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5.5 Multi-Degree-of-Freedom: The Response to Random Loads

Back

From the analysis of the Single-DoF system, we know that knowledge of the frequency response function

HðvÞ is crucial in obtaining the response statistics to random loads. For a system with more than one

DoF, there is more than one frequency response function HðvÞ: For an N-DoF system, the frequency

response functions will be given as an N £ N symmetric matrix HðtÞ:

Random Vibration 5-17

In the single-DoF system, the response statistics were given in terms of the mean response mX and

the spectral density of the response SXX ðvÞ: We will find here that the response statistics of an N-DoF

system are given in terms of the mean response for each DoF, mX1

; …; mXN

; and a matrix of spectral

densities of the response SðvÞ; which is also of dimension N £ N:

We present a priori the results which are derived in the subsequent section; the mean response vector

mX and the matrix spectral density of the output SXX ðvÞ:

mX ¼ HXX ð0ÞmF

SXX ðvÞ ¼ Hp

XX ðvÞSFF ðvÞHT

XX ðvÞ

Here, we will first obtain the frequency response function of a typical two-DoF system. We will then

consider a two-DoF system subjected to a single random load. Finally, we will expand our analysis to an

N-DoF system subjected to N random loads.

5.5.1 Response due to a Single Random Force

When a body vibrates, points on the body move relative to each other. To describe this relative

motion of two points on a body subjected to random loading, we must know both their crossspectral

densities and cross-correlations, and also the spectral densities and the autocorrelation

function of the individual motions of each point. This requirement is true even though there is

only one force acting on the system. This is similar to the problem of the single mass loaded by

two random loads.

Consider a body excited by a single randomly varying force PðtÞ with autocorrelation RTT ðtÞ and

spectral density SPP ðvÞ. Owing to this loading, two points on the body have displacements XðtÞ and Y ðtÞ

with means mX and mY ; autocorrelations RXX ðtÞ and RYY ðtÞ; and spectral densities SXX ðvÞ and SYY ðvÞ;

respectively. Note again that upper case letters are used to signify random processes.

We know the following from our earlier studies

E{XðtÞ} ¼ HXP ð0ÞmP

E{Y ðtÞ} ¼ HYP ð0ÞmP

and

SXX ðvÞ ¼ lHXP ðvÞl2SPP ðvÞ ð5:23Þ

SYY ðvÞ ¼ lHYP ðvÞl2SPP ðvÞ ð5:24Þ

where

HXP ðvÞ ¼

ð1

gXP ðtÞe2ivt dt

HYP ðvÞ ¼

ð1

gYP ðtÞe2ivt dt

and

XðtÞ ¼

ð1

gXP ðtÞPðt 2 tÞdt

Y ðtÞ ¼

ð1

gYP ðtÞPðt 2 tÞdt

5-18 Vibration and Shock Handbook

With these relations, we can now find the expressions for the cross-correlation RXY ðtÞ and cross-spectral

density SXY ðvÞ: Using the definition of a cross-correlation function, we obtain

RXY ðtÞ ¼ E{XðtÞY ðt þ tÞ} ¼ E

ð1

gXP ðt1ÞPðt 2 t1Þdt1

ð1

gYP ðt2ÞPðt þ t 2 t2Þdt2

􀀘 􀀙

ð1

gXP ðt1Þ

ð1

gYP ðt2ÞE{Pðt 2 t1ÞPðt þ t 2 t2Þ}dt2

􀀒 􀀓

dt1

ð1

gXP ðt1Þ

ð1

gYP ðt2ÞRPP ðt þ t1 2 t2Þdt2

􀀒 􀀓

dt1

Even though we have an explicit expression, it is generally not easy to interpret.

The cross-spectral density can be derived by using the Fourier transform relation between RXY ðtÞ and

SXY ðvÞ: Thus,

SXY ðvÞ¼

ð1

RXY ðtÞe2ivt dt ¼

ð1

gXP ðt1Þ

ð1

gYP ðt2ÞRPP ðt þt1 2t2Þdt2 dt1 e2ivt dt

ð1

gXP ðt1Þeþivt1

ð1

gYP ðt2Þe2ivt2

ð1

RPP ðt þt1 2t2Þe2ivðtþt1 2t2 Þdt dt2 dt1 ð5:25Þ

In order to integrate this last expression, it is easier to proceed by transforming the variables in the

innermost integral according to

l ;t þt1 2t2 and dl ¼dt

where t1 and t2 are dummy variables. Then, Equation 5.25 becomes

SXY ðvÞ¼

ð1

gXP ðt1Þeþivt1 dt1

ð1

gYP ðt2Þe2ivt2 dt2

ð1

RPP ðlÞe2ivl dl

¼Hp

XP ðvÞHYP ðvÞSPP ðvÞ ð5:26Þ

Where points X and Y coincide, this result reduces to the classical fundamental relation

SXX ðvÞ¼lHXP l2SPP ðvÞ:

Equation 5.23, Equation 5.24 and Equation 5.26 can be written in matrix form

SXX ðvÞ SXY ðvÞ

SYX ðvÞ SYY ðvÞ

" #

XP ðvÞ

YP ðvÞ

" #

SPP ðvÞ

􀀑

HXP ðvÞ HYP ðvÞ

􀀜

ð5:27Þ

SXX ðvÞ SXY ðvÞ

SYX ðvÞ SYY ðvÞ

" #

XP ðvÞSXX ðvÞHXP ðvÞ Hp

XP ðvÞSXY ðvÞHYP ðvÞ

YP ðvÞSYX ðvÞHXP ðvÞ Hp

YP ðvÞSYY ðvÞHYP ðvÞ

" #

Example 5.5 Response Spectra for a Two-Degree-of-Freedom System Subjected

to a Single Random Force

Consider the mass – spring – damper system shown in Figure 5.9. Assume that the random force F1ðtÞ is a

white noise random process with SF1 F1 ðvÞ ¼ S0 and F2ðtÞ ¼ 0: (i) Find the frequency response functions

H11ðvÞ and H21ðvÞ; and (ii) find the spectral densities SX1 X1 ðvÞ; SX1 X2 ðvÞ and SX2 X2 ðvÞ:

Random Vibration 5-19

Solution

The equation of motion for this system has been obtained as Equation 5.14 with F2ðtÞ ¼ 0: We found

previously that the frequency response function for this system is given by

½HðvÞ􀀉 ¼

det½Z􀀉

2m2v2 þ ivc2 þ k2 ivc2 þ k2

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

" #

where

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

2 v2ðm1k2 þ c1c2 þ k1m2 þ k2m2Þ þ k1k2

Since F2 ¼ 0; the frequency response functions due to F2; H12ðvÞ and H22ðvÞ; are zero. The frequency

response functions due to F1 are given by

H11ðvÞ ¼

2m2v2 þ ivc2 þ k2

det½Z􀀉

H21ðvÞ ¼

ivc2 þ k2

det½Z􀀉

Using Equation 5.27, the response spectra are given by

SX1 X1 ðvÞ ¼ Hp

11ðvÞH11ðvÞSF1 F1 ðvÞ ¼ S0 ðk2 2 m2v2Þ2 þ ðvc2Þ2

ðdet½Z􀀉Þ2

SX2 X2 ðvÞ ¼ Hp

22ðvÞH22ðvÞSF1 F1 ðvÞ ¼ S0

k22

þ ðvc2Þ2

ðdet½Z􀀉Þ2

SX1 X2 ðvÞ ¼ Hp

11ðvÞH21ðvÞSF1 F1 ðvÞ ¼ S0 ð2m2v2 2 ivc2 þ k2Þðivc2 þ k2Þ

ðdet½Z􀀉Þ2

SX2 X1 ðvÞ ¼ Hp

21ðvÞH11ðvÞSF1 F1 ðvÞ ¼ S0 ð2m2v2 þ ivc2 þ k2Þð2ivc2 þ k2Þ

ðdet½Z􀀉Þ2

Note that SX2 X1 ðvÞ ¼ SX1 X2 ð2vÞ ¼ Sp

X1 X2 ðvÞ:

5.5.2 Response to Multiple Random Forces

Consider a multi-DoF system subjected to multiple random forces. The goal is to obtain the relation

equivalent to Equation 5.27 for the general response for N DoF by first considering the response of a

Single-DoF to N forces, and then generalizing to N DoF. This will be done in two ways. First, by extending

the impulse-response method leading to the convolution integral, and then by utilizing the modal

F1(t)

X1(t) X2(t)

m1 m2

FIGURE 5.9 Two-DoF system excited by a single random force.

5-20 Vibration and Shock Handbook

analysis approach where the coupled physical equations of motion are decoupled in a new modal

coordinate system.

5.5.3 Impulse-Response Approach

Let us start by expressing the response XkðtÞ using the impulse-response functions gkiðtÞ: Recall that gkiðtÞ

is the impulse response of mass mk to force FiðtÞ: By linear superposition, the total response Xk ðtÞ of mass

mk is equal to the sum of the individual responses to each of the N forces

XkðtÞ ¼

i¼1

XkiðtÞ ¼

i¼1

ð1

gkiðtÞFiðt 2 tÞdt

Assuming that the forces are stationary with respective means mFi and cross-correlations RFi Fj ðtÞ; the

mean and correlation of the response can be found as follows:

E{XkðtÞ} ¼

i¼1

ð1

gkiðtÞE{Fiðt 2 tÞ}dt

mXk ¼

i¼1

mFi

ð1

gkiðtÞdt ¼

i¼1

mFi Hkið0Þ

In matrix form

mX ¼ HXX ð0ÞmF

This is a static response and, therefore, can be ignored here and added on at the end of the computations.

We next evaluate the expressions for the response correlations and spectral densities, from which we can

evaluate the mean-square response. By definition, the correlations are given by

RXk Xj ðtÞ ¼ E{XkðtÞXjðt þ tÞ} ¼ E

m¼1

XkmðtÞ

n¼1

Xjnðt þ tÞ

( )

¼ E

m¼1

ð1

gkmðz ÞFmðt 2 z Þdz

n¼1

ð1

gjnðj ÞFnðt þ t 2 j Þdj

( )

m¼1

n¼1

ð1

gkmðz Þgjnðj ÞE{Fmðt 2 z ÞFnðt þ t 2 j Þ}dz dj

m¼1

n¼1

ð1

gkmðz Þgjnðj ÞRFm Fn ðt 2 j þ z Þdz dj

The response spectral density is by definition

SXk Xj ðvÞ ¼

ð1

RXk Xj ðtÞexpð2ivtÞdt

Before substituting the correlation function into this equation, multiply it by expð2iv½z 2 j 􀀉Þ

£ expðiv½z 2 j 􀀉Þ: Also, define n ¼ t 2 j þ z with dn ¼ dt: All of these manipulations allow us to put

the spectral density in the following form:

SXk Xj ðvÞ ¼

m¼1

n¼1

ð1

gkmðzÞexpðivzÞdz

ð1

gjnðjÞexpð2ivjÞdj £

ð1

RFm Fn ðnÞexpð2ivnÞdn

m¼1

n¼1

kmðvÞHjnðvÞSFm Fn ðvÞ

Random Vibration 5-21

where the star denotes complex conjugate. In matrix form

SXk Xj ðvÞ ¼ Hp

k ðvÞSFF ðvÞHT

j ðvÞ ð5:28Þ

In this notation, Hp

k ðvÞ is a row vector of dimension 1 £ N

k ðvÞ ¼

􀀑

k1ðvÞ · · · Hp

kN ðvÞ

􀀜

HTj ðvÞ is a column vector of dimension N £ 1

j ðvÞ ¼

Hj1ðvÞ

.. .

HjN ðvÞ

66664

77775

and SFF ðvÞ is a matrix of dimension N £ N

SFF ðvÞ ¼

SF1 F1 ðvÞ · · · SF1 FN ðvÞ

.. .

. .

. .. .

SFN F1 ðvÞ · · · SFN FN ðvÞ

66664

77775

Equation 5.28 can now be generalized for any Xk and Xj

SXX ðvÞ ¼ HpðvÞSFF ðvÞHTðvÞ ð5:29Þ

where

HðvÞ ¼

H11ðvÞ · ·· H1N ðvÞ

.. .

. .

. .. .

HN1ðvÞ · ·· HNN ðvÞ

6664

7775

and

SXX ðvÞ ¼

SX1 X1 ðvÞ · · · SX1 XN ðvÞ

.. .

. .

. .. .

SXN X1 ðvÞ · · · SXN XN ðvÞ

66664

77775

Let us recall Equation 5.6, Equation 5.9, and Equation 5.27. Equation 5.6 is the response spectral density

of a single-DoF system to a single random force, Equation 5.9 is the response spectral density of a single-

DoF system to two random forces, and Equation 5.27 is the response spectral densities of a two-DoF

system due to a single force. Comparing them with Equation 5.29, we realize that they are all special cases

of this equation.

For a zero mean, the mean-square response of the jth coordinate is given by the relation

Xj ¼

ð1

SXj Xj ðvÞdv ð5:30Þ

Example 5.6 Spacecraft Design Problem (based on Wirsching et al., p. 228)

A rocket mounting is to be designed. The rocket of mass m1 ¼ 181 kg is mounted on a supporting

structure of mass m2 ¼ 362 kg using structural brackets that can be modeled as springs of constant

k ¼ 1:40 MN/m. There is also damping with coefficient c ¼ 3:50 kNsec/m. When the engine fires,

the thrust can be interpreted as being composed of a mean value thrust and a random fluctuating thrust.

5-22 Vibration and Shock Handbook

The random variations are due to factors such as fluctuations in fuel mix and eccentricities in thrust

direction. The mean thrust is given by mT ¼ 16 kN and the random component is given by T, which is

a random function of time. T is characterized as white noise of intensity 1980 N2/Hz over the frequency

range 0 – 50 Hz. The mean thrust can be viewed as a static load while the random thrust as

the dynamic load.

The design must consider that components in the supporting structure will be damaged if the

acceleration is too large. Also, the structural brackets will fail if the relative displacement between the

rocket and the supporting structure is too large. Both these design issues are difficult because the input is

random. Therefore, there is a trade-off between the cost of making a component stronger and making it

more reliable. The probabilistic framework allows the analyst to quantitatively understand the trade-off.

Solution

The goal of the analysis is to determine the acceleration of the supporting mass X€ 2 and the relative

displacement Z ¼ X2 2 X1:

First we solve the mean value problem. The mean value acceleration is

a ¼

m1 þ m2 ¼

16;000

181 þ 362 ¼ 29:46 m=sec2 ¼

29:46

9:81 ¼ 3:00g

The mean force in the bracket is found by applying the equation of motion to m2

mFb ¼ m2a2 ¼ 362 £ 29:46 ¼ 10665 N ¼ 10:66 kN

The mean bracket deflection is

mZ ¼

mFb

k ¼

10:66 £ 103

1:40 £ 106 ¼ 7:61 £ 1023 m ¼ 7:61 mm

While the mean value is the single most important descriptor of randomness, the variance is important

because it is a measure of the scatter of possible results. In order to calculate the variance, we need to

derive the response spectral density and the frequency response function (see Equation 5.30). This we

compute next. The equations of motion are given by

X€ 1 2 kðX2 2 X1Þ 2 cð X_ 2 2 X_ 1Þ ¼ FðtÞ

X€ 2 þ kðX2 2 X1Þ þ cð X_ 2 2 X_ 1Þ ¼ 0

Given the problem statement, the variables of interest are ðZ; X2Þ: Transform variables using

Z ¼ X2 2 X1: The equations of motion become

X€ 2 2 m1

Z€ 2 kZ 2 cZ_ ¼ FðtÞ

X€ 2 þ kZ þ cZ_ ¼ 0

To derive the frequency response function, assume harmonic excitation

FðtÞ ¼

" #

eivt

and harmonic response of the form

ZðtÞ

X€ 2

" #

X€ 0

" #

eivt

Substitute these equations into the matrix equation of motion to find

X€ 0 þ m1v2Z0 2 kZ0 2 icvZ0 ¼ F0

X€ 0 þ kZ0 þ icvZ0 ¼ 0

Random Vibration 5-23

The frequency response function is then found to be

HðvÞ ¼

DðvÞ

m2 2m1

2k 2 icv m1v2 2 k 2 icv

" #

DðvÞ ¼ m1m2v2 2 kðm1 þ m2Þ 2 icvðm1 þ m2Þ

There is only one applied force, and the spectral density function matrix is

SF ðvÞ ¼

S0 0

0 0

" #

Apply Equation 5.29 to find

SXX ðvÞ ¼

SZZ SZX€

SX€Z SX€X€

" #

lDðvÞl2

m22

2km2 þ icm2v

2km2 þ icm2v k2 þ ðcvÞ2

" #

To determine the root mean square of relative displacement ZðtÞ; note that the spectral density function

of ZðtÞ is the top term on the diagonal of SXX ðvÞ:

SZZ ðvÞ ¼ S0

m22

lDðvÞl2

The mean-square value of ZðtÞ is then

Z ¼

ð1

SZZ ðvÞ ¼

pS0m22

ckðm1 þ m2Þ2

The above are for a two-sided spectrum. The given spectrum is one-sided, therefore, S0 ¼ W0=4p: Then,

with appropriate substitutions, we find sZ . 0:212 mm: In summary, the relative displacement has

mean and root mean-square

mZ ¼ 7:61 mm; sZ ¼ 0:212 mm

The force in the spring has mean and root mean-square

mFb ¼ kmZ ¼ 1:40 £ 106 £ 7:61 ¼ 10:65 kN

sFb ¼ ksZ ¼ 1:40 £ 106 £ 0:212 ¼ 0:297 kN

Finally, the spectral density of the acceleration is the second diagonal term of SX ðvÞ:

SX€X€ ¼ S0

k2 þ ðcvÞ2

lDðvÞl2

X€ ¼

ð1

SX€X€ ðvÞdv

and

X€ ¼ pS0

cðm1 þ m2Þ2 þ

ðm1 þ m2Þm1m2

􀀒 􀀓

Upon making the appropriate substitutions, we find

sX€ ¼ 0:852 m=sec2 ¼ 0:087g

In summary, the mean and standard deviation of X€ 2 are

mX€ 2 ¼ 3g; sX€ 2 ¼ 0:087g

See Example 5.1 for a discussion on the above nonstandard integrals.

A designer would now use these mean and standard deviation numbers to decide how reliable to make

the components. For example, ifwe assume that Fb andX€ 2 are normal density,we can specify a design based

5-24 Vibration and Shock Handbook

on m ^ 3s, which tells us the probability of

exceeding this 6s range is less than 0.1%. If this

design is too expensive, then the designer will try a

design based on m ^ 3s; one that will have a higher

probability of failure. In this way, a trade-off study is

performed weighing cost against probability of

failure. Figure 5.10 demonstrates a design based on

5% failure probability.

5.5.4 Modal Analysis Approach

The response of a multi-DoF discrete system loaded by random forces is generally a very complicated

problem to formulate and solve. Here, to demonstrate one possible approach, we study the modal

analysis of such a structure and work through some of the intricacies.

We will start the analysis at the point where the modal equations of motion have been formulated and

the assumption of proportional damping has been made. From Equation 5.22, for a random vibration,

the modal equations in indicial notation are

Z€ i þ 2zivi

Z_ i þ v2i

Zi ¼ QiðtÞ; i ¼ 1; 2; …; N ð5:31Þ

where the parameters are familiar. We further assume that the modal forces QiðtÞ are ergodic random

excitations. The transformation between physical and modal spaces is

XjðtÞ ¼

i¼1

ujiZiðtÞ

and the transformation between physical and modal forces is

QjðtÞ ¼

i¼1

uijFiðtÞ ¼ uT

j F ð5:32Þ

where ui is the modal vector for the ith DoF and

ui ¼

u1i

.. .

uNi

6664

7775

In matrix form, we can write

X ¼ PZ

Q ¼ PTF

where P is the modal matrix given by

P ¼ ½ u1 · · · uN 􀀉

For a two-DoF system, we can be specific

XjðtÞ ¼

i¼1

ujiZiðtÞ ¼ uj1Z1ðtÞ þ uj2Z2ðtÞ; j ¼ 1; 2

QjðtÞ ¼

i¼1

uijFiðtÞ ¼ u1jF1ðtÞ þ u2jF2ðtÞ; j ¼ 1; 2

ð5:33Þ

mX −1.96σX mX + 1.96σX

0.95

fX (x)

FIGURE 5.10 Normal density with 5% failure

probability.

Random Vibration 5-25

for each DoF j of a two-DoF structure. Our goal in this analysis is to evaluate the statistics of the response,

that is, to find autocorrelations and cross-correlations, RXk Xj ðtÞ; and its Fourier transform, the power

spectrum, SXk Xj ðvÞ:

Begin with the definition of the autocorrelation and substitute Equation 5.33 for XjðtÞ:

RXk Xj ðtÞ ¼ E{XjðtÞXjðt þ tÞ} ¼ E

l¼1

m¼1

uklujmZlðtÞZmðt þ tÞ

( )

ð5:34Þ

where ZiðtÞ is the solution to Equation 5.31:

ZiðtÞ ¼

ðt

QiðtÞgiðt 2 tÞdt ð5:35Þ

giðtÞ ¼

vdi

expð2zivitÞsin vdi t ð5:36Þ

vdi ¼ við1 2 z2i

Þ1=2 ð5:37Þ

Note that, since the impulse response function gðtÞ is zero for t , 0; the lower limit on the integral

defining ZðtÞ can be made 21 without changing the value of the integral.

Substituting Equation 5.35 to Equation 5.37 into Equation 5.34 and letting the expectation operate

only on the stochastic terms, we have

RXk Xj ðtÞ ¼

l¼1

m¼1

ðtþt

ðt

uklujmE{Qlðu1ÞQm ðu2Þ}glðt 2 u1Þgmðt þ t 2 u2Þdu1 du2 ð5:38Þ

where u1 and u2 are dummy time variables, and

RQl Qm ðu2 2 u1Þ ¼ E{Qlðu1ÞQmðu2Þ}

owing to the assumed ergodicity (and thus stationarity) of the forcing. If the system is lightly damped

and has well-separated modal frequencies, as is the case in many engineering structures, the response due

to QlðtÞ is almost statistically independent of the response due to QmðtÞ: The cross-correlation terms that

arise in Equation 5.38 are then almost zero, with the only nonzero terms arising for m ¼ l :

RQl Ql ðu2 2 u1Þ ¼ E{Qlðu1ÞQlðu2Þ}

Now that we have the correlation function for the response in terms of the correlation function for the

random forcing, we can proceed to evaluate the response spectral density, from which we can evaluate

probabilities of occurrence. To do this, the following transformation of variables is first necessary:

u1 ; t 2 u1; u2 ; t þ t 2 u2;

du1 ¼ 2du1; du2 ¼ 2du2

resulting in the response correlation

RXk Xj ðtÞ ¼

l¼1

m¼1

ð1

uklujmRQl Qm ðu1 2 u2 þ tÞglðu1Þgmðu2Þdu1 du2

The power spectral density for response XðtÞ is equal to the Fourier transform of the correlation function

SXk Xj ð21 vÞ ¼

ð1

RXk Xj ðtÞe2ivt dt

ð1

l¼1

m¼1

ð1

uklujmRQl Qm ðu1 2 u2 þ tÞglðu1Þgmðu2Þdu1 du2

( )

e2ivt dt

5-26 Vibration and Shock Handbook

or, recalling and utilizing the assumption that the processes are ergodic, and, therefore, by averaging

in time

SXk Xj ðvÞ ¼

l¼1

m¼1

uklujm lim

T!1

ðT

glðu1Þdu1 lim

T!1

ðT

gmðu2Þdu2

􀀘

􀀐 lim

T!1

ðT

RQl Qm ðu1 2 u2 þ tÞe2ivt dt

􀀙

where the lower limits on the integrals were set to 2T since gðtÞ is zero for t , 0; and thus the change in

lower limit does not affect the values of the integrals. Using the change of variables

g ; u1 2 u2 þ t; dg ¼ dt

we obtain

SXk Xj ðvÞ ¼

l¼1

m¼1

uklujm lim

T!1

ðT

glðu1Þexpðivu1Þdu1

􀀘

􀀐 lim

T!1

ðT

gmðu2Þexpð2ivu2Þdu2 lim

T!1

ðT2u2 þu1

2T2u2 þu1

RQl Qm ðgÞe2ivg dg

􀀙

In the last integral, we make the physical argument that RQj Qm ðgÞ ! 0 as lgl increases and, therefore, the

limits can be replaced by 2T and T; respectively.10 Then,

Hlð2vÞ ¼ lim

T!1

ðT

glðu1Þexpðivu1Þdu1

HmðvÞ ¼ lim

T!1

ðT

gmðu2Þexpð2ivu2Þdu2

SQl Qm ðvÞ ¼ lim

T!1

ðT2u2 þu1

2T2u2 þu1

RQl Qm ðgÞe2ivg dg

with the resulting response spectral density

SXk Xj ðvÞ ¼

l¼1

m¼1

uklujmHlð2vÞHmðvÞSQl Qm ðvÞ

For a two-DoF system, we would have

SXk Xj ðvÞ ¼

􀀑

uk1 uk2

􀀜 Hp

1 ðvÞ 0

0 Hp

2 ðvÞ

" #

SQ1 Q1 ðvÞ SQ1 Q2 ðvÞ

SQ2 Q1 ðvÞ SQ2 Q2 ðvÞ

" #

H1ðvÞ 0

0 H2ðvÞ

" #

uj1

uj2

" #

¼ uk1uj1Hp

1 ðvÞ____________H1ðvÞSQ1 Q1 ðvÞ þ uk1uj2Hp

1 ðvÞH2ðvÞSQ1 Q2 ðvÞ þ uk2uj1Hp

2 ðvÞH1ðvÞSQ2 Q1 ðvÞ

þ uk2uj2Hp

2 ðvÞH2ðvÞSQ2 Q2 ðvÞ

Suppose that we wish to find SXX for a two-DoF system. Then

SXX ¼

u11 u12

u21 u22

" #

1 ðvÞ 0

0 Hp

2 ðvÞ

" #

SQ1 Q1 ðvÞ SQ1 Q2 ðvÞ

SQ2 Q1 ðvÞ SQ2 Q2 ðvÞ

" #

H1ðvÞ 0

0 H2ðvÞ

" #

u11 u21

u12 u22

" #

10Physically, we are stating that, as time difference (trial mode) increases, there will be an exponentially decaying

correlation. This is borne out by experiments on physical systems.

Random Vibration 5-27

In general, we can write

SXX ðvÞ ¼ PHpðvÞSQQðvÞHðvÞPT ð5:39Þ

where

HðvÞ ¼

H1ðvÞ 0 0

0 . .

0 0 HN ðvÞ

6664

7775

HiðvÞ ¼

2 v2 þ i2ziviv þ v2i

The spectral densities of the modal forces SQl Qm ðvÞ can be obtained from SFF ðvÞ using Equation 5.32.

The autocorrelation RQl Qm ðtÞ is defined as

RQl Qm ðtÞ ¼ E{QlðtÞQmðt þ tÞ} ¼ E

i¼1

uilFiðtÞ

j¼1

ujmFjðt þ tÞ

; ¼

i¼1

j¼1

uilujmE{FiðtÞFjðt þ tÞ}

i¼1

j¼1

uilujmRFi Fj ðtÞ

Taking the Fourier transform

SQl Qm ðvÞ ¼

i¼1

j¼1

uilujmSFi Fj ðvÞ

SQl Qm ðvÞ ¼ uT

l SFF um

SQQðvÞ ¼ PTSFF P ð5:40Þ

Substituting Equation 5.40 into Equation 5.39, we obtain

SXX ðvÞ ¼ PHpðvÞPTSFF ðvÞPHðvÞPT

Comparing this result with Equation 5.29, we find that

XX ðvÞ ¼ PHpðvÞPT

HXX ðvÞ ¼ PHðvÞPT

which is a fully populated transfer function matrix of XðtÞ:

For lightly damped systems with well-spaced modal frequencies, the cross-terms in the

dÐouble summation, those where l – m; contribute very little to the mean-square response given by

1 SXi Xj ðvÞdv: In this case, we can use the approximation

SXj Xj ðvÞ .

l¼1

jllHlðvÞl2SQl Ql ðvÞ

where

lHlðvÞl2 ¼

1=v2l

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

ð1 2 v2=v2l

Þ2 þ ð2zlv=vlÞ2

5-28 Vibration and Shock Handbook

Suppose that we only account for the first three modes out of 1000 DoF. Then, the spectral density SXX

based on three modes can be obtained by

SXX ðvÞ ¼ Pl1000£3HpðvÞl3£3PTl3£1000SFF l1000£1000Pl1000£3HðvÞl3£3PTl3£1000

where

Pl1000£3 ¼ ½ u1 u2 u3 􀀉

HðvÞl3£3 ¼

H1ðvÞ 0 0

0 H2ðvÞ 0

0 0 H3ðvÞ

664

775

Thus, we can easily see that the modal approach allows the analyst and designer to retain only those

modes that are key to the problem at hand. If, as above, a thousand-DoF system can be adequately

modeled as a three-DoF system because that is where all the energy is located, then a great deal of

unnecessary computation is avoided.

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