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C. Y. YANG
Univer.ity of Del.w.re New.rk. DeI,wlre
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A Wney.lnteraclence Publication JOHN WILEY" SONS
New York • Chlchuter • Brl.b.n. • Toronto • Slng.por.
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PREFA.CE
than the ever damping) and f~esent assumptions on structural models (mass, stiffness, and vibration an I a~l~re criteria (yield and fatigue). The product of such a random terms. 1'bi ~ YSIS IS a description of the safety of the structure in probabilistic This vot IS v~luable for structural design and planning. mind and h~~s prepared with the aforementioned engineering motivation in research ex . en developed during more than eighteen years of teaching and Delaware. ';flence in the Civil Engineering Department of the University of advanced en c:cau~ of the required backgr~)Und in structural dynamics, is'suitable pr:e~flng mathematics, and probability and statistics, the course Senior stud ardy for graduate students in civil and mechanical engineering. ents w'th with that of I good relevant background can make progress comparable individual g g.~duat~ students in small classes of ten to fifteen students, where provides a UI ance IS practical. For practicing structural engineers, the text random vi~ u~derstanding of the basic concepts and some applications of to follow t~atton of structures to enable them to tackle practical problems and placed on edresearch literature. Throughout the text, emphasis has been mathematicU~ erstan~ing the engineering significance involved rather than interesting aa /nalysls and rigor. Those who find the subject particularly retical WOrk ~ ,!seful can use this text as a springboard for either more theo n problems in tlh ~he statistical area or more complex research and engineering I' • The text' elrpro,esslon. subject of rlneludes ten chapters. Chapters 1-3 contain an introduction to the tools. Most andom vibration and the necessary mathematical and statistical of technica~~~ presentations start from basic definitions with no presumption engineering ackground beyond the senior level in civil and mechanical for roadwa;:nd are very brief. Chapter 4 introduces random excitation models most impOrt ' earthq~ake motion, and ocean waves, all centered around the and 7 prese~nt descrtptor, the power spectral density function. Chapters 5, 6, (SDOF) syste the exclt~tjon-response solutions for single-degree-of-freedom systems, res m~, multldegree-of-freedom systems (MDFS), and continuous in relatiOn t~c~lvely. Chapter 8 deals with the central issue Of structural safety response an~l YI~ld and fatigue type failures. Chapter 9 extends the excitation important ~ YSls from stationary to nonstationary cases, which are particularly gives a briO: ~tructural design for earthquake type excitations. Chapter 10 vibrations. ,e IOtroduction to the difficult problem of nonlinear random ' The text is r with Chapters ecommended for a regular two-semester academic year of study, examples and 1-3, 5, and 8 for the first semester. A sufficient number of students in the.home~ork problems are provided fo~ this material to h~lp study. ' Ir learnIng process. References are prOVIded for more extensIve It is my pIe given by PrO!i aSUre to acknowledge the superb course on Random Vibration edgeofattend~ssor Step~en S. Crandall of MIT in 1965 which I had the privil ible for my ~n~. The stImulating experience of tha,t course is partially respons e Olee of Subjects in teaching, research, and in writing this text
PREFA.CE
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on random vibration. Professor Y. K .. Lin of the University of Illinois has been a constant source of encouragement. My colleague at the University of Delaware, Professor Robert M. Stark, has provided an atmosphere of fruitful intellectual exchange of ideas in the broad area of probabilistic engineering. Professor John R. Zimmerman. also at Delaware, has been a most cordial . co-worker in teaching the course and has contributed problems and Appendix B in the text. A year of sabaticalleave u.nder the sponsorship of Professor Joseph Penzien at the University of California at Berkeley provided the much needed time for completing the first draft of the manuscript. Mrs. Betty Bramble's ever present cheerful attitude while typing the enormous amount of technical symbols and equations is also acknowledged with appreciation.
C. Y. YANG Newark, Delaware
December 1985
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CONTENTS
tJ ~ 1 INTRODUCTION
1
1.1 Random Vibration 1.2 Random Processes, Probability, and Statistics 2
1..2.1 Frequency Definition of Probability and Probability
3
Density 1.2.2 Joint Probability Density p(x I ' X2) 7
\.2.3 Conditional Probability and Independence 10 1.2.4 Statistics 12 1.2.5 Important Statistics in Practice 14
1.2.6 Rigid Body Analogy 18
Problems 20
2 STATIONARY RANDOM PROCESS,
AUTOCORRELATION, AND SPECTRAL DENSITY 2.1 2.2 2.3 2.4
3.1 3.2 3.3 3.4 3.5
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44
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Stationary Process 23 Autocorrelation Function R(f) 25
Fourier Series and Fourier Integral 29
Power Spectral Density S(ro) 36
Problems 42
3 ERGODIC PROCESSES AND TEMPORAL STATISTICS Ergodic Process 44 Temporal Autocorrelation 4l{t) 47
53 Temporal Spectral Density W(w) Alternative Definition of Temporal Spectral Density W(J) Equivalence of Two Definitions of W
Concluding Remarks 57
58 ," Problems
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STRUCTURES WITH SINGLE DEGREE OF FREEDOM (SDOF) Deterministic Transfer Relations 85 5.1.1 Frequency Domain Solution 86 5.1.2 Time Domain Solution 92 5.1.3 Time Domain versus Frequency Domain Solutions 5.2 Random ,Excitation and Response 97 5.2.1 Tiine Domain Approach 97 5.2.2 Frequency Domain Approach 104 5.2.3 A Direct Statistical Solution 101 Problems 109
7.2 7.3
7.4
85
5.1
6
RESPONSE OF LINEAR MULTIDEGREE-OF FREEDOM (MDOF) SYSTEMS Two-Degree-of-Freedom Systems (TDOF) 114
t 14 6.1.1 Deterministic Vibration 6.1 .2 Random Excitations 119 6.1.3 Response Autocorrelation 120 6.1.4 Response Spectral Density 121 6.1.5 Deterministic Damped Vibration 123 6.1.6 Damped Response Autocorrelation 126 ' '6.1.7 Damped Response Spectral Density 121 6.2 MDOF Systems 129 6.2.1 Deterministic Vibration .129 6.2.2 Stationary Random Vibration 130 6.3 An Alternative Solution Procedure 132 6.3.1 Deterministic Vibration 133 6.3.2 Complex Frequency Response H(w) 133 6.3.3 Impulse Response h(t) 134 6.3.4 Stationary Random Vibration 135 Problems 139
RESPONSE OF CONTINUOUS SYSTEMS 7.1
Random Roadway (Stationary Model) 61 4.1.1 Discrete Model 62 4.1.2 Continuous Model 65 4.2 Random Earthquake Motion (Non~ationary Model) 66 4.2.1 Discrete Model 68 4.2.2 Continuous Model 69 4.3 Random Ocean Waves (Multivariable Stationary Model) 11 4.3.1 Discrete Model 14 4.3.2 Continuous Model 76 4.3.3 Spectral Density of Wave Force 18 Problems 80
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MODELS OF RANDOM EXCITATIONS 4.1
III ill
xi
CONTENTS
CONTENTS
7.5 96
7.6
114
144
Shear Beams 144 7.1.1 Deterministic Vibration 144 7.1.2 Stationary Random Vibration 146 7.1.3 Concentrated Random Excitation 148 Flexural Beams 152 7.2.1 Deterministic Vibration 152 7.2.2 Stationary Random Vibration 154 'Thin Plates 158 7.3.1 Deterministic Vibration 158 159 7.3.2 Stationary Random Vibration Alternative Solution (Shear Beams) 161 7.4.1 Deterministic Vibration 162 163 7.4.2 Impulse Response hF(x, t) 7.4.3 Complex Frequency Response HF(x, w) . 7.4.4 Stationary Random Vibration 169 Dam-Reservoir (Vertical Excitation) 171 7.5. r Formulation 171 172 7.5.2 ImpUlse Response Function 178 7.5.3 Frequency Response Function 7.5.4 Response Power Spectral Density 182 184 7.5.5 Response Mean Square Dam-Reservoir (Horizontal Excitation) 187 7.6.1 Formulation 187 7.6.2 Complex Frequency Response 190 1.6.3 Deterministic Vibration 196 7.6.4 Random Vibmtion 196 Problems 199
168
6.1
8
DESIGN OF STRUCTURES FOR RANDOM
EXCITATIONS 8.1 8.2 8.3 8.4 8.5 8.6
9
201
Stationary Gaussian Process 201 Probability of Up-Crossing 205 Probability Density of the Peaks 209 Probability Density of the Envelope 210 213 Structural Design against Yield Failure Structural Design against Fatigue 224 224 8.6.1 Palmgren and Miner's Deterministic Hypothesis 8.6.2 Stationary Narrow-Band Random Loading 225 Problems 229
232
NONSTATIONARY RESPONSE 9.1
SDOF Systems with Stationary Excitation 9.. 1.1 Zero Damping System 234
232
.11 xii
CONTENTS
9.1.2 Lightly Damped Systems 234
Dam-Reservoir Systems with Stationary Excitation 236
9.2.1 Vertical Acceleration Excitation 236
9.2.2 Horizontal Acceleration Excitation 240
9.3 SDOF Systems with Nonstationary Excitation 241
9.3.1 Priestley's Model 241
9.3.2 Response ofSOOF Systems 243
9.3.3 Zero Damped SDOF Systems 245
9.3.4 Lightly Damped SOOF-Systems 246. 9.3.5 Bend'lt and Piersol's Model 251
9.4 Dam· Reservoir Systems with NOllstationary Excitation 253
9.4.1 Vertical Acceleration Excitation 253
9.4.2 Horizontal Acceleration Excitation 256
Problems 262
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RANDOM VIBRATION
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OF STRUCTURES
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NONLINEAR RANDOM VIBRATION
265
10.1 Derivation of the Random Walk Model' 266
10.1.1 Basic Probability Definitions 266
10.1.2 Chapman··Komogorov-Smoluchowski Equation 10.1.3 Random Walk Model 267
10.1.4 One.Step Transition Probability p 270
10.2 Applications of the Random Walk Model 271
10.3 Fokker-Planck Equation 276
10.4 Solution of the Fokker-Planck Equation 276
Problems 279
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282
Fast Fourier Transform (FFf) in Random Vibration 282
A. t Basic Concepts 282
A.2 Use of FFf Computer Subroutine 283
B. Monte Carlo Simulation 288
B.l Synthesis of a Sample Function by a Monte Carlo Simulation 288
B.2 Generating Random Numbers On a Computer 289
8.3 Simple Bubble Sort 289
C References 291
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INDEX
293
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CHAPTER
1
INTRODUCTION
1.1.
RANDOM VIBRATION
Structural engineers today are accustomed to dealing with the analysis and design of structures for dynamic loads, such as wind forces on tall buildings and bridges, strong earthquake motion effects on buildings and dams, ocean wave forces on offshore oil drilling platforms, vibration of ships in rough seas, air planes in flight and during taxiing, and vibrations of high speed trains on rails and automobiles on highways. In the dynamic analysis and design the basic difference from the static consideration is the inclusion of time dependent excitations from externally imposed dynamic effects and internal inertia loads. For ordinary 'structures and familiar loading conditions and for preliminary design purposes, these dynamic loads are often treated as pseudostalic loads in which inertia forces equal to the product of the estimated acceleration and the associated structural masses are added to the design process. For the final dynamic design of major structures, particularly with unusual structures and severe loading conditions, a pseudostatic approach is inadequate. A dynamic approach must be undertaken. In a dynamic analysis and design the first important step is to determine the dynamic load. Dynamic loads are generally divided into two categories, a transient or shock load of short duration and a steady-state load of relatively long duration. The characterization of each of these dynamic loads is accom plished completely by a definite function of time with controlling parameters such as amplitudes, frequency, period, and phase. Once the dynamic load or input excitation is defined, tben the response dynamic force, displacement, and so on, of the-structure can be determined. These dynamic responses are com bined with the static counterparts to form the basis of the complete structural design. 1
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RANDOM PROCeSSes. PROBABILITY. AND STATISTICS
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The fundamental difficulty in the structural analysis and design for dynamic loads. as briefly described here, lies in our inability to determine the design dynamic loads accurately. T.lking the case of strong earthquake motion as an example. our knowledge of ground motion is limited to acceleration measure ments such as the one illustrated in Figure 1.1. The peak acceleration of O.5g (one half of the gravitational acceleration) and a total duration of 10 sec gives an indication of the design acceleration in the high earthquake risk areas such as San Francisco. California. This acceleration record illustrates the irregularity and complexity of the time function. Furthermore, a rational dynamic design must consider the inherent uncertainty of possible future earthquakes at a structural site. To account for the large uncertainty in peak acceleration. vibration characteristics (frequency contents and phase angles), and the time duration, the classical approach where the ground acceleration is treated as a d~flnite timefunction, must be modified in a fundamental way. The uncertainties of ground acceleration must be built into a new class of indefinite or random lime fUl/ctiollS, known as random processes. For a random process, the charac teristics are no longer definite but are in statistical and probabilistic terms. When the input dynamic excitations are treated as random time functions, the struc tural responses are' naturally also random time functions. In this manner we are led to the study of random vibrations of structures, which mainly deals with the characterization of input random excitations to structures, determination of random structural responses, and assessment of structural safety under such random excitations.
acceleration records XII '(I). Xll'(l}, ... , which we call the ensemble. We display three typical samples in Figure 1.2 in the ensemble of the so-called random acceleration process X(t). These three samples illustrate the uncertainties of the acceleration process. As a convention, Capital letters are used for random processes and random variables and corresponding lower' case letters for sample functions and sample values. To give a rational characterization of the ensemble of acceleration records as a whole, we must make use of the methods and concepts of probability and statistics. We begin by looking at the values of X(t) from each of the three samples X(l)(t), X(ZI(t), and x(3)(t) at a particular time t = tl' The different sample values x(ll(tt), xl 21(tt), and x(3)(t\) are among the theoretically infinitely many in the ensemble representing the random variable X(t,) == XI' To characterize this random variable X I or any other single continuous random variable X I', we use the following probability density function p(x) together with the frequency definition of probability.
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We define the probability that (x < X < x + dx) as the fraction of samples for which (x < X < x + dx) in the ensemble of infinite samples, Thus, n P(x < X < x + dx) = lim -N
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(1.1)
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where N is the total· number of samples in the ensemble and n is the number of favorable samples, meaning those with (x < X < x + dx).
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INTRODUCTION
according to the frequency definition of probability. Random events of equal chance are usually called equally likely events. .
We further define the probability density function p(x) by the equality P(x
< X
+ dx) = p(x) dx
(1.2)
which is the area under the p(x) versus x curve between x and x + dx, as shown in Figure 1.3. . As consequences of the probability density definition, we find that
Pea < X < b) =
P(-oo < X < (0)
s: = f;",
p(x)dx
EXAMPLE 1.2. What is the probability of a 7 in a throw of a pair of dice? An examination of Figure 1.4 shows that there are a total of 36 equally likely random events for the sum of2 dice and that the number of favorable events is 6. Thus, P(the random sum X 7) 6/36 1/6. Similarly, P(the random sum
X
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From Equation (1.2) we see that once the probability density function p(x) is obtained, we have a complete knowledge about the uncertainty characteristics of the random variable X. We should point out that in the frequency definition of probability, Equation (1.1), the expression N ...... 00 reflects the concept of an ensemble of infini~ely many samples. In practice, of course, when we must work with a finite mlmber of samples and even just a few samples, we can obtain only an approximate probability. The case with infinite samples, how ever, can be achieved through a conceptual understanding of the underlying physical problem represented by the random variable, as will be illustrated in Example 1.1.
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EXAMPLE 1.1. We all know that for a fair die the probability of getting a certain number, say one, from a throw of the die is 1/6. This answer is arrived at by using the fact that the chance of each of the six numbers in a die is the same and the total probability is unity. Thus the probability is obtained through an understanding of the problem. The probability of 1/6, however, can be verified experimentally by counting the number of ones in a large number of throws and by using the frequency definition of probability, Equation (1.1). Thus we see that the concept of infinite samples is used here to draw the conclusion of equal chance for each number, and this conclusion can only be verified approximately
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2
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3 4 5 4 5 5 6 6 7 7 8
6 7
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7 7 8 8 '9 9 10 10 II
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Equ:llly likely mndolll even!"s in the sum of Iwo dice.
EXAMPLE 1.3. Throw a needle of length 2L on a table marked with parallel lines of equal spacing 2a, as shown in Figure 1.5. If it is assumed that L < a so that only one line can be covered by the needle, what is the probability that the needle will lie on a line? Let X be the random distance from the center of the needle to the nearest line and 0 the random angle that the needle makes with the lines. The condition ofcovering a line is L sin 11 > x. The range ofx is 0 ~ x .;;; a and of is 0 .;;; ~ 1£. Based on the equally likely assumption. the probability is proportional to area in the x - 0 plane, thus
e
P{covering line)
= area under arc an
This is shown in Figure 1.6.
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5
RANDOM PROCESSES. PROBABILITY. AND STATISTICS
FIG. 1.6.
Random pusiliull "f a needle.
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. III INTRODUCTION
6
1.2.
1.2.2.
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Joint Probability Density p(x l ' x 2 )
Section 1.2.1 begins with the quantitative description ofa random time function, or a random process X (I), by examining its value at a single time instant t = t I' The description is now extended to a pair of random variables X(t l ) :; X! at t = 1I and X(t l ) :; X:I at t = tl' A complete description of the pair comes from the joint probability density function P(x!, Xl), which is defined by the following equation, stating that the probability that Xl < Xl < XI + dx and Xl < X 2 < X2 + dxz simultaneously is equal to the volume p(x., X2)dxI dX2 under the P(Xt, X2) surface, Figure 1.8.
Lsin/l
a
1,
7
RANOOM PROCESSES. PROBABILITY. AND STATISTICS
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P(X, < XI <
XI
+ dXl and
X2
< Xl <
Xl
+ dX2)
= p(x t>x 2)dxl dx 1 (1.5)
Consider only the random angle 0 in Example 1.3. If we aSSUme the angle has a range of values from 0 to re with equal chance. what is the prob ability density function P(O)? What is the probability that 0 lies between 0 and 1£14? Since the total area under the density curve must be one and the density . Irve is rectangular (uniform chance) with the busc n. it follows that the height c~ 1'(0) Illusl be lIn as shown in Figure 1.7. () Having determined P(O) we can calculate the required probability by Equa tion (1.2). Thus
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4
=
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!!
Jor~ p(O) (10 = .4I
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slightly change the problem by assuming that the chance of the «(/ 0) 10 a maximum value at (/ :::: 1£ as shown in Figure 1.7b. Then,
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and
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P 0< (I <
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= p(.q. x21dxl dx2 = probability
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Relation between the density P(XI) in (a) and the joint density P(XI. Xl) in (b).
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8
INTRODUCTION
1.2.
This definition leads immediately to the following two equations:
P(a
(1.6)
c
Xl) dXI dXl = 1.0
RANDOM PROCESSES. PROBABILITY. AND STATISTICS
9
variables, which is analogous to the joint probability density p(x i , x 2) for two continuous random variables, For a pair of fair dice, each of the 36 random events represented by the plotted points on the XI - X2 plane has the same probability of 1/36. Further more, the probability that the first die has a value of one regardless of the number of the second is
(1.7)
P(X.
L
I)
P(I,x 2 )=P(I,I)+P(1,2)+"'+P(I,6}
allx;z
When the joint probability density P(XI, Xl) is known, we can derive the one-dimensional marginal densities P(XI) and P(-Xl) from it as follows: P(XI
il <
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regardless of X 2)
1:= "
p(Xt,x2)dx t dX2
J""- ." p(Xt,
Xl)
I)
1/6
which is obtained by using the discrete form of Equation (1.8). Note that the numbers of the two dice in this example are treated as two discrete random variables, whereas in Example 1.2 only a single random variable, the sum or the two random numbers, is considered.
(1.8)
dXl
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A probability is, of course, a dimens.onlas quantity. Thus the one dimensional probability density function ",xtjsimply has the dimensions of xII, whereas the joint probability density p(X!, Xl) has dimensions Of(X I X2 )-I.
• EXAMPLE 1.5. Each die ofa pair of d ice is thrown sequentially. Let the random values ofthe first and second be X I and X:1 re!lpectively.ln Figure 1.9 we can plot the joint probability mass function P(Xt, Xl) for the two discrete random
EXAMPLE 1.6. In Example 1.3 let us consider that there is an equal chance that the random center distance X has a possible magnitude between 0 and a, which' we cull a uniformly distributed continuolls mndom varinblc. Assume that the random anglc 0, however, is distributed linearly from 0 to its m
pili, xl
P(XI' %2)
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=
+ dXt
XI
=p(xddxl =
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< XI <
:. P(X.
'%2(secood
die)
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V
X2}
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%, (first
die)
for two discrete random variables X I and X 2'
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fiG. 1.10. Join! probability density function ortworandolll variables (hllld :\,
, IIiII'l
10
INTRODUCTION
1.2.
Equation (l.8),
( 4'x) = Jo
r~14
P 8<
==
samples. The probability P(XI < X t < XI for which (X2 < X.2 < Xl + dX2) is P(8) d8
i "14[i" o
0
]
Tilis is the volume of the smaller wedge with triangular shaded side in Figure 1.10. 1.2.3.
Consider two random variables X I and X 2, under the frequency definition of probability, Equation (t.1), we have XI
+ dX 1 and
X2 < Xl <
X2
+ dX2) =
lim nNI2
.'" I)d I',XI X2 XI
/.,m -IN n121N == N~
(U2)
I I IJ
',; I 1:1 i.1
H
Substituting Equations (1.9) and (1.10) into Equation (1.12) and using the definitions of density functions given by Equation (1.2) and (1.5) we obtain
J~
:~
P(Xt/X 2> ==
(1.9)
N~",
(1.11)
n2
which is known as the conditional probability of XI on X2 and is denoted by p(xtlxl)dx •• Thus we have
Conditional Probability and Independence
P(XI < XI <
+ dxtl among the selected samples
. nil II m
N~
lit- -Ia = I p(8,x)dx d8 = 242alt 16
11
RANDOM PROCESSES. PROBABILITY. AND STATISTICS
p(X., x 2 ) P( X 2)
(1.l3)
·~
W
II }
P(X2 < Xl < x 2
or
+ dX2) = lim nN1
(1.10)
N~",
p(XJ, X2)= P(XJ!X2)P(X2) = p(x2Ixtlp(xi)
where N represents the total number of samples, nIl is the number of samples falling in thc I1l1rmW uJ:ca dclincd hy x I < X I < .\" I .,. €Ix I nnd Xl < X! < Xl + dXl, and til is tbe number of samples in the narrow band X2 < Xl < X2 + dXl as shown in Figure I.tl. Sometimes we find it useful to consider the probability defined by the fraction of samples in the area in Figure 1.11 from a selected group of the total X2
which states thll! thc joint prohnhility density of two; random variables is equal to the prutim;l of the dellsity fUllctioll of olle ntlidOIll variuhlc aud its
1
In IU
COIl
ditional density on the other. The term conditional indicates clearly that the density function P(Xt/X2) depends on the occurrence frequency or chance of the variable X 2 as shown in Equation (tI2). However, in cases where XI and X 2 represent independent random events so that the occurrence of one does not influence the chance of occurrence of the other, then from Equation (1.12) and (1.13), p(xt/x2)dx1
---r~~~ dx2
(1.14)
== lim ~!. N~
N
= p(x1)dx l ;
P(XI, X2)
= P(XdP(X2)
(1.15)
which states that for two independent random variables, the joint probability density is equal to the product of the two individual densities.
X2
~I
!»"'~)o
FIG. 1.11. Number orsamples II, in Iheband X2 < X 2 < < x, + dXI and X, < X, < X2 + dX"
X,
+
dX2
XI
and nil in the areas x, < X,
EXAMPLE 1.7. A structural engineer estimates that, based on past observations, the probability of annual occurrence of strong earthquakes (with Richter magnitude greater than 4.0) at a site is peA) = 0.15. Furthermore, among the strong earthquakes that occurred in the past at the site, the probability of an extremely strong earthquake (with magnitude greater than 7.0) is estimated to be P(EIA) = 0.10. What is the probability P(Eand AI of an extremely strong
lin
~
lu
I 12
I ~
~. ;.j'l·
f:1· F·
lr
it
l iil
~~
m"
I
"nl.
~~
;···i.
m\~ ~
;
~~
.
,.I•. ; ,
5Jn~ Ilt~Jlm
10
~~
+ to + 20
;~
~~
20m = 12.5
L (value of x)(rrequency of X
""'
.~
sum of all sample values total number of samples
= 10(i) +
Up to this point, we have used a probability density functionp(x) to completely characterize the continuous random variable X and a probability mass function P(x) in the case or a discrete random variable. In most practical applications, however, it is neither easy nor necessary to have such a complete characteriza tion of the random variable. Instead, a. number of gross descriptions that we know as statistics, 'such as the mean value and the standard deviation, are sufficient. As gross descriptors, the statistics are intimately related to the probability mass and density functions. To illustrate this relationship, assume that we have a total of four samples of the discrete random variable X. We use four dots to represent these samples. Thus in Figure 1.12a we assume that there are three samples with X = 10 and one sample of X = 20. This simple illustration gives the approximate probability mass function plotted in Figure 1.l2b. It is approximate in the sense of a very small number of samples in the frequency definition of probability, Equat\on (1.1). First we find the most
E(X)
L
xIP(X
=
Xi)
x)
( 1.16)
allxJ
This statistical mean, which comes from the theoretically infinite samples, is called the ensemble mean, ensemble average, or expectation of X. Theexpecta tion of a continuous random variable is related to its probability density function p(x) in an analogous form E(X) =
f~a> xp(x) dx
(1.17)
Equation (L17) states that the expectation of X, which is an ensemble statistic of the continuous random variable, is equal to the product of X = x and the associated probability p(x) dx, integrated Over all x. Extending from this notion, we have the expectation of a function or X and of a function or two random variables X I and X 2 as
Pix) No. of samples
I
E[f(X)]
E{x}--,
•
10
20
)I
x
I I I I 0
(a)
FIG. 1.12. PIx) in (b).
( 1.18)
3/41
• • o
= f~oo f(x)p(x) dx
E[f(X 1, X 2)] =
1/4
I
10
•
x
20
f~"" f~"" I(x., X2)P(X t , X2) dXt dX2
(1.19)
The simple geometric relation between the expectation of X and its prob ability density function p(x) is illustrated in Figure 1.13.
(b)
Sample value distribution of X in (a) and its approximate probability mass function
,
.~.
Thus, in the limit when the number of samples approaches infinity, we have the statistical mean on one side, and the products of Xi and the associated probability mass function P(x;) on the other
Statistics
mi.
IJ'·FIij
= 10 +
~
EXAMPLE 1.8. Suppose a meteorologist estimates that the probability of the annual rainfall in a region, being less than to in., is P(R < to) = 0.50. What is the probability P(R m .. < to) that the maximum annual rainfall in 3 years will be less t.han 10 in., assuming that the yearly rainfall values are independent random variables. . P(R m.. < 10) P(rainfall offirst year < 10, rainfall ofthe second year < 10, and rainfall ofthe third year < 10) = [peR < 10)]3 = (0.5W = 0.125.
{/
It.! i U\),~~
arithmetic mean of X
~
IJ ~;.;. ~i. ~
13
'\. peE and A) = P(EjA)P(A) = 0.10(0.15) = O.ot5
1.01 '
RANDOM PROCESSES. PR08A8Il1TY. AND STATISTICS
familiar arithmetic mean of X as
earthquake occurring annually at the site? We write, .
1.2.4.
~
1.2.
INTRODUCT10N
EXAMPLE 1.9. Show that the expected value or the sum of two runctions or random variables X I and X2 is equal to the sum of expected values of the two
14
INTROOUCTION
RANDOM PROCESSES. PROBABIUTY, AND STATISTICS
1.2; "(x)
15
pix)
ul 111
-loX
-1
..
m
x
"
FIG. 1.13.
Expectation o( X and the associated probability density (unction
p(X),
FIG. 1.14.
'll
m,,
Mean EIX) and standard deviation (1.
"
fum.;tions. By deHnition, p(x)
E(fI(X I' X 2)
+ .f~(X I. X2)]
f, r',
[f1(X I' X 2)
l
+ .f~(X I' X l)]P(X I , x 2 )dx 1 dX2
I'
:i
= EUI(X I, X 2)]
1.2.5.
~
Narrow
..il,
+ E[f2(X I' X 2)]
Important Statistics in Practice
~
For" random vlIriable X. the important statistics are the mean E(X), the mean square 1\(:\2). the vari:lI1ce (12. and the standard deviation G. For two random variables X I and X 2, the additional statistics are the correlation E(X I, X 2), the covariance cov(X t. X 2), and the correlation coefficient p. We define these statistics next. Variance. G = E[(X - E(X))2] tions from the mean. l
= average of the squares
FIG. 1.15.
['i.,!, .
x
)IIi
,
Distributions with narrow and wide spread.
cov(X I • X 2 ) = E(X1X z) - E(X 1)E(X 2 ) E(X 1) and E(X l ) are zero.
= correlation
of the devia Correlation Coefficient.
Normalized covariance
when the means
= p = cov(X 1>
(12
= E[X2 - 2XE(X)
= .t.::( X 2)
+ (E(X))2]
[.t.::( X)]2 = mcan square minus the square of the mean
Standard Deviation. (1 = .Jvar X = a measure of the spread of the ran dom variuble about the mean as shown in Figurcs 1.14 and 1.15. Correlation.
.t.::(X I X 2)
= average across the ensemble of all products XI X 2
E(X t X 2) =
f
"" f'x, -0(. ' _ ""
XI X1P(X I ,
The correlation coefficient p is a measure of the knowledge of one random variable when another is known or simply of the statistical relation among the two variables. This relation is often displayed in a so-called scatter diagram, which is a plot of all sample values of the two random variables in Cartesian coordinates as those shown in Figures Ll6 and 1.17. EXAMPLE 1.10. If X I = are linearly correlated
± X 2, show that p = ±
cov(X I, X 2) =
X2) "XI '/:%:1
(1i
Covariance. cov(X I, X 2) = E((X I - E(X d](X 2 across the ensemble of all products of deviations.
-
E(X 2)]l
= average
1. We say that Xl and Xl
It!
IIIm ;l!
I
± E(Xi) '+ [E(X Il]l
= E(Xi) -
[E(X d]l=
_ ± E(Xn + [E(X 1)]2 p -
~
ili
X 2)/
(11(12
var X =
rJ
E(xi) - [E(X 1)]2
ui _
+
-
i
I
III
I
II
16
1.2.
INTRODUCTION
~I
~
RANDOM PROCESSES. PROBABILITY. AND STATISTICS
17
Let YI and }'z be two random variables and 0 a parameter, then
Xz
E[(OY1 - Y2)2] ;?!; 0
1II
fl
or
T
[E(nW 2
_______.l~y~~l~j;--------~ . q
-
2E(Y1 Y2 W + E(YW ~ 0
Consider this as a quadratic function of 0 and call it
~
a0 2 + bO
flU)
I
FIG. 1.16.
A scatter diagram with + 45° line corresponding to X I = + X 2 and p
~
=
+ I.
+ c~ 0
[(til
Xz
-~
~
I
, ~.
~~+--
.
'"
IIr
)
Xl (u)
{(O)
~ , ;..
~
FIG. 1.17.
Scatter diagram for two independent random va,riables.
\' ~
II
EXAMPLE
1.11. If X t and X z are independent, show that p = O. Since
P( Xt> X2) = P(Xt)p(xz) and E(X 1 X 2) =
J~oo J~.., XI X1P(XI X2)dxl dX2
,!
1;
~
.
(I
(b) ({/I)
= E(X dE(X 2)
it follows thatcov(X I, X 2) = 0 and p = O.
F .~
:;;>---L
III
iI; . ~ 1\,' t
~. ~
___~
Note that in the case of two independent variables, the correlation is equal to the product of the two mean values, wyich in general are not equal to zero. The scatter diagram is illustrated in Figure 1.17. In practical cases where one random Nariable is related somewhat to the other - 1 ~ p ~ + 1 and the scatter diagram looks somewhat between the two extremes given in Figures 1.16 and 1./7. 1.12.. Show that the correlatiob coefficient p' always lies between 1 and - 1.
EXAMPLE
+
~ ... 1/
(c)
FIG. 1.18. Three cases of the function [(0). Case (a): The function [(0) 0 has repeated roots 8 = 0,; Case (b): The function [(0) = Ohaslwodislinct roolsO = 0, and{il:Case(d:Thefunclion flO) 0 has no real rOOIS.
.1 18
INTRODUCTION
1.2.
From Figure U8a, 1.18b, and l.\ 8c we see that in order to guarantee frO) we must rule out case b. From algebra, we know that for cases a and c,
0=
- b
± ,jb" - 4ac 2a
,
~ 0,
RANDOM PROCESSES. PROBABILITY. AND STATISTICS
X I and X 2, Figure l.l9b, with its probability and statistics are given on the right for comparison.
b2 ~4ac
Total Mass
Total Probability
or
= f f dm
M
[E(Y. y2 }]2 ~ E(Yt>E(Y~)
=
f f P dXI dxz
Y.
= XI
E(X f), Y2 = X" - E(X 2)
=
X1
then
ff
11
Mean
E(Xd
XI dm/M
I'
I:
1 = f f p(x., X2) dXI dX2
Centroidal distance in XI direction
Let
19
= JJxlpdxl dX 2
!
~ ~ i
I1t E(Yi) = ai E(YI Y2 ) = E{[X.
f f pdXl dX2
E(n) =
Moment of inertia in
- E(X 1 )][X 2
COV(X,. XZ)]2
1.2.6.
Jl :.;
i ~ j
\ .;". .J
11,112
E(X,,)]}
O'tai
:. [cov(X .. X z)]2 ~ [
-
I
-1~{I~+1
~+,
_
direction
I~
'; Mean Square
13
x~p dx. dX2
11
Rigid Body Analogy
-rr-t2"' t
I I
~
,"I
~.
r
r1,
=
ff
IXt x • = f
f.'(X,)
..-l~1
Variance
(XI -
i
l )2
dm/M
0': = ff
..
xI
f
XI X2dm
[XI -
E(X d]2p dXI dXl
Correlation
E(X IX 2 ) = f f XI X2P dx l dX2
Product of inertia about centroid
w
plane.
Ij,Ji,
=f
f
(XI -
id(x2 - iz)dm
i~j
I· I .,;:j .:
I .
'X2
Radius of gyration in
XI direction about centroid
E(X1l-----l
M
. the x,
E(Xn = f f
Product of inertia
~~'~~;Jt:!J~~~late of variable tbickness. (b) Boundary of tbe two-dimensional probability .. '. - .'
= 1,;JM
pd"1 d"'2
"2
x1---J I
r;,
:,
I
xtdm
Radius of gyration (squared) in XI direction
There is a mathematical analogy between the characterization of random variables and that of rigid bodies in solid mechanics. Since structural engineers are familiar with rigid body descriptions, such an analogy should be helpful in understanding the characteristics of random variables, In the following we consider a thin plate of variable thickness: Figure 1.I9a with its mechanical properties on the left. The analogous distribution of two random variables
..
ff
lx,
X.
.,
',·i
~
COV(X"Xl)
= ff(XI-E(X 1»
x (X2 - E(X2))P dXI dX2
I: fi
I I I
I
20
I
I
II
tm.• [I
I ~ I ~.
'~' l)
.
~
;;
1'\ 'I
~l
PROBLEMS
INTRODUCTION
From this we see that the mass density p, total mass M, centroidal distance and product of inertia I::,x, are analogous to the probability density p, mean E(X t), mean square E(X f), and correlation E( X I X 2), respectively. To carry the analogy further, we need to know only the gross descriptions of a rigid body such as the total mass, the centroid. the moment of inertia, and product of inertia when we want to study the general rigid body motion. If we want to study the detailed internal force of the rigid body. however, we must kno.v its mass density function p. For random variables, the partial and gross descriptions are the statistics, such as the mean, the variance. and the covariance, which are sufficient in the majority of random vibration applica tions. There are some important applications, however, when we need to know the more complete description by the probability density function p.
Xt, radius of gyration
r;,.
~, 11 .:
IiI·'" ff1
PROBLEMS
1.1. A civil engineer has conducted a total of 605 concrete strength tests in a material testing laboratory. An examination of the test data shows that 125 tests have strength above 40 megapascal (MPa), 350 tests have strength betw~en 30 and 40 MPa; and the rest have strength below 30 MPa. Based on these test data and the frequency definition of prob ability, estimate the probabilities in each of the above strength intervals, that is, P(S> 40), P(30 < S < 40). and P(S < 30), where (S < 30) represents the event that the random COncrete strength S is less than 30 MPa. obtained either through test data (such as Problem 1.1) or through an understanding ofthe problem (such as the die in Example 1.1). To estimate the probabilities of other relevant engineering information, it is generally helpful first to establish the relation among the input random event and the other output random event or events through the so-called Venn diagram, and then to obtain the desired probability. Consider the concrete tests Problem 1.1. A Venn diagram is shown in Figure 1.20 for
tiJ~1 !\'~
------,
m·~~, j!,!{
t
~.--
' tii ;-,(
, !
Ir~
'J'~ I
I,Ill
the two random events (S <: 30) and (S > 40). It is clear from this Venn diagram that the random event (S > 30) is the so-called union (V) of the two random events (30 < S < 40) and (S > 40), Thus we may write (S > 30) = (30 < S < 40) V (S > 40) and obtain P(S > 30) = P(30 < S < 40) + peS > 40). This is known as a probability axion. Using this approach, obtain peS < 40). Referring to the data given in Problem 1.1 and the idea of the Venn diagram in Problem 1.2, we may define the random event (30 < S < 40) as the intersection «(1) of two random events (S < 40) and (S> 30), that is, (30 < S < 40)
(S < 40) (1 (S > 30)
which is different from the union (S < 40) V (S > 30) of the same two events. What is the relation between the two probabilities of the random events, P[(S < 40) (1 (S > 30)] and P[(S < 40) V (S > 30)]?
1.2. In a typical problem, the input probabilistic information is usually
I' I
21
(S
< 30)
r~-----
I I
30 MPa
I
_ _ _ _ _ _ .J
I I
40 MPa
(S > 40)
r IL. _ _ _ _ _ _
FIG. 1.20. A Venn diagram.
.
S
A useful probability function F(x) for the random variable X is defined as the cumulative frequency function that is the probability that the random variable X is less than or equal to x, . F(x) = P(X
~ x) = J:", p(x)dx
Based on this definition, find the inverse relation for p(x) expressed as a function of F(x). The velocity v of an accelerating car is equal to the initial velocity 1'0 plus the product of the acceleration a and time t, that is, v 1'0 + at. If Vo = 5 km/h, t = 2 h, and the acceleration (/ is a random vuriablc uni· formly distributed between 1 and 2 km/h 2• derive the probability density function p(v) for the random velocity ['. Consider a sine wave form X A sin 6 with a constant and nonrandom amplitude A and a random angle 6, which is uniformly distributed in the interval 0 ~ 8 ~ 2n. Determine the probability density function p(x) for the random wave X: In the design of a concrete gravity dam, the engineer is concerned with two random events that could cause failure of the dam. These are A is the random evertt that the shear force exceeds the shearing strength of the dam and the other B is the random event that the overturning moment exceeds the corresponding moment resistance capacity of the dam. If the probabilities are estimated as P(A) = 0.001 and P(B) = 0.002 and the conditional probability P(A/B) = 0.950, find the probability of failure of the dam Pc.
I!l;'IJ
~31
22 INTRODUCTION 1.8. For the dam failure Problem 1.7, consider the perfectly balanced design where failure by excessive force occurs simultaneously with that byover turning moment. In this case, derive the equation for the probability of dam failure Pc in terms of the probabilities P(A) and P(B), and the con ditional probability P(A/B). What conclusion can we draw on the value of P(A/B) and the relation between P(A) and P(B) as a direct consequence o(the balanced design assumption. 1.9. The joint probability density function p(x, x) of two uncorrelated random variables X and its time derivative X is given by
[1(X2 + 2:x2)]
1 - exp - - 2: p(x, x) = - 2
21tO"xO",;;
0""
O"x
Based on this density function, obtain the probability P(A (\ B) of the joint event A and B, that is defined by A = (x < a) and B = [x dt > (a - x)], with a positive constant a and the differential time dt.
'\.
CHAPTER
ill.
I'
mJ
STATIONARY RANDOM PROCESS, AUTOCORRELATION,
AND SPECTRAL DENSITY
II ~f
I
i
"
·'l
'1
;H! ·ll
W ~
,~
III III
Xl)
x exp( ---2 p( x-)2
a
I!J
2
1.10. For the following well-known probability density functions, obtain the ensemble meanE(x) and variance var(x~ (a) The Rayleigh distribution for random ocean wave height X,
-
~:I
2(1
2.1. STATIONARY PROCESS (b) The Poisson distribution for the random number of occurrences of strong earthquake in a design period, p(x) = a"e
4
x! (c) The Gaussian distribution for automobile vibration
1 (1 X2)
p(x) = 2~a exp - 2 ~2
A random process such as the earthquake ground acceleration shown in Figure 1.1, which has different statistics at different times, is called a nonstationary process. On the other hand, random excitations such as ocean wave oscillations and wind pressures quite often have small time variations in their statistics. As an important simplification we can treat these hi.tter random processes as stationary. A sample function of a stationary process is shown in Figure 2.1.
WI
~~! 111
;nl)
1.11. Verif:y that the correlation coefficient p for the two random variables X and X whose joint probability density function is given in Problem 1.9 is indeed zero.
FIG. 2.1. A sample function of a stati(lnary process.
23
I
D Ii
II
I I I
~..
E[x(to)y(to)] =
I fl
o
In general two harmonic functions of time will be correlated if they move in phase or anti-phase. and uncorrelated if they are in quadrature to each other. This conclusion is important in understanding the form of the autocorrelation function. to which we turn next.
(3.10)
the double integral of (2.8) reducing to a single integral since to is the only random variable. In this case we need only consider to varying from zero to 21t/co to cover a single fun cycle of the periodic motion, so that the probability
p
I~ .
(~)I
Autocorrelation The autocorrelation function for a random process x(t) is defined as the average value of the product x(t)x(t + f). The process is sampled at time t and then again at time t + f, Fig. 3.6, and the average value of the product, E[ x(t)x(t + calculated for the ensemble.
Tn,
o
!Iell
I
J:"" XoYo sin coto sin (coto + q,)P(to) dt
21t (.U
to
"('~
Fig. 3.4 Probability density function . for the random time of sampling to density function for to will be as shown in Fig. 3.4. Substituting this into (3.10) gives
E[x(to)y(t o)]
25
since Ox = xo/./i and Oy = Yo/J2 The two sine waves may therefore be said to be perfectly correlated when their phase difference is 0 or 1800 but uncorrelated when their phase difference is 900 or 270°, Fig. 3.5.
Suppose that tbese two waves are sampled at an arbitrary time to. and we calculate the average value of the product x(to)y(to). According to (2.8) the average value is given by
I I
Autocorrelation
t.,orrelaUan
co) Jo (2"'''' sin coto sin (coto + q,) dto = xoYo (21t co) Jo r'' 'lJ) {sin 2 cotocosq, + sincotocoscotosinq,} dt o = XoYo (21t (3.11)
= tXoYo cos q,
'. ('~ . t\~/A /'
\
I
/
II
"JV
I
\/1
V\,JI
V
/
VJrV
\ ,1
and the correlation coefficient is therefore
E[xy] = cos q,
Pxy
ihM&A
n
''i\{''
(3.12)
= (f,lly
I
X1(O+
I
I
I /
HI
/ -
- -
I
HI.
/
I ~
I ]TIl
/t'
/
/
/
/
Phase lag
/
¢
I
~
T
.. /
Fig. 3.6 Calculation of autocorrelation
J
Provided that the process is stationary, the value of E[x(t)x(t + r)] will be independent of absolute time t and will depend only on the time separation T
Fig. 3.5 Correlation coefficient for two sine waves with pbase difference
\
.
26
Correlation
Autocorrelation
so that we may put E[x(t)x(t
+
r)]
= !(r) =
Rx(r)
(3.13)
(say)
E[x(t)]
= E[x(t +
r)]
____ ~~~~~:.+.!'l~
I:
=m
The correlation coefficient for x(t) and x(t given by p=
E[{x(t) - m}{x(t
+ r)
u2 E[x(t)x(t
= Rx(t) U
Hence R,,(t) that
I
o UX(I+<)
+ t)]
m
U.
+
Fig. 3.7 Illustrating properties of the autocorrelation function R,,kr) of a stationary random process x(t)
- m}]
.
+ m2
Example
(3.14)
2
= u 2 p + m2 and, since the limiting values of pare ± l,t it follows I
+ m2
~ Rx(r) ~ u 2
+ m 2.
Calculate the autocorrelation function for an ergodic random process x(t) each of whose sample functions, Fig. 3.8(a), is a square wave of amplitude a and period T, but whose phase (that is the time of first switching after t = 0) is a random variable uniformly distributed between and T.
o
= E[X(t)2] = E[x1 ]
(3.16)
and is just equal to the mean square value for the process. At very large time intervals, t -+ 00, a random process will be uncorrelated since there will not be a coherent relationship between the two values x(t) and x(t + r) and so the correlation coefficient p -+ O. In this case, from (3.14),
RAt
-+
2
(0) -+ m
(3.17)
•
Finally, since, for a stationary process, Rx(t) depends only on the separation time t and not on absolute time t, Rx(t)
= E[x(t)x(t + t)] = E[x(t)x(t
- t)] = Rx( -t)
so that Rx(t).is an even function of t.
t For proof of this statement see, for instance, Meyer [48], p.. 131.
(3.18)
T
1,2'12'1,2'.1.
II
o __~~~~~Lm__
(bl
T (II)
P"l=L
OTto
II -L~~~I
to
II
IJ
I
1 1
:!
T
..,-----r-+--.-r----t
_u 2 + m 2 •
When the time interval t separating the two measuring points is zero then
T
x(t)
(3.15)
The value of the autocorrelation function can therefore never be greater than the mean square value E[x 2 ] = u 2 + m2 and it can never be less than
Rx(t = 0)
I
-o2+m 2
2
_U
T
r), defined by (3.8), is therefore
- mE[x(t + t)] - mE[x(t)] u2
I
~ I;
Rx (T)
and UX(I)
27
All these properties are illustrated in a typical graph of an autocorrelation function Rx(t) against separation time t shown in Fig. 3.7.
where Rx(r) is the autocorrelation function for x(t). We can deduce at once some of the properties of Rx(r). Firstly, if x(t) is stationary, the mean and standard deviation will be independent of t, so that
""",.
to +T
Fig. 3.8 Square wave sampled at arbitrary time to and at to+t to calculate its autocorrelation function There are two ways of making this calculation. The first is to calculate. an ensemble average, by looking across the ensemble of sample functions at two fixed times to and to + t (say) - see problem 3.2 The second way makes use of the fact that -the process is ergodic, when, as we have seen, any 'one sample function is completely representative of the process as a whole. This method is to average along a single sample function, by thinking of the measuring time to as a random variable uniformly
Il ~
II
I I I
I
I
~
i
,
i
Il~,
1m·
Ifig:
28
STATIONARY RANOOM PROCESS
2.3.
x(t)
.~
",< Il
/,'«i'////~»71
8=0 JI
",
fl
II
'1 I I I I I ·1 lI
J;
~
.
~
~~~
FOURIER SERIES AND FOURIER INTEGRAL
29
the autocorrelation function. We are working basically on the random function of time. An alternate approach in the evaluation of the two-time statistic is to shift our attention from the time domain to the frequency domain. This comes as no surprise, since we know in deterministic vibration analysis the comple mentary nature of the dual approaches. We begin this alternate approach by introducing the concepts of Fourier series and Fourier integral.
2.3.
FOURIER SERIES AND FOURIER INTEGRAL
Fourier Series... If f(t) is a periodic function of period T, it is at least piece wise continuous and the integral TI1.
f
_ Til
prO)
If(t)1 dt
exists, then f(t) has a series expansion that takes the following form of identity 00
f(t)
o-t FIG. 2.7. Three samples of the process XVI
}
= sin(wl
= L If=-QI)
(2.6)
1>8
+ (II. and the prohahilitydcnsilY functioll
1(10).
The Fourier coefficients en can be evaluated by multiplying both sides of this equation with cxp( - inwJot) and integrating ov~r a period. Thus 1'i2 f-1'12 f(t)e-im'U.f dt = L
1
1
= 2 cos OJ T
n=
,
-rt)
Cnei(n-ml"'ot dt
\:
7"
'\
>'
;)II
T
= cos 0 + i sin 0
(2.7)
this becomes .
TIZ
f
[cos(n - mlwot
+ i sin(n -
mlwot] dt = 0
-Til
for m FIG. 2.8.
1= n and equal to en T for m = n
The autocorrelation R~(,) for the periodic process XCt) = sin(lllt + 0).
:. en is not realistic because R(O) is the mean square value of the process. Conse quently, it must be finite. Nonetheless, such an idealization is useful in practice if proper restrictions are observed as we shall discuss in Example 5.12, Section 5.2.2. So far our attention has been focused on one time statistics such as the en semble mean, mean square, and variance and then on the two-time statistic,
I
f1'12
T
f(t)e-inWj)t dt
(2.8)
-1'/2
The Fourier coefficients Cn are generally complex numbers. When f(t) is a real and even function, C. are real. For a real function j(t), the integrand of Equation (2.8) for Cn is the complex conjugate of that for C _ft. That is, Cn = C!.n
I
.~.
II . _ ~ l·~
I;\;
.)~
1'/2
Using the identity that eiB
\\~ I
.~~
HZ
w
Rx(T)
21t
wO=T
CneR"'ot;
(2.9)
30
2.3. FOURIER SERIES AND FOURIER INTEGRAL
STATIONARV RANDOM PROCESS
Parseval's Formula for Periodic Functions. The time average of
ell
!
the square of the funclion f(t) or simply the temporal mean square value of
f(t)
= -1 fT'2 p(t) dt = -1 fTI2 ( Lao T
,= -1
T
=
1
T
T
-T12
fT I2 [ L 00
.=
-1'/2
fT/2
(
Clfeln"",' "=-00
-Tl2
(C"el."""
)2 dt
o
00
)
FIG. 2.10.
= n=~'" IC,,1 2 C~ + n~1 21C l2 ""
""
Then the Fourier series expansion is f(t)
2.2. Find the Fourier series expansion of the function f(t) given by Figure 2.9. The Fourier coefficients according to Equation (2.8)
(1'/,/
4
A cos nWol dt +
\
0
i
1 12 '
= Cli""" + C.e-i""" + C3 ei3""" + C 3 e- i3""" + ... =
n;
for odd n
o
The Fourier coefficients CII are all real and are displayed in Figure 2.10 along the discrete axis n, which is associated with the nth frequency component exp(inwot) in the series, Equation (2.6).
f(t)
~:
..
b -T12
FIG. 2.9.
-T14
. ) 2A( + I.sm wot + - cos wot n
A real, even, and periodic funclion !(t) of period T.
IW
~~
)"
,. \
I " fl
+ ...
3n;
Af-.
~
TI2
.. sm Wot)
I
+ (- 2A) - (2 cos 3wot) + ...
d
TI4
-
Thus the square toothed periodic function f(t), as shown in Figure 2.9, is decomposed into an infinite series of harmonic functions with decreasing amplitudes; 4A/n, 4A/3n, and so on, associated with increasing frequencies; wo, 3wo • and so on. It is instructive to compare the original function with approximations of one frequency (fundamental) component and two frequency components, respectively. These are shown in Figures 2.1l and 2.12. It is apparent that a good approximation can be obtained for this case with just two frequency components.
}.
A
2A( cos wot n
2A'(2 cos Wot) = -
2A. A cos nU}ot dt ) = -sm tilt -,
nn 2
T/4
2A (-1)1" 1J/2 = nn { for even n
Fourier coefficienls C•.
n
EXAMPLE
2 T
n
I
which is the sum of the squares of the absolute Fourier coefficients.
-
•
3
2
-2AI31T
(2.10)
Cn
2A1",
+ C:e-··.. , ) + Co J2 dt
"~I 2C"C: + q dt
-T/2
31
0
TI4 \,
I 1/ 3TI4
1 TI2
/
...... /
FIG. 2.11.
Approximation of !(/I by the fundamental frequency component (4A/x) cos wol.
I I I I I I I I I I I
I
I
ill ~~I
32
STATIONARY RANDOM PROCESS
2.3. FOURIER SERIES AND FOURIER INTEGRAL
33
'I
~. I II
I I I I I I I I I
f(t)
(ttl
"
~
A '-""
,...
II
!!
\
n
0
TI4
1
\
TI2
13TI4
-
~ FIG. 2.14. A nonperiodic function
Itt) plotted in the shaded area.
EXAMPLE 2.3, Determine the frequency distribution of the temporal mean square of the periodic function f(t) shown in Figure 2.9. Based on Equation (2.10) and the result of Example 2.2, the temporal mean square
TI fTI2 _T/2
f2(t) dl = 2
(4A2)
f(t)
'" =L
Cne1nWo ';
Wo =
2n = clrcu . Iar frequency T
-'"
and
(4A2) + 2 (4A2) 9n2 + 2 25n 2 +.;.
7
can be considered as the limit of a periodic function of period T as T approaches infinity. As a periodic function, from the Fourier series expansion, we can write
Cn = -I JT/2 f(t)e-InWo'dt
T -T/2
<
Thus. the temporal mean square can be considered as distributed to the fundamental frequency component (n = ± 1) with the portion 8A2/n2, to the third frequency component (n = ± 3) with 8A2/9n 2, and to the fifth frequency component (11= ± 5) with 8A2/25:n 2• This distribution is displayed in Figure 2.13. The sum of all contributions is, of course. A 2.
Multiplying both sides of the last equation by T, denoting Wo by liw, nwo by m, and then increasing T to approach infinity yields ;
Fourier Integral. A non periodic function f(t) such as the one with the shaded area in Figure 2.14, satisfying the absolutely integrable condition
f:",
If(t)1 dt <
(2.1 I)
00
lim TC n = lim
T-w
l'-t(}
f
T/2 f(t)e-iro'dt
".2
f"" -
~~~
t
f'"
lru (C. T)e '
(~)(;n)
I _ "" F(w)eiCO ' dw = 2n
•
•
-3
-2
•
-I
o
1
2
>-
Il
3
FIG. 2.13. Distribution of the temporal mean square of J(t) over discrete frequency c;:omponenls nwo·
f(t)e-iru'dt = F(w) (2.12)
t'IJ
and
f(t) = A2
=
- Ttl
(2.13)
This pair of Equations (2.12) and (2.13) are known as the Fourier transform (or integral) pair for a nonperiodic function f(t). Equation (2.13) gives the frequency decomposition of f(t) in a continuous frequency domain, which is analogous to that ofa periodic function in a discrete frequency domain. Equation (2.12) transforms the nonperiodic function f(t) of time into a function F(w) of frequency w, whereas Equation (2.13) transforms the function F(w) from the frequency domain back to the time domain.
~. r
I
I I f
t"
34
STATIONARY RANDOM PROCESS
2.3.
2.4. Determine and display the Fourier transform F(w) of the non periodic function f(t) shown in Figure 2.15 with duration T •. Since f(t) is absolutely integrable, real and even, its Fourier transform F(w) is also real and even.
EXAMPLE
F(w)
=
J'"
2
f(t)e-iUJ'dt
iT'"
, f'
1-",
T.ll cos wt dt == 2A sin wt fTIIl
= 2A f o
w
0
=
2A . wT, w
SIn-
2
In Figure 2.16, we see that F(w) = A Tj as w --> 0 by using L'HospitaJ's rule. As the time duration TI increases, the waves in F(w) tend to squeeze towards the zero frequency origin. Thus, a widely spread time function has a narrowly concentrated Fourier transform near the origin and vice versa.
, if
! \
?f r·
Parseval's Formula for' Nonperiodic Function. From the previous case of a periodic "function of period T the temporal mean square given by Equation (2.10) is 1 T
-
JTll -Til
f~(t) dt
L'" IC.1 2 = -T1 L
(T)
n=-Q)
•
n=
'"
TCnC: . 2n
-TlI2
= f""_
~2" dw
nT
'" f
IF(wW
_'"
A
1.
F
Ii
TI12
IF(w)1 2
~
2nT
t
f;
FIG. 2.15. A nonperiodic and absolutely integrable funclion f(I).
ft;
dm
A2Tj
-
T
where
K
1 4A 2 2n1'
-
.
2
Sin
Figure 2.17 shows the distribution of the total temporal mean square (A 2 T1IT total shaded area) along the frequency axis. The ordinate at any /J)
r
nOll sin
IF("12
ATI
21fT
~\ 'i
A2TlI21fT
~ ro
41f
p FIG. 2.16.. Fourier Iransform F(w) of Ihe funclion fit) shown in Figure 2.15.
~
(2.14)
EXAMPW 2.5. Determine the temporal mean square of the nonperiodic func tion f(t) shown in Figure 2.15 in terms of the. fictitious and infinitely large period T. Display the frequency distribution of the temporal mean square. The temporal mean square based on the Parseval's formula, Equation (2.14) is
Wo
Ii
f
.[2(t)dt
where the integrand IF(wW /2nT represents the contribution to the temporal mean square per unit frequency by continuous frequency components with center frequency w. Since the period T --> 00, the integrand [\F(wW/2xT] --> O. We note, how ever, that in Equation (2.14) the left-hand side also contains the infinitcsmal term liT, so that ParsevaJ's formula has finite integrands 12(r) and IF(w)l l /ln on both sides. When we apply the formula to the temporal mean square of a nonperiodic function, we must keep in mind that the limits of the time integral go to ± ct.). Consequently, only approximations ean be determined by using large but finite '/:
(ttl
!
I 12 ' -1"fl
dt
0
-00
35
As the period T --> 00, nwo --> w, Wo --> dw, Cn T --> F(w), C:T --> F*(w), so that the temporal mean square for a nonperiodic function can be expressed by !im -T'
Ae- i""
FOURIER SERIES AND FOURIER INTEGRAL
f
TJ
FIG. 2.1'1
I I I I
I I I I I I I Ii.! I
Tl
Frequencydistribulion of the lotal temporal mean square A'T,/T.
I
I!
,
I
I
I I I I I I I I I I I I
"'I "'~
~ r~ ,}
j,
~
b
;;) I'?"' 1-" B',
•
36
2.4.
STATIONARY RANDOM PROCESS
f,2
POWER SPECTRAL DENSITY S( w)
In random vibration problems, we find that very often an alternate form of the seconlt-order statistic, the so-called power spectral density S(w), is more useful than the autocorrelation function RCr). Let the Fourier transform of R(t)j21t ofa stationary random process X(t) be defined as the power spectral density Sew)
J'"
Sew) = 1 21t -
R(t}e-iru'dr
r
r;;.
=
J:
J:
=
x(t)e-i.., dt
x(t)e- i"" dt
1: 1: dt
0
x(s)e+i..• ds
dse-i..(I-S)R(t - s)
Plot the region of integration in the t versus s plane and then map the region on the t versus r plane with t defined as t - s. (See Figure 2.18.) By mapping the four corner points OABC from the t-s plane to the (-r plane the double integral can be carried out in the t-t plane first covering the upper triangle OBC and the lower triangle OAB as follows:
(2.15)
Then
~(t) = f~", S(w)e ... dw
I
E(e 2 )
00
i
=
.
E(e 2) = (216)
=
We define R(t)/21t and Sew) as a Fourier transform pair, which is valid since R(t)j21t is absolutely integrable as long as X(t) cont'4lins no purely periodic components and its mean value is zero or filtered out. Because the power spectral density function is generally regarded as the most important descriptor in random vibration, it is therefore essf;!ntiai for us to understand its properties and engineering significance. These are sum marized as follows:
Jofl' [ J.1' e-iwtR(t)dt] dt+ fO [ Jorl' + -T
t
1:
(T- t)e-i""R(t)dt
+
{:T
(T
t
e-i'U'RCr)dt ]
+ r)e-iMR(t)dr
Dividing both sides by 2nT and combining the two integrals gives E(e ) = ~ 21tT 21t 2
1. By definition,
JT -T
R(t)e- i ..t
(T -T Itl)
dt
s R(O)
= E(X2)
=
J~", S(w)dw
(2.17) Y
so that Sew) represents the distribution of the ensemble mean square in the frequency domain. When X(t) represents random current in a electric system, then the mean square value represents the power of the system with a unit electric resistance. Thus the term power spectral density is adopted. 2. S(w) is a real and even function of w because R(t) is a even and real function of t. 3.. There is no physical meaning for negative frequency, although S( - w) is defined together with its symmetric counterpart S( + w) to represent the ensemble mean square per unit frequency at the frequency + w. The function S(w) so defined is known as two sided, meaning the two sides of the origin w = O. 4. S(w);;';' O. As it represents the ensemble mean square per unit frequency, it must be greater or equal to zero for all frequencies. That this is indeed so
37
can be shown as follows: Consider
gives one haIr of the temporal mean square per unit frequency at that frequency w. Negative frequency has no physical meaning. It is used here simply for mathe matical convenience.
2.4.
POWER SPECTRAL DENSITY 5(w)
,ltP
'----.,Ie
:a B
~. r..,
T
=T B
~
FIG. 2.18. ~I:'
<.
Boundaries ofintegrations,
tlr
., 38
STATIONARY RANDOM PROCESS
Taking the limit as T ..... Figure 2.19 gives
00
2.4.
sew}
and observing the behavior or(T - ItllT) from
POWER SPECTRAL DENSITY S((j))
S(wl
S(.,l
2 r1'~"! £(e 1 f'" ) imt 21tT = 211: _'" R(t)e- dt = Sew) ~ 0
.. w
A typical spectral density function S(w) for a stationary random process is shown in Figure 2.20, which covers both sides of the origin w = 0, is non negative, and is symmetrical about the origin. The fact that the spectral density exists under the definition of Equations (2.15) and (2.16) and has no singularity indicates that the stationary random process has a zero ensemble mean value and has no purely periodic component. ' We see from Figure 2.20 that the spectral density is distributed smoothly over the frequency range from zero up to a certain maximum value W2' Within this frequency rangc, the ordinate h~ls maximum values somewhere near ± WI' When the spectral density is distributed over a narrow range of frequencies such as the one shown in Figure 2.2Ia, the associated sample functions of the random process appear to be similar to a simple harmonic time function. For a more irregular time fUllction, as shown in Figure 2.21/1, the corresponding spectral density has a wide range of frequencies. These two classes of processes are generally called narrow-band and wide-band processes respectively. T
.r
.........
FIG. 2.19.
)I
T
l'
Limiling behavior or Ihe rllnclion (1'
"it)
x(t)
x(l)
«t~
(t.)
(d
FIG. 2.21. Three classes er spectral densities corresponding to three classes of stationary random processes: (al a narrow-band process. (b) a wide-band process, and (e) an idealized white noise process.
An idealized spectral density with a uniform ordinate over the total fre quency range is shown in Figure 2.2lc along with the associated illustrative sample function X(t). This idealized spectrum is called a white noise spectrum because, as in the case of white light, it covers the entire frequency range approxi mately uniformly. The case is idealized because it corresponds to a stationary random process of unHmited mean square value, according to Equation (2.17). This idealized case, however, is important since it leads to analytieal simplicity in dealing with random vibration of structures and the idealized results are useful approximations under carefully observed limitations. This will be dis cussed later in the response of structures, Example 5.12 in Chapter 5.
T-x
-T
ItlllT as l' -cpo
•
s(w)
", -"'2
FIG. 2.20.
'-j.<'" -WI
...... "'I
"'2
A typical spectral density runction S(w}.
39
~
m
EXAMPLE 2.6. Show that the autocorrelation function R(t) corresponding to the idealized white noise ppectrum with constant magnitude So is 2nSob{t) where b(r) is the Dirac delta function. ' Although the autocorrelation function 2nSob(t) can be obtained by directly applying tne integration fprmula, Equat,ion (2.16), with S(wl = So, it is far simpler to approach the problem in a reversed manner by showing that the spectral density corresponding to R(t) = 2n:Sob(t) is So. Thus, from Equation (2.15),
S«(O)
= So
f:",
O(t)e-i""dt = So
I I I I
I
I I I I I
I I I
I
il I
~
l~, ; I
" I
,
:~
11
'
il
~.
:1 'I
II
II
I
I I I I I I I ~
I
fEil! '
40
2.4.
STATIONARY RANDOM PROCESS
according to the definition of the Dirac delta function. We can visualize this integral by observing that, the exponential function of t or any other function 4>('r) of t, when multiplied by oCt) gives a zero product except at the origin t = O. Thus, the relevant part of the integral covers only the narrow range of t near the origin, at which' 4>(0) is a constant and can be taken outside of the integra1. Consequently, the integral siITfpty becomes So4>(O) since the integral of the Direac delta function is unity by definition. The idealized white spectrum So and the corresponding autocorrelation function 21tSoo(t) are plotted in Figure t.22. EXAMPLE 2.7. A more realistic spectrum is shown in Figure 223a, which has a uniform magnitude So over the band oftrequency between WI and W2' For this so-called band-limited white spectrum, find the corresponding autocorrelation function. From Equation (2.16), since S(w) is real and even in w,
i
•
R(0) = I1m , .... 0
2S0(W2 cos wlt 1
•
WI
t
(sin W2t
-
'
1. The derivative of RJr) with respect to
tit 2.
a; =
HIT)
',vi J
II
o
==
t
at
t
= 0 is
(2.18)
R~(O) = 0
RAO) =
'"
f~
(2.19)
S(
3. The variance of the time derivative of X(t) with zero mean is
2lTSoli(T)
(I'
O'f = RAO) = (a)
)
The variance of X(t) with zero mean is
!
So
WI
Additional Properties of R(t) ~md Sew). For a stationary process X(t), additional properties of its autocorrelation function Rx(t) and spectral density Sew) are given in the following. The subscript x is added here for RAt) to clarify different derivatives. We use the prime sign to denote the derivative with respect to the argument of the function and use the dot exclusively for time derivation.
sin Wit)
S(wi
= 2S 0'(Wl
cos Wit)
This autocorrelation function is plotted in Figure 2.23b.
dRx(t)
cos wt dw =
41
which has a maximum value at t = 0 equal to the ensemble mean square of the process,
"'l
R(r) = 2S o
POWER SPECTRAL DENSITY S(ILI}
- R;(O)
L~ w l S ( w ) d w ' L 2 0 ,
4. The correlation between X(t) and X(t)
(b)
FIG. 2.22. A white spectrum So and the corresponding autocorrelation function 2n:S o.5(f) are plotted in (a) and (b), respectively.
E[X(t)X(t)]
dX(t)/dt is zero. That is.
(2.21)
0
These results are derived as follows: For Equation (2.18) we have S(..)
2SO("'2 -
SQ
-''''2
w,
-"'I
= B[X(t)X(t + t)] = £[X(t - t)X(t)) R~(t) = £[X(t)X'(r + t) = - £[X'(t t)X(t)] = - £[X'(t)X(t + r)] R~(O) = E[X(t)X'(t)] = - E[X'(t)X(t)] = 0 Rx(t)
R(f)
~
"'2
w
..A 1\"
'"
Wj)
""IV.... .
7
:.'
For Equation (2.20) we consider the second derivative of R,,( t) with respect to t, la)
fiG. 2.23.
(b)
A band-limited spectrum and Ihe corresponding autocorrelation function.
R~(t)
=-
£[X'(t)X'(t
:. R;(O)
=
Rx(O)
+ t)] = -
R.(t)
= E[X"(t -
r)X(t)]
PROBLEMS
42 STATIONARY RANDOM PROCESS
43
versus frequency w. For a complex F(w) the modulus and phase are defined by
Since by definition, R,Jr)
= J~a> S(w)e- i"'. d~
F(w)
then
= IF(w)le-i!(
f(t) = exp ( -
R~(t) J~", :.
R~(O)
2
-w S(w)e-
i ""
at)
u(t)
dw
J~", W2S(W) dw = -
R..(O)
Finally, from the definition for R.,(t) we also have the result of Equation (2.19). FIG. 2.25. The runction fit) "" exp( -al)U(I). where Ut,) is the unit step runction and a is a positive constant.
PROBLEMS
2.5. Verify that the Fourier transform of the time shifted function I(t - to)
2.1. Show that for a stationary random process the autocorrelation function R(t) is bounded by the inequality m1 - (11 ~ R(t) ~ m1 + 0'1, where m and (12 are the mean and variance, respectively.
is the phase shifted function F(w)e- itou, where F(w) is the Fourier trans form of I(t).
2.6. Show that the power spectral density function S«(I) is a real and even function of w.
2.2. Determine the correlation coefficient II for (a) the idealized white noise process of Example 2.6 and (b) the more realistic band-limited rrocess of Example 2.7.
x(1) + y(t) where the stationary random process xCI) = A sin(wt t + 6) witl'l constant amplitude A, constant frequency WI, and random phase angle (} is distributed uniformly in the interval (0, 2lt). The random process y(1) is stationary with a constant power spectral
density So in the iqterval (- Wo, + wo). For the case where Wo ~ WI>
obtain and sketch (a) the power spectral density S.(w) and (b) the auto
correlation functiod R.(t) for the random process z(t). Assume that the
processes x(t) and Y\t) are independent and with zero mean.
2.7. Let zIt) =
2.3. For the periodic function f(t) with period 2n shown in Figure 2.24, obtain its Fourier series exrunsion and riot the urpmximations corre sponding to one, two, and three terms of the series, respectively. Also plot the temporal mean square of I(t) versus the number n of the 11th frequency component. (t)
2.8. Obtain the ensemble mean and mean square of the random processes x(t) = A cos(wot + 0) and y(t) = te- al COs(Wlt +
the intervals (0, 1t) and (0, 2lt), respectively. Are these proce:;scs stationary?
2.9. Obtain and sketch the power spectral density functions S.(w) for the stationary process z(t) XI(t) + X2(t) where XIV) = AI COs(Wlt + ( 1) FIG. 2.24.
and xl(t) = Al COs(W2t + ( 2), Assume that AI' AI, WI> and W2( = 2wI) are constants and 6 1 and 6 2 are two different and independent random variables with identical probability density functions, that is, pWS> = P(02) = 1/2lt for 0 ~ 0 1 ~ 2lt and 0 ~ 01 ~ 2lt and P(01) = P(01) == 0 outside this interval. What kind of complications will be involved if 8 1 and O2 are not independent ']
A periodic runction I!,j·
2.4. For the non periodic function I(t) shown in Figure 2.25, obtain its Fourier transform F(w) and plot the modulus IF(w)l and phase ¢(w)
I I I
I
I I I I I I I' w
I
.'';: l>~
I~ ,j ••
II
~ I
l~
!,
I
i~.d .~
3.1.
11
!I 5~ 'i
11
CHAPTER
ERGODIC PROCESS
45
.. (t~3)
3
xltl)!3)
'J
I
I
I 1
I
I
I
I I I I I I r
\,
I·,
1,
,l.\ll
:1 i; i I~
ERGODIC PROCESSES AND TEMPORAL STATISTICS
FIG. 3.1. Thfl.'C samples X(lj'", X(lr 2l• and X(I)UI of a stationary random process X(I) muollg inlinilcly 1Il;IIlY samples or Ihe ensemble.
3.1.
Up to this point we have studied the probabtlistic and statistical properties of a random process X{t) based on an infinitely many imaginary sample functions, which we call the ensemble as shown in Figure 3.1. Thus, at a particular lime t., the random variable X(t.) = X I has a probability density function p(xd, which can be determined by examining each sample value at t, in the ensemble, using the frequency definition of probability. In Figure 3.1, three samples x(t)(I), x(t){2), and x(t){3) are shown with corresponding values x(td{l), x(t J )(2 1, and X(td{3) of the random variable X(t.) at the particular time t,. By examining all the values, x(t ,)(1), x(t d(21, ••• in the ensemble we can theoretically determine the probability density function p(xd. Similarly,.we can determine the ensemble mean value E[X(tt» and the mean square E[X 2(t.)] of the random variable X(tt) from the same infinitely many sample values in the ensemble. The same concept applies to the determination of the joint probability density function P(Xt, X2) and the autocorrelation function E[X(t .)X(t 2)] for two random variables X(tt) and X(t 2 ) of the random process X(t). In practice, however, we can never obtain an infinitely large number of samples. Usually, we can get hold of several sample functions and sometimes we have to work with a single sample. Thus, the practical problem confronting us is to approximately estimate the probabilistic-and statistical properties of the random process X(t) from the available single sample function. Fortunately for certain stationary processes, the statistics extracted from a single but long sample function of time, which we call temporal statistics, approximate the ensemble statistics quite well. Such a property of the stationary processes are
,
\l:m
44
called ergodic. Thus, for ergodic stationary ,processes, the temporal mean value (X(t» and the temporal correlation Cl>(t) defined by
ERGODIC p,ROCESS ,
I
(X(t» == lim T T ... ",
fT /2 x(t) dt -T12
and cfI(t)
= (X(t)X(t + tn =
lim T-oo
I
T
fTI2 -T12
x(t)x(t
(3.1 )
+ t)dt
(3.2)
respectively, have the property that from a sample function of infinite record length in time T,
(X(t» cfI(r) .j'
=
E[X(t»
= R(t)
(3.3) .(3.4)
where the symbol (.) is used for temporal mean value. From Equation (3.3) we see that an ergodic process has a constant mean value and therefore must be stationary. A stationary process, however, does not necessarily possess the ergodic property, meaning that the temporal statistics of one sample may be significantly ditTerent from that of another sample. A graphical illustration of an ergodic process is given in Figure 3.2 from which one can visualize the equality of the theoretical ensemble statistics and the practical temporal statistics. The former is based on' sample values across the ensemble axis whereas the latter is based on the single sample function along the time axis.
46
3.2.
ERGODIC PROCESSES AND TEMPORAL STATISTICS
TEMPORAL AUTOCORRELATION fIl(t)
47
The ensemble mean
f.uscl'loln ilKIS
E[X(t)) = E[A sin(wt
+ 8)]
[211
== A Jo sin(wt + 8)P(8) dO = 0
(3.7)
The ensemble mean square
E[X 2(t)]
= A2
f2. .
Jo sin 2(wt + 0)P(8)dO =
A2 (3.8)
The temporal mean Time axis
I (X (I)) •= Iim1'~", T
IT A sin(u)t + 0) dt = 0
(3.9)
0
•
The temporal mean squaGe fIG. 3.2. An ergodic process X(t) showing a single sample X(t)(I, as a function of time and an imag inary curve connecting all sample values at I I, across the ensemble.
3.1 Consider the simple but fundamentally important stationary process modeled by the sine wave
EXAMPLE
X(t)
= A sin(wt + 0),
oo
0,
(3.6)
otherwise
I ~im 'T-
1"'00
i'l' ' 0
A2 sin 2 (wl
+ O)dt =
A2
(3.10)
Thus the process satisfies the necessary conditions for being stationary and' ergodic. \
3.2.
o ~ 1I ~ 2lt
t
(35)
where the amplitude A and frequency ware deterministic, but the phase angle (J is assumed to be a random variable. The probability density function for 0 is distributed uniformly between 0 and 211: as shown in Figure 3.3. That is, p(O) = {1/2lt,
(X 2(rI)
TEMPORAL AUTOCORRELATION
When only one sample runction of a presumed stationary and ergodic process is available with a sufficiently long record length in time T, the ensemble autocorrelation function R(r) can be determined by the temporal autocorrela tion. X(t)
$( r) =
lim
~ C" X(t)X(1
T~"" T Jo
+ r) dl
(3.11 )
p((ll
oI
I
I'
1., ~
FIG. 3.3.
'
I 2~
Probability density function P{O).
ad
The temporal autocorrelation
I I I
I I I I I I I I I I
I
I
~A
:H
jll
"
:1
48
I I I I I I I tl
1 $(0) = lim -T T-
~:
2
(3.12)
0
which is the temporal mean square of x(t). When the sample function is purely periodic. as the one shown in Figure 3.4a and given by
+ 0),
-oo
(3.13)
T-oo
=
~ T
JofT A sin(wt + 8}A sin(wt + ('ot + 8) dt (3.14)
2
~2
11:
+ (A2 2" sin W!
)(2
~
)
(3.161
cos 2mt
i
2A K/2 . 2A E[X(t)] = sin(wr + 8) dO = - (cos Wt non
+ sin wt)
(3.171
is also different from the temporal mean
J!
;r cos Wt
2
+ O)A sin(wt + Wt + U)- dO
which is different from the temporal autocorrelation given by Equation (3.14). Thus, the process X(t) is no longer ergodic. It isn't even stationary anymore. Note also that for this case, the ensemble mean
Al -COSW!
(X(t» = lim -1 T-a;, T
iT
A sin(wt
.
+ 0) dt = 0
(3.18)
0
(3.15)
The ensemble mean square from Equation (3.16)
• X(t) =- A sin(wt + 0) with
Consequently, the stationary process a uniformly distributed phase angle f) satisfies the necessary condition for being an ergodic • process. It should be noted that the ergodicity"Of this process X(t) depends on the probability distribution of the random phase angle f).lf 8 has a different distribu
X(I)
= A sin (,.t + OJ:
H= 0
A2
e\l(T)
= "2COSWT
m
1
fj"
t~
o
2"
$(t) = R(r) =
li,
!
i
A sin(wt
= Al cos wt
which is also periodic with the same frequency as that of x(t) but with the amplitude of A 2/2. In fact, this temporal autocorrelation $(,,) is identical to the ensemble aut9correlation R(t) for a purely periodic stationary process X(t) with a unifotmly distributed phase angle 0 as given by Example 2.1 in Chapter 2 and is shown in Figure 3.4. Thus, We have
Ir
'I
In
=
the temporal autocorrelation $(t) = lim
49
+ t)]
R(t) = E[X(t)X(t nl2
x(t) = A sin(wt
II
,Ii
iT x (t) dt
TEMPORAL AUTOCORRELATION ¢>( 1:)
tion other than the, one used here and displayed in Figure 3.3, the result may be different. To illustrate this point, let us consider the case where 0 has a uniform distribution in the range 0 ~ 0 ~ n/2 instead of the previous range 0 ~ 0 ~ 2n. In such a case, P(8) = 2/n for 0 ~ 0 ~ n/2 and zero outside the range. The ensemble autocorrelation
When t is set equal to zero,
:;
n
3.2.
ERGODIC PROCESSES AND TEMPORAL STATISTICS
FIG. 3.4.
A purely periodic sample x(11 and its temporal autocOffdation function
R(O) = E[X(t)2] =
A2
"'2
(3.19)
and the temporal mean square
(X2(t)}
=
I lim T
T-a;,
iT A2 sin (wt + 0) de = -A2 2
0
(3.20)
2
however, are identical. To see the effect of the random phase angle 8 on the ergodicity of the process more clearly, we consider the limiting case of a deterministic phase angle 0 where all samples in the ensemble are identical. Its ensemble mean and mean square are then clearly time dependent or nonstationary and therefore the process cannot be ergodic. Let us now consider a typical sample of a stationary ergodic process such as the one displayed in Figure 3.2. We focus our attention lirst on a finite segment
50
ERGODIC PROCESSES AND TEMPORAL STATISTICS
3.2.
of the sample that lies in the interval 0 ~ I ~ 1: For this segment we can expand it into a Fourier series. Using an alternate form of the series with real co efficients a. and b. defined by
TEMPORAL AUTOCORRELATION 41(t)
Note that this temporaf mean square is identical to that given by the Parseval's formula, Equation (2.10), when T -+ 00, since
21C"l2 = a;
aZ
1 C e=...E.
ail = Co
(3.21)
C.
(3.22)
L"'. ( a. cos 27tnt + h. sm. -27tnt) T
(3.23)
2
tea. -
ib,,) =
we have
ao
x(t) = ·2-
+
n= I
whcre
an
-2 T
i1' x(t) cos 27tnt dt
2
h"
(3.24)
T Jo
fl. 'j'
x(t)
sm
and
ao ~ A" SIO . (27tnt T + ,,1;:1 T + 0")
2
+ '\'." [ a
.1;:,"
x(t)
An =
lit
cos 27t1~1 + T
t' + I, .
hll(1
+
t)]
Sill .._ ..........._.
•
T
The temporal autocorrelation fun<..1.ion for the process can be obtained by considering that the length of this segment increases towards infinity. That is, as T -+ 00
1 <1>(,) = I,im -1'
iT
1-'",
a2
=; +
I· I :) I ·i
!~j
11 1\
11
II I;.' ~
.I
.Ja; + h:
0" = tan II()
(3.28)
where the amplitude A. and phase angle Oft are given by
and =
2
(3.25)
dt
" + r)
+ b;
Equation (3.27) for the &emporal mean square of the process X(t), which is constructed from the sample function x(t) of infinite length, reveals the funda mental relationship betweetn the temporal mean square value and its frequency components. It states tha, the total mean temporal square is equal to the superposition of the mean squares of its frequency' components. Thus for the zero-frequency component with constant magnitude ao/2 the contribution to the total mean square is a~/4. Note that the zero frequency component is also the constant tern ral mean aal2. For the first frequency component with amplitude ai + bi, the mean square contribution is therefore (ai + bi)/2, since the sample function has the third alternate form of Fourier series ex pansion
x(t) =
,2nllt
2
(/11
~
0
fl'
51
x(t)x(t
0
+ r) dt
L t(a~ + b;) cos 2/tnt «>
from which we obtain the temporal mean square 1,2
+
co
L i(ll; + 11;)
.=
I
~
bIt
(3.30)
This interpretation of the temporal mean square leads directly to the intro duction of the temporal spectral density function W(f) or W(co). We shall treat this important topic systematically in Section 3.3.
. Before leaving the topic temporal autocorrelation function, we wish to
point out th~t for an ergodic stationary process, its temporal autocorrelation function
Chapter 2. Since a stationary process need not be ergodic, prop.erties of
from an ergodic process do not necessarily apply to R(r). An important illustra
tion is given in Example 3.2.
(3.26)
"= I
(x 2(t) = <1>(0) ;, '4'~
-1
(3.29)
(3.27)
EXAMPLE 3.2. Consider the idealized case where the stationary but not ergodic process X(t) consists of constants with random magnitudes as shown by two sample functions in Figure 3.5. (a) For this process, determine its en
semble auto.correlation function R(t). (b) Take one sample function x(t) = a
constant c for - 00 ~ t ~ 00 and assume that this function x(t) is a sample
~ Ii
I')
;
:'"..
.~
~ I.;;~j
!W;
I
r
1I
I
"I
I
I
II
;~ 1~
52
ERGODIC PROCESSES AND TEMPORAL STATISTICS
3.3. TEMPORAL SPECTRAL DENSITY W(w)
3.3.
53
TEMPORAL SPECTRAL DENSITY W(w)
One approach to introduce the temporal spectral density function W(w) is analogous to that for the ensemble spectral density Sew). By this approach we define W(w) as the Fourier transform of
':,
:,:'
W(w)
= -1
211
f""
,
(3.3\)
-OC!
and the inverse transform
f~", W(W)ei"" dw
Two samples of a stationary (but not ergodic) process
(3.32)
X(I),
When r = 0
I I I
I I.
]I
~ I
B!~" I it ,i '
!~
!
"
I
;
\
function of an ergodic stationary process X.(r). For this X .(1), determine its autocorrelation function R.(t). (c) Compare the ensemble mean square E(X2) with the square of ensemble mean [E(X)]~ in both cases (a) and (b).
!
1ft
=
f~
W(w)
~/w
(3.33)
I
(a) R(,) = E[X(t)X(t
+ ,)]
E[Xl(t)] = ensemble mean square (b)
R.{t}
= ~(t) =
.: x(t) = x(t
:. R.(,)
=e
2
1 .lim -T T-f'lJ
IT x(t)x(t + t} dt (J
+ ,) = c = R.(O) = ensemble mean square
(c) In case (a) the ensemble mean square E[X2(t)] is different from {E[X(t)]}2, the square of ensemble mean. This may be seen by con sidering the two sample functions in Figure 3.5. For E[X2(t)] we have (e 2 + b2)/2, whereas for {E[X(t)]}2 we have [(e + b)/2]1.
,
so that the temporal spectral density W(w) represents the frequency distribution of the temporal mean square (x 2 (t». The temporal spectral density W(w) defined indirectly through the temporal autocorrelation
In case (b), using the ergodic property, we have
E[Xl(t)] = R,(O)
and
•
f:..,
= el
{E[X(t)]P = (~(tW = c
2
;
~i
2
Thus the ensemble mean square is equal to the square of the ensemble mean in the ergodic case (b). The underlying idea here is that in the ergodic case the single sample function represents the ensemble, whiCh is not true in case (a).
(x(t)}2a(w) dw = (x(tW
Note that for a zero frequency component with constant magnitude of Ao/2, its temporal mean square is A5/4, which is identical to the square of its temporal mean (x(tj)2. That is, lim -1 T
T-",
iT(A-.l!.2 )2 dt = ( lim -T1 IT --..!! A2 de)2 = A2 ~ 4 0
T-oo
0
54
ERGODIC PROCESSES AND TEMPORAL STATISTICS
3.4.
Similarly, a purely periodic component of frequency Wo and amplitude x(t) has a temporal mean square of Al/2 and hence corresponds to two Dirac delta spike spectra of A2~(W - wo)/4 at the frequencies ± Wo in W(w~ This is shown in Figure 3.6. The temporal spectral density W(w) for an ergodic stationary process, being identical to the ensemble spectral density Sew) of the process, has the same properties as those of sew) given in Section 2.4 of Chapter 2. In addition. we wish to point out that since both W(w) and ~(t) are real and even functions of their respective argument, it follows that
AlTERNA~VE DEFINtTlqN OF TEMPORAL SPECTRAL DENSITY WIt)
55
Substituting W(w) from Equation (3.36) into Equation (3.35) gives
A in
1
W(w) = ;
roo
Jo
(3.34)
«lI(t) cos Wt dt
and inversely, <1>( t)
=
21,'" W(w) cos Wt dw
(3.35)
Moreover, the one-sided temporal spectral density W(j) is often used as a function of the frequency J = w/2n in cycles per second instead of the two sided spectral density W(w). The terms one-sided and two-sided refer to one or two sides of the origin for the frequency coordinate. The relation between Wen and W(w) can be established by consideration of the temporal mean squares contributed by components with frequencies between f and J + df lind between m and (I) + dm, respectively. Thus W(ndJ
= 2W(w)dw =
1<10 W(n cos 2nJ-c dJ
(t) =
(3.38)
Equation (3.37) and (3.38) give the relation between W(f) and «lI(t).
3.4. AlTERNAnVE DEFINITION OF TEMPORAL SPECTRAL DENSITY W(f) So far we have studied the spectral density functions Sew), W(w), and W(n, all through the Fourier transform of the corresponding autocorrelation functions R(t)/2n and ~(t)/2n. An alternative approach to introduce the temporal spectral density W(f) is through the Fourier transform of the sample function x(t) directly. We assume that the sample x(t) has zero mean and no purely periodic com ponent. In taking the Fourier transform of x(t) we adopt the strategy of first taking a finite segment of x(t), which lies in 0 ::::; t ::::; 1; treating the sample as a periodic function of period 1; and then taking the limit as the length of the segment T --+ 00, As a periodic function again we can expand it into,a Fourier series as x(t)
=:
4nW(w)dJ
Lon
(
.=1
met + b. sm. -met)
a. cos -T-
I
I
I
I,
I ,I
I
~i
;t;
:~
I
(3.39)
T
where
Hence,
(3.36)
W(n = 41tW(w)
a. =
Substituting W(w) from Equation (3.34) into Equation (3.36) gives W(n = 4
f'
(i'(r)
(3.37)
cos 21t/t dt
W("')
,.\'(t) >211( •• )
-(~O
b.
2
P' ,
n1tt x(t) cos T dl
(3.40)
2
fT x(t) sm. nnt dt
(3.41)
T Jo
=f Jo
+'·0
mw _ "'0)14
.. '"
FIG. 3.6. Temporal spectrum W(w) for a nonzero mean cOI:nponent (x(t)) and for a periodiC component of A sin (wot + 0) contained in x(t). '
x
(~)
=
f
(a. - ibn)
T[2 = 2 T =
rJoT
x(l)
2nnt cos T dt
JofT x(t) exp [i21tn~ - -;:r]dt
2
(T
iT Jo
.
2nnt
x(t) sm T
dt
I
I
1'
Now let us combine the Fourier coefficients to form , ,.\2
1,1'"
I~
]
(3.42)
I
II
I
I ~I
56
"
W(fl lim
:"~ <,
T~co
;1
;~ ~,~
"{
I I I I I I\ I
I I
1.1
11ftA, I III
,~
.i.
X(!!..) T -_T 2' (a.
+(~)I'
x(~)=f(a.-ibo)
?I
~
x( )
12
=
ibn),
n=foO
T
We see that when T -+ 00, liT = /0 -+ df, nlo = niT -+ I and thus in Equa tion (3.42), we have x(nIT) -+ x(f} which is tbe Fourier transform of the sample (unction x(t). The physical meaning of W(f), so defined, ~an be seen from Equation (3.42) , starting before the limiting process, since
~
57
and that its temporal mean is zero.
We then define the temporal spectral density
rn ".,
:11
CONCLUDING REMARKS
ERGODIC PROCESSES AND TEMPORAL STATISTICS
(n)
21 12 T1 = 2 1 (a; + b;), TXT
n=foO
'Substituting this into Equation (3.26) with 00 = 0, we have $(r) = I~m T
L -T21 x (n)12( cos -2nn<)(I) . T T T
co 0=1
a;.; b; T
then
-+
00,
I
T -+ df,
/I
T
ndl
-+
I
b
.~.
\
. W(f) = lim W(n/o) .fo~f
=
lim
T~",
(A 2/2)
Thus the temporal spectral density WU) is shown to be the limit as T -+ 00 of the temporal mean square A;/2 of the nth frequency component per unit frequency. Finally, we show in Section 3.5 that the two definitions of W(/) are equivalent as they should be.
3.5.
the definition of W(f) given by Equation (3.45) and changing the sum into integration in this equation, we then have
AnI u
i
$(r) 1'" WU) cos 2n/r d/ agrees completely with Equation (3.38) where WU) was defined not from the Fourier transform x(t) but from the Fourier transform of
EQUIVALENCE OF TWO DEFINITIONS OF W(f)
In this section we show that the temporal spectral density function W(f)", defined directly through the Fourier transform of x(t) by the equation W{f) =
{J!?!' 2IX~W 0,
1?t0
/<0
In Section 3.4. satisfies Equation (3.38). Thus, it is the same spectral function W(f), which is defined through the Fourier transform or the temporal autocorrelation function in Section 3.3. Let us again first consider a finite segment of x(t) lying in 0 ~ t ~ T expand this segment of x(t) into a Fourier series with coefficients a. and Note that we consider that there is no purely periodic component in the sample
$(r)
DING REMARKS this chapter we have presented the solution for the practical problem where ensemble statistics cannot be determined because it is impossible to obtain many sample functions. The solution is leaning on tlte assumption of t'Rodicity, so that a single sample function can be used instead of infinitely It is important to make sure that the single sample runction is indeed It'eSentative of the ensemble as illustrated in Figure 3.2. Furthermore, we point out that throughout the chapter we have always assumed that the or.this representative sample runction is infinite. This, of course, is still to realize in practice. Therefore, we must keep in mind that when we on a sample function of finite length, the results are necessarily approxi to the corresponding limiting case presented herein.
.
'\
-;.'"""
~.),
58
ERGODIC PROCESSES AND TEMPORAL STATISTICS
. Another comment at this time has something to do with various forms of the Fourier series and of the spectral density functions 8(w), W(w), and W(f). Among the three forms of the Fourier series, the complex form is simple in mathematical manipulations, the separate sine and cosine form is the elementary form, which is most familiar to beginners. and the combined trigonometrical form. ill terms of amplitude and phase angle, is most explicit in physical inter pretation. Among the three forms of the spectral density, the two-sided spectra S(w) and W(w) as functions of the circular frequency ware most convenient in analytical work. The one-sided temporal spectrum W(f) as a function of frequency f, with f = w/2'1t, is the natural form for the measurement and analysis of real data. Obviously, when we mix various forms, we generate a great deal of confusion. To minimize such confusion we shall from now on concentrate on the complex form ofthe Fourier series and Fourier integral and on the two-sided ensemble spectrum 8(w). The brief introduction in this chapter on other forms should provide the reader with a better understanding and a necessary link for con version.
PROBLEMS
59
in which the only random variable is the phase angle O. All others are nonrandom. (a) Determine the ensemble mean E[X(t)] and mean square E[X2(t)],
when 0 is uniformly distributed in the interval (0, 2'1t).
(b) Determine the ensemble mean E[X(t)] when 0 is uniformly distribu
ted in the interval (0, 'It/4).
(c) If a sample of X(t) is generated by using a randomly selected phase
angle 0 = 'It, determine the temporal mean (X(t)} and mean square
(X2(t)) based on this sample.
3.4. Use the information given in the Appendix A for Fast Fourier Transform (FFT) in structural dynamics. Compute the Fourier transform F(w) of I(t) given in Example 2.4 numerically by a computer library routine.
Plot F(w) versus w. Compare F(w) with the theoretical transform F(w)
in Example 2.4. Use F(w) so obtained to compute the inversion !(t).
Plot i(t) versus t and compare it with the original J(t). Use A = I,
11 = I for simplicity. . 3.5. Numerically obtain the inverse transform hIt) where, = wnt from the complex frequency response function
PROBLEMS
3.1. Based on the Fourier series expansion of a periodic function of period 1', as given by Equation (3.23). (a) Derive the alternate trigonometric form of Equation (3.28) and the complex form x(t)
= !;, en exp (in'ltt) T «J
(b) Derive the associate equation in the complex form for (n'
3.2. A nonrandom forcing function I(t) is given by
lUI
=
F cos(wol/2)
+ F cos (1)ot + F cos(3wol/2)
(a) Determine the temporal statistics (f(t)} and (f2(t)}. (b) Plot the one-sided, discrete amplitude IXn versus frequency Wn diagram for f(t), the so-called discrete amplitude spectrum. (c) Plot the two-sided, continuous, temporal mean square density W(w) versus frequency w diagram for I(t), the so-called temporal spectral density.
3.3. Consider a random process model X(t)
= IX COS(wf
0)
H(W)= Wn
I
(II(J),,) • (w/(J)n)l + 2ie(w/wnl'
e = O.ot, (J)n = ~
Plot h(,) versus r and compare it with the theoretical inversion. See Equation (5.49) and (5.51). Use the FFT computer information given in the Appendix A.
I I I I I I I I I I I I I I
I
I
il I l'jl
4.1.
" 1
1 ll'
Ii"
CHAPTER
111
~1~ij
Ij 11
I ;1
ii ~i
I I I
4
MODELS OF RANDOM EXCITATIONS
:~
I II
I I I I I I I
1
\
D.
61
model by superposition. From the final model we can generate sample functions . of the process. Three models of random excitations to structural systems are presented. The first is a roadway model, which is the simplest stationary random process X(t) of a single time variable t. It is commonly used for vibration studies of aulo mobiles, high speed trains, ships, and airplanes. The second is an earthquake ground motion mofiel. which is a nonstationary random process X(t), again of a single time variable t. Such a model is currently introduced into the earthquake •engineering field for tall buildings. bridges, high dams, and power plants. we present an ocean wave model, which is a stationary random process the rough ocean surface ,,(x, y) of two space coordinates x and y. Such an wave model is the basis for analysis and design of offshort structures such as oil drilling platforms.
RANDOM ROADWAY (STATIONARY MODEL)
.;
i
RANDOM ROADWAY (STATIONARY MODEL)
In the previou~ three chapters we have studied random excitations by either assuming that we have a large number of sample functions of a random process l or at least that we have a single representative sample function of sufficient' length or time duration. When we have a large number of samples, either real or; imagined, we proceed to analyze these samples, which we call the ensemble,. and determine the ensemble statistics such as the ensemble mean, mean square, autocorrelation function, spectral density function, and probability didrlhn. lions. When we have only a single sample function, we assume that the is ergodic and proceed to estimate its temporal statistics as approximations the ensemble counterparts. In summary, the problem is to analyze and mine the random characteristics from a set of given sample functions. In this chapter on random models we treat the same problem but in a manner. Starting with a simple but fundamental mathematical model deterministic and probabilistic characteristics are completely specified, we construct different models by superposition of the simple fundamental with various deterministic parameters and random variables. Since we pletely specify the characteristics of the basic building block and the manner in which these blocks are superimposed to form the final can therefore determine the statistical and probabilistic characteristics of final model. As in the previous chapters, the characterization of the final models is focused on· the autocorrelation and the spectral density runction~ because these are the most useful in practice. Once the final random model constructed, we can use it to generate the so-called artificial sample functions. In summary, in this chapter we begin with basic mathematical models completely specified probabilistic characteristics, and then construct a
60
first consider an idealized vehicle model of mass Ill, spring constaut k, aud damoing coefficient c moving with constant speed V ,along a rough road with elevation x(z) with z = Vt, as shown in Figure 4.1. The equation of motion for the vehicle model in terms of vehicle displace U, velocity and acceleration is given as
u,
u
- c(u - x) - k(u
my + cy + ky = x(z)
= x(Vt),
x) = mii
(4.1)
- mX
(4.2)
the equation of motion (4.2) falls into the typical form
Ii
t '..
FIG. 4.1.
• z == Vt
An idealized vehicle model moving on a random roadway.
.U
62
4.1.
MODELS OF RANDOM EXCITATIONS
uf u lineur SDOF (single-degrce-of-frecdom) system. The spring force k(u - x) = ky and the vibration of the vehicle u = x + y can be easily deter mined once the response y is solved for the given forcing function X(I). Let us consider the function X(I), which represents the rough roadway prolile x(z) with z = VI, as a stationary random process. For convenience let this stationary random process be denoted by XC!). We proceed to construct a random model for X(I) beginning with a simple periodic cosine function oftime with amplitude IX, circular frequency w, and phase angle O. Thus cos(wt - 0)
x(1) = IX
N
E[X(t)]
IX.E[cos(ru.t - 6.)] = 0
(4.8)
since E(cos
On)
=
I"
(2~) dO" = 0
(cos 0.)
(4.9)
(4.3)
The ensemble mean square. E [ .~I a. cos(w.t - 8n ) m~1 a", COs(Wml l i N
E[x2(1)]
0.)
=
an integral representation
Om)
J
I
N
(4.4)
11=1
L - a; .=1 2
(4.10)
since 0. and 0.. are assumed to be independent random variables and
f'",
X(I) =
a(ru) cos[rul - O(w)] dru
(4.5)
E[a.
cos(ru"t - B.lam cos(w",t - Om)]
= IX"a.. E[cos(runt -
and linally a complex integral representation x(t)
=
f:",
= IX;E[cos
a(ru)e i (o,,-6(o.1I dru
. (4.6)
where X(I) in Equation (4.6) takes a general complex form with the practically relevant real part given by Equation (4.5). Up to this point the functions X(I) are all deterministic. We now assume that the phase angles lI" in Equation (4.4) and O(w) in Equation (4.6) are inde pendent random variubles each uniformly distributed in the range from 0 to 2n;. Under this assumption, the functions in Equations (4.4) and (4.6) become the discrete and complex continuous form, respectively, of a random process X(I).
.
0,
·0 ~ 0" ~ 2n:
otherwise
(4.7)
.
«(J)"t - 0.)]
=
al 2~
(2X
Jo
cos
2
m:/: n
'i
«(J).t - (}.. ) dO.
m =n
EXAMPLE 4.lpetermine the temporal mean (x(t)} and temporal mean square (x 2(t» from a s!lmplc function X(I) given by the random process X(t) of Equation (4.4) with a particular sct of values 0 1 , O2 , •••• 0•.
Discrete Model
p(O,,) = { 21n;'
0... )] = 0,
O.)]E[cos(ru",t
Equutions (4.9) and (4.10) show that the process X(I) deHncd by Equation (4.4) satislles the necessary stationary condition.
lim -1 T
T-a>
First consider the discrete model, Equation (4.4), of the random process X(t). We can determine the ensemble mean E[X(t)] and mean square E[Xl(I)], using the probability density functions for O. with n 1,2, ... , N,
2
10:;,
=
(x(t) =
4.1.1.
= L
,,=1
N
L a. cos(ru.t -
63
Thus the ensemble mean
where IX, (t), and 0 are all deterministic and real valued quantities. With this simple function, we proceed to consider a finite sum of N discrete functions, x(1) =
RANDOM ROADWAY (STATIONARY MODEL)
=
I
N
.=1
IT[ L a" cos(ru.£ N
0
I
IX. lim -T T-..,
IT cos(ru.t)
IlT[NL
(X 2(t» = lim -. T-a> T
0
.
0.)] dt
n=\
0
.=1
IX.
On) dt
cos(ru.t 0.)
=0
(4.11)
L am cos(ru.. t- 0... ) N
]
dt
",=1
N
L !et; n=l
(4.12)
I I I I
I I I I I I I I I I
I 64
I ; ~ I ".J
~
MODELS OF RANDOM EXCITATIONS
1j
il I
II
. -T 1 hm T"'""
iT IXn cos(ront -
On)ctm cos(romt - 8m) dt
.
I
;~
1\ ,
I
\
'I I ~ ~ Ill!!
=:
0,
m =1= n
0
= IX; lim -T T-oo
Jo(T COsl(wnt -
8n ) dt
= llX;,
m =n
From Equations (4.8), (4.1 0), (4.11), and (4.12), we see that the discrete random model defined by Equation (4.4) also satisfies the ergodicity condition, which requires that the ensemble mean and mean square be equal to the corresponding temporal mean and mean square.
Furthermore, based on the concept of the temporal two-sided spectral
density W(w) defined by Equation (3.33) and on the spectral density shown in
Figure 3.6 for purely periodic components we have
N
(x 2(t»
>
65
sew)
, I
I
RANDOM ROADWAY (STATIONARY MODEL)
since
1
~I l
4.1.
L 2W(w.)8ro n=1
with I1ro = Wn+ 1 - WO' Depending· on the choice of I1w, the frequency band width, we can asSUll;1e that W(w n ) is the meae square contribution from the frequency component with ron divided by the frequency bandwidth, or the mean square per frequency bandwidth. Thus we consider W(Wir) to be distributed in the entire intervall1w centered at ron and not concentrated at the single frequency " ron' In so doing, we avoid the Delta spike function and obtain instead a smoothed ' or average spectral density. Equating Equations (4.12) and (4.13) and repla~ing W(w.) by the analytically preferable ensemble spectral density S(ro..) gives IXn = ,J4S(w.) 110:)
Substituting IX. in Equation (4.14) into Equation (4.4) we find x(t) =
f .j4S(ron) 8ro cos(ro.t - 8.)
w WI
FIG. 4.2.
"'2
'''7
Subdivided ensemble spel..1ral density function S(w) ror N
= 7.
: For instance. in Figure 4.2 we determine rol' ro2, ... , ro7; S(ro l ), S(Wl), ... , . and 8ro = WI' Then we select, completely at random, a phase angle 0 1 the interval between 0 and h. Such a completely random choice corre to a uniform distribution of the random phase angle. Wecan make such a for example, from a roulette wheel marked with angles from 0 to 2n. Let call the outcome of the roulette wheel from the tirst turn 81t from the second and so on, up to 8 7 , Finally. substituting the set ofdata WI> W2,' •• , W7; S(rod, .... S(ro7); 8ro; and 01 , O2 , ... ,07 so obtained into Equation (4.15), we the desired sample function x(t) as a superposition of seven periodic
Continuous Model the complex continuous model, Equation (4.6), results similar to the discrete Equation (4.4), can be obtained as follows: ensemble mean
E[X(t)) =
f~", lX(ro)e+ 1""E[e- i8
(W
I]
dw
0
(4.16)
:.
no=1
which is the basis for generating the so-called artificial sample functions from a given ensemble spectral density function S(ron) of a stationary ergodic process.
Artificial Sample Generation. On the basis of a discrete series Equation (4.4), we can generate artificial sample functiQCls of the underlying random process X(t) if we have an ensemble spectral density function S(ro) or the process. To generate a sample, we simply divide the positive frequency axis of the function S(w) into N equal intervals and label the frequencies rot , rol, roN: with corresponding S(wd, S(W2),' .. , S(roN) as shown in Figure 4.2.
E[e- l(1(wl] = E[cos O(ro)] - iE[sin O(w)]
= J:w (cos 0)
(;n) dO - J:« (sin (1)(;n) dO = 0 i
ensemble autocorrelation function R(-r), when X(t) is considered com when we are interested only in its real part, can be expressed by R(f) = E[X(t)* X(t
+ T)]
(4.17)
66
MODELS OF RANDOM EXCITATIONS
4.2.
where X(t)* is the complex conjugate of X(t). Substituting X(t)* and X(t according to Equation (4.6) for X(t), into Equation (4.17) gives R(T) = E
=
=
{f~"" o:(w)e
f:", f'",
i
(8{... )-w'l
dw
f~", o:(w')e [w·{.+.,-6(... i
·)J
+ f).
dW'}
RANDOM EARTH.QUAKE MOTION (NONSTATIONARY MODEL)
67
(4.22)
I I I I,
(4.23)
I I
in Figure 4.3. in which all the mass is assumed to be concentrated on the top girder. The girder is assumed to be perfectly rigid with horizontal displacement x(t). The two columns of the model are assumed to be elastic but massless. The horizontal ground displacement, velocity, and acceleration are denoted by u(t), u(t), and ii(t). respectively. The equation of motion is
E{oc(w)oc(w')ei[8(UJ)-BlUJ'lI dw dw'}ei'I..·- ...,+i...•• k(x - u) = mx
- c(x - u)
r~ (oc(w) dw]2e iw ,
(4.18)
because as the integration with respect to w' is carried out. the expectation E{eiI8(OI-O(w'll} = 0 for w' 1: wand equals unity when w' = w. Recall from the basic definition of ensemble spectral density S(w) given in Chapter 2, Equation (2.16), that R(T) =
f~", S(w~(·< dw
(4.19)
A comparison of Equation (4.18) and (4.19) gives o:(w) dw =
.JS(in) d~,
(4.20)
Substituting Equation (4.20) into Equation (4.6) we obtain the integral analogy to Equation (4.1 5) as follows: x( t)
=
f'" .J
S(-;~j~/-;~e'l<'"
- 11l"'1I
(4.211
Let the relative displacement (x - u) = y; then we have
my + cy + dy = - mii =
- rna
The earthquake generated shear force
v
= ky
+ cy
can be determined once the relative displacement y is found from response solution based on Equation (4.22). We point out here that~uation (4.22) derived for the tall building model subjected to earthquake ground acceleration is seen to be identical to Equation (4.2) for the veh ide model. The fundamental difference betwcen the two engineer ing problems lies in the nature ofthe forcing function u(l,.ln the vehicle problem, we assume that the road profile is a stationary proceSs where the key prob abalistic and statistical properties are independent of time t. In the case of the earthquake ground acceleration, as we pointed out in Chapter 1, the underlying random process is clearly nonstationary. The time dependent character of the random process is~of fundamental importance and must not be ignored.
-00
--l I--x(t)
where the lower limit may be changed to zero with the insertion of a factor of two and the replacement of the exponential function by its real part. the cosine function. . The mathematica\significance of the unusual term ~ in Equation (4.21) can be heuristically understood by a comparison of the two Equations (4.21) and (4.15). Unlike the usual integration, the associated summation consists of the term .JI'!w instead of I'!w.
~
I I I
I
I
4.2. RANDOM EARTHQUAKE MOTION (NONSTATIONARY MODEL) We consider an idealized structural model for tall buildings subjected to earth quake generated ground acceleration ail) == ii(t), with total mass m. spring constant k, and damping coefficient c. This idealized structural model is shown
I I
....
I
I
-
m
k(x-U)+c(±-u}
I I I I I I
.~ .~
II
. 11 >:l
m'l~1
- . " / /. a(t) .::. tilt}
FIG. 4.3
I I I I I) I,
u(t)
Tall building model subjecled to earthquake-generated ground acceleration aft) a;; ii(t).
" :i ~. '.:
I1l I;
If:1 :
(;
;!
::;:
'i
i
II
I
I
I
68
4.2.1.
~
.
i~ II
We introduce the time dependency of the random process in its discrete form by inserting a deterministic function A(t, w) of time t. and frequency w into the previous stationary model, Equation (4.4). In so doing we obtain the discrete model for the nonstationary process X(t), characteristic of earthquake ground acceleration aft) as N
! \~~I
I
\. I
,l
.
<
II 1\ I
ill
'I'"
'v
L A(t, w.)cx,. cos(wnt - OJ
X(t) =
.
. "·.. ;.SL;. '. ;;~ ,;.. M·
. (4.24)
.'" I
Oi course we use the symbol X(t) here for a general nonstationary random process as a model for earthquake ground acceleration as well as any other time dependent random excitations such as tornado wind force on buildings or hurricane generated ocean wave motions. Returning to Equation (4.24), we can calculate the ensemble mean E[X(t)]
= L
A(t, (/).p.E[cos(w.t - 0.)] = 0
(4.25)
since the nonstatifilnary process X(t) of Equation (4.24) is the same as the stationary process given by Equation (4.4) except for the deterministic functions A(t, Co.), which plays the role of multiplying factors on the N terms of zero mean values. The ensemble mean square value is
L A(t,w.)
N
8.)
.'" I
L 111=
A(t, w",)
1
L
tA2(t, m.)
(4.26)
.=1
since the deterministic functions A(t, m.) can be associ~ed with IX. and con sequently, we can use the previous result givell by Equation (4.10). . Now extending the concept of ensemble spectral density S(w) given by Equation (2.17) and considering the time dependent spectral density S(t, w.) associated with the frequency Wn to be distributed in the interval Aw = w.+ I - w., we then have
(4.29)
Finally, we observe that in Equation (4.24) when (4.30)
A(t, w.) = 1
X(t) reduces to the stationary process and from Equation (4.28) we have for A(t, w,,) = 1 and the stationary S(w.) IX.
(4.31)
.j4S(w.) Aw
A comparison of Equations (4.31) and (4.28) gives the relation between the nonstationary S(t, mn ) and the associated stationary S(wn ) as
= A2(t, w,,)S(w
lI )
(4.32)
It should be pointed out that the nonstationary process Equation (4.24) is no longer ergodic. This is reflected by the ensemble mean square given by Equation (4.26) as a function of time t. . In this and the next section we see that a nonstationary random process X(t) is introduced as a modified stationary process by introducing the deter ministic function A(t, w.). In this way the nonstationary process is intimately related to the associated stationary process [A(t, wn) == 1] by Equation (4.24) in its sample function form and by Equation (4.32) in terms of ensemble spectral densities.
4.2.2.
N
=
.j4S(t, (v.) Aw COS«(I).I - (lu)
n=l
S(t, w.)
n=1
E[X2(t)] =
69
N
L
X(t)
N
N
L
RANDOM EARTHQUAKE MOTION (NONSTATIONARY MODEL)
Substituting Equation (4.28) into Equation (4.24), we find that
Discrete Model
I
.; I
4.2.
MODELS OF RANDOM EXCITATIONS
Continuous Model
We again introduce the time dependency to the associated stationary random process in its continuous form, Equation (4.6), by inserting a deterministic function A(t, w). The resulting nonstationary process is then X{t)
=
J:",
A(t, w)
(4.33)
When A(t, w) == I, the process degenerates to the associated stationary process. The ensemble mean
N
E[X2{t)] =
L
2S(t, m,,) Am
(4.27) E[X(t)]
0=1
=
J:",
A(t, w)a:(w)eI"'E[e- l 8(w)] dw = 0
(4.34)
A comparison of Equations (4.26) and (4.27) gives A(t, w.)a:. = .j4S(t;w.)Aw
(4.28)
The ensemble autocorrelation function R(t, t) is now time dependent. When X(t) is considered complex and when we are interested only in its real part, it
70
MODELS OF RANDOM EXCITATIONS
4.3.
can be expressed by
R(t, -r) = E[X(t)* X(t
+ -r)]
(4.35)
which has the same right-hand side as Equation (4.17). Substituting X(I)· and X(t + f), according to Equation (4.33), into Equation (4.35) gives R(I, -r) =
E
[f:",
X e'("'(I+r I -6(w'l!
f:",
f:",
A(t, w)a(w)e'[6("I-ml) dw
A(t, w')a(w')
i
Before leaving this section on nonstationary random processes, we should point out that \be associated stationary process corresponds'to the case where, in Equation (4.33), thedeterministic function is A(t, w) == 1 for all - 00 ~ t ~ 00 and - 00 ~ w ~ 00. This is quite different, of course, from the simplest non stationary process where
J,
t;;;,O
A(t, w) :; A(t) = { 0,
(4.42)
t
which is the familiar unit step function of time. Physically such a nonstationary
process is; in fact, a suddenly applied or turned on stationary process. For this
reason it is sometimes still considered a stationary process. Here we consider
this as a nonstationary process with a nonstationary ensemble spectral density
dW'J (4.3(,1
1..1(1. mlCl'(wl ,1",F.· ... ,
71
RANDOM OCEAN WAVES (MULTIVARtABLE STATIONARY MODEL)
S(t,
/I)
{~(m).
I
t~O
(4.43)
t
which is "gain idcnticul to Equation (4.18) exccpt for the deterministic function ,,/(t. wi.
A comparison or Equation (4.36) with the extended dellnition or nonstation ary ensemble spectral density S(t, w) given by R(t, f) =
r~ S(l, w)e i'.' dw
(4.37)
gives
4.3. RANDOM OCEAN WAVES (MUlTIVARIABlE STATIONARY MODEl) A wave staff located at a fixed point in an ocean wave field may be used to measure surface or gravity wave oscillations a~ a functiQn of time 11(1) measured from the mean water level (MWL) as shown in Figure 4.4. To construct an ocean wave surface model. we begin from the simplest and ideal single com ponent standing wave with ~amplitude
(4.38)
A(t, w)a(w) dw = .jS(I, w) clw
I/(t) = iX cos(wt - 0)
..
f~ .JSi.t, (II) clwe'I••
'-O(••
1I
" l '< - , -I~I,
v
•/
f
IV' ,
(4.40) {t)
I 1 I
(4.41)
which is the relation between nonstationary S(t, w) and its associated stationary
/
FIG. 4.4.
MWl ~(tJ=o
alt)
d
z = 0
S(d) and is identical to Equation (4.32).
I j
A comparison of Equations (4.38) and (4.40) gives S(l, w) = A2(t, w)S(w)
i)
1i '
(4.39)
Finally, when we set A(t, w) = I in Equations (4.33) and (4.38) we obtain the associated stationary process and the result is a(w) dw = .jS(w) dw
~i
(4.44)
Substituting Equation (4.38) into Equation (4.34) we obtain the integral similar to Equation (4.29) as follows: X(I) =
I I II I
An offthore structural model with a wave staff,
I I I
1'1
I I I
72
MODELS OF RANDOM EXCITATIONS
For a progressive or moving wave we consider '1(x', t) =
(X
cos(kx' - WI
+ 0)
(4.4~)
co
c=k:
c
=
collI 21t
(4.4H)
which is the product of wavelength 1 and wave frequency J in cycles per second with J = co/21t. The frequency co and the wave number k are not independent. They are related, according to the linear theory of wave theory, by the so-called dis persion relation.
co 2 = gk tanh kd
(4.49)
where 9 and d are the gravitational acceleration and the water depth, respec tively. From Figures 4.6a and 4.6b we see that the wavelength 1 in the x' direction
A = ~7C
(4.47)
k'
.Y 'I(x', t)
x'
I"""'"
I
<
I
1x
/
~'"
'I'~
)0
x
)px'
J.ky
x',k '1(..',1
+
At)
y
,I II !
'1
'x
7'
",,)I
'1....
\\1
~.
x.k,
X'
~ *,',
...,.
~
~
I~
:'i.i
(4.46)
and the wavelength in Figure 4.5 is
~
,~
73
so that
where the wave form is no longer standing as that of Equation (4.44) but moves in the x' direction with constant wave speed c as shown in Figure 4.5. Also present in Equation (4.45) are the amplitude (t, the wavelength 1, and the wave number k. which is defined as the number of waves in radians per unit length. The wave speed is
I
I
I
~
RANDOM OCEAN WAVES (MULTIVARIABLE STATIONARY MODEL)
"(:
I I I
4.3.
FIG. 4.6, apart.
Two snapshots or a progressive surrace wa,ye with wave speed c at a time interval AI
FIG. 4.8, (a) Top view or a progressive wave train in the x' direction making an angle .p with the x axis. (b) Top view oftwo wave crests A and B showing the vector charaCieristics of the wavc number k and the geometric relation between land lx.
74
4.3.
MODELS OF RANDOM EXCITATIONS
RANDOM OCEAN WAVES (MUlTIVARIABLE STATIONARY MODEL)
associates with the wavelength A.. in the x direction and that the progressive wave model, Equation (4.45), is unidirectional. To extend it to the practically more important two-dimensional surface wave, we introduce the two space coordinates x and }' along with the wave direction 4> and obtain the two dimensional wave model, 'I(X,
y, t) = a cos((k cos
+ (k sin
wt
+ 0]
75
n
¢
(2.1)
Q
(2.2)
k" and k sin
"
, ~!
-iii!)
r directions,
k" and wave speed
(4.52)
e" = w/k" = A"l
These geometric wave relations can be illustrated in the following. From
Figure 4.6b,
A = A" cos
(4.53)
By definition the wave number is k
= (.~)(21t)
-~--
and
k" = ~!X2_1t) ,1. "
(4.54)
Substituting ,1. from Equation (4.53) into the first part of Equation (4.54) and
then using the second part of Equation (4.54), we find that
k" = k cos
(4.55)
which verifies the vector nature of the wave number k as it has been specified
by Equation (4.5l).
4.3.1.
Discrete Model
A discrete random ocean wave model is made up by the superposition of M by
N wave components, each having the form of Equation (4.50) as
'N
1/(X, y, t)
.M
I I
• =1 m=1
am. cos(kmx cos
o
6 (2.11
(1.11 FIG. 4.7.
" m
Four combinations of wave number m and wave direction n.
(4.51 )
k cos c/)
wmt
+ Om.)
(4.56)
It ; I .,
(4.50)
where k cos 4>
I
where the only set of random variables are the phase angles 8 11 .8 12 , _" .8MN • These random phase angles are assumed to be independent and each angle has the same uniform probability density funclion
p(Omn) = {
2~ , O.
o ~ Om. ~ 21t otherwise
To see this double sum in Equation (4.56) more clearly we refer to Figure 4.7 where the index m for wave number km and the index n for wave direction
I I
I· I~ It
I 1:.\ 11ft
I I I t
[
t:::
......1'>,1
I III
II
76
MODELS Of RANDOM EXCITATIONS
;:
N
E['1(x,y, t)]
M
= L L lXmft E;[cos(k.. x cos 4>. + kmJ sin 4>. .= 1 m=
w.,t
+ Om.)]
= 0
I
'1(x, y, t) =
f~~
I
(4.57)
I
I
I
I
RANDOM OCEAN WAVES (MUlTIVARIABlE STATIONARY MODEL)
since we can consider the three deterministic terms kmx cos 4>. + kmy sin 4>. - w",t together in place of the single term W r ! in Equation (4,8). By the same token, we can determine the ensemble mean square E['1 2(x, y, e)] =
N
,.,
L
~ ~(X!.
(4.58)
f"
a(w,
4»eiIO «O•.pI+k:< cos.pHysin.p-,.I] dw d4>
The ensemble autocorrelation R(X, Y, t) is now a function of the usual time lag r plus the space separation X and Y in the corresponding x and y directions. Thus we have R(X, Y,t) = E[t/(x, y, t)*'1(x
"
N
E['1 2(x, y, I)]
2S(wm , (P.) flw llc/)
(4.59)
I
f.
jl
I
L
i;'';}
I:
11
.
I
" .,
~;
P.IIP:I'
+ X, y +
Y, t
+ t)]
(4.64)
Substituting '1(x, y, I) from Equation (4.62), with appropriate modification for the complex conjugate operation and for the time and space lags, into Equation (4.64) gives
At
L L
(4.63)
E['1(X, y, t)] = 0
. " 1m-I
In the next step. we introduce a new term, the so-called directional wave spectral density S(w"" 4>.) as a function of two deterministic variables, the usual frequency Wm and the new wave direction IjJn. This density function is again an ensemble statistic and is defined analogously as the one variable density S(w.) by the ensemble mean square as
(4.62)
The ensemble mean is
tt>
R(X, Y,r) =
n= I m= 1
["
f'''=_«) J.. =-" x e- i1kx
f""
[n
",'=_",
J.p'=-" a(w.4»a«(1J',4>')
~o.t/>+ky.ln.p-'·I/el(k·" ~o.t/>·+k·yslnt/>-<»·II
A comparison of Equations (4.59) and (4.58) gives x E{ei(II'..••.p·)-l1lw .t/>II}ei(k·X
I
I
I
77
number and wave direction to give the following model:
The ensemble mean is
~l
~I
4.3.
teL!. =
2S(wm ,
4>n) flw fl4>
Substituting IXm. of Equation (4.60) into Equation (4.56) gives the discrete two dimensional ocean wave model in terms of its directional spectral density S(w"" 4>.) as N
,,(x, y, t)
M
= L L
n= 1 m= I
wmt
4.3.2.
.j4S(wm, 4>.) flw fl4> cos(kmx cos
+ 8",.)
~O$.p'+k·r$int/>·-"'·()
dw d4> dw' (14)'
(4.65)
(4.60)
After carrying out the double integration with respect to w' and 4>' and using the same uniform probability distributions of the independent phase angles as before, we obtain R(X, Y,
4>. + k.y sin 4>. (4.61 )
Continuous Model
A continuous random ocean wave model in a general complex form is con structed in· a manner analogous to the complex roadway model given by Equation (4.6). We therefore derive the model here- in a concise form leaving out some details, which can be found in Section 4.1.2. Based on the discrete model. Equation (4.56), we replace the cosine function by the complex ex ponential function and the double summation by the double integral in wave
Now for X
r).= f~",
=Y= r
f"
[IX(W,
4»
dw d4>]2eiikX oost/>+krsint/>-wt)
(4.66)
0, Equation (4.64) gives
R(O, 0, 0) = E['1(x, y, t)*'1(x, y, t)] =
J~",
f"
Sew,
4»
dw d4>
(4.67)
in which the real part of the expectation is the ensemble mean square and therefore it leads to the directional spectral density sew, 1jJ). From a comparison of Equations (4.66) and (4.67) we obtain a(w, 1jJ) dw d4> = .jS(w, 4» tlw de/>
(4.61:\)
,
78
4.3.
MODELS OF RANDOM EXCITATIONS
Substituting Equation (4.68) into Equation (4.62) gives the final complex ocean wave model in terms of the directional spectral density of the process as I'/(x. y. t)
4.3.3.
=
f:"" r~
.jS(w.
1) dw d1ei(6("'•.p)+~ cos.p+ky.ln.p-",'l
(4.69)
(4.70)
g
The first part on the right-hand side of Equation (4.70) consists of the product term 1'\1'1 of the water particle velocity v and represents the drag force. The second part represents the inertia force associated linearly with the water particle acceleration a. Also in the equation are the drag coefficient CD' inertia co efficient C"h unit weight of water w, gravitational acceleration g, and the diameter of the pile D. Based on thc Morison equation, Equation (4.70), and solutions of the linear water wave theory, a sct or transfer relations can be derived among ensemble spectral density' rundions or the wave r.,lrce S",(w), water particlc vclocity S,,(w), water particle acceleration S.,(CI'), and the wave surface S~(w), These transfer relations are of practical importance in offshore structural engineering since they can be used to prcdict the statistics of thc dcsign wave force on strul:tures fmm ocean wave mCllliUrCtllCnts. First we present the relevant linear wave solutions, then the spectral transfer relations, and finally some derivations. The basic one-dimensional linear wave solutions are I/(X, I)
= ex cos(kx -
(4.75)
S"(w, z) = W4Z2(Z)S~(w)
(4.76)
S",(w, z) =
w rrD2 +CM a
I
== W2Z2(Z)Siw)
S.(w, z)
Forces on pile structures due to ocean waves are usually determined by a semiempirical formula known as the Morison equation, where the wave force per unit length of pile is w
79
The ensemble spectral transfer relations are
Spectral Density of Wave Force
= Co 2g
RANDOM OCEAN WAVES (MULTIVARIABlE STATIONARY MODEL)
q;(z)
=
w ( w rrD2)2 ;t8 (CD 2g Dqv)2Sv(W, z) + CM 9 4 S..(w, z)
(4.77)
5:..,
(4.78)
Sv(W, z) dw
EXAMPLE 4.2. Verify the spectral transrormations between wave surface" and water particle velocity v, Equation (4.75), and between '1 and a, Equation (4.76). We introduce first the discrete form of the ensemble spectral density by N
L 2S~(wft) Aw
t)] =
E[I'/2(X,
(4.79)
q~1
and N
Ef,,2(X,..;:,
L
til
2S.,«(I)•• z) A~n
(4.RO)
n"" •
Based on Equations (4.71) and (4.72) the number of components N are the same in 11, 1', and N
E[1/ 2(x,
t
I)]
= L lex;
(4.81)
0=1 N
(ot
(4.70
+ 0)
~(L12(X, z, I)] =
L !a;w; Z;
(4.82)
.=1
t(x.
z. t) =
exwZ(z) cos(kx - wi 2
a(x, z, I) = ct:<1.I Z(z) sin(kx - wt Z(z)
= cosh kz
+ 0)
+ 0)
(4.72)
From Equations (4.79)-(4.82~ we obtain
(4.73)
1 . 2 S~(w.) = 4Aw Cl:q
Ii I I I I I I
I Ii I
(4.74)
sinh kd in which 1'/, v, a, a, k, (t), 0, z, alldd ar~ respectively, the wave surrace. water particle velocity, acceleration, amplitude, wave number. frequen!.,")', phase. water elevation measured from ocean bottom, and water depth.
S,,(w., z)
1 a.ro. 2 2 2 4-:'Z. un1
:. S,.(w., z) = w; Z;S~(w..)
(4.83)
I I 8}
I SO
I I
I I
I I I
I
MODELS OF RANDOM EXCITATIONS
where
I I
Iii
I, t;
ili' !
II I ,j
1lI
'S1
relations: Z. = .cosh k.z sinh k.d
(a) E[ft(Otl + 12(02 )] = E(ft(OI)] + E[fz(Ol)]. (b) E(ft(OI)fz(02)] = E(ft(Od]E(f2(02)]. (c) Will these hold when 0 1 and O2 are no longer independent?
(4.84)
4.2. Consider a single component random wave process
Similarly, we obtain N.
N
E[a 2(x, z, t)]
=L
2S.(w., z) !J.w
:. S.(w., z)
I'(x, t)
L tlX;W!Z;
=
n= 1
If=
t
= w!Z;Sq(W.)
(Jl =
(!
Ca,
= C
~nD2
M
IX
sin(kx - wt
+ 0)
in which the wave number k and frequency ware not random, but the amplitude IX and phase 0 are independent random variables. If the probability densities are those plotted in Figure 4.8, (a) Find E(I'(). (b) Find E(1'(2). (c) Find Rq(r).
(4.85)
EXAMPLE 4.3. Verify the spectral transformation for the wave force (Jl and the water particle acceleration a. , From the inertia force part of the Morison formula (4.86)
g 4
p(lJ)
p(a)
we take the ensemble mean squares E«(Jl2)
= C 2 E(a 2 )
(4.87)
Introducing the ensemble spectral densities S",(w, z) and S.(w, z) in Equation (4.87) gives N
L
L
I
o
.. 6
2rr
FIG. 4.8.
N
2S",(wn , z) !J.w = C2
n= 1
I
PROBLEMS
o
..........
)
(C
2
Probability densities p(O) and p(a).
2S,,(w., z)!J.w
n= 1
.'. S",(w., z) = C 2 S.(w., z) = ( CM
wnD2)2
g4
S.(wn , z)
(4.88)
which coinsides with the acceleration part of Equation'(4.77). Obviously the key to this simple derivation is the linear relation between II) and a in the Morison formula. Equation (4;10). Such It simple deviation is not possible for the nonlinear relation between II) and the water particle velocity v. For the complicated derivation of the transformation between S.(w, z) and S",(w, z), the reader is referred to the origin~l work by Borgman (t 967).
. PROBLEMS
°
4.1. Show that, if 11(Otl and 12(02 ) are two functions of two independent random variables 0 1 and 2 , we can use the following ensemble statistical
4.3. Suppose you are given a two-sided, so·called target power spectral density (PSD) of a stationary random process SI(W) and a simple mathe· matical model of the process x(t) as N
x(t)
==
L .j4S(w.) /J,.w cos(wnt n;1
On)
where the only random variables are O. for n = 1,2, ... , N, and O. are independent and uniformly distributed in the interval (0, 2n). (a) Find, presumably by computer, the ensemble autocorrelation func tion R.{t) from the simulated sample functions based on the given mathematical model and the target PSD S,(w). (b) Compare the simulated R.(r)·with the target R,('r), which is defined by the formal Fourier integral formula, Equation (2.16), with the associated target PSD S,(w).
82 MOOELS OF RANOOM EXCITATIONS
PROBLEMS
83
4.4. Considcr the following random process model N-
Y(t)
where
(12
S,(£O)
=
(1
fl L
sin(£Okt +4>k)
-N
field of mean water depth d. The pile has a circular section with constant radius R. Assume the pile is fixed at the end and neglect the drag part of the wave force. (See Figure 4.9.)
1<=1
= variance or mean square =
J:",
S,(£O) d£O
= a given target PSD, two-sided
4>" for k = I, 2, ... ,N are independent random variables uni formly distributed in the interval (0, 21t)
m
for k = 1,2, ...• N are also independent random variables but with probability density function
<,b(z)
(Uk
p(£O,,)
==
I»
d
p(£O) = S,(£O) (f2
(
~
simulated sample funCtions, based on the given mathematical model with the target PSD S,(£O).
FIG. 4.9..
Wave force .p(zl on pile.
(b) Find the simulated ensemble autocorrelation function Rlr). (c) Compare the simulated R.(-r) with the target R,{t). which is defined by the formal Fourier integral formulation, Equation (2.16), with the associated target PSD S,(m).
4.5. Consider an earthquake ground acceleration model N
xo(t) =
L
4.7. A computer simulation problem by Dr. John Zimmerman and Mr. Samir Chettri at the University of Delaware. Use the information given ; ' in the Appendix B. (a) Synthesize a rand(lm function of time by the Monte Carlo technique
(see Appendix A), based on a given target power spectral density
function W(f) and given frequency range
tal!-·j' cOs(Wjt + 4>j)
K
j= I
x(t)
where the only random variables are the phase angles 4>j for j = 1,2, ... , N and q)j arc indcpcndcnt and uniformly distributcd in the interval (O,21t). (ll) Find thc ensemble mean B[.\\MU from the Illodel. (b) Find the ensemble mean square E[Xo(I)2] from the model. (c) Use the definition of a two-sided nonstationary PSD as defined by N
E[XO(t)l] = L 2S(t. Wj) 8.(1) j= I
4.6. Based on the given PSD of the wave surface, Sq(£O), determine the PSD of the maximum bending stress (1. S.. (w), at the bottom of a pile in a wave
j
=L
ak sin(2ltht
+
k;1
where the lk and 4>" are selected randomly. Use K =:' 50. Store the
timc, t; = (i - l)T/N. i = I to N. Here l' = 12 sec. Store the tl' XI'
i = I to N, values in a data file for later use.
The PSD function is
.I~ and
wen = (1.34 x 0,;;:;
to find the acceleration model xo(t) ill terms of S(t, Wj)' (d) Find the autocorrelation function R(t, .) from the model.
I I,
I
10- 4 )(10- 0 ,185/1
I,;;:; 120 Hz,
I I I I I Il ;JtI~
[W(f)] = ft2/Hz
(b) Use a Fast Fourier Transform (FIT) computer subroutine to find the PSD from this sample function. Store values of frequency and
PSD in a data file for later use. Refer to Appendix A for FIT.
(c) Use the PROPLT plotting program (simply type PRO after the
m
I [I
, I
I ;: I
:j !~
"'
~;
I
I
I
I
I
I
I
,I Ir
, "~
(I .•
j
"\ I
II
IIU \\ i i
i i\l."
84
MODELS OF RANDOM EXCITATIONS
$ prompt) to plot the sample function. It is interactive. It will ask for the name of your data file and for other necessary information. (d) Write a short program to calculate values of the frequency and PSO from the given W(f) functiott. Use the same frequency spacing as is used for FFT computations. Store these values in a data file. You will use this file later. (e) Using PRO, plot the original W(f) curve and the calculated PSO curve on the same sheet.
CHAPTER
5
STRUCTURES WITH SINGlE"DEGREE OF FREEDOM (SDOF)
In the previous four chapters, we havc presented the characterization of random excitations to structures. In this and in Chapters 6·10 we shall study the relation between input excitation and output response, sometimes referred to as the transfer or transmission relation of the random vibration of structures. We begin with the deterministic transfer relation of single-degree-of-freedom (SOOF) structures. For simplicity, we shall from now on drop the rigid rule of using capital letters for random variables. 5.1.
DETERMINISTIC TRANSFER RELATIONS
Consider the simple SOOF oscillator model as shown in Figure 5.1, with mass In, spring constant k, damping coefficient c, and forcing function The equation of motion in terms of the displacement y, velocity y, and accelera tion ji may be given in the form ji
+ 2~rolfY + ro; y
x(t)
(5.1)
in which the natural frequency ron' damping ratio ~. critical damping coefficient and the excitation x(r) are given by
Ce •
",2 "'If
Ce
k
m•
= 2km,
~=C Cc
x(t) = f(t)
m
(5.2)
85
86
STRUCTURES WITH SINGLE DEGREE OF fREEDOM (SDOf)
5.1.
Then th~ total response tS obtained by superposition as
.v
~t}»;~)lk////' iOl
.
.,.
yet) fIt)
= -1 fa)
271.' _""
A simple oscillator model.
y(tJ' = 1 For the solution of the input excitation and output response problem, we proceed in two paraliel routes: The so-called frequency domain approach and the time domain approach. 5.1.1.
f'"
X(w)i Oll dw
y'(t)
= H(w)elw.
Yew)
iJ
+ flv =
x(t)
I I
(5.9)
I
H(w) =
Ii + iw
(S.lO)
I
I
then I
IH(wW ei(U'
I
. (5.8)
Substituting the unit response v' H(w)eiu,t together with the unit frequency excitation x'(t) = ei ." into Equation (5.9) gives
be the response to each frequency component excitation of unit modulus as x'(t) =
H(w)X(w)
EXAMPLE S.l. For the SDOF system with mass m, damping coefficient c but with no spring element, determine the complex frequency response function H(w) for the response velocity v. The excitation force is mx(t). (See Figure 5.3.) Let fJ = elm, then the equation of motion is
(S.3)
(S.4)
(5.7)
in which H(w) is called the complex frequency ;esponse function introduced by Equation (S.4j as the response y'(t) to a unit frequency component x'(t) = ei"".
_00
and can be interpreted as the superposition of components with different frequency w, it follows that the total response y(t) can be obtained if the response to each individual frequency component is known. Let
f""_a) Y(w)i"" dw
• • t
The frequency domain approach is based on the superposition of frequency components of the forcing function x(t) such as the arbitrary transient excitation shown in Figure 5.2. Since such a transient function x(t) can be expressed in terms of the following Fourier integral with its Fourier transform X(w),
271.'
(5.6)
I I
we then obtain the transfer relation in the frequency domain,
Frequency Domain Solution
x(t) = - 1
X(w)H(w)eiOJ'dw
Since the response y(t) can also be expressed in terms of its Fourier integral
c
fiG. 5.1.
87
DETERMINISTIC TRANSfER RELATIONS
= fi! + w2
(5.11 )
(5.5)
and is plotted in Figure 5.4. x(i)
,I
I. ·1
v
I
b , ~»»JJ///. m
fiG. 5.2.
An arbitrary transient excitation xl').
FIG. 5.3.
...
mx(t)
A SDOF system with no spring elemenl
~ f:
~:
,I
.
~
~
I I
m
,
~ ~'I L
~
I.') I.
88
STRUCTURES WITH SINGLE DEGREE Of FREEOOM (SDOF)
jJ
Jlt·l
--
2i.... _-0'-=. : E';::":='::'
I
:1 .~. ~
-
~: ~-
<
~.
o
,I
\
g'
..
= a cos wol,
!;-:~:. Z-.::":
~·~;':::i..!
~
... ' ....
::-..=~.;-=.;....L.= ::;t:::.::~::..:'"
--:f.
a. 2
x () t = - (e''''o'
FIG. 5.4. Square of the modulus of the complex frequency response for velocity v. Example 5,1.
15.1,:11
. + e-""'o')
(5.15)
it follows that the steady-state velocity response is EXAMPLE 5.2. For the SDOF structural model shown in Figure 5.1, determine the complex frequency response function H(w) for the displacement y. The excitation is X(l), Substituting the unil response y' = H(w)eiWl and Ihe unit excitation accelera tion x'(r) = eiUJ' into Equation (5.1) gives
H(w) =
~-:wl.!---- + 2i~wwn
1'(1)
'"
See Figure 5.5.
-"'n
a
+ . 1/( .2
a 1 (fl cos wot +lOo
.
wok-""""
+ lOo sin lOot)
4»
(5.16)
EXAMPLE 5.4. Determine the transient velocity response of the SOOF system of Example 5.1 when the system is subject to an acceleration excitation x(t}, as shown in Figure 5.7a, assuming the system is initially at rest. The transient response v(l) due to the transient excitation x(t) can be obtained by using Equations (5.7) and (5.8) where we first need to calculate the Fourier
Im",F,
"""=
2
.
1I(lIIok"''''
where p = a!.JfP- +- cOg and 4> = tan -1(wo!Pl are the amplitude and phase, respectively, of the steady-state velocity response. See Figure 5.6 for the relation between the set of constants p, a, and Wo for the structural model and excitation, and the corresponding constants P. 4>. and Wo for the response. Note that the steady-state response has the same frequency Wo as that of the excitation.
(5.13)
.•" ,
a -
= P cos(wot -
then IH(wW ="
= =
(5.12)
'{
....,.,
"""".
)"
"'n
~
-r~~
."
W
~:,. ~;t
3"1
r
~~~...~::J..:"!
_!
~.:..~:.:.: ;:~:::-_: :~:':'
where a and Wo are the amplitude and frequency, respectively. By definition, when the excitation is x'(t) = ei «>' the unit velocity is v'(t) H(w)e i "". Now, since the steady-state excitation specified by Equation (5.12) can be decomposed as
i;
i
::'-::-=-=-=.:i X{lJ
,,~I
I I I I I I
I
:. I
II l~ l~ I
~
-=-1.!...::...?..! :-
89
DETERMINISTIC TRANSFER RELATIONS
5.1.
'FIG. 5.5. Square of the.modulus of the complex frequency displacement response due to unit acceleration excitation.
K· f:
"'0
P
FIG. 5.6. Relation between the response characteristics (p,,p, and wo) and those for the excitation (a, w.} and structural model (/J).
liB ~:
rr
I[ I
90 X(tl
a
5.1.
STRUCTURES WITH SINGLE DEGREE OF FREEDOM (SDOF)
DETERMINISTIC TRANSFER RELATIONS
91
Imaginery
VW
"",I- - - - - - - ,
, I· i.,"
1
II I \'.1 '"i
I I I I I I I I
o
o
T
..
l!
..
Real
(i»
(al
FIG. 5.7. Transient excitat.ion (a) and response from Example 5.1.
(b)
for the SI)OF syslem given in Figure 5.3
F
transform of the excitation. For. this problem, the Fourier transform fror the excitation x(t) is
" f
X(w) =
x(c)e- iWf dl
=
-w
.0
j
e!iw'l)
(1 _
[a / e",T) . ]
Yew) = H(w)X(w)uW- ' -:- (.~ p
+ IW
and the transient velocity response, by inverse
1t
a
(5.18) J2
~o'urier transforr4 is
I
= 2ni".
-w
IWf
e
iw{w -
../
rm dw -
f'"
e'w(r-/l
_'" iw{W,L. ifl) dw
= 21[/. [ -.-I. - + -e -fl'] = "2- 1t1. (1 I( - IP) - fJ fJ •
-
_
I
e fl)
2m i(
-
e- ptr -
-
i{J)
T )] _
{J
-
2ni[1 - e-P1'-Tl] fl
] (5.19)
These two integrals, denoted b7.and 12 • respectivel;, can be ca rried (Jut by applying the theory of residuals as follows: Referring to the w plane in Figure 5.8, we observe that there are two singu larities or the so-called poles, at w = 0 and w = i/J, respectively, for both integrals 11 and 12 , FUrthermore, for 11 , when c > 0, the integrand vanishes as the radius R of the upper half-circle approaches infinity. The contour integral along the boundary of the upper half-circular area, including the pole at w = 0, is therefore zero except for the part along the real axis from - 00 to + 00. This nonzero part, which is 11 by definition, is then ,:qual to the sum of the residues at the two poles. Thus, for r > 0, II
_..[_1_ +
For 12 when (I - iT) < 0 a lower half-": that the integrand:vanishes a~ .,.
-00
fOCI
plane.
For 12 when (I 'f) >.10 the integral clin be carried out in exactly the same way as that for 11' Thllt is, for t > T,
V(w)ei Wf dw
r
II)
i
IW:
.I
Integration in the complex
J
.I i
f'"
FIG,S/S.
(5.17)
IW'
The Fourier transform of the response v(t) is then
I vet) = -2
,
,l
j
i
since there is no p( Summarizing th, stituting these result o
-(.
2ni
v(c) =
a
fJ{(l o
(S.24)
",,':!l -;;;(;l \
.
< 0, y(t) = o. initial velocity . by the magnitude
:;,~
S'. '& ... :;;~
.1
~
9· PO;;; ~ c· "t> 0::>
\~
(S.25)
~~~. ~
~
This transient velocity is ph Examples 5.3 and 5.4 eI, requires simple algebraic ca
--0t)
~~
~.
i.,.
he general appearance of this ~!5.9c with a nonzero 'to Note that
I II
:!l;ij i!1
90
STRUCTURES WITH SINGLE DEGREE OF FREEDOM (SDOF)
XCt)
5.1.
DETERMINISTIC TRANSFER RELATIONS
VItI
91
Imaginery
{.
I " ;: I ir
li
~~:
l~'
.. = ip
aljJ
a' ... - - - - -...
F ,,~
:i;
::;
~
o
o
T
T - - - - - -.......' - -....- - - "
~I
(a)
~
• I
"
Real
(/,)
FIG. 6.7. Transient excitation
(a) and rcspunse (iI) fur Ihc· SDOF system givlln in Figure 5.3
from Example 5.1.
I
I
I
"I 'j
transform of the excitation. For this problem, the Fourier transform for the excitation x(t) is X(w)
I
x(t)e-iOO'dt =
~(l
_ e- iIDr)
(5.11)
IW
The Fourier transform of the response v(t) is then
,
V(w) = H(w)X(w) =
FIG. 5.S.
V(I) = - 1
foo
~ l~ (1 IW
eiWT )]
fJ + IW
_ .[_1_
(5.18)
plane.
- fJ
-
fJ
For 12 when (I - T) < 0 a lower half-circular contour is required to ensure that the integrand. vanishes as the radius goes to infinity. That is, for
V(w)e'fJ>I dw
a [fOO = -. . e'fJ)1 . dro - f'"21t1 Iw(W - 'fJ) 00
00
.
ew,(I-TI
iw(w -
. dw
,fJ)
]
• 21[/ e-/1,] + -" = -(1 fJ
e- P1 )
O
(5.19)
These two integrals, denoted by 11 and 11 , respectively, can be carried out by applying the theory of residuals as follows: Referring to the w plane in Figure 5.8, we observe that there are two singu larities or the so-called poles, at w = 0 and w = ifJ, respectively, for both integrals I, and 12 , Furthermore, for I" when t > 0, the integrand vanishes as the radius R of the upper half-circle approaches infinity. The contour integral along the boundary of the upper half-circular area, including the pole at w = 0, is therefore zero except Cor the part along the real axis from - 00 to + 00. This nonzero part, which is It by definition, is then equal to the sum of the residues at the two poles. , Thus, for t > 0,
- p
(U
e-I/{I-TI] _ 21[;[1 _ e-/l(,-n]
12 - 21[1 i( _ ifJ) +
-00
I 1\ = 21[/ [ -.- . ,( - l{:l)
Integration in the complex
For 12 when (I - T) > 0 the integral can be carried out in exactly the same way as that for I,. That is, for t > T,
and the transient velocity response, by inverse Fourier transform is
21t
I·
foo -00
i
I
I
I
=
12 = 0
since there is no pole within the contour. Summarizing these results of contour integration for 11 and 12 and sub stituting these results into Equation (5.19) we find the response velocity
a 21[i (I t v{t) =
~ ({1
o
.
-
12 ) =
ff (1
- e - p')
e- lIl ) -: [1 - e-/1(I-n]}
forO < t < T for t > T
(5.20)
for t < 0
This transient velocity is plotted in Figure 5.1b. Examples 5.3 and 5.4 clearly show that the frequency domain approach requires simple algebraic calculations for the steady-state problem. On the
I1l ~
•I
92
STRUCTURES WITH SINGLE DEGREE OF FREEDOM (SDOF)
other hand, it leads to complicated contour integrations for the transient problem.
I
5.1.
~;'
f~
(i:!
:11
-:'"
:,1
li
I I
5.1.2.
93
Dirac unit impulse b(t) such that
·1
~)<
DETERMINISTIC TRANSFER RELATIONS
b(t)
Time Domain Solution
-+ 00
b(t) = 0
Parallel to the frequency domain solution, in the time domain solution, we consider an excitation x(t) as a superposition of impulses x(r) dr such as the shaded strip in Figure S.9a. We first obtain the response of the structure due to a
as t ...... 0 for all other
t
(5.21)
f~«> b(t)dt = 1
,j
·1
'I
I ,I
r\
I
jl
II
I I I I I I I!!lR .
and then determine the total response by a superposition integral, known as the Duhamel's integral. Hence the total response
xlt)
y(t) =
r-T--1
~dT
f=
-00
h(t - r)x(r) dr
(5.22)
In applying the time domain solution, we first obtain the impulse response function h(t) such as the one shown in Figure 5.9c with a time shift of r and then use it along with the given excitation x(r) in Equation (5.22).
(al .0;'(1)
1>(1 -
T)
EXAMPLE 5.5. Obtain the unit impulse displacement response function lI(t) for the SDOF structural model shown in Figure 5.1. Applying the principle of impulse and momentum to the SDOF system we can write
mdv
r-T---i
:=:
dt,
dv = f(t) dt/m
(5.23)
We recaIl that in structural dynamics, the response of a SDOF system to an initial velocity Vo is y(t)
(Il)
y(t)
= voe-{"'" wn~ sin(wn .J]=12t)
(5.24)
e
for a lightly dampled structure with ~ 1.0 and for t ;?: O. For t < 0, y(t) = O. If we use the differential velocity dv given by Equation (5.23) as an initial v~locity in place of Vo and divide the right-hand side of Equation (5.24) by the magnitude of the impulse !(t)"dt, we obtain the unit impulse response h(t) Ie)
FIG. 6.9. A excitation .x{t) subdivided into impulses .x{t} d", in (a), a unit impUlse excitation x'(t} = 6(t - r) in (b), and the corresponding response y'(t) = hIt - r) in (e).
e
=
e-(·...f
mw"J"f=Ti sin(wn,Ji" ~21)
(5.25)
for ~ 1.0 and t;?: O. For t < 0, h(t) = O. The general appearance of this' impulse response function is shown in Figure 5.9c with a nonzero r. Note that in Figure 5.9c, y'(t) h(t - r).
I
94
I I I I I I
I
EXAMPLE 5.6. Obtain the unit impulse velocity response for the SDOF system of Example 5.1. Let us again use the impulse and momentum approach and first obtain the free vibration solution due to an initial velocity. To do this we sel x(t) == 0 in the equation of motion, Equation (5.9), to get
v + fJv == 0 . v(t) = Ae-/lt
Using the initial condition that v =
1'0
at t
I,
!'I
It
l 'Ir ~~.
[
(5.27)
i
l ~
:t
t"hM
h(t)
+ fJh(t) = .:5(t)
(5.32)
where the excitation acceleration x(t) is specified as the Dirac delta function .:5(t). To solve h(t) from Equation (5.32) we observe that for t > O,o(t) = 0 and then h(t)
voe-/lt
+ fJh(t)
=0
To determine the integration constant A we integrate Equation (5.32)
The impulse and momentum relation gives rtlx(t) dt
= I'll dv;
f
dl) = x(t) dt
Using dv in plaL'"C of VII in Equation (5.28) and dividing by the magnitude of the impulse mx(t, dt, we obtain the impulse response for the velocity as h(t) =
1
e- Pr
= o(t) 1
=-
.:5(t)
I'll
(5.30)
as the corresponding acceleration excitation. Therefore, when the acceleration is specified as the Dirac delta function .:5(t) without the factor 1/1'11, the corresponding velocity impulse response function, Equation (5.29), becomes h(t) = e-(l,
h(t) dt
+
f+<
(5.34)
_r.Ph(t) cIt = I
Since JJ(t) is bounded and is 0 for t < 0, it follows that as integral in Equation (5.34) vanishes. Consequently,
1;
-+
0, the scwnd
lim h( + s) = h(O)
(5.35)
£~o
Using Equation (5.35) in Equation (5.33) gives A. = I and hence h(t)
= e-/lt
(5.36)
which agrees with Equation (5.3\).
·and we have x(t)
+< _<
(5.29)
Note that this velocity impulse response function h(t) corresponds to a Dirac force excitation mx(t)
(5.33)
(5.28)
(5.31)
This velocity impulse response can be obtained by an alternative mathe matical method as follows without employing tile principle of impulse and
\' Ir
x(t)
we obtain, by definition, the following ordinary differential equation for the velocity impulse response
= 0 gives
~.
~.
v + fJv =
:. h(t) = Ae-(lt v(t)
95
DETERMINISTIC TRANSFER RELATIONS
momentum. Starting from the equation of motion, Equation (5.9),
(5.26)
which has the following general solution with constant A:
I'll
I I I[
5.1.
STRUCTURES WITH SINGLE DEGREE OF FREEDOM (SDOFI
EXAMPLE 5.7. Determine the transient velocity response of the SDOF system of Example 5.1 when the system is subject to an acceleratiQn excitation x(t), as shown in Figure 5.7a. Use the time domain solution. Substituting h(t) from Equation (5.36) and .the acceleration excitation x(t) displayed by Figure 5.7a into Equation (5.22) we obtain the total velocity response
v(t) =
f' e-/l(t -ria dr = ~ (1 - e-/l~
r
{
Jo
e-/l(I-r)a
forO < t < T
fJ
dr
= ~ {(I
-
e-/l t -
This solution is shown in Figure S.7b.
[I -
e-/l(t-T)]}
for t > T
(5.37)
__
Eil
, -!.>;' :':'i.'"·.;;~~;K'i1rjy,~7Y.jb~>~1~\~~~.(:/~(;;i
II 96
I J
!I • ~
I.
STRUCTURES WITH SINGLE DEGREE Of FREEDOM {SDOF}
5.2.
EXAMPLE 5.8. Solve the steady-state problem of Example 5.3 by the alternate time domain approach. Note that for the steady-state acceleration excitation x(t) = a cos wot, which starts from a very long time in the pas~ the lower limits in the Duhamel's superposition integral Equation (5.22) must be - 00 instead of the usual zero. Thus v(t) =
t
foo
e-P1'.-tla
cos Wot'
1 ':::
= 2n
II
I I I I I I I I I.
I
~I I' L
,
rm>
f~", efJ
= ae- PI [ {J2
cos Wot'
eP<
2
+ Wo
dt'
-'"
H ((}) le
iw,
(5.41 )
dw
f~", h(t)e-
i ""
dt
(5.42)
dt'
({J cos Wo!
+ Wo sin Wo!)
J' -00
(5.3!!)
= p cos(mot - 1/»
EXAMPLE 5.9. Show that for the SDOF system of Example 5.1 with accelera tion excitation x(t) and velocity response v, the complex frequency response H(w) given by Equation (5.10) is the Fourier transform of the impulse response function h(t) given by Equation (5.31). From Equation (5.31),
which agrees with Equation (5.6).
5.1.3.
J
which is recognized as the Fourier inverse transform. Consequently, we also have H(w) =
t
97
He:.:.;
I
= ae- P '
RANDOM EXCITATION AND RESPONSE
h(t) = {e-PI,
0, ..
Time Domain versus Frequency Domain Solutions
t;;.>!:O t
<0
Its Fourier transform is The previous examples show that for the simple structural model of Figure 5.2 we can solve the transient and the steady-state problem using either the time domain or the frequency domain solution. We observe, however, that when the excitation is transient, it is more convenient to use the time domain approach. On the other hand, when we have a steady-state excitation, the frequency domain approach is simpler to apply. In these two parallel approaches, the basic idea is the superposition of responses to unit excitations. In the time domain solution, the superposition is based on the impulse response function h(t), whereas in the frequency domain solution the basic building block is the complex frequency response function H(w). These two functions represent the same dynamic characteristics of the structural model in different forms. In fact they are related by the Fourier trans form as shown in the following. Let us consider the special case where the total excitation consists of a single Dirac delta function x(t) = 6(t). Using Equation (5.6) we find that
foo
y(t) = h(t) = 1 X(w)H(w)eicut dw 21t _'"
(5.39)
where X(w) is the Fourier transform of the excitation x{t). For this special case, with x(t) = o(t), we have . X(W) =
f:",
x(t)e-1a>.
at =
I
(5.40)
f-'"'"
h(t)e-I",t dt
=
f"'e-(p + i",)t dt
Jo
= _1_._ (J
+ IW
which is the complex frequency response fUllction H(w) given by Equations (5.10).
5.2.
RANDOM EXCITATION AND RESPONSE
The preceding presentation of excitation and response of linear vibratory systems with SDOF is entirely deterministic. In the following we continue to treat the vibratory systems as deterministic, but consider the input excitation to be a stationary random process. The output response is in general a non stationary random process. For simplicity, however, we shall' first consider stationary random response. Again, the problem will be presented along two parallel routes, the time domain and the frequency domain approach, respectively. From here on all statistics discussed are ensemble statistics unless otherwise specified.
5.2.1.
Time Domain Approach
First consider the time domain method on the basis of the Duhamel's integral. The response y(t) of a linear SDOF system with impulse response h(t) to the
I 98
I I I I I I I I I ~
STRUCTURES WITH SINGLE DEGREE OF FREE[:)OM (SDOF)
excitation x(t) is given by
~I r~ ~.
RANDOM EXCITATION AND RESPONSE
f~oo h(r)x(r -
99
Rx(~1
Sx(w)
yet) =
(5.43)
r}dr
So ,. w
This is a different form, but equivalent to the previously derived integral, and is used for the convenience of subsequent development. We consider both yet) and x(t) as stationary random processes. Taking expectation on both sides of this integral gives E[y(r)]
=
f~,.., h(t)E[x(r
(a)
t)] dt
(h)
FIG. 6.10. Stati()nary white excitation acceleration xII) with II unifnrm power spcctrum dcnsity S.(W) = So in (a) and the corresponding autocorrelation R.(t) = 21t,S(t) in (/.).
Let my and mx denote the mean value of y(t) and x(r), respectively. Then from this equation we obtain For t > 0, integrating with respect to
my ,;", m.,
f:oo h('t) dt
(5.44)
RrC t ) =
which is the simplest statistical relation between input stationary excitation and output stationat.,y response. Next, we consider the second-order autocorrelation Rx(t) for the excitation and Ry(t) for the response R,(t)
= E[y(t)y(t + t)) =
f:., f:,.., h(OI)h(O~)E[x(t
= f:oo
f'",
1'"
(}2
and then Ot gives
2nSoe-PlJte-P(t+O"dO,
= 2nSoe-P'
'" f o
e- 2p8, dOl
S
-p,
=~
(5.46)
(J
The mean square response is - Odx(t
+t
- O2 )) dOl d0 2
E[y2(t)] = Ry(O) =
h(Odh(02)R x(t + 01
-
O2 ) dOl d0 2
(5.45)
~So
(5.47)
(J
As an even function of the time lag t, the stationary response autocorrelation Ry(t) is plotted in Figure 5.11.
EXAMPLE 5.10. For the SDOF structural model of Example 5.1, determine the stationary response mean m, and autocorrelation function Ry(t) with y(t) v(t), when the excitation acceleration x(t) is assumed to be a stationary white noise with zero mean. (See Figure 5.10.) Since RJt) -+ 0 as t -+ 00, the mean value of x(r), mx = O. The stationary mean response
~ ~
5.2.
m,
= mx f~oo h(t) dt = 0
Substituting the impulse response h(t) from Equation (5.31) into Equation (5.45) gives the stationary response autocorrelation R,(t)
=
I" I"
Ry(rI
e-/lO'e-/lO'2nSoo(t + 0 1
-
O2 ) dOl d0 2
·:t
_ ,
...
T
FIG. 5.11. Stationary response autocorrelation R,kt) of the SDOF system in E.~amplc 5.1 with v = y,
I
92
I
~3 I I I
STRUCTURES WITH SINGLE DEGREE OF FREEDOM (SDOF)
~ !l~:
other hand, it leads to complicated contour integrations for the transient problem.
{~
5.1.2.
~~
Si
:1j .:~
'·:1
ji'I "·1 'I
I
b(t) .... co b(t)
Parallel to the frequency domain solution, in the time domain solution, we consider an excitation x(t) as a superposition of impulses x(t) dt such as the shaded strip in Figure 5.9a. We first obtain the response of the structure due to a
I I I I
=0
f~.., J(t) dt =
y(t)
as t .... 0 for all other t
(5.21)
1
=
f=-",
h(t - ,)x(t)dt
(5.22)
In applying the time domain solution, we first obtain the impulse response function h(t) such as the one shown .in Figure 5.9c with a time shift of t and then use it along with the given excitation x(t) in Equation (5.22). (a)
II
I
93
and then determine the total response by a superposition integral, known as the Duhamel's integraL Hence the total response
x(t)
I
I
I
DETERMINISTIC TRANSFER RELATIONS
Dirac unit impulse J(t) such that
Time Domain Solution
!I
;1
5.1.
II(t -
r)
EXAMPLE 5.5. Obtain the unit impulse displacement response function Ir(t) for the SDOF structural model shown in Figure 5.1. Applying the principle of impulse and momentum to the SOOF system we can write
m dv
t-
----...j
(11)
= f(l) dl,
dv = f(t) dt/m
(5.23)
We recall that in structural dynamics, the response of a SDOF system to an initial velocity Vo is y(t) =
sin(wn.J!=et)
(5.24)
y(tl
for a lightly dampled structure with ~ ~ 1.0 and for t ;?: O. For t < 0, y(t} = O. Ifwe use the differential velocity dv given by Equation (5.23) as an initial velocity in place of Vo and divide the right-hand side ofEquation (5.24) by the magnitude of the impulse f(trdt, we obtain the unit impulse response
t- ---1 r
(e)
FIG. 5.9. Aexcitation ;>:(1) subdivided into impulses x('r) dt in ((1), a unit impulse excitation x'(/) = 6(1 - t) in (b). and the corresponding response i(l) h(t - ,) in Ie).
h(t)
= mw.~ e-~"·l sin(wn.JI - ~2t)
(5.25)
for ~ ~ 1.0 and t ;?: O. For t < 0, h(t) = O. The general appearance of this impulse response function is shown in Figure 5.9c with a nonzero t. Note that in Figure 5.9c, y'(t) = h(l - t).
I 94
I II f
II
I
I
I I I I
I tl 11i: I
I Jim .
5.1.
STRUCTURES WITH SINGLE DEGREE OF FREEDOM (SOOf)
EXAMPLE 5.6. Obtain the unit impulse velocity response for the SDOF system of Example 5.1. Let us again use the impulse and momentum approach and first obtain the free vibration solution due to an initial velocity. To do this we set x(t) ;; 0 in the equation of motion, Equation (5.9), to get i;
+ pv = 0
.vet) = Ae-!1' Using the initial condition that v = V(I)
['0
at
I =
(5.27)
==
= voe-/lt
+ flv
= x(t)
we obtain, by definition, the following ordinary differential equation for the velocity impulse response h(t)
+ Ph(l) =
(5.32)
b(t)
where the excitation acceleration x(t) is specified as the Dirac delta function 0(1). To solve h(t) from Equation (5.32) we observe that for t > O,o(t) = and then
°
h(t) + Ph(t) = 0 :. h(t) == Ae- Ilt
(5.33)
(5.28)
To determine the integration constant A we integrate Equation (5.32)
+£ f+< f _< h(t) dt + - t fJh(t) tit =
d/i = x(l)dt
m d/i;
Using dv in place of /il) in Equation (5.2S) amI dividing by the magnitude of the impulse mx(t) dt, we obtain the impulse response for the velocity as h(t)
i;
0 gives
The impulse and momentum relation gives mX(I) dl
momentum. Starting from the equation of motion, Equation (5.9),
(5.26)
which has the following general solution with constant A:
95
DETERMINISTIC TRANSFER RELATIONS
= -1 e-!1t m
mx(!) = 0(1)
1
m
lim h( + e) = h(O) = 1
(5.30)
as the corresponding acceleration excitation. Therefore, when the acceleration is specified as the Dirac delta function bet) without the factor 11m, the corresponding velocity impulse response function, Equation (5.29), becomes
Using Equation (5.35) in Equation (5.33) gives A = I and hence
(5.36)
h(t) = e-/Jt
EXAMPLE 5.7. Determine the transient velocity response of the SDOF system of Example 5.1 when the system is subject to an acceleration excitation x(t), as shown in Figure 5.7a. Use the time domain solution. Substituting h(r) from Equation (5.36) and the acceleration excitation x(£) displayed by Figure 5.7a into Equation (5.22) we obtain the total velocity response
(5.31 )
This velocity impulse response can be obtained by an afternative mathe matical method as follows without employing the principle of impulse and
' {ir
a e-fl(l-tladt = -(I - e- p,)
v(t} = h(t) = e -lit
(5.35)
<-0
which agrees with Equation (5.31).
. and we have x(t) = - oCt)
Since lI(t) is bounded and is 0 for I < 0, it follows that as " --> 0, the second integral in Equation (5.34) vanishes. Consequently,
(5.29)
Note that this velocity impulse response function h(t) corresponds to a Dirac force excitation
(5.34)
I
P T e-/l"-f)a dt = ~ {(t -
forO < t < T
0
Jo
P
e-(ll -
This solution is shown in Figure 5,Jb.
[I - e-!1V-TlJ}.
.
for t > T
(5.37)
II 96
~ '~ ~
I I
I.
STRUCTURES WITH SINGLE DEGREE OF FREEDOM (SDOF)
5.2.
EXAMPLE 5.8. Solve the steady-state problem of Example 5.3 by the alternate time domain approach. Note that for the steady-state acceleration excitation x(t) = a cos wot, which starts from a very long time in the past, the lower limits in the Duhamel's superposition integral Equation (5.22) must be - co instead of the usual zero. Thus v(t) =
f~", e-fl(t.-')a cos wor dr
RANDOM EXCITATION AND RESPONSE
-.~
_............ H ~-
j
1 '" il\r) = 2n :x:, H (wle""" dw
I I I I I I I I II I ..
"
i
= ae-
= ae- fl '
H(w) =
f~ro eP' cos wor dr 2 (/1 cos Wor + Wo sin wor) [ ~2 eP' + Wo
= I' cos(mot
J'-'"
(M
(5.38)
r:ro
h(e)e-
iwt
de
(5.42)
EXAMPLE 5.9. Show that for the SDOF system df Example 5.1 with accelera tion excitation x(t) and velocity response v, the complex frequency response H(w) given by Equation (5.10) is the Fourier transform of the impUlse response function h(t) given by Equation (5.31). . From Equation (5.31),
which agrees with Equation (5.6). 5.1.3.
\5.41)
which is recognized as the Fourier inverse transform. Consequently, we also have
I
flt
97
h(t) = {e0,
Pt ,
t;;?;O
1<0
Time Domain versus Frequencv Domain Solutions Its Fourier transform is
The previous examples show that for the simple structural model of Figure 5.2 we can solve the transient and the steady-state problem using either the time domain or the frequency domain solution. We observe, however, that when the excitation is transient, it is more convenient to use the time domain approach. On the other hand, when we have a steady-state excitation, the frequency .domain approach is simpler to apply. In these two parallel approaches, the basic idea is the superposition of responses to unit excitations. In the time domain solution, the superposition is based on the impulse response function h(t), whereas in the frequency domain solution the basic building block is the complex frequency response function H(w). These two functions represent the same dynamic characteristics of the structural model in different forms. In fact they are related by the Fourier trans. form as shown in the following. Let us consider the special case where the total excitation consists of a single Dirac delta function x(t) = li(t). Using Equation (5.6) we find that
y(t) = h(t) = 2n 1
f'"
- a:>
X(w)H(W)e'OI'dw
(5.39)
where X(w) is the Fourier transform of the excitation x(t). For this special . case, with x(t) = li(t), we have X(w) =
J~., x(t)e- r"" at =
1
(5.40)
f
oo
-
h(t)e-i"" dt = f""e-(Il+ iOJ)' dt = _1_._ ~ + /w
Jo
which is the complex frequency response function H(w) given by Equations (5.10).
5.2.
RANDOM EXCITATION AND RESPONSE
The preceding presentation of excitation and response of linear vibratory systems with SOOF is entirely deterministic. In the following we continue to treat the vibratory systems as deterministic, but consider the input excitation to be a stationary random process. The output response is in general a non stationary random process. For simplicity, however, we shall first consider stationary random response. Again, the problem will be presented along two paranel routes, the time domain and the frequency domain approach, respectively. From here on all statistics discussed are ensemble statistics unless otherwise specified. 5.2.1.
Time Domain Approach
First consider the lime domain method on the basis of the Duhamel's integra!. The response y(e) of a linear SDOF system with impulse response h(t) to the
II 98
I 'il
STRUCTURES WITH SINGLE DEGREE OF FREEDOM (SDOF)
excitation x(t) is given by
I I I I I I ~ rl
yet)
~i
,_.
R,.(TI
f:",
(S.43)
h(.)x(t - .)dr
So )
This is a different form, but equivalent to the previously der-ived integral, and
is used for the convenience of subsequent development.
We consider both y(t) and x(t) as stationary random processes. Taking
expectation on both sides of this integral !ives E[y(t)) =
f~.., h(.)E[x(t -
(al
.)] d.
T
(b)
FIG. 6.10. Stationary white excilntion accderation .«Il with ;1 uniform power spct:lfllm density S.(wl = Su in (ul and the corresponding autocorrelation R.(rl = 2nil(r) in (/.).
Let my and mx denote the mean value of y(t) and x(t), respectively. Then from this equation we obtain my
~ mx
f:",
For r > 0, integrating with respect to O2 and then (J, gives (S.44)
h(.)dr
Ry(t) =
which is the simplest statistical relation between input stationary excitation and output stationar~ response. Next, we consider the second-order autocorrelation Rx(r) ror the excitation and R y (') for the response Ry(r)
= E[y(t)y(t +.))
=
=
f:", f:.., h(Odh(O~)E[x(t
f'", f~
my = mx
rlX)
Jo
2n:Soe-P6Ie-/l(t+Otl
= 2n:S oe- Pr
i
'"
dO
•
- OI)X(t+ r- O2)] dOl d0 2
E[y2(t)] = Ry(O) = ,,!So
p
h(Odh(02)R x (r
f:",
s
-p,
e- 2PO, dO. = ~ o {J
(5.46)
The mean square response is
+ O.
- (2)dO I d0 2
(S.4S)
(S.47)
As an even function of the time lag r, the stationary response autocorrelation Ry(r) is plotted in Figure 5.11.
EXAMPLE 5.1 O. For the SDOF structural model of Example 5.1, determine the stationary response mean my and autocorrelation function Ry(t) with y(l) = v(t), when the excitation acceleration x(t) is assumed to be tI stationary white noise with zero mean. (See Figure 5.10.) Since R",(r) --> 0 as r -> 00, the mean value of x(t), mx = O. The stationary mean response
~ ~
=
99
RANDOM EXCITATION AND RESPONSE
s.(w)
~
I.
5.2.
Ry(T)
h{t)dt = 0
. Substituting the impulse response h(t) from Equation (5.31) into Equation (S.4S) gives the stationary response autocorrelatioR Ry(') =
L" L'"
e-po'e-Po'2nSo(j(t
+ 01 -
02)dO I d0 2
FIG. 5.11. Stationary response autocorrelation Ry(~l of the SDaF system in Example 5.1 with
v = y. ,;.
I
I I I I I I I I I I I I I \ f'IIj
100
5.2.
STRUCTURES WITH SINGLE DEGREE OF FREEDOM (SDOF)
.
RANDOM exCITATION AND RESPONSE
I
EXAMPLE 5.11. For the building and, vehicle model shown in Figure 5.12 with mass m, spring constant k, and damping coefficient c, determine the frequency response function H(w) and the impulse response h(t) of the relative displacement (Xl - xo) corresponding to the acceleration excitation xo(t). For both models the equation of motion is • ji
+ 2';wn y +'w;y =
Xo
101
1f1(w)j2
(5.48)
where
Y=
XI -
Xo
$m =
Wn
= relative displacement
natural circular frequency
FIG. 5.13. Squared modulus of the complex frequency response function 1I(w) given by Equation (5.50).
..; = c/2wn m = damping ratio
Substituting xo(t)
=
i
e "" and y(t) = H(w)i"" into the equation of motion gives
To determine the impulse response function h(t) of the relative displacement y(t) corresponding to the acceleration excitation >::o(t), we note that according to
H(w)
W..2
w 2 + i2';W W
(5.48) and by definition we have
Ii + 2';wnh + w;h
n
with its modulus ~hown in Figure 5.13.
IH(w)1 2 ="
Equa~ion
(5.49)
For t > 0, (i(t) ,.,
. -,
,
(5.50)
,
Note that for a smaller damping ratio .;, the peaks in Figure 5.13 will be higher and in the limit for the idealized undamped case'; = 0, the peaks go to infinity atw = Wn'
= - 6(t)
0 and the free vibration solution is h(t) = cle-~"'"' sin Wdt
+ c2e-~"" cos Wdt
where Wd = wn~ is the damped natural circular frequency. As f -+ 0, the equation of motion requires that h(t) -+ - b(t), h(t) .... - I, and hIt) -+ O. Hence CI = l/Wd, C2 '= 0 and e-~"'·'
h(t) =
"'o(t}
H 1-"
{
"'I{t)
m
-
'" IU}
sin Wdt,
0,
t>O (5.51)
t
This impulse response function along with its special behavior as t in Figure 5.14.
I k
I I I
%
-+
0 is shown
EXAMPLE 5.12. For the problem in Example5.lI, determine.the frequency response HI(w) and the impulse response hl(t) for the relative velocity Y(t) = XI(t) - xo(t) corresponding to the same input excitation xo(t). By definition, when xo(t) = e1"". y(t) = H(w)e " ·', then
I
I
.,
'i.
H
Y(t)
= HI (W)e/WI = iwH(w)eiwr
:;;o
W
W FIG. 5.12.
A building model (a) and a vehicle model (b).
:. HI(w) = jwH(w) =
+ i2~wnw
(5.52)
I 102
6.2.
STRUCTURES WITH SINGLE DEGREE OF FREEDOM (SDOF)
I
hm
-
21t~
- (oj
II
Let wdt Then
I
il
11(/)
::=
X
then t =
Jof'" e-2~'.nl sin
21tSo E(y2) = r
,
0
21t~o Wd
[1/(4 + 4~2~J;)l( (tJd
(tJd )
~(On
1tSo
i
I I
I
I I I I I I
w; = wJ + elo;.
f"" e - 2{'"nxl... sin 2 x dx
(Od
hW
(5.54)
= 2~w~
-II-I- -
The response autocorrelation can be determined by using Etillation (5.45~. For r > 0,0 1 > 0, O2 > 0, and R,.,(r) = 21tS(l(~(r).
hW
R,.( r) = ("'. 2nSuil(O. lll( T
J.
f
.l:,'
211So FIG. 5,14. Impulse response function hell given by Equation (5.51) with special behavior as I
....
0,
t!-~tIJU()1
o
-
t > 0,
EXAMPLE 5.13.
+ sin Wdr.
E(y2) = R1(0) =
=
f~
f~C() f~C() h(O.)h(02)R;;(OI -
f:",
(2)dO I d0 2
h(Odh(02)21tSOO(OI - O2 ) dO. d0 2
= 21tSo f~
dOl
COSWd'C
cos wd(r
+ 20d] dO.
IX. e-2" ••o'cos2wdOl dOl 0
t'" e-2~w.6.
sin 2Wd IJ l
dOl)
Since
For the problem in Example 5.11, determine the mean square
E(y2) and autocorrelation R,(r) of the relative displacement y through a time
domain approach when the excitation xu(c) is an idealized white noise with autocorrelation Ri(r) = 21tSoo(r).
sin Wd
(tJd
2~(tJn
COd
(5.53)
c
t!-~(dnti+Ol.
r = 1tSo -2-e-~(··' (cos - -Wd- -
e-~"'nl (cos Wd t - ~1'i Sinwdl). h.(t) = 0,
sin
Jo
When xo(t) = oCt), yet) = h(t),
h(t) =
+ (1.)<10 1
= 1tS~ e-~(••' f'.(~-2~(OnO, [cos Wdr COd
Y(t) = h.(t) =
103
2 Wd tdc
and dc = dx/wd' Note also that
X/Wd
II
~
RANDOM EXCITATION AND RESPONSE
/
"i o
tl
e- U ' cos bx dx = -'--, a2 + b2
L" e-
and
u "
b
sin bx lIx =
:. R (r) = 11So e-~"'.' (cos Wd r _ 2ewn cos Wd~. , w~ 2ewn 4w~ _ 1tSo
e
-~W.f (~) 2" 2
(
-
Wd2
=
1tSoe-("'''' ( 3 cos Wd r 2ewn
.. w n
"
'I _
;,r+ /;2
+ 2Wd sin Wd!) 4w~
~2 cos Wdr + ~ sin wir)
~.) + ".-----;:Sin (/Jd r
,,1 -
e
(5.55)
l~ •
:~ I
t',~
\1
I
1;
II
~~
.';
.j
~
I I
II
I
'm
I
I
I
II I
,I
Iit, I.
104
terms gives
From Examples 5.10 and 5.13, we see that when an idealized white noise station ary random excitation with zero autocorrelation (except at r = 0) is applied to a lightly damped SDOF structural model, the response autocorrelation function becomes nonzero for the time lag r from the origin r = 0 to an upper bound that depends on the damping ratio ~ and tbe natural frequency Wn' (See Figure 5.15.) A smaller product of corresponds to a higher upper bound t. Further more, the response mean square, unlike that of the excitation, is no longer unbounded.
Sy(w) =
eWn
5.2.2.
f'"
=
f'"_'" •.Ry(r)e-''''' . dr
Sy(w)
' ,.
+ 01 -
02)e-i(rH,-O,1 d(r
+ 01
-
O2 )
= H(w)H( -
w)S,,(w)
=
IH(w)fSAw)
(5.57)
EXAMPLE 5.14. For the problem of Example 5.1, determine the power spectral density Sy(w) and the mean square E[ y2(l)] of the stationary velocity response y(t) v(t), when the excitation acceleration x(t) is assumed to be a stationary .white noise with a uniform spectral density of So. Substituting IH(w)1 2 from Equation (S.ll) and SAw) = So into Equation (5.57) gives
(5.56)
Substituting Ry(r) from Equation (S.4S) into Equation (S.56) and rearranging
Sy(w)
= IH(wWSo = /12
So
+w2
(5.58)
The mean square is ~ 2S••;
<
"SO
. S'_'"" Sy(w)dw = S'_'"" /12SO+ w2 dw =. T1tSo
.
2S",~ e-''''n'
...... .......
...... .......
E[y2(t)] = R,(O) =
...... ~.
~ ,
,
,
\ ~
_'" h(02)e- I,.O'd0 2
which is the transfer relation between power spectral density functions of the stationary random excitation x(t) and response y(t). The simplicity of this algebraic equation as compared to the integral equa tion in terms of autocorrelations Ry(t) and R..,(t) makes the frequency approach more attractive than the time domain approach. Furthermore, in the time domain approach, we need to know the complete impulse response function h(l) whereas in the frequency domain approach, we need to know only the modulus of the complex function H(w) and not its phase.
I'
~i
foc>
The last of the three integrals is clearly S..,(w) by definition and according to Equation (5.42) the first two are complex frequency response functions H(w) and H( - w), respectively. Therefore we obtain
In the time domain approach, Section 5.2.1. we derived the transfer relation between autocorrelation functions of the stationary random excitation and response for linear SDOF structural models charact.erized by their impulse response functions. In fact, such a transfer relation applies to any linear struc tural system with a given impulse response function, as we shall see in subseq lIent chapters. Parallel to the time domain approach. we now consider the alternate fre quency domain approach. We develophhe transfer relation between power spectral density functions of the stationary random excitation and response for linear structural systems, characterizetl by their complex frequency response functions as follo~s: By definition, the response power spectral density is Sy(w)
a.·
f _'" h(O,)e-I,.o. dOl x -1 RAr 21t _'"
Frequency Domain Approach
1
105
5.2. RANDOM EXCITATION AND RESPONSE
STRUCTURES WITH SINGLE DEGREE OF FREEDOM (SDOF)
J,.
FIG. 5.15. Response autocorrelation R(tl of u SDOF system to a wbile noise random excitation.
(5.59)
which agrees with Equation (S.47). The simple transfer relation, Equation (5.58), is displayed in Figures 5.16a, S.l6b, and S.16c for the excitation spectral density S,,(w), the transfer function IH(w)12, and the response spectral density Sy(w), respectively. Two dotted lines at frequencies - Wo and + Wo help to indicate that if the excitation process is not white but has a uniform spectral density limited to the range - (1)0 < (t) < IOu.
I~ I
I~I I
106
STRUCTURES WITH SINGLE DEGREE OF FREEDOM (SDOF)
5.2.
E~
'1\
.so
"
t
)0
f~", IH«(1))1 2 dw = (B~/Ao)A2 + Bi .
.
~
1
(~.{>ll
AlA!
(5.62)
where
:r--?Jo
;coB 1 + Bo H(co) = _ co2A2 + iwA[
4~\.(Cd)
'f
:":=+ __ ... (.,
-wo
(5.63)
Examples 5.14 and 5.15 illustrate first that the SDOF structural model
(h)
__v
+ Ao
J:ft
'M
~
2';;W n
)
(al
'II I
fl
J
-
which agrees with Equation (5.54). This last integration is not simple. The result has been worked out on the basis of the following general formula:*
W
I/{(wlj<
IIi
I
I
I
I ;, I I I ~I ~ .
71'S. S ·lw) ,/(() = '-;~3
x
\,2, =: •
I
.1
107
The mean square is
S,.(•• J
rI
J
RANDOM EXCITATION AND RESPONSE
+'·0 (e)
FIG. 5.16. Geometric relation among the cxcitatiqn sfl~"(;tral density S,("'I in '(til, the transfer function III(wl1 2 in (hi, and the response spectral denslty S,(ml in (c),
then the only difference in the response sl*Ctral density Sy(w) is the secondary portion outside of the frequency range, as-shown by the dotted tails in Figure 5.16c. This illustrates the validity of assuming a white 1)0ise excitation as an approximation to a realistic process with il spectral density of finite frequency range. . EXAMPLE 5.15. For the problem in Example 5.11, determine the power spectral density S,(m) and the mean square B( y2) ofthe relative displacement y when the excitation xo(t) is assumed to be an idealized white noise process with spectral density Si,,(m) = So. Using the transfer function IH(m)l l given by Equation (5.50) we have the response power spectral density.
5.2.3.
A Direct Statistical Solution
We have already shown that for a linear SDOF system subjected to a stationary random excitation, the stationary response mean value can be determined directly by taking mean values on both sides of the Duhamel superposition integral to obtain the solution, Equation (5.44). Consider the problem in Example 5.11 again with the governing equation of motion
y + 2emn y + m~y =
- xo(t)
(5.64)
where xo(t) is the acceleration excitation assumed to be a white noise stationary process with zero mean and y(t) is the response displacement. Applying Equation (5.44) to the mean value of the stationary response y(t)
i
~ ;
'I
::1 I'
~I II
fill'.
~
S,(m)
= IH(w)1
2
S...(w)
= (0);
_
So (J)l}2
+ 4e2(J):w'
(5.60)
·See H. M. James. N. B. Nichols. and R. S. Phillips, "TIleory of Servomechanisms." MIT Radimi()fI Lulmmior.l' Sed,<.'. Vol. 25. McGraw·llill. New York. 1947.• PI'. 333 3<.9.
I:I
108
STRUCTURES WITH SINGLE DEGREE OF fREEDOM (SDOfl
PROBLEMS :,~.,
Sio{W)
~~
::~\
~i; ;~l
'Ii,)
- a;
:;"
+ w~a: = -
(5.66)
E(xoY)
~:"
'H; ~~i
- a; and E(yj) = 0, we then have
,,;' ;}:
il~:;j".~:
109
Multiplying both sides of Equation (5.64) by v =
I
'"=
J
<'(j
""
'-.,.
E(vji
w
- E(xov)
But E(vy) = 0 and E(vy) = 0, so we have
(cI)
'.)
+ 2~wnvy + w;vy)
yand taking expectations gives
:~
:I ~' I ~11I I t
IH(",)12
2~wnO";
;~ !
E(xoY)
:;1
j;' !
::'\
+wu
-W II
~
~~
oS,,('.)
g ~
~ ~
l
l'
~'1
~
."
l
ill ;.1.
I"
;g. 1
(5.69)
"
(5.70)1 : J
2~w
2_
(5.71)
,a" - 2~wn
(e)
FIG. 6.17. Geometric relation among the excitation spectral density S...(w) = So in (a), the trans fer function IH(w)1 1 in (b). and the response spectral density S.(w) in (c).
Equation (5.70) agrees with Equation (5.54) as it should. Note that the variance
a; equals the mean square E[ y2] when E[ y] = o.
and v(t) = j(t) immediately gives the solution
E(y) = E(v)
~
=0
(5.65)
.
PROBLEMS
.~(
5.1. In addition to these mean value solutions, we show in the following that two important statistics, a, and a., can also be determined directly from the equa tion of motion, Equation (5.64). Multiplyina both sides of Equation (5.64) by y and taking expectations gives E(yji
+ 2~wnyj + w;y2)
F'
For the vehicle vibration problem of Section 4.1 in Chapter 4 with the equation of motion
'",:.
my + cj + ky
Since E( y) = E(v)
mx =
where = -
E[yxo(t)]
1
(I
!( 21tSo)
-Wn
;~
:1
(5.68)
-~(l)
' t . . l
:i
:!
=0
1tSo3 2 ____
~~
III i til
r)xo(t)] dr
h(t)E(xo(c - r)'>':o(1) £It
(iy -
.~
I
..
f~
(5.67)
Substituting Equation (5.68) and (5.69) into Equations (5.66) and (5.67), respec tively, gives the required·solution.
~
• ~
E(llXO) =
E(xov)
i,(o+) == - I, from Figure 5.14 we have
= f~ro h(r)E(xo(t
--~-----L----~~~-£----~--~~_w
(bl
I II I ,I ., t.
0 and
Furthermore, since 11(0)
=-
= 0 and for the stationary response ~, E( yy) = -
R.(O) =
f(t)
wot = F cos wot + F cos (dot + F cos 3 ._,--, 2
• ':::A~
.,
"" ~
~ -II
m
III I '.,i I i
110
STRUCTURES WITH SINGLE DEGREE OF FREEDOM (SDOF)
k~
let the damping coefficient
~
and the natural frequency
Ei
~ = Co C
~!
i:
ill
I I I I I
I I I I I ~
c_ = 2mwo
= 0.2;
Wo
Wo
be given by
In
=
H(w) = IH(w)le- i "
k
Let the excitation x(t) be a stationary random process with zero mean and a white spectrum So. (a) Determine the stationary mean square response E(y2) when'; = 0 by a time domain approach.
5.4. Set up the Duhamel's integral for the transcient displacement response lI(t) for a SDOF system and find the displacement I'(l) at t = 30 sec assuming the system is at rest at t = 0 and a dynamic load p(t) as shown in Figure 5.18 is applied. Also assume that ~ = 0, Wo = 10 rad/sec, and m = 5 kips/sec 2 in.
where
11
111
(b) Determine the stationary mean square response E[y2] when ~ = 0.1 by a time domain approach. (c) Determine the stationary mean square response E[ },2] when.; = 0.1 by a frequency domain approach.
(a) Show that the complex frequency response function
IH(w)! =
PROBLEMS
Jt> - (:oYJ -(2~ :J2;
-
~ = tan-I{2~ ~:/[t C~J2]}
I'(tl kips
(b) Using H(w) from 5.1 a and considering that cos we is the real part of ei
100
.........
20
",
I (sec)
5.2. For the vehicle vibration problem with the equation of motion
my + cy + ky
= -
mx
= J(t)
FIG. 5.18.
let the excitation f(t) be nonrandom and specified by
f(t) = F sin wot
5.5. A SDOF system has the following equation of motion
+ iF sin 2wot
mi.i
(a) Determine the steady state response y(t) when
Wo
=~
and
~
=
c. c
for
y(t)
p(t)
Find the complex frequency response function H v(w) and impulse response function h.(t) for the velocity response ri(t).
c = 0.1 = 2mwo ~.
(b) Determine the temporal mean square (y2(t)). (c) Plot the two-sided temporal spectral density W(w~)
+ cv + kv =
%I~
using
8W
= Wo
Hinge
I.
...1
"I
<)
%b)
(
Rigid and massless bar
5.3. Consider a SDOF system with natural frequency excitation x(t), and response y(t) governed by
.
Wo.
damping ratio ~.
c
~
k
~
7/T/T/T/T/T///T/7777T/T/7/77777777
ji
+ 2woej + wb = x(t)
FIG. 5.19.
I
112
I I I I I I I I I
STRUCTURES WITH SINGLE DEGREE OF FREEDOM (SDOF)
5.6. (a) Set up the equation of motion in terms of Z(f) for the structural model shown in Figure 5.19 by the method of virtual work. (b) Determine the natural frequency of damped oscillation Wo and the critical damping coefficient c•.
5.7. (a) Formulate the equation of motion for the structural model by the
I I '-" Ail .,
•
principle of virtual work. (b) Find the complex frequency response H,,(w) and impulse response hu(l) for the displacement v(t) of the SDOF system in Figure 5.20. PW
,
"'0
p(tl
jJllITI ~ l ' ---~v(x.tl
= concentrated poil1t mass
/ I-- Assumed deflection shape .p(x) I
v(x. tl
FIG. 5.22.
L
x
o
1;7//J;j///;}"ff////////h •
V(x. t)
= .pix! Z(t)
FIG. 5.20.
5.S. (a) Set up the equation of motion for the simple model by virtual work in terms of Z(t). Take the axial graVity force IV into consideration. (b) Find the complex frequency response Hm(w) and impulse response hm(t) for the rotational moment of the coil spring shown in Figure 5.21.
w
zm
Point massm
x
Elastic
coil spring
'l
FIG. 5.21.
moment rotation
k=-
113
5.9. (a) By the principle of virtual displacement, formulate the equation of motion for the uniform beum treated upproximately as u SDOF shown in Figure 5.22. (b) Find the complex frequency response H,,(w) for the displacement v(x, tl. (See Figure 5.22.)
m.E! _ _
P(I) ~"-------:1i
I I
PROBLEMS
r.
(Sin
7)z(t)
..
,: '.#1
;1
; ~.
,
~: ~
6.1.
!!
11
115
Y2
I
:~
TWO-OEGREE-OF-FREEOOM SYSTEMS (TOOFl
fzW
CHAPTER
6
m
fb1ibll k
~
I I
I
I
I· !r
Ii;
IIIi:
~i
t ~: ).:
s.~
~~
l.
~
I I I I I :\
RESPONSE OF LINEAR MUlTIDEGREE-OF-FREEDOM (MDOF) SYSTEMS
A rDOF siructural model wilh no damping,
First consider the case ofa normal mode vibration where /1(1) = Oand /2(1) = 0, and assume the motion of the two masses to be harmonic with unknown ampli tudes AI and A l • respectively. at the same unknown frequency III: Yt(r)
= AI sinwt;
Yl(t) = A2 sinwt
(6.3)
SubstilUtingEquation (6.3) into Equations (6.1) and (6.2) with the no-load condition leads to the following frequency equation: 4k
4
2
3k 2
w --w +-=0 m m2
(6.4)
Solving Equation (6.4) for w 2 gives the two natural frequencies
6.1.
TWO-DEGREE-OF-FREEDOM SYSTEMS (TDOF) WI
All basic concepts and analysis methodology for the response of linear M DOF systems to random excitation' can be adequately brought out by a thorough treatment of the problem for a linear TDOF system. This is so simply because in the deterministic or nonrandom case the problem of the TDOF systems carries all the important features of the general MDOF system, and the addition of the random considerations does not alter this fact. The simplicity of the TDOF system is most helpful in understanding the random response problem. Following the TDOF system, linear MDOF systems with compact matrix notations will be presented.
6.1.1.
and
(02
= .j3kjm
(6.5)
and the corresponding natural mode shapes
~!.=
1
and
A21
AI! = _ I
All
\6.6.)
where the second index denotes the mode number.
Next we introduce the normal coordinates defined by
~: ~-:
114
= Jkji1i
ql
= AllYl + A ll Y2
= Yt
+ Y2
(6.7)
q2
= A 21 YI + A ll Y2 = Yt
- Y2
(6.8)
Deterministic Vibration
We consider the simple TDOF structural model shown in Figure 6.1. The equations of motion in terms of the displacemt:;nt YJ, excitations h(t); j = 1,2, the spring constant k, and mass m are
~
:,dllll'
FIG. 6.1.
Substituting Yl and Y2 from Equations (6.7) and (6.8) into Equations (6.1) and (6.2), with the specification of All = At2 = 1. we obtain the uncoupled and forced equations of motion
l't •
my I + 2kYI
- kh
= fl(t)
(6.1)
mh + 2kY2
- kYI
= un
(6.2)
iiI
+ wiqt = ~ (fl(t) +
f2(t)] = YI(t)
(6.9)
II 116
I I I I I I g: I
I
e·
i,
;,it
RESPONSE OF LINEAR MULTIDEGREE-OF-FREEDOM (MDOF) SYSTEMS
ii2
qt(l) =
q2(t) =
I l:\ I "3'~ I .
I
hj(t) =
==
1
!~
:
~,
~I §j ~,
J~~
\
fl !t
IIt i
I
.. f~. . f~
q2(t) =
Yt(t) = Hq.(t) Yz(t)
_(6.l2)
h2(8)Yl(t - 8) dO
for
£>0
for
1<0
1a "
iW '
(6.18)
dt
",fI
J
d8 dw
(6.14)
It(t)
. 2 ;
}
=
l.2
gl(t) =
(6.20)
for 0~ otherwise
FO
= {0
t ~
T
1
and
ft(t)
m
gz(t)
I
= -m II(t)
Substituting the preceding g.(t) and 92(t) into Equations (6.11) and (6.12) along with the impulse response functions from Equation (6.13) gives
(6.16) ql(t) =
It is-important to note that because the original TDOF system is transformed
by the normal mode method into two uncoupled SDOF systems, Equations (6.9) ana (6.10), the solution technique presented.in the Equations (6.1 1) (6.16) are identical to that for the SDOF system. The unit solutions, hj(t) and Hj(w), are the same as those for the SOOF system. Thus, as in the SOOF system, hit) and H,(£0) are Fourier transform of each otb.er with
qz(t)]
and Iz(t) = 0, determine the response displacements y.(t) and Y2(1).
(6.15)
where H,(w) and H2(W) are the complex frequency response functions for the uncoupled systems Equations (6.9) and (6.10), and are given by
1
= t[q.(t) -
(6.19)
EXAMPl.E 6.1. A transient excitation force I.(t) is applied to the TDOF system shown in Figure 6.1. If
,~ .
£OJ - £0
+ q2(t)]
and the solution for the transient response of the linear TDOF system to nOI1 random excitations I.(t) and 12(t) is now complete. The key in the. solution technique lies in the decoupling of the TOOF system into two independent SOOF systems by means of the modal superposition method. All of the analysis with the SOOF system can be carried over to the TDOF system. The analysis that is new for the TDOF system involves the determination of the natural frequencies Wi and the natural mode shapes At dA 12, At 2/An, and the introduction of the normal coordinates £11(1), lJz(r)· Furthermore in the random analysis, as we shall show in Section 6.1.2, the extension from the SDOF systems to TDOF systems involves only simple algebraic operations once the system is decoupled.
(6.13)
gl(O)e- i",9 dO] dw
;1[ f~oo H 2(W)t"" [ f~oo g2(O)e- l
2
f~oo hj(t)e-
Hj(w) =
(6.17)
(6.1 1)
ht(O)gt(t - 0) dO
~ fOCl Ht(w)e [fOO 21[ -... -
=
Hj(w)eiw'dw
-00
After the notmal coordinates qt(t) and q2(t) are obtained, the response displace ments }'I(t) and Y2(t) can be easily determined from Equations (6.7) and (6.8) as
1r~sin ~j Wi
Hj(w)
= -21 foo I[
00
~!
l'
hj(t)
where j = 1.2.
Alternatively. the solutions for the response qt(t) and q2(t) can be obtained through superposition in the frequency domain and are given by the Fourier integrals
"
~:~
(6.10)
where h1((J) and h2(O) are the impulse response functions for the uncoupled systems Equations (6.9) and (6.10), and are given by
!,
;~
1 [fl(t) - !2(t)] = 92(t)
m
117
TWO-DEGREE-OF-FREEDOM SYSTEMS (TDOF)
As two separate single-degree-of-freedom SDOF systems, the solutions for the responses qt(t) and q2(t) to transient excitations gt(t) and g2(1) can be derived from superposition in the time domain and are given by the Duhamel's integrals
ql(t)
:l i'
+ Wiq2 =
6.1.
'f
-00
= _ F0 ~ali~
mW I
-
I
1 sin[w.(t - 0)] - /1(0) dO
Wt
i' '
m
sin[wl(t - 0)) dO
0
_ Fo COS[w.(t-O)]I' - moo! WI 0
= F0 (1-coswtt) mOOt2
for
0<1 < T
"f
,ll
II •
118
!~
HI II:
";1
6.1.
RESPONSE OF LINEAR MUlTIDEGREE·OF·FREEDOM (MDOF) SYSTEMS
ql(l) =
f~
fT sin(wt{t mWt Jo
0)] dO =
Fo
t\ {cOS[Wt(1 mw,
tll( I)
for I> T
t, "'-,: ~
~,i !,,, "
I
11
Fo
qz(l) "
mW2
"mw;z
"
-1 -Fo [1 -
~ I ~
T
o[
\;~?~}'1-9jl) ~ ~_
00 >l;~1;<::
cos (JIol,
"'''''';:"'''~-~-
wi
I - cos W21] + --..,,--"-'
-
wi
forO
~
t
~
for t
T
6.1.2.
J
-oo
cos wot,
Random Excitations
Let us now consider the two excitations fl(l) and f2(1) as stationary random processes with zero mean and with autocorrelation functions
<()
EXAMI'Ul6.2. J.\ steady-state excitation II(f) = 1"0 cos (j)ol is applied to the TDOF system shown in Figure 6.1. Determine the s(eady-state response dis placements YI(I) and Y2(1). Again, from Equations (6.9) and (6.10), we find that
= .!. l,(t)
YI(I) :. Y2(t)
± Wz2 -1 Wo2
Note that when the forcing frequency Wo ... Wt, the first term on the right hand side dominates so that YI(I) ~ Yz(I), and the TDOF system vibrates in its first mode with A II /A 21 ~ l. On the other hand when Wo -> Wz, the second term dominates so that yz(t) ~ - YI(t), and the system vibrates in its second mode.)n the steady-state vibration the frequency of the response is always the same as that of the excitation.
I fO{COS[(/II(t '1')) COS (t}t 1 cos[w2(1- n-COSW21]} "-".-"- "- .. -" """"i - - ---"- "± -------·,,·--"""-"2 """---."---"---" " 2 m (/II Wz for t > T
o
2
m(w2 - (J}o)
YI(t)} F I 2 2 { Y2(t) = 2m WI - Wo
for t > T
T)] - cos W21}
cos wtl
2m
Yt(I)} { Y2(1)
Fo
= ---.. Y"-
" -, !~~1
The steady-state response displacements from Equations (6.19) and (6.20) are forO
cos W2 1)
(1
The response displacements from Equations (6.\9) and (6.20) are
I
_ l;4.z.
f
--2
="1 ~2 F {COS[Wz(t -
I
II I I I I I I I I
}' l~:
Similarly
119
""..i.:": ~
f.~.j
T)] - COSWtl}
;;,
'I
TWO'DEGREE-OF-FREEDOMf~~~~~l
m
-oo
= "'0 cos Wol,
m
RJ<'r) = E(jj(t)jj(1
+ r)],
.; = 1,2
(6.21 )
and power spectral density runctions
'/
f'"
1 _'" Siwl = 2n:
Uj(rje
.
"'''If liT:,
j = 1,2
(6.22)
In addition, for excitations, the cross correlation functions are defined by
Since
Fo F_ -cos w ot = ~(e"">'
2m
m
R I2(r) = EUI(t)f2(t + r)]
(6.23)
+ r)]
(6.24)
+ e-'....·) Rzl(r) = EU2(t)fl(1
it follows that:
and the cross spectral density functions by ql(t) = 2Fo HI(wo)ei"""
"
m
+ 2Fo H,( -
wo)e-I"""
m
I
fCO R (r)e-t
(6,25)
fco
(6.26)
2n: - co
Substituting H I(W) from Equation (6.16) into the preceeding equation gives
e''''''t') Fo (e- w~ + 2m wi -w~
Fo ( ql(t) = 2m wi
SI.i(W) = -
! S21(W) = 2-
I
"",/
~
)
n:
R21 (r)e- i 'Ot dr
-
Note that based on the stationary property, Fo m(wi _ w~) cos wot,
-oo
R I2 (r) = EUI(t - r)fz(t)] = R21 (
-
r)
(6.27)
II
[;1·.' ·f~.
<~
,1 '~I
,ll,
ii
120
RESPONSE OF LINEAR MULTIDEGREE-OF-FREEOOM (MOOF) SYSTEMS
I
I
j
,
-00
1t
[I t
~',
I
R,l(r)
121
I
J
.~,
I
;:R:.: - ;;. - ::
J--s,
= !E{[ql(l)
.
+ q2(t)](qt(t + '1') + q2(t +
= m12 f""
= E
f
co
x
(6.29)
0 1) + f2(t - O)]hl(OI) dOl
1
-00; (fl(t + 'I' -
(
2) + f2(t
+ 'I'
-
02)]h\(62)d62}
+ 't')] , = m12 foo
-
feo
_00
[RI(r - O2 + 0 1)
+ Ru('t' ~ O2 + 0\) + R 2(t -
I Ry\(t) =, 4m 2 -co
6.1.4.
O2 + Ot)]h l (OI)h\(02)dO I d0 2 (6.30)
,~
+ 't')] = m12 f'" fco (RI(t -
+ R;u(r -
-
62
O2 + 01)]h\(Ol)h 2(OddO I d0 2
[Rt(t - O2 + Ot!- R 12(r - O2 + Ot!
- 00
+ 01) + R2(t -
+ 9 1)
-
Ruer - 6 2
+ 0 1)
O2 + Od - R 2('t' - O2 + 01)]h l (Odh 2(02) dOl d0 2 (6.31)
1 -
O2 + Od]h 2(Od /2(02}d{}1 d0 2 ' (6.33)
°2)[h\(Ot)h\(02)
+ h2(Odh l(02) + h 2(Odh 2(02)] dOl
+ R12(r - 62 + 61)
Similady, the other three terms on the right-hand side o( Equation (6.28) are E(qt(t)q2(t
f'"
foo f'"_'" RI(t + 0
Interchanging the order of expectation and integration and making use of Equation (6.21)-(6.24) further reduces the preceding equation to give E[ql(t)q\(t
::
Equation (6.29) together with Equations (6.30H6.33) constitute the input excitation-output response relations in terms of autocorrelation and cross corrclmion functions. Although a lot more terms (16) .lre illvolved in deter mining the autocorrelation Ryl(t) of the stationary displacement response YI(t) as compared to a single term for the case of a SDOF system, these terms are basiC'dl1y similar in construction and thus introduce no basic ditliculty. One half of the 16 terms involve cross correlation functions R 12(t) and R21 (r) of the random excitations fl(t) and f2(t) as a direct consequence of the multiple input excitations. For the special case of a single random excitation fl(t), only the first term on the righ~-hand side of each of the four Equations (6,30)-(6.33) remains nonzero. The 16 terms are then reduced to only 4 and there is no cross correia tion function. This is given as:
+ 't')] + E[qt(t)q2(t + 't')] + E[q2(I)qt(t + '1')]
{f~", ~ (ft(t -
_ 00
- R 2I (r - O2
The right-hand side of Equation (6.29) consists of two autocOrrelation functions for the normal coordinates q I (t) and q2(t), respectively, and two cross correlation functi6ns. Using Equations (6.9)-(6.12) to relate the nonnal coordinate ql(t) to the random excitations fl(t) and f2(t) gives the autocorrelation
+ '1')]
1\:;.:
'I'm
+ E[q2(1)q2(t + r)]}
E[ql(t)ql(t
+ Otl- R2(t -
1-
(6.32)
E[q2(t)q2(t + '1')]
= E[Yl(t)Yl(t + '1')]
'
i,·,.,_'
m
(6.28)
w)
S21( -
~
I
= ~
The autocorrelation function Ry1(r) for the qisplacement response y\(t), based on Equations (6.19) and (6.20), is
I'
f'
=
Response Autocorrelation
= HE[ql(t)qt(1
I
I
·1
I
I
II
:;J
R 21 ( - r)e-h." dr
fOO R21 (r)e'"" " dr.
I = -2
6.1.3.
11
I
_'"
£i.q;,\'I'.t;1l -
- R2\(r - O2
I
;j
f'"
= -2!..1t
SI2(W)
'I
Ii:;"
TWO·OEGREE·OF·FREEOOM SYSTEMS (TOOFl
and
"
"
6,1,
d0 2
+ 11 1«(/1)11 2(0 2 ) (6.34)
Response Spectral Density
An examination of the input-output autocorrelation relation in the previous section shows strong similarities to that of the SDOF system. In fact, the first term on the right-hand side of Equation (6.30) is identical to the input-output autocorrelation relation of the SDOF system. As the input-output spectral density reliHion for the SDOF system has already been derived through the link between autocorrelation and spectral density functions, such a developed package is naturally applicable here as follows: The spectral density S,I(W) of the response displacement YI(t) is by definition and from Equation (6.29)
I 122
I I I II
g
I S,I(W) = -2 1C
ilk'"
f'"
_,.,
f'"
TWO-OEGREE-OF-FREEOOM SYSTEMS (TOOF)
123
For the TDOF structural model shown in Figure 6.1 determine power spectral density function Syl(W) of the stationary response dis IlIcement y\(t) when the excitation is a single stationary random process with a white spectrum So. Determine also the stationary mean square
RYI(t)e-IWf dt
1 ~'" 4 I {E[ql(t)ql(t = 21t
+ t)] + E[ql(t)ql(t + t)]
E(yn·
From Equation (6.16) the complex frequency responses are
+ E[qz(t)ql(t + t)] + E[q2(t)q2(t + t)]}e- iWf dt
I IH .(wW = (wi _ WI)2
Using Equation (6.30), the first term on the right-hand side of Equation (6.3S~ denoted here by 11 for convenience, becomes
1( = - 182 'ltm
H l(w)H 2( - (I))
I'" I'" Irk _0')
+ Rzl(t II =
I I I I I I I I
6.1.
RESPONSE OF LINEAR MUlTIOEGREE-OF-FREEOOM (MOOF)
-00
O2
[RI(t - O2
-a
+ 01) + R 2(t
IHI(wW 4m 2 [SI(W)
+ Od + Rdt - Oz + OJl
- O2
~bstituting
+ SI2(W) + S21(W) + S2(W)]
,
/ _ H l(w)H z( - w) [S (w) - SI 2(W) 2 4m' I 13 = H I ( -.W~H2(W) [SI(W)
= IH.2(w)l .,
+ S21(W)
+ S\2(w) -
S2(W)]
+ S2(W)]
+ 11 + 13 + 14
, For the special case of a single random excitation f.(t), the spectral
of the response displacement y.(r) is
~~
Sy\(W) = 4m 2 [lH l(w)I
2
+ H l(w)Hl( -
(0)
'
+ H 2(w)H I( -
(0)
S (w) = _S_o2 [ I y\ 4m (W~ - ( 2)2
+
2
(wf - ( 2)(wi - ( 2 )
+ --:,---I---:--:-J 2 (w~ _ (
)2
.The mean square E( yf) is equal to the area under the curve of the response density SYI(W). In this case E[y.(t)2] =
f~"" Syl(w) dw ...... 00
. Deterministic Damped Vibration
S2l(w) -S2(W)]
Finally, the spectral density function S,I(W) of the response displacement from Equation (6.35) is SYI(W) = 11
the preceding equations into Equation (6.41) gives
unrealistic response mean square arises as the direct consequence of the 8lization of zero damping and stationary excitation.
l
[SI(W) - S12(W) - S21(W)
'w~ _ W
IH 2(WW
+ Ot)]e-iO"hl(Odht(Oz) dOl d0 2
where the complex frequency response H I(W) is defined by Equation (6.17) and (6.18), and the spectral and cross s'pectral density functions are defined by Equations (6.22), (6.25), and (6.26). A comparison with that of the SDOF shows the addition of three spectral density terms due to the second excuuuon f2(t) and the constant multiplier 1/4m2.through modal superposition. Similarly, using Equations (6.31)-(6.33), the other three terms on the hand side of Equation (6.35), denoted by 11 • 13 , and /4, respectively,
14
I 2)( 2 1) = 1I 2(w)1I,( - (0) WI - W W2 - W
= (-2
.
+ IH2(w)1
. now in this chapter damping is left out for simplicity and clarity in with the new problems unique to the TDOF system. When damping is the only new problem lies in the deterministic aspect of decoupling, show in the following. Of course, with the additional damping term, become lengthier. dmplify the derivations, we consider only one single excitation fl(£) problem with the multiple excitations in the damped case is basically to that in the undamped case. Furthermore, we shall arrange all terms format in anticipation of the subsequent matrix formulation and for the general MDOF systems.
equations of motion for the TDOF system of Figure 6.2 are
2
]
0J{~I} + [2C -CJ{~I} + [2k
m
Y2
-c
2c
Y2
-k
-k]{YI} = {II(t)} (6.42) 2k)'2 0
~
I
;;1;ii
I
.:H ;}1
':1
124
RESPONSE OF LINEAR MUlTIOEGREE-OF-FREEOOM (MOOF) SYSTEMS
I I I I
TWO-OEGREE-OF-FREEOOM SYSTEMS (TOO F)
Mj = {IjtAT[m]{t/Ij};
c
Cj
FIG. 6.2. A TOOl" system with damping.
Fj
From the analysis of the undamped Cree vibration problem we again obtain the undamped natural frequencies
~
and
Wl
MJ
= fljtj}T[C]{ljtj};
(6.43)
= J3k/m
= {ljtj}T{f}
= 2m, Ml
C I = 2e, C 2
K j = {t/IjY[k]fljtj};
~d/7/7~$r//7//$ffm//a;?;7&;~
(1.)1
125
where the scalar quantities
...... .Y2It)
..... YI(t)
'~
6.1.
2111
=
= 6e
(6.51)
K! = 2k, K 2 = 6k
= Ijtofl;
(6.50)
(6.52)
F! = fl , F 2 = fl
(6.53)
are the generalized mass, damping coefficient, stiffness, and excitation, respec tively. Dividing Equation (6.49) through by Mj gives ..
Y}
•
2
+ 2e j w j Y) + Wj 1] =
•
Gj ;
)
= 1,2
(6.54)
and the natural mode shapes
I I I I I I I I tl,1),_.,. "
{ljtd =
{~~:} = {:},
I}
(6.44)
{1jt21={1jt12}={ 1jt22 -I
Introduce the normal coordinates Y such that the response' displacements are
{y}
= (1jt]{Y}
= {ljtdYI
+ {ljti}Y2
=
{~} Y
j
+
{-aY1
+ (e][Ijt]{Y} + (k][Ijt]{Y}
EXAMI'U16.4. Determine the transient displacement response vector {.I': to a single excitation fl(t) for the damped TDOF system shown in Figure 6.2 by a time domain superposition method. From Equations (6.45), the response vector is
= {f}
(6.46)
Multiplying Equation (6.46) through by the transpose of the jth modal vector {ljtjf gives
{t/lif(m](t/I]Y} + {t/lj}T[C][Ijt]{Y} + {1jt}f(k](Ijt]{Y} == {1jt}}T{f}
(6.47)
Making use of the following orthogonality conditions of the normal modes with respect to the mass [m], the damping [cJ, and the stiffness [k], respectively,
{1jt}V(m]{Ijt.} = {ljtjf[e]{lh} = {1jt}}T(k]{Ijt.}
"=
0
for j
-+ k
j=1,2
{y} = {ljtd Y1
+ {1jt2} Y2 =
{a
Y1
+ {_
a
Y2
(6.55)
Using Duhamel's integral to express the transient response for Equation (6.54) gives the normal coordinates
1)(t) =
f:",
Gi t - O)hj(O)dO;
j = 1,2
(6.56)
(6.48)
where the generalized force Gj = FJlM) with
we find that Equation (6.47) simplifies to the uncoupled systems
MjYj+CJYj+KjYj=Fj;
e
(6.45)
Substituting Equation (6.45) into Equati6n (6.40) gives
(m](Ijt]{y}
where 2~jwj = cj/Mj , wJ = KJiM j , and Gj = FiMj. Note that Wj is the jth undamped natural frequency and j is the jth modal damping ratio. Having obtained the uncoupled systems, Equation (6.54), which are inde pendent SDOF systems, the remaining response analysis is identical to that of Section 6.1.1 except for the addition of the damping terms in appropriate places. This addition, of course, was included in our earlier treatment ofdamped SDOF systems.
(6.49)
GJ = G2 = fl(t)/2m
(6.57)
~'I
I I
iI..~ , ' :i I .r,'l; I::·
,
o-
j
I II ~
6.1.
RESPONSE OF LINEAR MULTIDEGREE-OF-FREEDOM (MDOF) SYSTEMS
~jWjt) •
= exp( -
wj.,)l _
~j
sm../l -
~JWjt
(6.58) E[Yj(t)y"(t
+ t)] = E[
f'",
GJ(t - Odhj(Ot) dOl
f"",
Gk(t
+t
-
°2)h k(02)d0 2
I
I lI I' , I \I ii l
2~IWt =
mf
,! II
\ II
!
I I
I
IlIt'H'
2~2W2
elm,
= kim,
= 3clm
From Equations (6.50), (6.53), and (6.54) the generalized excitations are
wi = 3klm
(6.59)
EXAMPLE 6.5. Determine th'e transient displacement response vector {y} to a single excitation f,(t) for the damped TDOF system shown in Figure 6.2 by a frequency domain superposition method. In terms of Fourier integrals, the transient response normal coordinates Yj(t) in Equation (6.54) are given by Yj(t)
= d1t f~oo Hiw)e
1w
where from Equation (6.57), GI response Hj(w) =
I
1
•
[f~", Gj(O)e-lwe dO] dw; = G2 = f,(t)/2m I
2
Wj
2
W
2';:
+ ''>JWjW
j
. ) = 1,2
= E[YI(t)Y1(t + t)]
= E{[t/l1I Y1(t) + t/l12 Y2(t)][\f!11 YI(t + t) + t/l12 Y2{t + t)]}
+ t) + t/l1lt/lll(t)Y2(t + t}
• t/l12t/1" Y2(t)YI (t + t} +.t/lfl Y2(t)Y2(t + t)]
=E[t/lrl YI(t)YI(t
2
=
L L
t/llJt/llkE[YJ{t)"ik(t
+ t)]
(6.62)
!{I
-ro
t/I
-M'.!2 ' ~ R I(t J Mk
- ()z
+ () dlrPJ1 )lIk (O 2) dO I 110 2
!{III = I,
!{lIZ
=I
MI = 2m,
Mz
=
(6.65)
2m
= 4m1 2
f'" f'" _ '"
_ '"
R1(t - O2
+
O&h1(Odhl(Oz)
+ hz(Odh l (02) + h2(0,)h 2(Oz)] dOl dOz
+ h l (01)lI z(02) (6.66)
Note that the result in Equation (6.66) is in exactly the same form as that for the undamped TDOF system given by Equation (6.34). The only difference. of course, lies in the impulse response functions hiO) for which the present damped problem is given by Equation (6.58).
6.1.1.
2
J=11:1
ff<'
a,
Substituting these constants into Equations (6.65) and (6.62) gives the required input-output relation as R,I(t)
+
f
EXAM"L!' 6.6. Determine the stationary input output rcilition for the auto correlation functions of the TDOF system shown in Figure 6.2. From Equations (6A4) and (6.50) we have
The autocorrelation function for the stationary response displacement YI(t) of a damped TDOF system, as shown in Figure 6.2, to a sin~le random excitation ft(t) is by definition and from Equation (6.45), R,I(t)
=
where RI(t - Oz + Od is the autocorrelation function of the input excitation fl(t) and is defined by Equation (6.21). Equations (6.62) and (6.65) constitute the stationary input-output relation for the autocorrelation functions.
Damped Response Autocorrelation
,
(6.64)
Substituting Equation (6.64) into Equation (6.63) and interchanging the order of expectation and integration gives
1,2 (6.60)
(6.61 )
!{It.fl(t) = _!.H1 __ _ {t/lj Il'[nt] {!{Ij I
Mj
J
-co
and the complex frequency
;
FN) G.(t) = ---
E[ lj(t) y"(t + r)]
where 2~l1JJ and wJ are given by Equation (6.59). The required displacement response vector is again given by Equation (6.55). As we have pointed out in Chapter 5, for a problem with a transient excitation and response, the time domain approach is superior to the frequency domain approach.
6.1.6.
J
(6.63)
,
I
121
where t/I Ij and !{Ilk are the first elements of the first and second modal vectors, respectively. From Equation (6.56),
The impulse response functions are equal to 0 for t < O. For t > 0 hj(t)
TWO-DEGREE-OF-FREEDOM SYSTEMS (TDOFI
where
f
I
126
Damped Response Spectral Density
The stationary spectral density S,I(W) of the response displacement YI(I) of the TDOF system as shown in Figur-;: 6.2 to a single stationary excitation 1'1(1)
128
6.2.
RESPONSE Of LINEAR MULTIOEGREE-Of·fREEOeM (MOOf) SYSTEMS
6.2.
is related to the autocorrelation Ryl(r) by the equation I fa:> Ryl(r)e-iW'dt SYI(W) = -2
"II
~I 1111 ~m I~.I
II III
n
2 2
JI k~1 ~~~: fro
x
-00
e- i""
-00
Rtf, - 01
-
+ °1)h j (Ot)hk(02)
6.2.1.
= St(w)Hj(w)H,,( - wl
[m]{ji}
Replacing the triple integral in Equation (6.68) by Ijk of Equation (6.69) we obtain the spectral
Syt(W) = Stew)
II
I!I 1'1 IiI
ill I ,'I' :);1,,8' '1,
,
2
L L
1/121/12 j
[m][I/I]{Y} + [e][I/I]{Y}
(6.70)
MtJMlk Hiw)Hk( - w)
j=lk=1
Mt,
{I/Ij}T[m]{t/td
2m.
U}
(6.71 )
+ [k][I/I]{Y}
=
U}
= {I/IiV[e]{I/Id
= {I/IJ}T[k]{I/Id = 0
.,
•
YJ + 2~jWJYJ
M2 == 2m
2
+ Wj Yj
= GJ
where Substituting these constants and the complex frequency response functions from Equation (6.16) into Equation (6.70) we obtain
2ei»j :j::
, 1 Syl(W) = SI(W) 4n?' [H l(w)H I(
w)
+ H t(w)H 2( -
w)
+ H 2(w)H I ( -
w)
~,~, ~
< r. ,~
+ H 2(w)H 2( -
.
w)]
=
Cj = {I/IJ}T[e]{I/Ij}
MJ
wJ = K
j
Mj
MJ =
{I/Ij}T[k]{I/IJ} Mj
':,:,:
SI(W) [ 1 2 = -4-m(wi - ( 2 )2
+ (wi
- (
2
which agrees with the result of Example 6.3.
2
-:--::-_1----,"""']
)(wi - ( 2 ) + (wi _
(
2 )2
(6.72)
for j
f
k (6.73)
we obtain the uncoupled n SDOF systems
1/112 = 1
= I.
=
Now multiplying Equation (6.72) through by the transpose of the modal vector {I/I} T and making use of the orthogonality conditions
k
EXAMPLE 6.7. Determine the stationary input -output relation for the spectral density functions of the TDOF system shown in Figure 6.2. ' Again, from Equations (6.44) and (6.50) we have
1/111
+ [e]{ y} + [k]{ y}
It will be assumed here that the natural frequencies {w} and the natural mode shapes [1/1] have been obtained by a deterministic frequency analysis. With {w} and [I/IJ available. we introduce the normal coordinate vector {Y} by {y} = [1/1] {Y} into Equation (6.71) to give
(6.69)
~1
,,~
Deterministic Vibration
The matrix equation of motion in terms of the displacement vector { y}, excita tion vector {f}. mass matrix [m], damping matrix [e], and stiffness matrix [k] is
The triple integral in Equation (6.68), denoted for convenience as ljb can be expressed in terms of the spectral density Sc(w) defined by Equation (6.22) and the complex frequency response function Hj(w) defined by Equation (6.18). The integral Ijk in terms of Stew), Hiw). and Hk(w) is given by Ijk
MDOF SYSTEMS
(6,68)
dOt d0 2 dr
~
;
-
fro fa:>
1 1 2 1/1 1/1
= 2n
129
Stationary response of MDOF systems to random excitations will be given now after the basic concepts and results have been derived in detail for the TDOF systems. Here we shall use the compact matrix notations to clearly bring out the functional relationships among the important physical quantities without being obscured by lengti:tY expressions. As in the TDOF system, we begin by assembling the necessary results in the deterministic modal analysis before consideration of the random excitation.
(6.67)
From Equations (6.62) and (6.65) we have the general expression of Ryt(t) for the TDOF system. Substituting,the Ryl(r) into Equation (6.67) gives
Syl(W)
MOOf SYSTEMS
G = FJ = {I/IJ}T{f} J Mj MJ M J = {I/IJV[m]{I/IJ}'
j = 1,2•... ,,/1
(6.74)
I
130
I
I II
i
'j
1
11
I ~
RESPONSE OF LINEAR MUlTIDEGREE.OF-FREEDOM (MDOF) SYSTEMS
The solutions for Equations (6.74) in terms 9f time domain superposition are
Yj(t) =
f~", GJ(t -
fe,
"
,
,
~
1
j = 1,2, ... , Jl
(6.75)
hj(t) = exp( - ~jWjt).
w)",}l _ ~f
Sin
-
JI - ew;
\l!"J ,t
~I
1'8
t
= MjM
(6.77)
n
=
+
I/IIJl/luE[lj(I)y"(t
r
II~}I :~1
~' ·71...
( 2 ) + 1/12d2(t + r
O2 + 01)
[t/lljt/luR,(r
t/I'jt/lnkR.(r - O2
( 2 ) + '" + I/Indn(t + r - O2 ]
+ t/lijl/l2k R d r -
O2
+ ( + ... 1)
+ ...
+ 8 1)]
1)
E(Glt - 0dG.(1
+ r)]
+r
- 02}]
1
= MjM/lil/llkRl(r -
O2
(6.81)
+ lid
wbich bas been used in Equation (6.65) and there is no need for the two summa tions with indexes I and m. Substituting Equation (6.81) into Equation {6.79} and then Equation (6.7i) into Equation (6.78), we obtain the input-output autocorrelation relation as follows:
(6.78)
Using Equation (6.75) for the normal coordinates Yit), the; cross correlation is
E[lj(t)Y,,(t+r)]=E[f~", Gj(t-Ol)hifJdd8, f~", Gk(t+r
8 2)hk(02)d0 2] (6.79)
where the generalized force is
=!i = M j
E[I/IIj!I(t - 0 1) + t/l2J!2(t - Od + .. , + t/lnj!n(t - Od]
};1
j=Ik=1
Gi t )
(J2)}]
+ r)]
E"
:fi!l
-
1 • • = MjM ,~ t/lkjl/lmkR'm(r - O2 + 8 k
Note that when there is only a single excitation !I(t), for ex.ample,
displacement YI is given by
i;
,fEI
+r
+I/I,jt/luR.,(r - O2 + Od + I/InJt/l2k Rn2(r - O2 + 8 1)
n
LL
- O2)]
+t/l2JI/I.tR2n(r - O2 + Od + ...
6.2.2. ,Stationary Random Vibration respons~
131
+ t/l2il/llkRll(r - O2 + 8,) + 1/12il/l2kR2(r - O2 + 0,) + ...
It should be pointed out that the key in the classic'll deterministic analysis lies in the solution of the natural frequencies and natural modes. Furthermore, the orthogonality condition with respect to the damping matrix [c] as given by Equation (6.72) imposes certain limitations on the damping coefficients of the MDOF systems. No such restrictions are imposed on the masses and spring constants since the corresponding orthogonality conditions can be identified with energy laws of elastic structural systems.
The autocorrelation function for the
k
MDOF SYSTEMS
+ I/IIJl/lnkR'n(r - 0], + 0,)
= [I/I]{Y}
;
~
k
x [I/IU!l(t + r
(6.76)
Having obtained solutions for the normal coordinates that represent the modal participations, the response displacement
{y}
+r
k
I = MjM
Ryl(r) = E[y,(t)y,(t
!
O,)Gk(t
= M,lM E[{I/IiV{f(t - 0dHl/lkV{f(t
\
1,
~
O)hj(O) dO;
E[GJ(t
where the impulse response functions
I
I I I ',I
6.2.
;.
Ryl(r)
= J~' t~1 t/l1JI/Ilk n.
x R'm(r - 8 2
f"' f'" -co
-00
1 • • MjM k J~1 m~1 t/llit/lm.
+ 81)hj(Odhk(82) dOl d0 2
(6.82)
where R,.. (r) are the cross correlation functions of the excitations {f} and are defined by
{I/IJVU} , M·]
(6.80)
Substituting Equation (6.80) into Equation (6.79), we obtain the cross correlation relation
R'm(r)
= E[!t(t)!m(t + r)]
(6.83)
For autocorrelation functions Ry/(r) corresponding to the response displacement YI(t), the input ·-output relationship is the same as Equation (6.82) when the
I
I
132
6.3.
RESPONSE OF LINEAR MULTIDEGREE-OF-FREEDOM (MDOF) SYSTEMS
'1 "
I.
II
11
I
II
iI l I
I II I ':
I
I fa:> R (r)e-iW'dr Syl (ro) -- -21t _'" yl
(6.84)
6.3.1. Substituting the result for Ry1 (r) from Equation (6.82) into "Equation (6.84) and making use of the definitions given by Equations (6.18), (6.25), and (6.26), we obtain It It • • I Syl(W) = j~1 t~1 t/Jljt/Ju MjMI< 1~1 m~1 t/J1it/JmkS'm(w)Hj(w)H k ( - (0)
{y}
(6.85)
fco
R'm(r)e-'a>r dr
= [H(w)]{a}ej(O'
(6.89)
where {a} are the amplitudes of the steady state excitations and w is the excita tion frequency. When the excitations are general functions of time defined by
1
1t
Deterministic Vibration
Consider the same matrix equation of motion for the linear M OOF system given by Equation (6.71). Let the matrix [H(w)] of complex frequency response displacements be defined in such a way that the stcady state displal.-'COlcnts arc
where S,.,(ro) are the cross spectral density functions of the excitations {J} and are defined by
1 S'm = -2
133
An alternative solution procedure, which we shall present here, is to analyze the problem directly by the same method used for the SDOF systems with no reference to the modal analysis. The only modification needed is to generalize the single complex frequency response function (or the equivalent impulse response function) to a matrix of functions.
coordinate 1 for the terms Ry1 (t), t/Jlj. and t/Ju are replaced by the appropriate coordinate Ry/(t), t/Jlj' and t/Jlt. respectively. The input-output spectral density relations can be obtained by the definition
~~
AN ALTERNATIVE SOLUTION PROCEDURE
(6.86)
{I(t)}
_00
1 f"'" = 2n _'"
{F(w)}eiW't/w
(6.90)
I
:
Again, for the specia.l case of a single random excitation fl(t), two summation operations with respect to the indexes I and m reduce to a single term and we have
{F(w)} = /I
Syl(ro) =
ii!i Ii
L :1
:~
:i·
\: j:
,Ii
ii; !
1\ &:
Ii,
ilm( ~~
L L
t/J2 t/J2
MIJMIk SI(w)Hj(w)H k ( - (0) j
SY1(ro) =
i i
)=1 «=1
i
""
dt
(6.91)
the response displacement is
{y(t)}
= -2 1 ]I;
J'"
{Y(w)}e"u, dw
(6.92)
_00
where t/Jljt/Jlkt/JlJ'P2k S2(ro)H j (w)H k ( - w) MjM t
(6.88)
{Y(w)} = [H(w)]{F(w)}
,
'6.3.
f~co {I(t)}e-
(6.87)
k
which is identical to Equation (6.70) for the TOOF system when the upper limits ofthe summations in Equation (6.70) are replaced by n. Note that when the single excitation is 12(£) instead of 1.(£)
,
\
It
j,:,lk=1
t
~
with their Fourier transforms
AN ALTERNATIVE SOLUTION PROCEDURE
Up to this point we have presented response analysis of linear MOOF systems to deterministic excitations and random excitations on the basis of the classical modal analysis. By such an approach, the MDOF problem is first transformed into a superposition of independent SOOF systems in tenns of normal modes and frequencies and then' solved by the method already developed for SDOF systems.
,i (~
~~.
~"'t
(6.93)
is the Fourier transform of {y(t)}. It is instructive to see the similarity between Equations (5.3)-(5.8) in Chapter 5 for the SDOF systems. Having outlined the solution procedure here, we then concentrate on the determination of the matrix [H(w)] for the complex frequency response dis placements, which are the only new set of quantities in this analysis.
6.3.2.
Complex Frequency Response [H{w)]
The physical meaning of [H(w)] can be clearly revealed from consideration of a simple TooF system for which Equation (6.89) gives the steady state displace
~,,' I
I lIS f.1
134
~ :1
1
' ,l\
RESPONSE OF LINEAR MULTIDEGREE-OF-FREEDOM (MDOF) SYSTEMS
6.3.
ment of the first coordinate
'j.1.
I'
YI(t)
r:
f~
= [HIl«u)a + HJ2(cu)az]e
~~
i
""
(6.94)
hjk(t) =
= 0, and
H II(CU)
.,
Substituting 1';(w) of Equation (6.101) into Equation (6.100), we find the impulse response is
l
Furthermore, setting at =1, a2
:,~
= PI I(ru)e18
11
''''1
(6.95)
;!
we have
111 .i,
Yl(t) = PI I e"""
,'
I
\1 I
.l
(6.96) H ;.(w) =
~
III ill
+8u I
The last two equations show that the element H II(CU) of the matrix [H(w)] is a complex function of frequency, representing the amplitude PI I(CU) and phase 8 11 (cu) of the steady state displacement response only at the first coordinate to the harmonic excitation of unit amplitude onty at the first coordinate. Similarly therefore, Hjk(cu) represents the amplitude p~({J}) and phase Ojk(W) of the steady state displacement response at the jth coordinate to the steady state excitation only at the kth coordinate. To determine [H(w)], we substitute {f(t)} = {a}ej,·, and {Y(I») = [H(cu)]{a}e'''' into the matrix equation of motion,
I
(m]{Y}
+ [c]U} + [kJ{y}
= {f(t)}
6.3.4.
~~ ~
+ iw[c] + [k])-I
iV; :
where the superscripted - 1 denotes matrix inversion.
II~I;
6.3.3.
~
(6.97)
i.,
~.
~i
~ ~}I
~:I ~ ~j~
(6.98) Yj(t) =
Analogous to the complex frequency response Hjk(w), the element hjt(t) of the
impulse response matrix h(t) is defined as the response displacement at the jth
coordinate to the unit impulse excitation only at the kth coordinate.
To evaluate hjk(t), we refer to Equations (6.9)-(6.93) and write
f~ ~(t)e-I"" dt = 1
~ ~:
, l~ 1 ~;
J~ ~ tr . ~rJ
I ~
~
I
+ t)]
(6.104)
f~", I(t -
O)hjk(O) dO
.f~(I).
the
(6.105)
Substituting Equation (6.104) into Equation (6.104) gives
Impulse Response [h(t)]
Fl(w) =
(6.103)
dl
'
In terms of the impulse response function iljk(1) and the excitation response displacement at coordinate j is
t ~
~
1W
Consider the stationary response displacement .l'i\l) at coordimtle; of il M DOF system to a stationary random excitation at the coordinate k. Again, the random excitation 1;'(1) is assumed to have zero mean value, autocorrelation function Rk(t), and spectral density Sk(W), Our prime interest here is to determine the autocorrelation function Rit) and spectral density Siw) for the stationary response displacement Yj(t). By definition,
+ iw[e] + [k])[H(w)]{a} = {a}
:.JH(cu)] = (- w 2[m] ~;
f~", hjk(t)e-
Rj(r) = E[YitlYj(1
(- w 2[m]
~ I~
(6. H(2)
Stationary Random Vibration
and obtain
I
1 So;' H jk(w)e"'" , do} -2 n _'"
The impulse response function hjk therefore is related to the complex frequency response Hjk(w) by exactly the same Fourier integral as that for the SDOF system. The other part of the Fourier pair is then
~:
.3
135
AN ALTERNATIVE SOLUTION PROCEDURE
I YJ(t) = hjl(t) = 2
11:
fe . ~{cu)e+i'" dw
(6.100) (6.101)
= E[
f~«. fit ~ 0dhjk(O.l dOl f~«
f:", f:",
=
(6.99)
_00
1';(w) = H jk(w)Fk(cu) = HJ(cu)
Rj(t)
.
fi.(t
+r
o
- () 2)/ljk(O 2) dOl
Rk(t - O2 + dhjk(Odhjk(02) dOl d0 2
J (6.106)
which is identical in form to the input-output autocorrelation relation for the SDOF system as given by Equation (5.45) in Chapter 5. For the spectral density of the response displacement YjU), ~.l
f'"
.
. S ,(w) = -1 R ,(r)e-' w , dr ) 2n '" }
(6.107)
11
ill
ill
I Ii ~~!i~
I I j,~: I I ill I";,· :11II 1 1'\11 i Ii II 'N
,
•
136
RESPONSE OF LINEAR MUlTIDEGREE-OF-FREEDOM (MDOF) SYSTEMS
by definition. Substituting Equation (6.106) into Equation (6.101) and making use of Equation (6.103) gives the inpwt-output spectral relation,
6.3.
In matrix form,
AN AlTERNATIVE SOlUTION PROCEDURE
:~!
Sj(W) = If j iw)H jk( - W)Sk(W)
"i:
{: 'I
~:
~!
.:
!j ~
: :1
(6.108)
This again is identical in form to Equation (5.51) of the SDOF system. The only difference lies in the unit response solutions, hJt(t) and Hjk(w), of the determi nistic part of the problem. For the MDOF system, these unit solutions given by Equations (6.98) and (6.102) involve no fundamental difficulty beyond mathematical calculations. It should be pointed out that the preceding analysis of stationary random vibration is restricted to a single excitation lk(t) at the kth coordinate. When ,more than One excitation is considered, as wedid in the model analysis approach in Section 6.1.2, the input-output relations, Equations (6.106) and (6.108), become considerably lengthier. Furthermore, a simple superposition of two responses corresponding to two separate excitations are not valid because of the product terms. Specifications on the excitations must include cross correla tion functions and cross spectral densities among the different excitation pro cesses in addition to their individual autocorrelations and spectral densities.
[noll,) + [cll,) + [kll y )
~
{,k,x,; ""
where the mass, damping, and stiffness matrixes are
OJ
Em] -- [ml 02 '
[k] ""
m
[(k l +k k2) _
- k2] k'
2
-
[c] "'"
2
[(C -
2. Substituting YI = Zl + Xo and Yl = Equation (6.111) gives equation,
ml [ 1n2
o J{~I} + 1»2 Z2
Z2
+ Zl + Xo
+ ell
1
-
C2
into the
[el - ~2}{~1} + fk l 0
(2
_0
Z2
i
1,\11 ' , ,~
I..
li1l1 I
IIIIf ~
1111Ii )~:I ,:!! l!
i
. ~"
r "
~
EXAMPLE 6.8:" A TDOF system riding on a vehicle model (Figure 6.3) is set in vibration thr06gh the acceleration xo(t} of the vehicle. For this system (I) obtain the stiffness matrix [k], damping matrix [c], mass matrix [m], and the matrix equation of motion in terms of the displacements {y}, and (2) obtain the matrix equation of motion in terms of the relative displacements.
1.
Equations of motion for masses ml and m2 , respectively, are
m1 ;1
:::0
mzyz
=
(klxo
+ CIXO) - (k ll YI + k12 Y2) - (k 21 YI
+ kn
- (CII YI
Y2) - (C 2I
YI
~b
'.
~m ·1
:to(tl
+ Cl l Y2)
(6.109)
+ C22Y2)
(6.110)
Y2(11
k2
hI
::
~~\
\.
t... ...:
:.i!;l: i!: g
~
'li!i t
1
=
C2
IA ==
/2Jk;m;.
m2 /m l
Z2
{a
l
+
2/t{2m2]{~I} + [_ wt0
- 2e 2m 2
Z2
+ Ilwr - w~
!fh
t ", Mil ~,il';l;~, 4;~:
0J{~I} + [- 02ew
=xo
li~'I.'
e
Di,iding all 'e'm, of Equ"io n <6.113) by on, and u,ing 'he P'
- 1
YI(I)
m~ = k 2 / m2,
Ct/2../k:rn;,
2
In
f,;
kt/m 1 ,
el = - 1
[ - 1
~r fi~r ,;:
. wi =
'/
FIG. 8.3.
wh...e the fin' equa'ion j, fu,the. divided by _ 1 and second by identify 11(t) = Xo and 1 2(1) = Xo in this example by referring to E (6.97). As these quantities represent the excitations and are included in E (6.98) Thus for the complex frequency response [H(m)]; we Can Use it direcl
A TOOF system riding on a vehicle model.
·Examples 6,8, 6.9; and 6.10 are based on original work by CrandaH (1963).
- [
[H.(m)] -
)
""2<,"', - "'ll <""21'< x"'d ""'l) l' (m iw2c.,,, , .~, 2 _
I 138
I I I I il I I I
,
;1 It t
I
\I il ~
t ~ ill l tl I;
I-
I! ~
RESPONSE OF LINEAR MULTIDEOAEE-OF-FREEDOM (MDOF) SYSTEMS
PROBLEMS
• H u(W)
::=
(w 2
jW2~lW2 - w~)/A.
-
H 21 (W) = -w 2/A,
2 - w (wf
+
+ 2(1 +
1l)~2W2]
E(zr) = So
+ 2e1w;;wi] + wiw~ •
+ if 12(£0) =
H z2 (w) = H 21 (w)
+ Hu(w)
2
IHzl (w)1 dw
+ 2e2W2W!{[1
+ 1l)~2W2
[£02 - iw(l
= (- iw2e 1w 1
-
- (1
+ ll)wWA
- (I
+ Il(l + 1l)2(wzlw,)2] + Il)2(W2/W.)2]2 + 11(1 + It)2(W2/(O,)2}
+. 8e~w~(l + Jt)z[(l + It) + (~t!~2)2]] (6.117)
wf)/A
EXAMPLE 6.10. Determine the mean square relative displacement E(z~) for the TDOF system shown in Figure 6.3, when the acceleration xo(t) of the vehicle is an idealized stationary process with a while power spectrum of So. The mean square value is obtained by iritegrating the area under the appro priate spectral density curve. For a white noise excitation, the spectral density function S'I(W), according to Equation (6.108), is a constant multiplier of the complex response function IH z1 (w)l2, since in this problem the two excitations ltV) and h(t) are identical and hence are accounted for through the complex frequency response function H dw) in Equation (6.117). The integration with respect to £0 can be obtained from Equations (6.118) and (6.119)
- iW3B3 - w2B1 H(w) = w4A4 _ iw3.-4;--
IH(wl1
+ s.; •.;iw,wj(l + 1/)2[1 + (I + JI)(W2!W.)2]
H,,(w) = H lI(W)
C
f:
= (1tSo/A)[2~lwlwi[IlZ
(6.116)
Finally, by definition,
1
+ Jt)2~2W2
Using these constants in Equation (6.119) we obtain
+ (1 + Il)wi + 4~telWIW2]
iw[2e,w , wi
.8 1 = - (1
B3 = 0
B2 = 1,
H n(W) = (- iw21le2W2 - ILwil/A
H 22(w) = (w 2 .. iw2elwi - willA
A = w 4 - iw3[2e1wI
+ Il)wi,
Bo = - (1
By carrying out the inversion operation in Equation (6.115), we obtain
+ iwB, + Bo + iwA I + Ao
w1A2
. (6.118)
See the footnote given in Example 5.15 for source of the formulas.
PROBLEMS 6.1.
~
- !~
Find [K] for the rigid frame with coordinates
VI' V2'
~ ~ _\~ r~",
EI. L
~
•
FlO. 6.4.
"""_
m
_ _ ''2
5m
_ _ "1
A,A.) + A,(Bf - 2BoBo)+ A,(B~ - 28, 8,)+ {BiIA.)(A,A, - AoA,) A,(A1A,
A,A.)- AoAi
(6.119) For our problem with H'I(w) given by Equation (6.117), we have
Ao = wiwi, A2 = w~
+
(1
Al = 2~lwtwi
+ 2~2W2Wi
+ Il)wi + 4~1~2WtW1'
A3 = 2~twI
+ 2(1 + IMzwz /
A4 = 1
FIG.6.S.
(See Figure 6.4.)
2E{~~
l~
dw (B~/Ao){A2A~
139
HII
H
\~ I I
140
RESPONSE OF LINEAR MULTIDEGREE·OF·FREEOOM (MDOF) SYSTEMS
6.2.
Formulate the matrix equation of motion for the free vibration of the two-story building model. AssulT'-e rigid girders but massless elastic columns. (See Figure 6.5).
I !~
!: :
~
6.3.
[m]{ii}
111
+ [k]{v}
1\11
: .: ~ ~
".
[k]
f 1·t"
';
6.7.
0;
= 500 [
G~J
[nI]
_!
1
1.
kips/sec in.'
[,,'1]
f (.
.P OJ
_I
:0-1
=
~Wj; .... hen
pet)
I} Po
For the system in Problem 6.4, calculate the natural frequencieS and mode shapes for the case k, = 2k 1 •
6.6.
Determine the matrix equation of motion for the vibration model with a nonuniform rigid bar such that the center-or-mass has an eccentricity of e as shown in Figure 6.7.
= {~} poet)
1
J;
LI2
Mid point A
~~~j' II
Em]
=
[1.0 0 0] 0 2.0 0 kips/see 2 in. o 0 3.0
[k] - 500 [
-l.s
-1.5
3
-n
kips"n.
-3
PoCt) (kips)
VI
...
'~t--e~ Center of mass
.....
of rigid bar of IotaI mass m
20 ~
FIG. 6.7.
m:.:.d~
~m vJl
For an undamped three-degree-or-rreedom system with .
6.6.
a
{II .
'"
FIG. 6.6.
,
I WI ,!J
(~.70i
is applied. Use Po = 10 kips and w = 20 rad/sec. (b) Find the displacement {I(t)} at t = to sec in rree vibration when the system is disturbed by initial displacement v,(O) = 0.02 at (It only, that is, V2(0) = 0, 61(0) = 0, and 62(0) = O. (e) Find the transient displacement response o(t) when an impulsive load vector
6.S.
~!)
J~
\,
the steady state displacemem reo.pons,e state harmonic load vector
-~] kips/in.
lJ2
=
I pS
is applied where poet) is shown in Figure 6.8.
!{~,.
t~:
{12.t1 :'9.2j rad so!\:'.
Center of mass
~'.
I
[k] = 500 _ 1 .
I'll Fmd
VI ~I.
[k]{o}
2
Determine the matrix equation Jf motion Cor the vibration model with two coordinates VI and 02 showQ in Figure 6.6.
I
I
I
+
[ I - 2IJ k' /.m.
= [~ ~J kips/sec 2 in.;
•(;). =
Use A. = ro 2/500 for simplicity.
6.4.
For the undamped TOOF system with given [m]{v}
[m]
Find the natural frequencies rot and rol, and the natural mode shapes {Vd and {Vl} for the system with
\. ;j I
141
PROBLEMS
FIG.6.S.
•
I (sec)
~~,.
_ _ _---..LIL.
~'t "I\.:.
t';r
it.;,.li~ :II ,~
1
I:,~j I
t~:i
,:):l
I,i.i IIJ :li.;
142 RESPONSE OF LINEAR MULTIOEGREE-OF-FREEDOM (MOOF) SYSTEMS
"'
Find the natural frequencies {<.o} and mode shapes [rfr] by a simple computer program using a standard eigenvalue program routine and check the results by hand computations. 6.9. For Problem 6.7. find the complex frequency response vcctor {Hv(w)} and impulse response vector {hv(t}}ror the generalized displacement {Y(t)} due to the corresponding generalized load {P(t)}. Note that each of the three vectors has three modal components and
.
~~
~.
~l
{vet)}
I~J~II
= [4>]{ Y(t)};
M"Yn + K·nY. =
Pn(l)
:l.
where the generalized mass M n, stiffness Kn. and force p.(t) are given by Mn = {rfr.}T[m]{4>.}, K. = {rfr.}l,[k]{4>.}, and p.{t) = {rfrnrr{p(t)}.
I} I
~I
It
Ii';, !tl' t
I~m
'~ l~D
Ij IflGI '~
.'
t
6.10. For the undamped TDOF system of Problem 6.7, consider the excita tion vector
{pel)}
= {~} Po(t)
and
Po!l)
as a stationary random process with zero mean and white spectral density So. '; (a) Find the autocorrelation vector {Re(r)} for the response displace ' ment {V(I)}. (b) Find the spectral density vectol {S.(w)} for the response displace ment {vet)}. 6.11. For the undamped TDOF system of Problem 6.7.; consider the excita tion vector
{p(t)}
=+{ P.\{f)}
pz(t)
:k:
~:~ i~~
't'l I}~" "' ~~
hi
~<~~ml ~:~nl .' 'l~
,.!!J
:li <'Ill
I '';"ri
where PI(t) and pz(t) are stationary random processes with zero mean and white spectral densities SI and S2. respectively. In addition, the cross spectral density S; 2(W) is assumed to be a given function of w. (a) Find the spectral density vector {S.(w)} for the response displace . , ment {vet}}. (b) Find the autocorrelation vector {Rv(t)l for the response displace ment {v(t)}. 6.12. For the TDOF system of Problem 6.7., consider the damping ratio ~ = 10% in each normal mode and the excitation vector
{p(t)}
= {~} poet)
\
'!};.
PROBLEMS
143
where poet) is a stationary random process with zero mean and white spectral density So (a) Find the autocorrelation vector {R.(t)} for the response displace ment {vet)}, (b) Find the spectral density vector {{.(<.oj} for the response displace ment {I)(t)}.
II
I
I l, I I~. I
7.1.
CHAPTER
7
p Yo
s + ~dx ilx
--
e:JL...~ p 121
RESPONSE OF CONTINUOUS SYSTEMS
1
1~'I
'1
:~t
~I ,
fiG. 7.1.
+ ,Yo) dx
Shear beam with ground motion )'.(1) and excitation p(x. I),
We scek " modnl solution u:;
sum of products of normal modes", l\:) and
normal coordinates YN)
7.1,
SHEAR BEAM.S
y(x. t) =
I'" "'Ax) Y(1l
(7.3)
j
j= I
Parallel to our development of the multidegree-of-freedom (MOOF) systems
in Chapter 6, we begin here with the simplest continuous system, the shear beam, for which all basic concepts and analysis methodology for continuous systems can be adequately brought out with a minimum amount of mathe matical complexity. Furthermore, the selection of the shear beams has the advantage over other equally simple systems, such as a taut string, in that it simulates a tall building, which is of primary interest to structural engineers.
7.1.1.
Deterministic
a2(y + Yo)
at 2
+ CI
a.,
a2y
at - k ax" =
p(x,
t)
(7.1)
ClOY
'). 02y
_
p(x, t)
-+ ot2 -m-at- C ax"= -Yo+-m
144
(7.4)
nc = (2j - I) 2L
(75)
c
with the natural frequencies
satisfy the free vibration relations and orthogonality conditions c2 t/1j(x)
+ wjt/I(x) =
0
LL Joft/lj(x)t/lk(x)dx = 2,lijk
(7.6) (7.7)
where the delta function lijk = I for j = k and is 0 for j +- k. Substituting Equa tion (7.3) into Equation (7.2), multiplying through by '" i' integrating over x, and using Equations (7.6) and (7.7) we obtain the following single-degree-of-freedom (SDOF) systems ••
(7.2)
= sin WjX
t/lj(x)
wJ
Consider a uniform shear beam of length L, shear stiffness k. damping co efficient Cl, and mass per unit length m as shown in Figure 7.1. The beam is fixed at the base with coordinate x = 0 ,nd is free at the end x = L. When the base is given an arbitrary' transient acceleration Yo, the relative transverse displacement y(x, t) is governed by the simple wave equation (7.1).
oly
The normal modes for the shear beam
Vibrati~n
Rearranging terms in Equation (7.1), we.obtain the standard form, with wave speed C = Jk[m
tIm
ty
-c iJy ai dx 1
:%
m
!h
a at2
S = k ayiiix
~
~ ;1
2
111
__
~'!
~!
145
x
,wI
~~, I
SHEAR BEAMS
Yj(t)
' + PjYM + Wj1 Yj(t) = Git)
(7.8)
146
I
RESPONSE OF CONTINUOUS SYSTEMS
7.1.
fol. [
I
.. ( )
- Yo t
p(x, t)] +- !/ij(x) dx
(7.9)
pj = m
(7.10)
Solutions for the uncoupled SDOF systems, Equation (7.8), in terms of Duhamel's integrals are Yj(t) =
f", G)(~
(7.11 )
- O)hj(O) dO'
hit)
=
I
e- II / Z"J}. sin[wjtJI -
< () and for
I
y(x. t) =
.2:co
!/ij(X)
J~ I
m
fl.0 !/i~(Ixd dXI foo 11.0 p(Xlf J
=
I
!/iiX)[-
J= I
-00
(7.13)
(7.17)
!/iAxl!/ik(.x)_ ..._._____ .
'"
1
rr
U)
f~", f~""
x
(/x2
E[p(x 1 • t - 0dp(X2. 1 + r - (}2)]
x tftAXtl!/ik(Xz)llj(O.lhk(02) c/O I (10 2 c/Xl
(7.18)
dX2
In the frequency domain analysis, the spectral density function for the relative displacement y(x, t) is by definition,
= -2I f"" 1t
Jofl. !/i)(X)dX/ Jofl. !/i](X)dxJf' jio(t~O)hj(O)dO
O)!/ij(xl)dXt hj «(}) (I{)
.L L ;;;-fl. 1/,2(X I) tf;;n;-if,f(xzl
J O• I k
Sy(X, w)
=
0.16)
( 1 )]
-«)
00
R/x. t)
(7.12)
Equations (7.3)-(7.5) and (7.9)-(7.12) constitute the required solution for the relative displacement y(x, t) of the shear beam subject to a ground acceleration jio(t) and a distributed pressure p(x, C). For the special case where p(x, t) = 0, by substituting Equation (7.9) into Equation (7.11) and then Equation (7.11) into Equation (7.3), we obtain
+r
The autocorrelation function for the n.:httivc displm.:cmcnt rCx. II ill this case is
> () arc
ffJJ/4wJ)]
- 0lljio(1
For the case of only a distributed random excitation p(x, I), as shown in Figure 7.1, we use the term p(x, t)lm instead of - yo(t) in Equation (7.9) for the general ized force Gj(t) so that the relative displacement y(x, t) of Equation (7.13) becomes
.
where the impulse response functions hj(1) = () for
+ Od = E[h(t
Ro(t -0 2
m J~ !/iJ(x) dx
Git) =
7.1.2.
147
where the autocorrelation of the input ground acceleration j;o(t) is given by
where
y(x,t)
SHEAR BEAMS
.
(7.19)
Ry(x. l')e-w, 1/"1.'
-00
Substituting RJ.(x, t) of Equation (7.15) into Equation (7.19) we obtain, for the case of ground acceleration excitation jiO(I) only,
Stationary Random Vibration
S,(x, w)
'"
00
L L j%
=
So(w)
(7.20)
ajk!/ij(x)!/idx)Hiw)H k ( - cul
I k= I
The autocorrelation function Rv(:<. t) for the relative displacement y(x, t) at any location x is by definition, . Ry(x. t) = E[y(x. t)y(x, t
+ t)]
(7.14)
where
So(W)
= spectral density of the input ground acceleration j;o(t)
L
In the case ofonly a random ground acceleration yo(t), the relative displacement is given by Equation (7.1 3). Using Equation FJ 3) in Equation (7.14), we obtain
~. ~ R1(x, t ) -_ L... L... .
)=1 k=1
x
.1. ( ),1. ( ) '1') X 'l'k X
f~ !/iJ(Xt) dX 1 j~ !/ik(X2)dx;z I. ]. I.;Z
fo, !/i) (XI) dXI fo 2
!/ij(x)
!/ik(Xl)dxl
f~ r~ Ro(t .. ° + 0tlhiOj)hk(Ol)dO j dOl
alk = constants
= r
!/ij(Xt)dXl
L
(I. !/ik(X l)dX 2! r !/if(xtldxl rl. !/ir(x 2)dx 2
J o ' Jo
Jo
Jo
= jth normal mode
Hj(w) = frequency response function for the jth normal coordinate Yj(t)
(7.15)
I 2-
Wj
+
.
P
lW j -
W
2
I
148
RESPONSE OF CONTINUOUS SYSTEMS
7.1.
For the case of only a distributed random excitation p(x. t). the spectral density function for the relative displacement y(x. t) ean be derived in the same way by combining Equations (7.18) and (7.19).
:(
so that
J: J:
S"" {"" J~I k~1 bjk!Jtj(x)th(x)
. (0) = 2n 1 _"" Sy(x. x
-GO
-""
[L [L Rp(xl> xz: r
Jo Jo
=
_ O2 + 0tle-1w(.-6,+0.1
x !Jt lXI)!Jtk(Xz)hj(Ot\ei'dO. dO I hk(Ol)e -i
ll.1C } Z
{J(r - ()2
+0
00
co
j=
k=
I I I I bjk!Jtj(X)!Jtk(X)Hj(w)H x
f: f:
Sp(Xl>
Xl,
k( -
J: lL
= E[P(t)P(t
+ r)]
b(x, - a)o(x2 - a)Rp(r) dX1 dxz = Rp(r)
I)
RJI<{r) = E[GJ(t)Gk(t
[L fL
.
Jo Jo !Jtj(xl)!Jt(xl)E[p(XI, t)P(Xl, t + t)] dx , dX2
== bjk
w)
= bjk!JtJ(a)!Jtk(a)Rp(r)
(7.22)
w)!Jtj(Xtl!Jtk(XZ) dX 1 dX2
(7.25)
where
4
bJk = m 2 L2 Rp(X 1 ,
Xl' r)
= space-time correlation for p(x, t) = E[p(xl" r)p(x2.
SI'(X 1 , x2> (0)
t
+ r)] Ry(x, t) = E[y(x, t)}'(x, t
= space-frequency spectral density for p(x, t)
= constants ~
1/ J: 2
m
!Jt](XI) dx l
=E
J:
Lt ""
=
Consider now the stationary random excitation of the shear beam by a con centrated force located at X = a. The space-time correlation of p(x, t) is speci fied as
+
~
t)] = O(XI - a)b(xz - a)Rp(t)
!Jtj(x)Yj(t)
(0
Jl k~t co
co
i
JI !Jt~(X)Yk(t JeO-co J""
_00
1:::> • ':::.l* .
,
.&'"
+ r)]
= J~I k~1 !JtJ(X)!Jtk(X)
!Jtt(Xl) dX 1
Concentrated Random Excitation
E[p(x l , t)P(X2, t
(7.26)
The autocorrelation function for the relative displacement y(x, I) is then
1 SOO I = 211: _00 Rp(XI, X2, T)e- .,. dt
bjk
(7.24)
+ r)]
where
7.1.3.
+ t)] dXI dxz
The cross correlation of the modal forces given by Equation (7.9) is then evalu ated using Equation (7.23) as follows:
(7.21) S,(x, (0) =
E[P(XI, t)p(xz, t
GO
foo SeO
149
SHEAR SEAMS
+ r>}
(7.271
~
*t~
.),
hJ(01)lI k(02)R jk (r - O2
+ °ddO I {/O2
4 ) m2L2 !Jtix)!Jtk(x)!Jtj(a)!Jtk(a)I jk
(7.28)
(
where IJk denotes the double integral Ijk =
f~"" f~eO h (1I 1)hk (02)R j
p (r
+ 01
-
02) dOl d0 2
(7.29)
(7.23)
where Rp(r) is the autocorrelation function of the concentrated random force P at X = a. Note that b(x , - a) is the Dirac delta function with argument XI - a
Equations (7.28) and (7.29) together constitute the input -output autocorrelation
relation between the relative displacement y(x, t) and the concentrated force P
at location x = a on the shear beam.
,.
"I
150
~ .~ ~
, t, I 1 t
(ON
and
WN+I·
1 -2·
Sy(X, w)
1t
f'"
Ry(x, ,)e-I"" dr
(7.30)
-00
Since for Ry(x, r) in Equation (7.28) the only term which -contains the argument r is ljb it follows that when Ry(x, r) is substituted in Equation (7.30), the integral with r effects only ljk. This integral can be carried out as 1 21t
f'"
ljke-I""
dr
= Sp(w)Hj(w)H k( -
First, a review of the analysis leading to the autocorrelation fUllction for the displacement y(x, c), Equations (7.28) and (7.29), shows that when the velocity vex, f) is considered in place of y(x, f), the only modification needed to find the autocorrelation function for t'(x, t) is to change the impulse response functions from hj«(J) to its time derivative nj(O). Thus the mean square velocity, from Equation (7.28), is
E[112(X)) = Rv(x, O) w)
(7.31)
-u;,
""
=
ru
4
L L m2L2 t/I A>::)t/lk(X)t/I j(a)t/lk(a)I
jk
(7.34)
1/(/2
(7.35)
j=1 k=l
where
= f~", hj(t)e- I."
HAw)
where dt Ijk
Combining Equation (7.31) with the rest of Ry(x, r) from Equation (7.28), which is independent of r, we obtain the spectral density function from Equation (7.30) as
=
L~. L~ il j (Od i'lW2)U/.(OI JOC'
Ir•N) = (I hj(t) = _.I l/t 21t
_~,
- (l2l {/OI
'. iwH j(w)e"'"
I/tO
4
J! k~l m2L2 00
Sy(x, w) = Sp(w)
co
t/lj(x)t/lk(x)t/lAa)t/lk(a)Hiw)H k ( - w)
(7.32)
EXAMPLE 7.1. Determine the ensemble mean square velocity E[v 2(x)) at the location x of the shear beam shown in Figure 7.2 when the excitation is a random concentrated force P applied at x = a, with a zero mean and band-limited white spectral density Sp(w), defined by
So,
Sp(w) = { 0,
-
We
We
We
iwHj(w) =
"'. hNk-i!U' J -ru
(7.36)
l/e
Substituting into Equation (7.35) Rp(r)
= J~
Sp(w)e
00
iW '
(7.37)
dw
and using Equation (7.36) we obtain
SI"')
%
and
(7.33)
ljk'=
p
,
,,
151
SHEAR BEAMS
where the cut-off frequency w, lies between the natural frequencies
The spectral density function of the relative displacement y(x, c) is
,
~
1.1.
RESPONSE OF CONTINUOUS SYSTEMS
f:oo Sp(w)w H (w)Hk(- w)dw 2
j
(7.38)
Equutions (7.34) und (7.38) together give thc mean square velocity lit l~lCaliot\ x on the shear beam when a random concentrated force P is applied at x = d. To carry the solution further, we assume that the modal overlap ratio
-"'e
"e
;>
'"
FIG. 7.2. A shear beam subjected to a concentrated random load P at x = a with zero mean and a spectral density with cut·off frequency w,.
r
pj
= modal bandwidth modal spacing
Wj -
W)-1
elL
mnc
(7.39)
:;~t"'"
,. :1
,I
,.
, ~I
h ~
,
152
, , , ,
al
7.2.
is sufficiently small to permit the following approximation in the sum in Equa tion (7.34): Ijk
for j
= 0
1- k
FLEXURAL BEAMS
153
We.seek a solution of the form IX)
(7.40)
y(x, t)
=L
Yit)1jI j(x)
(7.46)
j= I
In view of the limited bandwidth of the spectral density function of the concentrated random force P as specified by Equation (7.33) we make another approximation I jj
0
for j > N
I
_ rrSo _ rrSom
jj
-P; -
-c.-'
j=1,2, ... ,N
2
rcSom (x a ) = ~B I'I' N
~
xa)
7.2.1.
m
2L2 ' -
j=1
.
sm
OJj
2
(2j - l)rrx . 2 '(2j 1)rra 2L sm 2L
(7.44)
vX
d lYi
. a2y
+ C 1 Tvi + pA TI" = f(x, t) ul
(iY J!i
(7.45)
m¢jk
2)
dy)
+ PrJ( + OJjYj = fit)
(7.48)
(7.49)
(7.50)
(7.51)
where the modal exciting force is
fit) = and the modal bandwidth
foL f(x, t)t/lj(x) dx
(7.52)
Pi is given by CI
where EI is the flexural rigidity.
J
where m pAL is the total mass of the beam and ¢jk is the Kronecker delta function. We substitute the modal solution, Equation (7.46), into Equation (7.45), multiply through by t/I}, and integrate over x usi~g Equations (7.48) and (7.50) to obtain the uncoupled equations for the Yj(e) m ( dt 2
Consider a simply supported uniform beam of length L and mass per unit length pA. If the beam is subjected to a viscous damping force C I per unit length per unit velocity and is excited by a transverse force f(x, t) per unit length,the transverse displacement y(x, t) satisfies the following partial differential equa tion: oy
=
f:PAljljt/lkdX
Deterministic Vibration
o4y
}
The normal modes tPJ{x) satisfy the orthogonality conditions
FLEXURAL BEAMS
EI ~
2
OJ· flAIjI.
Here OJ} are the natural frequencies
422
__ 4_ ~
7.2.
tJ4t/1j
El -dx -4
/;'lm2L2t/1j(x1t/lj(a)
-
(7.47)
satisfy the free vibration relations
(7.43)
where B(xIL, aiL, N) stands for the sum B ( I'I,N
J2 sin jnx L
i
IjI x )
(7.42)
which is independent of the index j and when inscrted in Equation (7.34) yields the mean square velocity E[v (x)]
where the normal modes
(7.41 )
To evaluate the remaining values of IjJ we use Equation (7.38) with Sp(OJ) = So for the entire frequency range - co < OJ < co as still another approximation. The result is
~
~
RESPONSE OF CONTINUOUS SYSTEMS
Pj = pA
(7.53)
154
RESPONSE OF CONTINUOUS SYSTEMS
7.2.
The formal solution to Equation (7.51) is given by the convolution integral
Yj(t)
= f~", hj(O)!j(t
(7.54)
i i"
+ r)]
L
=
o
where the impulse response function is
nXdtJ!k(Xz)E[f(xl> t)!(X2, t
0
+ r)] dXI dX 2
'
(7.60)
= tJ!j(a)tJ!k(a)Rp(r)
Il .(t)
}
=
O, { e
- 1/2(Pill.
t
sm Pj,
mpj
t
<0
(7.55)
I~O
As a measure of the local time average energy density we consider the mean square velocity at the location x E(1'2)
with
=E
pJ = wJ - fIJ/ 4
J!
tJ!j(X)
r~ Il (l1)Jj(1 j
Jl'"
tJ!ix)
f'"_'" hAO)!j(t
"') k~1
r
(7.56)
O)dO
(7.57)
0) dO
I I i
E[J(xl> t l )!(X2, (2)] = ~(XI - a)l5(x2 - a)RI'(t z
fL
}
a"
(7.61 )
tJ! j(X)tJ!k(X)tJ!j(£l)tJ!k(a)1 jk
+ t)] dXl dX2 = E[P(t)P(t + T)] =
",- II j(t)
t.)
(7.58)
Rp(r)
(7.59)
t
tt
( 2 ) dOl
(7.62)
tlO2
iwHiw)
=
f
Rp(r)
=
f:",
= 2'71:
'
_to
hj(t)e-
iror
.
iw/J j(Wk""f tIm
_ ".
dt
and Sp(w)e
iwt
dw
the double integral ljk can be expressed in terms of the spectral density S,.(w) of the exciting force as Ijk
OSee Crandall and Willig (1971).
( 2) ilOa
L
l
.
E[J(XI, t)!(X2, t
f'"., .,. l;k((11)f;,(t -
d t fOO
"i(t)
where Rp(r) is the autocorrelation function of the force process. Note that ~(Xl - a) is the Dirac delta function with argument (Xl - a), so that
(L
if1k(X)
Since
Stationary Random Vibration*
Jo Jo
Od dOl
f'", f~tO hj(Od/~d02)Rp(01
Ijk
'We now consider a stationary random excitation of the beam by a concentrated force P located at x = a. The space-time correlation of !tf~ t) is taken as
I
r~ i'PJ1J.lj(t
where ljk denotes the double integral
.
7.2.2.
tJ!ix)
j= I k= I
The corresponding transverse velocity oy/ot is
vex, t) =
Lt
X
The displacement response of the beam is thus
~(x, II =
155
The cross correlation of the modal forces is then evaluated as Rjdt) = E[JP)!k(t
0) dO
FLEXURA.L BEA.MS
f~,. f~,. [hi «(}. )hk(02) f~"
S/.(wjei",(O.-O,)
elm] (10 1 d0
2
156
RESPONSE OF CONTINUOUS SYSTEMS
7.2.
Interchanging the order ofintegration and using the complex frequency response function H(w) in place of the impulse resp
=
G(
f~", Sp(w>[ f~", hAOI)e- 0)8 dO~ f~", h (02)e-;.,e, d0 2] dw i
f:",
1
2
X
)
-
N
}-I
!f;j(x)!f;y(a) =
" IN 4 sin 2 Jrex sin 2 J1ra j=
G(
(7.63)
So,
-
We
a ) =9 (xL,N)+9 (aL,N) -"2I 9 (x----r.-,N -a )-2I 9 (x - LL + a,N\)
X L'L,N
(7.69)
<
W
<
where
g(~, N) = N + !
W
Wj_1
w] -
g2(2j
for j
-+ k
IlJ = 0
for j
>
c 1 L2 1).jEllpA
N
~
(7.65)
(al
To evaluate the remaining values of IlJ we use Equation (7.63) with Sp(w) = So for - 00 < W < 00. This is a good approximation since the contribution to the integral in the range Iwi > We is small corresponding to small Hj(w) for j ~ N. The result is 7[ So 1£So I jj = m1fJ) = mc,L'
j = 1,2, ... ,N
(7.66)
,,.,<
which when inserted in Equation (7.61) yields
S 2 1[ o E(v ) = mclL G
(xIff:a N )
r
o
(7.67)
(bl
~ .~ a.
• L.-a
1 L
(el
i.;: where G(xIL, aIL, N) stands for the sum
(7.70)
The spatial distribution of the ensemble mean square velocity according to Equations (7.67)-(7.70) is shown for three values of N in Figure 7.3. The three response distributions have been normalized to have the same average (lengthwise) ensemble mean square velocity. Note the local regions of 50% excess energy density at x := a and x = L - a. The high energy density at the a is anticipated, but the same high energy density at the loaded point x symmetrical location x = L - a comes as quite a surprise. In addition to this
. is sufficiently small to permit the following approximations in the sum of Equation (7.61): Ijk = 0
1)1t~
1[~
(7.64)
We take the cut-olT frequency we to lie somewhere b~tween the natural frequencies WN and WN + I . Furthermore we assume that the modal overlap ratio modal bandwidth r = modal-spacing
_ sin(2N + 2 sin
Iwi
we <
(7.68)
ILL
• EXAMPLB 7.2. Determine the ensemble mean square velocity E[u 2(x)] of a simply supported beam when the excitation is a random concentrated force P applied at x = a. The random force P has a zero mean and a band-limiled white noise spectrum Sp(w) = { 0,
157
This sum can be evaluated in closed form. We find
k
Sp(w)w Hj( - :w)Hk(w) dw •
a
L' L' N =.~
FLEXURAL BEAMS
FIG. 7.3. Sra1ial variations of the ensemble lUcan square velucity for excitation handwidth corresponding 10 (u) N 10. (b) N = 20, .... nd «.) N = 40.
158
RESPONSE OF CONTINUOUS SYSTEMS
1.3.
interesting result we note that the number of participating modes goes to N = 40, which is a great deal more than the usual deterministic case where a few modes are generally sufficient to determine the dynamic response.
7.3. 1.3.1.
D/',.4t/Jjp
(Ojp
Consider a simply supported uniform rectangular plate with dimensions L" by Ly by i and mass per unit area pt, If the plate is subjected to a viscous damping force C 1 per unit area per unit velocity aQd is excited by a transverse force f(x, y, c) per unit area, the transverse displacement l\I(x, y, I) satisfies the p'drtial differential equation •
D /1
a~w
Ow
4
W
+ C1 at + pc iff = f(x, y, c)
4
a4
- 4 /1 -- i'Jx
a4
(7.71)
(7.7S)
=
[(tY + (~:rJ~
(7,76}
These .frequencies are not uniformly. spaced. The average modal densit (' . per rad'Ian per second) I's I Y /I Ie. " average number of naturaI rrequencles ' lOwever independent of frequency. We may thus spea k of an average Spacing b ' etween resona.nces /',.w
=
1 4n f~ ~ = LxLy ,{pc
(7.77)
Subst~tuting Equation (7.72) into Equation (7.71) We obtain the uncoupled equations
a4
+ l)x - 2-l)y2 - + ti-4
m(
y
where 0 < x < Lx, 0"< y < L y, and D is the flexural rigidity. The procedure we follow to obtain the response of the plate is parallel to that given for the flexural beam in Section 7.2. The only differences are associated with the doubly infinite array of modes for the two-dimensional plate as compared with the singly infinite set of modes for the one-dimensional beam. We seek a solution of the form
= ptWJpt/Jjp
159
where the Wjp are the natural frequencies
THIN PlATES* Deterministic Vibration
THIN PLATES
2.) .Ii/IV) .
2
7 W jp + {ljp dWjp -di- + (Ojll\Vjp
d
=
(7.7g)
where the modal exciting force is
fL, fl.,
fjp(l) =
Jo Jo
I(x, y, r)t/Jjp dx dy
(7.79)
and the modal bandwidth is w(x, y, c) =
.., '"
L L
J= I
p~
Wjp(c)!/tjp(x, y)
1
(7.72)
CI
PiP
= pt
(7,80}
where the !/tjp are the natural modes of the simply supported plates .h
'l'jp
=
2' jxx . pny sm-smLy Lx
)
(7.73)
Since Equation (7.78) is identical in form to Equation . (7.51) forthe fi exuralbeam we can proceed by analogy to the formal solutIOn for the transv ) v == OW/(It on the plate erse velocity
which satisfy the orthogonality condition
fL. fL,
,
Jo Jo ptt/J jpt/Jkq dx dy
j~l p~1 t/J}p(X, y) IX)
v(x, y, c) =
= m6}p 6kq
(7.74)
1.3.2.
'"
fro
_'" h}p(O)iJp(C - 0) dO
Stationary Random Vibration
(7.81)
6
Here m =pcL"Ly is the total mass of the plate that also satisfies the free vibra tion equation
We now consider stationary random excit~tion of the plate by a conCentrated force located at x = a, y = h. To model thiS we lake the Space-lime correlation
·See Crandall and Wittig (1971),
·SeeCrandall and Wiuig(19111.
160
'I
RESPONSE OF CONTINUOUS SYSTEMS
7.4.
161
ALTERNATIVE SOLUTION (SHEAR BEAMS)
of the excitation to be
y
E[f(X1'Yb IdI(x2' Y2' (2)] = S(XI - a)S(X2 - a}6(y! - b)6(Y2 - b)Rp(12 _ II)
I I I I
:1
(7.82) which implies that the cross correlation of the modal forces is
7'
/'
Lyl3
+ r)]
E[!jp(t).hq(t
= IfJ jp(a; b)lfJkq(a, b)Rph)
/' Ly/3?
(7.83)
E[V2(X, y)]
=
«J
LI?
I I I I
j= I p= I k= I q= I
IfJ }p(X, y)lfJkq(X, y)IfJJV
,=
ljpkq
f.~,. I~,. ilill(O, 'J~kq(:'2)RI,«(l1
(2)
,lOl llOl
(7.XS)
i
I-:
2
Sp(w)w H Jp(w)H kq ( - w) dw
(7.86)
spec~rum ~e
As before, we take the input to band-limited white noise as given by Equation (7.64). We also assume that the average modal overlap ratio r=
modal bandwidth = Pjp c. LxLy _= _ _~ average modal spacing Aw 4rc.Jiiiij
(7.87)
is sufficiently small to permit approximations analogous to Equation (7.65),
that is, we only include in the summation in Equation (7.84) those terms of
I)p4q for which j = k, p = q, and the associated ~atural frequency w jp is less than
the cut-ofT frequency. For al\ such terms nSo nSo [ ---=-_. m 2p,p mC L L ,pjp -
1 x
y
(7.88)
Evaluation ofthe plate summation in Equation (7.84) is, however, fundamentally more difficult than that of the flexural beam summation in Equation (7:.61)
<
(5/
~
t-- L)2+L)2-1
x
FIG. 7.4. Distribulion of ensemble mean square velocity for wide-band excitation P applied the point x == 1.,/2 and y = 1.,/3.
,II
because of the irregular limits for.i and p, which are required to include only those modes whose natural frequencies lie below thc cut -off frcquency. We write
or in terms of the input spectral density. and the modal frequency response functions
Ijpkq =
('
(7.84)
when [jPkq is evaluated in terms of the input autocorrebtiol1 function iHld the modal impulse respnnsc functions
//
/'
w
(',()
--y/'
/'
By analogy with Equation (7.61) the mean square velocity at the location
(x, y) due to the excitation at the location (a, b) is 00
Pj--/-r//-
2
rcSo
E[v (x, y)] = mctLxLy
(jjjll<~t
I
IfJjp(X, y)lfJlv(a, b)
(7.89)
i,l>
The result of considerable computational skill, which we shall not discuss here, leads to an ensemble mean square velocity solution exactly similar to that of the flexural beam. The distribution of mean square velocity, as a function ofxand ywhen the random load P is applied atthejoint x = Lx/2andy Ly /3, has been obtained. It is extremely interesting to see a tic-tac-toe pattern in the two-dimensional sketch of Figure 7.4, which can be obtained experimentally by putting sand on a rectangular plate with a single random force P. Crossing lines in the plate represent the lines of high energy density.
7.4.
ALTERNATIV~
SOLUTION (SHEAR BEAMS)
In the previous analysis for the shear beam problem, we take advantage of the classical model analysis method to first simplify the problems by transforming them into uncoupled SDOF systems and then solve these SDOF systcms by either the time domain or the frequency domain superposition method, using impulse response and frequency response functions, respectively. Clearly, we may reverse this solution procedure by using the time domain or frequency domain superposition idea first and express the solution in terms of either the impulse response function or the frequency response function of the
l
rI
162
I !I
7.4.
:1
.j :~
z
7.4.1.
,
"
'.
I I
H~.(x.w)='
Consider the simplified shear beam problem with no damping. When the beam is subjected to a horizontal ground acceleration yo(tl. the total transverse displacement yo(t) + y(x, tl, as shown in Figure 7.1, is governed by the simple wave equation 2
ti ( y +_~)
ot 2
= k l)2( Y
2
(7.96)
hr(X.ljc-i""dl
(1.97)
Yo(t) = 15(1)
The associated term H ,,(x, role/"" is the response steady state shear force to a steady state excitation
+ ~)!
ox
-,.,
163
The subscript F indicates shear force so that, as usual, hf(x, t) is the response shear force of the beam to a unit impulse excitation; that is, for the special case where
Deterministic Vibration
m
ALTERNATIVE SOLUTION (SHEAR BEAMS)
'" f
original system (as versus the uncoupled SDOF systems). Then the task of evaluating the impulse response or frequency response functions follows. In this procedure, we are not restricted by the existence of modal solutions to the problem.
H
I I I I I
I I
RESPONSE OF CONTINUOUS SYSTEMS
Yo(t) =
(7.90)
ei(Uf
(7.98)
Note the difference between Ihis impulse response function Itt,(x, r) and ITPI from where the ground displacement yo(t) is added 10 the relative disphlccment y(x, I) in the derivative with respect 10 x bee.lUs\: yo(1l is imlepenliellt 01' the location x. Let the total displacement (Yo + y) be denoted by z(x, f), then
iJ2 z .
2
iJ2 z
-= 2 c -
ot
iJx 2
Equatioll (7.12), The lil'st is a function of x whereas the second is ass()t'iated with the ,ith mode of the problem.
F(x, I) ==
(7.91)
where the wave speed is
fa> j;o(1
(7.99)
t)lId x , tl tiT
or from the frequency superposition is c=
Jfn
(7.92)
F(x, t)
The boundary conditions are zero shear force at the free end on top of the beam
(Figure 7.1) and specified transverse acceleration at the fixed end at the bottom,
,
that is,
..-/
\
In terms of the unit solutions, the response shear force is
F(L, t) = k oz(L, t)
~=O
a~ z(O, t)
=
7.4.2.
(7.93)
2
yo(t)'
(7.94)
The shear beam is assumed to be initially at rest. For the solution of the shear beam problem defined by Equations (7.90) (7.94), let us introduce for the shear force F the impulse response function and frequency response function, respectively, as follows: hl'(x, I) =
2'~'1t S'_"
0'..
H,.{x, w)e'W'dw
(7.95)
r,.:;"
= ...... 1
f'"
2:n: - w
, [fa>
Ht·(x, ro)e"'"
-
00
'1
Yo(t)e- UUr tit dw
(7.100)
Impulse Response hdx. t)
Unlike the pair of unit solutions hj(t) and Hj(w), which are governed by the uncoupled SDOF systems and hence are already available from analysis of SDOF systems in Chapter 5, this pair, hF(x, t) and HF(x, w), must be evaluated from the formulation of the problem as follows: First consider thl! unit impulse response hF(x,t) to the unit impulse accelera tion excitation YoU) = a(O, t) = o(t), as shown in Figure 7.5 together with the velocity v(0, t). To solve for hF(x, t)= F(x, t) under this special excitation we begin by examining the governing equation, Equation (7.91). As a standard one-dimensional wave equation of the hyperbolic type, it is well known that there exist two characteristic lines defined by dx = dt
±
c
which represent straight lines in the x-t plane with slopes
(7.101)
± c. Sincc c = .Jk/III
~~ I
I:;'~ I
164
7.4.
RESPONSE OF CONTINUOUS SYSTEMS
HI
. f~"
,;;'')J:1 il
1:1
F4
LI
:.,
165
'"
alO, tl
I! I il
",'
oft)
I
ALTERNATIVE SOLUTION (SHEAR BEAMS)
P
VUU<~
F3
o
I'
wI v{O,l)
p«
r4'l~«.qz~
t-I- - - - - . . , . - - - -
:1
i
Ii
I.
FIG. 7.5.
Ground '~cceleration excitation and its associated ground velocity excitation.
1: rN/Z(UJ'l.(/AS\.~//~ o~~-+-~~
is known as the wave speed, the characteristic lines represent the trajectories of the leading wave front. In the x-I plane shown in Figure 7.6, the characteristic lines are inclined lines that are drawn starling at the origin x = 0 and I = O. For the wave solution behind the wave front we find that Equation (7.9l) admits the well-known solution of the form
z=
I(t ±~)
tal VelOcities,
(hl Shear torces, Jo'
IJ
FIG. 7.6. Waves of horizontal velocity v(x. t) = h.(x, t) and shear force F(x, II impulse ground acceleration a(O, t) = .5(1).
(7.102) V
in which 1 is any function that can be differentiated twice. The shear force is found by its relation to the slope of the beam
F=k5
ox
(7.103)
f'
and the solution, Equation (7.102).
~i
F= .
±~f'(d:~) c c
The velocity of the beam has the solution
.
f
(7.104) .
.
'-~:
oz = f' (x) t ±~
= at
"f(X, tldue tp an
(7.105)
Inspection of Equations (7.102), (7.104), and (7.105) show that c is the velocity of the wave of deformation as well as force. Furthermore, since the simple wave equation, Equation (7.91t is linear and homogeneous in z, any number of waves represented by Equations (7.102), (7.104), and (7.105) may be combined by superposition. The criterion for combining the wave solutions is to satisfy the specified boundary conditions given by Equations (7.93) and (7.94). For the unit impulse response problem the specified ground acceleration is
02 Z
iJt 2 (0, t) = a(O, t) = o(t)
(7.106)
and 0(1) is the Dirac delta function shown in Figure 7.5 l!long with the llssocilltcd ground velocity v(O, t).
I
I I I I I I I I I I ,I .;
~
:1
-t 1
i~ I
166
RESPONSE OF CONTINUOUS SYSTEMS
7.4.
In Figure 7.6 the ground moves according to the curve Vo shown as a unit boxcar function of t at the bottom. Then from EquatioQ (7.105),
v = Vi!
=;
f'(t)
= Vet) - U (t -
L 2e )
1'1
. /( x) ( x) ( = V
= ft."
c
t -
-.
c
-
V t - x.... - 2L) - c (.
(7.1 08)
L/4,
L/2, 3L/4,
f)
=- ~r c
(I - x.) = - ~ V(t. -:) + ~ V(r _2L) c
c
c
("
C
l:
I
Therefore, no new wave is needed at the time interval. This may be considered as requiring a new wave of zero magnitude. A review of the waves of horizontal velocity v(x, I) and shear force f(x, t) in Figure 7.6 reveals that the waves are periodic with a period of 4L/c Con sequently the wave solution for the velocity v(x, t) = II,.(X, I) and shear force f(x, t) = hF(x, t) are, respectively, hu(x, t)
+ f2
= 0
=
2L) = V (t + -x e2L) x -c-4L) - - V (t + -
x -c f •( t + -
+
f2 = +
Figure 7.6 shows block waves representing V2 and F2 and the combined values + V2 and f 1 + f 2 at different elevations. It is noted that V = 21'2 at the top. When the wave V2 returns to the bottom we have, from Equation (7.111),
VI
= V
(I _2~)
(t - ~~) - V (c - ~~)
V( _
V4
V
+ ...
(I _~ _2L) c
c
4L)
x -
1+-(-.
~.~~.~~) _
V ( _ x
~. ~~:)
(
8L) + ...
x -cV t +-
Jo (_1)n[v(t - 2n\+ x) + v(t _
21lL
hf·(x, t)
= f I + F2 +
k[
(7.112)
= V2(0, t) = r
(t _~)c
+
f
3
+V (
+(~L
-X)l
(7.114)
+ f 4 + ...
( cx) +V ( xc - c2L) 2L) -V (t+-x - 4L) t + --.'
=c -V t -
k ( + -x -c2L) k ( x - 2L) k ( x - 4L) crt - = cV t + - e - - cV r + - e -
v(O, t)
V3
- 6L) + V ( 1 + X--~--
(7.110)
(7.111)
+
2L) - V (
These requirements are satisfied by the solution V2
V2
X + V ( t+-,-.-
and is shown in Figure 7.6h. When the wave"l and 1-") reaches the top of the beam, u new wave I' = V2 and f f 2 is created, which travels downward. The V2 and f 2 must be func tions of I + (x/c). Furthermore, the boundary oondition at the top of the beam requires that
fl
= VI +
=V
(7.1 09)
I-
X -
c
c
- V(c - x : 4L) + V(I _x +c 6L) 6L) -V.(t +x -- - 8L) +,..
X +V ( t + -c
I
;:?,
4L 2L -~t~e c
Curves representing 1'1 arc shown in Figure 7.6<1 at different elevations, x = and L. These curves lie within a belt that has a slope c relative to the axis of t. The corresponding shear force according to Equation (7.104) is
;
iI
which satisfies the required boundary condition that v(O, t) = I in the time interval
(7.107)
where V is the Heaviside unit step function, and 2L/c is the time required for the wave to travel from the bottom of the beam x = 0 to the top x = Land back. The movement creates a wave that travels upward and according to Equa tion (7.105) contributes the velocity
167
ALTERNATIVE SOLUTION (SHEAR BEAMS)
(7.113)
~ (- 1)n [V(1 = kC n~o
e
-
2uL + 2L.'I:) --c----
V
+ x)] (7.115) ----c-·--·
( 1 - 211L
r
I
I
I I
~ ,. 'I
1
II
II i j
11 11 ] 11
168
Note that in Equation (7.114) a total of eight terms are given in detail corre sponding to Vl-V4' Of these eight terms, the first and the third correspond to 11 = 0 in the summation, whereas the second and fourth correspond to 11 = I. In Equation (7.115) a similar combination of terms is used. 7.4.3.
:1
02z(0, t)
~
yo(t)
f.
I~
I!
wL + cos wL -- cos (J)X)/ cos-·
- 1 ( sin wL. sm wx _.
c
C
C"
C
169 {7.1231
C
From Equation (7.119) the complex frequency response for the shear force is H p(x, w)
7.4.4.
=e
iWf
(7.116)
= we -k ( cos wL cos wx c e
-
. wL .
((IX)
e
c
Sill -
sm -
I
cos wL e
(7.124)
F(x, t)
= H p(x,
(1)
)ei ."
(7.117)
This shear force is related to the steady state displacement z by Equation (7.103). Using this relation and the definition Z(X, t)
= H .(x. W)ei""
Stationary Random Vibration
Now consider that the horizontal ground acceleration YoU) is a stationary random process with zero mean, spectral density function Sew), and auto correlation function R(t). The mean value, spectral density function, and the autocorrelation function for the response shear force can be determined in terms of the excitation process and the structural properties by exactly the same formulas, which have been developed for the SDOF systems. The stationary mean shear force is
and the steady state response shear force by
E(F) =
(7.118)
we find that
t",
(7.125)
E[jio(t)]hp(t - t;dt = 0
The spectral density for the shear force is Hp(x, (0)
I I ,I
ALTERNATIVE SOLUTION (SHEAR BEAMS)
Let the steady state ground acceleration excitation be specified by
'I
I
HJx, w) ==
Complex Frequency Response HF(x, ro)
1 1
7.4.
RESPONSE OF CONTINUOUS SYSTEMS
= k oH.(x, (0) ox
Sp(X, w)
(7.119)
= H~(x,
00)
Thus HAx, (0) can be obtained once H%(x, is determined. Substituting z(x, t) of Equation (7.118) into Equation (7.91) gives
2
d H. dx 2
+ (00)2' _ Hi<
0
e
(7.120) RF(x, t)
at
= I
x=o
dx
at
x=f.,
(7.12)
(7.122)
The solution for H.(x, w) that satisfies the governing Equation (7.120) and the boundary conditions, Equations (7.121) and (7.l22),is
=
f:", f:..,
R(t
+ 01
-
02)h p (x, O\)lrf(x, 02) (WI
(lOl
(7.127)
where the impulse response hp(x, t) is given by Equation (7.1 To gain some insight into the stationary random response of the shear beam, let us consider the idealized case where the excitation process is a white noise with uniform spectrum So and autocorrelation 2n:So()(t). The response mean square for the shear force from Equation (7.127) becomes
and from Equation (7.93) as dH, =0
26)
where the complex frequency response HF(x, w) for the shear force is given by Equation (7.124). The autocorrelation function for the shear force is
which is an ordinary differential equation for H.(x. (0), considered here as a function of x only with parameter w. The boundary conditions of H.(x, w) can be evaluated from Equation (7.116) as
- w 2 H.
= H F(X, w)H;(x, w)S(w)
'-,,',
E(F2)
f:", f:", = f:",
= RF(x, 0) =
2n: So
2n:Sob(8\ - 8 2 )h p (x, 8\)h f (x, 8 2 ) d8 1 d0 2
hi(x,OI)dO I
(7.128)
;" .'
~ ~i :llj
at1
fj :=;
:
I
Iii. I I I' I III I I
I I
I I
i
.,
170
7.5.
This last integral can be carried out most efficiently when we consider the graphical presentation of Equation (7.115) for the impulse response hf·(x, t) in . Figure 7.6b. Due to the symmetry and periodicity of hf·(x, tl, the relevant part of the integral in Equation (7.128) covers only one quarter of a period In the time axis. For the first quarter period Lie, the impulse response function hF(x, t) is zero at location x before the wave arrives at x and jumps to a value of kle according to Figure 1.6b. Thus at location x, the relevant part of the integral ill Equation (7.128) is simply LIC
f
xlc
k2 dt k e
=
E(F2)
k2
e
~
!
~
t
~ l
"
I K :
7.5.
DAM-RESERVOIR (VERTICAL EXCITATION)
Formulation
As shown in Figure 7.7 we consider here the random vibration of a dam reservoir system subjected to a vertical ground acceleration. Assuming a onc dimensional motion, a rigid dam, and an elastic reservoir foundation, the vertical displacement u at time t and location y for the foundation and ttie water, respec tively, are governed by the simple wave equations (,2.,
f~
f'
H;(O, w)So dw
_....2 = al
2 (,2"
Co .........
iJy2
2 iPu _=c 2 i)_ u
ot 2
oy2
for y
~
(7.IJ.1)
0
forO < y
~
H
p(H, t) = 0, p(O-, t) = p(O +, t)
i
oo
.
0
(k)2 - 2dw ..... co
i)u
(7.136)
p = - l i)y where l is the bulk modulus of water. The system is initially at rest. y
(1.132)
C.W
We have selected the undamped shear beam problem to introduce the alternate method of analyzing random vibration because of its simplicity. The idealization leads to an unbounded response statistic, which must be modified if the response analysis is to be practically useful. To achieve this objective, two improvements on the formulation of the problem can be made. One is to introduce a damping mechanism. The other is to consider a nonstationary
(7.135)
The pressure is related to the displacement by the compressibility law
.
E[F2(0)] = 2So
(7.134)
where c~ = lo/Po is the wave speed in the foundation and c is that in the water. Neglecting the surface wave effect the boundary conditions for the hydro dynamic pressure pare
(7.131)
Using the complex frequency response of Equation (1.124) in Equation (1.131) ~~
11'1
ILl
I I
(7.130)
- x)
where n is the number of wave passages. Eaeh walle passage is represented by a characteristic line in Figure 1.6b. As the number of waves approaches infinity in the case of an undamped shear beam, the mean square response shear force of Equation (7.130) also approaches infinity. Such an idealized answer is analogous to that of the un damped SDOF system where the mean square response to B white noise process tends to be unbounded because the stationary excitation has an idealized infinite duration and there is no energy disicipation mechanism. It is also instructive to see that the unbounded mean square force response, Equation (7.130), can be obtained through a frequency domain analysis using Equation (1.126). Thus, for simplicity. we consider the mean square shear response aUhe base of the beam x = O.
E[F 2(0)] =
171
(7.129)
(L - x)
= 2nSo 3' n(L
DAM-RESERVOIR (VERTICAL EXCITATION)
random excitation that only lasts for a finite duration. First we shall introduce a damping meclulllism 110t on the shear beam problem but on a dam rescrvoir problem which, however, is mathematically analogous to the shear beam problem. For nonstationary excitation and response, we shall discuss the subject in Chapter 9.
7.5.1.
2
2
The mean square value of the shear force is therefore
,
I' I
RESPONSE OF CONTINUOUS SYSTEMS
y=H
H
y=O FIG. 7.7.
Dam reservoir system.
H
I
~
172
~
I I " III il I II Il I ~
t '.
'I t
I III ~ ;~,
i,.... , I :
~
2
a u(o, t) = iJt 2
7.5.2.
~
I 115
uo( t)
(7.137)
Impulse Response Function*
f'
(t ± t)
(7.138)
for y ,;;; 0
are used first to derive laws of wave reflection and refraction as it encounters the interface as follows: Refer to Figure 7.8a. First we consider an upward moving particle velocity wave VI' After encountering the interface, a reflected wave t'2 in the rati.o of lX and a refractcd wavel'l in the ratio of 1 + lX are generated Sli Ihal lit the intt;rfan' r- n. ", I I'~
..
(I I .lIFtt) "-. I'J
17·139)
In the meantime the associated pressure waves PI' Pl' and P3 are determined
U3
y=o
=
(l
+ alr(t
- +)
P3 =(1
+
'
-
~)
~
Interlace
PI(O-, t)
lo (1 Co
=r(l- ~)
U2
= ar(1 + ~)
(a)
FIG. 7.8.
PI
P2 = -a~r(1 + !) Co c
AO ( /-y' ) =-r 'Co co
(6)
.
+ Pl(O-, t) = P3(O+, t)
=
~ (i + lX)
=
loc - leo CoPo fP = loc + lco CoPo + cp
C
(7.140)
where the constant ex representing the ratio between the reflected wave and the incident wave is called the reflection coefficient. When the foundation is rigid. Co approaches infinity and the rcflection coemdent becomes unity. Note that for an upward moving particle velocity wave, the reflection ratio is + lX whcrclis the ratio is - lX for an upward moving pressure wave (Figure 7.8). For a down ward moving wave it can be shown in a similar manner that the reflection ratio is - lX for particle velocity and + lX for pressure. Now we consider the impulse response function h,.(}'. r) due to an input Dirac (idea accch:rati(l\l incident wave. which is equivalent to a Heaviside ullit step particle velocity wave as shown in Figure 7.5 for the shear beam problem. To construct the wave solution, we refer again to the }'--t plane with inclined characteristic lines as shown in Figure 7.9 and superimpose basic wave solutions in such a way as to satisfy the specified input particle velocity at the interface and the boundary conditions, Equation (7.135). For particle velocity waves, the condition at the water surface y = H is to have 100% positive reflection. At the bottom interface the downward moving wave of particle velocity reflects in the ratio of - IX. The input particle velocity is a Heaviside function of time U(t) at the interface. First consider a boxcar function of particle velocity wave VI with unit magni tude and width 2H/c moving upward in Figure 7.9 in exactly the same manner as that for the shear beam of Figure 7.6. = U
(t - ~) - (t _~ ~ 2:) U
when the wave III reaches the top water surface y moving downward is created with
Reflection and refraction of upward' moving waves allhe interfaces.
·See Newark and Roscnhluclh (1911).
lX)
•• lX
VI UI
173
or
A comparison of the formulation for the dam-reservoir problem with that of the shear beam problem in Section 7.4 shows a mathematical analogy where the hydrodynamic pressure P here corresponds to the previous shear force F. The only difference is the additional boundary condition of continuity of pressure at the water-foundation interface, Equation (7.135). This condition and the wave equation for the elastic foundation, Equation (7.133), with its basiosolution V(y, t) =
DAM-RESERVOIR (VERTICAL EXCITATION)
according to Equation (7.1 36) and are shown in Figure 7.8b. Finally the bound ary condition, Equation (7.135) at the interface requires that
The excitation of the system is the vertical ground acceleration
HI ;\
7.5.
RESPONSE OF CONTINUOUS SYSTEMS
t'2 =
(7.141)
H, a reflected wave
y 2H) - U ( r + -(-.y - 4H) tT ( r + -(-:-
t'l
(7.142)
~., >,
I
I,
1;1 II I.
I.
II III ~I ~I ~II ~I
174
RESPONSE OF CONTINUOUS SYSTEMS
1.5.
y
175
DAM-RESERVOIR (VERTICAL EXCITATION)
created with Vs = (I - (X) [ V ( 1
JI 4H) + -c-'-
- V
(.+ y--(-~ - 6H)] 1
(7.145)
In this way. the impulse response function for the water particle velocity is
~~~Wff{{{{$~~«<~'''?';;~?~)o
y=HI
hv(y, t) = t'. +
+ ...
112
= U (/ -
~) -
(I _t _ 2H)
V
('
(
2H)
('
(
Y +V ( 1+-(-.-
2H)
.+ (l - (X) [V(t - -y - ('
+ (I
~ ~mmmi'Jm77J~j.1W4(~~,-W//./'J???0'//.AJ 1
Ty
=O
.::LI
-.
,
ItT
I
~'HC+'HC+'I/"'+
-
a[V
(t - ~ - 2:) - (t - ~ - 4:)] U
(X
+ (1
- (X
= V (/ -
~c - ~~) c
V
(I _!' - ~l!.) . c
c
+ a2V
+ (X)2 [V(I
y
- 'Y -
r
4H) c
(
611)J ('
I - 'V -
V
--
o
r
y - 8H)] + ... + (X2) [( V t + Y- c6H) - - V (I + -,-:-
.2
C
C
(t _Ec _ 4H) _ V (t _c c
:!:'. _
(
Y 6H) + rx V
- a V t - ~ -:- -;;-
(7.143)
2
(
t
y- 2H) (t - ~y) + V (t + --c- a.V ( + y -c 4H) + ...
= V
(7.144)
to form the upward moving wave V3 + V4' When this combined wave reaches the top water surface y = H, again a reflected wave Vs moving downward is
C
1 -
f!:=
This reflected wave now superimposes on the second boxcar input wave V4
C
y
C
= -
4H)]
-.Y -
y- 2H) ( Cy) + V (t + --c - (XV t - Y - - 2H) - (XV (y t + --4H) (
~tl
V3
(t -
V
-
C
a>[ V (t + -c 4H) - V (t + -c 6H)]
+ (I
V
FIG. 7.9. Waves of water particle \
tion u.,(/1 = <'i(/) in the vertical direction.
When the wave Vl reaches the bottom interface y = 0, a reflected wave v) moving upward is created with
4H)
J' V t+-(-.
r
hv(Y, t) = "~O ( - a)" V t
[(
2nH + c
6ft) + aV c
(t-
:!:'. _
c
611) c
y - 6H) + --c+ ...
(yC- -;;2H)
rxV t -
y) + V (t -
2nH +
2H - y)],(7.146)
c
•I
176
RESPONSE OF CONTINUOUS SYSTEMS
The impulse response function for the pressure is related to that for the water particle velocity by
I
111
I) I
p(Y, t)
=
OU = +- -,1,( f' t ± -Y)
A -0 y
c
=
- c
+- Ac v(y, t)
(7.147)
as shown in Figure 7.10 for each individual wave in the superposition process. To conform to the boundary condition at ,the top of the water surface, the pressure wave must have 100% negative reflection. At Ute bottom interface, a downward moving pressure wave reflects in the ratio of + IX. Referring to Figure 7.10 we have
~
" ~
"0
+
"
C :::I
hp(Y, t) = PI + P2 + ...
1'1 Ii I I~ I It I Iii
=
.j
III
~
~l
t
~ [ U (t - ~) - U (t - ~ - 2:)]
+ (l
::!
-
l -
,l [ (t - (I - IX) c U
Y -c411) +- -
(
U t
c
til
:::I
-
"'I"
"ii
611)J
l:!
a
E. ·u E ...
0.15 .~
611)J
Y t-~-c
~
.!:
c."
.g~
i'E .<:: >
-.,
'OJ:!
., c:: ... > .
~~
IX)"
C
::.
....
""II
15:
Y c+-
(t + Y -c 6H}_ U (t + Y -c 811)J+ ... = t ~.[U(t - 2nH + Y) - U (t - 2nH + 2H Y)]
"'"
::.:.
::!
(I - IX + a: 2 ) ~ [ U
hp{Y, t)
0.
.5
...
-
c c411) . U (
2 A[ +(I-a:+IX)C U(t- Y
""
~
2
.A[( Y 7' 211) - U ( Y IX); U ~ ~ 41I)] 7 l
2
:;
Ie
I
,
III
,.11
"ii
-cA[U(l +Y-- c2H) - -U ( l +Y-- c411)J
'f ~
"I'"
C
(7.148) Note that h,(Y, t) differs from h.(y, t) only by one factor Ale and one sign for the downward moving wave components, which are repreStmted by the second term in the summation. Note also that A = c"p so that 1ft .;" pc. lt is important to note thilt, as shown in Figure 7.9, the iaput particle velocity from y = 0- to Y = 0+ across the interfacqhali a constant magnitude of unity for 0 '" t, as required. The resultant particle velocity at y = 0 +, however,. consists of the unit magnitude input, the wave coming down from above, and the reflected wave. It is in general different from unity in magnitude except for
~
.
• 11 o~
!":-!!.
~
:t::
"
H~
":-3
ci g
ii: .;:
-
177
I
178
I
I
I~·.·I
Ii:
7.5.
RESPONSE OF CONTINUOUS SYSTEMS
the case of a rigid foundation where (7.146).
IX
I. The resultant is, from Equation'
211H) + U (t - 2nH c+ 2H)J
. IX, [( v(O, t) = 1t,,(O, t) = n~o (- et)" U t - -c-
i
il,..(y, II
~1;
(7.149)
h"
ll
--r
h,,(Y. I) dy
~
~I
Q
11
5
~
(7.150) .
II~'(J', I) dy
)'
,
t V&'/Il
I" ,,,1
,,:r~ ( \
vac////&'A
i, , , , , , ,~
V'//// i
1
11
1'1
I)
f
)'
I'MU', t) ""
179
hl'{l/.I)
For the other important response functions, namely the shear force F and overturning moment M on the dam due to hydrodynamic pressure, the respec tive impulSe response functions can be derived by the foflowing simple relations:
111
DAM-RESERVOIR (VERTICAL EXCITATION)
I) given hy EqualiOli (7.14!!). To I:arry nul these intcgnltions. it is convcnicnt h) lakc advantage ofLhc periodicity alld symmetry of hp(Y, t) as shown in Figure 7.11 such that only one quarter of a period in time is needed for intc~ration. Thus for 0 ::;; I.::;; HIe and for loc,itions behind the leading wave, that is,
I
alld the solution for il,.CI·,
for 0 ::;; y::;; el,
I
---.l.
'
+ IX) hp(Y, I) = pc ( '"-2
"
--'r
= Jyf
(7.151)
el
hf,(y, t)
hM(y, t) =
h,,(y, t) dy
(I
et)
+ (et - y) pc -2-
+ IX) {el J,fd hf·(y, t) dy = 2pc (1-2-'
y)2
(7.152)
"'-1
(7.153)
These are displayed in Figures 7.11-7.13, respectively.
~
~
7.5,3.
Although the complex frequency response function Hp(Y, (0) can be obtained by solving the steady state problem for pressure due to an input steady state ground acceleration ei"" in the vertical direction at the interface y = 0, it is more convenient here to use the Fourier transform relation. Thus H p(y, (0)
=
f:",
FIG. 7.11. Unil impuls1! re'pollse "I' pressure (fwill Newark autl Ru,-;cnhlucth. 1971 \. rh" l'IIlIS!
",r:."
Substituting 111'(.1', I) from Equation (7.148) into E(jcation (7.154) and currying out the integral we obtain ~'. "i"
:}
I~L"
i~-'
Frequency Response Function
H (v w) I' ,.,
hp(Y' t)e-;"" dt
(1 + IX) sin[w(H - .rl/e) = -pc -----------.----.-....
P.l55)
(7.154) Note that when the rdledion I:odlicicnl
(X '"
1. whidl I:orn:sponds Itl a 1 if:!.id
:7i
7.S.
I
I
I
II
III
~II
DAM-RESERVOIR (VERTICAL EXCITATION)
181
hA/(}l, I)
hp.H, I)
l/~
ID
~
to
I I hF(!!.' t) I ~ I \.
\
I
,
I I
I I
I
I
I
TI~I -I I
\
III
t~ ~
'YI
a/3
f
r
f
\
I J
\ \
I
I
I
I I
I
I I I
I I I
\\ I 1
\ \
\
I
\
\
\
,
1'n2J:'l1 ~ \
illl!
~I 1'1..1
E.
FIG. 7.13. Unit impulse response of moment (from Newark and Rosenblueth. 1971 ). The constant multiplier WeH'(1 + 11.)/19 is deleted.
;~
I~I
~I
foundation, Hp(Y, (0) simplifies to
,1
Hp(Y, (0) = pc sin £O«H - y)/c](l £0 cos(£OH/c)
;
til
~f.1
"r~:
FIG. 7.12. Unit impulse response of shear force (from Newark and Rosenblueth. 1971). The constant multiplier WCH(I + a.)/19 is deleted.
:-1
180
1
t:1!t1
+ IX)
(7.156)
which is analogous to HF(x. (0) given by Equation (7.124) for the shear beam problem. For the shear force F and overturning moment M, the respective complex frequency response functions can be derived from basic dclinitions
ttl 182
I
t. tl
t. ~
~
-, -, , , ,, ,
RESPONSE Of CONTINUOUS SYSTEMS
7.5.
(II Jy Hp(y,w)dy
.
(C)2
= f,(l
HM(y,w) =
+ (X) ~;;
I
cos[w(1I - 1')/(']
(1+- (X)-~;~(~,~H/<~)+i{T- 1)!,'~i~l(II;/I/(i
(7.157)
fft Hf·(y,w)dy
Jy
= p(1
7.5.4.
n
Il
.
+ (X) (~)2 ~::..")- (l{~:ill[
(7.158)
Q;
~
Now consider the problem of a v1:rtical ground acceleration input modeled by a stationary white noise process with zero mean. Sueh an idealized random motion is defined by a uniform power spectral density So with the associated autocorrelation function R,,(r) 2nSon(t), where n(r) is the Dirac delta function of the time lag T•. For such a random input, the stafionary response power spectral densities ~'SI'(Y' (II), Sr(V, w), and SMCV, (I) for pressure, shear, and moment, respectively, Call be readily determined by the following, well-kllown relation:
Sty.
(II)
= lI(y. wl/r"(y, m)Su
(7.159)
Power spectral density 01 pressure, shear, and moment
.:.
7
SF(Y, w) =
wAc2 H2 -2-2-
SoG(a, 4') sin2[w(U - r}/cJ
gw
SA-/(y, w)
W2 c 2
H2
(7.160)
C
., ... !'! ::J
(7.161)
I !,
'. r.\"
30
I I I I I I I I I I 1 I I I
- y)e] - sin[w(H
)')/C]}2
(7.162)
)2 [[(,+-;)2 cos 2(wfiTc) ++(1':'-;)2 sin2(Wii/c) (I
aJl
0 b
'u; c:
'"
'0
20
~
~
'"
J
depth.
rl
~
cos[w(H - y)fC]}2
where
G(a, w) = ( wH
I,
Wllkr
I I II Ire--- Approximation I I
Q.
SoG(a, wI{ 1
= - 2 - 2 - SoG«(X, w){[w(H gw
40
'0 al~ 8,,-~
SI'(Y' w) =
Variution of hydrodynamic r~Sp()llSC with
fiG. 7.14.
Thus
w1 H2
SlAv. 1)181'(0, !)
:;,
Response Power Spectral Density
~
\
mental circular frequency and is shown in Figure 7.14 at the reson.mt frequency = 1.0. Power spectral density for the stationary response pressure tit the dam base}' = 0 for the deformable reservoir base with IX = 0.815 is shown in Figure 7.15, where Po = wH is the corresponding hydrostatic pressure. Thc sharp resonant peak at w = WI = 1T.c/2H and the diminishing response at high frequencies are clearly noted.
and from Hp(Y, w) as follows: H,,(y,w) =
183
DAM-RESERVOIR (VERTICAL EXCITATION)
10
It
0.815
I.
,t'i
]
(7.163)
The variation of the power spectral density with the water depth ralio JI IH depends on the frequency ratio n = w/w l where WI = m:/211 is the funda
01
.., I
.......
o
2 Frequency,
fiG. 7.15.
'
3
n '"
Power spectral density
I
4 wlwi
"r pressure at dalll base.
..
5
r:,
184 7.5.5.
7.5,
RESPONSE OF CONTINUOUS SYSTEMS
•
Response Mean Square
with IX
185
DAM-RESERVOIR (VERTICAL EXCITATION)
= 0.815 are given as follows:
~
Since the mean square value of the response is equal to the area under the power spectral density curve, a straightforward integration operation should do the job, Unfortunately, a closed form solution has not been found for the integrals
involved and thus a numerical procedure must be used. For the present problem
the alternative approach using the time domain integration has been found to be more convenient. because of the periodicity and symmetry of the unit impulse functions presented in Section 7.5.2. In terms of the autocorrelation
functions, the response mean square pressure is
,I
,.
~
,
E[p2(y)]
= Rp(O)
L"" fo""Ra(t , -
E[ F2(O))
~
~
, ~ , , , ,. ~
'
1'"
(1p(O)
(I
+ IX)2(H -
'" L
y)
(X2n
(2F~tI! S~) g2
(7.171)
(7.172)
,-s-(F°'l./7 rs;;;;)
(7,174)
2'1./~
With the help of F~gure 7.11 and Equation (7.148) for the unit impulse response
h,,(y. t) this integral is carried to give the following closed form solution:
E[p2(y)) =
= 15.88
~(Po '1./7 ~) = 2 '1./1=7
(7.165)
h;(y, t) dt
(7.1 70)
The effect of the deformability of the reservoir base on the response standard deviation (J at the dam base is shown in Figure 7.16 using the following results:
t2)h p (Y, t,)h,,(y, t 2 ) dtl dt2 (7.164)
)
2nSo
Cpij;ISO)
E[M 2(0)] = 21.44 eM!~1 So)
Since the input autocorrelation has been specified as 2n:Sol5(t) from a stationary
white noise process, Equation (7,164) simplifies to
E[p2(y)]
= 11.91
«AI
(0)
(Mll !~:I)
2
(7.175)
The variation of the standard deviations with the random ground motion is shown in Figure 7.17 for dams with a deformable reservoir base. The standard
(7.166)
ft=o11
where 00
L
(7.167)
n=O 0:1
Similarly, for shear and moment from Figures 7.12 and 7.13 and Equations (7.152) and (7.153), the response mean squares are, respectively,
E[F2(y))
= n:S3g ow c (l + (Xf~ 2 1 _ /X2 (H -
.l:!
1
2
y)3
(7.168)
E[M2(y)] = nSow2c (I + «)2 , 20g2 1 \X2 (H - y)S
(7.169)
1.0
..:: ~ 0.815
i
0,5
!
0"
)r
I
Root mean square pressure, shear, and moment:
In terms of the hydrostatic pressure Po = wH, shear F{) = lwH2, and over turning moment Mo = iwH3, and the natural circular frequency WI = n:cj2H, the mean square responses at the base of the dam for an elastic reservoir base
(s.
I2S(;;:;;- , op(OllR
"p(OI/poV ~--2 g
FIG. 7.16.
0
Il.8so
"MCO)/Mo V ---==--2 ~
ElTecl of reservoir base rigidity on hydrodynamic response.
II;.1 h ,. ~I
~ ,.
186
, ,. , ~I
DAM-RESERVOIR (HORIZONTAL EXCITATION)
187
(7.160),
.'"
N
~
t
c:
C[/)-(O)]
" .:1 0.815
1.0
.z:
2
i'
1···· o
(21)(~SIlWl) \.(0. SI'(O, wI l l w · --,' ,. H)1 (/~! !J" (J (fJoSo/~J I
(7.176)
..g"'.
~"" ~.~
In Equation (7.176), the integral represents the
0'0
"'
"O~
~i (!)::
0.5
~
7.6.. DAM-RESERVOIR (HORIZONTAL EXCITATION) 00
2.0
1.0
7.6.1.
Root mean square pressu~, shear, and moment:
~
-,
7.6.
RESPONSE OF CONTINUOUS SYSTEMS
"1'(Ol/PQV2.
.FIG. 7.17.
,,~{0)IFoV67j. "M(O)IMo ~~
Referring to Figure 7.20 we consider a rig.id dam with a reservoir of water of depth 1/. Our main interest is to determine the hydrodynamic df,xts 011 Ih.: dam when the ground is given a hOl'iZOlllal acederati.m 1/(1) Sil1111lal iug an earthllua ke evcn\. To rOrlllulat.: this dynami..: pwbkm we assllmc lil;lllallhc displa~'~Ill"::tlls
5
EITL'Ct of vertical ground acceleration on
hyd~odynamic
ForrtlUlation
rcsponS().
deviations of the, hydrodynamic pressure, shear, and moment vary with i, l, and! power of the water depth ratio y/H, respectively. These are shown in Figure 7.18. Note that these response variations with depth are independent of the refraction coefficient ex. Having obtained the mean square responses througb a time domain integra tion approach. it is now instructive to make a very cmde check by integrating the area under the power spectral density curve given in Figure 7.15 as follows: For the mean square pressure at the dam base, by definttion and from Equation
ta
A :.-- cis ex
p..t (Jx
(c~/I -;;'-i
(1
(7.177)
lim
11
~
, , ~
,
1
IL·
1.0M::K--_
~ s: a
clX~
x
~
i
Wdl
0.5
~"
3::
I o I :::f:::::::;;ti;,J o
0.2
0.4
0.6
0.5
1.0
Root mean square pressure, shear, and I11()mMf FIG. 7.18.
Variation of hydrodynamic response with water depth.
..
I·A ,ix
__ "
"~lT+ FIG. 7.19. Ouc-uimcnsinn:lt slrcss wave ill lin d:t,tj.: hal'.
I iii
I
I. I
, I
II
I, II ~
I ~ I
188
RESPONSE OF CONTINUOUS SYSTEMS
where u is the displacement of the element. As an elastic bar, the stress related to the strain au/ax by the Yong's modulus E as: (J
ox 2 =
I I
IJ•
II.
III
~,~,
V2f)
E au
1 c2
02U
c2
E p
rao}
2
0
p at2 {u}
=
2
I 0 0
c2 p
(7.178)
ax
= (l + Jl) lax +
189
or
is
1+
(7.184)
p
where cp is known as the speed of the dilatational wave in elastic solids. Further more, since the pressure p is related to the dilatation 0 by
(7.179)
-
'\.;~
so that
.,
(7.185)
."
.~.
where c is the elastic wave speed. In a completely analogous manner, th~ three-dimensional wave equations can be derived. Here we shall present alld discuss these equations without derivations as follows: In a matrix form, the three-dimensional wave equations are
II I
DAM-RESERVOIR (HORIZONTAL EXCITATION)
P = -AO
a2u
~
II.
(J
Differentiating Equation (7.178) with respect to x and then substituting into Equation (7.177) gives the well-known simple wave equatjon
~
II.
7.6.
j {J2p
2
V p = c2" ot2 p and cp is therefore known as the speed of the p wave. For a non viscous fluid,lt = 0 so
(7.l80)
Cp
=
t=c
""' .~
.\<;;
i
(7.186)
::~
4)i (7.187)
which is known as the speed of a sound wave in a fluid. The combined wave equation for the fluid is therefore
where
8
=
au + ov + aw ax ov oz
(7.1SI)
V 20 _
= unit vol~me change or dilatation V2 =
02
a2
iJ2
--+-+_. iJx 2 iJy2 {JZ2
= Laplacian operator = bulk modulus K when /1 "" 0
Jl
= Lame's constant =
V2¢
a2 (J
ot
2
= {l
+ 2/1)V 28
=..!.. 02¢ cl
0 for non viscous fluid
(7.188)
7fi2
(7.189)
~'.
The preceding matrix equation has three scalar components, one in each of the three space directions x, Y. and z. These three component equations can be combined by differentiating the first with respect to x, the second with respect to y, and the third with respect to z and summing up the three to give the fol lowing single wave equations: p
02(J
It is instructive to point out that the derivation of Equation (7.188) follows the traditional approach in solid mechanics. The standard wave equation for non viscous fluid in fluid mechanics, however, is usually expressed in terms of a potential function ¢ as
(7.182)
l
j
- c2 ot l
','E.
where
{:x}¢ = :t The two forms of the wave equation for a non viscous fluid, Equations (7.188) and (7.189), are equivalent, as shown in the following. From the preceding wave equation in terms of ¢ and its definition, Equations
(7.183)
L
I 190
I I \~ I
(7.189) and (7.190)
!II
I'. I' I 1,1
DAM-RESERVOIR (HORIZONTAL EXCITATlONI )'
p=o
(12 iJ2u
,~-o
il2w)
(}2"
X'- 00
-}" ( iiic1x + (ilA), + (II ()~
(7.191)
al
aJ,:oy
Integrating with rcspc(;t to I givcs
1\ I i\1,, \
;
iJ
f)
(I -:;-
+ constant = p
=
~~
fI
=0
FIG. 7.20.
since p = 0 at I = O. Applying the Laplacian operator V 2 on Equation (7.192) and then making usc of Equation (7.19 I) gives
A dam· reservoir system subjected 10 a horizontal ground acceleration.
where ao is the constant amplitude and (() is the frequency. Thcn by dctinilil)ll. the steady state response of the potentiale/> is given in terms of its complex frequency response function H",(x, y, w) by
2 jill' i! 2 fl (Fp /IV -.- = P .• V q, = ._....
,
V-I')
(It
I
vi
)" Oil
,:2 lilt
(7.193)
which leads to Equation (7.188) becall~e p' - }"O. In summary, we have shown that there are four choices on the form of the general nonviscous three-dimensional prdblem. These are given by Equation (7.180) in terms of the displacement vector ~u}. by Equation (7.183) in dilatation 8, by Equation (7.186) in pressure p, or by EqujItion (7.189) in potential . The boundary conditions are p = 0 at the top water surface, vertical velocity o4>/iJy = 0 at the reservoir bottom, horizontal acceleration 024>/ot ox = - a(t) at the vertical face of the dam, and 4> -+ 0 as x .... 00. Finally the so-called radiation condition is imposed. This requires that waves propagate away from the dam. that is, in the + x direction.
7.6.2.
·See Chopra (1961).
(7.196)
= 0
2 ax
() H>
~'H + (W)2 ~ _(. H..
-~2+!l2
= 0
('J'
(7.197)
To solve for Hoi> in Equation (7.197) we first summarize the boundary conditions
. oH -iJ x01> = - 1
at x
oH4>=O
atr=O
(7.199)
H",=O
aty = H
(7.200)
H",
asx-> 00
110
a.I'
(7.194)
+ (~r H",
which reduces to the two-dimensional form
Complex Frequency Response*
aCt) = aot""
(7.195)
Substituting
j)lp
=
x
(7.192)
Let the input acceleration 'be specified by
~ !II.. '
,"
'/
2
~
r
- u(/)
V2
jJ() eJp }".-- = -
(Ie
a >ll)lo:< 2
V H>
11\ I t.~ dll
191
p (11 2
-
l! 11
7.6.
RESPONSE OF CONTINUOUS SYSTEMS
=0
and --+
0
and
0 ~ .\' ~ H
(7.198)
'III
III IIt
:5\.
,
Ir~
I
~~:
ot:l u
I I~r. I' L
192
RESPONSE OF CONTINUOUS SYSTEMS'
Let w/e
= k = wave number in radians per foot and H~(x, y, k)
}.',
1\: I
III rI ~I
DAM-RESERVOIR (HORIZONTAL EXCITATION)
= f!" cos, k~y
(7,201 )
4
+ k2 = 0
k;
ff" cos knH
at y = 0
(7.203)
0
t)Jk[
(7.210)
k2
With the constants an evaluated in Equation (7.210), the complex frequency response H.,(x, y, k), as given by Equation (7.207) is eompletcly determined. For the hydrodynamic pressure p, the corresponding complex frequency response function H p is defined by
(7.202)
- eax sin kn y = 0
(- I)n-I
iwn(2n -
Substituting Equation (7.201) into Equations (7.197), (7.199), and (7.200) gives (X2 -
193
Carrying out the preceding integration and making use of Equation (7.206) gives Un
,
~;
7.6.
(7.204)
p(x, y, r) = H p(x, y, w)uoei«lt
(7.205)
Since the pressure p is related to the potential> by Equation (7.192) it follows that
(7.2\1)
From Equations (7.202) and (7.204) we find that
(X=±.jk;'=P
n¢
and that
P = I' /It
k.
=
(2n - l)n 2H '
n =.1,2, ...
(7.206)
.
= piwH",aoe"ot
.
(7.212)
= HrGoe"ot
(7.213)
iwpH",(x,J',k)
:. Hp(x,jl,k)
11
I I 1\ I
I
~
.
~
II.. I
~
J
\~~
I' ;
g
N
H",(x, y, k)
II
IiI
=
L an cos kny exp[
ix.Jk 2
k;]
-
n; 1 to
+
to
L
H~(x, y, k)
2
an cos k.y dXp[ - x.Jk; - k ]
(7.207)
n==l
to
L
•
a".Jk;' - k2 cos k"y = 1
an
COS
kny exp[ - x.Jk?; - P]
(7.214)
where the quantities under the radicals are now all positive. The preceding shows that the response potential function, as defined by Equation (7.194), consists of two sums. The second sum approaches zero as x -> 00, which satisfies the required boundary condition of a decaying WLlVC. The first sum takes the form
where an are yet undetermined constants. Substituting the preceding H~ into Equallon (7.198) gives
iw
L N+I
~
II,.
1'1
which is determined once H> is solved. As u final notc for the complex frequency response functions, we point out that since k. = (2n 1)n:/2H and k = w/c, for each k value the quantity (k~ - k 2 ) may be negative for n up to some maximum N. Hence from Equation (7.207),
Note that k. depends dnly on the water depth H and represents the natural wave number of the system. Furthermore, since the governing ~uation (7.197), and the boundary conditions, Equations (7.199) and (7.200), are all homogeneous in H~, a superposition of solutions is also a solution. The ooundary condition at x = 0, Equation (7.198), however, is not homogeneous and thus the total solution, not any component, must satisfy this condition. The superimposed total solution
(7.208)
n= 1
To evaluate the constants a" we multiply both sroes of Equation (7.202) by the cos kmy and integrate over the water depth H, The orthogonality condition of the functions cos kn y leads to
>(x, y, t) = ,~:
:;;: .
.
2
2
iwan.Jk; - k = H
[II
Jo
cos kny dy
(7.209)
Ie ."
L N
";\
•
~.( 2 k; X
an cos k"y exp[ - l"k -
-
.J
WI)]
2
2
k - k.
(7.215)
which represents outgoing waves in the + x direction as required by the radia tion condition. The main interest' here, however, is the total hydrodynamic force and moment on the dam, the variation of H p with frequency w, and the resonance frequencies.
194
i
RESPONSE OF CONTINUOUS SYSTEMS
From Equation (7.210) it is clear that an ~
~ f'
2
I i~ I
lil. ~ I '\i;I I
~.2
ill II
'\ D
;1
11m
c
=
kj
=
A
WI ::::<
I~~.t[Ii
III
III _,
m _.
i
1-'(1) = iJollAwk ""
-!:::<
p
2:
ft 4120sec
(7.217)
= first wave number in radians per foot (rad/ft) .
74 rad
WI
or .f I
sec
-21t
::::<
IF(t)ly
'I
1"0
1
11
The complex frequency response function for the hydrodynamic pressure H p and the total force H F on the dam, x = 0, are related to Fl.,p by
IH/··(QlIo
(7.221 )
=Fo-
This dimensionless hydrodynamic force on the dam during a steady state ground acceleration is plotted in Figure 7.21 as a function of the frequency ratio Q = wlw i • Figure 7.21 shows the dynamic alllplific~ltion :IS a funclion of the frequency ratio, the diminishing effeet of response
l~(I)1
= -;- ::::< 0.08 sec
IH ~'«(/)Iy Fo
for II = 100 ft
12 cps
an<.lthe natural period
(f!)
= t.01!55
0.222,
Uo
It dearly indicates the' importance {)f the compressibility of water when I he loading frequency is near the fundamental frequency WI of the system. On the other hand when the gi'ound acceleration has a very low frequency stich that OJ -l> 0 and Q -l> 0, the hydrodynamic amplification iF(t)I'I/Fo(/o = 1.0l{55. This means that when the amplitude of the ground acceleration tlo is equal to g, for example, the amplitude of the hydrodynamic force on the dam W(tll is equal to 1.0855 times the hydrostatic force Fo.
W
Hp = (iw)- H
HF
=
Jo[" Hpdy
=4w _
rtg
i
ll
L ,-<,
~'o
(_1)"-1
0 n= I
= 1tg --;-
IH,,{Wi !L
(7.218)
cos k.y dy
(2n _
4W(2H) "',
~
I
(7.216)
ck,
WI =
f
II, I i
III III
e = k;
195
Let the earthquake excitation be a(t) = (/oei{UI with 'amplitude !1o, then the total hydrodynamic force on the dam in dimensionless form is
when
Thus the resonance or natural frequencies of the dam-reservoir system are = ck. and the first or fundamental natural frequency 1S
iii':
~
00
DAM-RESERVOIR (HO!\IZONTAl EXCITATION)
Wn
f~
iii
=
c
7.S.
.~I (2n -
c .g (ij .\,1
1 1)2 Jk';
=- (0/)c 2)
(7.219)
"= Ci
., .,Ec
E
2
Incompressible water -----------------_ .
<.>
.'
Using the dimensionless ratio Q = wjwl aDd the static water force F 0
=
WH2/2 in Equation (7.219) gives*
~
t HF(Q) _ ~ ~
Fo 'See Chopra (1967).
I
- rt 3g .=1 (2n _ IlJ(2n _
1)2 _ Q2
(7.220)
0
2
3 Frequency ratio
4
5
6
w
!! = -
Wl
FIG.7.21. Sleady-state hydrodynamic response force on a dam due 10 a steady-slale Iwrizontal ground acceleration.
I 196
I I I I I I
II
II I
7.6.3.
l
~...'
fl
7.6.
a(t) = - I
21[
32F o
fro
f"" A(w)Hr(w)e 'It _au
IWl
dw
which is a superposition in the frequency domain. Alternatively, in the time domain, the impulse response function is I -2 'It
f""_au H r{w)e''''' dw
8wcH ~
h .(1) = - f 1[2g
J o(k.ct)
.= I (2n _ 1)2
L. - _..-.
f", ~(r)hr(t
- r) dr
a·,
(7.227)
Random Vibration
where HF(W) is the complex frequency response (lUlction for shear and
(7.228)
1}2 _ Q2
(7.229) \
and
WI
1lc/2H
In terms of 0
SrlO)
1
,I
Siw)(32F o!n: 3!/)2:=
1
(21l - l)1Jpll
=-
1)2-'--n i
12
(7.2.~(l1
When S.(w) is assumed to be a white spectrum of constant magnitude So, the response shear spectrum Sf(Q) from Equation (7.230) varies with Q in a similar manner to IHF(O)III!F o , as shown in Figure 7.21. Note the singularities at resonant frequencies 0 == 1,3,5, and so on. The mean square value of the response shear force is equal to the area under the spectral curve. Thus
o}
=2
f'
(7.231)
Sr-(w) dw
Substituting Sf'(W) from Equation (7.230) into Equation (7.231) and putting dw = Wt dO gives
2=
(1r
2Wl
( -32F o)2 So 1"" I I 3-
00
1[ g
0
.:1
I (21t - l)2J(2n -
1)2
0
2
12 tlO (7.232)
In terms of 0, the dimensionless mean square is rex>
2 ___ ..--:
(1r
L
.•.
= Jo
(7.233)
I(O)dQ
where
1(0) =
Now consider that the horizontal ground acceleration is a stationary.random process with zero mean and· spectral density Sa(w). The spectral density of the stationary response shear force becomes .
I.~l (211 _ 1)2J(211 _
2
I
'"
O=W!WI
(7.226)
where J 0 is the Bessel function of order zero. Using hf·(t), the hydrodynamic force on the dam due \0 a general earthquake ground acceleration a(t) may be determined through the Duhamel's super position integral in the time domain. F(t) =
(1[3 g )
(7.225)
From Equation (7.220) for Hr(Q), the integral in Equation (1.225) can be carried out to give
Sr(w) = IH,,(w)12Sa(w)
I
(7.224)
2
197
with the frequency ratio
(7.223)
A(w)i"'· dw
F(t) = "2.!..
hF(I)
IHF(w)fZ =
_
so that a(t) can be considered as a superposition of steady-state excitations of various frequencies. Since for each of these frequency components the response hydrodynamic force on the dam is obtained in the form of the com plex frequency response Hf·(w), it follows that the total response is
7.6.4.
DAM-RESERVOIR (HORIZONTAL EXCITATION)
Deterministic Vibration
Let a general ground acceleration excitation a(t) be expressed in a Fourier integral
~ (I
'I
RESPONSE Of CONTINUOUS SYSTEMS
IJI
(21t _
l)2J(2~1 -
r
1)2 - 0 2
(7.234)
Some insight can be obtained by first considering a single term in the series in Equation (7.234). The one-term approximation of the integral in Equation (7.233) is 1 Jor"'\ r=-ili
I 2
dU
-+ CXJ
(7.235)
II
~ ~ U"lII
ill I li1 I Ii>! I If I :>',"
"'::·l
~·:l
ill
' t:;
,
I I! I It I lill It I Wit.1 i' III II II III I·,
198
RESPONSE OF CONTINUOUS SYSTEMS
PROBLEMS
This unbounded solution shows that the mean square shear response IT; analogourly to that of a linear SDOF system with zero damping. The stationary excitation feeds energy indefinitely to the system that has 110 damping mechanism. Therefore as the time increases the response of the system becomes infinitely large. This observation implies that the dam-reservoir problem must be modified before a practical solution can be obtained under the random stationary input and stationary output condition. One modilication is to assume that the water. is incompressible ;;0 that (' .... 00, W .... ct:l, 11 ..... 0, and the complex frequency response function II t·(wl in Equation (7.220) degenerates into a constant. Conseqllently the inlinite resonant magnilkation for the response spectrum S/,({lI) in Equation (7.22X) is 110 longcr present. For this case the series in Equation (7.229) hccomes behav~s
Ii
PROBLEMS Show that for the shear beam of Section 7.1 the natuntl freqllel1l:ies and mode shapes are, respectively,
(7.236)
fi3 = 1.0518
Itt"
(2; - I) --
(u. J
so that
(!:0)2 r'" S~(lII) elm II Ju
(7.237)
This c1e.arly shows that the mean square shear oJ is again unbounded if the excitation spectrum is an ideal white noise with a·uniform spectral density So indicating that when the excitation has an infinitely large mean square value, ·the response mean square will also be infinitely large. This, of course, is a structural behavior under static loads and not the dynamic tesonant behavior. For a band-limited white excitation with cut-off frequency We such that
w. = .
J
(i~)2 \
L
El
pA
I
I
I
11(51
So
{ 0
for 0 < Iwl < otherwise
c;r
= sin wi:': (.
t/J}x)
= sin 2ltx
7.5. For the shear beam Example 7.1, calculate the terms Ijk of Equation (7.38) for the various values of the modal overlapping ratio
elL
,. = W.i -: Wj_1
= 1tme = 0.1,0.0\, and 0.00\
Assume the stationary random concentrated load process has zero mean and a white spectral density Sp(w) = So. 7.6.
For the shear beam example of Problem 7.1, estimate the mean square velocity response E[v 2(L)) at x = L when the modal overlapping ratio r of Equation (7.39) is sufficiently small to neglect lJk terms for j -+ k.
7.7.
Derive the simple wave equations for the dam-reservoir problem of Section 7.5 in the solid and in water.
(7.238)
From this result, we see that when the standard deviation (I" of the horizontal
and
7.4. Estimate the spectral density function Sy(L, w) for lhe response dellec tion y(L, i) of the shear beam problem of Section 7.1, when p(x, i) = 0 and the ground acceleration yo(t) is a stationary random process with zero mean and a white spectral density So(w) = So.
We
the excitation acceleration has a mean square valueof IT; = 2S ow,. and the mean square response shear force on the darn
(J;' = 1.17M3( "',,)2
t/J l(x)
7.3. Estimate the deflection y(L. t) for the shear beam problem of Section 7.1 when p(x, il = 0 and Hi) = V(i), a unit step function of time.
I
S.(w)
and
frequencies and mode shapes are, respectively,
Substituting the 'const~nt 111/,01 2 into Equation p.2lS) for Sj."((II) lind then into Equation (7.231) yields the mean square response shear as
n} = 2.3567
2L
.
7.2. Show that for the flexural beam problem of Section 7.2 the natural
o)2
32F (1.05U1)-• ( -_ .. nly
IHt·(m)1 2
~
f
ground acceleration is, say 10% ~J, then the standard deviation of the hydro dynamic shear force on the dam should be 0.1 JU783F 0 or about 1O.9/.'. uf the hydrostatic shear force F o' Finally, we point out here that another way of extracting useful information from the dam -reservoir problem is to consider the transient response rather than the stationary response. In so doing the resonant build up remains finite for finite time i. This nonstationary response problem will be treated along with others in Chapter 9.
7.1.
II"" (211 -I
199
•I I I I I I I I I I
\1 t
I
I
. ilii
200
RESPONSE OF CONTINUOUS SYSTEMS
7.8.
Derive the pressure wave solutions PI. P2, and P3 in Figure 7.8b associ ated with the velocity waves VI' V2, and "3, resp~tively.
7.9.
For the dam-reservoir problem of Section 7.6 and Figure 7.20, consider the dam to be elastic with the deflection approximated by the assumed first mode shape ~(y) = O.l8(y/H) + 0.78(y/H)z. when a horiz{)ntal ground acceleration ug(t) is applied. (a) Formulate the equation of motion. for the dam in terms of the generalized coordinate yet) defined by th~ dam displacement u(y, t) = Y(t~(y) using the princip!eof virtual work. (b) Obtain the complex frequency response function Hy(w) for the generalized coordinate yet) of the elastic dam.
7.10.
For the dam-reservoir Problem 7.9, when the horizontal ground acceleration is assumed to be a stationary random process with zero mean and a white spectral density of So, (a) Find the root mean square displacement response a, at the top of the dam for the case of incompressible water in the reservoir. (b) Find (J, for the case of compressible water.
CHAPTER
8
DESIGN OF STRUCTURES FOR RANDOM EXCITATIONS
The fundamental problem of structural design for random excitation is based on two primary failure criteria. These are the yield failure and the fatigue failure criteria. To develop solutions to the design problem we need first to introduce the stationary Gaussian process, the famous work of Rice (1945) on the probability of up-crossing, and of Powell (1958) on the probability distribu tion of the peaks.
8.1. STATIONARY GAUSSIAN PROCESS By definition, a Gaussian distributed random variable x has the following probability density function: p(x) = _1_ exp [_ (x
J2i"r.a
m)2]
2a
l
(8.11
where m and a are the ensemble mean and standard deviation, respectively, of the random variable x. This famous probability density function is displayed in Figure 8.L For two random variables XI and X2, the joint probability density function
201
1~ • 111
J j.
;!. .J
;i n
f.!~
j
III
202
DESIGN OF STRUCTURES FOR RANDOM EXCITATIONS
.1
p(x)
STATIONARY GAUSSIAN PROCESS
203
/'(.'1. xl)
'i""
I I
I
I
~
II I II
-r I. FIG. 8.1.
1
I. I I
!I I I I ~I
I 'I i!~p •I. ~
.,g
iI !:".,_.
~
t:0'/O'h'/ffiW'pVh'A
:.x
"'---"'"I
....... '"
-;;'-X2
Gaussian probability density runction orlhc random variable X.
is defined by
1
.if~
8.1.
P(XI,
Xl)
=
{t
- IIIlf exp - - - - [ " [(XI ---2-. 21f.(J I (J2JI=lif2 2(1 - IJIl) (Jt 1
2'
Xl (0) Xi'
2Pt2(X, - 1II,)(X2 - 1112)
(X2 - m2)2]}
- "i·-·--··-·-(Jt;2----·---- + -
"~r'-"
(K2)
where fill is the correlation cocflicicnt or XI and X2' This so-called second order probability density function p(x I, X2) is shown in Figure 8.2. A random process x(t) with random variables XI = x(t l ), X2 = x(t 2 ), ••. is defined as a Gaussian random process if all orders of probability density functions are Gaussian. The higher-order Gaussian probability density func tions can be completely defined if only the parameters mit (J" and Pij; i, j = 1,2, ... , n are specified. Consequently, for a stationary Gaussian process where mit (fit and Pi) are independent of time, we need oflly to obtain the ensemble mean m, and the autocorrelation function R(r) or its equivalent. the spectral density function S(w), to completely specify the process. This is clear because (12
=
R(O) - m 2
m2
-0
(8.3) ,.
p(r) = R(r) - m
Xl
JIll
2
(8.4)
(b)
FIG. 8.2.
Gaussian stationary processes are important in studying random vibration problems because (I) many important engineering excitations can be approxi mately modeled by a Gaussian process, (2) linear operations on a Gaussian process produce a process that is also Gaussian so that when the input excitation to a linear vibratory system is Gaussian the output response is also Gaussian, and (3) analytical manipulations of Gaussian processes are relatively easy. The
Joint G:lUssl:m prohahilily density rUllction I~Xt. x,).
idea of retaining the Gaussian property after linear transformation can be illustrated by the following example: EXAMPl.E 8.1. Show that if Xl and X2 are Gaussian distributed random vari ables, then y = ax. + bX2 with constants a and /1 is also Gaussiml distriouted.
I 202
~ :~ '.~
DESIGN OF STRUCTURES FOR RANDOM EXCITATIONS
.1 I
p(x)
STATIONARY GAUSSIAN PROCESS
203
/I(XI'
I
r--
I "---f- ,,--!
I I
II
I
I I
I
I~I
1~A7'~1
-____-==!~~=q ____~~~?2c2/Z~~~c2?~/.~?~?~K~?262Z~?~X~V~?~Z~/'~Z~A~::~~~~~-------
II
,... FIG. 8.1.
I I I I ,I :1
8.1.
i~
'"
)
-- __ --
x'
-j
I
--,--I "'2--
/--,-----
1 I
--
·---:;;;t-.\"2
__ J/
Gaussian probability density funclion of the random variable X.
defined by
p(x., X2)
1
{I
f-(X , - md 2 21f.Ij~lj2.J1 _ Pi2 exp - 2(J---/~-2i-- Ij~-
XI
lal Xi*
2/112(X\ - 1II,)(X2 -
'7\-'---'
1112)
'-~-;-a;'''----'--
+
(X2 -
2)1]} ';r--1/1
(K2)
where 1112 is the correlation coellicient of x I and Xl' Thi~ w-callcd ~ccolld order probability density function p(x I , xJ is shown in Figure 8.2. A random process x(t) with random variables x I = X(ll), X2 X(12), ... is defined as a Gaussian random process if all orders of probability density functions are Gaussian. The higher-order Gaussian probability density func tions can be completely defined if only the parameters mi, Ij" and Pij; i, j = 1,2, ... , n are specified. Consequently, for a stationary Gaussian process where m i , 0'" and Pij are independent of time, we need only to obtain the ensemble mean ml and the autocorrelation function R(t) or its equivalent, the spectral density function Sew), to completely specify the process. This is clear because (1'2
R(O) _ m 2
"'2
-0
(8.3) ml
pIt) = R(t) (/2' -
m
XI
1
(8.4)
(b)
FIG. 8.2.
Gaussian stationary processes are important in studying random vibration problems because (1) many important engineering excitations can be approxi mately modeled by a Gaussian process, (2) linear operations on a Gaussian process produce a process that is also Gaussian so that when the input excitation to a linear vibratory system is Gaussian the output response is. also Gaussian, and (3) analytical manipulations of Gaussian processes are relatively easy. The
Joinl Gaussian prnhahilily densily rtiIlClinlll~\'I' x,l.
idea of retaining the Gaussian property after linear transformation can be illustrated by the following example: EXAMPLE 8.1. Show that if XI and -"2 are Gaussian distributed random vari ables, then y ax. + bX 2 with constants u and b is also Gaussiall dislribulcd.
,.B,_,., n ¥il)f 204
DESIGN OF STRUCTURES FOR RANDOM EXCITATIONS
8.2.
The probability p(y) dy is represented geometrically in Figure 8.3 by the volume under the p(x l , X2) surface with the strip dy as the base. Thus p(y) dy =
'f" [lIY+dy-ax,ul> -ax, - ao
J
exp - (y -
dXI
oo
V 21t(1y
f nf y
20'2 y
'"
-00
dy
Let x
:=
+ bm2'
ml
= alaf + 2abplZal(12 + "lai :=
var XI,
ai
:=
m2
E(XI),
a;
at
ami
var Xl,
-trl
P cos 0, Y = P sin 0, dx dy
(f
e-
ro
:=
foo
dxfoo e-""dy=f
-IXI
e-""+Y"dXdy=(f"" e- x 'dx)2
x2
dx
-""
= E(X2)
)2
1i2"
2
0
i
;~
I
:~
I
g
I
= P dO dp, then
m
= -1
I
-00
-I't;
)2
where
my
205
are the ensemble mean and the ensemble root mean square or standard deviation of the random variable x. Verify also that the total probability is indeed unity. First we consider the integral
1/1>
(y
1 Fe.
P(XI' X2) dXl
PROBABILITY OF UP-CROSSING
e- P ' dO d(pl) = rc
9=0
;1
··,i.,l
.
.' i:j
,
';1 :~
roo e-· du := 1t Jo
i
= j;r
e-·'du
COV(Xl, Pl2
=
Xl)
Next we verify the total probability. To do this we let .J2(}'u =
0'10'2
X -
~
~
m so, that
'" fO:' r--e-·\lirTdu=l 1 f -"" p(x)dx= -y, ,,21trT
Thus the random variable J' is also Gaussian distributed. 8.2. Show that the two parameters m and (}' in the following Gaussian probability density function
I
. EXAMPLE
p(X) =
~exp[- (x .J2rca
Finally, we calculate the ensemble statistics
- m)2]
f_' '
£(x)
0)
2(}'1
Xz
£(x
2
)
= m
xp(x) dx
I = -r=-'
'" 2rc rT
2
J""
(111
= 2
If"" _", + .J2au)e-· du 2
(m
J'"
u 2 3- ul (/u
= m 2 + (}'2
-1")'.1
var x = E[(x - £(X)2] = E(x 2) - (E(X»2
= (11.
= square of the standard deviation 8.2.
~
~
y + dy y
FIG. 8.3.
= axl
+
XI
PROBABILITY OF UP-CROSSING"
A sample function x(t) ofa rapdom process is shown in Figure 8.4 with a specified level x = a. We see in this illustrative figure that the function x(t) stays below the level line most of the time but there are several peaks exceeding the line. In particular we note that tile random even C of crossing the line x = a from
bX2
Geometric relation among x b
X2,
and y.
*Scc Rice (1945).
.
1I;\f,'
+ 2.JirTltlu + 2a 2 ul )e-·'.J2rT du
''Or,
2(12 + 0 + -;: ...Jrr.
Ir;.;•,. IiIi "t',;
= In
:t
II 206
I I I I
DESIGN Of' STRUCTURES FOR RP.NDOM EXCITATIONS
·_f
- -1-1 .tl"*-:--:I
I I I I I
A sample function xII) of a random process with a specilied level,
PROBABILITY OF UP-CROSSING
207
area defined by the two specific events. This base area in the x - .x plane is displayed in Figure 8.6. The first even (x < a) is clearly displayed as the area to the left of the line x = a or the line BD. For the second event x dt > (a xl, consider Ihe poinl A located inside x < a and x > 0 with coordinates (x, xl. Draw the line AD, which makes an angle dO with the line BD. Since (a - xl = :( tan dll, then (a - x) = x £it represents the line AD if the lingle dO is so chosenllwl Ian ,/(J = d/. Thus the event ({I x) < X £II is represented by the urea to the right of line AD and the joint event C is therefore deiined by the wedge area extending from the point D upward to infinity. The up-crossing probability in terms of the joint probability density fU!ll:lion p(x, x)is therefore
x(ll
FIG. 8,4,
I I I
8,2.
•
(/.
P(C) =
below during the time interval (It by the sample function x(1) depends on x(1) and x(t) at the beginning of the interval. Clearly from the enlarged Figlln: a.s, it is observed that X(I) at I must be below the line x = u and the slope ."'(lIllllls{ be sufficiently sleep. These conditions on the random even C can be expressed by the equation
C'" CD J;;:o Jx:a-xdl p(x,x) ,Ix elx
(H.6)
which, because of the differential dl simplifies to P(C) = dt
i:o
(8.7)
xp(a, x) eli
\
i:'
(Ix < (I)
lind
(:( til >
x)l]
(II
This up-crossing prob
(a.5)
so that the probability P( C) of an up-crossing in de by x(e) is equal 10 Ihe prob ability of the joint event of (x < If) and .X (/1 > la x). Tile calculation of 1his probability with two random variables x and .x follows tile lIslIal techniqllc of integration of the volume under the probability density slH'face over the b~se
x
x(t)
B
a
(a
_1_
x
xl
=i:dt
l:
--~~~---------X---------~r---------~~----------~X
t
+
at FIG. 8.6. Wedge area (shaded) to define the up-crossing random event in the plane of the two
FIG. 8.6.
I
Conditions of an up-crossing of the level x
random variables x and X.
(J.
I
208
DESIGN OF STRUCTURES FOR RANDOM EXCITATIONS
8.3.
is dt. The time rate of up-crossing probability 11:, however, is finite and can be calculated if p(x, x) is specified. This probability is also the mean value of the random number N of up crossings in de since, in the time interval dt, the possible values of N are zero and one only. Thus
L
£(N) =
+ (I)PN(l)
tI,PN(tli) = (O)PN(O)
= P(C)
(8.8)
PROBABIUTY DENSITY OF THE PEAKS
209
spectral moment, respectively, of the process x(t) about the origin w = O. Finally, we point out that for a stationary random process it has been shown that the correlation between x and i. is always zero. See Equation (2.21) in Chapter 2. 8.3.
PROBABILITY DENSITY OF THE PEAKS·
aU./
Dividing both sides by elt gives the result that tlie time rate of an up-crossing probability 110+ is equal to the mean number of up-crossings per unit time. Furthermore, under the restriction of a narrow-band random process, the mean up-crossing rate I'~ of the level x = 0 is an approximation of the mean (requency and the mean number of peaks per unit time. This is clear because .for a narrow-band process, the number of zero up-crossings, vibratory cycles, and peaks are approximately equal. EXAMPLE 8.3. Determine the time rate up-crossing probability v.+ of the level = a for a random process where x and x are Gaussian distrrbuted with zero mean and zero correlation. Substituting
Since typical structures are designed to withstand maximum loads or peak loads, it is therefore useful to examine all peak values of a random process x(t) and derive the probability density function among the peaks. Let us consider the entire time history of a sample x(t), such as the one shown in Figure 8.7 under the restriction of a stationary narrow-band process. For such a sample with zero mean, denote the number of peaks per unit time with magnitude lying between z and z + dz by np: and denote the total number of peaks per unit time by nr Then the probability that a peak with magnitude lying between z and z + dz among a total population of peaks can be estimated by
x
2 X2)} t {I p(a, :c) = ---exp - _. (a" 2 +"2 21t(1.•(1"
2
(J"
(1 x
p(z) dz
P(C) dt
=-I - exp (t12 - - 2) 211:0",,(1x 2(1x
i"" 0
x cxp
E[lIp.]
(-
-2- exp (a - -2
X2).dx
+
•• 11.
=
(1,i;
1t(Jx
2
)
= £[tI<:
(8.12)
11<:7 +dz]
Similarly. E[II,.] is approximately equal to the average time ratc or zcro up
-'-2
crossings I'(~ . Thererore,
2".,i;
2
.
(8.11)
£(tl p )
Again, under the assumption of a narrow-band process, HI': in Equation (8.11) is approximately equal to the difference between the number of up-crossings of the level z per unit time Hcz and that of the level z + dz per unit time tI<,:+d:' Th~ ,
into Equation (8.7) and carrying out the integral yields \'a+
~ £[Hp:]
-
.
v+ -
p(z) dz i = :
(8.9)
(j"
+
dv. +V:+d: = ---~. Vo
Note that in order to determine 11; from the result in Equation (8.9), we need to know the standard deviations (jx and (j" of the two random processes x(t) and xU). When the processes are stationary, however, then
(8.13)
v;
x(t)
a; = RAO) = f~oo SAw) dw 0"1 =
R..(O) = - R;(O) =
f~oo w S,,(w)dw 2
(8.10)
so that I',: can be dCkrmincd if thc spcctrul density function Sx(w) is known. The two integrals in Equation (8.10) are known as the zeroth and second
FIG. 8.7. 'Sec Powell 11958),
1\ samplc rUllclion with scvcwl peaks loc;atcd in the interval (z, 2
+ '/21·
i
I
I 210
I "I
i
DESIGN OF STRUCTURES FOR RANDOM EXCITATIONS
8.4.
PROBABIliTY DENSITY OF THE ENVELOPE
Xf"'O
pz(z)
211
xl",O "-\)I
,~
I'" " I "'" x
II I
11
I
II11
tL II
_ _ _ _-;---:~=.
a
.... x
1£'
o
J~.
kl
oX
J':t.
t
FIG. 8.8.
Rayleigh probability density function pz{z) given by Equation (8.14).
Recall that for x(t) and i(t) with uncofl:,ell\ted Gaussian distribution and zero means, the average time rate of zero up-crossings v;j and z level up-crossings are, respectively, .
!
I II I
,.........
z
II
vci
,
:. p(z)
Ux 21tU x'
= :; exp
vi =
exp ( -
, .......,,'
a(l)
2~;)
(Z2) - 2u;
(8.14)
,!
which is the well-known Rayleigh distributioniil ocean wave height as illustrated in Figure 8.8. Note that pz(z) = 0 for Z < O. (II)
;1
I 'I I
'I
III
I
hi
~ ill
I '~1
I
~i
8.4.
PROBABILITY DENSITY OF THE ENVELOPE
The concept of the envelope function of a random process x(t) is introduced through a natural generalization of the envelope for a simple deterministic sinusoid x(t) = a sin wot. In Figure 8.9 let this sinusoid be represented by the horizontal projection of a radius vector of magnitude a rotating with constant circular frequency Wo from the + i.jwo axis in the phase plane. In the x versus t plot, the two dash lines x = ± a geometrically represent the envelope of the sinusoid. Thus if a(t) is used to define the envelope as a general function of time itis therefore equal to the radius of the moving point P.ln the case ofthe sinusoid this radius is a constant so that the plot of a(t) versus tis the pair of dashed lines in Figures 8.9a. Applying this definition to a sample of a narrow-band random process x(t). the envelope function a(t) is no longer a constant. It is defined by the equation
f,
lllll\
2(
a t) = x2(t)
+ X'2(t) ~io
(I»
FIG. 8.9. Phase-plane diagrams of (a) simple harmonic motion and random process, from Crandall and Mark (1963),
(8.15)
(h)
sample of narrow·band
The plot of a(t) versus t is shown by the dashed curves in Figure 8.9b, exhibiting the geometric characteristic of the envelope to the sample function x(t). Having defined the envelope function a(t), it is then possible to derive ils probability density function from the joint density of x(t) and xV). Thus
p(a)da
=
ff
P(X,:JdXd(:J
(8.16)
The integral in Equation (8.16) covers the area defined by the boundary (a, a + da) in the x - X/wo plane as shown in Figure 8.10. The geometry ofthe envelope equation suggests, as usual, that it is advantageous to use polar coordinates. Furthermore, to change the ordinate x/wo to i, it is necessary to
212
DESIGN OF STRUCTURES FOR RANDOM EXCITATIONS
8.5.
Let x
i}tdO
213
STRUCTURAL DESIGN AGAINST YIELD FAILURE
= a sin 0, i/wo = a cos 0, then P(x, xl = _1_ exp {2naxa x
;:2}
(8.21)
x
Since roo = ai/a". Substituting Equation (S.21) into Equation (S.20) gives
r1
V-~
;>x
p(a)
= az exp {- 2:2.} a"
. (8.22)
..
which again is the Rayleigh probability density function as shown in Figure 8.S.
8.5. FIG. 8.10.
Envelope a(t) in the plane or x and x/woo
relate the corresponding probabilities
P(X,~)dXd(X) = p(x,x)dxdx Wo wo
(8.17)
so that
p
(x. ~)
=
(8.18)
(Oop(x, x)
(1)0
f21t
p(a) da = Je~ 0 (1)oP(X, x)a dO da
STRUCTURAL QESIGN AGAINST YielD FAILURE
Consider an offshore structure subject to a strong random wave force. The maximum stress x(t) at a critical location on the structure is examined against the allowable yield strength a. Since the ocean wave is treated as random, the maximum stress x(t) is therefore also random. Consequently, there is a prob ability that after a certain time T during the pounding of the wave force, the maximum stress x(t) will exceed the yield strength q of the structure. Such a sample is shown in Figure 8.11. This random time Tto failure of the structure by yielding. when failure is defined as the exceedance of the strength by the maximum stress, is directly ~elated to the probability of structural failure Pro The probability of survival or simply the reliability of the structure R is defined as 1 - Pc. Both P r and R ard functions of time t as shown in Figure 8.11. To derive the probabiiity density function of this random time 1: the so-
(8.19) ",(t)
p(a) = a(1)o where x and
flO
Jo
(8.20)
p(x, x) dO
I~
T
~I
I· I I I I I I
··~~
··t:
~':!
I,
I I
I
x are functions of a and O.
8.4. Determine the probability density function p(a) of the envelope process a(t) for a narrow-band stationary Gaussian process with zero mean. We have
EXAMPLE
1 {I-2 (x2ax + axX2)}
p(x, x) = -2--. exp
na"ax
-
"2
I 11
~
FIG. 8.11..
A sample runction x(t) that exceeds x
="
a at time T.
I ill
I" I
II
.y
:j'
I:
214
8.5.
DESIGN OF STRUCTURES FOR RANDOM EXCITATIONS
:i'i
:1
I
)1 I I I
+ dt)
= P(no exceedance in,(O, t)]P(one exceedance in (t, t
P(O; (0, t)]P[I; (t,
t
I I I I I '.I
I
'\ ;~~I'
+ dt)]
peT
(8.23)
where the occurrences of exceedance are assumed to be independent From the result of Rice (1945)
+ dt)) = v: de
(S.24)
> t)
= R(t)
fO?
pet) dt
J'" v:e- :< de = e-':<
Pr(t) = 1 - R(t) = I -
prO; (t, t
+ dt)]
= I -
v: dt
but &
pro; (0, t
+ dt))
= P[O; (0, t)]P(O; (t, t
,
+ dtn •
= P[O; (0, t)](l - v4+ dt)
tit •
(0, I)]V:
f
dP[O; (0, e)] P[O, (0, t)]
Assume that the mean exceedance rate
...:
f v;;
i de
+ 2ew ox + w~x
f(l)
m
+c p(tl
When t = 0, P[O; (0, 0)]= 1, c = O. Thus P(O; (0, t)] =
e-':'
The excitation f(t) is a Gaussian stationary random process with zero mean and a white spectrum So. Under the assumption of a Poisson distribution for the first passage time,
v: is constant, then
In P[O; (0, t)) = - v;;t
e-':'
(8.25)
Equation (8.25) states thiu the probability that no exceedance of the level a throughout the time interval (0, e) is which is by definition the reliability function, that is,
e-·:·,
reliability
= R(t) = e-'!'
(8.26)
Substituting Equations (8.24) and (8.25) into (8.23) yields the probability density function p(t) ofthe random first passage time T, known as the Poisson distribu tion p(t) =
v+ e-·:· Q
(S.28)
EXAMPLE 8.5. A single-degree-of-freedom (SDOF) system is subjected to a stationary Gaussian force excitation f(t) with zero mean and uniform spectral density So. Determine the reliability R{t) as a function of the dimensionless product wot of the Wldamped natural circular frequency Wo and the time t. The reliability R(t) is defined as the probability that the response displacement x(t) has not crossed a barrier a = 2ux after time t . The governing equation of motion for the SDOF system in terms of !he displacement x(t) is
Rearranging terms and integrating both sides gives
•
v
which agrees with Equation (8.26) and is plotted in Figure 8.l3a along with the probability of failure Pr in Figure S.l3b.
then
dl'[O; (0, t)] 'T -
I
+ de)]
P[l;(t, t
215
The probability density function for T is shown in Figure 8.12. Note that the area under the curve to the right of t is peT > t). Since the event (T > t) is the same as that of no exceedance in (0, t), it follows that
called first passage time, it is observed that
.,
p(t) dt = pet < T < t
STRUCTURAL DESIGN AGAINST YIELD FAILURE
(8.27)
FIG. 8.12.
Probability density function P(ll for the random first passage lime T.
216
DESIGN OF STRUCTURES FOR RANDOM EXCITATIONS,
8.5.
STRUCTURAL DESIGN AGAINST YIELD FAILURE
217
where
RIt)
=
hAt)
1.0
o.
t
e-("'ot
{
.
/0,
t
t;:::o
Sin ""d.
-illd
I
Carrying out this integral we obtain 1
ax
=
__ nSo 2e(J)~
I
lal
Similarly P (1l
r
O''f
-----------
~~--
1"h1( dt I)
where
hAt) =
(b)
11:- = O.0431.J(-;;~ e- v !'
+
Rtt) =
nO'" '
= 2n:So
e-·;' =
e-O.0431"'o'
e.
(ax) -
In R(tl
0'",
The standard deviations of the response displacement x and velocity determined by integrating the respective impulse response functions. 0';
0.0431 (J)o
which is independent of the damping ratio To plot this result, let us take the logarithm on each side. (See Figure 8.14.)
'itO' x
tx> h;(t) dt
x can be
I I I I
I
= a= 20'x
0'i _(a'/2,,', --e •
= -O'i- e -2 = 0.0431
=
Then the reliability is
v;;
where is the frequency of crossings of the displacement levellxl in this problem lIa
t;:::O
Substituting ax and a" intfl the equation for v:, we obtain
the reliability is R(t) =
t
2emo
ax Reliability R(II in (al and probability offailure P, in Ih}.
oJ {d[h,,(/))/dt, nS_ _ o
2
o FIG. 8.13.
2nSo
- 0.0431wol
=
2.303 log R(t)
or log R(I) =
0.01872wol = .V
dy d(wo/) =
0.01872
Ii II I I I
II
I I
218
DESIGN OF STRUCTURES FOR RANDOM EXCITATIONS
wo
5
8.5.
15
20
~~
.
~,o
R(y, t)
= exp{ -
v;t} 2
t
""-
0.81 \-
taF(Y) [ -a (y) -ex { ---exp -]} p 1toAy) 2o}(y)
y
.'
219
hydrodynamic shear force, we can estimate the dam reliability by Equation
l
10
K
,_
I I I I I I I I II I
STRUCTURAL DESIGN AGAINST VIELD FAILURE
'"
I
t
0.651-
RII)
""
(8.29)
where the standard deviation of the hydrodynamic shear force from Equation
-1-0,19
m. tl
hi-
0.52
0.42'
.
211 ) h~. ( 3.1
Y -0.38
FIG. 8.14.
Reliability R(I} as a function of Wof.
y
wot
0 5 10 15 20
-
0 0.0936 0.1872 0.2808 0.3744
I,
R(t)
I I )fUid ? hI' ( 3' t I
.1.0000
0.8061
,0.6499
0.5239
'0.4223
8.6. Consider the problem in Section 7.5 of the dam-reservoir system subjected to a vertical ground acceleration, modeled as a stationary Gaussian process with a zero mean, and a white noise spectral density So. It is assumed that failure occurs when the response hydrodynamic shear force F at the dam base exceeds a barrier level of a = pal' where P= 2.0,2.5,3.0, and 3.5, respectively,and aF is the standard deviation of the response hydrodynamic force. Assuming equal resistance of the dam to two opposite directions of the
fLAm]
_.
I I I I
,-;1,I,--'J~~lV~7~7~7~J~---j.~7~A7~71f/~/2'2/a,~--~r~/~C~<~'~L??77~~~------------V euA
Note that wot = 21t(tlto) so that for wot = 21t.;:::. 6, tlto = 1, which means that we are considering the time one natural period to after loading.
1»Y4
-.."nj
..
,I;
EXAMPLE
I
.•
' '
vu,c'Z/1777777777777A :I J 1/r,«i"/UZVUZf )'777777 j
FIG.8.15. Unit impulse response of derivative of shear. The constant multiplier we'll + a:){2g is deleted.
220
DESIGN OF STRUCTURES FOR RANDOM exCITATIONS
8.5.
(7.168) is O}(y)
=;
(y)3
1 + 0: 2 E[F (y)] = 3(1 _ 0:) 7r.SQp 2cH3 1 - H
(S.30)
Based on the unit impulse response hy(Y, t) of the shear force, as shown in Figure 7.12, and the simple relation d
hj(y, t) = dt hf(y, t)
STRUCTURAL DESIGN AGAINST YIELD FAILURE
221
we can determine the unit impulse response hj(y, t) of the time derivative ofthe shear force. This is shown in Figure 8.15. With hy(y, t) known we can then determine the square of ths standard deviation 1
•
•
O'j(Y) = E[F2(y)] = 1nSo
•
JofOOh1(y, t) dt =
l'
+ IXIX nSoP 1c3 H (1.-
1 .....,
y)
H
(8.31)
Substituting O'y(y) from pquation (S.20) and O'i'(y) from Equation (S.31) into
.
3.5
0.8
1
0.7
t
g :0 .,
~
:0 .~
~ 0.6
i!
..
~
ci ~
ci
~
0.5
0.4 1
o
1 ! ! ! ) J 20 30 40 50 61
10 01\
t = 2tdl1'j = time _
FIG. 8.16. Variation of shear reliability at dam baSe y. 0 with time w,1 and strength ratio a(O)/O".,{O).
10 "'II '" 2'ITtJtl
= time
20
30
_
FIG. 8.17. Comparison of reliabilities for shear F, moment M at base of dam, and for SDOr system at 20" strength level.
I
I
I
I
I
I
I
I
I
I
111:lr J
II
1
Ii
"if
f;·
I
l~
i~ I
222
8.5.
DESIGN OF STRUCTuRES FOR RANDOM EXCITATIONS
223
STRUCTURAL DESIGN AGAINST YIELD FAILURE
Equation (8.29) yields
.1
•
i:
I .] I
r#,i
~.~
R(y. t) = exp
,I
11 ~·t
.:{
,I
2./3 wit . - a 2(y) ] [ - 7 1 _ (ylH) exp 2[1 _ (J'IH)]3G~0)
(8.32)
where WI = 1tc/2H is the fundamental circular frequency of the dam-reservoir system. Note that the shear reliability in Equation (8.32) depends on the refrac tion coefficient (1. implicitly through GF(O) in Equation (8.30). To develop the reliability analysis further it is necessary to specify the available shear strength aCyl as a junction of y and then the ratio between the
WI
= 10
1.0
1
II I
§.a .,
0.9
~
II.
Wit
s
= 10
~
I I
'II
I .! ;I
!
I .
I
I
tl
L
t 0.7'~------~~
....
o
S
____~~____~~____~L-______~______~ ylH ~ lop
tt:
AG.8.19. Variation of shear reliability with elevation (ylm and time (W,t) for the triangular section with 0(0)/0',,(0) = 3.0.
r.
!
Jr
0.81
~
I
I
......
j 0
0.2
0.4
0.6
strength and the shear response GF(O). For the specification of the strength response ratio, first consider the critical location at the base of the dam y O. The reliability is R(O, t) = exp 0.8
1.0
ylH _ l o p
FIG. 8.18. Variation of shear reliability with elevation (YIH) and time (w, t) for the uniform section with a(0)/0'''(0) 3.0.
[ x2 exp {- 2J3w,t
a
2
(0)]}
2O'i-(0)
(8.33)
This shear reliability function depends on two parameters, the strength-response parameter a(O)/O'F(O) and the time parameter lOll. The function is plotted in Figure 8.16 as a family of straight lines in a semilog scale for a(O)/O'F(O) = 2.0, 2.5,3.0, and 3.5. It is interesting to note that R{O, t) for shear in the dam problem
I
224
DESIGN OF STRUCTURES FOR RANDOM EXCITATIONS
8.6.
differs from that for a SOOF systcm with an undampcd natural circular fre quency (J)Q = WI only in the constant multiplier, which is 0,3510 for the dam versus 0.3183 for the SDOF system. The resulting difference in the reliability functions is shown in Figure 8.1 7 for the strength-response ratio of 2.0. Also included in Figure 8.17 is the moment reliability function for which the deriva tion is completely analogous to the shear case. The details will not be given here. To examine the variation of the reliability function with the elevation y, two basic and simple dam configurations are considered, the uniform dam section with a(y) = a(O) and the triangular dam section where a(y) = [I - (y/H}]a(O). For each configuration, the shear reliability R(y, t) is calcu lated for the typical strength-response ratio of a(O)/O'F(O) = 3.0 using Equation (8.32). The result is shown in Figures 8.18 and 8.19, respectively, for two typical values of the time parameter (I)ll = to and 60. In both dam configurations, the shear reliability approaches 100% towards the top of the dam indicating that the bottom section is critical in design as expected. A comparison of Figures 11.111 and 8.19 also gives a quantitative measure for the effect of reducing dam sectional area near its top on the system safety.
STRUCTURAL DESIGN AGAINST FATIGUE
Palmgren and Miner's Deterministic Hypothesis
From standard fatigue tests of structural elements, it is observed that the most important parameter that governs the fatigue failure is the amplitude S of the cyclic load. If N(s) denotes the number of load cycles with constant amplitude S to cause fatigue failure or simply the so-called fatigue life at amplitude S, then a linear approximation can be obtained from experimental data for typical structural materials between N(s) and S on a log-log scale as shown in Figure 8.20. That is log N = a - b(logS) log (NSh)
b
100.- 2
~1 alb
10
10gN " a 10" = N
1- 1
10
C
~
~
100
'" N
10gN
Idealized raligue lire N versus stress amplitude S.
II,
....
D
(8.35)
= N(Sj)
where N(Sj) is the fatigue life with cyclic load of constant amplitude St. Further more, for groups of different load amplitudes, the total percentage of damage is "1 D = N(S;j
nl
+ N(S2) + ...
=
LI N(SI) "I
(8.36)
Clearly fatigue failure is defined by 100% damage. Extending the discrete loading groups to a continuously variable amplitUde S gives the result D
= roo
Js=o
n(S) dS N(S)
(8.34)
where constants band cdepend only on material strength. Based on this idealized experimental law, Palmgren and Miner's hypothesis states that the percentage of damage D due to loadings with n, cycles of ampli-
8.6.2.
I I I I I E
,I I}
1 :
~ \:
:{
Ii1 ~ "
.
(8.37)
where n(S) dS represents the number of cyclic loadings with amplitude in the interval (S, S + dS) and N(S) is the fatigue life at amplitude S.
=a
to" = NSb =
225
tude S, is accumulated linearly. That is,
The basic result for the estimation of fatigue failure of structures under random loadings is again derived from a generalization of the deterministic theory. This dctcrministic thcory is bascd 011 thc wcll-known hypothcsis of Palmgrcn (llJ24) and Miner (1945) as seen in Section 8.6.1. 8.6.1.
•
SlogS
FIG. 8.20.
8.6.
STRUCTURAL DESIGN AGAINST FATIGUE
Stationary Narrow-Band Random Loading
j
,~ ~
i; IIH;. ~
~J
,11
1 ~l I JI
,{
'f, I.,i~
~\i
Consider now that the cyclic loading is a stationary narrow-band random
load and n(S) dS represents the random number of load cycles with amplitUdes in the interval (S, S + dS). From the precedinp ;'i,cgral equation it is seen 'that
; .l
'.'.!
jl ~ J [~" ~I.
~i
~ tIl, l~ i:~
•I"
i.
I
I
I
19 ,~
h
fl
f.~
'ij
7~
~
" ,.
.~ i 'I
~
226
DESIGN OF STRUCTURES FOR RANDOM EXCITATIONS
the percentage of damage D must also be a random variable. Thus the mean value of D is given by E(D)
I
.,
!~
I i I
peS) dS = _ dV.+
Substituting the last two equations into the mean damage integral gives E(D)
I
'i tt
I
I. '.
I
= roo
Js=o
~
,.
SbN
p(S)
:~
. N(S)Sb
I'..
il
S = -zexp (1JC
=c
On the basis of the Palmgren-Miner hypothesis what should the amplitude A be of the equivalent excitation I
X.q
=
A sin wot
if this excitation is to do the same damage in the same time as the average damage due to the random excitation x(t), which has the spectral density
{~o(W~ + w ) l
S,,(w) =
for Wo - e < elsewhere
< Wo
+e
This spectral density function is shown in Figure 8.21. Fatigue damage from nonrandom loading is
~ = (wot) S:
1.'6 t P(S) dS
N
(8.41 )
N(S)
Finally making use of the Rayleigh distribution of the load peaks, the experi mental relation for fatigue life, and 1.'6. = wo/21t, the integral in Equation (8.4l) can be carried out as
1
+ j(w/wo'
= •
A fatigue sensitive element is exposed to the stress process y(t), The S-N curve for this element is
(8.39)
(8.40)
-;r
I
III
!'·I
(- dV$+)(t)
(S2)
-2 2
21t
C
S...(tuj
I"""-
r
I II
III
(1"
=C
I,
E(D)
~
"~.I
il
'I !~Irr.';i! l~ fiB
!:.,
it· lU' :l ~!
I' Itt ,:~ ~il"
I ... t),
wt 1 foo Sb+l exp (S2 = ~--z - - 2) ds 21t
C
(1"
Wo r;; )br ( I - -t (-../2(1 21t C x where r(t
+ x) =
-"-1:1
2(1".
0
b)
+ -2
"-1:1
--+H
(8.42)
J~ e-Yy" dy is the Gamma function.
EXAMPLE 8.7. A stationary narrow-band Gaussian 'excitation x(t) acts on a system whose response yet) is determined by the frequency response function
227
I
H(w)
(8.38)
N(S)
where 1.'.+ is the mean up-crossing rate of the level x = Sand t is the time dura tion of load application. Recan that the probability density function of the load amplitude or peak has been derived before as
~
I
i
Jo
since the fatigue life N(s) is a deterministic or nonrandom function of amplitude S. The mean number of load cydes with amplitude or peak in the interval (S, S + dS) is denoted by E[n(S) dS]. In view of the work by Powell (1958), it is found to be the difference of the mean number of up-crossings of two peaks, that is,
jl
I
= roo E(n(S)]dS
E[n(S)dS] = (1.'.+ - v,++d.)t
;l
8,&.STRUCTUAAL DESIGN AGAINST FATIGUE
FIG. 8.21.
l>
'"
.1 228
PROBLEMS
DESIGN OF STRUCTURES FOR RANDOM EXCITATIONS
229
\:,
which ilithis problem is made equal to the mean damage from random loading given as·
!(..[iasrr(l +~) C
wot
211:
A =
Equating the two fatigue damage expressions yields the stress amplitude
..[ias
Sa =
[r (I + ~)J'b
(8.43)
The remaining task now is to relate the stress amplitude Sa to the amplitude A of the equivalent excitation X cq • and to relate the standard deviation of the random stress a. to the given spectral density S,,(w) oC the random excitation X(I). These arc worked out as follows: For the first part,
= A sin wot = ~ (e l"""
excitation X cq
response y =
A.
2i [H(wo)e"'''' Ii
.
i (Sill (/lui
-
Substituting Sa from Equation (8.44) and a. from Equation (8.45) into Equation (8.43) yields
-
8.1. Find the probability density Cunction p(x, y) in the circular shaded area as shown, if it is known that for any point (x, y) in the plane of x-y outside of the area, the corresponding probability is zero, and inside the arca tbe probability is uniformly distributed. (Sec Figure 8.22.) y
.
cos wot) =
.
ji sm(/IJ()I
(I
)
Xeq.
PROBLEMS
- lI( - UJo)e- W ",,] A
(1+ n]~
which is the required amplitude of the equivalent nonrandom excitation
t
e - i"'o')
4wo.[eSo[r
.r
"
0
FIG. 8.22.
response stress ky =
~1 sin(wol -
0)
8.2. Show that if Y = g(x), and Pr(Y) and px(x) are the probability density functions fOf the two random variables' Y and X, respectively, then
where k is the spring 'constant and the stress amplitude
kA
(8.44)
../i
Sa = •
py(y) =
Idg~:(y)\ PX[g-l(y)]
where x = g-l(y)
8.3. Show that if X and Yare two random variables with the joint prob
For the second part,
ability density fu~ion Px.r(x, y) and that Z = X + Y, then
(1; =
E(S2)
= k 2E(y2) = k2
S:",
pz(z) =
J.,.-. IH(wWS,,(w)dw
:. a; :::: 4ek2Sow~
I I I I I
1 II
1
f-"'", Px.y(z -
y, y) dy
8.4. Find the probability density function pz(z) if Z (8.45)
= X + Y and X and Y are two independent random variables with the same uniform distribu tion in the interv;u (- 1 to + I).
'i i
', W
:]
j!
11\ ill!
<
I
I I: II
Sy(W) dw
f"o+£
= 2kl
I
Ii· §:
$:,1
Ill:i1
1111 1/
~ !1~t
I';
lij
;',,;11 m
f,l 'll' ...I b
I:I
.':~1 ;1. i: ,' 1
230 DESIGN OF STRUCTURES FOR RANDOM EXCITATIONS
t~
tl ~j
"
-,
:1"
";~
"i
I ]
"~I
;1
:~
I
I
I ! I ' 11 I
8.8.
densities
py(y)
0,
x
= {fJe-JlJ',
y~O
0,
,,~
~~.
(b) Find the probability density function pea) of aCt) in terms of the joint probability density p(x, x) in an integral form.
8.9.
y
find the probability density function pz(z) where Z = X + Y.
8.6. Given: X(t) is a stationary random process and E(x) (a) Show that E(xx) = O.
0 and E(x)
=0
(b) Show that R,t(-r) = - (d 2 /d-r 2 )RA-r);;: - R~(-r). (c) Find (f,t
Given:
A SDOF system is subjected to a stationary narrow-band random excitation x(r) with E(X(l)] = 0 and spectral density Sx(w) given by the plot in Figure 8.24. Estimate the time 'If to fatigue faiiure under this excitation. Assume that the complex frequency response for the critical stress yet) is
'
Hy(w) = 1 +
k
i2~(w/wo) _
~ =
.
(OO/Wo)
2'
<.0 0
k
S:co Sx(oo)e- iwr doo
Rx(-r) =
= a(ncosOJ!
For a narrow-band random process x(t) and ware random,
(a) Define the random envelope process aft) in terms of the random process X(l) and i(I).
x~o
Px(x) = {ae- ......
I
11
.,
",.,,:~~
8.5. Given two independent random variables X and Yand their probability
"
~"
PR
=
damping ratio natural, frequency a constant
Also the fatigue life Curve is NSb = C with constants band C.
and SAw) is given as a band-limited whit~ noise as shown in Figure 8.21
S"ln)
= solwii
S,,(w)
'I- .,2)
8,.{w)
i
lao
Wo
10"//#fffl:!
J
l:I0'/I'!'ffffA
~
0
I
1/1(:1
wo
,.. .,
-1 u
(oJ
2.
\dvI
FIG. 8.23.
,
8.7. A stationary narrow-band process X(t). with E[x]
= 0 and E[x] = x at all times. - 1~ x~ 1
p(x, x)
~
I
,~
ri5i1
= {~
for - 1 ~ x , otherwise
~
1 and
FIG. 8.24.
0,
has the following joint probability density for x and
(a) Find (fx and (f,t. (b) Find the mean number of up-crossings of the level x = a (with a = 1/2) per unit time III in Hertz. (c) Find the probability of zero up-crossings per unit time P(c)/dt.
8.10. Suppose yell is a stationary, narrow-band, Gaussian stress process at a critical location of a structure. E(y) = 0 and the central frequency of Y(l) is <.00 = 30 rad/sec.
(a) Estimate the probability v;; per unit time that the stress yet) crosses the design yield stress level of a = 4(f)/. (b) Estimate the probability P that the random time T of first up crossings of the 4(fy level will be within the interval, (\ sec
~
T
~
2 sec)
....
9.1.
CHAPTER
SDOF SYSTEMS WITH STATIONARY EXCITATION
233
It is assumed that x(t) = 0 for I < O. For I ~ 0 it is a stationary Gaussian process with zero mean and a white spectrum So. The solution to Equation (9.1) . with zero initial displacement and velocity is
9
y(/) =:'
f~ 11(1 -
0);(01 dO
(9.2)
where the unit impulse response function is
NONSTATIONARV RESPONSE
e-(Wo'
h(t) = {
wo~ sin wo~t,
t~O
O.
t
(9.3)
The mean value of the response displacement can be obtained from Equation (9.2).
£[y(t)] =
In this chapter we consider the time-dependent response of structures. This is important in the structural problems dealing with dynamic effects from earth quake motion, strong wind, and ocean waves, where both the excitation and the response are essentially transient in nature. The presentation is made in two parts. The first considers a suddenly applied stationary random excitation, which may be called a pseudononstationary excitation. The second part deals with a more practical case where the excitation not only has a zero starting condition but is also transient during the load application. 9.1.
Consider single-degree-of-freedom (SDOF) linear systems governed by the equation of motion
where
Wo =:
+ 2woey + w~,
= X(/)
= damping ratio, y = displacement response, and x(tl = excitation force divided by the mass. 'See emil-hey and Stump£j196l).
232
=0
(9.4)
since the excitation process x(t) is assumed to have zero mean. The variance of yet) when yet) has zero mean is equal to tbe mean square. From Equation (9.2), var yet) = 0';(1) = £[y2(r)] =
f~ f~ h(t -
0dll(t - 02)£[x(Odx(02)] dOl d0 2 (9.5)
(9.1)
£[x(O, )x(02)] = R.,,(O, - O2 ) = 2n;So<>(O, - O2 )
(9.6)
is the autocorrelation function of the stationary excitation x(t) with a white spectrum So. Substituting the autocorrelation function of Equation (9.6) into Equation (95) and carrying out the integration with respect to O2 yields
natural circular frequency,
~
O)£[x(O)] dO
where
SDOF SYSTEMS WITH STATIONARY EXCITATION *
ji
f~ h(t -
11';(1) = 2n;So
£
h2 (t
-
Od dOl
(9.7)
This simple result is obtained on the basis of the key assumption of a white excitation with the corresponding Dirac delta autocorrelation function, Equa tion (9.6).
~ I
~ I
I! I
U I
:11
'1,
234
NONSTATIONARY RESPONSE
9.1.1.
Zero Damping System
a yet - 2n:S0 2 ) _
;f!q
l
II i
II
i II 1 I
1 I
f I: I
i)
which approaches infinity as stationary solution.
9.1.2.
o
,
;~
,
.; ,
t -+ 00.
2
(ly(t)
Let
~
a;(t) =
. sm 2wot)
(9.8)
This result agrees with the unbounded
sin 2Wdt _ O(~o) O(e n ~ 2
wo~(c
2nSo
i'
IJ) =
z, then dO
2nS~ (2wot
(9.12a)
sin 2wol)
Wo
This agrees with Equation (9.8) as expected. The mean square a:(rl, as given by Equation (9.10), is plotted in Figure 9.1 for = 0,0.025, and 0.1. It is observed that ror = 0.1 the system response approaches to the corresponding stationary case in about three cycles of natural vibration when wor = 27tfol ~ 6n so that t ~ 3'0, three times the natural period When t -+ oc, q;(t) from Equation (9.10) reduces to
e
e
'0-
For the practical case with light dampingO < .
from Equations (9.3) and (9.7) is
e < l.0, the mean square response
n:So
q;(t) =
e-2~"'o('-O) sin wo~(t - 0)
2( 2 - dO o Wo I e )
(9.l2b)
2wJ~
2
=
(9.9)
dz/wo~ and
which agrees with the well-known stationary solution. For ~ a;(t)wMnSo = 1/2e = 5.0 in Figure 9.1.
= 0.1 0 the ordinate
~
o}(t) = w~(l2nSo _ ..1'2)3/2
iWo~l e-2~;~2 0
i,
• 2 zdz sm
-'-')
"so''''o
q;(t) =
2~oe [1 - e-2~Wo' (1 + 1 :~2 sin 2 wdt + ...
h
sin 2Wdt)]
I,
or
(9.10)
\.
;}
- 2ewot +
(1)0
!t l :~
= [1
235
the last term in the approximation Equation (9.11) must be retained. For a zero-damped system,
2
lightly Damped Systems
i
I
e- {o,<>, sin 2Wdt 1
eis zero in Equation
sin wo(t2 - 0) dO
= -2 3 ( 2w ot Wo
I
!I
!
i'
n:So
{\
.
2
For the idealized but fundamental case of zero damping
: l'l
II.
:j
SDOF SYSTEMS WITH STATIONARY EXCITATION
(9.3) so that the mean square response from Equation (9.7) becomes
!' ,~
!~
9.1.
~
nSo [
2
O',,(t) = 2~w~ I -
~i 8!
e- 2~"'ot --wr(Wd + WOWde sm 2Wdt + 2woe sm 2
•
2 2 • 2
Wdt)
. For small
"
]
".
U'
t~
20
15
e 10
n:S [1 2(t) "'" _ 0 -(1
0',
-2w~
e
e- 2{"'o') -
e-2~WoI.
- - - s m 2w"t
~
]
f
0.10
(9.11 )
t~
x
5
. Since I~~
e-2~Wo' = I
e! (l -
2ew ot +
o
O(e 2)
e- 2~Wot) = 2wol
+ O(e) =
I
3"
/6"
I
:>-'·ot=2"fot
9..
Three cycles of natural vibration
FIG. 9.1. Nonstationary n:sponse of linear SDOF systems under stationary random excitation
O(eo)
with a white spectrum So (Caughey and Stumpf. 19611.
.-.----------~~-
.11 .
236
9.2.
NONSTATIONARY RESPONSE
DAM-RESERVOIR SYSTEMS WITH STATIONARY EXCITATION
237
9.2. DAM-RESERVOIR SYSTEMS WITH STATIONARY EXCITATION
modified by a scale factor of a2to - lI • The effective starting time depends on the elevation y and is equal to y/c so that the mean square shear is
9.2.1.
E[F2(t)]
Vertical Acceleration Excitation
= 271:So
[f
81c
ylc
We consider now the nonstationary response of the dam-reservoir system to vertical ground acceleration excitation, which is modeled by a stationary process x(t) with a white spectrum So and a zero mean value. Again we assume x(t) = 0 for t < O. As in the case of SDOF systems in Section 9.1, the response mean value is again zero. The mean square response for the shear force F is, from Equation (9.7),
~[F2(t)] =
u}(t)
= 271:So f~ h}(O) dO
(1
+ ex)2 p. 2C2(CO 2
J
[1 + ex 2 + ... + a ltn -
_ y)2 dO
_ 1tSo(l + ex)1.p~CH3 (1 _ .t)3 [1 + ex2 + ... + a2(O-1I]
-
3
H
I ,]
(9.15)
when t ... 00, n ... 00 according to Equation (9.14), so that the mean square in Equation (9.15) becomes lim E[F2(t)]
(9.13)
r~'"
1.)3
= 3~
+ a) 1tSop1cH3 (1 _ 1 - a) H
(9.16)
I: I
I
I
I" I
I
I
i
"
in which hp(t) is the unit impulse response function given in Section 7.5.2 and plotted in Figure 9.2. An examination of the figure reveals that the integral in Equation (9.13) changes with time t in a simple manner when t is limited to a multiple of the half-period 2H/c. Let n be defined such that
which agrees with the stationary solution, Equation (7.168), as expected. In a similar manner, the mean square response for the overturning moment Mis E[M2(t)]
2H
= 2nSo
c
=
is lhe restricted time t. Then from Figure 9.2. we see that the integral in Equation (9.13) equals to n times the basic integral within the range of 0 ~ t ~ 2H/c, "~·l\'.
"IC
ylc
(9.14)
t=n
[l
(1 + a)2
p 2C2(cO _
and when t ...
]
d6 [1 + a 2 + ... + ex 2(n-I,]
1_)S [I + II
(X2
+, ... + ex 2(n-1)]
(y)S
,"'..,
1- -2 '" "
peH ( 1 - ij)
o
~
I", " hp=-2 pcict-yl
._ _ _ Half [leriod
1+ 2
I > •
a
1'cl.2H _ h - cel = hF
V
A
a(~)peH(l 2
Half
W'
-!)
P
.
FIG. 9.2. Impulse response (or hydrodynamic shear (orce on dam due to vertical ground accelera tion.
(9.17)
00,
. (I + ex) hm E[M2(t)] = - - - - 1tSop2cHs 1 - -
()
8
71:50 (1 +3~.~!!lCH~ (I _
---20
y)4
20(1 - a)
H
(9.18)
Note that both the stationary response shear of Equation (9.16) and moment of Equation (9.18) become unbounded for dam-reservoir systems with a rigid foundation when a = 1.0. This behavior is analogous to that of an undamped linear SDOF system for which the unbounded mean square response to a stationary excitation process with a white spectrum is well known. To study the transient behavior of the mean square response, consider the response at the base of the dam y = 0 and for a deformable foundation of 'ex = 0.815. From Equation (9.15), using the hydrostatic force Fo = WH2/2 and the fundamental circular frequency WI = 1tc/2H, we have
= F~C;~I)n(l + a)2(1 + ex 2 + ... + a2(n-Il)]
i
11
1
m~
II ~
'"
\.t
,~~
t
!\ "~l
-!: ~
"
:~! ,!~
(9.19)
.<.
Note that the right-hand side of Equation (9.19) d~pends on the time t implicitly through the half-period 2H/c by the definition t = n(2H/c).
!:
E[F2(t)]
r
1 ::1"
1m Ii
IIDH ,.: w:
I ' ll:;", ' ; i .13 W
.,1..;
;!~
.t' . ~I
I
~
IIill~ I fll!
I II I II ~~:
U ~I
f!
238
NONSTATIONARY RESPONSE
The nondimensional mean square shear E[Fl(t)]/(FASowt/gl) is plotted against the number of half-natural periods n = t/(2H/c) for 0: = 0.815 in Figure 9.3. It shows clearly that the transient response approaches a stationary state in about eight natural periods (n ::::: 16~ For the mean square moment, using the hydrostatic moment Mo = wH 3 /6, y = 0, and WI = 1I'.c/2H in Equation (9.17) gives
!I
,I
E(M2(t)] =
II t! II
I
9.2.
II
M~ e:~l)[ 158 (I -+ OC)2(1
+
(X2 + ... + a 2 l]
(9.20)
(n-l
DAM-RESERVOIR SYSTEMS WITH STATIONARY EXCITATION
239
A plot of the nondimensional mean square moment E[M2(t)]/(M~Sowd~/2} against II t/(2H/c:) for a = 0.815 in Figure 9.4 shows again that the transient moment approaches a stationary state in about eight natural periods (II :::::. 16) as in the case of shear. Also'plotted in Figure 9.3 and 9.4 are the idealized case of a rigid foundation with a = 1.0, clearly showing the effect of damping generated by the deformable foundation. Note that the dimensional mean square shear (or moment) in Equation (9.19) [or Equation (9.20)] depends on the time t, the natural period 4H/c, the
70
56
II
60 48
.
E
:11 I
Si
E E
.c
0
!go 40
'"'"0 '"
'" l!!
l!! 50
.
c:
"ill
'"E
E
~
r
m
: I!
;! J
1:
,!
I I
I
.I
Stationary case
r
Stationary case
~
I~
I
N 40r .!:!>
N 32
0
Zb
'"r;:
~
~ 30
;:..
S 24
~
;!'oil
101 20
16 .
•
t; ,"
"
,
:~!
::i:, :·1
.
f·:
-,
~~
:
~~
.
I!H.
~;
fU
}
,<
l .
".~ I ~~f
:U ",mJ
6 n = 1112Hel
= time
14
16
FIG. 9.3. Nonstationary response mean square shear at'
16 6 n tJ(2H1el = time 'FIG. 9.4. Nonstationary response mean square moment at dam base versus time (exacl at integer 2
II).
4
II 240
NONSTATIONARV RESPONSE
9.3.
refraction coellicient.a, the excitation acceleration parameter Sowdg 2, and the hydrostatic force Fo = WH2/2 (or the hydrostatic moment Mo = wIl J /6). When the spectral density of the vertical ground acceleration So has a magnitude of g2/WI' meaning that the mean square acceleration in the frequency range (0, WI) is 2g2, then the stationary response root mean square (RMS) shear is equal to .J6.541 (or 2.558) times the hydrostatic shear F Q for ex = 0.815 from Equation (9.19) or Figure 9.3. In the case of overturning moment, the stationary response RMS moment is equal to ..)35.319 (or 5.943) times the hydrostatic moment Mo for (X = 0.815 from Equation (9.20). or Figure 9.4. These RMS responses are at the base of the dam y = o.
9.2.2.
Horizontal Acceleration Excitation
Consider the nonstationary response of tile dam-reservoir system to a suddenly applied horizontal ground acceleration excitation, which is modeled by a stationary process with a zero mean and with a white spectrum So. For such a system. the impulse response function for the hydrodynamic shear force F(t) given by Equation (7.226) is
where J II is the Bessel fUllction of the IIrst kind and of ord~r zero, f 0 = (l)H 2/2 is the hydrostatic shear force, and kn = n(211 I )f2U is /I the wave number of the vibrating water in the reservoir. The mean value of the response shear force is zero and the mean square v
E[fl(l}]
= a}(t) = 2nSo
0
t
)2 i' [WI 0
n= I
.
1
o}(t)
elK.
(4096/1r4")(f~)(2Sowtlg2) ~ 41r Jo J~x) dx
Jof
2
".
J~x) dx ~ 1.6
for
t =
1
o}(l)
(4096/1r4}(f~}(2SoWl/g2) ~ 0.1 The first term approximation indicates that when the excitation acceleration has a mean square value equal to O.oli! in the frequency range 0 ~ W ~ Wt, the mean square response shear force 0}(1) ~ (O.1)(42.05)(O.Ol)Ft = O,042f~ after one natural period of time. The standard deviation (IF ~ O.20F0 or 20% of the hydrostatic shear force f o.
SDOf SYSTEMS WITH NONSTATIONARY EXCITATION
As pointed out in the beginning of the chapter, we treat the nonstationary response problem in two parts. Sections 9.1 and 9.2 deal with the so-called zero start stationary or pseudononstationary excitation. Ih this and the following section, Sections 9.3 and 9.4, we treat the excitation as a zero start nonstationary 'process with time dependent statistics for t > O. 9.3.1.
Priestley's Model
~
,
First consider a stationary random process x(t) with zero mean, using Priestley's complex integral model (Priestley, 1967)
J (k "--2 tt) -~-.
(2/1 -
I
I
D
I I I I
;· ' I
,. '.
I I ,
11;'(0) dO
Substituting I',,(t) in the preceding equation 16Cf aNt) = 2nSo ( 2n ~/H
or
9.3.
16c 00 J (k 11,._(1) = - - F L _I!. -ncr} n2011 o. = I (211-=-,)i
241
SOOF SYSTEMS WITH NONSTATIONARV EXCITATION
I)
J2 dt
As a IIrst approximation, consider only one term in the preceding convergent series of Bessel functions. Using kl = n/2H. (01 = nc/2H, 1: w l t/2n, and k I cl = 2nt in the approximation gives
x(t)
= f~., e
."
dX(w)
(9.21)
where dX(w) is defined as a zero mean random orthogonal process such that E[dX(w)] = 0 and
E[dX(w)*dX(w)] (2n)( 16<:) f' P(2 ) (2n) d (F~)(SOWt!g2) ~ WI n 2H. Jo . 0 nt WI t
i
=0
for w :/: w'
(9.22)
I::
~
I
i
~
I
,{
~
,
I
;;: I m
h: ;: ! ri ~ ;:~
o}(t)
~,
As usual w is the frequency arid dX(w)* is the complex conjugate of dX(w).
1 I
' !II mii <,
~~ f.;;; v.J ~t
IR!
q
fi
Ii
I~H i~;
III 242
I I I I I I I I I 'II " I
t
!
~
!:.
Ii
NONSTATfONARY RESPONSE •
9.3.
The autocorrelation function for x(t),in terms ofthe lime lag r, is by definition Rx(r)
real part of E[x(l)*X{l
= Re
IA(t, wWEldX(w)!2 = SAt, w) dw +I",'f
too", ei"'tEldX(wW
= Re f~ S,,(w)ej
(9.24)
(9.25)
must also be an even function of wand consequently the integral in Equation (9.25) is always a real function of the time lag t. Thus we can drop the real part symbol in Equation (9.25) to give
J:oo S..(w)e
i 6>t
dw
The last equation is clearly recognized as the standard relation between auto correlation function R.Jr) and the power spectral density function S..(w) of the " stationary process x{t). Now let us extend the preceding development to a nonstationary random process x{t) with zero mean by introducing a deterministic modulation function A(t, w) in Equation (9.21) as x(t)
=
f~"" A(t, w)e",·t dX(w)
!
.
I
,
~I
(9.21)
The autocorrelation function in this case is then a funl;:tion of both t and r.
;,'
I,
.~
Rx{t, r)
= Re J:"" IA(t, w)1 2ei"'tEldX(wW
t
I '"
La
= r~ SAt, w)ei.,,, dw
Sx(t,w)
(9.30)
(9.31)
= IA(t,wWSAw)
(9.32)
For the special case ofa frequency independent modulation function A(r, (I) A(t), Equations (9.31) and (9.32) lead to RAt, t)
IA(tW
I
SX(W)e'Wf dw
(9.33)
'" Using Equation (9.26) in Equation (9.33) gives RAt, t)
= IA(tWRAt)
(9.34)
In summary, we see that based on the integral representations of a stationary and nonstationary random processes, Equations (9.21) and (9.21), respectively, we have derived for the two processes a spectral relation, Equation"(9.32), and for the case where A(t,"w) A(t}, a correlation relation, Equation (9.34). 9.3.2.
Response of SDOF Systems*
Let the impulse response and complex frequency response of a SDOF linear system be denoted by h(t) and H(w1 respectively. When the system is excited by
(9.28) ·See Shinozuka (1970).
I
f~"" SAt, Wk"'f dw
A comparison of Equations (9.26) and (9.31) justifies the definitions of the nonstationary autocorrelation function for Rx(t, r) and the nonstationary spectrum for SAt, w). Furthermore, from Equations (9.24) and (9.29), we obtain the relation
(9.26)
L=.
!
RAt, t)
At this point, let us restrict x(l) to real functions of t .so that from Equation (9.21), dX(w) must: be even functions of w. Then, from Equation (9.24), Sx(w)
R..{t) =
= Re
Based on the same argument of the even function S.(t, w) of w, we can drop the real part symbol in Equation (9.30) and obtain
Substituting Equation (9.24) into Equation (9.23) gives Rx(r)
Rx(t, t)
(9.23)
8,,(w) dw
(9.29)
Substituting this into Equation (9.28) gives
dX(W 1 ) ]
as a result of the orthogonality property, Equation (9.22). Now let "EldX(W!2
243
Let
+ r)]
fOoo f:"" e-i"'t dX{w)*eioJ't
= Re E [
SDOF SYSTEMS WITH NONSTATIONARY EXCITATION
244
9.3.
NONSTATIONARY RESPONSE
a nonstationary random process x(t), as defined by Equation (9.27), the response y(t) is also a nonstationary random process with zero mean and has the modula tion function B(t, w), integral representation. spectral densities S,(t, w), S,(w), and correlation functions R,(t, -r), R,(t) as follows: yet) = S,(!, w)
J:",
(9.35)
B(t, w)e·"" dY(w)
= IB(t. wWS,(w)
R,(t. t) =
J:oo S,(t, w)e
i ""
(9.36)
=
J~ h(t -
Letting t -
t
=
f;o[J~",
y(t) =
f: _'.
e ,.'
A(T. w)e'''' dX(W)] Ii(t -
[J~=(J A(t -
= J:oo
II
f'"
O. w)e-i.,°h(O) dO] dX«(JJ)
S..(W)ei(l)' dw
(9.44)
R..(V)e-i(l)" dv
Zero Damped SDOF Systems
her) =
(9.39)
(9.40)
(9.41)
f0,
.
dT
(9.42)
Finally, substituting B(t, w) of Equation (9.42) and S,(w) of Equation (9.41) into Equation (9.36) gives the following input-output spectral relation due to
t;:;o 0
Wo
(9.46)
t <0
Let the excitation x(t) of the oscillator be a simple nonstationary random process with zero mean and with a nonstationary spectrum that is made up of the product of a unit step time function U(t) and a stationary white spectrum So. Thus A(t, wJ = U(t), S..(w) = So. and from Equation (9.32) • Sx(t, w)
t, w)e-'''''h(T)
(9.45)
For an undamped linear oscillator, the unit impulse response function is
Substituting dX(w) of Equation (9.40) into Equation (9.39) and comparing the
resulting Equation (9.39) with Equation (9.35) gives
,=0 A(t -
r
_'"
sin wot ,
= IH(w)l2S..(w)
f'
(9.43)
w)e:,iO"h(t) dTr SAw)
A(t - U, w)ei"'"h(u) du
1 S..(w) = -2
9.3.3.
T) dT
dY(w) = H(ctl) dX(ctl)
1 B(t, w) = H(w)
t,
When A(t, w) = 1 and t...... 00 again Equation (9.44) reduces to the Fourier transform between Ry{t}/2lt and S,(w) in the stationary case.
Recall that in the case of stationary excitation and response we have
S,(w)
R,(t, t)
It
= 0 and interchanging the order of integration gives i
= 11'=0 A(t -
Note that when A(t, w) = I and t -+ 00 the integral in Equation (9.43) reduces to H(w~ which is the well-known transfer function in the stationary case. The input-output correlation relation can be obtained from Equations (9.37) and (9.43)
(9.38)
T)X(t) dt
Substituting x(t) of Equation (9.27) into Equation (9.38) gives yet)
S,(t, w)
where
When the system is initially at rest. the response is yell
Shinozuka (1970).
(9.37)
dw
245
SDOf SYSTEMS WITH NONSTAnONARY EXCITATION
= U(t)So
(9.47)
The response y(r) is also It nonstationary random process with zero mean. Its nonstationary spectrum can be obtained by substituting h(t) from Equation (9.46), S..(w) = So and A(t, w)·= U(t) into Equation (9.43) and carrying out the integral as S,(t, w) =
If'
<=0
'. U(t -t r)e-''''' sm Wot dT 12 -So = II(t, (tI)1 2 -So Wo
Wo
(9.48)
,II
246
(~
I I I I I I
NONSTATIONARY RESPONSE
9.3.
For t > 0 the unit step function in Equation (9.48) is unity within the range of integration. Consequently, it can be deleted and the integral is I(t, w)
=
I
h(t)
(9.49)
e-;"" sin Wot dt
So (w~
(
W
2
)2
and U(t)
Ij
Ii 1i1
. 1iJ·
t
(9.53)
(J)o
where the integral Wo
The mean square response can be obtained by setting t and is Ry(t,O) = o-;(t) =
(9.50)
0 in Equation (9.44)
I
2
nSo
-3
f~", S,(t, w) dw
2(/)0
.
(2w ot - sm 2wot)
Let Wdt
f'
e-(;w+{WoI,
sin
Wd'!: d(Wd t
"'"' ~
~,
)
=
I Wd
(9.54)
~~
= 0, then I(t, w)
f"'d! e-(I"'+~""'16/UJd .' sin 0 dO 0
1 + e""''''(a sin Wdt - cos wat)
2 Wd(1 + a )
(9.55)
where (9.52)
This result agrees with that given by Equation (9.8) as expected, since the non stationary spectrum S..(t, w) for the excitation x(t) is so specified here that it is
exactly the same as before, namely S..(t, (/) = 0 for t < 0 and Sx(t, w) = So for
t ;;::: O. Note that this approach is basically different from that used before
since one is through a frequency domain superposition and the other through a
time domain.
9.3.4.
.,
~ ~";
WdJO
The integral in Eql,lation (9.51) represents the total area under the nonstationary spectral curve, which is analogous to the stationary case. Substituting S,(t, w) from Equation (9.50) into Equation (9.51) and carrying out the contour integration, the following transient mean square response can be obtained (Caughey and Stumpf, 1961) O"y(t)
b,
I e-i..r-~""'t sin Wdt dt
I(t, w)
iw
+ ,;wo
';wo. W
----I Wn
Wd
and
1+ a
w"
2= (w~ (2) + -2
WJ
j
2';wwo
WJ
Let
b = Pr + ibj = .- e
lightly Damped SDOF Systems*
•
= ea""'(a sin Wdt
Parallel to the undamped system presented in Section 9.3.3, the nonstationary response spectrum S,(t, co) and mean square q;(t) to a simple excitation process x(t) with zero mean and nonstationary spectrum S ..(t, w) = U(t)So are derived as follows.
cos Wdt)
-~"",t '(-. ;wo. . • cos wt sm Wdt + cos wt cos Wdt + -( j )sm wt sm Wd
. (w
- Ie _~""" t
-
Wd
.
cos wt sm Wdt
Wd
,;wo .wt . . wt cos Wdt ) - sm sm Wdt -sm Wd
then I/(t, w)1 2 = 1(1. w)l(t,
·See Corolis and Vanmarcke (1975).
~
[11
(1)*
~ ! + 2b,2 + b; +2 bf (I + a )(J + a )*
wJ
(9.56)
I I I I
Iii
It
t;;::: 0
S1(t, w) = S02 II(t, wW , Wd
a
Il, [15
{
e-~""" t 0 Wd sin Wd
= A(t) into Equation (9.43) gives
sin 2 wot - 2 cos wt cos (l)ot
. wot ] - -2w. sm wt sm
I 11
=
2
+ cos 2 wot + 2
247
Substituting the impulse response function
Carrying out the simple integral in Equation (9,49) and then substituting the result into Equation (9.48) gives
Sl.t,
SDOF SYSTEMS WITH NONSTATIONARY EXCITATION
'"
!1llI
, .
~
1 ,.
248
k..
II'
w"
1
+ a 2 )(t + a 2). = w: [(w~
2
- (
)2
+ 4.;2W~W2]
=
...
(9.571
c(t)
(02
+2
Wd
.
I/(t,
•
sm 2 wt 8m 2 w.t
•
Wd
+ wd(t)sinwt]}
(9.59)
Wd.
2 So = 21/(t, w)1
S yet, w)
(9.60)
w"
The nonstationary spectral density Sy(t, (I) degenerates to the stationary response spectrum
2
Wd
wW = wJIH(wW{l + e- 2(".'[l + aCt) + w 2 b(t)]
The nonstationary spectrum solution is now complete since
Wd
bl =e- 2~""" [(';2~~ sin 2Wd t + COS 2W,.,t + ,;wo sin 2W.t) sin 2wt +2
W.
- e-~"""[c(t)coswt
Wa
2
.
= - - sm Wdt + 2 cos w"t
then
.. . Wd! + cos W.,t )] + 2 -W sm wt sm Wat cos wt (.;wo - - sm
W
2~wo
d(t). = 2 sin Wd t
Wd
b; = e-2~""" [(';2~~ sin 2 Wd t + cos 2 Wa t + ~wo sin 2wat) cos1wt Wd Wd
'·1
249
• 2
and
L
SDOF SYSTEMS WITH NONSTATIONARY EXCITATION
sm Wd t b() t =--2 (l
Ii ,.
I I
I I: I I I I I
9.3.
where
~
~
NONSTATIONARY RESPONSE
2
cos wt sin w.t
Sl'(W) = Il1(w)12So
- 2
W cos -.
Wd
wt
• sm
Wdt
• sm
wt (.;wo. - - sm Wdt
+
cos Wdt)]
when
W"
:. b; + bl = e-~"""
';2W2 . Wd
2 ( ---; sin Wdt
I ......
ce, as expected. For small damping,
ai')
+ cos 2 Wdt + ';W _ 0 sin 2w.t + 2 Wd
Wd
• 2
(9.58)
Substituting Equations (9.57) and (9.58) into Equation (9.56) gives
'1/(t, w)1 2 = wiIH(w)1 2
{l - 2e-~""" e
i t
::. (,;wo cos wt
('(t)~
2
+ W sin rot)
)~ S y( t, w
-
t
(.;
2
Wo2
' ';Wo sm . 2wdt]}
+ W2) + cos 2 Wdt + --
w"
+2coswot;
When';
(2
e-~""'I
So
Wo -
[2
2)2 (I)
e .
cos wot cos wt
eW2 _W2) ( Wd . sin 0 2
d
2
Wd t
';W
+ _ 0 sin 2Wdt
W.
• 2 +w 2 sm wot] Wo 2
cos wot
+ ~: sin wot sin wt]}
(9.61)
=0 Wo -
aCt) =
d(t) ~ 2 sin wot Wo
{t + -2~"",t [2
'[ Syet, W) = (2 S o 2)2 t
Let
wo
so that
+ cos Wdt cos wt] + e- 2~"",I [sin Wd 2 Wd
bet) ~~ot 2
aCt) ~ - sin 2 wot;
sin 2 Wdt
W
+ cos 2 wot + w 2
2
Wo
• 2
sm wot
.. - 2 cos wt cos wot - -2 w sm wt sm wot Wo
J
(9.62)
250
NONSTATIONARY RESPONSE
9.3.
which agrees with Equation (9.50) for the zero-damped case derived in the previous section, The mean square response is O',(t) = Ry(t, O) =
n;S { I =--;
2ewo
f:,.., Sy(t, w) dw e-2~"'o' [2eW2 1 + ~sin2
Wd t
';w 0 sin 2Wdt]} + __
Wd
9.3.5.
SDOF SYSTEMS WITH NONSTATIONARY EXCITATION
251
I I
Bendat and Piersol's Model
An alternate approach to the nonstationary excitation and response problem is due to Bendat and Piersol (1971) and is presented here for comparison. Let the time dependent autocorrelation function fora zero mean excitation process x(t) be defined by RAt, 1") = E[x(t)x(t
(9.63)
+ 1")]
(9.64)
.
(9.65)
m
Wd
The Fourier transform of (l/2n)R x (t. 1") is which agrees with Equation (9.11) derived in Section 9.1.2 through a time domain analysis. Equation (9.61) for the nonstationary response spectrum is plotted in Figure 9.5 for = 0.10, om, and O. These plots show that as the time wot increases, the peak spectra in all cases increase in magnitude. Furthermore, the widths of the spectra become narrower as damping decreases. In all cases, the main contribution to the mean square value comes from the spectrum distributed near the undamped natural frequency wo' This concentration near Wo is strongest for systems with the lightest damping.
(9.68)
I I I I I I
is also a time dependent random process with zero mean. According to the same definition given by Equation (9.64) for the excitation process x(t), the time dependent autocorrelation function for the response y(t) is
II II
1 SAt, W) = 2-
fan
1t
e
_..,
R",(t, 1')-"'" dt
which is defined as the nonstationary spectrum of x(t) with the inverse transform
f:..,
RAt, 1')
SAt, w)e;"" dw
(9.66)
Consider now a linear SDOF system with unit impulse response function
Ii(t) and with the excitation process x(t) specified in Equation (9.66). Assume
26
, '" 0.10
24
2400
22
-"'01
20 18 16 S
(t.
,.1
14
='"
that the system is initially at rest and
260,000
2600
240'OOO~ ,= 0
, = 0.01
2200
I
wot "" 1000
h(t)
220,000
2000
"'0
1800
,/"
x
200,000
{
e-~"'ol
= -;;;;- sin w"t; 0;
t~O
t
(9.67)
'
180,000
1600
160,000
"'01,",201400
140.000
So',.,g
The respons~ y(t) =
f~ h(t -
,)x(r) dr
120,000
12
100,000
10
"'01
8
100
800
80,000
6 600
6
I "'01 "
60,oooL
500
Ry(t, i) = E[y(t)y(t
"'0'" 50
4
400 200 1.0 OJ
1.5
2.0
0
= 0
0.5
1.0
1.5
2.0
0
0.5
w
"0
wo
(al
(h)
FIG. 9,5. Nonslalionary response spectra (u) ~ = 0.10, (h) ~ = 0.01, and (e) ~ Vanmarcke. 1975).
1.0
1.5
+ 1")] .
Jof' Jof'+' h(t -
u)h(t
+
1" - v)E[x(u)x(v)] du du· (9.69)
2.0
w
Fa (e)
Now consider that the excitation process x(t) can be separated as the product of a detenninistic time function A(t) and a zero mean stationary process f(t). That is,
0 (Corolis and
x(t) = A(t)f(t)
(9.70)
I I I I
II
i'l lt
_,
~
252
,'. "
I'
(9.71)
E(x(u)x(v)] = A(u)A(v)Rj(u - v)
S,(t,w) = S/(w)
where the autocorrelation Rj{u - d for the associated stationary excitation J(t) is related to the spectrum by the usual integral Rf(u - v) =
f~", Sf(w)eiW(U-V) dw
(9.72)
Substituting RAu - v) of Equation (9.72) into Equation (9.71) and then the expectation of Equation (9.71) into Equation (9.69) gives
,1'+< A(u)A(ll)h(t 1o
1l)It(t
+r
- v)
0
J'"
= Sf(w)
= Sf(W)
(9.73)
,, 11.+' Atu)A(v)h(t 1o
«)/I(t
+r
MUltiplying both sides of Equation (9.74) by
e- i""
v)iw(U-.)+ du tlv
0
A(t - 8de- iw6'h(8d d8 1
r
J~-o
A(t
8 l )e'tJJ8 2 h(8 l ) d8l
J~ A(v)e-iW('-V)h(t -
t
J~ A(u)A(v)h(t -
v)dv
S~ A(u)e''''('-·)h(t -
u)h(t - v)e-'''(u-o, du dv
«)du
(9.77)
After removing the imaginary part from the exponential function on account of the real valued function, Equation (9.77) is indeed identical to Equation (9.76) as expected.
SAw)e'W(.-O' dw d« dv
-'"
Again, according to the same definition given by Equation (9.66) fOr the time dependent autocorrelation of the excitation x(t), the autocorrelation Rit , r) of the response yet) is related to its nonstationary spectrum S,(t,W) by the same integral. A comparison of that integral relation, Equation (9.66) (or yet), and Equation (9.73) yields ' e''''S,(r, w)
r
J~.o
Let t - 8 1 = v and t - 8 2 = u. Then S,(t,w) = Sj(w)
Ry(t. r) =
253
DAM-RESERVOIR SYSTEMS WITH NONSTATIONARY EXCITATION
square of the absolute value of the integral in Equation (9.43) by the product of the integral and its complex conjugate yields
Under this restriction, the expectation in Equation (9.69) becomes
~
II I:' I I I I I I I I
9.4.
NONSTATIONARY RESPONSE
9.4. DAM-RESERVOIR SYSTEMS WITH NONSTATIONARY EXCITATION 9.4.1.
Vertical Acceleration Excitation
We consider the nonstationary response of a dam -reservoir system to vertical ground acceleration modeled by a zero mean nonstationary process. This is the same problem as treated in Section 9.2 with the only exception being a non stationary excitation. Following the derivations in Sections 9.3.1 and 9.3.2, the excitation acceleration has a nonstationary spectrum
(9.74)
•
gives th~ nonstationary
spectrum of ytl) as S).{t, (tI) = S f«(tJ)
' J'+f A(Il)Alv)l1(t Jo 0
Sa(t,
u)/l(t
+t
-
0
(1)e'w(u- 1 du,
dv
f' f'. A(<<)A(v)h(t -
Jo Jo
'
«)h(t- v) cos (m(<< - v)] du dv
= A 2 (t)S.(w)
(9.78)
(9.75)
Referring to Equations (9.64)-(9.66) for the response yet) we see that Ry(t. or) is an even function of r. Then from Equation (9.65), S,,(t,w) must be real and even in w. Furthermore, since S,(t,W) in Equation (9.75) is indepel)dent of the time lag r. it can then be simplified by setting r = O. Thus Sy(t, w) = Sj(w)
(I)
r
(9.7 6
It is instructive to show in the following equation that the response spectrum Sy(t,w) of Equation (9.76) and of Equation (9.43) obtained previously ate identical. Putting A(t - r, w) = A(t - r) and S"Jw) = S jew) and replacing the
where A(t) is limited to being a real valued function of time t and SQ(w) is the associated spectrum of the stationary acceleration. Using Equation (9.43) for the response shear F(t) the nonstationary spectrum for the shear force is then SF(t,W)
=
If.o
=
J
L
1
A(t - t)e- ""hAr) dtr S.(w)
e-1wthF(t)
dtr
S..(w)
(9.79)
where A(t) is assumed to be a Heaviside unit step function. Fot the dam problem with a vertical ground acceleration excitation, the unit impulse response shear hF(t) is given in Section 7.5.2 and plotted here for
254
NONSTATIONARY RESPONSE
9.4.
y = 0 in Figure 9.6. The integral in Equation (9.79) in the first-quarter period 0:;:;; t:;:;; H/ds
, (I + C<) I o
e -iwB - 2
2
pc 0 dO =
I)
iW1 e-;<41 + a)pc 2 (te- - + - - __._
(1
-2m
i
DAM-RESERVOIR SYSTEMS WITH NON STATIONARY EXCITATION
then the nonstationary spectral density for the shear force at t = Hie is
The mean square
O'~(t) =
elm')
te - 10" I e- i(o/)(tei<01 I +w- - -w- - +w- -w ( -i -i =
+ a)2 (F~)(SOWI) w! 7 0I4 (0 2 - 2 sin 8 + 2 - 2 cos 0)
H w ) = (I SF ( -;;-'
J:i (U}2 t W
-
wt2 sin wt
+2-
(9.80)
-w-w
Since
2
255
=
(I
+ a)2 4
2 cos rut)
xJoroo (w t
2 2
the nonstationary spectrum from Equation (9.79) is, for 0:;:;; t:;:;; Hlc and for
r a
2 LOOS.,(t: w) dw = 2w!
Sf(t, ru) dQ
(4)(24) (F~SOWI) 1l
WI
-
2M sin M
-4-
--2-
9
2w! dQ
+ 2 - 2 cos twt) Q4
(9.81)
SQ(w) = So S,,(t, w) =
+ a)2p 2c4S o
(l
.
d
(W
2 2
t
-
2wt sin wt
+2-
_
At t = fIle lhc~
2 cos wt)
(H) _5 +
O'F2 -<1 -
Let
Q=~ w' I F~
=
2
w H4 4
llC H WI = 2H' t =-, SOW! SollC
and C
wt =
Q
114
Let
a)2
I)
(F~SoW - - Ji'OO g2
0
[(Qll)2 . Qll - -Qll2 s m 2
2
2
QllJdQ
+2
llC H Q1t 2H~= 2 = 0
7=2Hg 2
2 (1
(9.82)
2cosT Q4
Qll/2 = O. Then
O'~(H)
c
= 4(1 + a)2 (F~SoWI) 1t
g2
r"" 02 -
Jo
20 sin 0
+ (2 - 2 cos 8) dO 0
4
(9.83)
hF'.O. t)
Note that hF'.O. t)
~
(1; a)
, 02 - 20 sin 0 + 2 - 2 cos 0 I1m - - - - - - 4: - - - - 0
pelt
8-+0
.. (l+a)pcH 2
FIG. B.S. Impulse response function (or shear h,(O, t)at the dam base y acceleration.
= Odue to vertical ground
4
A crude numerical integration shows that for a = 0.815
O'i (~) ~ 4.19 (F~!~WI) (0.915) =
3.84 (F~:~WI)
(9.84)
which roughly agrees with Figure 9.3 (Section 9.2,0. A comparison of the frequency domain integration approach here with the time domain integration in Section 9.2.1 shows clearly the relative simplicity in the time domain approach. Therefore, if only the mean square response to the special unit step modulated excitation is of interest, then the time domain approach should be followed. The present frequency domain, however, gives
j ~ I
' ::} fI "I
t~,
•
:11
i~ ~
256
II
'! I I
9.4.
more information on the distribution of the mean square resf'(lnse in the frequency domain. When the associated stationary excitation has a frequency dependent spectral density, the present approach may be more effective. Further more, when the time modulation function is not a simple unit step function but a general time function, the excitation is a true nonstationary process and is no longer a suddenly applied stationary process. In such a case, of course, the previous time domain approach does not apply. 9.4.2.
[I I
NONSTATIONARV RESPONSE
Horizontal Acceleration Excitation"
We consider the nonstationary response of a dam-reservoir system to horizontal ground acceleration modeled by a zero mean nonstationary proc{ . Again, following the derivation in Sections 9.3.1 and 9.3.2, the nonstationary spectral density of the excitation Sa(t, w) is equal to the product of the square of the modulating function A2(t) and the associated stationary spectrum S.(w) Sa(t, w)
1.1
I
= A 2(t)S.(w)
t
1
A(t - O)e-;",OhF(O) dO
r
S,,(w)
(9.86)
I I
where IIF{/) is the impulse· response function for shear force F(t). Wh-en the time modulation function A(t) is further limited to be a Heaviside unit step function and the stationary spectrum S.(w) = So then Equation (9.86) can be simplified as
I I I
where the impulse response function from Section 9.2.2 is
\
I I'
SF(t, w)
hF(t)
I
= f~ e-1OJIhF(t) dt
'" = -16c .F ~ 2 1t gH
0,,';:1
r
So
J oCknct)
(9.87)
(9.88)
Let the non dimensional frequency a and the time t be defined \)y 0= w W' I
-I it -
t=O
e
-1«11<1
~ J o(21t1:(2n - 1)] d
L...
"=1
(2n -1) 1
T
11
(9.89)
It is useful to keep in mind that the dimension of SF(t, 0) is the force squared per unit width of dam squared per frequency or simply F2 T/U, that of So is the acceleration squared per frequency of L2/T 3, and that of H4 W 2/g2 is F2T4/L4.. Thus the ratio on the left-hand side of Equation (9.89) is dimensionless. Equa tion (9.89) applies for t > O. For t < 0, SF(t, 0) = O. The nondimensional mean square force on the dam, becomes o-;(r) So(1024H3w2c/n3g2) =
f""
Jo
[(t, 0) dO
(9.90)
in which f(t, 0) represents the right-hand side of Equation (9.89) and is the non dimensional, nonstationary spectral density of the hydrodynamic force on the dam. Note in Equation (9.90) the one-sided integral due to symmetry of l(t,O) in a and the change of some terms in the denominator on the left-hand side according to a = W/WI' Equation (9.90) applies for t ;;;:, O. For t < O,o}(t) = O. Analytic Study of Solution Equation (9.89). Although the primary solution for the random response force on the dam in terms of the nonstationary power spectral density S~T, n) and the mean square function oJ(t) have been obtained in integral forms, a general analytic solution, unfortunately, cannot be found. Special cases, however, can be solved analytically so that the validity of the solution can be checked with known specialized solutions in the literature before numerical integration computations. This assurance is needed here because of the expected sharp peaks in the force spectra, the singularities in the limiting stationary spectrum. and the infinite series of Bessel functions involved. First, consider the stationary case when the upper limit t approaches in finity in Equation (9.89). As infinite integrals they are known as the Weber Schafheitlin integrals and are tabulated in the Handbook of Mathematical Functions (Abramowitz and Stegum, 1964). The right-hand side of Equation (9.89) with t -+ co then becomes
iiI
= TI
where the fundamental circular frequency WI = ftc/2H and period ~ ·See Yang and Charita (1981).
SF(t, 0) 4. i 4. 1 So(t024Hw/1tg)
t
t
257
Then kftct = 21t(2n - l)t, wt = 211:Ot, and the nondimensional, nonstationary power speCtral density of the hYdrodynamic force on the dam from Equation (9.87) and (9.88) becomes
(9.85)
The mean value of the response shear force .F(t) is zero. The nonstationary spectral density of the response shear force can be obtained by using Equation (9.43) SF(t, w) =
DAM-RESERVOIR SYSTEMS WITH NONSTATIONARY EXCITATION
[
JI 21t(2n -
= 2x/w p +
L:t+
1
J2
1)2 • J02' - (2n _ 1)2
=-=n~2J
1 I 21t(2n - 1)2 • -">==(2=n=_=1=::1);;=2
258
NONSTATIONARY RESPONSE
9.4.
DAM-RESERVOIR SYSTEMS WITH NONSTATIONARY EXCITATION
259
where N is the largest value of n such that [(2n - 1)2 - 0 2] < O. The two terms in the preceding equation can be combined to give
In~1 co,
I /2 2lt(2n - 1)2 J(2n - 1)2 _ 0 2
Thus Equation (9.89) for the stationary case with
)2/
~.
32WH2)2 ( -1 S 0 = (-' F( ) It 2g 2lt
~
l
I
.-
---.
'..\1'"... ~'t
is reduced to
1 12 So JI (2n _ 1)2 J(2n _ 1)2 _ 01 co
Sf'( 00,0)
J ~
I(
1024H4W2So) lt 4g 1
=
9.9 I) (
0.0280
(9.92)
Second, consider the case of almost incompressible water but without the assumption of a stationary state, in such a way that the nonstationary spectrum is almost independent of w. Then setting 0 = 0 in the integral in Equation (9.89) we obtain
J;
[J o(2Itt)
+ ~ J 0(6ltt) + 2~ J o(lOltt) + .. -J dt
These integrations can be carried out analytically. A fifsHerm approximation and a change of variable reduce this to
1 2It
f2nr Jo(t)dt = 0
I '"
-
L
J O + 2 k+l(2ltT)
(9.93)
ltk=O
This analytic approximation is evaluated with the help of the tabulated Bessel functions and the nonstationary force spectrum is plotted in Figure 9.7. Note that since Tl is very small, the actual time t for the transient duration t ~ 10 ' is almost instantaneous. Last, for the special case of 0 = I, the right·hand side of Equation (9.89) reduces to the following first-term analytic approximation 2ltJo
]2 + [12ltJof21<' sintJo(t)dtJ2
costJo(t)dt
=t
2
[J5(2ltt)+JH2ltT)] (9.94)
g~
I
2 --'-' a ~
10 '
FIG; 9.7. NORstationary spectrum of hydrodynamic force SF(t, O)/(!024d"w'So/n 4 g') versus time t = tiT"~ analytic approximation. incompressible water.
This analytic approximation equals 0 and 0.093 for c = 0 and I, respectively. These results and the analytic approximation for the case of incompressible water (a = 0), together with the well-known stationary solution Equation (9.91) are used as checks for the subsequent numerical computations.
Numerical Results. It is first noted that the series of Bessel functions in Equation (9.89) converges at least as fast as l/n2 and the range of integration is finite. Consequently. SF(r, a) is a well·behaved function of T and a. Furthermore, when the right-hand side of Equation (9.89) is expressed as a sum of squares of the real and imaginary parts, the sum approaches 0 when a --- 00 because for a finite t and a finite m, lim 0-",
[1f1.1«
1°·0280
..g
in which the right·hand side equals \H Aa)\lSo and checks with the well-known stationary solution. Furthermore. if the water is assumed to be incompressible. then the sound speed c and the fundamental frequency WI approach infinity. Consequently. the frequency ratio a approaches zero and the force spectrum from Equation (9.89) degenerates to a constant as the stationary response in incompressible water
~
i
t -+ 00
Exact stationary solulion
0.03
I
Jof' sin 2ltazJo(2ltmz) dz = 0
(9.95)
This is based on the observation that the integral in Equation (9.95) is a Fourier coefficient for the series expansion of the Bessel function J o(2ltmz) in the finite period of (0, T). When J o(2ltmz) satisfies the Dirichlet's conditions of finite number of maxima, minima, and discontinuities in (0, r) then the convergence. of the Fourier infinite series guarantees the validity of Equation(9.95~ Note that only small m needs to be considered here because of the fast convergence of the series of Bessel functions as pointed out before in Equation (9.89). From a physical point of view it is expected that the dam-reservoir system filters out high frequencies of the input excitation so that the output response consists of low frequency components only.
I
;~:
[I
"tf'
.~:, ;;7'
I I I.i I'
260
9A DAM-RESERVOIR SYSTEMS WITH NONSTATIONARY EXCITATION
NONSTATIONARY RESPONSE
frequency ratio 0 > 4, spectral response of the force is negligible except for the idealized stationary case, where singularities exist at resonant frequencies. Good agreement exists between the analytic approximation plotted in Figure 9.7 for the case of incompressible water (0 = 0) and the corresponding numerical computations in Figure 9.8. For 0 = 1 and t = 1 the analytic approximation of 0.093 from Equation (9.103) also checks with the numerical results. The overall response force spectra as shown in Figure 9.8 display the time variation as they approach the stationary limit and the narrowness of the frequency band because of the sharply resonant linear system. Figure 9.9 shows the nondimensional, nonstationary mean square hydro dynamic force on the dam
Having established the vanishing condition at high frequepcy for the non stationary power spectral density SF(,r,O) in Equation (9.89), it follows from Equation (9.90) that the integral and hence the nonstationary mean square value oJ(t) is finite for a finite t. A simple computer program based on Simpson's rule of numerical integra tion is used to carry out the computations of Equations (9.90) and (9.91). Numerical results are plotted in Figure 9.8 for the nondimensional, non stationary power spectral density of the hydrodynamic force on the dam
'
I(1024HgwlSo) 4
SF(t, 0)
:,'11
1r:
:'11
4 2
~.!'l....
as a function of the nondimensional frequency 0 :::;;; 4 within the figure. For the
I I I I
tl I I t: ~
I I If
5.0
r
~ Stationary to" T ,.
1.0
0
'" 'leN .. ::t; a ,. V
I
II
I
3 Z
(1024H3 22 CS o) 7[ g
as a function of the nondimensional time t = t/Tt up to t = tOO. For a dam having loo-ft depth of water, the fundamental period 11 is of the order of 0.1 sec, so that t = 100 corresponds to an earthquake with a typical duration of 10 sec. The rapid initial rise in response is noted in Figure 9.9 at t = 10 to about 75% of that at t == 100. The contribution to the response is mainly
I I I
I
0,5
I l-
l':t
N
z ) O'F(t
I I I I I
100
261
Sl
I I I I
T
I
= 10
I I I I
,:
~ 0,\
r:ii '"~I'"::c: ~ '1
I t, -->0-\1 = 100
0,05
8
,i.
I
0,1
(!
I I I
o,od
'\
I I
===:d
1 J 0 2 3
-
Ib
I
01
o
4
u FIG. 9.8. Nonstationary spectra of hydrodynamic force SF(t. !l)f(lb24Jow2 SoIn'gz) versus time t = lIT. and frequency!l = wlwlo all dimensionless.
!
-
I
I
.-
-
80
100
T
AG.9.9. t
= I/~.
Mean square hydrodynamic force on dam o}(t)J(1024d'w2cSo/lt'g2) versus lime
262
PROBLEMS
NONSTATIONARY RESPONSE
from the frequency interval 0 ~ 0 ~ 2 surrounding the fundamental frequency of the dam-reservoir system.
9.2. Consitier a SDOF system governed by the standard equation of motion wi.th ercitation x(r) and response y(r) as ji
From Equation (9.90), the mean square response
Application of Results.
can be rearranged in the form
42.0SF~(2Sowtlg2j
=
Jo
/("1:,0) dO
(9.96) p(x)
=
Then at t
=:
x(t)
m,y].
._1_._ exp [- (x 2(1;'
.j21C(1x
where the integral is plotted as a function of t in Figure 9.9. Consider the case where the mean square acceleration excitation is uniformly distributed in the frequency range 0 ~ W ~ Wl so that
E[a 2(t)] = 2Sowj
+ 2woey + w~y =
If x(r) = 0 for t ~ 0 and is a stationary Gaussian process with the prob ability density function
roo
oXr)
263
t>O
A(t)
= 0.01g2
100 we find from Figure 9.9 that
L"" 1(100,0) dO = 0.221 \
Hence the RMS hy4rodynamic response shear force (11'(100)
~---'------:!:-------'::~-----.... I (sec)
= O.30SF 0
Sx(w)
which is about 30% of the corresponding hydrostatic shear.
(al
So
PROBLEMS
9.1. Consider a SDOF system with natural frequency WOo damping ratio excitation x(t), response y(t), and the governing equation ji
+ 2wo~Y + w~y =
~.
Let the excitation x(t) be a stationary random process with zero mean and a white spectral density So. The system is at rest at t ~ 0 and is subjected to the excitation x(t) at t > O. (a) Determine the nonstationary mean square response E[y2] when 0, by a time
e= e
.
1'\
x(t)
I
2
3
I
4
;>0
(bl
FIG. 9.10. (a) Time modulalion function A(l) and (b) stationary spectral density by Kanai (1957) aod Tajimi (1960) for earthquake acceleration. . I + 4,j{w/wl) . So S.(w) - [I _ (W/W/)l]l + 4{j{w/wf)2
'I
where WI
5.1 - 51.7 rad/sec,
= 0.1 - 0.9, and
So = while spectral density.
i i I I I I I I I I I I I I
I
I I I ",.
:1·
I I I I I I~I
I I I fi
264
NONSTATIONAAY RESPONSE
determine the mean response E(y(t)] in terms of m", (1", Woo and integral form.
~
in
9.3. (a) For Problem 9.1, determine the nonstationary spectral density
function for the response displacement S,,(t. w) when'; = 0.10. (b) Determine the mean square response E(yl) by integrating S,,(t. w) and verify the result by comparing with the time domain solution from Problem 9.l(b).
9.4. For Problem 9.1, determine the nonstationary spectral density function
e
S,(/, w) when = 0.10 and when the excitation x(t) is a nonstationary process according to Priestley's model with the time modulation function A(t) and the associated stationary spectral density shown in Figure 9.10. Use wI = 5.7 rad/sec and (.r = 0.1. This is an earthquake acceleration model by Priestley (1967), Kanai (1957), and Tajimi (1960).
CHAPTER
10
NONLINEAR RANDOM VIBRATION
9.5. For Problem 9.1, determine the nonstationary response spectral density S,,(t, w) when { = 0.10, but the stationary excitation x(t) does not have a white spectral density function. Instead x(t) has the Kanai-Tajimi spectral density function given in Problem 9.4.
9.6 .. For Problem 9.1, determine the nonstationary response mean square E(yl(t)] when ~ = 0.10, but the stationary excitation x(t) dGes not have a white spectral density function. Instead x(t) has the Kanai-Tajimi spectral density function given in Problem 9.4. 9.7. A nonstationary random model for earthquake ground acceleration is given by
. _{.t
xo(t) -
J~ I
0,
tClje-:·jI
cos(Wj r + cPj);
r~o
t <0
where ai' lXi' and Wj are given sets of real positive numbers with Wt < W2 < ... < Wn and 4>] are n independent random variables uniformly distributed in (0, 21t). (a) Find E[xo(t» and (1;(t). (b) Use aJ I, Wj = j, IXj = 1, and n = 20 to compute a sample function of xo(t) for 0 ~ t ~ 10.0 by the computer simulation method .of Chapter 4. (c) Use a] 1, Wj = j, IX) = 1, and n = 4 to generate 1000 sa~le functions of io(t) for 0 ",:; t ",:; 5.0 and obtain the variance al(t) from the generated samples. Plot (1j(t) for 0 ",:; t ",:; 5.0 from the simulated result of Problem 9.7(c) and compare it with the theoretical one obtained in Problem 9.1(a).
From the standpoint of practical structural engineering, it is important to consider the random vibration of structures beyond the linear elastic range. Unfortunately when nonlinear structures are consid~red, the fundamental solution approach, which is either a time domain superposition or a frequency domain superposition, no longer applies. One method of analyzing nonlinear random vibration problems is simulation. The basic concept of the simulation method consists of (a) generating a large number of sample input excitation functions of time based on the specified random excitation process, (b) deter mining the nonlinear response functions corresponding to each sample excita tion, and (c) analyzing the statistics and probability characteristics of the nonlinear structural response on the basis of all the sample solutions. Because of its general applicability and of the increasing efficiency of the digital com puter, the simulation method is becoming a very dependable and popular tool (Shinozuka,1971). Another approach to the nonlinear random vibration problem is centered around the formulation and solution of the Fokker-Planck equation (Caughey, 1963). This is a second-order nonlinear partial differential equation for the time dependent probability density function of the structural response. It applies to linear and nonlinear structures and to stationary and nonstationary response. It is, however, restricted to stationary white excitation and analytic solutions are available for only a few special cases. In this chapter we present the basic concepts of the Fokker-Planck approach by a detaiied derivation from a random walk model and then present two solutions. The first is a nurrierical step-by-step solution for an idealized non linear structure and the second is an analytic solution for the stationary response.
265
I
I
2 266
NONLINEAR RANDOM VIBRATION
10.1.
DERIVATION OF THE RANDOM WALK MODEL *
10.1.1 a.
10.1.
different times, the random and transient state of the structural response follows the probability relation
Basic Probability Definitions
P(Y;, Yj, tk) = L L P(Yi, Yj, t k ; Ym, Y., tk -
= x]
P[X = x and Y = y]
PMF
L
= Px.r(x, y)
px.y(x, y;) = marginal PMF of
Yi =
b.
Px(l)
= 0.5,
P x (2) = 0.2
Conditional Probability P(x, y) . P(x/y) = P(y) ,
PU;. )'j, (kiO, 0, 0)
tk
= kAt
=
Lm L P(i'm,)'n, Ik - .1°,0, O)P(Yi' Yj' Ikl.lim' Y., Ik
I; 0, 0, 0)
It
P(x/y)P(y) = P(x, y)
Here the sUbscripts denoting the random variables for the PMFs are deleted for clarity.
Finally, the one-step time transition probability for the random response vector from tk - I to tk, given as the second term on the right-hand side of the previous equation, is assumed to be independent of the initial condition and thus
10.1.2.
P(y;, Yi' tklO, 0, 0) =
Chapman-Komogorov-Smoluchowski Equation
Extending the marginal probability 90ncept from one random variable to the response vector with components Y (velocity) and Y (displacement) at two
PX,y{x,y)
~~
t
.. .
L L P(Y.. , y., tk_110, 0, 0) x
P(Yi, Yj, tk!Ym, Y., tk-I) (10.2)
This is known as the Chapman-Komogorov-Smoluchowski (CKS) equation o.n the basis of the aforementioned assumption, which is known as the Markov assumption.
10.1.3.
Random Walk Model
To further simplify the CKS equa~ion, assume that during a single small time interval At, the change in displacement is completely determined by the initially known and determined velocity y according to the simple relation
\ \
Yj = jAy,
The time t is attached to each random response vector indicating different vectors at different times. Next, the joint probability mass function for two different response vectors U;,)'» tk) and Um, Y., tk-I) is replaced by the product of the conditional probabilities P(Y;. Yj, tk/Ym, Y., t k - I ) and PUm, )'n. It .. ,). In addition, all probabilities are subject to the initial conditions (0, 0, 0). Thus the preceding equation becomes
For example, in Figure 10.1,
= 0.3,
lAy,
X
allYl
Px(o)
(10.1)
In Equation (10.1), subscripts i,j, and k are used as indexes such that with the incremel1ts Ay, Ay, and At,
= Px(x)
= joint PMF of X and Y
Px(x) =
tl
m "
Marginal Probability Mass Function (PMF) P[X
267
DERIVATION OF THE RANDOM WALK MODEL
0.3
Ay = yAt
\
0'"
:Y
,1£..
.. x
In other words, given the velocity Ym and displacement Y. at tk-I, the displace ment YJ at the next time incr.ement tk is no longer random. It is de.termined by FIG. 10.1.
An illustration of marginal probabitity mass function P..{x).
·See Toland. Yang. and Hsu (1972).
Yj= Y.
+ Ym At
with certainty or with 100% probability according to. this crucial assumptio.n.
I f
I ~
I
I I I I I I I I I I I
I I Illli
268
NONLINEAR RANDOM VIBRATION
10.1.
Rearranging as Yft = YJ -
and carrying out the summation on
11
Ym lll
in Equation (lO.2) yields the single sum
P(ji> Yi' ttlO, 0, 0) =
L. PUm, YJ - Ym At, tt-IIO, 0, O)PCV., Yi' t~IY.., Yi - Y". Ill, '''-1) .
(to.3)
Since the only term with nonzero probability is the one with y.,. = YJ - \~'.. ~I, the sum with the index n therefore gives only this particular term. Next, it is assumed that during a single small interval Ill, the change in velocity is limited to either an increase of one small increment Ily with yet undetermined probability P or a decrease ofonelly with probability q = (l - pl. Under this assumption, the summation on m given by Equation (lO.3) can be carried out to give the following result:
DERIVATION OFTHE RANDOM WALK MODEL
decrease. Such a diffusion ~echanism in the two-dimensional phase plane is identified with the classical random walk model. As shown in Figure lO.2, the state of the structural response walks to the point A at t" in two paths. Either it was at point B at time t"_1 and went up one step with probability P or it was at point C at t,,_ 1 and dropped down one step with probability q. An alternate view point may be provided for the diffusion mechanism, which is instructive and useful in practical computations. Instead of examination of the probability masses arriving at point A from points Band C in the previous case, the dispersion or diffusion of probability mass at point A to points Band C in the next time step is considered. As shown in Figure lO.3, the probability mass at point A walks to the point B with probability p and to point C with
y
P(ji' Yi' ttlO, 0, 0)
= PUr-I> Y1 -
Yl-!' t,,_ dO, 0, O)p + P(YI+ h Yj -
Yi+ 1 Ilt,t,,_ dO, 0,0)q (10,4)
where p =
(lO.S)
P(y" Yj, ("IYi-1t Yj - Yi-I Ill, ("-I)
• y
and
FIG. 10.2.
q = PUb Yi' ("IYh I, YJ - YI+ 1 Ill, t"-I)
Having carried out the double summations or m and Equation (10.4). Equation (10.2) becomes P(yj, Yi' tt) = P(YI-I, Yi - Yi-I At, ("-I)P
+ P(ji+1> YJ -
Random walk in Ihe phase plane (backward).
y /I
to arrive at the
Yi+t
Ilt,tt-I)q (10.6)
where the initial zero conditions are deleted for clarity. This is. a recurrence equation that governs the diffusion of the probability masses in the phase plane (y- Yplane)in a very simple manner. Its physical meaning can be explained as follows: The probability that the structural response will be (jh YJ) at tt is equal to the sum of two probabilities. One is the probability that its response was (Yi-I' YJ - Yi-I At) at t"_1 multiplied by the probability P of a one-step velocity increase. The other is the probability that its response was (Yi+ I' Yj - Yi + 1 Ill) at 1,,-1 multiplied by the probability q of a one-step velocity
269
B
c
.. y
FIG, 10.3.
Random walk in the phase plane (forward).
270
NONLINEAR RANDOM VIBRATION
probability q. This alternate viewpoint selects a point A and examines the diffusion mechanism in the next forward step timewise. Thus it is referred to as the forward mechanism as opposed to the previous backward mechanism, in which, at the selected point A a backward step is examined. Experience seems to indicate a preference of the backward model in derivation but forward model in computation.
10.2.
APPLICATIONS OF THE RANDOM WALK MODEL
'1+4Y
dy
y
f'+41 H(y, y) d~
+
I
Ai'
=
271
f'+41 F(l;) dl; 1
+ H(Y, y) At = F(t) At E[Ay] = - H(y, Y) At
(10.11)
Furthermore, 10.1.4.
One-Step Transition Probability p
Ay
The random walk model, Equation (10.6), is complete except for the one step transition probability p defined by Equation (10.5). Note that by definition q = 1 - p. The probability p depends on the state of the structural response (Y, y) and the random applied load at tk ~ I' Indeed the structure and loading characteristics have yet to be specified. Consider a nonlinear structure in random vibration governed by the follow ing equation of motion and initial conditions ji
+
H{Ji, y)
F(t);
y(O)
y(O)
E[Ay]2
where the excitation /-'(t) is a stationary Gaussian process with zero mean and white power spectr,um density of intensity So. That is, E[F(t)]
0,
R F (.)
= E[F(t)F(t + .)] =
2rcSo15(.)
= A*p + (- A*)q =
where ,1* is the constant grid size in the response velocity phase plane. Similarly,
y coordinate in the
A*)2q = (,1*)2
(10.\0)
On the other hand, these expected values are related to the structure and loading by the following derivation: From Equations (10.7) and (10.8) it is clear that
: + H(y, y)
= F(t)
I
F(l;) dl;
Jf + F(l;) dl;,1'+~ F(q) till + (L1t)2 '
'+&'
f
=,
f'+4' J,. E[FR)F('1)] dl; d'1 f'+M f1+4'
2rcSo
J,
= 2rcSo
J,
J/
15(l; - 11) dl; dll
dq = 2rcSo At
( 10.12)
From the rombination of Equations (10.9)-(10.12) it follows that:
(to.9)
= (A*)2p + (_
f'+41
+.
(10.8)
A*(2p - 1)
E[(Ay)2]
1 ,
H(y, y) dl;
f/+&1
where RF (.) and 15(.) are the autocorrelation and Dirac delta function, respec tively. For this nonlinear structure with initial conditions and random loading we can determine the one-step transition probability p as follows: First, by definition and the assumption of one-step up or down, the expecta tion of the velocity response increment during At at each point in the phase plane at any instant t is E(Ay)
,+41
I
4
(Ay)2
(l0.7)
0
=
p= ,1*
~[l - H(Y. y) !!]
(10.13)
..j21tSo At
(10.14)
The random walk model for the time dependent, joint probability P(Yj, Yj' tk) Qf the response of nonlinear structures subject to stationary Gaussian loading is now completely specified by Equations (10.6), (10.13), and (10.14). The last two equations imply physically that the magnitude of the velocity response increment ,1* is directly related to the intensity of the random load and the time increment. The probability of increasing velocity response p depends oil the current structural characteristics in addition to the applied random load and the time increment. 10.2.
APPLICATIONS OF THE RANDOM WALK MODEL
For computational efficiency and clarity, the model given by Equation (10.6) can be further simplified.
i I I I I I I I
•
II
I I I I I I I I I I I" I I ..:Ill '
272
10.2. APPLICATIONS OF THE RANDOM WALK MODEL
NONLINEAR RANDOM VIBRATION
Let )\ = ; fly = ; /l*, Yj = j /ly, and rk = kilt. Then YI_I = (i - 1)1l*, YJ - YI-l /It
Select fly
=j
273
sgny
fly - (i - I)/l* Ilt
= /l* Ilt. Then Yj = j /l* Ilt, lk-l
= (j -i + I) /l* Ilt
YJ - Yi-\Ilt
= (k - 1)1lt,
YJ - Yi+11lt = j /l* III - (i
YI+I = (i
+
+
1.0
1)1l*
10
... y
I)/l* Ilt = (j - ; - 1)/l* Ilt -1.0
Thus if the grid size of /l* is selected for the Y coordinate, /l* Ilt for the Y co ordinate in addition to Ilt for the time step, then all the subscripted variables can be replaced by the corresponding subscripts and the grid size as specified in the preceding equations. Furthermore, keeping in mind that i,j, and k associ ate with y, y, and t, respectively, then the respective increments /l*, /l* Ilt, and Ilt can be deleted in the random walk model. After making these simplifications, the model, Equation (10.6), becomes P(; /l*,j /l* Ilt, kilt)
p(i,j, k/i - I,j - i
= P(i,j, k) = P(i - I,j - ;
+ P(; + I,j -
+ 1, k -
1)p(;,j, kl; - I,j - ;
i -1,k-l)q(;,j,kli
+ 1, k -
+ I,j- i -
I)
I,k - I) (10.15)
Let K = C /It and L
EXAMPLE 10.1. motion is
+
~-
+
I,k - I)
I,j - i-I, k - I)
Ilt
= 0.05, =
2nSo
+
The special nonlinear function sgn y is displayed in Figure 10.4. Note that sgn y = 0 at y = O. This particular function HU, y) with constants C and D specifics a linearly damped structure with nonlinear stiffness. From Equations (10.1 J) and (10.14) Il* = J2nSo Ilt
~:: sgnU - ; + 1)
K(i - I) - L"sgn(j - i
=! + K(i +
= I,
F(r)
D sgn y
=! -
.j2nSo/lt = 0.224,
/ltD
= C/lt = 0.05,
L
with the structural property function = 2Cy
I)C /It -
I)
+
+ Lsgn (j - ; -
I)
(10.16)
l)
"I I I I
K
HU, y)
(; -
= DIlt/2/l*, then
then /l*
+ HU, y) =
1, k - I) =
+
For numerical computation of response probabilities let
Consider a nonlinear structure for which the equation of
ji
The sgn y function.
p(i,j,kl; - I,j - ; q(i,j, k/;
As pointed out previously, this gives the backward random walk model witll a recurrence mechanism illustrated in Figure 10.2.
FIG. 10.4.
= 2.1*
C = 1, /l*/lt
D= 1
= 0.0112
0.05 = 2(0.224)
= 0.1 12
For this particular nonlinear structure and random load, the probability that the structure will have the response velocity Yi and displacement Yj at time r k after it starts from rest may be detennined by the following random walk model, incorporating Equations (10.15) and (10.16), P(;,j, k) = P(; - 1,j - ;
+
I, k - 1)[0.5 - 0.05(; _ I)
+ I)) + P(; + I,j + 0.112 sgn U - i-I)]
- 0.112 sgnU - ;
+ 0.05(; +
I)
i-I, k - 1)[0.5
1>2
9.3. (iv) 0·21 < X~ < 4·6 (v) Prob(xio> y)
{R,} = (~O Ro, .!pRJ'
= e-y/l{l + iy + jy2 + lsy3 + Jhy4 }
{R_,}
0'15 for chi-square cf.0·16 for Gaussian.
9.4. 0·987So < m < l-(H3So;
10.1. (i) (0, (ii) (0,
a 2a
-ii, 0, 0, 0, 0, 0, to
{Vn
(0, -i~ 0, 0, 0, ...• 0,0,0,
to
10.2. (O,t - i!<.j2 + O,o,t - i!<.j2
11.3.
l~O.t
(.!jRo,lj!R,), !j!-Rs.
>=
2a
(i) So
(ii) Sf( =
2. 2. 2.) (0, -;",0, - 3x" 0, 3rt" 0,;"
I 2048 cm 2
for
O.:s;; k .:s;; 1023
4~96 cm 2
for
k
S512
10.4. (i) x, = 1for all r x = { 1 for r even
11.4. X(w) "
(iii) {x,} = (1, 1, -I, -I, t, l, -1, -I, ..., -I, -I) (iv) x, sin 2xr/N for all r
where 0 .:s;; r .:s;; (N - 1) in each case.
2n:m
. 2xm
10.7. R, = 2(am2
N
i )
=m
= 0, 1,2, .... (N -
(2~8 + O.OOiS) cm
~sin(wo
-
2L
=
N -
for which Sm =
In,
SN-m
+ XIY1). to
= (R o, ljR 1• lj-R 2,!.pR J , 0, 0, 0,. 0, 0, OJ} = (R o, ljR 9.lj-Rs, !.pR 7 , 0,0,0,0,0,0)
- w)T (Wo - w)T
+
A 2 wou
2'
(Wo
S = a f{Sin(2W OA k 64 . sm 4
a2 . 2
+ W)TJb(W _
+ dJ)T
..
Xk) T
a2
2
. 2
woA, TCOS woAstn
4"S1O
1tk)}2 + {Sin(2WoA + Xk)JJ . (2woA4+ Xk) sm
12.1. x == { I for r even ' - I for r odd 12.3. {}j} where }j = 0 for alii except
y. _ 1
f
sin(wo
(2woA - 1tk)
-fo( XIYO + XlYl + X3Yl»
{R,} {iL,}
2
2
1)
11.1. {R,} =to
I
II.S. (TCOS woAcos
and k
=
S3584
2
Since there is not a complete number of cycles of the cosine function on the record, E[Xl] +- ta2. IXkl would be the same, although alternate terms of tXt} have opposite sign. a2
r
2n:m for r + b2;,)cosNr
St = 0 for all k except k = (a; + b!).
=
t=
-lfor r odd
10.5. 2amcosNr + 2bmsln
=
8 -
or part (11) Y24 =
41
4-
.(ft +
l) _ Yo'" 56
.(ft -
I)
1
1
4 •
=
I I I
= 3072
due to aliasing. Values of Sk for k > 2048 (i.e. for frequencies above the Nyquist frequency) have no practical value.
Xk) .. k 0 dd 1 cot 64 lor
III _.
,
k
1024 and
(iii) St = as above except
0 for keven
32\
3073.:s;; k .:s;; 4095
and
1025.:s;; k .:s;; 3071
.0 for
{
)
...
= ~rt cm 2 s :
2.
(this result is obtained by extending the result of problem 10.1 (iii) - see also problem 10.5) but the higher harmonics distort the calculated spectrum at lower frequencies due to aliasing.
for part (iv)
\0 R 7 , 0, O. 0, 0, 0, 0)
= (a, o. 0, 0, O. 0, ...)
+ i!<.jJ. - 1),0,1 + i!<.j2 + I»
If only the first two harmonics of the square wave were present, the OFT would be .
10.3. X k
2a
11.2. {Vk} = (2' x2' 0, 9x 2 ' 0,
0'974So < m < I·026So
i, 0, 0, 0, 0, 0, i)
\0 RJ • O. O. 0, 0, 0, O)}
... Y 40
2 a . 2 )
4 sm
woA
I I I
I I I I I I II
fW
til ,.. 276
I , I
Answers to problems
13.1. R..{ t)
=
2(
{
i
~:
j
2a
2
0~
2
+
a
N2
for
11'1
~
/N 2 elsewhere
Sx(ro) = ;2 (ro)
t5
+
M}.
(1 _ ~)fl.t(sinrofl.t/2)2
a2
N
~~
N
= (2n - I)
II
roAt/2
2n
References
Reversing the bias would not alter these results.
~
r:
t:
I )( 11'1) a I - N2 I - fl.l
'I:,,'
fli: II II
133 S ( ') = ~~() •• x ro N2f.1 ro
2(N.+ 1)(SinroAt/2)2 ~ N2 roAt/2 k=L.. a:>
a,
,
I,
Nfl.t
ro,
I
k~O
14.1. (i) SI(ro) = S2(ro)
I;
I
S13(ro)
I I I I I II jI
= S3(ro) = .
I. BEAUCHAMP, K. G.
I
S21(ro)
= S34(ro) =
Random Data: Analysis and Measurement Procedures, John Wiley, New York,
1971.
4. BINGHAM, C., GoDFREY, M. D. and TUKEY, J. W.
r
Sdro)
-I-fa:>
R(J(b 2
+
p2»e- i (wP/Vldp
-00
= S%I(ro) = S23(ro) = S12(ro) = _1-
foo
21tV _a:>
R(J{b 2 + (I - pfZ}re-I(p/V)dp
t-
Mil + 2ev + 2kv = 2eu + 2ku
+
(
!eb 2 ()
+ !kb 2 0
So
(m) S" ro) = 2 V'
(
S. ro)
E[v2] = 1tSoro l(1
= !eb 2
q, + tkb 2 ¢
2So
= b2 V
t
+ 4(i) E[8 2 ] = 1£Soro 2(1 + 4(n b 2(2V
4(IV' 1£So{ro 1 4V Yt'"(l
+ 2
where
2
4(d
2k,
rol = M'
_ 2e -
(v) 3·6 per cent approx. (vi) There is no change
'Modem Techniques of Power Spectrum Estimation', IEEE Trans. Audio and
' 5. BLACKMAN, R. B. and TUKEY, J. W. , The Measurement of Power Spectra, Dover, New York, 1959. • 6.BIlIGGS, P. A. N., HAMMOND, P. H., HUGHES, M. T. G. and PLUMB, G. O. 'Correlation Analysis of Process Dynamics using Pseudo-Random Binary Test Perturbations', Proc. I. Mech. E., Vol. 179, 1964-65, Part 3H, 53-67. 7. BRIGHAM, E. O. The Fast Fourier Transform, Prentice-Hall, Englewood Cliffs, New Jersey, 1974. tf'S'I'-l' \ 8. CHURCHILL, R. V. Fourier Series and Boundory Value Problems. McGraw-Hili, New York, 1941. 9. CooLEY, J. W. and TUKEY, J. W. 'An Algorithm for the Machine Calculation of Complex Fourier Series',
Mathematics of Computation, Vol. 19, 1965, 297-301. (Reprinted in A. V.
Oppenheim (ed.) Papers on Digital Signal Processing, MIT Press, Cambridge,
Mass., 1969.) 10. CooLEY. J. W., LEwiS, P. A. W. and WELCH, P. D. The Application of the Fast Fourier Transform Algorithm to the Estimation of Spectra and Cross Spectra. IBM Research Paper, IBM Watson Research Center,
Yorktown Heights, N.Y., 1967. This work is included in a chapter entitled 'The Fast Fourier Transform and its Applications to Time Series Analysis' in Enslein
ro2 2} + (;(1 + 4(2) 2( l ro l
Measurement and Analysis of Random Data, John Wiley, New York, 1966.
3. BENDAT, J. S. and PIERSOL. A. G.
Electroacoustics, Vol. AU-IS, 1967, No. 2, 56-66.
21£ V
...
London, 1973. 2. BENDAT. J. S. and PiERSOL, A. G.
21tV _a:>
=
(ii)
Signal Processing Using Analog and Digital Techniques. George Allen & Unwin,
S~I(ro) = S24(ro) = S%2(ro) = 2~Vf~a:> R(l- p)e-1(W"W1dp
Sdro)
SI4(ro)
fIX> R(p)e-I(IIJP/Y1dp
S4(ro} = - I
M'
et al. (26). 2 ro 2
14.4. (iv) I 3 5 6 5 3 1 . 4!' 4!' 4!' 4!' 4!' 4! and respectively
j
~( _ 21tk)
f.I
The difference between this answer and that to problem 13.1 is that the
latter is for genuinely random noise, whereas the peculiar shape of R..kr) (and therefore of Sx(ro» in 13.3 arises from the repetitive properties of pseudo random sequences.
"
';'
+
=
2 kb -
21 '
2(2 ro 2 =
cb 2
2I
l
II. CooLEY, J. W., LEwIs, P. A. W. and WELCH, P. D. 'Application of the Fast Fourier Transform to Computation of Fourier Integrals, Fourier Series and Convolution Integrals', IEEE Trans. Audj(j and Electro acoustics, Vol. AU-IS, 1967, No. 2, 79-84. 12. CooLEY, 1. W., Ll!Wls, P. A. W. and WELCH, P. D. 'Historical Notes on the Fast Fourier Transform', IEEE Trans. Audio and Electro acoiistics, Vol. AU-IS, 1967, No.2, 76-9. 13. CooLEY, J. W., Ll!Wls, P. A. W. and WELCH, P. D. 'The Fast Fourier Transform and its Applications', IEEE Transactions on Education, Vol. 12, 1969, No. I, 27-34.
14. CooLEY, J. W., LEwIS, P. A. W. and WELCH, P. D. 'The Finite Fourier Transfonn', IEEE Trans. Audio and Electroacoustics, Vol. AU·17, 1969, No.2, 77-85. 15. CooLEY, J. W., LEWIS, P. A. W. and WELCH, P. D. 'The Fast Fourier Transfonn Algorithm: Programming Considerations in the Calculation of Sine, Cosine and Laplace Transfonns', J. Sound Vib., Vol. 12, 1970, 315-37. 16. CRANDALL, S. H. (ed.) Random Vibration, M.I.T. Press, Cambridge, Mass. and John Wiley, New York, 1958. 17. CRANDALL, S. H. (ed.) Random Vibration - Vol. 2, M.I.T. Press, Cambridge, Mass., 1963. 18. CRANDALL, S. H. and MARK, W. D. Random Vibration in Mechanical Systems, Academic Press, New York, 1963. 19. CRANDALL, S. H., CHANDIRAMANI, K. L. and CooK, R. G. 'Some First·Passage Problems in Random Vibration', J. Appl. Mech., Trans. ASME., Vol. 33, 1966, 532-8. 20. CRANDALL, S. H. 'First Crossing Probabilities of the Linear Oscillator'. J. Sound Vib., Vol. 12, 1970, No.3, 285-99. . 21. CRANDALL, S. H. 'Distribution of Maxima in the Response of an Oscillator to Random Excitation', J. Acoust. Soc. Am., Vol. 47, 1970, No.3 (Part 2). 838-45. 22. DAVENPORT, W. B. Jr. and ROOT, W. L. An Introchlction to the Theory of Random Signals and Noise, McGraw-Hili, New York, 1958. 23. D(mDS, C. J. 'The Laboratory Simulation of Vehicle Service Stress', J. Engng. Ind.• Trans. ASME, Vol. 96,1974, No.3, 391-8. . 24. DODDS, C. J. and ROBSON, 1. D. 'The Description of Road Surface Roughness', J. Sound Vib., Vol. 31, 1973, No. 22, 175-83. 25. DURRANI, T. S. and NIGHTINGALE, J. M. 'Data Windows for Digital Spectral Analysis', Prcc. I/::£" Vol. 119, 1972, No.3, 343-52. 26. ENSLElN, K., RALSTON, A. and WILF, H. S. (eds) Statistical Methods for Digital Computers (Vol. 3 of Mathematical Methods for Digital Computers), John Wiley, New York, 1975. 27. FELLER, W. Probability Theory and its Applications, John Wiley, New York, 1950. 28. FISHER, R. A. and YATES, F. Statistical Tables for Biological. Agricultural and Medical Research, Sixth edition, Oliver and Boyd, ~inburgh, 1963. ,9. GOLD, B. and RADER, C. M. Digital Processing of Signals, McGraw·HilI, New York, 1969. 30. GoLOMB, S. W.
Shift Register Sequences, Holden-Day, San Francisco, 1967.
31. GREEN, B. F., SMITH, J. E. J(. and KLEM, L. 'Empirical Tests of an Additive Random Number Generator', J. Assoc. Computer Machinery, Vol. 6, 1959, No.4, 527-37. 32. HALFMAN, R. L. Dynamics (2 Vols.), Addison-Wesley, Reading, Mass., 1962. 33. HARTLEY, M. G. 'Development, Design and Test Procedures for Random Generators using Chaincodes', Proc. lEE., Vol. 116, 1969, 22-6.
~
34. HARTLEY, M. G. 'Evaluation of Perfonnance of Random Generators Employing Chaincodes', Proc. lEE., Vol. 116, 1969, 27-34. 35. HEAl'H. F. G. and GRIBBLE, M. W. 'Chaincodes and their Electronic Applications', Proc. lEE.• Vol. 108C, 1961, 50-7. 36. HOE'SCHELE, D. F. Jr. Analog-to-Digital/Digital.(o-Analog Conversion Techniques, John Wiley, New York,1968. 37. HUFFMAN, P. A. 'The Synthesis of Linear Sequential Coding Networks', in C. Cherry (ed.), In/ormation Theory, Butterworth, 1956. 38. JAME'S, H. M., NICHOLS, N. B. and PHILLIPS, R. S. Theory ofServomechanisms, MIT Radiation Laboratory Series, Vol. 25, McGraw Hill, New York, 1947. 39. JENKINS, G. M. and WATfS, D. G. Spectral Analysis and its Applications, Holden-Day, San Francisco, 1968. 40. JOLLEY,.L. B. W. Summation of Series, Second edition, Dover, New York, 1961. 41. KENDALL, M. G. and STIJART, A. The Advanced Theory of Statistics, Charles Griffin, London, (a) Vol. I (2nd edition) 1963, (b) Vol. 2 1961, (c) Vol. 3 1966. 42. KENDALL, M. G.
Time·Series, Charles Griffin, London, 1973.
4.1 KLINE, MORRIS Mathematical Thought from Ancient (0 Modern Times, Oxford University Press, New York, 1972. 44. KOOPMANS, L. H. The Spectral Analysis of Time Series, Academic Press, New York, 1974. 45. KREYSZI,G, E. Advanced Engineering Mathematics, Second edition, John Wiley, New York, 196-7. 46. LIGHTHILL, M. J. Intt~uction to Fourier Analysis and Generalised Functions, Cambridge University Press, 1962. 47. MACLAREN, M. D. and MARSAGLlA, G. 'Unifonn Random Number Generators', J. Assoc. Computer Machinery, Vol. 12, 1965,83-9. 48. MEYER, P. L. Introchlctory Probability and Statistical Applications, Addison-Wesley, Reading, Mass., 1965. 49. MIDDLETON, D. An Introchlctjon to Statistical Communication Theory, McGraw-Hill, New York, 1960. 50. MINER, M. A. 'Cumulative Damage in Fatigue', J. Appl. Mech., Trans. ASME, Vol. 12, 1945, A 159-64. .51. MusTER" D. and CRENWELGE, O. E. Jr. 'Simulation of Complex Excitation of Structures in Random Vibration by One Point Excitation', Proc. Soc. of Environmental Engineers Symposium - Vibra tio~ in Environmental Engineering, London, 1973, MI":M13. 52. OTNE'S, R. K .•and ENOCHSON, L.
Digital Time Series Analysis, John Wiley, New York, 1972.
53. PALMGREN, it 'Die Lebensdauer von Kugellagem', Ver. Deut. Ingr., Vol. 68, 1924,339-41.
~,;~<~
~
I
I fI II
I
I
(I
I
I
I
I
I
I
I I
I
280
List ofreferences
54. PARZAN,E. Modem Probability Theory and its Applications,Iohn Wiley, New York, 1960. 55. PIERSOL, A. G.
'Power Spectra Measurements for Spacecraft Vibration Data', J. Spacecraft and
Rockets, Vol. 4,1967,1613.
56. PoWELL, ALAN I
'On the Fatigue Failure of Structures due to Vibrations Excited by Random
. Pressure Fields', J. Acoust. Soc. Am., Vol. 30, 1958, No. 12.. 1130-5.
57. PRIcE, W. G. and BISHOP, R. E. D. Probabilistic Theory of Ship Dynamics, Chapman & Hall, lAndon, 1974. 58. RICE, S. O.
'Mathematical Analysis of Random Noise', Bell System Tec". J., Vol. 23, 1944.
282-332, and Vol. 24, 1945,46-156, both of which are reprinted in N. Wax (ed.),
Selected Papers on Noise and Stochastic Processes, Dover, New York, 1954. 59. ROBSON, J. D. An Introduction to Random Vibration, Edinburgh University Press, 1963. 60. ROBSON,.1. D. and ROBERTS, 1. W.
'A Theoretical Basis for the Practical Simulation of Random Motions', J. Mech.
£ngng. Sci., Vol. 7, 1965, No.3, 246-5L 61. SHANNON, C. E. 'Communication in the PresenCe of Noise', Proc. IRE.• Vol. 37, 1949, 10-21.
.62. SmNOZUKA, M. and IAN, C. M.
'Digital Simulation of Random Processes and its Applications', J. Sound. Vih., Vol. 25,1972,111-28. ·63. SmNOZUKA. M.
'Simulation of Multivariate and Multidimensional Random ProcesSes', J. Acoust.
Soc. Am., Vol. 49, 1971,357-67. • 64. SINGLETON, R. C.
'A Method for Computing the Fast Fourier Transfonn with Auxiliary Memory
and Limited High-Speed Storage" IEEE Trans. Audio and Electroacoustics, Vol.
AU-IS, June, 1967, No. 2, 91-8.
• 65. SLOANE, E. A.
'Comparison of Linearly and Quadratically Modified Spectral Estimates of
Gaussian Signals', IEEE Trans. Audio and Electroacoustics, Vol. AU-17, 1969,
No.2, 133-7.
66. VIRcms, V. J. and ROBSON, 1. D.
'The Response of an Accelerating Vehicle to Random Road Undulation', J.
Sound vih., Vol. 18, 1971,423-7.
67. WATIS, D. G. 'A General Theory of Amplitude Quantization with Application to Correlation
Detennination', Proc. lEE., Vot. I09C, 1962, 209-18.
68. WELCH, P. D. The Use of Fast Fourier Transfonn for the Estimation of Power Spectra: A Method Based on Time Averaging Over Short, Modified Periodograms', IEEE Trans. Audio and Electroacoustics, Vol. AU-15, Iune 1967, No. 2, 10-3. 69. WIIINBR, N. .
The Fourier Integral and Certain ofIts Applications, CAmbridge University Press,
1933.
70. WIENER, N.
Extrapolation, Interpolation and Smoothiilg ofStationary Time Series,Iohn Wiley,
New York, 1949.
Index
Accuracy of
measurements, 95
spectral analysis, example calculation, 102
spectral estimates, 138
Addition of zeros to sample record, 139
Aliased spectral window, 256
Aliasing, 118
distortion, 120
Alias spectra, 120
Analogue
digital conversion, 149
error analysis, 149
spectrum analysis, 95
spectrum analyzer, 95
Aperiodic functions, Fourier transform of a
train of, 129
A utocol'relation, 25
function
definition of, 25
for a pseudo random binary signal. 173
for a random binary process, 168
for white noise, 46
properties of, 26, 27
spatial, 198
. input-output relation for a linear system, 69
Average frequency of crossings, 86
Averages, second order, 14
Averaging
.spectral estimates, 138 .
time of a spectrum analyzer, 95
Average value of a function of random
variables, 14
Bandwidth
half-power, 191
mean square, 191
of analogue spectrum aualyzer, 96
ofspectral window, 108
Bias error of spectral measurement, 103
Binary
random process, 168
sequence, full length, 262
Binomial probability distribution. 175
Broad band random process, 44
Calculation
ofaverages, 7 .
of spectral estimates from discrete data, 120
Central limit theorem, 80
Chain code, 183
Chi-square
random variable. 109
distribution of, 110, 246
probability density function. 246
probability distribution.
application to spectral analysis, 138
dependence on degrees-or-freedom. ItO
tabulated values of distribution, 224
Circular correlation function, 122
interpretation in terms of linear correlation
function estimate, t 29
relationship with linear correlation function,
126,129
Clock frequency of random binary process,
170 .
Coherence functions, 207
digital calculation of, 210
mUltiple, 207, 209
ordinary. 209
partial. 267
Complex frequency
response function, 56
measurement of, 206
Computer program for the fast Fourier
transform, 220
Conditional probability, 17
Confidence limits
of spectral measurements, 108
sample calculation. 112
Continuous time series, 113
Convolution, 64, 241
integral. 64
Correction for slow trend in spectral analysis,
147 .
Correlated random noise, synthesis of, 185
Correlation, 21·
coefficient, 23
function (see also autocorrelation, cross
correlation and circular correlation)
calculation from data extended by
additional zeros, 139
circular, 122
consistent estimate for, 127
error due to "wrap around", 125
1iil
A.2.
S.
APPENDIX
Using F(w) from Equation (A.S) in Equation (A.6)
ret) - = 2"'f"''' F(w)i"" dw
A
n -"'''
= -
If"''' [N-IL fer At)e-""'4f 'J At .
e''''' dw
21£ -."'''
FAST FOURIER TRANSFORM (FFT) IN RANDOM 'VIBRATION*
t
f [N-I L 2.. "
0
J
fer fl.t)e -i,.r4' tl.t e"O,41 lim
,=0
21n; fer' At) tl.t(2WN)
Exact Fourier transform pair 'F(W)
=
f(t) =
f:",
= f(r'fl.l)
for r'
f'" x
I
-2'
(A.I)
!(r,tl.t)
f(t) = 0
for -
WN ;;;. W ;;;. WN;
for t < 0 and
L
F(kAw)eikAWfAW;
k=O,I, ... ,N-I;r=O,l, ... ,N-1
A.2.
Use of FFT Computer Subroutine
(A.2)
F(w)e''''' dw
1.
Fourier Transform Pair
WN
= x/Ill =
f(t)e-i'.f
F(w) =
Nuquist frequency
> T
t
(A.3) (A.4)
de;
I
f(l)
2n
F
•fW :
Fe,..): ..3..sin~· 2
transient in t
N-l
L
f(r
Ilt)e- i",r4.
Ilt;
T
= N Ilt
(AS)
I 2x
f(t) = ·See Yang and Shinozuka (1969).
282
fWN
-0.5
F(w)e'w'dw
= transient in (IJ
til
r=O
4. From Equations (A.2) and (A.3),
«())eiM Iho
For example, if! Figure A.l
3. Discrete approximation F(w)
F(w) =
(A.8)
k=O
2. Limitations on F(w) and f(t) in typical structural dynamics problems
°
- I
N-\
f(t)e- iCD• dt
-00
F(w) ==
= 0, 1,2, ... , N
This shows that the continuous approximation l(t) f(t) is the true inversion at r' fl.t discrete points. 7. Discrete approximate inversion let) based on Equation (A.7).
Basic Concepts
1.
(A.7)
r=O
6. Since F(w) of Equation (AS) is periodic in w with period of 2WN and = r' At for r' = 0, 1,2..... N I, then
I ]'(r' Ilt) = --. 2n
A.1.
283
USE OF FFT COMPUTER SUBROUTINE
'\/
0,5
(A.6)
-«IN
FIG. A.1.
',/
a-w
I I
I I I I I I I I I I I I
...~
"'1
: .......Y'"
,,1
';1
:t~ 11f,j ~. I
't.;1,/ :-
I I '
284 2.
FAST FOURIER TRANSFORM (FFT) IN RANDOM VIBRATION
A.2.
USE OF FFT COMPUTER SUB.ROUTINE
Discrete Approximation F(w) and let)
F(w)
Ilt
L'"
285
f(ll
fer llt)e-i""AI
(A.9)
r= -ro
"
-0,5
:1
let) = 11
[I
~
I I I I I I D'
F(k Ilw)e-
Ilw
0.5
(A.lO)
f{t)
-0.5 00 p =
(A.H)
2n:/1lt
L..
I
-I_
Tp
I
r;, =
(A.l2)
2n:/Aw
'. Wh
.,..+
I' J \.
"'p
(A.l3)
4. Relation between F(w) and F(w). In an analogous manner, F(w) is periodic in 00 with period wp 21t/At. The principal part lies in the interval ( - w,,/2, + (/),,/2). 5. Distortion of F(w) (Aliasing). In case the highest frequency Wh of the true Fouri~r transform F(w) is less than half of the common period wp as is the case shown, then there is no distortion due to the periodic superposition or folding elTect. If, however, Wh is greater than w p /2, then distortion is introduced as shown in Figure A.3. Thus to avoid distortion due to folding use the criterion 00 p =
,
F{",)
, I.
21t Aw = Tp
< 00 p/2;
I
F(",)
For the same example, in Figure A.2 3. Relation between f(t) and let). The time function ..c(t) is usually limited to a transient aperiodic function such as the one in the example. The approximate time function 1(1) is always periodic with period Tp = 21t/Aw, in addition to being approximately equal to f(t) in one period. Thus the first step in using the FIT is to select r;" and construct the periodic function let) in the interval (0, Tp). Once Tp is selected, then the frequency increment can be deter mined.
Wh
Tp = 211:111",
""'I
Similarly Equation (A.IO) represents an infinite sum of harmonic functions of time I with periods 2n:/k Aw. The common and the largest pefiod is 21t/llw, Let the common period be r;, then
j
II
L
1U ",'
Equation (A.9) represents an infinite sum of harmonic functions of frequency 00 with real' amplitudes fer At) Ilt and periods 2n:/r At. The common periOd, which is also the largest, is 2n:/At. Thus, F(w) is periodic in 00 with period 2n:/At. Let this common period be w p , then
~j
1.
""
2n: k=-",
" m
I
.]1
I
2n:/llt
< n:/Ilt = WN = Nuquist frequency
(A.l4)
= 211:111/
-I
FIG. A.2.
6. Basic Discrete Formulas. Let the period T, in time t and period wp in frequency co be divided into N equal intervals of Ilt and Ilw, respectively, then
r" =
N Ilt;
wp
= N Aw
(A.IS)
Using Equation (A.15) and the dummy index k for discrete frequencies such that to = k L\w in Equation (A.9) gives F(w) == Fk
= III
N-I
L
,=0
/'e-{2nk'IN
for
k
= 0, I, 2, .... N
- 1 (A. 16)
1:
286
A.2.
FAST FOURIER TRANSFORM (FFT) IN RANDOM VIBRATION
USE OF FFT COMPUTER SUBROUTINE
287
5. Output
F(,.)
-I
1
N-l
N
r=9
Fk = -
I
fre-iZttkr/N;
6. Corrected Ft = TpFk ; Tp 7. Output Fk corresponds to W
Fl,.}
= k Aw,
k = 0, I, ... ,N
= NAt
Aw = 21tITp;
k = 0, I, 2, ...• N - I
8. For inversion, change - ito + i. divide the output l~ by 11M to get the approximate inversion 1. for r = 0, 1, 2•... , N - 1. -...."
. Distortion
:.7£
•
w
FIG.A.3.
Similarly, since Aw121t _ _ 1 f(t) == J.. = N
= 1/7;. =
N-l
liN At, Equation (A.lO) becomes
1
I - F.. e+ibkr/N
&=0
At
for
k
= 0, 1, 2, ... , N -
1 (A.l7)
7. Computer Program Application for Fourier Transform and Invasion. To calculate the approximate Fourier transform F.. by Equation (A.l6); 1. 2. 3. 4.
i ~
I m
F(,.)
'''''"'"'"""
I
Select 7;,. Select N, which must be equal to integer power of 2, that is, N == 2'. At = TpIN, check that Wh < WN = wI2 = 1t/At. Read input f(At) == J..at discrete times covering a complete period (0, Tp).
I I I I I I I I I
18
I I 11 I
I
I
I
I I
I I I I I
B.3. SIMPLE BUBBLE SORT
9. Fori
APPENDIX
289
1 to N:
B
T
tl
1) N
(i
-
K
L Ok sin(27tll t
Xi -
i
+
k= 1
MONTE CARLO SIMUlATION*
10. Store the results: B.2.
t/> Xi;
i
Generating Random Numbers on a Computer
1. Basic Rule.
The equation Y
B.1. Synthesis of a Sample Function by a Monte Carlo Simulation
3. 4. 5. 6. 7. 8.
II - Imin; IK+I - Imax
Select 12 to !K. from random uniform distribution over (fmiq. lmax)·
Sort the fi, k= 1 to K + 1. into ascending order.
];. filk + 1, k = 1 to K; (find midfrequency).
Calculate »I: - W(h), k = 1 to K. A. - .j2»1:Ui,+I - A), k = 1 to K. Store the results: 1., Wb A ••
.J
c
c c
10 W{f)
0+ (b -
o)r
where r is a random number from a uniform distribution in (0, l) and generates a random number y from a uniform distribution in (a, b)~ 2. Example
1. (See Figure A.4.) Select
= 1 to N
real phi (50)
write (6,*) 'type in a large odd integer'
read (5:) ix .
once ix is initialized, it shall be
left alone. It is the property
of the random number generator.
pi=3.1415027
twopi=2.*pi
do 10 j'=1, 50
phi(i)==ran(ix) * twopi
write (6:) i. phi (i)
continue
stop
end
Atl2
B.3. Simple Bubble sort
Sort the array xCi), i = I to N into ascending order.
,..f
VI [min
I; 1;+1
fro••
FIG.A.4, .*From Zimmermam and Cheltri.
288
subroutine sort (x,
real x(50)
do 2 j 1. n - 1
ji == i +1
do 3 j == ji, n
f(x(j).It.x(i» then
temp '= x(i)
290
MONTE CARLO SIMULATION
xCi)
x(j)
x(j) = temp
3
2
else end if continue continue return end
The first time through the outer loop, the smallest number rises like a bubble to first position in the array. The second time through, the smallest of the remaining numbers bubbles up to the second position, and so on.
APPENDIX
C
~ "-::).
I
:~.
I .,
REFERENCES
l.~ t
[iJ
Abramowitz, M., and I. A. Stegun (Eds), National Bureau of Standards. /I(wdblltJk q( Mwlll'/lIII/ica( Futlcliani, AMS 55. 1964, p. 358. Bendat, J. S., and A. G. Piersol. Random Data: Allal}'sl, alld Measuremellt PI'OL'l!dures. Wiley Interscience, New York, 1971. Borgman. L E., "Spectral Analysis of Ocean Wave forces on Piling." J. lVtlfl!f\l'Uys /larbo" DI." ASCE93, WW2,I29 ,156(1967). Caughey. T. K., "Derivation and Application ofthe Fokker· Planck Equation to Discr.:te Nonlinear Dynamic Systems Subjected to White Random Excitation." J, Acmes/. Soc, Alii .• 35(11).1683 1692 (1963). Caughey. T. K., and H. 1. Stumpf, "Transient Response of a Dynamic System under Random Excitation," J. Appl. Me"'... 28(41. 563 {l961l. Chopra, A. K" "Hydrodynamic Pr~ssures ill Dm"s During Eal'th4uakc": J. /;II!I, Mccll, Oil". ASCE, 93, EM6, 205-224 (1967). Corotis, R. B~ and E. H. Vanmarke, "Time-Dependent Spectral Contem of System Response." J. Eng. Mech. Div. Proc. ASCE. tOt, 623·637 (1975). Crandall. S. H., and W. D. Mark, RWldolll Vibration in M",'Iulllicul Sr.• It'III.'_ 1963;
Acad~lI1i,.
New York.
Crandall.S. H .• and L. E. Wittig, "Chladni's Patterns for Random Vibration of Plate: in G. Her rmann and N. Perrone, eds., Dynamic ResponseafSrruc/ures, Pergamon. New York. 1971. James, H. M.• N. B. Nichols, and R. S. Phillips, "Theory of Servomechanisms," MIT Radiation LabOratory Series 333-369, McGraw Hill, New York, 1947. Kanai, K., "Semi-empirical Formula for the Seismic Characteristics of the Ground." VIIii'. 1ilkyo Bilil. Earthq. Res. Insl., 35, 309-325 (1957). Miner. M. A., "Cumulative Damage in Fatigue," J. Appl. Mech" 12. AI59-AI640945). Newark. N. M., and E. Rosenblueth, Fundamenlals of earil.qltuke englll«·rinil. Prentice Hall. Englewood Cliffs. NJ. 1971. Palmgren. A.• "Die lebensdauer von Kugellagern." H'r. delli. 11111...,68, 339 341 (l9!41. Powell, A., "On the Fatigue Failure of Structures Due to Vibrations Excited by Random Process fields," J. AcOU$/. Soc. Am., 30. 1130 1135 (1958). Priestley, M. B~ "Power Spectral Analysis of Non-Stationary Random Processes." J, S",md ViIlJ'll lion, 6(1), 86-97 (1967). Rice. S. 0., "Mathematical Analysis of Random Noise," Bell S}",wn Tet·h. J" 2.1.282 332 (1944):
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REFERENCES
24, 46,156 (19451; also in N, Wax, ed., Selected Papers on Noise alld Slochastic Processes,
Dover, New York, 1954.
Shinozuka, M., "Random Processes with Evolutionary Power," J. Eng. Mech, Div., ASCE, 96,
EM4. 543-545 (1970).
Shinozuki, M~ "Simulation of Multivariate and Multidimensional Random Processcs,~ J, ACOUSl. SO(', Alii" 49(1), 357 ·,367 (19711, Tajimi, H~ "A Statistical Method of Determining the Maximum Response of a Building Structure
During an Earthquake." Proc. 2nd World Conf. Earrhq, Eng., Tokyo and Kyoto, fl, 781-198
(1960),
Toland, R. H., C. Y. Yang. and C. Hsu, "Nonstationary Random Vibration of Nonlinear Structures," 1111. J. No/llil/ear Me('h .. 7. 395-409 (1972). Yang. c.Y., and V, Charito, "Random Hydrodynamic FOrce on Dams from Earthquakes,," J. Eng. Merit. Div.Proc. ASCE, 107. EMI. 717-722 (19811. Yang. 1. N.• and M. Shinozuka. "Numerical Fourier Transform in Random Vibration: J. Eng. Meclt. Dl!,. Pr"c. ASCE, 95, EM3. 731-746 (1969).
INDEX
Acceleration. ground. 2, 147
Aliasing, 285
Autocorrelation. 25.26, 119, 130
d,mlpcd ~'flI>nse, 126
ensemhle, 47
resJIOnsc. 120
temporal. 47,48,56
Beams: IIcxural.
1~2
I~O, t61 Bendat and Piersol. 25 I
Blacklllall·Tukcy method. 53
shear. 144.
n"J'~"'i1n. Iluildill~,
'''''"1. gU
ilK)
tall. 144
Chupman-KOlnogorov-Smohlchowski. 266
Caughey. T. K. and H. I. Stumpf. 232
Chamclerislic ,lines. 163
ChOprd, A. 1( .• 190
Coordinates. normal. 115, 124. 129
Correlation. 14
Correlation cocflident. 15
Covariance, 14
Cmndall. S. H.and L E. Wittig. 154
Crosscorrclation. 25. 27, 119. 120. 130
Cumulalivc frC
Dam-reservoirsystclIIs. 171. 187.191,223. 224.253
Dc,i~l1. slnKlun.'. 211
Diulll'.,llI:
scalier. 16. 27
Ve.nn.2()
Distribution:
Gaussian, 22
Poisson, 22
Rayleigh. 22
Duhamel's integral, 9.1, 116
Eal1hquake:
ground aceeler-tl;on. 67
motioll. I
Envelope. 210. 212
Ex.citatiuns:
models or I"dndonl, 60
random. flO, I I')
Expeetatinll. D, 14
Failure:
fatigue. 224
yield. 213
Fast Fourier transform. 59. 83. 282
Filter. narrow.band. 107
Fokker-Planck. 265, 276
Fourier:
integral. 29
series. 29
Function:
311iticial sample. 64
c.·omple" frequency resflIIOsc. 59. 87. I 10.
\11
133, 137
Dirac delta. 27. 94
harmonic. 31. 39
impulse rcsflIlnsc. 93, II h. 134
transfer. 10K
[mtelltial. 1119
Gaussi.m process. 201. 202
293
294
INDEX
Hydrodynamic response, 185, 186, 195
Imerseetion, 21
loint probability density, 7
loint probability mass function, 8
,
I
.\ \ \
,
\
{ \
Marginal probability density, 8
Matrix, 123
damping, 129, 137
mass, 129, 137
stiffness, 129, 137
Mean square, 14, 24
ensemble, 25·
tempordl, 30, 32
Models:
continuous, 65, 69, 76
discrete, 62, 68, 74
multivariable stationary, 71
nonstationary. 66
offshore structural. 71
of random excitations. 60
stationary. 60
structural. 104
two-dimensional wave. 75
vehicle. 61, 100. 136 ..
Newmark. N. M. and E. Rosenblueth, 172
Nonlinear ralldom vibralions. 265
Nonstationary excitation. 241, 253
Nonslationary response. 232, 238. 250
Normal mode. 116
Offshore structuml model. 71
Orthogonalily conditions. 124, 129. 145
Palmgren and Miner, 224
Parseva!' s formula:
for nonperiodic function, 34
for periodic function. 30
Phase angle, 27
Plates:
rectangular, 158
thin. 158
Poisson distribution, 215
Powell, A., 201, 209
Power spectral density, 36, 39, 104
cross, 132
damped response, 127
directional, 76
nonstationary ensemble, 70
one-sided temp
targel, HI
INDEX temporal, 53, 56. 110
two-sided temp
Priestley, 241
Probability:
conditional, 10
density function, 4
frequency definition of, 3
mass function, 8, 12
Processes:
ergodic, 44, 46
ergodic stationary, 45
Systems:
cnntinu(lYs, t44
mliltidegrcc of freedom, 114. 123, 129
sin~~~ degree. of freedom, 94. III, 112, 117.
Random excilations:
eoncelltrated. 14H
distributed. 147
Random processes. 2
narrow-band. 39
nonstationary. 2
periodic. 27
stationary. 22. 26
Variance, 14
Vibration:
detenninistic, 114, 129, 133
nonnal mode, 115
random, I·
wide-hand. ]()
Randolll vari~llJlcs:
continuous, 4
discfCte. 8
independent. II, 16
Random vibration, I
stationary. 130. 136. 146
Random walk, 267.271
Rayleigh probability density, 213
Reliability. 214. 216. 218
Response:
complex frequency. I23
steady-state. 118
Rice, S. 0 .. 201
Rigid body anology. 18
Shinozuka. M., 243
Simulation:
computer. H3
Monte Carlo. 287
Solution:
frequency domain, 86, 96
modal, 145
time domain. 92. 96
Spectral moment. 209
Spectrum:
band-limited. 40
white, 40
Standard deviation, 14
Statislics:
ensemble, 45
temporal. 44, 45, 110
Structures:
design of. 20 I
single degree of freedolll. H5
two degree offreedom, 114, 115, 117. 123,
136
uncoupled, I
i4
Up-crossing, 205, 206, 207, 208, 209
Union, 21
Wave:
dilatational, I H'l
direction, 75
equation, 163, 188
force, 7~
~umber, 75
random ocean, 71
speed, 164
surface, 72
White noise, 39
Yang, 1. N. and M. Shinozuka, 282
Yield failure, 213
295
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