Stochastic Structural Dynamics Lecture-40 · 2017. 8. 4. · 1 Stochastic Structural Dynamics...

transcript

Stochastic Structural Dynamics

Lecture-40

Dr C S ManoharDepartment of Civil Engineering

Professor of Structural EngineeringIndian Institute of ScienceBangalore 560 012 India

manohar@civil.iisc.ernet.in

Problem solving session-4

Problem 40A sdof system driven by a filteredGaussian excitation is governed by the equations

2 ; 0 0; 0

Here is zero mean Gaussian white noise process

such that 2

x x x f t x x x

Set up the equations for time evolution of firsttwo order moments using Markov process approach.Consider the response in the steady state and evaluate

the response moments

1 2 2 1

2 2 1 3

2 ; 0 0; 0

x x x f t x x x

t t t D t t

x xx xx f

x x x xx x t

22 2 1 3

, , ; 0; 0

, ~ 1; ~ 1; ~

n n nt

j ijj i jj i j

dx x dt

dx x x x dt

dx x dB t

dB t dB t dB t D

dX t f X t t dt G X t t dB t t X X

X t f n dB t m G n md h X t tdt

h h hf X t GDGt X X X

Recall

22 2 1 3

3 23 3

n n nt

j ijj i jj i j

d h X t tdt

h h hf X t GDGt X X X

d h hh X t t X X X Xdt X X

h D hXX X

22 2 1 3

d X Xdtd X X X Xdtd X Xdt

21 1 2

2 22 2 1 3 2

23 3 3

2 21 2 2 2 1 3 1

1 3 2 3 3 1

22 3 2 1 3 3 3 2

d X X Xdtd X X X X Xdtd X X X Ddtd X X X X X X Xdtd X X X X X Xdtd X X X X X X X Xdt

2 2 21 2 3 1 2 1 3 2 3

Initial conditionsAssume that , & 0 = are all deterministic.

, , , , , 0@ 0

x x f f

X X X X X X X X X t

22 2 1 3 3 2

1 2 3 3

22 2 1 3

Steady state response analysissteady state

d h X t tdt

h h h D hX X X X XX X X X

d X Xdtd X X X Xdtd X Xdt

21 1 2

2 22 2 1 3 2

23 3 3

2 21 2 2 2 1 3 1

1 3 2 3 3 1

22 3 2 1 3 3 3 2

d X X Xdtd X X X X Xdtd X X X Ddtd X X X X X X Xdtd X X X X X Xdtd X X X X X X X Xdt

0 0 0 1 0 0 00 4 0 2 0 2 00 0 4 0 0 0

1 0 2 1 0 00 0 0 0 1 00 0 1 0 2 0

0 0 0 1 0 00 4

X XX XX X

00 2 0 2 0

0 0 4 0 0 01 0 2 1 0 0

0 0 0 0 1 00 0 1 0 2 0

Problem 41In a study on reliability analysis of a cracked plate ithas become necessary to simulate a vector of six non-Gaussian random variables. The specification of theserandom variables is limited to the description of the 1storder pdf-s and the matrix of correlation coefficients.Develop a simulation procedure based on the Nataftransformation to simulate 5000 samples of the randomvariables. Estimate the 1st order PDF-s from the simulatedsample and perform the Kolmogorov-Smirnov test to verifyif the simulations have been performed satisfactorily.

4 5 6 2

The distribution of the basic random variables are as follows:~ 60,10 ; ~ 1,0.2 ; ~ 2,0.1 ;

~ 1 ; , ~ 33.0,0.47,3.5,0.3, 0.9

1.0 0 0 0 0 00 1.0 0 0 0 00 0 1.0 0 0 00 0 0 1.0 0 00 0 0 0 1.0 0.8350 0 0 0 0.835 1.0

X N X LN X LN

X EX X X N

Partially specified non-Gaussian RVsNataf’s transformation

Let X1 and X2 be two random variables such that

•X1 and X2 are not completely specified•Knowledge on X1 and X2 is limited to first order pdfs and the covariance matrix.

Question: How to transform X to standard normal space?

*1 2 1 2 12

1 11 1 1

2 22 2 2

1 1 1 2

2 1 1 2 2

Let( ) ( )

with ~ (0,1), ~ (0,1) &

( ) ( ); 0

( ) 0( ) ( ) ( )

( ) ( ) ( )0( )

U N U N U Udx dxp x udu du

dx dxp x udu duu

p x u uJu p x p x

1 21 2 1 1 2 2

1 11 1 2 2

1 12 1 2 2

1 1 2 21 11 2

12 1 1 2 2 1 2 1 2

( , )( , ) ( ) ( )

( ) ( )

@ ( ) , ( )

( ) , ( )( ) ( )

( ) ( )

( )( ) ( , )

U UX X

p u up x x p x p x

u P x u P x

P x u P xp x p x

P x P x

x x p x x dx dx

1 12 1 2 2

1 1 2 2 1 11 2

1 1 2 2 1 2

( ) , ( )( )( )

( ) ( )

P x u P xx x

P x P x

p x p x dx dx

1 1 2 1

1 2 X 1 X 1 1 1 1 2

1 1 *12 1 1 1 2 2 1 2 12 1 2

( )( ) , ,

P x z P x z

dx dx p x p x z z dz dz

P z P z z z dz dz

Substitute

1 1 *12 1 1 2 2 2 1 2 12 1 2

( )( ) , ,

(1) Divide the range [-1,1] of into L divisions.

(2) For each value of solve the

P z P z z z dz dz

Strategy for the determination of the unknown

*12 121

above equation

(numerically) and obtain the

corresponding values of . Note that

-1 1 1,2, , .

(3) Interpolate to obtain the value of

for which the target value of is r

ealized.

12specified

*12 determined

1 1 *12 1 1 2 2 2 1 2 12 1 2

solve for by solving

( )( ) , ,

Simulate ~ 0, .

Simulate and us

X XP z P z z z dz dz

Steps for simulation of 2 - dimensional Nataf random variables

Step 1

Step 2

Step 3

; 1,2i X iX P U i

1.6 1.7 1.8 1.9 2 2.1 2.2 2.3 2.4 2.50

simulation

target lognormal cdf

0 2 4 6 8 10 120

simulation

target exponential cdf

Rho_equivalent_gaussianrho_t =1.0000 0 0 0 0 0

0 1.0000 0 0 0 00 0 1.0000 0 0 00 0 0 1.0000 0 00 0 0 0 1.0000 -0.88330 0 0 0 -0.8833 1.0000

Msim=59.8787 1.0019 1.9990 1.0044 5.1723e-015 3.4961

Stdsim= 9.9726 0.1986 0.0998 1.0104 2.5838e-015 0.2982

Rhosim=1.0000 -0.0154 0.0093 -0.0135 -0.0013 0.0126-0.0154 1.0000 0.0206 0.0025 0.0014 -0.00680.0093 0.0206 1.0000 -0.0016 0.0018 -0.0058-0.0135 0.0025 -0.0016 1.0000 -0.0180 0.0116-0.0013 0.0014 0.0018 -0.0180 1.0000 -0.83450.0126 -0.0068 -0.0058 0.0116 -0.8345 1.0000

Problem 42A two dof system with cubic and hysteretic nonlinearstiffness characteristics is shown in the figure.

k1 k2 k3

c1 c2 c3

p1(t) p2(t)

u1(t) u2(t)

Left support

Right support

R1(t) R2(t)

Springs and have cubic force displacementcharacteristics and spring k is an inelasticspring modeled using Bouc's approach.

31 1 1 1 2 1 2 1 1 1 1 2 1 2

32 1 2 1 1

32 2 2 2 1 3 2 2 2 1 2 2 1

3 2 3 2 2

12 2 2

The system is taken to be governed by the equations

| | | | | |n n

m u c u c u u k u u k u u

u u p t w t

m u c u u c u k u u u u

k u k z p t w t

z u z z u z Au

1 1 1 4

2 2 2 5

p p w t

1 2 1 2

are given to be a set of independent

white noise processes with .

The various system paramters are as follows:1.0kg, 1.5kg, 0.1kN / m, 0.2kN/m,0.15kN/m, 0.05,

w t w t

m m k kk

1 2 3 4

4, 0.05, 0.5, 0.5, 1, 3,10, 20, 15.6 , 2 , 0.00420.01N, 0.02N, 0.01N, 10.0N and 20.0N

Using 1.5 order strong Taylor's scheme, develop a procedure to simulate samples of the

A nT s t s s

sytem response.Hence estimate the response moments (up to second order)and the first order pdf-s.

, 0 21

Recall

2 1 2 ,

1 1 1 ,12 2

mn n n n j j nk k k k k

mj n j n j j j

j jj j j j

j i p j pi

p k iki i k

d mi j lk k

x x a b W L a

L a Z b W Zt

W Z a a

L ati x

j j i jki l i

ik k k

L bx x x

1 1 2 2 1 2

1 1 1 0 1 2 1 1 1 2 1 2 3 1 31

4 1 4 5 1 5

2 2 2 1 0 2 2 1 2 1 2 2 211

13 2 3 4 2 4 5 2 5

3 3 3 0 3 2 11

k k k k k k k

k k k k k k

k k k k

x t u t u t u t u t z t p t p t

x x a L a L a Z L a Z L a Z

L a Z L a Z

x x a W L a L a Z L a Zm

L a Z L a Z L a Z

x x a L a L a

3 1 2 3 2 3 3 3

4 3 4 5 3 5

Z L a Z L a Z

L a Z L a Z

4 4 4 2 0 4 2 1 4 1 2 4 221

23 4 3 4 4 4 5 4 5

5 5 5 3 0 5 2 1 5 1 2 5 21 3

3 5 3 4 5 4 5 5 5

6 6 6 4 0 6 2 1 6 1 2 6 21 4

k k k k k k

x x a W L a L a Z L a Zm

L a Z L a Z L a Z

x x a W L a L a Z L a Z

L a Z L a Z L a Z

3 6 3 4 6 4 5 6 5

7 7 7 5 0 7 2 1 7 1 2 7 21 5

3 7 3 4 7 4 5 7 5

k k k k k k

L a Z L a Z L a Z

0 1 2 1 1 2 1 3 1 4 1 5 11

1 2 20 2 1 1 31 2 1 2

2 3 4 521 31 2 2 2 2

1 1 1 1

1 2 2 2 3 21 21 2 22

, , 0, 0, 0, 0

k k k k k k k

kk k k k

k k k kk k

L a a L a L a L a L a L am

aL a k k x x xm

a a a ac c k x x cm m m m

L a c c L a c L am m m

4 2 5 24

, 0k kL a L am

0 3 4 1 3 2 3 3 3 4 3 5 32

1 220 4 3 12 2 2

3 4 523 12 3 2 2 3 3

1 4 2 41 22 2 32

, 0, , 0, 0, 0

k k k k k k k

k kk k k

k k kk k

L a a L a L a L a L a L am

a aL a k x x cm m

a a ak k x x c c km m m

L a c L a c cm m m

4 4 4 5 43 53

1 , 0,k k kk L a L am m

0 5 4 5 5 1 4 5

5 4 5 1 4 5 5 2 5 4 5 1 5

24 5 5 12

4 5 2 5 4 5 5 3

| | sgn | |

| || | 1 | | | | sgn | | sgn

2 | |2

1 | || | sgn 1 2 | | | | sgn

n nk k k k k k

n n nk k k k k k k k k k

nk k k

n nk k k k k k k

L a a x x x x A

a x x n x x x x nx x x

x x xm

n x x x n n x x x x

4 5 2 5 4 5 5 2 5

24 5 2 5 4 5 1 5

1 | || | sgn 2 1 | | | |

1 | | sgn 2 | |

n nk k k k k k k

n nk k k k k k

n x x x n x x x x

n n x x x nx x x

1 5 2 5 5 5 1 4 5222

3 5 4 5 1 4 5 5 2 5 4 5 2 53

4 5 5 5

0 6 6 1 6 2 6 3 6 4 6 51 4 1

0, | | sgn | | ,

| || | 1 | | | | sgn | | sgn ,

, 0, 0, 0, ,

n nk k k k k k

n n nk k k k k k k k k

k k k k k k

L a L a x x x x Am

L a x x n x x x x nx x x

L a L a

L a a L a L a L a L a L a

0 7 7 1 7 2 7 3 7 4 7 5 72 4 2

, 0, 0, 0, 0,

k k k k k k kL a a L a L a L a L a L a

time s

Stochastic structural dynamics

Uncertainty modeling•Probability•Random variables•Random processes

Propagation of uncertainty•Analytical procedures forLTI systems (time/frequency)•Markov vector approach

Reliability analysis•First passage•Extremes•Fatigue

Monte Carlo simulations•Gaussian/non-GaussianRandom variables & random processes•Variance reduction

Applications•Earthquake•Fatigue•High frequency analysis

What next?

•Structural system identification•Reliability model updating•Structural health monitoring

Further applications•Wind, waves, guideway uneveneness,…•Hazard and risk analysis

•Earthquake•Wind

•Performance based design•Design code development…..

Sensing + Computing

Discussion on mean square estimation

Introductory comments

Let X and Y be two random variables with a known jpdf. Assuming that in a particular experiment, the random variable Y can be measured and takes the value y. What can we say about the corresponding value, say x, of the unobservable variable X?

Suppose we make an estimate, say, x*, of the value of Xwhen Y=y, according to the rule x*=h(y). h(y)=unspecified function of y. The error of our estimate e=x-h(y).We can never hope to make e=0. Can we select h such that we minimize the expected value of some function of e?

Let x(t) and y(t) be two Gaussian random processes with a known joint pdf.Let it be assumed that we can observe y(t) and not x(t).Given the observation of a sample of y(t) for t in 0 to T, how to estimate the value of x(t) for some value of t?

Typical problem in dealing with random processes

Problem 1

Let Y be a random variable and c be a constant. We wish to estimateY by a constant.Find c such that E[(Y-c)2] is minimized.

YEdyyypcce

dyypcycYEe

Problem 2Let X and Y be two random variables. We wish to estimate Y by a function c(X).To find c(X) such that e=E{[Y-c(X)]2} is minimized.

( ) ( ) ( , )

( ) ( | ) ( )

( ) ( ) ( | )

would be a minimum if ( ) minimizes

( ) ( | ) for every fixed value of .

e E Y c X y c x p x y dxdy

y c x p y X x p x dxdy

p x y c x p y X x dy dx

y c x p y X x dy x

From solution of Problem 1 we have

( ) ( | ) |Yc x yp y X x dy E Y X x

Remarks•If Y=g(X), c(x)=E[g(X)|X=x]=g(x) & e=0.•If X and Y are independent, c(x)=<Y>=constant

Let ( )

Y X Y X

XY XY X Y XY Y

Y X XY YXY Y

c X AX B

e E Y AX B

e E Y AX B B AB

e E Y AX A E Y A X

e E Y X A XA

re E Y r X

Linear MS estimation

YY XY X Y XY

E Y r X r

22min 1 XYY

YXYXYXY

Let X and Y be Gaussian

YXY rxrxXYExc

For normal random variables, linearand nonlinear ms estimation lead to identical results.

The orthogonality principle

Errordata

e E Y AX B

e E Y AX B XA

Y AX BX

Data is orthogonal to error

1 1 2 2

2 21 1 2 2

Let be a random variable &ˆ

be an estimatior of .

Select 0 1, 2, ,

S a X a X a X a XS

P E S S E S a X a X a X

Pa i na

General case of linear ms estimation

21 1 1 2 1 2 1 1

22 1 1 2 2 2 1 2

21 1 2 2 1

01 11 12 1 1

02 21 22 2 2

n n n n n

n n n nn n

SX a X a X X a X X

SX a X X a X a X X

SX a X X a X X a X

R R R R aR R R R aR

R R R R a

R RA A R

1 1 2 2

1 01 2 02 0

ˆWe have 0 1,

ˆ ˆ ˆ ˆ0

ˆ ˆ ˆ ˆ ˆ

E S S X i n

E S S a X a X a X

E S S S S S S

P E S S S S E S S S E S S S

P E S S S E S E SS

E SS E S a X a X a X

a R a R a R

tP E S A R

( , , , )

( , , , ) ( , )

( , , , ) ( | ) ( )

( ) ( , , , ) ( | )

P E S g X X X

s g x x x p s x dsdx

s g x x x p s X x p x dsdx

p x s g x x x p s X x ds dx

Nonlinear estimation

is minimum when the second integrand is minimum for any

( , , , ) |n

g x x x E S X x

1 1 2 2

We have

( ) 0 1,

ˆi.e., 0 1,

ˆ 0 for any 1,

E S a X a X a X X i n

E S S X i n

E S S c X c X c X c i n

General orthogonality principle

If ( ) is the nonlinear ms estimator of , the estimation error ( ) is orthogonal to any function ,linear or nonlinear function of data.

( ) ( ) ( ) ( ) ( , )

g X SS g X w X

E S g X w X s g x w x p s x dsdx

) ( ) ( | ) ( )

( ) ( ) ( | ) ( )

( ) ( ) | ( )

( ) | ( ) ( )

( ) ( ) ( ) ( ) 0

w x p s x p x dsdx

w x s g x p s x ds p x dx

w x E S g X X x p x dx

w x E S X x g x p x dx

w x g x g x p x dx QED

If S and X are jointly normal it can be shown that linear and nonlinearestimations of S are equal.

Estimation of a random process

Let S(t) & X(ξ) be two random processes with .ba

Consider the problem of estimating S(t) for a fixed t in terms of X(ξ)specified for every ξ in an interval of finite or infinite length. X(ξ) = data available.

ˆ( ) ( ) ( )

( ) ( ) ( )

Select ( ) such that is minimized.

ˆ( ) ( )

error ( ) ( )

Orthogonality principle

( ) ( ) ( )

k k jk

S t X h d

P E S t X h d

S t h X

E S t h X X

( , ) ( ) ( , ) 1,m

SX j k XX k jk

R t h R j m

0lim ( , ) ( ) ( , )

Smoothing: Prediction: ; , Forward prediction: Backward prediction: Filtering:

SX XXa

R t h R d

a t bX t S t t a b

t bt a

X t S t

Problem: Let ( ) be a stationary random process. Estimate ( ) in terms of .ˆ( ) ( )

( ) ( )

0 ( ) ( ) ( ) 0

( , ) ( )( , )

S tS t S t

S t aS t

P E S t aS t

P E S t aS t S ta

R t t RaR t t

( ) ( ) ( ) ( )

( ) ( ) ( )

(0) ( )

( )(0)(0)

Let ( ) exp | |

exp | |exp | |

P E S t aS t S t aS t

E S t aS t S t

ˆExample : Let ( ) ( ) ( )

Note: ( ) ( ) ( ) ( ), ( )

( ) ( ) 0

( ) ( ) ( )

0 ( , ) ( , ) 0

( , ) ( )( , )

S t a S t a S t

S t a S t a S t S t S t

E S t S t

P E S t a S t a S t

P R t t a R t ta

R t t RaR t t

( , ), ) ( , ) 0

( , )SS

R t tt t a R t t a

1 2 1 2

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

(0) ( ) ( , )SS SS SS

P E S t a S t a S t S t a S t a S t

E S t a S t a S t S t

R a R a R t t

Filteringˆ( ) ( )

( ) ( )

(0)0(0)

( ) ( ) ( )

(0)(0)(0)

S t aX t

P E S t aX t

RP aa R

P E S t aX t S t

Interpolation

t-NT t+NTλt t+T

To estimate s(t+ λ) in the interval t to t+T in terms of samples of S(t), S(t+kT), k=-N,-(N-1),…,0,1,2,…,N

ˆ( ) ( ); 0

( ) ( )

0 ( ) ( ) ( ) ,

Set of 2 1 equations for

Nk k N

S t a S t kT T

P E S t a S t kT

P E S t a S t kT S t jT j N Na

Quadrature

ˆ (0) ( ) ( );

( ) (0) ( ) ( )

0 ( ) ( ) (0) ( )

( ) ( ) ( ) ( ) 0 0,1 equations in 1

Z S t dt

bZ a S a S T a S NT TN

P E S t dt a S a S T a S NT

P E S t S kT dt a E S S kTa

a E S T S kT a E S nT S kT j NN N

unknowns

Estimate present value of ( ) in terms of values of( ) for

( ) ( ) ( )

ˆ( ) ( ) ( )

ˆ( ) ( ) ( ) ,

( ) ( ) ( ) ( ) 0

( ) ( ) (SX XX

X t S t t

S t h X t d

S t S t X

E S t h X t d X t

Smoothing

Probelm of dynamic state estimationProcess equation

Measurement equation

: 1 state evector: 1 process noise; iid sequence N(0,1): state transistion matrix: 1 measurem

k k k k

z H xx nw n

n nz m

ent vector: relates states to measurements

: 1 measurement noise; iid N(0,1)k

H m nm

| | | 1:

Probelm of dynamic state estimationDetermine

k k k k

k k k k k k k k k

x a x a z

Kalman filter provides the exact solution to this problem

Stochastic Structural Dynamics Lecture-40 · 2017. 8. 4. · 1 Stochastic Structural Dynamics...

Documents