111

Stochastic Structural Dynamics

Lecture-15

Dr C S ManoharDepartment of Civil Engineering

Professor of Structural EngineeringIndian Institute of ScienceBangalore 560 012 India

manohar@civil.iisc.ernet.in

Random vibration analysis of MDOF systems-3

2

12

2 21 2

Nrn sn

n n n n

H M i C ki

1

1 exp sinN

rs rn sn n n dnn dn

h t h t t t

exp

1 exp2

ij ij

ij ij

H h t i t d

h t H i t d

1 2

1 2 1 1 2 1 2 2 1 20 0

,t t

tXX FFR t t h t R h t d d

*tXX FFS H S H

Recall

3

Structures under differential support motions

Tu t

6

24gN

N

4

0

~ 1, ~ 1;

, , ~, , ~

, , ~

Pseudo-dynamic response

T T Tg g g

t t tg gg g gg g gg gg g g

T

g g g T g

g g g g

gg gg gg g g

g

M M C C K Ku u uM M C C K K p tu u u

u Nu p t N N N N

M C K N NM C K N N

M C K N N

K K

1

1

1

0

0

p

t pg gg gg

p pg g g g g

g

p t p tg g gg g g g gg g

uK K p tu

Ku K u u K K u t u t

K K

p t K u K u K K K K u t

5

Total response= pseudo-dynamic response + dynamic response

0

( )

0

( )

T p

g g

p pg g

t tg gg g ggg g

pg

tg gg gg

eff

peff g g

u u t u tu u t

M M C Cu u t u uM M C Cu u

K K u uK K p tu

Mu Cu Ku p t

p t Mu t M u

( )

( ) ( )

( ) ( )

pg g

g g g g g g

g g g g

Cu t C u

M u t M u Cu t C u

M M u t C C u t

6

1

( ) ( )

Special CaseMass matrix is diagonal M 0

is proportional to ( )

0

( )

~

( ) ~ 1

Note:

g g g g

g

g g g g

g

g

g g

Mu Cu Ku M M u t C C u t

C K C K

C C K K KK K K

Mu Cu Ku M u t

N N

u t N

If all supports suffer the same motion, 1

1 1 1

g

t

N

7

*

( ) ( ) ( )

( ) ~ 1: vector of stationary random process with zero mean

and PSD matrix1lim

( )

g g g g

g g

tgg gT gTT

Mu Cu Ku M M u t C C u t p t

u t N

S U UT

p t

Random vibration analysis in frequency domain

2

2

* * 2

( ) ( )g g g g

T g gT g gT

g g gT

t t t tT gT g g

M M u t C C u t

P M M U i C C U

M M i C C U

P U M M i C C

8

2

* 2

2

2

*

1limpp g g gTT

t t tgT g g

pp g g

t tgg g g

tUU pp

S M M i C C UT

U M M i C C

S M M i C C

S M M i C C

S H S H

12

2 21 2

Nrn sn

n n n n

H M i C ki

9

1

1

Pseudo-dynamic response

p p

pg g g

g

tggu u

u K K u t u t

K K

S S

2

2

Total response( )T p

g

TT gT T

gT T

T g g gT

TT g g gT

u t u t u t

u t u t

U U U

U H P

P M M i C C U

U M M i C C U

10

2

2

2

Total response

Variance of total response=variance of pseudo-dynamic response+variance of dyna

TT g g gT

TT g g gg

t

g g

U M M i C C U

S M M i C C S

M M i C C

mic response+contributions due to correlation betweenpseudo-dynamic and dynamic responses

11

1 2 1 2

( ) ( ) ( )

( ) ~ 1: vector of stationary random process with zero mean

and auto-covariance matrix

, ( ) ( )

g g g g

g g

tgg g g

Mu Cu Ku M M u t C C u t p t

u t N

R t t u t u t

Random vibration analysis in time domain

1 2

1 1 1

2 2 2

( ) ( ) ( )

( ) ( )

gg

g g g g

t t t t tg g g g

R t t

p t M M u t C C u t

p t u t M M u t C C

12

1 2 1 2

1 2

1 2

1 2

1 2

1 2 1 1 2 1 2 2

( ) ,

( ) ( )

( ) ( )

( ) ( )

( ) ( )

,

tpp

t tg g g g

t tg g g g

t tg g g g

t tg g g g

UU pp

p t p t R t t

M M u t u t M M

M M u t u t C MC

C C u t u t M M

C C u t u t C MC

R t t h t R h t

1 2

1 20 0

t ttd d

1

1 exp sinN

rs rn sn n n dnn dn

h t h t t t

13

1 2

1 2

1 2

,

1

n mn mXX

n m n mt t t t

n m n mm XY

n m n m

R t td X d Xdt dt t t

d X t d Y t d Rdt dt d

Recall

14

0

Total response= pseudo-dynamic response+dynamic response

( )

( ) ( )

T p

g

t

g g g g g

u t u t u t

u t u t

u t h t M M u C C u d

Variance of total response=variance of pseudo-dynamic response+variance of dynamic response+contributions due to correlation betweenpseudo-dynamic and dynamic responses

15

RANDOM VIBRATION ANALYSIS OF CONTINUOUS MDOF SYSTEMS

Chimney underwind and earthqukae loads

16

Vehicle structure interactionGuideway unevennessVehicle motionVehicle parameters

17

Re view of dynamics of Euler-Bernoulli beams under deterministic excitations

18

2 2

2 2

2

2

0 0; ( ) 0

0 0; ( ) 0

d d yEI x q xdx dx

y y l

d yy EI ldx

Statically loaded beam

19

2 2 3 2

2 2 2 2

2

2

0 0

,

0, 0; ( , ) 0

0, 0; ( , ) 0

,0 ; ,0

y y y yEI x a x m x c x q x tx x x t t t

BCSy t y l t

yy t EI l tx

ICSyy x y x x y xt

Dynamically loaded beam

Strain rate dependentviscous damping

velocity dependentviscous damping

20

2 2 2

2 2 2

2

2

2

2

0

0, 0; ( , ) 0

0, 0; ( , ) 0

Seek the solution in the form,

0

&

y yEI x m xx x t

BCSy t y l t

yy t EI l tx

y x t x T t

EI x T t x m x T t

NoteTT t

x t

Undamped free vibration analysis

21

2

2

2

2

2

0

0

constant=

0 cos sin

0

0, 0; ( , ) 0; 0, 0; ( , ) 0

0 0; 0; 0 0

EI x T t x m x T t

EI x T t x m x T tx T t x T t

EI x T tx m x T t

T t T t T t A t B t

EI x m x x

yy t y l t y t EI l tx

l

2

2; ( ) 0dEI ldx

22

2

2

2

2

4

Eigenvalue problem

0

0 0; 0; 0 0; ( ) 0

Special case; ( )

0

0cos cosh cos cosh

sin sinh sin sinh

sin sinh si

iv

iv

EI x x m x x

dl EI ldx

EI x EI m x m

EI x m x

x a x x b x x

c x x d x x

x a x x b

2 2

2 2

n sinh

cos cosh cos cosh

cos cosh cos cosh

sin sinh sin sinh

x x

c x x d x x

x a x x b x x

c x x d x x

23

2

2

2 2

2

0 0; 0; 0 0; ( ) 0

cos cosh cos cosh

sin sinh sin sinh

sin sinh sin sinh

cos cosh cos cosh

cos cosh cos cosh

sin

dl EI ldx

x a x x b x x

c x x d x x

x a x x b x x

c x x d x x

x a x x b x x

c x

2

2 2

sinh sin sinh

0 0 0

0 0 0

0 cos cosh sin sinh

0 cos cosh sin sinh

x d x x

a

c

l b l l d l l

l b l l d l l

24

2 2

2 2

2 2

0 cos cosh sin sinh

0 cos cosh sin sinh

cos cosh sin sinh0

cos cosh sin sinh

cos cosh sin sinh0

cos cosh sin sinh

l b l l d l l

l b l l d l l

l l l l bl l l l d

l l l ll l l l

1tan tanh : characteristic values

cosh cos sinh sincos coshsin sinh

n n

n n n n n n

n nn

n n

l l

x x x x xl ll l

25

24

5

15

1

;

15.4118, 49.9648,104.2477,178.2697, 272.0309

1.000777,1.000001,1.000000,1.00000,1.00000

n n n n

n n

n n

EIC C lml

C

26

2 2

2 1, 2,

( ) sin

n

n

n EI nL m

n xxL

Homogeneous simply supported beam

Exercise

27

n=12 3 4

0 0.2 0.4 0.6 0.8-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

x/l

mod

e sh

ape

mode shapes for a simply supported beam

28

Orthogonality conditions

L

nkk

L

kn

L

nkn

L

nk

kkk

nnn

dxmdxEI

dxmdxEI

mEI

mEI

0

2

0

0

2

0

2

2

)4(][

)3(][

)2(][

)1(][

29

dxEIEIEI

dxEIEIdxEI

L

knL

knL

kk

L

knL

kk

L

kn

000

00

0

][][

][][][

30

knfordxxxxm

knfordxxxxEI

k

L

n

k

L

n

0)()()(

0)()()(

0

0

31

Forced response analysis using eigenfunction expansion

2 2 3

2 2 2

0 0

( ) ( ) ( ) ( , )

ICS: ( ) ( ,0) ( ) ( ,0) & BCS as appropriate.

y yEI x x m x y c x y f x tx x x t

y x y x y x y x

1

2

0 0

( , ) ( ) ( )

( )

0 0

n nn

n n n

L L

n k n k

y x t a t x

EI m x

EI dx n k m dx n k

32

1 1 1 1

1 1 1 1

2

1 1

[ ( ) ( ) ] ( ) ( ) ( ) ( ) ( , )

( )[ ] ( )[ ] ( ) ( ) ( ) ( ) ( , )

( ) ( )[

n n n n n n n nn n n n

n n n n n n n nn n n n

n n n nn n

EI a t a t c x a t m x a t f x t

a t EI a t x c x a t m x a t f x t

a t m a t x

1 1

] ( ) ( ) ( ) ( ) ( , )n n n n nn n

c x a t m x a t f x t

( ) (Damping force proportional to time rate of bending strain)

( )(Damping force is proportional to velocity)

x EI x

c x m x

Proportional damping model

33

2

1 1 1 1

2 2

1 1 1 1

( ) ( )[ ] ( ) ( ) ( ) ( ) ( , )

( ) ( ) ( ) ( ) ( ) ( ) ( , )

n n n n n n n n nn n n n

n n n n n n n n n nn n n n

a t m a t EI x m x a t m x a t f x t

a t m a t m x m x a t m x a t f x t

dxtxfxdxxmtadxxcta

dxxmtavdxxmta

l

kn

n

L

knn

n

L

kn

nn

L

knnnn

L

knn

),()()()()()(

)()()()(

01 01 0

1 0

2

1 0

2

34

)()()(

)()(

22

22

tpm

tpaaa

tpamamam

nn

nnnnnn

nnnnnnnnn

2

2

0

2

0

2 ( );

2 ( );

( ) ( , )( ) 1, 2,

( ) ( )

n n n n n n n

n n nL

n

n L

n

a a a p t

x f x t dxp t n

x m x dx

35

dxxyxxmadxxyxxma

xaxy

xtatxy

n

L

nn

L

n

kkk

kkk

)0,()()()0()0,()()()0(

)()0()0,(

)()(),(

00

1

1

Initial conditions

36

1 0

)()(]sincos)[exp()(),(n

t

nndnndnnnnn dpthtBtAtxtxy

Final solution

Remark:

Once the displacement field is found, we can easily find stress resultants (BM and SF) and the required bending/shear stresses

37

Example 1: An undamped simply supported beam carries an udl. The load is suddenlyremoved. Determine the ensuing vibrations.

Solution:The initial deflection of the beam satisfies the equation

qEIyiv 0

Assume that the initial velocity of the beam is zero.

38

2

1

1

01

4 4

0 5 51

( , ) ( ) ( ) 0

( ) cos sin

( ,0) 0 (0) 0 ( , ) cos sin

( ,0) sin ( )

4sin

n n n n nn

n n n n n

n n nn

nn

ivn n

n

y x t a t x a a

a t A t B tn xy x a y x t A tL

n xy x A y xL

n n x L qEIy q EI A q AL L n EI

Lxnt

EIq

nLtxy

nn

,5,3,155

4

sincos4),(

39

Example 2: An udl is suddenly applied on an undamped simply supported beam.Determine the ensuing vibrations. Assume that the beam is at rest at t=0.

1

2

0

( )

( , ) ( )sin

( ) sin

iv

nn

L

n n nn

EIy my qU tn xy x t a tL

q n xa a U t q dxm L

,5,3,1

2 sin)cos1(2),(n

nnn L

xntmnqLtxy

40

Seismic wave amplification through soil layers

2 2 3

2 2 2

0, exp ; , 0

u u uGt z z t

uu t i t L tz

41

zLzxL

zBzAziG

iGtiitiGti

tiztzu

sintancos)(0)(1)0(

sincos)()(

;0

0)()exp()exp()exp(

)exp()(),(

222

2

2

42

)21()1(

cos

1cos

1)(

**

*

iGiGG

LLL

43

xi=0xi=0xi=0

0 20 40 60 8010

-1

100

101

102

frequency rad/s

Am

plifi

catio

n fa

ctor