Seismic Analysis of Structures - II
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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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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.. ..
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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)
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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..
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
;
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..
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
-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..
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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.
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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..
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
-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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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..
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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:
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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..
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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:
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
Displacement for d. o. f. 2
Contd..
Fig3.16
5/5
Time histories of displacements are shown in Fig 3.16 - 3.17
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
)
Displacement for d. o. f. 1
Displacement for d. o. f. 2
Contd..5/6
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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:
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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.
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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.
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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Response Analysis for specified ground motion
Lec-1/74
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
Chapter -4
FREQUENCY DOMAIN SPECTRAL ANALYSIS
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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 -
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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)
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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)
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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Response Analysis for specified ground motion
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
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Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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)
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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)
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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)
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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)
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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.
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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)
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
)
Displacement for d. o. f. 1
Displacement for d. o. f. 2
Fig 4.16 (a-b)
4/4
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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)
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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).
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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T.K. Datta Department Of Civil Engineering, IIT Delhi
Response Analysis for specified ground motion
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
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Response Analysis for specified ground motion
Lec-1/74