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ME 535 - Vibrations

Systems with Two or More DOF

Introduction

An N-DOF system will have N natural frequencies

Each natural frequency has a natural state of vibration called its normal mode.

Eigenvalues and eigenvectors are related to the normal modes

Normal mode vibrations are undamped free vibrations that depend only on the mass and

stiffness distribution in the system.

Damping limits amplitude and forces the free vibrations to decay.

Normal Mode Analysis

Example (5.1.1 pg. 127):

( ) ( )

( ) ( )

mx kx k x x mx kx k x x

mx k x x kx mx k x x kx

&& &&

&& &&

1 1 1 2 1 1 2 1

2 1 2 2 2 2 1 2

0

2 0 2

+ + = = +

+ = =

For normal mode vibration, each mass undergoes harmonic motion at the same frequency,

passing through the equilibrium position simultaneously, leading to the following solutions:

x A t A ex A t A e

i t

i t

1 1 1

1 2 2

==

sinsin

oror

substituting into the EOMs:


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( )

( )

& &

&& &&

x iA e x iA e

x A e x A e

mA e kA e kA e kA e

mA e kA e kA e kA e

mA kA kA e

mA kA kA e

m

i t i t

i t i t

i t i t i t i t

i t i t i t i t

i t

i t

1 1 2 2

1 1

2

2 2

2

1

2

1 2 1

2

2

1 2 2

1

2

1 2

2

2

1 2

0

2

2 0

2 2 0

= =

= =

+ + =

+ +

+ =

+ =

( )

( )

2

1 2

1

2

2

2 0

2 2 0

+ =

+ + =

k A kA

kA m k A

or:

2

2 2 0

2

2

1

2

k m k

k k m

A

A

=

which is satisfied if:

2

2 20

2

2

k m k

k k m

=

if we let then: 2 =

2

2 20

k m k

k k m

=

or:


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( )( )

( ) ( )

2 2 2 0

4 4 2 2 0

2 6 3 0

33

20

3 9 6

2

3

2

1

2

3

0634

2 366

0634

2

2 2 2 2

2 2 2

2

2

1 2

2 2

1

2

1 1

k m k m k

k km km m k

m km k

k

m

k

m

k

mk

m

k

m

k

m

km

km

km

=

+ =

+ =

+

=

=

=

=

=

= =

,

.

.

.

eigenvalues of the system

2 2 2 366= =

.

k

m

natural frequencies of the system

From:

( )

( )

+ =

+ + =

m k A kA

kA m k A

2

1 2

1

2

2

2 0

2 2 0

we can form the following relationship:

A

A

k

k m

k m

k

1

2

2

2

2

2 2

=

=

Since this is a 2-DOF system, there are two natural frequencies. Substituting in for each, gives


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the following:

for :1

A

A

k

k m km

1

2

1

2 634

1

2 6340732

=

=

=

( )

(. ) ..

for :2

AA

kk m km

1

2

2

2 2 3661

2 2 3662 732 = = =

( )

( . ) ..

If one of the amplitudes is chosen = 1, then the amplitude ratios are normalized. These are called

the normal modes, , or mode shapes for the system:i x( )

1 20732

100

2 73

100( )

.

.( )

.

.x x=

=

The normal modes are also the eigenvectors for the system, and the normal mode oscillations aregiven by:

( )

( )

x

xA t

x

xA t

1

2

1

1 1 1

1

2

2

2 2 2

0732

100

2 73

100

=

+

=

+

( )

( )

.

.sin

.

.sin

The mode shapes can be represented graphically as follows:


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Example (5.1.2 pg. 129):


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Example (5.1.3 pg. 130):

( )

( )

ml mgl ka

ml mgl ka

2

1 1

2

1 2

2

2 2

2

1 2

&&

&&

=

= +


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Initial Conditions

Once we know the modal frequencies and mode shapes, we can determine the free vibration of

the system for any initial conditions as a sum of the normal modes.

1 1

2 2

06340732

1000

2 3662 732

1000

= =

= =

..

.

..

.

km

km

For free vibrations in either of the normal modes:

( )xx

c t i

i

i i i i

1

2

1 2

= + =( )

sin ; ,

where:

come from initial conditionsci iand

assures that the correct amplitude ratio is maintainedi

For initial conditions in general, both modes are present:

( ) ( )x

xc t c t

1

21 1 1 2 2 2

0732

1000

2 732

1000

=

+ +

+.

.sin

.

.sin

Note that there are four constants and two equations. The four constants are:


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c c1 2

1 2

,

,

define the contribution of each mode, and

describe the phase relationships

We can differentiate to get two more equations:

( ) ( )&

&

.

.cos

.

.cos

x

xc t c t

2

21 1 1 1 2 2 2 2

0732

1000

2 732

1000

=

+ +

+

We can then let t= 0 and solve the four equations for the four unknowns.

Example (5.2.1 pg. 132):

x

x

x

x1

2

1

2

0

0

2 0

4 0

0

0

0

0

( )

( )

.

.

& ( )

& ( ) = =


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Coordinate Coupling

The equations for a 2 DOF system, in general form are:

m x m x k x k x

m x m x k x k x

11 1 12 2 11 1 12 2

21 1 22 2 21 1 22 2

0

0

&& &&

&& &&

+ + + =+ + + =

or:

m m

m m

x

x

k k

k k

x

x

11 12

21 22

1

2

11 12

21 22

1

2

0

0

+

=

&&

&&

If the mass matrix is non-diagonal then mass or dynamical coupling is present.

If the stiffness matrix is non-diagonal then stiffness or static coupling is present.

If principle (normal) coordinates are used then neither type of coupling is present. This is

always possible for an undamped system, but not always for a damped system.

The following system is coupled through the damping matrix:

m

m

x

x

c c

c c

x

x

k

k

x

x

11

22

1

2

11 12

21 22

1

2

11

22

1

2

0

0

0

0 0

+

+

=

&&

&&

&

&

Example (5.3.1 pg. 135):


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The choice of coordinates has an effect on the coupling.


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Example (5.3.2 pg. 136):


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Forced Harmonic Vibrations

For a general 2 DOF system, with a sinusoidal forcing function, the equations of motion are:

m m

m m

x

x

k k

k k

x

x

Ft

11 12

21 22

1

2

11 12

21 22

1

2

1

0

+

=

&&

&&sin

by assuming a solution of the form:

x

x

X

Xt

1

2

1

2

=

sin

and substituting into the equations of motion, we get:

k m k m

k m k m

X

X

F11 11

2

12 12

2

21 21

2

22 22

2

1

2

1

0

=

or:

[ ]z XX

F( )

1

2

1

0

=

The normal mode frequencies are the eigenvalues of :[ ]z( )

( )( )

1 211 22 22 11 12 21 21 12

11 22 12 212, =

+

k m k m k m k m R

m m m m


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where:

( )( )

R

k m k m k m k m

k k m m k k m m k k m m k k m m

k k m m k k m m

m m m m k k k k

=

+ + +

+

+

12

2

21

2

21

2

12

2

11

2

22

2

22

2

11

2

12 21 12 21 11 12 21 22 12 22 11 21 11 21 12 22

21 22 11 12 11 22 11 22

11 22 12 21 11 22 12 21

2

4

it can also be shown that the amplitudes are given by:

( )( ) ( )( )[ ]

( )

( ) ( )( )[ ]

Xk m F

m m m m

Xk m F

m m m m

1

22 222

1

11 22 12 21 1

2 2

2

2 2

2

21 21

2

1

11 22 12 21 1

2 2

2

2 2

=

=

Vibration Absorber

( )( )

( )

( )

m x k x k x x F t

m x k k x k x F t

m x k x x

m

m

x

x

k k k

k k

x

x

Ft

Xk m F

m m

1 1 1 1 2 1 2 0

1 1 1 2 1 2 2 0

2 2 2 1 2

1

2

1

2

1 2 2

2 2

1

2

0

1

2 2

2

1

1 2 1

0

0

0 0

&& sin

&& sin

&&

&&

&&sin

+ + =+ + =

=

++

=

=

( )( )2 2 22 2


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We can tune the system such that which will make .

k

m

2

2

2

= x1 0=Note: From the following figure, it is clear that there are resonant frequencies on either side of

.

Example (Problem 5.7 pg. 152):

Determine the natural frequency of the torsional system shown, and draw the normal mode curve.

G X= 115 106. psi

k GIl

I l= =; 4

32

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