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5.4 Multi-Degree-of-Freedom Vibration

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In this section, a review of deterministic vibration is provided. These results are used in the next section,

where the forcing is taken to be a random process.

5.4.1 Deterministic Vibration

Consider a case where there is more than one DoF.

The equations of motion can be written as

½m􀀉{x€ðtÞ} þ ½c􀀉{x_ðtÞ} þ ½k􀀉{xðtÞ} ¼ {FðtÞ}

For an N-DoF system, the matrices ½m􀀉; ½c􀀉 and ½k􀀉

are of dimension N £ N; and the response {xðtÞ}

and force {FðtÞ} vectors are of dimension N £ 1:

For purposes of demonstration and discussion,

the necessary concepts will be introduced primarily

by working through the solution of a two-DoF

system. All these ideas transfer to larger systems, but with the two-DoF model we can demonstrate the

key ideas without the complications of the major algebraic and numerical demands made by the larger

systems.

For the system shown in Figure 5.7, we can derive the coupled equations of motion using either

Newton’s Second Law of motion applied to a free body diagram for each mass, as shown in Figure 5.8, or

by Lagrange’s equation. In either case, we find the governing equations to be

m1x€1ðtÞþðc1 þ c2Þx_1ðtÞ þ ðk1 þ k2Þx1ðtÞ 2 c2x_2ðtÞ 2 k2x2ðtÞ ¼ F1ðtÞ;

m2x€2ðtÞ þ c2x_2ðtÞ 2 c2x_1ðtÞ 2 k2x1ðtÞ þ k2x2ðtÞ ¼ F2ðtÞ

or in matrix form

m1 0

0 m2

" #

x€1ðtÞ

x€2ðtÞ

( )

c1 þ c2 2c2

2c2 c2

" #

x_1ðtÞ

x_2ðtÞ

( )

k1 þ k2 2k2

2k2 k2

" #

x1ðtÞ

x2ðtÞ

( )

F1ðtÞ

F2ðtÞ

( )

ð5:14Þ

F2(t)

x1(t)

F1(t)

x2(t)

FIGURE 5.7 Two-DoF mass – spring – damper system.

5-12 Vibration and Shock Handbook

In this case, the mass, damping, and stiffness matrices are given by

½m􀀉 ¼

m1 0

0 m2

" #

; ½c􀀉 ¼

c1 þ c2 2c2

2c2 c2

" #

; ½k􀀉 ¼

k1 þ k2 2k2

2k2 k2

" #

5.4.2 Solution by Frequency Response Function

Begin by taking the Fourier transform of the equations of motion, obtaining

ð2v2½m􀀉 þ iv½c􀀉 þ ½k􀀉Þ{XðvÞ} ¼ {FðvÞ}

where {XðvÞ} and {FðvÞ} are Fourier transforms of {xðtÞ} and {FðtÞ}; respectively. The matrix

ð2v2½m􀀉 þ iv½c􀀉 þ ½k􀀉Þ is denoted as ½ZðvÞ􀀉: Then,

{XðvÞ} ¼ ½ZðvÞ􀀉21{FðvÞ}

The matrix ½ZðvÞ􀀉21 is identical to the frequency response matrix denoted as ½HðvÞ􀀉:

{XðvÞ} ¼ ½HðvÞ􀀉{FðvÞ} ð5:15Þ

For the two-DoF system in Figure 5.7

½HðvÞ􀀉 ¼

det½Z􀀉

2m2v2 þ ivc2 þ k2 ivc2 þ k2

ivc2 þ k2 2m1v2 þ ivðc1 þ c2Þ þ ðk1 þ k2Þ

" #

det½Z􀀉 ¼ v4m1m2 2 iv3ðm1c2 þ c1m2 þ c2m2Þ þ ivðc1k2 þ k1c2Þ

2 v2ðm1k2 þ c1c2 þ k1m2 þ k2m2Þ þ k1k2

Note that the frequency response function of a two-DoF system is given as a matrix with dimension of

2 £ 2. To clarify the meaning of each element, let us expand Equation 5.15:

X1ðvÞ ¼ H11ðvÞF1ðvÞ þ H12ðvÞF2ðvÞ

X2ðvÞ ¼ H21ðvÞF1ðvÞ þ H22ðvÞF2ðvÞ

Now, the meaning of each element is clear. Each element HijðvÞ is the frequency response function for

coordinate i due to a force at j: In general, ½HðvÞ􀀉 is a symmetric matrix since ½m􀀉; ½c􀀉 and ½k􀀉 are

symmetric.

For a deterministic response, recall that the impulse response function is related to the frequency

response function by

gijðtÞ ¼

ð1

HijðvÞexpðivtÞdv

F1(t)

x1(t)

F2(t)

x2(t)

k2(x2−x1)

c2(x2−x1)

k1 x1

c1 x1

FIGURE 5.8 Free body diagram of a two-DoF system.

Random Vibration 5-13

which is written for each element in the matrix. For the two-DoF system

g11ðtÞ ¼

ð1

2m2v2 þ ivc2 þ k2

det½Z��

expðivtÞdv

g12ðtÞ ¼

ð1

ivc2 þ k2

det½Z􀀉

expðivtÞdv

g21ðtÞ ¼

ð1

ivc2 þ k2

det½Z􀀉

expðivtÞdv ¼ g12ðtÞ

g22ðtÞ ¼

ð1

2m1v2 þ ivðc1 þ c2Þ þ ðk1 þ k2Þ

det½Z􀀉

expðivtÞdv

Again, ½gðtÞ􀀉 is a symmetric matrix.

Using the convolution integral, the response xiðtÞ due to Fj is

xiðtÞ ¼

ð1

gijðtÞFjðt 2 tÞdt

The total response xiðtÞ of mass mi is then equal to the sum of the individual responses to each of the

forces. In our case, the responses are

x1ðtÞ ¼

ð1

21 ½g11ðtÞF1ðt 2 tÞ þ g12ðtÞF2ðt 2 tÞ􀀉dt

x2ðtÞ ¼

ð1

21 ½g21ðtÞF1ðt 2 tÞ þ g22ðtÞF2ðt 2 tÞ􀀉dt

We can extend our analysis to a system with N DoF. The equations of motion are written in matrix form

½m􀀉{x€ðtÞ} þ ½c􀀉{x_ðtÞ} þ ½k􀀉{xðtÞ} ¼ {FðtÞ}

where {xðtÞ} and {FðtÞ} are response and force vectors and the frequency response matrix is given by

½HðvÞ􀀉 ¼ ð2v2½m􀀉 þ iv½c􀀉 þ ½k􀀉Þ21

The impulse response matrix ½gðtÞ􀀉 is given by

½gðtÞ􀀉 ¼

ð1

21 ½HðvÞ􀀉expðivtÞdv

and the response matrix is given by

{xðtÞ} ¼

ð1

21 ½gðtÞ􀀉{Fðt 2 tÞ}dt

Note that the impulse response vector is not trivial to evaluate, much less the convolution integrals. More

importantly, this method becomes much harder for each additional DoF. Therefore, this method is rarely

used, and we often rely on modal analysis.

5.4.3 Modal Analysis

Let us first consider an undamped system in free vibration with equations of motion

½m􀀉{x€ðtÞ} þ ½k􀀉{xðtÞ} ¼ 0

If we assume that {xðtÞ} is harmonic with frequency v;

{xðtÞ} ¼ A0{u}eivt

the equations of motion are reduced to the eigenvalue problem given by

ð2v2½m􀀉 þ ½k􀀉Þ{u} ¼ {0} ð5:16Þ

5-14 Vibration and Shock Handbook

The goal is to find the nonzero vector {u}: Therefore, equations must be linearly dependent,8 or there

should be no inverse for ð2v2½m􀀉 þ ½k􀀉Þ: This can be satisfied if the determinant of ð2v2½m􀀉 þ ½k􀀉Þ is

zero:

l 2 v2½m􀀉 þ ½k􀀉l ¼ 0

There are several values of v2 which can make the determinant equal zero, and these values of v2 are

called eigenvalues. For each nonrepeating eigenvalue, there is a corresponding vector {u} that satisfies

Equation 5.16. This vector is called the eigenvector. Physically, the eigenvalues of this problem are the

natural frequencies squared, and the eigenvectors are the mode shapes, since they represent the respective

motion of each modal coordinate.

By solving this eigenvalue problem, we can obtain N sets of eigenvalues v2i

with corresponding

eigenvectors {u}i for an N-DoF system.9 It is customary that the eigenvalues are arranged in ascending

order

v21

, v22

, · · · , v2N

The eigenvectors can be normalized with respect to the mass matrix,

{u}T

i ½m􀀉{u}i ¼ 1; i ¼ 1; 2; …; N

where

{u}i ¼

u1i

.. .

uNi

8>>><

>>>:

9>>>=

>>>;

For the ith and jth set of eigenvalues and eigenvectors, we can write

2v2i

½m􀀉{u}i þ ½k􀀉{u}i ¼ {0}

2v2j

½m􀀉{u}j þ ½k􀀉{u}j ¼ {0}

Multiplying the first equation by {u}T

j and the second equation by {u}T

i ; we obtain

2v2i

{u}T

j ½m􀀉{u}i þ {u}T

j ½k􀀉{u}i ¼ {0} ð5:17Þ

2v2j

{u}T

i ½m􀀉{u}j þ {u}T

i ½k􀀉{u}j ¼ {0} ð5:18Þ

Take the transpose of Equation 5.18,

2v2j

{u}T

j ½m􀀉T{u}i þ {u}T

j ½k􀀉T{u}i ¼ {0}

Since ½m􀀉 and ½k􀀉 are symmetric, ½m􀀉 ¼ ½m􀀉T and ½k􀀉 ¼ ½k􀀉T; and

2v2j

{u}T

j ½m􀀉{u}i þ {u}T

j ½k􀀉{u}i ¼ {0} ð5:19Þ

8An example of a linear dependent equation is

1 2

2 4

" #

( )

u1 þ 2u2 ¼ 0

2u1 þ 4u2 ¼ 0

In this case, u1 and u2 need not be zeros. Instead, the equations are satisfied as long as u1 ¼ 2u2: Therefore, there is an infinite

number of solutions.

9An eigenvalue problem will be demonstrated in the next example.

Random Vibration 5-15

Subtracting Equation 5.19 from Equation 5.17, we obtain

ðv2j

2 v2i

Þ{u}T

j ½m􀀉{u}i ¼ 0

If the eigenvalues are unique ðv2j

– v2i

Þ; then

{u}T

j ½m􀀉{u}i ¼ 0; i – j

From Equation 5.19, similarly

{u}T

j ½k􀀉{u}i ¼ 0; i – j

That is, the eigenvectors are orthogonal (or normal) to each other with respect to the mass and the stiffness

matrices. If the eigenvectors are normalized with respect to the mass matrix, Equation 5.19 with i ¼ j can

be written as

2v2i

{u}T

i ½m􀀉{u}i þ {u}T

i ½k􀀉{u}i ¼ {0}

and therefore

{u}T

i ½k􀀉{u}i ¼ v2i

In summary, we just found that the eigenvectors have the following properties:

{u}T

i ½m􀀉{u}j ¼

1; i ¼ j

0; i – j

(

and

{u}T

i ½k􀀉{u}j ¼

v2i

; i ¼ j

0; i – j

(

If we construct a matrix composed of eigenvectors

½P􀀉 ¼ ½ {u}1 · · · {u}N􀀉 ð5:20Þ

we can write

½p􀀉T½m􀀉½p􀀉 ¼ ½I􀀉

½p􀀉T½k􀀉½p􀀉 ¼

v21

0 0

0 . .

0 0 v2N 2

66664

77775

¼ diagðv2Þ

where ½I􀀉 is the identity matrix and [P] is called the modal matrix.

Now, let us return to our original equations of motion given by

½m􀀉{x€ðtÞ} þ ½c􀀉{x_ðtÞ} þ ½k􀀉{xðtÞ} ¼ {FðtÞ}

We restrict our problem to the proportional damping case, where ½c􀀉 is a linear combination of ½m􀀉

and ½k􀀉:

½c􀀉 ¼ a½m􀀉 þ b½k􀀉

with constant a and b:

Let us define a new set of coordinates {zðtÞ}; so that

{xðtÞ} ¼ ½P􀀉{zðtÞ} ð5:21Þ

where ½p􀀉 is the modal matrix defined in Equation 5.20. The equations of motion then become

½m􀀉½P􀀉{z€ðtÞ} þ ½c􀀉½P􀀉{zðtÞ} þ ½k􀀉½P􀀉{zðtÞ} ¼ {FðtÞ}

5-16 Vibration and Shock Handbook

Multiplying by ½P􀀉T;

½P􀀉T½m􀀉½P􀀉{z€ðtÞ} þ ½P􀀉T½c􀀉½P􀀉{zðtÞ} þ ½P􀀉T½k􀀉½P􀀉{zðtÞ} ¼ ½P􀀉T{FðtÞ}

which reduces to

{z€ðtÞ} þ ða½I􀀉 þb £ diagðv2ÞÞ{zðtÞ} þ diagðv2Þ{zðtÞ} ¼ ½P􀀉T{FðtÞ}

z€1ðtÞ þ

􀀄

a þ bv21

􀀅

z1ðtÞ þv21

zðtÞ ¼ {u}T

1 {FðtÞ}

.. .

z€NðtÞ þ

􀀄

a þ bv2N

􀀅

zNðtÞ þv2N

zðtÞ ¼ {u}T

N {FðtÞ}

It is clear now why proportional damping is considered. The damping matrix ½c􀀉 becomes diagonalized

so that the equations of motion are decoupled. The coordinates ziðtÞ are called the modal coordinates, and

{u}T

i {FðtÞ} are called the modal forces. We often let

􀀄

a þ bv2i

􀀅

¼ 2vizi

{u}T

i {FðtÞ} ¼ qiðtÞ

to simplify the equations to

z€1ðtÞ þ 2v1z1z1ðtÞ þv21

z1ðtÞ ¼ q1ðtÞ

.. .

z€NðtÞ þ 2vNzNzNðtÞ þv2N

zðtÞ ¼ qN ðtÞ

ð5:22Þ

The solution to these equations can be obtained using the impulse response function and the convolution

integral for each coordinate

ziðtÞ ¼

ð1

giðtÞgiðt 2 tÞdt; i ¼ 1; 2; …; N

where

giðtÞ ¼

vdi

e2zvi t sin vdi t; t $ 0

0; t , 0

8><

and

vdi ¼ vi

ffiffiffiffiffiffiffiffi

1 2 z2i

Once we find all the ziðtÞ; we recover the physical coordinates by using the transformation in Equation

5.21 or

{xðtÞ} ¼ ½P􀀉{zðtÞ}

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