown in figure For a viscous damped system the damping force is

where c is a constant of proportionality It is called damping coefficient or coefficient of viscous damping

Symbolically it is designated by a dashpot as shown in Fig 3

x xampx

39

At any instant of time let the mass be displaced by x from the equilibrium position Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

Okxxcxm =minusminus= ampampamp

Okxxcxm =++ ampampamp ----------------------------------------------------------(1)

or

0kxdtdx

Cdt

xdM

2

2

=++

Assuming the solution of equation as

The first term in the above equation m2Ct

eminus

is an exponentially decaying function of time The behavior of the terms in the parenthesis however depends on whether numerical value within the radicals is positive zero or negative

CASE 1 if 0mk

m2c

2

=minus

40

CASE 2 if 0mk

m2c

2

=gt

CASE 3 if 0mk

m2c

2

=lt

41

Critically damped coefficient Critically damped coefficient is defined as the value of the damping coeffient for which radical in equation

mk

m2c

m2c

S2

21 minus

plusmn

= becomes zero

Damping Ratio ( Damping factor) ( ξ ) It is defined as the ratio of damping coefficient to the critical damping coefficient Note The damping coefficient c is defined as the actual amount of damping while the damping coefficient is a constant depending upon the mass and stiffness of the system The roots of characteristics equation can be now written as Note if c=cc or ξ =1 then system is critically damped if cgt cc or ξ gt1 then system is over damped I if clt cc or ξ lt1 then system is under damped

42

Take home quiz Derive General solution of different damped system

1 Critically damped system 2 Under damped system 3 Over damped system

Logarithmic Decrement It is used to determine the amount of damping present in system This measure The rate of decay of free vibration The larger the damping the greater will be rate of decay Logarithmic decrement is defined as the natural logarithm of ratio of any two successive amplitudes

=δ

2

1

xx

ln

Expression for logarithmic decrement (δ ) Consider a damped free vibration which is expressed by

( ) ( ) φ+ξminusω= ξω t 1sinXex 2n

tn

x is the displacement at any instant of time X and φ are arbitrary constants

21

2

ξminus

πξ=δ

43

Tutorial problems 1 Derive an expression for damping ratio in terms of logarithmic decrement

2 Show that logarithmic decrement

=δ

n

o

xx

lnn1

where xo is the initial amplitude xn

is amplitude after n cycles 3 A mass of 1 kg is to be supported on a spring having stiffness K= 9800 Nm The damping constant c= 59 Nsm Determine the natural frequency of system Find also logarithmic decrement and amplitude after 3 cycles if initial displacement is 03 cm 4 A vibrating system is defined by the following parameters M = 3kg k= 100 Nm damping coefficient = 3 Nsm determine Damping factor natural frequency damped vibration logarithmic decrement ratio of consecutive amplitude no of cycles after which the original amplitude is reduced to 20 5A single degree damped vibrating system a suspended mass of 18kg makes 15 oscillations in 003 seconds The amplitude decreases to 025 of the initial value after 5 oscillations Determine the stiffness of spring Logarithmic decrement damping factor and damping coefficient 6 A damped vibration record of a spring mass system shows the following data Amplitude at the end of 2nd cycle is 9mm amplitude at eh end of 3rd cycle is 6mm amplitude at the end of 4th cycle is 4mm stiffness of spring =8000 Nm weight =20N Find the logarithmic decrement damping force at unit velocity periodic time of vibration 7 For the system shown in figure the characteristics of dashpot is such that the constant force of 49 N is applied to the piston it velocity is found to be constant at 012 ms Determine the value of C would you expect the complete system to be periodic or aperiodic 8 For a torsional system shown in the figuref1 is natural frequency in air and f2 is the natural frequency when immersed in oil Determine the damping coefficient

Viscous fluid

Disc of mass moment of Inertia J

2a

C

m M

b

a a

k

9 A thin plate of area A and weigh w is fixed to the end of the spring and allow to oscillate in a viscous fluid as shown in figure It f1 is the frequency of oscillation in air and f2 in liquid Show that

22

21 ff

gA2

minusπω

=micro where the damping force on the plate

AV2Fd micro= V being the velocity of vibration 10 A torsional pendulum has a natural frequency of 175 cyclessecond vibrating in vacuum The mass moment of inertia is 2 kgm2 It is immersed in oil and it is observed that the natural frequency is 142 cyclessecond Find the damping torque If the disc is replaced by 34 deg When in oil find the displacement at the end of the first complete cycle 11 An automobile can be modeled as a mass placed on 4 shock absorbers each consisting of a spring and a damper Such that each spring is equally loaded Determine the stiffness damping constant of each shock absorber So that natural frequency is 2Hz and system is critically damped The mass of vehicle is 200kg 12 Write the differential equation of motion for the system shown in the figure below Determine the natural frequency of the damped oscillation and critical damping coefficient

Determine the differential equation of motion and find the critical cdamped natural frequency for system shown in figure

a

C

m M

M =200kg

44

200kg kgkg

oefficient and

45

Derive the equation of motion and find the critical damping coefficient and damped natural frequency for the system shown below Write the differential equation of motion and determine the expression for critical damping coefficient and damped natural frequency of system shown in figure

Set up a differential equation of motion for system shown in figure and determine the following Undamped natural frequencies critical damping coefficient damping ratio damped natural frequency

b

C

kM

a

m

46

A mass of 300N is resting on two springs of stiffness 3000 Nm each and a dashpot of damping coefficient 150 Nsm as shown in figure If the initial velocity of 10cms is given to mass at its equilibrium position what will be the displacement from the equilibrium position at the end of first second Derive the formula used A gun barrel of mass 600kg has a recoil spring of stiffness 294 knm if the barrel recoils 13m on firing determine i) The initial recoil velocity of barrel ii) The critical damping coefficient of dashpot which is engaged at the end of recoil stroke Iii) Time required for the barrel to return at the end of recoil stroke Iv) time required for the barrel to return to a position 5cm from the initial position

W= 3000 N

FIRING

CHAPTER 4

FORCED VIBRATION LEARNING OBJECTIVES

] Introduction to forced free vibration

] Terminologies used in forced vibration

] Concept of magnification factor

] Study of reciprocating and rotating unbalance

] Concept of base excitation and support excitation

] Concept of Transmissibility ratio

] Problems related to forced vibration

The vibration that takes place under the excitation of external forces are called

forced vibration When a system is subjected to harmonic excitation it is forced to vibrate at the same frequency as that of excitation Most of the mechanical systems are prone to steady state forced vibration which cause fatigue failure The vibration problem also becomes very important when any of excitation frequencies coincide with one of natural frequencies of a system Under this condition resonance occurs during which large amplitude of vibrations are observed To avoid resonance external frequency (operating speed) may be changed or properties of the system may be changed to alter the natural frequency In some cases sufficient amount of damping may be provided to avoid large amplitude during resonance Thus the problem of force vibration is very important in mechanical design

FORCED VIBRATIONS OF SINGLE DOF SYSTEMS UNDER HARMONIC EXCITATION

` [ ] tsin FF o ω=

x

sin FF o=

Fo

xampamp

[ ] tω

xamp

47

48

Consider a spring mass system having a viscous damping excited by a harmonic force [ ] tsin FF o ω= as shown in figure The various forces acting on the system is shown in figure Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

O)tsin(Fkxxcxm o =ω+minusminus= ampampamp

)tsin(Fkxxcxm o ω=++ ampampamp ----------------------------------------------------------(1)

or

)tsin(Fkxdtdx

Cdt

xdM o2

2

ω=++

Equation (1) represents the differential equation of motion of the system The solution to this equation consists of two parts namely complementary function and a particular integral

The first of solution is also termed as transient solution of homogeneous solution and can be obtained from the homogeneous differential equation 0kxxcxm =++ ampampamp

The second part of solution particular integral is also termed as steady state solution ad is of form )tsin(Xx φminusω= Where X= steady state amplitude φ is phase difference between external force and displacement

)tsin(Xx φminusω=

The vector representation of the shown

49

The force polygon for the above vector diagram can be written as follows

( )φminusωt O

KX

xm 2ω

Fo

xCω

tω

φ

50

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

is the governing equation

Zero frequency deflection (Xst) It is deflection observed at zero frequency ( when the external frequency 0=ω

( ) kF

001

kF

x o

22

o

st =

+minus=

Magnification factor (M) It is defined as the ratio of steady state amplitude to zero frequency deflection xst

M = arX

X=

A plot of magnification factor versus frequency ratio

damping factor ( )ξ is called frequency plot of a harmon All the frequency terms starts from magnification factor of damping ratio ( )ξ The amplitude of vibration at any on the value of damping The amplitude will be higher for higher damping The amplitude of vibration at reso

amplitude tends to zero as

ωω

n tends to infinity

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

Freq

ζ =

ζ = 1

Magnification Factor M = XXst

kFo

51

ωω

nfor varying values of

ically excite vibrating system

M=1 what ever may be value particular frequency depends

for lesser damping and lesser nance will be very high The

uency Ratio r = (ωωn)

ζ = 0 0

ζ = 025 ζ = 0375 ζ = 0 5

ζ = 0707

ζ = 2

52

Plot of phase angle φ versus frequency ratio

ωω

n is shown in the

figure The phase angle φ is zero for all the values of damping ratio

at

ωω

n=0 The phase

angle φ is 90 for all the values of damping ratio

at

ωω

n=1 ie at

resonance The phase angle between o to 90 deg for all the values of

ωω

n= 0 to 1 and

between 90 deg to 180 deg for all the values of

ωω

ngt1

Class work Derive an expression for frequency at peak amplitude

2np 21 ξminusω=ω=ω is the required equation

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

53

Derive expression amplitude at resonance ( In dimensional form) In dimensional form We know that steady state amplitude

( ) ( )

ω+ωminus=

222

o

Cmk

Fx Equation in dimensional form

We know that at resonance mk

n =ω=ω

( )

ω+

minus

=2

n

2

oR

Cmk

mk

Fx

( )

ω=

2n

oR

C

Fx

( )

ω=

n

oR C

Fx

Class work Derive an expression for amplitude at resonance ( in non dimensional form)

Tutorial Problems on Forced Vibration 1A weight of 60N is suspended by a spring stiffness 12 kNm is forced to vibrate by a harmonic force of 10N assuming viscous damping of 0086 kN-sm Find i) resonance frequency ii) amplitude at resonance iii) phase angle at resonance iv) frequency corresponding to peak amplitude 2 A periodic torque having a maximum value of 05888Nm at a frequency of 4 rads is impressed upon a flywheel suspended from a wire The wheel has mass moment of inertia 012 kgm2 and the wire has stiffness 1176 Nmrad A viscous dashpot applies damping couple of 0784 Nm an angular velocity of 2 rads calculate i)maximum angular displacement from rest position ii) maximum couple applied to dashpot Iii) angle by which the angular displacement lags the torque A mass of 1019 kg is suspended from a spring stiffness 20 Nmm viscous damping causes the amplitude to decrease to 110th of initial value in 4 complete cycle If a periodic force of 20 cos(50 t) is applied to the mass Find the amplitude of forced vibration What would be the amplitude of oscillation if the applied force coincides with the natural period

m=1019 kg

[ ] t50cos 30F =

54

55

ROTATING UNBALANCE Unbalance in rotating machinery is a common source of vibration excitation Consider a machine of mass m having mounted on a foundation of stiffness k and damping coefficient C as shown in figure The unbalance is represented by an eccentric rotating mass m with eccentricity (e) and having a angular velocity (ω) let x be the displacement of non rotating mass from the static equilibrium position The displacement of rotating mass m from the static equilibrium is ( ) tsin( ex ω+ Applying Newtonrsquos law of motion we have

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

56

RECIPROCATING UNBALANCE Unbalance in reciprocating machinery is a common source of vibration excitation Consider a machine of mass M having mounted on a foundation of stiffness K and damping coefficient C as shown in figure The unbalance is represented by a reciprocating mass (m) having crank radus (e) and connecting rod of length (l) Let ωωωω be angular velocity of the crank Let x be the displacement of non reciprocating mass from the static equilibrium position at any instant of time The displacement of reciprocating mass m from the static equilibrium position is

)2sin(sin +++ tle

tex ωωωωωωωω

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

57

Note The equations obtained in reciprocating unbalance is same as the rotating unbalance and having the following differences Rotating Unbalance Reciprocating Unbalance Here mass of the rotating element is considered

Here mass of reciprocating element is considered

Here the eccentricity (e) which is the distance between the centre of rotation and CG

Here the eccentricity e is the crank radius and crank radius = stroke length2

The frequency plot of non dimensional factor( ( )me

MX versus frequency ratio

ωω

n for various values of ( )ξξξξ is shown in figure This frequency plot is same for

the rotating and reciprocating unbalance with the following points are noted from the diagram

1 Damping ratio plays very important role during resonance in controlling the amplitude

2 When

ωω

n=0 then ( )me

MX =0 for all the values of damping ratio ( )ξξξξ

3 When

ωω

nis higher than ( )me

MX tends to unity

4 At very high speeds ie at very high frequency ratio

ωω

n there is no effect

of damping The frequency plot of phase angle versus the frequency ratio is given as shown in figure The explanation is same as in forced harmonic vibration

58

VIBRATION ISOLATION

Vibratory forces and motions generated by machine and other caused are unavoidable Howeever their effect on the other dynamical systems can be minimized by proper isolator design An isolation system is either to protect o delicate machine from excessive vibration transmitted to it from its supporting structure or to prevent vibratory forces and motions generated by machines from being transmitted to its surroundings The effectiveness of isolation may be measured in terms of forces and motion transmitted The first type is known as force isolation and the second type is known as motion isolation

Force isolation (Force Transmissibility) When vibratory forces generated by machine are isolated the effectiveness of forced isolation can be made by defining the term transmissibility The term transmissibility is defined as the ratio of force transmitted to the foundation to that of the external impressed force

Force Transmissibility = o

tr

FF

force Impressed Externalfoundation toted transmitForce

=

Consider viscous damped spring mass system as shown in figure Upon this system an external harmonic force ) t (Sin Fo ω is applied The difference forces acting on system are inertia forces damping force spring force and external force All these forces are represented on force polygon shown below Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these two forces

( ) ( )22 xckx ω+=

59

This is the expression for force transmissibility in dimensional form To represent same in the non dimensional form divide numerator and denominator by k

2

n

22

n

2

n

o

tr

21

21

FF

TR

ωω

ξ+

ωω

minus

ωω

ξ+

==

PHASE DIFFERNECE BETWEEN EXTERNAL FORCE AND DISPLACEMENT Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these

ωω

minus

ωω

ξ=

ωminusω

=φ minusminus2

n

n1

2

1

1

2tan

mKC

tan

The phase difference between force transmitted to the foundation Ftr to the displacement

kc

KxxC

tanω

=ω

=α or

ωω

ξ=

ω=α minusminus

n

11 2tanKx

xCtan

The phase difference between external force Fo and force transmitted to foundationis β

60

β= αminusφ =

ωω

minus

ωω

ξminus

2

n

n1

1

2tan -

ωω

ξminus

n

1 2tan

Displacement Transmissibility It is the ratio of amplitude of steady state vibration of machine to that of the base motion Forced vibration due to support motion Consider a spring mass damper system subjected to harmonic excitation )tsin(Yy ω= at the base as shown in figure Let x be the absolute displacement of the mass at any instant of time t The different forces acting on system are shown in free body diagram Applying Newtonrsquos Law of motion Ma = sumF

)yx(k)yx(cxm minusminusminusminus= ampampampamp

0)yx(k)yx(cxm =minus+minus+ ampampampamp

61

Frequency plot of transmissibility versus frequency ratio

62

The plot of transmissibility versus frequency ratio is as shown in figure The following points can be noted from it

i) The value of transmissibility is unity at frequency ratio

ωω

n=0 and 2

ii) For an undamped system ( )ξ =0 transmissibility tends to infin at resonance iii) The value of transmissibility is less than unity for the value of frequency ratio

greater than 2

iv) For frequency ratio

ωω

nlt 2 smaller damping ratio lead to larger values of

transmissibility on the other hand for frequency ratio

ωω

ngt 2 smaller damping

ratio leads to smaller value of transmissibility

The frequency plot of transmissibility (TR) versus frequency ratio

ωω

n can be

broadly divided into two zones

1

ωω

n= 0 to 2

2

ωω

n= 2 to infin

CASE 1

When

ωω

n= 0 to 2

In this zone the value of transmissibility is always greater than 1 for all values of damping ratio In this zone the transmissibility decreases with increase in damping Hence we can conclude that in this zone damping is useful in isolating vibration forces being transmitted to the support CASE 2

When

ωω

n= 2 to infin

In this zone the value of transmissibility is less than 1 for all values of damping ratio In this zone the transmissibility increases with increase in damping Hence we can conclude that in this zone damping will adversely affect in isolating vibratory forces

63

Tutorial problems on Forced Vibration 1 Machine of mass 75kg is mounted on spring of stiffness 12000 Nm with an assumed damping factor ξ =02 piston within the machine of mass 2 kg has a reciprocating motion with a stroke of 75 cm and speed of 15 cyclessecond Assuming the motion of piston to be harmonic determine amplitude of machine ( X) Transmissibility and force transmitted to foundation phase angle of transmitted force wrt to exciting force 2 Motor weighs 600N is mounted on a simple beam that has a spring rate of 400 Ncm The motor armature base 160 N and has an eccentricity of 0005cm What will be the amplitude of motion when it runs at 1760 RPM Neglect damping the deflection of armature shaft and weight of the beam 3 A Heavy machine of mass 300 kg is supported on resilient foundation the static deflection of foundation due to weight of machine is found to be 75mm It is observed that the machine vibrates with an amplitude of 10mm when the base of the foundation is subjected to harmonic oscillation at the undampeds natural frequency of system with an amplitude of 25mm Find the damping constant of foundation Ii) dynamic force amplitude on the base iii) amplitude of the displacement of machine relative to base 4 An air compressor of mass 450 kg operates at a constant speed of 1750 RPM The reciprocating masses are 10 kg the crank radius is 100mm If the damper for mounting introduces a damping factor 015 i) specify the springs of mounting such that only 20 of the unbalanced force is transmitted to the foundation ii) determine the magnitude of the transmitted force bull Repeat the above problem by assuming that there is no damping in system

5 A body of a car having mass 1500 kg and is mounted on 4 equal springs which deflect through 225 cm under the weight of the body A total damping forces the 4 shock absorbers is 46 N at a velocity of 1cmsec The car is placed with all the four wheels on a test platform which is move up and down at resonant speed With an amplitude of 25cm Find the amplitude of the car body on its springs assuming its CG to be in the centre of wheel base 6A 75 kg machine which is mounted on springs of stiffness 116x105 Nm with the damping of 20=ξ A 2kg piston within the machine as reciprocating motion with a stroke of 008m and speed of 3000 RPM Assuming the motion of piston to be harmonic determine the amplitude of vibration of machine

7 The spring of an automobile are compressed 10cm under its own weight Find the critical speed when a trailer is traveling over road with a people approximaterly by a sine wave of amplitude 75 cm and wave length 15m What will be the amplitude of vibration at 64kmhour 8 An eccentric mass excites us used to determine vibratory characteristics of a structure of mass 25kg as given in figure At a speed of 1000 RPM the eccentric mass to beat top position at the instant the structure was moved upwards through its static equilibrium position and corresponding amplitude was 20mm The eccentric mass is 05kg at an eccentricity of 10mm Determine i) Undamped natural frequency of the system ii) Damping ratio iii) Steady state amplitude and phase angle at 1500 RPM iv) Force transmitted to the ground at 1200 RPM

M=25kg

C

K

Eccentric mass

64

65

CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Support motion Consider a spring mass damper system as shown in figure Let it excited by motion

of support let y be the harmonic displacement of support From Newtonrsquos second law of motion

)yx(k)yx(cxm minusminusminusminus= ampampampamp

)1(0)yx(k)yx(cxm minusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminus=minus+minus+ ampampampamp

Let z be the relative displacement of the mass with respect to support then

Z=x-y

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

27

11 A compound pendulum which is rigid body of mass m and it is pivoted at O The point of pivot is at distance d from the centre of gravity It is free to rotate about its axis Find the frequency of oscillation of such pendulum 12 A connecting rod shown in fig is supported at the wrist pin end It is displaced and allowed to oscillate The mass of rod is 5kg and centre of gravity is 20 cm from the pivot point O If the frequency of oscillation is 40 cyclesminute calculate the mass moment of inertia about its CG

13 A semi circular homogenous disc of radius r and mass m is pivoted freely about its centre as shown in figure Determine the natural frequency of oscillation 14A simply supported beam of square cross section 5mmx5mm and length 1m carrying a mass of 0575 kg at the middle is found to have natural frequency of 30 radsec Determine youngrsquos modulus of elasticity of beam 15 A spring mass system k1 and m have a natural frequency f1 Determine the value of k2 of another spring in terms of k1 which when placed in series with k1 lowers the natural frequency to 23 f1

COMPLETE SOLUTION OF SYSTEM EXECUTING SHM

The equation of motion of system executing SHM can be represented by

0)( =+ xkxm ampamp ----------------------------------------(1)

022

2=+

dtdx

dt

dx ω

The general solution of equation (1) can be expressed as

------------------------(2)

Where A and B are arbitrary constant which can be determined from the initial conditions of the system Two initial conditions are to be specified to evaluate these

constants x=x0 at t=0 and xamp = Vo at t=0 substituting in the equation (2) Energy method In a conservative system the total energy is constant The dwell as natural frequency can be determined by the princienergy For free vibration of undamped system at any instant and partly potential The kinetic energy T is stored in the masswhere as the potential energy U is stored in the form of sdeformation or work done in a force field such as gravity The total energy being constant T+U = constant Its rate of cha

t)(sin Bt)cos( AX ω+ω=

t)(sin V

t)cos( xx o0 ω

ω+ω=

Is the required complete

solution

28

ifferential equation as ple of conservation of of time is partly kinetic by virtue of its velocity train energy in elastic

nge

Is given by [ ] 0UTdtd

=+

From this we get a differential equation of motion as well as natural frequency of the system Determine the natural frequency of spring mass system using energy method

Determine the natural frequency of system shown in figure Determine the natural frequency of the system shown in figure Is there any limitation on the value of K Discuss Determine the natural frequency of s

m

l k

m

a

θ

m

a

θ

l

k

29

k

k

ystem shown below Neglect the mass of ball

l

m

30

A string shown in figure is under tension T which can be assumed to remain constant for small displacements Find the natural frequency of vertical vibrations of spring An acrobat 120kg walks on a tight rope as shown in figure The frequency of vibration in the given position is vertical direction is 30 rads Find the tension in the rope A manometer has a uniform bore of cross section area A If the column of liquid of length L and Density ρ is set into motion as shown in figure Find the frequency of the resulting oscillation

m

l

a

T T

36m 8m

31

k

R

k

m

r

m

Find the expression for natural frequency of system shown in the figure Neglect the mass of the cantilever beam Study the special case i) k=Infinity ii) I = infinity Determine the expression for the natural frequency of a system shown in figure The two discs are keyed to a common shaft and have a combined mass moment of inertia about centre of oscillation O It is assumed that the card attached to mass m does not stretch and is always under tension Determine the natural frequency of a system shown

m

l

32

Determine the expression for the natural frequency of the system shown in figure Assume that the wires connecting the masses do not stretch and are always in tension

Determine the natural frequency of spring mass system taking the MASS OF SPRING (ms ) into account

M3 M1

M2

k2 k1

m

k

33

RAYLEIGHrsquoS METHOD

This is the extension of energy method Here we can determine the natural frequency of a conservative system without finding the equation of motion The natural frequency is a function of the rate of change of Kinetic energy and potential energy of the system From the principle of conservation of energy we have T+U = Constant T= KE and U =PE We can write the above equation as T1+U1 = --------------------------(1) Where 1 and 2 represents tow instances of time Let 1 be the time when the mass is passing through static equilibrium position Let 2 be the time corresponding to the mass displacement of the mass At this instant the velocity f the mass is zero and hence Substituting in equation (1) we get ( )Max = ( )Max Above equation leads directly to natural frequency Determine the natural frequency of spring mass system using Rayleighrsquos method m

34

r

B

K K

Home work

Determine the natural frequency of pendulum using Rayleighrsquos method

A Cylinder of mass m and moment of inertia Jo is to role without slipping but is sustained by spring K as shown in figure Determine the natural frequency of oscillation

Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping

a

Determine the natural frequency of system shown where in cylindrical disc rolls over the inextensible spring A cylinder of radius r rolls without slipping on cylindrical the equation of motion and hence determine the naoscillations about the lowest point shown Repeat the above problem ndash Instead of cylinder assume swithout slipping on concave surface as shown above

r

K

35

portion of radius R Derive tural frequency of small

phere of radius with r rolls

The torsional pendulum has a natural frequency of 5 Hz What length of steel wire of diameter 2 mm should be used for pendulum The inertia of mass fixed at the free end is 00098 kgm2 Take g= 083x1011 Nm2 Determine the torsional natural frequency of the system shown in figure Neglect the mass moment of inertia of the shaft ( Figure not proportionate) Determine the natural frequency of simple pendulum considering the mass of the rod into account

Using the energy method find tThe chord may be assumed ine

Kt1

Kt2

20cm

06m 08m 04m

20cm 20cm 20cm

m

M

he natural frxtensible in

k

M

36

equency of the system shown in figure the spring mass pulley system

A rod of length l and depth H having a mass m rests on a semicylidrical surface os radius r It is tipped slightly and let go Find the frequency of oscillation

A thin bar of l having a mass m rests on a semicylidrical surface radius r It is tipped slightly and let go Find the frequency of oscillation

l

h

R

l

37

38

CHAPTER 3

DAMPED FREE VIBRATION OF SINGLE DEGREE FREEDOM SYSTEM LEARNING OBJECTIVES

] Introduction to damped free vibration

] Terminologies used in damped free vibration

] Three different types of damping

] Concept of critical damping and its importance

] Study of response of viscous damped system for cases of under damping

critical and over damping

] Concept of logarithmic decrement

] Problems related to damped free vibration problems

Viscously damped free vibration of single degree freedom sytem Consider a single degree freedom system with viscous damping as shown in figure For a viscous damped system the damping force is

where c is a constant of proportionality It is called damping coefficient or coefficient of viscous damping

Symbolically it is designated by a dashpot as shown in Fig 3

x xampx

39

At any instant of time let the mass be displaced by x from the equilibrium position Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

Okxxcxm =minusminus= ampampamp

Okxxcxm =++ ampampamp ----------------------------------------------------------(1)

or

0kxdtdx

Cdt

xdM

2

2

=++

Assuming the solution of equation as

The first term in the above equation m2Ct

eminus

is an exponentially decaying function of time The behavior of the terms in the parenthesis however depends on whether numerical value within the radicals is positive zero or negative

CASE 1 if 0mk

m2c

2

=minus

40

CASE 2 if 0mk

m2c

2

=gt

CASE 3 if 0mk

m2c

2

=lt

41

Critically damped coefficient Critically damped coefficient is defined as the value of the damping coeffient for which radical in equation

mk

m2c

m2c

S2

21 minus

plusmn

= becomes zero

Damping Ratio ( Damping factor) ( ξ ) It is defined as the ratio of damping coefficient to the critical damping coefficient Note The damping coefficient c is defined as the actual amount of damping while the damping coefficient is a constant depending upon the mass and stiffness of the system The roots of characteristics equation can be now written as Note if c=cc or ξ =1 then system is critically damped if cgt cc or ξ gt1 then system is over damped I if clt cc or ξ lt1 then system is under damped

42

Take home quiz Derive General solution of different damped system

1 Critically damped system 2 Under damped system 3 Over damped system

Logarithmic Decrement It is used to determine the amount of damping present in system This measure The rate of decay of free vibration The larger the damping the greater will be rate of decay Logarithmic decrement is defined as the natural logarithm of ratio of any two successive amplitudes

=δ

2

1

xx

ln

Expression for logarithmic decrement (δ ) Consider a damped free vibration which is expressed by

( ) ( ) φ+ξminusω= ξω t 1sinXex 2n

tn

x is the displacement at any instant of time X and φ are arbitrary constants

21

2

ξminus

πξ=δ

43

Tutorial problems 1 Derive an expression for damping ratio in terms of logarithmic decrement

2 Show that logarithmic decrement

=δ

n

o

xx

lnn1

where xo is the initial amplitude xn

is amplitude after n cycles 3 A mass of 1 kg is to be supported on a spring having stiffness K= 9800 Nm The damping constant c= 59 Nsm Determine the natural frequency of system Find also logarithmic decrement and amplitude after 3 cycles if initial displacement is 03 cm 4 A vibrating system is defined by the following parameters M = 3kg k= 100 Nm damping coefficient = 3 Nsm determine Damping factor natural frequency damped vibration logarithmic decrement ratio of consecutive amplitude no of cycles after which the original amplitude is reduced to 20 5A single degree damped vibrating system a suspended mass of 18kg makes 15 oscillations in 003 seconds The amplitude decreases to 025 of the initial value after 5 oscillations Determine the stiffness of spring Logarithmic decrement damping factor and damping coefficient 6 A damped vibration record of a spring mass system shows the following data Amplitude at the end of 2nd cycle is 9mm amplitude at eh end of 3rd cycle is 6mm amplitude at the end of 4th cycle is 4mm stiffness of spring =8000 Nm weight =20N Find the logarithmic decrement damping force at unit velocity periodic time of vibration 7 For the system shown in figure the characteristics of dashpot is such that the constant force of 49 N is applied to the piston it velocity is found to be constant at 012 ms Determine the value of C would you expect the complete system to be periodic or aperiodic 8 For a torsional system shown in the figuref1 is natural frequency in air and f2 is the natural frequency when immersed in oil Determine the damping coefficient

Viscous fluid

Disc of mass moment of Inertia J

2a

C

m M

b

a a

k

9 A thin plate of area A and weigh w is fixed to the end of the spring and allow to oscillate in a viscous fluid as shown in figure It f1 is the frequency of oscillation in air and f2 in liquid Show that

22

21 ff

gA2

minusπω

=micro where the damping force on the plate

AV2Fd micro= V being the velocity of vibration 10 A torsional pendulum has a natural frequency of 175 cyclessecond vibrating in vacuum The mass moment of inertia is 2 kgm2 It is immersed in oil and it is observed that the natural frequency is 142 cyclessecond Find the damping torque If the disc is replaced by 34 deg When in oil find the displacement at the end of the first complete cycle 11 An automobile can be modeled as a mass placed on 4 shock absorbers each consisting of a spring and a damper Such that each spring is equally loaded Determine the stiffness damping constant of each shock absorber So that natural frequency is 2Hz and system is critically damped The mass of vehicle is 200kg 12 Write the differential equation of motion for the system shown in the figure below Determine the natural frequency of the damped oscillation and critical damping coefficient

Determine the differential equation of motion and find the critical cdamped natural frequency for system shown in figure

a

C

m M

M =200kg

44

200kg kgkg

oefficient and

45

Derive the equation of motion and find the critical damping coefficient and damped natural frequency for the system shown below Write the differential equation of motion and determine the expression for critical damping coefficient and damped natural frequency of system shown in figure

Set up a differential equation of motion for system shown in figure and determine the following Undamped natural frequencies critical damping coefficient damping ratio damped natural frequency

b

C

kM

a

m

46

A mass of 300N is resting on two springs of stiffness 3000 Nm each and a dashpot of damping coefficient 150 Nsm as shown in figure If the initial velocity of 10cms is given to mass at its equilibrium position what will be the displacement from the equilibrium position at the end of first second Derive the formula used A gun barrel of mass 600kg has a recoil spring of stiffness 294 knm if the barrel recoils 13m on firing determine i) The initial recoil velocity of barrel ii) The critical damping coefficient of dashpot which is engaged at the end of recoil stroke Iii) Time required for the barrel to return at the end of recoil stroke Iv) time required for the barrel to return to a position 5cm from the initial position

W= 3000 N

FIRING

CHAPTER 4

FORCED VIBRATION LEARNING OBJECTIVES

] Introduction to forced free vibration

] Terminologies used in forced vibration

] Concept of magnification factor

] Study of reciprocating and rotating unbalance

] Concept of base excitation and support excitation

] Concept of Transmissibility ratio

] Problems related to forced vibration

The vibration that takes place under the excitation of external forces are called

forced vibration When a system is subjected to harmonic excitation it is forced to vibrate at the same frequency as that of excitation Most of the mechanical systems are prone to steady state forced vibration which cause fatigue failure The vibration problem also becomes very important when any of excitation frequencies coincide with one of natural frequencies of a system Under this condition resonance occurs during which large amplitude of vibrations are observed To avoid resonance external frequency (operating speed) may be changed or properties of the system may be changed to alter the natural frequency In some cases sufficient amount of damping may be provided to avoid large amplitude during resonance Thus the problem of force vibration is very important in mechanical design

FORCED VIBRATIONS OF SINGLE DOF SYSTEMS UNDER HARMONIC EXCITATION

` [ ] tsin FF o ω=

x

sin FF o=

Fo

xampamp

[ ] tω

xamp

47

48

Consider a spring mass system having a viscous damping excited by a harmonic force [ ] tsin FF o ω= as shown in figure The various forces acting on the system is shown in figure Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

O)tsin(Fkxxcxm o =ω+minusminus= ampampamp

)tsin(Fkxxcxm o ω=++ ampampamp ----------------------------------------------------------(1)

or

)tsin(Fkxdtdx

Cdt

xdM o2

2

ω=++

Equation (1) represents the differential equation of motion of the system The solution to this equation consists of two parts namely complementary function and a particular integral

The first of solution is also termed as transient solution of homogeneous solution and can be obtained from the homogeneous differential equation 0kxxcxm =++ ampampamp

The second part of solution particular integral is also termed as steady state solution ad is of form )tsin(Xx φminusω= Where X= steady state amplitude φ is phase difference between external force and displacement

)tsin(Xx φminusω=

The vector representation of the shown

49

The force polygon for the above vector diagram can be written as follows

( )φminusωt O

KX

xm 2ω

Fo

xCω

tω

φ

50

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

is the governing equation

Zero frequency deflection (Xst) It is deflection observed at zero frequency ( when the external frequency 0=ω

( ) kF

001

kF

x o

22

o

st =

+minus=

Magnification factor (M) It is defined as the ratio of steady state amplitude to zero frequency deflection xst

M = arX

X=

A plot of magnification factor versus frequency ratio

damping factor ( )ξ is called frequency plot of a harmon All the frequency terms starts from magnification factor of damping ratio ( )ξ The amplitude of vibration at any on the value of damping The amplitude will be higher for higher damping The amplitude of vibration at reso

amplitude tends to zero as

ωω

n tends to infinity

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

Freq

ζ =

ζ = 1

Magnification Factor M = XXst

kFo

51

ωω

nfor varying values of

ically excite vibrating system

M=1 what ever may be value particular frequency depends

for lesser damping and lesser nance will be very high The

uency Ratio r = (ωωn)

ζ = 0 0

ζ = 025 ζ = 0375 ζ = 0 5

ζ = 0707

ζ = 2

52

Plot of phase angle φ versus frequency ratio

ωω

n is shown in the

figure The phase angle φ is zero for all the values of damping ratio

at

ωω

n=0 The phase

angle φ is 90 for all the values of damping ratio

at

ωω

n=1 ie at

resonance The phase angle between o to 90 deg for all the values of

ωω

n= 0 to 1 and

between 90 deg to 180 deg for all the values of

ωω

ngt1

Class work Derive an expression for frequency at peak amplitude

2np 21 ξminusω=ω=ω is the required equation

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

53

Derive expression amplitude at resonance ( In dimensional form) In dimensional form We know that steady state amplitude

( ) ( )

ω+ωminus=

222

o

Cmk

Fx Equation in dimensional form

We know that at resonance mk

n =ω=ω

( )

ω+

minus

=2

n

2

oR

Cmk

mk

Fx

( )

ω=

2n

oR

C

Fx

( )

ω=

n

oR C

Fx

Class work Derive an expression for amplitude at resonance ( in non dimensional form)

Tutorial Problems on Forced Vibration 1A weight of 60N is suspended by a spring stiffness 12 kNm is forced to vibrate by a harmonic force of 10N assuming viscous damping of 0086 kN-sm Find i) resonance frequency ii) amplitude at resonance iii) phase angle at resonance iv) frequency corresponding to peak amplitude 2 A periodic torque having a maximum value of 05888Nm at a frequency of 4 rads is impressed upon a flywheel suspended from a wire The wheel has mass moment of inertia 012 kgm2 and the wire has stiffness 1176 Nmrad A viscous dashpot applies damping couple of 0784 Nm an angular velocity of 2 rads calculate i)maximum angular displacement from rest position ii) maximum couple applied to dashpot Iii) angle by which the angular displacement lags the torque A mass of 1019 kg is suspended from a spring stiffness 20 Nmm viscous damping causes the amplitude to decrease to 110th of initial value in 4 complete cycle If a periodic force of 20 cos(50 t) is applied to the mass Find the amplitude of forced vibration What would be the amplitude of oscillation if the applied force coincides with the natural period

m=1019 kg

[ ] t50cos 30F =

54

55

ROTATING UNBALANCE Unbalance in rotating machinery is a common source of vibration excitation Consider a machine of mass m having mounted on a foundation of stiffness k and damping coefficient C as shown in figure The unbalance is represented by an eccentric rotating mass m with eccentricity (e) and having a angular velocity (ω) let x be the displacement of non rotating mass from the static equilibrium position The displacement of rotating mass m from the static equilibrium is ( ) tsin( ex ω+ Applying Newtonrsquos law of motion we have

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

56

RECIPROCATING UNBALANCE Unbalance in reciprocating machinery is a common source of vibration excitation Consider a machine of mass M having mounted on a foundation of stiffness K and damping coefficient C as shown in figure The unbalance is represented by a reciprocating mass (m) having crank radus (e) and connecting rod of length (l) Let ωωωω be angular velocity of the crank Let x be the displacement of non reciprocating mass from the static equilibrium position at any instant of time The displacement of reciprocating mass m from the static equilibrium position is

)2sin(sin +++ tle

tex ωωωωωωωω

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

57

Note The equations obtained in reciprocating unbalance is same as the rotating unbalance and having the following differences Rotating Unbalance Reciprocating Unbalance Here mass of the rotating element is considered

Here mass of reciprocating element is considered

Here the eccentricity (e) which is the distance between the centre of rotation and CG

Here the eccentricity e is the crank radius and crank radius = stroke length2

The frequency plot of non dimensional factor( ( )me

MX versus frequency ratio

ωω

n for various values of ( )ξξξξ is shown in figure This frequency plot is same for

the rotating and reciprocating unbalance with the following points are noted from the diagram

1 Damping ratio plays very important role during resonance in controlling the amplitude

2 When

ωω

n=0 then ( )me

MX =0 for all the values of damping ratio ( )ξξξξ

3 When

ωω

nis higher than ( )me

MX tends to unity

4 At very high speeds ie at very high frequency ratio

ωω

n there is no effect

of damping The frequency plot of phase angle versus the frequency ratio is given as shown in figure The explanation is same as in forced harmonic vibration

58

VIBRATION ISOLATION

Vibratory forces and motions generated by machine and other caused are unavoidable Howeever their effect on the other dynamical systems can be minimized by proper isolator design An isolation system is either to protect o delicate machine from excessive vibration transmitted to it from its supporting structure or to prevent vibratory forces and motions generated by machines from being transmitted to its surroundings The effectiveness of isolation may be measured in terms of forces and motion transmitted The first type is known as force isolation and the second type is known as motion isolation

Force isolation (Force Transmissibility) When vibratory forces generated by machine are isolated the effectiveness of forced isolation can be made by defining the term transmissibility The term transmissibility is defined as the ratio of force transmitted to the foundation to that of the external impressed force

Force Transmissibility = o

tr

FF

force Impressed Externalfoundation toted transmitForce

=

Consider viscous damped spring mass system as shown in figure Upon this system an external harmonic force ) t (Sin Fo ω is applied The difference forces acting on system are inertia forces damping force spring force and external force All these forces are represented on force polygon shown below Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these two forces

( ) ( )22 xckx ω+=

59

This is the expression for force transmissibility in dimensional form To represent same in the non dimensional form divide numerator and denominator by k

2

n

22

n

2

n

o

tr

21

21

FF

TR

ωω

ξ+

ωω

minus

ωω

ξ+

==

PHASE DIFFERNECE BETWEEN EXTERNAL FORCE AND DISPLACEMENT Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these

ωω

minus

ωω

ξ=

ωminusω

=φ minusminus2

n

n1

2

1

1

2tan

mKC

tan

The phase difference between force transmitted to the foundation Ftr to the displacement

kc

KxxC

tanω

=ω

=α or

ωω

ξ=

ω=α minusminus

n

11 2tanKx

xCtan

The phase difference between external force Fo and force transmitted to foundationis β

60

β= αminusφ =

ωω

minus

ωω

ξminus

2

n

n1

1

2tan -

ωω

ξminus

n

1 2tan

Displacement Transmissibility It is the ratio of amplitude of steady state vibration of machine to that of the base motion Forced vibration due to support motion Consider a spring mass damper system subjected to harmonic excitation )tsin(Yy ω= at the base as shown in figure Let x be the absolute displacement of the mass at any instant of time t The different forces acting on system are shown in free body diagram Applying Newtonrsquos Law of motion Ma = sumF

)yx(k)yx(cxm minusminusminusminus= ampampampamp

0)yx(k)yx(cxm =minus+minus+ ampampampamp

61

Frequency plot of transmissibility versus frequency ratio

62

The plot of transmissibility versus frequency ratio is as shown in figure The following points can be noted from it

i) The value of transmissibility is unity at frequency ratio

ωω

n=0 and 2

ii) For an undamped system ( )ξ =0 transmissibility tends to infin at resonance iii) The value of transmissibility is less than unity for the value of frequency ratio

greater than 2

iv) For frequency ratio

ωω

nlt 2 smaller damping ratio lead to larger values of

transmissibility on the other hand for frequency ratio

ωω

ngt 2 smaller damping

ratio leads to smaller value of transmissibility

The frequency plot of transmissibility (TR) versus frequency ratio

ωω

n can be

broadly divided into two zones

1

ωω

n= 0 to 2

2

ωω

n= 2 to infin

CASE 1

When

ωω

n= 0 to 2

In this zone the value of transmissibility is always greater than 1 for all values of damping ratio In this zone the transmissibility decreases with increase in damping Hence we can conclude that in this zone damping is useful in isolating vibration forces being transmitted to the support CASE 2

When

ωω

n= 2 to infin

In this zone the value of transmissibility is less than 1 for all values of damping ratio In this zone the transmissibility increases with increase in damping Hence we can conclude that in this zone damping will adversely affect in isolating vibratory forces

63

Tutorial problems on Forced Vibration 1 Machine of mass 75kg is mounted on spring of stiffness 12000 Nm with an assumed damping factor ξ =02 piston within the machine of mass 2 kg has a reciprocating motion with a stroke of 75 cm and speed of 15 cyclessecond Assuming the motion of piston to be harmonic determine amplitude of machine ( X) Transmissibility and force transmitted to foundation phase angle of transmitted force wrt to exciting force 2 Motor weighs 600N is mounted on a simple beam that has a spring rate of 400 Ncm The motor armature base 160 N and has an eccentricity of 0005cm What will be the amplitude of motion when it runs at 1760 RPM Neglect damping the deflection of armature shaft and weight of the beam 3 A Heavy machine of mass 300 kg is supported on resilient foundation the static deflection of foundation due to weight of machine is found to be 75mm It is observed that the machine vibrates with an amplitude of 10mm when the base of the foundation is subjected to harmonic oscillation at the undampeds natural frequency of system with an amplitude of 25mm Find the damping constant of foundation Ii) dynamic force amplitude on the base iii) amplitude of the displacement of machine relative to base 4 An air compressor of mass 450 kg operates at a constant speed of 1750 RPM The reciprocating masses are 10 kg the crank radius is 100mm If the damper for mounting introduces a damping factor 015 i) specify the springs of mounting such that only 20 of the unbalanced force is transmitted to the foundation ii) determine the magnitude of the transmitted force bull Repeat the above problem by assuming that there is no damping in system

5 A body of a car having mass 1500 kg and is mounted on 4 equal springs which deflect through 225 cm under the weight of the body A total damping forces the 4 shock absorbers is 46 N at a velocity of 1cmsec The car is placed with all the four wheels on a test platform which is move up and down at resonant speed With an amplitude of 25cm Find the amplitude of the car body on its springs assuming its CG to be in the centre of wheel base 6A 75 kg machine which is mounted on springs of stiffness 116x105 Nm with the damping of 20=ξ A 2kg piston within the machine as reciprocating motion with a stroke of 008m and speed of 3000 RPM Assuming the motion of piston to be harmonic determine the amplitude of vibration of machine

7 The spring of an automobile are compressed 10cm under its own weight Find the critical speed when a trailer is traveling over road with a people approximaterly by a sine wave of amplitude 75 cm and wave length 15m What will be the amplitude of vibration at 64kmhour 8 An eccentric mass excites us used to determine vibratory characteristics of a structure of mass 25kg as given in figure At a speed of 1000 RPM the eccentric mass to beat top position at the instant the structure was moved upwards through its static equilibrium position and corresponding amplitude was 20mm The eccentric mass is 05kg at an eccentricity of 10mm Determine i) Undamped natural frequency of the system ii) Damping ratio iii) Steady state amplitude and phase angle at 1500 RPM iv) Force transmitted to the ground at 1200 RPM

M=25kg

C

K

Eccentric mass

64

65

CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Support motion Consider a spring mass damper system as shown in figure Let it excited by motion

of support let y be the harmonic displacement of support From Newtonrsquos second law of motion

)yx(k)yx(cxm minusminusminusminus= ampampampamp

)1(0)yx(k)yx(cxm minusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminus=minus+minus+ ampampampamp

Let z be the relative displacement of the mass with respect to support then

Z=x-y

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

COMPLETE SOLUTION OF SYSTEM EXECUTING SHM

The equation of motion of system executing SHM can be represented by

0)( =+ xkxm ampamp ----------------------------------------(1)

022

2=+

dtdx

dt

dx ω

The general solution of equation (1) can be expressed as

------------------------(2)

Where A and B are arbitrary constant which can be determined from the initial conditions of the system Two initial conditions are to be specified to evaluate these

constants x=x0 at t=0 and xamp = Vo at t=0 substituting in the equation (2) Energy method In a conservative system the total energy is constant The dwell as natural frequency can be determined by the princienergy For free vibration of undamped system at any instant and partly potential The kinetic energy T is stored in the masswhere as the potential energy U is stored in the form of sdeformation or work done in a force field such as gravity The total energy being constant T+U = constant Its rate of cha

t)(sin Bt)cos( AX ω+ω=

t)(sin V

t)cos( xx o0 ω

ω+ω=

Is the required complete

solution

28

ifferential equation as ple of conservation of of time is partly kinetic by virtue of its velocity train energy in elastic

nge

Is given by [ ] 0UTdtd

=+

From this we get a differential equation of motion as well as natural frequency of the system Determine the natural frequency of spring mass system using energy method

Determine the natural frequency of system shown in figure Determine the natural frequency of the system shown in figure Is there any limitation on the value of K Discuss Determine the natural frequency of s

m

l k

m

a

θ

m

a

θ

l

k

29

k

k

ystem shown below Neglect the mass of ball

l

m

30

A string shown in figure is under tension T which can be assumed to remain constant for small displacements Find the natural frequency of vertical vibrations of spring An acrobat 120kg walks on a tight rope as shown in figure The frequency of vibration in the given position is vertical direction is 30 rads Find the tension in the rope A manometer has a uniform bore of cross section area A If the column of liquid of length L and Density ρ is set into motion as shown in figure Find the frequency of the resulting oscillation

m

l

a

T T

36m 8m

31

k

R

k

m

r

m

Find the expression for natural frequency of system shown in the figure Neglect the mass of the cantilever beam Study the special case i) k=Infinity ii) I = infinity Determine the expression for the natural frequency of a system shown in figure The two discs are keyed to a common shaft and have a combined mass moment of inertia about centre of oscillation O It is assumed that the card attached to mass m does not stretch and is always under tension Determine the natural frequency of a system shown

m

l

32

Determine the expression for the natural frequency of the system shown in figure Assume that the wires connecting the masses do not stretch and are always in tension

Determine the natural frequency of spring mass system taking the MASS OF SPRING (ms ) into account

M3 M1

M2

k2 k1

m

k

33

RAYLEIGHrsquoS METHOD

This is the extension of energy method Here we can determine the natural frequency of a conservative system without finding the equation of motion The natural frequency is a function of the rate of change of Kinetic energy and potential energy of the system From the principle of conservation of energy we have T+U = Constant T= KE and U =PE We can write the above equation as T1+U1 = --------------------------(1) Where 1 and 2 represents tow instances of time Let 1 be the time when the mass is passing through static equilibrium position Let 2 be the time corresponding to the mass displacement of the mass At this instant the velocity f the mass is zero and hence Substituting in equation (1) we get ( )Max = ( )Max Above equation leads directly to natural frequency Determine the natural frequency of spring mass system using Rayleighrsquos method m

34

r

B

K K

Home work

Determine the natural frequency of pendulum using Rayleighrsquos method

A Cylinder of mass m and moment of inertia Jo is to role without slipping but is sustained by spring K as shown in figure Determine the natural frequency of oscillation

Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping

a

Determine the natural frequency of system shown where in cylindrical disc rolls over the inextensible spring A cylinder of radius r rolls without slipping on cylindrical the equation of motion and hence determine the naoscillations about the lowest point shown Repeat the above problem ndash Instead of cylinder assume swithout slipping on concave surface as shown above

r

K

35

portion of radius R Derive tural frequency of small

phere of radius with r rolls

The torsional pendulum has a natural frequency of 5 Hz What length of steel wire of diameter 2 mm should be used for pendulum The inertia of mass fixed at the free end is 00098 kgm2 Take g= 083x1011 Nm2 Determine the torsional natural frequency of the system shown in figure Neglect the mass moment of inertia of the shaft ( Figure not proportionate) Determine the natural frequency of simple pendulum considering the mass of the rod into account

Using the energy method find tThe chord may be assumed ine

Kt1

Kt2

20cm

06m 08m 04m

20cm 20cm 20cm

m

M

he natural frxtensible in

k

M

36

equency of the system shown in figure the spring mass pulley system

A rod of length l and depth H having a mass m rests on a semicylidrical surface os radius r It is tipped slightly and let go Find the frequency of oscillation

A thin bar of l having a mass m rests on a semicylidrical surface radius r It is tipped slightly and let go Find the frequency of oscillation

l

h

R

l

37

38

CHAPTER 3

DAMPED FREE VIBRATION OF SINGLE DEGREE FREEDOM SYSTEM LEARNING OBJECTIVES

] Introduction to damped free vibration

] Terminologies used in damped free vibration

] Three different types of damping

] Concept of critical damping and its importance

] Study of response of viscous damped system for cases of under damping

critical and over damping

] Concept of logarithmic decrement

] Problems related to damped free vibration problems

Viscously damped free vibration of single degree freedom sytem Consider a single degree freedom system with viscous damping as shown in figure For a viscous damped system the damping force is

where c is a constant of proportionality It is called damping coefficient or coefficient of viscous damping

Symbolically it is designated by a dashpot as shown in Fig 3

x xampx

39

At any instant of time let the mass be displaced by x from the equilibrium position Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

Okxxcxm =minusminus= ampampamp

Okxxcxm =++ ampampamp ----------------------------------------------------------(1)

or

0kxdtdx

Cdt

xdM

2

2

=++

Assuming the solution of equation as

The first term in the above equation m2Ct

eminus

is an exponentially decaying function of time The behavior of the terms in the parenthesis however depends on whether numerical value within the radicals is positive zero or negative

CASE 1 if 0mk

m2c

2

=minus

40

CASE 2 if 0mk

m2c

2

=gt

CASE 3 if 0mk

m2c

2

=lt

41

Critically damped coefficient Critically damped coefficient is defined as the value of the damping coeffient for which radical in equation

mk

m2c

m2c

S2

21 minus

plusmn

= becomes zero

Damping Ratio ( Damping factor) ( ξ ) It is defined as the ratio of damping coefficient to the critical damping coefficient Note The damping coefficient c is defined as the actual amount of damping while the damping coefficient is a constant depending upon the mass and stiffness of the system The roots of characteristics equation can be now written as Note if c=cc or ξ =1 then system is critically damped if cgt cc or ξ gt1 then system is over damped I if clt cc or ξ lt1 then system is under damped

42

Take home quiz Derive General solution of different damped system

1 Critically damped system 2 Under damped system 3 Over damped system

Logarithmic Decrement It is used to determine the amount of damping present in system This measure The rate of decay of free vibration The larger the damping the greater will be rate of decay Logarithmic decrement is defined as the natural logarithm of ratio of any two successive amplitudes

=δ

2

1

xx

ln

Expression for logarithmic decrement (δ ) Consider a damped free vibration which is expressed by

( ) ( ) φ+ξminusω= ξω t 1sinXex 2n

tn

x is the displacement at any instant of time X and φ are arbitrary constants

21

2

ξminus

πξ=δ

43

Tutorial problems 1 Derive an expression for damping ratio in terms of logarithmic decrement

2 Show that logarithmic decrement

=δ

n

o

xx

lnn1

where xo is the initial amplitude xn

is amplitude after n cycles 3 A mass of 1 kg is to be supported on a spring having stiffness K= 9800 Nm The damping constant c= 59 Nsm Determine the natural frequency of system Find also logarithmic decrement and amplitude after 3 cycles if initial displacement is 03 cm 4 A vibrating system is defined by the following parameters M = 3kg k= 100 Nm damping coefficient = 3 Nsm determine Damping factor natural frequency damped vibration logarithmic decrement ratio of consecutive amplitude no of cycles after which the original amplitude is reduced to 20 5A single degree damped vibrating system a suspended mass of 18kg makes 15 oscillations in 003 seconds The amplitude decreases to 025 of the initial value after 5 oscillations Determine the stiffness of spring Logarithmic decrement damping factor and damping coefficient 6 A damped vibration record of a spring mass system shows the following data Amplitude at the end of 2nd cycle is 9mm amplitude at eh end of 3rd cycle is 6mm amplitude at the end of 4th cycle is 4mm stiffness of spring =8000 Nm weight =20N Find the logarithmic decrement damping force at unit velocity periodic time of vibration 7 For the system shown in figure the characteristics of dashpot is such that the constant force of 49 N is applied to the piston it velocity is found to be constant at 012 ms Determine the value of C would you expect the complete system to be periodic or aperiodic 8 For a torsional system shown in the figuref1 is natural frequency in air and f2 is the natural frequency when immersed in oil Determine the damping coefficient

Viscous fluid

Disc of mass moment of Inertia J

2a

C

m M

b

a a

k

9 A thin plate of area A and weigh w is fixed to the end of the spring and allow to oscillate in a viscous fluid as shown in figure It f1 is the frequency of oscillation in air and f2 in liquid Show that

22

21 ff

gA2

minusπω

=micro where the damping force on the plate

AV2Fd micro= V being the velocity of vibration 10 A torsional pendulum has a natural frequency of 175 cyclessecond vibrating in vacuum The mass moment of inertia is 2 kgm2 It is immersed in oil and it is observed that the natural frequency is 142 cyclessecond Find the damping torque If the disc is replaced by 34 deg When in oil find the displacement at the end of the first complete cycle 11 An automobile can be modeled as a mass placed on 4 shock absorbers each consisting of a spring and a damper Such that each spring is equally loaded Determine the stiffness damping constant of each shock absorber So that natural frequency is 2Hz and system is critically damped The mass of vehicle is 200kg 12 Write the differential equation of motion for the system shown in the figure below Determine the natural frequency of the damped oscillation and critical damping coefficient

Determine the differential equation of motion and find the critical cdamped natural frequency for system shown in figure

a

C

m M

M =200kg

44

200kg kgkg

oefficient and

45

Derive the equation of motion and find the critical damping coefficient and damped natural frequency for the system shown below Write the differential equation of motion and determine the expression for critical damping coefficient and damped natural frequency of system shown in figure

Set up a differential equation of motion for system shown in figure and determine the following Undamped natural frequencies critical damping coefficient damping ratio damped natural frequency

b

C

kM

a

m

46

A mass of 300N is resting on two springs of stiffness 3000 Nm each and a dashpot of damping coefficient 150 Nsm as shown in figure If the initial velocity of 10cms is given to mass at its equilibrium position what will be the displacement from the equilibrium position at the end of first second Derive the formula used A gun barrel of mass 600kg has a recoil spring of stiffness 294 knm if the barrel recoils 13m on firing determine i) The initial recoil velocity of barrel ii) The critical damping coefficient of dashpot which is engaged at the end of recoil stroke Iii) Time required for the barrel to return at the end of recoil stroke Iv) time required for the barrel to return to a position 5cm from the initial position

W= 3000 N

FIRING

CHAPTER 4

FORCED VIBRATION LEARNING OBJECTIVES

] Introduction to forced free vibration

] Terminologies used in forced vibration

] Concept of magnification factor

] Study of reciprocating and rotating unbalance

] Concept of base excitation and support excitation

] Concept of Transmissibility ratio

] Problems related to forced vibration

The vibration that takes place under the excitation of external forces are called

forced vibration When a system is subjected to harmonic excitation it is forced to vibrate at the same frequency as that of excitation Most of the mechanical systems are prone to steady state forced vibration which cause fatigue failure The vibration problem also becomes very important when any of excitation frequencies coincide with one of natural frequencies of a system Under this condition resonance occurs during which large amplitude of vibrations are observed To avoid resonance external frequency (operating speed) may be changed or properties of the system may be changed to alter the natural frequency In some cases sufficient amount of damping may be provided to avoid large amplitude during resonance Thus the problem of force vibration is very important in mechanical design

FORCED VIBRATIONS OF SINGLE DOF SYSTEMS UNDER HARMONIC EXCITATION

` [ ] tsin FF o ω=

x

sin FF o=

Fo

xampamp

[ ] tω

xamp

47

48

Consider a spring mass system having a viscous damping excited by a harmonic force [ ] tsin FF o ω= as shown in figure The various forces acting on the system is shown in figure Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

O)tsin(Fkxxcxm o =ω+minusminus= ampampamp

)tsin(Fkxxcxm o ω=++ ampampamp ----------------------------------------------------------(1)

or

)tsin(Fkxdtdx

Cdt

xdM o2

2

ω=++

Equation (1) represents the differential equation of motion of the system The solution to this equation consists of two parts namely complementary function and a particular integral

The first of solution is also termed as transient solution of homogeneous solution and can be obtained from the homogeneous differential equation 0kxxcxm =++ ampampamp

The second part of solution particular integral is also termed as steady state solution ad is of form )tsin(Xx φminusω= Where X= steady state amplitude φ is phase difference between external force and displacement

)tsin(Xx φminusω=

The vector representation of the shown

49

The force polygon for the above vector diagram can be written as follows

( )φminusωt O

KX

xm 2ω

Fo

xCω

tω

φ

50

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

is the governing equation

Zero frequency deflection (Xst) It is deflection observed at zero frequency ( when the external frequency 0=ω

( ) kF

001

kF

x o

22

o

st =

+minus=

Magnification factor (M) It is defined as the ratio of steady state amplitude to zero frequency deflection xst

M = arX

X=

A plot of magnification factor versus frequency ratio

damping factor ( )ξ is called frequency plot of a harmon All the frequency terms starts from magnification factor of damping ratio ( )ξ The amplitude of vibration at any on the value of damping The amplitude will be higher for higher damping The amplitude of vibration at reso

amplitude tends to zero as

ωω

n tends to infinity

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

Freq

ζ =

ζ = 1

Magnification Factor M = XXst

kFo

51

ωω

nfor varying values of

ically excite vibrating system

M=1 what ever may be value particular frequency depends

for lesser damping and lesser nance will be very high The

uency Ratio r = (ωωn)

ζ = 0 0

ζ = 025 ζ = 0375 ζ = 0 5

ζ = 0707

ζ = 2

52

Plot of phase angle φ versus frequency ratio

ωω

n is shown in the

figure The phase angle φ is zero for all the values of damping ratio

at

ωω

n=0 The phase

angle φ is 90 for all the values of damping ratio

at

ωω

n=1 ie at

resonance The phase angle between o to 90 deg for all the values of

ωω

n= 0 to 1 and

between 90 deg to 180 deg for all the values of

ωω

ngt1

Class work Derive an expression for frequency at peak amplitude

2np 21 ξminusω=ω=ω is the required equation

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

53

Derive expression amplitude at resonance ( In dimensional form) In dimensional form We know that steady state amplitude

( ) ( )

ω+ωminus=

222

o

Cmk

Fx Equation in dimensional form

We know that at resonance mk

n =ω=ω

( )

ω+

minus

=2

n

2

oR

Cmk

mk

Fx

( )

ω=

2n

oR

C

Fx

( )

ω=

n

oR C

Fx

Class work Derive an expression for amplitude at resonance ( in non dimensional form)

Tutorial Problems on Forced Vibration 1A weight of 60N is suspended by a spring stiffness 12 kNm is forced to vibrate by a harmonic force of 10N assuming viscous damping of 0086 kN-sm Find i) resonance frequency ii) amplitude at resonance iii) phase angle at resonance iv) frequency corresponding to peak amplitude 2 A periodic torque having a maximum value of 05888Nm at a frequency of 4 rads is impressed upon a flywheel suspended from a wire The wheel has mass moment of inertia 012 kgm2 and the wire has stiffness 1176 Nmrad A viscous dashpot applies damping couple of 0784 Nm an angular velocity of 2 rads calculate i)maximum angular displacement from rest position ii) maximum couple applied to dashpot Iii) angle by which the angular displacement lags the torque A mass of 1019 kg is suspended from a spring stiffness 20 Nmm viscous damping causes the amplitude to decrease to 110th of initial value in 4 complete cycle If a periodic force of 20 cos(50 t) is applied to the mass Find the amplitude of forced vibration What would be the amplitude of oscillation if the applied force coincides with the natural period

m=1019 kg

[ ] t50cos 30F =

54

55

ROTATING UNBALANCE Unbalance in rotating machinery is a common source of vibration excitation Consider a machine of mass m having mounted on a foundation of stiffness k and damping coefficient C as shown in figure The unbalance is represented by an eccentric rotating mass m with eccentricity (e) and having a angular velocity (ω) let x be the displacement of non rotating mass from the static equilibrium position The displacement of rotating mass m from the static equilibrium is ( ) tsin( ex ω+ Applying Newtonrsquos law of motion we have

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

56

RECIPROCATING UNBALANCE Unbalance in reciprocating machinery is a common source of vibration excitation Consider a machine of mass M having mounted on a foundation of stiffness K and damping coefficient C as shown in figure The unbalance is represented by a reciprocating mass (m) having crank radus (e) and connecting rod of length (l) Let ωωωω be angular velocity of the crank Let x be the displacement of non reciprocating mass from the static equilibrium position at any instant of time The displacement of reciprocating mass m from the static equilibrium position is

)2sin(sin +++ tle

tex ωωωωωωωω

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

57

Note The equations obtained in reciprocating unbalance is same as the rotating unbalance and having the following differences Rotating Unbalance Reciprocating Unbalance Here mass of the rotating element is considered

Here mass of reciprocating element is considered

Here the eccentricity (e) which is the distance between the centre of rotation and CG

Here the eccentricity e is the crank radius and crank radius = stroke length2

The frequency plot of non dimensional factor( ( )me

MX versus frequency ratio

ωω

n for various values of ( )ξξξξ is shown in figure This frequency plot is same for

the rotating and reciprocating unbalance with the following points are noted from the diagram

1 Damping ratio plays very important role during resonance in controlling the amplitude

2 When

ωω

n=0 then ( )me

MX =0 for all the values of damping ratio ( )ξξξξ

3 When

ωω

nis higher than ( )me

MX tends to unity

4 At very high speeds ie at very high frequency ratio

ωω

n there is no effect

of damping The frequency plot of phase angle versus the frequency ratio is given as shown in figure The explanation is same as in forced harmonic vibration

58

VIBRATION ISOLATION

Vibratory forces and motions generated by machine and other caused are unavoidable Howeever their effect on the other dynamical systems can be minimized by proper isolator design An isolation system is either to protect o delicate machine from excessive vibration transmitted to it from its supporting structure or to prevent vibratory forces and motions generated by machines from being transmitted to its surroundings The effectiveness of isolation may be measured in terms of forces and motion transmitted The first type is known as force isolation and the second type is known as motion isolation

Force isolation (Force Transmissibility) When vibratory forces generated by machine are isolated the effectiveness of forced isolation can be made by defining the term transmissibility The term transmissibility is defined as the ratio of force transmitted to the foundation to that of the external impressed force

Force Transmissibility = o

tr

FF

force Impressed Externalfoundation toted transmitForce

=

Consider viscous damped spring mass system as shown in figure Upon this system an external harmonic force ) t (Sin Fo ω is applied The difference forces acting on system are inertia forces damping force spring force and external force All these forces are represented on force polygon shown below Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these two forces

( ) ( )22 xckx ω+=

59

This is the expression for force transmissibility in dimensional form To represent same in the non dimensional form divide numerator and denominator by k

2

n

22

n

2

n

o

tr

21

21

FF

TR

ωω

ξ+

ωω

minus

ωω

ξ+

==

PHASE DIFFERNECE BETWEEN EXTERNAL FORCE AND DISPLACEMENT Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these

ωω

minus

ωω

ξ=

ωminusω

=φ minusminus2

n

n1

2

1

1

2tan

mKC

tan

The phase difference between force transmitted to the foundation Ftr to the displacement

kc

KxxC

tanω

=ω

=α or

ωω

ξ=

ω=α minusminus

n

11 2tanKx

xCtan

The phase difference between external force Fo and force transmitted to foundationis β

60

β= αminusφ =

ωω

minus

ωω

ξminus

2

n

n1

1

2tan -

ωω

ξminus

n

1 2tan

Displacement Transmissibility It is the ratio of amplitude of steady state vibration of machine to that of the base motion Forced vibration due to support motion Consider a spring mass damper system subjected to harmonic excitation )tsin(Yy ω= at the base as shown in figure Let x be the absolute displacement of the mass at any instant of time t The different forces acting on system are shown in free body diagram Applying Newtonrsquos Law of motion Ma = sumF

)yx(k)yx(cxm minusminusminusminus= ampampampamp

0)yx(k)yx(cxm =minus+minus+ ampampampamp

61

Frequency plot of transmissibility versus frequency ratio

62

The plot of transmissibility versus frequency ratio is as shown in figure The following points can be noted from it

i) The value of transmissibility is unity at frequency ratio

ωω

n=0 and 2

ii) For an undamped system ( )ξ =0 transmissibility tends to infin at resonance iii) The value of transmissibility is less than unity for the value of frequency ratio

greater than 2

iv) For frequency ratio

ωω

nlt 2 smaller damping ratio lead to larger values of

transmissibility on the other hand for frequency ratio

ωω

ngt 2 smaller damping

ratio leads to smaller value of transmissibility

The frequency plot of transmissibility (TR) versus frequency ratio

ωω

n can be

broadly divided into two zones

1

ωω

n= 0 to 2

2

ωω

n= 2 to infin

CASE 1

When

ωω

n= 0 to 2

In this zone the value of transmissibility is always greater than 1 for all values of damping ratio In this zone the transmissibility decreases with increase in damping Hence we can conclude that in this zone damping is useful in isolating vibration forces being transmitted to the support CASE 2

When

ωω

n= 2 to infin

In this zone the value of transmissibility is less than 1 for all values of damping ratio In this zone the transmissibility increases with increase in damping Hence we can conclude that in this zone damping will adversely affect in isolating vibratory forces

63

Tutorial problems on Forced Vibration 1 Machine of mass 75kg is mounted on spring of stiffness 12000 Nm with an assumed damping factor ξ =02 piston within the machine of mass 2 kg has a reciprocating motion with a stroke of 75 cm and speed of 15 cyclessecond Assuming the motion of piston to be harmonic determine amplitude of machine ( X) Transmissibility and force transmitted to foundation phase angle of transmitted force wrt to exciting force 2 Motor weighs 600N is mounted on a simple beam that has a spring rate of 400 Ncm The motor armature base 160 N and has an eccentricity of 0005cm What will be the amplitude of motion when it runs at 1760 RPM Neglect damping the deflection of armature shaft and weight of the beam 3 A Heavy machine of mass 300 kg is supported on resilient foundation the static deflection of foundation due to weight of machine is found to be 75mm It is observed that the machine vibrates with an amplitude of 10mm when the base of the foundation is subjected to harmonic oscillation at the undampeds natural frequency of system with an amplitude of 25mm Find the damping constant of foundation Ii) dynamic force amplitude on the base iii) amplitude of the displacement of machine relative to base 4 An air compressor of mass 450 kg operates at a constant speed of 1750 RPM The reciprocating masses are 10 kg the crank radius is 100mm If the damper for mounting introduces a damping factor 015 i) specify the springs of mounting such that only 20 of the unbalanced force is transmitted to the foundation ii) determine the magnitude of the transmitted force bull Repeat the above problem by assuming that there is no damping in system

5 A body of a car having mass 1500 kg and is mounted on 4 equal springs which deflect through 225 cm under the weight of the body A total damping forces the 4 shock absorbers is 46 N at a velocity of 1cmsec The car is placed with all the four wheels on a test platform which is move up and down at resonant speed With an amplitude of 25cm Find the amplitude of the car body on its springs assuming its CG to be in the centre of wheel base 6A 75 kg machine which is mounted on springs of stiffness 116x105 Nm with the damping of 20=ξ A 2kg piston within the machine as reciprocating motion with a stroke of 008m and speed of 3000 RPM Assuming the motion of piston to be harmonic determine the amplitude of vibration of machine

7 The spring of an automobile are compressed 10cm under its own weight Find the critical speed when a trailer is traveling over road with a people approximaterly by a sine wave of amplitude 75 cm and wave length 15m What will be the amplitude of vibration at 64kmhour 8 An eccentric mass excites us used to determine vibratory characteristics of a structure of mass 25kg as given in figure At a speed of 1000 RPM the eccentric mass to beat top position at the instant the structure was moved upwards through its static equilibrium position and corresponding amplitude was 20mm The eccentric mass is 05kg at an eccentricity of 10mm Determine i) Undamped natural frequency of the system ii) Damping ratio iii) Steady state amplitude and phase angle at 1500 RPM iv) Force transmitted to the ground at 1200 RPM

M=25kg

C

K

Eccentric mass

64

65

CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Support motion Consider a spring mass damper system as shown in figure Let it excited by motion

of support let y be the harmonic displacement of support From Newtonrsquos second law of motion

)yx(k)yx(cxm minusminusminusminus= ampampampamp

)1(0)yx(k)yx(cxm minusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminus=minus+minus+ ampampampamp

Let z be the relative displacement of the mass with respect to support then

Z=x-y

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

Is given by [ ] 0UTdtd

=+

From this we get a differential equation of motion as well as natural frequency of the system Determine the natural frequency of spring mass system using energy method

Determine the natural frequency of system shown in figure Determine the natural frequency of the system shown in figure Is there any limitation on the value of K Discuss Determine the natural frequency of s

m

l k

m

a

θ

m

a

θ

l

k

29

k

k

ystem shown below Neglect the mass of ball

l

m

30

A string shown in figure is under tension T which can be assumed to remain constant for small displacements Find the natural frequency of vertical vibrations of spring An acrobat 120kg walks on a tight rope as shown in figure The frequency of vibration in the given position is vertical direction is 30 rads Find the tension in the rope A manometer has a uniform bore of cross section area A If the column of liquid of length L and Density ρ is set into motion as shown in figure Find the frequency of the resulting oscillation

m

l

a

T T

36m 8m

31

k

R

k

m

r

m

Find the expression for natural frequency of system shown in the figure Neglect the mass of the cantilever beam Study the special case i) k=Infinity ii) I = infinity Determine the expression for the natural frequency of a system shown in figure The two discs are keyed to a common shaft and have a combined mass moment of inertia about centre of oscillation O It is assumed that the card attached to mass m does not stretch and is always under tension Determine the natural frequency of a system shown

m

l

32

Determine the expression for the natural frequency of the system shown in figure Assume that the wires connecting the masses do not stretch and are always in tension

Determine the natural frequency of spring mass system taking the MASS OF SPRING (ms ) into account

M3 M1

M2

k2 k1

m

k

33

RAYLEIGHrsquoS METHOD

This is the extension of energy method Here we can determine the natural frequency of a conservative system without finding the equation of motion The natural frequency is a function of the rate of change of Kinetic energy and potential energy of the system From the principle of conservation of energy we have T+U = Constant T= KE and U =PE We can write the above equation as T1+U1 = --------------------------(1) Where 1 and 2 represents tow instances of time Let 1 be the time when the mass is passing through static equilibrium position Let 2 be the time corresponding to the mass displacement of the mass At this instant the velocity f the mass is zero and hence Substituting in equation (1) we get ( )Max = ( )Max Above equation leads directly to natural frequency Determine the natural frequency of spring mass system using Rayleighrsquos method m

34

r

B

K K

Home work

Determine the natural frequency of pendulum using Rayleighrsquos method

A Cylinder of mass m and moment of inertia Jo is to role without slipping but is sustained by spring K as shown in figure Determine the natural frequency of oscillation

Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping

a

Determine the natural frequency of system shown where in cylindrical disc rolls over the inextensible spring A cylinder of radius r rolls without slipping on cylindrical the equation of motion and hence determine the naoscillations about the lowest point shown Repeat the above problem ndash Instead of cylinder assume swithout slipping on concave surface as shown above

r

K

35

portion of radius R Derive tural frequency of small

phere of radius with r rolls

The torsional pendulum has a natural frequency of 5 Hz What length of steel wire of diameter 2 mm should be used for pendulum The inertia of mass fixed at the free end is 00098 kgm2 Take g= 083x1011 Nm2 Determine the torsional natural frequency of the system shown in figure Neglect the mass moment of inertia of the shaft ( Figure not proportionate) Determine the natural frequency of simple pendulum considering the mass of the rod into account

Using the energy method find tThe chord may be assumed ine

Kt1

Kt2

20cm

06m 08m 04m

20cm 20cm 20cm

m

M

he natural frxtensible in

k

M

36

equency of the system shown in figure the spring mass pulley system

A rod of length l and depth H having a mass m rests on a semicylidrical surface os radius r It is tipped slightly and let go Find the frequency of oscillation

A thin bar of l having a mass m rests on a semicylidrical surface radius r It is tipped slightly and let go Find the frequency of oscillation

l

h

R

l

37

38

CHAPTER 3

DAMPED FREE VIBRATION OF SINGLE DEGREE FREEDOM SYSTEM LEARNING OBJECTIVES

] Introduction to damped free vibration

] Terminologies used in damped free vibration

] Three different types of damping

] Concept of critical damping and its importance

] Study of response of viscous damped system for cases of under damping

critical and over damping

] Concept of logarithmic decrement

] Problems related to damped free vibration problems

Viscously damped free vibration of single degree freedom sytem Consider a single degree freedom system with viscous damping as shown in figure For a viscous damped system the damping force is

where c is a constant of proportionality It is called damping coefficient or coefficient of viscous damping

Symbolically it is designated by a dashpot as shown in Fig 3

x xampx

39

At any instant of time let the mass be displaced by x from the equilibrium position Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

Okxxcxm =minusminus= ampampamp

Okxxcxm =++ ampampamp ----------------------------------------------------------(1)

or

0kxdtdx

Cdt

xdM

2

2

=++

Assuming the solution of equation as

The first term in the above equation m2Ct

eminus

is an exponentially decaying function of time The behavior of the terms in the parenthesis however depends on whether numerical value within the radicals is positive zero or negative

CASE 1 if 0mk

m2c

2

=minus

40

CASE 2 if 0mk

m2c

2

=gt

CASE 3 if 0mk

m2c

2

=lt

41

Critically damped coefficient Critically damped coefficient is defined as the value of the damping coeffient for which radical in equation

mk

m2c

m2c

S2

21 minus

plusmn

= becomes zero

Damping Ratio ( Damping factor) ( ξ ) It is defined as the ratio of damping coefficient to the critical damping coefficient Note The damping coefficient c is defined as the actual amount of damping while the damping coefficient is a constant depending upon the mass and stiffness of the system The roots of characteristics equation can be now written as Note if c=cc or ξ =1 then system is critically damped if cgt cc or ξ gt1 then system is over damped I if clt cc or ξ lt1 then system is under damped

42

Take home quiz Derive General solution of different damped system

1 Critically damped system 2 Under damped system 3 Over damped system

Logarithmic Decrement It is used to determine the amount of damping present in system This measure The rate of decay of free vibration The larger the damping the greater will be rate of decay Logarithmic decrement is defined as the natural logarithm of ratio of any two successive amplitudes

=δ

2

1

xx

ln

Expression for logarithmic decrement (δ ) Consider a damped free vibration which is expressed by

( ) ( ) φ+ξminusω= ξω t 1sinXex 2n

tn

x is the displacement at any instant of time X and φ are arbitrary constants

21

2

ξminus

πξ=δ

43

Tutorial problems 1 Derive an expression for damping ratio in terms of logarithmic decrement

2 Show that logarithmic decrement

=δ

n

o

xx

lnn1

where xo is the initial amplitude xn

is amplitude after n cycles 3 A mass of 1 kg is to be supported on a spring having stiffness K= 9800 Nm The damping constant c= 59 Nsm Determine the natural frequency of system Find also logarithmic decrement and amplitude after 3 cycles if initial displacement is 03 cm 4 A vibrating system is defined by the following parameters M = 3kg k= 100 Nm damping coefficient = 3 Nsm determine Damping factor natural frequency damped vibration logarithmic decrement ratio of consecutive amplitude no of cycles after which the original amplitude is reduced to 20 5A single degree damped vibrating system a suspended mass of 18kg makes 15 oscillations in 003 seconds The amplitude decreases to 025 of the initial value after 5 oscillations Determine the stiffness of spring Logarithmic decrement damping factor and damping coefficient 6 A damped vibration record of a spring mass system shows the following data Amplitude at the end of 2nd cycle is 9mm amplitude at eh end of 3rd cycle is 6mm amplitude at the end of 4th cycle is 4mm stiffness of spring =8000 Nm weight =20N Find the logarithmic decrement damping force at unit velocity periodic time of vibration 7 For the system shown in figure the characteristics of dashpot is such that the constant force of 49 N is applied to the piston it velocity is found to be constant at 012 ms Determine the value of C would you expect the complete system to be periodic or aperiodic 8 For a torsional system shown in the figuref1 is natural frequency in air and f2 is the natural frequency when immersed in oil Determine the damping coefficient

Viscous fluid

Disc of mass moment of Inertia J

2a

C

m M

b

a a

k

9 A thin plate of area A and weigh w is fixed to the end of the spring and allow to oscillate in a viscous fluid as shown in figure It f1 is the frequency of oscillation in air and f2 in liquid Show that

22

21 ff

gA2

minusπω

=micro where the damping force on the plate

AV2Fd micro= V being the velocity of vibration 10 A torsional pendulum has a natural frequency of 175 cyclessecond vibrating in vacuum The mass moment of inertia is 2 kgm2 It is immersed in oil and it is observed that the natural frequency is 142 cyclessecond Find the damping torque If the disc is replaced by 34 deg When in oil find the displacement at the end of the first complete cycle 11 An automobile can be modeled as a mass placed on 4 shock absorbers each consisting of a spring and a damper Such that each spring is equally loaded Determine the stiffness damping constant of each shock absorber So that natural frequency is 2Hz and system is critically damped The mass of vehicle is 200kg 12 Write the differential equation of motion for the system shown in the figure below Determine the natural frequency of the damped oscillation and critical damping coefficient

Determine the differential equation of motion and find the critical cdamped natural frequency for system shown in figure

a

C

m M

M =200kg

44

200kg kgkg

oefficient and

45

Derive the equation of motion and find the critical damping coefficient and damped natural frequency for the system shown below Write the differential equation of motion and determine the expression for critical damping coefficient and damped natural frequency of system shown in figure

Set up a differential equation of motion for system shown in figure and determine the following Undamped natural frequencies critical damping coefficient damping ratio damped natural frequency

b

C

kM

a

m

46

A mass of 300N is resting on two springs of stiffness 3000 Nm each and a dashpot of damping coefficient 150 Nsm as shown in figure If the initial velocity of 10cms is given to mass at its equilibrium position what will be the displacement from the equilibrium position at the end of first second Derive the formula used A gun barrel of mass 600kg has a recoil spring of stiffness 294 knm if the barrel recoils 13m on firing determine i) The initial recoil velocity of barrel ii) The critical damping coefficient of dashpot which is engaged at the end of recoil stroke Iii) Time required for the barrel to return at the end of recoil stroke Iv) time required for the barrel to return to a position 5cm from the initial position

W= 3000 N

FIRING

CHAPTER 4

FORCED VIBRATION LEARNING OBJECTIVES

] Introduction to forced free vibration

] Terminologies used in forced vibration

] Concept of magnification factor

] Study of reciprocating and rotating unbalance

] Concept of base excitation and support excitation

] Concept of Transmissibility ratio

] Problems related to forced vibration

The vibration that takes place under the excitation of external forces are called

forced vibration When a system is subjected to harmonic excitation it is forced to vibrate at the same frequency as that of excitation Most of the mechanical systems are prone to steady state forced vibration which cause fatigue failure The vibration problem also becomes very important when any of excitation frequencies coincide with one of natural frequencies of a system Under this condition resonance occurs during which large amplitude of vibrations are observed To avoid resonance external frequency (operating speed) may be changed or properties of the system may be changed to alter the natural frequency In some cases sufficient amount of damping may be provided to avoid large amplitude during resonance Thus the problem of force vibration is very important in mechanical design

FORCED VIBRATIONS OF SINGLE DOF SYSTEMS UNDER HARMONIC EXCITATION

` [ ] tsin FF o ω=

x

sin FF o=

Fo

xampamp

[ ] tω

xamp

47

48

Consider a spring mass system having a viscous damping excited by a harmonic force [ ] tsin FF o ω= as shown in figure The various forces acting on the system is shown in figure Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

O)tsin(Fkxxcxm o =ω+minusminus= ampampamp

)tsin(Fkxxcxm o ω=++ ampampamp ----------------------------------------------------------(1)

or

)tsin(Fkxdtdx

Cdt

xdM o2

2

ω=++

Equation (1) represents the differential equation of motion of the system The solution to this equation consists of two parts namely complementary function and a particular integral

The first of solution is also termed as transient solution of homogeneous solution and can be obtained from the homogeneous differential equation 0kxxcxm =++ ampampamp

The second part of solution particular integral is also termed as steady state solution ad is of form )tsin(Xx φminusω= Where X= steady state amplitude φ is phase difference between external force and displacement

)tsin(Xx φminusω=

The vector representation of the shown

49

The force polygon for the above vector diagram can be written as follows

( )φminusωt O

KX

xm 2ω

Fo

xCω

tω

φ

50

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

is the governing equation

Zero frequency deflection (Xst) It is deflection observed at zero frequency ( when the external frequency 0=ω

( ) kF

001

kF

x o

22

o

st =

+minus=

Magnification factor (M) It is defined as the ratio of steady state amplitude to zero frequency deflection xst

M = arX

X=

A plot of magnification factor versus frequency ratio

damping factor ( )ξ is called frequency plot of a harmon All the frequency terms starts from magnification factor of damping ratio ( )ξ The amplitude of vibration at any on the value of damping The amplitude will be higher for higher damping The amplitude of vibration at reso

amplitude tends to zero as

ωω

n tends to infinity

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

Freq

ζ =

ζ = 1

Magnification Factor M = XXst

kFo

51

ωω

nfor varying values of

ically excite vibrating system

M=1 what ever may be value particular frequency depends

for lesser damping and lesser nance will be very high The

uency Ratio r = (ωωn)

ζ = 0 0

ζ = 025 ζ = 0375 ζ = 0 5

ζ = 0707

ζ = 2

52

Plot of phase angle φ versus frequency ratio

ωω

n is shown in the

figure The phase angle φ is zero for all the values of damping ratio

at

ωω

n=0 The phase

angle φ is 90 for all the values of damping ratio

at

ωω

n=1 ie at

resonance The phase angle between o to 90 deg for all the values of

ωω

n= 0 to 1 and

between 90 deg to 180 deg for all the values of

ωω

ngt1

Class work Derive an expression for frequency at peak amplitude

2np 21 ξminusω=ω=ω is the required equation

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

53

Derive expression amplitude at resonance ( In dimensional form) In dimensional form We know that steady state amplitude

( ) ( )

ω+ωminus=

222

o

Cmk

Fx Equation in dimensional form

We know that at resonance mk

n =ω=ω

( )

ω+

minus

=2

n

2

oR

Cmk

mk

Fx

( )

ω=

2n

oR

C

Fx

( )

ω=

n

oR C

Fx

Class work Derive an expression for amplitude at resonance ( in non dimensional form)

Tutorial Problems on Forced Vibration 1A weight of 60N is suspended by a spring stiffness 12 kNm is forced to vibrate by a harmonic force of 10N assuming viscous damping of 0086 kN-sm Find i) resonance frequency ii) amplitude at resonance iii) phase angle at resonance iv) frequency corresponding to peak amplitude 2 A periodic torque having a maximum value of 05888Nm at a frequency of 4 rads is impressed upon a flywheel suspended from a wire The wheel has mass moment of inertia 012 kgm2 and the wire has stiffness 1176 Nmrad A viscous dashpot applies damping couple of 0784 Nm an angular velocity of 2 rads calculate i)maximum angular displacement from rest position ii) maximum couple applied to dashpot Iii) angle by which the angular displacement lags the torque A mass of 1019 kg is suspended from a spring stiffness 20 Nmm viscous damping causes the amplitude to decrease to 110th of initial value in 4 complete cycle If a periodic force of 20 cos(50 t) is applied to the mass Find the amplitude of forced vibration What would be the amplitude of oscillation if the applied force coincides with the natural period

m=1019 kg

[ ] t50cos 30F =

54

55

ROTATING UNBALANCE Unbalance in rotating machinery is a common source of vibration excitation Consider a machine of mass m having mounted on a foundation of stiffness k and damping coefficient C as shown in figure The unbalance is represented by an eccentric rotating mass m with eccentricity (e) and having a angular velocity (ω) let x be the displacement of non rotating mass from the static equilibrium position The displacement of rotating mass m from the static equilibrium is ( ) tsin( ex ω+ Applying Newtonrsquos law of motion we have

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

56

RECIPROCATING UNBALANCE Unbalance in reciprocating machinery is a common source of vibration excitation Consider a machine of mass M having mounted on a foundation of stiffness K and damping coefficient C as shown in figure The unbalance is represented by a reciprocating mass (m) having crank radus (e) and connecting rod of length (l) Let ωωωω be angular velocity of the crank Let x be the displacement of non reciprocating mass from the static equilibrium position at any instant of time The displacement of reciprocating mass m from the static equilibrium position is

)2sin(sin +++ tle

tex ωωωωωωωω

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

57

Note The equations obtained in reciprocating unbalance is same as the rotating unbalance and having the following differences Rotating Unbalance Reciprocating Unbalance Here mass of the rotating element is considered

Here mass of reciprocating element is considered

Here the eccentricity (e) which is the distance between the centre of rotation and CG

Here the eccentricity e is the crank radius and crank radius = stroke length2

The frequency plot of non dimensional factor( ( )me

MX versus frequency ratio

ωω

n for various values of ( )ξξξξ is shown in figure This frequency plot is same for

the rotating and reciprocating unbalance with the following points are noted from the diagram

1 Damping ratio plays very important role during resonance in controlling the amplitude

2 When

ωω

n=0 then ( )me

MX =0 for all the values of damping ratio ( )ξξξξ

3 When

ωω

nis higher than ( )me

MX tends to unity

4 At very high speeds ie at very high frequency ratio

ωω

n there is no effect

of damping The frequency plot of phase angle versus the frequency ratio is given as shown in figure The explanation is same as in forced harmonic vibration

58

VIBRATION ISOLATION

Vibratory forces and motions generated by machine and other caused are unavoidable Howeever their effect on the other dynamical systems can be minimized by proper isolator design An isolation system is either to protect o delicate machine from excessive vibration transmitted to it from its supporting structure or to prevent vibratory forces and motions generated by machines from being transmitted to its surroundings The effectiveness of isolation may be measured in terms of forces and motion transmitted The first type is known as force isolation and the second type is known as motion isolation

Force isolation (Force Transmissibility) When vibratory forces generated by machine are isolated the effectiveness of forced isolation can be made by defining the term transmissibility The term transmissibility is defined as the ratio of force transmitted to the foundation to that of the external impressed force

Force Transmissibility = o

tr

FF

force Impressed Externalfoundation toted transmitForce

=

Consider viscous damped spring mass system as shown in figure Upon this system an external harmonic force ) t (Sin Fo ω is applied The difference forces acting on system are inertia forces damping force spring force and external force All these forces are represented on force polygon shown below Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these two forces

( ) ( )22 xckx ω+=

59

This is the expression for force transmissibility in dimensional form To represent same in the non dimensional form divide numerator and denominator by k

2

n

22

n

2

n

o

tr

21

21

FF

TR

ωω

ξ+

ωω

minus

ωω

ξ+

==

PHASE DIFFERNECE BETWEEN EXTERNAL FORCE AND DISPLACEMENT Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these

ωω

minus

ωω

ξ=

ωminusω

=φ minusminus2

n

n1

2

1

1

2tan

mKC

tan

The phase difference between force transmitted to the foundation Ftr to the displacement

kc

KxxC

tanω

=ω

=α or

ωω

ξ=

ω=α minusminus

n

11 2tanKx

xCtan

The phase difference between external force Fo and force transmitted to foundationis β

60

β= αminusφ =

ωω

minus

ωω

ξminus

2

n

n1

1

2tan -

ωω

ξminus

n

1 2tan

Displacement Transmissibility It is the ratio of amplitude of steady state vibration of machine to that of the base motion Forced vibration due to support motion Consider a spring mass damper system subjected to harmonic excitation )tsin(Yy ω= at the base as shown in figure Let x be the absolute displacement of the mass at any instant of time t The different forces acting on system are shown in free body diagram Applying Newtonrsquos Law of motion Ma = sumF

)yx(k)yx(cxm minusminusminusminus= ampampampamp

0)yx(k)yx(cxm =minus+minus+ ampampampamp

61

Frequency plot of transmissibility versus frequency ratio

62

The plot of transmissibility versus frequency ratio is as shown in figure The following points can be noted from it

i) The value of transmissibility is unity at frequency ratio

ωω

n=0 and 2

ii) For an undamped system ( )ξ =0 transmissibility tends to infin at resonance iii) The value of transmissibility is less than unity for the value of frequency ratio

greater than 2

iv) For frequency ratio

ωω

nlt 2 smaller damping ratio lead to larger values of

transmissibility on the other hand for frequency ratio

ωω

ngt 2 smaller damping

ratio leads to smaller value of transmissibility

The frequency plot of transmissibility (TR) versus frequency ratio

ωω

n can be

broadly divided into two zones

1

ωω

n= 0 to 2

2

ωω

n= 2 to infin

CASE 1

When

ωω

n= 0 to 2

In this zone the value of transmissibility is always greater than 1 for all values of damping ratio In this zone the transmissibility decreases with increase in damping Hence we can conclude that in this zone damping is useful in isolating vibration forces being transmitted to the support CASE 2

When

ωω

n= 2 to infin

In this zone the value of transmissibility is less than 1 for all values of damping ratio In this zone the transmissibility increases with increase in damping Hence we can conclude that in this zone damping will adversely affect in isolating vibratory forces

63

Tutorial problems on Forced Vibration 1 Machine of mass 75kg is mounted on spring of stiffness 12000 Nm with an assumed damping factor ξ =02 piston within the machine of mass 2 kg has a reciprocating motion with a stroke of 75 cm and speed of 15 cyclessecond Assuming the motion of piston to be harmonic determine amplitude of machine ( X) Transmissibility and force transmitted to foundation phase angle of transmitted force wrt to exciting force 2 Motor weighs 600N is mounted on a simple beam that has a spring rate of 400 Ncm The motor armature base 160 N and has an eccentricity of 0005cm What will be the amplitude of motion when it runs at 1760 RPM Neglect damping the deflection of armature shaft and weight of the beam 3 A Heavy machine of mass 300 kg is supported on resilient foundation the static deflection of foundation due to weight of machine is found to be 75mm It is observed that the machine vibrates with an amplitude of 10mm when the base of the foundation is subjected to harmonic oscillation at the undampeds natural frequency of system with an amplitude of 25mm Find the damping constant of foundation Ii) dynamic force amplitude on the base iii) amplitude of the displacement of machine relative to base 4 An air compressor of mass 450 kg operates at a constant speed of 1750 RPM The reciprocating masses are 10 kg the crank radius is 100mm If the damper for mounting introduces a damping factor 015 i) specify the springs of mounting such that only 20 of the unbalanced force is transmitted to the foundation ii) determine the magnitude of the transmitted force bull Repeat the above problem by assuming that there is no damping in system

5 A body of a car having mass 1500 kg and is mounted on 4 equal springs which deflect through 225 cm under the weight of the body A total damping forces the 4 shock absorbers is 46 N at a velocity of 1cmsec The car is placed with all the four wheels on a test platform which is move up and down at resonant speed With an amplitude of 25cm Find the amplitude of the car body on its springs assuming its CG to be in the centre of wheel base 6A 75 kg machine which is mounted on springs of stiffness 116x105 Nm with the damping of 20=ξ A 2kg piston within the machine as reciprocating motion with a stroke of 008m and speed of 3000 RPM Assuming the motion of piston to be harmonic determine the amplitude of vibration of machine

7 The spring of an automobile are compressed 10cm under its own weight Find the critical speed when a trailer is traveling over road with a people approximaterly by a sine wave of amplitude 75 cm and wave length 15m What will be the amplitude of vibration at 64kmhour 8 An eccentric mass excites us used to determine vibratory characteristics of a structure of mass 25kg as given in figure At a speed of 1000 RPM the eccentric mass to beat top position at the instant the structure was moved upwards through its static equilibrium position and corresponding amplitude was 20mm The eccentric mass is 05kg at an eccentricity of 10mm Determine i) Undamped natural frequency of the system ii) Damping ratio iii) Steady state amplitude and phase angle at 1500 RPM iv) Force transmitted to the ground at 1200 RPM

M=25kg

C

K

Eccentric mass

64

65

CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Support motion Consider a spring mass damper system as shown in figure Let it excited by motion

of support let y be the harmonic displacement of support From Newtonrsquos second law of motion

)yx(k)yx(cxm minusminusminusminus= ampampampamp

)1(0)yx(k)yx(cxm minusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminus=minus+minus+ ampampampamp

Let z be the relative displacement of the mass with respect to support then

Z=x-y

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

30

A string shown in figure is under tension T which can be assumed to remain constant for small displacements Find the natural frequency of vertical vibrations of spring An acrobat 120kg walks on a tight rope as shown in figure The frequency of vibration in the given position is vertical direction is 30 rads Find the tension in the rope A manometer has a uniform bore of cross section area A If the column of liquid of length L and Density ρ is set into motion as shown in figure Find the frequency of the resulting oscillation

m

l

a

T T

36m 8m

31

k

R

k

m

r

m

Find the expression for natural frequency of system shown in the figure Neglect the mass of the cantilever beam Study the special case i) k=Infinity ii) I = infinity Determine the expression for the natural frequency of a system shown in figure The two discs are keyed to a common shaft and have a combined mass moment of inertia about centre of oscillation O It is assumed that the card attached to mass m does not stretch and is always under tension Determine the natural frequency of a system shown

m

l

32

Determine the expression for the natural frequency of the system shown in figure Assume that the wires connecting the masses do not stretch and are always in tension

Determine the natural frequency of spring mass system taking the MASS OF SPRING (ms ) into account

M3 M1

M2

k2 k1

m

k

33

RAYLEIGHrsquoS METHOD

This is the extension of energy method Here we can determine the natural frequency of a conservative system without finding the equation of motion The natural frequency is a function of the rate of change of Kinetic energy and potential energy of the system From the principle of conservation of energy we have T+U = Constant T= KE and U =PE We can write the above equation as T1+U1 = --------------------------(1) Where 1 and 2 represents tow instances of time Let 1 be the time when the mass is passing through static equilibrium position Let 2 be the time corresponding to the mass displacement of the mass At this instant the velocity f the mass is zero and hence Substituting in equation (1) we get ( )Max = ( )Max Above equation leads directly to natural frequency Determine the natural frequency of spring mass system using Rayleighrsquos method m

34

r

B

K K

Home work

Determine the natural frequency of pendulum using Rayleighrsquos method

A Cylinder of mass m and moment of inertia Jo is to role without slipping but is sustained by spring K as shown in figure Determine the natural frequency of oscillation

Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping

a

Determine the natural frequency of system shown where in cylindrical disc rolls over the inextensible spring A cylinder of radius r rolls without slipping on cylindrical the equation of motion and hence determine the naoscillations about the lowest point shown Repeat the above problem ndash Instead of cylinder assume swithout slipping on concave surface as shown above

r

K

35

portion of radius R Derive tural frequency of small

phere of radius with r rolls

The torsional pendulum has a natural frequency of 5 Hz What length of steel wire of diameter 2 mm should be used for pendulum The inertia of mass fixed at the free end is 00098 kgm2 Take g= 083x1011 Nm2 Determine the torsional natural frequency of the system shown in figure Neglect the mass moment of inertia of the shaft ( Figure not proportionate) Determine the natural frequency of simple pendulum considering the mass of the rod into account

Using the energy method find tThe chord may be assumed ine

Kt1

Kt2

20cm

06m 08m 04m

20cm 20cm 20cm

m

M

he natural frxtensible in

k

M

36

equency of the system shown in figure the spring mass pulley system

A rod of length l and depth H having a mass m rests on a semicylidrical surface os radius r It is tipped slightly and let go Find the frequency of oscillation

A thin bar of l having a mass m rests on a semicylidrical surface radius r It is tipped slightly and let go Find the frequency of oscillation

l

h

R

l

37

38

CHAPTER 3

DAMPED FREE VIBRATION OF SINGLE DEGREE FREEDOM SYSTEM LEARNING OBJECTIVES

] Introduction to damped free vibration

] Terminologies used in damped free vibration

] Three different types of damping

] Concept of critical damping and its importance

] Study of response of viscous damped system for cases of under damping

critical and over damping

] Concept of logarithmic decrement

] Problems related to damped free vibration problems

Viscously damped free vibration of single degree freedom sytem Consider a single degree freedom system with viscous damping as shown in figure For a viscous damped system the damping force is

where c is a constant of proportionality It is called damping coefficient or coefficient of viscous damping

Symbolically it is designated by a dashpot as shown in Fig 3

x xampx

39

At any instant of time let the mass be displaced by x from the equilibrium position Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

Okxxcxm =minusminus= ampampamp

Okxxcxm =++ ampampamp ----------------------------------------------------------(1)

or

0kxdtdx

Cdt

xdM

2

2

=++

Assuming the solution of equation as

The first term in the above equation m2Ct

eminus

is an exponentially decaying function of time The behavior of the terms in the parenthesis however depends on whether numerical value within the radicals is positive zero or negative

CASE 1 if 0mk

m2c

2

=minus

40

CASE 2 if 0mk

m2c

2

=gt

CASE 3 if 0mk

m2c

2

=lt

41

Critically damped coefficient Critically damped coefficient is defined as the value of the damping coeffient for which radical in equation

mk

m2c

m2c

S2

21 minus

plusmn

= becomes zero

Damping Ratio ( Damping factor) ( ξ ) It is defined as the ratio of damping coefficient to the critical damping coefficient Note The damping coefficient c is defined as the actual amount of damping while the damping coefficient is a constant depending upon the mass and stiffness of the system The roots of characteristics equation can be now written as Note if c=cc or ξ =1 then system is critically damped if cgt cc or ξ gt1 then system is over damped I if clt cc or ξ lt1 then system is under damped

42

Take home quiz Derive General solution of different damped system

1 Critically damped system 2 Under damped system 3 Over damped system

Logarithmic Decrement It is used to determine the amount of damping present in system This measure The rate of decay of free vibration The larger the damping the greater will be rate of decay Logarithmic decrement is defined as the natural logarithm of ratio of any two successive amplitudes

=δ

2

1

xx

ln

Expression for logarithmic decrement (δ ) Consider a damped free vibration which is expressed by

( ) ( ) φ+ξminusω= ξω t 1sinXex 2n

tn

x is the displacement at any instant of time X and φ are arbitrary constants

21

2

ξminus

πξ=δ

43

Tutorial problems 1 Derive an expression for damping ratio in terms of logarithmic decrement

2 Show that logarithmic decrement

=δ

n

o

xx

lnn1

where xo is the initial amplitude xn

is amplitude after n cycles 3 A mass of 1 kg is to be supported on a spring having stiffness K= 9800 Nm The damping constant c= 59 Nsm Determine the natural frequency of system Find also logarithmic decrement and amplitude after 3 cycles if initial displacement is 03 cm 4 A vibrating system is defined by the following parameters M = 3kg k= 100 Nm damping coefficient = 3 Nsm determine Damping factor natural frequency damped vibration logarithmic decrement ratio of consecutive amplitude no of cycles after which the original amplitude is reduced to 20 5A single degree damped vibrating system a suspended mass of 18kg makes 15 oscillations in 003 seconds The amplitude decreases to 025 of the initial value after 5 oscillations Determine the stiffness of spring Logarithmic decrement damping factor and damping coefficient 6 A damped vibration record of a spring mass system shows the following data Amplitude at the end of 2nd cycle is 9mm amplitude at eh end of 3rd cycle is 6mm amplitude at the end of 4th cycle is 4mm stiffness of spring =8000 Nm weight =20N Find the logarithmic decrement damping force at unit velocity periodic time of vibration 7 For the system shown in figure the characteristics of dashpot is such that the constant force of 49 N is applied to the piston it velocity is found to be constant at 012 ms Determine the value of C would you expect the complete system to be periodic or aperiodic 8 For a torsional system shown in the figuref1 is natural frequency in air and f2 is the natural frequency when immersed in oil Determine the damping coefficient

Viscous fluid

Disc of mass moment of Inertia J

2a

C

m M

b

a a

k

9 A thin plate of area A and weigh w is fixed to the end of the spring and allow to oscillate in a viscous fluid as shown in figure It f1 is the frequency of oscillation in air and f2 in liquid Show that

22

21 ff

gA2

minusπω

=micro where the damping force on the plate

AV2Fd micro= V being the velocity of vibration 10 A torsional pendulum has a natural frequency of 175 cyclessecond vibrating in vacuum The mass moment of inertia is 2 kgm2 It is immersed in oil and it is observed that the natural frequency is 142 cyclessecond Find the damping torque If the disc is replaced by 34 deg When in oil find the displacement at the end of the first complete cycle 11 An automobile can be modeled as a mass placed on 4 shock absorbers each consisting of a spring and a damper Such that each spring is equally loaded Determine the stiffness damping constant of each shock absorber So that natural frequency is 2Hz and system is critically damped The mass of vehicle is 200kg 12 Write the differential equation of motion for the system shown in the figure below Determine the natural frequency of the damped oscillation and critical damping coefficient

Determine the differential equation of motion and find the critical cdamped natural frequency for system shown in figure

a

C

m M

M =200kg

44

200kg kgkg

oefficient and

45

Derive the equation of motion and find the critical damping coefficient and damped natural frequency for the system shown below Write the differential equation of motion and determine the expression for critical damping coefficient and damped natural frequency of system shown in figure

Set up a differential equation of motion for system shown in figure and determine the following Undamped natural frequencies critical damping coefficient damping ratio damped natural frequency

b

C

kM

a

m

46

A mass of 300N is resting on two springs of stiffness 3000 Nm each and a dashpot of damping coefficient 150 Nsm as shown in figure If the initial velocity of 10cms is given to mass at its equilibrium position what will be the displacement from the equilibrium position at the end of first second Derive the formula used A gun barrel of mass 600kg has a recoil spring of stiffness 294 knm if the barrel recoils 13m on firing determine i) The initial recoil velocity of barrel ii) The critical damping coefficient of dashpot which is engaged at the end of recoil stroke Iii) Time required for the barrel to return at the end of recoil stroke Iv) time required for the barrel to return to a position 5cm from the initial position

W= 3000 N

FIRING

CHAPTER 4

FORCED VIBRATION LEARNING OBJECTIVES

] Introduction to forced free vibration

] Terminologies used in forced vibration

] Concept of magnification factor

] Study of reciprocating and rotating unbalance

] Concept of base excitation and support excitation

] Concept of Transmissibility ratio

] Problems related to forced vibration

The vibration that takes place under the excitation of external forces are called

forced vibration When a system is subjected to harmonic excitation it is forced to vibrate at the same frequency as that of excitation Most of the mechanical systems are prone to steady state forced vibration which cause fatigue failure The vibration problem also becomes very important when any of excitation frequencies coincide with one of natural frequencies of a system Under this condition resonance occurs during which large amplitude of vibrations are observed To avoid resonance external frequency (operating speed) may be changed or properties of the system may be changed to alter the natural frequency In some cases sufficient amount of damping may be provided to avoid large amplitude during resonance Thus the problem of force vibration is very important in mechanical design

FORCED VIBRATIONS OF SINGLE DOF SYSTEMS UNDER HARMONIC EXCITATION

` [ ] tsin FF o ω=

x

sin FF o=

Fo

xampamp

[ ] tω

xamp

47

48

Consider a spring mass system having a viscous damping excited by a harmonic force [ ] tsin FF o ω= as shown in figure The various forces acting on the system is shown in figure Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

O)tsin(Fkxxcxm o =ω+minusminus= ampampamp

)tsin(Fkxxcxm o ω=++ ampampamp ----------------------------------------------------------(1)

or

)tsin(Fkxdtdx

Cdt

xdM o2

2

ω=++

Equation (1) represents the differential equation of motion of the system The solution to this equation consists of two parts namely complementary function and a particular integral

The first of solution is also termed as transient solution of homogeneous solution and can be obtained from the homogeneous differential equation 0kxxcxm =++ ampampamp

The second part of solution particular integral is also termed as steady state solution ad is of form )tsin(Xx φminusω= Where X= steady state amplitude φ is phase difference between external force and displacement

)tsin(Xx φminusω=

The vector representation of the shown

49

The force polygon for the above vector diagram can be written as follows

( )φminusωt O

KX

xm 2ω

Fo

xCω

tω

φ

50

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

is the governing equation

Zero frequency deflection (Xst) It is deflection observed at zero frequency ( when the external frequency 0=ω

( ) kF

001

kF

x o

22

o

st =

+minus=

Magnification factor (M) It is defined as the ratio of steady state amplitude to zero frequency deflection xst

M = arX

X=

A plot of magnification factor versus frequency ratio

damping factor ( )ξ is called frequency plot of a harmon All the frequency terms starts from magnification factor of damping ratio ( )ξ The amplitude of vibration at any on the value of damping The amplitude will be higher for higher damping The amplitude of vibration at reso

amplitude tends to zero as

ωω

n tends to infinity

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

Freq

ζ =

ζ = 1

Magnification Factor M = XXst

kFo

51

ωω

nfor varying values of

ically excite vibrating system

M=1 what ever may be value particular frequency depends

for lesser damping and lesser nance will be very high The

uency Ratio r = (ωωn)

ζ = 0 0

ζ = 025 ζ = 0375 ζ = 0 5

ζ = 0707

ζ = 2

52

Plot of phase angle φ versus frequency ratio

ωω

n is shown in the

figure The phase angle φ is zero for all the values of damping ratio

at

ωω

n=0 The phase

angle φ is 90 for all the values of damping ratio

at

ωω

n=1 ie at

resonance The phase angle between o to 90 deg for all the values of

ωω

n= 0 to 1 and

between 90 deg to 180 deg for all the values of

ωω

ngt1

Class work Derive an expression for frequency at peak amplitude

2np 21 ξminusω=ω=ω is the required equation

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

53

Derive expression amplitude at resonance ( In dimensional form) In dimensional form We know that steady state amplitude

( ) ( )

ω+ωminus=

222

o

Cmk

Fx Equation in dimensional form

We know that at resonance mk

n =ω=ω

( )

ω+

minus

=2

n

2

oR

Cmk

mk

Fx

( )

ω=

2n

oR

C

Fx

( )

ω=

n

oR C

Fx

Class work Derive an expression for amplitude at resonance ( in non dimensional form)

Tutorial Problems on Forced Vibration 1A weight of 60N is suspended by a spring stiffness 12 kNm is forced to vibrate by a harmonic force of 10N assuming viscous damping of 0086 kN-sm Find i) resonance frequency ii) amplitude at resonance iii) phase angle at resonance iv) frequency corresponding to peak amplitude 2 A periodic torque having a maximum value of 05888Nm at a frequency of 4 rads is impressed upon a flywheel suspended from a wire The wheel has mass moment of inertia 012 kgm2 and the wire has stiffness 1176 Nmrad A viscous dashpot applies damping couple of 0784 Nm an angular velocity of 2 rads calculate i)maximum angular displacement from rest position ii) maximum couple applied to dashpot Iii) angle by which the angular displacement lags the torque A mass of 1019 kg is suspended from a spring stiffness 20 Nmm viscous damping causes the amplitude to decrease to 110th of initial value in 4 complete cycle If a periodic force of 20 cos(50 t) is applied to the mass Find the amplitude of forced vibration What would be the amplitude of oscillation if the applied force coincides with the natural period

m=1019 kg

[ ] t50cos 30F =

54

55

ROTATING UNBALANCE Unbalance in rotating machinery is a common source of vibration excitation Consider a machine of mass m having mounted on a foundation of stiffness k and damping coefficient C as shown in figure The unbalance is represented by an eccentric rotating mass m with eccentricity (e) and having a angular velocity (ω) let x be the displacement of non rotating mass from the static equilibrium position The displacement of rotating mass m from the static equilibrium is ( ) tsin( ex ω+ Applying Newtonrsquos law of motion we have

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

56

RECIPROCATING UNBALANCE Unbalance in reciprocating machinery is a common source of vibration excitation Consider a machine of mass M having mounted on a foundation of stiffness K and damping coefficient C as shown in figure The unbalance is represented by a reciprocating mass (m) having crank radus (e) and connecting rod of length (l) Let ωωωω be angular velocity of the crank Let x be the displacement of non reciprocating mass from the static equilibrium position at any instant of time The displacement of reciprocating mass m from the static equilibrium position is

)2sin(sin +++ tle

tex ωωωωωωωω

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

57

Note The equations obtained in reciprocating unbalance is same as the rotating unbalance and having the following differences Rotating Unbalance Reciprocating Unbalance Here mass of the rotating element is considered

Here mass of reciprocating element is considered

Here the eccentricity (e) which is the distance between the centre of rotation and CG

Here the eccentricity e is the crank radius and crank radius = stroke length2

The frequency plot of non dimensional factor( ( )me

MX versus frequency ratio

ωω

n for various values of ( )ξξξξ is shown in figure This frequency plot is same for

the rotating and reciprocating unbalance with the following points are noted from the diagram

1 Damping ratio plays very important role during resonance in controlling the amplitude

2 When

ωω

n=0 then ( )me

MX =0 for all the values of damping ratio ( )ξξξξ

3 When

ωω

nis higher than ( )me

MX tends to unity

4 At very high speeds ie at very high frequency ratio

ωω

n there is no effect

of damping The frequency plot of phase angle versus the frequency ratio is given as shown in figure The explanation is same as in forced harmonic vibration

58

VIBRATION ISOLATION

Vibratory forces and motions generated by machine and other caused are unavoidable Howeever their effect on the other dynamical systems can be minimized by proper isolator design An isolation system is either to protect o delicate machine from excessive vibration transmitted to it from its supporting structure or to prevent vibratory forces and motions generated by machines from being transmitted to its surroundings The effectiveness of isolation may be measured in terms of forces and motion transmitted The first type is known as force isolation and the second type is known as motion isolation

Force isolation (Force Transmissibility) When vibratory forces generated by machine are isolated the effectiveness of forced isolation can be made by defining the term transmissibility The term transmissibility is defined as the ratio of force transmitted to the foundation to that of the external impressed force

Force Transmissibility = o

tr

FF

force Impressed Externalfoundation toted transmitForce

=

Consider viscous damped spring mass system as shown in figure Upon this system an external harmonic force ) t (Sin Fo ω is applied The difference forces acting on system are inertia forces damping force spring force and external force All these forces are represented on force polygon shown below Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these two forces

( ) ( )22 xckx ω+=

59

This is the expression for force transmissibility in dimensional form To represent same in the non dimensional form divide numerator and denominator by k

2

n

22

n

2

n

o

tr

21

21

FF

TR

ωω

ξ+

ωω

minus

ωω

ξ+

==

PHASE DIFFERNECE BETWEEN EXTERNAL FORCE AND DISPLACEMENT Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these

ωω

minus

ωω

ξ=

ωminusω

=φ minusminus2

n

n1

2

1

1

2tan

mKC

tan

The phase difference between force transmitted to the foundation Ftr to the displacement

kc

KxxC

tanω

=ω

=α or

ωω

ξ=

ω=α minusminus

n

11 2tanKx

xCtan

The phase difference between external force Fo and force transmitted to foundationis β

60

β= αminusφ =

ωω

minus

ωω

ξminus

2

n

n1

1

2tan -

ωω

ξminus

n

1 2tan

Displacement Transmissibility It is the ratio of amplitude of steady state vibration of machine to that of the base motion Forced vibration due to support motion Consider a spring mass damper system subjected to harmonic excitation )tsin(Yy ω= at the base as shown in figure Let x be the absolute displacement of the mass at any instant of time t The different forces acting on system are shown in free body diagram Applying Newtonrsquos Law of motion Ma = sumF

)yx(k)yx(cxm minusminusminusminus= ampampampamp

0)yx(k)yx(cxm =minus+minus+ ampampampamp

61

Frequency plot of transmissibility versus frequency ratio

62

The plot of transmissibility versus frequency ratio is as shown in figure The following points can be noted from it

i) The value of transmissibility is unity at frequency ratio

ωω

n=0 and 2

ii) For an undamped system ( )ξ =0 transmissibility tends to infin at resonance iii) The value of transmissibility is less than unity for the value of frequency ratio

greater than 2

iv) For frequency ratio

ωω

nlt 2 smaller damping ratio lead to larger values of

transmissibility on the other hand for frequency ratio

ωω

ngt 2 smaller damping

ratio leads to smaller value of transmissibility

The frequency plot of transmissibility (TR) versus frequency ratio

ωω

n can be

broadly divided into two zones

1

ωω

n= 0 to 2

2

ωω

n= 2 to infin

CASE 1

When

ωω

n= 0 to 2

In this zone the value of transmissibility is always greater than 1 for all values of damping ratio In this zone the transmissibility decreases with increase in damping Hence we can conclude that in this zone damping is useful in isolating vibration forces being transmitted to the support CASE 2

When

ωω

n= 2 to infin

In this zone the value of transmissibility is less than 1 for all values of damping ratio In this zone the transmissibility increases with increase in damping Hence we can conclude that in this zone damping will adversely affect in isolating vibratory forces

63

Tutorial problems on Forced Vibration 1 Machine of mass 75kg is mounted on spring of stiffness 12000 Nm with an assumed damping factor ξ =02 piston within the machine of mass 2 kg has a reciprocating motion with a stroke of 75 cm and speed of 15 cyclessecond Assuming the motion of piston to be harmonic determine amplitude of machine ( X) Transmissibility and force transmitted to foundation phase angle of transmitted force wrt to exciting force 2 Motor weighs 600N is mounted on a simple beam that has a spring rate of 400 Ncm The motor armature base 160 N and has an eccentricity of 0005cm What will be the amplitude of motion when it runs at 1760 RPM Neglect damping the deflection of armature shaft and weight of the beam 3 A Heavy machine of mass 300 kg is supported on resilient foundation the static deflection of foundation due to weight of machine is found to be 75mm It is observed that the machine vibrates with an amplitude of 10mm when the base of the foundation is subjected to harmonic oscillation at the undampeds natural frequency of system with an amplitude of 25mm Find the damping constant of foundation Ii) dynamic force amplitude on the base iii) amplitude of the displacement of machine relative to base 4 An air compressor of mass 450 kg operates at a constant speed of 1750 RPM The reciprocating masses are 10 kg the crank radius is 100mm If the damper for mounting introduces a damping factor 015 i) specify the springs of mounting such that only 20 of the unbalanced force is transmitted to the foundation ii) determine the magnitude of the transmitted force bull Repeat the above problem by assuming that there is no damping in system

5 A body of a car having mass 1500 kg and is mounted on 4 equal springs which deflect through 225 cm under the weight of the body A total damping forces the 4 shock absorbers is 46 N at a velocity of 1cmsec The car is placed with all the four wheels on a test platform which is move up and down at resonant speed With an amplitude of 25cm Find the amplitude of the car body on its springs assuming its CG to be in the centre of wheel base 6A 75 kg machine which is mounted on springs of stiffness 116x105 Nm with the damping of 20=ξ A 2kg piston within the machine as reciprocating motion with a stroke of 008m and speed of 3000 RPM Assuming the motion of piston to be harmonic determine the amplitude of vibration of machine

7 The spring of an automobile are compressed 10cm under its own weight Find the critical speed when a trailer is traveling over road with a people approximaterly by a sine wave of amplitude 75 cm and wave length 15m What will be the amplitude of vibration at 64kmhour 8 An eccentric mass excites us used to determine vibratory characteristics of a structure of mass 25kg as given in figure At a speed of 1000 RPM the eccentric mass to beat top position at the instant the structure was moved upwards through its static equilibrium position and corresponding amplitude was 20mm The eccentric mass is 05kg at an eccentricity of 10mm Determine i) Undamped natural frequency of the system ii) Damping ratio iii) Steady state amplitude and phase angle at 1500 RPM iv) Force transmitted to the ground at 1200 RPM

M=25kg

C

K

Eccentric mass

64

65

CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Support motion Consider a spring mass damper system as shown in figure Let it excited by motion

of support let y be the harmonic displacement of support From Newtonrsquos second law of motion

)yx(k)yx(cxm minusminusminusminus= ampampampamp

)1(0)yx(k)yx(cxm minusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminus=minus+minus+ ampampampamp

Let z be the relative displacement of the mass with respect to support then

Z=x-y

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

31

k

R

k

m

r

m

Find the expression for natural frequency of system shown in the figure Neglect the mass of the cantilever beam Study the special case i) k=Infinity ii) I = infinity Determine the expression for the natural frequency of a system shown in figure The two discs are keyed to a common shaft and have a combined mass moment of inertia about centre of oscillation O It is assumed that the card attached to mass m does not stretch and is always under tension Determine the natural frequency of a system shown

m

l

32

Determine the expression for the natural frequency of the system shown in figure Assume that the wires connecting the masses do not stretch and are always in tension

Determine the natural frequency of spring mass system taking the MASS OF SPRING (ms ) into account

M3 M1

M2

k2 k1

m

k

33

RAYLEIGHrsquoS METHOD

This is the extension of energy method Here we can determine the natural frequency of a conservative system without finding the equation of motion The natural frequency is a function of the rate of change of Kinetic energy and potential energy of the system From the principle of conservation of energy we have T+U = Constant T= KE and U =PE We can write the above equation as T1+U1 = --------------------------(1) Where 1 and 2 represents tow instances of time Let 1 be the time when the mass is passing through static equilibrium position Let 2 be the time corresponding to the mass displacement of the mass At this instant the velocity f the mass is zero and hence Substituting in equation (1) we get ( )Max = ( )Max Above equation leads directly to natural frequency Determine the natural frequency of spring mass system using Rayleighrsquos method m

34

r

B

K K

Home work

Determine the natural frequency of pendulum using Rayleighrsquos method

A Cylinder of mass m and moment of inertia Jo is to role without slipping but is sustained by spring K as shown in figure Determine the natural frequency of oscillation

Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping

a

Determine the natural frequency of system shown where in cylindrical disc rolls over the inextensible spring A cylinder of radius r rolls without slipping on cylindrical the equation of motion and hence determine the naoscillations about the lowest point shown Repeat the above problem ndash Instead of cylinder assume swithout slipping on concave surface as shown above

r

K

35

portion of radius R Derive tural frequency of small

phere of radius with r rolls

The torsional pendulum has a natural frequency of 5 Hz What length of steel wire of diameter 2 mm should be used for pendulum The inertia of mass fixed at the free end is 00098 kgm2 Take g= 083x1011 Nm2 Determine the torsional natural frequency of the system shown in figure Neglect the mass moment of inertia of the shaft ( Figure not proportionate) Determine the natural frequency of simple pendulum considering the mass of the rod into account

Using the energy method find tThe chord may be assumed ine

Kt1

Kt2

20cm

06m 08m 04m

20cm 20cm 20cm

m

M

he natural frxtensible in

k

M

36

equency of the system shown in figure the spring mass pulley system

A rod of length l and depth H having a mass m rests on a semicylidrical surface os radius r It is tipped slightly and let go Find the frequency of oscillation

A thin bar of l having a mass m rests on a semicylidrical surface radius r It is tipped slightly and let go Find the frequency of oscillation

l

h

R

l

37

38

CHAPTER 3

DAMPED FREE VIBRATION OF SINGLE DEGREE FREEDOM SYSTEM LEARNING OBJECTIVES

] Introduction to damped free vibration

] Terminologies used in damped free vibration

] Three different types of damping

] Concept of critical damping and its importance

] Study of response of viscous damped system for cases of under damping

critical and over damping

] Concept of logarithmic decrement

] Problems related to damped free vibration problems

Viscously damped free vibration of single degree freedom sytem Consider a single degree freedom system with viscous damping as shown in figure For a viscous damped system the damping force is

where c is a constant of proportionality It is called damping coefficient or coefficient of viscous damping

Symbolically it is designated by a dashpot as shown in Fig 3

x xampx

39

At any instant of time let the mass be displaced by x from the equilibrium position Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

Okxxcxm =minusminus= ampampamp

Okxxcxm =++ ampampamp ----------------------------------------------------------(1)

or

0kxdtdx

Cdt

xdM

2

2

=++

Assuming the solution of equation as

The first term in the above equation m2Ct

eminus

is an exponentially decaying function of time The behavior of the terms in the parenthesis however depends on whether numerical value within the radicals is positive zero or negative

CASE 1 if 0mk

m2c

2

=minus

40

CASE 2 if 0mk

m2c

2

=gt

CASE 3 if 0mk

m2c

2

=lt

41

Critically damped coefficient Critically damped coefficient is defined as the value of the damping coeffient for which radical in equation

mk

m2c

m2c

S2

21 minus

plusmn

= becomes zero

Damping Ratio ( Damping factor) ( ξ ) It is defined as the ratio of damping coefficient to the critical damping coefficient Note The damping coefficient c is defined as the actual amount of damping while the damping coefficient is a constant depending upon the mass and stiffness of the system The roots of characteristics equation can be now written as Note if c=cc or ξ =1 then system is critically damped if cgt cc or ξ gt1 then system is over damped I if clt cc or ξ lt1 then system is under damped

42

Take home quiz Derive General solution of different damped system

1 Critically damped system 2 Under damped system 3 Over damped system

Logarithmic Decrement It is used to determine the amount of damping present in system This measure The rate of decay of free vibration The larger the damping the greater will be rate of decay Logarithmic decrement is defined as the natural logarithm of ratio of any two successive amplitudes

=δ

2

1

xx

ln

Expression for logarithmic decrement (δ ) Consider a damped free vibration which is expressed by

( ) ( ) φ+ξminusω= ξω t 1sinXex 2n

tn

x is the displacement at any instant of time X and φ are arbitrary constants

21

2

ξminus

πξ=δ

43

Tutorial problems 1 Derive an expression for damping ratio in terms of logarithmic decrement

2 Show that logarithmic decrement

=δ

n

o

xx

lnn1

where xo is the initial amplitude xn

is amplitude after n cycles 3 A mass of 1 kg is to be supported on a spring having stiffness K= 9800 Nm The damping constant c= 59 Nsm Determine the natural frequency of system Find also logarithmic decrement and amplitude after 3 cycles if initial displacement is 03 cm 4 A vibrating system is defined by the following parameters M = 3kg k= 100 Nm damping coefficient = 3 Nsm determine Damping factor natural frequency damped vibration logarithmic decrement ratio of consecutive amplitude no of cycles after which the original amplitude is reduced to 20 5A single degree damped vibrating system a suspended mass of 18kg makes 15 oscillations in 003 seconds The amplitude decreases to 025 of the initial value after 5 oscillations Determine the stiffness of spring Logarithmic decrement damping factor and damping coefficient 6 A damped vibration record of a spring mass system shows the following data Amplitude at the end of 2nd cycle is 9mm amplitude at eh end of 3rd cycle is 6mm amplitude at the end of 4th cycle is 4mm stiffness of spring =8000 Nm weight =20N Find the logarithmic decrement damping force at unit velocity periodic time of vibration 7 For the system shown in figure the characteristics of dashpot is such that the constant force of 49 N is applied to the piston it velocity is found to be constant at 012 ms Determine the value of C would you expect the complete system to be periodic or aperiodic 8 For a torsional system shown in the figuref1 is natural frequency in air and f2 is the natural frequency when immersed in oil Determine the damping coefficient

Viscous fluid

Disc of mass moment of Inertia J

2a

C

m M

b

a a

k

9 A thin plate of area A and weigh w is fixed to the end of the spring and allow to oscillate in a viscous fluid as shown in figure It f1 is the frequency of oscillation in air and f2 in liquid Show that

22

21 ff

gA2

minusπω

=micro where the damping force on the plate

AV2Fd micro= V being the velocity of vibration 10 A torsional pendulum has a natural frequency of 175 cyclessecond vibrating in vacuum The mass moment of inertia is 2 kgm2 It is immersed in oil and it is observed that the natural frequency is 142 cyclessecond Find the damping torque If the disc is replaced by 34 deg When in oil find the displacement at the end of the first complete cycle 11 An automobile can be modeled as a mass placed on 4 shock absorbers each consisting of a spring and a damper Such that each spring is equally loaded Determine the stiffness damping constant of each shock absorber So that natural frequency is 2Hz and system is critically damped The mass of vehicle is 200kg 12 Write the differential equation of motion for the system shown in the figure below Determine the natural frequency of the damped oscillation and critical damping coefficient

Determine the differential equation of motion and find the critical cdamped natural frequency for system shown in figure

a

C

m M

M =200kg

44

200kg kgkg

oefficient and

45

Derive the equation of motion and find the critical damping coefficient and damped natural frequency for the system shown below Write the differential equation of motion and determine the expression for critical damping coefficient and damped natural frequency of system shown in figure

Set up a differential equation of motion for system shown in figure and determine the following Undamped natural frequencies critical damping coefficient damping ratio damped natural frequency

b

C

kM

a

m

46

A mass of 300N is resting on two springs of stiffness 3000 Nm each and a dashpot of damping coefficient 150 Nsm as shown in figure If the initial velocity of 10cms is given to mass at its equilibrium position what will be the displacement from the equilibrium position at the end of first second Derive the formula used A gun barrel of mass 600kg has a recoil spring of stiffness 294 knm if the barrel recoils 13m on firing determine i) The initial recoil velocity of barrel ii) The critical damping coefficient of dashpot which is engaged at the end of recoil stroke Iii) Time required for the barrel to return at the end of recoil stroke Iv) time required for the barrel to return to a position 5cm from the initial position

W= 3000 N

FIRING

CHAPTER 4

FORCED VIBRATION LEARNING OBJECTIVES

] Introduction to forced free vibration

] Terminologies used in forced vibration

] Concept of magnification factor

] Study of reciprocating and rotating unbalance

] Concept of base excitation and support excitation

] Concept of Transmissibility ratio

] Problems related to forced vibration

The vibration that takes place under the excitation of external forces are called

forced vibration When a system is subjected to harmonic excitation it is forced to vibrate at the same frequency as that of excitation Most of the mechanical systems are prone to steady state forced vibration which cause fatigue failure The vibration problem also becomes very important when any of excitation frequencies coincide with one of natural frequencies of a system Under this condition resonance occurs during which large amplitude of vibrations are observed To avoid resonance external frequency (operating speed) may be changed or properties of the system may be changed to alter the natural frequency In some cases sufficient amount of damping may be provided to avoid large amplitude during resonance Thus the problem of force vibration is very important in mechanical design

FORCED VIBRATIONS OF SINGLE DOF SYSTEMS UNDER HARMONIC EXCITATION

` [ ] tsin FF o ω=

x

sin FF o=

Fo

xampamp

[ ] tω

xamp

47

48

Consider a spring mass system having a viscous damping excited by a harmonic force [ ] tsin FF o ω= as shown in figure The various forces acting on the system is shown in figure Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

O)tsin(Fkxxcxm o =ω+minusminus= ampampamp

)tsin(Fkxxcxm o ω=++ ampampamp ----------------------------------------------------------(1)

or

)tsin(Fkxdtdx

Cdt

xdM o2

2

ω=++

Equation (1) represents the differential equation of motion of the system The solution to this equation consists of two parts namely complementary function and a particular integral

The first of solution is also termed as transient solution of homogeneous solution and can be obtained from the homogeneous differential equation 0kxxcxm =++ ampampamp

The second part of solution particular integral is also termed as steady state solution ad is of form )tsin(Xx φminusω= Where X= steady state amplitude φ is phase difference between external force and displacement

)tsin(Xx φminusω=

The vector representation of the shown

49

The force polygon for the above vector diagram can be written as follows

( )φminusωt O

KX

xm 2ω

Fo

xCω

tω

φ

50

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

is the governing equation

Zero frequency deflection (Xst) It is deflection observed at zero frequency ( when the external frequency 0=ω

( ) kF

001

kF

x o

22

o

st =

+minus=

Magnification factor (M) It is defined as the ratio of steady state amplitude to zero frequency deflection xst

M = arX

X=

A plot of magnification factor versus frequency ratio

damping factor ( )ξ is called frequency plot of a harmon All the frequency terms starts from magnification factor of damping ratio ( )ξ The amplitude of vibration at any on the value of damping The amplitude will be higher for higher damping The amplitude of vibration at reso

amplitude tends to zero as

ωω

n tends to infinity

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

Freq

ζ =

ζ = 1

Magnification Factor M = XXst

kFo

51

ωω

nfor varying values of

ically excite vibrating system

M=1 what ever may be value particular frequency depends

for lesser damping and lesser nance will be very high The

uency Ratio r = (ωωn)

ζ = 0 0

ζ = 025 ζ = 0375 ζ = 0 5

ζ = 0707

ζ = 2

52

Plot of phase angle φ versus frequency ratio

ωω

n is shown in the

figure The phase angle φ is zero for all the values of damping ratio

at

ωω

n=0 The phase

angle φ is 90 for all the values of damping ratio

at

ωω

n=1 ie at

resonance The phase angle between o to 90 deg for all the values of

ωω

n= 0 to 1 and

between 90 deg to 180 deg for all the values of

ωω

ngt1

Class work Derive an expression for frequency at peak amplitude

2np 21 ξminusω=ω=ω is the required equation

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

53

Derive expression amplitude at resonance ( In dimensional form) In dimensional form We know that steady state amplitude

( ) ( )

ω+ωminus=

222

o

Cmk

Fx Equation in dimensional form

We know that at resonance mk

n =ω=ω

( )

ω+

minus

=2

n

2

oR

Cmk

mk

Fx

( )

ω=

2n

oR

C

Fx

( )

ω=

n

oR C

Fx

Class work Derive an expression for amplitude at resonance ( in non dimensional form)

Tutorial Problems on Forced Vibration 1A weight of 60N is suspended by a spring stiffness 12 kNm is forced to vibrate by a harmonic force of 10N assuming viscous damping of 0086 kN-sm Find i) resonance frequency ii) amplitude at resonance iii) phase angle at resonance iv) frequency corresponding to peak amplitude 2 A periodic torque having a maximum value of 05888Nm at a frequency of 4 rads is impressed upon a flywheel suspended from a wire The wheel has mass moment of inertia 012 kgm2 and the wire has stiffness 1176 Nmrad A viscous dashpot applies damping couple of 0784 Nm an angular velocity of 2 rads calculate i)maximum angular displacement from rest position ii) maximum couple applied to dashpot Iii) angle by which the angular displacement lags the torque A mass of 1019 kg is suspended from a spring stiffness 20 Nmm viscous damping causes the amplitude to decrease to 110th of initial value in 4 complete cycle If a periodic force of 20 cos(50 t) is applied to the mass Find the amplitude of forced vibration What would be the amplitude of oscillation if the applied force coincides with the natural period

m=1019 kg

[ ] t50cos 30F =

54

55

ROTATING UNBALANCE Unbalance in rotating machinery is a common source of vibration excitation Consider a machine of mass m having mounted on a foundation of stiffness k and damping coefficient C as shown in figure The unbalance is represented by an eccentric rotating mass m with eccentricity (e) and having a angular velocity (ω) let x be the displacement of non rotating mass from the static equilibrium position The displacement of rotating mass m from the static equilibrium is ( ) tsin( ex ω+ Applying Newtonrsquos law of motion we have

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

56

RECIPROCATING UNBALANCE Unbalance in reciprocating machinery is a common source of vibration excitation Consider a machine of mass M having mounted on a foundation of stiffness K and damping coefficient C as shown in figure The unbalance is represented by a reciprocating mass (m) having crank radus (e) and connecting rod of length (l) Let ωωωω be angular velocity of the crank Let x be the displacement of non reciprocating mass from the static equilibrium position at any instant of time The displacement of reciprocating mass m from the static equilibrium position is

)2sin(sin +++ tle

tex ωωωωωωωω

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

57

Note The equations obtained in reciprocating unbalance is same as the rotating unbalance and having the following differences Rotating Unbalance Reciprocating Unbalance Here mass of the rotating element is considered

Here mass of reciprocating element is considered

Here the eccentricity (e) which is the distance between the centre of rotation and CG

Here the eccentricity e is the crank radius and crank radius = stroke length2

The frequency plot of non dimensional factor( ( )me

MX versus frequency ratio

ωω

n for various values of ( )ξξξξ is shown in figure This frequency plot is same for

the rotating and reciprocating unbalance with the following points are noted from the diagram

1 Damping ratio plays very important role during resonance in controlling the amplitude

2 When

ωω

n=0 then ( )me

MX =0 for all the values of damping ratio ( )ξξξξ

3 When

ωω

nis higher than ( )me

MX tends to unity

4 At very high speeds ie at very high frequency ratio

ωω

n there is no effect

of damping The frequency plot of phase angle versus the frequency ratio is given as shown in figure The explanation is same as in forced harmonic vibration

58

VIBRATION ISOLATION

Vibratory forces and motions generated by machine and other caused are unavoidable Howeever their effect on the other dynamical systems can be minimized by proper isolator design An isolation system is either to protect o delicate machine from excessive vibration transmitted to it from its supporting structure or to prevent vibratory forces and motions generated by machines from being transmitted to its surroundings The effectiveness of isolation may be measured in terms of forces and motion transmitted The first type is known as force isolation and the second type is known as motion isolation

Force isolation (Force Transmissibility) When vibratory forces generated by machine are isolated the effectiveness of forced isolation can be made by defining the term transmissibility The term transmissibility is defined as the ratio of force transmitted to the foundation to that of the external impressed force

Force Transmissibility = o

tr

FF

force Impressed Externalfoundation toted transmitForce

=

Consider viscous damped spring mass system as shown in figure Upon this system an external harmonic force ) t (Sin Fo ω is applied The difference forces acting on system are inertia forces damping force spring force and external force All these forces are represented on force polygon shown below Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these two forces

( ) ( )22 xckx ω+=

59

This is the expression for force transmissibility in dimensional form To represent same in the non dimensional form divide numerator and denominator by k

2

n

22

n

2

n

o

tr

21

21

FF

TR

ωω

ξ+

ωω

minus

ωω

ξ+

==

PHASE DIFFERNECE BETWEEN EXTERNAL FORCE AND DISPLACEMENT Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these

ωω

minus

ωω

ξ=

ωminusω

=φ minusminus2

n

n1

2

1

1

2tan

mKC

tan

The phase difference between force transmitted to the foundation Ftr to the displacement

kc

KxxC

tanω

=ω

=α or

ωω

ξ=

ω=α minusminus

n

11 2tanKx

xCtan

The phase difference between external force Fo and force transmitted to foundationis β

60

β= αminusφ =

ωω

minus

ωω

ξminus

2

n

n1

1

2tan -

ωω

ξminus

n

1 2tan

Displacement Transmissibility It is the ratio of amplitude of steady state vibration of machine to that of the base motion Forced vibration due to support motion Consider a spring mass damper system subjected to harmonic excitation )tsin(Yy ω= at the base as shown in figure Let x be the absolute displacement of the mass at any instant of time t The different forces acting on system are shown in free body diagram Applying Newtonrsquos Law of motion Ma = sumF

)yx(k)yx(cxm minusminusminusminus= ampampampamp

0)yx(k)yx(cxm =minus+minus+ ampampampamp

61

Frequency plot of transmissibility versus frequency ratio

62

The plot of transmissibility versus frequency ratio is as shown in figure The following points can be noted from it

i) The value of transmissibility is unity at frequency ratio

ωω

n=0 and 2

ii) For an undamped system ( )ξ =0 transmissibility tends to infin at resonance iii) The value of transmissibility is less than unity for the value of frequency ratio

greater than 2

iv) For frequency ratio

ωω

nlt 2 smaller damping ratio lead to larger values of

transmissibility on the other hand for frequency ratio

ωω

ngt 2 smaller damping

ratio leads to smaller value of transmissibility

The frequency plot of transmissibility (TR) versus frequency ratio

ωω

n can be

broadly divided into two zones

1

ωω

n= 0 to 2

2

ωω

n= 2 to infin

CASE 1

When

ωω

n= 0 to 2

In this zone the value of transmissibility is always greater than 1 for all values of damping ratio In this zone the transmissibility decreases with increase in damping Hence we can conclude that in this zone damping is useful in isolating vibration forces being transmitted to the support CASE 2

When

ωω

n= 2 to infin

In this zone the value of transmissibility is less than 1 for all values of damping ratio In this zone the transmissibility increases with increase in damping Hence we can conclude that in this zone damping will adversely affect in isolating vibratory forces

63

Tutorial problems on Forced Vibration 1 Machine of mass 75kg is mounted on spring of stiffness 12000 Nm with an assumed damping factor ξ =02 piston within the machine of mass 2 kg has a reciprocating motion with a stroke of 75 cm and speed of 15 cyclessecond Assuming the motion of piston to be harmonic determine amplitude of machine ( X) Transmissibility and force transmitted to foundation phase angle of transmitted force wrt to exciting force 2 Motor weighs 600N is mounted on a simple beam that has a spring rate of 400 Ncm The motor armature base 160 N and has an eccentricity of 0005cm What will be the amplitude of motion when it runs at 1760 RPM Neglect damping the deflection of armature shaft and weight of the beam 3 A Heavy machine of mass 300 kg is supported on resilient foundation the static deflection of foundation due to weight of machine is found to be 75mm It is observed that the machine vibrates with an amplitude of 10mm when the base of the foundation is subjected to harmonic oscillation at the undampeds natural frequency of system with an amplitude of 25mm Find the damping constant of foundation Ii) dynamic force amplitude on the base iii) amplitude of the displacement of machine relative to base 4 An air compressor of mass 450 kg operates at a constant speed of 1750 RPM The reciprocating masses are 10 kg the crank radius is 100mm If the damper for mounting introduces a damping factor 015 i) specify the springs of mounting such that only 20 of the unbalanced force is transmitted to the foundation ii) determine the magnitude of the transmitted force bull Repeat the above problem by assuming that there is no damping in system

5 A body of a car having mass 1500 kg and is mounted on 4 equal springs which deflect through 225 cm under the weight of the body A total damping forces the 4 shock absorbers is 46 N at a velocity of 1cmsec The car is placed with all the four wheels on a test platform which is move up and down at resonant speed With an amplitude of 25cm Find the amplitude of the car body on its springs assuming its CG to be in the centre of wheel base 6A 75 kg machine which is mounted on springs of stiffness 116x105 Nm with the damping of 20=ξ A 2kg piston within the machine as reciprocating motion with a stroke of 008m and speed of 3000 RPM Assuming the motion of piston to be harmonic determine the amplitude of vibration of machine

7 The spring of an automobile are compressed 10cm under its own weight Find the critical speed when a trailer is traveling over road with a people approximaterly by a sine wave of amplitude 75 cm and wave length 15m What will be the amplitude of vibration at 64kmhour 8 An eccentric mass excites us used to determine vibratory characteristics of a structure of mass 25kg as given in figure At a speed of 1000 RPM the eccentric mass to beat top position at the instant the structure was moved upwards through its static equilibrium position and corresponding amplitude was 20mm The eccentric mass is 05kg at an eccentricity of 10mm Determine i) Undamped natural frequency of the system ii) Damping ratio iii) Steady state amplitude and phase angle at 1500 RPM iv) Force transmitted to the ground at 1200 RPM

M=25kg

C

K

Eccentric mass

64

65

CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Support motion Consider a spring mass damper system as shown in figure Let it excited by motion

of support let y be the harmonic displacement of support From Newtonrsquos second law of motion

)yx(k)yx(cxm minusminusminusminus= ampampampamp

)1(0)yx(k)yx(cxm minusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminus=minus+minus+ ampampampamp

Let z be the relative displacement of the mass with respect to support then

Z=x-y

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

32

Determine the expression for the natural frequency of the system shown in figure Assume that the wires connecting the masses do not stretch and are always in tension

Determine the natural frequency of spring mass system taking the MASS OF SPRING (ms ) into account

M3 M1

M2

k2 k1

m

k

33

RAYLEIGHrsquoS METHOD

This is the extension of energy method Here we can determine the natural frequency of a conservative system without finding the equation of motion The natural frequency is a function of the rate of change of Kinetic energy and potential energy of the system From the principle of conservation of energy we have T+U = Constant T= KE and U =PE We can write the above equation as T1+U1 = --------------------------(1) Where 1 and 2 represents tow instances of time Let 1 be the time when the mass is passing through static equilibrium position Let 2 be the time corresponding to the mass displacement of the mass At this instant the velocity f the mass is zero and hence Substituting in equation (1) we get ( )Max = ( )Max Above equation leads directly to natural frequency Determine the natural frequency of spring mass system using Rayleighrsquos method m

34

r

B

K K

Home work

Determine the natural frequency of pendulum using Rayleighrsquos method

A Cylinder of mass m and moment of inertia Jo is to role without slipping but is sustained by spring K as shown in figure Determine the natural frequency of oscillation

Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping

a

Determine the natural frequency of system shown where in cylindrical disc rolls over the inextensible spring A cylinder of radius r rolls without slipping on cylindrical the equation of motion and hence determine the naoscillations about the lowest point shown Repeat the above problem ndash Instead of cylinder assume swithout slipping on concave surface as shown above

r

K

35

portion of radius R Derive tural frequency of small

phere of radius with r rolls

The torsional pendulum has a natural frequency of 5 Hz What length of steel wire of diameter 2 mm should be used for pendulum The inertia of mass fixed at the free end is 00098 kgm2 Take g= 083x1011 Nm2 Determine the torsional natural frequency of the system shown in figure Neglect the mass moment of inertia of the shaft ( Figure not proportionate) Determine the natural frequency of simple pendulum considering the mass of the rod into account

Using the energy method find tThe chord may be assumed ine

Kt1

Kt2

20cm

06m 08m 04m

20cm 20cm 20cm

m

M

he natural frxtensible in

k

M

36

equency of the system shown in figure the spring mass pulley system

A rod of length l and depth H having a mass m rests on a semicylidrical surface os radius r It is tipped slightly and let go Find the frequency of oscillation

A thin bar of l having a mass m rests on a semicylidrical surface radius r It is tipped slightly and let go Find the frequency of oscillation

l

h

R

l

37

38

CHAPTER 3

DAMPED FREE VIBRATION OF SINGLE DEGREE FREEDOM SYSTEM LEARNING OBJECTIVES

] Introduction to damped free vibration

] Terminologies used in damped free vibration

] Three different types of damping

] Concept of critical damping and its importance

] Study of response of viscous damped system for cases of under damping

critical and over damping

] Concept of logarithmic decrement

] Problems related to damped free vibration problems

Viscously damped free vibration of single degree freedom sytem Consider a single degree freedom system with viscous damping as shown in figure For a viscous damped system the damping force is

where c is a constant of proportionality It is called damping coefficient or coefficient of viscous damping

Symbolically it is designated by a dashpot as shown in Fig 3

x xampx

39

At any instant of time let the mass be displaced by x from the equilibrium position Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

Okxxcxm =minusminus= ampampamp

Okxxcxm =++ ampampamp ----------------------------------------------------------(1)

or

0kxdtdx

Cdt

xdM

2

2

=++

Assuming the solution of equation as

The first term in the above equation m2Ct

eminus

is an exponentially decaying function of time The behavior of the terms in the parenthesis however depends on whether numerical value within the radicals is positive zero or negative

CASE 1 if 0mk

m2c

2

=minus

40

CASE 2 if 0mk

m2c

2

=gt

CASE 3 if 0mk

m2c

2

=lt

41

Critically damped coefficient Critically damped coefficient is defined as the value of the damping coeffient for which radical in equation

mk

m2c

m2c

S2

21 minus

plusmn

= becomes zero

Damping Ratio ( Damping factor) ( ξ ) It is defined as the ratio of damping coefficient to the critical damping coefficient Note The damping coefficient c is defined as the actual amount of damping while the damping coefficient is a constant depending upon the mass and stiffness of the system The roots of characteristics equation can be now written as Note if c=cc or ξ =1 then system is critically damped if cgt cc or ξ gt1 then system is over damped I if clt cc or ξ lt1 then system is under damped

42

Take home quiz Derive General solution of different damped system

1 Critically damped system 2 Under damped system 3 Over damped system

Logarithmic Decrement It is used to determine the amount of damping present in system This measure The rate of decay of free vibration The larger the damping the greater will be rate of decay Logarithmic decrement is defined as the natural logarithm of ratio of any two successive amplitudes

=δ

2

1

xx

ln

Expression for logarithmic decrement (δ ) Consider a damped free vibration which is expressed by

( ) ( ) φ+ξminusω= ξω t 1sinXex 2n

tn

x is the displacement at any instant of time X and φ are arbitrary constants

21

2

ξminus

πξ=δ

43

Tutorial problems 1 Derive an expression for damping ratio in terms of logarithmic decrement

2 Show that logarithmic decrement

=δ

n

o

xx

lnn1

where xo is the initial amplitude xn

is amplitude after n cycles 3 A mass of 1 kg is to be supported on a spring having stiffness K= 9800 Nm The damping constant c= 59 Nsm Determine the natural frequency of system Find also logarithmic decrement and amplitude after 3 cycles if initial displacement is 03 cm 4 A vibrating system is defined by the following parameters M = 3kg k= 100 Nm damping coefficient = 3 Nsm determine Damping factor natural frequency damped vibration logarithmic decrement ratio of consecutive amplitude no of cycles after which the original amplitude is reduced to 20 5A single degree damped vibrating system a suspended mass of 18kg makes 15 oscillations in 003 seconds The amplitude decreases to 025 of the initial value after 5 oscillations Determine the stiffness of spring Logarithmic decrement damping factor and damping coefficient 6 A damped vibration record of a spring mass system shows the following data Amplitude at the end of 2nd cycle is 9mm amplitude at eh end of 3rd cycle is 6mm amplitude at the end of 4th cycle is 4mm stiffness of spring =8000 Nm weight =20N Find the logarithmic decrement damping force at unit velocity periodic time of vibration 7 For the system shown in figure the characteristics of dashpot is such that the constant force of 49 N is applied to the piston it velocity is found to be constant at 012 ms Determine the value of C would you expect the complete system to be periodic or aperiodic 8 For a torsional system shown in the figuref1 is natural frequency in air and f2 is the natural frequency when immersed in oil Determine the damping coefficient

Viscous fluid

Disc of mass moment of Inertia J

2a

C

m M

b

a a

k

9 A thin plate of area A and weigh w is fixed to the end of the spring and allow to oscillate in a viscous fluid as shown in figure It f1 is the frequency of oscillation in air and f2 in liquid Show that

22

21 ff

gA2

minusπω

=micro where the damping force on the plate

AV2Fd micro= V being the velocity of vibration 10 A torsional pendulum has a natural frequency of 175 cyclessecond vibrating in vacuum The mass moment of inertia is 2 kgm2 It is immersed in oil and it is observed that the natural frequency is 142 cyclessecond Find the damping torque If the disc is replaced by 34 deg When in oil find the displacement at the end of the first complete cycle 11 An automobile can be modeled as a mass placed on 4 shock absorbers each consisting of a spring and a damper Such that each spring is equally loaded Determine the stiffness damping constant of each shock absorber So that natural frequency is 2Hz and system is critically damped The mass of vehicle is 200kg 12 Write the differential equation of motion for the system shown in the figure below Determine the natural frequency of the damped oscillation and critical damping coefficient

Determine the differential equation of motion and find the critical cdamped natural frequency for system shown in figure

a

C

m M

M =200kg

44

200kg kgkg

oefficient and

45

Derive the equation of motion and find the critical damping coefficient and damped natural frequency for the system shown below Write the differential equation of motion and determine the expression for critical damping coefficient and damped natural frequency of system shown in figure

Set up a differential equation of motion for system shown in figure and determine the following Undamped natural frequencies critical damping coefficient damping ratio damped natural frequency

b

C

kM

a

m

46

A mass of 300N is resting on two springs of stiffness 3000 Nm each and a dashpot of damping coefficient 150 Nsm as shown in figure If the initial velocity of 10cms is given to mass at its equilibrium position what will be the displacement from the equilibrium position at the end of first second Derive the formula used A gun barrel of mass 600kg has a recoil spring of stiffness 294 knm if the barrel recoils 13m on firing determine i) The initial recoil velocity of barrel ii) The critical damping coefficient of dashpot which is engaged at the end of recoil stroke Iii) Time required for the barrel to return at the end of recoil stroke Iv) time required for the barrel to return to a position 5cm from the initial position

W= 3000 N

FIRING

CHAPTER 4

FORCED VIBRATION LEARNING OBJECTIVES

] Introduction to forced free vibration

] Terminologies used in forced vibration

] Concept of magnification factor

] Study of reciprocating and rotating unbalance

] Concept of base excitation and support excitation

] Concept of Transmissibility ratio

] Problems related to forced vibration

The vibration that takes place under the excitation of external forces are called

forced vibration When a system is subjected to harmonic excitation it is forced to vibrate at the same frequency as that of excitation Most of the mechanical systems are prone to steady state forced vibration which cause fatigue failure The vibration problem also becomes very important when any of excitation frequencies coincide with one of natural frequencies of a system Under this condition resonance occurs during which large amplitude of vibrations are observed To avoid resonance external frequency (operating speed) may be changed or properties of the system may be changed to alter the natural frequency In some cases sufficient amount of damping may be provided to avoid large amplitude during resonance Thus the problem of force vibration is very important in mechanical design

FORCED VIBRATIONS OF SINGLE DOF SYSTEMS UNDER HARMONIC EXCITATION

` [ ] tsin FF o ω=

x

sin FF o=

Fo

xampamp

[ ] tω

xamp

47

48

Consider a spring mass system having a viscous damping excited by a harmonic force [ ] tsin FF o ω= as shown in figure The various forces acting on the system is shown in figure Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

O)tsin(Fkxxcxm o =ω+minusminus= ampampamp

)tsin(Fkxxcxm o ω=++ ampampamp ----------------------------------------------------------(1)

or

)tsin(Fkxdtdx

Cdt

xdM o2

2

ω=++

Equation (1) represents the differential equation of motion of the system The solution to this equation consists of two parts namely complementary function and a particular integral

The first of solution is also termed as transient solution of homogeneous solution and can be obtained from the homogeneous differential equation 0kxxcxm =++ ampampamp

The second part of solution particular integral is also termed as steady state solution ad is of form )tsin(Xx φminusω= Where X= steady state amplitude φ is phase difference between external force and displacement

)tsin(Xx φminusω=

The vector representation of the shown

49

The force polygon for the above vector diagram can be written as follows

( )φminusωt O

KX

xm 2ω

Fo

xCω

tω

φ

50

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

is the governing equation

Zero frequency deflection (Xst) It is deflection observed at zero frequency ( when the external frequency 0=ω

( ) kF

001

kF

x o

22

o

st =

+minus=

Magnification factor (M) It is defined as the ratio of steady state amplitude to zero frequency deflection xst

M = arX

X=

A plot of magnification factor versus frequency ratio

damping factor ( )ξ is called frequency plot of a harmon All the frequency terms starts from magnification factor of damping ratio ( )ξ The amplitude of vibration at any on the value of damping The amplitude will be higher for higher damping The amplitude of vibration at reso

amplitude tends to zero as

ωω

n tends to infinity

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

Freq

ζ =

ζ = 1

Magnification Factor M = XXst

kFo

51

ωω

nfor varying values of

ically excite vibrating system

M=1 what ever may be value particular frequency depends

for lesser damping and lesser nance will be very high The

uency Ratio r = (ωωn)

ζ = 0 0

ζ = 025 ζ = 0375 ζ = 0 5

ζ = 0707

ζ = 2

52

Plot of phase angle φ versus frequency ratio

ωω

n is shown in the

figure The phase angle φ is zero for all the values of damping ratio

at

ωω

n=0 The phase

angle φ is 90 for all the values of damping ratio

at

ωω

n=1 ie at

resonance The phase angle between o to 90 deg for all the values of

ωω

n= 0 to 1 and

between 90 deg to 180 deg for all the values of

ωω

ngt1

Class work Derive an expression for frequency at peak amplitude

2np 21 ξminusω=ω=ω is the required equation

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

53

Derive expression amplitude at resonance ( In dimensional form) In dimensional form We know that steady state amplitude

( ) ( )

ω+ωminus=

222

o

Cmk

Fx Equation in dimensional form

We know that at resonance mk

n =ω=ω

( )

ω+

minus

=2

n

2

oR

Cmk

mk

Fx

( )

ω=

2n

oR

C

Fx

( )

ω=

n

oR C

Fx

Class work Derive an expression for amplitude at resonance ( in non dimensional form)

Tutorial Problems on Forced Vibration 1A weight of 60N is suspended by a spring stiffness 12 kNm is forced to vibrate by a harmonic force of 10N assuming viscous damping of 0086 kN-sm Find i) resonance frequency ii) amplitude at resonance iii) phase angle at resonance iv) frequency corresponding to peak amplitude 2 A periodic torque having a maximum value of 05888Nm at a frequency of 4 rads is impressed upon a flywheel suspended from a wire The wheel has mass moment of inertia 012 kgm2 and the wire has stiffness 1176 Nmrad A viscous dashpot applies damping couple of 0784 Nm an angular velocity of 2 rads calculate i)maximum angular displacement from rest position ii) maximum couple applied to dashpot Iii) angle by which the angular displacement lags the torque A mass of 1019 kg is suspended from a spring stiffness 20 Nmm viscous damping causes the amplitude to decrease to 110th of initial value in 4 complete cycle If a periodic force of 20 cos(50 t) is applied to the mass Find the amplitude of forced vibration What would be the amplitude of oscillation if the applied force coincides with the natural period

m=1019 kg

[ ] t50cos 30F =

54

55

ROTATING UNBALANCE Unbalance in rotating machinery is a common source of vibration excitation Consider a machine of mass m having mounted on a foundation of stiffness k and damping coefficient C as shown in figure The unbalance is represented by an eccentric rotating mass m with eccentricity (e) and having a angular velocity (ω) let x be the displacement of non rotating mass from the static equilibrium position The displacement of rotating mass m from the static equilibrium is ( ) tsin( ex ω+ Applying Newtonrsquos law of motion we have

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

56

RECIPROCATING UNBALANCE Unbalance in reciprocating machinery is a common source of vibration excitation Consider a machine of mass M having mounted on a foundation of stiffness K and damping coefficient C as shown in figure The unbalance is represented by a reciprocating mass (m) having crank radus (e) and connecting rod of length (l) Let ωωωω be angular velocity of the crank Let x be the displacement of non reciprocating mass from the static equilibrium position at any instant of time The displacement of reciprocating mass m from the static equilibrium position is

)2sin(sin +++ tle

tex ωωωωωωωω

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

57

Note The equations obtained in reciprocating unbalance is same as the rotating unbalance and having the following differences Rotating Unbalance Reciprocating Unbalance Here mass of the rotating element is considered

Here mass of reciprocating element is considered

Here the eccentricity (e) which is the distance between the centre of rotation and CG

Here the eccentricity e is the crank radius and crank radius = stroke length2

The frequency plot of non dimensional factor( ( )me

MX versus frequency ratio

ωω

n for various values of ( )ξξξξ is shown in figure This frequency plot is same for

the rotating and reciprocating unbalance with the following points are noted from the diagram

1 Damping ratio plays very important role during resonance in controlling the amplitude

2 When

ωω

n=0 then ( )me

MX =0 for all the values of damping ratio ( )ξξξξ

3 When

ωω

nis higher than ( )me

MX tends to unity

4 At very high speeds ie at very high frequency ratio

ωω

n there is no effect

of damping The frequency plot of phase angle versus the frequency ratio is given as shown in figure The explanation is same as in forced harmonic vibration

58

VIBRATION ISOLATION

Vibratory forces and motions generated by machine and other caused are unavoidable Howeever their effect on the other dynamical systems can be minimized by proper isolator design An isolation system is either to protect o delicate machine from excessive vibration transmitted to it from its supporting structure or to prevent vibratory forces and motions generated by machines from being transmitted to its surroundings The effectiveness of isolation may be measured in terms of forces and motion transmitted The first type is known as force isolation and the second type is known as motion isolation

Force isolation (Force Transmissibility) When vibratory forces generated by machine are isolated the effectiveness of forced isolation can be made by defining the term transmissibility The term transmissibility is defined as the ratio of force transmitted to the foundation to that of the external impressed force

Force Transmissibility = o

tr

FF

force Impressed Externalfoundation toted transmitForce

=

Consider viscous damped spring mass system as shown in figure Upon this system an external harmonic force ) t (Sin Fo ω is applied The difference forces acting on system are inertia forces damping force spring force and external force All these forces are represented on force polygon shown below Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these two forces

( ) ( )22 xckx ω+=

59

This is the expression for force transmissibility in dimensional form To represent same in the non dimensional form divide numerator and denominator by k

2

n

22

n

2

n

o

tr

21

21

FF

TR

ωω

ξ+

ωω

minus

ωω

ξ+

==

PHASE DIFFERNECE BETWEEN EXTERNAL FORCE AND DISPLACEMENT Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these

ωω

minus

ωω

ξ=

ωminusω

=φ minusminus2

n

n1

2

1

1

2tan

mKC

tan

The phase difference between force transmitted to the foundation Ftr to the displacement

kc

KxxC

tanω

=ω

=α or

ωω

ξ=

ω=α minusminus

n

11 2tanKx

xCtan

The phase difference between external force Fo and force transmitted to foundationis β

60

β= αminusφ =

ωω

minus

ωω

ξminus

2

n

n1

1

2tan -

ωω

ξminus

n

1 2tan

Displacement Transmissibility It is the ratio of amplitude of steady state vibration of machine to that of the base motion Forced vibration due to support motion Consider a spring mass damper system subjected to harmonic excitation )tsin(Yy ω= at the base as shown in figure Let x be the absolute displacement of the mass at any instant of time t The different forces acting on system are shown in free body diagram Applying Newtonrsquos Law of motion Ma = sumF

)yx(k)yx(cxm minusminusminusminus= ampampampamp

0)yx(k)yx(cxm =minus+minus+ ampampampamp

61

Frequency plot of transmissibility versus frequency ratio

62

The plot of transmissibility versus frequency ratio is as shown in figure The following points can be noted from it

i) The value of transmissibility is unity at frequency ratio

ωω

n=0 and 2

ii) For an undamped system ( )ξ =0 transmissibility tends to infin at resonance iii) The value of transmissibility is less than unity for the value of frequency ratio

greater than 2

iv) For frequency ratio

ωω

nlt 2 smaller damping ratio lead to larger values of

transmissibility on the other hand for frequency ratio

ωω

ngt 2 smaller damping

ratio leads to smaller value of transmissibility

The frequency plot of transmissibility (TR) versus frequency ratio

ωω

n can be

broadly divided into two zones

1

ωω

n= 0 to 2

2

ωω

n= 2 to infin

CASE 1

When

ωω

n= 0 to 2

In this zone the value of transmissibility is always greater than 1 for all values of damping ratio In this zone the transmissibility decreases with increase in damping Hence we can conclude that in this zone damping is useful in isolating vibration forces being transmitted to the support CASE 2

When

ωω

n= 2 to infin

In this zone the value of transmissibility is less than 1 for all values of damping ratio In this zone the transmissibility increases with increase in damping Hence we can conclude that in this zone damping will adversely affect in isolating vibratory forces

63

Tutorial problems on Forced Vibration 1 Machine of mass 75kg is mounted on spring of stiffness 12000 Nm with an assumed damping factor ξ =02 piston within the machine of mass 2 kg has a reciprocating motion with a stroke of 75 cm and speed of 15 cyclessecond Assuming the motion of piston to be harmonic determine amplitude of machine ( X) Transmissibility and force transmitted to foundation phase angle of transmitted force wrt to exciting force 2 Motor weighs 600N is mounted on a simple beam that has a spring rate of 400 Ncm The motor armature base 160 N and has an eccentricity of 0005cm What will be the amplitude of motion when it runs at 1760 RPM Neglect damping the deflection of armature shaft and weight of the beam 3 A Heavy machine of mass 300 kg is supported on resilient foundation the static deflection of foundation due to weight of machine is found to be 75mm It is observed that the machine vibrates with an amplitude of 10mm when the base of the foundation is subjected to harmonic oscillation at the undampeds natural frequency of system with an amplitude of 25mm Find the damping constant of foundation Ii) dynamic force amplitude on the base iii) amplitude of the displacement of machine relative to base 4 An air compressor of mass 450 kg operates at a constant speed of 1750 RPM The reciprocating masses are 10 kg the crank radius is 100mm If the damper for mounting introduces a damping factor 015 i) specify the springs of mounting such that only 20 of the unbalanced force is transmitted to the foundation ii) determine the magnitude of the transmitted force bull Repeat the above problem by assuming that there is no damping in system

5 A body of a car having mass 1500 kg and is mounted on 4 equal springs which deflect through 225 cm under the weight of the body A total damping forces the 4 shock absorbers is 46 N at a velocity of 1cmsec The car is placed with all the four wheels on a test platform which is move up and down at resonant speed With an amplitude of 25cm Find the amplitude of the car body on its springs assuming its CG to be in the centre of wheel base 6A 75 kg machine which is mounted on springs of stiffness 116x105 Nm with the damping of 20=ξ A 2kg piston within the machine as reciprocating motion with a stroke of 008m and speed of 3000 RPM Assuming the motion of piston to be harmonic determine the amplitude of vibration of machine

7 The spring of an automobile are compressed 10cm under its own weight Find the critical speed when a trailer is traveling over road with a people approximaterly by a sine wave of amplitude 75 cm and wave length 15m What will be the amplitude of vibration at 64kmhour 8 An eccentric mass excites us used to determine vibratory characteristics of a structure of mass 25kg as given in figure At a speed of 1000 RPM the eccentric mass to beat top position at the instant the structure was moved upwards through its static equilibrium position and corresponding amplitude was 20mm The eccentric mass is 05kg at an eccentricity of 10mm Determine i) Undamped natural frequency of the system ii) Damping ratio iii) Steady state amplitude and phase angle at 1500 RPM iv) Force transmitted to the ground at 1200 RPM

M=25kg

C

K

Eccentric mass

64

65

CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Support motion Consider a spring mass damper system as shown in figure Let it excited by motion

of support let y be the harmonic displacement of support From Newtonrsquos second law of motion

)yx(k)yx(cxm minusminusminusminus= ampampampamp

)1(0)yx(k)yx(cxm minusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminus=minus+minus+ ampampampamp

Let z be the relative displacement of the mass with respect to support then

Z=x-y

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

33

RAYLEIGHrsquoS METHOD

This is the extension of energy method Here we can determine the natural frequency of a conservative system without finding the equation of motion The natural frequency is a function of the rate of change of Kinetic energy and potential energy of the system From the principle of conservation of energy we have T+U = Constant T= KE and U =PE We can write the above equation as T1+U1 = --------------------------(1) Where 1 and 2 represents tow instances of time Let 1 be the time when the mass is passing through static equilibrium position Let 2 be the time corresponding to the mass displacement of the mass At this instant the velocity f the mass is zero and hence Substituting in equation (1) we get ( )Max = ( )Max Above equation leads directly to natural frequency Determine the natural frequency of spring mass system using Rayleighrsquos method m

34

r

B

K K

Home work

Determine the natural frequency of pendulum using Rayleighrsquos method

A Cylinder of mass m and moment of inertia Jo is to role without slipping but is sustained by spring K as shown in figure Determine the natural frequency of oscillation

Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping

a

Determine the natural frequency of system shown where in cylindrical disc rolls over the inextensible spring A cylinder of radius r rolls without slipping on cylindrical the equation of motion and hence determine the naoscillations about the lowest point shown Repeat the above problem ndash Instead of cylinder assume swithout slipping on concave surface as shown above

r

K

35

portion of radius R Derive tural frequency of small

phere of radius with r rolls

The torsional pendulum has a natural frequency of 5 Hz What length of steel wire of diameter 2 mm should be used for pendulum The inertia of mass fixed at the free end is 00098 kgm2 Take g= 083x1011 Nm2 Determine the torsional natural frequency of the system shown in figure Neglect the mass moment of inertia of the shaft ( Figure not proportionate) Determine the natural frequency of simple pendulum considering the mass of the rod into account

Using the energy method find tThe chord may be assumed ine

Kt1

Kt2

20cm

06m 08m 04m

20cm 20cm 20cm

m

M

he natural frxtensible in

k

M

36

equency of the system shown in figure the spring mass pulley system

A rod of length l and depth H having a mass m rests on a semicylidrical surface os radius r It is tipped slightly and let go Find the frequency of oscillation

A thin bar of l having a mass m rests on a semicylidrical surface radius r It is tipped slightly and let go Find the frequency of oscillation

l

h

R

l

37

38

CHAPTER 3

DAMPED FREE VIBRATION OF SINGLE DEGREE FREEDOM SYSTEM LEARNING OBJECTIVES

] Introduction to damped free vibration

] Terminologies used in damped free vibration

] Three different types of damping

] Concept of critical damping and its importance

] Study of response of viscous damped system for cases of under damping

critical and over damping

] Concept of logarithmic decrement

] Problems related to damped free vibration problems

Viscously damped free vibration of single degree freedom sytem Consider a single degree freedom system with viscous damping as shown in figure For a viscous damped system the damping force is

where c is a constant of proportionality It is called damping coefficient or coefficient of viscous damping

Symbolically it is designated by a dashpot as shown in Fig 3

x xampx

39

At any instant of time let the mass be displaced by x from the equilibrium position Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

Okxxcxm =minusminus= ampampamp

Okxxcxm =++ ampampamp ----------------------------------------------------------(1)

or

0kxdtdx

Cdt

xdM

2

2

=++

Assuming the solution of equation as

The first term in the above equation m2Ct

eminus

is an exponentially decaying function of time The behavior of the terms in the parenthesis however depends on whether numerical value within the radicals is positive zero or negative

CASE 1 if 0mk

m2c

2

=minus

40

CASE 2 if 0mk

m2c

2

=gt

CASE 3 if 0mk

m2c

2

=lt

41

Critically damped coefficient Critically damped coefficient is defined as the value of the damping coeffient for which radical in equation

mk

m2c

m2c

S2

21 minus

plusmn

= becomes zero

Damping Ratio ( Damping factor) ( ξ ) It is defined as the ratio of damping coefficient to the critical damping coefficient Note The damping coefficient c is defined as the actual amount of damping while the damping coefficient is a constant depending upon the mass and stiffness of the system The roots of characteristics equation can be now written as Note if c=cc or ξ =1 then system is critically damped if cgt cc or ξ gt1 then system is over damped I if clt cc or ξ lt1 then system is under damped

42

Take home quiz Derive General solution of different damped system

1 Critically damped system 2 Under damped system 3 Over damped system

Logarithmic Decrement It is used to determine the amount of damping present in system This measure The rate of decay of free vibration The larger the damping the greater will be rate of decay Logarithmic decrement is defined as the natural logarithm of ratio of any two successive amplitudes

=δ

2

1

xx

ln

Expression for logarithmic decrement (δ ) Consider a damped free vibration which is expressed by

( ) ( ) φ+ξminusω= ξω t 1sinXex 2n

tn

x is the displacement at any instant of time X and φ are arbitrary constants

21

2

ξminus

πξ=δ

43

Tutorial problems 1 Derive an expression for damping ratio in terms of logarithmic decrement

2 Show that logarithmic decrement

=δ

n

o

xx

lnn1

where xo is the initial amplitude xn

is amplitude after n cycles 3 A mass of 1 kg is to be supported on a spring having stiffness K= 9800 Nm The damping constant c= 59 Nsm Determine the natural frequency of system Find also logarithmic decrement and amplitude after 3 cycles if initial displacement is 03 cm 4 A vibrating system is defined by the following parameters M = 3kg k= 100 Nm damping coefficient = 3 Nsm determine Damping factor natural frequency damped vibration logarithmic decrement ratio of consecutive amplitude no of cycles after which the original amplitude is reduced to 20 5A single degree damped vibrating system a suspended mass of 18kg makes 15 oscillations in 003 seconds The amplitude decreases to 025 of the initial value after 5 oscillations Determine the stiffness of spring Logarithmic decrement damping factor and damping coefficient 6 A damped vibration record of a spring mass system shows the following data Amplitude at the end of 2nd cycle is 9mm amplitude at eh end of 3rd cycle is 6mm amplitude at the end of 4th cycle is 4mm stiffness of spring =8000 Nm weight =20N Find the logarithmic decrement damping force at unit velocity periodic time of vibration 7 For the system shown in figure the characteristics of dashpot is such that the constant force of 49 N is applied to the piston it velocity is found to be constant at 012 ms Determine the value of C would you expect the complete system to be periodic or aperiodic 8 For a torsional system shown in the figuref1 is natural frequency in air and f2 is the natural frequency when immersed in oil Determine the damping coefficient

Viscous fluid

Disc of mass moment of Inertia J

2a

C

m M

b

a a

k

9 A thin plate of area A and weigh w is fixed to the end of the spring and allow to oscillate in a viscous fluid as shown in figure It f1 is the frequency of oscillation in air and f2 in liquid Show that

22

21 ff

gA2

minusπω

=micro where the damping force on the plate

AV2Fd micro= V being the velocity of vibration 10 A torsional pendulum has a natural frequency of 175 cyclessecond vibrating in vacuum The mass moment of inertia is 2 kgm2 It is immersed in oil and it is observed that the natural frequency is 142 cyclessecond Find the damping torque If the disc is replaced by 34 deg When in oil find the displacement at the end of the first complete cycle 11 An automobile can be modeled as a mass placed on 4 shock absorbers each consisting of a spring and a damper Such that each spring is equally loaded Determine the stiffness damping constant of each shock absorber So that natural frequency is 2Hz and system is critically damped The mass of vehicle is 200kg 12 Write the differential equation of motion for the system shown in the figure below Determine the natural frequency of the damped oscillation and critical damping coefficient

Determine the differential equation of motion and find the critical cdamped natural frequency for system shown in figure

a

C

m M

M =200kg

44

200kg kgkg

oefficient and

45

Derive the equation of motion and find the critical damping coefficient and damped natural frequency for the system shown below Write the differential equation of motion and determine the expression for critical damping coefficient and damped natural frequency of system shown in figure

Set up a differential equation of motion for system shown in figure and determine the following Undamped natural frequencies critical damping coefficient damping ratio damped natural frequency

b

C

kM

a

m

46

A mass of 300N is resting on two springs of stiffness 3000 Nm each and a dashpot of damping coefficient 150 Nsm as shown in figure If the initial velocity of 10cms is given to mass at its equilibrium position what will be the displacement from the equilibrium position at the end of first second Derive the formula used A gun barrel of mass 600kg has a recoil spring of stiffness 294 knm if the barrel recoils 13m on firing determine i) The initial recoil velocity of barrel ii) The critical damping coefficient of dashpot which is engaged at the end of recoil stroke Iii) Time required for the barrel to return at the end of recoil stroke Iv) time required for the barrel to return to a position 5cm from the initial position

W= 3000 N

FIRING

CHAPTER 4

FORCED VIBRATION LEARNING OBJECTIVES

] Introduction to forced free vibration

] Terminologies used in forced vibration

] Concept of magnification factor

] Study of reciprocating and rotating unbalance

] Concept of base excitation and support excitation

] Concept of Transmissibility ratio

] Problems related to forced vibration

The vibration that takes place under the excitation of external forces are called

forced vibration When a system is subjected to harmonic excitation it is forced to vibrate at the same frequency as that of excitation Most of the mechanical systems are prone to steady state forced vibration which cause fatigue failure The vibration problem also becomes very important when any of excitation frequencies coincide with one of natural frequencies of a system Under this condition resonance occurs during which large amplitude of vibrations are observed To avoid resonance external frequency (operating speed) may be changed or properties of the system may be changed to alter the natural frequency In some cases sufficient amount of damping may be provided to avoid large amplitude during resonance Thus the problem of force vibration is very important in mechanical design

FORCED VIBRATIONS OF SINGLE DOF SYSTEMS UNDER HARMONIC EXCITATION

` [ ] tsin FF o ω=

x

sin FF o=

Fo

xampamp

[ ] tω

xamp

47

48

Consider a spring mass system having a viscous damping excited by a harmonic force [ ] tsin FF o ω= as shown in figure The various forces acting on the system is shown in figure Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

O)tsin(Fkxxcxm o =ω+minusminus= ampampamp

)tsin(Fkxxcxm o ω=++ ampampamp ----------------------------------------------------------(1)

or

)tsin(Fkxdtdx

Cdt

xdM o2

2

ω=++

Equation (1) represents the differential equation of motion of the system The solution to this equation consists of two parts namely complementary function and a particular integral

The first of solution is also termed as transient solution of homogeneous solution and can be obtained from the homogeneous differential equation 0kxxcxm =++ ampampamp

The second part of solution particular integral is also termed as steady state solution ad is of form )tsin(Xx φminusω= Where X= steady state amplitude φ is phase difference between external force and displacement

)tsin(Xx φminusω=

The vector representation of the shown

49

The force polygon for the above vector diagram can be written as follows

( )φminusωt O

KX

xm 2ω

Fo

xCω

tω

φ

50

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

is the governing equation

Zero frequency deflection (Xst) It is deflection observed at zero frequency ( when the external frequency 0=ω

( ) kF

001

kF

x o

22

o

st =

+minus=

Magnification factor (M) It is defined as the ratio of steady state amplitude to zero frequency deflection xst

M = arX

X=

A plot of magnification factor versus frequency ratio

damping factor ( )ξ is called frequency plot of a harmon All the frequency terms starts from magnification factor of damping ratio ( )ξ The amplitude of vibration at any on the value of damping The amplitude will be higher for higher damping The amplitude of vibration at reso

amplitude tends to zero as

ωω

n tends to infinity

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

Freq

ζ =

ζ = 1

Magnification Factor M = XXst

kFo

51

ωω

nfor varying values of

ically excite vibrating system

M=1 what ever may be value particular frequency depends

for lesser damping and lesser nance will be very high The

uency Ratio r = (ωωn)

ζ = 0 0

ζ = 025 ζ = 0375 ζ = 0 5

ζ = 0707

ζ = 2

52

Plot of phase angle φ versus frequency ratio

ωω

n is shown in the

figure The phase angle φ is zero for all the values of damping ratio

at

ωω

n=0 The phase

angle φ is 90 for all the values of damping ratio

at

ωω

n=1 ie at

resonance The phase angle between o to 90 deg for all the values of

ωω

n= 0 to 1 and

between 90 deg to 180 deg for all the values of

ωω

ngt1

Class work Derive an expression for frequency at peak amplitude

2np 21 ξminusω=ω=ω is the required equation

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

53

Derive expression amplitude at resonance ( In dimensional form) In dimensional form We know that steady state amplitude

( ) ( )

ω+ωminus=

222

o

Cmk

Fx Equation in dimensional form

We know that at resonance mk

n =ω=ω

( )

ω+

minus

=2

n

2

oR

Cmk

mk

Fx

( )

ω=

2n

oR

C

Fx

( )

ω=

n

oR C

Fx

Class work Derive an expression for amplitude at resonance ( in non dimensional form)

Tutorial Problems on Forced Vibration 1A weight of 60N is suspended by a spring stiffness 12 kNm is forced to vibrate by a harmonic force of 10N assuming viscous damping of 0086 kN-sm Find i) resonance frequency ii) amplitude at resonance iii) phase angle at resonance iv) frequency corresponding to peak amplitude 2 A periodic torque having a maximum value of 05888Nm at a frequency of 4 rads is impressed upon a flywheel suspended from a wire The wheel has mass moment of inertia 012 kgm2 and the wire has stiffness 1176 Nmrad A viscous dashpot applies damping couple of 0784 Nm an angular velocity of 2 rads calculate i)maximum angular displacement from rest position ii) maximum couple applied to dashpot Iii) angle by which the angular displacement lags the torque A mass of 1019 kg is suspended from a spring stiffness 20 Nmm viscous damping causes the amplitude to decrease to 110th of initial value in 4 complete cycle If a periodic force of 20 cos(50 t) is applied to the mass Find the amplitude of forced vibration What would be the amplitude of oscillation if the applied force coincides with the natural period

m=1019 kg

[ ] t50cos 30F =

54

55

ROTATING UNBALANCE Unbalance in rotating machinery is a common source of vibration excitation Consider a machine of mass m having mounted on a foundation of stiffness k and damping coefficient C as shown in figure The unbalance is represented by an eccentric rotating mass m with eccentricity (e) and having a angular velocity (ω) let x be the displacement of non rotating mass from the static equilibrium position The displacement of rotating mass m from the static equilibrium is ( ) tsin( ex ω+ Applying Newtonrsquos law of motion we have

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

56

RECIPROCATING UNBALANCE Unbalance in reciprocating machinery is a common source of vibration excitation Consider a machine of mass M having mounted on a foundation of stiffness K and damping coefficient C as shown in figure The unbalance is represented by a reciprocating mass (m) having crank radus (e) and connecting rod of length (l) Let ωωωω be angular velocity of the crank Let x be the displacement of non reciprocating mass from the static equilibrium position at any instant of time The displacement of reciprocating mass m from the static equilibrium position is

)2sin(sin +++ tle

tex ωωωωωωωω

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

57

Note The equations obtained in reciprocating unbalance is same as the rotating unbalance and having the following differences Rotating Unbalance Reciprocating Unbalance Here mass of the rotating element is considered

Here mass of reciprocating element is considered

Here the eccentricity (e) which is the distance between the centre of rotation and CG

Here the eccentricity e is the crank radius and crank radius = stroke length2

The frequency plot of non dimensional factor( ( )me

MX versus frequency ratio

ωω

n for various values of ( )ξξξξ is shown in figure This frequency plot is same for

the rotating and reciprocating unbalance with the following points are noted from the diagram

1 Damping ratio plays very important role during resonance in controlling the amplitude

2 When

ωω

n=0 then ( )me

MX =0 for all the values of damping ratio ( )ξξξξ

3 When

ωω

nis higher than ( )me

MX tends to unity

4 At very high speeds ie at very high frequency ratio

ωω

n there is no effect

of damping The frequency plot of phase angle versus the frequency ratio is given as shown in figure The explanation is same as in forced harmonic vibration

58

VIBRATION ISOLATION

Vibratory forces and motions generated by machine and other caused are unavoidable Howeever their effect on the other dynamical systems can be minimized by proper isolator design An isolation system is either to protect o delicate machine from excessive vibration transmitted to it from its supporting structure or to prevent vibratory forces and motions generated by machines from being transmitted to its surroundings The effectiveness of isolation may be measured in terms of forces and motion transmitted The first type is known as force isolation and the second type is known as motion isolation

Force isolation (Force Transmissibility) When vibratory forces generated by machine are isolated the effectiveness of forced isolation can be made by defining the term transmissibility The term transmissibility is defined as the ratio of force transmitted to the foundation to that of the external impressed force

Force Transmissibility = o

tr

FF

force Impressed Externalfoundation toted transmitForce

=

Consider viscous damped spring mass system as shown in figure Upon this system an external harmonic force ) t (Sin Fo ω is applied The difference forces acting on system are inertia forces damping force spring force and external force All these forces are represented on force polygon shown below Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these two forces

( ) ( )22 xckx ω+=

59

This is the expression for force transmissibility in dimensional form To represent same in the non dimensional form divide numerator and denominator by k

2

n

22

n

2

n

o

tr

21

21

FF

TR

ωω

ξ+

ωω

minus

ωω

ξ+

==

PHASE DIFFERNECE BETWEEN EXTERNAL FORCE AND DISPLACEMENT Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these

ωω

minus

ωω

ξ=

ωminusω

=φ minusminus2

n

n1

2

1

1

2tan

mKC

tan

The phase difference between force transmitted to the foundation Ftr to the displacement

kc

KxxC

tanω

=ω

=α or

ωω

ξ=

ω=α minusminus

n

11 2tanKx

xCtan

The phase difference between external force Fo and force transmitted to foundationis β

60

β= αminusφ =

ωω

minus

ωω

ξminus

2

n

n1

1

2tan -

ωω

ξminus

n

1 2tan

Displacement Transmissibility It is the ratio of amplitude of steady state vibration of machine to that of the base motion Forced vibration due to support motion Consider a spring mass damper system subjected to harmonic excitation )tsin(Yy ω= at the base as shown in figure Let x be the absolute displacement of the mass at any instant of time t The different forces acting on system are shown in free body diagram Applying Newtonrsquos Law of motion Ma = sumF

)yx(k)yx(cxm minusminusminusminus= ampampampamp

0)yx(k)yx(cxm =minus+minus+ ampampampamp

61

Frequency plot of transmissibility versus frequency ratio

62

The plot of transmissibility versus frequency ratio is as shown in figure The following points can be noted from it

i) The value of transmissibility is unity at frequency ratio

ωω

n=0 and 2

ii) For an undamped system ( )ξ =0 transmissibility tends to infin at resonance iii) The value of transmissibility is less than unity for the value of frequency ratio

greater than 2

iv) For frequency ratio

ωω

nlt 2 smaller damping ratio lead to larger values of

transmissibility on the other hand for frequency ratio

ωω

ngt 2 smaller damping

ratio leads to smaller value of transmissibility

The frequency plot of transmissibility (TR) versus frequency ratio

ωω

n can be

broadly divided into two zones

1

ωω

n= 0 to 2

2

ωω

n= 2 to infin

CASE 1

When

ωω

n= 0 to 2

In this zone the value of transmissibility is always greater than 1 for all values of damping ratio In this zone the transmissibility decreases with increase in damping Hence we can conclude that in this zone damping is useful in isolating vibration forces being transmitted to the support CASE 2

When

ωω

n= 2 to infin

In this zone the value of transmissibility is less than 1 for all values of damping ratio In this zone the transmissibility increases with increase in damping Hence we can conclude that in this zone damping will adversely affect in isolating vibratory forces

63

Tutorial problems on Forced Vibration 1 Machine of mass 75kg is mounted on spring of stiffness 12000 Nm with an assumed damping factor ξ =02 piston within the machine of mass 2 kg has a reciprocating motion with a stroke of 75 cm and speed of 15 cyclessecond Assuming the motion of piston to be harmonic determine amplitude of machine ( X) Transmissibility and force transmitted to foundation phase angle of transmitted force wrt to exciting force 2 Motor weighs 600N is mounted on a simple beam that has a spring rate of 400 Ncm The motor armature base 160 N and has an eccentricity of 0005cm What will be the amplitude of motion when it runs at 1760 RPM Neglect damping the deflection of armature shaft and weight of the beam 3 A Heavy machine of mass 300 kg is supported on resilient foundation the static deflection of foundation due to weight of machine is found to be 75mm It is observed that the machine vibrates with an amplitude of 10mm when the base of the foundation is subjected to harmonic oscillation at the undampeds natural frequency of system with an amplitude of 25mm Find the damping constant of foundation Ii) dynamic force amplitude on the base iii) amplitude of the displacement of machine relative to base 4 An air compressor of mass 450 kg operates at a constant speed of 1750 RPM The reciprocating masses are 10 kg the crank radius is 100mm If the damper for mounting introduces a damping factor 015 i) specify the springs of mounting such that only 20 of the unbalanced force is transmitted to the foundation ii) determine the magnitude of the transmitted force bull Repeat the above problem by assuming that there is no damping in system

5 A body of a car having mass 1500 kg and is mounted on 4 equal springs which deflect through 225 cm under the weight of the body A total damping forces the 4 shock absorbers is 46 N at a velocity of 1cmsec The car is placed with all the four wheels on a test platform which is move up and down at resonant speed With an amplitude of 25cm Find the amplitude of the car body on its springs assuming its CG to be in the centre of wheel base 6A 75 kg machine which is mounted on springs of stiffness 116x105 Nm with the damping of 20=ξ A 2kg piston within the machine as reciprocating motion with a stroke of 008m and speed of 3000 RPM Assuming the motion of piston to be harmonic determine the amplitude of vibration of machine

7 The spring of an automobile are compressed 10cm under its own weight Find the critical speed when a trailer is traveling over road with a people approximaterly by a sine wave of amplitude 75 cm and wave length 15m What will be the amplitude of vibration at 64kmhour 8 An eccentric mass excites us used to determine vibratory characteristics of a structure of mass 25kg as given in figure At a speed of 1000 RPM the eccentric mass to beat top position at the instant the structure was moved upwards through its static equilibrium position and corresponding amplitude was 20mm The eccentric mass is 05kg at an eccentricity of 10mm Determine i) Undamped natural frequency of the system ii) Damping ratio iii) Steady state amplitude and phase angle at 1500 RPM iv) Force transmitted to the ground at 1200 RPM

M=25kg

C

K

Eccentric mass

64

65

CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Support motion Consider a spring mass damper system as shown in figure Let it excited by motion

of support let y be the harmonic displacement of support From Newtonrsquos second law of motion

)yx(k)yx(cxm minusminusminusminus= ampampampamp

)1(0)yx(k)yx(cxm minusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminus=minus+minus+ ampampampamp

Let z be the relative displacement of the mass with respect to support then

Z=x-y

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

34

r

B

K K

Home work

Determine the natural frequency of pendulum using Rayleighrsquos method

A Cylinder of mass m and moment of inertia Jo is to role without slipping but is sustained by spring K as shown in figure Determine the natural frequency of oscillation

Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping Determine the natural frequency of system shown in figure If the cylinder is free to roll without slipping

a

Determine the natural frequency of system shown where in cylindrical disc rolls over the inextensible spring A cylinder of radius r rolls without slipping on cylindrical the equation of motion and hence determine the naoscillations about the lowest point shown Repeat the above problem ndash Instead of cylinder assume swithout slipping on concave surface as shown above

r

K

35

portion of radius R Derive tural frequency of small

phere of radius with r rolls

The torsional pendulum has a natural frequency of 5 Hz What length of steel wire of diameter 2 mm should be used for pendulum The inertia of mass fixed at the free end is 00098 kgm2 Take g= 083x1011 Nm2 Determine the torsional natural frequency of the system shown in figure Neglect the mass moment of inertia of the shaft ( Figure not proportionate) Determine the natural frequency of simple pendulum considering the mass of the rod into account

Using the energy method find tThe chord may be assumed ine

Kt1

Kt2

20cm

06m 08m 04m

20cm 20cm 20cm

m

M

he natural frxtensible in

k

M

36

equency of the system shown in figure the spring mass pulley system

A rod of length l and depth H having a mass m rests on a semicylidrical surface os radius r It is tipped slightly and let go Find the frequency of oscillation

A thin bar of l having a mass m rests on a semicylidrical surface radius r It is tipped slightly and let go Find the frequency of oscillation

l

h

R

l

37

38

CHAPTER 3

DAMPED FREE VIBRATION OF SINGLE DEGREE FREEDOM SYSTEM LEARNING OBJECTIVES

] Introduction to damped free vibration

] Terminologies used in damped free vibration

] Three different types of damping

] Concept of critical damping and its importance

] Study of response of viscous damped system for cases of under damping

critical and over damping

] Concept of logarithmic decrement

] Problems related to damped free vibration problems

Viscously damped free vibration of single degree freedom sytem Consider a single degree freedom system with viscous damping as shown in figure For a viscous damped system the damping force is

where c is a constant of proportionality It is called damping coefficient or coefficient of viscous damping

Symbolically it is designated by a dashpot as shown in Fig 3

x xampx

39

At any instant of time let the mass be displaced by x from the equilibrium position Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

Okxxcxm =minusminus= ampampamp

Okxxcxm =++ ampampamp ----------------------------------------------------------(1)

or

0kxdtdx

Cdt

xdM

2

2

=++

Assuming the solution of equation as

The first term in the above equation m2Ct

eminus

is an exponentially decaying function of time The behavior of the terms in the parenthesis however depends on whether numerical value within the radicals is positive zero or negative

CASE 1 if 0mk

m2c

2

=minus

40

CASE 2 if 0mk

m2c

2

=gt

CASE 3 if 0mk

m2c

2

=lt

41

Critically damped coefficient Critically damped coefficient is defined as the value of the damping coeffient for which radical in equation

mk

m2c

m2c

S2

21 minus

plusmn

= becomes zero

Damping Ratio ( Damping factor) ( ξ ) It is defined as the ratio of damping coefficient to the critical damping coefficient Note The damping coefficient c is defined as the actual amount of damping while the damping coefficient is a constant depending upon the mass and stiffness of the system The roots of characteristics equation can be now written as Note if c=cc or ξ =1 then system is critically damped if cgt cc or ξ gt1 then system is over damped I if clt cc or ξ lt1 then system is under damped

42

Take home quiz Derive General solution of different damped system

1 Critically damped system 2 Under damped system 3 Over damped system

Logarithmic Decrement It is used to determine the amount of damping present in system This measure The rate of decay of free vibration The larger the damping the greater will be rate of decay Logarithmic decrement is defined as the natural logarithm of ratio of any two successive amplitudes

=δ

2

1

xx

ln

Expression for logarithmic decrement (δ ) Consider a damped free vibration which is expressed by

( ) ( ) φ+ξminusω= ξω t 1sinXex 2n

tn

x is the displacement at any instant of time X and φ are arbitrary constants

21

2

ξminus

πξ=δ

43

Tutorial problems 1 Derive an expression for damping ratio in terms of logarithmic decrement

2 Show that logarithmic decrement

=δ

n

o

xx

lnn1

where xo is the initial amplitude xn

is amplitude after n cycles 3 A mass of 1 kg is to be supported on a spring having stiffness K= 9800 Nm The damping constant c= 59 Nsm Determine the natural frequency of system Find also logarithmic decrement and amplitude after 3 cycles if initial displacement is 03 cm 4 A vibrating system is defined by the following parameters M = 3kg k= 100 Nm damping coefficient = 3 Nsm determine Damping factor natural frequency damped vibration logarithmic decrement ratio of consecutive amplitude no of cycles after which the original amplitude is reduced to 20 5A single degree damped vibrating system a suspended mass of 18kg makes 15 oscillations in 003 seconds The amplitude decreases to 025 of the initial value after 5 oscillations Determine the stiffness of spring Logarithmic decrement damping factor and damping coefficient 6 A damped vibration record of a spring mass system shows the following data Amplitude at the end of 2nd cycle is 9mm amplitude at eh end of 3rd cycle is 6mm amplitude at the end of 4th cycle is 4mm stiffness of spring =8000 Nm weight =20N Find the logarithmic decrement damping force at unit velocity periodic time of vibration 7 For the system shown in figure the characteristics of dashpot is such that the constant force of 49 N is applied to the piston it velocity is found to be constant at 012 ms Determine the value of C would you expect the complete system to be periodic or aperiodic 8 For a torsional system shown in the figuref1 is natural frequency in air and f2 is the natural frequency when immersed in oil Determine the damping coefficient

Viscous fluid

Disc of mass moment of Inertia J

2a

C

m M

b

a a

k

9 A thin plate of area A and weigh w is fixed to the end of the spring and allow to oscillate in a viscous fluid as shown in figure It f1 is the frequency of oscillation in air and f2 in liquid Show that

22

21 ff

gA2

minusπω

=micro where the damping force on the plate

AV2Fd micro= V being the velocity of vibration 10 A torsional pendulum has a natural frequency of 175 cyclessecond vibrating in vacuum The mass moment of inertia is 2 kgm2 It is immersed in oil and it is observed that the natural frequency is 142 cyclessecond Find the damping torque If the disc is replaced by 34 deg When in oil find the displacement at the end of the first complete cycle 11 An automobile can be modeled as a mass placed on 4 shock absorbers each consisting of a spring and a damper Such that each spring is equally loaded Determine the stiffness damping constant of each shock absorber So that natural frequency is 2Hz and system is critically damped The mass of vehicle is 200kg 12 Write the differential equation of motion for the system shown in the figure below Determine the natural frequency of the damped oscillation and critical damping coefficient

Determine the differential equation of motion and find the critical cdamped natural frequency for system shown in figure

a

C

m M

M =200kg

44

200kg kgkg

oefficient and

45

Derive the equation of motion and find the critical damping coefficient and damped natural frequency for the system shown below Write the differential equation of motion and determine the expression for critical damping coefficient and damped natural frequency of system shown in figure

Set up a differential equation of motion for system shown in figure and determine the following Undamped natural frequencies critical damping coefficient damping ratio damped natural frequency

b

C

kM

a

m

46

A mass of 300N is resting on two springs of stiffness 3000 Nm each and a dashpot of damping coefficient 150 Nsm as shown in figure If the initial velocity of 10cms is given to mass at its equilibrium position what will be the displacement from the equilibrium position at the end of first second Derive the formula used A gun barrel of mass 600kg has a recoil spring of stiffness 294 knm if the barrel recoils 13m on firing determine i) The initial recoil velocity of barrel ii) The critical damping coefficient of dashpot which is engaged at the end of recoil stroke Iii) Time required for the barrel to return at the end of recoil stroke Iv) time required for the barrel to return to a position 5cm from the initial position

W= 3000 N

FIRING

CHAPTER 4

FORCED VIBRATION LEARNING OBJECTIVES

] Introduction to forced free vibration

] Terminologies used in forced vibration

] Concept of magnification factor

] Study of reciprocating and rotating unbalance

] Concept of base excitation and support excitation

] Concept of Transmissibility ratio

] Problems related to forced vibration

The vibration that takes place under the excitation of external forces are called

forced vibration When a system is subjected to harmonic excitation it is forced to vibrate at the same frequency as that of excitation Most of the mechanical systems are prone to steady state forced vibration which cause fatigue failure The vibration problem also becomes very important when any of excitation frequencies coincide with one of natural frequencies of a system Under this condition resonance occurs during which large amplitude of vibrations are observed To avoid resonance external frequency (operating speed) may be changed or properties of the system may be changed to alter the natural frequency In some cases sufficient amount of damping may be provided to avoid large amplitude during resonance Thus the problem of force vibration is very important in mechanical design

FORCED VIBRATIONS OF SINGLE DOF SYSTEMS UNDER HARMONIC EXCITATION

` [ ] tsin FF o ω=

x

sin FF o=

Fo

xampamp

[ ] tω

xamp

47

48

Consider a spring mass system having a viscous damping excited by a harmonic force [ ] tsin FF o ω= as shown in figure The various forces acting on the system is shown in figure Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

O)tsin(Fkxxcxm o =ω+minusminus= ampampamp

)tsin(Fkxxcxm o ω=++ ampampamp ----------------------------------------------------------(1)

or

)tsin(Fkxdtdx

Cdt

xdM o2

2

ω=++

Equation (1) represents the differential equation of motion of the system The solution to this equation consists of two parts namely complementary function and a particular integral

The first of solution is also termed as transient solution of homogeneous solution and can be obtained from the homogeneous differential equation 0kxxcxm =++ ampampamp

The second part of solution particular integral is also termed as steady state solution ad is of form )tsin(Xx φminusω= Where X= steady state amplitude φ is phase difference between external force and displacement

)tsin(Xx φminusω=

The vector representation of the shown

49

The force polygon for the above vector diagram can be written as follows

( )φminusωt O

KX

xm 2ω

Fo

xCω

tω

φ

50

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

is the governing equation

Zero frequency deflection (Xst) It is deflection observed at zero frequency ( when the external frequency 0=ω

( ) kF

001

kF

x o

22

o

st =

+minus=

Magnification factor (M) It is defined as the ratio of steady state amplitude to zero frequency deflection xst

M = arX

X=

A plot of magnification factor versus frequency ratio

damping factor ( )ξ is called frequency plot of a harmon All the frequency terms starts from magnification factor of damping ratio ( )ξ The amplitude of vibration at any on the value of damping The amplitude will be higher for higher damping The amplitude of vibration at reso

amplitude tends to zero as

ωω

n tends to infinity

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

Freq

ζ =

ζ = 1

Magnification Factor M = XXst

kFo

51

ωω

nfor varying values of

ically excite vibrating system

M=1 what ever may be value particular frequency depends

for lesser damping and lesser nance will be very high The

uency Ratio r = (ωωn)

ζ = 0 0

ζ = 025 ζ = 0375 ζ = 0 5

ζ = 0707

ζ = 2

52

Plot of phase angle φ versus frequency ratio

ωω

n is shown in the

figure The phase angle φ is zero for all the values of damping ratio

at

ωω

n=0 The phase

angle φ is 90 for all the values of damping ratio

at

ωω

n=1 ie at

resonance The phase angle between o to 90 deg for all the values of

ωω

n= 0 to 1 and

between 90 deg to 180 deg for all the values of

ωω

ngt1

Class work Derive an expression for frequency at peak amplitude

2np 21 ξminusω=ω=ω is the required equation

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

53

Derive expression amplitude at resonance ( In dimensional form) In dimensional form We know that steady state amplitude

( ) ( )

ω+ωminus=

222

o

Cmk

Fx Equation in dimensional form

We know that at resonance mk

n =ω=ω

( )

ω+

minus

=2

n

2

oR

Cmk

mk

Fx

( )

ω=

2n

oR

C

Fx

( )

ω=

n

oR C

Fx

Class work Derive an expression for amplitude at resonance ( in non dimensional form)

Tutorial Problems on Forced Vibration 1A weight of 60N is suspended by a spring stiffness 12 kNm is forced to vibrate by a harmonic force of 10N assuming viscous damping of 0086 kN-sm Find i) resonance frequency ii) amplitude at resonance iii) phase angle at resonance iv) frequency corresponding to peak amplitude 2 A periodic torque having a maximum value of 05888Nm at a frequency of 4 rads is impressed upon a flywheel suspended from a wire The wheel has mass moment of inertia 012 kgm2 and the wire has stiffness 1176 Nmrad A viscous dashpot applies damping couple of 0784 Nm an angular velocity of 2 rads calculate i)maximum angular displacement from rest position ii) maximum couple applied to dashpot Iii) angle by which the angular displacement lags the torque A mass of 1019 kg is suspended from a spring stiffness 20 Nmm viscous damping causes the amplitude to decrease to 110th of initial value in 4 complete cycle If a periodic force of 20 cos(50 t) is applied to the mass Find the amplitude of forced vibration What would be the amplitude of oscillation if the applied force coincides with the natural period

m=1019 kg

[ ] t50cos 30F =

54

55

ROTATING UNBALANCE Unbalance in rotating machinery is a common source of vibration excitation Consider a machine of mass m having mounted on a foundation of stiffness k and damping coefficient C as shown in figure The unbalance is represented by an eccentric rotating mass m with eccentricity (e) and having a angular velocity (ω) let x be the displacement of non rotating mass from the static equilibrium position The displacement of rotating mass m from the static equilibrium is ( ) tsin( ex ω+ Applying Newtonrsquos law of motion we have

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

56

RECIPROCATING UNBALANCE Unbalance in reciprocating machinery is a common source of vibration excitation Consider a machine of mass M having mounted on a foundation of stiffness K and damping coefficient C as shown in figure The unbalance is represented by a reciprocating mass (m) having crank radus (e) and connecting rod of length (l) Let ωωωω be angular velocity of the crank Let x be the displacement of non reciprocating mass from the static equilibrium position at any instant of time The displacement of reciprocating mass m from the static equilibrium position is

)2sin(sin +++ tle

tex ωωωωωωωω

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

57

Note The equations obtained in reciprocating unbalance is same as the rotating unbalance and having the following differences Rotating Unbalance Reciprocating Unbalance Here mass of the rotating element is considered

Here mass of reciprocating element is considered

Here the eccentricity (e) which is the distance between the centre of rotation and CG

Here the eccentricity e is the crank radius and crank radius = stroke length2

The frequency plot of non dimensional factor( ( )me

MX versus frequency ratio

ωω

n for various values of ( )ξξξξ is shown in figure This frequency plot is same for

the rotating and reciprocating unbalance with the following points are noted from the diagram

1 Damping ratio plays very important role during resonance in controlling the amplitude

2 When

ωω

n=0 then ( )me

MX =0 for all the values of damping ratio ( )ξξξξ

3 When

ωω

nis higher than ( )me

MX tends to unity

4 At very high speeds ie at very high frequency ratio

ωω

n there is no effect

of damping The frequency plot of phase angle versus the frequency ratio is given as shown in figure The explanation is same as in forced harmonic vibration

58

VIBRATION ISOLATION

Vibratory forces and motions generated by machine and other caused are unavoidable Howeever their effect on the other dynamical systems can be minimized by proper isolator design An isolation system is either to protect o delicate machine from excessive vibration transmitted to it from its supporting structure or to prevent vibratory forces and motions generated by machines from being transmitted to its surroundings The effectiveness of isolation may be measured in terms of forces and motion transmitted The first type is known as force isolation and the second type is known as motion isolation

Force isolation (Force Transmissibility) When vibratory forces generated by machine are isolated the effectiveness of forced isolation can be made by defining the term transmissibility The term transmissibility is defined as the ratio of force transmitted to the foundation to that of the external impressed force

Force Transmissibility = o

tr

FF

force Impressed Externalfoundation toted transmitForce

=

Consider viscous damped spring mass system as shown in figure Upon this system an external harmonic force ) t (Sin Fo ω is applied The difference forces acting on system are inertia forces damping force spring force and external force All these forces are represented on force polygon shown below Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these two forces

( ) ( )22 xckx ω+=

59

This is the expression for force transmissibility in dimensional form To represent same in the non dimensional form divide numerator and denominator by k

2

n

22

n

2

n

o

tr

21

21

FF

TR

ωω

ξ+

ωω

minus

ωω

ξ+

==

PHASE DIFFERNECE BETWEEN EXTERNAL FORCE AND DISPLACEMENT Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these

ωω

minus

ωω

ξ=

ωminusω

=φ minusminus2

n

n1

2

1

1

2tan

mKC

tan

The phase difference between force transmitted to the foundation Ftr to the displacement

kc

KxxC

tanω

=ω

=α or

ωω

ξ=

ω=α minusminus

n

11 2tanKx

xCtan

The phase difference between external force Fo and force transmitted to foundationis β

60

β= αminusφ =

ωω

minus

ωω

ξminus

2

n

n1

1

2tan -

ωω

ξminus

n

1 2tan

Displacement Transmissibility It is the ratio of amplitude of steady state vibration of machine to that of the base motion Forced vibration due to support motion Consider a spring mass damper system subjected to harmonic excitation )tsin(Yy ω= at the base as shown in figure Let x be the absolute displacement of the mass at any instant of time t The different forces acting on system are shown in free body diagram Applying Newtonrsquos Law of motion Ma = sumF

)yx(k)yx(cxm minusminusminusminus= ampampampamp

0)yx(k)yx(cxm =minus+minus+ ampampampamp

61

Frequency plot of transmissibility versus frequency ratio

62

The plot of transmissibility versus frequency ratio is as shown in figure The following points can be noted from it

i) The value of transmissibility is unity at frequency ratio

ωω

n=0 and 2

ii) For an undamped system ( )ξ =0 transmissibility tends to infin at resonance iii) The value of transmissibility is less than unity for the value of frequency ratio

greater than 2

iv) For frequency ratio

ωω

nlt 2 smaller damping ratio lead to larger values of

transmissibility on the other hand for frequency ratio

ωω

ngt 2 smaller damping

ratio leads to smaller value of transmissibility

The frequency plot of transmissibility (TR) versus frequency ratio

ωω

n can be

broadly divided into two zones

1

ωω

n= 0 to 2

2

ωω

n= 2 to infin

CASE 1

When

ωω

n= 0 to 2

In this zone the value of transmissibility is always greater than 1 for all values of damping ratio In this zone the transmissibility decreases with increase in damping Hence we can conclude that in this zone damping is useful in isolating vibration forces being transmitted to the support CASE 2

When

ωω

n= 2 to infin

In this zone the value of transmissibility is less than 1 for all values of damping ratio In this zone the transmissibility increases with increase in damping Hence we can conclude that in this zone damping will adversely affect in isolating vibratory forces

63

Tutorial problems on Forced Vibration 1 Machine of mass 75kg is mounted on spring of stiffness 12000 Nm with an assumed damping factor ξ =02 piston within the machine of mass 2 kg has a reciprocating motion with a stroke of 75 cm and speed of 15 cyclessecond Assuming the motion of piston to be harmonic determine amplitude of machine ( X) Transmissibility and force transmitted to foundation phase angle of transmitted force wrt to exciting force 2 Motor weighs 600N is mounted on a simple beam that has a spring rate of 400 Ncm The motor armature base 160 N and has an eccentricity of 0005cm What will be the amplitude of motion when it runs at 1760 RPM Neglect damping the deflection of armature shaft and weight of the beam 3 A Heavy machine of mass 300 kg is supported on resilient foundation the static deflection of foundation due to weight of machine is found to be 75mm It is observed that the machine vibrates with an amplitude of 10mm when the base of the foundation is subjected to harmonic oscillation at the undampeds natural frequency of system with an amplitude of 25mm Find the damping constant of foundation Ii) dynamic force amplitude on the base iii) amplitude of the displacement of machine relative to base 4 An air compressor of mass 450 kg operates at a constant speed of 1750 RPM The reciprocating masses are 10 kg the crank radius is 100mm If the damper for mounting introduces a damping factor 015 i) specify the springs of mounting such that only 20 of the unbalanced force is transmitted to the foundation ii) determine the magnitude of the transmitted force bull Repeat the above problem by assuming that there is no damping in system

5 A body of a car having mass 1500 kg and is mounted on 4 equal springs which deflect through 225 cm under the weight of the body A total damping forces the 4 shock absorbers is 46 N at a velocity of 1cmsec The car is placed with all the four wheels on a test platform which is move up and down at resonant speed With an amplitude of 25cm Find the amplitude of the car body on its springs assuming its CG to be in the centre of wheel base 6A 75 kg machine which is mounted on springs of stiffness 116x105 Nm with the damping of 20=ξ A 2kg piston within the machine as reciprocating motion with a stroke of 008m and speed of 3000 RPM Assuming the motion of piston to be harmonic determine the amplitude of vibration of machine

7 The spring of an automobile are compressed 10cm under its own weight Find the critical speed when a trailer is traveling over road with a people approximaterly by a sine wave of amplitude 75 cm and wave length 15m What will be the amplitude of vibration at 64kmhour 8 An eccentric mass excites us used to determine vibratory characteristics of a structure of mass 25kg as given in figure At a speed of 1000 RPM the eccentric mass to beat top position at the instant the structure was moved upwards through its static equilibrium position and corresponding amplitude was 20mm The eccentric mass is 05kg at an eccentricity of 10mm Determine i) Undamped natural frequency of the system ii) Damping ratio iii) Steady state amplitude and phase angle at 1500 RPM iv) Force transmitted to the ground at 1200 RPM

M=25kg

C

K

Eccentric mass

64

65

CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Support motion Consider a spring mass damper system as shown in figure Let it excited by motion

of support let y be the harmonic displacement of support From Newtonrsquos second law of motion

)yx(k)yx(cxm minusminusminusminus= ampampampamp

)1(0)yx(k)yx(cxm minusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminus=minus+minus+ ampampampamp

Let z be the relative displacement of the mass with respect to support then

Z=x-y

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

Determine the natural frequency of system shown where in cylindrical disc rolls over the inextensible spring A cylinder of radius r rolls without slipping on cylindrical the equation of motion and hence determine the naoscillations about the lowest point shown Repeat the above problem ndash Instead of cylinder assume swithout slipping on concave surface as shown above

r

K

35

portion of radius R Derive tural frequency of small

phere of radius with r rolls

The torsional pendulum has a natural frequency of 5 Hz What length of steel wire of diameter 2 mm should be used for pendulum The inertia of mass fixed at the free end is 00098 kgm2 Take g= 083x1011 Nm2 Determine the torsional natural frequency of the system shown in figure Neglect the mass moment of inertia of the shaft ( Figure not proportionate) Determine the natural frequency of simple pendulum considering the mass of the rod into account

Using the energy method find tThe chord may be assumed ine

Kt1

Kt2

20cm

06m 08m 04m

20cm 20cm 20cm

m

M

he natural frxtensible in

k

M

36

equency of the system shown in figure the spring mass pulley system

A rod of length l and depth H having a mass m rests on a semicylidrical surface os radius r It is tipped slightly and let go Find the frequency of oscillation

A thin bar of l having a mass m rests on a semicylidrical surface radius r It is tipped slightly and let go Find the frequency of oscillation

l

h

R

l

37

38

CHAPTER 3

DAMPED FREE VIBRATION OF SINGLE DEGREE FREEDOM SYSTEM LEARNING OBJECTIVES

] Introduction to damped free vibration

] Terminologies used in damped free vibration

] Three different types of damping

] Concept of critical damping and its importance

] Study of response of viscous damped system for cases of under damping

critical and over damping

] Concept of logarithmic decrement

] Problems related to damped free vibration problems

Viscously damped free vibration of single degree freedom sytem Consider a single degree freedom system with viscous damping as shown in figure For a viscous damped system the damping force is

where c is a constant of proportionality It is called damping coefficient or coefficient of viscous damping

Symbolically it is designated by a dashpot as shown in Fig 3

x xampx

39

At any instant of time let the mass be displaced by x from the equilibrium position Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

Okxxcxm =minusminus= ampampamp

Okxxcxm =++ ampampamp ----------------------------------------------------------(1)

or

0kxdtdx

Cdt

xdM

2

2

=++

Assuming the solution of equation as

The first term in the above equation m2Ct

eminus

is an exponentially decaying function of time The behavior of the terms in the parenthesis however depends on whether numerical value within the radicals is positive zero or negative

CASE 1 if 0mk

m2c

2

=minus

40

CASE 2 if 0mk

m2c

2

=gt

CASE 3 if 0mk

m2c

2

=lt

41

Critically damped coefficient Critically damped coefficient is defined as the value of the damping coeffient for which radical in equation

mk

m2c

m2c

S2

21 minus

plusmn

= becomes zero

Damping Ratio ( Damping factor) ( ξ ) It is defined as the ratio of damping coefficient to the critical damping coefficient Note The damping coefficient c is defined as the actual amount of damping while the damping coefficient is a constant depending upon the mass and stiffness of the system The roots of characteristics equation can be now written as Note if c=cc or ξ =1 then system is critically damped if cgt cc or ξ gt1 then system is over damped I if clt cc or ξ lt1 then system is under damped

42

Take home quiz Derive General solution of different damped system

1 Critically damped system 2 Under damped system 3 Over damped system

Logarithmic Decrement It is used to determine the amount of damping present in system This measure The rate of decay of free vibration The larger the damping the greater will be rate of decay Logarithmic decrement is defined as the natural logarithm of ratio of any two successive amplitudes

=δ

2

1

xx

ln

Expression for logarithmic decrement (δ ) Consider a damped free vibration which is expressed by

( ) ( ) φ+ξminusω= ξω t 1sinXex 2n

tn

x is the displacement at any instant of time X and φ are arbitrary constants

21

2

ξminus

πξ=δ

43

Tutorial problems 1 Derive an expression for damping ratio in terms of logarithmic decrement

2 Show that logarithmic decrement

=δ

n

o

xx

lnn1

where xo is the initial amplitude xn

is amplitude after n cycles 3 A mass of 1 kg is to be supported on a spring having stiffness K= 9800 Nm The damping constant c= 59 Nsm Determine the natural frequency of system Find also logarithmic decrement and amplitude after 3 cycles if initial displacement is 03 cm 4 A vibrating system is defined by the following parameters M = 3kg k= 100 Nm damping coefficient = 3 Nsm determine Damping factor natural frequency damped vibration logarithmic decrement ratio of consecutive amplitude no of cycles after which the original amplitude is reduced to 20 5A single degree damped vibrating system a suspended mass of 18kg makes 15 oscillations in 003 seconds The amplitude decreases to 025 of the initial value after 5 oscillations Determine the stiffness of spring Logarithmic decrement damping factor and damping coefficient 6 A damped vibration record of a spring mass system shows the following data Amplitude at the end of 2nd cycle is 9mm amplitude at eh end of 3rd cycle is 6mm amplitude at the end of 4th cycle is 4mm stiffness of spring =8000 Nm weight =20N Find the logarithmic decrement damping force at unit velocity periodic time of vibration 7 For the system shown in figure the characteristics of dashpot is such that the constant force of 49 N is applied to the piston it velocity is found to be constant at 012 ms Determine the value of C would you expect the complete system to be periodic or aperiodic 8 For a torsional system shown in the figuref1 is natural frequency in air and f2 is the natural frequency when immersed in oil Determine the damping coefficient

Viscous fluid

Disc of mass moment of Inertia J

2a

C

m M

b

a a

k

9 A thin plate of area A and weigh w is fixed to the end of the spring and allow to oscillate in a viscous fluid as shown in figure It f1 is the frequency of oscillation in air and f2 in liquid Show that

22

21 ff

gA2

minusπω

=micro where the damping force on the plate

AV2Fd micro= V being the velocity of vibration 10 A torsional pendulum has a natural frequency of 175 cyclessecond vibrating in vacuum The mass moment of inertia is 2 kgm2 It is immersed in oil and it is observed that the natural frequency is 142 cyclessecond Find the damping torque If the disc is replaced by 34 deg When in oil find the displacement at the end of the first complete cycle 11 An automobile can be modeled as a mass placed on 4 shock absorbers each consisting of a spring and a damper Such that each spring is equally loaded Determine the stiffness damping constant of each shock absorber So that natural frequency is 2Hz and system is critically damped The mass of vehicle is 200kg 12 Write the differential equation of motion for the system shown in the figure below Determine the natural frequency of the damped oscillation and critical damping coefficient

Determine the differential equation of motion and find the critical cdamped natural frequency for system shown in figure

a

C

m M

M =200kg

44

200kg kgkg

oefficient and

45

Derive the equation of motion and find the critical damping coefficient and damped natural frequency for the system shown below Write the differential equation of motion and determine the expression for critical damping coefficient and damped natural frequency of system shown in figure

Set up a differential equation of motion for system shown in figure and determine the following Undamped natural frequencies critical damping coefficient damping ratio damped natural frequency

b

C

kM

a

m

46

A mass of 300N is resting on two springs of stiffness 3000 Nm each and a dashpot of damping coefficient 150 Nsm as shown in figure If the initial velocity of 10cms is given to mass at its equilibrium position what will be the displacement from the equilibrium position at the end of first second Derive the formula used A gun barrel of mass 600kg has a recoil spring of stiffness 294 knm if the barrel recoils 13m on firing determine i) The initial recoil velocity of barrel ii) The critical damping coefficient of dashpot which is engaged at the end of recoil stroke Iii) Time required for the barrel to return at the end of recoil stroke Iv) time required for the barrel to return to a position 5cm from the initial position

W= 3000 N

FIRING

CHAPTER 4

FORCED VIBRATION LEARNING OBJECTIVES

] Introduction to forced free vibration

] Terminologies used in forced vibration

] Concept of magnification factor

] Study of reciprocating and rotating unbalance

] Concept of base excitation and support excitation

] Concept of Transmissibility ratio

] Problems related to forced vibration

The vibration that takes place under the excitation of external forces are called

forced vibration When a system is subjected to harmonic excitation it is forced to vibrate at the same frequency as that of excitation Most of the mechanical systems are prone to steady state forced vibration which cause fatigue failure The vibration problem also becomes very important when any of excitation frequencies coincide with one of natural frequencies of a system Under this condition resonance occurs during which large amplitude of vibrations are observed To avoid resonance external frequency (operating speed) may be changed or properties of the system may be changed to alter the natural frequency In some cases sufficient amount of damping may be provided to avoid large amplitude during resonance Thus the problem of force vibration is very important in mechanical design

FORCED VIBRATIONS OF SINGLE DOF SYSTEMS UNDER HARMONIC EXCITATION

` [ ] tsin FF o ω=

x

sin FF o=

Fo

xampamp

[ ] tω

xamp

47

48

Consider a spring mass system having a viscous damping excited by a harmonic force [ ] tsin FF o ω= as shown in figure The various forces acting on the system is shown in figure Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

O)tsin(Fkxxcxm o =ω+minusminus= ampampamp

)tsin(Fkxxcxm o ω=++ ampampamp ----------------------------------------------------------(1)

or

)tsin(Fkxdtdx

Cdt

xdM o2

2

ω=++

Equation (1) represents the differential equation of motion of the system The solution to this equation consists of two parts namely complementary function and a particular integral

The first of solution is also termed as transient solution of homogeneous solution and can be obtained from the homogeneous differential equation 0kxxcxm =++ ampampamp

The second part of solution particular integral is also termed as steady state solution ad is of form )tsin(Xx φminusω= Where X= steady state amplitude φ is phase difference between external force and displacement

)tsin(Xx φminusω=

The vector representation of the shown

49

The force polygon for the above vector diagram can be written as follows

( )φminusωt O

KX

xm 2ω

Fo

xCω

tω

φ

50

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

is the governing equation

Zero frequency deflection (Xst) It is deflection observed at zero frequency ( when the external frequency 0=ω

( ) kF

001

kF

x o

22

o

st =

+minus=

Magnification factor (M) It is defined as the ratio of steady state amplitude to zero frequency deflection xst

M = arX

X=

A plot of magnification factor versus frequency ratio

damping factor ( )ξ is called frequency plot of a harmon All the frequency terms starts from magnification factor of damping ratio ( )ξ The amplitude of vibration at any on the value of damping The amplitude will be higher for higher damping The amplitude of vibration at reso

amplitude tends to zero as

ωω

n tends to infinity

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

Freq

ζ =

ζ = 1

Magnification Factor M = XXst

kFo

51

ωω

nfor varying values of

ically excite vibrating system

M=1 what ever may be value particular frequency depends

for lesser damping and lesser nance will be very high The

uency Ratio r = (ωωn)

ζ = 0 0

ζ = 025 ζ = 0375 ζ = 0 5

ζ = 0707

ζ = 2

52

Plot of phase angle φ versus frequency ratio

ωω

n is shown in the

figure The phase angle φ is zero for all the values of damping ratio

at

ωω

n=0 The phase

angle φ is 90 for all the values of damping ratio

at

ωω

n=1 ie at

resonance The phase angle between o to 90 deg for all the values of

ωω

n= 0 to 1 and

between 90 deg to 180 deg for all the values of

ωω

ngt1

Class work Derive an expression for frequency at peak amplitude

2np 21 ξminusω=ω=ω is the required equation

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

53

Derive expression amplitude at resonance ( In dimensional form) In dimensional form We know that steady state amplitude

( ) ( )

ω+ωminus=

222

o

Cmk

Fx Equation in dimensional form

We know that at resonance mk

n =ω=ω

( )

ω+

minus

=2

n

2

oR

Cmk

mk

Fx

( )

ω=

2n

oR

C

Fx

( )

ω=

n

oR C

Fx

Class work Derive an expression for amplitude at resonance ( in non dimensional form)

Tutorial Problems on Forced Vibration 1A weight of 60N is suspended by a spring stiffness 12 kNm is forced to vibrate by a harmonic force of 10N assuming viscous damping of 0086 kN-sm Find i) resonance frequency ii) amplitude at resonance iii) phase angle at resonance iv) frequency corresponding to peak amplitude 2 A periodic torque having a maximum value of 05888Nm at a frequency of 4 rads is impressed upon a flywheel suspended from a wire The wheel has mass moment of inertia 012 kgm2 and the wire has stiffness 1176 Nmrad A viscous dashpot applies damping couple of 0784 Nm an angular velocity of 2 rads calculate i)maximum angular displacement from rest position ii) maximum couple applied to dashpot Iii) angle by which the angular displacement lags the torque A mass of 1019 kg is suspended from a spring stiffness 20 Nmm viscous damping causes the amplitude to decrease to 110th of initial value in 4 complete cycle If a periodic force of 20 cos(50 t) is applied to the mass Find the amplitude of forced vibration What would be the amplitude of oscillation if the applied force coincides with the natural period

m=1019 kg

[ ] t50cos 30F =

54

55

ROTATING UNBALANCE Unbalance in rotating machinery is a common source of vibration excitation Consider a machine of mass m having mounted on a foundation of stiffness k and damping coefficient C as shown in figure The unbalance is represented by an eccentric rotating mass m with eccentricity (e) and having a angular velocity (ω) let x be the displacement of non rotating mass from the static equilibrium position The displacement of rotating mass m from the static equilibrium is ( ) tsin( ex ω+ Applying Newtonrsquos law of motion we have

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

56

RECIPROCATING UNBALANCE Unbalance in reciprocating machinery is a common source of vibration excitation Consider a machine of mass M having mounted on a foundation of stiffness K and damping coefficient C as shown in figure The unbalance is represented by a reciprocating mass (m) having crank radus (e) and connecting rod of length (l) Let ωωωω be angular velocity of the crank Let x be the displacement of non reciprocating mass from the static equilibrium position at any instant of time The displacement of reciprocating mass m from the static equilibrium position is

)2sin(sin +++ tle

tex ωωωωωωωω

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

57

Note The equations obtained in reciprocating unbalance is same as the rotating unbalance and having the following differences Rotating Unbalance Reciprocating Unbalance Here mass of the rotating element is considered

Here mass of reciprocating element is considered

Here the eccentricity (e) which is the distance between the centre of rotation and CG

Here the eccentricity e is the crank radius and crank radius = stroke length2

The frequency plot of non dimensional factor( ( )me

MX versus frequency ratio

ωω

n for various values of ( )ξξξξ is shown in figure This frequency plot is same for

the rotating and reciprocating unbalance with the following points are noted from the diagram

1 Damping ratio plays very important role during resonance in controlling the amplitude

2 When

ωω

n=0 then ( )me

MX =0 for all the values of damping ratio ( )ξξξξ

3 When

ωω

nis higher than ( )me

MX tends to unity

4 At very high speeds ie at very high frequency ratio

ωω

n there is no effect

of damping The frequency plot of phase angle versus the frequency ratio is given as shown in figure The explanation is same as in forced harmonic vibration

58

VIBRATION ISOLATION

Vibratory forces and motions generated by machine and other caused are unavoidable Howeever their effect on the other dynamical systems can be minimized by proper isolator design An isolation system is either to protect o delicate machine from excessive vibration transmitted to it from its supporting structure or to prevent vibratory forces and motions generated by machines from being transmitted to its surroundings The effectiveness of isolation may be measured in terms of forces and motion transmitted The first type is known as force isolation and the second type is known as motion isolation

Force isolation (Force Transmissibility) When vibratory forces generated by machine are isolated the effectiveness of forced isolation can be made by defining the term transmissibility The term transmissibility is defined as the ratio of force transmitted to the foundation to that of the external impressed force

Force Transmissibility = o

tr

FF

force Impressed Externalfoundation toted transmitForce

=

Consider viscous damped spring mass system as shown in figure Upon this system an external harmonic force ) t (Sin Fo ω is applied The difference forces acting on system are inertia forces damping force spring force and external force All these forces are represented on force polygon shown below Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these two forces

( ) ( )22 xckx ω+=

59

This is the expression for force transmissibility in dimensional form To represent same in the non dimensional form divide numerator and denominator by k

2

n

22

n

2

n

o

tr

21

21

FF

TR

ωω

ξ+

ωω

minus

ωω

ξ+

==

PHASE DIFFERNECE BETWEEN EXTERNAL FORCE AND DISPLACEMENT Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these

ωω

minus

ωω

ξ=

ωminusω

=φ minusminus2

n

n1

2

1

1

2tan

mKC

tan

The phase difference between force transmitted to the foundation Ftr to the displacement

kc

KxxC

tanω

=ω

=α or

ωω

ξ=

ω=α minusminus

n

11 2tanKx

xCtan

The phase difference between external force Fo and force transmitted to foundationis β

60

β= αminusφ =

ωω

minus

ωω

ξminus

2

n

n1

1

2tan -

ωω

ξminus

n

1 2tan

Displacement Transmissibility It is the ratio of amplitude of steady state vibration of machine to that of the base motion Forced vibration due to support motion Consider a spring mass damper system subjected to harmonic excitation )tsin(Yy ω= at the base as shown in figure Let x be the absolute displacement of the mass at any instant of time t The different forces acting on system are shown in free body diagram Applying Newtonrsquos Law of motion Ma = sumF

)yx(k)yx(cxm minusminusminusminus= ampampampamp

0)yx(k)yx(cxm =minus+minus+ ampampampamp

61

Frequency plot of transmissibility versus frequency ratio

62

The plot of transmissibility versus frequency ratio is as shown in figure The following points can be noted from it

i) The value of transmissibility is unity at frequency ratio

ωω

n=0 and 2

ii) For an undamped system ( )ξ =0 transmissibility tends to infin at resonance iii) The value of transmissibility is less than unity for the value of frequency ratio

greater than 2

iv) For frequency ratio

ωω

nlt 2 smaller damping ratio lead to larger values of

transmissibility on the other hand for frequency ratio

ωω

ngt 2 smaller damping

ratio leads to smaller value of transmissibility

The frequency plot of transmissibility (TR) versus frequency ratio

ωω

n can be

broadly divided into two zones

1

ωω

n= 0 to 2

2

ωω

n= 2 to infin

CASE 1

When

ωω

n= 0 to 2

In this zone the value of transmissibility is always greater than 1 for all values of damping ratio In this zone the transmissibility decreases with increase in damping Hence we can conclude that in this zone damping is useful in isolating vibration forces being transmitted to the support CASE 2

When

ωω

n= 2 to infin

In this zone the value of transmissibility is less than 1 for all values of damping ratio In this zone the transmissibility increases with increase in damping Hence we can conclude that in this zone damping will adversely affect in isolating vibratory forces

63

Tutorial problems on Forced Vibration 1 Machine of mass 75kg is mounted on spring of stiffness 12000 Nm with an assumed damping factor ξ =02 piston within the machine of mass 2 kg has a reciprocating motion with a stroke of 75 cm and speed of 15 cyclessecond Assuming the motion of piston to be harmonic determine amplitude of machine ( X) Transmissibility and force transmitted to foundation phase angle of transmitted force wrt to exciting force 2 Motor weighs 600N is mounted on a simple beam that has a spring rate of 400 Ncm The motor armature base 160 N and has an eccentricity of 0005cm What will be the amplitude of motion when it runs at 1760 RPM Neglect damping the deflection of armature shaft and weight of the beam 3 A Heavy machine of mass 300 kg is supported on resilient foundation the static deflection of foundation due to weight of machine is found to be 75mm It is observed that the machine vibrates with an amplitude of 10mm when the base of the foundation is subjected to harmonic oscillation at the undampeds natural frequency of system with an amplitude of 25mm Find the damping constant of foundation Ii) dynamic force amplitude on the base iii) amplitude of the displacement of machine relative to base 4 An air compressor of mass 450 kg operates at a constant speed of 1750 RPM The reciprocating masses are 10 kg the crank radius is 100mm If the damper for mounting introduces a damping factor 015 i) specify the springs of mounting such that only 20 of the unbalanced force is transmitted to the foundation ii) determine the magnitude of the transmitted force bull Repeat the above problem by assuming that there is no damping in system

5 A body of a car having mass 1500 kg and is mounted on 4 equal springs which deflect through 225 cm under the weight of the body A total damping forces the 4 shock absorbers is 46 N at a velocity of 1cmsec The car is placed with all the four wheels on a test platform which is move up and down at resonant speed With an amplitude of 25cm Find the amplitude of the car body on its springs assuming its CG to be in the centre of wheel base 6A 75 kg machine which is mounted on springs of stiffness 116x105 Nm with the damping of 20=ξ A 2kg piston within the machine as reciprocating motion with a stroke of 008m and speed of 3000 RPM Assuming the motion of piston to be harmonic determine the amplitude of vibration of machine

7 The spring of an automobile are compressed 10cm under its own weight Find the critical speed when a trailer is traveling over road with a people approximaterly by a sine wave of amplitude 75 cm and wave length 15m What will be the amplitude of vibration at 64kmhour 8 An eccentric mass excites us used to determine vibratory characteristics of a structure of mass 25kg as given in figure At a speed of 1000 RPM the eccentric mass to beat top position at the instant the structure was moved upwards through its static equilibrium position and corresponding amplitude was 20mm The eccentric mass is 05kg at an eccentricity of 10mm Determine i) Undamped natural frequency of the system ii) Damping ratio iii) Steady state amplitude and phase angle at 1500 RPM iv) Force transmitted to the ground at 1200 RPM

M=25kg

C

K

Eccentric mass

64

65

CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Support motion Consider a spring mass damper system as shown in figure Let it excited by motion

of support let y be the harmonic displacement of support From Newtonrsquos second law of motion

)yx(k)yx(cxm minusminusminusminus= ampampampamp

)1(0)yx(k)yx(cxm minusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminus=minus+minus+ ampampampamp

Let z be the relative displacement of the mass with respect to support then

Z=x-y

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

The torsional pendulum has a natural frequency of 5 Hz What length of steel wire of diameter 2 mm should be used for pendulum The inertia of mass fixed at the free end is 00098 kgm2 Take g= 083x1011 Nm2 Determine the torsional natural frequency of the system shown in figure Neglect the mass moment of inertia of the shaft ( Figure not proportionate) Determine the natural frequency of simple pendulum considering the mass of the rod into account

Using the energy method find tThe chord may be assumed ine

Kt1

Kt2

20cm

06m 08m 04m

20cm 20cm 20cm

m

M

he natural frxtensible in

k

M

36

equency of the system shown in figure the spring mass pulley system

A rod of length l and depth H having a mass m rests on a semicylidrical surface os radius r It is tipped slightly and let go Find the frequency of oscillation

A thin bar of l having a mass m rests on a semicylidrical surface radius r It is tipped slightly and let go Find the frequency of oscillation

l

h

R

l

37

38

CHAPTER 3

DAMPED FREE VIBRATION OF SINGLE DEGREE FREEDOM SYSTEM LEARNING OBJECTIVES

] Introduction to damped free vibration

] Terminologies used in damped free vibration

] Three different types of damping

] Concept of critical damping and its importance

] Study of response of viscous damped system for cases of under damping

critical and over damping

] Concept of logarithmic decrement

] Problems related to damped free vibration problems

Viscously damped free vibration of single degree freedom sytem Consider a single degree freedom system with viscous damping as shown in figure For a viscous damped system the damping force is

where c is a constant of proportionality It is called damping coefficient or coefficient of viscous damping

Symbolically it is designated by a dashpot as shown in Fig 3

x xampx

39

At any instant of time let the mass be displaced by x from the equilibrium position Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

Okxxcxm =minusminus= ampampamp

Okxxcxm =++ ampampamp ----------------------------------------------------------(1)

or

0kxdtdx

Cdt

xdM

2

2

=++

Assuming the solution of equation as

The first term in the above equation m2Ct

eminus

is an exponentially decaying function of time The behavior of the terms in the parenthesis however depends on whether numerical value within the radicals is positive zero or negative

CASE 1 if 0mk

m2c

2

=minus

40

CASE 2 if 0mk

m2c

2

=gt

CASE 3 if 0mk

m2c

2

=lt

41

Critically damped coefficient Critically damped coefficient is defined as the value of the damping coeffient for which radical in equation

mk

m2c

m2c

S2

21 minus

plusmn

= becomes zero

Damping Ratio ( Damping factor) ( ξ ) It is defined as the ratio of damping coefficient to the critical damping coefficient Note The damping coefficient c is defined as the actual amount of damping while the damping coefficient is a constant depending upon the mass and stiffness of the system The roots of characteristics equation can be now written as Note if c=cc or ξ =1 then system is critically damped if cgt cc or ξ gt1 then system is over damped I if clt cc or ξ lt1 then system is under damped

42

Take home quiz Derive General solution of different damped system

1 Critically damped system 2 Under damped system 3 Over damped system

Logarithmic Decrement It is used to determine the amount of damping present in system This measure The rate of decay of free vibration The larger the damping the greater will be rate of decay Logarithmic decrement is defined as the natural logarithm of ratio of any two successive amplitudes

=δ

2

1

xx

ln

Expression for logarithmic decrement (δ ) Consider a damped free vibration which is expressed by

( ) ( ) φ+ξminusω= ξω t 1sinXex 2n

tn

x is the displacement at any instant of time X and φ are arbitrary constants

21

2

ξminus

πξ=δ

43

Tutorial problems 1 Derive an expression for damping ratio in terms of logarithmic decrement

2 Show that logarithmic decrement

=δ

n

o

xx

lnn1

where xo is the initial amplitude xn

is amplitude after n cycles 3 A mass of 1 kg is to be supported on a spring having stiffness K= 9800 Nm The damping constant c= 59 Nsm Determine the natural frequency of system Find also logarithmic decrement and amplitude after 3 cycles if initial displacement is 03 cm 4 A vibrating system is defined by the following parameters M = 3kg k= 100 Nm damping coefficient = 3 Nsm determine Damping factor natural frequency damped vibration logarithmic decrement ratio of consecutive amplitude no of cycles after which the original amplitude is reduced to 20 5A single degree damped vibrating system a suspended mass of 18kg makes 15 oscillations in 003 seconds The amplitude decreases to 025 of the initial value after 5 oscillations Determine the stiffness of spring Logarithmic decrement damping factor and damping coefficient 6 A damped vibration record of a spring mass system shows the following data Amplitude at the end of 2nd cycle is 9mm amplitude at eh end of 3rd cycle is 6mm amplitude at the end of 4th cycle is 4mm stiffness of spring =8000 Nm weight =20N Find the logarithmic decrement damping force at unit velocity periodic time of vibration 7 For the system shown in figure the characteristics of dashpot is such that the constant force of 49 N is applied to the piston it velocity is found to be constant at 012 ms Determine the value of C would you expect the complete system to be periodic or aperiodic 8 For a torsional system shown in the figuref1 is natural frequency in air and f2 is the natural frequency when immersed in oil Determine the damping coefficient

Viscous fluid

Disc of mass moment of Inertia J

2a

C

m M

b

a a

k

9 A thin plate of area A and weigh w is fixed to the end of the spring and allow to oscillate in a viscous fluid as shown in figure It f1 is the frequency of oscillation in air and f2 in liquid Show that

22

21 ff

gA2

minusπω

=micro where the damping force on the plate

AV2Fd micro= V being the velocity of vibration 10 A torsional pendulum has a natural frequency of 175 cyclessecond vibrating in vacuum The mass moment of inertia is 2 kgm2 It is immersed in oil and it is observed that the natural frequency is 142 cyclessecond Find the damping torque If the disc is replaced by 34 deg When in oil find the displacement at the end of the first complete cycle 11 An automobile can be modeled as a mass placed on 4 shock absorbers each consisting of a spring and a damper Such that each spring is equally loaded Determine the stiffness damping constant of each shock absorber So that natural frequency is 2Hz and system is critically damped The mass of vehicle is 200kg 12 Write the differential equation of motion for the system shown in the figure below Determine the natural frequency of the damped oscillation and critical damping coefficient

Determine the differential equation of motion and find the critical cdamped natural frequency for system shown in figure

a

C

m M

M =200kg

44

200kg kgkg

oefficient and

45

Derive the equation of motion and find the critical damping coefficient and damped natural frequency for the system shown below Write the differential equation of motion and determine the expression for critical damping coefficient and damped natural frequency of system shown in figure

Set up a differential equation of motion for system shown in figure and determine the following Undamped natural frequencies critical damping coefficient damping ratio damped natural frequency

b

C

kM

a

m

46

A mass of 300N is resting on two springs of stiffness 3000 Nm each and a dashpot of damping coefficient 150 Nsm as shown in figure If the initial velocity of 10cms is given to mass at its equilibrium position what will be the displacement from the equilibrium position at the end of first second Derive the formula used A gun barrel of mass 600kg has a recoil spring of stiffness 294 knm if the barrel recoils 13m on firing determine i) The initial recoil velocity of barrel ii) The critical damping coefficient of dashpot which is engaged at the end of recoil stroke Iii) Time required for the barrel to return at the end of recoil stroke Iv) time required for the barrel to return to a position 5cm from the initial position

W= 3000 N

FIRING

CHAPTER 4

FORCED VIBRATION LEARNING OBJECTIVES

] Introduction to forced free vibration

] Terminologies used in forced vibration

] Concept of magnification factor

] Study of reciprocating and rotating unbalance

] Concept of base excitation and support excitation

] Concept of Transmissibility ratio

] Problems related to forced vibration

The vibration that takes place under the excitation of external forces are called

forced vibration When a system is subjected to harmonic excitation it is forced to vibrate at the same frequency as that of excitation Most of the mechanical systems are prone to steady state forced vibration which cause fatigue failure The vibration problem also becomes very important when any of excitation frequencies coincide with one of natural frequencies of a system Under this condition resonance occurs during which large amplitude of vibrations are observed To avoid resonance external frequency (operating speed) may be changed or properties of the system may be changed to alter the natural frequency In some cases sufficient amount of damping may be provided to avoid large amplitude during resonance Thus the problem of force vibration is very important in mechanical design

FORCED VIBRATIONS OF SINGLE DOF SYSTEMS UNDER HARMONIC EXCITATION

` [ ] tsin FF o ω=

x

sin FF o=

Fo

xampamp

[ ] tω

xamp

47

48

Consider a spring mass system having a viscous damping excited by a harmonic force [ ] tsin FF o ω= as shown in figure The various forces acting on the system is shown in figure Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

O)tsin(Fkxxcxm o =ω+minusminus= ampampamp

)tsin(Fkxxcxm o ω=++ ampampamp ----------------------------------------------------------(1)

or

)tsin(Fkxdtdx

Cdt

xdM o2

2

ω=++

Equation (1) represents the differential equation of motion of the system The solution to this equation consists of two parts namely complementary function and a particular integral

The first of solution is also termed as transient solution of homogeneous solution and can be obtained from the homogeneous differential equation 0kxxcxm =++ ampampamp

The second part of solution particular integral is also termed as steady state solution ad is of form )tsin(Xx φminusω= Where X= steady state amplitude φ is phase difference between external force and displacement

)tsin(Xx φminusω=

The vector representation of the shown

49

The force polygon for the above vector diagram can be written as follows

( )φminusωt O

KX

xm 2ω

Fo

xCω

tω

φ

50

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

is the governing equation

Zero frequency deflection (Xst) It is deflection observed at zero frequency ( when the external frequency 0=ω

( ) kF

001

kF

x o

22

o

st =

+minus=

Magnification factor (M) It is defined as the ratio of steady state amplitude to zero frequency deflection xst

M = arX

X=

A plot of magnification factor versus frequency ratio

damping factor ( )ξ is called frequency plot of a harmon All the frequency terms starts from magnification factor of damping ratio ( )ξ The amplitude of vibration at any on the value of damping The amplitude will be higher for higher damping The amplitude of vibration at reso

amplitude tends to zero as

ωω

n tends to infinity

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

Freq

ζ =

ζ = 1

Magnification Factor M = XXst

kFo

51

ωω

nfor varying values of

ically excite vibrating system

M=1 what ever may be value particular frequency depends

for lesser damping and lesser nance will be very high The

uency Ratio r = (ωωn)

ζ = 0 0

ζ = 025 ζ = 0375 ζ = 0 5

ζ = 0707

ζ = 2

52

Plot of phase angle φ versus frequency ratio

ωω

n is shown in the

figure The phase angle φ is zero for all the values of damping ratio

at

ωω

n=0 The phase

angle φ is 90 for all the values of damping ratio

at

ωω

n=1 ie at

resonance The phase angle between o to 90 deg for all the values of

ωω

n= 0 to 1 and

between 90 deg to 180 deg for all the values of

ωω

ngt1

Class work Derive an expression for frequency at peak amplitude

2np 21 ξminusω=ω=ω is the required equation

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

53

Derive expression amplitude at resonance ( In dimensional form) In dimensional form We know that steady state amplitude

( ) ( )

ω+ωminus=

222

o

Cmk

Fx Equation in dimensional form

We know that at resonance mk

n =ω=ω

( )

ω+

minus

=2

n

2

oR

Cmk

mk

Fx

( )

ω=

2n

oR

C

Fx

( )

ω=

n

oR C

Fx

Class work Derive an expression for amplitude at resonance ( in non dimensional form)

Tutorial Problems on Forced Vibration 1A weight of 60N is suspended by a spring stiffness 12 kNm is forced to vibrate by a harmonic force of 10N assuming viscous damping of 0086 kN-sm Find i) resonance frequency ii) amplitude at resonance iii) phase angle at resonance iv) frequency corresponding to peak amplitude 2 A periodic torque having a maximum value of 05888Nm at a frequency of 4 rads is impressed upon a flywheel suspended from a wire The wheel has mass moment of inertia 012 kgm2 and the wire has stiffness 1176 Nmrad A viscous dashpot applies damping couple of 0784 Nm an angular velocity of 2 rads calculate i)maximum angular displacement from rest position ii) maximum couple applied to dashpot Iii) angle by which the angular displacement lags the torque A mass of 1019 kg is suspended from a spring stiffness 20 Nmm viscous damping causes the amplitude to decrease to 110th of initial value in 4 complete cycle If a periodic force of 20 cos(50 t) is applied to the mass Find the amplitude of forced vibration What would be the amplitude of oscillation if the applied force coincides with the natural period

m=1019 kg

[ ] t50cos 30F =

54

55

ROTATING UNBALANCE Unbalance in rotating machinery is a common source of vibration excitation Consider a machine of mass m having mounted on a foundation of stiffness k and damping coefficient C as shown in figure The unbalance is represented by an eccentric rotating mass m with eccentricity (e) and having a angular velocity (ω) let x be the displacement of non rotating mass from the static equilibrium position The displacement of rotating mass m from the static equilibrium is ( ) tsin( ex ω+ Applying Newtonrsquos law of motion we have

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

56

RECIPROCATING UNBALANCE Unbalance in reciprocating machinery is a common source of vibration excitation Consider a machine of mass M having mounted on a foundation of stiffness K and damping coefficient C as shown in figure The unbalance is represented by a reciprocating mass (m) having crank radus (e) and connecting rod of length (l) Let ωωωω be angular velocity of the crank Let x be the displacement of non reciprocating mass from the static equilibrium position at any instant of time The displacement of reciprocating mass m from the static equilibrium position is

)2sin(sin +++ tle

tex ωωωωωωωω

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

57

Note The equations obtained in reciprocating unbalance is same as the rotating unbalance and having the following differences Rotating Unbalance Reciprocating Unbalance Here mass of the rotating element is considered

Here mass of reciprocating element is considered

Here the eccentricity (e) which is the distance between the centre of rotation and CG

Here the eccentricity e is the crank radius and crank radius = stroke length2

The frequency plot of non dimensional factor( ( )me

MX versus frequency ratio

ωω

n for various values of ( )ξξξξ is shown in figure This frequency plot is same for

the rotating and reciprocating unbalance with the following points are noted from the diagram

1 Damping ratio plays very important role during resonance in controlling the amplitude

2 When

ωω

n=0 then ( )me

MX =0 for all the values of damping ratio ( )ξξξξ

3 When

ωω

nis higher than ( )me

MX tends to unity

4 At very high speeds ie at very high frequency ratio

ωω

n there is no effect

of damping The frequency plot of phase angle versus the frequency ratio is given as shown in figure The explanation is same as in forced harmonic vibration

58

VIBRATION ISOLATION

Vibratory forces and motions generated by machine and other caused are unavoidable Howeever their effect on the other dynamical systems can be minimized by proper isolator design An isolation system is either to protect o delicate machine from excessive vibration transmitted to it from its supporting structure or to prevent vibratory forces and motions generated by machines from being transmitted to its surroundings The effectiveness of isolation may be measured in terms of forces and motion transmitted The first type is known as force isolation and the second type is known as motion isolation

Force isolation (Force Transmissibility) When vibratory forces generated by machine are isolated the effectiveness of forced isolation can be made by defining the term transmissibility The term transmissibility is defined as the ratio of force transmitted to the foundation to that of the external impressed force

Force Transmissibility = o

tr

FF

force Impressed Externalfoundation toted transmitForce

=

Consider viscous damped spring mass system as shown in figure Upon this system an external harmonic force ) t (Sin Fo ω is applied The difference forces acting on system are inertia forces damping force spring force and external force All these forces are represented on force polygon shown below Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these two forces

( ) ( )22 xckx ω+=

59

This is the expression for force transmissibility in dimensional form To represent same in the non dimensional form divide numerator and denominator by k

2

n

22

n

2

n

o

tr

21

21

FF

TR

ωω

ξ+

ωω

minus

ωω

ξ+

==

PHASE DIFFERNECE BETWEEN EXTERNAL FORCE AND DISPLACEMENT Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these

ωω

minus

ωω

ξ=

ωminusω

=φ minusminus2

n

n1

2

1

1

2tan

mKC

tan

The phase difference between force transmitted to the foundation Ftr to the displacement

kc

KxxC

tanω

=ω

=α or

ωω

ξ=

ω=α minusminus

n

11 2tanKx

xCtan

The phase difference between external force Fo and force transmitted to foundationis β

60

β= αminusφ =

ωω

minus

ωω

ξminus

2

n

n1

1

2tan -

ωω

ξminus

n

1 2tan

Displacement Transmissibility It is the ratio of amplitude of steady state vibration of machine to that of the base motion Forced vibration due to support motion Consider a spring mass damper system subjected to harmonic excitation )tsin(Yy ω= at the base as shown in figure Let x be the absolute displacement of the mass at any instant of time t The different forces acting on system are shown in free body diagram Applying Newtonrsquos Law of motion Ma = sumF

)yx(k)yx(cxm minusminusminusminus= ampampampamp

0)yx(k)yx(cxm =minus+minus+ ampampampamp

61

Frequency plot of transmissibility versus frequency ratio

62

The plot of transmissibility versus frequency ratio is as shown in figure The following points can be noted from it

i) The value of transmissibility is unity at frequency ratio

ωω

n=0 and 2

ii) For an undamped system ( )ξ =0 transmissibility tends to infin at resonance iii) The value of transmissibility is less than unity for the value of frequency ratio

greater than 2

iv) For frequency ratio

ωω

nlt 2 smaller damping ratio lead to larger values of

transmissibility on the other hand for frequency ratio

ωω

ngt 2 smaller damping

ratio leads to smaller value of transmissibility

The frequency plot of transmissibility (TR) versus frequency ratio

ωω

n can be

broadly divided into two zones

1

ωω

n= 0 to 2

2

ωω

n= 2 to infin

CASE 1

When

ωω

n= 0 to 2

In this zone the value of transmissibility is always greater than 1 for all values of damping ratio In this zone the transmissibility decreases with increase in damping Hence we can conclude that in this zone damping is useful in isolating vibration forces being transmitted to the support CASE 2

When

ωω

n= 2 to infin

In this zone the value of transmissibility is less than 1 for all values of damping ratio In this zone the transmissibility increases with increase in damping Hence we can conclude that in this zone damping will adversely affect in isolating vibratory forces

63

Tutorial problems on Forced Vibration 1 Machine of mass 75kg is mounted on spring of stiffness 12000 Nm with an assumed damping factor ξ =02 piston within the machine of mass 2 kg has a reciprocating motion with a stroke of 75 cm and speed of 15 cyclessecond Assuming the motion of piston to be harmonic determine amplitude of machine ( X) Transmissibility and force transmitted to foundation phase angle of transmitted force wrt to exciting force 2 Motor weighs 600N is mounted on a simple beam that has a spring rate of 400 Ncm The motor armature base 160 N and has an eccentricity of 0005cm What will be the amplitude of motion when it runs at 1760 RPM Neglect damping the deflection of armature shaft and weight of the beam 3 A Heavy machine of mass 300 kg is supported on resilient foundation the static deflection of foundation due to weight of machine is found to be 75mm It is observed that the machine vibrates with an amplitude of 10mm when the base of the foundation is subjected to harmonic oscillation at the undampeds natural frequency of system with an amplitude of 25mm Find the damping constant of foundation Ii) dynamic force amplitude on the base iii) amplitude of the displacement of machine relative to base 4 An air compressor of mass 450 kg operates at a constant speed of 1750 RPM The reciprocating masses are 10 kg the crank radius is 100mm If the damper for mounting introduces a damping factor 015 i) specify the springs of mounting such that only 20 of the unbalanced force is transmitted to the foundation ii) determine the magnitude of the transmitted force bull Repeat the above problem by assuming that there is no damping in system

5 A body of a car having mass 1500 kg and is mounted on 4 equal springs which deflect through 225 cm under the weight of the body A total damping forces the 4 shock absorbers is 46 N at a velocity of 1cmsec The car is placed with all the four wheels on a test platform which is move up and down at resonant speed With an amplitude of 25cm Find the amplitude of the car body on its springs assuming its CG to be in the centre of wheel base 6A 75 kg machine which is mounted on springs of stiffness 116x105 Nm with the damping of 20=ξ A 2kg piston within the machine as reciprocating motion with a stroke of 008m and speed of 3000 RPM Assuming the motion of piston to be harmonic determine the amplitude of vibration of machine

7 The spring of an automobile are compressed 10cm under its own weight Find the critical speed when a trailer is traveling over road with a people approximaterly by a sine wave of amplitude 75 cm and wave length 15m What will be the amplitude of vibration at 64kmhour 8 An eccentric mass excites us used to determine vibratory characteristics of a structure of mass 25kg as given in figure At a speed of 1000 RPM the eccentric mass to beat top position at the instant the structure was moved upwards through its static equilibrium position and corresponding amplitude was 20mm The eccentric mass is 05kg at an eccentricity of 10mm Determine i) Undamped natural frequency of the system ii) Damping ratio iii) Steady state amplitude and phase angle at 1500 RPM iv) Force transmitted to the ground at 1200 RPM

M=25kg

C

K

Eccentric mass

64

65

CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Support motion Consider a spring mass damper system as shown in figure Let it excited by motion

of support let y be the harmonic displacement of support From Newtonrsquos second law of motion

)yx(k)yx(cxm minusminusminusminus= ampampampamp

)1(0)yx(k)yx(cxm minusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminus=minus+minus+ ampampampamp

Let z be the relative displacement of the mass with respect to support then

Z=x-y

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

A rod of length l and depth H having a mass m rests on a semicylidrical surface os radius r It is tipped slightly and let go Find the frequency of oscillation

A thin bar of l having a mass m rests on a semicylidrical surface radius r It is tipped slightly and let go Find the frequency of oscillation

l

h

R

l

37

38

CHAPTER 3

DAMPED FREE VIBRATION OF SINGLE DEGREE FREEDOM SYSTEM LEARNING OBJECTIVES

] Introduction to damped free vibration

] Terminologies used in damped free vibration

] Three different types of damping

] Concept of critical damping and its importance

] Study of response of viscous damped system for cases of under damping

critical and over damping

] Concept of logarithmic decrement

] Problems related to damped free vibration problems

Viscously damped free vibration of single degree freedom sytem Consider a single degree freedom system with viscous damping as shown in figure For a viscous damped system the damping force is

where c is a constant of proportionality It is called damping coefficient or coefficient of viscous damping

Symbolically it is designated by a dashpot as shown in Fig 3

x xampx

39

At any instant of time let the mass be displaced by x from the equilibrium position Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

Okxxcxm =minusminus= ampampamp

Okxxcxm =++ ampampamp ----------------------------------------------------------(1)

or

0kxdtdx

Cdt

xdM

2

2

=++

Assuming the solution of equation as

The first term in the above equation m2Ct

eminus

is an exponentially decaying function of time The behavior of the terms in the parenthesis however depends on whether numerical value within the radicals is positive zero or negative

CASE 1 if 0mk

m2c

2

=minus

40

CASE 2 if 0mk

m2c

2

=gt

CASE 3 if 0mk

m2c

2

=lt

41

Critically damped coefficient Critically damped coefficient is defined as the value of the damping coeffient for which radical in equation

mk

m2c

m2c

S2

21 minus

plusmn

= becomes zero

Damping Ratio ( Damping factor) ( ξ ) It is defined as the ratio of damping coefficient to the critical damping coefficient Note The damping coefficient c is defined as the actual amount of damping while the damping coefficient is a constant depending upon the mass and stiffness of the system The roots of characteristics equation can be now written as Note if c=cc or ξ =1 then system is critically damped if cgt cc or ξ gt1 then system is over damped I if clt cc or ξ lt1 then system is under damped

42

Take home quiz Derive General solution of different damped system

1 Critically damped system 2 Under damped system 3 Over damped system

Logarithmic Decrement It is used to determine the amount of damping present in system This measure The rate of decay of free vibration The larger the damping the greater will be rate of decay Logarithmic decrement is defined as the natural logarithm of ratio of any two successive amplitudes

=δ

2

1

xx

ln

Expression for logarithmic decrement (δ ) Consider a damped free vibration which is expressed by

( ) ( ) φ+ξminusω= ξω t 1sinXex 2n

tn

x is the displacement at any instant of time X and φ are arbitrary constants

21

2

ξminus

πξ=δ

43

Tutorial problems 1 Derive an expression for damping ratio in terms of logarithmic decrement

2 Show that logarithmic decrement

=δ

n

o

xx

lnn1

where xo is the initial amplitude xn

is amplitude after n cycles 3 A mass of 1 kg is to be supported on a spring having stiffness K= 9800 Nm The damping constant c= 59 Nsm Determine the natural frequency of system Find also logarithmic decrement and amplitude after 3 cycles if initial displacement is 03 cm 4 A vibrating system is defined by the following parameters M = 3kg k= 100 Nm damping coefficient = 3 Nsm determine Damping factor natural frequency damped vibration logarithmic decrement ratio of consecutive amplitude no of cycles after which the original amplitude is reduced to 20 5A single degree damped vibrating system a suspended mass of 18kg makes 15 oscillations in 003 seconds The amplitude decreases to 025 of the initial value after 5 oscillations Determine the stiffness of spring Logarithmic decrement damping factor and damping coefficient 6 A damped vibration record of a spring mass system shows the following data Amplitude at the end of 2nd cycle is 9mm amplitude at eh end of 3rd cycle is 6mm amplitude at the end of 4th cycle is 4mm stiffness of spring =8000 Nm weight =20N Find the logarithmic decrement damping force at unit velocity periodic time of vibration 7 For the system shown in figure the characteristics of dashpot is such that the constant force of 49 N is applied to the piston it velocity is found to be constant at 012 ms Determine the value of C would you expect the complete system to be periodic or aperiodic 8 For a torsional system shown in the figuref1 is natural frequency in air and f2 is the natural frequency when immersed in oil Determine the damping coefficient

Viscous fluid

Disc of mass moment of Inertia J

2a

C

m M

b

a a

k

9 A thin plate of area A and weigh w is fixed to the end of the spring and allow to oscillate in a viscous fluid as shown in figure It f1 is the frequency of oscillation in air and f2 in liquid Show that

22

21 ff

gA2

minusπω

=micro where the damping force on the plate

AV2Fd micro= V being the velocity of vibration 10 A torsional pendulum has a natural frequency of 175 cyclessecond vibrating in vacuum The mass moment of inertia is 2 kgm2 It is immersed in oil and it is observed that the natural frequency is 142 cyclessecond Find the damping torque If the disc is replaced by 34 deg When in oil find the displacement at the end of the first complete cycle 11 An automobile can be modeled as a mass placed on 4 shock absorbers each consisting of a spring and a damper Such that each spring is equally loaded Determine the stiffness damping constant of each shock absorber So that natural frequency is 2Hz and system is critically damped The mass of vehicle is 200kg 12 Write the differential equation of motion for the system shown in the figure below Determine the natural frequency of the damped oscillation and critical damping coefficient

Determine the differential equation of motion and find the critical cdamped natural frequency for system shown in figure

a

C

m M

M =200kg

44

200kg kgkg

oefficient and

45

Derive the equation of motion and find the critical damping coefficient and damped natural frequency for the system shown below Write the differential equation of motion and determine the expression for critical damping coefficient and damped natural frequency of system shown in figure

Set up a differential equation of motion for system shown in figure and determine the following Undamped natural frequencies critical damping coefficient damping ratio damped natural frequency

b

C

kM

a

m

46

A mass of 300N is resting on two springs of stiffness 3000 Nm each and a dashpot of damping coefficient 150 Nsm as shown in figure If the initial velocity of 10cms is given to mass at its equilibrium position what will be the displacement from the equilibrium position at the end of first second Derive the formula used A gun barrel of mass 600kg has a recoil spring of stiffness 294 knm if the barrel recoils 13m on firing determine i) The initial recoil velocity of barrel ii) The critical damping coefficient of dashpot which is engaged at the end of recoil stroke Iii) Time required for the barrel to return at the end of recoil stroke Iv) time required for the barrel to return to a position 5cm from the initial position

W= 3000 N

FIRING

CHAPTER 4

FORCED VIBRATION LEARNING OBJECTIVES

] Introduction to forced free vibration

] Terminologies used in forced vibration

] Concept of magnification factor

] Study of reciprocating and rotating unbalance

] Concept of base excitation and support excitation

] Concept of Transmissibility ratio

] Problems related to forced vibration

The vibration that takes place under the excitation of external forces are called

forced vibration When a system is subjected to harmonic excitation it is forced to vibrate at the same frequency as that of excitation Most of the mechanical systems are prone to steady state forced vibration which cause fatigue failure The vibration problem also becomes very important when any of excitation frequencies coincide with one of natural frequencies of a system Under this condition resonance occurs during which large amplitude of vibrations are observed To avoid resonance external frequency (operating speed) may be changed or properties of the system may be changed to alter the natural frequency In some cases sufficient amount of damping may be provided to avoid large amplitude during resonance Thus the problem of force vibration is very important in mechanical design

FORCED VIBRATIONS OF SINGLE DOF SYSTEMS UNDER HARMONIC EXCITATION

` [ ] tsin FF o ω=

x

sin FF o=

Fo

xampamp

[ ] tω

xamp

47

48

Consider a spring mass system having a viscous damping excited by a harmonic force [ ] tsin FF o ω= as shown in figure The various forces acting on the system is shown in figure Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

O)tsin(Fkxxcxm o =ω+minusminus= ampampamp

)tsin(Fkxxcxm o ω=++ ampampamp ----------------------------------------------------------(1)

or

)tsin(Fkxdtdx

Cdt

xdM o2

2

ω=++

Equation (1) represents the differential equation of motion of the system The solution to this equation consists of two parts namely complementary function and a particular integral

The first of solution is also termed as transient solution of homogeneous solution and can be obtained from the homogeneous differential equation 0kxxcxm =++ ampampamp

The second part of solution particular integral is also termed as steady state solution ad is of form )tsin(Xx φminusω= Where X= steady state amplitude φ is phase difference between external force and displacement

)tsin(Xx φminusω=

The vector representation of the shown

49

The force polygon for the above vector diagram can be written as follows

( )φminusωt O

KX

xm 2ω

Fo

xCω

tω

φ

50

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

is the governing equation

Zero frequency deflection (Xst) It is deflection observed at zero frequency ( when the external frequency 0=ω

( ) kF

001

kF

x o

22

o

st =

+minus=

Magnification factor (M) It is defined as the ratio of steady state amplitude to zero frequency deflection xst

M = arX

X=

A plot of magnification factor versus frequency ratio

damping factor ( )ξ is called frequency plot of a harmon All the frequency terms starts from magnification factor of damping ratio ( )ξ The amplitude of vibration at any on the value of damping The amplitude will be higher for higher damping The amplitude of vibration at reso

amplitude tends to zero as

ωω

n tends to infinity

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

Freq

ζ =

ζ = 1

Magnification Factor M = XXst

kFo

51

ωω

nfor varying values of

ically excite vibrating system

M=1 what ever may be value particular frequency depends

for lesser damping and lesser nance will be very high The

uency Ratio r = (ωωn)

ζ = 0 0

ζ = 025 ζ = 0375 ζ = 0 5

ζ = 0707

ζ = 2

52

Plot of phase angle φ versus frequency ratio

ωω

n is shown in the

figure The phase angle φ is zero for all the values of damping ratio

at

ωω

n=0 The phase

angle φ is 90 for all the values of damping ratio

at

ωω

n=1 ie at

resonance The phase angle between o to 90 deg for all the values of

ωω

n= 0 to 1 and

between 90 deg to 180 deg for all the values of

ωω

ngt1

Class work Derive an expression for frequency at peak amplitude

2np 21 ξminusω=ω=ω is the required equation

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

53

Derive expression amplitude at resonance ( In dimensional form) In dimensional form We know that steady state amplitude

( ) ( )

ω+ωminus=

222

o

Cmk

Fx Equation in dimensional form

We know that at resonance mk

n =ω=ω

( )

ω+

minus

=2

n

2

oR

Cmk

mk

Fx

( )

ω=

2n

oR

C

Fx

( )

ω=

n

oR C

Fx

Class work Derive an expression for amplitude at resonance ( in non dimensional form)

Tutorial Problems on Forced Vibration 1A weight of 60N is suspended by a spring stiffness 12 kNm is forced to vibrate by a harmonic force of 10N assuming viscous damping of 0086 kN-sm Find i) resonance frequency ii) amplitude at resonance iii) phase angle at resonance iv) frequency corresponding to peak amplitude 2 A periodic torque having a maximum value of 05888Nm at a frequency of 4 rads is impressed upon a flywheel suspended from a wire The wheel has mass moment of inertia 012 kgm2 and the wire has stiffness 1176 Nmrad A viscous dashpot applies damping couple of 0784 Nm an angular velocity of 2 rads calculate i)maximum angular displacement from rest position ii) maximum couple applied to dashpot Iii) angle by which the angular displacement lags the torque A mass of 1019 kg is suspended from a spring stiffness 20 Nmm viscous damping causes the amplitude to decrease to 110th of initial value in 4 complete cycle If a periodic force of 20 cos(50 t) is applied to the mass Find the amplitude of forced vibration What would be the amplitude of oscillation if the applied force coincides with the natural period

m=1019 kg

[ ] t50cos 30F =

54

55

ROTATING UNBALANCE Unbalance in rotating machinery is a common source of vibration excitation Consider a machine of mass m having mounted on a foundation of stiffness k and damping coefficient C as shown in figure The unbalance is represented by an eccentric rotating mass m with eccentricity (e) and having a angular velocity (ω) let x be the displacement of non rotating mass from the static equilibrium position The displacement of rotating mass m from the static equilibrium is ( ) tsin( ex ω+ Applying Newtonrsquos law of motion we have

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

56

RECIPROCATING UNBALANCE Unbalance in reciprocating machinery is a common source of vibration excitation Consider a machine of mass M having mounted on a foundation of stiffness K and damping coefficient C as shown in figure The unbalance is represented by a reciprocating mass (m) having crank radus (e) and connecting rod of length (l) Let ωωωω be angular velocity of the crank Let x be the displacement of non reciprocating mass from the static equilibrium position at any instant of time The displacement of reciprocating mass m from the static equilibrium position is

)2sin(sin +++ tle

tex ωωωωωωωω

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

57

Note The equations obtained in reciprocating unbalance is same as the rotating unbalance and having the following differences Rotating Unbalance Reciprocating Unbalance Here mass of the rotating element is considered

Here mass of reciprocating element is considered

Here the eccentricity (e) which is the distance between the centre of rotation and CG

Here the eccentricity e is the crank radius and crank radius = stroke length2

The frequency plot of non dimensional factor( ( )me

MX versus frequency ratio

ωω

n for various values of ( )ξξξξ is shown in figure This frequency plot is same for

the rotating and reciprocating unbalance with the following points are noted from the diagram

1 Damping ratio plays very important role during resonance in controlling the amplitude

2 When

ωω

n=0 then ( )me

MX =0 for all the values of damping ratio ( )ξξξξ

3 When

ωω

nis higher than ( )me

MX tends to unity

4 At very high speeds ie at very high frequency ratio

ωω

n there is no effect

of damping The frequency plot of phase angle versus the frequency ratio is given as shown in figure The explanation is same as in forced harmonic vibration

58

VIBRATION ISOLATION

Vibratory forces and motions generated by machine and other caused are unavoidable Howeever their effect on the other dynamical systems can be minimized by proper isolator design An isolation system is either to protect o delicate machine from excessive vibration transmitted to it from its supporting structure or to prevent vibratory forces and motions generated by machines from being transmitted to its surroundings The effectiveness of isolation may be measured in terms of forces and motion transmitted The first type is known as force isolation and the second type is known as motion isolation

Force isolation (Force Transmissibility) When vibratory forces generated by machine are isolated the effectiveness of forced isolation can be made by defining the term transmissibility The term transmissibility is defined as the ratio of force transmitted to the foundation to that of the external impressed force

Force Transmissibility = o

tr

FF

force Impressed Externalfoundation toted transmitForce

=

Consider viscous damped spring mass system as shown in figure Upon this system an external harmonic force ) t (Sin Fo ω is applied The difference forces acting on system are inertia forces damping force spring force and external force All these forces are represented on force polygon shown below Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these two forces

( ) ( )22 xckx ω+=

59

This is the expression for force transmissibility in dimensional form To represent same in the non dimensional form divide numerator and denominator by k

2

n

22

n

2

n

o

tr

21

21

FF

TR

ωω

ξ+

ωω

minus

ωω

ξ+

==

PHASE DIFFERNECE BETWEEN EXTERNAL FORCE AND DISPLACEMENT Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these

ωω

minus

ωω

ξ=

ωminusω

=φ minusminus2

n

n1

2

1

1

2tan

mKC

tan

The phase difference between force transmitted to the foundation Ftr to the displacement

kc

KxxC

tanω

=ω

=α or

ωω

ξ=

ω=α minusminus

n

11 2tanKx

xCtan

The phase difference between external force Fo and force transmitted to foundationis β

60

β= αminusφ =

ωω

minus

ωω

ξminus

2

n

n1

1

2tan -

ωω

ξminus

n

1 2tan

Displacement Transmissibility It is the ratio of amplitude of steady state vibration of machine to that of the base motion Forced vibration due to support motion Consider a spring mass damper system subjected to harmonic excitation )tsin(Yy ω= at the base as shown in figure Let x be the absolute displacement of the mass at any instant of time t The different forces acting on system are shown in free body diagram Applying Newtonrsquos Law of motion Ma = sumF

)yx(k)yx(cxm minusminusminusminus= ampampampamp

0)yx(k)yx(cxm =minus+minus+ ampampampamp

61

Frequency plot of transmissibility versus frequency ratio

62

The plot of transmissibility versus frequency ratio is as shown in figure The following points can be noted from it

i) The value of transmissibility is unity at frequency ratio

ωω

n=0 and 2

ii) For an undamped system ( )ξ =0 transmissibility tends to infin at resonance iii) The value of transmissibility is less than unity for the value of frequency ratio

greater than 2

iv) For frequency ratio

ωω

nlt 2 smaller damping ratio lead to larger values of

transmissibility on the other hand for frequency ratio

ωω

ngt 2 smaller damping

ratio leads to smaller value of transmissibility

The frequency plot of transmissibility (TR) versus frequency ratio

ωω

n can be

broadly divided into two zones

1

ωω

n= 0 to 2

2

ωω

n= 2 to infin

CASE 1

When

ωω

n= 0 to 2

In this zone the value of transmissibility is always greater than 1 for all values of damping ratio In this zone the transmissibility decreases with increase in damping Hence we can conclude that in this zone damping is useful in isolating vibration forces being transmitted to the support CASE 2

When

ωω

n= 2 to infin

In this zone the value of transmissibility is less than 1 for all values of damping ratio In this zone the transmissibility increases with increase in damping Hence we can conclude that in this zone damping will adversely affect in isolating vibratory forces

63

Tutorial problems on Forced Vibration 1 Machine of mass 75kg is mounted on spring of stiffness 12000 Nm with an assumed damping factor ξ =02 piston within the machine of mass 2 kg has a reciprocating motion with a stroke of 75 cm and speed of 15 cyclessecond Assuming the motion of piston to be harmonic determine amplitude of machine ( X) Transmissibility and force transmitted to foundation phase angle of transmitted force wrt to exciting force 2 Motor weighs 600N is mounted on a simple beam that has a spring rate of 400 Ncm The motor armature base 160 N and has an eccentricity of 0005cm What will be the amplitude of motion when it runs at 1760 RPM Neglect damping the deflection of armature shaft and weight of the beam 3 A Heavy machine of mass 300 kg is supported on resilient foundation the static deflection of foundation due to weight of machine is found to be 75mm It is observed that the machine vibrates with an amplitude of 10mm when the base of the foundation is subjected to harmonic oscillation at the undampeds natural frequency of system with an amplitude of 25mm Find the damping constant of foundation Ii) dynamic force amplitude on the base iii) amplitude of the displacement of machine relative to base 4 An air compressor of mass 450 kg operates at a constant speed of 1750 RPM The reciprocating masses are 10 kg the crank radius is 100mm If the damper for mounting introduces a damping factor 015 i) specify the springs of mounting such that only 20 of the unbalanced force is transmitted to the foundation ii) determine the magnitude of the transmitted force bull Repeat the above problem by assuming that there is no damping in system

5 A body of a car having mass 1500 kg and is mounted on 4 equal springs which deflect through 225 cm under the weight of the body A total damping forces the 4 shock absorbers is 46 N at a velocity of 1cmsec The car is placed with all the four wheels on a test platform which is move up and down at resonant speed With an amplitude of 25cm Find the amplitude of the car body on its springs assuming its CG to be in the centre of wheel base 6A 75 kg machine which is mounted on springs of stiffness 116x105 Nm with the damping of 20=ξ A 2kg piston within the machine as reciprocating motion with a stroke of 008m and speed of 3000 RPM Assuming the motion of piston to be harmonic determine the amplitude of vibration of machine

7 The spring of an automobile are compressed 10cm under its own weight Find the critical speed when a trailer is traveling over road with a people approximaterly by a sine wave of amplitude 75 cm and wave length 15m What will be the amplitude of vibration at 64kmhour 8 An eccentric mass excites us used to determine vibratory characteristics of a structure of mass 25kg as given in figure At a speed of 1000 RPM the eccentric mass to beat top position at the instant the structure was moved upwards through its static equilibrium position and corresponding amplitude was 20mm The eccentric mass is 05kg at an eccentricity of 10mm Determine i) Undamped natural frequency of the system ii) Damping ratio iii) Steady state amplitude and phase angle at 1500 RPM iv) Force transmitted to the ground at 1200 RPM

M=25kg

C

K

Eccentric mass

64

65

CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Support motion Consider a spring mass damper system as shown in figure Let it excited by motion

of support let y be the harmonic displacement of support From Newtonrsquos second law of motion

)yx(k)yx(cxm minusminusminusminus= ampampampamp

)1(0)yx(k)yx(cxm minusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminus=minus+minus+ ampampampamp

Let z be the relative displacement of the mass with respect to support then

Z=x-y

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

38

CHAPTER 3

DAMPED FREE VIBRATION OF SINGLE DEGREE FREEDOM SYSTEM LEARNING OBJECTIVES

] Introduction to damped free vibration

] Terminologies used in damped free vibration

] Three different types of damping

] Concept of critical damping and its importance

] Study of response of viscous damped system for cases of under damping

critical and over damping

] Concept of logarithmic decrement

] Problems related to damped free vibration problems

Viscously damped free vibration of single degree freedom sytem Consider a single degree freedom system with viscous damping as shown in figure For a viscous damped system the damping force is

where c is a constant of proportionality It is called damping coefficient or coefficient of viscous damping

Symbolically it is designated by a dashpot as shown in Fig 3

x xampx

39

At any instant of time let the mass be displaced by x from the equilibrium position Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

Okxxcxm =minusminus= ampampamp

Okxxcxm =++ ampampamp ----------------------------------------------------------(1)

or

0kxdtdx

Cdt

xdM

2

2

=++

Assuming the solution of equation as

The first term in the above equation m2Ct

eminus

is an exponentially decaying function of time The behavior of the terms in the parenthesis however depends on whether numerical value within the radicals is positive zero or negative

CASE 1 if 0mk

m2c

2

=minus

40

CASE 2 if 0mk

m2c

2

=gt

CASE 3 if 0mk

m2c

2

=lt

41

Critically damped coefficient Critically damped coefficient is defined as the value of the damping coeffient for which radical in equation

mk

m2c

m2c

S2

21 minus

plusmn

= becomes zero

Damping Ratio ( Damping factor) ( ξ ) It is defined as the ratio of damping coefficient to the critical damping coefficient Note The damping coefficient c is defined as the actual amount of damping while the damping coefficient is a constant depending upon the mass and stiffness of the system The roots of characteristics equation can be now written as Note if c=cc or ξ =1 then system is critically damped if cgt cc or ξ gt1 then system is over damped I if clt cc or ξ lt1 then system is under damped

42

Take home quiz Derive General solution of different damped system

1 Critically damped system 2 Under damped system 3 Over damped system

Logarithmic Decrement It is used to determine the amount of damping present in system This measure The rate of decay of free vibration The larger the damping the greater will be rate of decay Logarithmic decrement is defined as the natural logarithm of ratio of any two successive amplitudes

=δ

2

1

xx

ln

Expression for logarithmic decrement (δ ) Consider a damped free vibration which is expressed by

( ) ( ) φ+ξminusω= ξω t 1sinXex 2n

tn

x is the displacement at any instant of time X and φ are arbitrary constants

21

2

ξminus

πξ=δ

43

Tutorial problems 1 Derive an expression for damping ratio in terms of logarithmic decrement

2 Show that logarithmic decrement

=δ

n

o

xx

lnn1

where xo is the initial amplitude xn

is amplitude after n cycles 3 A mass of 1 kg is to be supported on a spring having stiffness K= 9800 Nm The damping constant c= 59 Nsm Determine the natural frequency of system Find also logarithmic decrement and amplitude after 3 cycles if initial displacement is 03 cm 4 A vibrating system is defined by the following parameters M = 3kg k= 100 Nm damping coefficient = 3 Nsm determine Damping factor natural frequency damped vibration logarithmic decrement ratio of consecutive amplitude no of cycles after which the original amplitude is reduced to 20 5A single degree damped vibrating system a suspended mass of 18kg makes 15 oscillations in 003 seconds The amplitude decreases to 025 of the initial value after 5 oscillations Determine the stiffness of spring Logarithmic decrement damping factor and damping coefficient 6 A damped vibration record of a spring mass system shows the following data Amplitude at the end of 2nd cycle is 9mm amplitude at eh end of 3rd cycle is 6mm amplitude at the end of 4th cycle is 4mm stiffness of spring =8000 Nm weight =20N Find the logarithmic decrement damping force at unit velocity periodic time of vibration 7 For the system shown in figure the characteristics of dashpot is such that the constant force of 49 N is applied to the piston it velocity is found to be constant at 012 ms Determine the value of C would you expect the complete system to be periodic or aperiodic 8 For a torsional system shown in the figuref1 is natural frequency in air and f2 is the natural frequency when immersed in oil Determine the damping coefficient

Viscous fluid

Disc of mass moment of Inertia J

2a

C

m M

b

a a

k

9 A thin plate of area A and weigh w is fixed to the end of the spring and allow to oscillate in a viscous fluid as shown in figure It f1 is the frequency of oscillation in air and f2 in liquid Show that

22

21 ff

gA2

minusπω

=micro where the damping force on the plate

AV2Fd micro= V being the velocity of vibration 10 A torsional pendulum has a natural frequency of 175 cyclessecond vibrating in vacuum The mass moment of inertia is 2 kgm2 It is immersed in oil and it is observed that the natural frequency is 142 cyclessecond Find the damping torque If the disc is replaced by 34 deg When in oil find the displacement at the end of the first complete cycle 11 An automobile can be modeled as a mass placed on 4 shock absorbers each consisting of a spring and a damper Such that each spring is equally loaded Determine the stiffness damping constant of each shock absorber So that natural frequency is 2Hz and system is critically damped The mass of vehicle is 200kg 12 Write the differential equation of motion for the system shown in the figure below Determine the natural frequency of the damped oscillation and critical damping coefficient

Determine the differential equation of motion and find the critical cdamped natural frequency for system shown in figure

a

C

m M

M =200kg

44

200kg kgkg

oefficient and

45

Derive the equation of motion and find the critical damping coefficient and damped natural frequency for the system shown below Write the differential equation of motion and determine the expression for critical damping coefficient and damped natural frequency of system shown in figure

Set up a differential equation of motion for system shown in figure and determine the following Undamped natural frequencies critical damping coefficient damping ratio damped natural frequency

b

C

kM

a

m

46

A mass of 300N is resting on two springs of stiffness 3000 Nm each and a dashpot of damping coefficient 150 Nsm as shown in figure If the initial velocity of 10cms is given to mass at its equilibrium position what will be the displacement from the equilibrium position at the end of first second Derive the formula used A gun barrel of mass 600kg has a recoil spring of stiffness 294 knm if the barrel recoils 13m on firing determine i) The initial recoil velocity of barrel ii) The critical damping coefficient of dashpot which is engaged at the end of recoil stroke Iii) Time required for the barrel to return at the end of recoil stroke Iv) time required for the barrel to return to a position 5cm from the initial position

W= 3000 N

FIRING

CHAPTER 4

FORCED VIBRATION LEARNING OBJECTIVES

] Introduction to forced free vibration

] Terminologies used in forced vibration

] Concept of magnification factor

] Study of reciprocating and rotating unbalance

] Concept of base excitation and support excitation

] Concept of Transmissibility ratio

] Problems related to forced vibration

The vibration that takes place under the excitation of external forces are called

forced vibration When a system is subjected to harmonic excitation it is forced to vibrate at the same frequency as that of excitation Most of the mechanical systems are prone to steady state forced vibration which cause fatigue failure The vibration problem also becomes very important when any of excitation frequencies coincide with one of natural frequencies of a system Under this condition resonance occurs during which large amplitude of vibrations are observed To avoid resonance external frequency (operating speed) may be changed or properties of the system may be changed to alter the natural frequency In some cases sufficient amount of damping may be provided to avoid large amplitude during resonance Thus the problem of force vibration is very important in mechanical design

FORCED VIBRATIONS OF SINGLE DOF SYSTEMS UNDER HARMONIC EXCITATION

` [ ] tsin FF o ω=

x

sin FF o=

Fo

xampamp

[ ] tω

xamp

47

48

Consider a spring mass system having a viscous damping excited by a harmonic force [ ] tsin FF o ω= as shown in figure The various forces acting on the system is shown in figure Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

O)tsin(Fkxxcxm o =ω+minusminus= ampampamp

)tsin(Fkxxcxm o ω=++ ampampamp ----------------------------------------------------------(1)

or

)tsin(Fkxdtdx

Cdt

xdM o2

2

ω=++

Equation (1) represents the differential equation of motion of the system The solution to this equation consists of two parts namely complementary function and a particular integral

The first of solution is also termed as transient solution of homogeneous solution and can be obtained from the homogeneous differential equation 0kxxcxm =++ ampampamp

The second part of solution particular integral is also termed as steady state solution ad is of form )tsin(Xx φminusω= Where X= steady state amplitude φ is phase difference between external force and displacement

)tsin(Xx φminusω=

The vector representation of the shown

49

The force polygon for the above vector diagram can be written as follows

( )φminusωt O

KX

xm 2ω

Fo

xCω

tω

φ

50

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

is the governing equation

Zero frequency deflection (Xst) It is deflection observed at zero frequency ( when the external frequency 0=ω

( ) kF

001

kF

x o

22

o

st =

+minus=

Magnification factor (M) It is defined as the ratio of steady state amplitude to zero frequency deflection xst

M = arX

X=

A plot of magnification factor versus frequency ratio

damping factor ( )ξ is called frequency plot of a harmon All the frequency terms starts from magnification factor of damping ratio ( )ξ The amplitude of vibration at any on the value of damping The amplitude will be higher for higher damping The amplitude of vibration at reso

amplitude tends to zero as

ωω

n tends to infinity

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

Freq

ζ =

ζ = 1

Magnification Factor M = XXst

kFo

51

ωω

nfor varying values of

ically excite vibrating system

M=1 what ever may be value particular frequency depends

for lesser damping and lesser nance will be very high The

uency Ratio r = (ωωn)

ζ = 0 0

ζ = 025 ζ = 0375 ζ = 0 5

ζ = 0707

ζ = 2

52

Plot of phase angle φ versus frequency ratio

ωω

n is shown in the

figure The phase angle φ is zero for all the values of damping ratio

at

ωω

n=0 The phase

angle φ is 90 for all the values of damping ratio

at

ωω

n=1 ie at

resonance The phase angle between o to 90 deg for all the values of

ωω

n= 0 to 1 and

between 90 deg to 180 deg for all the values of

ωω

ngt1

Class work Derive an expression for frequency at peak amplitude

2np 21 ξminusω=ω=ω is the required equation

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

53

Derive expression amplitude at resonance ( In dimensional form) In dimensional form We know that steady state amplitude

( ) ( )

ω+ωminus=

222

o

Cmk

Fx Equation in dimensional form

We know that at resonance mk

n =ω=ω

( )

ω+

minus

=2

n

2

oR

Cmk

mk

Fx

( )

ω=

2n

oR

C

Fx

( )

ω=

n

oR C

Fx

Class work Derive an expression for amplitude at resonance ( in non dimensional form)

Tutorial Problems on Forced Vibration 1A weight of 60N is suspended by a spring stiffness 12 kNm is forced to vibrate by a harmonic force of 10N assuming viscous damping of 0086 kN-sm Find i) resonance frequency ii) amplitude at resonance iii) phase angle at resonance iv) frequency corresponding to peak amplitude 2 A periodic torque having a maximum value of 05888Nm at a frequency of 4 rads is impressed upon a flywheel suspended from a wire The wheel has mass moment of inertia 012 kgm2 and the wire has stiffness 1176 Nmrad A viscous dashpot applies damping couple of 0784 Nm an angular velocity of 2 rads calculate i)maximum angular displacement from rest position ii) maximum couple applied to dashpot Iii) angle by which the angular displacement lags the torque A mass of 1019 kg is suspended from a spring stiffness 20 Nmm viscous damping causes the amplitude to decrease to 110th of initial value in 4 complete cycle If a periodic force of 20 cos(50 t) is applied to the mass Find the amplitude of forced vibration What would be the amplitude of oscillation if the applied force coincides with the natural period

m=1019 kg

[ ] t50cos 30F =

54

55

ROTATING UNBALANCE Unbalance in rotating machinery is a common source of vibration excitation Consider a machine of mass m having mounted on a foundation of stiffness k and damping coefficient C as shown in figure The unbalance is represented by an eccentric rotating mass m with eccentricity (e) and having a angular velocity (ω) let x be the displacement of non rotating mass from the static equilibrium position The displacement of rotating mass m from the static equilibrium is ( ) tsin( ex ω+ Applying Newtonrsquos law of motion we have

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

56

RECIPROCATING UNBALANCE Unbalance in reciprocating machinery is a common source of vibration excitation Consider a machine of mass M having mounted on a foundation of stiffness K and damping coefficient C as shown in figure The unbalance is represented by a reciprocating mass (m) having crank radus (e) and connecting rod of length (l) Let ωωωω be angular velocity of the crank Let x be the displacement of non reciprocating mass from the static equilibrium position at any instant of time The displacement of reciprocating mass m from the static equilibrium position is

)2sin(sin +++ tle

tex ωωωωωωωω

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

57

Note The equations obtained in reciprocating unbalance is same as the rotating unbalance and having the following differences Rotating Unbalance Reciprocating Unbalance Here mass of the rotating element is considered

Here mass of reciprocating element is considered

Here the eccentricity (e) which is the distance between the centre of rotation and CG

Here the eccentricity e is the crank radius and crank radius = stroke length2

The frequency plot of non dimensional factor( ( )me

MX versus frequency ratio

ωω

n for various values of ( )ξξξξ is shown in figure This frequency plot is same for

the rotating and reciprocating unbalance with the following points are noted from the diagram

1 Damping ratio plays very important role during resonance in controlling the amplitude

2 When

ωω

n=0 then ( )me

MX =0 for all the values of damping ratio ( )ξξξξ

3 When

ωω

nis higher than ( )me

MX tends to unity

4 At very high speeds ie at very high frequency ratio

ωω

n there is no effect

of damping The frequency plot of phase angle versus the frequency ratio is given as shown in figure The explanation is same as in forced harmonic vibration

58

VIBRATION ISOLATION

Vibratory forces and motions generated by machine and other caused are unavoidable Howeever their effect on the other dynamical systems can be minimized by proper isolator design An isolation system is either to protect o delicate machine from excessive vibration transmitted to it from its supporting structure or to prevent vibratory forces and motions generated by machines from being transmitted to its surroundings The effectiveness of isolation may be measured in terms of forces and motion transmitted The first type is known as force isolation and the second type is known as motion isolation

Force isolation (Force Transmissibility) When vibratory forces generated by machine are isolated the effectiveness of forced isolation can be made by defining the term transmissibility The term transmissibility is defined as the ratio of force transmitted to the foundation to that of the external impressed force

Force Transmissibility = o

tr

FF

force Impressed Externalfoundation toted transmitForce

=

Consider viscous damped spring mass system as shown in figure Upon this system an external harmonic force ) t (Sin Fo ω is applied The difference forces acting on system are inertia forces damping force spring force and external force All these forces are represented on force polygon shown below Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these two forces

( ) ( )22 xckx ω+=

59

This is the expression for force transmissibility in dimensional form To represent same in the non dimensional form divide numerator and denominator by k

2

n

22

n

2

n

o

tr

21

21

FF

TR

ωω

ξ+

ωω

minus

ωω

ξ+

==

PHASE DIFFERNECE BETWEEN EXTERNAL FORCE AND DISPLACEMENT Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these

ωω

minus

ωω

ξ=

ωminusω

=φ minusminus2

n

n1

2

1

1

2tan

mKC

tan

The phase difference between force transmitted to the foundation Ftr to the displacement

kc

KxxC

tanω

=ω

=α or

ωω

ξ=

ω=α minusminus

n

11 2tanKx

xCtan

The phase difference between external force Fo and force transmitted to foundationis β

60

β= αminusφ =

ωω

minus

ωω

ξminus

2

n

n1

1

2tan -

ωω

ξminus

n

1 2tan

Displacement Transmissibility It is the ratio of amplitude of steady state vibration of machine to that of the base motion Forced vibration due to support motion Consider a spring mass damper system subjected to harmonic excitation )tsin(Yy ω= at the base as shown in figure Let x be the absolute displacement of the mass at any instant of time t The different forces acting on system are shown in free body diagram Applying Newtonrsquos Law of motion Ma = sumF

)yx(k)yx(cxm minusminusminusminus= ampampampamp

0)yx(k)yx(cxm =minus+minus+ ampampampamp

61

Frequency plot of transmissibility versus frequency ratio

62

The plot of transmissibility versus frequency ratio is as shown in figure The following points can be noted from it

i) The value of transmissibility is unity at frequency ratio

ωω

n=0 and 2

ii) For an undamped system ( )ξ =0 transmissibility tends to infin at resonance iii) The value of transmissibility is less than unity for the value of frequency ratio

greater than 2

iv) For frequency ratio

ωω

nlt 2 smaller damping ratio lead to larger values of

transmissibility on the other hand for frequency ratio

ωω

ngt 2 smaller damping

ratio leads to smaller value of transmissibility

The frequency plot of transmissibility (TR) versus frequency ratio

ωω

n can be

broadly divided into two zones

1

ωω

n= 0 to 2

2

ωω

n= 2 to infin

CASE 1

When

ωω

n= 0 to 2

In this zone the value of transmissibility is always greater than 1 for all values of damping ratio In this zone the transmissibility decreases with increase in damping Hence we can conclude that in this zone damping is useful in isolating vibration forces being transmitted to the support CASE 2

When

ωω

n= 2 to infin

In this zone the value of transmissibility is less than 1 for all values of damping ratio In this zone the transmissibility increases with increase in damping Hence we can conclude that in this zone damping will adversely affect in isolating vibratory forces

63

Tutorial problems on Forced Vibration 1 Machine of mass 75kg is mounted on spring of stiffness 12000 Nm with an assumed damping factor ξ =02 piston within the machine of mass 2 kg has a reciprocating motion with a stroke of 75 cm and speed of 15 cyclessecond Assuming the motion of piston to be harmonic determine amplitude of machine ( X) Transmissibility and force transmitted to foundation phase angle of transmitted force wrt to exciting force 2 Motor weighs 600N is mounted on a simple beam that has a spring rate of 400 Ncm The motor armature base 160 N and has an eccentricity of 0005cm What will be the amplitude of motion when it runs at 1760 RPM Neglect damping the deflection of armature shaft and weight of the beam 3 A Heavy machine of mass 300 kg is supported on resilient foundation the static deflection of foundation due to weight of machine is found to be 75mm It is observed that the machine vibrates with an amplitude of 10mm when the base of the foundation is subjected to harmonic oscillation at the undampeds natural frequency of system with an amplitude of 25mm Find the damping constant of foundation Ii) dynamic force amplitude on the base iii) amplitude of the displacement of machine relative to base 4 An air compressor of mass 450 kg operates at a constant speed of 1750 RPM The reciprocating masses are 10 kg the crank radius is 100mm If the damper for mounting introduces a damping factor 015 i) specify the springs of mounting such that only 20 of the unbalanced force is transmitted to the foundation ii) determine the magnitude of the transmitted force bull Repeat the above problem by assuming that there is no damping in system

5 A body of a car having mass 1500 kg and is mounted on 4 equal springs which deflect through 225 cm under the weight of the body A total damping forces the 4 shock absorbers is 46 N at a velocity of 1cmsec The car is placed with all the four wheels on a test platform which is move up and down at resonant speed With an amplitude of 25cm Find the amplitude of the car body on its springs assuming its CG to be in the centre of wheel base 6A 75 kg machine which is mounted on springs of stiffness 116x105 Nm with the damping of 20=ξ A 2kg piston within the machine as reciprocating motion with a stroke of 008m and speed of 3000 RPM Assuming the motion of piston to be harmonic determine the amplitude of vibration of machine

7 The spring of an automobile are compressed 10cm under its own weight Find the critical speed when a trailer is traveling over road with a people approximaterly by a sine wave of amplitude 75 cm and wave length 15m What will be the amplitude of vibration at 64kmhour 8 An eccentric mass excites us used to determine vibratory characteristics of a structure of mass 25kg as given in figure At a speed of 1000 RPM the eccentric mass to beat top position at the instant the structure was moved upwards through its static equilibrium position and corresponding amplitude was 20mm The eccentric mass is 05kg at an eccentricity of 10mm Determine i) Undamped natural frequency of the system ii) Damping ratio iii) Steady state amplitude and phase angle at 1500 RPM iv) Force transmitted to the ground at 1200 RPM

M=25kg

C

K

Eccentric mass

64

65

CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Support motion Consider a spring mass damper system as shown in figure Let it excited by motion

of support let y be the harmonic displacement of support From Newtonrsquos second law of motion

)yx(k)yx(cxm minusminusminusminus= ampampampamp

)1(0)yx(k)yx(cxm minusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminus=minus+minus+ ampampampamp

Let z be the relative displacement of the mass with respect to support then

Z=x-y

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

39

At any instant of time let the mass be displaced by x from the equilibrium position Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

Okxxcxm =minusminus= ampampamp

Okxxcxm =++ ampampamp ----------------------------------------------------------(1)

or

0kxdtdx

Cdt

xdM

2

2

=++

Assuming the solution of equation as

The first term in the above equation m2Ct

eminus

is an exponentially decaying function of time The behavior of the terms in the parenthesis however depends on whether numerical value within the radicals is positive zero or negative

CASE 1 if 0mk

m2c

2

=minus

40

CASE 2 if 0mk

m2c

2

=gt

CASE 3 if 0mk

m2c

2

=lt

41

Critically damped coefficient Critically damped coefficient is defined as the value of the damping coeffient for which radical in equation

mk

m2c

m2c

S2

21 minus

plusmn

= becomes zero

Damping Ratio ( Damping factor) ( ξ ) It is defined as the ratio of damping coefficient to the critical damping coefficient Note The damping coefficient c is defined as the actual amount of damping while the damping coefficient is a constant depending upon the mass and stiffness of the system The roots of characteristics equation can be now written as Note if c=cc or ξ =1 then system is critically damped if cgt cc or ξ gt1 then system is over damped I if clt cc or ξ lt1 then system is under damped

42

Take home quiz Derive General solution of different damped system

1 Critically damped system 2 Under damped system 3 Over damped system

Logarithmic Decrement It is used to determine the amount of damping present in system This measure The rate of decay of free vibration The larger the damping the greater will be rate of decay Logarithmic decrement is defined as the natural logarithm of ratio of any two successive amplitudes

=δ

2

1

xx

ln

Expression for logarithmic decrement (δ ) Consider a damped free vibration which is expressed by

( ) ( ) φ+ξminusω= ξω t 1sinXex 2n

tn

x is the displacement at any instant of time X and φ are arbitrary constants

21

2

ξminus

πξ=δ

43

Tutorial problems 1 Derive an expression for damping ratio in terms of logarithmic decrement

2 Show that logarithmic decrement

=δ

n

o

xx

lnn1

where xo is the initial amplitude xn

is amplitude after n cycles 3 A mass of 1 kg is to be supported on a spring having stiffness K= 9800 Nm The damping constant c= 59 Nsm Determine the natural frequency of system Find also logarithmic decrement and amplitude after 3 cycles if initial displacement is 03 cm 4 A vibrating system is defined by the following parameters M = 3kg k= 100 Nm damping coefficient = 3 Nsm determine Damping factor natural frequency damped vibration logarithmic decrement ratio of consecutive amplitude no of cycles after which the original amplitude is reduced to 20 5A single degree damped vibrating system a suspended mass of 18kg makes 15 oscillations in 003 seconds The amplitude decreases to 025 of the initial value after 5 oscillations Determine the stiffness of spring Logarithmic decrement damping factor and damping coefficient 6 A damped vibration record of a spring mass system shows the following data Amplitude at the end of 2nd cycle is 9mm amplitude at eh end of 3rd cycle is 6mm amplitude at the end of 4th cycle is 4mm stiffness of spring =8000 Nm weight =20N Find the logarithmic decrement damping force at unit velocity periodic time of vibration 7 For the system shown in figure the characteristics of dashpot is such that the constant force of 49 N is applied to the piston it velocity is found to be constant at 012 ms Determine the value of C would you expect the complete system to be periodic or aperiodic 8 For a torsional system shown in the figuref1 is natural frequency in air and f2 is the natural frequency when immersed in oil Determine the damping coefficient

Viscous fluid

Disc of mass moment of Inertia J

2a

C

m M

b

a a

k

9 A thin plate of area A and weigh w is fixed to the end of the spring and allow to oscillate in a viscous fluid as shown in figure It f1 is the frequency of oscillation in air and f2 in liquid Show that

22

21 ff

gA2

minusπω

=micro where the damping force on the plate

AV2Fd micro= V being the velocity of vibration 10 A torsional pendulum has a natural frequency of 175 cyclessecond vibrating in vacuum The mass moment of inertia is 2 kgm2 It is immersed in oil and it is observed that the natural frequency is 142 cyclessecond Find the damping torque If the disc is replaced by 34 deg When in oil find the displacement at the end of the first complete cycle 11 An automobile can be modeled as a mass placed on 4 shock absorbers each consisting of a spring and a damper Such that each spring is equally loaded Determine the stiffness damping constant of each shock absorber So that natural frequency is 2Hz and system is critically damped The mass of vehicle is 200kg 12 Write the differential equation of motion for the system shown in the figure below Determine the natural frequency of the damped oscillation and critical damping coefficient

Determine the differential equation of motion and find the critical cdamped natural frequency for system shown in figure

a

C

m M

M =200kg

44

200kg kgkg

oefficient and

45

Derive the equation of motion and find the critical damping coefficient and damped natural frequency for the system shown below Write the differential equation of motion and determine the expression for critical damping coefficient and damped natural frequency of system shown in figure

Set up a differential equation of motion for system shown in figure and determine the following Undamped natural frequencies critical damping coefficient damping ratio damped natural frequency

b

C

kM

a

m

46

A mass of 300N is resting on two springs of stiffness 3000 Nm each and a dashpot of damping coefficient 150 Nsm as shown in figure If the initial velocity of 10cms is given to mass at its equilibrium position what will be the displacement from the equilibrium position at the end of first second Derive the formula used A gun barrel of mass 600kg has a recoil spring of stiffness 294 knm if the barrel recoils 13m on firing determine i) The initial recoil velocity of barrel ii) The critical damping coefficient of dashpot which is engaged at the end of recoil stroke Iii) Time required for the barrel to return at the end of recoil stroke Iv) time required for the barrel to return to a position 5cm from the initial position

W= 3000 N

FIRING

CHAPTER 4

FORCED VIBRATION LEARNING OBJECTIVES

] Introduction to forced free vibration

] Terminologies used in forced vibration

] Concept of magnification factor

] Study of reciprocating and rotating unbalance

] Concept of base excitation and support excitation

] Concept of Transmissibility ratio

] Problems related to forced vibration

The vibration that takes place under the excitation of external forces are called

forced vibration When a system is subjected to harmonic excitation it is forced to vibrate at the same frequency as that of excitation Most of the mechanical systems are prone to steady state forced vibration which cause fatigue failure The vibration problem also becomes very important when any of excitation frequencies coincide with one of natural frequencies of a system Under this condition resonance occurs during which large amplitude of vibrations are observed To avoid resonance external frequency (operating speed) may be changed or properties of the system may be changed to alter the natural frequency In some cases sufficient amount of damping may be provided to avoid large amplitude during resonance Thus the problem of force vibration is very important in mechanical design

FORCED VIBRATIONS OF SINGLE DOF SYSTEMS UNDER HARMONIC EXCITATION

` [ ] tsin FF o ω=

x

sin FF o=

Fo

xampamp

[ ] tω

xamp

47

48

Consider a spring mass system having a viscous damping excited by a harmonic force [ ] tsin FF o ω= as shown in figure The various forces acting on the system is shown in figure Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

O)tsin(Fkxxcxm o =ω+minusminus= ampampamp

)tsin(Fkxxcxm o ω=++ ampampamp ----------------------------------------------------------(1)

or

)tsin(Fkxdtdx

Cdt

xdM o2

2

ω=++

Equation (1) represents the differential equation of motion of the system The solution to this equation consists of two parts namely complementary function and a particular integral

The first of solution is also termed as transient solution of homogeneous solution and can be obtained from the homogeneous differential equation 0kxxcxm =++ ampampamp

The second part of solution particular integral is also termed as steady state solution ad is of form )tsin(Xx φminusω= Where X= steady state amplitude φ is phase difference between external force and displacement

)tsin(Xx φminusω=

The vector representation of the shown

49

The force polygon for the above vector diagram can be written as follows

( )φminusωt O

KX

xm 2ω

Fo

xCω

tω

φ

50

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

is the governing equation

Zero frequency deflection (Xst) It is deflection observed at zero frequency ( when the external frequency 0=ω

( ) kF

001

kF

x o

22

o

st =

+minus=

Magnification factor (M) It is defined as the ratio of steady state amplitude to zero frequency deflection xst

M = arX

X=

A plot of magnification factor versus frequency ratio

damping factor ( )ξ is called frequency plot of a harmon All the frequency terms starts from magnification factor of damping ratio ( )ξ The amplitude of vibration at any on the value of damping The amplitude will be higher for higher damping The amplitude of vibration at reso

amplitude tends to zero as

ωω

n tends to infinity

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

Freq

ζ =

ζ = 1

Magnification Factor M = XXst

kFo

51

ωω

nfor varying values of

ically excite vibrating system

M=1 what ever may be value particular frequency depends

for lesser damping and lesser nance will be very high The

uency Ratio r = (ωωn)

ζ = 0 0

ζ = 025 ζ = 0375 ζ = 0 5

ζ = 0707

ζ = 2

52

Plot of phase angle φ versus frequency ratio

ωω

n is shown in the

figure The phase angle φ is zero for all the values of damping ratio

at

ωω

n=0 The phase

angle φ is 90 for all the values of damping ratio

at

ωω

n=1 ie at

resonance The phase angle between o to 90 deg for all the values of

ωω

n= 0 to 1 and

between 90 deg to 180 deg for all the values of

ωω

ngt1

Class work Derive an expression for frequency at peak amplitude

2np 21 ξminusω=ω=ω is the required equation

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

53

Derive expression amplitude at resonance ( In dimensional form) In dimensional form We know that steady state amplitude

( ) ( )

ω+ωminus=

222

o

Cmk

Fx Equation in dimensional form

We know that at resonance mk

n =ω=ω

( )

ω+

minus

=2

n

2

oR

Cmk

mk

Fx

( )

ω=

2n

oR

C

Fx

( )

ω=

n

oR C

Fx

Class work Derive an expression for amplitude at resonance ( in non dimensional form)

Tutorial Problems on Forced Vibration 1A weight of 60N is suspended by a spring stiffness 12 kNm is forced to vibrate by a harmonic force of 10N assuming viscous damping of 0086 kN-sm Find i) resonance frequency ii) amplitude at resonance iii) phase angle at resonance iv) frequency corresponding to peak amplitude 2 A periodic torque having a maximum value of 05888Nm at a frequency of 4 rads is impressed upon a flywheel suspended from a wire The wheel has mass moment of inertia 012 kgm2 and the wire has stiffness 1176 Nmrad A viscous dashpot applies damping couple of 0784 Nm an angular velocity of 2 rads calculate i)maximum angular displacement from rest position ii) maximum couple applied to dashpot Iii) angle by which the angular displacement lags the torque A mass of 1019 kg is suspended from a spring stiffness 20 Nmm viscous damping causes the amplitude to decrease to 110th of initial value in 4 complete cycle If a periodic force of 20 cos(50 t) is applied to the mass Find the amplitude of forced vibration What would be the amplitude of oscillation if the applied force coincides with the natural period

m=1019 kg

[ ] t50cos 30F =

54

55

ROTATING UNBALANCE Unbalance in rotating machinery is a common source of vibration excitation Consider a machine of mass m having mounted on a foundation of stiffness k and damping coefficient C as shown in figure The unbalance is represented by an eccentric rotating mass m with eccentricity (e) and having a angular velocity (ω) let x be the displacement of non rotating mass from the static equilibrium position The displacement of rotating mass m from the static equilibrium is ( ) tsin( ex ω+ Applying Newtonrsquos law of motion we have

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

56

RECIPROCATING UNBALANCE Unbalance in reciprocating machinery is a common source of vibration excitation Consider a machine of mass M having mounted on a foundation of stiffness K and damping coefficient C as shown in figure The unbalance is represented by a reciprocating mass (m) having crank radus (e) and connecting rod of length (l) Let ωωωω be angular velocity of the crank Let x be the displacement of non reciprocating mass from the static equilibrium position at any instant of time The displacement of reciprocating mass m from the static equilibrium position is

)2sin(sin +++ tle

tex ωωωωωωωω

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

57

Note The equations obtained in reciprocating unbalance is same as the rotating unbalance and having the following differences Rotating Unbalance Reciprocating Unbalance Here mass of the rotating element is considered

Here mass of reciprocating element is considered

Here the eccentricity (e) which is the distance between the centre of rotation and CG

Here the eccentricity e is the crank radius and crank radius = stroke length2

The frequency plot of non dimensional factor( ( )me

MX versus frequency ratio

ωω

n for various values of ( )ξξξξ is shown in figure This frequency plot is same for

the rotating and reciprocating unbalance with the following points are noted from the diagram

1 Damping ratio plays very important role during resonance in controlling the amplitude

2 When

ωω

n=0 then ( )me

MX =0 for all the values of damping ratio ( )ξξξξ

3 When

ωω

nis higher than ( )me

MX tends to unity

4 At very high speeds ie at very high frequency ratio

ωω

n there is no effect

of damping The frequency plot of phase angle versus the frequency ratio is given as shown in figure The explanation is same as in forced harmonic vibration

58

VIBRATION ISOLATION

Vibratory forces and motions generated by machine and other caused are unavoidable Howeever their effect on the other dynamical systems can be minimized by proper isolator design An isolation system is either to protect o delicate machine from excessive vibration transmitted to it from its supporting structure or to prevent vibratory forces and motions generated by machines from being transmitted to its surroundings The effectiveness of isolation may be measured in terms of forces and motion transmitted The first type is known as force isolation and the second type is known as motion isolation

Force isolation (Force Transmissibility) When vibratory forces generated by machine are isolated the effectiveness of forced isolation can be made by defining the term transmissibility The term transmissibility is defined as the ratio of force transmitted to the foundation to that of the external impressed force

Force Transmissibility = o

tr

FF

force Impressed Externalfoundation toted transmitForce

=

Consider viscous damped spring mass system as shown in figure Upon this system an external harmonic force ) t (Sin Fo ω is applied The difference forces acting on system are inertia forces damping force spring force and external force All these forces are represented on force polygon shown below Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these two forces

( ) ( )22 xckx ω+=

59

This is the expression for force transmissibility in dimensional form To represent same in the non dimensional form divide numerator and denominator by k

2

n

22

n

2

n

o

tr

21

21

FF

TR

ωω

ξ+

ωω

minus

ωω

ξ+

==

PHASE DIFFERNECE BETWEEN EXTERNAL FORCE AND DISPLACEMENT Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these

ωω

minus

ωω

ξ=

ωminusω

=φ minusminus2

n

n1

2

1

1

2tan

mKC

tan

The phase difference between force transmitted to the foundation Ftr to the displacement

kc

KxxC

tanω

=ω

=α or

ωω

ξ=

ω=α minusminus

n

11 2tanKx

xCtan

The phase difference between external force Fo and force transmitted to foundationis β

60

β= αminusφ =

ωω

minus

ωω

ξminus

2

n

n1

1

2tan -

ωω

ξminus

n

1 2tan

Displacement Transmissibility It is the ratio of amplitude of steady state vibration of machine to that of the base motion Forced vibration due to support motion Consider a spring mass damper system subjected to harmonic excitation )tsin(Yy ω= at the base as shown in figure Let x be the absolute displacement of the mass at any instant of time t The different forces acting on system are shown in free body diagram Applying Newtonrsquos Law of motion Ma = sumF

)yx(k)yx(cxm minusminusminusminus= ampampampamp

0)yx(k)yx(cxm =minus+minus+ ampampampamp

61

Frequency plot of transmissibility versus frequency ratio

62

The plot of transmissibility versus frequency ratio is as shown in figure The following points can be noted from it

i) The value of transmissibility is unity at frequency ratio

ωω

n=0 and 2

ii) For an undamped system ( )ξ =0 transmissibility tends to infin at resonance iii) The value of transmissibility is less than unity for the value of frequency ratio

greater than 2

iv) For frequency ratio

ωω

nlt 2 smaller damping ratio lead to larger values of

transmissibility on the other hand for frequency ratio

ωω

ngt 2 smaller damping

ratio leads to smaller value of transmissibility

The frequency plot of transmissibility (TR) versus frequency ratio

ωω

n can be

broadly divided into two zones

1

ωω

n= 0 to 2

2

ωω

n= 2 to infin

CASE 1

When

ωω

n= 0 to 2

In this zone the value of transmissibility is always greater than 1 for all values of damping ratio In this zone the transmissibility decreases with increase in damping Hence we can conclude that in this zone damping is useful in isolating vibration forces being transmitted to the support CASE 2

When

ωω

n= 2 to infin

In this zone the value of transmissibility is less than 1 for all values of damping ratio In this zone the transmissibility increases with increase in damping Hence we can conclude that in this zone damping will adversely affect in isolating vibratory forces

63

Tutorial problems on Forced Vibration 1 Machine of mass 75kg is mounted on spring of stiffness 12000 Nm with an assumed damping factor ξ =02 piston within the machine of mass 2 kg has a reciprocating motion with a stroke of 75 cm and speed of 15 cyclessecond Assuming the motion of piston to be harmonic determine amplitude of machine ( X) Transmissibility and force transmitted to foundation phase angle of transmitted force wrt to exciting force 2 Motor weighs 600N is mounted on a simple beam that has a spring rate of 400 Ncm The motor armature base 160 N and has an eccentricity of 0005cm What will be the amplitude of motion when it runs at 1760 RPM Neglect damping the deflection of armature shaft and weight of the beam 3 A Heavy machine of mass 300 kg is supported on resilient foundation the static deflection of foundation due to weight of machine is found to be 75mm It is observed that the machine vibrates with an amplitude of 10mm when the base of the foundation is subjected to harmonic oscillation at the undampeds natural frequency of system with an amplitude of 25mm Find the damping constant of foundation Ii) dynamic force amplitude on the base iii) amplitude of the displacement of machine relative to base 4 An air compressor of mass 450 kg operates at a constant speed of 1750 RPM The reciprocating masses are 10 kg the crank radius is 100mm If the damper for mounting introduces a damping factor 015 i) specify the springs of mounting such that only 20 of the unbalanced force is transmitted to the foundation ii) determine the magnitude of the transmitted force bull Repeat the above problem by assuming that there is no damping in system

5 A body of a car having mass 1500 kg and is mounted on 4 equal springs which deflect through 225 cm under the weight of the body A total damping forces the 4 shock absorbers is 46 N at a velocity of 1cmsec The car is placed with all the four wheels on a test platform which is move up and down at resonant speed With an amplitude of 25cm Find the amplitude of the car body on its springs assuming its CG to be in the centre of wheel base 6A 75 kg machine which is mounted on springs of stiffness 116x105 Nm with the damping of 20=ξ A 2kg piston within the machine as reciprocating motion with a stroke of 008m and speed of 3000 RPM Assuming the motion of piston to be harmonic determine the amplitude of vibration of machine

7 The spring of an automobile are compressed 10cm under its own weight Find the critical speed when a trailer is traveling over road with a people approximaterly by a sine wave of amplitude 75 cm and wave length 15m What will be the amplitude of vibration at 64kmhour 8 An eccentric mass excites us used to determine vibratory characteristics of a structure of mass 25kg as given in figure At a speed of 1000 RPM the eccentric mass to beat top position at the instant the structure was moved upwards through its static equilibrium position and corresponding amplitude was 20mm The eccentric mass is 05kg at an eccentricity of 10mm Determine i) Undamped natural frequency of the system ii) Damping ratio iii) Steady state amplitude and phase angle at 1500 RPM iv) Force transmitted to the ground at 1200 RPM

M=25kg

C

K

Eccentric mass

64

65

CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Support motion Consider a spring mass damper system as shown in figure Let it excited by motion

of support let y be the harmonic displacement of support From Newtonrsquos second law of motion

)yx(k)yx(cxm minusminusminusminus= ampampampamp

)1(0)yx(k)yx(cxm minusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminus=minus+minus+ ampampampamp

Let z be the relative displacement of the mass with respect to support then

Z=x-y

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

40

CASE 2 if 0mk

m2c

2

=gt

CASE 3 if 0mk

m2c

2

=lt

41

Critically damped coefficient Critically damped coefficient is defined as the value of the damping coeffient for which radical in equation

mk

m2c

m2c

S2

21 minus

plusmn

= becomes zero

Damping Ratio ( Damping factor) ( ξ ) It is defined as the ratio of damping coefficient to the critical damping coefficient Note The damping coefficient c is defined as the actual amount of damping while the damping coefficient is a constant depending upon the mass and stiffness of the system The roots of characteristics equation can be now written as Note if c=cc or ξ =1 then system is critically damped if cgt cc or ξ gt1 then system is over damped I if clt cc or ξ lt1 then system is under damped

42

Take home quiz Derive General solution of different damped system

1 Critically damped system 2 Under damped system 3 Over damped system

Logarithmic Decrement It is used to determine the amount of damping present in system This measure The rate of decay of free vibration The larger the damping the greater will be rate of decay Logarithmic decrement is defined as the natural logarithm of ratio of any two successive amplitudes

=δ

2

1

xx

ln

Expression for logarithmic decrement (δ ) Consider a damped free vibration which is expressed by

( ) ( ) φ+ξminusω= ξω t 1sinXex 2n

tn

x is the displacement at any instant of time X and φ are arbitrary constants

21

2

ξminus

πξ=δ

43

Tutorial problems 1 Derive an expression for damping ratio in terms of logarithmic decrement

2 Show that logarithmic decrement

=δ

n

o

xx

lnn1

where xo is the initial amplitude xn

is amplitude after n cycles 3 A mass of 1 kg is to be supported on a spring having stiffness K= 9800 Nm The damping constant c= 59 Nsm Determine the natural frequency of system Find also logarithmic decrement and amplitude after 3 cycles if initial displacement is 03 cm 4 A vibrating system is defined by the following parameters M = 3kg k= 100 Nm damping coefficient = 3 Nsm determine Damping factor natural frequency damped vibration logarithmic decrement ratio of consecutive amplitude no of cycles after which the original amplitude is reduced to 20 5A single degree damped vibrating system a suspended mass of 18kg makes 15 oscillations in 003 seconds The amplitude decreases to 025 of the initial value after 5 oscillations Determine the stiffness of spring Logarithmic decrement damping factor and damping coefficient 6 A damped vibration record of a spring mass system shows the following data Amplitude at the end of 2nd cycle is 9mm amplitude at eh end of 3rd cycle is 6mm amplitude at the end of 4th cycle is 4mm stiffness of spring =8000 Nm weight =20N Find the logarithmic decrement damping force at unit velocity periodic time of vibration 7 For the system shown in figure the characteristics of dashpot is such that the constant force of 49 N is applied to the piston it velocity is found to be constant at 012 ms Determine the value of C would you expect the complete system to be periodic or aperiodic 8 For a torsional system shown in the figuref1 is natural frequency in air and f2 is the natural frequency when immersed in oil Determine the damping coefficient

Viscous fluid

Disc of mass moment of Inertia J

2a

C

m M

b

a a

k

9 A thin plate of area A and weigh w is fixed to the end of the spring and allow to oscillate in a viscous fluid as shown in figure It f1 is the frequency of oscillation in air and f2 in liquid Show that

22

21 ff

gA2

minusπω

=micro where the damping force on the plate

AV2Fd micro= V being the velocity of vibration 10 A torsional pendulum has a natural frequency of 175 cyclessecond vibrating in vacuum The mass moment of inertia is 2 kgm2 It is immersed in oil and it is observed that the natural frequency is 142 cyclessecond Find the damping torque If the disc is replaced by 34 deg When in oil find the displacement at the end of the first complete cycle 11 An automobile can be modeled as a mass placed on 4 shock absorbers each consisting of a spring and a damper Such that each spring is equally loaded Determine the stiffness damping constant of each shock absorber So that natural frequency is 2Hz and system is critically damped The mass of vehicle is 200kg 12 Write the differential equation of motion for the system shown in the figure below Determine the natural frequency of the damped oscillation and critical damping coefficient

Determine the differential equation of motion and find the critical cdamped natural frequency for system shown in figure

a

C

m M

M =200kg

44

200kg kgkg

oefficient and

45

Derive the equation of motion and find the critical damping coefficient and damped natural frequency for the system shown below Write the differential equation of motion and determine the expression for critical damping coefficient and damped natural frequency of system shown in figure

Set up a differential equation of motion for system shown in figure and determine the following Undamped natural frequencies critical damping coefficient damping ratio damped natural frequency

b

C

kM

a

m

46

A mass of 300N is resting on two springs of stiffness 3000 Nm each and a dashpot of damping coefficient 150 Nsm as shown in figure If the initial velocity of 10cms is given to mass at its equilibrium position what will be the displacement from the equilibrium position at the end of first second Derive the formula used A gun barrel of mass 600kg has a recoil spring of stiffness 294 knm if the barrel recoils 13m on firing determine i) The initial recoil velocity of barrel ii) The critical damping coefficient of dashpot which is engaged at the end of recoil stroke Iii) Time required for the barrel to return at the end of recoil stroke Iv) time required for the barrel to return to a position 5cm from the initial position

W= 3000 N

FIRING

CHAPTER 4

FORCED VIBRATION LEARNING OBJECTIVES

] Introduction to forced free vibration

] Terminologies used in forced vibration

] Concept of magnification factor

] Study of reciprocating and rotating unbalance

] Concept of base excitation and support excitation

] Concept of Transmissibility ratio

] Problems related to forced vibration

The vibration that takes place under the excitation of external forces are called

forced vibration When a system is subjected to harmonic excitation it is forced to vibrate at the same frequency as that of excitation Most of the mechanical systems are prone to steady state forced vibration which cause fatigue failure The vibration problem also becomes very important when any of excitation frequencies coincide with one of natural frequencies of a system Under this condition resonance occurs during which large amplitude of vibrations are observed To avoid resonance external frequency (operating speed) may be changed or properties of the system may be changed to alter the natural frequency In some cases sufficient amount of damping may be provided to avoid large amplitude during resonance Thus the problem of force vibration is very important in mechanical design

FORCED VIBRATIONS OF SINGLE DOF SYSTEMS UNDER HARMONIC EXCITATION

` [ ] tsin FF o ω=

x

sin FF o=

Fo

xampamp

[ ] tω

xamp

47

48

Consider a spring mass system having a viscous damping excited by a harmonic force [ ] tsin FF o ω= as shown in figure The various forces acting on the system is shown in figure Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

O)tsin(Fkxxcxm o =ω+minusminus= ampampamp

)tsin(Fkxxcxm o ω=++ ampampamp ----------------------------------------------------------(1)

or

)tsin(Fkxdtdx

Cdt

xdM o2

2

ω=++

Equation (1) represents the differential equation of motion of the system The solution to this equation consists of two parts namely complementary function and a particular integral

The first of solution is also termed as transient solution of homogeneous solution and can be obtained from the homogeneous differential equation 0kxxcxm =++ ampampamp

The second part of solution particular integral is also termed as steady state solution ad is of form )tsin(Xx φminusω= Where X= steady state amplitude φ is phase difference between external force and displacement

)tsin(Xx φminusω=

The vector representation of the shown

49

The force polygon for the above vector diagram can be written as follows

( )φminusωt O

KX

xm 2ω

Fo

xCω

tω

φ

50

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

is the governing equation

Zero frequency deflection (Xst) It is deflection observed at zero frequency ( when the external frequency 0=ω

( ) kF

001

kF

x o

22

o

st =

+minus=

Magnification factor (M) It is defined as the ratio of steady state amplitude to zero frequency deflection xst

M = arX

X=

A plot of magnification factor versus frequency ratio

damping factor ( )ξ is called frequency plot of a harmon All the frequency terms starts from magnification factor of damping ratio ( )ξ The amplitude of vibration at any on the value of damping The amplitude will be higher for higher damping The amplitude of vibration at reso

amplitude tends to zero as

ωω

n tends to infinity

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

Freq

ζ =

ζ = 1

Magnification Factor M = XXst

kFo

51

ωω

nfor varying values of

ically excite vibrating system

M=1 what ever may be value particular frequency depends

for lesser damping and lesser nance will be very high The

uency Ratio r = (ωωn)

ζ = 0 0

ζ = 025 ζ = 0375 ζ = 0 5

ζ = 0707

ζ = 2

52

Plot of phase angle φ versus frequency ratio

ωω

n is shown in the

figure The phase angle φ is zero for all the values of damping ratio

at

ωω

n=0 The phase

angle φ is 90 for all the values of damping ratio

at

ωω

n=1 ie at

resonance The phase angle between o to 90 deg for all the values of

ωω

n= 0 to 1 and

between 90 deg to 180 deg for all the values of

ωω

ngt1

Class work Derive an expression for frequency at peak amplitude

2np 21 ξminusω=ω=ω is the required equation

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

53

Derive expression amplitude at resonance ( In dimensional form) In dimensional form We know that steady state amplitude

( ) ( )

ω+ωminus=

222

o

Cmk

Fx Equation in dimensional form

We know that at resonance mk

n =ω=ω

( )

ω+

minus

=2

n

2

oR

Cmk

mk

Fx

( )

ω=

2n

oR

C

Fx

( )

ω=

n

oR C

Fx

Class work Derive an expression for amplitude at resonance ( in non dimensional form)

Tutorial Problems on Forced Vibration 1A weight of 60N is suspended by a spring stiffness 12 kNm is forced to vibrate by a harmonic force of 10N assuming viscous damping of 0086 kN-sm Find i) resonance frequency ii) amplitude at resonance iii) phase angle at resonance iv) frequency corresponding to peak amplitude 2 A periodic torque having a maximum value of 05888Nm at a frequency of 4 rads is impressed upon a flywheel suspended from a wire The wheel has mass moment of inertia 012 kgm2 and the wire has stiffness 1176 Nmrad A viscous dashpot applies damping couple of 0784 Nm an angular velocity of 2 rads calculate i)maximum angular displacement from rest position ii) maximum couple applied to dashpot Iii) angle by which the angular displacement lags the torque A mass of 1019 kg is suspended from a spring stiffness 20 Nmm viscous damping causes the amplitude to decrease to 110th of initial value in 4 complete cycle If a periodic force of 20 cos(50 t) is applied to the mass Find the amplitude of forced vibration What would be the amplitude of oscillation if the applied force coincides with the natural period

m=1019 kg

[ ] t50cos 30F =

54

55

ROTATING UNBALANCE Unbalance in rotating machinery is a common source of vibration excitation Consider a machine of mass m having mounted on a foundation of stiffness k and damping coefficient C as shown in figure The unbalance is represented by an eccentric rotating mass m with eccentricity (e) and having a angular velocity (ω) let x be the displacement of non rotating mass from the static equilibrium position The displacement of rotating mass m from the static equilibrium is ( ) tsin( ex ω+ Applying Newtonrsquos law of motion we have

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

56

RECIPROCATING UNBALANCE Unbalance in reciprocating machinery is a common source of vibration excitation Consider a machine of mass M having mounted on a foundation of stiffness K and damping coefficient C as shown in figure The unbalance is represented by a reciprocating mass (m) having crank radus (e) and connecting rod of length (l) Let ωωωω be angular velocity of the crank Let x be the displacement of non reciprocating mass from the static equilibrium position at any instant of time The displacement of reciprocating mass m from the static equilibrium position is

)2sin(sin +++ tle

tex ωωωωωωωω

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

57

Note The equations obtained in reciprocating unbalance is same as the rotating unbalance and having the following differences Rotating Unbalance Reciprocating Unbalance Here mass of the rotating element is considered

Here mass of reciprocating element is considered

Here the eccentricity (e) which is the distance between the centre of rotation and CG

Here the eccentricity e is the crank radius and crank radius = stroke length2

The frequency plot of non dimensional factor( ( )me

MX versus frequency ratio

ωω

n for various values of ( )ξξξξ is shown in figure This frequency plot is same for

the rotating and reciprocating unbalance with the following points are noted from the diagram

1 Damping ratio plays very important role during resonance in controlling the amplitude

2 When

ωω

n=0 then ( )me

MX =0 for all the values of damping ratio ( )ξξξξ

3 When

ωω

nis higher than ( )me

MX tends to unity

4 At very high speeds ie at very high frequency ratio

ωω

n there is no effect

of damping The frequency plot of phase angle versus the frequency ratio is given as shown in figure The explanation is same as in forced harmonic vibration

58

VIBRATION ISOLATION

Vibratory forces and motions generated by machine and other caused are unavoidable Howeever their effect on the other dynamical systems can be minimized by proper isolator design An isolation system is either to protect o delicate machine from excessive vibration transmitted to it from its supporting structure or to prevent vibratory forces and motions generated by machines from being transmitted to its surroundings The effectiveness of isolation may be measured in terms of forces and motion transmitted The first type is known as force isolation and the second type is known as motion isolation

Force isolation (Force Transmissibility) When vibratory forces generated by machine are isolated the effectiveness of forced isolation can be made by defining the term transmissibility The term transmissibility is defined as the ratio of force transmitted to the foundation to that of the external impressed force

Force Transmissibility = o

tr

FF

force Impressed Externalfoundation toted transmitForce

=

Consider viscous damped spring mass system as shown in figure Upon this system an external harmonic force ) t (Sin Fo ω is applied The difference forces acting on system are inertia forces damping force spring force and external force All these forces are represented on force polygon shown below Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these two forces

( ) ( )22 xckx ω+=

59

This is the expression for force transmissibility in dimensional form To represent same in the non dimensional form divide numerator and denominator by k

2

n

22

n

2

n

o

tr

21

21

FF

TR

ωω

ξ+

ωω

minus

ωω

ξ+

==

PHASE DIFFERNECE BETWEEN EXTERNAL FORCE AND DISPLACEMENT Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these

ωω

minus

ωω

ξ=

ωminusω

=φ minusminus2

n

n1

2

1

1

2tan

mKC

tan

The phase difference between force transmitted to the foundation Ftr to the displacement

kc

KxxC

tanω

=ω

=α or

ωω

ξ=

ω=α minusminus

n

11 2tanKx

xCtan

The phase difference between external force Fo and force transmitted to foundationis β

60

β= αminusφ =

ωω

minus

ωω

ξminus

2

n

n1

1

2tan -

ωω

ξminus

n

1 2tan

Displacement Transmissibility It is the ratio of amplitude of steady state vibration of machine to that of the base motion Forced vibration due to support motion Consider a spring mass damper system subjected to harmonic excitation )tsin(Yy ω= at the base as shown in figure Let x be the absolute displacement of the mass at any instant of time t The different forces acting on system are shown in free body diagram Applying Newtonrsquos Law of motion Ma = sumF

)yx(k)yx(cxm minusminusminusminus= ampampampamp

0)yx(k)yx(cxm =minus+minus+ ampampampamp

61

Frequency plot of transmissibility versus frequency ratio

62

The plot of transmissibility versus frequency ratio is as shown in figure The following points can be noted from it

i) The value of transmissibility is unity at frequency ratio

ωω

n=0 and 2

ii) For an undamped system ( )ξ =0 transmissibility tends to infin at resonance iii) The value of transmissibility is less than unity for the value of frequency ratio

greater than 2

iv) For frequency ratio

ωω

nlt 2 smaller damping ratio lead to larger values of

transmissibility on the other hand for frequency ratio

ωω

ngt 2 smaller damping

ratio leads to smaller value of transmissibility

The frequency plot of transmissibility (TR) versus frequency ratio

ωω

n can be

broadly divided into two zones

1

ωω

n= 0 to 2

2

ωω

n= 2 to infin

CASE 1

When

ωω

n= 0 to 2

In this zone the value of transmissibility is always greater than 1 for all values of damping ratio In this zone the transmissibility decreases with increase in damping Hence we can conclude that in this zone damping is useful in isolating vibration forces being transmitted to the support CASE 2

When

ωω

n= 2 to infin

In this zone the value of transmissibility is less than 1 for all values of damping ratio In this zone the transmissibility increases with increase in damping Hence we can conclude that in this zone damping will adversely affect in isolating vibratory forces

63

Tutorial problems on Forced Vibration 1 Machine of mass 75kg is mounted on spring of stiffness 12000 Nm with an assumed damping factor ξ =02 piston within the machine of mass 2 kg has a reciprocating motion with a stroke of 75 cm and speed of 15 cyclessecond Assuming the motion of piston to be harmonic determine amplitude of machine ( X) Transmissibility and force transmitted to foundation phase angle of transmitted force wrt to exciting force 2 Motor weighs 600N is mounted on a simple beam that has a spring rate of 400 Ncm The motor armature base 160 N and has an eccentricity of 0005cm What will be the amplitude of motion when it runs at 1760 RPM Neglect damping the deflection of armature shaft and weight of the beam 3 A Heavy machine of mass 300 kg is supported on resilient foundation the static deflection of foundation due to weight of machine is found to be 75mm It is observed that the machine vibrates with an amplitude of 10mm when the base of the foundation is subjected to harmonic oscillation at the undampeds natural frequency of system with an amplitude of 25mm Find the damping constant of foundation Ii) dynamic force amplitude on the base iii) amplitude of the displacement of machine relative to base 4 An air compressor of mass 450 kg operates at a constant speed of 1750 RPM The reciprocating masses are 10 kg the crank radius is 100mm If the damper for mounting introduces a damping factor 015 i) specify the springs of mounting such that only 20 of the unbalanced force is transmitted to the foundation ii) determine the magnitude of the transmitted force bull Repeat the above problem by assuming that there is no damping in system

5 A body of a car having mass 1500 kg and is mounted on 4 equal springs which deflect through 225 cm under the weight of the body A total damping forces the 4 shock absorbers is 46 N at a velocity of 1cmsec The car is placed with all the four wheels on a test platform which is move up and down at resonant speed With an amplitude of 25cm Find the amplitude of the car body on its springs assuming its CG to be in the centre of wheel base 6A 75 kg machine which is mounted on springs of stiffness 116x105 Nm with the damping of 20=ξ A 2kg piston within the machine as reciprocating motion with a stroke of 008m and speed of 3000 RPM Assuming the motion of piston to be harmonic determine the amplitude of vibration of machine

7 The spring of an automobile are compressed 10cm under its own weight Find the critical speed when a trailer is traveling over road with a people approximaterly by a sine wave of amplitude 75 cm and wave length 15m What will be the amplitude of vibration at 64kmhour 8 An eccentric mass excites us used to determine vibratory characteristics of a structure of mass 25kg as given in figure At a speed of 1000 RPM the eccentric mass to beat top position at the instant the structure was moved upwards through its static equilibrium position and corresponding amplitude was 20mm The eccentric mass is 05kg at an eccentricity of 10mm Determine i) Undamped natural frequency of the system ii) Damping ratio iii) Steady state amplitude and phase angle at 1500 RPM iv) Force transmitted to the ground at 1200 RPM

M=25kg

C

K

Eccentric mass

64

65

CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Support motion Consider a spring mass damper system as shown in figure Let it excited by motion

of support let y be the harmonic displacement of support From Newtonrsquos second law of motion

)yx(k)yx(cxm minusminusminusminus= ampampampamp

)1(0)yx(k)yx(cxm minusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminus=minus+minus+ ampampampamp

Let z be the relative displacement of the mass with respect to support then

Z=x-y

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

41

Critically damped coefficient Critically damped coefficient is defined as the value of the damping coeffient for which radical in equation

mk

m2c

m2c

S2

21 minus

plusmn

= becomes zero

Damping Ratio ( Damping factor) ( ξ ) It is defined as the ratio of damping coefficient to the critical damping coefficient Note The damping coefficient c is defined as the actual amount of damping while the damping coefficient is a constant depending upon the mass and stiffness of the system The roots of characteristics equation can be now written as Note if c=cc or ξ =1 then system is critically damped if cgt cc or ξ gt1 then system is over damped I if clt cc or ξ lt1 then system is under damped

42

Take home quiz Derive General solution of different damped system

1 Critically damped system 2 Under damped system 3 Over damped system

Logarithmic Decrement It is used to determine the amount of damping present in system This measure The rate of decay of free vibration The larger the damping the greater will be rate of decay Logarithmic decrement is defined as the natural logarithm of ratio of any two successive amplitudes

=δ

2

1

xx

ln

Expression for logarithmic decrement (δ ) Consider a damped free vibration which is expressed by

( ) ( ) φ+ξminusω= ξω t 1sinXex 2n

tn

x is the displacement at any instant of time X and φ are arbitrary constants

21

2

ξminus

πξ=δ

43

Tutorial problems 1 Derive an expression for damping ratio in terms of logarithmic decrement

2 Show that logarithmic decrement

=δ

n

o

xx

lnn1

where xo is the initial amplitude xn

is amplitude after n cycles 3 A mass of 1 kg is to be supported on a spring having stiffness K= 9800 Nm The damping constant c= 59 Nsm Determine the natural frequency of system Find also logarithmic decrement and amplitude after 3 cycles if initial displacement is 03 cm 4 A vibrating system is defined by the following parameters M = 3kg k= 100 Nm damping coefficient = 3 Nsm determine Damping factor natural frequency damped vibration logarithmic decrement ratio of consecutive amplitude no of cycles after which the original amplitude is reduced to 20 5A single degree damped vibrating system a suspended mass of 18kg makes 15 oscillations in 003 seconds The amplitude decreases to 025 of the initial value after 5 oscillations Determine the stiffness of spring Logarithmic decrement damping factor and damping coefficient 6 A damped vibration record of a spring mass system shows the following data Amplitude at the end of 2nd cycle is 9mm amplitude at eh end of 3rd cycle is 6mm amplitude at the end of 4th cycle is 4mm stiffness of spring =8000 Nm weight =20N Find the logarithmic decrement damping force at unit velocity periodic time of vibration 7 For the system shown in figure the characteristics of dashpot is such that the constant force of 49 N is applied to the piston it velocity is found to be constant at 012 ms Determine the value of C would you expect the complete system to be periodic or aperiodic 8 For a torsional system shown in the figuref1 is natural frequency in air and f2 is the natural frequency when immersed in oil Determine the damping coefficient

Viscous fluid

Disc of mass moment of Inertia J

2a

C

m M

b

a a

k

9 A thin plate of area A and weigh w is fixed to the end of the spring and allow to oscillate in a viscous fluid as shown in figure It f1 is the frequency of oscillation in air and f2 in liquid Show that

22

21 ff

gA2

minusπω

=micro where the damping force on the plate

AV2Fd micro= V being the velocity of vibration 10 A torsional pendulum has a natural frequency of 175 cyclessecond vibrating in vacuum The mass moment of inertia is 2 kgm2 It is immersed in oil and it is observed that the natural frequency is 142 cyclessecond Find the damping torque If the disc is replaced by 34 deg When in oil find the displacement at the end of the first complete cycle 11 An automobile can be modeled as a mass placed on 4 shock absorbers each consisting of a spring and a damper Such that each spring is equally loaded Determine the stiffness damping constant of each shock absorber So that natural frequency is 2Hz and system is critically damped The mass of vehicle is 200kg 12 Write the differential equation of motion for the system shown in the figure below Determine the natural frequency of the damped oscillation and critical damping coefficient

Determine the differential equation of motion and find the critical cdamped natural frequency for system shown in figure

a

C

m M

M =200kg

44

200kg kgkg

oefficient and

45

Derive the equation of motion and find the critical damping coefficient and damped natural frequency for the system shown below Write the differential equation of motion and determine the expression for critical damping coefficient and damped natural frequency of system shown in figure

Set up a differential equation of motion for system shown in figure and determine the following Undamped natural frequencies critical damping coefficient damping ratio damped natural frequency

b

C

kM

a

m

46

A mass of 300N is resting on two springs of stiffness 3000 Nm each and a dashpot of damping coefficient 150 Nsm as shown in figure If the initial velocity of 10cms is given to mass at its equilibrium position what will be the displacement from the equilibrium position at the end of first second Derive the formula used A gun barrel of mass 600kg has a recoil spring of stiffness 294 knm if the barrel recoils 13m on firing determine i) The initial recoil velocity of barrel ii) The critical damping coefficient of dashpot which is engaged at the end of recoil stroke Iii) Time required for the barrel to return at the end of recoil stroke Iv) time required for the barrel to return to a position 5cm from the initial position

W= 3000 N

FIRING

CHAPTER 4

FORCED VIBRATION LEARNING OBJECTIVES

] Introduction to forced free vibration

] Terminologies used in forced vibration

] Concept of magnification factor

] Study of reciprocating and rotating unbalance

] Concept of base excitation and support excitation

] Concept of Transmissibility ratio

] Problems related to forced vibration

The vibration that takes place under the excitation of external forces are called

forced vibration When a system is subjected to harmonic excitation it is forced to vibrate at the same frequency as that of excitation Most of the mechanical systems are prone to steady state forced vibration which cause fatigue failure The vibration problem also becomes very important when any of excitation frequencies coincide with one of natural frequencies of a system Under this condition resonance occurs during which large amplitude of vibrations are observed To avoid resonance external frequency (operating speed) may be changed or properties of the system may be changed to alter the natural frequency In some cases sufficient amount of damping may be provided to avoid large amplitude during resonance Thus the problem of force vibration is very important in mechanical design

FORCED VIBRATIONS OF SINGLE DOF SYSTEMS UNDER HARMONIC EXCITATION

` [ ] tsin FF o ω=

x

sin FF o=

Fo

xampamp

[ ] tω

xamp

47

48

Consider a spring mass system having a viscous damping excited by a harmonic force [ ] tsin FF o ω= as shown in figure The various forces acting on the system is shown in figure Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

O)tsin(Fkxxcxm o =ω+minusminus= ampampamp

)tsin(Fkxxcxm o ω=++ ampampamp ----------------------------------------------------------(1)

or

)tsin(Fkxdtdx

Cdt

xdM o2

2

ω=++

Equation (1) represents the differential equation of motion of the system The solution to this equation consists of two parts namely complementary function and a particular integral

The first of solution is also termed as transient solution of homogeneous solution and can be obtained from the homogeneous differential equation 0kxxcxm =++ ampampamp

The second part of solution particular integral is also termed as steady state solution ad is of form )tsin(Xx φminusω= Where X= steady state amplitude φ is phase difference between external force and displacement

)tsin(Xx φminusω=

The vector representation of the shown

49

The force polygon for the above vector diagram can be written as follows

( )φminusωt O

KX

xm 2ω

Fo

xCω

tω

φ

50

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

is the governing equation

Zero frequency deflection (Xst) It is deflection observed at zero frequency ( when the external frequency 0=ω

( ) kF

001

kF

x o

22

o

st =

+minus=

Magnification factor (M) It is defined as the ratio of steady state amplitude to zero frequency deflection xst

M = arX

X=

A plot of magnification factor versus frequency ratio

damping factor ( )ξ is called frequency plot of a harmon All the frequency terms starts from magnification factor of damping ratio ( )ξ The amplitude of vibration at any on the value of damping The amplitude will be higher for higher damping The amplitude of vibration at reso

amplitude tends to zero as

ωω

n tends to infinity

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

Freq

ζ =

ζ = 1

Magnification Factor M = XXst

kFo

51

ωω

nfor varying values of

ically excite vibrating system

M=1 what ever may be value particular frequency depends

for lesser damping and lesser nance will be very high The

uency Ratio r = (ωωn)

ζ = 0 0

ζ = 025 ζ = 0375 ζ = 0 5

ζ = 0707

ζ = 2

52

Plot of phase angle φ versus frequency ratio

ωω

n is shown in the

figure The phase angle φ is zero for all the values of damping ratio

at

ωω

n=0 The phase

angle φ is 90 for all the values of damping ratio

at

ωω

n=1 ie at

resonance The phase angle between o to 90 deg for all the values of

ωω

n= 0 to 1 and

between 90 deg to 180 deg for all the values of

ωω

ngt1

Class work Derive an expression for frequency at peak amplitude

2np 21 ξminusω=ω=ω is the required equation

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

53

Derive expression amplitude at resonance ( In dimensional form) In dimensional form We know that steady state amplitude

( ) ( )

ω+ωminus=

222

o

Cmk

Fx Equation in dimensional form

We know that at resonance mk

n =ω=ω

( )

ω+

minus

=2

n

2

oR

Cmk

mk

Fx

( )

ω=

2n

oR

C

Fx

( )

ω=

n

oR C

Fx

Class work Derive an expression for amplitude at resonance ( in non dimensional form)

Tutorial Problems on Forced Vibration 1A weight of 60N is suspended by a spring stiffness 12 kNm is forced to vibrate by a harmonic force of 10N assuming viscous damping of 0086 kN-sm Find i) resonance frequency ii) amplitude at resonance iii) phase angle at resonance iv) frequency corresponding to peak amplitude 2 A periodic torque having a maximum value of 05888Nm at a frequency of 4 rads is impressed upon a flywheel suspended from a wire The wheel has mass moment of inertia 012 kgm2 and the wire has stiffness 1176 Nmrad A viscous dashpot applies damping couple of 0784 Nm an angular velocity of 2 rads calculate i)maximum angular displacement from rest position ii) maximum couple applied to dashpot Iii) angle by which the angular displacement lags the torque A mass of 1019 kg is suspended from a spring stiffness 20 Nmm viscous damping causes the amplitude to decrease to 110th of initial value in 4 complete cycle If a periodic force of 20 cos(50 t) is applied to the mass Find the amplitude of forced vibration What would be the amplitude of oscillation if the applied force coincides with the natural period

m=1019 kg

[ ] t50cos 30F =

54

55

ROTATING UNBALANCE Unbalance in rotating machinery is a common source of vibration excitation Consider a machine of mass m having mounted on a foundation of stiffness k and damping coefficient C as shown in figure The unbalance is represented by an eccentric rotating mass m with eccentricity (e) and having a angular velocity (ω) let x be the displacement of non rotating mass from the static equilibrium position The displacement of rotating mass m from the static equilibrium is ( ) tsin( ex ω+ Applying Newtonrsquos law of motion we have

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

56

RECIPROCATING UNBALANCE Unbalance in reciprocating machinery is a common source of vibration excitation Consider a machine of mass M having mounted on a foundation of stiffness K and damping coefficient C as shown in figure The unbalance is represented by a reciprocating mass (m) having crank radus (e) and connecting rod of length (l) Let ωωωω be angular velocity of the crank Let x be the displacement of non reciprocating mass from the static equilibrium position at any instant of time The displacement of reciprocating mass m from the static equilibrium position is

)2sin(sin +++ tle

tex ωωωωωωωω

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

57

Note The equations obtained in reciprocating unbalance is same as the rotating unbalance and having the following differences Rotating Unbalance Reciprocating Unbalance Here mass of the rotating element is considered

Here mass of reciprocating element is considered

Here the eccentricity (e) which is the distance between the centre of rotation and CG

Here the eccentricity e is the crank radius and crank radius = stroke length2

The frequency plot of non dimensional factor( ( )me

MX versus frequency ratio

ωω

n for various values of ( )ξξξξ is shown in figure This frequency plot is same for

the rotating and reciprocating unbalance with the following points are noted from the diagram

1 Damping ratio plays very important role during resonance in controlling the amplitude

2 When

ωω

n=0 then ( )me

MX =0 for all the values of damping ratio ( )ξξξξ

3 When

ωω

nis higher than ( )me

MX tends to unity

4 At very high speeds ie at very high frequency ratio

ωω

n there is no effect

of damping The frequency plot of phase angle versus the frequency ratio is given as shown in figure The explanation is same as in forced harmonic vibration

58

VIBRATION ISOLATION

Vibratory forces and motions generated by machine and other caused are unavoidable Howeever their effect on the other dynamical systems can be minimized by proper isolator design An isolation system is either to protect o delicate machine from excessive vibration transmitted to it from its supporting structure or to prevent vibratory forces and motions generated by machines from being transmitted to its surroundings The effectiveness of isolation may be measured in terms of forces and motion transmitted The first type is known as force isolation and the second type is known as motion isolation

Force isolation (Force Transmissibility) When vibratory forces generated by machine are isolated the effectiveness of forced isolation can be made by defining the term transmissibility The term transmissibility is defined as the ratio of force transmitted to the foundation to that of the external impressed force

Force Transmissibility = o

tr

FF

force Impressed Externalfoundation toted transmitForce

=

Consider viscous damped spring mass system as shown in figure Upon this system an external harmonic force ) t (Sin Fo ω is applied The difference forces acting on system are inertia forces damping force spring force and external force All these forces are represented on force polygon shown below Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these two forces

( ) ( )22 xckx ω+=

59

This is the expression for force transmissibility in dimensional form To represent same in the non dimensional form divide numerator and denominator by k

2

n

22

n

2

n

o

tr

21

21

FF

TR

ωω

ξ+

ωω

minus

ωω

ξ+

==

PHASE DIFFERNECE BETWEEN EXTERNAL FORCE AND DISPLACEMENT Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these

ωω

minus

ωω

ξ=

ωminusω

=φ minusminus2

n

n1

2

1

1

2tan

mKC

tan

The phase difference between force transmitted to the foundation Ftr to the displacement

kc

KxxC

tanω

=ω

=α or

ωω

ξ=

ω=α minusminus

n

11 2tanKx

xCtan

The phase difference between external force Fo and force transmitted to foundationis β

60

β= αminusφ =

ωω

minus

ωω

ξminus

2

n

n1

1

2tan -

ωω

ξminus

n

1 2tan

Displacement Transmissibility It is the ratio of amplitude of steady state vibration of machine to that of the base motion Forced vibration due to support motion Consider a spring mass damper system subjected to harmonic excitation )tsin(Yy ω= at the base as shown in figure Let x be the absolute displacement of the mass at any instant of time t The different forces acting on system are shown in free body diagram Applying Newtonrsquos Law of motion Ma = sumF

)yx(k)yx(cxm minusminusminusminus= ampampampamp

0)yx(k)yx(cxm =minus+minus+ ampampampamp

61

Frequency plot of transmissibility versus frequency ratio

62

The plot of transmissibility versus frequency ratio is as shown in figure The following points can be noted from it

i) The value of transmissibility is unity at frequency ratio

ωω

n=0 and 2

ii) For an undamped system ( )ξ =0 transmissibility tends to infin at resonance iii) The value of transmissibility is less than unity for the value of frequency ratio

greater than 2

iv) For frequency ratio

ωω

nlt 2 smaller damping ratio lead to larger values of

transmissibility on the other hand for frequency ratio

ωω

ngt 2 smaller damping

ratio leads to smaller value of transmissibility

The frequency plot of transmissibility (TR) versus frequency ratio

ωω

n can be

broadly divided into two zones

1

ωω

n= 0 to 2

2

ωω

n= 2 to infin

CASE 1

When

ωω

n= 0 to 2

In this zone the value of transmissibility is always greater than 1 for all values of damping ratio In this zone the transmissibility decreases with increase in damping Hence we can conclude that in this zone damping is useful in isolating vibration forces being transmitted to the support CASE 2

When

ωω

n= 2 to infin

In this zone the value of transmissibility is less than 1 for all values of damping ratio In this zone the transmissibility increases with increase in damping Hence we can conclude that in this zone damping will adversely affect in isolating vibratory forces

63

Tutorial problems on Forced Vibration 1 Machine of mass 75kg is mounted on spring of stiffness 12000 Nm with an assumed damping factor ξ =02 piston within the machine of mass 2 kg has a reciprocating motion with a stroke of 75 cm and speed of 15 cyclessecond Assuming the motion of piston to be harmonic determine amplitude of machine ( X) Transmissibility and force transmitted to foundation phase angle of transmitted force wrt to exciting force 2 Motor weighs 600N is mounted on a simple beam that has a spring rate of 400 Ncm The motor armature base 160 N and has an eccentricity of 0005cm What will be the amplitude of motion when it runs at 1760 RPM Neglect damping the deflection of armature shaft and weight of the beam 3 A Heavy machine of mass 300 kg is supported on resilient foundation the static deflection of foundation due to weight of machine is found to be 75mm It is observed that the machine vibrates with an amplitude of 10mm when the base of the foundation is subjected to harmonic oscillation at the undampeds natural frequency of system with an amplitude of 25mm Find the damping constant of foundation Ii) dynamic force amplitude on the base iii) amplitude of the displacement of machine relative to base 4 An air compressor of mass 450 kg operates at a constant speed of 1750 RPM The reciprocating masses are 10 kg the crank radius is 100mm If the damper for mounting introduces a damping factor 015 i) specify the springs of mounting such that only 20 of the unbalanced force is transmitted to the foundation ii) determine the magnitude of the transmitted force bull Repeat the above problem by assuming that there is no damping in system

5 A body of a car having mass 1500 kg and is mounted on 4 equal springs which deflect through 225 cm under the weight of the body A total damping forces the 4 shock absorbers is 46 N at a velocity of 1cmsec The car is placed with all the four wheels on a test platform which is move up and down at resonant speed With an amplitude of 25cm Find the amplitude of the car body on its springs assuming its CG to be in the centre of wheel base 6A 75 kg machine which is mounted on springs of stiffness 116x105 Nm with the damping of 20=ξ A 2kg piston within the machine as reciprocating motion with a stroke of 008m and speed of 3000 RPM Assuming the motion of piston to be harmonic determine the amplitude of vibration of machine

7 The spring of an automobile are compressed 10cm under its own weight Find the critical speed when a trailer is traveling over road with a people approximaterly by a sine wave of amplitude 75 cm and wave length 15m What will be the amplitude of vibration at 64kmhour 8 An eccentric mass excites us used to determine vibratory characteristics of a structure of mass 25kg as given in figure At a speed of 1000 RPM the eccentric mass to beat top position at the instant the structure was moved upwards through its static equilibrium position and corresponding amplitude was 20mm The eccentric mass is 05kg at an eccentricity of 10mm Determine i) Undamped natural frequency of the system ii) Damping ratio iii) Steady state amplitude and phase angle at 1500 RPM iv) Force transmitted to the ground at 1200 RPM

M=25kg

C

K

Eccentric mass

64

65

CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Support motion Consider a spring mass damper system as shown in figure Let it excited by motion

of support let y be the harmonic displacement of support From Newtonrsquos second law of motion

)yx(k)yx(cxm minusminusminusminus= ampampampamp

)1(0)yx(k)yx(cxm minusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminus=minus+minus+ ampampampamp

Let z be the relative displacement of the mass with respect to support then

Z=x-y

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

42

Take home quiz Derive General solution of different damped system

1 Critically damped system 2 Under damped system 3 Over damped system

Logarithmic Decrement It is used to determine the amount of damping present in system This measure The rate of decay of free vibration The larger the damping the greater will be rate of decay Logarithmic decrement is defined as the natural logarithm of ratio of any two successive amplitudes

=δ

2

1

xx

ln

Expression for logarithmic decrement (δ ) Consider a damped free vibration which is expressed by

( ) ( ) φ+ξminusω= ξω t 1sinXex 2n

tn

x is the displacement at any instant of time X and φ are arbitrary constants

21

2

ξminus

πξ=δ

43

Tutorial problems 1 Derive an expression for damping ratio in terms of logarithmic decrement

2 Show that logarithmic decrement

=δ

n

o

xx

lnn1

where xo is the initial amplitude xn

is amplitude after n cycles 3 A mass of 1 kg is to be supported on a spring having stiffness K= 9800 Nm The damping constant c= 59 Nsm Determine the natural frequency of system Find also logarithmic decrement and amplitude after 3 cycles if initial displacement is 03 cm 4 A vibrating system is defined by the following parameters M = 3kg k= 100 Nm damping coefficient = 3 Nsm determine Damping factor natural frequency damped vibration logarithmic decrement ratio of consecutive amplitude no of cycles after which the original amplitude is reduced to 20 5A single degree damped vibrating system a suspended mass of 18kg makes 15 oscillations in 003 seconds The amplitude decreases to 025 of the initial value after 5 oscillations Determine the stiffness of spring Logarithmic decrement damping factor and damping coefficient 6 A damped vibration record of a spring mass system shows the following data Amplitude at the end of 2nd cycle is 9mm amplitude at eh end of 3rd cycle is 6mm amplitude at the end of 4th cycle is 4mm stiffness of spring =8000 Nm weight =20N Find the logarithmic decrement damping force at unit velocity periodic time of vibration 7 For the system shown in figure the characteristics of dashpot is such that the constant force of 49 N is applied to the piston it velocity is found to be constant at 012 ms Determine the value of C would you expect the complete system to be periodic or aperiodic 8 For a torsional system shown in the figuref1 is natural frequency in air and f2 is the natural frequency when immersed in oil Determine the damping coefficient

Viscous fluid

Disc of mass moment of Inertia J

2a

C

m M

b

a a

k

9 A thin plate of area A and weigh w is fixed to the end of the spring and allow to oscillate in a viscous fluid as shown in figure It f1 is the frequency of oscillation in air and f2 in liquid Show that

22

21 ff

gA2

minusπω

=micro where the damping force on the plate

AV2Fd micro= V being the velocity of vibration 10 A torsional pendulum has a natural frequency of 175 cyclessecond vibrating in vacuum The mass moment of inertia is 2 kgm2 It is immersed in oil and it is observed that the natural frequency is 142 cyclessecond Find the damping torque If the disc is replaced by 34 deg When in oil find the displacement at the end of the first complete cycle 11 An automobile can be modeled as a mass placed on 4 shock absorbers each consisting of a spring and a damper Such that each spring is equally loaded Determine the stiffness damping constant of each shock absorber So that natural frequency is 2Hz and system is critically damped The mass of vehicle is 200kg 12 Write the differential equation of motion for the system shown in the figure below Determine the natural frequency of the damped oscillation and critical damping coefficient

Determine the differential equation of motion and find the critical cdamped natural frequency for system shown in figure

a

C

m M

M =200kg

44

200kg kgkg

oefficient and

45

Derive the equation of motion and find the critical damping coefficient and damped natural frequency for the system shown below Write the differential equation of motion and determine the expression for critical damping coefficient and damped natural frequency of system shown in figure

Set up a differential equation of motion for system shown in figure and determine the following Undamped natural frequencies critical damping coefficient damping ratio damped natural frequency

b

C

kM

a

m

46

A mass of 300N is resting on two springs of stiffness 3000 Nm each and a dashpot of damping coefficient 150 Nsm as shown in figure If the initial velocity of 10cms is given to mass at its equilibrium position what will be the displacement from the equilibrium position at the end of first second Derive the formula used A gun barrel of mass 600kg has a recoil spring of stiffness 294 knm if the barrel recoils 13m on firing determine i) The initial recoil velocity of barrel ii) The critical damping coefficient of dashpot which is engaged at the end of recoil stroke Iii) Time required for the barrel to return at the end of recoil stroke Iv) time required for the barrel to return to a position 5cm from the initial position

W= 3000 N

FIRING

CHAPTER 4

FORCED VIBRATION LEARNING OBJECTIVES

] Introduction to forced free vibration

] Terminologies used in forced vibration

] Concept of magnification factor

] Study of reciprocating and rotating unbalance

] Concept of base excitation and support excitation

] Concept of Transmissibility ratio

] Problems related to forced vibration

The vibration that takes place under the excitation of external forces are called

forced vibration When a system is subjected to harmonic excitation it is forced to vibrate at the same frequency as that of excitation Most of the mechanical systems are prone to steady state forced vibration which cause fatigue failure The vibration problem also becomes very important when any of excitation frequencies coincide with one of natural frequencies of a system Under this condition resonance occurs during which large amplitude of vibrations are observed To avoid resonance external frequency (operating speed) may be changed or properties of the system may be changed to alter the natural frequency In some cases sufficient amount of damping may be provided to avoid large amplitude during resonance Thus the problem of force vibration is very important in mechanical design

FORCED VIBRATIONS OF SINGLE DOF SYSTEMS UNDER HARMONIC EXCITATION

` [ ] tsin FF o ω=

x

sin FF o=

Fo

xampamp

[ ] tω

xamp

47

48

Consider a spring mass system having a viscous damping excited by a harmonic force [ ] tsin FF o ω= as shown in figure The various forces acting on the system is shown in figure Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

O)tsin(Fkxxcxm o =ω+minusminus= ampampamp

)tsin(Fkxxcxm o ω=++ ampampamp ----------------------------------------------------------(1)

or

)tsin(Fkxdtdx

Cdt

xdM o2

2

ω=++

Equation (1) represents the differential equation of motion of the system The solution to this equation consists of two parts namely complementary function and a particular integral

The first of solution is also termed as transient solution of homogeneous solution and can be obtained from the homogeneous differential equation 0kxxcxm =++ ampampamp

The second part of solution particular integral is also termed as steady state solution ad is of form )tsin(Xx φminusω= Where X= steady state amplitude φ is phase difference between external force and displacement

)tsin(Xx φminusω=

The vector representation of the shown

49

The force polygon for the above vector diagram can be written as follows

( )φminusωt O

KX

xm 2ω

Fo

xCω

tω

φ

50

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

is the governing equation

Zero frequency deflection (Xst) It is deflection observed at zero frequency ( when the external frequency 0=ω

( ) kF

001

kF

x o

22

o

st =

+minus=

Magnification factor (M) It is defined as the ratio of steady state amplitude to zero frequency deflection xst

M = arX

X=

A plot of magnification factor versus frequency ratio

damping factor ( )ξ is called frequency plot of a harmon All the frequency terms starts from magnification factor of damping ratio ( )ξ The amplitude of vibration at any on the value of damping The amplitude will be higher for higher damping The amplitude of vibration at reso

amplitude tends to zero as

ωω

n tends to infinity

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

Freq

ζ =

ζ = 1

Magnification Factor M = XXst

kFo

51

ωω

nfor varying values of

ically excite vibrating system

M=1 what ever may be value particular frequency depends

for lesser damping and lesser nance will be very high The

uency Ratio r = (ωωn)

ζ = 0 0

ζ = 025 ζ = 0375 ζ = 0 5

ζ = 0707

ζ = 2

52

Plot of phase angle φ versus frequency ratio

ωω

n is shown in the

figure The phase angle φ is zero for all the values of damping ratio

at

ωω

n=0 The phase

angle φ is 90 for all the values of damping ratio

at

ωω

n=1 ie at

resonance The phase angle between o to 90 deg for all the values of

ωω

n= 0 to 1 and

between 90 deg to 180 deg for all the values of

ωω

ngt1

Class work Derive an expression for frequency at peak amplitude

2np 21 ξminusω=ω=ω is the required equation

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

53

Derive expression amplitude at resonance ( In dimensional form) In dimensional form We know that steady state amplitude

( ) ( )

ω+ωminus=

222

o

Cmk

Fx Equation in dimensional form

We know that at resonance mk

n =ω=ω

( )

ω+

minus

=2

n

2

oR

Cmk

mk

Fx

( )

ω=

2n

oR

C

Fx

( )

ω=

n

oR C

Fx

Class work Derive an expression for amplitude at resonance ( in non dimensional form)

Tutorial Problems on Forced Vibration 1A weight of 60N is suspended by a spring stiffness 12 kNm is forced to vibrate by a harmonic force of 10N assuming viscous damping of 0086 kN-sm Find i) resonance frequency ii) amplitude at resonance iii) phase angle at resonance iv) frequency corresponding to peak amplitude 2 A periodic torque having a maximum value of 05888Nm at a frequency of 4 rads is impressed upon a flywheel suspended from a wire The wheel has mass moment of inertia 012 kgm2 and the wire has stiffness 1176 Nmrad A viscous dashpot applies damping couple of 0784 Nm an angular velocity of 2 rads calculate i)maximum angular displacement from rest position ii) maximum couple applied to dashpot Iii) angle by which the angular displacement lags the torque A mass of 1019 kg is suspended from a spring stiffness 20 Nmm viscous damping causes the amplitude to decrease to 110th of initial value in 4 complete cycle If a periodic force of 20 cos(50 t) is applied to the mass Find the amplitude of forced vibration What would be the amplitude of oscillation if the applied force coincides with the natural period

m=1019 kg

[ ] t50cos 30F =

54

55

ROTATING UNBALANCE Unbalance in rotating machinery is a common source of vibration excitation Consider a machine of mass m having mounted on a foundation of stiffness k and damping coefficient C as shown in figure The unbalance is represented by an eccentric rotating mass m with eccentricity (e) and having a angular velocity (ω) let x be the displacement of non rotating mass from the static equilibrium position The displacement of rotating mass m from the static equilibrium is ( ) tsin( ex ω+ Applying Newtonrsquos law of motion we have

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

56

RECIPROCATING UNBALANCE Unbalance in reciprocating machinery is a common source of vibration excitation Consider a machine of mass M having mounted on a foundation of stiffness K and damping coefficient C as shown in figure The unbalance is represented by a reciprocating mass (m) having crank radus (e) and connecting rod of length (l) Let ωωωω be angular velocity of the crank Let x be the displacement of non reciprocating mass from the static equilibrium position at any instant of time The displacement of reciprocating mass m from the static equilibrium position is

)2sin(sin +++ tle

tex ωωωωωωωω

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

57

Note The equations obtained in reciprocating unbalance is same as the rotating unbalance and having the following differences Rotating Unbalance Reciprocating Unbalance Here mass of the rotating element is considered

Here mass of reciprocating element is considered

Here the eccentricity (e) which is the distance between the centre of rotation and CG

Here the eccentricity e is the crank radius and crank radius = stroke length2

The frequency plot of non dimensional factor( ( )me

MX versus frequency ratio

ωω

n for various values of ( )ξξξξ is shown in figure This frequency plot is same for

the rotating and reciprocating unbalance with the following points are noted from the diagram

1 Damping ratio plays very important role during resonance in controlling the amplitude

2 When

ωω

n=0 then ( )me

MX =0 for all the values of damping ratio ( )ξξξξ

3 When

ωω

nis higher than ( )me

MX tends to unity

4 At very high speeds ie at very high frequency ratio

ωω

n there is no effect

of damping The frequency plot of phase angle versus the frequency ratio is given as shown in figure The explanation is same as in forced harmonic vibration

58

VIBRATION ISOLATION

Vibratory forces and motions generated by machine and other caused are unavoidable Howeever their effect on the other dynamical systems can be minimized by proper isolator design An isolation system is either to protect o delicate machine from excessive vibration transmitted to it from its supporting structure or to prevent vibratory forces and motions generated by machines from being transmitted to its surroundings The effectiveness of isolation may be measured in terms of forces and motion transmitted The first type is known as force isolation and the second type is known as motion isolation

Force isolation (Force Transmissibility) When vibratory forces generated by machine are isolated the effectiveness of forced isolation can be made by defining the term transmissibility The term transmissibility is defined as the ratio of force transmitted to the foundation to that of the external impressed force

Force Transmissibility = o

tr

FF

force Impressed Externalfoundation toted transmitForce

=

Consider viscous damped spring mass system as shown in figure Upon this system an external harmonic force ) t (Sin Fo ω is applied The difference forces acting on system are inertia forces damping force spring force and external force All these forces are represented on force polygon shown below Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these two forces

( ) ( )22 xckx ω+=

59

This is the expression for force transmissibility in dimensional form To represent same in the non dimensional form divide numerator and denominator by k

2

n

22

n

2

n

o

tr

21

21

FF

TR

ωω

ξ+

ωω

minus

ωω

ξ+

==

PHASE DIFFERNECE BETWEEN EXTERNAL FORCE AND DISPLACEMENT Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these

ωω

minus

ωω

ξ=

ωminusω

=φ minusminus2

n

n1

2

1

1

2tan

mKC

tan

The phase difference between force transmitted to the foundation Ftr to the displacement

kc

KxxC

tanω

=ω

=α or

ωω

ξ=

ω=α minusminus

n

11 2tanKx

xCtan

The phase difference between external force Fo and force transmitted to foundationis β

60

β= αminusφ =

ωω

minus

ωω

ξminus

2

n

n1

1

2tan -

ωω

ξminus

n

1 2tan

Displacement Transmissibility It is the ratio of amplitude of steady state vibration of machine to that of the base motion Forced vibration due to support motion Consider a spring mass damper system subjected to harmonic excitation )tsin(Yy ω= at the base as shown in figure Let x be the absolute displacement of the mass at any instant of time t The different forces acting on system are shown in free body diagram Applying Newtonrsquos Law of motion Ma = sumF

)yx(k)yx(cxm minusminusminusminus= ampampampamp

0)yx(k)yx(cxm =minus+minus+ ampampampamp

61

Frequency plot of transmissibility versus frequency ratio

62

The plot of transmissibility versus frequency ratio is as shown in figure The following points can be noted from it

i) The value of transmissibility is unity at frequency ratio

ωω

n=0 and 2

ii) For an undamped system ( )ξ =0 transmissibility tends to infin at resonance iii) The value of transmissibility is less than unity for the value of frequency ratio

greater than 2

iv) For frequency ratio

ωω

nlt 2 smaller damping ratio lead to larger values of

transmissibility on the other hand for frequency ratio

ωω

ngt 2 smaller damping

ratio leads to smaller value of transmissibility

The frequency plot of transmissibility (TR) versus frequency ratio

ωω

n can be

broadly divided into two zones

1

ωω

n= 0 to 2

2

ωω

n= 2 to infin

CASE 1

When

ωω

n= 0 to 2

In this zone the value of transmissibility is always greater than 1 for all values of damping ratio In this zone the transmissibility decreases with increase in damping Hence we can conclude that in this zone damping is useful in isolating vibration forces being transmitted to the support CASE 2

When

ωω

n= 2 to infin

In this zone the value of transmissibility is less than 1 for all values of damping ratio In this zone the transmissibility increases with increase in damping Hence we can conclude that in this zone damping will adversely affect in isolating vibratory forces

63

Tutorial problems on Forced Vibration 1 Machine of mass 75kg is mounted on spring of stiffness 12000 Nm with an assumed damping factor ξ =02 piston within the machine of mass 2 kg has a reciprocating motion with a stroke of 75 cm and speed of 15 cyclessecond Assuming the motion of piston to be harmonic determine amplitude of machine ( X) Transmissibility and force transmitted to foundation phase angle of transmitted force wrt to exciting force 2 Motor weighs 600N is mounted on a simple beam that has a spring rate of 400 Ncm The motor armature base 160 N and has an eccentricity of 0005cm What will be the amplitude of motion when it runs at 1760 RPM Neglect damping the deflection of armature shaft and weight of the beam 3 A Heavy machine of mass 300 kg is supported on resilient foundation the static deflection of foundation due to weight of machine is found to be 75mm It is observed that the machine vibrates with an amplitude of 10mm when the base of the foundation is subjected to harmonic oscillation at the undampeds natural frequency of system with an amplitude of 25mm Find the damping constant of foundation Ii) dynamic force amplitude on the base iii) amplitude of the displacement of machine relative to base 4 An air compressor of mass 450 kg operates at a constant speed of 1750 RPM The reciprocating masses are 10 kg the crank radius is 100mm If the damper for mounting introduces a damping factor 015 i) specify the springs of mounting such that only 20 of the unbalanced force is transmitted to the foundation ii) determine the magnitude of the transmitted force bull Repeat the above problem by assuming that there is no damping in system

5 A body of a car having mass 1500 kg and is mounted on 4 equal springs which deflect through 225 cm under the weight of the body A total damping forces the 4 shock absorbers is 46 N at a velocity of 1cmsec The car is placed with all the four wheels on a test platform which is move up and down at resonant speed With an amplitude of 25cm Find the amplitude of the car body on its springs assuming its CG to be in the centre of wheel base 6A 75 kg machine which is mounted on springs of stiffness 116x105 Nm with the damping of 20=ξ A 2kg piston within the machine as reciprocating motion with a stroke of 008m and speed of 3000 RPM Assuming the motion of piston to be harmonic determine the amplitude of vibration of machine

7 The spring of an automobile are compressed 10cm under its own weight Find the critical speed when a trailer is traveling over road with a people approximaterly by a sine wave of amplitude 75 cm and wave length 15m What will be the amplitude of vibration at 64kmhour 8 An eccentric mass excites us used to determine vibratory characteristics of a structure of mass 25kg as given in figure At a speed of 1000 RPM the eccentric mass to beat top position at the instant the structure was moved upwards through its static equilibrium position and corresponding amplitude was 20mm The eccentric mass is 05kg at an eccentricity of 10mm Determine i) Undamped natural frequency of the system ii) Damping ratio iii) Steady state amplitude and phase angle at 1500 RPM iv) Force transmitted to the ground at 1200 RPM

M=25kg

C

K

Eccentric mass

64

65

CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Support motion Consider a spring mass damper system as shown in figure Let it excited by motion

of support let y be the harmonic displacement of support From Newtonrsquos second law of motion

)yx(k)yx(cxm minusminusminusminus= ampampampamp

)1(0)yx(k)yx(cxm minusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminus=minus+minus+ ampampampamp

Let z be the relative displacement of the mass with respect to support then

Z=x-y

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

43

Tutorial problems 1 Derive an expression for damping ratio in terms of logarithmic decrement

2 Show that logarithmic decrement

=δ

n

o

xx

lnn1

where xo is the initial amplitude xn

is amplitude after n cycles 3 A mass of 1 kg is to be supported on a spring having stiffness K= 9800 Nm The damping constant c= 59 Nsm Determine the natural frequency of system Find also logarithmic decrement and amplitude after 3 cycles if initial displacement is 03 cm 4 A vibrating system is defined by the following parameters M = 3kg k= 100 Nm damping coefficient = 3 Nsm determine Damping factor natural frequency damped vibration logarithmic decrement ratio of consecutive amplitude no of cycles after which the original amplitude is reduced to 20 5A single degree damped vibrating system a suspended mass of 18kg makes 15 oscillations in 003 seconds The amplitude decreases to 025 of the initial value after 5 oscillations Determine the stiffness of spring Logarithmic decrement damping factor and damping coefficient 6 A damped vibration record of a spring mass system shows the following data Amplitude at the end of 2nd cycle is 9mm amplitude at eh end of 3rd cycle is 6mm amplitude at the end of 4th cycle is 4mm stiffness of spring =8000 Nm weight =20N Find the logarithmic decrement damping force at unit velocity periodic time of vibration 7 For the system shown in figure the characteristics of dashpot is such that the constant force of 49 N is applied to the piston it velocity is found to be constant at 012 ms Determine the value of C would you expect the complete system to be periodic or aperiodic 8 For a torsional system shown in the figuref1 is natural frequency in air and f2 is the natural frequency when immersed in oil Determine the damping coefficient

Viscous fluid

Disc of mass moment of Inertia J

2a

C

m M

b

a a

k

9 A thin plate of area A and weigh w is fixed to the end of the spring and allow to oscillate in a viscous fluid as shown in figure It f1 is the frequency of oscillation in air and f2 in liquid Show that

22

21 ff

gA2

minusπω

=micro where the damping force on the plate

AV2Fd micro= V being the velocity of vibration 10 A torsional pendulum has a natural frequency of 175 cyclessecond vibrating in vacuum The mass moment of inertia is 2 kgm2 It is immersed in oil and it is observed that the natural frequency is 142 cyclessecond Find the damping torque If the disc is replaced by 34 deg When in oil find the displacement at the end of the first complete cycle 11 An automobile can be modeled as a mass placed on 4 shock absorbers each consisting of a spring and a damper Such that each spring is equally loaded Determine the stiffness damping constant of each shock absorber So that natural frequency is 2Hz and system is critically damped The mass of vehicle is 200kg 12 Write the differential equation of motion for the system shown in the figure below Determine the natural frequency of the damped oscillation and critical damping coefficient

Determine the differential equation of motion and find the critical cdamped natural frequency for system shown in figure

a

C

m M

M =200kg

44

200kg kgkg

oefficient and

45

Derive the equation of motion and find the critical damping coefficient and damped natural frequency for the system shown below Write the differential equation of motion and determine the expression for critical damping coefficient and damped natural frequency of system shown in figure

Set up a differential equation of motion for system shown in figure and determine the following Undamped natural frequencies critical damping coefficient damping ratio damped natural frequency

b

C

kM

a

m

46

A mass of 300N is resting on two springs of stiffness 3000 Nm each and a dashpot of damping coefficient 150 Nsm as shown in figure If the initial velocity of 10cms is given to mass at its equilibrium position what will be the displacement from the equilibrium position at the end of first second Derive the formula used A gun barrel of mass 600kg has a recoil spring of stiffness 294 knm if the barrel recoils 13m on firing determine i) The initial recoil velocity of barrel ii) The critical damping coefficient of dashpot which is engaged at the end of recoil stroke Iii) Time required for the barrel to return at the end of recoil stroke Iv) time required for the barrel to return to a position 5cm from the initial position

W= 3000 N

FIRING

CHAPTER 4

FORCED VIBRATION LEARNING OBJECTIVES

] Introduction to forced free vibration

] Terminologies used in forced vibration

] Concept of magnification factor

] Study of reciprocating and rotating unbalance

] Concept of base excitation and support excitation

] Concept of Transmissibility ratio

] Problems related to forced vibration

The vibration that takes place under the excitation of external forces are called

forced vibration When a system is subjected to harmonic excitation it is forced to vibrate at the same frequency as that of excitation Most of the mechanical systems are prone to steady state forced vibration which cause fatigue failure The vibration problem also becomes very important when any of excitation frequencies coincide with one of natural frequencies of a system Under this condition resonance occurs during which large amplitude of vibrations are observed To avoid resonance external frequency (operating speed) may be changed or properties of the system may be changed to alter the natural frequency In some cases sufficient amount of damping may be provided to avoid large amplitude during resonance Thus the problem of force vibration is very important in mechanical design

FORCED VIBRATIONS OF SINGLE DOF SYSTEMS UNDER HARMONIC EXCITATION

` [ ] tsin FF o ω=

x

sin FF o=

Fo

xampamp

[ ] tω

xamp

47

48

Consider a spring mass system having a viscous damping excited by a harmonic force [ ] tsin FF o ω= as shown in figure The various forces acting on the system is shown in figure Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

O)tsin(Fkxxcxm o =ω+minusminus= ampampamp

)tsin(Fkxxcxm o ω=++ ampampamp ----------------------------------------------------------(1)

or

)tsin(Fkxdtdx

Cdt

xdM o2

2

ω=++

Equation (1) represents the differential equation of motion of the system The solution to this equation consists of two parts namely complementary function and a particular integral

The first of solution is also termed as transient solution of homogeneous solution and can be obtained from the homogeneous differential equation 0kxxcxm =++ ampampamp

The second part of solution particular integral is also termed as steady state solution ad is of form )tsin(Xx φminusω= Where X= steady state amplitude φ is phase difference between external force and displacement

)tsin(Xx φminusω=

The vector representation of the shown

49

The force polygon for the above vector diagram can be written as follows

( )φminusωt O

KX

xm 2ω

Fo

xCω

tω

φ

50

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

is the governing equation

Zero frequency deflection (Xst) It is deflection observed at zero frequency ( when the external frequency 0=ω

( ) kF

001

kF

x o

22

o

st =

+minus=

Magnification factor (M) It is defined as the ratio of steady state amplitude to zero frequency deflection xst

M = arX

X=

A plot of magnification factor versus frequency ratio

damping factor ( )ξ is called frequency plot of a harmon All the frequency terms starts from magnification factor of damping ratio ( )ξ The amplitude of vibration at any on the value of damping The amplitude will be higher for higher damping The amplitude of vibration at reso

amplitude tends to zero as

ωω

n tends to infinity

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

Freq

ζ =

ζ = 1

Magnification Factor M = XXst

kFo

51

ωω

nfor varying values of

ically excite vibrating system

M=1 what ever may be value particular frequency depends

for lesser damping and lesser nance will be very high The

uency Ratio r = (ωωn)

ζ = 0 0

ζ = 025 ζ = 0375 ζ = 0 5

ζ = 0707

ζ = 2

52

Plot of phase angle φ versus frequency ratio

ωω

n is shown in the

figure The phase angle φ is zero for all the values of damping ratio

at

ωω

n=0 The phase

angle φ is 90 for all the values of damping ratio

at

ωω

n=1 ie at

resonance The phase angle between o to 90 deg for all the values of

ωω

n= 0 to 1 and

between 90 deg to 180 deg for all the values of

ωω

ngt1

Class work Derive an expression for frequency at peak amplitude

2np 21 ξminusω=ω=ω is the required equation

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

53

Derive expression amplitude at resonance ( In dimensional form) In dimensional form We know that steady state amplitude

( ) ( )

ω+ωminus=

222

o

Cmk

Fx Equation in dimensional form

We know that at resonance mk

n =ω=ω

( )

ω+

minus

=2

n

2

oR

Cmk

mk

Fx

( )

ω=

2n

oR

C

Fx

( )

ω=

n

oR C

Fx

Class work Derive an expression for amplitude at resonance ( in non dimensional form)

Tutorial Problems on Forced Vibration 1A weight of 60N is suspended by a spring stiffness 12 kNm is forced to vibrate by a harmonic force of 10N assuming viscous damping of 0086 kN-sm Find i) resonance frequency ii) amplitude at resonance iii) phase angle at resonance iv) frequency corresponding to peak amplitude 2 A periodic torque having a maximum value of 05888Nm at a frequency of 4 rads is impressed upon a flywheel suspended from a wire The wheel has mass moment of inertia 012 kgm2 and the wire has stiffness 1176 Nmrad A viscous dashpot applies damping couple of 0784 Nm an angular velocity of 2 rads calculate i)maximum angular displacement from rest position ii) maximum couple applied to dashpot Iii) angle by which the angular displacement lags the torque A mass of 1019 kg is suspended from a spring stiffness 20 Nmm viscous damping causes the amplitude to decrease to 110th of initial value in 4 complete cycle If a periodic force of 20 cos(50 t) is applied to the mass Find the amplitude of forced vibration What would be the amplitude of oscillation if the applied force coincides with the natural period

m=1019 kg

[ ] t50cos 30F =

54

55

ROTATING UNBALANCE Unbalance in rotating machinery is a common source of vibration excitation Consider a machine of mass m having mounted on a foundation of stiffness k and damping coefficient C as shown in figure The unbalance is represented by an eccentric rotating mass m with eccentricity (e) and having a angular velocity (ω) let x be the displacement of non rotating mass from the static equilibrium position The displacement of rotating mass m from the static equilibrium is ( ) tsin( ex ω+ Applying Newtonrsquos law of motion we have

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

56

RECIPROCATING UNBALANCE Unbalance in reciprocating machinery is a common source of vibration excitation Consider a machine of mass M having mounted on a foundation of stiffness K and damping coefficient C as shown in figure The unbalance is represented by a reciprocating mass (m) having crank radus (e) and connecting rod of length (l) Let ωωωω be angular velocity of the crank Let x be the displacement of non reciprocating mass from the static equilibrium position at any instant of time The displacement of reciprocating mass m from the static equilibrium position is

)2sin(sin +++ tle

tex ωωωωωωωω

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

57

Note The equations obtained in reciprocating unbalance is same as the rotating unbalance and having the following differences Rotating Unbalance Reciprocating Unbalance Here mass of the rotating element is considered

Here mass of reciprocating element is considered

Here the eccentricity (e) which is the distance between the centre of rotation and CG

Here the eccentricity e is the crank radius and crank radius = stroke length2

The frequency plot of non dimensional factor( ( )me

MX versus frequency ratio

ωω

n for various values of ( )ξξξξ is shown in figure This frequency plot is same for

the rotating and reciprocating unbalance with the following points are noted from the diagram

1 Damping ratio plays very important role during resonance in controlling the amplitude

2 When

ωω

n=0 then ( )me

MX =0 for all the values of damping ratio ( )ξξξξ

3 When

ωω

nis higher than ( )me

MX tends to unity

4 At very high speeds ie at very high frequency ratio

ωω

n there is no effect

of damping The frequency plot of phase angle versus the frequency ratio is given as shown in figure The explanation is same as in forced harmonic vibration

58

VIBRATION ISOLATION

Vibratory forces and motions generated by machine and other caused are unavoidable Howeever their effect on the other dynamical systems can be minimized by proper isolator design An isolation system is either to protect o delicate machine from excessive vibration transmitted to it from its supporting structure or to prevent vibratory forces and motions generated by machines from being transmitted to its surroundings The effectiveness of isolation may be measured in terms of forces and motion transmitted The first type is known as force isolation and the second type is known as motion isolation

Force isolation (Force Transmissibility) When vibratory forces generated by machine are isolated the effectiveness of forced isolation can be made by defining the term transmissibility The term transmissibility is defined as the ratio of force transmitted to the foundation to that of the external impressed force

Force Transmissibility = o

tr

FF

force Impressed Externalfoundation toted transmitForce

=

Consider viscous damped spring mass system as shown in figure Upon this system an external harmonic force ) t (Sin Fo ω is applied The difference forces acting on system are inertia forces damping force spring force and external force All these forces are represented on force polygon shown below Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these two forces

( ) ( )22 xckx ω+=

59

This is the expression for force transmissibility in dimensional form To represent same in the non dimensional form divide numerator and denominator by k

2

n

22

n

2

n

o

tr

21

21

FF

TR

ωω

ξ+

ωω

minus

ωω

ξ+

==

PHASE DIFFERNECE BETWEEN EXTERNAL FORCE AND DISPLACEMENT Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these

ωω

minus

ωω

ξ=

ωminusω

=φ minusminus2

n

n1

2

1

1

2tan

mKC

tan

The phase difference between force transmitted to the foundation Ftr to the displacement

kc

KxxC

tanω

=ω

=α or

ωω

ξ=

ω=α minusminus

n

11 2tanKx

xCtan

The phase difference between external force Fo and force transmitted to foundationis β

60

β= αminusφ =

ωω

minus

ωω

ξminus

2

n

n1

1

2tan -

ωω

ξminus

n

1 2tan

Displacement Transmissibility It is the ratio of amplitude of steady state vibration of machine to that of the base motion Forced vibration due to support motion Consider a spring mass damper system subjected to harmonic excitation )tsin(Yy ω= at the base as shown in figure Let x be the absolute displacement of the mass at any instant of time t The different forces acting on system are shown in free body diagram Applying Newtonrsquos Law of motion Ma = sumF

)yx(k)yx(cxm minusminusminusminus= ampampampamp

0)yx(k)yx(cxm =minus+minus+ ampampampamp

61

Frequency plot of transmissibility versus frequency ratio

62

The plot of transmissibility versus frequency ratio is as shown in figure The following points can be noted from it

i) The value of transmissibility is unity at frequency ratio

ωω

n=0 and 2

ii) For an undamped system ( )ξ =0 transmissibility tends to infin at resonance iii) The value of transmissibility is less than unity for the value of frequency ratio

greater than 2

iv) For frequency ratio

ωω

nlt 2 smaller damping ratio lead to larger values of

transmissibility on the other hand for frequency ratio

ωω

ngt 2 smaller damping

ratio leads to smaller value of transmissibility

The frequency plot of transmissibility (TR) versus frequency ratio

ωω

n can be

broadly divided into two zones

1

ωω

n= 0 to 2

2

ωω

n= 2 to infin

CASE 1

When

ωω

n= 0 to 2

In this zone the value of transmissibility is always greater than 1 for all values of damping ratio In this zone the transmissibility decreases with increase in damping Hence we can conclude that in this zone damping is useful in isolating vibration forces being transmitted to the support CASE 2

When

ωω

n= 2 to infin

In this zone the value of transmissibility is less than 1 for all values of damping ratio In this zone the transmissibility increases with increase in damping Hence we can conclude that in this zone damping will adversely affect in isolating vibratory forces

63

Tutorial problems on Forced Vibration 1 Machine of mass 75kg is mounted on spring of stiffness 12000 Nm with an assumed damping factor ξ =02 piston within the machine of mass 2 kg has a reciprocating motion with a stroke of 75 cm and speed of 15 cyclessecond Assuming the motion of piston to be harmonic determine amplitude of machine ( X) Transmissibility and force transmitted to foundation phase angle of transmitted force wrt to exciting force 2 Motor weighs 600N is mounted on a simple beam that has a spring rate of 400 Ncm The motor armature base 160 N and has an eccentricity of 0005cm What will be the amplitude of motion when it runs at 1760 RPM Neglect damping the deflection of armature shaft and weight of the beam 3 A Heavy machine of mass 300 kg is supported on resilient foundation the static deflection of foundation due to weight of machine is found to be 75mm It is observed that the machine vibrates with an amplitude of 10mm when the base of the foundation is subjected to harmonic oscillation at the undampeds natural frequency of system with an amplitude of 25mm Find the damping constant of foundation Ii) dynamic force amplitude on the base iii) amplitude of the displacement of machine relative to base 4 An air compressor of mass 450 kg operates at a constant speed of 1750 RPM The reciprocating masses are 10 kg the crank radius is 100mm If the damper for mounting introduces a damping factor 015 i) specify the springs of mounting such that only 20 of the unbalanced force is transmitted to the foundation ii) determine the magnitude of the transmitted force bull Repeat the above problem by assuming that there is no damping in system

5 A body of a car having mass 1500 kg and is mounted on 4 equal springs which deflect through 225 cm under the weight of the body A total damping forces the 4 shock absorbers is 46 N at a velocity of 1cmsec The car is placed with all the four wheels on a test platform which is move up and down at resonant speed With an amplitude of 25cm Find the amplitude of the car body on its springs assuming its CG to be in the centre of wheel base 6A 75 kg machine which is mounted on springs of stiffness 116x105 Nm with the damping of 20=ξ A 2kg piston within the machine as reciprocating motion with a stroke of 008m and speed of 3000 RPM Assuming the motion of piston to be harmonic determine the amplitude of vibration of machine

7 The spring of an automobile are compressed 10cm under its own weight Find the critical speed when a trailer is traveling over road with a people approximaterly by a sine wave of amplitude 75 cm and wave length 15m What will be the amplitude of vibration at 64kmhour 8 An eccentric mass excites us used to determine vibratory characteristics of a structure of mass 25kg as given in figure At a speed of 1000 RPM the eccentric mass to beat top position at the instant the structure was moved upwards through its static equilibrium position and corresponding amplitude was 20mm The eccentric mass is 05kg at an eccentricity of 10mm Determine i) Undamped natural frequency of the system ii) Damping ratio iii) Steady state amplitude and phase angle at 1500 RPM iv) Force transmitted to the ground at 1200 RPM

M=25kg

C

K

Eccentric mass

64

65

CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Support motion Consider a spring mass damper system as shown in figure Let it excited by motion

of support let y be the harmonic displacement of support From Newtonrsquos second law of motion

)yx(k)yx(cxm minusminusminusminus= ampampampamp

)1(0)yx(k)yx(cxm minusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminus=minus+minus+ ampampampamp

Let z be the relative displacement of the mass with respect to support then

Z=x-y

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

2a

C

m M

b

a a

k

9 A thin plate of area A and weigh w is fixed to the end of the spring and allow to oscillate in a viscous fluid as shown in figure It f1 is the frequency of oscillation in air and f2 in liquid Show that

22

21 ff

gA2

minusπω

=micro where the damping force on the plate

AV2Fd micro= V being the velocity of vibration 10 A torsional pendulum has a natural frequency of 175 cyclessecond vibrating in vacuum The mass moment of inertia is 2 kgm2 It is immersed in oil and it is observed that the natural frequency is 142 cyclessecond Find the damping torque If the disc is replaced by 34 deg When in oil find the displacement at the end of the first complete cycle 11 An automobile can be modeled as a mass placed on 4 shock absorbers each consisting of a spring and a damper Such that each spring is equally loaded Determine the stiffness damping constant of each shock absorber So that natural frequency is 2Hz and system is critically damped The mass of vehicle is 200kg 12 Write the differential equation of motion for the system shown in the figure below Determine the natural frequency of the damped oscillation and critical damping coefficient

Determine the differential equation of motion and find the critical cdamped natural frequency for system shown in figure

a

C

m M

M =200kg

44

200kg kgkg

oefficient and

45

Derive the equation of motion and find the critical damping coefficient and damped natural frequency for the system shown below Write the differential equation of motion and determine the expression for critical damping coefficient and damped natural frequency of system shown in figure

Set up a differential equation of motion for system shown in figure and determine the following Undamped natural frequencies critical damping coefficient damping ratio damped natural frequency

b

C

kM

a

m

46

A mass of 300N is resting on two springs of stiffness 3000 Nm each and a dashpot of damping coefficient 150 Nsm as shown in figure If the initial velocity of 10cms is given to mass at its equilibrium position what will be the displacement from the equilibrium position at the end of first second Derive the formula used A gun barrel of mass 600kg has a recoil spring of stiffness 294 knm if the barrel recoils 13m on firing determine i) The initial recoil velocity of barrel ii) The critical damping coefficient of dashpot which is engaged at the end of recoil stroke Iii) Time required for the barrel to return at the end of recoil stroke Iv) time required for the barrel to return to a position 5cm from the initial position

W= 3000 N

FIRING

CHAPTER 4

FORCED VIBRATION LEARNING OBJECTIVES

] Introduction to forced free vibration

] Terminologies used in forced vibration

] Concept of magnification factor

] Study of reciprocating and rotating unbalance

] Concept of base excitation and support excitation

] Concept of Transmissibility ratio

] Problems related to forced vibration

The vibration that takes place under the excitation of external forces are called

forced vibration When a system is subjected to harmonic excitation it is forced to vibrate at the same frequency as that of excitation Most of the mechanical systems are prone to steady state forced vibration which cause fatigue failure The vibration problem also becomes very important when any of excitation frequencies coincide with one of natural frequencies of a system Under this condition resonance occurs during which large amplitude of vibrations are observed To avoid resonance external frequency (operating speed) may be changed or properties of the system may be changed to alter the natural frequency In some cases sufficient amount of damping may be provided to avoid large amplitude during resonance Thus the problem of force vibration is very important in mechanical design

FORCED VIBRATIONS OF SINGLE DOF SYSTEMS UNDER HARMONIC EXCITATION

` [ ] tsin FF o ω=

x

sin FF o=

Fo

xampamp

[ ] tω

xamp

47

48

Consider a spring mass system having a viscous damping excited by a harmonic force [ ] tsin FF o ω= as shown in figure The various forces acting on the system is shown in figure Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

O)tsin(Fkxxcxm o =ω+minusminus= ampampamp

)tsin(Fkxxcxm o ω=++ ampampamp ----------------------------------------------------------(1)

or

)tsin(Fkxdtdx

Cdt

xdM o2

2

ω=++

Equation (1) represents the differential equation of motion of the system The solution to this equation consists of two parts namely complementary function and a particular integral

The first of solution is also termed as transient solution of homogeneous solution and can be obtained from the homogeneous differential equation 0kxxcxm =++ ampampamp

The second part of solution particular integral is also termed as steady state solution ad is of form )tsin(Xx φminusω= Where X= steady state amplitude φ is phase difference between external force and displacement

)tsin(Xx φminusω=

The vector representation of the shown

49

The force polygon for the above vector diagram can be written as follows

( )φminusωt O

KX

xm 2ω

Fo

xCω

tω

φ

50

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

is the governing equation

Zero frequency deflection (Xst) It is deflection observed at zero frequency ( when the external frequency 0=ω

( ) kF

001

kF

x o

22

o

st =

+minus=

Magnification factor (M) It is defined as the ratio of steady state amplitude to zero frequency deflection xst

M = arX

X=

A plot of magnification factor versus frequency ratio

damping factor ( )ξ is called frequency plot of a harmon All the frequency terms starts from magnification factor of damping ratio ( )ξ The amplitude of vibration at any on the value of damping The amplitude will be higher for higher damping The amplitude of vibration at reso

amplitude tends to zero as

ωω

n tends to infinity

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

Freq

ζ =

ζ = 1

Magnification Factor M = XXst

kFo

51

ωω

nfor varying values of

ically excite vibrating system

M=1 what ever may be value particular frequency depends

for lesser damping and lesser nance will be very high The

uency Ratio r = (ωωn)

ζ = 0 0

ζ = 025 ζ = 0375 ζ = 0 5

ζ = 0707

ζ = 2

52

Plot of phase angle φ versus frequency ratio

ωω

n is shown in the

figure The phase angle φ is zero for all the values of damping ratio

at

ωω

n=0 The phase

angle φ is 90 for all the values of damping ratio

at

ωω

n=1 ie at

resonance The phase angle between o to 90 deg for all the values of

ωω

n= 0 to 1 and

between 90 deg to 180 deg for all the values of

ωω

ngt1

Class work Derive an expression for frequency at peak amplitude

2np 21 ξminusω=ω=ω is the required equation

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

53

Derive expression amplitude at resonance ( In dimensional form) In dimensional form We know that steady state amplitude

( ) ( )

ω+ωminus=

222

o

Cmk

Fx Equation in dimensional form

We know that at resonance mk

n =ω=ω

( )

ω+

minus

=2

n

2

oR

Cmk

mk

Fx

( )

ω=

2n

oR

C

Fx

( )

ω=

n

oR C

Fx

Class work Derive an expression for amplitude at resonance ( in non dimensional form)

Tutorial Problems on Forced Vibration 1A weight of 60N is suspended by a spring stiffness 12 kNm is forced to vibrate by a harmonic force of 10N assuming viscous damping of 0086 kN-sm Find i) resonance frequency ii) amplitude at resonance iii) phase angle at resonance iv) frequency corresponding to peak amplitude 2 A periodic torque having a maximum value of 05888Nm at a frequency of 4 rads is impressed upon a flywheel suspended from a wire The wheel has mass moment of inertia 012 kgm2 and the wire has stiffness 1176 Nmrad A viscous dashpot applies damping couple of 0784 Nm an angular velocity of 2 rads calculate i)maximum angular displacement from rest position ii) maximum couple applied to dashpot Iii) angle by which the angular displacement lags the torque A mass of 1019 kg is suspended from a spring stiffness 20 Nmm viscous damping causes the amplitude to decrease to 110th of initial value in 4 complete cycle If a periodic force of 20 cos(50 t) is applied to the mass Find the amplitude of forced vibration What would be the amplitude of oscillation if the applied force coincides with the natural period

m=1019 kg

[ ] t50cos 30F =

54

55

ROTATING UNBALANCE Unbalance in rotating machinery is a common source of vibration excitation Consider a machine of mass m having mounted on a foundation of stiffness k and damping coefficient C as shown in figure The unbalance is represented by an eccentric rotating mass m with eccentricity (e) and having a angular velocity (ω) let x be the displacement of non rotating mass from the static equilibrium position The displacement of rotating mass m from the static equilibrium is ( ) tsin( ex ω+ Applying Newtonrsquos law of motion we have

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

56

RECIPROCATING UNBALANCE Unbalance in reciprocating machinery is a common source of vibration excitation Consider a machine of mass M having mounted on a foundation of stiffness K and damping coefficient C as shown in figure The unbalance is represented by a reciprocating mass (m) having crank radus (e) and connecting rod of length (l) Let ωωωω be angular velocity of the crank Let x be the displacement of non reciprocating mass from the static equilibrium position at any instant of time The displacement of reciprocating mass m from the static equilibrium position is

)2sin(sin +++ tle

tex ωωωωωωωω

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

57

Note The equations obtained in reciprocating unbalance is same as the rotating unbalance and having the following differences Rotating Unbalance Reciprocating Unbalance Here mass of the rotating element is considered

Here mass of reciprocating element is considered

Here the eccentricity (e) which is the distance between the centre of rotation and CG

Here the eccentricity e is the crank radius and crank radius = stroke length2

The frequency plot of non dimensional factor( ( )me

MX versus frequency ratio

ωω

n for various values of ( )ξξξξ is shown in figure This frequency plot is same for

the rotating and reciprocating unbalance with the following points are noted from the diagram

1 Damping ratio plays very important role during resonance in controlling the amplitude

2 When

ωω

n=0 then ( )me

MX =0 for all the values of damping ratio ( )ξξξξ

3 When

ωω

nis higher than ( )me

MX tends to unity

4 At very high speeds ie at very high frequency ratio

ωω

n there is no effect

of damping The frequency plot of phase angle versus the frequency ratio is given as shown in figure The explanation is same as in forced harmonic vibration

58

VIBRATION ISOLATION

Vibratory forces and motions generated by machine and other caused are unavoidable Howeever their effect on the other dynamical systems can be minimized by proper isolator design An isolation system is either to protect o delicate machine from excessive vibration transmitted to it from its supporting structure or to prevent vibratory forces and motions generated by machines from being transmitted to its surroundings The effectiveness of isolation may be measured in terms of forces and motion transmitted The first type is known as force isolation and the second type is known as motion isolation

Force isolation (Force Transmissibility) When vibratory forces generated by machine are isolated the effectiveness of forced isolation can be made by defining the term transmissibility The term transmissibility is defined as the ratio of force transmitted to the foundation to that of the external impressed force

Force Transmissibility = o

tr

FF

force Impressed Externalfoundation toted transmitForce

=

Consider viscous damped spring mass system as shown in figure Upon this system an external harmonic force ) t (Sin Fo ω is applied The difference forces acting on system are inertia forces damping force spring force and external force All these forces are represented on force polygon shown below Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these two forces

( ) ( )22 xckx ω+=

59

This is the expression for force transmissibility in dimensional form To represent same in the non dimensional form divide numerator and denominator by k

2

n

22

n

2

n

o

tr

21

21

FF

TR

ωω

ξ+

ωω

minus

ωω

ξ+

==

PHASE DIFFERNECE BETWEEN EXTERNAL FORCE AND DISPLACEMENT Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these

ωω

minus

ωω

ξ=

ωminusω

=φ minusminus2

n

n1

2

1

1

2tan

mKC

tan

The phase difference between force transmitted to the foundation Ftr to the displacement

kc

KxxC

tanω

=ω

=α or

ωω

ξ=

ω=α minusminus

n

11 2tanKx

xCtan

The phase difference between external force Fo and force transmitted to foundationis β

60

β= αminusφ =

ωω

minus

ωω

ξminus

2

n

n1

1

2tan -

ωω

ξminus

n

1 2tan

Displacement Transmissibility It is the ratio of amplitude of steady state vibration of machine to that of the base motion Forced vibration due to support motion Consider a spring mass damper system subjected to harmonic excitation )tsin(Yy ω= at the base as shown in figure Let x be the absolute displacement of the mass at any instant of time t The different forces acting on system are shown in free body diagram Applying Newtonrsquos Law of motion Ma = sumF

)yx(k)yx(cxm minusminusminusminus= ampampampamp

0)yx(k)yx(cxm =minus+minus+ ampampampamp

61

Frequency plot of transmissibility versus frequency ratio

62

The plot of transmissibility versus frequency ratio is as shown in figure The following points can be noted from it

i) The value of transmissibility is unity at frequency ratio

ωω

n=0 and 2

ii) For an undamped system ( )ξ =0 transmissibility tends to infin at resonance iii) The value of transmissibility is less than unity for the value of frequency ratio

greater than 2

iv) For frequency ratio

ωω

nlt 2 smaller damping ratio lead to larger values of

transmissibility on the other hand for frequency ratio

ωω

ngt 2 smaller damping

ratio leads to smaller value of transmissibility

The frequency plot of transmissibility (TR) versus frequency ratio

ωω

n can be

broadly divided into two zones

1

ωω

n= 0 to 2

2

ωω

n= 2 to infin

CASE 1

When

ωω

n= 0 to 2

In this zone the value of transmissibility is always greater than 1 for all values of damping ratio In this zone the transmissibility decreases with increase in damping Hence we can conclude that in this zone damping is useful in isolating vibration forces being transmitted to the support CASE 2

When

ωω

n= 2 to infin

In this zone the value of transmissibility is less than 1 for all values of damping ratio In this zone the transmissibility increases with increase in damping Hence we can conclude that in this zone damping will adversely affect in isolating vibratory forces

63

Tutorial problems on Forced Vibration 1 Machine of mass 75kg is mounted on spring of stiffness 12000 Nm with an assumed damping factor ξ =02 piston within the machine of mass 2 kg has a reciprocating motion with a stroke of 75 cm and speed of 15 cyclessecond Assuming the motion of piston to be harmonic determine amplitude of machine ( X) Transmissibility and force transmitted to foundation phase angle of transmitted force wrt to exciting force 2 Motor weighs 600N is mounted on a simple beam that has a spring rate of 400 Ncm The motor armature base 160 N and has an eccentricity of 0005cm What will be the amplitude of motion when it runs at 1760 RPM Neglect damping the deflection of armature shaft and weight of the beam 3 A Heavy machine of mass 300 kg is supported on resilient foundation the static deflection of foundation due to weight of machine is found to be 75mm It is observed that the machine vibrates with an amplitude of 10mm when the base of the foundation is subjected to harmonic oscillation at the undampeds natural frequency of system with an amplitude of 25mm Find the damping constant of foundation Ii) dynamic force amplitude on the base iii) amplitude of the displacement of machine relative to base 4 An air compressor of mass 450 kg operates at a constant speed of 1750 RPM The reciprocating masses are 10 kg the crank radius is 100mm If the damper for mounting introduces a damping factor 015 i) specify the springs of mounting such that only 20 of the unbalanced force is transmitted to the foundation ii) determine the magnitude of the transmitted force bull Repeat the above problem by assuming that there is no damping in system

5 A body of a car having mass 1500 kg and is mounted on 4 equal springs which deflect through 225 cm under the weight of the body A total damping forces the 4 shock absorbers is 46 N at a velocity of 1cmsec The car is placed with all the four wheels on a test platform which is move up and down at resonant speed With an amplitude of 25cm Find the amplitude of the car body on its springs assuming its CG to be in the centre of wheel base 6A 75 kg machine which is mounted on springs of stiffness 116x105 Nm with the damping of 20=ξ A 2kg piston within the machine as reciprocating motion with a stroke of 008m and speed of 3000 RPM Assuming the motion of piston to be harmonic determine the amplitude of vibration of machine

7 The spring of an automobile are compressed 10cm under its own weight Find the critical speed when a trailer is traveling over road with a people approximaterly by a sine wave of amplitude 75 cm and wave length 15m What will be the amplitude of vibration at 64kmhour 8 An eccentric mass excites us used to determine vibratory characteristics of a structure of mass 25kg as given in figure At a speed of 1000 RPM the eccentric mass to beat top position at the instant the structure was moved upwards through its static equilibrium position and corresponding amplitude was 20mm The eccentric mass is 05kg at an eccentricity of 10mm Determine i) Undamped natural frequency of the system ii) Damping ratio iii) Steady state amplitude and phase angle at 1500 RPM iv) Force transmitted to the ground at 1200 RPM

M=25kg

C

K

Eccentric mass

64

65

CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Support motion Consider a spring mass damper system as shown in figure Let it excited by motion

of support let y be the harmonic displacement of support From Newtonrsquos second law of motion

)yx(k)yx(cxm minusminusminusminus= ampampampamp

)1(0)yx(k)yx(cxm minusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminus=minus+minus+ ampampampamp

Let z be the relative displacement of the mass with respect to support then

Z=x-y

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

45

Derive the equation of motion and find the critical damping coefficient and damped natural frequency for the system shown below Write the differential equation of motion and determine the expression for critical damping coefficient and damped natural frequency of system shown in figure

Set up a differential equation of motion for system shown in figure and determine the following Undamped natural frequencies critical damping coefficient damping ratio damped natural frequency

b

C

kM

a

m

46

A mass of 300N is resting on two springs of stiffness 3000 Nm each and a dashpot of damping coefficient 150 Nsm as shown in figure If the initial velocity of 10cms is given to mass at its equilibrium position what will be the displacement from the equilibrium position at the end of first second Derive the formula used A gun barrel of mass 600kg has a recoil spring of stiffness 294 knm if the barrel recoils 13m on firing determine i) The initial recoil velocity of barrel ii) The critical damping coefficient of dashpot which is engaged at the end of recoil stroke Iii) Time required for the barrel to return at the end of recoil stroke Iv) time required for the barrel to return to a position 5cm from the initial position

W= 3000 N

FIRING

CHAPTER 4

FORCED VIBRATION LEARNING OBJECTIVES

] Introduction to forced free vibration

] Terminologies used in forced vibration

] Concept of magnification factor

] Study of reciprocating and rotating unbalance

] Concept of base excitation and support excitation

] Concept of Transmissibility ratio

] Problems related to forced vibration

The vibration that takes place under the excitation of external forces are called

forced vibration When a system is subjected to harmonic excitation it is forced to vibrate at the same frequency as that of excitation Most of the mechanical systems are prone to steady state forced vibration which cause fatigue failure The vibration problem also becomes very important when any of excitation frequencies coincide with one of natural frequencies of a system Under this condition resonance occurs during which large amplitude of vibrations are observed To avoid resonance external frequency (operating speed) may be changed or properties of the system may be changed to alter the natural frequency In some cases sufficient amount of damping may be provided to avoid large amplitude during resonance Thus the problem of force vibration is very important in mechanical design

FORCED VIBRATIONS OF SINGLE DOF SYSTEMS UNDER HARMONIC EXCITATION

` [ ] tsin FF o ω=

x

sin FF o=

Fo

xampamp

[ ] tω

xamp

47

48

Consider a spring mass system having a viscous damping excited by a harmonic force [ ] tsin FF o ω= as shown in figure The various forces acting on the system is shown in figure Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

O)tsin(Fkxxcxm o =ω+minusminus= ampampamp

)tsin(Fkxxcxm o ω=++ ampampamp ----------------------------------------------------------(1)

or

)tsin(Fkxdtdx

Cdt

xdM o2

2

ω=++

Equation (1) represents the differential equation of motion of the system The solution to this equation consists of two parts namely complementary function and a particular integral

The first of solution is also termed as transient solution of homogeneous solution and can be obtained from the homogeneous differential equation 0kxxcxm =++ ampampamp

The second part of solution particular integral is also termed as steady state solution ad is of form )tsin(Xx φminusω= Where X= steady state amplitude φ is phase difference between external force and displacement

)tsin(Xx φminusω=

The vector representation of the shown

49

The force polygon for the above vector diagram can be written as follows

( )φminusωt O

KX

xm 2ω

Fo

xCω

tω

φ

50

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

is the governing equation

Zero frequency deflection (Xst) It is deflection observed at zero frequency ( when the external frequency 0=ω

( ) kF

001

kF

x o

22

o

st =

+minus=

Magnification factor (M) It is defined as the ratio of steady state amplitude to zero frequency deflection xst

M = arX

X=

A plot of magnification factor versus frequency ratio

damping factor ( )ξ is called frequency plot of a harmon All the frequency terms starts from magnification factor of damping ratio ( )ξ The amplitude of vibration at any on the value of damping The amplitude will be higher for higher damping The amplitude of vibration at reso

amplitude tends to zero as

ωω

n tends to infinity

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

Freq

ζ =

ζ = 1

Magnification Factor M = XXst

kFo

51

ωω

nfor varying values of

ically excite vibrating system

M=1 what ever may be value particular frequency depends

for lesser damping and lesser nance will be very high The

uency Ratio r = (ωωn)

ζ = 0 0

ζ = 025 ζ = 0375 ζ = 0 5

ζ = 0707

ζ = 2

52

Plot of phase angle φ versus frequency ratio

ωω

n is shown in the

figure The phase angle φ is zero for all the values of damping ratio

at

ωω

n=0 The phase

angle φ is 90 for all the values of damping ratio

at

ωω

n=1 ie at

resonance The phase angle between o to 90 deg for all the values of

ωω

n= 0 to 1 and

between 90 deg to 180 deg for all the values of

ωω

ngt1

Class work Derive an expression for frequency at peak amplitude

2np 21 ξminusω=ω=ω is the required equation

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

53

Derive expression amplitude at resonance ( In dimensional form) In dimensional form We know that steady state amplitude

( ) ( )

ω+ωminus=

222

o

Cmk

Fx Equation in dimensional form

We know that at resonance mk

n =ω=ω

( )

ω+

minus

=2

n

2

oR

Cmk

mk

Fx

( )

ω=

2n

oR

C

Fx

( )

ω=

n

oR C

Fx

Class work Derive an expression for amplitude at resonance ( in non dimensional form)

Tutorial Problems on Forced Vibration 1A weight of 60N is suspended by a spring stiffness 12 kNm is forced to vibrate by a harmonic force of 10N assuming viscous damping of 0086 kN-sm Find i) resonance frequency ii) amplitude at resonance iii) phase angle at resonance iv) frequency corresponding to peak amplitude 2 A periodic torque having a maximum value of 05888Nm at a frequency of 4 rads is impressed upon a flywheel suspended from a wire The wheel has mass moment of inertia 012 kgm2 and the wire has stiffness 1176 Nmrad A viscous dashpot applies damping couple of 0784 Nm an angular velocity of 2 rads calculate i)maximum angular displacement from rest position ii) maximum couple applied to dashpot Iii) angle by which the angular displacement lags the torque A mass of 1019 kg is suspended from a spring stiffness 20 Nmm viscous damping causes the amplitude to decrease to 110th of initial value in 4 complete cycle If a periodic force of 20 cos(50 t) is applied to the mass Find the amplitude of forced vibration What would be the amplitude of oscillation if the applied force coincides with the natural period

m=1019 kg

[ ] t50cos 30F =

54

55

ROTATING UNBALANCE Unbalance in rotating machinery is a common source of vibration excitation Consider a machine of mass m having mounted on a foundation of stiffness k and damping coefficient C as shown in figure The unbalance is represented by an eccentric rotating mass m with eccentricity (e) and having a angular velocity (ω) let x be the displacement of non rotating mass from the static equilibrium position The displacement of rotating mass m from the static equilibrium is ( ) tsin( ex ω+ Applying Newtonrsquos law of motion we have

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

56

RECIPROCATING UNBALANCE Unbalance in reciprocating machinery is a common source of vibration excitation Consider a machine of mass M having mounted on a foundation of stiffness K and damping coefficient C as shown in figure The unbalance is represented by a reciprocating mass (m) having crank radus (e) and connecting rod of length (l) Let ωωωω be angular velocity of the crank Let x be the displacement of non reciprocating mass from the static equilibrium position at any instant of time The displacement of reciprocating mass m from the static equilibrium position is

)2sin(sin +++ tle

tex ωωωωωωωω

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

57

Note The equations obtained in reciprocating unbalance is same as the rotating unbalance and having the following differences Rotating Unbalance Reciprocating Unbalance Here mass of the rotating element is considered

Here mass of reciprocating element is considered

Here the eccentricity (e) which is the distance between the centre of rotation and CG

Here the eccentricity e is the crank radius and crank radius = stroke length2

The frequency plot of non dimensional factor( ( )me

MX versus frequency ratio

ωω

n for various values of ( )ξξξξ is shown in figure This frequency plot is same for

the rotating and reciprocating unbalance with the following points are noted from the diagram

1 Damping ratio plays very important role during resonance in controlling the amplitude

2 When

ωω

n=0 then ( )me

MX =0 for all the values of damping ratio ( )ξξξξ

3 When

ωω

nis higher than ( )me

MX tends to unity

4 At very high speeds ie at very high frequency ratio

ωω

n there is no effect

of damping The frequency plot of phase angle versus the frequency ratio is given as shown in figure The explanation is same as in forced harmonic vibration

58

VIBRATION ISOLATION

Vibratory forces and motions generated by machine and other caused are unavoidable Howeever their effect on the other dynamical systems can be minimized by proper isolator design An isolation system is either to protect o delicate machine from excessive vibration transmitted to it from its supporting structure or to prevent vibratory forces and motions generated by machines from being transmitted to its surroundings The effectiveness of isolation may be measured in terms of forces and motion transmitted The first type is known as force isolation and the second type is known as motion isolation

Force isolation (Force Transmissibility) When vibratory forces generated by machine are isolated the effectiveness of forced isolation can be made by defining the term transmissibility The term transmissibility is defined as the ratio of force transmitted to the foundation to that of the external impressed force

Force Transmissibility = o

tr

FF

force Impressed Externalfoundation toted transmitForce

=

Consider viscous damped spring mass system as shown in figure Upon this system an external harmonic force ) t (Sin Fo ω is applied The difference forces acting on system are inertia forces damping force spring force and external force All these forces are represented on force polygon shown below Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these two forces

( ) ( )22 xckx ω+=

59

This is the expression for force transmissibility in dimensional form To represent same in the non dimensional form divide numerator and denominator by k

2

n

22

n

2

n

o

tr

21

21

FF

TR

ωω

ξ+

ωω

minus

ωω

ξ+

==

PHASE DIFFERNECE BETWEEN EXTERNAL FORCE AND DISPLACEMENT Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these

ωω

minus

ωω

ξ=

ωminusω

=φ minusminus2

n

n1

2

1

1

2tan

mKC

tan

The phase difference between force transmitted to the foundation Ftr to the displacement

kc

KxxC

tanω

=ω

=α or

ωω

ξ=

ω=α minusminus

n

11 2tanKx

xCtan

The phase difference between external force Fo and force transmitted to foundationis β

60

β= αminusφ =

ωω

minus

ωω

ξminus

2

n

n1

1

2tan -

ωω

ξminus

n

1 2tan

Displacement Transmissibility It is the ratio of amplitude of steady state vibration of machine to that of the base motion Forced vibration due to support motion Consider a spring mass damper system subjected to harmonic excitation )tsin(Yy ω= at the base as shown in figure Let x be the absolute displacement of the mass at any instant of time t The different forces acting on system are shown in free body diagram Applying Newtonrsquos Law of motion Ma = sumF

)yx(k)yx(cxm minusminusminusminus= ampampampamp

0)yx(k)yx(cxm =minus+minus+ ampampampamp

61

Frequency plot of transmissibility versus frequency ratio

62

The plot of transmissibility versus frequency ratio is as shown in figure The following points can be noted from it

i) The value of transmissibility is unity at frequency ratio

ωω

n=0 and 2

ii) For an undamped system ( )ξ =0 transmissibility tends to infin at resonance iii) The value of transmissibility is less than unity for the value of frequency ratio

greater than 2

iv) For frequency ratio

ωω

nlt 2 smaller damping ratio lead to larger values of

transmissibility on the other hand for frequency ratio

ωω

ngt 2 smaller damping

ratio leads to smaller value of transmissibility

The frequency plot of transmissibility (TR) versus frequency ratio

ωω

n can be

broadly divided into two zones

1

ωω

n= 0 to 2

2

ωω

n= 2 to infin

CASE 1

When

ωω

n= 0 to 2

In this zone the value of transmissibility is always greater than 1 for all values of damping ratio In this zone the transmissibility decreases with increase in damping Hence we can conclude that in this zone damping is useful in isolating vibration forces being transmitted to the support CASE 2

When

ωω

n= 2 to infin

In this zone the value of transmissibility is less than 1 for all values of damping ratio In this zone the transmissibility increases with increase in damping Hence we can conclude that in this zone damping will adversely affect in isolating vibratory forces

63

Tutorial problems on Forced Vibration 1 Machine of mass 75kg is mounted on spring of stiffness 12000 Nm with an assumed damping factor ξ =02 piston within the machine of mass 2 kg has a reciprocating motion with a stroke of 75 cm and speed of 15 cyclessecond Assuming the motion of piston to be harmonic determine amplitude of machine ( X) Transmissibility and force transmitted to foundation phase angle of transmitted force wrt to exciting force 2 Motor weighs 600N is mounted on a simple beam that has a spring rate of 400 Ncm The motor armature base 160 N and has an eccentricity of 0005cm What will be the amplitude of motion when it runs at 1760 RPM Neglect damping the deflection of armature shaft and weight of the beam 3 A Heavy machine of mass 300 kg is supported on resilient foundation the static deflection of foundation due to weight of machine is found to be 75mm It is observed that the machine vibrates with an amplitude of 10mm when the base of the foundation is subjected to harmonic oscillation at the undampeds natural frequency of system with an amplitude of 25mm Find the damping constant of foundation Ii) dynamic force amplitude on the base iii) amplitude of the displacement of machine relative to base 4 An air compressor of mass 450 kg operates at a constant speed of 1750 RPM The reciprocating masses are 10 kg the crank radius is 100mm If the damper for mounting introduces a damping factor 015 i) specify the springs of mounting such that only 20 of the unbalanced force is transmitted to the foundation ii) determine the magnitude of the transmitted force bull Repeat the above problem by assuming that there is no damping in system

5 A body of a car having mass 1500 kg and is mounted on 4 equal springs which deflect through 225 cm under the weight of the body A total damping forces the 4 shock absorbers is 46 N at a velocity of 1cmsec The car is placed with all the four wheels on a test platform which is move up and down at resonant speed With an amplitude of 25cm Find the amplitude of the car body on its springs assuming its CG to be in the centre of wheel base 6A 75 kg machine which is mounted on springs of stiffness 116x105 Nm with the damping of 20=ξ A 2kg piston within the machine as reciprocating motion with a stroke of 008m and speed of 3000 RPM Assuming the motion of piston to be harmonic determine the amplitude of vibration of machine

7 The spring of an automobile are compressed 10cm under its own weight Find the critical speed when a trailer is traveling over road with a people approximaterly by a sine wave of amplitude 75 cm and wave length 15m What will be the amplitude of vibration at 64kmhour 8 An eccentric mass excites us used to determine vibratory characteristics of a structure of mass 25kg as given in figure At a speed of 1000 RPM the eccentric mass to beat top position at the instant the structure was moved upwards through its static equilibrium position and corresponding amplitude was 20mm The eccentric mass is 05kg at an eccentricity of 10mm Determine i) Undamped natural frequency of the system ii) Damping ratio iii) Steady state amplitude and phase angle at 1500 RPM iv) Force transmitted to the ground at 1200 RPM

M=25kg

C

K

Eccentric mass

64

65

CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Support motion Consider a spring mass damper system as shown in figure Let it excited by motion

of support let y be the harmonic displacement of support From Newtonrsquos second law of motion

)yx(k)yx(cxm minusminusminusminus= ampampampamp

)1(0)yx(k)yx(cxm minusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminus=minus+minus+ ampampampamp

Let z be the relative displacement of the mass with respect to support then

Z=x-y

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

46

A mass of 300N is resting on two springs of stiffness 3000 Nm each and a dashpot of damping coefficient 150 Nsm as shown in figure If the initial velocity of 10cms is given to mass at its equilibrium position what will be the displacement from the equilibrium position at the end of first second Derive the formula used A gun barrel of mass 600kg has a recoil spring of stiffness 294 knm if the barrel recoils 13m on firing determine i) The initial recoil velocity of barrel ii) The critical damping coefficient of dashpot which is engaged at the end of recoil stroke Iii) Time required for the barrel to return at the end of recoil stroke Iv) time required for the barrel to return to a position 5cm from the initial position

W= 3000 N

FIRING

CHAPTER 4

FORCED VIBRATION LEARNING OBJECTIVES

] Introduction to forced free vibration

] Terminologies used in forced vibration

] Concept of magnification factor

] Study of reciprocating and rotating unbalance

] Concept of base excitation and support excitation

] Concept of Transmissibility ratio

] Problems related to forced vibration

The vibration that takes place under the excitation of external forces are called

forced vibration When a system is subjected to harmonic excitation it is forced to vibrate at the same frequency as that of excitation Most of the mechanical systems are prone to steady state forced vibration which cause fatigue failure The vibration problem also becomes very important when any of excitation frequencies coincide with one of natural frequencies of a system Under this condition resonance occurs during which large amplitude of vibrations are observed To avoid resonance external frequency (operating speed) may be changed or properties of the system may be changed to alter the natural frequency In some cases sufficient amount of damping may be provided to avoid large amplitude during resonance Thus the problem of force vibration is very important in mechanical design

FORCED VIBRATIONS OF SINGLE DOF SYSTEMS UNDER HARMONIC EXCITATION

` [ ] tsin FF o ω=

x

sin FF o=

Fo

xampamp

[ ] tω

xamp

47

48

Consider a spring mass system having a viscous damping excited by a harmonic force [ ] tsin FF o ω= as shown in figure The various forces acting on the system is shown in figure Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

O)tsin(Fkxxcxm o =ω+minusminus= ampampamp

)tsin(Fkxxcxm o ω=++ ampampamp ----------------------------------------------------------(1)

or

)tsin(Fkxdtdx

Cdt

xdM o2

2

ω=++

Equation (1) represents the differential equation of motion of the system The solution to this equation consists of two parts namely complementary function and a particular integral

The first of solution is also termed as transient solution of homogeneous solution and can be obtained from the homogeneous differential equation 0kxxcxm =++ ampampamp

The second part of solution particular integral is also termed as steady state solution ad is of form )tsin(Xx φminusω= Where X= steady state amplitude φ is phase difference between external force and displacement

)tsin(Xx φminusω=

The vector representation of the shown

49

The force polygon for the above vector diagram can be written as follows

( )φminusωt O

KX

xm 2ω

Fo

xCω

tω

φ

50

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

is the governing equation

Zero frequency deflection (Xst) It is deflection observed at zero frequency ( when the external frequency 0=ω

( ) kF

001

kF

x o

22

o

st =

+minus=

Magnification factor (M) It is defined as the ratio of steady state amplitude to zero frequency deflection xst

M = arX

X=

A plot of magnification factor versus frequency ratio

damping factor ( )ξ is called frequency plot of a harmon All the frequency terms starts from magnification factor of damping ratio ( )ξ The amplitude of vibration at any on the value of damping The amplitude will be higher for higher damping The amplitude of vibration at reso

amplitude tends to zero as

ωω

n tends to infinity

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

Freq

ζ =

ζ = 1

Magnification Factor M = XXst

kFo

51

ωω

nfor varying values of

ically excite vibrating system

M=1 what ever may be value particular frequency depends

for lesser damping and lesser nance will be very high The

uency Ratio r = (ωωn)

ζ = 0 0

ζ = 025 ζ = 0375 ζ = 0 5

ζ = 0707

ζ = 2

52

Plot of phase angle φ versus frequency ratio

ωω

n is shown in the

figure The phase angle φ is zero for all the values of damping ratio

at

ωω

n=0 The phase

angle φ is 90 for all the values of damping ratio

at

ωω

n=1 ie at

resonance The phase angle between o to 90 deg for all the values of

ωω

n= 0 to 1 and

between 90 deg to 180 deg for all the values of

ωω

ngt1

Class work Derive an expression for frequency at peak amplitude

2np 21 ξminusω=ω=ω is the required equation

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

53

Derive expression amplitude at resonance ( In dimensional form) In dimensional form We know that steady state amplitude

( ) ( )

ω+ωminus=

222

o

Cmk

Fx Equation in dimensional form

We know that at resonance mk

n =ω=ω

( )

ω+

minus

=2

n

2

oR

Cmk

mk

Fx

( )

ω=

2n

oR

C

Fx

( )

ω=

n

oR C

Fx

Class work Derive an expression for amplitude at resonance ( in non dimensional form)

Tutorial Problems on Forced Vibration 1A weight of 60N is suspended by a spring stiffness 12 kNm is forced to vibrate by a harmonic force of 10N assuming viscous damping of 0086 kN-sm Find i) resonance frequency ii) amplitude at resonance iii) phase angle at resonance iv) frequency corresponding to peak amplitude 2 A periodic torque having a maximum value of 05888Nm at a frequency of 4 rads is impressed upon a flywheel suspended from a wire The wheel has mass moment of inertia 012 kgm2 and the wire has stiffness 1176 Nmrad A viscous dashpot applies damping couple of 0784 Nm an angular velocity of 2 rads calculate i)maximum angular displacement from rest position ii) maximum couple applied to dashpot Iii) angle by which the angular displacement lags the torque A mass of 1019 kg is suspended from a spring stiffness 20 Nmm viscous damping causes the amplitude to decrease to 110th of initial value in 4 complete cycle If a periodic force of 20 cos(50 t) is applied to the mass Find the amplitude of forced vibration What would be the amplitude of oscillation if the applied force coincides with the natural period

m=1019 kg

[ ] t50cos 30F =

54

55

ROTATING UNBALANCE Unbalance in rotating machinery is a common source of vibration excitation Consider a machine of mass m having mounted on a foundation of stiffness k and damping coefficient C as shown in figure The unbalance is represented by an eccentric rotating mass m with eccentricity (e) and having a angular velocity (ω) let x be the displacement of non rotating mass from the static equilibrium position The displacement of rotating mass m from the static equilibrium is ( ) tsin( ex ω+ Applying Newtonrsquos law of motion we have

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

56

RECIPROCATING UNBALANCE Unbalance in reciprocating machinery is a common source of vibration excitation Consider a machine of mass M having mounted on a foundation of stiffness K and damping coefficient C as shown in figure The unbalance is represented by a reciprocating mass (m) having crank radus (e) and connecting rod of length (l) Let ωωωω be angular velocity of the crank Let x be the displacement of non reciprocating mass from the static equilibrium position at any instant of time The displacement of reciprocating mass m from the static equilibrium position is

)2sin(sin +++ tle

tex ωωωωωωωω

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

57

Note The equations obtained in reciprocating unbalance is same as the rotating unbalance and having the following differences Rotating Unbalance Reciprocating Unbalance Here mass of the rotating element is considered

Here mass of reciprocating element is considered

Here the eccentricity (e) which is the distance between the centre of rotation and CG

Here the eccentricity e is the crank radius and crank radius = stroke length2

The frequency plot of non dimensional factor( ( )me

MX versus frequency ratio

ωω

n for various values of ( )ξξξξ is shown in figure This frequency plot is same for

the rotating and reciprocating unbalance with the following points are noted from the diagram

1 Damping ratio plays very important role during resonance in controlling the amplitude

2 When

ωω

n=0 then ( )me

MX =0 for all the values of damping ratio ( )ξξξξ

3 When

ωω

nis higher than ( )me

MX tends to unity

4 At very high speeds ie at very high frequency ratio

ωω

n there is no effect

of damping The frequency plot of phase angle versus the frequency ratio is given as shown in figure The explanation is same as in forced harmonic vibration

58

VIBRATION ISOLATION

Vibratory forces and motions generated by machine and other caused are unavoidable Howeever their effect on the other dynamical systems can be minimized by proper isolator design An isolation system is either to protect o delicate machine from excessive vibration transmitted to it from its supporting structure or to prevent vibratory forces and motions generated by machines from being transmitted to its surroundings The effectiveness of isolation may be measured in terms of forces and motion transmitted The first type is known as force isolation and the second type is known as motion isolation

Force isolation (Force Transmissibility) When vibratory forces generated by machine are isolated the effectiveness of forced isolation can be made by defining the term transmissibility The term transmissibility is defined as the ratio of force transmitted to the foundation to that of the external impressed force

Force Transmissibility = o

tr

FF

force Impressed Externalfoundation toted transmitForce

=

Consider viscous damped spring mass system as shown in figure Upon this system an external harmonic force ) t (Sin Fo ω is applied The difference forces acting on system are inertia forces damping force spring force and external force All these forces are represented on force polygon shown below Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these two forces

( ) ( )22 xckx ω+=

59

This is the expression for force transmissibility in dimensional form To represent same in the non dimensional form divide numerator and denominator by k

2

n

22

n

2

n

o

tr

21

21

FF

TR

ωω

ξ+

ωω

minus

ωω

ξ+

==

PHASE DIFFERNECE BETWEEN EXTERNAL FORCE AND DISPLACEMENT Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these

ωω

minus

ωω

ξ=

ωminusω

=φ minusminus2

n

n1

2

1

1

2tan

mKC

tan

The phase difference between force transmitted to the foundation Ftr to the displacement

kc

KxxC

tanω

=ω

=α or

ωω

ξ=

ω=α minusminus

n

11 2tanKx

xCtan

The phase difference between external force Fo and force transmitted to foundationis β

60

β= αminusφ =

ωω

minus

ωω

ξminus

2

n

n1

1

2tan -

ωω

ξminus

n

1 2tan

Displacement Transmissibility It is the ratio of amplitude of steady state vibration of machine to that of the base motion Forced vibration due to support motion Consider a spring mass damper system subjected to harmonic excitation )tsin(Yy ω= at the base as shown in figure Let x be the absolute displacement of the mass at any instant of time t The different forces acting on system are shown in free body diagram Applying Newtonrsquos Law of motion Ma = sumF

)yx(k)yx(cxm minusminusminusminus= ampampampamp

0)yx(k)yx(cxm =minus+minus+ ampampampamp

61

Frequency plot of transmissibility versus frequency ratio

62

The plot of transmissibility versus frequency ratio is as shown in figure The following points can be noted from it

i) The value of transmissibility is unity at frequency ratio

ωω

n=0 and 2

ii) For an undamped system ( )ξ =0 transmissibility tends to infin at resonance iii) The value of transmissibility is less than unity for the value of frequency ratio

greater than 2

iv) For frequency ratio

ωω

nlt 2 smaller damping ratio lead to larger values of

transmissibility on the other hand for frequency ratio

ωω

ngt 2 smaller damping

ratio leads to smaller value of transmissibility

The frequency plot of transmissibility (TR) versus frequency ratio

ωω

n can be

broadly divided into two zones

1

ωω

n= 0 to 2

2

ωω

n= 2 to infin

CASE 1

When

ωω

n= 0 to 2

In this zone the value of transmissibility is always greater than 1 for all values of damping ratio In this zone the transmissibility decreases with increase in damping Hence we can conclude that in this zone damping is useful in isolating vibration forces being transmitted to the support CASE 2

When

ωω

n= 2 to infin

In this zone the value of transmissibility is less than 1 for all values of damping ratio In this zone the transmissibility increases with increase in damping Hence we can conclude that in this zone damping will adversely affect in isolating vibratory forces

63

Tutorial problems on Forced Vibration 1 Machine of mass 75kg is mounted on spring of stiffness 12000 Nm with an assumed damping factor ξ =02 piston within the machine of mass 2 kg has a reciprocating motion with a stroke of 75 cm and speed of 15 cyclessecond Assuming the motion of piston to be harmonic determine amplitude of machine ( X) Transmissibility and force transmitted to foundation phase angle of transmitted force wrt to exciting force 2 Motor weighs 600N is mounted on a simple beam that has a spring rate of 400 Ncm The motor armature base 160 N and has an eccentricity of 0005cm What will be the amplitude of motion when it runs at 1760 RPM Neglect damping the deflection of armature shaft and weight of the beam 3 A Heavy machine of mass 300 kg is supported on resilient foundation the static deflection of foundation due to weight of machine is found to be 75mm It is observed that the machine vibrates with an amplitude of 10mm when the base of the foundation is subjected to harmonic oscillation at the undampeds natural frequency of system with an amplitude of 25mm Find the damping constant of foundation Ii) dynamic force amplitude on the base iii) amplitude of the displacement of machine relative to base 4 An air compressor of mass 450 kg operates at a constant speed of 1750 RPM The reciprocating masses are 10 kg the crank radius is 100mm If the damper for mounting introduces a damping factor 015 i) specify the springs of mounting such that only 20 of the unbalanced force is transmitted to the foundation ii) determine the magnitude of the transmitted force bull Repeat the above problem by assuming that there is no damping in system

5 A body of a car having mass 1500 kg and is mounted on 4 equal springs which deflect through 225 cm under the weight of the body A total damping forces the 4 shock absorbers is 46 N at a velocity of 1cmsec The car is placed with all the four wheels on a test platform which is move up and down at resonant speed With an amplitude of 25cm Find the amplitude of the car body on its springs assuming its CG to be in the centre of wheel base 6A 75 kg machine which is mounted on springs of stiffness 116x105 Nm with the damping of 20=ξ A 2kg piston within the machine as reciprocating motion with a stroke of 008m and speed of 3000 RPM Assuming the motion of piston to be harmonic determine the amplitude of vibration of machine

7 The spring of an automobile are compressed 10cm under its own weight Find the critical speed when a trailer is traveling over road with a people approximaterly by a sine wave of amplitude 75 cm and wave length 15m What will be the amplitude of vibration at 64kmhour 8 An eccentric mass excites us used to determine vibratory characteristics of a structure of mass 25kg as given in figure At a speed of 1000 RPM the eccentric mass to beat top position at the instant the structure was moved upwards through its static equilibrium position and corresponding amplitude was 20mm The eccentric mass is 05kg at an eccentricity of 10mm Determine i) Undamped natural frequency of the system ii) Damping ratio iii) Steady state amplitude and phase angle at 1500 RPM iv) Force transmitted to the ground at 1200 RPM

M=25kg

C

K

Eccentric mass

64

65

CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Support motion Consider a spring mass damper system as shown in figure Let it excited by motion

of support let y be the harmonic displacement of support From Newtonrsquos second law of motion

)yx(k)yx(cxm minusminusminusminus= ampampampamp

)1(0)yx(k)yx(cxm minusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminus=minus+minus+ ampampampamp

Let z be the relative displacement of the mass with respect to support then

Z=x-y

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

CHAPTER 4

FORCED VIBRATION LEARNING OBJECTIVES

] Introduction to forced free vibration

] Terminologies used in forced vibration

] Concept of magnification factor

] Study of reciprocating and rotating unbalance

] Concept of base excitation and support excitation

] Concept of Transmissibility ratio

] Problems related to forced vibration

The vibration that takes place under the excitation of external forces are called

forced vibration When a system is subjected to harmonic excitation it is forced to vibrate at the same frequency as that of excitation Most of the mechanical systems are prone to steady state forced vibration which cause fatigue failure The vibration problem also becomes very important when any of excitation frequencies coincide with one of natural frequencies of a system Under this condition resonance occurs during which large amplitude of vibrations are observed To avoid resonance external frequency (operating speed) may be changed or properties of the system may be changed to alter the natural frequency In some cases sufficient amount of damping may be provided to avoid large amplitude during resonance Thus the problem of force vibration is very important in mechanical design

FORCED VIBRATIONS OF SINGLE DOF SYSTEMS UNDER HARMONIC EXCITATION

` [ ] tsin FF o ω=

x

sin FF o=

Fo

xampamp

[ ] tω

xamp

47

48

Consider a spring mass system having a viscous damping excited by a harmonic force [ ] tsin FF o ω= as shown in figure The various forces acting on the system is shown in figure Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

O)tsin(Fkxxcxm o =ω+minusminus= ampampamp

)tsin(Fkxxcxm o ω=++ ampampamp ----------------------------------------------------------(1)

or

)tsin(Fkxdtdx

Cdt

xdM o2

2

ω=++

Equation (1) represents the differential equation of motion of the system The solution to this equation consists of two parts namely complementary function and a particular integral

The first of solution is also termed as transient solution of homogeneous solution and can be obtained from the homogeneous differential equation 0kxxcxm =++ ampampamp

The second part of solution particular integral is also termed as steady state solution ad is of form )tsin(Xx φminusω= Where X= steady state amplitude φ is phase difference between external force and displacement

)tsin(Xx φminusω=

The vector representation of the shown

49

The force polygon for the above vector diagram can be written as follows

( )φminusωt O

KX

xm 2ω

Fo

xCω

tω

φ

50

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

is the governing equation

Zero frequency deflection (Xst) It is deflection observed at zero frequency ( when the external frequency 0=ω

( ) kF

001

kF

x o

22

o

st =

+minus=

Magnification factor (M) It is defined as the ratio of steady state amplitude to zero frequency deflection xst

M = arX

X=

A plot of magnification factor versus frequency ratio

damping factor ( )ξ is called frequency plot of a harmon All the frequency terms starts from magnification factor of damping ratio ( )ξ The amplitude of vibration at any on the value of damping The amplitude will be higher for higher damping The amplitude of vibration at reso

amplitude tends to zero as

ωω

n tends to infinity

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

Freq

ζ =

ζ = 1

Magnification Factor M = XXst

kFo

51

ωω

nfor varying values of

ically excite vibrating system

M=1 what ever may be value particular frequency depends

for lesser damping and lesser nance will be very high The

uency Ratio r = (ωωn)

ζ = 0 0

ζ = 025 ζ = 0375 ζ = 0 5

ζ = 0707

ζ = 2

52

Plot of phase angle φ versus frequency ratio

ωω

n is shown in the

figure The phase angle φ is zero for all the values of damping ratio

at

ωω

n=0 The phase

angle φ is 90 for all the values of damping ratio

at

ωω

n=1 ie at

resonance The phase angle between o to 90 deg for all the values of

ωω

n= 0 to 1 and

between 90 deg to 180 deg for all the values of

ωω

ngt1

Class work Derive an expression for frequency at peak amplitude

2np 21 ξminusω=ω=ω is the required equation

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

53

Derive expression amplitude at resonance ( In dimensional form) In dimensional form We know that steady state amplitude

( ) ( )

ω+ωminus=

222

o

Cmk

Fx Equation in dimensional form

We know that at resonance mk

n =ω=ω

( )

ω+

minus

=2

n

2

oR

Cmk

mk

Fx

( )

ω=

2n

oR

C

Fx

( )

ω=

n

oR C

Fx

Class work Derive an expression for amplitude at resonance ( in non dimensional form)

Tutorial Problems on Forced Vibration 1A weight of 60N is suspended by a spring stiffness 12 kNm is forced to vibrate by a harmonic force of 10N assuming viscous damping of 0086 kN-sm Find i) resonance frequency ii) amplitude at resonance iii) phase angle at resonance iv) frequency corresponding to peak amplitude 2 A periodic torque having a maximum value of 05888Nm at a frequency of 4 rads is impressed upon a flywheel suspended from a wire The wheel has mass moment of inertia 012 kgm2 and the wire has stiffness 1176 Nmrad A viscous dashpot applies damping couple of 0784 Nm an angular velocity of 2 rads calculate i)maximum angular displacement from rest position ii) maximum couple applied to dashpot Iii) angle by which the angular displacement lags the torque A mass of 1019 kg is suspended from a spring stiffness 20 Nmm viscous damping causes the amplitude to decrease to 110th of initial value in 4 complete cycle If a periodic force of 20 cos(50 t) is applied to the mass Find the amplitude of forced vibration What would be the amplitude of oscillation if the applied force coincides with the natural period

m=1019 kg

[ ] t50cos 30F =

54

55

ROTATING UNBALANCE Unbalance in rotating machinery is a common source of vibration excitation Consider a machine of mass m having mounted on a foundation of stiffness k and damping coefficient C as shown in figure The unbalance is represented by an eccentric rotating mass m with eccentricity (e) and having a angular velocity (ω) let x be the displacement of non rotating mass from the static equilibrium position The displacement of rotating mass m from the static equilibrium is ( ) tsin( ex ω+ Applying Newtonrsquos law of motion we have

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

56

RECIPROCATING UNBALANCE Unbalance in reciprocating machinery is a common source of vibration excitation Consider a machine of mass M having mounted on a foundation of stiffness K and damping coefficient C as shown in figure The unbalance is represented by a reciprocating mass (m) having crank radus (e) and connecting rod of length (l) Let ωωωω be angular velocity of the crank Let x be the displacement of non reciprocating mass from the static equilibrium position at any instant of time The displacement of reciprocating mass m from the static equilibrium position is

)2sin(sin +++ tle

tex ωωωωωωωω

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

57

Note The equations obtained in reciprocating unbalance is same as the rotating unbalance and having the following differences Rotating Unbalance Reciprocating Unbalance Here mass of the rotating element is considered

Here mass of reciprocating element is considered

Here the eccentricity (e) which is the distance between the centre of rotation and CG

Here the eccentricity e is the crank radius and crank radius = stroke length2

The frequency plot of non dimensional factor( ( )me

MX versus frequency ratio

ωω

n for various values of ( )ξξξξ is shown in figure This frequency plot is same for

the rotating and reciprocating unbalance with the following points are noted from the diagram

1 Damping ratio plays very important role during resonance in controlling the amplitude

2 When

ωω

n=0 then ( )me

MX =0 for all the values of damping ratio ( )ξξξξ

3 When

ωω

nis higher than ( )me

MX tends to unity

4 At very high speeds ie at very high frequency ratio

ωω

n there is no effect

of damping The frequency plot of phase angle versus the frequency ratio is given as shown in figure The explanation is same as in forced harmonic vibration

58

VIBRATION ISOLATION

Vibratory forces and motions generated by machine and other caused are unavoidable Howeever their effect on the other dynamical systems can be minimized by proper isolator design An isolation system is either to protect o delicate machine from excessive vibration transmitted to it from its supporting structure or to prevent vibratory forces and motions generated by machines from being transmitted to its surroundings The effectiveness of isolation may be measured in terms of forces and motion transmitted The first type is known as force isolation and the second type is known as motion isolation

Force isolation (Force Transmissibility) When vibratory forces generated by machine are isolated the effectiveness of forced isolation can be made by defining the term transmissibility The term transmissibility is defined as the ratio of force transmitted to the foundation to that of the external impressed force

Force Transmissibility = o

tr

FF

force Impressed Externalfoundation toted transmitForce

=

Consider viscous damped spring mass system as shown in figure Upon this system an external harmonic force ) t (Sin Fo ω is applied The difference forces acting on system are inertia forces damping force spring force and external force All these forces are represented on force polygon shown below Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these two forces

( ) ( )22 xckx ω+=

59

This is the expression for force transmissibility in dimensional form To represent same in the non dimensional form divide numerator and denominator by k

2

n

22

n

2

n

o

tr

21

21

FF

TR

ωω

ξ+

ωω

minus

ωω

ξ+

==

PHASE DIFFERNECE BETWEEN EXTERNAL FORCE AND DISPLACEMENT Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these

ωω

minus

ωω

ξ=

ωminusω

=φ minusminus2

n

n1

2

1

1

2tan

mKC

tan

The phase difference between force transmitted to the foundation Ftr to the displacement

kc

KxxC

tanω

=ω

=α or

ωω

ξ=

ω=α minusminus

n

11 2tanKx

xCtan

The phase difference between external force Fo and force transmitted to foundationis β

60

β= αminusφ =

ωω

minus

ωω

ξminus

2

n

n1

1

2tan -

ωω

ξminus

n

1 2tan

Displacement Transmissibility It is the ratio of amplitude of steady state vibration of machine to that of the base motion Forced vibration due to support motion Consider a spring mass damper system subjected to harmonic excitation )tsin(Yy ω= at the base as shown in figure Let x be the absolute displacement of the mass at any instant of time t The different forces acting on system are shown in free body diagram Applying Newtonrsquos Law of motion Ma = sumF

)yx(k)yx(cxm minusminusminusminus= ampampampamp

0)yx(k)yx(cxm =minus+minus+ ampampampamp

61

Frequency plot of transmissibility versus frequency ratio

62

The plot of transmissibility versus frequency ratio is as shown in figure The following points can be noted from it

i) The value of transmissibility is unity at frequency ratio

ωω

n=0 and 2

ii) For an undamped system ( )ξ =0 transmissibility tends to infin at resonance iii) The value of transmissibility is less than unity for the value of frequency ratio

greater than 2

iv) For frequency ratio

ωω

nlt 2 smaller damping ratio lead to larger values of

transmissibility on the other hand for frequency ratio

ωω

ngt 2 smaller damping

ratio leads to smaller value of transmissibility

The frequency plot of transmissibility (TR) versus frequency ratio

ωω

n can be

broadly divided into two zones

1

ωω

n= 0 to 2

2

ωω

n= 2 to infin

CASE 1

When

ωω

n= 0 to 2

In this zone the value of transmissibility is always greater than 1 for all values of damping ratio In this zone the transmissibility decreases with increase in damping Hence we can conclude that in this zone damping is useful in isolating vibration forces being transmitted to the support CASE 2

When

ωω

n= 2 to infin

In this zone the value of transmissibility is less than 1 for all values of damping ratio In this zone the transmissibility increases with increase in damping Hence we can conclude that in this zone damping will adversely affect in isolating vibratory forces

63

Tutorial problems on Forced Vibration 1 Machine of mass 75kg is mounted on spring of stiffness 12000 Nm with an assumed damping factor ξ =02 piston within the machine of mass 2 kg has a reciprocating motion with a stroke of 75 cm and speed of 15 cyclessecond Assuming the motion of piston to be harmonic determine amplitude of machine ( X) Transmissibility and force transmitted to foundation phase angle of transmitted force wrt to exciting force 2 Motor weighs 600N is mounted on a simple beam that has a spring rate of 400 Ncm The motor armature base 160 N and has an eccentricity of 0005cm What will be the amplitude of motion when it runs at 1760 RPM Neglect damping the deflection of armature shaft and weight of the beam 3 A Heavy machine of mass 300 kg is supported on resilient foundation the static deflection of foundation due to weight of machine is found to be 75mm It is observed that the machine vibrates with an amplitude of 10mm when the base of the foundation is subjected to harmonic oscillation at the undampeds natural frequency of system with an amplitude of 25mm Find the damping constant of foundation Ii) dynamic force amplitude on the base iii) amplitude of the displacement of machine relative to base 4 An air compressor of mass 450 kg operates at a constant speed of 1750 RPM The reciprocating masses are 10 kg the crank radius is 100mm If the damper for mounting introduces a damping factor 015 i) specify the springs of mounting such that only 20 of the unbalanced force is transmitted to the foundation ii) determine the magnitude of the transmitted force bull Repeat the above problem by assuming that there is no damping in system

5 A body of a car having mass 1500 kg and is mounted on 4 equal springs which deflect through 225 cm under the weight of the body A total damping forces the 4 shock absorbers is 46 N at a velocity of 1cmsec The car is placed with all the four wheels on a test platform which is move up and down at resonant speed With an amplitude of 25cm Find the amplitude of the car body on its springs assuming its CG to be in the centre of wheel base 6A 75 kg machine which is mounted on springs of stiffness 116x105 Nm with the damping of 20=ξ A 2kg piston within the machine as reciprocating motion with a stroke of 008m and speed of 3000 RPM Assuming the motion of piston to be harmonic determine the amplitude of vibration of machine

7 The spring of an automobile are compressed 10cm under its own weight Find the critical speed when a trailer is traveling over road with a people approximaterly by a sine wave of amplitude 75 cm and wave length 15m What will be the amplitude of vibration at 64kmhour 8 An eccentric mass excites us used to determine vibratory characteristics of a structure of mass 25kg as given in figure At a speed of 1000 RPM the eccentric mass to beat top position at the instant the structure was moved upwards through its static equilibrium position and corresponding amplitude was 20mm The eccentric mass is 05kg at an eccentricity of 10mm Determine i) Undamped natural frequency of the system ii) Damping ratio iii) Steady state amplitude and phase angle at 1500 RPM iv) Force transmitted to the ground at 1200 RPM

M=25kg

C

K

Eccentric mass

64

65

CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Support motion Consider a spring mass damper system as shown in figure Let it excited by motion

of support let y be the harmonic displacement of support From Newtonrsquos second law of motion

)yx(k)yx(cxm minusminusminusminus= ampampampamp

)1(0)yx(k)yx(cxm minusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminus=minus+minus+ ampampampamp

Let z be the relative displacement of the mass with respect to support then

Z=x-y

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

48

Consider a spring mass system having a viscous damping excited by a harmonic force [ ] tsin FF o ω= as shown in figure The various forces acting on the system is shown in figure Then different forces acting on system is shown in the free body diagram By Newtonrsquos second law of motion

O)tsin(Fkxxcxm o =ω+minusminus= ampampamp

)tsin(Fkxxcxm o ω=++ ampampamp ----------------------------------------------------------(1)

or

)tsin(Fkxdtdx

Cdt

xdM o2

2

ω=++

Equation (1) represents the differential equation of motion of the system The solution to this equation consists of two parts namely complementary function and a particular integral

The first of solution is also termed as transient solution of homogeneous solution and can be obtained from the homogeneous differential equation 0kxxcxm =++ ampampamp

The second part of solution particular integral is also termed as steady state solution ad is of form )tsin(Xx φminusω= Where X= steady state amplitude φ is phase difference between external force and displacement

)tsin(Xx φminusω=

The vector representation of the shown

49

The force polygon for the above vector diagram can be written as follows

( )φminusωt O

KX

xm 2ω

Fo

xCω

tω

φ

50

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

is the governing equation

Zero frequency deflection (Xst) It is deflection observed at zero frequency ( when the external frequency 0=ω

( ) kF

001

kF

x o

22

o

st =

+minus=

Magnification factor (M) It is defined as the ratio of steady state amplitude to zero frequency deflection xst

M = arX

X=

A plot of magnification factor versus frequency ratio

damping factor ( )ξ is called frequency plot of a harmon All the frequency terms starts from magnification factor of damping ratio ( )ξ The amplitude of vibration at any on the value of damping The amplitude will be higher for higher damping The amplitude of vibration at reso

amplitude tends to zero as

ωω

n tends to infinity

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

Freq

ζ =

ζ = 1

Magnification Factor M = XXst

kFo

51

ωω

nfor varying values of

ically excite vibrating system

M=1 what ever may be value particular frequency depends

for lesser damping and lesser nance will be very high The

uency Ratio r = (ωωn)

ζ = 0 0

ζ = 025 ζ = 0375 ζ = 0 5

ζ = 0707

ζ = 2

52

Plot of phase angle φ versus frequency ratio

ωω

n is shown in the

figure The phase angle φ is zero for all the values of damping ratio

at

ωω

n=0 The phase

angle φ is 90 for all the values of damping ratio

at

ωω

n=1 ie at

resonance The phase angle between o to 90 deg for all the values of

ωω

n= 0 to 1 and

between 90 deg to 180 deg for all the values of

ωω

ngt1

Class work Derive an expression for frequency at peak amplitude

2np 21 ξminusω=ω=ω is the required equation

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

53

Derive expression amplitude at resonance ( In dimensional form) In dimensional form We know that steady state amplitude

( ) ( )

ω+ωminus=

222

o

Cmk

Fx Equation in dimensional form

We know that at resonance mk

n =ω=ω

( )

ω+

minus

=2

n

2

oR

Cmk

mk

Fx

( )

ω=

2n

oR

C

Fx

( )

ω=

n

oR C

Fx

Class work Derive an expression for amplitude at resonance ( in non dimensional form)

Tutorial Problems on Forced Vibration 1A weight of 60N is suspended by a spring stiffness 12 kNm is forced to vibrate by a harmonic force of 10N assuming viscous damping of 0086 kN-sm Find i) resonance frequency ii) amplitude at resonance iii) phase angle at resonance iv) frequency corresponding to peak amplitude 2 A periodic torque having a maximum value of 05888Nm at a frequency of 4 rads is impressed upon a flywheel suspended from a wire The wheel has mass moment of inertia 012 kgm2 and the wire has stiffness 1176 Nmrad A viscous dashpot applies damping couple of 0784 Nm an angular velocity of 2 rads calculate i)maximum angular displacement from rest position ii) maximum couple applied to dashpot Iii) angle by which the angular displacement lags the torque A mass of 1019 kg is suspended from a spring stiffness 20 Nmm viscous damping causes the amplitude to decrease to 110th of initial value in 4 complete cycle If a periodic force of 20 cos(50 t) is applied to the mass Find the amplitude of forced vibration What would be the amplitude of oscillation if the applied force coincides with the natural period

m=1019 kg

[ ] t50cos 30F =

54

55

ROTATING UNBALANCE Unbalance in rotating machinery is a common source of vibration excitation Consider a machine of mass m having mounted on a foundation of stiffness k and damping coefficient C as shown in figure The unbalance is represented by an eccentric rotating mass m with eccentricity (e) and having a angular velocity (ω) let x be the displacement of non rotating mass from the static equilibrium position The displacement of rotating mass m from the static equilibrium is ( ) tsin( ex ω+ Applying Newtonrsquos law of motion we have

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

56

RECIPROCATING UNBALANCE Unbalance in reciprocating machinery is a common source of vibration excitation Consider a machine of mass M having mounted on a foundation of stiffness K and damping coefficient C as shown in figure The unbalance is represented by a reciprocating mass (m) having crank radus (e) and connecting rod of length (l) Let ωωωω be angular velocity of the crank Let x be the displacement of non reciprocating mass from the static equilibrium position at any instant of time The displacement of reciprocating mass m from the static equilibrium position is

)2sin(sin +++ tle

tex ωωωωωωωω

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

57

Note The equations obtained in reciprocating unbalance is same as the rotating unbalance and having the following differences Rotating Unbalance Reciprocating Unbalance Here mass of the rotating element is considered

Here mass of reciprocating element is considered

Here the eccentricity (e) which is the distance between the centre of rotation and CG

Here the eccentricity e is the crank radius and crank radius = stroke length2

The frequency plot of non dimensional factor( ( )me

MX versus frequency ratio

ωω

n for various values of ( )ξξξξ is shown in figure This frequency plot is same for

the rotating and reciprocating unbalance with the following points are noted from the diagram

1 Damping ratio plays very important role during resonance in controlling the amplitude

2 When

ωω

n=0 then ( )me

MX =0 for all the values of damping ratio ( )ξξξξ

3 When

ωω

nis higher than ( )me

MX tends to unity

4 At very high speeds ie at very high frequency ratio

ωω

n there is no effect

of damping The frequency plot of phase angle versus the frequency ratio is given as shown in figure The explanation is same as in forced harmonic vibration

58

VIBRATION ISOLATION

Vibratory forces and motions generated by machine and other caused are unavoidable Howeever their effect on the other dynamical systems can be minimized by proper isolator design An isolation system is either to protect o delicate machine from excessive vibration transmitted to it from its supporting structure or to prevent vibratory forces and motions generated by machines from being transmitted to its surroundings The effectiveness of isolation may be measured in terms of forces and motion transmitted The first type is known as force isolation and the second type is known as motion isolation

Force isolation (Force Transmissibility) When vibratory forces generated by machine are isolated the effectiveness of forced isolation can be made by defining the term transmissibility The term transmissibility is defined as the ratio of force transmitted to the foundation to that of the external impressed force

Force Transmissibility = o

tr

FF

force Impressed Externalfoundation toted transmitForce

=

Consider viscous damped spring mass system as shown in figure Upon this system an external harmonic force ) t (Sin Fo ω is applied The difference forces acting on system are inertia forces damping force spring force and external force All these forces are represented on force polygon shown below Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these two forces

( ) ( )22 xckx ω+=

59

This is the expression for force transmissibility in dimensional form To represent same in the non dimensional form divide numerator and denominator by k

2

n

22

n

2

n

o

tr

21

21

FF

TR

ωω

ξ+

ωω

minus

ωω

ξ+

==

PHASE DIFFERNECE BETWEEN EXTERNAL FORCE AND DISPLACEMENT Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these

ωω

minus

ωω

ξ=

ωminusω

=φ minusminus2

n

n1

2

1

1

2tan

mKC

tan

The phase difference between force transmitted to the foundation Ftr to the displacement

kc

KxxC

tanω

=ω

=α or

ωω

ξ=

ω=α minusminus

n

11 2tanKx

xCtan

The phase difference between external force Fo and force transmitted to foundationis β

60

β= αminusφ =

ωω

minus

ωω

ξminus

2

n

n1

1

2tan -

ωω

ξminus

n

1 2tan

Displacement Transmissibility It is the ratio of amplitude of steady state vibration of machine to that of the base motion Forced vibration due to support motion Consider a spring mass damper system subjected to harmonic excitation )tsin(Yy ω= at the base as shown in figure Let x be the absolute displacement of the mass at any instant of time t The different forces acting on system are shown in free body diagram Applying Newtonrsquos Law of motion Ma = sumF

)yx(k)yx(cxm minusminusminusminus= ampampampamp

0)yx(k)yx(cxm =minus+minus+ ampampampamp

61

Frequency plot of transmissibility versus frequency ratio

62

The plot of transmissibility versus frequency ratio is as shown in figure The following points can be noted from it

i) The value of transmissibility is unity at frequency ratio

ωω

n=0 and 2

ii) For an undamped system ( )ξ =0 transmissibility tends to infin at resonance iii) The value of transmissibility is less than unity for the value of frequency ratio

greater than 2

iv) For frequency ratio

ωω

nlt 2 smaller damping ratio lead to larger values of

transmissibility on the other hand for frequency ratio

ωω

ngt 2 smaller damping

ratio leads to smaller value of transmissibility

The frequency plot of transmissibility (TR) versus frequency ratio

ωω

n can be

broadly divided into two zones

1

ωω

n= 0 to 2

2

ωω

n= 2 to infin

CASE 1

When

ωω

n= 0 to 2

In this zone the value of transmissibility is always greater than 1 for all values of damping ratio In this zone the transmissibility decreases with increase in damping Hence we can conclude that in this zone damping is useful in isolating vibration forces being transmitted to the support CASE 2

When

ωω

n= 2 to infin

In this zone the value of transmissibility is less than 1 for all values of damping ratio In this zone the transmissibility increases with increase in damping Hence we can conclude that in this zone damping will adversely affect in isolating vibratory forces

63

Tutorial problems on Forced Vibration 1 Machine of mass 75kg is mounted on spring of stiffness 12000 Nm with an assumed damping factor ξ =02 piston within the machine of mass 2 kg has a reciprocating motion with a stroke of 75 cm and speed of 15 cyclessecond Assuming the motion of piston to be harmonic determine amplitude of machine ( X) Transmissibility and force transmitted to foundation phase angle of transmitted force wrt to exciting force 2 Motor weighs 600N is mounted on a simple beam that has a spring rate of 400 Ncm The motor armature base 160 N and has an eccentricity of 0005cm What will be the amplitude of motion when it runs at 1760 RPM Neglect damping the deflection of armature shaft and weight of the beam 3 A Heavy machine of mass 300 kg is supported on resilient foundation the static deflection of foundation due to weight of machine is found to be 75mm It is observed that the machine vibrates with an amplitude of 10mm when the base of the foundation is subjected to harmonic oscillation at the undampeds natural frequency of system with an amplitude of 25mm Find the damping constant of foundation Ii) dynamic force amplitude on the base iii) amplitude of the displacement of machine relative to base 4 An air compressor of mass 450 kg operates at a constant speed of 1750 RPM The reciprocating masses are 10 kg the crank radius is 100mm If the damper for mounting introduces a damping factor 015 i) specify the springs of mounting such that only 20 of the unbalanced force is transmitted to the foundation ii) determine the magnitude of the transmitted force bull Repeat the above problem by assuming that there is no damping in system

5 A body of a car having mass 1500 kg and is mounted on 4 equal springs which deflect through 225 cm under the weight of the body A total damping forces the 4 shock absorbers is 46 N at a velocity of 1cmsec The car is placed with all the four wheels on a test platform which is move up and down at resonant speed With an amplitude of 25cm Find the amplitude of the car body on its springs assuming its CG to be in the centre of wheel base 6A 75 kg machine which is mounted on springs of stiffness 116x105 Nm with the damping of 20=ξ A 2kg piston within the machine as reciprocating motion with a stroke of 008m and speed of 3000 RPM Assuming the motion of piston to be harmonic determine the amplitude of vibration of machine

7 The spring of an automobile are compressed 10cm under its own weight Find the critical speed when a trailer is traveling over road with a people approximaterly by a sine wave of amplitude 75 cm and wave length 15m What will be the amplitude of vibration at 64kmhour 8 An eccentric mass excites us used to determine vibratory characteristics of a structure of mass 25kg as given in figure At a speed of 1000 RPM the eccentric mass to beat top position at the instant the structure was moved upwards through its static equilibrium position and corresponding amplitude was 20mm The eccentric mass is 05kg at an eccentricity of 10mm Determine i) Undamped natural frequency of the system ii) Damping ratio iii) Steady state amplitude and phase angle at 1500 RPM iv) Force transmitted to the ground at 1200 RPM

M=25kg

C

K

Eccentric mass

64

65

CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Support motion Consider a spring mass damper system as shown in figure Let it excited by motion

of support let y be the harmonic displacement of support From Newtonrsquos second law of motion

)yx(k)yx(cxm minusminusminusminus= ampampampamp

)1(0)yx(k)yx(cxm minusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminus=minus+minus+ ampampampamp

Let z be the relative displacement of the mass with respect to support then

Z=x-y

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

49

The force polygon for the above vector diagram can be written as follows

( )φminusωt O

KX

xm 2ω

Fo

xCω

tω

φ

50

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

is the governing equation

Zero frequency deflection (Xst) It is deflection observed at zero frequency ( when the external frequency 0=ω

( ) kF

001

kF

x o

22

o

st =

+minus=

Magnification factor (M) It is defined as the ratio of steady state amplitude to zero frequency deflection xst

M = arX

X=

A plot of magnification factor versus frequency ratio

damping factor ( )ξ is called frequency plot of a harmon All the frequency terms starts from magnification factor of damping ratio ( )ξ The amplitude of vibration at any on the value of damping The amplitude will be higher for higher damping The amplitude of vibration at reso

amplitude tends to zero as

ωω

n tends to infinity

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

Freq

ζ =

ζ = 1

Magnification Factor M = XXst

kFo

51

ωω

nfor varying values of

ically excite vibrating system

M=1 what ever may be value particular frequency depends

for lesser damping and lesser nance will be very high The

uency Ratio r = (ωωn)

ζ = 0 0

ζ = 025 ζ = 0375 ζ = 0 5

ζ = 0707

ζ = 2

52

Plot of phase angle φ versus frequency ratio

ωω

n is shown in the

figure The phase angle φ is zero for all the values of damping ratio

at

ωω

n=0 The phase

angle φ is 90 for all the values of damping ratio

at

ωω

n=1 ie at

resonance The phase angle between o to 90 deg for all the values of

ωω

n= 0 to 1 and

between 90 deg to 180 deg for all the values of

ωω

ngt1

Class work Derive an expression for frequency at peak amplitude

2np 21 ξminusω=ω=ω is the required equation

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

53

Derive expression amplitude at resonance ( In dimensional form) In dimensional form We know that steady state amplitude

( ) ( )

ω+ωminus=

222

o

Cmk

Fx Equation in dimensional form

We know that at resonance mk

n =ω=ω

( )

ω+

minus

=2

n

2

oR

Cmk

mk

Fx

( )

ω=

2n

oR

C

Fx

( )

ω=

n

oR C

Fx

Class work Derive an expression for amplitude at resonance ( in non dimensional form)

Tutorial Problems on Forced Vibration 1A weight of 60N is suspended by a spring stiffness 12 kNm is forced to vibrate by a harmonic force of 10N assuming viscous damping of 0086 kN-sm Find i) resonance frequency ii) amplitude at resonance iii) phase angle at resonance iv) frequency corresponding to peak amplitude 2 A periodic torque having a maximum value of 05888Nm at a frequency of 4 rads is impressed upon a flywheel suspended from a wire The wheel has mass moment of inertia 012 kgm2 and the wire has stiffness 1176 Nmrad A viscous dashpot applies damping couple of 0784 Nm an angular velocity of 2 rads calculate i)maximum angular displacement from rest position ii) maximum couple applied to dashpot Iii) angle by which the angular displacement lags the torque A mass of 1019 kg is suspended from a spring stiffness 20 Nmm viscous damping causes the amplitude to decrease to 110th of initial value in 4 complete cycle If a periodic force of 20 cos(50 t) is applied to the mass Find the amplitude of forced vibration What would be the amplitude of oscillation if the applied force coincides with the natural period

m=1019 kg

[ ] t50cos 30F =

54

55

ROTATING UNBALANCE Unbalance in rotating machinery is a common source of vibration excitation Consider a machine of mass m having mounted on a foundation of stiffness k and damping coefficient C as shown in figure The unbalance is represented by an eccentric rotating mass m with eccentricity (e) and having a angular velocity (ω) let x be the displacement of non rotating mass from the static equilibrium position The displacement of rotating mass m from the static equilibrium is ( ) tsin( ex ω+ Applying Newtonrsquos law of motion we have

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

56

RECIPROCATING UNBALANCE Unbalance in reciprocating machinery is a common source of vibration excitation Consider a machine of mass M having mounted on a foundation of stiffness K and damping coefficient C as shown in figure The unbalance is represented by a reciprocating mass (m) having crank radus (e) and connecting rod of length (l) Let ωωωω be angular velocity of the crank Let x be the displacement of non reciprocating mass from the static equilibrium position at any instant of time The displacement of reciprocating mass m from the static equilibrium position is

)2sin(sin +++ tle

tex ωωωωωωωω

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

57

Note The equations obtained in reciprocating unbalance is same as the rotating unbalance and having the following differences Rotating Unbalance Reciprocating Unbalance Here mass of the rotating element is considered

Here mass of reciprocating element is considered

Here the eccentricity (e) which is the distance between the centre of rotation and CG

Here the eccentricity e is the crank radius and crank radius = stroke length2

The frequency plot of non dimensional factor( ( )me

MX versus frequency ratio

ωω

n for various values of ( )ξξξξ is shown in figure This frequency plot is same for

the rotating and reciprocating unbalance with the following points are noted from the diagram

1 Damping ratio plays very important role during resonance in controlling the amplitude

2 When

ωω

n=0 then ( )me

MX =0 for all the values of damping ratio ( )ξξξξ

3 When

ωω

nis higher than ( )me

MX tends to unity

4 At very high speeds ie at very high frequency ratio

ωω

n there is no effect

of damping The frequency plot of phase angle versus the frequency ratio is given as shown in figure The explanation is same as in forced harmonic vibration

58

VIBRATION ISOLATION

Vibratory forces and motions generated by machine and other caused are unavoidable Howeever their effect on the other dynamical systems can be minimized by proper isolator design An isolation system is either to protect o delicate machine from excessive vibration transmitted to it from its supporting structure or to prevent vibratory forces and motions generated by machines from being transmitted to its surroundings The effectiveness of isolation may be measured in terms of forces and motion transmitted The first type is known as force isolation and the second type is known as motion isolation

Force isolation (Force Transmissibility) When vibratory forces generated by machine are isolated the effectiveness of forced isolation can be made by defining the term transmissibility The term transmissibility is defined as the ratio of force transmitted to the foundation to that of the external impressed force

Force Transmissibility = o

tr

FF

force Impressed Externalfoundation toted transmitForce

=

Consider viscous damped spring mass system as shown in figure Upon this system an external harmonic force ) t (Sin Fo ω is applied The difference forces acting on system are inertia forces damping force spring force and external force All these forces are represented on force polygon shown below Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these two forces

( ) ( )22 xckx ω+=

59

This is the expression for force transmissibility in dimensional form To represent same in the non dimensional form divide numerator and denominator by k

2

n

22

n

2

n

o

tr

21

21

FF

TR

ωω

ξ+

ωω

minus

ωω

ξ+

==

PHASE DIFFERNECE BETWEEN EXTERNAL FORCE AND DISPLACEMENT Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these

ωω

minus

ωω

ξ=

ωminusω

=φ minusminus2

n

n1

2

1

1

2tan

mKC

tan

The phase difference between force transmitted to the foundation Ftr to the displacement

kc

KxxC

tanω

=ω

=α or

ωω

ξ=

ω=α minusminus

n

11 2tanKx

xCtan

The phase difference between external force Fo and force transmitted to foundationis β

60

β= αminusφ =

ωω

minus

ωω

ξminus

2

n

n1

1

2tan -

ωω

ξminus

n

1 2tan

Displacement Transmissibility It is the ratio of amplitude of steady state vibration of machine to that of the base motion Forced vibration due to support motion Consider a spring mass damper system subjected to harmonic excitation )tsin(Yy ω= at the base as shown in figure Let x be the absolute displacement of the mass at any instant of time t The different forces acting on system are shown in free body diagram Applying Newtonrsquos Law of motion Ma = sumF

)yx(k)yx(cxm minusminusminusminus= ampampampamp

0)yx(k)yx(cxm =minus+minus+ ampampampamp

61

Frequency plot of transmissibility versus frequency ratio

62

The plot of transmissibility versus frequency ratio is as shown in figure The following points can be noted from it

i) The value of transmissibility is unity at frequency ratio

ωω

n=0 and 2

ii) For an undamped system ( )ξ =0 transmissibility tends to infin at resonance iii) The value of transmissibility is less than unity for the value of frequency ratio

greater than 2

iv) For frequency ratio

ωω

nlt 2 smaller damping ratio lead to larger values of

transmissibility on the other hand for frequency ratio

ωω

ngt 2 smaller damping

ratio leads to smaller value of transmissibility

The frequency plot of transmissibility (TR) versus frequency ratio

ωω

n can be

broadly divided into two zones

1

ωω

n= 0 to 2

2

ωω

n= 2 to infin

CASE 1

When

ωω

n= 0 to 2

In this zone the value of transmissibility is always greater than 1 for all values of damping ratio In this zone the transmissibility decreases with increase in damping Hence we can conclude that in this zone damping is useful in isolating vibration forces being transmitted to the support CASE 2

When

ωω

n= 2 to infin

In this zone the value of transmissibility is less than 1 for all values of damping ratio In this zone the transmissibility increases with increase in damping Hence we can conclude that in this zone damping will adversely affect in isolating vibratory forces

63

Tutorial problems on Forced Vibration 1 Machine of mass 75kg is mounted on spring of stiffness 12000 Nm with an assumed damping factor ξ =02 piston within the machine of mass 2 kg has a reciprocating motion with a stroke of 75 cm and speed of 15 cyclessecond Assuming the motion of piston to be harmonic determine amplitude of machine ( X) Transmissibility and force transmitted to foundation phase angle of transmitted force wrt to exciting force 2 Motor weighs 600N is mounted on a simple beam that has a spring rate of 400 Ncm The motor armature base 160 N and has an eccentricity of 0005cm What will be the amplitude of motion when it runs at 1760 RPM Neglect damping the deflection of armature shaft and weight of the beam 3 A Heavy machine of mass 300 kg is supported on resilient foundation the static deflection of foundation due to weight of machine is found to be 75mm It is observed that the machine vibrates with an amplitude of 10mm when the base of the foundation is subjected to harmonic oscillation at the undampeds natural frequency of system with an amplitude of 25mm Find the damping constant of foundation Ii) dynamic force amplitude on the base iii) amplitude of the displacement of machine relative to base 4 An air compressor of mass 450 kg operates at a constant speed of 1750 RPM The reciprocating masses are 10 kg the crank radius is 100mm If the damper for mounting introduces a damping factor 015 i) specify the springs of mounting such that only 20 of the unbalanced force is transmitted to the foundation ii) determine the magnitude of the transmitted force bull Repeat the above problem by assuming that there is no damping in system

5 A body of a car having mass 1500 kg and is mounted on 4 equal springs which deflect through 225 cm under the weight of the body A total damping forces the 4 shock absorbers is 46 N at a velocity of 1cmsec The car is placed with all the four wheels on a test platform which is move up and down at resonant speed With an amplitude of 25cm Find the amplitude of the car body on its springs assuming its CG to be in the centre of wheel base 6A 75 kg machine which is mounted on springs of stiffness 116x105 Nm with the damping of 20=ξ A 2kg piston within the machine as reciprocating motion with a stroke of 008m and speed of 3000 RPM Assuming the motion of piston to be harmonic determine the amplitude of vibration of machine

7 The spring of an automobile are compressed 10cm under its own weight Find the critical speed when a trailer is traveling over road with a people approximaterly by a sine wave of amplitude 75 cm and wave length 15m What will be the amplitude of vibration at 64kmhour 8 An eccentric mass excites us used to determine vibratory characteristics of a structure of mass 25kg as given in figure At a speed of 1000 RPM the eccentric mass to beat top position at the instant the structure was moved upwards through its static equilibrium position and corresponding amplitude was 20mm The eccentric mass is 05kg at an eccentricity of 10mm Determine i) Undamped natural frequency of the system ii) Damping ratio iii) Steady state amplitude and phase angle at 1500 RPM iv) Force transmitted to the ground at 1200 RPM

M=25kg

C

K

Eccentric mass

64

65

CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Support motion Consider a spring mass damper system as shown in figure Let it excited by motion

of support let y be the harmonic displacement of support From Newtonrsquos second law of motion

)yx(k)yx(cxm minusminusminusminus= ampampampamp

)1(0)yx(k)yx(cxm minusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminus=minus+minus+ ampampampamp

Let z be the relative displacement of the mass with respect to support then

Z=x-y

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

50

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

is the governing equation

Zero frequency deflection (Xst) It is deflection observed at zero frequency ( when the external frequency 0=ω

( ) kF

001

kF

x o

22

o

st =

+minus=

Magnification factor (M) It is defined as the ratio of steady state amplitude to zero frequency deflection xst

M = arX

X=

A plot of magnification factor versus frequency ratio

damping factor ( )ξ is called frequency plot of a harmon All the frequency terms starts from magnification factor of damping ratio ( )ξ The amplitude of vibration at any on the value of damping The amplitude will be higher for higher damping The amplitude of vibration at reso

amplitude tends to zero as

ωω

n tends to infinity

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

Freq

ζ =

ζ = 1

Magnification Factor M = XXst

kFo

51

ωω

nfor varying values of

ically excite vibrating system

M=1 what ever may be value particular frequency depends

for lesser damping and lesser nance will be very high The

uency Ratio r = (ωωn)

ζ = 0 0

ζ = 025 ζ = 0375 ζ = 0 5

ζ = 0707

ζ = 2

52

Plot of phase angle φ versus frequency ratio

ωω

n is shown in the

figure The phase angle φ is zero for all the values of damping ratio

at

ωω

n=0 The phase

angle φ is 90 for all the values of damping ratio

at

ωω

n=1 ie at

resonance The phase angle between o to 90 deg for all the values of

ωω

n= 0 to 1 and

between 90 deg to 180 deg for all the values of

ωω

ngt1

Class work Derive an expression for frequency at peak amplitude

2np 21 ξminusω=ω=ω is the required equation

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

53

Derive expression amplitude at resonance ( In dimensional form) In dimensional form We know that steady state amplitude

( ) ( )

ω+ωminus=

222

o

Cmk

Fx Equation in dimensional form

We know that at resonance mk

n =ω=ω

( )

ω+

minus

=2

n

2

oR

Cmk

mk

Fx

( )

ω=

2n

oR

C

Fx

( )

ω=

n

oR C

Fx

Class work Derive an expression for amplitude at resonance ( in non dimensional form)

Tutorial Problems on Forced Vibration 1A weight of 60N is suspended by a spring stiffness 12 kNm is forced to vibrate by a harmonic force of 10N assuming viscous damping of 0086 kN-sm Find i) resonance frequency ii) amplitude at resonance iii) phase angle at resonance iv) frequency corresponding to peak amplitude 2 A periodic torque having a maximum value of 05888Nm at a frequency of 4 rads is impressed upon a flywheel suspended from a wire The wheel has mass moment of inertia 012 kgm2 and the wire has stiffness 1176 Nmrad A viscous dashpot applies damping couple of 0784 Nm an angular velocity of 2 rads calculate i)maximum angular displacement from rest position ii) maximum couple applied to dashpot Iii) angle by which the angular displacement lags the torque A mass of 1019 kg is suspended from a spring stiffness 20 Nmm viscous damping causes the amplitude to decrease to 110th of initial value in 4 complete cycle If a periodic force of 20 cos(50 t) is applied to the mass Find the amplitude of forced vibration What would be the amplitude of oscillation if the applied force coincides with the natural period

m=1019 kg

[ ] t50cos 30F =

54

55

ROTATING UNBALANCE Unbalance in rotating machinery is a common source of vibration excitation Consider a machine of mass m having mounted on a foundation of stiffness k and damping coefficient C as shown in figure The unbalance is represented by an eccentric rotating mass m with eccentricity (e) and having a angular velocity (ω) let x be the displacement of non rotating mass from the static equilibrium position The displacement of rotating mass m from the static equilibrium is ( ) tsin( ex ω+ Applying Newtonrsquos law of motion we have

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

56

RECIPROCATING UNBALANCE Unbalance in reciprocating machinery is a common source of vibration excitation Consider a machine of mass M having mounted on a foundation of stiffness K and damping coefficient C as shown in figure The unbalance is represented by a reciprocating mass (m) having crank radus (e) and connecting rod of length (l) Let ωωωω be angular velocity of the crank Let x be the displacement of non reciprocating mass from the static equilibrium position at any instant of time The displacement of reciprocating mass m from the static equilibrium position is

)2sin(sin +++ tle

tex ωωωωωωωω

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

57

Note The equations obtained in reciprocating unbalance is same as the rotating unbalance and having the following differences Rotating Unbalance Reciprocating Unbalance Here mass of the rotating element is considered

Here mass of reciprocating element is considered

Here the eccentricity (e) which is the distance between the centre of rotation and CG

Here the eccentricity e is the crank radius and crank radius = stroke length2

The frequency plot of non dimensional factor( ( )me

MX versus frequency ratio

ωω

n for various values of ( )ξξξξ is shown in figure This frequency plot is same for

the rotating and reciprocating unbalance with the following points are noted from the diagram

1 Damping ratio plays very important role during resonance in controlling the amplitude

2 When

ωω

n=0 then ( )me

MX =0 for all the values of damping ratio ( )ξξξξ

3 When

ωω

nis higher than ( )me

MX tends to unity

4 At very high speeds ie at very high frequency ratio

ωω

n there is no effect

of damping The frequency plot of phase angle versus the frequency ratio is given as shown in figure The explanation is same as in forced harmonic vibration

58

VIBRATION ISOLATION

Vibratory forces and motions generated by machine and other caused are unavoidable Howeever their effect on the other dynamical systems can be minimized by proper isolator design An isolation system is either to protect o delicate machine from excessive vibration transmitted to it from its supporting structure or to prevent vibratory forces and motions generated by machines from being transmitted to its surroundings The effectiveness of isolation may be measured in terms of forces and motion transmitted The first type is known as force isolation and the second type is known as motion isolation

Force isolation (Force Transmissibility) When vibratory forces generated by machine are isolated the effectiveness of forced isolation can be made by defining the term transmissibility The term transmissibility is defined as the ratio of force transmitted to the foundation to that of the external impressed force

Force Transmissibility = o

tr

FF

force Impressed Externalfoundation toted transmitForce

=

Consider viscous damped spring mass system as shown in figure Upon this system an external harmonic force ) t (Sin Fo ω is applied The difference forces acting on system are inertia forces damping force spring force and external force All these forces are represented on force polygon shown below Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these two forces

( ) ( )22 xckx ω+=

59

This is the expression for force transmissibility in dimensional form To represent same in the non dimensional form divide numerator and denominator by k

2

n

22

n

2

n

o

tr

21

21

FF

TR

ωω

ξ+

ωω

minus

ωω

ξ+

==

PHASE DIFFERNECE BETWEEN EXTERNAL FORCE AND DISPLACEMENT Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these

ωω

minus

ωω

ξ=

ωminusω

=φ minusminus2

n

n1

2

1

1

2tan

mKC

tan

The phase difference between force transmitted to the foundation Ftr to the displacement

kc

KxxC

tanω

=ω

=α or

ωω

ξ=

ω=α minusminus

n

11 2tanKx

xCtan

The phase difference between external force Fo and force transmitted to foundationis β

60

β= αminusφ =

ωω

minus

ωω

ξminus

2

n

n1

1

2tan -

ωω

ξminus

n

1 2tan

Displacement Transmissibility It is the ratio of amplitude of steady state vibration of machine to that of the base motion Forced vibration due to support motion Consider a spring mass damper system subjected to harmonic excitation )tsin(Yy ω= at the base as shown in figure Let x be the absolute displacement of the mass at any instant of time t The different forces acting on system are shown in free body diagram Applying Newtonrsquos Law of motion Ma = sumF

)yx(k)yx(cxm minusminusminusminus= ampampampamp

0)yx(k)yx(cxm =minus+minus+ ampampampamp

61

Frequency plot of transmissibility versus frequency ratio

62

The plot of transmissibility versus frequency ratio is as shown in figure The following points can be noted from it

i) The value of transmissibility is unity at frequency ratio

ωω

n=0 and 2

ii) For an undamped system ( )ξ =0 transmissibility tends to infin at resonance iii) The value of transmissibility is less than unity for the value of frequency ratio

greater than 2

iv) For frequency ratio

ωω

nlt 2 smaller damping ratio lead to larger values of

transmissibility on the other hand for frequency ratio

ωω

ngt 2 smaller damping

ratio leads to smaller value of transmissibility

The frequency plot of transmissibility (TR) versus frequency ratio

ωω

n can be

broadly divided into two zones

1

ωω

n= 0 to 2

2

ωω

n= 2 to infin

CASE 1

When

ωω

n= 0 to 2

In this zone the value of transmissibility is always greater than 1 for all values of damping ratio In this zone the transmissibility decreases with increase in damping Hence we can conclude that in this zone damping is useful in isolating vibration forces being transmitted to the support CASE 2

When

ωω

n= 2 to infin

In this zone the value of transmissibility is less than 1 for all values of damping ratio In this zone the transmissibility increases with increase in damping Hence we can conclude that in this zone damping will adversely affect in isolating vibratory forces

63

Tutorial problems on Forced Vibration 1 Machine of mass 75kg is mounted on spring of stiffness 12000 Nm with an assumed damping factor ξ =02 piston within the machine of mass 2 kg has a reciprocating motion with a stroke of 75 cm and speed of 15 cyclessecond Assuming the motion of piston to be harmonic determine amplitude of machine ( X) Transmissibility and force transmitted to foundation phase angle of transmitted force wrt to exciting force 2 Motor weighs 600N is mounted on a simple beam that has a spring rate of 400 Ncm The motor armature base 160 N and has an eccentricity of 0005cm What will be the amplitude of motion when it runs at 1760 RPM Neglect damping the deflection of armature shaft and weight of the beam 3 A Heavy machine of mass 300 kg is supported on resilient foundation the static deflection of foundation due to weight of machine is found to be 75mm It is observed that the machine vibrates with an amplitude of 10mm when the base of the foundation is subjected to harmonic oscillation at the undampeds natural frequency of system with an amplitude of 25mm Find the damping constant of foundation Ii) dynamic force amplitude on the base iii) amplitude of the displacement of machine relative to base 4 An air compressor of mass 450 kg operates at a constant speed of 1750 RPM The reciprocating masses are 10 kg the crank radius is 100mm If the damper for mounting introduces a damping factor 015 i) specify the springs of mounting such that only 20 of the unbalanced force is transmitted to the foundation ii) determine the magnitude of the transmitted force bull Repeat the above problem by assuming that there is no damping in system

5 A body of a car having mass 1500 kg and is mounted on 4 equal springs which deflect through 225 cm under the weight of the body A total damping forces the 4 shock absorbers is 46 N at a velocity of 1cmsec The car is placed with all the four wheels on a test platform which is move up and down at resonant speed With an amplitude of 25cm Find the amplitude of the car body on its springs assuming its CG to be in the centre of wheel base 6A 75 kg machine which is mounted on springs of stiffness 116x105 Nm with the damping of 20=ξ A 2kg piston within the machine as reciprocating motion with a stroke of 008m and speed of 3000 RPM Assuming the motion of piston to be harmonic determine the amplitude of vibration of machine

7 The spring of an automobile are compressed 10cm under its own weight Find the critical speed when a trailer is traveling over road with a people approximaterly by a sine wave of amplitude 75 cm and wave length 15m What will be the amplitude of vibration at 64kmhour 8 An eccentric mass excites us used to determine vibratory characteristics of a structure of mass 25kg as given in figure At a speed of 1000 RPM the eccentric mass to beat top position at the instant the structure was moved upwards through its static equilibrium position and corresponding amplitude was 20mm The eccentric mass is 05kg at an eccentricity of 10mm Determine i) Undamped natural frequency of the system ii) Damping ratio iii) Steady state amplitude and phase angle at 1500 RPM iv) Force transmitted to the ground at 1200 RPM

M=25kg

C

K

Eccentric mass

64

65

CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Support motion Consider a spring mass damper system as shown in figure Let it excited by motion

of support let y be the harmonic displacement of support From Newtonrsquos second law of motion

)yx(k)yx(cxm minusminusminusminus= ampampampamp

)1(0)yx(k)yx(cxm minusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminus=minus+minus+ ampampampamp

Let z be the relative displacement of the mass with respect to support then

Z=x-y

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

Zero frequency deflection (Xst) It is deflection observed at zero frequency ( when the external frequency 0=ω

( ) kF

001

kF

x o

22

o

st =

+minus=

Magnification factor (M) It is defined as the ratio of steady state amplitude to zero frequency deflection xst

M = arX

X=

A plot of magnification factor versus frequency ratio

damping factor ( )ξ is called frequency plot of a harmon All the frequency terms starts from magnification factor of damping ratio ( )ξ The amplitude of vibration at any on the value of damping The amplitude will be higher for higher damping The amplitude of vibration at reso

amplitude tends to zero as

ωω

n tends to infinity

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

Freq

ζ =

ζ = 1

Magnification Factor M = XXst

kFo

51

ωω

nfor varying values of

ically excite vibrating system

M=1 what ever may be value particular frequency depends

for lesser damping and lesser nance will be very high The

uency Ratio r = (ωωn)

ζ = 0 0

ζ = 025 ζ = 0375 ζ = 0 5

ζ = 0707

ζ = 2

52

Plot of phase angle φ versus frequency ratio

ωω

n is shown in the

figure The phase angle φ is zero for all the values of damping ratio

at

ωω

n=0 The phase

angle φ is 90 for all the values of damping ratio

at

ωω

n=1 ie at

resonance The phase angle between o to 90 deg for all the values of

ωω

n= 0 to 1 and

between 90 deg to 180 deg for all the values of

ωω

ngt1

Class work Derive an expression for frequency at peak amplitude

2np 21 ξminusω=ω=ω is the required equation

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

53

Derive expression amplitude at resonance ( In dimensional form) In dimensional form We know that steady state amplitude

( ) ( )

ω+ωminus=

222

o

Cmk

Fx Equation in dimensional form

We know that at resonance mk

n =ω=ω

( )

ω+

minus

=2

n

2

oR

Cmk

mk

Fx

( )

ω=

2n

oR

C

Fx

( )

ω=

n

oR C

Fx

Class work Derive an expression for amplitude at resonance ( in non dimensional form)

Tutorial Problems on Forced Vibration 1A weight of 60N is suspended by a spring stiffness 12 kNm is forced to vibrate by a harmonic force of 10N assuming viscous damping of 0086 kN-sm Find i) resonance frequency ii) amplitude at resonance iii) phase angle at resonance iv) frequency corresponding to peak amplitude 2 A periodic torque having a maximum value of 05888Nm at a frequency of 4 rads is impressed upon a flywheel suspended from a wire The wheel has mass moment of inertia 012 kgm2 and the wire has stiffness 1176 Nmrad A viscous dashpot applies damping couple of 0784 Nm an angular velocity of 2 rads calculate i)maximum angular displacement from rest position ii) maximum couple applied to dashpot Iii) angle by which the angular displacement lags the torque A mass of 1019 kg is suspended from a spring stiffness 20 Nmm viscous damping causes the amplitude to decrease to 110th of initial value in 4 complete cycle If a periodic force of 20 cos(50 t) is applied to the mass Find the amplitude of forced vibration What would be the amplitude of oscillation if the applied force coincides with the natural period

m=1019 kg

[ ] t50cos 30F =

54

55

ROTATING UNBALANCE Unbalance in rotating machinery is a common source of vibration excitation Consider a machine of mass m having mounted on a foundation of stiffness k and damping coefficient C as shown in figure The unbalance is represented by an eccentric rotating mass m with eccentricity (e) and having a angular velocity (ω) let x be the displacement of non rotating mass from the static equilibrium position The displacement of rotating mass m from the static equilibrium is ( ) tsin( ex ω+ Applying Newtonrsquos law of motion we have

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

56

RECIPROCATING UNBALANCE Unbalance in reciprocating machinery is a common source of vibration excitation Consider a machine of mass M having mounted on a foundation of stiffness K and damping coefficient C as shown in figure The unbalance is represented by a reciprocating mass (m) having crank radus (e) and connecting rod of length (l) Let ωωωω be angular velocity of the crank Let x be the displacement of non reciprocating mass from the static equilibrium position at any instant of time The displacement of reciprocating mass m from the static equilibrium position is

)2sin(sin +++ tle

tex ωωωωωωωω

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

57

Note The equations obtained in reciprocating unbalance is same as the rotating unbalance and having the following differences Rotating Unbalance Reciprocating Unbalance Here mass of the rotating element is considered

Here mass of reciprocating element is considered

Here the eccentricity (e) which is the distance between the centre of rotation and CG

Here the eccentricity e is the crank radius and crank radius = stroke length2

The frequency plot of non dimensional factor( ( )me

MX versus frequency ratio

ωω

n for various values of ( )ξξξξ is shown in figure This frequency plot is same for

the rotating and reciprocating unbalance with the following points are noted from the diagram

1 Damping ratio plays very important role during resonance in controlling the amplitude

2 When

ωω

n=0 then ( )me

MX =0 for all the values of damping ratio ( )ξξξξ

3 When

ωω

nis higher than ( )me

MX tends to unity

4 At very high speeds ie at very high frequency ratio

ωω

n there is no effect

of damping The frequency plot of phase angle versus the frequency ratio is given as shown in figure The explanation is same as in forced harmonic vibration

58

VIBRATION ISOLATION

Vibratory forces and motions generated by machine and other caused are unavoidable Howeever their effect on the other dynamical systems can be minimized by proper isolator design An isolation system is either to protect o delicate machine from excessive vibration transmitted to it from its supporting structure or to prevent vibratory forces and motions generated by machines from being transmitted to its surroundings The effectiveness of isolation may be measured in terms of forces and motion transmitted The first type is known as force isolation and the second type is known as motion isolation

Force isolation (Force Transmissibility) When vibratory forces generated by machine are isolated the effectiveness of forced isolation can be made by defining the term transmissibility The term transmissibility is defined as the ratio of force transmitted to the foundation to that of the external impressed force

Force Transmissibility = o

tr

FF

force Impressed Externalfoundation toted transmitForce

=

Consider viscous damped spring mass system as shown in figure Upon this system an external harmonic force ) t (Sin Fo ω is applied The difference forces acting on system are inertia forces damping force spring force and external force All these forces are represented on force polygon shown below Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these two forces

( ) ( )22 xckx ω+=

59

This is the expression for force transmissibility in dimensional form To represent same in the non dimensional form divide numerator and denominator by k

2

n

22

n

2

n

o

tr

21

21

FF

TR

ωω

ξ+

ωω

minus

ωω

ξ+

==

PHASE DIFFERNECE BETWEEN EXTERNAL FORCE AND DISPLACEMENT Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these

ωω

minus

ωω

ξ=

ωminusω

=φ minusminus2

n

n1

2

1

1

2tan

mKC

tan

The phase difference between force transmitted to the foundation Ftr to the displacement

kc

KxxC

tanω

=ω

=α or

ωω

ξ=

ω=α minusminus

n

11 2tanKx

xCtan

The phase difference between external force Fo and force transmitted to foundationis β

60

β= αminusφ =

ωω

minus

ωω

ξminus

2

n

n1

1

2tan -

ωω

ξminus

n

1 2tan

Displacement Transmissibility It is the ratio of amplitude of steady state vibration of machine to that of the base motion Forced vibration due to support motion Consider a spring mass damper system subjected to harmonic excitation )tsin(Yy ω= at the base as shown in figure Let x be the absolute displacement of the mass at any instant of time t The different forces acting on system are shown in free body diagram Applying Newtonrsquos Law of motion Ma = sumF

)yx(k)yx(cxm minusminusminusminus= ampampampamp

0)yx(k)yx(cxm =minus+minus+ ampampampamp

61

Frequency plot of transmissibility versus frequency ratio

62

The plot of transmissibility versus frequency ratio is as shown in figure The following points can be noted from it

i) The value of transmissibility is unity at frequency ratio

ωω

n=0 and 2

ii) For an undamped system ( )ξ =0 transmissibility tends to infin at resonance iii) The value of transmissibility is less than unity for the value of frequency ratio

greater than 2

iv) For frequency ratio

ωω

nlt 2 smaller damping ratio lead to larger values of

transmissibility on the other hand for frequency ratio

ωω

ngt 2 smaller damping

ratio leads to smaller value of transmissibility

The frequency plot of transmissibility (TR) versus frequency ratio

ωω

n can be

broadly divided into two zones

1

ωω

n= 0 to 2

2

ωω

n= 2 to infin

CASE 1

When

ωω

n= 0 to 2

In this zone the value of transmissibility is always greater than 1 for all values of damping ratio In this zone the transmissibility decreases with increase in damping Hence we can conclude that in this zone damping is useful in isolating vibration forces being transmitted to the support CASE 2

When

ωω

n= 2 to infin

In this zone the value of transmissibility is less than 1 for all values of damping ratio In this zone the transmissibility increases with increase in damping Hence we can conclude that in this zone damping will adversely affect in isolating vibratory forces

63

Tutorial problems on Forced Vibration 1 Machine of mass 75kg is mounted on spring of stiffness 12000 Nm with an assumed damping factor ξ =02 piston within the machine of mass 2 kg has a reciprocating motion with a stroke of 75 cm and speed of 15 cyclessecond Assuming the motion of piston to be harmonic determine amplitude of machine ( X) Transmissibility and force transmitted to foundation phase angle of transmitted force wrt to exciting force 2 Motor weighs 600N is mounted on a simple beam that has a spring rate of 400 Ncm The motor armature base 160 N and has an eccentricity of 0005cm What will be the amplitude of motion when it runs at 1760 RPM Neglect damping the deflection of armature shaft and weight of the beam 3 A Heavy machine of mass 300 kg is supported on resilient foundation the static deflection of foundation due to weight of machine is found to be 75mm It is observed that the machine vibrates with an amplitude of 10mm when the base of the foundation is subjected to harmonic oscillation at the undampeds natural frequency of system with an amplitude of 25mm Find the damping constant of foundation Ii) dynamic force amplitude on the base iii) amplitude of the displacement of machine relative to base 4 An air compressor of mass 450 kg operates at a constant speed of 1750 RPM The reciprocating masses are 10 kg the crank radius is 100mm If the damper for mounting introduces a damping factor 015 i) specify the springs of mounting such that only 20 of the unbalanced force is transmitted to the foundation ii) determine the magnitude of the transmitted force bull Repeat the above problem by assuming that there is no damping in system

5 A body of a car having mass 1500 kg and is mounted on 4 equal springs which deflect through 225 cm under the weight of the body A total damping forces the 4 shock absorbers is 46 N at a velocity of 1cmsec The car is placed with all the four wheels on a test platform which is move up and down at resonant speed With an amplitude of 25cm Find the amplitude of the car body on its springs assuming its CG to be in the centre of wheel base 6A 75 kg machine which is mounted on springs of stiffness 116x105 Nm with the damping of 20=ξ A 2kg piston within the machine as reciprocating motion with a stroke of 008m and speed of 3000 RPM Assuming the motion of piston to be harmonic determine the amplitude of vibration of machine

7 The spring of an automobile are compressed 10cm under its own weight Find the critical speed when a trailer is traveling over road with a people approximaterly by a sine wave of amplitude 75 cm and wave length 15m What will be the amplitude of vibration at 64kmhour 8 An eccentric mass excites us used to determine vibratory characteristics of a structure of mass 25kg as given in figure At a speed of 1000 RPM the eccentric mass to beat top position at the instant the structure was moved upwards through its static equilibrium position and corresponding amplitude was 20mm The eccentric mass is 05kg at an eccentricity of 10mm Determine i) Undamped natural frequency of the system ii) Damping ratio iii) Steady state amplitude and phase angle at 1500 RPM iv) Force transmitted to the ground at 1200 RPM

M=25kg

C

K

Eccentric mass

64

65

CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Support motion Consider a spring mass damper system as shown in figure Let it excited by motion

of support let y be the harmonic displacement of support From Newtonrsquos second law of motion

)yx(k)yx(cxm minusminusminusminus= ampampampamp

)1(0)yx(k)yx(cxm minusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminus=minus+minus+ ampampampamp

Let z be the relative displacement of the mass with respect to support then

Z=x-y

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

52

Plot of phase angle φ versus frequency ratio

ωω

n is shown in the

figure The phase angle φ is zero for all the values of damping ratio

at

ωω

n=0 The phase

angle φ is 90 for all the values of damping ratio

at

ωω

n=1 ie at

resonance The phase angle between o to 90 deg for all the values of

ωω

n= 0 to 1 and

between 90 deg to 180 deg for all the values of

ωω

ngt1

Class work Derive an expression for frequency at peak amplitude

2np 21 ξminusω=ω=ω is the required equation

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

53

Derive expression amplitude at resonance ( In dimensional form) In dimensional form We know that steady state amplitude

( ) ( )

ω+ωminus=

222

o

Cmk

Fx Equation in dimensional form

We know that at resonance mk

n =ω=ω

( )

ω+

minus

=2

n

2

oR

Cmk

mk

Fx

( )

ω=

2n

oR

C

Fx

( )

ω=

n

oR C

Fx

Class work Derive an expression for amplitude at resonance ( in non dimensional form)

Tutorial Problems on Forced Vibration 1A weight of 60N is suspended by a spring stiffness 12 kNm is forced to vibrate by a harmonic force of 10N assuming viscous damping of 0086 kN-sm Find i) resonance frequency ii) amplitude at resonance iii) phase angle at resonance iv) frequency corresponding to peak amplitude 2 A periodic torque having a maximum value of 05888Nm at a frequency of 4 rads is impressed upon a flywheel suspended from a wire The wheel has mass moment of inertia 012 kgm2 and the wire has stiffness 1176 Nmrad A viscous dashpot applies damping couple of 0784 Nm an angular velocity of 2 rads calculate i)maximum angular displacement from rest position ii) maximum couple applied to dashpot Iii) angle by which the angular displacement lags the torque A mass of 1019 kg is suspended from a spring stiffness 20 Nmm viscous damping causes the amplitude to decrease to 110th of initial value in 4 complete cycle If a periodic force of 20 cos(50 t) is applied to the mass Find the amplitude of forced vibration What would be the amplitude of oscillation if the applied force coincides with the natural period

m=1019 kg

[ ] t50cos 30F =

54

55

ROTATING UNBALANCE Unbalance in rotating machinery is a common source of vibration excitation Consider a machine of mass m having mounted on a foundation of stiffness k and damping coefficient C as shown in figure The unbalance is represented by an eccentric rotating mass m with eccentricity (e) and having a angular velocity (ω) let x be the displacement of non rotating mass from the static equilibrium position The displacement of rotating mass m from the static equilibrium is ( ) tsin( ex ω+ Applying Newtonrsquos law of motion we have

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

56

RECIPROCATING UNBALANCE Unbalance in reciprocating machinery is a common source of vibration excitation Consider a machine of mass M having mounted on a foundation of stiffness K and damping coefficient C as shown in figure The unbalance is represented by a reciprocating mass (m) having crank radus (e) and connecting rod of length (l) Let ωωωω be angular velocity of the crank Let x be the displacement of non reciprocating mass from the static equilibrium position at any instant of time The displacement of reciprocating mass m from the static equilibrium position is

)2sin(sin +++ tle

tex ωωωωωωωω

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

57

Note The equations obtained in reciprocating unbalance is same as the rotating unbalance and having the following differences Rotating Unbalance Reciprocating Unbalance Here mass of the rotating element is considered

Here mass of reciprocating element is considered

Here the eccentricity (e) which is the distance between the centre of rotation and CG

Here the eccentricity e is the crank radius and crank radius = stroke length2

The frequency plot of non dimensional factor( ( )me

MX versus frequency ratio

ωω

n for various values of ( )ξξξξ is shown in figure This frequency plot is same for

the rotating and reciprocating unbalance with the following points are noted from the diagram

1 Damping ratio plays very important role during resonance in controlling the amplitude

2 When

ωω

n=0 then ( )me

MX =0 for all the values of damping ratio ( )ξξξξ

3 When

ωω

nis higher than ( )me

MX tends to unity

4 At very high speeds ie at very high frequency ratio

ωω

n there is no effect

of damping The frequency plot of phase angle versus the frequency ratio is given as shown in figure The explanation is same as in forced harmonic vibration

58

VIBRATION ISOLATION

Vibratory forces and motions generated by machine and other caused are unavoidable Howeever their effect on the other dynamical systems can be minimized by proper isolator design An isolation system is either to protect o delicate machine from excessive vibration transmitted to it from its supporting structure or to prevent vibratory forces and motions generated by machines from being transmitted to its surroundings The effectiveness of isolation may be measured in terms of forces and motion transmitted The first type is known as force isolation and the second type is known as motion isolation

Force isolation (Force Transmissibility) When vibratory forces generated by machine are isolated the effectiveness of forced isolation can be made by defining the term transmissibility The term transmissibility is defined as the ratio of force transmitted to the foundation to that of the external impressed force

Force Transmissibility = o

tr

FF

force Impressed Externalfoundation toted transmitForce

=

Consider viscous damped spring mass system as shown in figure Upon this system an external harmonic force ) t (Sin Fo ω is applied The difference forces acting on system are inertia forces damping force spring force and external force All these forces are represented on force polygon shown below Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these two forces

( ) ( )22 xckx ω+=

59

This is the expression for force transmissibility in dimensional form To represent same in the non dimensional form divide numerator and denominator by k

2

n

22

n

2

n

o

tr

21

21

FF

TR

ωω

ξ+

ωω

minus

ωω

ξ+

==

PHASE DIFFERNECE BETWEEN EXTERNAL FORCE AND DISPLACEMENT Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these

ωω

minus

ωω

ξ=

ωminusω

=φ minusminus2

n

n1

2

1

1

2tan

mKC

tan

The phase difference between force transmitted to the foundation Ftr to the displacement

kc

KxxC

tanω

=ω

=α or

ωω

ξ=

ω=α minusminus

n

11 2tanKx

xCtan

The phase difference between external force Fo and force transmitted to foundationis β

60

β= αminusφ =

ωω

minus

ωω

ξminus

2

n

n1

1

2tan -

ωω

ξminus

n

1 2tan

Displacement Transmissibility It is the ratio of amplitude of steady state vibration of machine to that of the base motion Forced vibration due to support motion Consider a spring mass damper system subjected to harmonic excitation )tsin(Yy ω= at the base as shown in figure Let x be the absolute displacement of the mass at any instant of time t The different forces acting on system are shown in free body diagram Applying Newtonrsquos Law of motion Ma = sumF

)yx(k)yx(cxm minusminusminusminus= ampampampamp

0)yx(k)yx(cxm =minus+minus+ ampampampamp

61

Frequency plot of transmissibility versus frequency ratio

62

The plot of transmissibility versus frequency ratio is as shown in figure The following points can be noted from it

i) The value of transmissibility is unity at frequency ratio

ωω

n=0 and 2

ii) For an undamped system ( )ξ =0 transmissibility tends to infin at resonance iii) The value of transmissibility is less than unity for the value of frequency ratio

greater than 2

iv) For frequency ratio

ωω

nlt 2 smaller damping ratio lead to larger values of

transmissibility on the other hand for frequency ratio

ωω

ngt 2 smaller damping

ratio leads to smaller value of transmissibility

The frequency plot of transmissibility (TR) versus frequency ratio

ωω

n can be

broadly divided into two zones

1

ωω

n= 0 to 2

2

ωω

n= 2 to infin

CASE 1

When

ωω

n= 0 to 2

In this zone the value of transmissibility is always greater than 1 for all values of damping ratio In this zone the transmissibility decreases with increase in damping Hence we can conclude that in this zone damping is useful in isolating vibration forces being transmitted to the support CASE 2

When

ωω

n= 2 to infin

In this zone the value of transmissibility is less than 1 for all values of damping ratio In this zone the transmissibility increases with increase in damping Hence we can conclude that in this zone damping will adversely affect in isolating vibratory forces

63

Tutorial problems on Forced Vibration 1 Machine of mass 75kg is mounted on spring of stiffness 12000 Nm with an assumed damping factor ξ =02 piston within the machine of mass 2 kg has a reciprocating motion with a stroke of 75 cm and speed of 15 cyclessecond Assuming the motion of piston to be harmonic determine amplitude of machine ( X) Transmissibility and force transmitted to foundation phase angle of transmitted force wrt to exciting force 2 Motor weighs 600N is mounted on a simple beam that has a spring rate of 400 Ncm The motor armature base 160 N and has an eccentricity of 0005cm What will be the amplitude of motion when it runs at 1760 RPM Neglect damping the deflection of armature shaft and weight of the beam 3 A Heavy machine of mass 300 kg is supported on resilient foundation the static deflection of foundation due to weight of machine is found to be 75mm It is observed that the machine vibrates with an amplitude of 10mm when the base of the foundation is subjected to harmonic oscillation at the undampeds natural frequency of system with an amplitude of 25mm Find the damping constant of foundation Ii) dynamic force amplitude on the base iii) amplitude of the displacement of machine relative to base 4 An air compressor of mass 450 kg operates at a constant speed of 1750 RPM The reciprocating masses are 10 kg the crank radius is 100mm If the damper for mounting introduces a damping factor 015 i) specify the springs of mounting such that only 20 of the unbalanced force is transmitted to the foundation ii) determine the magnitude of the transmitted force bull Repeat the above problem by assuming that there is no damping in system

5 A body of a car having mass 1500 kg and is mounted on 4 equal springs which deflect through 225 cm under the weight of the body A total damping forces the 4 shock absorbers is 46 N at a velocity of 1cmsec The car is placed with all the four wheels on a test platform which is move up and down at resonant speed With an amplitude of 25cm Find the amplitude of the car body on its springs assuming its CG to be in the centre of wheel base 6A 75 kg machine which is mounted on springs of stiffness 116x105 Nm with the damping of 20=ξ A 2kg piston within the machine as reciprocating motion with a stroke of 008m and speed of 3000 RPM Assuming the motion of piston to be harmonic determine the amplitude of vibration of machine

7 The spring of an automobile are compressed 10cm under its own weight Find the critical speed when a trailer is traveling over road with a people approximaterly by a sine wave of amplitude 75 cm and wave length 15m What will be the amplitude of vibration at 64kmhour 8 An eccentric mass excites us used to determine vibratory characteristics of a structure of mass 25kg as given in figure At a speed of 1000 RPM the eccentric mass to beat top position at the instant the structure was moved upwards through its static equilibrium position and corresponding amplitude was 20mm The eccentric mass is 05kg at an eccentricity of 10mm Determine i) Undamped natural frequency of the system ii) Damping ratio iii) Steady state amplitude and phase angle at 1500 RPM iv) Force transmitted to the ground at 1200 RPM

M=25kg

C

K

Eccentric mass

64

65

CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Support motion Consider a spring mass damper system as shown in figure Let it excited by motion

of support let y be the harmonic displacement of support From Newtonrsquos second law of motion

)yx(k)yx(cxm minusminusminusminus= ampampampamp

)1(0)yx(k)yx(cxm minusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminus=minus+minus+ ampampampamp

Let z be the relative displacement of the mass with respect to support then

Z=x-y

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

53

Derive expression amplitude at resonance ( In dimensional form) In dimensional form We know that steady state amplitude

( ) ( )

ω+ωminus=

222

o

Cmk

Fx Equation in dimensional form

We know that at resonance mk

n =ω=ω

( )

ω+

minus

=2

n

2

oR

Cmk

mk

Fx

( )

ω=

2n

oR

C

Fx

( )

ω=

n

oR C

Fx

Class work Derive an expression for amplitude at resonance ( in non dimensional form)

Tutorial Problems on Forced Vibration 1A weight of 60N is suspended by a spring stiffness 12 kNm is forced to vibrate by a harmonic force of 10N assuming viscous damping of 0086 kN-sm Find i) resonance frequency ii) amplitude at resonance iii) phase angle at resonance iv) frequency corresponding to peak amplitude 2 A periodic torque having a maximum value of 05888Nm at a frequency of 4 rads is impressed upon a flywheel suspended from a wire The wheel has mass moment of inertia 012 kgm2 and the wire has stiffness 1176 Nmrad A viscous dashpot applies damping couple of 0784 Nm an angular velocity of 2 rads calculate i)maximum angular displacement from rest position ii) maximum couple applied to dashpot Iii) angle by which the angular displacement lags the torque A mass of 1019 kg is suspended from a spring stiffness 20 Nmm viscous damping causes the amplitude to decrease to 110th of initial value in 4 complete cycle If a periodic force of 20 cos(50 t) is applied to the mass Find the amplitude of forced vibration What would be the amplitude of oscillation if the applied force coincides with the natural period

m=1019 kg

[ ] t50cos 30F =

54

55

ROTATING UNBALANCE Unbalance in rotating machinery is a common source of vibration excitation Consider a machine of mass m having mounted on a foundation of stiffness k and damping coefficient C as shown in figure The unbalance is represented by an eccentric rotating mass m with eccentricity (e) and having a angular velocity (ω) let x be the displacement of non rotating mass from the static equilibrium position The displacement of rotating mass m from the static equilibrium is ( ) tsin( ex ω+ Applying Newtonrsquos law of motion we have

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

56

RECIPROCATING UNBALANCE Unbalance in reciprocating machinery is a common source of vibration excitation Consider a machine of mass M having mounted on a foundation of stiffness K and damping coefficient C as shown in figure The unbalance is represented by a reciprocating mass (m) having crank radus (e) and connecting rod of length (l) Let ωωωω be angular velocity of the crank Let x be the displacement of non reciprocating mass from the static equilibrium position at any instant of time The displacement of reciprocating mass m from the static equilibrium position is

)2sin(sin +++ tle

tex ωωωωωωωω

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

57

Note The equations obtained in reciprocating unbalance is same as the rotating unbalance and having the following differences Rotating Unbalance Reciprocating Unbalance Here mass of the rotating element is considered

Here mass of reciprocating element is considered

Here the eccentricity (e) which is the distance between the centre of rotation and CG

Here the eccentricity e is the crank radius and crank radius = stroke length2

The frequency plot of non dimensional factor( ( )me

MX versus frequency ratio

ωω

n for various values of ( )ξξξξ is shown in figure This frequency plot is same for

the rotating and reciprocating unbalance with the following points are noted from the diagram

1 Damping ratio plays very important role during resonance in controlling the amplitude

2 When

ωω

n=0 then ( )me

MX =0 for all the values of damping ratio ( )ξξξξ

3 When

ωω

nis higher than ( )me

MX tends to unity

4 At very high speeds ie at very high frequency ratio

ωω

n there is no effect

of damping The frequency plot of phase angle versus the frequency ratio is given as shown in figure The explanation is same as in forced harmonic vibration

58

VIBRATION ISOLATION

Vibratory forces and motions generated by machine and other caused are unavoidable Howeever their effect on the other dynamical systems can be minimized by proper isolator design An isolation system is either to protect o delicate machine from excessive vibration transmitted to it from its supporting structure or to prevent vibratory forces and motions generated by machines from being transmitted to its surroundings The effectiveness of isolation may be measured in terms of forces and motion transmitted The first type is known as force isolation and the second type is known as motion isolation

Force isolation (Force Transmissibility) When vibratory forces generated by machine are isolated the effectiveness of forced isolation can be made by defining the term transmissibility The term transmissibility is defined as the ratio of force transmitted to the foundation to that of the external impressed force

Force Transmissibility = o

tr

FF

force Impressed Externalfoundation toted transmitForce

=

Consider viscous damped spring mass system as shown in figure Upon this system an external harmonic force ) t (Sin Fo ω is applied The difference forces acting on system are inertia forces damping force spring force and external force All these forces are represented on force polygon shown below Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these two forces

( ) ( )22 xckx ω+=

59

This is the expression for force transmissibility in dimensional form To represent same in the non dimensional form divide numerator and denominator by k

2

n

22

n

2

n

o

tr

21

21

FF

TR

ωω

ξ+

ωω

minus

ωω

ξ+

==

PHASE DIFFERNECE BETWEEN EXTERNAL FORCE AND DISPLACEMENT Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these

ωω

minus

ωω

ξ=

ωminusω

=φ minusminus2

n

n1

2

1

1

2tan

mKC

tan

The phase difference between force transmitted to the foundation Ftr to the displacement

kc

KxxC

tanω

=ω

=α or

ωω

ξ=

ω=α minusminus

n

11 2tanKx

xCtan

The phase difference between external force Fo and force transmitted to foundationis β

60

β= αminusφ =

ωω

minus

ωω

ξminus

2

n

n1

1

2tan -

ωω

ξminus

n

1 2tan

Displacement Transmissibility It is the ratio of amplitude of steady state vibration of machine to that of the base motion Forced vibration due to support motion Consider a spring mass damper system subjected to harmonic excitation )tsin(Yy ω= at the base as shown in figure Let x be the absolute displacement of the mass at any instant of time t The different forces acting on system are shown in free body diagram Applying Newtonrsquos Law of motion Ma = sumF

)yx(k)yx(cxm minusminusminusminus= ampampampamp

0)yx(k)yx(cxm =minus+minus+ ampampampamp

61

Frequency plot of transmissibility versus frequency ratio

62

The plot of transmissibility versus frequency ratio is as shown in figure The following points can be noted from it

i) The value of transmissibility is unity at frequency ratio

ωω

n=0 and 2

ii) For an undamped system ( )ξ =0 transmissibility tends to infin at resonance iii) The value of transmissibility is less than unity for the value of frequency ratio

greater than 2

iv) For frequency ratio

ωω

nlt 2 smaller damping ratio lead to larger values of

transmissibility on the other hand for frequency ratio

ωω

ngt 2 smaller damping

ratio leads to smaller value of transmissibility

The frequency plot of transmissibility (TR) versus frequency ratio

ωω

n can be

broadly divided into two zones

1

ωω

n= 0 to 2

2

ωω

n= 2 to infin

CASE 1

When

ωω

n= 0 to 2

In this zone the value of transmissibility is always greater than 1 for all values of damping ratio In this zone the transmissibility decreases with increase in damping Hence we can conclude that in this zone damping is useful in isolating vibration forces being transmitted to the support CASE 2

When

ωω

n= 2 to infin

In this zone the value of transmissibility is less than 1 for all values of damping ratio In this zone the transmissibility increases with increase in damping Hence we can conclude that in this zone damping will adversely affect in isolating vibratory forces

63

Tutorial problems on Forced Vibration 1 Machine of mass 75kg is mounted on spring of stiffness 12000 Nm with an assumed damping factor ξ =02 piston within the machine of mass 2 kg has a reciprocating motion with a stroke of 75 cm and speed of 15 cyclessecond Assuming the motion of piston to be harmonic determine amplitude of machine ( X) Transmissibility and force transmitted to foundation phase angle of transmitted force wrt to exciting force 2 Motor weighs 600N is mounted on a simple beam that has a spring rate of 400 Ncm The motor armature base 160 N and has an eccentricity of 0005cm What will be the amplitude of motion when it runs at 1760 RPM Neglect damping the deflection of armature shaft and weight of the beam 3 A Heavy machine of mass 300 kg is supported on resilient foundation the static deflection of foundation due to weight of machine is found to be 75mm It is observed that the machine vibrates with an amplitude of 10mm when the base of the foundation is subjected to harmonic oscillation at the undampeds natural frequency of system with an amplitude of 25mm Find the damping constant of foundation Ii) dynamic force amplitude on the base iii) amplitude of the displacement of machine relative to base 4 An air compressor of mass 450 kg operates at a constant speed of 1750 RPM The reciprocating masses are 10 kg the crank radius is 100mm If the damper for mounting introduces a damping factor 015 i) specify the springs of mounting such that only 20 of the unbalanced force is transmitted to the foundation ii) determine the magnitude of the transmitted force bull Repeat the above problem by assuming that there is no damping in system

5 A body of a car having mass 1500 kg and is mounted on 4 equal springs which deflect through 225 cm under the weight of the body A total damping forces the 4 shock absorbers is 46 N at a velocity of 1cmsec The car is placed with all the four wheels on a test platform which is move up and down at resonant speed With an amplitude of 25cm Find the amplitude of the car body on its springs assuming its CG to be in the centre of wheel base 6A 75 kg machine which is mounted on springs of stiffness 116x105 Nm with the damping of 20=ξ A 2kg piston within the machine as reciprocating motion with a stroke of 008m and speed of 3000 RPM Assuming the motion of piston to be harmonic determine the amplitude of vibration of machine

7 The spring of an automobile are compressed 10cm under its own weight Find the critical speed when a trailer is traveling over road with a people approximaterly by a sine wave of amplitude 75 cm and wave length 15m What will be the amplitude of vibration at 64kmhour 8 An eccentric mass excites us used to determine vibratory characteristics of a structure of mass 25kg as given in figure At a speed of 1000 RPM the eccentric mass to beat top position at the instant the structure was moved upwards through its static equilibrium position and corresponding amplitude was 20mm The eccentric mass is 05kg at an eccentricity of 10mm Determine i) Undamped natural frequency of the system ii) Damping ratio iii) Steady state amplitude and phase angle at 1500 RPM iv) Force transmitted to the ground at 1200 RPM

M=25kg

C

K

Eccentric mass

64

65

CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Support motion Consider a spring mass damper system as shown in figure Let it excited by motion

of support let y be the harmonic displacement of support From Newtonrsquos second law of motion

)yx(k)yx(cxm minusminusminusminus= ampampampamp

)1(0)yx(k)yx(cxm minusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminus=minus+minus+ ampampampamp

Let z be the relative displacement of the mass with respect to support then

Z=x-y

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

Tutorial Problems on Forced Vibration 1A weight of 60N is suspended by a spring stiffness 12 kNm is forced to vibrate by a harmonic force of 10N assuming viscous damping of 0086 kN-sm Find i) resonance frequency ii) amplitude at resonance iii) phase angle at resonance iv) frequency corresponding to peak amplitude 2 A periodic torque having a maximum value of 05888Nm at a frequency of 4 rads is impressed upon a flywheel suspended from a wire The wheel has mass moment of inertia 012 kgm2 and the wire has stiffness 1176 Nmrad A viscous dashpot applies damping couple of 0784 Nm an angular velocity of 2 rads calculate i)maximum angular displacement from rest position ii) maximum couple applied to dashpot Iii) angle by which the angular displacement lags the torque A mass of 1019 kg is suspended from a spring stiffness 20 Nmm viscous damping causes the amplitude to decrease to 110th of initial value in 4 complete cycle If a periodic force of 20 cos(50 t) is applied to the mass Find the amplitude of forced vibration What would be the amplitude of oscillation if the applied force coincides with the natural period

m=1019 kg

[ ] t50cos 30F =

54

55

ROTATING UNBALANCE Unbalance in rotating machinery is a common source of vibration excitation Consider a machine of mass m having mounted on a foundation of stiffness k and damping coefficient C as shown in figure The unbalance is represented by an eccentric rotating mass m with eccentricity (e) and having a angular velocity (ω) let x be the displacement of non rotating mass from the static equilibrium position The displacement of rotating mass m from the static equilibrium is ( ) tsin( ex ω+ Applying Newtonrsquos law of motion we have

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

56

RECIPROCATING UNBALANCE Unbalance in reciprocating machinery is a common source of vibration excitation Consider a machine of mass M having mounted on a foundation of stiffness K and damping coefficient C as shown in figure The unbalance is represented by a reciprocating mass (m) having crank radus (e) and connecting rod of length (l) Let ωωωω be angular velocity of the crank Let x be the displacement of non reciprocating mass from the static equilibrium position at any instant of time The displacement of reciprocating mass m from the static equilibrium position is

)2sin(sin +++ tle

tex ωωωωωωωω

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

57

Note The equations obtained in reciprocating unbalance is same as the rotating unbalance and having the following differences Rotating Unbalance Reciprocating Unbalance Here mass of the rotating element is considered

Here mass of reciprocating element is considered

Here the eccentricity (e) which is the distance between the centre of rotation and CG

Here the eccentricity e is the crank radius and crank radius = stroke length2

The frequency plot of non dimensional factor( ( )me

MX versus frequency ratio

ωω

n for various values of ( )ξξξξ is shown in figure This frequency plot is same for

the rotating and reciprocating unbalance with the following points are noted from the diagram

1 Damping ratio plays very important role during resonance in controlling the amplitude

2 When

ωω

n=0 then ( )me

MX =0 for all the values of damping ratio ( )ξξξξ

3 When

ωω

nis higher than ( )me

MX tends to unity

4 At very high speeds ie at very high frequency ratio

ωω

n there is no effect

of damping The frequency plot of phase angle versus the frequency ratio is given as shown in figure The explanation is same as in forced harmonic vibration

58

VIBRATION ISOLATION

Vibratory forces and motions generated by machine and other caused are unavoidable Howeever their effect on the other dynamical systems can be minimized by proper isolator design An isolation system is either to protect o delicate machine from excessive vibration transmitted to it from its supporting structure or to prevent vibratory forces and motions generated by machines from being transmitted to its surroundings The effectiveness of isolation may be measured in terms of forces and motion transmitted The first type is known as force isolation and the second type is known as motion isolation

Force isolation (Force Transmissibility) When vibratory forces generated by machine are isolated the effectiveness of forced isolation can be made by defining the term transmissibility The term transmissibility is defined as the ratio of force transmitted to the foundation to that of the external impressed force

Force Transmissibility = o

tr

FF

force Impressed Externalfoundation toted transmitForce

=

Consider viscous damped spring mass system as shown in figure Upon this system an external harmonic force ) t (Sin Fo ω is applied The difference forces acting on system are inertia forces damping force spring force and external force All these forces are represented on force polygon shown below Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these two forces

( ) ( )22 xckx ω+=

59

This is the expression for force transmissibility in dimensional form To represent same in the non dimensional form divide numerator and denominator by k

2

n

22

n

2

n

o

tr

21

21

FF

TR

ωω

ξ+

ωω

minus

ωω

ξ+

==

PHASE DIFFERNECE BETWEEN EXTERNAL FORCE AND DISPLACEMENT Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these

ωω

minus

ωω

ξ=

ωminusω

=φ minusminus2

n

n1

2

1

1

2tan

mKC

tan

The phase difference between force transmitted to the foundation Ftr to the displacement

kc

KxxC

tanω

=ω

=α or

ωω

ξ=

ω=α minusminus

n

11 2tanKx

xCtan

The phase difference between external force Fo and force transmitted to foundationis β

60

β= αminusφ =

ωω

minus

ωω

ξminus

2

n

n1

1

2tan -

ωω

ξminus

n

1 2tan

Displacement Transmissibility It is the ratio of amplitude of steady state vibration of machine to that of the base motion Forced vibration due to support motion Consider a spring mass damper system subjected to harmonic excitation )tsin(Yy ω= at the base as shown in figure Let x be the absolute displacement of the mass at any instant of time t The different forces acting on system are shown in free body diagram Applying Newtonrsquos Law of motion Ma = sumF

)yx(k)yx(cxm minusminusminusminus= ampampampamp

0)yx(k)yx(cxm =minus+minus+ ampampampamp

61

Frequency plot of transmissibility versus frequency ratio

62

The plot of transmissibility versus frequency ratio is as shown in figure The following points can be noted from it

i) The value of transmissibility is unity at frequency ratio

ωω

n=0 and 2

ii) For an undamped system ( )ξ =0 transmissibility tends to infin at resonance iii) The value of transmissibility is less than unity for the value of frequency ratio

greater than 2

iv) For frequency ratio

ωω

nlt 2 smaller damping ratio lead to larger values of

transmissibility on the other hand for frequency ratio

ωω

ngt 2 smaller damping

ratio leads to smaller value of transmissibility

The frequency plot of transmissibility (TR) versus frequency ratio

ωω

n can be

broadly divided into two zones

1

ωω

n= 0 to 2

2

ωω

n= 2 to infin

CASE 1

When

ωω

n= 0 to 2

In this zone the value of transmissibility is always greater than 1 for all values of damping ratio In this zone the transmissibility decreases with increase in damping Hence we can conclude that in this zone damping is useful in isolating vibration forces being transmitted to the support CASE 2

When

ωω

n= 2 to infin

In this zone the value of transmissibility is less than 1 for all values of damping ratio In this zone the transmissibility increases with increase in damping Hence we can conclude that in this zone damping will adversely affect in isolating vibratory forces

63

Tutorial problems on Forced Vibration 1 Machine of mass 75kg is mounted on spring of stiffness 12000 Nm with an assumed damping factor ξ =02 piston within the machine of mass 2 kg has a reciprocating motion with a stroke of 75 cm and speed of 15 cyclessecond Assuming the motion of piston to be harmonic determine amplitude of machine ( X) Transmissibility and force transmitted to foundation phase angle of transmitted force wrt to exciting force 2 Motor weighs 600N is mounted on a simple beam that has a spring rate of 400 Ncm The motor armature base 160 N and has an eccentricity of 0005cm What will be the amplitude of motion when it runs at 1760 RPM Neglect damping the deflection of armature shaft and weight of the beam 3 A Heavy machine of mass 300 kg is supported on resilient foundation the static deflection of foundation due to weight of machine is found to be 75mm It is observed that the machine vibrates with an amplitude of 10mm when the base of the foundation is subjected to harmonic oscillation at the undampeds natural frequency of system with an amplitude of 25mm Find the damping constant of foundation Ii) dynamic force amplitude on the base iii) amplitude of the displacement of machine relative to base 4 An air compressor of mass 450 kg operates at a constant speed of 1750 RPM The reciprocating masses are 10 kg the crank radius is 100mm If the damper for mounting introduces a damping factor 015 i) specify the springs of mounting such that only 20 of the unbalanced force is transmitted to the foundation ii) determine the magnitude of the transmitted force bull Repeat the above problem by assuming that there is no damping in system

5 A body of a car having mass 1500 kg and is mounted on 4 equal springs which deflect through 225 cm under the weight of the body A total damping forces the 4 shock absorbers is 46 N at a velocity of 1cmsec The car is placed with all the four wheels on a test platform which is move up and down at resonant speed With an amplitude of 25cm Find the amplitude of the car body on its springs assuming its CG to be in the centre of wheel base 6A 75 kg machine which is mounted on springs of stiffness 116x105 Nm with the damping of 20=ξ A 2kg piston within the machine as reciprocating motion with a stroke of 008m and speed of 3000 RPM Assuming the motion of piston to be harmonic determine the amplitude of vibration of machine

7 The spring of an automobile are compressed 10cm under its own weight Find the critical speed when a trailer is traveling over road with a people approximaterly by a sine wave of amplitude 75 cm and wave length 15m What will be the amplitude of vibration at 64kmhour 8 An eccentric mass excites us used to determine vibratory characteristics of a structure of mass 25kg as given in figure At a speed of 1000 RPM the eccentric mass to beat top position at the instant the structure was moved upwards through its static equilibrium position and corresponding amplitude was 20mm The eccentric mass is 05kg at an eccentricity of 10mm Determine i) Undamped natural frequency of the system ii) Damping ratio iii) Steady state amplitude and phase angle at 1500 RPM iv) Force transmitted to the ground at 1200 RPM

M=25kg

C

K

Eccentric mass

64

65

CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Support motion Consider a spring mass damper system as shown in figure Let it excited by motion

of support let y be the harmonic displacement of support From Newtonrsquos second law of motion

)yx(k)yx(cxm minusminusminusminus= ampampampamp

)1(0)yx(k)yx(cxm minusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminus=minus+minus+ ampampampamp

Let z be the relative displacement of the mass with respect to support then

Z=x-y

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

55

ROTATING UNBALANCE Unbalance in rotating machinery is a common source of vibration excitation Consider a machine of mass m having mounted on a foundation of stiffness k and damping coefficient C as shown in figure The unbalance is represented by an eccentric rotating mass m with eccentricity (e) and having a angular velocity (ω) let x be the displacement of non rotating mass from the static equilibrium position The displacement of rotating mass m from the static equilibrium is ( ) tsin( ex ω+ Applying Newtonrsquos law of motion we have

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

56

RECIPROCATING UNBALANCE Unbalance in reciprocating machinery is a common source of vibration excitation Consider a machine of mass M having mounted on a foundation of stiffness K and damping coefficient C as shown in figure The unbalance is represented by a reciprocating mass (m) having crank radus (e) and connecting rod of length (l) Let ωωωω be angular velocity of the crank Let x be the displacement of non reciprocating mass from the static equilibrium position at any instant of time The displacement of reciprocating mass m from the static equilibrium position is

)2sin(sin +++ tle

tex ωωωωωωωω

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

57

Note The equations obtained in reciprocating unbalance is same as the rotating unbalance and having the following differences Rotating Unbalance Reciprocating Unbalance Here mass of the rotating element is considered

Here mass of reciprocating element is considered

Here the eccentricity (e) which is the distance between the centre of rotation and CG

Here the eccentricity e is the crank radius and crank radius = stroke length2

The frequency plot of non dimensional factor( ( )me

MX versus frequency ratio

ωω

n for various values of ( )ξξξξ is shown in figure This frequency plot is same for

the rotating and reciprocating unbalance with the following points are noted from the diagram

1 Damping ratio plays very important role during resonance in controlling the amplitude

2 When

ωω

n=0 then ( )me

MX =0 for all the values of damping ratio ( )ξξξξ

3 When

ωω

nis higher than ( )me

MX tends to unity

4 At very high speeds ie at very high frequency ratio

ωω

n there is no effect

of damping The frequency plot of phase angle versus the frequency ratio is given as shown in figure The explanation is same as in forced harmonic vibration

58

VIBRATION ISOLATION

Vibratory forces and motions generated by machine and other caused are unavoidable Howeever their effect on the other dynamical systems can be minimized by proper isolator design An isolation system is either to protect o delicate machine from excessive vibration transmitted to it from its supporting structure or to prevent vibratory forces and motions generated by machines from being transmitted to its surroundings The effectiveness of isolation may be measured in terms of forces and motion transmitted The first type is known as force isolation and the second type is known as motion isolation

Force isolation (Force Transmissibility) When vibratory forces generated by machine are isolated the effectiveness of forced isolation can be made by defining the term transmissibility The term transmissibility is defined as the ratio of force transmitted to the foundation to that of the external impressed force

Force Transmissibility = o

tr

FF

force Impressed Externalfoundation toted transmitForce

=

Consider viscous damped spring mass system as shown in figure Upon this system an external harmonic force ) t (Sin Fo ω is applied The difference forces acting on system are inertia forces damping force spring force and external force All these forces are represented on force polygon shown below Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these two forces

( ) ( )22 xckx ω+=

59

This is the expression for force transmissibility in dimensional form To represent same in the non dimensional form divide numerator and denominator by k

2

n

22

n

2

n

o

tr

21

21

FF

TR

ωω

ξ+

ωω

minus

ωω

ξ+

==

PHASE DIFFERNECE BETWEEN EXTERNAL FORCE AND DISPLACEMENT Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these

ωω

minus

ωω

ξ=

ωminusω

=φ minusminus2

n

n1

2

1

1

2tan

mKC

tan

The phase difference between force transmitted to the foundation Ftr to the displacement

kc

KxxC

tanω

=ω

=α or

ωω

ξ=

ω=α minusminus

n

11 2tanKx

xCtan

The phase difference between external force Fo and force transmitted to foundationis β

60

β= αminusφ =

ωω

minus

ωω

ξminus

2

n

n1

1

2tan -

ωω

ξminus

n

1 2tan

Displacement Transmissibility It is the ratio of amplitude of steady state vibration of machine to that of the base motion Forced vibration due to support motion Consider a spring mass damper system subjected to harmonic excitation )tsin(Yy ω= at the base as shown in figure Let x be the absolute displacement of the mass at any instant of time t The different forces acting on system are shown in free body diagram Applying Newtonrsquos Law of motion Ma = sumF

)yx(k)yx(cxm minusminusminusminus= ampampampamp

0)yx(k)yx(cxm =minus+minus+ ampampampamp

61

Frequency plot of transmissibility versus frequency ratio

62

The plot of transmissibility versus frequency ratio is as shown in figure The following points can be noted from it

i) The value of transmissibility is unity at frequency ratio

ωω

n=0 and 2

ii) For an undamped system ( )ξ =0 transmissibility tends to infin at resonance iii) The value of transmissibility is less than unity for the value of frequency ratio

greater than 2

iv) For frequency ratio

ωω

nlt 2 smaller damping ratio lead to larger values of

transmissibility on the other hand for frequency ratio

ωω

ngt 2 smaller damping

ratio leads to smaller value of transmissibility

The frequency plot of transmissibility (TR) versus frequency ratio

ωω

n can be

broadly divided into two zones

1

ωω

n= 0 to 2

2

ωω

n= 2 to infin

CASE 1

When

ωω

n= 0 to 2

In this zone the value of transmissibility is always greater than 1 for all values of damping ratio In this zone the transmissibility decreases with increase in damping Hence we can conclude that in this zone damping is useful in isolating vibration forces being transmitted to the support CASE 2

When

ωω

n= 2 to infin

In this zone the value of transmissibility is less than 1 for all values of damping ratio In this zone the transmissibility increases with increase in damping Hence we can conclude that in this zone damping will adversely affect in isolating vibratory forces

63

Tutorial problems on Forced Vibration 1 Machine of mass 75kg is mounted on spring of stiffness 12000 Nm with an assumed damping factor ξ =02 piston within the machine of mass 2 kg has a reciprocating motion with a stroke of 75 cm and speed of 15 cyclessecond Assuming the motion of piston to be harmonic determine amplitude of machine ( X) Transmissibility and force transmitted to foundation phase angle of transmitted force wrt to exciting force 2 Motor weighs 600N is mounted on a simple beam that has a spring rate of 400 Ncm The motor armature base 160 N and has an eccentricity of 0005cm What will be the amplitude of motion when it runs at 1760 RPM Neglect damping the deflection of armature shaft and weight of the beam 3 A Heavy machine of mass 300 kg is supported on resilient foundation the static deflection of foundation due to weight of machine is found to be 75mm It is observed that the machine vibrates with an amplitude of 10mm when the base of the foundation is subjected to harmonic oscillation at the undampeds natural frequency of system with an amplitude of 25mm Find the damping constant of foundation Ii) dynamic force amplitude on the base iii) amplitude of the displacement of machine relative to base 4 An air compressor of mass 450 kg operates at a constant speed of 1750 RPM The reciprocating masses are 10 kg the crank radius is 100mm If the damper for mounting introduces a damping factor 015 i) specify the springs of mounting such that only 20 of the unbalanced force is transmitted to the foundation ii) determine the magnitude of the transmitted force bull Repeat the above problem by assuming that there is no damping in system

5 A body of a car having mass 1500 kg and is mounted on 4 equal springs which deflect through 225 cm under the weight of the body A total damping forces the 4 shock absorbers is 46 N at a velocity of 1cmsec The car is placed with all the four wheels on a test platform which is move up and down at resonant speed With an amplitude of 25cm Find the amplitude of the car body on its springs assuming its CG to be in the centre of wheel base 6A 75 kg machine which is mounted on springs of stiffness 116x105 Nm with the damping of 20=ξ A 2kg piston within the machine as reciprocating motion with a stroke of 008m and speed of 3000 RPM Assuming the motion of piston to be harmonic determine the amplitude of vibration of machine

7 The spring of an automobile are compressed 10cm under its own weight Find the critical speed when a trailer is traveling over road with a people approximaterly by a sine wave of amplitude 75 cm and wave length 15m What will be the amplitude of vibration at 64kmhour 8 An eccentric mass excites us used to determine vibratory characteristics of a structure of mass 25kg as given in figure At a speed of 1000 RPM the eccentric mass to beat top position at the instant the structure was moved upwards through its static equilibrium position and corresponding amplitude was 20mm The eccentric mass is 05kg at an eccentricity of 10mm Determine i) Undamped natural frequency of the system ii) Damping ratio iii) Steady state amplitude and phase angle at 1500 RPM iv) Force transmitted to the ground at 1200 RPM

M=25kg

C

K

Eccentric mass

64

65

CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Support motion Consider a spring mass damper system as shown in figure Let it excited by motion

of support let y be the harmonic displacement of support From Newtonrsquos second law of motion

)yx(k)yx(cxm minusminusminusminus= ampampampamp

)1(0)yx(k)yx(cxm minusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminus=minus+minus+ ampampampamp

Let z be the relative displacement of the mass with respect to support then

Z=x-y

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

56

RECIPROCATING UNBALANCE Unbalance in reciprocating machinery is a common source of vibration excitation Consider a machine of mass M having mounted on a foundation of stiffness K and damping coefficient C as shown in figure The unbalance is represented by a reciprocating mass (m) having crank radus (e) and connecting rod of length (l) Let ωωωω be angular velocity of the crank Let x be the displacement of non reciprocating mass from the static equilibrium position at any instant of time The displacement of reciprocating mass m from the static equilibrium position is

)2sin(sin +++ tle

tex ωωωωωωωω

minus

= minus2

1

1

2tan

n

n

ωωωωωωωω

ωωωωωωωωξξξξ

φφφφ

57

Note The equations obtained in reciprocating unbalance is same as the rotating unbalance and having the following differences Rotating Unbalance Reciprocating Unbalance Here mass of the rotating element is considered

Here mass of reciprocating element is considered

Here the eccentricity (e) which is the distance between the centre of rotation and CG

Here the eccentricity e is the crank radius and crank radius = stroke length2

The frequency plot of non dimensional factor( ( )me

MX versus frequency ratio

ωω

n for various values of ( )ξξξξ is shown in figure This frequency plot is same for

the rotating and reciprocating unbalance with the following points are noted from the diagram

1 Damping ratio plays very important role during resonance in controlling the amplitude

2 When

ωω

n=0 then ( )me

MX =0 for all the values of damping ratio ( )ξξξξ

3 When

ωω

nis higher than ( )me

MX tends to unity

4 At very high speeds ie at very high frequency ratio

ωω

n there is no effect

of damping The frequency plot of phase angle versus the frequency ratio is given as shown in figure The explanation is same as in forced harmonic vibration

58

VIBRATION ISOLATION

Vibratory forces and motions generated by machine and other caused are unavoidable Howeever their effect on the other dynamical systems can be minimized by proper isolator design An isolation system is either to protect o delicate machine from excessive vibration transmitted to it from its supporting structure or to prevent vibratory forces and motions generated by machines from being transmitted to its surroundings The effectiveness of isolation may be measured in terms of forces and motion transmitted The first type is known as force isolation and the second type is known as motion isolation

Force isolation (Force Transmissibility) When vibratory forces generated by machine are isolated the effectiveness of forced isolation can be made by defining the term transmissibility The term transmissibility is defined as the ratio of force transmitted to the foundation to that of the external impressed force

Force Transmissibility = o

tr

FF

force Impressed Externalfoundation toted transmitForce

=

Consider viscous damped spring mass system as shown in figure Upon this system an external harmonic force ) t (Sin Fo ω is applied The difference forces acting on system are inertia forces damping force spring force and external force All these forces are represented on force polygon shown below Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these two forces

( ) ( )22 xckx ω+=

59

This is the expression for force transmissibility in dimensional form To represent same in the non dimensional form divide numerator and denominator by k

2

n

22

n

2

n

o

tr

21

21

FF

TR

ωω

ξ+

ωω

minus

ωω

ξ+

==

PHASE DIFFERNECE BETWEEN EXTERNAL FORCE AND DISPLACEMENT Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these

ωω

minus

ωω

ξ=

ωminusω

=φ minusminus2

n

n1

2

1

1

2tan

mKC

tan

The phase difference between force transmitted to the foundation Ftr to the displacement

kc

KxxC

tanω

=ω

=α or

ωω

ξ=

ω=α minusminus

n

11 2tanKx

xCtan

The phase difference between external force Fo and force transmitted to foundationis β

60

β= αminusφ =

ωω

minus

ωω

ξminus

2

n

n1

1

2tan -

ωω

ξminus

n

1 2tan

Displacement Transmissibility It is the ratio of amplitude of steady state vibration of machine to that of the base motion Forced vibration due to support motion Consider a spring mass damper system subjected to harmonic excitation )tsin(Yy ω= at the base as shown in figure Let x be the absolute displacement of the mass at any instant of time t The different forces acting on system are shown in free body diagram Applying Newtonrsquos Law of motion Ma = sumF

)yx(k)yx(cxm minusminusminusminus= ampampampamp

0)yx(k)yx(cxm =minus+minus+ ampampampamp

61

Frequency plot of transmissibility versus frequency ratio

62

The plot of transmissibility versus frequency ratio is as shown in figure The following points can be noted from it

i) The value of transmissibility is unity at frequency ratio

ωω

n=0 and 2

ii) For an undamped system ( )ξ =0 transmissibility tends to infin at resonance iii) The value of transmissibility is less than unity for the value of frequency ratio

greater than 2

iv) For frequency ratio

ωω

nlt 2 smaller damping ratio lead to larger values of

transmissibility on the other hand for frequency ratio

ωω

ngt 2 smaller damping

ratio leads to smaller value of transmissibility

The frequency plot of transmissibility (TR) versus frequency ratio

ωω

n can be

broadly divided into two zones

1

ωω

n= 0 to 2

2

ωω

n= 2 to infin

CASE 1

When

ωω

n= 0 to 2

In this zone the value of transmissibility is always greater than 1 for all values of damping ratio In this zone the transmissibility decreases with increase in damping Hence we can conclude that in this zone damping is useful in isolating vibration forces being transmitted to the support CASE 2

When

ωω

n= 2 to infin

In this zone the value of transmissibility is less than 1 for all values of damping ratio In this zone the transmissibility increases with increase in damping Hence we can conclude that in this zone damping will adversely affect in isolating vibratory forces

63

Tutorial problems on Forced Vibration 1 Machine of mass 75kg is mounted on spring of stiffness 12000 Nm with an assumed damping factor ξ =02 piston within the machine of mass 2 kg has a reciprocating motion with a stroke of 75 cm and speed of 15 cyclessecond Assuming the motion of piston to be harmonic determine amplitude of machine ( X) Transmissibility and force transmitted to foundation phase angle of transmitted force wrt to exciting force 2 Motor weighs 600N is mounted on a simple beam that has a spring rate of 400 Ncm The motor armature base 160 N and has an eccentricity of 0005cm What will be the amplitude of motion when it runs at 1760 RPM Neglect damping the deflection of armature shaft and weight of the beam 3 A Heavy machine of mass 300 kg is supported on resilient foundation the static deflection of foundation due to weight of machine is found to be 75mm It is observed that the machine vibrates with an amplitude of 10mm when the base of the foundation is subjected to harmonic oscillation at the undampeds natural frequency of system with an amplitude of 25mm Find the damping constant of foundation Ii) dynamic force amplitude on the base iii) amplitude of the displacement of machine relative to base 4 An air compressor of mass 450 kg operates at a constant speed of 1750 RPM The reciprocating masses are 10 kg the crank radius is 100mm If the damper for mounting introduces a damping factor 015 i) specify the springs of mounting such that only 20 of the unbalanced force is transmitted to the foundation ii) determine the magnitude of the transmitted force bull Repeat the above problem by assuming that there is no damping in system

5 A body of a car having mass 1500 kg and is mounted on 4 equal springs which deflect through 225 cm under the weight of the body A total damping forces the 4 shock absorbers is 46 N at a velocity of 1cmsec The car is placed with all the four wheels on a test platform which is move up and down at resonant speed With an amplitude of 25cm Find the amplitude of the car body on its springs assuming its CG to be in the centre of wheel base 6A 75 kg machine which is mounted on springs of stiffness 116x105 Nm with the damping of 20=ξ A 2kg piston within the machine as reciprocating motion with a stroke of 008m and speed of 3000 RPM Assuming the motion of piston to be harmonic determine the amplitude of vibration of machine

7 The spring of an automobile are compressed 10cm under its own weight Find the critical speed when a trailer is traveling over road with a people approximaterly by a sine wave of amplitude 75 cm and wave length 15m What will be the amplitude of vibration at 64kmhour 8 An eccentric mass excites us used to determine vibratory characteristics of a structure of mass 25kg as given in figure At a speed of 1000 RPM the eccentric mass to beat top position at the instant the structure was moved upwards through its static equilibrium position and corresponding amplitude was 20mm The eccentric mass is 05kg at an eccentricity of 10mm Determine i) Undamped natural frequency of the system ii) Damping ratio iii) Steady state amplitude and phase angle at 1500 RPM iv) Force transmitted to the ground at 1200 RPM

M=25kg

C

K

Eccentric mass

64

65

CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Support motion Consider a spring mass damper system as shown in figure Let it excited by motion

of support let y be the harmonic displacement of support From Newtonrsquos second law of motion

)yx(k)yx(cxm minusminusminusminus= ampampampamp

)1(0)yx(k)yx(cxm minusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminus=minus+minus+ ampampampamp

Let z be the relative displacement of the mass with respect to support then

Z=x-y

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

57

Note The equations obtained in reciprocating unbalance is same as the rotating unbalance and having the following differences Rotating Unbalance Reciprocating Unbalance Here mass of the rotating element is considered

Here mass of reciprocating element is considered

Here the eccentricity (e) which is the distance between the centre of rotation and CG

Here the eccentricity e is the crank radius and crank radius = stroke length2

The frequency plot of non dimensional factor( ( )me

MX versus frequency ratio

ωω

n for various values of ( )ξξξξ is shown in figure This frequency plot is same for

the rotating and reciprocating unbalance with the following points are noted from the diagram

1 Damping ratio plays very important role during resonance in controlling the amplitude

2 When

ωω

n=0 then ( )me

MX =0 for all the values of damping ratio ( )ξξξξ

3 When

ωω

nis higher than ( )me

MX tends to unity

4 At very high speeds ie at very high frequency ratio

ωω

n there is no effect

of damping The frequency plot of phase angle versus the frequency ratio is given as shown in figure The explanation is same as in forced harmonic vibration

58

VIBRATION ISOLATION

Vibratory forces and motions generated by machine and other caused are unavoidable Howeever their effect on the other dynamical systems can be minimized by proper isolator design An isolation system is either to protect o delicate machine from excessive vibration transmitted to it from its supporting structure or to prevent vibratory forces and motions generated by machines from being transmitted to its surroundings The effectiveness of isolation may be measured in terms of forces and motion transmitted The first type is known as force isolation and the second type is known as motion isolation

Force isolation (Force Transmissibility) When vibratory forces generated by machine are isolated the effectiveness of forced isolation can be made by defining the term transmissibility The term transmissibility is defined as the ratio of force transmitted to the foundation to that of the external impressed force

Force Transmissibility = o

tr

FF

force Impressed Externalfoundation toted transmitForce

=

Consider viscous damped spring mass system as shown in figure Upon this system an external harmonic force ) t (Sin Fo ω is applied The difference forces acting on system are inertia forces damping force spring force and external force All these forces are represented on force polygon shown below Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these two forces

( ) ( )22 xckx ω+=

59

This is the expression for force transmissibility in dimensional form To represent same in the non dimensional form divide numerator and denominator by k

2

n

22

n

2

n

o

tr

21

21

FF

TR

ωω

ξ+

ωω

minus

ωω

ξ+

==

PHASE DIFFERNECE BETWEEN EXTERNAL FORCE AND DISPLACEMENT Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these

ωω

minus

ωω

ξ=

ωminusω

=φ minusminus2

n

n1

2

1

1

2tan

mKC

tan

The phase difference between force transmitted to the foundation Ftr to the displacement

kc

KxxC

tanω

=ω

=α or

ωω

ξ=

ω=α minusminus

n

11 2tanKx

xCtan

The phase difference between external force Fo and force transmitted to foundationis β

60

β= αminusφ =

ωω

minus

ωω

ξminus

2

n

n1

1

2tan -

ωω

ξminus

n

1 2tan

Displacement Transmissibility It is the ratio of amplitude of steady state vibration of machine to that of the base motion Forced vibration due to support motion Consider a spring mass damper system subjected to harmonic excitation )tsin(Yy ω= at the base as shown in figure Let x be the absolute displacement of the mass at any instant of time t The different forces acting on system are shown in free body diagram Applying Newtonrsquos Law of motion Ma = sumF

)yx(k)yx(cxm minusminusminusminus= ampampampamp

0)yx(k)yx(cxm =minus+minus+ ampampampamp

61

Frequency plot of transmissibility versus frequency ratio

62

The plot of transmissibility versus frequency ratio is as shown in figure The following points can be noted from it

i) The value of transmissibility is unity at frequency ratio

ωω

n=0 and 2

ii) For an undamped system ( )ξ =0 transmissibility tends to infin at resonance iii) The value of transmissibility is less than unity for the value of frequency ratio

greater than 2

iv) For frequency ratio

ωω

nlt 2 smaller damping ratio lead to larger values of

transmissibility on the other hand for frequency ratio

ωω

ngt 2 smaller damping

ratio leads to smaller value of transmissibility

The frequency plot of transmissibility (TR) versus frequency ratio

ωω

n can be

broadly divided into two zones

1

ωω

n= 0 to 2

2

ωω

n= 2 to infin

CASE 1

When

ωω

n= 0 to 2

In this zone the value of transmissibility is always greater than 1 for all values of damping ratio In this zone the transmissibility decreases with increase in damping Hence we can conclude that in this zone damping is useful in isolating vibration forces being transmitted to the support CASE 2

When

ωω

n= 2 to infin

In this zone the value of transmissibility is less than 1 for all values of damping ratio In this zone the transmissibility increases with increase in damping Hence we can conclude that in this zone damping will adversely affect in isolating vibratory forces

63

Tutorial problems on Forced Vibration 1 Machine of mass 75kg is mounted on spring of stiffness 12000 Nm with an assumed damping factor ξ =02 piston within the machine of mass 2 kg has a reciprocating motion with a stroke of 75 cm and speed of 15 cyclessecond Assuming the motion of piston to be harmonic determine amplitude of machine ( X) Transmissibility and force transmitted to foundation phase angle of transmitted force wrt to exciting force 2 Motor weighs 600N is mounted on a simple beam that has a spring rate of 400 Ncm The motor armature base 160 N and has an eccentricity of 0005cm What will be the amplitude of motion when it runs at 1760 RPM Neglect damping the deflection of armature shaft and weight of the beam 3 A Heavy machine of mass 300 kg is supported on resilient foundation the static deflection of foundation due to weight of machine is found to be 75mm It is observed that the machine vibrates with an amplitude of 10mm when the base of the foundation is subjected to harmonic oscillation at the undampeds natural frequency of system with an amplitude of 25mm Find the damping constant of foundation Ii) dynamic force amplitude on the base iii) amplitude of the displacement of machine relative to base 4 An air compressor of mass 450 kg operates at a constant speed of 1750 RPM The reciprocating masses are 10 kg the crank radius is 100mm If the damper for mounting introduces a damping factor 015 i) specify the springs of mounting such that only 20 of the unbalanced force is transmitted to the foundation ii) determine the magnitude of the transmitted force bull Repeat the above problem by assuming that there is no damping in system

5 A body of a car having mass 1500 kg and is mounted on 4 equal springs which deflect through 225 cm under the weight of the body A total damping forces the 4 shock absorbers is 46 N at a velocity of 1cmsec The car is placed with all the four wheels on a test platform which is move up and down at resonant speed With an amplitude of 25cm Find the amplitude of the car body on its springs assuming its CG to be in the centre of wheel base 6A 75 kg machine which is mounted on springs of stiffness 116x105 Nm with the damping of 20=ξ A 2kg piston within the machine as reciprocating motion with a stroke of 008m and speed of 3000 RPM Assuming the motion of piston to be harmonic determine the amplitude of vibration of machine

7 The spring of an automobile are compressed 10cm under its own weight Find the critical speed when a trailer is traveling over road with a people approximaterly by a sine wave of amplitude 75 cm and wave length 15m What will be the amplitude of vibration at 64kmhour 8 An eccentric mass excites us used to determine vibratory characteristics of a structure of mass 25kg as given in figure At a speed of 1000 RPM the eccentric mass to beat top position at the instant the structure was moved upwards through its static equilibrium position and corresponding amplitude was 20mm The eccentric mass is 05kg at an eccentricity of 10mm Determine i) Undamped natural frequency of the system ii) Damping ratio iii) Steady state amplitude and phase angle at 1500 RPM iv) Force transmitted to the ground at 1200 RPM

M=25kg

C

K

Eccentric mass

64

65

CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Support motion Consider a spring mass damper system as shown in figure Let it excited by motion

of support let y be the harmonic displacement of support From Newtonrsquos second law of motion

)yx(k)yx(cxm minusminusminusminus= ampampampamp

)1(0)yx(k)yx(cxm minusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminus=minus+minus+ ampampampamp

Let z be the relative displacement of the mass with respect to support then

Z=x-y

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

58

VIBRATION ISOLATION

Vibratory forces and motions generated by machine and other caused are unavoidable Howeever their effect on the other dynamical systems can be minimized by proper isolator design An isolation system is either to protect o delicate machine from excessive vibration transmitted to it from its supporting structure or to prevent vibratory forces and motions generated by machines from being transmitted to its surroundings The effectiveness of isolation may be measured in terms of forces and motion transmitted The first type is known as force isolation and the second type is known as motion isolation

Force isolation (Force Transmissibility) When vibratory forces generated by machine are isolated the effectiveness of forced isolation can be made by defining the term transmissibility The term transmissibility is defined as the ratio of force transmitted to the foundation to that of the external impressed force

Force Transmissibility = o

tr

FF

force Impressed Externalfoundation toted transmitForce

=

Consider viscous damped spring mass system as shown in figure Upon this system an external harmonic force ) t (Sin Fo ω is applied The difference forces acting on system are inertia forces damping force spring force and external force All these forces are represented on force polygon shown below Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these two forces

( ) ( )22 xckx ω+=

59

This is the expression for force transmissibility in dimensional form To represent same in the non dimensional form divide numerator and denominator by k

2

n

22

n

2

n

o

tr

21

21

FF

TR

ωω

ξ+

ωω

minus

ωω

ξ+

==

PHASE DIFFERNECE BETWEEN EXTERNAL FORCE AND DISPLACEMENT Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these

ωω

minus

ωω

ξ=

ωminusω

=φ minusminus2

n

n1

2

1

1

2tan

mKC

tan

The phase difference between force transmitted to the foundation Ftr to the displacement

kc

KxxC

tanω

=ω

=α or

ωω

ξ=

ω=α minusminus

n

11 2tanKx

xCtan

The phase difference between external force Fo and force transmitted to foundationis β

60

β= αminusφ =

ωω

minus

ωω

ξminus

2

n

n1

1

2tan -

ωω

ξminus

n

1 2tan

Displacement Transmissibility It is the ratio of amplitude of steady state vibration of machine to that of the base motion Forced vibration due to support motion Consider a spring mass damper system subjected to harmonic excitation )tsin(Yy ω= at the base as shown in figure Let x be the absolute displacement of the mass at any instant of time t The different forces acting on system are shown in free body diagram Applying Newtonrsquos Law of motion Ma = sumF

)yx(k)yx(cxm minusminusminusminus= ampampampamp

0)yx(k)yx(cxm =minus+minus+ ampampampamp

61

Frequency plot of transmissibility versus frequency ratio

62

The plot of transmissibility versus frequency ratio is as shown in figure The following points can be noted from it

i) The value of transmissibility is unity at frequency ratio

ωω

n=0 and 2

ii) For an undamped system ( )ξ =0 transmissibility tends to infin at resonance iii) The value of transmissibility is less than unity for the value of frequency ratio

greater than 2

iv) For frequency ratio

ωω

nlt 2 smaller damping ratio lead to larger values of

transmissibility on the other hand for frequency ratio

ωω

ngt 2 smaller damping

ratio leads to smaller value of transmissibility

The frequency plot of transmissibility (TR) versus frequency ratio

ωω

n can be

broadly divided into two zones

1

ωω

n= 0 to 2

2

ωω

n= 2 to infin

CASE 1

When

ωω

n= 0 to 2

In this zone the value of transmissibility is always greater than 1 for all values of damping ratio In this zone the transmissibility decreases with increase in damping Hence we can conclude that in this zone damping is useful in isolating vibration forces being transmitted to the support CASE 2

When

ωω

n= 2 to infin

In this zone the value of transmissibility is less than 1 for all values of damping ratio In this zone the transmissibility increases with increase in damping Hence we can conclude that in this zone damping will adversely affect in isolating vibratory forces

63

Tutorial problems on Forced Vibration 1 Machine of mass 75kg is mounted on spring of stiffness 12000 Nm with an assumed damping factor ξ =02 piston within the machine of mass 2 kg has a reciprocating motion with a stroke of 75 cm and speed of 15 cyclessecond Assuming the motion of piston to be harmonic determine amplitude of machine ( X) Transmissibility and force transmitted to foundation phase angle of transmitted force wrt to exciting force 2 Motor weighs 600N is mounted on a simple beam that has a spring rate of 400 Ncm The motor armature base 160 N and has an eccentricity of 0005cm What will be the amplitude of motion when it runs at 1760 RPM Neglect damping the deflection of armature shaft and weight of the beam 3 A Heavy machine of mass 300 kg is supported on resilient foundation the static deflection of foundation due to weight of machine is found to be 75mm It is observed that the machine vibrates with an amplitude of 10mm when the base of the foundation is subjected to harmonic oscillation at the undampeds natural frequency of system with an amplitude of 25mm Find the damping constant of foundation Ii) dynamic force amplitude on the base iii) amplitude of the displacement of machine relative to base 4 An air compressor of mass 450 kg operates at a constant speed of 1750 RPM The reciprocating masses are 10 kg the crank radius is 100mm If the damper for mounting introduces a damping factor 015 i) specify the springs of mounting such that only 20 of the unbalanced force is transmitted to the foundation ii) determine the magnitude of the transmitted force bull Repeat the above problem by assuming that there is no damping in system

5 A body of a car having mass 1500 kg and is mounted on 4 equal springs which deflect through 225 cm under the weight of the body A total damping forces the 4 shock absorbers is 46 N at a velocity of 1cmsec The car is placed with all the four wheels on a test platform which is move up and down at resonant speed With an amplitude of 25cm Find the amplitude of the car body on its springs assuming its CG to be in the centre of wheel base 6A 75 kg machine which is mounted on springs of stiffness 116x105 Nm with the damping of 20=ξ A 2kg piston within the machine as reciprocating motion with a stroke of 008m and speed of 3000 RPM Assuming the motion of piston to be harmonic determine the amplitude of vibration of machine

7 The spring of an automobile are compressed 10cm under its own weight Find the critical speed when a trailer is traveling over road with a people approximaterly by a sine wave of amplitude 75 cm and wave length 15m What will be the amplitude of vibration at 64kmhour 8 An eccentric mass excites us used to determine vibratory characteristics of a structure of mass 25kg as given in figure At a speed of 1000 RPM the eccentric mass to beat top position at the instant the structure was moved upwards through its static equilibrium position and corresponding amplitude was 20mm The eccentric mass is 05kg at an eccentricity of 10mm Determine i) Undamped natural frequency of the system ii) Damping ratio iii) Steady state amplitude and phase angle at 1500 RPM iv) Force transmitted to the ground at 1200 RPM

M=25kg

C

K

Eccentric mass

64

65

CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Support motion Consider a spring mass damper system as shown in figure Let it excited by motion

of support let y be the harmonic displacement of support From Newtonrsquos second law of motion

)yx(k)yx(cxm minusminusminusminus= ampampampamp

)1(0)yx(k)yx(cxm minusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminus=minus+minus+ ampampampamp

Let z be the relative displacement of the mass with respect to support then

Z=x-y

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

59

This is the expression for force transmissibility in dimensional form To represent same in the non dimensional form divide numerator and denominator by k

2

n

22

n

2

n

o

tr

21

21

FF

TR

ωω

ξ+

ωω

minus

ωω

ξ+

==

PHASE DIFFERNECE BETWEEN EXTERNAL FORCE AND DISPLACEMENT Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation Hence the force transmitted to the foundation Ftr is the vector sum of these

ωω

minus

ωω

ξ=

ωminusω

=φ minusminus2

n

n1

2

1

1

2tan

mKC

tan

The phase difference between force transmitted to the foundation Ftr to the displacement

kc

KxxC

tanω

=ω

=α or

ωω

ξ=

ω=α minusminus

n

11 2tanKx

xCtan

The phase difference between external force Fo and force transmitted to foundationis β

60

β= αminusφ =

ωω

minus

ωω

ξminus

2

n

n1

1

2tan -

ωω

ξminus

n

1 2tan

Displacement Transmissibility It is the ratio of amplitude of steady state vibration of machine to that of the base motion Forced vibration due to support motion Consider a spring mass damper system subjected to harmonic excitation )tsin(Yy ω= at the base as shown in figure Let x be the absolute displacement of the mass at any instant of time t The different forces acting on system are shown in free body diagram Applying Newtonrsquos Law of motion Ma = sumF

)yx(k)yx(cxm minusminusminusminus= ampampampamp

0)yx(k)yx(cxm =minus+minus+ ampampampamp

61

Frequency plot of transmissibility versus frequency ratio

62

The plot of transmissibility versus frequency ratio is as shown in figure The following points can be noted from it

i) The value of transmissibility is unity at frequency ratio

ωω

n=0 and 2

ii) For an undamped system ( )ξ =0 transmissibility tends to infin at resonance iii) The value of transmissibility is less than unity for the value of frequency ratio

greater than 2

iv) For frequency ratio

ωω

nlt 2 smaller damping ratio lead to larger values of

transmissibility on the other hand for frequency ratio

ωω

ngt 2 smaller damping

ratio leads to smaller value of transmissibility

The frequency plot of transmissibility (TR) versus frequency ratio

ωω

n can be

broadly divided into two zones

1

ωω

n= 0 to 2

2

ωω

n= 2 to infin

CASE 1

When

ωω

n= 0 to 2

In this zone the value of transmissibility is always greater than 1 for all values of damping ratio In this zone the transmissibility decreases with increase in damping Hence we can conclude that in this zone damping is useful in isolating vibration forces being transmitted to the support CASE 2

When

ωω

n= 2 to infin

In this zone the value of transmissibility is less than 1 for all values of damping ratio In this zone the transmissibility increases with increase in damping Hence we can conclude that in this zone damping will adversely affect in isolating vibratory forces

63

Tutorial problems on Forced Vibration 1 Machine of mass 75kg is mounted on spring of stiffness 12000 Nm with an assumed damping factor ξ =02 piston within the machine of mass 2 kg has a reciprocating motion with a stroke of 75 cm and speed of 15 cyclessecond Assuming the motion of piston to be harmonic determine amplitude of machine ( X) Transmissibility and force transmitted to foundation phase angle of transmitted force wrt to exciting force 2 Motor weighs 600N is mounted on a simple beam that has a spring rate of 400 Ncm The motor armature base 160 N and has an eccentricity of 0005cm What will be the amplitude of motion when it runs at 1760 RPM Neglect damping the deflection of armature shaft and weight of the beam 3 A Heavy machine of mass 300 kg is supported on resilient foundation the static deflection of foundation due to weight of machine is found to be 75mm It is observed that the machine vibrates with an amplitude of 10mm when the base of the foundation is subjected to harmonic oscillation at the undampeds natural frequency of system with an amplitude of 25mm Find the damping constant of foundation Ii) dynamic force amplitude on the base iii) amplitude of the displacement of machine relative to base 4 An air compressor of mass 450 kg operates at a constant speed of 1750 RPM The reciprocating masses are 10 kg the crank radius is 100mm If the damper for mounting introduces a damping factor 015 i) specify the springs of mounting such that only 20 of the unbalanced force is transmitted to the foundation ii) determine the magnitude of the transmitted force bull Repeat the above problem by assuming that there is no damping in system

5 A body of a car having mass 1500 kg and is mounted on 4 equal springs which deflect through 225 cm under the weight of the body A total damping forces the 4 shock absorbers is 46 N at a velocity of 1cmsec The car is placed with all the four wheels on a test platform which is move up and down at resonant speed With an amplitude of 25cm Find the amplitude of the car body on its springs assuming its CG to be in the centre of wheel base 6A 75 kg machine which is mounted on springs of stiffness 116x105 Nm with the damping of 20=ξ A 2kg piston within the machine as reciprocating motion with a stroke of 008m and speed of 3000 RPM Assuming the motion of piston to be harmonic determine the amplitude of vibration of machine

7 The spring of an automobile are compressed 10cm under its own weight Find the critical speed when a trailer is traveling over road with a people approximaterly by a sine wave of amplitude 75 cm and wave length 15m What will be the amplitude of vibration at 64kmhour 8 An eccentric mass excites us used to determine vibratory characteristics of a structure of mass 25kg as given in figure At a speed of 1000 RPM the eccentric mass to beat top position at the instant the structure was moved upwards through its static equilibrium position and corresponding amplitude was 20mm The eccentric mass is 05kg at an eccentricity of 10mm Determine i) Undamped natural frequency of the system ii) Damping ratio iii) Steady state amplitude and phase angle at 1500 RPM iv) Force transmitted to the ground at 1200 RPM

M=25kg

C

K

Eccentric mass

64

65

CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Support motion Consider a spring mass damper system as shown in figure Let it excited by motion

of support let y be the harmonic displacement of support From Newtonrsquos second law of motion

)yx(k)yx(cxm minusminusminusminus= ampampampamp

)1(0)yx(k)yx(cxm minusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminus=minus+minus+ ampampampamp

Let z be the relative displacement of the mass with respect to support then

Z=x-y

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

60

β= αminusφ =

ωω

minus

ωω

ξminus

2

n

n1

1

2tan -

ωω

ξminus

n

1 2tan

Displacement Transmissibility It is the ratio of amplitude of steady state vibration of machine to that of the base motion Forced vibration due to support motion Consider a spring mass damper system subjected to harmonic excitation )tsin(Yy ω= at the base as shown in figure Let x be the absolute displacement of the mass at any instant of time t The different forces acting on system are shown in free body diagram Applying Newtonrsquos Law of motion Ma = sumF

)yx(k)yx(cxm minusminusminusminus= ampampampamp

0)yx(k)yx(cxm =minus+minus+ ampampampamp

61

Frequency plot of transmissibility versus frequency ratio

62

The plot of transmissibility versus frequency ratio is as shown in figure The following points can be noted from it

i) The value of transmissibility is unity at frequency ratio

ωω

n=0 and 2

ii) For an undamped system ( )ξ =0 transmissibility tends to infin at resonance iii) The value of transmissibility is less than unity for the value of frequency ratio

greater than 2

iv) For frequency ratio

ωω

nlt 2 smaller damping ratio lead to larger values of

transmissibility on the other hand for frequency ratio

ωω

ngt 2 smaller damping

ratio leads to smaller value of transmissibility

The frequency plot of transmissibility (TR) versus frequency ratio

ωω

n can be

broadly divided into two zones

1

ωω

n= 0 to 2

2

ωω

n= 2 to infin

CASE 1

When

ωω

n= 0 to 2

In this zone the value of transmissibility is always greater than 1 for all values of damping ratio In this zone the transmissibility decreases with increase in damping Hence we can conclude that in this zone damping is useful in isolating vibration forces being transmitted to the support CASE 2

When

ωω

n= 2 to infin

In this zone the value of transmissibility is less than 1 for all values of damping ratio In this zone the transmissibility increases with increase in damping Hence we can conclude that in this zone damping will adversely affect in isolating vibratory forces

63

Tutorial problems on Forced Vibration 1 Machine of mass 75kg is mounted on spring of stiffness 12000 Nm with an assumed damping factor ξ =02 piston within the machine of mass 2 kg has a reciprocating motion with a stroke of 75 cm and speed of 15 cyclessecond Assuming the motion of piston to be harmonic determine amplitude of machine ( X) Transmissibility and force transmitted to foundation phase angle of transmitted force wrt to exciting force 2 Motor weighs 600N is mounted on a simple beam that has a spring rate of 400 Ncm The motor armature base 160 N and has an eccentricity of 0005cm What will be the amplitude of motion when it runs at 1760 RPM Neglect damping the deflection of armature shaft and weight of the beam 3 A Heavy machine of mass 300 kg is supported on resilient foundation the static deflection of foundation due to weight of machine is found to be 75mm It is observed that the machine vibrates with an amplitude of 10mm when the base of the foundation is subjected to harmonic oscillation at the undampeds natural frequency of system with an amplitude of 25mm Find the damping constant of foundation Ii) dynamic force amplitude on the base iii) amplitude of the displacement of machine relative to base 4 An air compressor of mass 450 kg operates at a constant speed of 1750 RPM The reciprocating masses are 10 kg the crank radius is 100mm If the damper for mounting introduces a damping factor 015 i) specify the springs of mounting such that only 20 of the unbalanced force is transmitted to the foundation ii) determine the magnitude of the transmitted force bull Repeat the above problem by assuming that there is no damping in system

5 A body of a car having mass 1500 kg and is mounted on 4 equal springs which deflect through 225 cm under the weight of the body A total damping forces the 4 shock absorbers is 46 N at a velocity of 1cmsec The car is placed with all the four wheels on a test platform which is move up and down at resonant speed With an amplitude of 25cm Find the amplitude of the car body on its springs assuming its CG to be in the centre of wheel base 6A 75 kg machine which is mounted on springs of stiffness 116x105 Nm with the damping of 20=ξ A 2kg piston within the machine as reciprocating motion with a stroke of 008m and speed of 3000 RPM Assuming the motion of piston to be harmonic determine the amplitude of vibration of machine

7 The spring of an automobile are compressed 10cm under its own weight Find the critical speed when a trailer is traveling over road with a people approximaterly by a sine wave of amplitude 75 cm and wave length 15m What will be the amplitude of vibration at 64kmhour 8 An eccentric mass excites us used to determine vibratory characteristics of a structure of mass 25kg as given in figure At a speed of 1000 RPM the eccentric mass to beat top position at the instant the structure was moved upwards through its static equilibrium position and corresponding amplitude was 20mm The eccentric mass is 05kg at an eccentricity of 10mm Determine i) Undamped natural frequency of the system ii) Damping ratio iii) Steady state amplitude and phase angle at 1500 RPM iv) Force transmitted to the ground at 1200 RPM

M=25kg

C

K

Eccentric mass

64

65

CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Support motion Consider a spring mass damper system as shown in figure Let it excited by motion

of support let y be the harmonic displacement of support From Newtonrsquos second law of motion

)yx(k)yx(cxm minusminusminusminus= ampampampamp

)1(0)yx(k)yx(cxm minusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminus=minus+minus+ ampampampamp

Let z be the relative displacement of the mass with respect to support then

Z=x-y

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

61

Frequency plot of transmissibility versus frequency ratio

62

The plot of transmissibility versus frequency ratio is as shown in figure The following points can be noted from it

i) The value of transmissibility is unity at frequency ratio

ωω

n=0 and 2

ii) For an undamped system ( )ξ =0 transmissibility tends to infin at resonance iii) The value of transmissibility is less than unity for the value of frequency ratio

greater than 2

iv) For frequency ratio

ωω

nlt 2 smaller damping ratio lead to larger values of

transmissibility on the other hand for frequency ratio

ωω

ngt 2 smaller damping

ratio leads to smaller value of transmissibility

The frequency plot of transmissibility (TR) versus frequency ratio

ωω

n can be

broadly divided into two zones

1

ωω

n= 0 to 2

2

ωω

n= 2 to infin

CASE 1

When

ωω

n= 0 to 2

In this zone the value of transmissibility is always greater than 1 for all values of damping ratio In this zone the transmissibility decreases with increase in damping Hence we can conclude that in this zone damping is useful in isolating vibration forces being transmitted to the support CASE 2

When

ωω

n= 2 to infin

In this zone the value of transmissibility is less than 1 for all values of damping ratio In this zone the transmissibility increases with increase in damping Hence we can conclude that in this zone damping will adversely affect in isolating vibratory forces

63

Tutorial problems on Forced Vibration 1 Machine of mass 75kg is mounted on spring of stiffness 12000 Nm with an assumed damping factor ξ =02 piston within the machine of mass 2 kg has a reciprocating motion with a stroke of 75 cm and speed of 15 cyclessecond Assuming the motion of piston to be harmonic determine amplitude of machine ( X) Transmissibility and force transmitted to foundation phase angle of transmitted force wrt to exciting force 2 Motor weighs 600N is mounted on a simple beam that has a spring rate of 400 Ncm The motor armature base 160 N and has an eccentricity of 0005cm What will be the amplitude of motion when it runs at 1760 RPM Neglect damping the deflection of armature shaft and weight of the beam 3 A Heavy machine of mass 300 kg is supported on resilient foundation the static deflection of foundation due to weight of machine is found to be 75mm It is observed that the machine vibrates with an amplitude of 10mm when the base of the foundation is subjected to harmonic oscillation at the undampeds natural frequency of system with an amplitude of 25mm Find the damping constant of foundation Ii) dynamic force amplitude on the base iii) amplitude of the displacement of machine relative to base 4 An air compressor of mass 450 kg operates at a constant speed of 1750 RPM The reciprocating masses are 10 kg the crank radius is 100mm If the damper for mounting introduces a damping factor 015 i) specify the springs of mounting such that only 20 of the unbalanced force is transmitted to the foundation ii) determine the magnitude of the transmitted force bull Repeat the above problem by assuming that there is no damping in system

5 A body of a car having mass 1500 kg and is mounted on 4 equal springs which deflect through 225 cm under the weight of the body A total damping forces the 4 shock absorbers is 46 N at a velocity of 1cmsec The car is placed with all the four wheels on a test platform which is move up and down at resonant speed With an amplitude of 25cm Find the amplitude of the car body on its springs assuming its CG to be in the centre of wheel base 6A 75 kg machine which is mounted on springs of stiffness 116x105 Nm with the damping of 20=ξ A 2kg piston within the machine as reciprocating motion with a stroke of 008m and speed of 3000 RPM Assuming the motion of piston to be harmonic determine the amplitude of vibration of machine

7 The spring of an automobile are compressed 10cm under its own weight Find the critical speed when a trailer is traveling over road with a people approximaterly by a sine wave of amplitude 75 cm and wave length 15m What will be the amplitude of vibration at 64kmhour 8 An eccentric mass excites us used to determine vibratory characteristics of a structure of mass 25kg as given in figure At a speed of 1000 RPM the eccentric mass to beat top position at the instant the structure was moved upwards through its static equilibrium position and corresponding amplitude was 20mm The eccentric mass is 05kg at an eccentricity of 10mm Determine i) Undamped natural frequency of the system ii) Damping ratio iii) Steady state amplitude and phase angle at 1500 RPM iv) Force transmitted to the ground at 1200 RPM

M=25kg

C

K

Eccentric mass

64

65

CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Support motion Consider a spring mass damper system as shown in figure Let it excited by motion

of support let y be the harmonic displacement of support From Newtonrsquos second law of motion

)yx(k)yx(cxm minusminusminusminus= ampampampamp

)1(0)yx(k)yx(cxm minusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminus=minus+minus+ ampampampamp

Let z be the relative displacement of the mass with respect to support then

Z=x-y

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

62

The plot of transmissibility versus frequency ratio is as shown in figure The following points can be noted from it

i) The value of transmissibility is unity at frequency ratio

ωω

n=0 and 2

ii) For an undamped system ( )ξ =0 transmissibility tends to infin at resonance iii) The value of transmissibility is less than unity for the value of frequency ratio

greater than 2

iv) For frequency ratio

ωω

nlt 2 smaller damping ratio lead to larger values of

transmissibility on the other hand for frequency ratio

ωω

ngt 2 smaller damping

ratio leads to smaller value of transmissibility

The frequency plot of transmissibility (TR) versus frequency ratio

ωω

n can be

broadly divided into two zones

1

ωω

n= 0 to 2

2

ωω

n= 2 to infin

CASE 1

When

ωω

n= 0 to 2

In this zone the value of transmissibility is always greater than 1 for all values of damping ratio In this zone the transmissibility decreases with increase in damping Hence we can conclude that in this zone damping is useful in isolating vibration forces being transmitted to the support CASE 2

When

ωω

n= 2 to infin

In this zone the value of transmissibility is less than 1 for all values of damping ratio In this zone the transmissibility increases with increase in damping Hence we can conclude that in this zone damping will adversely affect in isolating vibratory forces

63

Tutorial problems on Forced Vibration 1 Machine of mass 75kg is mounted on spring of stiffness 12000 Nm with an assumed damping factor ξ =02 piston within the machine of mass 2 kg has a reciprocating motion with a stroke of 75 cm and speed of 15 cyclessecond Assuming the motion of piston to be harmonic determine amplitude of machine ( X) Transmissibility and force transmitted to foundation phase angle of transmitted force wrt to exciting force 2 Motor weighs 600N is mounted on a simple beam that has a spring rate of 400 Ncm The motor armature base 160 N and has an eccentricity of 0005cm What will be the amplitude of motion when it runs at 1760 RPM Neglect damping the deflection of armature shaft and weight of the beam 3 A Heavy machine of mass 300 kg is supported on resilient foundation the static deflection of foundation due to weight of machine is found to be 75mm It is observed that the machine vibrates with an amplitude of 10mm when the base of the foundation is subjected to harmonic oscillation at the undampeds natural frequency of system with an amplitude of 25mm Find the damping constant of foundation Ii) dynamic force amplitude on the base iii) amplitude of the displacement of machine relative to base 4 An air compressor of mass 450 kg operates at a constant speed of 1750 RPM The reciprocating masses are 10 kg the crank radius is 100mm If the damper for mounting introduces a damping factor 015 i) specify the springs of mounting such that only 20 of the unbalanced force is transmitted to the foundation ii) determine the magnitude of the transmitted force bull Repeat the above problem by assuming that there is no damping in system

5 A body of a car having mass 1500 kg and is mounted on 4 equal springs which deflect through 225 cm under the weight of the body A total damping forces the 4 shock absorbers is 46 N at a velocity of 1cmsec The car is placed with all the four wheels on a test platform which is move up and down at resonant speed With an amplitude of 25cm Find the amplitude of the car body on its springs assuming its CG to be in the centre of wheel base 6A 75 kg machine which is mounted on springs of stiffness 116x105 Nm with the damping of 20=ξ A 2kg piston within the machine as reciprocating motion with a stroke of 008m and speed of 3000 RPM Assuming the motion of piston to be harmonic determine the amplitude of vibration of machine

7 The spring of an automobile are compressed 10cm under its own weight Find the critical speed when a trailer is traveling over road with a people approximaterly by a sine wave of amplitude 75 cm and wave length 15m What will be the amplitude of vibration at 64kmhour 8 An eccentric mass excites us used to determine vibratory characteristics of a structure of mass 25kg as given in figure At a speed of 1000 RPM the eccentric mass to beat top position at the instant the structure was moved upwards through its static equilibrium position and corresponding amplitude was 20mm The eccentric mass is 05kg at an eccentricity of 10mm Determine i) Undamped natural frequency of the system ii) Damping ratio iii) Steady state amplitude and phase angle at 1500 RPM iv) Force transmitted to the ground at 1200 RPM

M=25kg

C

K

Eccentric mass

64

65

CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Support motion Consider a spring mass damper system as shown in figure Let it excited by motion

of support let y be the harmonic displacement of support From Newtonrsquos second law of motion

)yx(k)yx(cxm minusminusminusminus= ampampampamp

)1(0)yx(k)yx(cxm minusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminus=minus+minus+ ampampampamp

Let z be the relative displacement of the mass with respect to support then

Z=x-y

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

63

Tutorial problems on Forced Vibration 1 Machine of mass 75kg is mounted on spring of stiffness 12000 Nm with an assumed damping factor ξ =02 piston within the machine of mass 2 kg has a reciprocating motion with a stroke of 75 cm and speed of 15 cyclessecond Assuming the motion of piston to be harmonic determine amplitude of machine ( X) Transmissibility and force transmitted to foundation phase angle of transmitted force wrt to exciting force 2 Motor weighs 600N is mounted on a simple beam that has a spring rate of 400 Ncm The motor armature base 160 N and has an eccentricity of 0005cm What will be the amplitude of motion when it runs at 1760 RPM Neglect damping the deflection of armature shaft and weight of the beam 3 A Heavy machine of mass 300 kg is supported on resilient foundation the static deflection of foundation due to weight of machine is found to be 75mm It is observed that the machine vibrates with an amplitude of 10mm when the base of the foundation is subjected to harmonic oscillation at the undampeds natural frequency of system with an amplitude of 25mm Find the damping constant of foundation Ii) dynamic force amplitude on the base iii) amplitude of the displacement of machine relative to base 4 An air compressor of mass 450 kg operates at a constant speed of 1750 RPM The reciprocating masses are 10 kg the crank radius is 100mm If the damper for mounting introduces a damping factor 015 i) specify the springs of mounting such that only 20 of the unbalanced force is transmitted to the foundation ii) determine the magnitude of the transmitted force bull Repeat the above problem by assuming that there is no damping in system

5 A body of a car having mass 1500 kg and is mounted on 4 equal springs which deflect through 225 cm under the weight of the body A total damping forces the 4 shock absorbers is 46 N at a velocity of 1cmsec The car is placed with all the four wheels on a test platform which is move up and down at resonant speed With an amplitude of 25cm Find the amplitude of the car body on its springs assuming its CG to be in the centre of wheel base 6A 75 kg machine which is mounted on springs of stiffness 116x105 Nm with the damping of 20=ξ A 2kg piston within the machine as reciprocating motion with a stroke of 008m and speed of 3000 RPM Assuming the motion of piston to be harmonic determine the amplitude of vibration of machine

7 The spring of an automobile are compressed 10cm under its own weight Find the critical speed when a trailer is traveling over road with a people approximaterly by a sine wave of amplitude 75 cm and wave length 15m What will be the amplitude of vibration at 64kmhour 8 An eccentric mass excites us used to determine vibratory characteristics of a structure of mass 25kg as given in figure At a speed of 1000 RPM the eccentric mass to beat top position at the instant the structure was moved upwards through its static equilibrium position and corresponding amplitude was 20mm The eccentric mass is 05kg at an eccentricity of 10mm Determine i) Undamped natural frequency of the system ii) Damping ratio iii) Steady state amplitude and phase angle at 1500 RPM iv) Force transmitted to the ground at 1200 RPM

M=25kg

C

K

Eccentric mass

64

65

CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Support motion Consider a spring mass damper system as shown in figure Let it excited by motion

of support let y be the harmonic displacement of support From Newtonrsquos second law of motion

)yx(k)yx(cxm minusminusminusminus= ampampampamp

)1(0)yx(k)yx(cxm minusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminus=minus+minus+ ampampampamp

Let z be the relative displacement of the mass with respect to support then

Z=x-y

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

7 The spring of an automobile are compressed 10cm under its own weight Find the critical speed when a trailer is traveling over road with a people approximaterly by a sine wave of amplitude 75 cm and wave length 15m What will be the amplitude of vibration at 64kmhour 8 An eccentric mass excites us used to determine vibratory characteristics of a structure of mass 25kg as given in figure At a speed of 1000 RPM the eccentric mass to beat top position at the instant the structure was moved upwards through its static equilibrium position and corresponding amplitude was 20mm The eccentric mass is 05kg at an eccentricity of 10mm Determine i) Undamped natural frequency of the system ii) Damping ratio iii) Steady state amplitude and phase angle at 1500 RPM iv) Force transmitted to the ground at 1200 RPM

M=25kg

C

K

Eccentric mass

64

65

CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Support motion Consider a spring mass damper system as shown in figure Let it excited by motion

of support let y be the harmonic displacement of support From Newtonrsquos second law of motion

)yx(k)yx(cxm minusminusminusminus= ampampampamp

)1(0)yx(k)yx(cxm minusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminus=minus+minus+ ampampampamp

Let z be the relative displacement of the mass with respect to support then

Z=x-y

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

65

CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Support motion Consider a spring mass damper system as shown in figure Let it excited by motion

of support let y be the harmonic displacement of support From Newtonrsquos second law of motion

)yx(k)yx(cxm minusminusminusminus= ampampampamp

)1(0)yx(k)yx(cxm minusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminusminus=minus+minus+ ampampampamp

Let z be the relative displacement of the mass with respect to support then

Z=x-y

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

66

ωω

minus

ωω

ξ=φ minus

2

n

n1

1

2tan

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

67

From the frequency plot shown following points can be noted

bull At higher frequency ratio the amplitude ratio YZ

is almost equal to unity Then

relative amplitude z and support amplitude Y are equal

bull When YZ

is equal to 1 or Z=Y it means that the mass will not be having any

displacement or zero displacement bull For higher frequency ratios the amplitude ratio will not have any effect

VIBRATION MEASURING INSTRUMENTS

It measure displacement velocity and acceleration of a vibrating body These are also known as seismic instruments This instrument have mass spring and dashpot The construction of vibrating measuring instrument is as shown in above figure Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines The output of electrical signal of instrument will be proportional to the quantity which is to be measured There are two types of vibration measuring system

1 Vibrometer (seismometer) ndash an instrument with low natural frequency 2 Accelerometer

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

68

VIBROMETER This is a device to measure the displacement of vibrating body This is designed wit low natural frequency We know for vibration measuring instrument

( ) ( )

ξ+minus=

222

2

r2r1

rYZ

when the natural frequency of instrument is low then frequency ratio

ωω

n tends to

higher values for undamped system

( )

minus=

22

2

r1

rYZ which implies 1

YZ

= or Z= Y

so relative amplitude Z is equal to the amplitude of vibrating body Y for every values

of damping ratio This is shown in frequency plot of YZ

versus

ωω

n

One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency The seismic mass remains stationary while the frame moves with an vibrating body These instruments can be used to measure velocity and acceleration by incorporating differentiators ACEELEROMETER Accelerometer is used to measure the acceleration of vibratory body The natural frequency of accelerometer is high compare to the frequency which is to be measured We know for a vibration measuring instrument

2

n

22

n

2

n

21YZ

ωω

ξ+

ωω

minus

ωω

=

where Z = n

2

2Yωω

f

where f is a factor which remains constant for the useful range of accelerometer

Where f = ( ) ( )

ξ+minus 222 r2r1

1 with r =

ωω

n

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

69

The equation (1) Y2ω is the acceleration of the vibrating body It is clearly seen that

the acceleration is multiplied by factor n

2

1ω

To keep the value of factor f=1 for the

higher range of

ωω

n ξ should be high then amplitude Z becomes proportional to

amplitude of acceleration to be measured ie Zgt Y2ω or Z α acceleration where

nω is constant

From the equation (2) the following plot can be drawn The figure shows the response of the accelerometer It is seen that for ξ =07 there is complete linearity for accelerometer ie f=1 for frequency ratio less than 025 Since the natural frequency of accelerometer is high it is very light in construction Using integration circuits one can get the display of velocity and displacement the same instrument can be used to measure the velocity and displacement by incorporating integrator

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

70

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1A vibrometer has a natural frequency of 4 rads and ξ 02 is attached to a structure that perform the harmonic motion If the difference of maximum and minimum recorded value is 8mm Find the amplitude of motion of vibrating structure when the frequency is 40 radsecond

2 A damping in a vibrometer is observed to be negligible it gave a relative displacement of 01cm when used on vibrating structures The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cyclesminutes Find the magnitude of displacement and acceleration of the vibrating units 3A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0049 kgm2 is connected to shaft by a spiral spring having constant 098 Nmrad and a viscous damper having a constant of 011 Nm-srad When the shaft vibrates with a frequency of 15 cyclesminute The relative amplitude between ring and shaft is found to be 2 deg What is the maximum acceleration of the shaft

4A commercial vibration pick-up has a natural frequency of 575 Hz and damping factor of 065 What is the lowest frequency beyond which the amplitude can be measured with in (a) 1 error (b) 2 error

5An accelerometer is made with a crystal of natural frequency 20 kHz The damping ratio of accelerometer is found to be 071 Determine the upper cut off frequency of the accelerometer for 1 accuracy 6A vibrometer having a natural frequency of 4 radsec and ζ = 02 is attached to a structure that executes harmonic motion If the difference between the maximum and minimum recorded value is 8 mm find the amplitude of vibration of structure when its frequency is 40 radsec

7 A vibrometer has a natural frequency of 10 cps and has a damping ratio of 07 It is used by mistake to measure vibrations of a fan base at an exciting frequency of 180 rpm The measured vibration velocity of the fan base is 3 mms What is the actual velocity of the fan base 8 A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm If the natural frequency of the instrument is 5 Hz and if it shows 0004 cm determine the displacement velocity and acceleration assuming no damping

9A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz If the lowest frequency that can be measured is 40 Hz find the value of damping factor

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

71

CHAPTER 5 ( CONTD)

WHIRLING OF SHAFTS LEARNING OBJECTIVES

] Introduction to whirling of shafts

] Terminologies used

] Concept of critical speed and its importance in engineering practice

] Derivation of critical speeds with and without air damping

] Discussion of speeds above and below critical speeds

] Problems related to whirling of whirling of shafts

Introduction Critical speed occurs when the speed of the rotation of shaft is equal to the natural

frequency of the lateral vibration of shafts At this speed shaft starts to vibrate violently in the transverse direction Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage Example The rotor blades of a turbine may come in contact with stator blades Larger shaft deflections produce larger bearing reactionswhich may lead to bearing failure The amplitude build up is a time dependent phenomenon and therefore it is very dangerous to continue to run the shaft at it critical speed

This phenomenon results from the following causes 1Mass unbalance 2hystersis damping in shaft 3Gyroscope forces 4Fluid friction in bearing

O

C

G

O C

G

X

e

Bearing centre

Undeflected Position Deflected Position

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

72

The whirling motion of a shaft consists of two components of motion as shown

a Spinning of the shaft along with rotor about the bent up shaft axis b Rotation of plane A made by the centre line of the bearings and bent up-shaft

about the centre line of the bearings

The rotation of plane A which is generally referred as whirling may take place in the same sense as that of spinning of the shaft or in the opposite sense Further the speed of whirling may or may not be equal to the speed of spinning of the shaft When the whirling speed is equal to the speed of rotation of shaft it is called ldquosynchronous whirlrdquo

ωωωω

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Rotation of plane A

ωωωωplane A

ωωωω

Bent up shaft axis

Bearing centre line

Rotor or Disc

Bearing Bearing

Plane A Plane A

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

73

Critical speed of a shaft with a single rotor (with out damping) Consider a shaft on which a rotor in symmetrically located between two bearings The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions

1 Shaft is light and flexible 2 Gravity effects are negligible 3 Friction at shaft centre is small 4 Damping due to air is neglected

Let m mass of the disc ω Angular rotation of the disc (uniform angular velocity of shaft) e eccentricity of the disc radial distance of the mass centre of the disc from its

geometric centre- G K Stiffness of the shaft in transverse direction C Geometric centre of the disc G CG of disc (mass centre) X Lateral deflection of the shaft centre from 0 (OC) (deflection of the geometric

centre of the disc) ωc Critical speed of the shaft The rotor (disc) is in equilibrium under the action of two forces Centrifugal force which acts radially outwards through G = mω2 (x + e) Restoring force which act radially inwards through C = KX there4For equilibrium restoring force = Centrifugal force Xe = 1[(1r2) ndash1]

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

74

r lt 1

10

r gt1

ωωωωn

ωωωω

Xe

+ ve

- ve

Discussions The relation between Xe and ωcω can be plotted as shown below in Figure 513 Relation between Xe and ωωωωn ωωωω Case (i) When ωωωω =ωωωωn (r =1)

bull Forcing frequency coincides with the natural frequency of transverse vibration of the shaft Xe ndash approaches infinity ie the deflection of geometric centre of the disc tends to infinity

bull The disk has a tendency to fly out if the damping is insufficient There will be severe vibrations of the shaft thereby producing huge bearings reactions

bull At ω = ωn the above undesirable effects would occur and therefore ωωωω = ωωωωn = ωωωωc is called the critical speed of the shaft

Case (ii) ωωωω lt ωωωωc r lt 1 ω ltltlt ωn r lt 1

there4Xe = is positive The deflection x and eccentricity lsquoersquo are in the same sense This condition of disc is referred as ldquoHeavy side outsiderdquo ie The disc rotates with heavy side outside Thus C will lie between O and G Positive sign indicates that X is in phase with CF

Disk with Heavy side outside

O

C

G

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

75

Case (iii) When ωωωω gt ωωωωc r gt 1 ω gtgtgt ωn

Xe = negative the deflection x and the eccentricity e are in opposite sense This condition of the disc is referred as ldquoHeavy side insiderdquo G falls between O and C Negative sign indicates that X is out of phase with CF

Disk with Heavy side inside

When ωωωω is very large ωωωωωωωωn = r infininfininfininfin G tends to coincide with O The disc tends to rotate about its mass centre and hence

vibrations are very minimum This is the principle used for stabilization of aircrafts at high speeds

Dynamic force transmitted to the bearings Fd = KX ωn

2 = Km K = mωn2 at the critical speed

= mω2nX

= mω2X at ω Note

1 ω lt ωn r lt 1 Xe is positive Fd = mω2 (X + e) 2 ω gt ωn r gt 1 Xe is negative Fd = mω2 (X ndash e)

If the shaft is vertical dynamic load on each bearing FB = Fd2 If the shaft is horizontal dynamic load on each bearing = FB = (mg2 + Fd2)

O

C

G

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

76

Tutorial problems on Critical speeds with out damping 1 A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings The bearings are 1 m apart The shaft rotates at 1200 rpm The mass centre of the rotor is 011 mm away from the geometric centre of the rotor due to certain manufacturing errors Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GNm2 2 A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal

shaft supported at the ends by two bearings The bearing span is 400 mm Due to certain manufacturing in accuracies the CG of the disc is 002 mm away from the geometric centre of the rotor If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings Neglect damping Take E = 196 105 Nmm2

3 shaft of 14 mm φand the length 12 m is held in long bearings It carries a rotor of 16 Kgs at its midspan The eccentricity of the mass centre of the rotor from the shaft centre is 04 mm The shaft is made of steel for which E = 200 GNm2 and permissible stress is 70 MPa Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft Assume the shaft is mass less (a) When the shaft is horizontal (b) When the shaft if vertical 4 A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a

rotor of 20 Kgs midway between the bearings Determine the critical speed of the shaft if the shaft material has a density of 8000 Kgm3 and E = 21 105Nmm2

5 A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply

supports Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 25 cm from the centre line The shaft rotates at 700 RPM Neglect the weight of the shaft assume that the elastic limit of the shaft is not exceeded Determine the maximum bending and minimum bending stresses on the surface of shaft if the shaft is horizontal ASSUME NO DAMPING take E=200 GPA

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications rotors are subjected to air-resistance or structural damping However for analytical purposes equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ It has been shown that in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by tan φ = 2ξr(1-r2) r = ωωn ξ = damping ratio Due to damping the points O C and G no longer remain collinear and take up the configuration given below as shown in Figure 516 The point C is pulled back due to damping Thus the rotor will be in equilibrium under the action of the following forces

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

a Centrifugal force = meω2 due to the eccentricity of mass of the rotor b Spring force = KX c Centrifugal force mXω2 due to whirling d Damping force CωX

The above forces are shown both in magnitude and direction as given below in Figure

516 Resolve these forces in horizontal and vertical direction and for equilibrium Xe = r2 radic[(1-r2)2 + (2ζr)2] and tan φφφφ = Cωωωω(K-mωωωω2) = 2ξξξξr(1-r2) These expressions are very much similar to frequency response curve of

single DOF system subjected to harmonic excitation due to rotating unbalance The frequency response curves are as shown in Figure 517 and 518

e

0

1

5

0

ωn

KX O X

C

G

CωX

mXω2

meω2

meω2

CωX

(K -mω2)x

) φ ) φ

meω2 Cos φ

meω2 Sin φ

r = ω

ζ = 1

= 025 ζ = 0

ζ = 015 ζ

ζ = 0

ζ =

X

77

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

78

Discussions

a When ωωωωltltlt ωωωωn r ltltlt 1 tanφ infin φ 900 Disc rotates with heavier side outside ie G outside C as shown in figure (a)

b When ωωωω = ωωωωn r = 1 tanφ infin φ 900

ζ = 0

05 10 15 20 25 30

ζ = 025

ζ = 05 ζ = 0707

ζ = 10 ζ = 20

Frequency Ratio r = (ωωn)

Phase Angle φ

Figure 518

G

C φ

O

(a) φφφφ lt 90

G

C

φ

O

(b) φφφφ = 90

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

79

Resonance occurs Deflection X is maximum As damping increases deflection

reduces Severe lateral vibrations occurs

c ωωωωgtgtgtgt ωωωωn r gtgtgt 1 900 lt φ lt 1800

Disc rotates with heavy side inside d When φφφφ = 1800 Irrespective of amount of damping the point G approaches O The system tends to

be more stable and it is the desirable conditions Fig shows the phase at different rotational speeds

G

C φ

O

G

C

φ G C

φ G

C

φ

O O

(a) φφφφ lt 90

O

(b) φφφφ = 90 (c) 900 lt φφφφ lt 1800

G C

φ

O

(c) 900 lt φφφφ lt 1800

(d) φφφφ = 1800

G C

φ

O

(d) φφφφ = 1800

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

80

Tutorial problems on Critical speeds with out damping 1A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart on a horizontal steel shaft 9 mm in diameter The CG of the disc is displaced by 3 mm from its geometric centre Equivalent viscous damping at the centre of the disc is 49 Nsm If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft (b) What is the power required to drive the shaft at this speed (c) Also compare the maximum bending stress with the dead load stress in the shaft Also find the power required to drive the shaft at this speed 2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter The bearing span is 40cm It is known that C G of rotor is 025mm from its geometric centre If the system rotates at 1000 RPM and damping ratio is estimated to be 005 Find

1 amplitude of vibration 2 Dynamic load transmitted to bearings 3 Maximum stress induced in the shaft when the shaft is supports vertically

Neglect the weight of the shaft Assume the shaft to be simply supported and take E = 196 GPa 3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM is mounted at the centre of the shaft of span 08m and diameter 5cm The disc has an unbalance of 10-3 kg-m Assume the bearings to be rigid and the end conditions to be simply supported Damping is represented by an equivalent viscous damping ratio 008 Find the critical speed whirling amplitude at the critical speed Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) Power required to direve the disc at a speed of 25 more than critical speed Take E =196 GPa 4 A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft The span of shaft between the bearings is 1m The mass centre of the disc is 0005mm from the axis of the shaft Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends determine the critical speed of rotation Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa Take E=200 GPa

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system

] Terminologies used ndash principle modes and normal modes of vibrations

] Concept of co ordinate coupling

] Generalized and principal co ordinates

] Free vibration in terms of initial conditions

] Problems on Geared system

] Forced oscillations ndash Harmonic excitation applications

] Vehicle suspension

] Dynamic vibration absorber

] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = masseach ofmotion of typepossible Noofsystem in the Masses of No = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

bull bull bull bull

m1

K

m2

X1 X2

m1

K1

m2

K2

x1

x2

θ1 J1

θ2 J2

Kt

1

Kt2

Figure 7

gure 3

Figure 4

Fi

Figure 2

Figure 1

81

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

Two degrees of freedom will vmodes If masses m1 and m2 shof vibration is called first principain out of phase such mode of vib There are two equations of moteach mass ( more precisely focoupled differential equation ieassuming harmonic solution tofrequencies During free vibratiodefinite value of ratio of the ampratio of amplitude of vibration in aprincipal mode if one of amplitudeas normal mode UNDAMPED FREE VIBRATION

bull bull bull bull

K1

m1

K2 K3

m2

X1 X2

Figure 6

Figure 5

ibrate in two different modes called as principal own in figure (1) are vibrating in phase such mode l mode When the masses m1 and m2 are vibrating ration is called second mode of vibration

i degree of freedom system one for r

Figure 7

ons for a two

each DOF0 They are generally in the form of each equation involves all the coordinates After the equation of motion it gives two natural n at one of natural frequencies there exists a

82

litude which is referred to as principal mode ( the particular mode is known as principal mode) In the is taken as unity then principal mode is referred at

OF TWO DEGREES OF FREEDOM

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

83

1write the equation of motion for the following fig Determine the mode shape

natural frequency no of modes in different modes ( Figure 1)

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

2 Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2 3 Find the natural frequencies of system shown in figure Take J1 = 06 kgm2 J2 = 134 kgm2 G=083x1011 Nm2

Definite and semi-definite systems A system which is free from both the ends is referred as semi-definite system Ex Train Turbo-generator etc

re 2

Figure 1

θ1 J1

θ2 J2

Kt

Kt2

θ1

θ2

Disc-1

Disc-2

)θ(θK 122 minus

22θJ ampamp

11θJ ampamp11θK

Free body diagram

Figu

84

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

85

For the linear system shown below (i) derive the equations of motion (ii) setup the frequency equation and obtain the

fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system Class work 1 A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 Nm Determine the natural frequency of the system

bull bull bull bull

m1

K

m2

X1 X2

x2 gt x1

KX2 KX2

KX1 KX1

m1 m2

X1 X2

K (X2 ndashX1)

K (X2 ndashX1) m1 m2

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

86

2 Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm The length of the shaft is 3 m and its diameter = 10 cm Modulus of rigidity for shaft material of the shaft G = 083 1011 Nm2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm ( hint for the part II is given below) GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied For a general disturbance not satisfying amplitude ratios the motion cannot be harmonic The general solution includes the response containing harmonics as well as second natural frequencies of the system Since the differential euation for mass m1 and mass m2 both are of second order there will be two constants for each differential equations hence the general solution can be written as

J1 J2

Kt1 Kt2

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

87

For the system shown mass m1 is displaced 001 m from its static equilibrium position and released Determine the resulting displacements x1 and x2 of masses m1 and m2 if m1=m2=m Also determine the displacement of masses after 01 second If k=10 KNm and m=10kg Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

88

Determine the natural frequencies of the system shown below Take m1=m2=m Derive the

differential equation of motion for system shown Assume a chord passing over cylinder do not slip Determine an expression for natural frequencies of system

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

89

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes UNDAMPED DYNAMIC VIBRATION ABSORBER ( Forced harmonic vibration of two degree freedom system) When a single degree freedom system having large amplitudes of vibration under external excitation at resonance the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system This is the principle of undamped dynamic vibration absorber Here the excitation is finally transmitted to the auxiliary system bringing the main system to rest

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

90

Consider a two degree freedom system as shown in Figure The spring mass system

k1 m1 is called the main system which is excited by an external

harmonic force tsinωFF 01 = From Force equilibrium diagram of mass m1

tsinωF)x(xKxKxm 01221111 =minusminus+ampamp tsinωFxK)xK(Kxm 02212111 =minus++ampamp

(137) From Force equilibrium diagram of mass m2

0)x(xKxm 12222 =minus+ampamp 0xKxKxm 221222 =+minusampamp

(138) It is possible to have pure harmonic free vibration for both the masses Therefore Let us assume

tsinωAx 11 = tsinωAx 22 = where ω is the forcing frequency on the system

m1

K1x1

K2(x2-x1)

11xm ampamp

m2 22xm ampamp

tsinωFF 01 =

Force equilibrium diagram of the system

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

91

[ ]212

1222214

21

2o2 kk)mkmkmk(mm

KFx

+ω++minusω=

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

92

[ ][ ]21

2211212

421

222o

1 kk)mkmkmk(mmmkF

x+ω++minusω

ωminus=

The amplitude of vibration of main system can be reduced to zero if the numerator of equation (2) is zero

0ωmKF 2220 =minus )(

0ωmK 2

22 =minus

2

22

mK

ω =

srad2

2

mK

ω =

This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system The addition of a vibration absorber to main system is not much meaningful unless

the main system is operating under resonance Then 1nω=ω but absorber to be

effective 2nω=ω we can write 21 nn ω=ω=ω When this condition is fulfilled the absorber is known as tuned absorber

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES

93

• CHAPTER 1- INTRODUCTION TO VIBRATION
• • LEARNING OBJECTIVES
  • DESCRIBING MOTIONS OF VIBRATING SYSTEMS
  • • CHAPTER 2 UNDAMPED FREE VIBRATION
    • • LEARNING OBJECTIVES
      • • CHAPTER 3
        • • LEARNING OBJECTIVES
          • • CHAPTER 4
            • • LEARNING OBJECTIVES
              • • CHAPTER 5
                • • LEARNING OBJECTIVES
                  • • CHAPTER 5 ( CONTD)
                    • • LEARNING OBJECTIVES
                      • • CHAPTER 6
                        • • LEARNING OBJECTIVES