multi degree freedom systems

8/4/2019 Multi Degree Freedom Systems

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Multi-degree of Freedom Systems

Motivation: Many systems are too complex

to be represented by a single degree of

freedom model. Objective of this chapter: Understand

vibration of systems with more than one

degree of freedom.


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Free-vibration of undamped two-

degree of freedom system We learn how to analyze free vibration by

considering an example

k1 k2 k3

m1 m2

x1 x2


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Deriving equations of motion

0)(

0)(

2321222

2212111

!

!

xkkxkxm

xkxkkxm

m1

k1x1 k2(x1-x2)

m2k2(x2-x1)

k3x2

Interpretation of coefficients

First equation: k1+k2 is the force on m1 needed to move it slowly by one unit while m2 is

held stationary. k2 is the force on m1 need to hold it steady if m2 is displaced slowly by oneunit.

Second equation: k1+k2 is the force on m2 needed to move it slowly by one unit while m1 is

held stationary. k2 is the force on m2 need to hold is steady if m1 is displaced slowly by

one unit.

Maxwell reciprocity theorem: the force on m1 need to hold it steady if m2 is displaced slowly

by one unit=force on m2 need to hold is steady if m1 is displaced slowly by one unit.


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Special case

Let m1=m2=m and k1=k2=k3=k. Then:

02

02

212

211

!

!

kxkxx

kxkxx


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Solution of equations of motion

We know from experience that:

)sin()(

)sin()(

2

1

tBtx

tAtx

[

[

!

!

Substituting the above equation to eq. of motion, weobtain two eqs w.r.t. A, B :

0sin)2(

0sin)2(

2

2

!

!

tkAkBmB

tkBkAmA

[[

[[

The above two equations are satisfied for every t

02

02

2

2

!

!

kAkBmB

kBkAmA

[

[


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Trivial solution A, B=0. In order to have a nontrivial solution:

02

2det

2

2!

[

[

mkk

kmk

m

k

m

k 3, ss![

First natural frequency:

BAm

k!![

The displacement for the first natural frequency is:

A

A

This vector is called mode shape. Constant cannot be determined.

Usually assume first entry is 1. Therefore, mode shape is:

1

1


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Thus, both masses move in phase and the have

the same amplitudes.


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Second natural frequency:

BAm

k!!

3[

Mode shape for above frequency:

A

A

Usually assume first entry is 1. Therefore, mode shape is:

1

1

The two masses move in opposite directions.


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)3

sin(1

1)sin(

1

1

)(

)(2211

2

1 ]]

!

t

m

kct

m

kc

tx

tx

Displacements of two masses are sums of displacements

in the two modes:


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General expression for vibration of the

two-degree of freedom system

)sin()sin()(

)(22

2

2211

1

11

2

1][][

!

t

B

Act

B

Ac

tx

tx


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Observations:

System motion is superposition of twoharmonic (sinusoidal) motions withfrequencies [1 and [2.

Participation of each mode depends oninitial conditions.

Four unknowns (c1, c2, ]1, ]2) can befound using four initial conditions.

Can find specific initial conditions so thatonly one mode is excited.


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How to solve a free vibration problem

involving a two degree of freedom system

1) Write equations of motion for free vibration (no external force or moment)

2) Assume displacements are sinusoidal waves, and plug in equations ofmotion: Obtain equation:

where

3) Solve det[C()]=0, obtain two natural frequencies, 1, 2.

4) Solve,

Assume A=1, and obtain first mode shape. Obtain second mode shape in asimilar manner.

!

0

0*)]([

B

AC [

[ ( )] *CA

B[1

0

0

!

][][)]([ 2 MKC [[ !


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5) Find constants from initial conditions.

)sin()sin()()( 22

2

22111

112

1 ][][

!

t

B

ActB

Actx

tx

Free vibration response is:

multi degree freedom systems

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