seismic analysis of structures - ii

T.K. Datta Department Of Civil Engineering, IIT Delhi

Response Analysis for specified ground motion

Chapters – 3 & 4

Chapter -3

RESPONSE ANALYSIS

FOR

SPECIFIED GROUND MOTION1



Introduction

Time history analysis of structuures is carried out when input is in the form of specified time history of ground motion.

Time history analysis can be performed using direct integration methods or using fourier transform techniques.

In the direct integration methods, there are many integration schemes; two most popular methods used in earthquake engineering will be discussed here.

In addition, time history analysis using FFT will be presented.

Before they are described, several concepts used in dynamic analysis of structures under support motion will be summerised ( assuming that they are already known to the students).

1/1



1/2

Models for an SDOF

Spring-mass-dashpot systemFig: 3.1 a

k

c

x

m

gx..

Idealized single frameFig: 3.1b

Rigid beam

Lumped massAll membersare inextensible

gx gx.. ..



Equation of motion of an SDOF system can be written in three different ways:

For MDOF system, equations of motion are written for two cases; single & multi support excitations.

Contd..

g

t t tg g

g

t t

mx + cx + kx = -mx ( 3.1)

mx + cx + kx = cx + kx ( 3.3)

X =AX+f ( 3.4a)

0x 0 1X= A = f= ( 3.4b

-xx -k/m -c/m

X = AX + Ff ( 3.5a)

in which (3.5 )b

tgt

tg

x0 0xX = ; F = ; f =

xk/m c/mx

1/3



For single support single component excitation

For two component ground motion

For three component ground motion

Contd..

T

Tg g1 g2

1 0 1 0 - - - - - -I = ( 3.9a)

0 1 0 1 - - - - - -

X = x x ( 3.10a)

T

Tg g1 g2 g3

1 0 0 1 0 0 - - - -

I = 0 1 0 0 1 0 - - - - ( 3.9b)

0 0 1 0 0 1 - - - -

X = x x x ( 3.10b)

1/4

gMX + CX + KX = -MIX ( 3.8)



Contd..

Example 3.1: Determine for the following structures .Solution :

I

1 1 2 3 1 2 3T Tu v u u u u uI = I =

1 0 1 1 1 1 1

Bracket frame

3u

2u

1u

Shear building frame

u2

u3

u1

v1

gxgx

Fig3.4a

Fig3.4b

1/5



010010

001001

001001

T

T

I

I

Contd..

Case 1 Single component Case 2 Two component

11 & vu

2

1

v2

u2

v1

u1

3-D model of a shear building frame Fig3.4c

1/6



Example 3.2 : Find the mass and stiffness matrices for the two models of 3D frame shown in Fig 3.5.Solution :

Contd..

3u

2u

1u

L x y

C.M.

L

k k

gx

Model-1

4 -2 2 4 -1 3 0m

K = k -2 3 -2 ; M = -1 4 -3 ; I = 06

2 -2 3 3 -3 6 1

Fig3.5a

1/7

Teff g

mP ( model1)= - 3 -3 6 x

6



Contd..

gx

C.M. u C.R. v

L

k k

Model-2

For Model -2

2

3 0 0.5L 1 0 0 0

K = k 0 3 0.5L ; M = m 0 1 0 ; I= 1

0.5L 0.5L 1.5L L 00 0

6

Fig3.5b

1/8

Teff gP ( model2)= -m 0 1 0 x



Example 3.3 : All members are inextensible for the pitched roof portal; column & beam rigidities are K & 0.5K obtain mass matrix & force vector.

Solution:For Model 1

Contd..

2.5 1.67 1M = m I =

1.67 2.5 0 2u

l

m u1

m

L

L

gx

2m

3L

gx

1u 2u m

2m

m

Model-1 Model-2 Fig3.6a Fig3.6b

1/9



eff g

1.406 -0.156 1 1.25M = m I = P = -m x

-0.156 1.406 1 1.25

Contd..

For Model 2

C

B m m D

A E

2m Instantaneous Centre

C

l B

D

A E

m

2m

sec

m

Unit acceleration givento u1 (model-1)

Unit acceleration givento u1 (model-2)

Fig3.6c Fig3.6d

1/10



For multi support excitation, equation of motion

Contd..

)11.3(0

gg

t

gggs

sgss

g

t

gggs

sgss

g

t

gggs

sgss

PX

X

KK

KK

X

X

CC

CC

X

X

MM

MM

(3.12)

0 (3.13)

(3.14)

(3.15)

( ) ( )

( )

tg

t tss sg g ss sg g ss g

t t tss ss ss sg g sg g sg g

t t tss ss ss sg g

ss ss ss sg ss g sg ss g

sg ss g

X X rX

M X M X C X C X K X

or M X C X K X M X C X K X

M X C X K X K X

M X C X K X M M r X C C r X

K K r X

1

(3.16)

0 (3.17)

(3.18 )

0 (3.18 )

(3.19)

ss s sg g

s ss sg g g

ss sg

ss ss ss ss g

K X K X

X K K X rX a

K r K b

M X C X K X M rX

2/1



Contd.. Example 3.4 : Find the r matrices for the two frames shown in Fig 3.7 & 3.8.Solution : For the rectangular frame

k k k

1u1

u5

2k 2k 2k

xg1

2m

Fig3.7

2/2

ss

3 -3K = k

-3 9

sg

0 0 0K =

-2k -2k -2k

gg

2k 0 0

K = 0 2k 0

0 0 2k

-1ss sg

1 10 0 0 1 1 11 12 6r=-K K =- =

1 1 -2k -2k -2k 1 1 1k 36 6

.. ..xg2

..xg3

u3 u4

m

u2



Contd.. For the inclined leg portal frame

Unit rotation given at B

1 B

(1)

(2)

(3) 3L

D (5)

C

(4) B

A E

L (6) (7)Kinematic

D.O.F.

Fig3.8

2/3

2

38.4 12 0 6 20 6 0

12 48 12 0 0 0 03.6

0 12 38.4 6 12 6 6rr ru

EI EIK K

L L

3 3

24 16 12 12 8 5.53 8 4

16 181 0 16 5.53 52 5.33 6.33

12 0 12 0 8 5.33 8 4

12 16 0 12 4 6.33 4 3.69

uu

EI EIK

L L

3 3

1

19.56 10.51 4 8

10.51 129 5.33 22.3

0.0661 0.0054 4 8

0.0054 0.0082 5.33 22.31

0.2926 0.4074

0.654 0.1389

us usg

us usg

EI EIK K

L L

r K K



Contd..Example 3.5 : Obtain the r matrix for d.o.fs 1, 2 & 3 for the bridge shown in Fig 3.9.

Solution: To solve the problem, following values are assumed

2/4

Simplified model of a cable stayed bridge Fig 3.9

60 m (3)

Cables 1l

20 m (1)

(9)

240m

20 m (2)

(10)

3

(6) 120m

(7)

4

50m

30m (5)

2

(8)

120m

(4)

1

gx gx gx gx

t dEI =1.25EI =1.25EI;1

0.8480 deck cable

AE AE

L

12Cosθ=

135

13Sin

; ;3 3 3

1

AE 12EI 3EI 120m 12EI= = =400m

l 80120 80 120



2 211 3

i

21 31i

241 51 3

i

61 81 91 101

222 11; 32 31 71 52

1

62 51 72 41 82 92 102

33 31

3.75EI 2AEk = + cos θ=1.875m+800mcos θ

l80

AEk =0; k =- cosθsinθ;

l

3 AE 3.75EIk =- cos θ;k =-

2 l 80

k =k =k =k =0

AEk =k k =-k ; k =- cos θ; k =0

2L

k =k ;k =k ;k =k =k =0

24EI 2AEk = +

L120 2 2

43 53 63 73

sin θ=800m 1+sin θ ;

k =k =k =k =0

2/5Contd..



;

83 93 1032

44 42 71 74deck

54 64 84 94 104

55 65 75 85 95 1053

66 76 86 96 1063

77 87 97 107

88

6EIk =- =-24000m;k =0;k =24000m

LAE

k = =320m;k =k k =-320m480

k =k =k =k =k =0

3.75EIk = ;k =k =k =k =k =0

80

3.75EIk = ;k =k =k =k =0

80

k =320m;k =k =k =0

7EIk =

12 2

98 88 108

99 88 109 88 1010 88

7 2= ×400m×120 ;k = k ;k =0

0 12 77EI 8 2

k = = k ;k = k ;k =k120 7 7

2/6Contd..



-0.781 -0.003 0.002 -0.218

r=- -0.218 0.002 -0.003 -0.781

-0.147 -0.0009 0.0009 0.147

2/7

Using the above stifness coefficients,the condensed stifness matrix corresponding to the translational degees of freedom is obtained.

The first 3X3 sub matrix is the stifness matrix corresponding to the non support translational degrees of freedom.

The coupling matrix between the support and non support translational degree of freedom is the upper 3X4 matrix.

Using them the above matrices the r matrix is obtained as

Contd..



Contd..

Equation of Motion in state spaceZ= AZ + f ( 3.20)

in which

-1 -1

g ss ss ss ss

0 0 IxZ = ; f = ; A = ( 3.21)

-rx -K M -C Mx

t tZ = AZ + f ( 3.22)

in which

tt

-1tss ss g

0xZ = ; f = ( 3.23)

-M K xx

2/8

Example 3.6: Write equations of motion in state space for example problem 3.4 using both relative & absolute motions.



For relative motion of the structure

Contd..2/9

2 21 2

k k k mω =1.9 ; ω =19.1 ; α=0.105 ; β=0.017

m m m k

ss

1 0 3 -3 0.156 -0.051C =α m+β k= km

0 2 -3 9 -0.051 0.205

2 2

2 2

0 0 1 0

0 0 0 1A =

-3ρ 1.5ρ -0.156ρ 0.026ρ

3ρ -4.5ρ 0.051ρ -0.182ρ



2/10

For absolute motion of the structure

1 2 3

2

0

0

0

g g gx x x

f

1 2 3

1 2 3

g g g

g g g

0

0k

ρ= ; f = -0.33 x +x +xm

-0.33 x +x +x

Contd..



Both time & frequency domain solutions for SDOF system are presented first and then, they are extended to MDOF system.

Two methods are described here: duhamel integration &Newmark’s - - methods.

Duhamel integration treats the earthquake force as a series of impulses of short duration shown in the figure.

In Newmark’s method, the equation of motion is solved using a step by step numerical integration technique.

For both methods, a recursive relationship is derived to find responses at K+1 time station given those at K time station.

t

Response analysis2/11



Contd..

gx

d t

n-ξω t1 d 2 dx t =e C cos ω t+C sinω t ( 3.24)

n-ξω tn 1 2 d d d 1 n 2 dx t =e -ξω C +C ω cos ω t+ -ω C -ξω C sin ω t ( 3.25)

Fig3.10

Duhamel Integration:

Δt=t - t ( 3.34)k+1 k

k+1 kk

F -FF τ =F + τ ( 3.35)

Δt

2/12



Responses at the tk+1 is the sum of following :

• Response for initial condition

• Response due to Fk between tk and tk+1

• Response due to triangular variation of F

Response at tk+1 clearly depends upon

The constants etc can be evaluated from the three response analyses mentioned above.

Contd..

(0) 0 (0) 0x x

x =C x +C x +C F +C F ( 3.36)1 4k+1 k 2 k 3 k k+1

x =D x +D x +D F +D F ( 3.37)1 4k+1 k 2 k 3 k k+12x =-x - 2ξω x - ω x ( 3.38)n nk+1 gk+1 k+1 k+1

C ,C1 2

2/13



Contd..

k+1 k k+1 ( 3.49)q =Aq +HF

in which

i 4

i i 4

2in 4 n 4

x C

q = x H= D ( 3.50)

x 1-2ξω D -ω C

m

1 3 2 3 3

1 3 2 3 3

2 23n 1 3 n 2 3

n 3n 1 3 n 2 3

C +kC C +cC mC

A = D +kD D +cD mD ( 3.51)

-mC-ω C +kC -ω C +cC

-2ξ ω mD-2ξω D +k D -2ξω D +cD

2/14

Using expression for these constants, the respons at can be written in recursive form ast k+1



-ξω Δt ξnC =e cosω Δt+ sinω Δt ( 3.39a)1 d d21-ξ

-ξω Δt 1nC =e sinω Δt ( 3.39b)2 dωd

2-ξω Δt1 2ξ 2ξ 1-2ξ ξnC = +e - 1+ cosω Δt+ - sinω Δt3 d dk ω Δt ω Δt ω Δt 2n n 1-ξd

( 3.40)

2-ξω Δt1 2ξ 2ξ 2ξ -1nC = 1- +e cosω Δt+ sinω Δt ( 3.41)4 d dk ω Δt ω Δt ω Δtn n d

Contd..2/15



ω-ξω Δ t nnD =e - sinω Δt ( 3.42)1 d21-ξ

-ξω Δt ξnD =e cosω Δt- sinω Δt ( 3.43)2 d d21-ξ

ω-ξω Δt1 1 1 ξnnD = - +e cosω Δt+ + sinω Δt (3 d dk Δt Δt 2 21-ξ Δt 1-ξ

3.44)

-ξω Δt1 ξnD = 1-e cos ω Δt+ sinω Δt ( 3.45)4 d dkΔt 21-ξ

Contd..2/16



Contd..

Newmark’s - method:

With known displacement, velocity & acceleration at kth time, it calculates the corresponding quantities at k+1th time; Fk+1 is known.

Two relationships are used for this purpose; they mean that within time interval , the displacement is assumed to vary quadratically.

Substituting these relationships in the equation of motion & performing algebraic manipulation, following recursive relationship is obtained.

t

k+1 k k k+1

2 2

k+1 k k k k+1

x =x + 1-δ x Δt+x δΔt ( 3.52)

1x =x +x Δt+ -β Δt x +β Δt x ( 3.53)

2

3/1



Contd..

k+1 N k N k+1 ( 3.66)q =F q +H F

2

i

i i N

i

x β Δt1

q = x H = δΔt ( 3.67)mα

x 1

2 2 3 2 22 2n n n

22 2N n n n

2 2n n n

1α-ω α Δt Δt-2ξω β Δt -ω β Δt α Δt -β α+γ Δt

21

F = -ω δΔt α-2ξω δΔt-ω δ Δt αΔt-δ α+γ Δtα

-ω -2ξω -ω Δt -γ

( 3.68)

in which

3/2

221 2 n nt t



With known responses at kth time step, responses at k+1 th time step are obtained.

State space solution in time domain

0

0

k+1

k+1

k

g

gg

tA t-t At -As

0 gt

tAtAΔt -As

k+1 k gt

Z=AZ+f ( 3.69)

0 1 0 xA = f = Z= ( 3.70)k c -x x- -

m m

Z t =e Z t +e e f s ds ( 3.71)

Z =e Z +e e f s ds ( 3.72)

3/3



Eqn (3.73) is preferred since it does not require inversion.

Once Z is known, displacement and velocity are known.

Second order differential equation is solved to find the acceleration .

Contd..

AΔtAΔtZ =e Z +Δte f ( 3.73)k+1 k gk

AΔt -1 AΔtZ =e Z +A e -I f ( 3.74)k+1 k gk

At λt -1e =φe φ ( 3.75)

The integration in Eqn (3.72) can be performed in two ways.

3/4



It obtains steady state solution of the equation of motion & hence, not strictly valid for short duration excitations like, earthquake.

However, under many cases (as mentioned before) a good estimate of rms & peak values of response may be obtained.

For obtaining the response, ground motion is Fourier synthesized and responses are obtained by the use of the pair of Fourier integral (discussed before) .

Frequency domain analysis

α-iωt

g g-α

αiωt

g g-α

1x iω = x t e dt ( 3.76)

2π

x t = x iω e dω ( 3.77)

3/5



The FFT algorithm now available solves the integral using their discrete forms.

For linear systems, the well known response equation in complex frequency domain is used:

For earthquake excitation, the response due to the jth frequency component of excitation is given by

Contd..

N-1-i 2πkr N

gk grr=0

N-1i 2πkr N

gr gkk=0

1x = x e ( 3.78)

N

x = x e ( 3.79)

x( iω) = h( iω) p( iω)

jiω tj j gjx t =h iω x iω e ( 3.80)

3/6



Total response is given by

N/2

jj=0

x( t) = x( t)

Contd..

in which -1

2 2n j n jj

h iω = ω -ω +2iξω ω

The following steps may be used for programming the solution procedure.

• Sample at an interval of (N).

• Input in FFT.

• Consider first N/2+1 values of the output

• Obtain

( )gx t t

( 0... 1)grx r N

e

N

T t

j j

Nh( iω) ; ω =0... Δω

2

3/7



Contd..

• Obtain .

• Add to make input for IFFT.

• IFFT of gives .

j j gjx( iω)=h( iω) x( iω)

*jx( iω)

*jx( iω) j=0...N-1 ( )x t

Since both frequency & time domain ( Duhamel integration) can be used for the solution, there exists a relation between and .

This relationships forms Fourier transforms pair of integral.

)(h )(h

α-iωt

-α

αiωt

-α

h iω = h t e dt ( 3.83)

1h t = h iω e dt ( 3.84)

2π

3/8



Contd..Example 3.7:Single bay portal frame in Fig 3.11 is subjected to Elcentro earthquake. Find with

Solution: For Duhamel integral

n

d

Δt=0.02s

ω =12.24 rad/s

ω =12.23rad/s

0.0150 0.0312 0.0257 0.0098

0.5980 0.9696 0.0146 0.0060

1.5153 3.4804 0.0436 1.0960

A H

n n

0.9854 0.0196 0.0001 0.0001

F = -1.4601 0.9589 0.0097 H = 0.0097

-146.0108 -4.1124 -0.0265 0.9735

3/9

m m ,EI L u

,EI L

60o

k L3

10 rad/seckm

gx 60o

,EIL

For New Mark’s method

( )x t(0) 0; (0) 0; 0.05x x

Fig 3.11



Contd..

For state space solution

AΔt

0 1 -0.0161-0.16045i -0.0161+0.16045iA = φ =

-37.50 -1.2247 0.9869 0.9869

0.9926 0.0197e =

-0.7390 0.9684

Responses for first few time steps are given in Table 3.1 and 3.2.(in the book)

Time history of responses are shown in Fig.3.12

3/10



0 4 8 12 16 20 24 28 30-0.1

-0.05

0

0.05

0.1

Time (sec)

Dis

plac

emen

t (m

)

0 4 8 12 16 20 24 28 30-0.1

-0.05

0

0.05

0.1

Time (sec)

Dis

plac

emen

t (m

)

Newmark’s--method

Frequency domain FFT analysis

Fig3.12

Contd..3/11



Response of MDOF system Two types of analysis are possible: direct & modal analysis.

For direct analysis damping matrix is generated using Rayleigh damping,

For modal analysis, mode shapes & frequencies are used & modal damping is assumed the same for all modes.

Both time & frequency domain analyses using second order & state space equations can be carried out.

Frequency domain analysis uses FFT algorithm.

C = αM + βK

4/1



Direct analysis in time domain

Equation of motion takes the form.

The same two equations as used in SDOF are used by replacing x by vector X

Substituting those two equations in equation (3.85), the solution is put finally in the recursive form

MX +CX +KX =-MrX ( 3.85)k+1 k+1 k+1 gk+1

k+1 k k k+1

2 2

k+1 k k k k+1

x =x + 1-δ x Δt+x δΔt ( 3.86)

1x =x +x Δt+ -β Δt x +β Δt x ( 3.87)

2

k+1 N k N gk+1Q =F Q +H X 3.92

4/2Contd..



Contd..

2 2 2 2 2

2

N

2

S Δt Δt Δt QΔt SΔt ΔtI- IΔt- Q+S Δt I - +

4 4 4 2 4 4

S Δt Δt Δt QΔt SΔt ΔtF = - I- Q+S Δt I - + ( 3.93)

2 2 2 2 4 2

QΔt SΔt-S - Q+S Δt - +

2 4

2

-1 -1 -1N

Δt-T

4Δt

S =G K Q =G C T =G Mr H = -T ( 3.94)2

-T

4/3

2G M C t K t



Frequency domain analysis using FFT

Using the steps mentioned before for SDOF X(t) is obtained using FFT & IFFT

Note that the method requires inversion of a complex matrix Eq.(3.98)

Contd..

....

......

T

g 1 2 n g

T

g g 1 11 g1 12 g2 13 g3 2 21 g1 22 g2 23 g3

j j gj

-12j j j

P =- m m m x ( 3.95)

P =-M rX =- m r x +r x +r x ,m r x +r x +r x , ( 3.96)

X iω =H iω P i ω ( 3.97)

H iω = K -M ω +i C ω ( 3.98)

4/4



Contd..

Example 3.8 : For the portal frame shown in Fig 3.7 , find the displacements by Newmark’s method:time delay = 5 s; k/m = 100; = 5%; duration = 40s

• : Last 10s of record have zero values

• : First 5s & last 5s have zero values

• : First 10s have zero values

1gx

2gx

3gx

m

k k k 1u1

2u2

2k 2k 2k

2m

Fig3.7

4/5

Solution:

xg2 xg3xg1

.... ..u3 u4

u5



Equation of motion can be written as:

1 212.25 / ; 24.5 / ; 0.816; 0.0027; 0.02rad s rad s t s

Contd..

1

2

3

1 1 1

2 2 2

1 0 1 0 3 3 3 3

0 2 0 2 3 9 3 9

1 0 1 1 1

0 2 1 1 13

g

g

g

x x xm m k k

x x x

xm

x

x

0.9712 0.0272 0.0193 0.0006 0.0001 0.0

0.0132 0.9581 0.0003 0.0192 0.0 0.0001

2.8171 2.7143 0.9281 0.0611 0.0096 0.0003

1.3572 4.1751 0.0302 0.8973 0.0002 0.0094

281.7651 271.4872 7.1831 6.1402 0.0432 0.0343

135.7433 417.5

n

F

091 3.0701 10.2531 0.0171 0.0602

4/6



0 0 0T

x

Contd..

0.0 0.0 0.0

0.0 0.0 0.0

0.0033 0.0033 0.0033;

0.0032 0.0032 0.0032

0.3301 0.3301 0.3301

0.3182 0.3182 0.3182

n

H

Using recursive Eqn. (3.92), relative displacement, velocity and accelerations are obtained.

Time histories of displacements, (u1 and u2) are shown in Fig 3.13

4/7



Contd..

0 5 10 15 20 25 30 35 40-0.04

-0.02

0

0.02

0.04

Time (sec)

Dis

plac

emen

t (m

)

0 5 10 15 20 25 30 35 40-0.02

-0.01

0

0.01

0.02

Time (sec)

Dis

plac

emen

t (m

)

Displacement u1

Displacement u2

Fig3.13

4/8



Example 3.9: For the pitched roof portal frame shownin Fig 3.8, find displacements 4 & 5 for zero & 5s time delay between the supports.

Contd..

1

A E

1

Unit displacement given at A

Unit displacement given at E

Fig3.8

1/ 2

32 / ; 5%

EIrad s

mL

1 25.58 / ; 18.91 / ; 0.4311; 0.004; 0.02rad s rad s t

4/9

Solution:



For the first case , duration is 30s and excitation are same at all supports.

For the second case, duration = 35 s and excitations are different at different supports.

Contd..

3

2.50 1.67;

1.67 2.50

16 10.50

10.50 129

0.2926 0.4074

0.654 0.139

m

EI

L

M

K

r

Time histories of displacements are shown for the two cases in Fig3.14

4/10



4/11

0 4 8 12 16 20 24 28 32 36-0.1

-0.05

0

0.05

0.1

Time (sec)

Dis

plac

emen

t (m

)

0 4 8 12 16 20 24 28 32 36-2

-1

0

1

2 x 10-3

Time (sec)

Dis

plac

emen

t (m

)

Without time delay for the d.o.f. 4

Without time delay for the d.o.f. 5

Contd..



Contd..

0 5 10 15 20 25 30 35 40-0.05

0

0.05

Time (sec)

Dis

plac

emen

t (m

)

0 5 10 15 20 25 30 35 40-4

-2

0

2

4x 10

-3

Time (sec)

Dis

plac

emen

t (m

)

With time delay for the d.o.f. 4

With time delay for the d.o.f. 5

Fig3.14

4/12



The excitation vector fg is of size 2nX1 (single point)

The excitation vector fg is of size 2nX1 (multi point excitation).

The time domain analysis is performed in the same way as for SDOF system.

In frequency domain, state space solution can be performed as given below

g gp rx

0(3.99)g gx

fr

0(3.100)g

g

f

p

5/1

Contd..State Space Direct Analysis



Contd..

(3.101)g z Az f

in which

1 1

0; ; (3.102)g g g

g

xand

x

0

I

A f z p rxpKM CM

By using FFT of , jth component of is obtained.

jth frequency component of response is given by:

)( if gjgf

1

(3.103)

(3.104)

j j gj

j

i i i

i

z H f

H I A

By using IFFT, may be obtained (as before)( )z t

5/2



Contd..

Example 3.9: Find the displacement, responses corresponding to d.o.f 1,2 and 3 using frequency domain ordinary & state space solutions for the beam shown in Fig 3.15

3

316; 48 ; 0.6 ; 2%s s

EIk m C m

mL

5/3

A pipeline supported on soft soil (Exmp. 3.10)

(1) (2) (3)

2m m (4) 2m

sk sc gx

L L (5) (6) (7)

Wave propagation

sk sc gx

sc sk

gx



0.0 0.0 0.0 1.0 0.0 0.0 0

0.0 0.0 0.0 0.0 1.0 0.0 0

0.0 0.0 0.0 0.0 0.0 1.0 0

112.0 16.0 16.0 1.622 0.035 0.035 1

32.0 80.0 32.0 0.070 0.952 0.070 1

16.0 16.0 112.0 0.035 0.035 1.622 1

gm

A f gx

Contd..

For solution of second order differential equation, FFT & IFFT are used as before.

5/4

1 2 3

56 16 8 0.813 0.035 0.017

16 80 16 0.035 0.952 0.035

8 16 56 0.017 0.035 0.813

8.1 / , 9.8 / , 12.2 /

0.1761 0.0022

m m

rad s rad s rad s

K C

Solution:



0 5 10 15 20 25 30-0.1

-0.05

0

0.05

0.1

Time (sec)

Dis

plac

emen

t (m

)

0 5 10 15 20 25 30-0.2

-0.1

0

0.1

0.2

Time (sec)

Dis

pla

cem

ent

(m)

Displacement for d. o. f. 1


Contd..

Fig3.16

5/5

Time histories of displacements are shown in Fig 3.16 - 3.17



Fig3.17

15 20 25 30 35 40 45-0.1

-0.05

0

0.05

0.1

Time (sec)

Dis

plac

emen

t (m

)

15 20 25 30 35 40 45-0.2

-0.1

0

0.1

0.2

Time (sec)

Dis

plac

emen

t (m

)



Contd..5/6



Excitation load vector Pg is of the form

Generally is set to zero (in most cases)

The solution procedure remains the same.

and are of the order of n x s; s is the number of supports.

Solution requires time histories of and .

Solutions can be obtained both in time and frequency domains.

Solution for absolute displacements

(3.105)g sg g sg g P K x C x

sgCsgk

sgC

gx gx

5/7



Modal analysis In the modal analysis, equation of motion is decoupled into a set of N uncoupled equations of motion .

Normal mode theory stipulates that the response is a weighted summation of its undamped mode shapes.

....

Ti i i i i i i g

T 2 ii i i i i i i i

i

s2

i i i i i ik gkk=1

Ti k

ik Ti i

X =φz ( 3.106)

mz +cz +kz =-φ Mrx ( 3.111)

kk =φ Kφ ω = c =2ξωm ( 3.112)

m

z +2ξωz +ω z =- λ x i=1 m ( 3.113)

φ Mrλ = ( 3.114)

φ Mφ

5/8



For s=1, equation 3.113 represents the equation for single point excitation.

Each SDOF system can be solved in time or frequency domain as describes before.

Example 3.11: For the cable stayed bridge shown before , find the displacement responses of d.o.f 1,2,3 for Elcentro earthquake; 5s time delay is assumed between supports.

Contd..

684 0 149 20 0 0

0 684 149 0 20 0

149 149 575 0 0 60

m m

K M

5/9

Solution:



Contd..

1 2 3

T1

T2

T3

-0.781 -0.003 0.002 -0.218

r=- -0.218 0.002 -0.003 -0.781

-0.147 -0.0009 0.0009 0.147

ω =2.86rad/s ω =5.85rad/s ω =5.97rad/s

φ = -0.036 0.036 -0.125

φ = 0.158 0.158 0

φ = -0.154 0.154 0.030

First modal equation

g1

Tg22 1

1 1 1 1 1 Tg31 1

g4

x

xφ Mrz +2ηω z +ω z =-

xφ Mφx

5/10



1

2

13

4

1.474 0.008 0.0061 1.474

g

g

gg

g

x

xp

x

x

0.025t

• will have first 30s as the actual record & the last 15s as zeros.

• will have first 10s as zeroes followed by 30s of’ actual record & the last 5s as zeros.

• Time histories of generalized loads are shown in Fig3.18.

• Time histories of generalized displacement are show in Fig 3.19

1gx

3gx

Contd..5/11



0 5 10 15 20 25 30 35-1

-0.5

0

0.5

1

Time (sec) Seco

nd g

ener

aliz

ed f

orce

(g)

0 5 10 15 20 25 30 35-1

-0.5

0

0.5

1

Time (sec)

Fir

st g

ener

aliz

ed f

orce

(g)

First generalized force

Second generalized force (pg2)

Fig3.18

Contd..5/12



Contd..

0 5 10 15 20 25 30 35 40 45-0.04

-0.02

0

0.02

0.04

Time (sec)

Dis

plac

emen

t (m

)

0 5 10 15 20 25 30 35 40 45-0.04

-0.02

0

0.02

0.04

Time (sec)

Dis

plac

emen

t (m

)

First generalized displacement (Z1)

Second generalized displacement (Z2)

Fig3.19

5/13



Contd..

Solution Z1 Z2 Z3

Time history

rms(m)

peak(m)

rms(m)

peak(m)

rms(m)

peak(m)

0.00910.036

90.0048 0.0261 0.0044 0.0249

Frequency

domain

0.0090.036

80.0049 0.0265 0.0044 0.0250

1z2z3z

5/14

Table 3.4 Comparison of generalized displacement



One index which is used to determine number of modes required is called mass participation factor.

Number of modes m to be considered in the analysis is determined by

i r irr

i

m

M

Contd..

1

1m

ii

5/15

Number of modes depends upon the nature of excitation, dynamic characteristics of structure & response quantity of interest.



The approach provides a good estimate of response quantity with few number of modes.

)(tR

Mode acceleration approach

i i g i i i2 2i i

m mi

i g i i i i2 2i=1 i=1i i

m

i i i i2i=1 i

1 1z = -λx - z +2ξωz ( 3.117)

ω ω

φ 1R( t)=- λx - z +2ξωz φ ( 3.118)

ω ω

1=R t - z +2ξωz φ ( 3.119)

ω

is the quasi static response for which can be proved as below

gMrx

g

T Tg

KX t =-Mrx t ( 3.120)

φ Kφz t =-φ Mrx t ( 3.121)

6/1



The solution is obtained by:

• Find quasi static response for . • Find :

The second quantity is obtained from

1

12

m

i i i iii

z z

)(tR

Contd..

; n ni g i g

i i i i2 2i=1 i=1i i

-λx λxz = R t = φ z =- φ ( 3.122,123)

ω ω

gMrx

i g

i i i i2 2i i

-λx1z +2ξωz = -z ( 3.124)

ω ω

6/2

The contribution of the second part of the solution from higher modes is small; first part contributing maximum to the response consists of contribution from all the modes.



Computation of internal forces Determination of internal forces may be obtained in two ways:

Using the known displacements and member properties.

Using the mode shape coefficients of the internal forces.

For the latter , any response quantity of interest is obtained as a weighted summation of mode shapes; the mode shape coefficients correspond to those of the response quantities of interest.

The method is applicable where modal analysis is possible;number of modes to be considered depends upon the response quantity of interest.

6/3



Contd..6/4

These coefficients are obtained by solving the MDOF system for a static load given below & finding the response quantity of interest.

For the first approach member stiffness matrix is multiplied by displacement vector in member co-ordinate system.

Rotations which are condensed out are regenerated from the condensation relationships.

The second approach is widely used in softwarewhich deals with the solutions of framed structurein general.

2 ; 1 (3.125)i i i i m P M



z(t) is expressed as a weighted summation of mode shapes.

Equation of motion can be then written as

2n uncoupled equations are then obtained as

The initial condition is obtained from

State space analysis

Z t =φq ( 3.126)

g

-1 -1 -1g

φq=Aφq+f ( 3.127)

φ φq=φ Aφq+φ f ( 3.128)

i i i giq =λq +fi =1.......2n ( 3.129)

-10 0q =φ Z ( 3.130)

6/5



In frequency domain, Eq 3.129 is solved using the same FFT approach with

Contd..

1

1 (3.131)jh i

Example 3.12: For the frame shown in Fig3.20, find the base shear for the right column; k/m = 100; 5%

k k

k k

2k 2k

2k 2k

gx

4u

3u

2u

1u m

2m

2m

2m

Fig3.20

Solution:

2 2 0 0

2 4 2 0

0 2 6 4

0 0 4 8

k

K

1 0 0 0

0 2 0 0

0 0 2 0

0 0 0 2

m

M

1 25.06 rad/s; 12.57 rad/s

3 418.65rad/s; 23.85rad/s

6/6



1.0 0.871 0.520 0.278

1.0 0.210 0.911 0.752

1.0 0.738 0.090 0.347

1.0 0.843 0.268 0.145

T

T

T

T

1

2

3

4

Contd..

1.500 1.140 0.0 0.0

1.140 3.001 1.140 0.0

0.0 1.140 4.141 2.280

0.0 0.0 2.280 5.281

m

C M K

0.0 0.0 0.0 0.0 1.0 0.0 0.0 0.0

0.0 0.0 0.0 0.0 0.0 1.0 0.0 0.0

0.0 0.0 0.0 0.0 0.0 0.0 1.0 0.0

0.0 0.0 0.0 0.0 0.0 0.0 0.0 1.0

2.0 1.0 0.0 0.0 1.454 0.567 0.0 0.0

2.0 2.0 1.0 0.0 1.134 1.495 0.567 0.0

0.0 1.0 3.0 2.0 0.0 0.567 2.062 1.134

0.0 0.0 2.

A

0 4.0 0.0 0.0 1.134 2.630

6/7



Time histories of the base shear obtained by mode simulation, mode acceleration and modal state space analyses are shown in Fig 3.21.

Contd..

1 2

3 4

1.793 1.571 ; 1.793 1.571 ;

1.157 1.461 ; 1.157 1.461

i i

i i

0 5 10 15 20 25 30-0.1

-0.05

0

0.05

0.1

Time (sec)

Sh

ear

(in

ter

ms

of

mas

s m

)

0 5 10 15 20 25 30-0.1

-0.05

0

0.05

0.1

Time (sec)

Sh

ear

(in

ter

ms

of

mas

s m

)

Mode acceleration method

State space method

× 102

× 102

6/8

Fig 3.21



Computational steps for MATLAB programming are given in the book in section 3.5.7 for all types of analyses.

Steps are given in following sections:

Contd..

Computation of basic elements required for all types of analysis.

Time domain analysis covering direct analysis, modal analysis, mode acceleration approach, state space analysis (modal and direct).

Frequency domain analysis covering all cases considered for the time domain analysis

6/9



Lec-1/74



Chapter -4

FREQUENCY DOMAIN SPECTRAL ANALYSIS



Introduction Spectral analysis is a popular method for finding seismic response of structures for ground motions modeled as random process.

Since the analysis is performed in frequency domain it is known as frequency domain spectral analysis.

The analysis requires the knowledge of random vibration analysis which forms a special subject.

However, without the rigors of the theory of random vibration, spectral analysis is developed here.

It requires some simple concepts which will be explained first.

1/1



Stationary random process

x2 (t)

x1 (t)

t1t2

x3 (t)

x4 (t)

1( )x t1( )x t

Fig 4.1

Mean of is 2 2

1 1[{ ( ) ( )} ]X E x t x t Sample

2 21[ ( ) ( )]i

xi i iT x t x t dt Ergodic process 2 2

x xi

1/2



Fourier series & integral Fourier series decomposes any arbitrary function x(t) into Fourier components.

α0

k kk=1

T T2 2

k 0T T

- -2 2

T2

kT

-2

a 2πkt 2πktx( t)= + a cos +b sin ( 4.1)

2 T T

2 2πkt 2a = x( t) cos dt a = x t dt ( 4.2)

T T T

2 2πktb = x( t) sin dt ( 4.3)

T T

1/3

T2

ok k

T2

a Δωx( t)= + x( t) cos( ω t) dt cos( ω t)+

2 πk=1 -

∞

(4.4)

T2

k kT2

x( t) sin( t ) dt sin( t )k=1 -



α α

ω=0 ω=0

α α

-α -α

x( t)=2 A( ω)cos( ωt)dω+2 B( ω)sin( ωt)dω ( 4.7a)

x( t)= A ω cosωtdω+ B ω sin ωtdω ( 4.7b)

Contd.

The complex harmonic function is introduced to define the pair of Fourier integral.

α

iωt

-α

( 4.10)x( t)= x( ω) e dω

, 2 /T T d . It can be shown that (book)

1/4

α

-iωt

-α

1x( ω)=A( ω)-i B( ω)= x( t) e dt ( 4.9)

2π



Contd.

Discrete form of Fourier integral is given by

2πkrN-1 -iN

k rr=0

2πkrN-1 iN

r kk=0

1x( ω)= x e ( 4.12)

N

x( t)= x e ( 4.13)

FFT & IFFT are based on DFT.

From , Fourier amplitude is obtained. xk Ak

2 2k k k

0 0

NA =2 c +d k=1...... ( 4.14)

2A =c ( 4.15)

For the Fourier integral to be strictly valid

α

-α

x( t)dt<α ( 4.11)

1/5



In MATLAB, , is divided by N/2 (not N), then

Parsavel’s theorem is useful for finding mean square value

Note equation 4.18 pertains to Eq 4.1 & Eq 4.19 pertains to Eqs 4.12 – 4.13.

(X(t) is divided by N not N/2 as in MATLAB)

rx

Contd.

2 2k k k

00

A = c +d ( 4.16)

cA = ( 4.17)

2

T 22 2 20

k k0

T N-1 N-122 2

r kr=0 K=00

a1 1x( t) dt= + ( a +b ) ( 4.18)

T 4 2

1 1x( t) dt= x = x ( 4.19)

T N

1/6



Correlation functions Random values taken across the ensemble are different than those would take although and are the same.

How these two sets of random variables are different is denoted by auto correlation function

Obviously, mean square value of the process & the two sets are perfectly correlated.

Similarly, cross correlation between two random process is given by

1x( t )2 1x( t =t + τ)

21E [x( t ) ] 2

2E [x( t ) ]

XXR (τ=0)=

yx xyxy 1 1 ( )R ( τ)=E x( t )y( t +τ) ; R τ =R ( -τ) ( 4.21)

1/7

xx 1 1R ( τ)=E[x( t )x( t +τ) ] ( 4.20)



E [x2] = 2 + m2

xR

2m

o 2 + m2

0 o

xyR x y + mx my

mx my

x y + mx my

Fig 4.2

Fig 4.3

Contd.1/8



PSDFs & correlation functions Correlation functions & power spectral density functions form Fourier transform pairs

From equation 4.23, It follows

∞-iwτ

xx xx-∞

∞-iwτ

xy xy-∞

∞iwτ

xx xx-∞

∞iwτ

xy xy-∞

1S ( ω)= R ( τ) e dτ ( 4.22)

2π

1S ( ω)= R ( τ) e dτ ( 4.23)

2π

R ( τ)= S ( ω) e d ( 4.24)

R ( τ)= S ( ω) e d ( 4.25)

yx xyS ( ω)=comp.conjugS ( ω)

1/9



An indirect proof of the relationships may be given as

Contd.

∞2 2

xx x xx-∞

α

xy xy-α

R ( 0)=r = S ( ω) dω=E x ( 4.26a)

R ( 0)= S ( ω) dω=E xy ( 4.26b)

Eq. 4.26a provides a physical meaning of PSDF; distribution of mean square value of the process with frequency.

PSDF forms an ideal input for frequency domain analysis of structures for two reasons:

1/10



A second order random process is uniquely defined by its mean square value.

Distribution of mean square value with frequency helps in ascertaining the contribution of each frequency content to the overall response.

Contd.

S

d Fig 4.4

1/11



1/12Contd

Fig 4.5(b)Fig 4.5(a)

kS

2 T

k

S = | x | at kth frequency

k

k

S

Compactedordinates

S at closer frequencies

S

kkth ordinate of the power spectral density function

Fig 4.5(c)



Concept of PSDF becomes simple if ergodicity is assumed;for a single sample, mean square value

For large T , ordinates become more packed; kth ordinate is divided by ; sum of areas will result in variance; smooth curve passing through points is the PSDF.

This definition is useful in the development of spectral analysis technique for single point excitation and widely used in the random vibration analysis of structures.

d

Contd.

T 2 N-1

22 2 2 20x k k k

k=00

a1 1r = x( t) dt= + a +b = x ( 4.27a)

T 4 2

1/13



It is difficult to attach a physical significance to cross PSDF; however some physical significance can be understood from the problem below.

For different values of ǿ, degree of correlation varies; the responses will be different.

Contd.

1P1 = A sin t P2 = A sin (t + )

For random excitations & ,the response will be obviously different depending upon the degree of correlation & hence cross PSDF.

1p( t) 2p( t)

1/14



PSDF matrix is involved when more than one random processes are involved

PSDF matrix

1 1 2 2

1 2 2 1

1 1 2 2 1 2

11 1 2 2 1 2

2

2 2 2 2 21 1 2 2 1 2 1 2 2 1 2 1

2 21 2

1 2 2 1

2 21 2 1 2

[ ] (4.28)

(4.30)

yy x x x x

x x x x

yy x x x x x x

xy a x a x a a

x

E y E a x a x a a x x a a x x

S d a S d a S d

a a S d a a S d

S d a S a S a a S

2 1

1 1 2 2 1 2 2 1

2 1

1 1 1 2 12 21 2 1 2 2 1 1 2

2 1 2 2 2

yy

(4.31)

(4.32 )

S (4.32 )

x x

x x x xyy x x x x x x x x

x x x x

a a S d

S S aS a S a S a a S a a S a a a

S S a

b

TxxaS a

2/1



For single variable

Eq (4.32b) can be extended to establish two stochastic vectors

Contd.

2y xS =a S ( 4.33)

1 1 2 2 1 2 2 1

n×mn×1 m×1

Tyy xx

n×m 1 n×r 2m×1 r×1

T T T Tyy x x x x x x x x

xy xx

y t =A x t ( 4.34)

S =AS A ( 4.35)

y t =A x t +B x t ( 4.36)

S =AS A +BS B +AS B +BS A ( 4.37)

S =AS ( 4.38) PSDF of the derivatives of the process is required in many cases. It can be shown (book):

2x x

2 4x x x

S =ω S ( 4.48)

S =ω S =ω S ( 4.49)

2/2



It is assumed that p(t) is an ergodic process; then in frequency domain

From (4.51a), it is possible to write

Using Eq(4.19), mean square value may be written as

SISO

x(� ω)=h( ω) p( ω) ( 4.51a)

2 2 -1n n

g

h( ω)=( ω -ω +2iξω ω) ( 4.51c)

p( ω)=-x( ω) ( 4.51d)

* * *

2 2 2

x ω x ω =h ω h ω p ω p ω ( 4.53a)

x ω =h ω p ω ( 4.53b)

T N-1 N-1222 2

r Kr=0 K=00

T2 22

0

1 1x( t) dt= x = x = x( ω) ( 4.54)

T N

1x( t) dt= h( ω) p( ω) ( 4.55)

T

2/3



T

Contd.

T

∞ 22 2x p0

0

1x( t) dt=r = h( ω) S dω ( 4.56)

T

ω ω

2

x p0 0

2

x p

*x p

S( ω) dω= h( ω) S dω ( 4.57)

S( ω)=h( ω) S ( 4.58)

S( ω)=h( ω)S h( ω) ( 4.59)

2/4



2x x

xx x xx x

2

4x x

2 2xx x xx x

2x x

4x x

T

xx x xx x

T2 2xx x xx x

x ω =iωx ω ( 4.60)

S =ω S ( 4.61)

S =iωS S =-iωS ( 4.62)

x ω =-ω x ω ( 4.63)

S =ω S ( 4.64)

S =-ω S S =-ω S ( 4.65)

S =ω S ( 4.66)

S =ω S ( 4.67)

S =iωS ; S = iωS ( 4.68)

S =-ω S ; S = -ω S ( 4.69a)

xx xxS +S =0 ( 4.69b)

Contd.2/5



Example 4.1: For the problem in example3.7, findthe rms value of the displacement;

Solution: Digitized values of PSDF of Elcentro are given in Appendix 4A(book)

can be obtained using above equations.

0ω =12.24 rad/s ; Δω = 0.209 rad/s

Contd.

2 2 -1n n

2

x p

h( ω)=( ω -ω +2iξω ω) ( 4.51c)

S( ω)=h( ω) S ( 4.58)

Xh( w)& S

2/6



Contd. PSDF of Elcentro earthquake, absolute square of and PSDF of displacements are shown in the Figs 4.8-4.10 .

h( ω)

0 20 40 60 80 100 120 140 1600

0.005

0.01

0.015

0.02

Frequency (rad/sec)

PS

DF

of

acce

lera

tion

(m

2 sec–

3 /ra

d )

Fig 4.8

2/7



Contd.

0 4 8 12 16 20 240

2

4

6

8 x 10-4

Frequency (rad/sec)

|h(w

)|2

0 5 10 15 20 250

2

4

6

Frequency (rad/sec)

PS

DF

of

dis

plac

emen

t (m

2 sec

/rad

8 x 10-4

Fig 4.10

Fig 4.9

2/8



Single point excitation, P(t) is given by

Using Eqns.4.35 & 4.71, following equation can be written

Using Eqns.4.35 & 4.72

gP( t) = -MIx ( 4.70)

x ω =H ω P ω ( 4.71)

MDOF system

*T

xx ppS =H ω S H ω ( 4.72)

(4.74)gT T

pp xS =MII M S

2/9

(4.75)gT T *T

xx xS =HM II M H S

g gx xS = - HMIS (4.76)x



Example 4.2 : For example 3.8, find PSDFs of the displacement & for the same excitations at all supports. Solution:

1u 2u

Contd.

2 -1

T

1 2

1 0M= m

0 2

1 0 3 -3C =0.816 m+0.0027 k

0 2 -3 9

3 -3K = k

-3 9

H=[K -Mω +iCω]

k=100; I ={1 1}

mω =12.25rad/s; ω =24.49rad/s

2/10



Using Eqn.4.75, is obtained; PSDFs of & are shown in Fig.4.11

XXS

Contd.

1u 2u

0 5 10 15 20 2500.20.40.60.80.11.21.4

Frequency (rad/sec)

PS

DF

of

dis

pla

cem

ent

(m2 s

ec/r

ad)

(10-4

)

0 5 10 15 20 250

0.5

1

1.5

2

2.5

3

3.5

Frequency (rad/sec)

-

For u1

For u2Fig 4.11

2/11

u1

u2

σ =0.0154m

σ =0.0078mP

SD

F o

f d

isp

lace

men

t(m

2 sec

/rad

) (1

0-5)



Contd.

For multipoint excitation:

g g

g g

T T *Tx x x

x x x

S =HMrS r M H ( 4.77)

S =-HMrS ( 4.78)

gxS is of size s x s; r is of size n x s

If all displacements are not required, then a reduced is used of size m x n & is given by H( ω) XXS

gT T *T

xx m×n x n×mS =H MrS r M H (4.79)

Without the assumption of ergodicity, Eqns 4.75- 4.78 can be derived (1-3).

3/1



Contd.Example4.2: (a) For example 4.2, if a time lag of 5s is introduced between supports, find the PSDFs of u1 & u2 ; b) For example 3.9, find PSDFs of the degrees of freedom 4 & 5 for correlated and partially correlated excitations (with time lag 5s).

3/2

Solution: For example 4.2

ij

s

1 2

xg 1 1 xg 1 2

2 1

1 1 11r=

1 1 13

r ωcoh( i, j)=exp -

2πv

1 ρ ρ5ω 10ω

S = ρ 1 ρ S ρ =exp - ρ =exp -2π 2π

ρ ρ 1

PSDFs & cross PSDFs are shown in Fig4.12 (a-d)



Contd.

0 5 10 15 20 250

1

2

3

4

5x 10-5

Frequency (rad/sec)

0 5 10 15 20 25 300

0.2

0.4

0.6

0.8

1

1.2 x 10-5

Frequency (rad/sec)

Displacement u1

Displacement u2

Fig 4.12

3/3

Rms values of u1 & u2 are 0.0089m (0.0092m) & 0.0045m (0.0048m)

PS

DF

of

dis

pla

cem

ent

(m2 s

ec/r

ad)

PS

DF

of

dis

pla

cem

ent

(m2 s

ec/r

ad)



3/4

0 5 10 15 20 250

1

2

3

4

5 x 10-5

Frequency (rad/sec)

0 5 10 15 20 25 300

0.2

0.4

0.6

0.8

1

1.2x 10

-5

Frequency (rad/sec)

Displacement u1

Displacement u2

Contd.P

SD

F o

f d

isp

lace

men

t (m

2 sec

/rad

)

PS

DF

of

dis

pla

cem

ent

(m2 s

ec/r

ad)



For the problem of example 3.9, required matrices & values of other parameters are given below:

3/5

1 5.58 rad/s 2 18.91rad/s

3

19.56 10.5EIK =

10.5 129L3

EI=2

mLα=0.431

β=0.004

0.479 0.331r=

-0.131 0.146C =αM+βK

2.5 1.67M= m

1.67 2.5

Using Eq 4.77, PSDFs are obtained & are shown in Figs.4.13 & 4.14.



Contd. Rms values for d.o.f 4 & 5 are

0.0237( corr.) & 0.0168( partially corr.)

0.0005( corr.) & 0.0008( partially corr.)

Fig 4.13

0 2 4 6 8 10 12 14 16 18 200

2

4

6

8x 10

-4

Frequency (rad/sec)

PS

DF o

f d

isp

lacem

en

t (

m2 s

ec/r

ad

)

0 5 10 15 20 25 300

0.5

1

1.5

2

3

3.5x 10-4

Frequency (rad/sec)

PS

DF o

f d

isp

lacem

en

t (m

2 s

ec/r

ad

)Without time lag

With time lag

3/6



Contd.

0 2 4 6 8 10 12 14 16 18 200

1

2

3

4x 10

-7

Frequency (rad/sec)

PS

DF o

f d

isp

lacem

en

t (m

2 s

ec/r

ad

)

0 5 10 15 20 25 300

0.5

1

1.5

2x 10

-7

Frequency (rad/sec)

PS

DF o

f d

isp

lacem

en

t (m

2 s

ec/r

ad

)

Without time lag

With time lag Fig 4.14

3/7



PSDFs of absolute displacements Absolute displacement is obtained by adding ground displacement to the relative displacement

PSDF of is obtained using Eqn. 4.37

Example 4.4: For example 4.3, find the PSDFs of absolute displacement of d.o.f 4 & 5. Solution: Let & represent

x =Ix+rx ( 4.80)a g

ax

a g g g

g g g g

T T Tx xx x xx x x

2g g

2 Tx x x xx x x

S =S +rS r +IS r +rS I ( 4.81)

x ω =-HMrx ω =HMrω x ω ( 4.82)

S =HMrω S and S =S ( 4.83)

x gx

T T4 5 g g1 g2x = x x x = x x

3/8



Contd. Using Eqn. 4.83

Using Eqn. 4.81, is obtained.

The PSDFs of & are shown in Figs 4.15(a-b). rms values are given as 0.052m & 0.015m respectively.

g g g

g g

g1 4 g2 4

g

g1 5 g2 5

2 -2x x x x

11 21x x ij

12 22

x x x x

x xx x x x

S =HMrω S =HMrω S ( 4.84)

c cS = S c =coh( i, j) ( 4.85)

c c

S SS = ( 4.86)

S S

xaS

4x aS 5x aS

3/9



Contd.

0 2 4 6 8 10 12 14 16 18 200

0.2

0.4

0.6

0.8

1

1.2x 10-3

Frequency (rad/sec)

0 2 4 6 8 10 12 14 16 18 200

2

4

6

8 x 10-5

Frequency (rad/sec)

For d. o. f. 4

For d. o. f. 5Fig 4.15

3/10

PS

DF

of

dis

pla

cem

ent

(m2 s

ec/r

ad)

PS

DF

of

dis

pla

cem

ent

(m2 s

ec/r

ad)



PSDF of member forces Consider the frame in Fig 3.7; x is the vector of dyn. d.o.f. & θ as that of the condensed out d.o.f.

Consider column i-j; displacement at the ends of column are

PSDF matrix of the member end forces are

Tθθ xx

Txθ xx θx xθ

θ=Ax ( 4.87)

S =AS A ( 4.88)

S =AS & S =S ( 4.89)

T

i i j jδ= xθxθ ( 4.90)

f =Kδ ( 4.91)

Tff δδS =KS K ( 4.92)

4/1



Contd. If are in local coordinates, then

Example 4.5 : For the example 3.10, find PSDFs of displacemens1 & 2; also find the PSDF of bending moment at the centre.Solution:

T T Tff δδδδ

δ=Tδ ( 4.93)

S =KS K =KTS T K ( 4.94)

1 2 3

56 -16 8 0.813 -0.035 0.017

K = -16 80 -16 m C = -0.035 0.952 -0.035 m

8 -16 56 0.017 -0.035 0.811

ω =8.0rad/s; ω =9.8rad/s and ω =12.0rad/s

4/2



Following the same procedure , PSDFs of 1& 2 are obtained and are shown in Fig4.16(a-b).

For finding PSDF of bending moment, at the center is required

& are zero as is zero at the center.

Displacement vector is

Contd.

T1 2 3x = x x x

3θ= -1 0 1 x=Ax

4L

θθSxθS θ

δ

T21 1 2δ= x ,θ ,x ,θ

4/3



Contd.

0 2 4 6 8 10 12 14 16 18 200

2

4

6

8x 10-3

Frequency (rad/sec)

PS

DF o

f d

isp

lacem

en

t (m

2 s

ec/r

ad

)



Fig 4.16 (a-b)

4/4



Contd. Using modified stiffness for pinned end, is eliminated ; are abs.displacements.

PSDF of bending moment obtained using Eqn. 4.92 is shown in Fig 4.16c. rms values of are 0.0637m, 0.1192m & 1.104m .

1θ 1 2x &x

1 2x ,x & B.M.

4/5

0 2 4 6 8 10 12 14 16 18 200

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

Frequency (rad/sec)

PS

DF o

f B

.M.

x m

ass

m('

g'm

)2sec/r

ad

)

Fig 4.16(c)



Modal spectral analysis Advantages of modal analysis have been described before.

For multipoint excitation, the ith modal equation can be written as

r is the influence coefficient matrix of size mxn.

hi(w) for each modal equation can be easily obtained.

2 Ti i i i i i i i i i g

T2 i

i i i i i i g ii

Ti i i

mz +2ξωmz +mω z =-j Mrx ( 4.95)

-j Mrz +2ξωz +ω z = x =p i=1.....r ( 4.96)

m

m =j Mj

4/6



iΖ( ω) ip( ω) can be related to by

Contd.

i i i

Ti g

ii

z ω =h ω p ω i=1.....r ( 4.97a)

-j Mrx ωp ω = ( 4.97b)

m Elements of PSDF matrix of z are given by

i j g

*i j T T T

z z i x ji j

h hS = j MrS r M j i=1....r, j=1....r ( 4.98)

mm

Using modal transformation rule

Txx zz

x=φz

S =φS φ ( 4.99)

φ is m×r

4/7



Contd.Example 4.6: For problem example 3.11,find PSDF of d.o.f 1(tower top) and d.o.f 2(centre of deck). It is assumed that a uniform time lag of 5s between the supports exists.

684 0 149 20 0 0

0 684 149 , 0 20 0

149 149 575 0 0 60

m m

K M

0.781 0.003 0.002 0.218

0.218 0.002 0.003 0.781

0.147 0.009 0.0009 0.147

r

1 2 32.86 rad/s; 5.85rad/s ; 5.97 rad/s

0.036 0.036 0.125

0.158 0.158 0

0.154 0.154 0.030

T

T

T

1

2

3

Solution

4/8



Contd.

rms values of displacement for d.o.f 1 & 2 are:rms modal direct

d.o.f1 0.021m 0.021m

d.o.f2 0.015m 0.015m

value

PSDFs are calculated using equations 4.98-4.99 and shown in Fig 4.17 a-c.

4/9

It is seen that the values obtained by modal and direct analyses are the same because the number of modes taken are equal to the number of d.o.f. (ie all modes are considered).



0 1 2 3 4 5 6 7 8 9 100

1

2

3

4

5

6x 10

-4

Frequency (rad/sec)

PS

DF o

f d

isp

lacem

en

t (m

2 s

ec/r

ad

)

0 5 10 15 20 25 30 35 40 45 50-6

-4

-2

0

2

4x 10

-5

Frequency (rad/sec)

Real Part

Cro

ss P

SD

F b

etw

een

u1 a

nd

u2 (

m2 s

ec/r

ad

)

Fig 4.17b

4/10

Contd.

Fig 4.17a



Contd. 4/11

0 5 10 15 20 25 30 35 40

-1

0

1

2

3x 10

-20

Frequency (rad/sec)

Imaginary Part

Cro

ss P

SD

F b

etw

een

u1 a

nd

u2 (

m2 s

ec/r

ad

)

Fig 4.17c



4/12

0 1 2 3 4 5 6 7 8 9 100

1

x 10-4

Frequency (rad/sec)

PS

DF o

f d

isp

lacem

en

t (m

2 s

ec/r

ad

)

0 1 2 3 4 5 6 7 8 9 100

1

2

3

4

5

6x 10

-4

Frequency (rad/sec)

PS

DF o

f d

isp

lacem

en

t (m

2 s

ec/r

ad

)Top of the left tower

Centre of the deck

Contd.

Fig 4.18



Spectral analysis (state space) When state space equation is used, is obtained as:

g g

*Tzz ff

-1

S =HS H ( 100a)

H= Iω-A ( 100b)

zzS

g gf fS is the PSDF matrix of vector.

contains and terms; addition of these two terms becomes zero.

For modal spectral analysis, eigen values & eigen vector of matrix A are obtained.

Same equations as used before are utilized to find PSDF of responses.

gf

xx xxS +S =0

xxS xxSzzS

4/13



Contd.Example4.7: For the exercise problem 3.12, find the PSDF matrices of top & first floor of displacements; ground motion is perfectly correlated.

0.0 0.0 0.0 0.0 1 0.0 0.0 0.0

0.0 0.0 0.0 0.0 0.0 1 0.0 0.0

0.0 0.0 0.0 0.0 0.0 0.0 1 0.0

0.0 0.0 0.0 0.0 0.0 0.0 0.0 1

2.0 1.0 0.0 0.0 1.454 0.567 0.0 0.0

2.0 2.0 1.0 0.0 1.134 1.495 0.567 0.0

0.0 1.0 3.0 2.0 0.0 0.567 2.062 1.134

0.0 0.0 2.0 4.0 0.0

A

0.0 1.134 2.630

Using eqns 4.100(a-b), PSDFs are calculated and are shown in fig.4.19.

4/14

Solution



Contd. rms values are as below:

rms modal direct statespace

d.o.f1 0.0903m 0.0907m 0.0905m

d.o.f4 0.0263m 0.0259m 0.0264m

value

Steps for developing a program for spectral analysis of structures with multi-support excitations using MATLAB are given.

Steps cover all types of methods ie;modal , direct and state space formulations.

The program can easily make use of the standard MATLAB routines.

4/15



Lec-1/74

seismic analysis of structures - ii

Education

iitresponse analysis

specified ground motioncontd

specified ground motionexample

component ground motioncontd

equations of motion

single componentcase

t tu v u u u u ui

10b14 gmx cx kx