3.8 Forced Vibration

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The forced motion of a linear, n-DoF, undamped system is given by the nonhomogeneous equation of

motion 3.1:

My€ þ Ky ¼ f ðtÞ ð3:41Þ

Even though the following discussion is based on this undamped model, the results can be easily

extended to the damped case.

We have observed that the modal vectors form a basis for the configuration space. In other words, it is

possible to express the response y as a linear combination of the modal vectors ci:

y ¼ q1c1 þ q2c2 þ · · · þ qncn ð3:42Þ

The parameters qi are a set of generalized coordinates and are functions of time t: Equation 3.42 is written

in the vector– matrix form

y ¼ ½c1; c2; …; cn􀀉

q1

q2

.. .

qn

2

66666664

3

77777775

or

y ¼ Cq ð3:43Þ

and can be viewed as a coordinate transformation from the trajectory space to the canonical space of

generalized coordinates (principal coordinates or natural coordinates). Note that the inverse

transformation exists because the modal matrix C is nonsingular. On substituting Equation 3.43 into

Equation 3.41, we obtain

MCq€ þ KCq ¼ f ðtÞ

This result is premultiplied by CT and the orthogonality conditions (Equation 3.24 and Equation 3.25)

are substituted to obtain the canonical form of the system equation:

M􀀊 q€ þ K􀀊 q ¼ f􀀊ðtÞ ð3:44Þ

in which M􀀊 and K􀀊 are the diagonal matrices given by Equation 3.24 and Equation 3.25, and the

transformed forcing vector is given by

f􀀊ðtÞ ¼ CTf ðtÞ ð3:45Þ

3-28 Vibration and Shock Handbook

© 2005 by Taylor & Francis Group, LLC

Since M􀀊 and K􀀊 are diagonal matrices, Equation 3.44 corresponds to the set of n uncoupled simple

oscillator equations:

Miq€i þ Kiqi ¼ f􀀊iðtÞ for i ¼ 1; 2; …; n ð3:46Þ

In other words, the coordinate transformation 3.43 using the modal matrix has uncoupled the equations

of forced motion. It follows from Equation 3.46 that the natural frequencies of the system are given by

v2i

¼ Ki=Mi for i ¼ 1; 2; …; n ð3:47Þ

as we have noted before.

It is particularly convenient to employM-normal modal vectors. In this case,M􀀊 becomes the nth order

identity matrix, and 􀀊 K the diagonal matrix having v2i

as its diagonal elements; thus

M􀀊 ¼ I ð3:48Þ

K􀀊 ¼ diag

h

v21

; v22

; …; v2

n

i

ð3:49Þ

The corresponding uncoupled equations 3.46 take the form

q€i þv2i

qi ¼ f􀀊iðtÞ for i ¼ 1; 2; …; n ð3:50Þ

in which the forcing terms f􀀊iðtÞ are given by Equation 3.45, C being M-normal.

Typically, the initial conditions for the original system are provided as the initial position yð0Þ and the

initial velocity y_ð0Þ: The corresponding initial conditions for the transformed equations of motion are

obtained using Equation 3.43 as

qð0Þ ¼ C21yð0Þ ð3:51Þ

q_ ð0Þ ¼ C21 y_ð0Þ ð3:52Þ

The complete response of the original system can be conveniently obtained by first solving the simple

oscillator equation 3.50 and then transforming the results back into the trajectory space using

Equation 3.43.

The complete solution to this linear system can be viewed as the sum of the initial condition response

in the absence of the forcing function and the forced response with zero initial conditions, as discussed in

Chapter 2. For the simple oscillator equation 3.50, the initial condition response qiI is given by

qiI ¼ qið0Þ cos vit þ

q_ið0Þ

vi

sin vit ð3:53Þ

The impulse response function (i.e., response to a unit impulse excitation) for the undamped oscillator

equation is

hiðtÞ ¼

1

vi

sin vit for t $ 0 ð3:54Þ

The forced response qiF; with zero initial conditions, is obtained using the convolution integral method;

specifically

qiF ¼

ðt

0

f􀀊iðtÞhiðt 2 tÞdt ð3:55Þ

The complete solution in the canonical domain is

qi ¼ qiI þ qiF ¼ qið0Þcos vit þ

q_ið0Þ

vi

sin vit þ

1

vi

ðt

0

f􀀊iðtÞsin viðt 2 tÞdt i ¼ 1; 2; …; n ð3:56Þ

Each of the n responses is a modal response that is the contribution from that mode to the actual

response y. This approach is summarized in Box 3.4. Next, we will illustrate this approach of solving

for the forced vibration by means of an example.

Modal Analysis 3-29

© 2005 by Taylor & Francis Group, LLC

Box 3.4

MODAL APPROACH TO FORCED RESPONSE

Forced system: My€ þ Ky ¼ f ðtÞ

Number of DoFs ¼ n

Modal transformation: y ¼ Cq

where modal matrix C ¼ ½c1; c2; …; cn􀀉 ¼ matrix of mode shape vectors.

We obtain the diagonalized system

M􀀊 q€ þ K􀀊 q ¼ f􀀊ðtÞ

or

Miq€i þ Kiqi ¼ fiðtÞ; i ¼ 1; 2;…; n

where

M􀀊 ¼ CTMC ¼ diag½M1; M2; …; Mn􀀉

K􀀊 ¼ CTKC ¼ diag½K1; K2; …; Kn􀀉

f􀀊ðtÞ ¼ CTf ðtÞ ¼ ½f􀀊1; f􀀊2;…; f􀀊n􀀉T

Initial conditions:

qð0Þ ¼ C21yð0Þ

q_ ð0Þ ¼ C21 y_ð0Þ

Steps: Use M-normal case: Mi ¼ 1; Ki ¼ v2i

Then,

q€i þv2i

qi ¼ f􀀊iðtÞ; i ¼ 1; 2; …; n

1. Free, initial-condition response (zero-input response)

qiI ¼ qið0Þcos vit þ

q_ið0Þ

vi

sin vit

2. Forced, zero initial-condition response

qiF ¼

1

vi

ðt

0

f􀀊iðtÞsin viðt 2 tÞdt

3. qi ¼ qiI þ qiF

4. Transform back to y using y ¼ Cq

3-30 Vibration and Shock Handbook

© 2005 by Taylor & Francis Group, LLC

Example 3.7

Let us consider again the system shown in Figure 3.8. A step-input force f ðtÞ given by

f ðtÞ ¼

f0 for t $ 0

0 for t , 0

(

is applied to the left-hand mass (DoF y1). Assume that the system starts from rest ðyð0Þ ¼ 0 and

y_ð0Þ ¼ 0Þ:

As before, the M-normal modal matrix of the system is

C ¼

1ffiffiffiffi

2m p

1ffiffiffiffi

2m p

1ffiffiffiffi

2m p 2

1ffiffiffiffi

2m p

2

6664

3

7775

¼

1ffiffiffiffi

2m p

1 1

1 21

" #

No forcing input is applied to the second DoF ðy2Þ: Hence, the overall forcing input vector is

fðtÞ ¼

f0

0

" #

for t $ 0

From Equation 3.45, the transformed forcing input vector is obtained as

f􀀊ðtÞ ¼ CTf ðtÞ ¼

1ffiffiffiffi

2m p

1 1

1 21

" #

f0

0

" #

¼

fffi0ffiffiffi

2m p

1

1

" #

for t $ 0

From Equation 3.51 and Equation 3.52, the initial conditions for the modal (canonical) variables

are obtained as q_ ð0Þ ¼ 0 and qð0Þ ¼ 0: The modal responses q1ðtÞ and q2ðtÞ are obtained using

Equation 3.56.

3.8.1 First Mode (Rigid-Body Mode)

Note that

lim

vi !0

sin viðt 2 tÞ

vi ¼ t 2 t

It follows that

q1ðtÞ ¼

ðt

0

f0ðt ffi2ffiffiffitÞ

2m p dt ¼

f0t2

2

ffiffiffiffi

2m p

3.8.2 Second Mode (Oscillatory Mode)

q2ðtÞ ¼

1

v2

ðt

0

fffi0ffiffiffi

2m p sin v2ðt 2 tÞdt ¼

f0

v22

ffiffiffiffi

2m p ð1 2 cos v2tÞ

The overall response in the physical trajectory space is obtained by transforming the modal responses

using Equation 3.43; thus

y ¼

1ffiffiffiffi

2m p

1 1

1 21

" # f0t2

2

ffiffiffiffi

2m p

f0ð1 2 cos v2tÞ

v22

ffiffiffiffi

2m p

2

66664

3

77775

¼

f0

2m

t2

2 þ

1

v22

ð1 2 cos v2tÞ

t2

2

2

1

v22

ð1 2 cos v2tÞ

2

66664

3

77775

Modal Analysis 3-31

© 2005 by Taylor & Francis Group, LLC

with v22

¼ 2k=m: The response of both masses grows (unstable) quadratically in an oscillatory manner, as

shown in Figure 3.11.

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