opgthpskyqlmtnrgxlxsynqk%2fetd78euzzx57f7i7pm%3d

8/21/2019 OPgTHpSKyQLmtNrGXLXSynqK%2fETd78eUzzx57f7I7PM%3d

1/8

Lecture 19-2 dofForced

ResponseMEEM 3700 1

MEEM 3700MEEM 3700

Mechanical VibrationsMechanical Vibrations

Mohan D. Rao

Chuck Van Karsen

Mechanical Engineering-Engineering MechanicsMichigan Technological University

Copyright 2003


ResponseMEEM 3700 2

1K 2K 3K

1 x 2 x

1 M 2 M

1 f 2 f

Given a 2Given a 2--degree of freedom system with no damping the equations ofdegree of freedom system with no damping the equations of

motion in matrix form can be generalized as:motion in matrix form can be generalized as:

11 1 11 12 1 1

22 2 21 22 2 2

0

0

m x k k x f

m x k k x f

⎡ ⎤ ⎧ ⎫ ⎡ ⎤ ⎧ ⎫ ⎧ ⎫+ =⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎩ ⎭ ⎣ ⎦ ⎩ ⎭ ⎩ ⎭

&&

&&

2 DOF2 DOF ForcedForced VibrationVibration


2/8


ResponseMEEM 3700 3

Assuming harmonic forcing functions of the form:Assuming harmonic forcing functions of the form:

( )1 12 2

sin f F

t f F

ω

⎧ ⎫ ⎧ ⎫=⎨ ⎬ ⎨ ⎬

⎩ ⎭ ⎩ ⎭

The solution (motion of xThe solution (motion of x11 and xand x22) will be of the form:) will be of the form:

( )1 12 2

sin x X

t x X

ω ⎧ ⎫ ⎧ ⎫

=⎨ ⎬ ⎨ ⎬⎩ ⎭ ⎩ ⎭

2 DOF Forced Vibration2 DOF Forced Vibration

1K 2K 3K

1 x 2 x

1 M 2 M

1 f 2 f


ResponseMEEM 3700 4

2 DOF Forced Vibration: Solution2 DOF Forced Vibration: Solution

( ) ( )11 11 12 1 1222 21 22 2 2

0sin sin

0

m k k X F t t

m k k X F ω ω ω

⎛ ⎞⎡ ⎤ ⎡ ⎤ ⎧ ⎫ ⎧ ⎫− + =⎜ ⎟ ⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦ ⎩ ⎭ ⎩ ⎭⎝ ⎠

Substituting into the equation of motion:Substituting into the equation of motion:

and simplifyingand simplifying

11 11 12 1 12

22 21 22 2 2

0

0

m k k X F

m k k X F ω

⎛ ⎞⎡ ⎤ ⎡ ⎤ ⎧ ⎫ ⎧ ⎫− + =⎜ ⎟ ⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎩ ⎭ ⎩ ⎭⎝ ⎠


3/8


ResponseMEEM 3700 5

2 DOF Forced Vibration: Impedance Matrix2 DOF Forced Vibration: Impedance Matrix

Define:Define: ( ) 11 12 11221 22 22

00

k k m Z k k m

ω ω ⎡ ⎤ ⎡ ⎤⎡ ⎤ = −⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦ ⎣ ⎦

WhereWhere ( ) Z ω ⎡ ⎤⎣ ⎦ is called theis called the system system oror impedanceimpedance matrix.matrix.

Then the responseThen the response can be solved for as:can be solved for as:1

2

X

X

⎧ ⎫⎨ ⎬⎩ ⎭

( ) 11 1

2 2

X F Z

X F ω

−⎧ ⎫ ⎧ ⎫⎡ ⎤=⎨ ⎬ ⎨ ⎬⎣ ⎦⎩ ⎭ ⎩ ⎭ ( )

( )

( )

1

det

adj Z

Z Z

ω

ω ω

− ⎡ ⎤⎣ ⎦⎡ ⎤ =⎣ ⎦ ⎡ ⎤⎣ ⎦

( )2

11 11 12

2

21 22 22

k m k Z

k k m

ω

ω

ω

⎡ ⎤−⎡ ⎤ = ⎢ ⎥⎣ ⎦ −⎣ ⎦


ResponseMEEM 3700 6

Inverse of AInverse of A

[ ]11 12 13

21 22 23

31 32 33

a a a A a a a

a a a

⎛ ⎞⎜ ⎟= ⎜ ⎟

⎜ ⎟⎝ ⎠

[ ] [ ]

[ ]1

det

adjoint A A

A

−=

[ ] [ ]adjoint transpose of cofactor

cofactors minor determinants

A A=

= ±

[ ]( )( )12 33 13

22 33 23 32

32

...

...

...a acof Aa a

aa a

a−⎛ ⎞⎜ ⎟

= ⎜ ⎟⎜

+−

⎟⎝ ⎠

−O

Linear Algebra Review: Matrix OperationsLinear Algebra Review: Matrix Operations


4/8


ResponseMEEM 3700 7


The inverse of the impedance matrix is written as:The inverse of the impedance matrix is written as:

( )( )( ) ( )

2

22 22 12

21 21 11 11

2

11 11 22 22 12 21

k m k

k k m Z

k m k m k k

ω

ω

ω

ω ω

−

⎡ ⎤− −⎢ ⎥

− −⎣ ⎦⎡ ⎤ =⎣ ⎦− − −


2

X

X

⎧ ⎫⎨ ⎬⎩ ⎭

( ) 11 1

2 2

X F

Z X F ω

−⎧ ⎫ ⎧ ⎫

⎡ ⎤=⎨ ⎬ ⎨ ⎬⎣ ⎦⎩ ⎭ ⎩ ⎭


ResponseMEEM 3700 8


Therefore each response can be written as:Therefore each response can be written as:

( )( )( ) ( )

2

22 22 1 12 2

1 2 2

11 11 22 22 12 21

k m F k F X

k m k m k k

ω

ω ω

− −=

− − −

( )( ) ( ) ( )

2

11 11 2 21 1

2 2 2

11 11 22 22 12 21

k m F k F

X k m k m k k

ω

ω ω

− −

= − − −


5/8


ResponseMEEM 3700 9

2 DOF Forced Vibration: FRF Matrix2 DOF Forced Vibration: FRF Matrix

DefineDefine ( ) ( ) 1 H Z ω ω −⎡ ⎤ ⎡ ⎤=⎣ ⎦ ⎣ ⎦

as the Frequency Response Function Matrixas the Frequency Response Function Matrix

( )1 1 11 12 12 2 21 22 2

X F h h F H

X F h h F ω

⎧ ⎫ ⎧ ⎫ ⎡ ⎤ ⎧ ⎫⎡ ⎤= =⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎢ ⎥⎣ ⎦

⎩ ⎭ ⎩ ⎭ ⎣ ⎦ ⎩ ⎭


1 11 1 12 2 X h F h F = +

2 21 1 22 2 X h F h F = +


ResponseMEEM 3700 10

2 DOF Forced Vibration: FRF2 DOF Forced Vibration: FRF

The individualThe individual FRFFRF’’ss are defined as:are defined as:

( ) ( ) ( )

2

1 22 2211 2 2

1 11 11 22 22 12 21

X k mh

F k m k m k k

ω

ω ω

−= =

− − −

( )( ) ( )1 12

12 2 22 11 11 22 22 12 21

X k h

F k m k m k k ω ω

−= =

− − −

( )( ) ( )2 21

21 2 21 11 11 22 22 12 21

X k h

F k m k m k k ω ω

−= =

− − −

( )( ) ( )

22 11 11

22 2 22 11 11 22 22 12 21

X k mh

F k m k m k k

ω

ω ω

−= =− − −


6/8



0 0.5 1 1.5 2 2.5 3 3.5 4-20

-15

-10

-5

0

5

10

15

20

Frequency Hz

X 2 / F 1

X2/F1

0 0.5 1 1.5 2 2.5 3 3.5 4-20

-15

-10

-5

0

5

10

15

20

Frequency Hz

X 1 / F 1

X1/F1

0 0.5 1 1.5 2 2.5 3 3.5 4-20

-15

-10

-5

0

5

10

15

20

Frequency Hz

X 1 / F 2

X1/F2

0 0.5 1 1.5 2 2.5 3 3.5 4-20

-15

-10

-5

0

5

10

15

20

Frequency Hz

X 2 / F 2

X2/F2

Arrows indicate that

function goes to infinity



2 DOF Forced Vibration: Damping2 DOF Forced Vibration: Damping

If damping is considered the equations of motion become:If damping is considered the equations of motion become:

11 1 11 12 1 11 12 1 1

22 2 21 22 2 21 22 2 2

0

0

m x c c x k k x f

m x c c x k k x f

⎡ ⎤ ⎧ ⎫ ⎡ ⎤ ⎧ ⎫ ⎡ ⎤ ⎧ ⎫ ⎧ ⎫+ + =⎨ ⎬ ⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎩ ⎭ ⎣ ⎦ ⎩ ⎭ ⎣ ⎦ ⎩ ⎭ ⎩ ⎭

&& &

&& &

And if the forcing functions are harmonic functions (as before)And if the forcing functions are harmonic functions (as before) the solution is:the solution is:

11 11 12 11 12 1 12

22 21 22 21 22 2 2

0

0

m c c k k X F

jm c c k k X F ω ω

⎛ ⎞⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎧ ⎫ ⎧ ⎫

− + + =⎜ ⎟ ⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎩ ⎭ ⎩ ⎭⎝ ⎠


7/8



2 DOF Forced Vibration: FRF w/ Damping2 DOF Forced Vibration: FRF w/ Damping

Define:Define: ( ) 11 12 11 11 12221 22 22 21 22

0

0

k k m c c Z j

k k m c cω ω ω

⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎡ ⎤ = − +⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦ ⎣ ⎦ ⎣ ⎦

WhereWhere ( ) Z ω ⎡ ⎤⎣ ⎦ is called theis called the system system oror impedanceimpedance matrix.matrix.


2

X

X

⎧ ⎫⎨ ⎬⎩ ⎭

( ) 11 1

2 2

X F Z

X F ω

−⎧ ⎫ ⎧ ⎫⎡ ⎤=⎨ ⎬ ⎨ ⎬⎣ ⎦

⎩ ⎭ ⎩ ⎭

WhereWhere ( ) ( )1

Z H ω ω −

⎡ ⎤ ⎡ ⎤=⎣ ⎦ ⎣ ⎦ is the Frequency Response Function Matrixis the Frequency Response Function Matrix

( )1 12 2

X F H X F

ω ⎧ ⎫ ⎧ ⎫⎡ ⎤=⎨ ⎬ ⎨ ⎬⎣ ⎦⎩ ⎭ ⎩ ⎭




The FRF matrix can be written as:The FRF matrix can be written as:

( ) ( )

( )det

adj Z H

Z

ω

ω

ω

⎡ ⎤⎣ ⎦⎡ ⎤ =⎣ ⎦ ⎡ ⎤⎣ ⎦

This is expanded to:This is expanded to:

( )

( )( )

( )( ) ( )( )

2

22 22 22 12 12

2

21 21 11 11 11

2

11 11 11 22 22 22 12 12 21 21

k m j c j c k

j c k k m j c H

k m j c k m j c j c k j c k

ω ω ω

ω ω ω

ω

ω ω ω ω ω ω

⎡ ⎤− + − +⎢ ⎥

− + − +⎣ ⎦⎡ ⎤ =⎣ ⎦ − + − + − + +


1 11 1 12 2 X h F h F = +

2 21 1 22 2 X h F h F = +


8/8




( )( )( ) ( )( )

2

22 22 22111 2 2

1 11 11 22 22 12 12 21 21

k m j c X h

F k m k m j c k j c k

ω ω

ω ω ω ω

− += =

− − − + +

( )

( )( ) ( )( )12 121

12 2 22 11 11 22 22 12 12 21 21

k j c X h


ω

ω ω ω ω

− += =

− − − + +

( )

( ) ( ) ( )( )21 212

21 2 21 11 11 22 22 12 12 21 21

k j c X h


ω

ω ω ω ω

− += =

− − − + +

( )

( )( ) ( )( )

2

11 11 11222 2 2

2 11 11 22 22 12 12 21 21

k m j c X h


ω ω

ω ω ω ω

− += =

− − − + +

The individualThe individual FRFFRF’’ss are defined as:are defined as:



0 0.5 1 1.5 2 2.5 3 3.5 40

0.5

1

1.5

2

2.5

3

3.5

4

4.5Frequency Response Function

Frequency Hz

m a g n i t u d e

X1/F1

0 0.5 1 1.5 2 2.5 3 3.5 40

0.5

1

1.5

2

2.5

3

3.5

4

4.5Frequency Response Function

Frequency Hz

m a g n i t

u d e

X2/F2

0 0.5 1 1.5 2 2.5 3 3.5 40

1

2

3

4

5

6

7

8Frequency Response Function

Frequency Hz

M a g n i t u d e

X1/F2

0 0.5 1 1.5 2 2.5 3 3.5 40

1

2

3

4

5

6

7

8Frequency Response Function

Frequency Hz

M a g n i t u d e

X2/F1

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