台灣師範大學機電科技學系 C. R. Yang, NTNU MT -1- Chapter 14 Random Vibration 14.

台灣師範大學機電科技學系

C. R. Yang, NTNU MT

-1-

Chapter 14Random Vibration

14


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


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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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• 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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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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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


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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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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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• 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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• 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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• 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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• 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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14.7Stationary 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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• 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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• If x(t)≈x(t+τ), R(τ) will be constant.


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• 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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• 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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• 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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• 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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• 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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• 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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• 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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Example 14.2Complex Fourier Series Expansion

Find the complex Fourier series expansion of the functions shown below:


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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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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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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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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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Frequency spectrum is as shown:


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• 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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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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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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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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Example 14.3Fourier Transform of a Triangular Pulse

Find the Fourier transform of the triangular pulse shown below.


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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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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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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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• 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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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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• 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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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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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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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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• Impulse Response Approach

Let the forcing function be a series of impulses of varying magnitude as shown:


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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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• 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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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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Autocorrelation

212121

212121

212121

222111

ddhhR

ddhhtxtxE

tytyER

ddhhtxtx

dhtxdhtxtyty

x

y


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Power Spectral Density

1

2

1

2

1

2121

212121

iii

x

i

iyy

eee

ddhhR

de

deRS


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xi

x

ix

ix

iiy

SdeR

deR

deR

dehdehS

2

12

1

2

1

21

21

21

21

21

21

2211


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H(-ω) is the complex conjugate of H(ω).

Mean Square Response:

xy SHS2

dSHdS

ddhhRRyE

xy

xy

2

2111212

0


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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


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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


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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


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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.


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Example 14.6Design of the Columns of a BuildingSolution

Model the building as a single DOF system.

4403 64

,3

4 , iddIh

EIkgWm


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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


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When the base moves, equation of motion:

titi

titi

eeHm

k

m

ci

eHtzex

xzm

kz

m

cz

xmkzzczm

2

and


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kc

mSd

mk

mc

iSdSzE

mk

mc

iSSHS

mk

mc

iH

z

xz

2

02

02

2

20

2

2

1

1

1


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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


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14.14 Response of a Multi-DOF System

14.14


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• Equation of motion:

• Physical and generalized coordinates are related as:

nitQtqtqtq iiiiiii ,,2,1 ;2 2

n

jj

jii tqXtxtqXtx

1

or


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• 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


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• 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


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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


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• For stationary random process,

dSHH

dttHHT

dSdttT

t

sr

T

T srT

T

TT

2

2_______

2

2

1lim

2

1lim


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• 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


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• For lightly damped systems,

r

rr

rrr

S

dHSdSH

2

22

r

rrn

r r

rrii

SNXtx

214

22

_______2


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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.


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Example 14.7Response of a Building Frame Under an EarthquakeSolution

100

010

001

110

121

012

mm

kk


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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


C. R. Yang, NTNU MT

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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


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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


C. R. Yang, NTNU MT

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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


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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


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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


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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


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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.


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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

台灣師範大學機電科技學系 C. R. Yang, NTNU MT -1- Chapter 14 Random Vibration 14.

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Transcript of 台灣師範大學機電科技學系 C. R. Yang, NTNU MT -1- Chapter 14 Random Vibration 14.