Vibration and Shock Handbook 05

Vibration and Shock Handbook 05 Every so often, a reference book appears that stands apart from all others, destined to become the definitive work in its field. The Vibration and Shock Handbook is just such a reference. From its ambitious scope to its impressive list of contributors, this handbook delivers all of the techniques, tools, instrumentation, and data needed to model, analyze, monitor, modify, and control vibration, shock, noise, and acoustics. Providing convenient, thorough, up-to-date, and authoritative coverage, the editor summarizes important and complex concepts and results into “snapshot” windows to make quick access to this critical information even easier. The Handbook’s nine sections encompass: fundamentals and analytical techniques; computer techniques, tools, and signal analysis; shock and vibration methodologies; instrumentation and testing; vibration suppression, damping, and control; monitoring and diagnosis; seismic vibration and related regulatory issues; system design, application, and control implementation; and acoustics and noise suppression. The book also features an extensive glossary and convenient cross-referencing, plus references at the end of each chapter. Brimming with illustrations, equations, examples, and case studies, the Vibration and Shock Handbook is the most extensive, practical, and comprehensive reference in the field. It is a must-have for anyone, beginner or expert, who is serious about investigating and controlling vibration and acoustics.

Random Loads 5-2Formulation † Derivation of Equations † Response

Correlations † Response Spectral Density5.3 Response to Two Random Loads 5-75.4 Multi-Degree-of-Freedom Vibration 5-12Deterministic Vibration † Solution by Frequency Response

Function † Modal Analysis5.5 Multi-Degree-of-Freedom: The Response toRandom Loads 5-17Response due to a Single Random Force † Response to

Multiple Random Forces † Impulse-Response Approach †

Modal Analysis Approach5.6 Continuous System Random Vibration 5-29Transverse Vibration of Beams † Random Transverse

Vibration

Summary

This chapter summarizes the key ideas of linear random vibration This discipline focuses on determining theresponse statistics of an oscillator or structure to input forces that are definable only in terms of their statistics.Typical problems include the following: (1) given the power spectrum of the force, find the power spectrum of theresponse; (2) given the mean value and variance of the force, find the mean value and variance of the response Themethodology is built upon the linear theory of vibration for discrete single- and multi-degree-of-freedom (DoF)systems, and continuous systems The approaches are essentially the direct method and the modal analysis method.The direct method may also be called a transfer matrix method (see Chapter 2) Modal analysis (see Chapters 3 and4) has the same benefit in random vibration as is done in deterministic vibration studies: it can be computationallymore efficient A number of examples are given, as are a list of representative references

5.1 Random Vibration

The discipline of random vibration of structures was borne of the need to understand how structuresrespond to dynamic loads that are too complex to model deterministically Examples include aerodynamicloading on aircraft and earthquake loading of structures Essentially, the question that must be answered is:given the statistics (read: uncertainties) of the loading, what are the statistics (read: most likely values withbounds) of the response? Generally, for engineering applications the statistics of greatest concern are themean, or average value, and the variance, or scatter These concepts are discussed in detail subsequently.Suppose that we are aircraft designers currently working on the analysis and design of a wing for a newairplane As engineers, we are very familiar with the mechanics of solids and can size the wing for staticloads Also, we have vibration experience and can evaluate the response of the wing to a harmonic

5-1

or impulsive forcing However, this wing provides

lift to an airplane flying through a turbulent

atmosphere Even though we are not fluid

dynamicists, we know that turbulence is a very

complicated physical process In fact, the fluid

(air) motion is so complicated that probabilistic

models are required to model the behavior Here, a

plausibly deterministic but very complicated

dynamic process is taken to be random for the

purposes of modeling Wing design requires force

data resulting from the interaction between fluid

and structure Such data can be shown as the time history in Figure 5.1

The challenge is to make sense of such intricate fluctuations The analyst and designer must run scalemodel tests A wing section is set up in the wind tunnel and representative aerodynamic forces aregenerated Data on wind forces and structural response are gathered and analyzed With additional dataanalysis, it is possible to estimate the force magnitudes Estimates of the mean values of these forces can becalculated, as well as of the range of possible forces With these estimates, it is possible to study the behavior

of the wing under a variety of realistic loading scenarios using the tools of probability and statistics tomodel this complex physical problem This text introduces the use of probabilistic information inmechanical systems, primarily structural and dynamic systems These tools are applicable to all the sciencesand engineering, even though this text focuses on the mechanical sciences and engineering

5.2 Single Degree of Freedom: The Response to Random Loads

Consider the second-order differential equation1governing the linear motion of an oscillator

where the input force per unit mass is given by stationary random process FðtÞ and the outputdisplacement by random process XðtÞ: Note that it is customary to use upper case letters XðtÞ to representrandom processes, and lower case letters xðtÞ to represent the realizations of a random process.The notation here is standard: 2zvn¼ c=m where c ¼ viscous damping and m ¼ mass; andv2¼ k=mwhere k ¼ stiffness

We assume that the reader understands the concepts of impulse response and convolution We wouldalso like to present a priori the results which are derived in the subsequent section; the mean value of theresponsemX and the spectral density of the output SXXðvÞ:

mX¼ Hð0ÞmF

SXXðvÞ ¼ lHðvÞl2SFFðvÞ

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffið1 2v2=v2Þ2þ ð2zv=vnÞ2p

FIGURE 5.1 Turbulent force history.

1 In this instance, the system parameters z and v n are deterministic and, therefore, so is the governing equation of motion.

If z and v n are either random variables or processes then the governing equation is random The case of random parameters

is much more complicated because the system itself is random rather than just the forcing We would have to solve the problem for the “many randomly prescribed systems” rather than for just one system with randomly prescribed forces.

By previewing the end results, the reader will hopefully be able to better follow the mathematicalmanipulations which follow.

5.2.1 Formulation

Consider the linear system defined by Equation 5.1 and assume random process input FðtÞ to bestationary, with mean mF and power spectrum SFFðvÞ: The stationarity assumption for the forcingmeans that transient dynamic behavior cannot be directly considered here.2Thus, the initial loadingtransients of an earthquake, a wind gust, or an extreme ocean wave cannot be considered as stationary.Assuming that the character of the loading does not change, steady-state behavior can be assumed to bestatistically stationary

mX¼ Hð0ÞmF ¼ v12mF;

which shows us that sincemF is time independent, then so must be E{xðtÞ}:

In order to derive the output spectral density, we must work through the intermediate results involvingthe correlation function

5.2.3 Response Correlations

For a stationary process, the autocorrelation function is given by

RXXðtÞ ¼ð1

21xðtÞxðt þtÞfXðxÞdxwhere fXðxÞ is the probability density function of the process We cannot take the Fourier transform

of this equation to find the response spectral density because we do not know the response density

2 There are, however, clever ways by which stationary solutions can be utilized in nonstationary cases One possibility is to multiply the stationary process by a deterministic time function such that the product is an evolutionary, or nonstationary, process For example, use AðtÞFðtÞ as the forcing function where AðtÞ is a deterministic transient function and FðtÞ is stationary.

3 The force is stationary and has a constant mean value.

function fXðxÞ: There are two other equivalent ways to proceed: (i) utilize the ergodic definition of theautocorrelation,4or (ii) use Equation 5.2, utilizing available information on FðtÞ; as follows.

First, derive the cross-correlation between FðtÞ and XðtÞ: Multiply both sides of Equation 5.2 byFðt 2a1Þ and take the expected values of both sides:

E{XðtÞFðt 2a1Þ} ¼ð1

21gðt1ÞE{Fðt 2t1ÞFðt 2a1Þ}dt1where E{Fðt 2t1ÞFðt 2a1Þ} ¼ RFFðt12a1Þ and E{XðtÞFðt 2a1Þ} ¼ RXFða1Þ; the cross-correlationbetween loading FðTÞ and response XðtÞ: Thus,

5.2.4 Response Spectral Density

Begin with the Fourier transform relation between power spectrum and correlation function

It is emphasized here that the derivation of Equation 5.6 made use of the convolution equation that isvalid for linear systems and structures Any generalization for nonlinear behavior requires problem-specific approaches.

Example 5.1 Oscillator Response to White Noise

Consider a simple application of the above ideas to an oscillator What is the response of a dampedoscillator to a force with white noise density?

Solution

The governing equation of motion is

€XðtÞ þ 2zvn_XðtÞ þ v2

nXðtÞ ¼ FðtÞwhere FðtÞ is the external force per unit mass, the system transfer function is given by

HðvÞ ¼ v2þ i2zv1

nv þ ðivÞ2and the squared magnitude of HðvÞ is given by

ðv22v2Þ2þ ð2zvnvÞ2Therefore, given any input spectral density SFFðvÞ; the response spectral density is

SXXðvÞ ¼ lHðvÞl2SFFðvÞ ¼ ðv22v2SÞFF2ðvÞþ ð2zv

nvÞ2Suppose, for mathematical simplicity, that the forcing is white noise, SFFðvÞ ¼ S0: Then,

E{XðtÞ2} ¼ pm2S0

m2W04kcwhere the one-sided density W0is related to the two-sided density by S0¼ pW0=4:

This integral is not standard, but can be found in texts on random vibration.5Even though infinitemean-square energy is input to the system,6it responds with finite mean-square energy.See Figure 5.2forplots of the components of Equation 5.7 Only the positive frequency ranges are plotted as they are

5 For example, the integral of this example problem is a specialized version of

ð 1 21

A 0 ¼ v 2 ; A 1 ¼ 2 zv n ; A 2 ¼ 2; B 0 ¼ 1; B 1 ¼ 0:

6 The energy input equals to the area under the spectral density, which for white noise is

ð 1

21 S 0 d v ¼ 1:

symmetric about the abscissae White noise is

useful and frequently used, even though it is

nonphysical, because it leads to good approximate

results

Example 5.2 Response to Colored

Noise

Suppose the same system as in the last

example is subjected to more complex loading,

where the spectral density of the forcing is not a

constant, but a function ofv: How would the above

analysis change?

Solution

The output spectral density becomes a more

complex function of frequency, for example, if

the loading density is similar to those found for

wind loads Then, the mean-square response must

be evaluated numerically

The applied problems are always solved

numeri-cally, although hopefully after some significant

analytical exposition There are numerical

methods that are specifically geared to handling

uncertainties Of particular note is the group

known as Monte Carlo methods These methods

utilize the massive computational power available

today to account for uncertainties This is

accomplished by generating random numbers that are used to represent random parameters For each

of these generated random values, the program recalculates the problem After running enough cycles toensure the convergence of the statistics, these numerical realizations are averaged to find the mean valueand variance of the relevant parameters

We summarize the key results for the random vibration of a single-DoF linear oscillator in Table 5.1 to

Table 5.3 Figure 5.3shows some of the most important correlation and spectral density pairs

ð1 2 v 2 = v 2 Þ 2 þ ð2 zv = v n Þ 2 q

2 6

3 7 2

m F is known from force data

TABLE 5.2 Output Correlation Function/Variance

5.3 Response to Two Random Loads

Previously, the system responses were due to a single randomly varying force In general, the situation ismore complicated, because more than one load may act on a system and the resulting response dependsnot only on the properties of each force, but also on the correlation between the two forces

TABLE 5.3 Output Spectral Density

S XX ð v Þ ¼ lHð v Þl 2 S FF ð v Þ lHð v Þl 2 from structural parameters

S FF ð v Þ is the input force power spectrum

FIGURE 5.3 Correlation functions and corresponding power spectral densities.

Consider the response of the system to two

random loadings, PðtÞ and QðtÞ; acting

simul-taneously but at different points on the system, as

shown in Figure 5.4 We are interested in

calculating the response statistics of the

displace-ment XðtÞ at an arbitrary point on the system

Assume that E{PðtÞ} ¼ 0 and E{QðtÞ} ¼ 0: Also,

by utilizing available data, we are able to estimate

RPPðtÞ and RQQðtÞ: Our interest is in evaluating

RXXðtÞ and its Fourier transform SXXðvÞ: Using linear superposition and the convolution integral, theresponse due to both forces is given by

XðtÞ ¼ð1

21 gXPðt1ÞPðt 2t1Þ þ gXQðt1ÞQðt 2t1Þ dt1Similarly, for Xðt þtÞ:

Xðt þtÞ ¼ð1

21 gXPðt2ÞPðt þt 2 t2Þ þ gXQðt2ÞQðt þt 2 t2Þ dt2where gXPis the impulse response function at coordinate X due to force P and gXQis the impulse responsefunction at X due to Q: Then,

RXXðtÞ ¼ E{XðtÞXðt þ tÞ} ¼ E ð1

21 gXPðt1ÞPðt 2t1Þ þ gXQðt1ÞQðt 2t1Þ dt1

ð1

21 gXPðt2ÞPðt þt 2 t2Þ þ gXQðt2ÞQðt þt 2 t2Þ dt2Now expand the product, and then move the expectation operator to the random processes, as follows:

The importance of this result is primarily in the observation that we cannot derive RXXðtÞ unless thecross-correlations RPQðtÞ and RQPðtÞ are also known Using the Fourier transform relation between

SPQðvÞ ¼ 2p1 ð1

21RPQðtÞe2i vtdt

As expected, the evaluation of the output spectral density requires knowledge about the cross-spectra

SPQðvÞ and SQPðvÞ: If there are more than two forces, then we will have additional cross-spectra betweeneach pair of forces

Examining Equation 5.8 closely, we find that we can write SXXðvÞ as

Example 5.3 Conjugates of Cross Spectra

It was briefly mentioned that RPQðtÞ ¼ RQPð2tÞ: How are SPQðvÞ and SQPðvÞ related?

Solution

By the definition of spectral density

SPQðvÞ ¼ 2p1 ð1

21RPQðtÞexpð2ivtÞdtReplacing RPQðtÞ with RQPð2tÞ

SPQðvÞ ¼ 2p1 ð1

21RQPð2tÞexpð2ivtÞdtLetting 2t ¼ t

SPQðvÞ ¼ 2p1 ð1

21RPQðtÞexpðivtÞdtThen,

SPQðvÞ ¼ SQPð2vÞ ¼ Sp

QPðvÞ

Example 5.4 Response Spectrum due to Two Random Loads

Consider a mass–spring–damper system inFigure 5.5subject to two random forces PðtÞ and QðtÞ: Findthe response spectrum SXXðvÞ assuming that

SPPðvÞ ¼ SP; SPQðvÞ ¼ 0; SQQðvÞ ¼ SQ

The equation of motion for this system is given by

m €X þ c _X þ kX ¼ PðtÞ þ QðtÞ

First, assume that QðtÞ ¼ 0 in order to first obtain

HXPðvÞ: Taking the Fourier transform, the equation

HXQðvÞ ¼ mð2v2þ 2v1

nziv þ v2ÞThen, the spectral density of the response is given by

SXXðvÞ ¼h mv22v22 2vnziv 21 mv22v22 2vnziv 21i SPPðvÞ SPQðvÞ

SQPðvÞ SQQðvÞ

24

35ðmðv22v2þ 2vnzivÞÞ21

ðmðv22v2þ 2vnzivÞÞ21

24

1 PðtÞ and QðtÞ arise from independent sources and are, therefore, uncorrelated.7Then, RPQðtÞ ¼ 0;

RQPðtÞ ¼ 0 and SPQðvÞ ¼ 0; SQPðvÞ ¼ 0:

2 PðtÞ and QðtÞ are directly correlated; that is, QðtÞ ¼ kPðtÞ where k is a constant

3 PðtÞ and QðtÞ are exponentially correlated, E{PðtÞQðt þtÞ} ¼ kPQexp{2at} where kPQ is aconstant

4 PðtÞ and QðtÞ are correlated in a “simplified” exponential; that is, with a triangular correlationdefined by E{PðtÞQðt þtÞ} ¼ kPQð1 2t=t1Þ; 2t1#t # t1:

X(t)

P(t) m

k

Q(t) c

FIGURE 5.5 Single-DoF system subjected to two random loads.

7 Independence implies that

E{Pðt 1 ÞQðt 2 Þ} ¼ E{Pðt 1 Þ}E{Qðt 2 Þ}

They are uncorrelated if

CovðPðt 1 ÞQðt 2 ÞÞ ¼ E{Pðt 1 ÞQðt 2 Þ} 2 E{Pðt 1 Þ}E{Qðt 2 Þ} ¼ 0 Independent processes are always uncorrelated whereas uncorrelated processes may not be independent.

We consider in more detail the first two cases listed above Where the loads are independent, becausethe cross-correlations are identically zero, the output spectral density is just the sum of the two respectivespectral densities obtained with the forces acting separately:

SXXðvÞ ¼ HXPðvÞHp

XPðvÞSPPðvÞ þ HXQðvÞHp

XQðvÞSQQðvÞ

¼ lHXPðvÞl2SPPðvÞ þ lHXQðvÞl2SQQðvÞ ð5:10Þ

We note that the output spectral density for a linear system follows a strict interpretation of the principle

of linear superposition only when the forces are uncorrelated

For the case where QðtÞ ¼ kPðtÞ, we have

RPQðtÞ ¼ E{PðtÞkPðt þ tÞ} ¼ kRPPðtÞ

RQPðtÞ ¼ E{kPðtÞPðt þ tÞ} ¼ kRPPðtÞ

RQQðtÞ ¼ E{kPðtÞkPðt þ tÞ} ¼ k2RPPðtÞThen, we can obtain the respective spectral densities

SPQðvÞ ¼ SQPðvÞ ¼ kSPPðvÞ

SQQðvÞ ¼ k2SPPðvÞleading to the spectral density of the response

will be identical to those that are correlated where

cosf ¼ 0; that is, f ¼ ^p=2: This is when the two

vectors in Figure 5.6 are perpendicular to each

other For other values off, the spectral density in

the correlated case may have any value in the

range defined by ½lHXPl2^ lHXQl2 SðvÞ depending

on the value off:

If, at some frequency, HXP¼ 2HXQ, the spectral

density at that frequency for the correlated case

with k ¼ 1 will be zero For any case where

HXP¼ HXQ, the spectral density with correlation

will be twice that obtained without correlation

Another specialized result is where QðtÞ follows

PðtÞ after a lag of t0 so that QðtÞ ¼ Pðt þt0Þ:

RPQðtÞ ¼ E{PðtÞPðt þ t0þtÞ} ¼ RPPðt0þtÞwith the respective spectral density

SQPðvÞ ¼ e2i vt 0SPPðvÞAlso, SPPðvÞ ¼ SQQðvÞ: Then,

Consider a case where there is more than one DoF

The equations of motion can be written as

½m {€xðtÞ} þ ½c {_xðtÞ} 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

primar-ily 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 thekey ideas without the complications of the major algebraic and numerical demands made by the largersystems

For the system shown in Figure 5.7, we can derive the coupled equations of motion using eitherNewton’s Second Law of motion applied to a free body diagram for each mass, as shown inFigure 5.8,or

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

_x1ðtÞ_x2ðtÞ

þ k1þ k2 2k22k2 k2

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

5.4.2 Solution by Frequency Response Function

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

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

X1ðvÞ ¼ H11ðvÞF1ðvÞ þ H12ðvÞF2ðvÞ

X2ðvÞ ¼ H21ðvÞF1ðvÞ þ H22ðvÞF2ðvÞNow, the meaning of each element is clear Each element HijðvÞ is the frequency response function forcoordinate i due to a force at j: In general, ½HðvÞ is a symmetric matrix since ½m ; ½c and ½k aresymmetric

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

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

g11ðtÞ ¼ 1

2p

ð1 21

ivc2þ k2det½Z expðivtÞdv

g21ðtÞ ¼ 2p1 ð1

21

ivc2þ k2det½Z expðivtÞdv ¼ g12ðtÞ

g22ðtÞ ¼ 1

2p

ð1 21

2m1v2þ ivðc1þ c2Þ þ ðk1þ k2Þ

Again, ½gðtÞ is a symmetric matrix

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

xiðtÞ ¼ð1

21gijðtÞFjðt 2tÞdtThe total response xiðtÞ of mass miis then equal to the sum of the individual responses to each of theforces In our case, the responses are

½gðtÞ ¼ 1

2p

ð1

21½HðvÞ expðivtÞdvand the response matrix is given by

{xðtÞ} ¼ð1

21½gðtÞ {Fðt 2 tÞ}dtNote that the impulse response vector is not trivial to evaluate, much less the convolution integrals Moreimportantly, this method becomes much harder for each additional DoF Therefore, this method is rarelyused, and we often rely on modal analysis

The goal is to find the nonzero vector {u}: Therefore, equations must be linearly dependent,8or thereshould be no inverse for ð2v2½m þ ½k Þ: This can be satisfied if the determinant of ð2v2½m þ ½k Þ iszero:

l2v2½m þ ½kl¼ 0There are several values ofv2which can make the determinant equal zero, and these values ofv2arecalled eigenvalues For each nonrepeating eigenvalue, there is a corresponding vector {u} that satisfiesEquation 5.16 This vector is called the eigenvector Physically, the eigenvalues of this problem are thenatural frequencies squared, and the eigenvectors are the mode shapes, since they represent the respectivemotion of each modal coordinate

By solving this eigenvalue problem, we can obtain N sets of eigenvalues v2

i with correspondingeigenvectors {u}ifor an N-DoF system.9It is customary that the eigenvalues are arranged in ascendingorder

v2,v2, · · · ,v2

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

{u}T

i½m {u}i¼ 1; i ¼ 1; 2; …; Nwhere

{u}i¼

u1i

2v2

i½m {u}iþ ½k {u}i¼ {0}

2v2j½m {u}jþ ½k {u}j¼ {0}

Multiplying the first equation by {u}T

j and the second equation by {u}T

i; we obtain2v2

Take the transpose of Equation 5.18,

2v2j{u}Tj½mT{u}iþ {u}Tj½kT{u}i¼ {0}

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

or

u 1 þ 2u 2 ¼ 0 2u 1 þ 4u 2 ¼ 0

In this case, u 1 and u 2 need not be zeros Instead, the equations are satisfied as long as u 1 ¼ 2u 2 : Therefore, there is an infinite number of solutions.

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

Subtracting Equation 5.19 from Equation 5.17, we obtain

{u}Tj½k {u}i¼ 0; i – jThat is, the eigenvectors are orthogonal (or normal) to each other with respect to the mass and the stiffnessmatrices If the eigenvectors are normalized with respect to the mass matrix, Equation 5.19 with i ¼ j can

and

{u}Ti½k {u}j ¼ v2

i; i ¼ j0; i – j(

If we construct a matrix composed of eigenvectors

377

5¼ 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 {€xðtÞ} þ ½c {_xðtÞ} þ ½k {xðtÞ} ¼ {FðtÞ}

We restrict our problem to the proportional damping case, where ½c is a linear combination of ½mand ½k :

½c ¼a½m þ b½kwith constanta and b:

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

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

½m ½P {€zðtÞ} þ ½c ½P {_zðtÞ} þ ½k ½P {zðtÞ} ¼ {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 bv2

i ¼ 2vizi{u}T

i{FðtÞ} ¼ qiðtÞ

to simplify the equations to

€z1ðtÞ þ 2v1z1_z1ðtÞ þv21z1ðtÞ ¼ q1ðtÞ

Once we find all the ziðtÞ; we recover the physical coordinates by using the transformation in Equation5.21 or

{xðtÞ} ¼ ½P {zðtÞ}

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

From the analysis of the Single-DoF system, we know that knowledge of the frequency response functionHðvÞ is crucial in obtaining the response statistics to random loads For a system with more than oneDoF, there is more than one frequency response function HðvÞ: For an N-DoF system, the frequencyresponse functions will be given as an N £ N symmetric matrix HðtÞ: