III

1-_

-

... ,... l

~

."

...

"

"

\.

, .'

Ail/V \y

-.'

...

~~ .

:1 '

I

..

....

RANDOM VIBRATION OF STRUCTURES

"" ....

:1I I"

r/0-....

",x 2)dxl dx 1(1.5)

'1

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 114? 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). ThusP 0< (I 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 O 2 are not independent ']

III

I~ ,j

i~ .d.~

l~

~ I

!,

ICHAPTER

3.1.

ERGODIC PROCESS

45

11'J

!I5~ 'i

11

3

.. (t~3)

xltl)!3)

I

I

I1

\,

I,

I I I I I I I I I r1, ,l.\llI~

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.

ERGODIC p,ROCESS ,

called ergodic. Thus, for ergodic stationary ,processes, the temporal mean value (X(t and the temporal correlation Cl>(t) defined by(X(t == lim TT ... ",

:1 i; i,

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

I

fT /2 x(t) dt -T12I

(3.1 )

andcfI(t)

= (X(t)X(t + tn =

lim T-oo

fTI2-T12

T

x(t)x(t

+ t)dt

(3.2)

respectively, have the property that from a sample function of infinite record length in time T,

(X(tcfI(r).j'

=

E[X(t

(3.3) .(3.4)

= R(t)

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.

\ l:m

44

46

ERGODIC PROCESSES AND TEMPORAL STATISTICSf.uscl'loln ilKIS

3.2.

TEMPORAL AUTOCORRELATION fIl(t)

47

The ensemble meanE[X(t)) = E[A sin(wt

+ 8)](3.7)

== A Jo sin(wt + 8)P(8) dO = 0The ensemble mean square

[211

E[X 2(t)]The temporal meanTime axis

= A2

Jo sin 2(wt + 0)P(8)dO =

f2. .

A2(3.8)

I I I

I (X (I)) = Iim1'~", T

IT A sin(u)t + 0) dt = 00

(3.9)

t

The temporal mean squaGefIG. 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.

(X 2(rI)

1"'00

I ~im 'T-

i'l' '0

A2 sin 2 (wl

+ O)dt =

A2

(3.10)

EXAMPLE

3.1 Consider the simple but fundamentally important stationary process modeled by the sine waveX(t)

Thus the process satisfies the necessary conditions for being stationary and' ergodic. \

= A sin(wt + 0),

oo' and using the same uniform probability distributions of the independent phase angles as before, we obtainR(X, Y,

= L Lwmt

N

M

n= 1 m= I

, 4>.) flw fl4> cos(kmx cos .j4S(wm

4>. + k.y sin 4>.(4.61 )

r).= f~",

f"

[IX(W,

4

dw d4>]2eiikX oost/>+krsint/>-wt)

(4.66)

+ 8",.)

Now for X

=Y= r

0, Equation (4.64) gives

I

f.L

4.3.2.

Continuous Model

R(O, 0, 0) = E['1(x, y, t)*'1(x, y, t)] =

J~",

f"

Sew,

4

dw d4>

(4.67)

i;'';}I

jl I

I:. P.IIP:I'

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

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 obtaina(w, 1jJ) dw d4> = .jS(w, 4 tlw de/>

(4.61:\)

11~;

" .,

,

78

MODELS OF RANDOM EXCITATIONS

4.3.

RANDOM OCEAN WAVES (MULTIVARIABlE STATIONARY MODEL)

79

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 asI'/(x. y. t)

The ensemble spectral transfer relations areS.(w, z)

I(4.75) (4.76)

=

f:"" r~

.jS(w.

T for t < 0(5.20)

~ ({1

e- lIl ) -: [1 - e-/1(I-n]}

o

I 1\ = 21[/ [ -.- . ,( - l{:l)

21[/ e-/1,] + -" = -(1 - p

fJ

e- P1 )

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 ~1

I

I I

92

STRUCTURES WITH SINGLE DEGREE OF FREEDOM (SDOF)

other hand, it leads to complicated contour integrations for the transient problem.

I

5.1.

DETERMINISTIC TRANSFER RELATIONS

93

Dirac unit impulse b(t) such thatb(t)-+ 00

~) ;' :':'i.'".;;~~;K'i1rjy,~7Y.jb~>~1~\~~~.(:/~(;;i

II96STRUCTURES WITH SINGLE DEGREE Of FREEDOM {SDOF} 5.2. RANDOM EXCITATION AND RESPONSE 97

I.t

!I ~

I JI

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. Thusv(t) =

He:.:.;1 ':::

= 2n

J

-'"

H ((}) le

iw,

dw

(5.41 )

foo

which is recognized as the Fourier inverse transform. Consequently, we also haveH(w) =

e-P1'.-tla

cos Wot' cos Wot'

dt'

II

I I I I I I I II.I

' = ae- P

f~", efJeP 0, the second integral in Equation (5.34) vanishes. Consequently,

X(w)H(W)e'OI'dw

(5.39)

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

where X(w) is the Fourier transform of the excitation x(t). For this special . case, with x(t) = li(t), we haveX(w) =

J~., x(t)e- r"" at =

1

(5.40)

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

i

II98STRUCTURES WITH SINGLE DEGREE OF FREEDOM (SDOF) 5.2.s.(w)

RANDOM EXCITATION AND RESPONSE

99

I.

I I I I I I~ rl

I 'il~

excitation x(t) is given byyet)

R,.(TI

=

f:",

h(.)x(t - .)dr

(S.43)So) T

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 !ivesE[y(t)) =

f~.., h(.)E[x(t -

.)] d.

(al

(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 obtainmy

~ mx

f:",

For r > 0, integrating with respect to O 2 and then (J, givesh(.)dr (S.44)

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 responseRy(r)

Jo

rlX)

2n:Soe-P6Ie-/l(t+Otl

dO

= 2n:S oe- PrThe mean square response is

i

'"

o

e- 2PO, dO. = ~ {J

s

-p,

(5.46)

= E[y(t)y(t +.))

=

f:", f:.., h(Odh(O~)E[x(t

- OI)X(t+ r- O 2)] dOl d0 2

=

f'", f~

E[y2(t)] = Ry(O) = ,,!So

p

(S.47)

h(Odh(02)R x (r

+ O.

- (2)dO I d0 2

(S.4S)

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

Ry(T)

~~~iRy(') =

my = mx

f:",

h{t)dt = 0

. Substituting the impulse response h(t) from Equation (5.31) into Equation (S.4S) gives the stationary response autocorrelatioR

,_.

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

STRUCTURES WITH SINGLE DEGREE OF FREEDOM (SDOF)

5.2.

RANDOM exCITATION AND RESPONSE1f1(w)j2

.

I

101

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 where

+ 2';wn y +'w;y =

Xo

(5.48)

Y=Wn

XI -

Xo

= relative displacement

$m =i

natural circular frequency

..; = c/2wn m = damping ratio

FIG. 5.13. Squared modulus of the complex frequency response function 1I(w) given by Equation (5.50).

Substituting xo(t)

=

e "" and y(t) = H(w)i"" into the equation of motion gives H(w) w 2 + i2';W Wn

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 toEqua~ion

(5.48) and by definition we have

2 W..

(5.49)

Ii + 2';wnh + w;hFor t > 0, (i(t)

= - 6(t)

with its modulus ~hown in Figure 5.13.

0 and the free vibration solution ish(t) = cle-~"'"' sin Wdt

IH(w)1 2 ="

,.,

. -,

,

,

(5.50)

+ c2e-~"" cos Wdt

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'"'o(t}

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 andh(t) =-

m

H 1-""'I{t)

%

e-~"''

sin Wdt,

t>O

{'" IU}

0,

t}(7.271

~

*t~

.),

bjk

= constants ~

1/ J:m2

!Jt](XI) dx l

J:

!Jtt(Xl) dX 1

= J~I k~1 !JtJ(X)!Jtk(X)

2 hJ(01)lI k(02)R jk (r - O

+ ddO I {/O2(7.28)

7.1.3.

Concentrated Random Excitation

=

Jl k~tco co

(

4 ) m2L2 !Jtix)!Jtk(x)!Jtj(a)!Jtk(a)I jk

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 asE[p(x l , t)P(X2, t

where IJk denotes the double integralIjk =

f~"" f~eO h (1I 1)hk (02)Rj

p (r

+ 01

-

02) dOl d0 2

(7.29)

+

t)] = O(XI - a)b(xz - a)Rp(t)

(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

RESPONSE OF CONTINUOUS SYSTEMS

1.1.

SHEAR BEAMS(ON

151and

The spectral density function of the relative displacement y(x, c) isSy(X, w)

where the cut-off frequency w, lies between the natural frequenciesWN+I

1 -21t

f'"-00

Ry(x, ,)e-I"" dr

(7.30)

~ .~ ~

~

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 as1 21t

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

f'"

ljke-I""

dr

= Sp(w)Hj(w)H k( -

E[112(X)) = Rv(x, O)w)

(7.31)=j=1 k=l

-u;,

whereHAw)

L L m2L2 t/I A>::)t/lk(X)t/I j(a)t/lk(a)I

""

ru

4

jk

(7.34)

= f~", hj(t)e- I."

wheredtIjk

=

, t ,I1 t,

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) asSy(x, w) = Sp(w)

L~. L~ il j (Od i'lW2)U/.(OIJOC'_~,

- (l2l {/OI

1/(/2

(7.35)

N) = (I hj(t) = _. I Ir l/t 21t

'. iwH j(w)e"'"

I/tO

J! k~l m2L200

co

4

t/lj(x)t/lk(x)t/lAa)t/lk(a)Hiw)H k ( - w)

(7.32)

andiwHj(w) =

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 bySp(w) = { 0,

"'. hNk-i!U' J-ru

l/e

(7.36)

Substituting into Equation (7.35)Rp(r)

So,

We

We

00

Yo(t)e- UUr tit dw

'1

(7.100)

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,

7.4.2.

Impulse Response hdx. t)

..-/

F(L, t) = k oz(L, t)

~=O

(7.93)

a z(O, t) ~ =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) =

r,.:;"

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 bydx = dt

c

(7.101)

2'~'1t S' " _

H,.{x, w)e'W'dw

(7.95)

0'..

which represent straight lines in the x-t plane with slopes

c. Sincc c = .Jk/III

~~ I" . f~

I:; I '~HI

164

RESPONSE OF CONTINUOUS SYSTEMSoft)",'

7.4.

ALTERNATIVE SOLUTION (SHEAR BEAMS)

165

':1 ,;; ')J

il

I

'"F4

I! Iil

alO, tl

LI

VUU 00

~ !II.. '

'III III:5 \.

192

RESPONSE OF CONTINUOUS SYSTEMS'

7.6.

DAM-RESERVOIR (HORIZONTAL EXCITATION)

193

Ir~~~:

,

I,

It I

Let w/e

= k = wave number in radians per foot andH~(x, y, k)

Carrying out the preceding integration and making use of Equation (7.206) gives (7,201 )Un

= f!" cos, k~y

4

(- I)n-I

Substituting Equation (7.201) into Equations (7.197), (7.199), and (7.200) gives(X2 -

iwn(2n -

t)Jk[

k2

(7.210)

ot:l uL~;

I I ~r. I'}.',

k;

+ k2 = 0at y = 0

(7.202) (7.203) (7.204)

- eax sin kn y = 0ff" cos knH

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 p(x, y, r) = H p(x, y, w)uoeilt(7.2\1)

0

From Equations (7.202) and (7.204) we find that

1\: I III rI~I~

(X=.jk;'=Pand that

(7.205)

Since the pressure p is related to the potential by Equation (7.192) it follows that P = I' /Itn

= piwH",aoe"ot

.

= HrGoe"ot

.

(7.212) (7.213)

k.

=

(2n - l)n 2H '

n =.1,2, ...

(7.206)

:. Hp(x,jl,k)

iwpH",(x,J',k)

I I 1\ I

I

11

~

.

II. .I~J

\~ ~gI' ;~

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 solutionH~(x, y, k)

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),

H",(x, y, k)

=

L an cos kny exp[n; 1

N

ix.Jk 2

-

k;](7.214)

Ln==l

to

an cos k.y dXp[ - x.Jk; - k ]

2

(7.207)

+

LN+I

to

an

COS

kny exp[ - x.Jk?; - P]

where an are yet undetermined constants. Substituting the preceding H~ into Equallon (7.198) gives

II,.

iw

Ln= 1

to

a".Jk;' - k2 cos k"y = 1

(7.208)

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(x, y, t) =,~:

II IiI

1'1

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

";\

LN

an cos k"y exp[ - l"k -

~.( 2 k; X

-

.J

k - k.

2

WI)]2

(7.215)

:;;: .

iwan.Jk; - k = H

2

.

2

Jo

[IIcos kny dy

(7.209)

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.

Ie."

194

RESPONSE OF CONTINUOUS SYSTEMS00

7.S.

DAM-RESERVOIR (HO!\IZONTAl EXCITATION)

195

i

I i~ I

~ f'

From Equation (7.210) it is clear that an ~

when (7.216)

~. 2c

2

Let the earthquake excitation be a(t) = (/oei{UI with 'amplitude !1o, then the total hydrodynamic force on the dam in dimensionless form is1-'(1) = iJollAwk ""i

=

e = k;

lil. ~ I'\i; IIf~

Wn

Thus the resonance or natural frequencies of the dam-reservoir system are = ck. and the first or fundamental natural frequency 1SWI =

IF(t)ly

IH ~'(/)IyFo

=Fo-

IH/(QlIo

(7.221 )

ck,

c

==

A-!::: 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.

Hp = (iw)- H

I(O)dQ

(7.233)

f", ~(r)hr(t

- r) dr

(7.227)

where

'Il

~...'

7.6.4.

Random Vibration

1(0) =

IJI

(21t _

l)2J(2~1 -

1)2 - 0 2

r

(7.234)

flIa,

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 .Sr(w) = IH,,(w)12Sa(w)

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

(7.228)

where HF(W) is the complex frequency response (lUlction for shear and

r"'\ r=-ili 1 Jo

I2

dU

-+ CXJ

(7.235)

~ ~U"lII

II198RESPONSE OF CONTINUOUS SYSTEMS PROBLEMS

ill I li1 I Ii>! I If I:>',"

199

"'::l

~:l

ill

' t:;

,

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) hccomesbehav~s

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.

PROBLEMS 7.1. Show that for the shear beam of Section 7.1 the natuntl freqllel1l:ies and mode shapes are, respectively,(u. J

I I! I It I lill It I Wit.1 i' III II II IIII, ~

"" I II (211 so that

fi3 = 1.0518

(7.236) (2; - I) -.

Itt"

2L

and

t/J l(x)

= sin wi:': (.

IHt(m)1 2

32F ( -_ .. (1.05U1)nly

o)2

7.2. Show that for the flexural beam problem of Section 7.2 the naturalfrequencies and mode shapes are, respectively,w. =. J

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

(i~)2\

El

L

pA

and

t/J}x)

= sin 2ltx

n} = 2.3567

(!:0)2 r'" S~(lII) elm II Ju

(7.237)

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. 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. 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,. = = 1tme = 0.1,0.0\, and 0.00\

f

This c1e.arly shows that the mean square shear oJ is again unbounded if the excitation spectrum is an ideal white noise with auniform 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 thatS.(w)

Ii

I

{0

So

for 0 < Iwl < otherwise

elL

We

W.i -: Wj_1

I

the excitation acceleration has a mean square valueof IT; = 2S ow,. and the mean square response shear force on the darn

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. Derive the simple wave equations for the dam-reservoir problem of Section 7.5 in the solid and in water.

I

(J;' = 1.17M3( "',,)2

c;r

(7.238)7.7.

From this result, we see that when the standard deviation (I" of the horizontal

I

11(51

II I I I I I I I I

200 7.8. 7.9.

RESPONSE OF CONTINUOUS SYSTEMS

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. 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. 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

7.10.

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 PROCESSBy definition, a Gaussian distributed random variable x has the following probability density function:p(x) = _1_ exp [_ (x

\1t

I I

J2i"r.a

2a

l

m)2]

(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

. ilii

1~ 111

Jj.

.1I I

III

202

DESIGN OF STRUCTURES FOR RANDOM EXCITATIONSp(x)

8.1./'(.'1. xl)

STATIONARY GAUSSIAN PROCESS

203

;!..J

'i""

;in

I

~ f.!

j

I-rI.FIG. 8.1.

~

I I III1

t:0'/O'h'/ffiW'pVh'A

~

:.x....... '" -;;'-X2

"'---"'"IGaussian probability density runction orlhc random variable X.

is defined byP(XI,Xl)

1

!II I I

I. II

=

- IIIlf exp - - - - [ " [(XI ---2-. 21f.(J I (J2JI=lif2 2(1 - IJIl) (Jt12'

{t

Xl(0)Xi'

- "i-----(Jt;2-------- + -

2Pt2(X, - 1II,)(X2 - 1112)

(X2 - m2)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

m2

-0,. Xl

=

R(O) - m 22

(8.3)JIll

I 'I. ~ if~ I. p i!

~I~

p(r) = R(r) - m

(8.4)FIG. 8.2.

(b)

Joint G:lUssl:m prohahilily density rUllction I~Xt. x,).

g .,

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

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 I !:". ,_.

I202DESIGN OF STRUCTURES FOR RANDOM EXCITATIONSp(x)

8.1./I(XI'

STATIONARY GAUSSIAN PROCESS

203

IIIII I I I ,I :1

~ :~ '. ~

.1 I

II I

r--

I "---f- ,,--!

1~A7'~1

I~I

I I

II

-____-==!~~=q ____~~~?2c2/Z~~~c2?~/.~?~?~K~?262Z~?~X~V~?~Z~/'~Z~A~::~~~~~------,...FIG. 8.1.i~

)

x'

'"

-j

-- __ --

--,-I "'2-1 I

I

/--,-----

--

---:;;;t-.\"2

Gaussian probability density funclion of the random variable X.

__ J/XI

defined by

p(x., X2)

1 21f.Ij~lj2.J1'7\-'---'

_ Pi2 exp

{2(J---/~-2i-I f-(X , -Ij~- md 21112)

lalXi*

2/112(X\ - 1II,)(X2

'-~-;-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

"'2

-0ml XI(b)

R(O) _ m 2

(8.3)

pIt) = R(t)(/2' -

m

1

(8.4)FIG. 8.2.

Joinl Gaussian prnhahilily densily rtiIlClinlll~\'I' x,l.

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

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,_,., f n il)204DESIGN OF STRUCTURES FOR RANDOM EXCITATIONS 8.2. PROBABILITY OF UP-CROSSING

205

Ii

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. Thusp(y) dy =

' " [lIY+dy-ax,ul> f -ax,- ao(y

P(XI' X2) dXlnf y

J

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 integraloo

I I

;~:~;11 :~ ';

dXI

1/1>

1 Fe.V 21t(1y

f)2

dy

'"-IXI

dxfoo e-""dy=f-00 -trl

foo-I't;

e-""+Y"dXdy=(f"" e- x 'dx)2-00

;! :iii I 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

A sample function xII) of a random process with a specilied level,

(/.

I I II I I I I

P(C) =

C'" CD J;;:o Jx:a-xdl p(x,x) ,Ix elx

(H.6)

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\

which, because of the differential dl simplifies toP(C) = dt

i:o

xp(a, x) eli

(8.7)

i:'

(Ix < (I)

lind

(:( til >

(II

x)l]

(a.5)

This up-crossing prob 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~sex(t)

x

B

a

(a

xl

=i:dt

x

_1_t

l:

--~~~---------X---------~r---------~~----------~X

+

atFIG. 8.6. Wedge area (shaded) to define the up-crossing random event in the plane of the two(J.

FIG. 8.6.

Conditions of an up-crossing of the level x

random variables x and X.

I

I

208

DESIGN OF STRUCTURES FOR RANDOM EXCITATIONS

8.3.

PROBABIUTY DENSITY OF THE PEAKS

209

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(N) =

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

LaU./

tI,PN(tli) = (O)PN(O)

+ (I)PN(l)

= P(C)

(8.8)

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 byp(z) dz-

i

I

x

~ [Hp:](tl p )

(8.11)

2 X2)} t {I p(a, :c) = ---exp - _. (a " 2 +"221t(1.(1"2 (1 x

(J"

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 t) is the same as that of no exceedance in (0, t), it follows thatpeT

= P(no exceedance in,(O, t)]P(one exceedance in (t, t

> t)

= R(t)

)1I I I

P(O; (0, t)]P[I; (t,

t

+ de)]

fO?

pet) dt

J'" v:e- :< de = e-': YJ -

Yi+t

Ilt,tt-I)q(10.6)B

I I

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

c

.. y

FIG, 10.3.

Random walk in the phase plane (forward).

270

NONLINEAR RANDOM VIBRATION

10.2.

APPLICATIONS OF THE RANDOM WALK MODEL

271

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. Furthermore,10.1.4. One-Step Transition Probability p

4Y + ' 1y

dy

+

f'+41 H(y, y) d~I

=

f'+41 F(l;) dl;1

Ai'

+ H(Y, y) At = F(t) AtE[Ay] = - H(y, Y) At(10.11)

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 conditionsji

Ay(Ay)2

=

f + F(l;) dl;, 1'+~ F(q) till + (L1t)2 J4 '

1 ,I

,+41

H(y, y) dl;

+.

f'+41I

F(l;) dl;

E[Ay]2

=,

f

'+&'

f'+4' J,. E[FR)F('1)] dl; d'1

+

H{Ji, y)

F(t);

y(O)

y(O)

0

(l0.7)

2rcSo= 2rcSo

J,

f'+M f1+4'

J/

15(l; - 11) dl; dll( 10.12)

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 (.)

J,

f/+&1dq = 2rcSo At

= E[F(t)F(t + .)] =

2rcSo15(.)

(10.8)

From the rombination of Equations (10.9)-(10.12) it follows that:p=,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 isE(Ay)

~[l - H(Y. y) !!]..j21tSo At

(10.13) (10.14)

= A*p + (- A*)q =

A*(2p - 1)

(to.9)

where ,1* is the constant grid size in the response velocity phase plane. Similarly,E[(Ay)2]

y coordinate in the(10.\0)

= (A*)2p + (_

A*)2q = (,1*)2

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

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)

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

NONLINEAR RANDOM VIBRATION

10.2. APPLICATIONS OF THE RANDOM WALK MODELsgny

273

Let )\ = ; fly = ; /l*, Yj = j /ly, and rk = kilt. ThenYI_I = (i - 1)1l*, YJ - YI-l /It

=j

fly - (i - I)/l* Ilt

Select fly

= /l* Ilt. ThenYj = j /l* Ilt,lk-l

YJ - Yi-\Ilt YI+I = (i

= (j -i + I) /l* Ilt+1)1l*-1.0

1.0

= (k - 1)1lt,

10

... y

YJ - Yi+11lt = j /l* III - (i

+

I)/l* Ilt = (j - ; - 1)/l* Ilt

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), becomesP(; /l*,j /l* Ilt, kilt)

FIG. 10.4.

The sgn y function.

= P(i,j, k)= P(i - I,j - ;

p(i,j, k/i - I,j - i

+

1, k - I) =

~-

(; -

I)C /It -

~:: sgnU - ; + 1)

+ 1, k -

1)p(;,j, kl; - I,j - ;

+ 1, k -

I)

Let K = C /It and L

= DIlt/2/l*, then+I,k - I)

+ P(; + I,j -

i -1,k-l)q(;,j,kli

+ I,j- i -

I,k - I) (10.15)

p(i,j,kl; - I,j - ; q(i,j, k/;

=! -

K(i - I) - L"sgn(j - i

+

I)l)

+

I,j - i-I, k - I)

=! + K(i +

I)

+ Lsgn (j - ; -

(10.16)

"II II

As pointed out previously, this gives the backward random walk model witll a recurrence mechanism illustrated in Figure 10.2. EXAMPLE 10.1. motion is Consider a nonlinear structure for which the equation of

For numerical computation of response probabilities letIlt

= 0.05,=

2nSo

= I,

C = 1, /l*/lt

D= 1

then /l*ji

.j2nSo/lt = 0.224,

= 0.0112 = 0.1 12

+ HU, y) =

F(r)

K with the structural property function

= C/lt = 0.05,

/ltDL

0.05= 2(0.224)

= 2.1*

HU, y)

= 2Cy

+

D sgn y

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

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-I, k - 1)[0.5

- 0.112 sgnU - ;

+ 0.05(; +

I)

+ I)) + P(; + I,j + 0.112 sgn U - i-I)]

1>2

9.3. (iv) 021 < X~ < 46 (v) Prob(xio> y)

{R,} = (~O Ro, .!pRJ' {R_,}>=

\0 RJ O. O. 0, 0, 0, O)}

= e-y/l{l + iy + jy2 + lsy3 + Jhy4 }

0'974So < m < I026So

(.!jRo,lj!R,), !j!-Rs.2a 2a

\0 R 7 , 0, O. 0, 0, 0, 0)...

)

for part (iv)

0'15 for chi-square cf.016 for Gaussian. 9.4. 0987So < m < l-(H3So;

11.2. {Vk} = (2' x2' 0, 9x 2 ' 0,

a 2a

10.1. (i) (0,(ii) (0,

i, 0, 0, 0, 0, 0, i) (0, -i~ 0, 0, 0, ... 0,0,0,

-ii, 0, 0, 0, 0, 0, to

{Vn

= (a, o. 0, 0, O. 0, ...)

= ~rt cm 2 s :I 2048 cm 2for for O.:s;; k .:s;; 1023 and 3073.:s;; k .:s;; 4095

to

l~O.t

11.3.

(i) So

10.2. (O,t - i!