Almitani.com MENG470 Spring2011 Vibration of Two DOFS

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MENG-470 Spring 2011 Dr.Khalid Almatani 1

King Abdul Aziz University

Department of Production Engineering and Mechanical Systems Design

MENG-470

Spring 2011

Prepared and presented by

Dr. Khalid AlmataniAssistant professor of mechanical Engineering

Production Engineering and mechanical systems design department

College of Engineering

King Abdul Aziz University

Jeddah, Saudi ArabiaFebruary, 2011

INTRODUCTIONINTRODUCTION TO TO

LINEAR VIBRATIONSLINEAR VIBRATIONS



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Introduction to Linear Vibrations

UndampedUndamped Free Free

Vibration Vibration Of Of TwoTwo Degrees of Freedom SystemsDegrees of Freedom Systems



UNDAMPED FREE VIBRATION OF 2-D-O-F-S

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1x

1k

1m

2x

2k

2m

3k

1x

1m

2x

2m1 1

k x3 2

k x2 2 1( )k x x−




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First Mass:

Second Mass:

2 2 2 1 2 3 2( ) 0m x k x k k x− + + =&&

1 1 1 2 1 2 2( ) 0m x k k x k x+ + − =&&

Equations of Motion

Let 1 1cos( )x X tω φ= + & 2 2

cos( )x X tω φ= +




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2

1 1 2 1 2 2( ) 0m k k X k Xω − + + − =

&2

2 1 2 2 3 2( ) 0k X m k k Xω − + − + + =

Solution exists when:

2

1 1 2 2

2

2 2 2 3

( )det 0

( )

m k k k

k m k k

ω

ω

− + + − = − − + +




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[ ]4 2

1 2 1 2 2 2 3 1

2

1 2 2 3 2

( ) ( )

( )( ) 0

m m k k m k k m

k k k k k

ω ω− + + + +

+ + − =

2 1 2 2 2 3 211,2 2

1 2

2 1/ 22

1 2 2 2 3 2 1 2 2 3 212

1 2 1 2

( ) ( )

( ) ( ) ( )( )4

k k m k k m

m m

k k m k k m k k k k k

m m m m

ω + + +

=

+ + + + + −−

m

Solution:




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Values of X1 & X2:

2

2 1 1 22 2

1 2

1 2 2 1 2 31

( )

( )

m k kX kr

X k m k k

ω

ω

− + + = = =

− + +

&

2

2 2 1 22 2

2 2

1 2 2 2 2 32

( )

( )

m k kX kr

X k m k k

ω

ω

− + + = = = − + +




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For 1 1cos( )x X tω φ= + & 2 2

cos( )x X tω φ= +

1 1

2 2

x X

x X= Independent of time

Synchronous Motion at frequency ω

2 1 2 2 2 3 111,2 2

1 2

2 1/ 22

1 2 2 2 3 1 1 2 2 3 21

2

1 2 1 2

( ) ( )

( ) ( ) ( )( )4

k k m k k m

m m

k k m k k m k k k k k

m m m m

ω + + +

= ±

+ + + + + −−




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with2

2 1 1 22 2

1 2

1 2 2 1 2 31

( )

( )

m k kX kr

X k m k k

ω

ω

− + + = = =

− + + &

2

2 2 1 22 2

2 2

1 2 2 2 2 32

( )

( )

m k kX kr

X k m k k

ω

ω

− + + = = = − + +

{ } 1 1

1

2 1 11 1

X XX

X r X

= =

{ } 1 1

2

2 2 12 2

X XX

X r X

= =

&

1 2&ω ω Natural frequencies

{ } { }1 2

&X X Modes shapes or normal modes




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{ }1 1

1 1 1 1

1 1 1

2 1 1 1 1

cos( )

cos( )

x X tx

x r X t

ω φ

ω φ

+ = =

+

Final Solutions:

{ }2 2

1 1 2 2

2 2 2

2 2 1 2 2

cos( )

cos( )

x X tx

x r X t

ω φ

ω φ

+ = =

+ &

1 2

1 1 1 2, , , &X X φ φwhere

are determined from initial conditions




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Initial Conditions

The system can be made to vibrate in its ith mode by

subjecting it to specific initial conditions:

1 1 1

2 1 2

( 0) , ( 0) 0,

( 0) , ( 0) 0,

i

i

i

x t X x t

x t r X x t

= = = =

= = = =

&

&




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General Response { } { } { }1 2

1 2x c x c x= +

{ } { } { }1 2x x x= +

&

1 2

1 1 1 1 1 2 2cos( ) cos( )x X t X tω φ ω φ= + + +

1 2

2 1 1 1 1 2 1 2 2cos( ) cos( )x r X t r X tω φ ω φ= + + +

Without loss of generality

1 2

1 1 1 2, , , &X X φ φwhere

are determined from initial conditions




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0 0

0 0

1 1 1 1

2 2 2 2

( 0) , ( 0) ,

( 0) , ( 0) ,

x t x x t x

x t x x t x

= = = =

= = = =

& &

& &

Initial Conditions:

0

1 2

1 1 1 1 1 2( 0) cos cosx t x X Xφ φ= = = +

0

1 2

2 2 1 1 1 2 1 2( 0) cos cosx t x r X r Xφ φ= = = +

0

1 2

1 1 1 1 1 2 1 2( 0) sin sinx t x X Xω φ ω φ= = = − −& &

&

0

1 2

2 2 1 1 1 1 2 2 1 2( 0) sin sinx t x r X r Xω φ ω φ= = = − −& &




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Solution:

0 02 1 21

1 1

2 1

cosr x x

Xr r

φ−

= −

0 01 1 22

1 2

2 1

cosr x x

Xr r

φ− +

= −

0 02 1 21

1 1

1 2 1

sin( )

r x xX

r rφ

ω

− + =

−

& &&

,

0 02 1 22

1 2

2 2 1

sin( )

r x xX

r rφ

ω

− =

−

& &

{ } 0 0

0 0

1

2 22 2 1 21

1 2 1 2

2 1 1

1

( )

r x xX r x x

r r ω

− + = − +

−

& &

{ } 0 0

0 0

1

2 22 2 1 22

1 2 1 2

2 1 1

1

( )

r x xX r x x

r r ω

− = − + +

−

& &




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&

0 0

0 0

2 1 21

1

1 2 1 2

tan[ ]

r x x

r x xφ

ω

− − +

= −

& &

0 0

0 0

1 1 21

2

1 1 1 2

tan[ ]

r x x

r x xφ

ω

− −

= − +

& &

{ } 0 0

0 0

1

2 22 2 1 21

1 2 1 2

2 1 1

1

( )

r x xX r x x

r r ω

− + = − +

−

& &

{ } 0 0

0 0

1

2 22 2 1 22

1 2 1 2

2 1 1

1

( )

r x xX r x x

r r ω

− = − + +

−

& &




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m2

x2

k k=

3k k=

1x

1k k=

m

Equations of Motion

Let 1 1cos( )x X tω φ= +

& 2 2cos( )x X tω φ= +

1 1 22 0mx kx kx+ − =&&

2 1 22 0mx kx kx− + =&&&

EXAMPLE:




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( )

( )

2

2

2det 0

2

m k k

k m k

ω

ω

− + − = − − +

Conditions for Synchronous Motion:

2 4 2 24 3 0m km kω ω− + =

Solution:

1,2

3,

k k

m mω =




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2

12

1 2

1 11

21

2

m kX kr

X k m k

ω

ω

− + = = = = − +

2

22

2 2

1 22

21

2

m kX kr

X k m k

ω

ω

− + = = = = −

− + &

1

1 1

1

1

1 1

cos

{ }

cos

kX t

mx

kX t

m

φ

φ

+

=

+

Normal Modes:

2

1 2

2

2

1 2

3cos

{ }3

cos

kX t

mx

kX t

m

φ

φ

+

=

− +

&



1

1 1

1

1

1 1

cos

{ }

cos

kX t

mx

kX t

m

φ

φ

+

=

+

Normal Modes:

2

1 2

2

2

1 2

3cos

{ }3

cos

kX t

mx

kX t

m

φ

φ

+

=

− +

&


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m1 m2

Mode 1 Mode2m1

Node

m2




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1 2

1 1 1 1 2

3cos cos

k kx X t X t

m mφ φ

= + + +

General Response

&

1 2

2 1 1 1 2

3cos cos

k kx X t X t

m mφ φ

= + − +




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Redo the previous example taking k=1N/m and m=1kg

m2

x2

k k=

3k k=

1x

1k k=

m1 1 2

2 0x x x+ − =&&

2 1 22 0x x x− + =&&

&

Equations of Motion:

Or in a matrix form:




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Let 1 1cos( )x X tω φ= + & 2 2

cos( )x X tω φ= +

Which upon substitution yields:

( )

( )

2

2

2 1det 0

1 2

ω

ω

− + − = − − +

Conditions for Synchronous Motion




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4 24 3 0ω ω− + =

Solution:

2 2 2

1, 21 , 3 /rad sω =

{ } 1

1

12 1

1

1

XX

X

= =

{ } 1

2

22 2

1

1

XX

X

= =

− &Also




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>> k=[2 -1;-1 2];

>> m=eye(2);

>> A=inv(m)*k

>> [v,e]=eig(A)

A =

2 -1

-1 2

v =

-0.7071 -0.7071

-0.7071 0.7071

e =

1 0

0 3

MATLAB CODES

INPUT OUTPUT




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Coordinate Coupling

&

Principal Coordinates




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m2

x2

k k=

3k k=

1x

1k k=

m

1 1 22 0mx kx kx+ − =&&

2 1 22 0mx kx kx− + =&&

& Coupled Eqns.

1 2

1 1 1 1 2

3cos cos

k kx X t X t

m mφ φ

= + + +

1 2

2 1 1 1 2

3cos cos

k kx X t X t

m mφ φ

= + − +

Solution for Physical Coordinates:

&

EXAMPLE (Revisit):

Equations of Motion




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let1

1 1 1cos

kq X t

mφ

= +

2

2 1 2

3cos

kq X t

mφ

= +

&

Add the two equations

Subtract the two equations




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Un-Coupled Equations:

&

These are two un-coupled, linear and first degree of

freedom systems. So, one can solve for the Principal

Coordinates q1 & q2 independently, and then find the

solutions in “physical coordinates” as:

&

Where q1 & q2 are defined as the “Principal Coordinates”



FORCED VIBRATION OF 2-D-O-F-S

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SIMULINK OUTPUTSX1:




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SIMULINK OUTPUTSX2:



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Forced Forced Vibration Vibration Of Two Degrees of Freedom Of Two Degrees of Freedom

SystemsSystems




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1x

1k

1c

1m

2x

2k

2c

2m

3k

3c

1x

1m

2x

2m1 1

k x3 2

k x

1 1c x& 3 2

c x&2 2 1( )k x x−

2 2 1( )c x x−& &

FFFF1111 FFFF2222


11 12 1 11 12 1 12 12 1 1

21 22 2 21 21 2 21 22 2 2

m m x c c x k k x F

m m x c c x k k x F

+ + =

&& &

&& &




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1, 2o

i t

j jF F e jω= =Let

& 1, 2i t

j jx X e jω= =

0

0

2 2111 11 11 12 12 12 1

2 22 221 21 21 22 22 22

( ) ( )

( ) ( )

Fm i c k m i c k X

X Fm i c k m i c k

ω ω ω ω

ω ω ω ω

− + + − + + =

− + + − + +

2( ) , 1, 2

rs rs rs rsZ m i c k r sω ω= − + + =Let

0

0

111 12 1

21 22 2 2

FZ Z X

Z Z X F

=

or 0[ ]{ } { }Z X F= 1

0{ } [ ] { }X Z F

−=




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0 022 1 12 2

1 2

11 22 12

Z F Z FX

Z Z Z

−=

−

0 012 1 11 2

2 2

11 22 12

Z F Z FX

Z Z Z

− +=

−&




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m2

x2

k k=

3k k=

1x

1k k=

m

01cosF tω

EXAMPLE:

01 1 2 12 cosmx kx kx F tω+ − =&&


2 1 22 0mx kx kx− + =&&&

011 1

2 2

cos2

2 0

F tx xm o k k

o m x k k x

ω − + =

−

&&

&&

11 22 12 21, 0m m m m m= = = =

11 12 21 220c c c c= = = =

11 22 12 212 ,k k k k k k= = = =−




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cos 1, 2j jx X t jω= =Solution

2

11 22( 2 )Z Z m kω= = − + 12 21

Z Z k= = −&

( )

( )

( )( )( )

0 0

2 2

1 1

1 2 2 22 2

2 2

32

m k F m k FX

m k m km k k

ω ω

ω ωω

− + − += =

− + − +− + −

( ) ( )( )0 01 1

2 2 2 22 2 32

kF kFX

m k m km k k ω ωω= =

− + − +− + −&

0

2

1

1

1 2 2 2

2

1 1 1

2

1

F

X

k

ω

ω

ω ω ω

ω ω ω

−

=

− −

&01

2 2 2 2

2

1 1 1

1

FX

kω ω ω

ω ω ω

=

− −

2 2

1 2/ , 3 /k m k mω ω= =where




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Orthogonality

of Modes




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[ ]{ } [ ]{ }2

11 1K X M Xω=

[ ]{ } [ ]{ }2K X M Xω=

Stiffness Matrix Mass Matrix

As

Consider two different natural frequencies ω1 & ω2

Eigenvalue Eigenvector

& [ ]{ } [ ]{ }2

22 2K X M Xω=




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{ } [ ]{ } { } [ ]{ }2

12 1 2 1

T TX K X X M Xω=

Multiply 1st eqn by {X}2T & 2nd eqn by {X}1

T gives

& { } [ ]{ } { } [ ]{ }2

21 2 1 2

T TX K X X M Xω=

Subtract the above equations

2 2

2 1 2 1( ){ } [ ]{ } 0

TX M Xω ω− =




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2 1{ } [ ]{ } 0

TX M X =

Similarly,2 1

{ } [ ]{ } 0T

X K X =

Hence, the eigenvectors (mode shapes) {X}1

& {X}2 are orthogonal with respect to both

the mass & stiffness matrices




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When k=1N/m & m=1kg

m2

x2

k k=

3k k=

1x

1k k=

m

2 2 2

1,21, 3 /rad sω =

{ } 1

1

12 1

1

1

XX

X

= =

{ } 1

2

22 2

1

1

XX

X

= =

− &

1 12

2 2

2 1 1 0

1 2 0 1

X X

X Xω

− =

−





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{ }2 1

1 0 1{ } [ ]{ } 1 1 0

0 1 1

TX M X

= − =

Similarly,

{ }2 1

2 1 1{ } [ ]{ } 1 1 0

1 2 1

TX K X

− = − =

−

Hence, the eigenvectors (mode shapes) {X}1

& {X}2 are orthogonal with respect to both

the mass & stiffness matrices




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Stability of 2DOF

System




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11 12 1 11 12 1 12 12 1

21 22 2 21 21 2 21 22 2

0

0

m m x c c x k k x

m m x c c x k k x

+ + =

&& &

&& &

Equations of Motion

Let

2 2

11 11 11 12 12 12 1

2 2221 21 21 22 22 22

( ) ( ) 0

0( ) ( )

s m sc k s m sc k X

Xs m sc k s m sc k

+ + + + =

+ + + +

1,2st

j jx X e j= =




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Solution exists if

2 2

11 11 11 12 12 12

2 2

21 21 21 22 22 22

( ) ( )det 0

( ) ( )

s m sc k s m sc k

s m sc k s m sc k

+ + + +=

+ + + +

or4 3 2

0 1 2 3 40a s a s a s a s a+ + + + =

System is stable if ALL the roots of above

equations have negative real parts




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Modal Modal

UncouplingUncoupling




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Equation of Motion:

[ ]{ } [ ]{ } { }M x K x F+ =&&

Let 1 2[ ] [{ } { } .... { } ]

NX X X X=

be the matrix of the eigenvectors

And define { } [ ]{ }x X q=




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[ ][ }{ } [ ][ ]{ } { }M X q K X q F+ =&&

Pre-multiply both sides by [X]T gives

[ ] [ ][ ]{ } [ ] [ ][ ]{ } [ ] { }T T T

X M X q X K X q X F+ =&&

Consider [X]T [M] [X] for 2DOF system:

1 1

2 2

{ } [ ]{ } 0

0 { } [ ]{ }

T

T

X M X

X M X

=

1 1 1 1 2

1 2

2 2 1 2 2

{ } { } [ ]{ } { } [ ]{ }[ ] [ ][ ] [ ] { } { }

{ } { } [ ]{ } { } [ ]{ }

T T T

T

T T T

X X M X X M XX M X M X X

X X M X X M X

= =




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For 2DOF system

[X]T[M][X]{q¨}+ [X]T [K] [X]{q}=0

1

1 1 1 1 1 1

2 22 2 2 2

{ } [ ]{ } 0 { } [ ]{ } 00

0 { } [ ]{ } 0 { } [ ]{ }

T T

T T

q X M X X K X q

q qX M X X K X

−

+ =

&&

&&

1 1 1 1 1 1

2 22 2 2 2

{ } [ ]{ } 0 { } [ ]{ } 00

0 { } [ ]{ } 0 { } [ ]{ }

T T

T T

X M X q X K X q

q qX M X X K X

+ =

&&

&&




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For 2DOF system

[X]T[M][X]{q¨}+ [X]T [K] [X]{q}=0

1 1

1 1 1 1

2 22 2

2 2

{ } [ ]{ }0

{ } [ ]{ }0

{ } [ ]{ }0

{ } [ ]{ }

T

T

T

T

X K X

q X M X q

q qX K X

X M X

+ =

&&

&&




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1 1

1 1 1 1

2 22 2

2 2

{ } [ ]{ }0

{ } [ ]{ }0

{ } [ ]{ }0

{ } [ ]{ }

T

T

T

T

X K X

q X M X q

q qX K X

X M X

+ =

&&

&&

{ } [ ]{ }

{ } [ ]{ }2 1 1

1

1 1

T

T

X K X

X M Xω =

Rayleigh’s Quotient:

{ } [ ]{ }

{ } [ ]{ }2 2 2

2

2 2

T

T

X K X

X M Xω =&




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2

1 1 1

22 22

00

0

q q

q q

ω

ω

+ =

&&

&&

For N DOF:

2

11 1

2

2 22

2

0 ..... 0

0 ..... 00

.... ........ ..... ..... ....

0 ..... .....N NN

q q

q q

q q

ω

ω

ω

+ =

&&

&&

&&

21,..,i n i i iq q Q i Nω+ = =&&or Uncoupled




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Then, transform to the physical domain:

{ } [ ]{ }x X q=

Solve for qi

0

(0)(0)cos( ) sin( )

1( )sin ( ) 1,..,

ii i ni ni

ni

t

i ni

ni

qq q t t

Q t d i N

ω ωω

τ ω τ τω

= +

+ − =∫

&




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Rayleigh’s

Quotient




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[ ]{ } [ ]{ }2

11 1K X M Xω=

[ ]{ } [ ]{ }2K X M Xω=


As

At the natural frequencies ω1

Eigenvalue Eigenvector

{ } [ ]{ }

{ } [ ]{ }2 1 1

1

1 1

T

T

X K X

X M Xω =




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Called Rayleigh’s Quotient




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EXAMPLE: Estimate the first (fundamental) natural

frequency of the given system by means of Rayleigh’s Quotient

1k 3

k2

k

FFFF1111 FFFF2222 FFFF3333

Take:

Compare the result with the exact value




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Deriving the equations of motion of the given system, the mass and stiffness

matrices will be:

We need first to start with a “trail displacement vector”. A good initial guess

can be found from:

To use Rayleigh’s Quotient




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Hence, we will let

We can see that the estimated value is very close to the exact one!




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EXAMPLE: Estimate all eigenvalues (natural frequencies)

and all eigenvectors (mode shape vectors) of the given system

in the previous example by means of Rayleigh’s Quotient

To estimate all eigenvalues and eigenvectors vectors of the system, let us

assume a general eigenvector of the form

Where “a” and “b” are unknowns. Our job now is to find the values of “a”

and “b” at which minimize Rayleigh’s Quotient. In other words, we need to

solve the following two equations for “a” and “b” :

The following MATHEMATICATM will do this task!




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MATHEMATICAMATHEMATICATMTM CODESCODES

Thus the three eigenvectors (mode shapes) are:




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Thus the three eigenvectors (mode shapes) are:

And the three eigenvalues (natural frequencies) are:

2nd mode shape at 2nd natural frequency

3rd mode shape at 3rd natural frequency

1st mode shape at 1st natural frequency




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CHECK WITH MATHEMATICACHECK WITH MATHEMATICATMTM

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Almitani.com MENG470 Spring2011 Vibration of Two DOFS

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