memb343 mechanical vibrations presentation slides

FUNDAMENTALS OF

VIBRATIONSBASIC CONCEPTS

MEMB343 Mechanical Vibrations

LEARNING OBJECTIVES

Upon completion of this lecture, you should be able to:

Understand the concepts of degree-of-freedom, and discreteand continuous systems.

Compute the stiffness of some simple spring elements.

Determine the equivalent mass or inertia, and equivalentspring and damping constants of vibrating systems.

Understand the definitions of free and forced vibration,undamped and damped vibration, linear and non-linear

vibration, and deterministic and random vibration.


VIBRATION IN ENGINEERING

PRACTICE

Good

Machine condition monitoring.

Vibrating sieves, mixers and tools.

Electric massaging units, dentist drills, electrictoothbrushes.

Bad

Noise, vibration and harshness (NVH).

Machinery and structural failures.

Motion sickness, white finger syndrome, etc.


TACOMA NARROWS BRIDGE

This photograph shows the twisting motion of

the center span just prior to failure.

The nature and severity of the torsional

movement is revealed in this picture taken

from the Tacoma end of the suspension span.

When the twisting motion was at the

maximum, elevation of the sidewalk at the

right was 28 feet (8.5m) higher than the

sidewalk at the left.


TACOMA NARROWS BRIDGE (cont.)


POSITIVE DISPLACEMENT BLOWER

Shearing of shaft attributed to high torsional vibrations. Rotor operating

speed was within the vicinity of a torsional natural frequency.


POSITIVE DISPLACEMENT BLOWER

(cont.)


EXAMPLES OF VIBRATING

SYSTEM

Masses attached to springs.

Flexible Rods.

Pendulums.


BASIC CONCEPTS OF VIBRATION

When a particle or a rigid body in stable equilibrium isdisplaced by the application of an additional force,

mechanical vibration will result.

Some important concepts in mechanical vibration theorycan be categorized into the followings:

Elementary parts of vibrating systems.

Degree of freedom.

Discrete and continuous system.


ELEMENTARY PARTS OF VIBRATING

SYSTEMS

To be subject to vibration, a system must be able to storeenergy in two different forms and allow energy to be

transferred from one to the other.

In particular, for vibration to exist, there must be a transferof energy from potential to kinetic and vice-versa.

Potential energy is due to either gravity or the elasticity ofthe system, whilst the kinetic energy is due to the motion

of the mass.

The simplest mechanical oscillators are the pendulum andthe spring-mass system. The corresponding simplest

electrical oscillator is the capacitor-inductor system.


SPRING ELEMENTS

A spring is a mechanical link that is generally assumed tohave negligible mass and damping.

A force is developed in a spring whenever there is arelative motion between two ends of the spring.

Work done in deforming a spring is stored as potentialenergy in the spring.

kxFS

TS kM

Linear Spring

Torsional Spring


SPRING ELEMENTS (cont.)

A spring element is generally made of an elastic material.

The stiffness in a spring element can be related moredirectly to its material (elastic modulus) and geometric

properties.

A spring-like behavior results from a variety of motionconfigurations, including:

Longitudinal motion (vibration in the direction of thelength).

Transverse motion (vibration perpendicular to the length).

Torsional motion (vibration rotating around the length).


STIFFNESS OF SPRING ELEMENTS

Stiffness associated with the longitudinal vibration of aslender prismatic bar.



(cont.)

Stiffness associated with the torsional vibration of a shaft.



(cont.)

Stiffness associated with a helical spring.



(cont.)

Beam stiffness associated with the transverse vibration ofthe tip of a beam.


EQUIVALENT SPRING CONSTANT

Springs in Parallel Springs in Series

neq kkkk ....21neq kkkk

1....

111

21


MASS OR INERTIA ELEMENTS

The mass or inertia element is assumed to be a rigid body.

A rigid bodys inertia is responsible for the resistance toacceleration of a system.

Work done on a mass is stored in the form of kineticenergy of the mass.

Linear Motion

Rotational Motion

xmF

IM


EQUIVALENT MASS OF A SYSTEM

Translational Masses Connected by a Rigid Bar.

Pivot point

1l

2l

3l

3x2x1x

2m3m1

m

Original system

eqm

eqx

1l

Equivalent system

From trigonometric

relationship

Assume

eqxl

lx

1

22

eqxl

lx

1

33

1xxeq



(cont.)

Translational Masses Connected by a Rigid Bar.

22

33

2

22

2

112

1

2

1

2

1

2

1eqeq xmxmxmxm

Equate the kinetic energy of the three masses to that of the

equivalent systems mass

The equivalent systems mass is therefore obtained

3

2

1

32

2

1

21 m

l

lm

l

lmmeq



(cont.)Coupled Translational and Rotational Masses.

m

0J

r

x

From kinematics, the relationship

between the linear and the angular

velocity is

To obtain the equivalent translational

mass

To obtain the equivalent rotational

mass

r

x

xxeq

eq



(cont.)

Equivalent Translational

Mass

Equivalent Rotational Mass

22

02

2

1

2

1

2

1eqeq xm

r

xJxm

2

0

r

Jmmeq

2202

2

1

2

1

2

1eqeqJJrm

0

2 JmrJeq

Coupled Translational and Rotational Masses.


DAMPING ELEMENTS

A damper is generally assumed to have negligible massand stiffness.

A force is developed in a damper whenever there is arelative velocity between two ends of the damper.

The damper dissipates energy from a system in the form ofheat or sound.

xcFD

TD cM

Linear Damper

Torsional Damper


DAMPING ELEMENTS MODELS

Viscous Damping

The damping force is proportional to the velocity of the

vibrating body.

Coulomb Damping

The damping force is constant in amplitude but opposite

the direction to that of the motion of the vibrating body.

Hysteretic Damping

The energy dissipated per cycle is proportional to the

square of the vibration amplitude.


EQUIVALENT DAMPING CONSTANT

Dampers in Parallel Dampers in Series

neq cccc ....21neq cccc

1....

111

21


DEGREE OF FREEDOM

The minimum number of independent coordinates requiredto determine completely the positions of all part of a

system at any instant of times.

x

y

z

x

y

z

Unconstrained rigid body with 6 d.o.f.


DISCRETE AND CONTINUOUS SYSTEM

Systems with a finite number of degrees of freedom arecalled discrete or lumped parameter systems, and those

with an infinite number of degrees of freedom are called

continuous or distributed systems.

Discrete System Continuous System

Solution: 2nd Order Ordinary

Differential Equation

Solution: Partial Differential

Equation


CLASSIFICATION OF VIBRATION

Free and Forced Vibration.

Undamped and Damped Vibration.

Linear and Nonlinear Vibration.

Deterministic and RandomVibration.


FREE AND FORCED VIBRATION

Free Vibration

Oscillation occurring at a natural frequency, after an initial

force input.

Forced Vibration

Oscillation occurring at the frequency of a driving force

input.

0

0

kxxcxm

kxxm

)(

)(

tFkxxcxm

tFkxxm


UNDAMPED AND DAMPED VIBRATION

Undamped Vibration

No energy is lost or dissipated in friction or other

resistance during oscillation.

Damped Vibration

Energy is lost or dissipated during oscillation.

)(

0

tFkxxm

kxxm

)(

0

tFkxxcxm

kxxcxm


LINEAR AND NONLINEAR VIBRATION

Linear VibrationThe cause (force) and effect (response) are proportionallyrelated. Principle of superposition holds.

Nonlinear VibrationRelationship between cause and effect is no longerproportional.


DETERMINISTIC AND RANDOM

VIBRATION

Deterministic Vibration

The instantaneous values of the vibration amplitude at anytime (t) can be determined from mathematical expressions.

Random Vibration

Future instantaneous values of the vibration amplitudecannot be predicted in a deterministic sense.


EXAMPLE 1

A composite propeller shaft, which is made of steel andaluminium, is shown below. Determine the torsional

spring constant of the shaft The shear modulus G of steel

is 80 GPa and for aluminium 26 GPa.


EXAMPLE 1 (cont.)

The polar area moment of inertia for a hollow shaft isgiven by the following equation where D and d are the

outer and inner diameters, respectively.

Torsional stiffness for the shaft is:

32

44 dDIP

l

dDG

l

GIk Pt

32

44


EXAMPLE 1 (cont.)

For the steel shaft, torsional stiffness is:

For the aluminium shaft, torsional stiffness is:

Shafts are in parallel, therefore:

Nm/rad 1034.5

532

15.025.01080 6449

tSk

Nm/rad 10207.0

532

1.015.01026 6449

tAk

Nm/rad 10547.5 6 tAtSeq kkk


FUNDAMENTALS OF

VIBRATIONSSIMPLE HARMONIC MOTION


LEARNING OBJECTIVES


Determine if a vibration motion can be classified as simpleharmonic.

Represent simple harmonic motion in both trigonometric andcomplex forms.

Understand some basic terminology that are used to define avibration signal.


INTRODUCTION

Oscillatory motion may repeat itself regularly as in thecase of a simple pendulum, or it may display considerable

irregularity, as in earthquakes.

When the motion is repeated in equal intervals of time ,it is called periodic motion.

The repetition time is called the period of oscillation,and its reciprocal is called the frequency.

If the motion is designated by the time function , thenany periodic motion must satisfy the relationship:

1f

tx

txtx


SIMPLE HARMONIC MOTION

The simplest type of periodic motion is called a simpleharmonic motion.

It is essentially a periodic motion with a single frequency.

It can be demonstrated by a mass suspended from a lightspring.

If the mass is displaced from its rest position and released,it will oscillate up and down.


SIMPLE HARMONIC MOTION (cont.)

The motion of the mass can be expressed by the equation

is the amplitude of oscillation, measured from theequilibrium position of the mass.

is the period, and the motion is repeated when .

tAx 2sin

A

t



Harmonic motion is often represented as the projection ona straight line of a point that is moving on a circle at

constant speed.

With the angular speed of the line designated by ,the displacement can be written as . tAx sin

OP

x



The quantity is measured in radians per second, and isreferred to as the circular or angular frequency.

Since the motion repeat itself every radians, we havethe relationship .

and are the period and frequency of the harmonicmotion, usually measured in seconds and cycles per second

(Hz), respectively.

The velocity and acceleration of harmonic motion can besimply determined by differentiation of .

f

2

2

2

f

tAtAx

tAtAx

sinsin

sincos

22

2

tAx sin



Thus the velocity and acceleration are also harmonic withthe same frequency of oscillation but lead the displacement

by and radians, respectively.2/



The acceleration can be written as .

Therefore, in simple harmonic motion, the acceleration isproportional to the displacement and is directed towards

the origin.

As Newtons 2nd law states that the acceleration isproportional to the force, harmonic motion can be expected

with linear springs with force varying as .kx

xx 2


COMPLEX FORM OF SIMPLE

HARMONIC MOTION

The trigonometric functions of sine and cosine are relatedto the exponential function by Eulers equation:

A vector of amplitude rotating at constant angular speedcan be represented as a complex quantity in the

Argand diagram.

z

sincos iei

A


COMPLEX FORM OF SIMPLE HARMONIC

MOTION (cont.)

The following relationship holds:

The quantity is referred to as the complex sinusoid withand as the real and imaginary components.

z

iyx

tiAtA

Aez ti

sincos

x y


COMPLEX FORM OF SIMPLE HARMONIC

MOTION (cont.)

The above figure shows and its conjugate which isrotating in the negative direction with angular speed .

From this diagram, it is seen that the real component isexpressed in terms of and by the equation:

stands for the real part of the quantity .

z

tiAetAzzx Recos*21

x

*z

Re

z

*z

z


COMPLEX FORM OF SIMPLE

HARMONIC MOTION (cont.)

Some of the rules of exponential operations betweenand are:

Multiplication

Division

Powers

212121

ieAAzz

111

ieAz 222

ieAz

21

2

1

2

1

ie

A

A

z

z

ni

nn eAz

eAz innn

11


VIBRATION TERMINOLOGY

Since the position, velocity and acceleration changecontinually with time, several other quantities are used to

discuss vibration.

The peak value, defined as the maximum displacement, ormagnitude , usually indicates the maximum stress that

the vibrating part is undergoing.

A


VIBRATION TERMINOLOGY (cont.)

Another quantity useful in describing vibration is theaverage value, denoted as , and defined by:

The average value indicates a steady or static valuesomewhat like the DC level of an electrical current.

For example, the average value for a complete cycle of asine wave is zero.

Its average value for a half-cycle is, however:

AA

dttA

x 637.02

sin0

tAsin

T

Tdttx

Tx

0

1lim

x



Since the square of displacement is associated with asystems potential energy, the average of the displacement

squared is sometimes a useful vibration property to

discuss.

The mean square value of a time function , denotedby is found from the average of the squared values,

integrated over some time interval :

If , its mean square value is:

2x

T

Tdttx

Tx

0

22 1lim

tx

T

20

22

2

12cos1

2

1lim Adtt

T

Ax

T

T

tAtx sin



The root mean square (rms) value is the square root of themean square value.

From the previous example for , the rms ofthe sine wave of amplitude is:

Vibration instrumentation generally measures root meansquare vibration amplitudes (displacement, velocity or

acceleration).

AA

Arms 707.02

A

tAtx sin



Since the peak value of the velocity and acceleration aremultiples of the circular frequency times the displacement

amplitude, these three basic quantities often differ in value

by an order of magnitude.

For systems with circular frequency larger than 1 rad/s,the relative amplitude of the velocity response is larger

than that of the displacement by a multiple of , and the

acceleration response is larger by a multiple of .

For systems with circular frequency less than 1 rad/s, thevelocity and acceleration have smaller relative amplitudes

than the displacement.

2



A common unit of measurement for vibration amplitudesand rms values is the decibel (dB). As the decibel is a

logarithmic unit, it compresses or expands the scale.

The decibel was originally defined in terms of the base 10logarithm of the power ratio of two electrical signals, or as

the ratio of the square of the amplitudes of two signals.

The equation based on the ratio of the square of theamplitudes of two signals results from the fact that power

is proportional to the square of the amplitude or voltage.

2

110

2

2

110

2

110 log20log10log10

x

x

x

x

P

PdB



Cycle is the movement of a vibrating body from itsequilibrium position to its extreme position in one

direction, then to the equilibrium position, then to its

extreme position in other direction, and back to

equilibrium position. 1 revolution (angular displacement

of 2 radians) of the pin or one revolution of the vector

constitute a cycle.

P

OP



Amplitude, denoted by , is defined as the maximumdisplacement of a vibrating body from its equilibrium

position.

Period of oscillation, denoted by , is defined as the time(second) taken to complete one cycle of motion.

It is equal to the time required for the vector to rotatethrough an angle of 2 and therefore:

Frequency of oscillation, denoted by , is defined as thenumber of cycles per unit time. is measured in

cycles/second (Hz).

A

OP

2

f

2

1f

f



Phase angle ( ) is the difference in time between twoevents such as the zero crossing of two waveforms.

is expressed in radians as the time between two eventsdivided by the period, times 2.

2t


EXAMPLE 1

The maximum displacement and the maximumacceleration of the foundation of a centrifugal pump were

found to be 0.25 mm and 0.4g. Find the operating speed of

the pump. Assume the motion of the foundation is

harmonic.


EXAMPLE 1 (cont.)

For harmonic motion, the following relationship betweendisplacement and acceleration is valid:

From the above equation, we solve for the angularfrequency:

max

2

max xx

rad/s 28.1251025.0

81.94.03

max

max

x

x


EXAMPLE 1 (cont.)

The operating speed of the pump is related to the angularfrequency by the following equation:

rpm 3.11962

28.12560

2

60

N


FREE VIBRATION OF SINGLE-

DEGREE-OF-FREEDOM SYSTEMSUNDAMPED FREE VIBRATIONS


LEARNING OBJECTIVES


Determine the equation of motion of an undamped single-degree-of-freedom system.

Determine the undamped natural frequency of a single-degree-of-freedom system.

Solve the equation of motion of the undamped free response.

Understand the relationship between various solutions to theequation of motion.


INTRODUCTION

The vibration analysis procedure generally involves: Mathematical modeling.

Derivation of governing equations.

Solution of the governing equations.

Interpretation of the results.

This lecture deals with the undamped free vibrationresponse of a single-degree-of-freedom system.


UNDAMPED FREE VIBRATIONS

A free vibration is produced and maintained by forces suchas elastic and gravitational forces.

These forces depend only on the position and motion of thebody.

When forces that oppose the restoring force (friction, airresistance, etc.) are negligible, the vibration is called

undamped.

An undamped free vibration will repeat itself indefinitely.

In real systems, the frequency and period of vibrationobtained for a freely vibrating system are very close to the

values obtained for a system that has a small amount of

damping.


UNDAMPED FREE VIBRATIONS (cont.)

Consider a block of mass sliding on a frictionlesshorizontal surface.

Vibration is induced by displacing the block a distanceand then releasing it with an initial velocity of

.0x

00 vx

m



The elastic restoring force of the spring is alwaysdirected toward the equilibrium position, whereas the

acceleration acts in the direction of

positive displacement.

It is important to remember that since acceleration is thesecond time derivative of displacement, both the

displacement and the acceleration must be measured

positive in the same direction.

Applying Newtons 2nd law to the block gives thedifferential equation of motion of the block:

or

kxFs

xdtxda 22 /

x x

xmkx 0 kxxm



The general solution for the previous equation based onphysical observation is given by:

is the amplitude, or maximum value of the function.

, the angular natural frequency measured in rad/s,determines the interval in time that the function repeats

itself.

, the phase measured in radians, determines the initialvalue of the sine function.

It is standard to measure time in seconds.

A

tAtx nsin

n

t



The velocity and acceleration of the block are obtained bydifferentiating with time.

The velocity of the block is .

The acceleration of the block is .

Substitution of these equations into the equation of motiongiven by yields:

tAtx nn cos

tAtx nn sin2

tAtx nsin

0 kxxm

0sinsin2 tkAtAm nn



Dividing the previous equation by and yields thefact that this equation is satisfied if .

The constant characterizes the spring-mass system, aswell as the frequency at which the motion repeats itself,

and hence is called the systems natural frequency.

m

kn 2

n

A m



The constants and are determined by the initialstage of motion of the mass-spring system. If no energy is

imparted to the mass, it will stay at rest.

If, however, the mass is displaced to a position at time, the potential energy in the spring will result in

motion. Also, if the mass is given an initial velocity of

at time , motion will result.

These are called initial conditions and when substitutedinto the previous equations for displacement and velocity

yield:

cos0cos00 AAxv nnn

A

0x

sin0sin00 AAxx n

0v

0t

0t



Solving these two equations simultaneously for the twounknowns and yields:A

n

n vxA

202

0

2

0

01tanv

xn



Thus the solution of the equation of motion for the spring-mass system is given by:

0

01

2

0

2

0

2

tansinv

xt

vxtx nn

n

n



Since the solution repeats whenever their argumentincreases by an angle , the period of the oscillation is

given by:

The natural frequency of the oscillation in Hz (cycles persecond) is then:

The results above may be used to analyze the vibrationalmotion of a particle or rigid-body whenever the equations

of motion reduce to the form which characterizes simple

harmonic motion given by .

2

n

n

2

02 xx n

2

1 n

n

nf


ALTERNATIVE SOLUTIONS

Following the theory of elementary differential equations,we can assume the solution for the equation of motion as:

and are nonzero constants to be determined.

Upon successive differentiation of , we obtain:

Substituting the assumed exponential form into theequation of motion yields:

a taetx

02 tt kaeaem

tx

taetx 2

taetx


ALTERNATIVE SOLUTIONS (cont.)

Since the term is never zero, the previous expressioncan be divided by to yield:

Solving this algebraically results in:

is the imaginary number and is the naturalfrequency.

tae

02 km

1j

jjm

k

m

kn

tae

n



Note that there are two values for :

This results because the equation for is of second order,which implies that there must be two solutions of the

equation of motion.

Substituting these two values of into yields:

The principal of superposition for linear systems states thatthe sum of two solutions is also a solution, hence:

and are complex-valued constants of integration.

jn

tj naetx

tjtj nn eaeatx 21

taetx

jn

tj naetx

1a 2a



Another alternative solution for the equation of motion is:

and are real-valued constants of integration to bedetermined from the initial conditions:

The velocity and acceleration of the block are obtained bydifferentiating with time.

1A

0 when and 00 tvtxxtx

tx

tAtAtx nn sincos 21

tAtAtx nnnn cossin 21

2A

tAtAtx nnnn sincos 22

1

2


SOLUTIONS RELATIONSHIP

The solution of subject to non-zero initialconditions can be written in three equivalent ways.

First, the solution can be written as follows where andare real-valued constants.

Second, the solution can be written as follows whereand are complex-valued constants.

Third, the solution can be written as follows where andare real-valued constants.

1A

0 kxxm

tAtAtx nn sincos 21 2A

tjtj nn eaeatx 21

1a 2a

tAtx nsin

A


SOLUTIONS RELATIONSHIP (cont.)

Each set of two constants is determined by the initialconditions.

Following trigonometric identities and the Eulersformulas, the various constants are related by:

2

2

tan

212

211

212211

2

11-2

2

2

1

jAAa

jAAa

jaaAaaA

A

AAAA


DISPLACED EQUILIBRIUM POSITION

The deformation of the spring in the static equilibriumposition is , and the spring force is equal to the

gravitational force acting on the mass .W k

m

mgWk



(cont.)

Measuring the displacement from the static equilibriumposition, the forces acting on are and .

With chosen to be positive in the downward direction,so are all quantities (force, velocity and acceleration).

Applying Newtons 2nd law of motion to the mass :

Since , we obtain

The choice of the static equilibrium position as referencefor has eliminated , the force due to gravity, and the

static spring force , from the equation of motion.

W

m

xkWFxm

x

Wk

x

m

Wk 0 x

m

kx

x

k



(cont.)

The equation of motion may be written as

where .

The natural period of oscillation and the natural frequencyis respectively established from:

and

depend only on the mass and stiffness of thesystem, which are properties of the system.

For this case and any case where vibration is measuredfrom gravity equilibrium position, no gravity terms have to

be included in the equations of motion.

nnn f and ,

02 xx n

k

mn 2

mkn /2

m

kfn

2

1


EXAMPLE 1

A heavy ring of mass moment of inertia 1 kgm2 is attachedat the end of a two-layered hollow shaft of length 2 m. If

the two layers of the shaft are made of steel and brass,

repectively, determine the natural time period of torsional

vibration of the heavy ring. Shear modulus for steel and

brass are 80 GPa and 40 GPa, respectively.


EXAMPLE 1 (cont.)

Torsional stiffness for a hollow shaft is:

l

dDG

l

GIk Pt

32

44


EXAMPLE 1 (cont.)

For the steel shaft, torsional stiffness is:

For the brass shaft, torsional stiffness is:

Shafts are in parallel, therefore:

Nm/rad 6.14490

232

04.005.01080 449

tSk

Nm/rad 1.3436

232

03.004.01040 449

tBk

Nm/rad 7.17926 tBtSeq kkk


EXAMPLE 1 (cont.)

The torsional natural frequency is:

The torsional natural time period is:

rad/s 89.1331

7.17926

0

J

keqn

s 047.089.133

22

n

n



DEGREE-OF-FREEDOM SYSTEMSENERGY METHOD


LEARNING OBJECTIVES


Derive the equation of motion of a single-degree-of-freedomsystem using the energy method.

Determine the natural frequency of a single-degree-of-freedom system using the energy method.

Understand that the energy method is based on the principleof conservation of energy and therefore is only valid for

undamped vibration systems.


INTRODUCTION

Dynamic systems can be characterized in terms of one ormore natural frequencies.

The natural frequency is the frequency at which the systemwould vibrate if it were given an initial disturbance and

then allowed to vibrate freely.

There are many available methods for determining thenatural frequency. The most utilized include:

Newtons Law of Motion

Rayleighs Method

Energy Method

Lagranges Equation.


ENERGY METHOD

The total energy of a conservative system (damping andexternal forces equal zero) is constant. Thus

Kinetic energy is the energy of motion, as calculated fromthe velocity.

Potential energy has several forms. One is strain energy.Another is the work done against a gravity field.

0UTdt

d T : kinetic energy

U : potential energy


RAYLEIGHS METHOD

The principle of conservation of energy in the context ofan undamped vibrating system can be restated as

Subscript 1 denotes the time when the mass passes throughits static equilibrium position; therefore U1=0.

Subscript 2 denotes the time corresponding to themaximum displacement of the mass; therefore T2=0.

For a system undergoing harmonic motion T1 and U2denote the maximum values of T and U, respectively.

2211 UTUT 1,2 : denotes two different

instants of time

maxmax UT


EXAMPLE 1

A manometer used in a fluid mechanics laboratory has auniform bore of cross-section area A.

If a column of liquid of length l and density r is set intomotion as shown in the above figure, find the frequency of

the resulting motion.

xx


SOLUTION TO EXAMPLE 1

Energy Method.

sradl

gn /

2

22

22

2

2

1

2

1

2

1

2

1

xx

gAxkxU

xlAxmT

r

r

02

02

xl

gx

xgxAxxlAUTdt

d

rr


SOLUTION TO EXAMPLE 1 (cont.)

Newtons Law.

02

2

xl

gx

xlAAxg

xmF

rr

sradl

gn /

2


EXAMPLE 2

A circular cylinder of mass m and radius r is connected bya spring of modulus k.

If it is free to roll on the rough horizontal surface withoutslipping, finds its natural frequency.

x

r k

m



Energy Method.

2

220

2

2

1

4

3

2

1

2

1

kxU

xmJxmT

02

3

02

3

kxxm

xkxxxmUTdt

d

xr

xr

mrJ

20

2

1

sradm

kn /

3

2



Newtons Law.

fFkxxm

xmF

xmF

rFr

xmr

rFJ

JM

f

f

f

2

1

2

1 2

0

0

kx

Ff

(1)

(2)

Insert (2) into (1)0

2

3 kxxm

sradm

kn /

3

2


EXAMPLE 3

Determine the effect of the massof the spring on the natural

frequency of the spring-mass

system.

k : spring stiffness

m : mass density of the spring (mass/length)

l : length of spring

k l

y

dy

m

x



Kinetic energy of the mass:

Potential energy of the spring:

Kinetic energy of the spring:

Total kinetic energy of the system:

2

2

1xmTm

2

2

1kxU

2

6

1xlTs m

2

3

1

2

1xlmT

m



Assume a harmonic motion where X is the maximumamplitude and n the natural frequency:

The maximum potential and kinetic energies are:

Equating Tmax and Umax, we have

tXtx ncos

2max

22max

2

1

3

1

2

1

kXU

XlmT n

m

srad

lm

kn /

3

1m


RAYLEIGHS METHOD: EFFECTIVE

MASS

Rayleighs method can be used for multi-mass or fordistributed mass systems, provided the motion of every

point in the system is known.

In multi-mass systems which are joined by rigid links,levers or gears, the motion of the various masses can be

expressed in terms of the motion of some specific point,

and thus reducing the system to 1 DOF.

Kinetic energy can then be written as .

meff is the effective mass or an equivalent mass at thespecified point where motion of the various masses are

referred to.

2

2

1xmT eff



MASS (cont.)

If the stiffness at the specified point is also known, thenatural frequency can be calculated from:

effn

m

k



MASS (cont.)

In distributed mass systems such as springs and beams, aknowledge of the distribution of the vibration amplitude

becomes necessary before the kinetic energy can be

calculated.

Rayleigh showed that with a reasonable assumption for thevibration amplitude, it is possible to take into account

previously ignored masses and arrive at a better estimate

for the fundamental frequency.


EXAMPLE 4

A simply supportedbeam of total mass m has

a concentrated mass M at

mid-span.

Determine the effectivemass of the system at

mid-span and find its

natural frequency. M : mass of

concentrated load

m : total mass of beam

l : length of beam

l

m

x

l/2

M



Assume the deflection of the beam to be due to aconcentrated load at mid-span:

The velocity is therefore:

3

max 43

l

x

l

xyy

3

max 43

l

x

l

xyy



The maximum kinetic energy of the beam is given by:

2maxmax

2

0max

4857.02

1

22

1 2

ymT

dxxyl

mT

l



If meq denotes the equivalent mass of the beam acting atmid-span, its maximum kinetic energy is:

The effective mass at mid-span is expressed as:

mm

ymT

eq

eq

4857.0

2

1 2maxmax

mMmeff 4857.0



The natural frequency is:

effn

m

k

mMlEI

mM

k

n

n

4857.0

48

4857.0

3



DEGREE-OF-FREEDOM SYSTEMSDAMPED FREE VIBRATIONS


LEARNING OBJECTIVES


Determine the equation of motion of a damped single-degree-of-freedom system.

Determine if the system is underdamped, critically dampedor overdamped.

Solve the equation of motion of the damped free response.

Determine the damped natural frequency of an underdampedsingle-degree-of-freedom system.

Determine the logarithmic decrement of an underdampedsingle-degree-of-freedom system, and relate it to the

damping ratio.


INTRODUCTION

The analysis of undamped free vibrations is only anidealization of real systems because it does not account for

the energy lost to friction.

Once set into motion, such idealized systems vibrateforever with a constant amplitude.

All real systems, however, lose energy to friction and willeventually stop unless there is a source of energy to keep

them going.

When the amount of energy lost in the system is small, theresults of undamped free vibrations are often in good

agreement with real systems.


DAMPED FREE VIBRATIONS

Common types of friction forces that can removemechanical energy from vibrating systems include:

Fluid friction (viscous damping force), which arises whenbodies move through viscous fluids;

Dry friction (coulomb friction), which arises when a bodyslides across a dry surface; and

Internal friction, which arises when a solid body isdeformed.

Damping caused by fluid friction is quite common inengineering work, and only linear viscous damping is

considered in this lecture.


THE LINEAR VISCOUS DAMPER

Viscous damping is usually represented by a dashpot,which consists of a piston moving in a cylinder filled with

viscous fluid.

The viscous dampers considered are linear; the magnitudeof the viscous damping force is directly proportional to the

velocity with which the damper is being extended or

compressed .

The constant of proportionality is called the coefficientof viscous damping. Units in the SI system are .

The direction of the viscous damping force is alwaysopposite to the direction of the velocity.

xcF c

Ns/m


VISCOUS DAMPED FREE VIBRATIONS

Consider a block of mass sliding on a frictionlesshorizontal surface.

Vibration is induced by displacing the block a distanceand then releasing it with an initial velocity of

.0x

00 vx

m



(cont.)

The elastic restoring force of the spring is stilldirected toward the equilibrium position (negative

coordinate direction).

Since the positive directions for the velocity andacceleration are the same as the positive coordinate

direction, the damping force also acts in the

negative coordinate direction.

Applying Newtons 2nd law to the block gives thedifferential equation of motion of the block

or

The above equation is a second-order linear differentialequation with constant coefficients.

kxFs

xcFd

xx

xmxckx 0 kxxcxm



(cont.)

From the theory of ordinary differential equations, thesolution of any linear, ordinary differential equation with

constant coefficients is always of the form: .

and must be chosen to satisfy the differentialequation and the initial conditions.

Substituting the solution into the governing equation gives:

If the constant is zero, the trivial solution isobtained, which is of no interest. Since the exponential

is never zero, we have the characteristic equation:

a

0xa

taetx

02 kcm

02 kcmae t

te



(cont.)

The characteristic equation has roots

The displacement of the block is then given by

The constants are determined from the initial conditions.

The roots can be rewritten in terms of a non-dimensionalparameter known as the damping ratio.

tt eaeatx 21 21

0221121 and ;0 0 vaaxxaaxt

m

mkcc

2

42

2,1

nm

c

mk

c

22



(cont.)

In terms of the damping ratio and the natural frequency, the roots of the characteristic equation become:

The behavior of the system depends on whether thequantity under the radical is positive, zero or negative.

The value of that makes the radical zero is called thecritical damping coefficient .

The solution will have three distinct types of behaviordepending on whether the actual system damping is

greater than, equal to, or less than .

ccc

n

mkmc nc 22

122,1 nn

c

cc


OVERDAMPED SYSTEMS

When the damping coefficient is greater than , thenthe damping ratio , and the radical is real.

The two roots and are both real and are unequal,and since , both roots will be negative.

The solution then becomes:

c

1

12

1cc

2

121 nn

122 nn

ttt nnn eaeaetx

1

2

1

1

22


OVERDAMPED SYSTEMS (cont.)

The real-valued constants of integration and aredetermined by the initial conditions.

1a 2a

12

1

2

0

2

0

1

n

n xva

12

1

2

0

2

0

2

n

n xva


OVERDAMPED SYSTEMS (cont.)

Therefore the displacement simply decreases to zero asincreases, and the motion is non-vibratory.t


CRITICALLY DAMPED SYSTEMS

When the damping coefficient is equal to , then thedamping ratio , and the radical is zero.

The two roots and are equal and negative.

The solution in this case has the special form:

The real-valued constants of integration and aredetermined by the initial conditions.

c

tnetaatx 21

1cc

n 21

1a 2a

01 xa

002 xva n

1 2


CRITICALLY DAMPED SYSTEMS (cont.)

The displacement simply decreases to zero as increases,and the motion is non-vibratory.

Qualitatively the motion for critical damping is the same asthe motion for overdamped system.

t


CRITICALLY DAMPED SYSTEMS (cont.)

Critical damping is of special importance because it is thedividing point between non-vibratory motions and damped

oscillatory motions.

Critical damping is the smallest amount of damping forwhich a system will not oscillate.

A critically damped system will come to rest in less timethan any other system starting from the same initial

conditions (i.e. the value of damping that provides the

fastest return to zero without oscillation).


UNDERDAMPED SYSTEMS

When the damping coefficient is less than , then thedamping ratio , and the radical is imaginary.

The two roots are complex conjugates.

The constant is called the damped naturalcircular frequency.

The solution then becomes:

c

21 nd

1 dn j

1

cc

21

dn j 2

tjtjt ddn eaeaetx 21


UNDERDAMPED SYSTEMS (cont.)

By making use of the Euler formula ,the equation for the displacement can be written as:

The constants and are evaluated using the initialconditions.

sin0 0 Axx

cot0 000 dn xxvx

sincos je j

tAe

tAtAe

taajtaaetx

d

t

dd

t

dd

t

n

n

n

sin

sincos

sincos

21

2121

A



The constants and can be expressed as:

With this value of , the sine becomes:

Thus value of and are determined to be:

sin0xA

00

0tanxv

x

n

d

A

202

00

0sin

dn

d

xxv

x

2

2

0

2

00

d

dn xxvA

00

01tanxv

x

n

d

A



Motion of the underdamped system is called time-periodic.

The motion oscillates about the equilibrium position, butthe amplitude decreases because the exponent is

negative.

tnAe



Since the amplitude decreases monotonically with time,the oscillation will never repeat itself exactly, unlike the

case of a undamped free vibration.

The constant is called the damped naturalcircular frequency.

Since for underdamped vibration , the dampednatural frequency will always be less than the

undamped natural circular frequency .d

10

n

21 nd



By analogy with the undamped free vibration, a dampednatural frequency and a damped natural period

may be defined as

It is interesting to note that the damped natural periodand the damped natural frequency are constant

(independent of time) even though the amplitude is not.

2

2

1

22

2

1

2

nd

d

nddf

d

d


EXAMPLE 1

A torsional pendulum has a natural frequency of 200cycles/min when vibrating in vacuum. The mass moment

of inertia of the disc is 0.2 kgm2. It is then immersed in oil

and its natural frequency is found to be 180 cycles/min.

Determine the damping constant. If the disc, when placed

in oil, is given an initial displacement of 2, find its

displacement at the end of the first cycle.


EXAMPLE 1 (cont.)

Undamped and damped torsional natural frequencies are:

Damping ratio is obtained from:

rad/s 85.1860

18022

rad/s 94.2060

20022

dd

nn

f

f

4355.094.20

85.1811

22

n

d


EXAMPLE 1 (cont.)

Torsional damping coefficient is obtained from:

Initial conditions are:

Damped natural period is:

Nms/rad 65.34355.094.202.022 0 nt Jc

0

rad 035.02

0

0

s 33.085.18

22

d

d


EXAMPLE 1 (cont.)

Value of is determined from:

rad 039.0

85.18

85.18035.0035.094.204355.002

22

2

2

0

2

00

d

dn


EXAMPLE 1 (cont.)

Value of is determined from:

rad 2.1

6.68

035.085.184355.00

94.20035.0tan

tan

1

00

01

n

d


EXAMPLE 1 (cont.)

At the end of 1 cycle, . The displacement isobtained from:

s 33.0 dt

097.0

rad 0017.0

2.133.085.18sin039.0

sin

33.094.204355.0e

tet dtn


LOGARITHMIC DECREMENT

Viscous damping in an underdamped 1-DOF systemcauses the vibration to decay exponentially.

The exponent is a linear function of the damping ratio .

A convenient measure of the amount of damping isprovided by the extent to which the amplitude has

fallen during one complete cycle of vibration.

Logarithmic decrement is defined as the naturallogarithm of the ratio of two successive amplitudes, and is

defined as follows, where is the damped natural

period:

dtxtx

ln

d

tx


LOGARITHMIC DECREMENT (cont.)

Substitution of the analytical form of the underdampedresponse yields:

As , the expression for the logarithmicdecrement reduces to:

The circular damped natural frequency is substituted intothe above equation to give:

ddd

t

d

t

tAe

tAedn

n

sin

sinln

2dd

dndne ln

22 1

2

1

2

n

n



Solving the expression for yields:

Logarithmic decrement can be determined from a record ofthe displacement response of an underdamped system.

Let and represent the times correspondingto consecutive displacements and measured one

cycle apart.

The logarithmic decrement is given by:

1x

1t

2

1lnx

x

224

dtt 122x



The peak measurements can be used over any integermultiple of the period to increase the accuracy over the

measurements taken at adjacent peaks ( is an integer

which corresponds to the number of complete cycles).

1

1ln1

mx

x

m

m



For or , the log decrement and thedamping ratio can be approximated as:

2 2

3.0 2


EXAMPLE 2

The free vibration response of an electric motor of weight500 N mounted on a flexible foundation is shown in the

figure below. Find

the undamped and damped natural frequencies of the electricmotor - foundation system, and

the spring constant and damping coefficient of thefoundation.


EXAMPLE 2 (cont.)


EXAMPLE 2 (cont.)

From the time-response plot, the damped natural period is0.2 s. The damped natural frequency is:

Log decrement is also obtained from this plot from:

693.01

8ln

3

1ln

1

1

1 mx

x

m

rad/s 416.312.0

22

d

d


EXAMPLE 2 (cont.)

Since , damping ratio is obtained from:

Undamped natural frequency is then computed from:

rad/s 608.3111.01

416.31

1 22

dn

11.02

693.0

2

2


EXAMPLE 2 (cont.)

Spring constant of the foundation is obtained from:

Damping constant of the foundation is:

Ns/m 4.35411.0608.3181.9

50022 nmc

N/m 8.50920608.3181.9

500 22 nmk


REVIEW OF FREE VIBRATIONS

There are four cases of interest with regards to the freevibrations of a single degree-of-freedom-system.

Undamped system (0)

Underdamped system (0

REVIEW OF FREE VIBRATIONS (cont.)

Characteristics roots in the Argand (complex) plane.


HARMONICALLY EXCITED

VIBRATION OF SINGLE-DEGREE-

OF-FREEDOM SYSTEMSUNDAMPED FORCED VIBRATIONS


LEARNING OBJECTIVES


Compute the homogeneous (transient) and particular(steady-state) solutions for the undamped forced vibration of

a single-degree-of-freedom system.

Understand the concept of resonance and the definition ofmagnification factor.

Understand the concept of beat phenomena.


INTRODUCTION

Free vibration takes place when a system oscillates underthe forces inherent in the system itself, and when the

external forces are absent.

Forced vibration takes place when a system oscillatesunder the action of external forces. When the excitation

force is harmonic, the system is forced to vibrate at the

excitation frequency.

If the frequency of excitation coincides with one of thenatural frequencies, resonance is encountered.

Resonance is a phenomena in which the amplitude buildup to dangerously high levels, limited only by the degree

of damping.


EQUATION OF MOTION

The equation of motion for a viscously damped spring-mass system subjected to an external excitation force is

given by:

k

c

x

F(t)m

tFkxxcxm


EQUATION OF MOTION (cont.)

The general solution is given by the sum of thehomogeneous solution and the particular solution.

The homogeneous solution represents the free vibration ofthe system; the free vibration eventually dies out with time.

The general solution therefore reduces to the particularsolution which represents the steady state motion of the

same frequency as that of the excitation.



Homogeneous, particular and general solutions for anunderdamped forced vibration system.


UNDAMPED FORCED HARMONIC

MOTION

The equation of motion for a spring-mass system subjectedto an external harmonic force excitation is given by:

The homogeneous solution of this equation is givenby

Because the exciting force is harmonic the particularsolution is also harmonic and has the same

frequency .

txp

txh

tFkxxm cos0

tAtAtx nnh sincos 21

tXtxp cos



MOTION (cont.)

Solving for , we obtain:

The total solution therefore is the sum of the homogeneoussolution and the particular solution . txp txh

X

2

0

mk

FX

tmk

FtAtAtx nn

cossincos

2

021



MOTION (cont.)

Using the initial conditions and :

The total solution therefore is given by:

00 xx 00 vx

2

001

mk

FxA

n

vA

0

2

tmk

Ft

vt

mk

Fxtx n

n

n

cossincos2

00

2

00



MOTION (cont.)

1n

1n


MAGNIFICATION FACTOR

The maximum amplitude can also be expressed as

has a physical significance: it is the static deflectionof the spring under the constant load .

kF /0

st

X

0F

2

0

1

n

k

F

X


MAGNIFICATION FACTOR (cont.)

The previous equation can be expressed in the followingform:

The quantity represents the ratio of the dynamicto the static amplitude of motion and is called the

magnification factor, amplification factor or amplitude

ratio.

stX /

2

1

1

n

st

X


X/st vs. /n

n /

stX / Variation of the amplitude ratio with the frequencyratio .


X/st vs. /n (cont.)

For , the response of the system is in phasewith the excitation force.

For , the response is 180 degrees out of phasewith the external force.

1/0 n

1/ n

1/0 n 1/ n


X/st vs. /n (cont.)

As approaches infinity, the vibration responseapproaches 0.

For , the vibration response becomes infinite.

This condition, for which the forcing frequency isequal to the natural frequency of the system , is called

resonance.

At resonance, the vibration response increases linearlywith time.

1/ n

n

n /


RESONANCE

As the forcing frequency becomes exactly equal to thesystems natural frequency, , the solution becomes:

Evaluating the initial displacement and velocityyields:

n

ttf

tAtAtx nn

nn

sin2

sincos 021

m

Ff 00

0x

0v

ttf

tv

txtx nn

n

n

n

sin2

sincos 000


RESONANCE (cont.)

grows without bound, and this defines the phenomenaof resonance.

The amplitude of vibration becomes unbounded at:

mkn

tx


BEAT PHENOMENA

A very important phenomena occur when the drivingfrequency becomes close, but not exactly, to the systems

natural frequency.

This phenomena is known as beat.

In this kind of vibration, the amplitude builds up and thendiminishes in a regular pattern.

For zero initial conditions , the general solutionfor an undamped forced vibration system can be written as:

000 vx

ttmF

tx nn

coscos22

0


BEAT PHENOMENA (cont.)

The previous equation may be expressed as:

Let the forcing frequency be slightly less than the naturalfrequency where is a small positive

quantity.

Then and .

Multiplication of and yields:

tt

mFtx nn

n 2sin

2sin2

22

0

2n

n 2n

2n 2n

422 n



Substituting , andinto the general solution equation, we obtain:

If is small, the function varies slowly; its period,equal to , is large.

The above equation therefore represents a vibrationresponse with period and of variable amplitude

equal to:

2n 2n 422 n

ttmF

tx

sinsin2

0

tsin

2

2

tmF

sin2

0



It can be observed that the curve will go throughseveral cycles, while the wave goes through a singlecycle.

Thus the amplitude builds up and dies down continuously.

The time between the points of zero amplitude or thepoints of maximum amplitude is called the beat period.

The period of the beat is given by:

tsin

n

b

2

2

2

tsin



The beat frequency is given by:

2 nbeat


EXAMPLE 1

A mass m is suspended from a spring of stiffness 4000N/m and is subjected to a harmonic force having an

amplitude of 100 N and a frequency of 5 Hz. The

amplitude of the forced motion of the mass is observed to

be 20 mm. Find the value of m.


EXAMPLE 1 (cont.)

Forcing frequency is given by:

Static deflection is obtained from:

rad/s 42.31522 f

m 025.04000

1000 k

Fst


EXAMPLE 1 (cont.)

Use the following relationship to relate the magnificationfactor with the ratio of forcing to natural frequency.

Ratio of forcing to natural frequency is then solved.

8.0025.0

1020

1

1 3

2

st

n

X

5.1n


EXAMPLE 1 (cont.)

Natural frequency is determined from the previousequation.

The mass is obtained from:

rad/s 95.205.1

42.31

5.1

n

kg 1.995.20

400022

n

km


EXAMPLE 2

A spring-mass system is subjected to a harmonic forcewhose frequency is close to the natural frequency of the

system. If the forcing frequency is 39.8 Hz and the natural

frequency is 40.0 Hz, determine the period of beating.


EXAMPLE 2 (cont.)

Period of beating is obtained from:

s 5

8.39402

22

n

b




OF-FREEDOM SYSTEMSDAMPED FORCED VIBRATIONS


LEARNING OBJECTIVES


Compute the homogeneous (transient) and particular(steady-state) solutions for the damped forced vibration of a

single-degree-of-freedom system.

Appreciate the role of damping in forced vibration.

Determine the quality factor from the damped forcedresponse of a single-degree-of-freedom system and relate it

to the damping ratio.


EQUATION OF MOTION

The equation of motion for a viscously damped spring-mass system subjected to an external excitation force is

given by:

k

c

x

F(t)m

tFkxxcxm



The general solution is given by the sum of thehomogeneous solution and the particular solution.

The homogeneous solution represents the free vibration ofthe system; the free vibration eventually dies out with time.

The general solution therefore reduces to the particularsolution which represents the steady state motion of the

same frequency as that of the excitation.



Homogeneous, particular and general solutions for anunderdamped forced vibration system.


DAMPED FORCED HARMONIC

VIBRATION

The equation of motion for a viscously damped spring-mass system subjected to an external harmonic force

excitation is given by:

tFkxxcxm cos0



VIBRATION (cont.)

As the exciting force is harmonic, is also harmonicand has the same frequency .

Taking the appropriate derivatives of and insertingthem into the differential equation gives:

txp

sinsincoscos

cos

tXtX

tXtxp

tFtcmkX

tcmkX

cossincossin

cossincos

0

2

2

txp



VIBRATION (cont.)

As the solution is supposed to satisfy the differentialequation at every instant of time, the above equation must

hold for every instant of time.

Since the equation holds for all time, the constantcoefficients of and may be equated to

give:

tsin tcos

0cossin

sincos

2

0

2

cmkX

FcmkX



VIBRATION (cont.)

Solving the previous equations and taking intoconsideration the relationship between and

with gives:

2mk c

2220

cmk

FX

2

1tan

mk

c



VIBRATION (cont.)

Graphical representation of forcing function and response.



VIBRATION (cont.)

222

21

1

nn

st

X

2

1

1

2

tan

n

n

undamped

natural

frequency

critical

damping

damping

ratio

m

kn

nc mc 2

cc

c

Dividing the numerator and denominator of the previousequations by yields the non-dimensional equations:k



VIBRATION (cont.)

n / Variation of amplitude ratio and phase angle with .



VIBRATION (cont.)

In summary, we can write the differential equation and itscomplete solution, including the transient term as:

22

00

2

0

21

cos1sin

nn

n

t t

k

FteXtx n


VECTOR RELATIONSHIP IN FORCED

VIBRATION

For , both the inertia and damping forces aresmall, which results in a small phase angle.

The magnitude of the impressed force is then nearly equalto the spring force.

1n


VECTOR RELATIONSHIP IN FORCED

VIBRATION (cont.)

For the case of , the phase angle is 90.

The inertia force, which is now larger is balanced by thespring force, whereas the impressed force overcomes the

damping force.

The amplitude at resonance can be found from:

For large values of , approaches 180.

The impressed force is expended almost entirely inovercoming the large inertia force.

1n

k

F

c

FX

n 200

1n


QUALITY FACTOR AND BANDWIDTH

When the damping ratio is increased, two phenomena isobserved in the vicinity of the resonance frequency:

the amplitude tends to decrease, and

the amplitude peak tends to shift to the left of .

The maximum value of occurs at a frequency of:

The maximum value of the amplitude ratio is given by:

1n

2max 12

1

st

X

stX

221 n



(cont.)

For light damping , the curves are approximatelysymmetric about the vertical through .

The peak value of occurs in the immediate vicinityof and is given by:

The value of the amplitude ratio at resonance is called QFactor or Quality Factor.

The Q Factor is a measure of damping in a system.

1n

QX

st

2

1

05.0

stX

1n



(cont.)

Harmonic response curve showing half power points andbandwidth.



(cont.)

Points and where the amplitude ofreduces to are called the half power points.

The difference in the frequencies of points and iscalled the Bandwidth (or 3 dB Bandwidth)of the system.

For light damping , it can be shown that:

and are the frequencies at points and ,respectively.

1R

122

1

nQ

05.0

stX

2Q

2R

1R 2R

1 2 1R 2R


EXAMPLE 1

Find the response of a single-degree-of-freedom systemwith m=10 kg, c=20 Ns/m, k=4000 N/m, x0=0.01 m, and

released from rest at time t=0 s, under the following

conditions:

Free vibration with F(t)=0, and

An external force F(t)=F0cos(t) with F0=100 N and =10rad/s (neglect the transient response of the system).


EXAMPLE 1 (cont.)

Undamped natural frequency is given by:

Critical damping is given by:

Damping ratio is given by:

rad/s 2010

4000

m

kn

Ns/m 400201022 nc mc

05.0400

20

cc

c


EXAMPLE 1 (cont.)

Damped natural frequency is given by:

Ratio of forcing frequency to undamped natural frequencyis given by:

Free response of an underdamped system is given by:

rad/s 97.1905.01201 22 nd

00 sin teXtx d

t

hn

5.020

10

n

r


EXAMPLE 1 (cont.)

Determine X0 and 0 from:

m 01.0

97.19

97.1901.001.02005.002

22

2

2

0

2

000

d

dn xxvX

rad 52.101.02005.00

97.1901.0tantan 1

00

01

0

xv

x

n

d


EXAMPLE 1 (cont.)

Free response is therefore:

Forced response is given by:

2220

21

cos

rr

t

k

Ftxp

m 52.197.19sin01.0

52.197.19sin01.0

sin

2005.0

00

te

te

teXtx

t

t

d

t

hn


EXAMPLE 1 (cont.)

Determine from:

Forced response is then:

rad 067.05.01

5.005.02tan

1

2tan

2

1

2

1

r

r

m 067.010cos033.0

5.005.025.01

067.010cos

4000

100

21

cos

222

222

0

t

t

rr

t

k

Ftxp




OF-FREEDOM SYSTEMSRESPONSE DUE TO BASE EXCITATION


LEARNING OBJECTIVES


Differentiate between the two cases of damped forcedvibration: (1) the mass is acted upon by a harmonic force

and (2) the support or base is subjected to harmonic

excitation.

Compute the motion of a mass when its base or support issubjected to harmonic displacement.

Compute the force transmitted to the base or the support dueto the reactions from the systems spring and dashpot.


INTRODUCTION

This lecture is concern with a special case of dampedforced vibration which is of significant importance.

The damped forced vibration response is due to theharmonic excitation of the base or support of a system.

The forcing in this system, unlike the case of a forcingfunction of acting on the mass, is due to

the displacement of the base or support.

tFtF sin0 tYty sin


RESPONSE TO BASE EXCITATION

The base or support of a spring-mass-damper systemsometimes undergoes harmonic motion.

Forced vibration may therefore arise from the harmonicexcitation of the base or support of a system.


RESPONSE TO BASE EXCITATION (cont.)

Let denote the displacement of the base and thedisplacement of the mass from its static equilibrium

position at time .

The net elongation of the spring is and the relativevelocity between the two ends of the damper is .

In the displaced position the unbalanced forces are due tothe damper and the springs, and the equation of motion

becomes:

0 yxkyxcxm

ty tx

yx yx

t



Making the substitution and assuming thesolution , the previous equation becomes:

The solution to the above equation is ,which yields the steady-state amplitude and phase for the

relative motion:

yxz

tYmymkzzczm sin2

tYty sin

tZtz sin

2222

cmk

YmZ

2

1tan

mk

c



If the steady-state absolute motion of the mass, , isdesired, we can solve for .

The solution is made easier if we use the exponential formof the harmonic motion given by:

yzx

tx

tjjtj

tjjtj

tj

eXeXetx

eZeZetz

Yety



Substituting the exponential form of the harmonic motioninto the solution of the equation of motion yields:

From the relationship , we have:

jcmk

YmZe j

2

2

yzx

tjjtjj eYZeeXe



Substituting the exponential form into theexponential form of the relationship , we obtain:

jZe

yzx

tjtjj Yejcmk

cjkeXe

2



The ratio of the amplitude of the response to that ofthe base motion , is the Displacement

Transmissibility. Y

X

tx

22222

cmk

ck

Y

X

ty

22

31tan

cmkk

mc



dT n Variation of and with .



The value of is unity at , and close to unity forsmall values of .

For an undamped system ( ), at resonance.

The value of is less than unity ( ) for any amountof damping for .

The value of for all values of at .

dT

0

1dT

1dT

2n

2n

0n

n

dT

dT



For , smaller damping ratios lead to largervalues of .

For , smaller values of damping lead to smallervalues of .

The displacement transmissibility attains a maximumfor at the frequency ratio given by:

dT

10

2n

2n

n

dT

dT

21

1812

1 2

n


FORCE TRANSMITTED

A force is transmitted to the base or the support due to thereactions from the spring and the dashpot.

The force can be determined as:

The following relationship is obtained by assuming asolution in the form of : tjXetx

xmyxcyxkF

tjTtj eFXemF 2


FORCE TRANSMITTED (cont.)

is the amplitude or maximum value of the forcetransmitted to the base given by:

The ratio is known as the Force Transmissibility.

The transmitted force is in phase with the motion ofthe mass . tx

kY

FT

TF

21

222

222

cmk

ck

kY

F

n

T

TF


FORCE TRANSMITTED (cont.)

kY

FT n Variation of with for different values of .


EXAMPLE 1

A 3000 N heavy machine is supported on a resilientfoundation. The static deflection of the foundation due to

the weight of the machine is 0.075 m. It is observed that

the machine vibrates with an amplitude of 0.01 m when the

base of the foundation is subjected to harmonic oscillation

at the undamped natural frequency of the system with an

amplitude of 0.0025 m. Find the:

Damping constant of the foundation.

Dynamic force amplitude on the base.

Amplitude of displacement of machine relative to the base.


EXAMPLE 1 (cont.)

The stiffness of the foundation is given by:

The undamped natural frequency is given by:

The displacement transmissibility is given by:

N/m 40000075.0

3000

st

mgk

rad/s 11.4481.9/3000

40000

m

kn

400250

010

.

.

Y

XTD


EXAMPLE 1 (cont.)

It is given that . With this information, we solvefor the damping ratio using the following equation.

n

129.0

4164

1211

1214

22

222

2


EXAMPLE 1 (cont.)

The damping coefficient is obtained from:

The dynamic force amplitude on the base is:

Ns/m 6.902129.044.1181.9

300022 nc mcc

N 400

40025.0140000 2

2

Y

XkYF

n

T


EXAMPLE 1 (cont.)

The displacement of the machine relative to the base isgiven by:

m 0097.0

44.116.90244.1181.9

300040000

0025.044.1181.9

3000

2

2

2

2

222

2

cmk

YmZ




OF-FREEDOM SYSTEMSRESPONSE DUE TO ROTATING

UNBALANCE


LEARNING OBJECTIVES


Differentiate between the two cases of damped forcedvibration: (1) the mass is acted upon by a constant harmonic

force and (2) the mass is acted upon by a frequency-

dependent harmonic force.

Compute the motion of a mass when it is subjected toharmonic force due to rotating unbalance.


INTRODUCTION

This lecture is concern with a special case of dampedforced vibration which is of significant importance.

The damped forced vibration response is due to rotorimbalance, which is common in rotating machinery.

The forcing in this system, unlike the case of a forcingfunction of constant amplitude , is due to

imbalance force in rotating machinery whose force

amplitude is dependent on frequency .

tFtF sin0

tmetF sin2


RESPONSE TO ROTATING UNBALANCE

Unbalance in rotating machines is a common source ofvibration excitation.

The unbalance is represented by an eccentric mass witheccentricity which is rotating with angular velocity .e

m



(cont.)

Let be the displacement of the non-rotating massfrom the static equilibrium position.

The displacement of is therefore given by:

The equation of motion can therefore be written as:

x mM

tex sin

m

xckxtexdt

dmxmM sin

2

2



(cont.)

The previous equation may be rearranged to give:

This equation is similar to the case of a damped forcedvibration, where the force is given by , if we

replace with .

The steady-state solution is therefore given by:

2me

tFtF sin00F

tmekxxcxM sin2

2222

cMk

meX

2

1tan

Mk

c



(cont.)

The steady-state solution can be expressed in non-dimensional form as:

22

2

2

21

nn

n

me

MX

2

1

1

2

tan

n

n



(cont.)

Variation of with for different values of .me

MXn



(cont.)

All the curves begin at zero amplitude.

The amplitude near resonance is significantlyaffected by damping.

At very high speeds , is almost unity, andthe effect of damping is negligible.

For , the maximum occurs when:

1n me

MX

n

210 me

MX

121

1

2

n

0

me

MX

d

d

n



(cont.)

The corresponding maximum value of is given by:

The peaks occur to the right of the resonance value.

For , does not attain a maximum.

Its value grows from 0 at to 1 as .

me

MX

21me

MX

n0n

1n

2max 12

1

me

MX


EXAMPLE 1

A centrifugal pump weighing 600 N and operating at 1000rpm is mounted on six springs of stiffness 6000 N/m each.

Find the maximum permissible unbalance in order to limit

the steady-state deflection to 5 mm peak-to-peak.


EXAMPLE 1 (cont.)

The mass of the centrifugal pump is given by:

The angular frequency is given by:

The equivalent spring stiffness is given by:

kg 16.6181.9

600

g

Wm

rad/s 104.7260

10002

60

2

N

N/m 3600060006 eqk


EXAMPLE 1 (cont.)

The natural frequency is:

The ratio of the forcing frequency to the natural frequencyis:

rad/s 26.2416.61

36000

m

keqn

66.1832.4

32.426.24

72.104

22

r

rn


EXAMPLE 1 (cont.)

The unbalance can be found using the following equation:

The unbalance is therefore:

m-kg 145.072.104

166.180025.0360001

22

2

0

rkXem

12

2

0

rk

emX


TWO-DEGREE-OF-FREEDOM

SYSTEMSEQUATIONS OF MOTION AND FREE

VIBRATION RESPONSE


LEARNING OBJECTIVES


Understand the difference between generalized and principalcoordinates.

Determine the equations of motion of a two-degree-of-freedom system.

Express the equations of motion in matrix form.

Determine the characteristics equation, undamped naturalfrequencies and associated mode shapes.


INTRODUCTION

Most engineering systems are continuous and have aninfinite number of degrees of freedom.

For simplicity of analysis, continuous systems are oftenapproximated as multi-degree of freedom systems, which

requires the solution of a set of ordinary differential

equations.

There is one equation of motion for each degree offreedom.

There are n natural frequencies, each associated with itsown mode shape, for a system having n degrees of

freedom.


MATHEMATICAL MODELING

A continuous sys