mechanical vibrations

© S.Adhikari 2012 1 of 18

EG-260 DYNAMICS 1 COURSE HANDBOOK

1 VIBRATION AND THE FREE RESPONSE 2

1.1 Simple Harmonic Motion (SHM) 2

1.2 Single degree of freedom system (SDOF) 2

1.3 DAMPING 3

1.4 The idealised dashpot 3

1.5 Characteristic equation 3

1.6 Critical damping coefficient and the damping ratio 4

1.7 The damped solution 4

2 MEASUREMENT 5 2.1 The log decrement method 5

2.1.1 Under damped system 5 2.1.2 Measurement over more than one cycle. 5

2.2 Added mass method 5

3 STIFFNESS 5 3.1 Combination of springs 6

3.1.1 Springs in parallel 6 3.1.2 Springs in series 6

4 MODELLING AND ENERGY METHODS 6 4.1 A vertical spring-mass system 6

5 STABILITY 7 5.1 Stable 7

5.2 Unstable, divergent 8

5.3 Flutter 8

6 HARMONIC EXCITATION 8 6.1 Equation of motion 8

6.2 Harmonic excitation of undamped SDOF system 8

6.3 The phenomenon of beats 9

6.4 Resonance 10

6.5 Harmonic excitation of damped SDOF system 10

6.5.1 Equation of motion 11 6.5.2 Particular Solution 11 6.5.3 The Complementary Solution 11 6.5.4 The Complete Solution 11 6.5.5 Transient and steady state responses 11

7 RESONANCE IN DAMPED SYSTEMS 12 8 ALTERNATIVE solution methods 12

8.1 The geometrical method. 12

8.2 The frequency response method 12

8.3 The transform method 12

9 HARMONIC EXCITATION OF THE BASE 13 9.1 Force transmitted to the base 13

10 ROTATING IMBALANCE 14 11 MEASUREMENT 14

11.1 Transducers 14

11.2 Accelerometers 14

12 OTHER FORMS OF DAMPING 15 12.1 Fundamental definition of damping. 15

12.2 Viscous damping 15

12.3 Coulomb damping 15

12.4 Structural Damping 16

13 FOURIER TRASFORM 16 13.1 Fourier Series 16

14 MULTIPLE DEGREES OF FREEDOM 17 15 LAGRANGE’S METHOD 17

15.1 An overview of the procedure. 18

15.2 Advantages of Lagrange’s method 18


mk

n =ω

( )tx

1 VIBRATION AND THE FREE RESPONSE 1.1 Simple Harmonic Motion (SHM) Simple harmonic motion is defined as oscillating motion of an object about a fixed point, such that the acceleration, a, is: • Proportional to the displacement, x from the fixed point. • Always directed towards the fixed point. Equation for SHM is :

(1) is the acceleration (m/s2), x(t) is the displacement (m) and wn is the angular natural frequency (rads/s) and the negative sign indicates that the acceleration is always directed towards the fixed point. Newton’s 2nd Law

(2) Thus restoring force must be proportional to the displacement from a fixed point. ANY OSCILLATING SYSTEM FOR WHICH THE RESTORING FORCE MEETS THESE CONDITIONS WILL MOVE IN SHM Solution of the SHM equation Compact form of the solution is :

)tsin(A)t(x n φω += (3) where is the phase angle and is the natural frequency. Remember thee are other forms of the solution.

1.2 Single degree of freedom system (SDOF) Simplest form of oscillatory motion - a single mass attached to a light spring Considering the only the horizontal forces in Figure 1

( ) ( )tkxtxm −= (4) Dividing through by m and re-arranging the equation of motion is:

( ) ( ) 0=+ txmktx

(5) Let natural frequency then equation of motion is:

( ) ( ) 02 =+ txtxn

ω (6)

( ) ( )txtx n2ω−=

( ) x( -xmmaF n constant)2 =−== ω

Equilibrium

+ -

0

x Mass, m

Friction-free surface

k

-kx m

g N

Figure 1 Spring - Mass system

Free body diagram


( )( )

000

0

0 =!"#

=

=t

vvxx

��

If the mass is displaced from equilibrium and then released from rest, the initial conditions are: Using

)sin()( φω += tAtxn (7)

and substituting for the initial conditions gives:

!!"

#$$%

&=

+= −

0

01

2

00

2

tanvxvx

A n

n

n ωφ

ωω ��

(8) Thus the solution is

( )!"#

$%&

''(

)**+

,+

+= −

0

01

2

00

2

tansinvxt

vxtx n

n

n

n ωω

ωω

(9) where the period T is

ωπ2

=T (10)

1.3 DAMPING Equation of motion

( ) ( ) ( ) ( )tftkxtxctxm =++ (11) The constant c is the damping coefficient and has units Ns/m or kg/s and the damping force is:

( ) ( )txctfc = (12) Damping dissipates energy and the physical representation for dissipating energy is the dashpot or damper.

1.4 The idealised dashpot The dashpot is modelled as a piston inside an oil filled cylinder as seen in Figure 2 Schematically the dashpot is indicated by For the homogeneous equation:

( ) ( ) ( ) 0=++ tkxtxctxm (13) Assume solutions of form

( ) taetx λ= (14) Successive integration gives: 1.5 Characteristic equation

02 =++ kcm λλ (15) which and has two solutions:

Mounting point Mounting point Seal

Casing

Piston Oil

Orifice x(t)

Figure 2 Idealised Dashpot

fc

Mass, m


k

c

x(t)

N

mg

fk

Figure 3 Schematic for SDOF system with viscous damping

Free body diagram

Martin Hayes

Martin Hayes

2

Martin Hayes

2

Martin Hayes


mmkc

mc

, 24

2

2

21−

±−=λ (16)

The roots will be either real or complex depending on the value of the discriminant, c2 - 4mk. 1.6 Critical damping coefficient and the damping

ratio The critical damping coefficient, ccr is defined as:

ncr

cr

mmkmc

kmc

ω22

2

==⇒

=

(17) And the non-dimensional damping ratio, �, is defined as:

ncr mc

kmc

cc

ωζ

22===

(18) 1.7 The damped solution The solution to the characteristic equation (16)may be expressed as:

1221 −±−= ζωζωλ nn, (19)

• Overdamped

041 2 >−⇒> mkcζ 1 and 2 will be distinct real numbers. Solution is of the form:

( ) tt eaeatx 2121

λλ += (20)

Does not oscillate but returns to rest position exponentially.

• Underdamped

041 2 <−⇒< mkcζ 1 and 2 will be a complex conjugate pair with negative real part. Solution is of the form:

( ) ( )φωζω += − tsinAetx dtn

(21) where the damped natural frequency, wd in rads/s, is given by:

( )21 ζωω −= nd (22) Oscillatory motion with decaying amplitude.

• Critically damped

041 2 =−⇒= mkcζ

1 and 2 will be equal negative numbers, such that �n. Solution is of the form:

( ) ( ) tnetaatx ω−+= 21 (23) Does not oscillate but returns to rest position. Critical damping provides the fastest return to zero without oscillation.


2 MEASUREMENT Mass and stiffness are measured statically but damping must be measured from a dynamic test. Hence, damping measurements are often the least reliable coefficient in the equations of motion. 2.1 The log decrement method Undamped system The logarithmic decrement method measures damping from the transient response of the system and represents the rate at which the amplitude decays. The logarithmic decrement is defined as:

(24) 2.1.1 Under damped system Displacement given by:

( ) ( )φωζω += − tsinAetx dtn

(25)

After substitution, the log decrement for an under-damped system is:

21

2

ζ

πζζωδ

−== Tn

(26) Thus the damping ratio is given by:

224 δπ

δζ

+=

(27) and once is known � may be calculated. If k and m are known then

the critical value of damping given by: and the damping coefficient c is

related to critical damping by thus:

2242

δπδ

+=

kmc (28)

2.1.2 Measurement over more than one cycle. If the amplitudes are measures over n cycles then the logarithmic decrement is given by:

(29)

2.2 Added mass method Mass and stiffness are usually measured from static tests, but occasionally this is not possible. In such cases the frequency of the system is measured before and after an additional mass m2 has been applied Given two values of frequency 1 and 2, then:

212

11 mm

kandmk

+== ωω

(30) which may be solved for m1 and k 3 STIFFNESS The stiffness of a body is directly related to the geometry and material properties of the body. Young’s modulus, E, commonly called the

!!"

#$$%

&

+=

)()(lnTtxtx

δ

kmccr 2=

crcc ζ=

!!"

#$$%

&

+=

)()(ln1nTtxtx

nδ


elastic modulus, units pascal (Pa), i.e. newtons per square metre, N/m2, as does G, the shear modulus or modulus of rigidity 3.1 Combination of springs A number of springs may be combined in many practical applications to form a single equivalent spring. There are two ways that the springs may be combined 3.1.1 Springs in parallel If n springs in parallel, the equivalent stiffness, keq is:

∑ ==

n

i ieq kk1 (31)

3.1.2 Springs in series For n springs in series, the equivalent stiffness, keq is:

∑=

=n

i ieq kk 1

11 (32)

4 MODELLING AND ENERGY METHODS The Energy Method is an alternative method to the Method of Forces to derive the equations of motion. If the total potential energy is U and the total kinetic energy of the system is T then:

T + U = constant (33) When T is a maximum U must be a minimum and vice versa, thus the equation of motion may be obtained by equating Tmax with Umax

4.1 A vertical spring-mass system The PE of the system is due to gravity and the spring. From above:

( )221

xkUmgxU springgrav +Δ=−= (34)

where is the static equilibrium position and x is the displacement from equilibrium. The negative sign for gravity PE indicates that the mass is below the reference point. The KE of the system is:

2

21xmT = (35)

Substituting in equation (33)gives:

( ) constant 21

21 22 =+Δ+− xkmgxxm (36)

Mass, m

Figure 4 Vertical Spring Mass System

mg

k

- static equilibrium

x(t)

k Ο - rest position

Free body diagram


Differentiating with respect to time: ( ) ( ) 0 =−Δ++ mgkxkxxmx (37)

But from the free body diagram in Figure 3

mgk =Δ (38) hence

( ) 0 =+ kxxmx (39) but the velocity can not be zero for all time, so

0 =+ kxxm (40)

which is the same equation as obtained using the summation of forces method. 5 STABILITY A system may be either: • Stable

Stable finite • Asymptotically stable • Unstable,

• Flutter

This is illustrated in Figure 5

The response equations in Figure 5 are:

• Stable ( ) ( )tsintx = a finite constant

• Asymptotically stable

( ) ( ) ( )( )02 →∞→

⇒= − txtLim

tsinetx t

• Unstable, divergent ( ) ( ) ( )tcoshtsinhtx += grows without bound

• Flutter ( ) ( )tsinetx t2= oscillates as it grows without

bound 5.1 Stable For the undamped case

( ) ( )ϕω += tsinAtx (41)

( )( )0→∞→ txtLim

unstable

stable

asymptotically stable

Figure 5 Response

( ) constant finite a≤tx

( ) bound without grows divergent,≤tx

( ) bound without grows as oscillates≤tx


Thus for all values of t ( ) ( ) AtsinAtx =+= ϕω (42)

Hence is bounded by the finite number A. In the damped case the motion may be unstable in one of two ways 5.2 Unstable, divergent The motion may grow without bound and does not oscillate, i.e. divergent instability 5.3 Flutter The motion may grow without bound oscillates, i.e. flutter. or self-excited vibration. In all such cases a source of energy is required to maintain the instability 6 HARMONIC EXCITATION Harmonic excitation refers to the excitation due to a sinusoidal external force of a single frequency 6.1 Equation of motion A force F (t) acts on a viscously damped spring-mass system is shown in Figure 6

Figure 6 Viscously damped SDOF spring-mass-damper system The equation of motion is:

( )tFkxxcxm =++ (43) The homogenous equation:

0=++ kxxcxm (44) represents the free vibration of the system, and dies out due to the effect of damping, i.e. underdamped, overdamped or critically damped Thus the general solution to equation (43)eventually reduces over time to particular solution, xp (t), i.e. to the steady state vibration. 6.2 Harmonic excitation of undamped SDOF

system An undamped system is harmonically excited by a horizontal force:

F0 cos(ω t) The equation of motion is:

( )tx

F(t)


k

c

x(t)

m

fc

mg

N fk F

Free body diagram


( )( )tcosfxx

tcosFkxxm

n ωω

ω

02

0

=+

=+

(45)

where mFf 00 = and mkn =2ω

First find xcf the solution to the complementary equation:

02 =+ xx nω (46) then consider the right hand side of equation. The particular solution will be of the form:

( )tcosXxp ω= (47) where X is the amplitude of the forced response, since the system is undamped .

if nωω ≠ : 220

ωω −=

n

fX (48)

and the particular solution is

( ) ( )tcosf

txn

p ωωω 220

−= (49)

Thus the complete solution is ( ) ( ) ( )txtxtx pcf += , Given initial conditions 00 vx the full solution is:

( ) ( ) ( ) ( )tcosf

tcosf

xtsinv

txn

nn

nn

ωωω

ωωω

ωω 22

022

00

0

−+##

$

%&&'

(

−−+=

(50)

6.3 The phenomenon of beats For initial conditions 00 00 == xandv equation (50) gives:

( ) ( ) ( )

( ) ( ) ( )( )tcostcosf

tx

tcosf

tcosf

tx

nn

nn

n

ωωωω

ωωω

ωωω

−−

=

−+

−−=

220

220

220

(51)

Equation (51) may be written as

( ) !"

#$%

& +!"

#$%

& −

−= tsintsin

ftx nn

n 222

220 ωωωωωω

(52)

If forcing frequency is close to natural frequency then ( )ωω −n is

small and ( )ωω +n is large so that !"

#$%

& − tsin n

2ωω has a much longer

period than !"

#$%

& + tsin n

2ωω i.e. phenomenon of BEATS

The beat frequency is the time between two successive maximums, i.e. half the time for one complete oscillation.: The beat frequency is (53) The period the beat is (54)

ωωω −= nbeat

( )ωωπ −= nbeatT 2


6.4 Resonance If n the particular solution ( )tcosXxp ω= to equation (50) is not valid (identical roots, see revision).

The particular solution is ( )tsintf

xp ωω20= (55)

The general solution is of the form:

( ) ( ) ( ) ( )tsintf

tcosAtsinAtx ωω

ωω20

21 ++= (56)

Given initial conditions 00 vx

( ) ( ) ( ) ( )tsintf

tcosxtsinv

tx ωω

ωωω 2

00

0 ++= (57)

The type of response given by equation (57) for a spring-mass system is shown in Figure 8

Figure 8 shows that the response x(t) grows without bound, which is the definition of resonance. For the spring-mass system this occurs if

mk

n ==ωω , in which case the spring would fail

6.5 Harmonic excitation of damped SDOF system

!"

#$%

& −

−tsinf n

n 22

220 ωωωω

ωπ2

( )ωωπ−n

4

Figure 7 Phenomenon of Beats

Figure 8 Resonance - Spring mass system

fc F(t)


k

c

x(t)

mg

N

fk

F

Free body diagram

Figure 9 Damped SDOF sy

Figure 9 Damped SDOF system


6.5.1 Equation of motion (58) Dividing by m gives

( )tcosfxmk

xmc

x ω0=++ (59)

where mFf 0

0 = . Equation (59) may be written in terms of the natural

frequency and damping ratio, giving: ( )tcosfxxx nn ωωζω 0

22 =++ (60) 6.5.2 Particular Solution The particular solution is assumed to be of the form ( ) ( )θω −= tcosXtx p (61)

The phase shift is due to the effect of the damping force The particular solution is given by:

( )( ) ( )

( )θωωζωωω

−+−

= tcosf

txnn

p 2222

0

2 (62)

The forcing term amplitude is given by:

( ) ( )2222

0

2 ωζωωω nn

fX+−

= (63)

The forcing term phase angle is given by:

!!"

#$$%

&

−= −

221 2

ωωωζω

θn

ntan (64)

6.5.3 The Complementary Solution For the under-damped case (0< � <1), the solution to the complementary function, i.e. the case of free vibration, is:

( ) ( )φωζω += − tAetx d

tn sin (65) 6.5.4 The Complete Solution The complete solution is:

( ) ( ) ( )θωφωζω −++= − tcosXtsinAetx dtn (66)

6.5.5 Transient and steady state responses Equation (66) may be considered as a combination of a transient response and a steady state response as below:

( ) ( ) ( ) response statesteady responsetransient

θωφωζω −++= − tcosXtsinAetx dtn (67)

Assuming that the system has relatively large damping, i.e. transient

term may be ignored, the frequency ratio r is defined asn

rωω

= , which

is a dimensionless quantity. The frequency ratio may be used in equations (63) and (64) to give the forcing term amplitude as:

( ) ( )222

20

21 rr

fX n

ζ

ω

+−= (68)

The forcing term phase angle as !"

#$%

&−

= −2

1

12rrtan ζ

θ (69)

( )tFkxxcxm =++


If the system is lightly damped, i.e. ζ very small, the transient term may last a sufficiently long time that it may be significant. 7 RESONANCE IN DAMPED SYSTEMS Resonance is defined to occur if nωω = i.e. r = 1. Equation (68) may be expressed in terms of the normalised amplitude as:

( ) ( )22200

2

21

1

rrFXk

fX n

ζ

ω

+−== (70)

The maximum value of 0FXk occurs at different values of r depending on

the value of the damping ratio � 8 ALTERNATIVE SOLUTION METHODS Alternatives to the method of undetermined coefficients are 8.1 The geometrical method. Uses the fact that position, velocity and acceleration will be out of phase with each other by /2 radians. Therefore express displacement, velocity and acceleration as a vector Compute X in terms of F0 via vector addition 8.2 The frequency response method If the forcing function for a harmonically excited system is expressed in complex form as:

( ) tjeFtF ω0= (71)

Then the equation of motion for the spring-mass-damper system may be expressed as:

tjeFkxxcxm ω0=++ (72)

But the excitation is real, so the response will also be the real part of x(t), where x(t) is a complex function that by definition must satisfy the equation of motion (72) and the particular solution must also be a complex number

( ) tjp Xetx ω= (73)

The solution proceeds using the complementary function and the particular solution using complex algebra to give:

( )( ) ( )

( ) ( )( )θωθωωω

−+−+−

= tsinjtcoscmk

Ftxp

222

0 (74)

Comparing the real part of equation (74) with equation (62), both methods yield the same result 8.3 The transform method The equation of motion for a spring-mass damper system subject to harmonic excitation is given by:

tcosFkxxcxm ω0=++ (75) which may be transformed using the Laplace transform:

dte)t(x))t(x()s(X st−∞

∫==0

L (76)

To give )s)(kcsms(

sF)s(X 2220

ω+++= : (77)

Then an inverse transformation must be performed to obtain the solution in the time domain


9 HARMONIC EXCITATION OF THE BASE Figure 10 shows a spring- mass-damper SDOF system undergoing harmonic excitation of the base. The displacement of the base is denoted by y(t) and the displacement of the mass from its static equilibrium position at time t is denoted by x(t). The equation of motion is:

( ) ( ) 0=−+−+ yxkyxcxm (78) The base is moving under harmonic motion such that ( ) ( )tsinYty ω= then equation (78) gives:

( ) ( )tcosYctsinkYkxxcxm ωωω +=++ (79) The particular solution is :

( ) ( )φω −= tsinXtxp (80) where

( )( ) ( )222

22

ωω

ω

cmkck

YX

+−

+= (81)

and

( ) ( ) !!"

#$$%

&

+−= −

22

31

ωω

ωφ

cmkkmctan

(82) Equation (81)may be expressed as the ratio of the maximum response amplitude and the input displacement amplitude:

( )( ) ( )222

2

2121

rrr

YX

ζ

ζ

+−

+=

(83) is the displacement transmissibility. Describes how motion is transmitted from the base to the mass of the system. is the frequency of the base and is frequency ratio 9.1 Force transmitted to the base A force F(t) is transmitted to the mass due to the reaction from the spring and dashpot. Must balance the inertial force, hence:

( ) ( ) ( ) xmyxcyxktF −=−+−= (84) The steady state solution is:

( ) ( )φω −= tsinFtF Tp (85) where FT is the amplitude or maximum value of the force transmitted and is given by:

Figure 10 Base excitation spring-mass-damper SDOF system

+ x

k c

m

y(t)= Ysin(t)

+ y

m

( )yxc −( )yxk −

+x

x+

Free body diagram

dTYX =

nr ωω=


( )( ) ( )222

22

2121

rrrr

kYFT

ζ

ζ

+−

+=

(86) The ratio is the force transmissibility. 10 ROTATING IMBALANCE A schematic of rotating imbalance of mass m0 a distance e, eccentricity, from the centre of rotation is shown in Figure 11 Assuming no motion in the horizontal direction due to t presence guides

Considering the forces for the rotating imbalance and the machine gives:

00 =+++ kxxcxmxm r (87) The particular solution is:

)tsin(X)t(x rp φω −= (88) The magnitude and the phase of the steady state motion of the mass m due to the rotating imbalance of mass m0

( )!"

#$%

&−

=+−

= −2

1

222

2

12

21 rr

tanandr)r(

rmem

X o ζφ

ζ (89) 11 MEASUREMENT 11.1 Transducers Transducers converts mechanical vibration into an electric signal such as voltage proportional to acceleration, Used to measure vibration. 11.2 Accelerometers Accelerometers are based on piezoelectric elements and are spring-mass-damper systems. Convert the acceleration into an electric signal

a b

Figure 11Free body diagram of the imbalance (a), and machine (b)

kYFT


OTHER FORMS OF DAMPING 11.3 Fundamental definition of damping. By definition the change in energy, E, is equal to the work done by damping forces over a cycle, so that:

(90)

Loss factor, , is defined as:

max2 UE

π

Δ=η (91)

11.4 Viscous damping Energy dissipated in a viscously damped system per cycle, with a viscous damping coefficient, c is

(92) Assuming harmonic motion

(93)

For harmonic motion 2

max 21 kXU =

(94) If resonance then nωω = so

(95)

11.5 Coulomb damping For Coulomb or dry friction damping force is constant in magnitude but opposite to the direction of motion of the vibrating body. Caused when components of a system slide relative to one another. Friction force, fc is given by:

(96) The equation of motion is:

(97) where sgn(y) is the signum function, where

(98) fc always opposes the direction of motion therefore equation of motion is nonlinear which may be solved as piecewise linear. Harmonic motion

(99) The change in energy for viscous damper is and the equivalent damping coefficient for Coulomb damping as:

∫=Δ dxFE d

∫=Δωπ2

0

2visc dtxcE

cXcXE 222visc ω

ωω π=#

$

%&'

( π=Δ

kc

UE ω

η =π

Δ=

2 max

viscvisc

ζω

η 2visc ==kc

!"

!#

$

<

=

>−

=

0)(0)(00)(

)(txNtxtxN

xfc

µ

µ

0)sgn( =++ kxxmgxm µ

( )( )( ) 0for1sgn

0for0sgn0for1sgn

>=

==

<−=

yyyyyy

gXmE

dtttXgmE

clmb

clmb

µ

ωωωµ

4

)cossgn(cos

=Δ

=Δ ∫

cXEvisc2ωπ=Δ

XgmcgXmcX eqeq ω

µµω

π=⇒=π442


(100) The equivalent damping ratio for a system with Coulomb damping is:

(101) If driving force is large with respect to the friction force, Coulomb damping can be approximated by viscous damping, using the effective relationships 11.6 Structural Damping Generated by friction, rubbing or impact of two surfaces at a joint. Common to assume some amplitude-dependent loss mechanism. For harmonic excitation the change in energy is defined as:

(102) where is a frequency dependent multiplier. An equivalent viscous damping coefficient is given by:

(103) The equivalent damping ratio for a system with structural damping is:

(104) The equation of motion with complex forcing function is:

(105)

Assuming the solution is ( ) tiXetx ω= gives:

(106) leads to the concept of “complex” stiffness. This analysis is only valid for harmonic excitation. This analysis is only valid for harmonic excitation. 12 FOURIER TRASFORM Any function which is periodic, can be expressed as a series of sinusoids. Given the solution for sinusoidal forcing function, the solution may be found of the dynamics of a system to any periodic input 12.1 Fourier Series A periodic function may be expressed as a sum of an infinite series of sine and cosine functions.

( ) ( )tnbtnaa

tf Tnn

Tnn

o ωω sincos2

)(11∑∑∞

=

∞

=

++= (107)

where :

∫=T

o dttfT

a0

)(2 (108)

( ) .....,.........3,2,1cos)(2

0== ∫ mdttmtf

Ta

T

Tn ω (109)

( ) ...............3,2,1sin)(20

== ∫ mdttmtfT

bT

Tn ω (110)

Xg

neq ωω

µζ

π=

2

( ) 2XkEstruc βπ=Δ

( )ωβ

πωβkcXcXk eqeq =⇒=π 22

eqn

eq mk

ζβωβ

ζ 22 2 =⇒=

tiFekxxk

xm ω

ωβ

=++

( ) tieFxikxm ωβ 01 =++


13 MULTIPLE DEGREES OF FREEDOM (MDOF) Number of degrees of freedom of a system dependant on the number of moving parts and the directions in which they may move. Each degree of freedom will have an associated natural frequency thus increasing the opportunity for the occurrence of resonance. Illustrated for two degrees of freedom.

Figure 12 MDOF example

TO SOLVE YOU MUST DRAW A FORCE DIAGRAM See Figure 13 MDOF Force diagram Only motion in horizontal direction so considering the forces on each mass in the horizontal direction only.

( )1221111 xxkxkxm −+−= (111)

( )12222 xxkxm −−= (112)

Expressed in matrix form as: 0=+KxxM (113)

where

!"

#$%

&=

2

1

00m

mM (114)

!"

#$%

&

−

−+=

22

221

kkkkk

K (115)

For non-trivial solution ( ) 02 =+− KMωdet (116)

Equation (116) is quadratic in so solve for the two frequencies (Eigenfrequencies) 1 and 2 To obtain the mode shapes (Eigenmodes) u 1 u 2 solve:

( ) 0121 =+− uKMω (117)

( ) 0222 =+− uKMω (118)

14 LAGRANGE’S METHOD Lagrangian methods use an energy approach to obtain the equations of motion in terms of generalised co-ordinates. The equation of motion in

x2

k1 k2

x1 m1 m2

m1

m2

k2(x2 - x1)

k2(x2 – x1) k1

x1

x1

x2

Figure 13 MDOF Force diagram


terms of the kinetic energy (T), potential energy (U) and generalised co-ordinated qi for an n degree of freedom system is given by:

( ) niQqU

qT

qT

dtd n

iiii

,.......2,1==∂∂

+∂∂

−##$

%&&'

(

∂∂ (119)

where is the generalised velocity and Qi

(n) is the non-conservative force corresponding to the generalised co-ordinates qi

If Fxk, Fyk and Fzk represent the external forces acting on the kth mass if the system in the x, y and z directions, respectively, the generalised force is given by:

( ) ∑ ""#

$%%&

'

∂

∂+

∂

∂+

∂

∂=

i

kzk

i

kyk

i

kxk

ni q

zF

qy

Fqx

FQ (120)

where xk, yk, zk are the displacements of the kth. mass in the x , y and z directions, respectively. For a torsional system the force Fxk is replaced by the moment Mxk about the x axis and the displacement xk is replaced by the angular displacement qxk about the x axis If the Lagrangian, L, is defined as L = T- U, then equation (119) may be expressed as:

iii

QqL

qL

dtd

=∂

∂−##$

%&&'

(

∂

∂

(121)

14.1 An overview of the procedure.

• Determine a set of generalized co-ordinates, denoted by q, which are an independent set of variables to record position.

• Express the kinetic and potential energy in terms of the generalised co-ordinates q.

• Express the work done by forces and moments. • Derive the equation of motion using equations of the type

expressed by equations (119) and (121) 14.2 Advantages of Lagrange’s method

• Equations contain only scalar quantities • One equation for each degree of freedom. • The resulting equations are independent of the choice of

coordinate system since kinetic and potential energy does not depend on coordinates

tqqi ∂∂=

mechanical vibrations

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