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台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-1-
Chapter 14Random Vibration
14
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-2-
Chapter Outline
14.1 Introduction14.2 Random Variables and Random Processes14.3 Probability Distribution14.4 Mean Value and Standard Deviation14.5 Joint Probability Distribution of Several Random Variables14.6 Correlation Functions of a Random Process14.7 Stationary Random Process14.8 Gaussian Random Process14.9 Fourier Analysis14.10 Power Spectral Density14.11 Wide-Band and Narrow-Band Processes14.12 Response of a Single DOF system14.13 Response Due to Stationary Random Excitations14.14 Response of a Multi-DOF System
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-3-
14.1Introduction
14.1
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14.1 Introduction
• Random processes has parameters that cannot be precisely predicted.
• E.g. pressure fluctuation on the surface of a flying aircraft
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14.2Random Variables and Random Processes
14.2
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14.2 Random Variables and Random Processes
• Any quantity whose magnitude cannot be precisely predicted is known as a random variable (R.V)
• Experiments conducted to find the value of the random variable will give an outcome that is not a function of any parameter
• If n experiments are conducted, the n outcomes form the sample space of the random variable.
• Random processes produces outcomes that is a function of some parameters.
• If n experiments are conducted, the n sample functions form the ensemble of the random variable.
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14.3
14.3Probability Distribution
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14.3 Probability Distribution
• Consider a random variable x.
n
nxP
xxn
xxxxn
xn
nxx
n
i
n
~
~~
21
~
~
~
limfunction on distributiy Probabilit
toequalor smaller ofnumber theis
,, as available are of valuesalexperiment
valuespecified some is
Prob
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14.3 Probability Distribution
• Consider a random time function as shown:
1Prob
0Prob
1lim
1
Prob
i
i~
Ptx
Ptx
tt
xP
tt
xtx
in
i
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14.3 Probability Distribution
1
lim
xdxpP
xdxpxP
x
xPxxP
dx
xdPxp
n
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14.4Mean Value and Standard Deviation
14.4
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14.4 Mean Value and Standard Deviation
• Expected value of f(x) =μf
• The positive square root of σ(x) is the standard deviation of x.
2____
22__2__
2
2__
222
__
______
of Variance
, If
, If
2
xxdxxpxxxxE
x
dxxpxxxExxf
dxxxpxxExxf
dxxpxfxfxfE
x
x
x
f
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14.4 Mean Value and Standard Deviation
Example 14.1Probabilistic Characteristics of Eccentricity of a Rotor
The eccentricity of a rotor (x), due to manufacturing errors, is found to have the following distribution
where k is a constant. Find the mean, standard deviation and the mean square value of the eccentricity and the probability of realizing x less than or equal to 2mm.
elsewhere ,0
mm5x0 ,2kxxp
台灣師範大學機電科技學系
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-14-
14.4 Mean Value and Standard Deviation
Example 14.1Probabilistic Characteristics of Eccentricity of a RotorSolution
Normalize the probability density function:
Mean value of x:
Standard deviation of x:
125
3 i.e. 1
3 i.e. 1
5
0
35
0
2
k
xkdxkxdxxp
mm75.34
5
0
45
0
xkxdxxpx
mm9682.0
9375.075.35
3125
5
2
22
5
0
525
0
4
5
0
225
0
22
x
x
kxx
kxdxkx
dxxpxxxxdxxpxx
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14.4 Mean Value and Standard Deviation
Example 14.1Probabilistic Characteristics of Eccentricity of a RotorSolution
The mean square value of x is
064.0125
8
3
2Prob
mm155
3125
2
0
3
2
0
2
0
2
2___
2
x
k
dxxkdxxpx
kx
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14.5Joint Probability Distribution of Several RV
14.5
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14.5 Joint Probability Distribution of Several RV
• Joint behavior of 2 or more RV is determined by joint probability distribution function
• Joint pdf of single RV is called univariate distributions
• Joint pdf of 2 RVs is called bivariate distributions
• Joint pdf of more than one RV is called multivariate distributions
• Bivariate density function of RV x1 and x2:
222211112121 ,Prob, dxxxxdxxxxdxdxxxp
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14.5 Joint Probability Distribution of Several RV
• Joint pdf of x1 and x2:
• Marginal density functions:
1, 2121 dxdxxxp
1 2
2121
221121
,
,Prob,x
-
xxdxdxxp
xxxxxxP
-dyyxpyp
dyyxpxp
,
,
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14.5 Joint Probability Distribution of Several RV
• Variances of x and y:
dyypyyE
dxxpxxE
yyy
xxx
222
222
yx
yxx
y
yxxy
yx
yxxy
xyE
dxdyyxpdxdyyxyp
dxdyyxxpdxdyyxxyp
dxdyyxpyxxy
dxdyyxpyx
yxE
,,
,,
,
,
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14.5 Joint Probability Distribution of Several RV
• Correlation coefficient between x and y:
11
xy
yx
xyxy
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14.6Correlation Functions of a Random Process
14.6
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14.6 Correlation Functions of a Random Process
• Form products of RV x1, x2, …
• Average the products over the set of all possibilities to obtain a sequence of functions:
• These functions are called correlation functions
on... so and
,,
,
321321321
212121
xxxEtxtxtxEtttK
xxEtxtxEttK
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C. R. Yang, NTNU MT
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14.6 Correlation Functions of a Random Process
• E[x1x2] is also known as the autocorrelation function, designated as
R(t1,t2)
• Experimentally we can find R(t1,t2) by multiplying x(i)(t1) and x(i)(t2)
and averaging over the ensemble:
21212121 ,, dxdxxxpxxttR
n
i
ii txtxn
ttR1
2121
1,
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14.6 Correlation Functions of a Random Process
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14.7Stationary Random Process
14.7
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14.7 Stationary Random Process
• Probability distribution remain invariant under shift of time scale
• Pdf p(x1) becomes universal density function p(x) independent of time
• Joint density function p(x1,x2) becomes p(t,t+τ)
• Expected value of stationary random processes
• Autocorrelation function depend only on the separation time τ where τ=t2-t1
tttxEtxE any for 11
tRtxtxExxEttR any for , 2121
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14.7 Stationary Random Process
• R(0)=E[x2]
• If the process has zero mean and is extremely irregular as shown, R(τ) will be small.
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14.7 Stationary Random Process
• If x(t)≈x(t+τ), R(τ) will be constant.
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14.7 Stationary Random Process
• If x(t) is stationary, its mean and standard deviations will be independent of t: txtxtxEtxE and
2222
22
2
2
2
2
2
,1 Since
i.e.
:tcoefficienn Correlatio
R
R
R
txEtxEtxtxE
txtxE
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14.7 Stationary Random Process
• R(τ) is an even function of τ.
• When τ∞, ρ0, R(τ∞)μ2
• A typical autocorrelation function is shown:
RtxtxEtxtxER
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14.7 Stationary Random Process
• Ergodic Process
We can obtain all the probability info from a single sample function and assume it applies to the entire ensemble.
x(i)(t) represents the temporal average of x(t)
2
2
2
2
222
2
2
1lim
1lim
1lim
T
T
ii
T
T
T
i
T
T
T
i
T
dttxtxT
txtxR
dttxT
txxE
dttxT
txxE
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14.8Gaussian Random Process
14.8
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14.8 Gaussian Random Process
• Most commonly used distribution for modeling physical random processes
• The forms of its probability distribution are invariant wrt linear operations
• Standard normal variable:
2
2
1
2
1
x
xx
x
exp
x
xxz
2
2
1
2
1 zexp
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14.8 Gaussian Random Process
• The graph of a Gaussian probability density function is as shown:
c
x
c
c
z
dxectx
dxectxc
2
2
2
2
1
2
1
2
2Prob
2
1Prob
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14.8 Gaussian Random Process
• Some typical values are shown below:
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14.9Fourier Analysis
14.9
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14.9 Fourier Analysis
• Fourier Series
Any periodic function x(t) of period τ can be express as a complex Fourier series
Multiply both side with e-imω0t and integrating:
2
where 00
n
tinnectx
-nn
-n
tmnin
tim
dttmnitmnc
dtecdtetx
2
2 00
2
2
2
2
sincos
00
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14.9 Fourier Analysis
• Fourier Series
x(t) can be expressed as a sum of infinite number of harmonics
Difference between any 2 consecutive frequencies:
If x(t) is real, the integrand of cn is the complex conjugate of that of
c-n
2
2
01
dtetxc tin
n
0001
21
nnnn
*nn cc
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14.9 Fourier Analysis
• Fourier Series
Mean square value of x(t):
nn
nn
nnn
n
tinn
tinn
n
tinn
n
tinn
n
tinn
ccc
dtccc
dtececc
dteccec
dtecdttxtx
2
1
220
2
21
*20
2
2
2
1
*0
2
2
2
10
1
2
2
22
2
2_______
2
2
21
1
1
11
00
00
0
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14.9 Fourier Analysis
Example 14.2Complex Fourier Series Expansion
Find the complex Fourier series expansion of the functions shown below:
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14.9 Fourier Analysis
Example 14.2Complex Fourier Series ExpansionSolution
2
0
0
2
2
2
0
00
0
111
1
tscoefficienFourier
2 and 2 where
20 ,1
02
,1
dtea
tAdte
a
tA
dtetxc
aa
ta
tA
ta
tA
tx
tintin
tinn
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14.9 Fourier Analysis
Example 14.2Complex Fourier Series ExpansionSolution
2
0
020
2
00
0
2
020
0
20
2
1
11
1
0
0
0
0
inin
e
a
Ae
in
A
inin
e
a
Ae
in
Ac
ktk
edtte
tintin
tintin
n
ktkt
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14.9 Fourier Analysis
Example 14.2Complex Fourier Series ExpansionSolution
The equation can be reduced to
inin
inin
ininn
einna
Aein
na
A
ena
Ae
na
A
ein
A
na
Ae
in
Ac
20
220
2
20
220
2
020
20
11
11
121
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14.9 Fourier Analysis
Example 14.2Complex Fourier Series ExpansionSolution
Note that
,6,4,2 ,0
,5,3,1 ,24
0 ,2
,6,4,2 ,1
,5,3,1 ,1
0 ,1
or
2220
2
n
nn
A
na
A
nA
c
n
n
n
ee
n
inin
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14.9 Fourier Analysis
Example 14.2Complex Fourier Series ExpansionSolution
Frequency spectrum is as shown:
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14.9 Fourier Analysis
• Fourier Integral
A non periodic function as shown can be treated as a periodic function with τ∞
2
where 00
n
tinnectx
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14.9 Fourier Analysis
• Fourier Integral
As τ∞,
deX
ec
ectx
dtetxcX
dtetxdtetxc
ti
n
tin
n
tin
tin
titin
2
1
2
12lim
2
2lim
lim Define
limlim2
2
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14.9 Fourier Analysis
• Fourier Integral
Integral Fourier Transform pair
Mean square value of x(t):
dtetxX
deXtx
ti
ti
2
1
nnn
nnn
nnn
nn
cccc
cccdttx
2
12
1
0*0*
0
0*22
2
2
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14.9 Fourier Analysis
• Fourier Integral
This is known as Parseval’s formula for nonperiodic functions.
d
Xdttxtx
dXcXc nn
2
1lim
as and ,2
2
2
2_______
2
0**
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14.9 Fourier Analysis
Example 14.3Fourier Transform of a Triangular Pulse
Find the Fourier transform of the triangular pulse shown below.
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14.9 Fourier Analysis
Example 14.3Fourier Transform of a Triangular PulseSolution
0
011
1
otherwise ,0
,1
dtea
tAdte
a
tA
dtea
tAX
ata
tA
tx
titi
ti
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14.9 Fourier Analysis
Example 14.3Fourier Transform of a Triangular PulseSolution
222
0
20
0
0
2
1
11
a
Ae
a
Ae
a
A
tii
e
a
Ae
i
A
dtea
tAdte
a
tAX
titi
atia
ti
titi
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14.9 Fourier Analysis
Example 14.3Fourier Transform of a Triangular PulseSolution
2sin
4cos1
2
sincos
sincos2
222
2
22
a
a
Aa
a
A
aiaa
A
aiaa
A
a
AX
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14.10Power Spectral Density
14.10
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14.10 Power Spectral Density
• Power spectral density S(ω) is the Fourier transform of R(τ)/2π
• If the mean is zero,
• R(0) is the average energy
dSxER
deSR
deRS
i
i
20
2
1
dSRx 02
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14.10 Power Spectral Density
• S(-ω)=S(ω)
• Only positive frequencies are counted in an equivalent one-sided spectrum Wx(f)
0
2 dffWdSxE x
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14.10 Power Spectral Density
xxxx
xx
Sd
dS
df
dSfW
dffWdS
42
22
2
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14.11Wide-Band and Narrow-Band Processes
14.11
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14.11 Wide-Band and Narrow-Band Processes
• Wide-band random process:
• E.g. pressure fluctuations on surface of rocket
• Narrow-band random process:
• A process whose power spectral density is constant over a frequency range is called white noise.
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14.11 Wide-Band and Narrow-Band Processes
• Ideal white noise – band of frequencies is infinitely wide
• Band-limited white noise – band of frequencies has finite cut off frequencies.
• Mean square value is the total area under the spectrum: 2S0(ω2 –
ω1)
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14.11 Wide-Band and Narrow-Band Processes
Example 14.4Autocorrelation and Mean Square Value of a Stationary Process
The power spectral density of a stationary random process x(t) is shown below. Find its autocorrelation function and the mean square value.
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14.11 Wide-Band and Narrow-Band Processes
Example 14.4Autocorrelation and Mean Square Value of a Stationary ProcessSolution
We have
2sin
2cos
4
sinsin2
sin1
2
cos2cos2
21210
120
0
00
2
1
2
1
S
SS
dSdSR xx
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14.11 Wide-Band and Narrow-Band Processes
Example 14.4Autocorrelation and Mean Square Value of a Stationary ProcessSolution
Mean Square Value 12002 22
SdSdSxE x
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14.12Response of a Single DOF System
14.12
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14.12 Response of a Single DOF System
mkc
c
c
m
k
m
tFtx
txyyy
cc
n
nn
2,,, where
2 2
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14.12 Response of a Single DOF System
• Impulse Response Approach
Let the forcing function be a series of impulses of varying magnitude as shown:
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14.12 Response of a Single DOF System
• Impulse Response Approach
y(t)=h(t-τ) is the impulse response function
Total response can be found by superposing the responses.
Response to total excitation:
tdthxty
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14.12 Response of a Single DOF System
• Frequency Response Approach
Transient function:
, H(ω) is the complex frequency response function.
Total response of the system:
titi eHtyetx ~~
, If
deXtx ti
2
1
XHY
deY
deXH
deXHtxHty
ti
ti
ti
2
1
2
1
2
1
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14.12 Response of a Single DOF System
• Frequency Response Approach
Since h(t- )=0 when t< or >t,
Change the variable from to θ=t- ,
Both the superposition integral and the Fourier integral can be used to find system response
dthxty
dhtxty
dtethHdeHth
dtetdtetxX
deHXthty
titi
titi
ti
,2
1
1
2
1
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14.13Response Due to Stationary Random Excitations
14.13
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14.13 Response Due to Stationary Random Excitations
• When excitation is a stationary random process, the response is also a stationary random process
dtthH
dhtxE
dhtxE
dhtxEtyE
dhtxty
0
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14.13 Response Due to Stationary Random Excitations
• Impulse Response Approach
Autocorrelation
212121
212121
212121
222111
ddhhR
ddhhtxtxE
tytyER
ddhhtxtx
dhtxdhtxtyty
x
y
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-73-
14.13 Response Due to Stationary Random Excitations
• Frequency Response Approach
Power Spectral Density
1
2
1
2
1
2121
212121
iii
x
i
iyy
eee
ddhhR
de
deRS
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-74-
14.13 Response Due to Stationary Random Excitations
• Frequency Response Approach
xi
x
ix
ix
iiy
SdeR
deR
deR
dehdehS
2
12
1
2
1
21
21
21
21
21
21
2211
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-75-
14.13 Response Due to Stationary Random Excitations
• Frequency Response Approach
H(-ω) is the complex conjugate of H(ω).
Mean Square Response:
xy SHS2
dSHdS
ddhhRRyE
xy
xy
2
2111212
0
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-76-
14.13 Response Due to Stationary Random Excitations
Example 14.5Mean Square Value of Response
A single DOF system is subjected to a force whose spectral density is a white noise Sx(ω)=S0. Find the following:
a)Complex frequency response function of the systemb)Power spectral density of the responsec)Mean square value of the response
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-77-
14.13 Response Due to Stationary Random Excitations
Example 14.5Mean Square Value of ResponseSolution
a)Substitute input as eiωt and corresponding response as y(t)=H(ω)eiωt
kicm
H
eeHkicm
txkyycymtiti
2
2
1
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-78-
14.13 Response Due to Stationary Random Excitations
Example 14.5Mean Square Value of ResponseSolution
b) We have
c) Mean square value
2
20
2 1
kicmSSHS xy
kc
Sd
kicmS
dSyE y
0
2
20
2
_
1
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-79-
14.13 Response Due to Stationary Random Excitations
Example 14.6Design of the Columns of a Building
A single-storey building is modeled by 4 identical columns of Young’s modulus E and height h and a rigid floor of weight W. The columns act as cantilevers fixed at the ground. The damping in the structure can be approximated by a constant spectrum S0. If each column has a
tubular cross section with mean diameter d and wall thickness t=d/10, find the mean diameter of the columns such that the standard deviation of the displacement of the floor relative to the ground does not exceed a specified value δ.
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-80-
14.13 Response Due to Stationary Random Excitations
Example 14.6Design of the Columns of a BuildingSolution
Model the building as a single DOF system.
4403 64
,3
4 , iddIh
EIkgWm
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-81-
14.13 Response Due to Stationary Random Excitations
Example 14.6Design of the Columns of a BuildingSolution
3
4
3
4
44
22
22
0044
0
10
47592.003966.012
03966.08000
101,10With
8
64
64
,
h
Ed
h
dEk
ddIdt
tddt
tdtdtdtdtdtd
ddddddI
tddtdd
iii
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-82-
14.13 Response Due to Stationary Random Excitations
Example 14.6Design of the Columns of a BuildingSolution
When the base moves, equation of motion:
titi
titi
eeHm
k
m
ci
eHtzex
xzm
kz
m
cz
xmkzzczm
2
and
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-83-
14.13 Response Due to Stationary Random Excitations
Example 14.6Design of the Columns of a BuildingSolution
kc
mSd
mk
mc
iSdSzE
mk
mc
iSSHS
mk
mc
iH
z
xz
2
02
02
2
20
2
2
1
1
1
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-84-
14.13 Response Due to Stationary Random Excitations
Example 14.6Design of the Columns of a BuildingSolution
42
3202
42
32
02
47592.0
47592.0
cEdg
hWSzE
Edcg
hWSzE
z
41
22
320
22
3204
242
320
47592.0
47592.0or
47592.0, Since
cEg
hWSd
cEg
hWSd
cEdg
hWSz
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-85-
14.14 Response of a Multi-DOF System
14.14
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-86-
14.14 Response of a Multi-DOF System
• Equation of motion:
• Physical and generalized coordinates are related as:
nitQtqtqtq iiiiiii ,,2,1 ;2 2
n
jj
jii tqXtxtqXtx
1
or
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-87-
14.14 Response of a Multi-DOF System
• Physical and generalized forces are related as:
tftF
tFXtQtFXtQ
jj
n
jj
jii
T
Let
or 1
ii
i
i
ii
ii
ti
n
jj
iji
i
n
jj
jii
i
H
tHN
tqet
fXN
tNtfXtQ
21
1 where
, Assume
where
2
2
1
1
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-88-
14.14 Response of a Multi-DOF System
• Mean square value of physical displacement:
21
222
1 1
222
2_______
2
21
2
1lim
2
1lim
rr
rr
irr
n
r
n
s
T
T srT
s
s
r
rsi
ri
T
T it
i
H
eHH
dttHHT
NNXX
dttxT
tx
r
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-89-
14.14 Response of a Multi-DOF System
T
T srT
T
T srT
r
rr
r
dttHHT
dttHHT
2
2
21
2
1lim
2
1lim
angles. phase eNeglect th
1
2
tan
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-90-
14.14 Response of a Multi-DOF System
• For stationary random process,
dSHH
dttHHT
dSdttT
t
sr
T
T srT
T
TT
2
2_______
2
2
1lim
2
1lim
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-91-
14.14 Response of a Multi-DOF System
• Mean square value of xi(t)
n
r
n
ssr
s
s
r
rsi
rii dSHH
NNXXtx
1 122
_______2
dSHN
Xtx r
n
r r
rrii
2
14
22
_______2
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-92-
14.14 Response of a Multi-DOF System
• For lightly damped systems,
r
rr
rrr
S
dHSdSH
2
22
r
rrn
r r
rrii
SNXtx
214
22
_______2
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-93-
14.14 Response of a Multi-DOF System
Example 14.7Response of a Building Frame Under an Earthquake
A 3-storey building is subjected to an earthquake. The ground acceleration during the earthquake can be assumed to be a stationary random process with a power spectral density S(y)=0.05(m2/s4)/(rad/s). Assuming a modal damping ratio of 0.02 in each mode, determine the mean square values of the responses of the various floors of the building frame under the earthquake.
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-94-
14.14 Response of a Multi-DOF System
Example 14.7Response of a Building Frame Under an EarthquakeSolution
100
010
001
110
121
012
mm
kk
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-95-
14.14 Response of a Multi-DOF System
Example 14.7Response of a Building Frame Under an EarthquakeSolution
Compute eigenvalues and eigenvectors using k=106N/m and m=1000kg
rad/s 0001.578025.1
rad/s 4368.392471.1
rad/s 0734.1444504.0
3
2
1
m
k
m
k
m
k
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-96-
14.14 Response of a Multi-DOF System
Example 14.7Response of a Building Frame Under an EarthquakeSolution
01036.0
02330.0
01869.0
5544.0
2468.1
0000.15991.0
01869.0
01037.0
02331.0
8020.0
4450.0
0000.17370.0
02330.0
01869.0
01037.0
2470.2
8019.1
0000.13280.0
3
2
1
mZ
mZ
mZ
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-97-
14.14 Response of a Multi-DOF System
Example 14.7Response of a Building Frame Under an EarthquakeSolution
Relative displacements of the floors: zi(t)=xi(t)-y(t), i=1,2,3
Equation of motion:
where [Z] denotes the modal matrix.
qZz
ymzkzczm
zkzcxm
0
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-98-
14.14 Response of a Multi-DOF System
Example 14.7Response of a Building Frame Under an EarthquakeSolution
Assume damping ratio ζi = 0.02
Uncoupled equations of motion:
tytmmf
tftymtymtF
tFZQ
iqq
jj
jjj
n
jj
iji
iiiii
,
and
where
3,2,1 ;Q2
1
i2
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-99-
14.14 Response of a Multi-DOF System
Example 14.7Response of a Building Frame Under an EarthquakeSolution
Mean square values
35.5205235.01000
37.5205237.01000
36.5205236.01000
2
3
1
33
1
33
3
1
23
1
22
3
1
13
1
11
3
13
22
_______2
ji
jji
ji
jji
ji
jji
rrr r
rrii
ZmfZN
ZmfZN
ZmfZN
SN
Ztz
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-100-
14.14 Response of a Multi-DOF System
Example 14.7Response of a Building Frame Under an EarthquakeSolution
Mean square values of relative displacements of various floors of the building frame:
2_______
23
2_______
22
2_______
21
m 00216455.0
m 00139957.0
m 00053132.0
tz
tz
tz
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-101-
14.14 Response of a Multi-DOF System
Example 14.8Probability of Relative Displacement Exceeding a Specified Value
Find the probability of the magnitude of the relative displacement of the various floors exceeding 1,2,3, and 4 standard deviations of the corresponding relative displacement for the building frame of Example 14.7
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-102-
14.14 Response of a Multi-DOF System
Example 14.8Probability of Relative Displacement Exceeding a Specified Value Solution
Assume ground acceleration to be normally distributed random process with zero mean.
Relative displacements of various floors can also be assumed to be normally distributed.
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-103-
14.14 Response of a Multi-DOF System
Example 14.8Probability of Relative Displacement Exceeding a Specified Value Solution
4for 00006.0
3for 00270.0
2for 04550.0
1for 31732.0
3,2,1 ;_______
21
p
p
p
p
ptzP
itz
zii
zi