Chapters – 5 & 6
Chapter -5
RESPONSE SPECTRUM
METHOD OF ANALYSIS
T.K. Datta
Department Of Civil Engineering, IIT Delhi
Response Spectrum Method Of Analysis
1/1
Introduction
Response spectrum method is favoured by
earthquake engineering community because of:
 It provides a technique for performing an
equivalent static lateral load analysis.
 It allows a clear understanding of the
contributions of different modes of vibration.
 It offers a simplified method for finding the
design forces for structural members for
earthquake.
 It is also useful for approximate evaluation
of seismic reliability of structures.
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Department Of Civil Engineering, IIT Delhi
Response Spectrum Method Of Analysis
1/2
Contd…
 The concept of equivalent lateral forces for earthquake is a unique concept because it converts a
dynamic analysis partly to dynamic & partly to
static analysis for finding maximum stresses.
 For seismic design, these maximum stresses are
of interest, not the time history of stress.
 Equivalent lateral force for an earthquake is
defined as a set of lateral force which will
produce the same peak response as that
obtained by dynamic analysis of structures .
 The equivalence is restricted to a single mode of
vibration.
T.K. Datta
Department Of Civil Engineering, IIT Delhi
Response Spectrum Method Of Analysis
1/3
Contd…
 The
response spectrum method of analysis is
developed using the following steps.
 A modal analysis of the structure is carried out
to obtain mode shapes, frequencies & modal
participation factors.
 Using the acceleration response spectrum, an
equivalent static load is derived which will
provide the same maximum response as that
obtained in each mode of vibration.
 Maximum modal responses are combined to
find total maximum response of the structure.
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Department Of Civil Engineering, IIT Delhi
Response Spectrum Method Of Analysis
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Contd…
 The first step is the dynamic analysis while , the
second step is a static analysis.
 The first two steps do not have approximations,
while the third step has some approximations.
 As a result, response spectrum analysis is
called an approximate analysis; but applications
show that it provides mostly a good estimate of
peak responses.
 Method is developed for single point, single
component excitation for classically damped
linear systems. However, with additional
approximations it has been extended for multi
point-multi component excitations & for nonclassically damped systems.
T.K. Datta
Department Of Civil Engineering, IIT Delhi
Response Spectrum Method Of Analysis
1/5
Development of the method
 Equation of motion for MDOF system under
single point excitation
(5.1)
Mx  Cx  Kx   MIx
g
 Using modal transformation, uncoupled sets of
equations take the form
zi  2ii zi  i2 zi  i xg ; i  1
iT MI
i  T
i Mi

m
(5.2)
i is the mode shape; ωi is the natural frequency
λ is the more participation factor; ξ is the
i
i
modal damping ratio.
T.K. Datta
Department Of Civil Engineering, IIT Delhi
Response Spectrum Method Of Analysis
1/6
Contd…
 Response of the system in the ith mode is
(5.3)
x i = φi z i
 Elastic force on the system in the ith mode
fsi = Kx i = Kφi zi
(5.4)
 As the undamped mode shape i satisfies
Kφi = ωi2Mφi
(5.5)
 Eq 5.4 can be written as
fsi = ωi2Mφi z i
(5.6)
 The maximum elastic force developed in the ith
mode
fsimax = Mφiωi2 z imax
(5.7)
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Department Of Civil Engineering, IIT Delhi
Response Spectrum Method Of Analysis
1/7
Contd…
 Referring to the development of displacement
response spectrum
(5.8)
zi max  i Sdi i , i 
2
 Using Sa   Sd , Eqn 5.7 may be written as
f s i max  i M i S ai  Pei
(5.9)
 Eq 5.4 can be written as
xi max  K 1 f si max  K 1 Pei
(5.10)
 Pe i is the equivalent static load for the ith mode
of vibration.
 Pe i is the static load which produces structural
displacements same as the maximum modal
displacement.
T.K. Datta
Department Of Civil Engineering, IIT Delhi
Response Spectrum Method Of Analysis
1/8
Contd…
 Since both response spectrum & mode shape
properties are required in obtaining Pe i , it is known
as modal response spectrum analysis.
 It is evident from above that both the dynamic &
static analyses are involved in the method of
analysis as mentioned before.
 As the contributions of responses from different
modes constitute the total response, the total
maximum response is obtained by combining modal
quantities.
 This combination is done in an approximate manner
since actual dynamic analysis is now replaced by
partly dynamic & partly static analysis.
T.K. Datta
Department Of Civil Engineering, IIT Delhi
Response Spectrum Method Of Analysis
2/1
Contd…
Modal combination rules
 Three different types of modal combination rules
are popular
 ABSSUM
 SRSS
 CQC
 ABSSUM stands for absolute sum of maximum
values of responses; If x is the response quantity
of interest
m
x

i 1
xi
(5.11)
max
xi max is the absolute maximum value of
response in the ith mode.
T.K. Datta
Department Of Civil Engineering, IIT Delhi
Response Spectrum Method Of Analysis
2/2
Contd…
 The combination rule gives an upper bound to the
computed values of the total response for two
reasons:
 It assumes that modal peak responses occur at
the same time.
 It ignores the algebraic sign of the response.
 Actual time history analysis shows modal peaks
occur at different times as shown in Fig. 5.1;further
time history of the displacement has peak value at
some other time.
 Thus, the combination provides a conservative
estimate of response.
T.K. Datta
Department Of Civil Engineering, IIT Delhi
Response Spectrum Method Of Analysis
Top floor displacement (m)
2/3
0.4
0.2
0
-0.2
-0.4
0
5 t=6.15
10
15
20
25
30
First generalized displacement (m)
(a) Top storey displacement
0.4
0.2
0
-0.2
-0.4
0
5 t=6.1
10
15
Time (sec)
20
25
30
(b) First generalized displacement
Fig
T.K. Datta
Department Of Civil Engineering, IIT Delhi
5.1
Response Spectrum Method Of Analysis
Contd…
2/3
Second generalized displacement (m)
0.06
0.04
0.02
0
-0.02
-0.04
-0.06
0
t=2.5
5
10
15
20
25
30
Time (sec)
(c) Second generalized displacement
Fig 5.1 (contd.)
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Department Of Civil Engineering, IIT Delhi
Response Spectrum Method Of Analysis
2/4
Contd…
 SRSS combination rule denotes square root of sum
of squares of modal responses
 For structures with well separated frequencies, it
provides a good estimate of total peak response.
x
m
x
i 1
2
i max
(5.12)
 When frequencies are not well separated, some
errors are introduced due to the degree of
correlation of modal responses which is ignored.
 The CQC rule called complete quadratic
combination rule takes care of this correlation.
T.K. Datta
Department Of Civil Engineering, IIT Delhi
Response Spectrum Method Of Analysis
2/5
Contd…
 It is used for structures having closely spaced
frequencies:
x
m
m
m
2
x
 i   ij xi x j
i 1
(5.13)
i 1 j 1
 Second term is valid for i
of degree of correlation.
j
& includes the effect
 Due to the second term, the peak response may be
estimated less than that of SRSS.
 Various expressions for  i j are available; here
only two are given :
T.K. Datta
Department Of Civil Engineering, IIT Delhi
Response Spectrum Method Of Analysis
2/6
Contd…
ij 
1   
1     4

2
2
ij
2
2
ij
ij 
1    
  4  1   
8
1  
3
2
ij
2
ij
(Rosenblueth & Elordy) (5.14)
 ij
2
ij
2
ij
(Der Kiureghian)
(5.15)
2
ij
 Both SRSS & CQC rules for combining peak modal
responses are best derived by assuming
earthquake as a stochastic process.
 If the ground motion is assumed as a stationary
random process, then generalized coordinate in
each mode is also a random process & there
should exist a cross correlation between
generalized coordinates.
T.K. Datta
Department Of Civil Engineering, IIT Delhi
Response Spectrum Method Of Analysis
2/7
Contd…
 Because of this,  i j exists between two modal
peak responses.
 Both CQC & SRSS rules provide good estimates of
peak response for wide band earthquakes with
duration much greater than the period of structure.
 Because of the underlying principle of random
vibration in deriving the combination rules, the
peak response would be better termed as mean
peak response.
Fig 5.2 shows the variation of  i j with frquency
ratio.  i j rapidly decreases as frequency ratio
increases.
T.K. Datta
Department Of Civil Engineering, IIT Delhi
Response Spectrum Method Of Analysis
2/8
Contd…
Fig 5.2
T.K. Datta
Department Of Civil Engineering, IIT Delhi
Response Spectrum Method Of Analysis
2/9
Contd…
As both response spectrum & PSDF represent
frequency contents of ground motion, a relationship
exists between the two.
 This relationship is investigated for the smoothed
curves of the two.
Here a relationship proposed by Kiureghian is
presented
 2 4   D  ,   
S xg     
 

 
   ff  
   p0   

 2
p0 ( ) 
T.K. Datta
Department Of Civil Engineering, IIT Delhi
2
 2.8 
2 ln 

 2 
(5.16 a)
(5.16 b)
Response Spectrum Method Of Analysis
2/10
Contd…
2 -3
PSDF of acceleration
sec
/rad)
(m
Example 5.1 : Compare between PSDFs obtained
from the smoothed displacement RSP and FFT of
Elcentro record.
0.05
Unsmoothed PSDF from Eqn 5.16a
Raw PSDF from fourier spectrum
0.04
0.03
0.02
0.01
00
10
20
30
40
Frequency (rad/sec)
Eqn.5.16a
Fourier spectrum of El Centro
0.02
2
-3
60
Unsmoothed
0.025
(m
PSDFs of acceleration
sec/rad)
50
0.015
0.01
0.005
00
10
20
30
40
50
60
Frequency (rad/sec)
70
80
90
100
5 Point smoothed
T.K. Datta
Department Of Civil Engineering, IIT Delhi
Fig5.3
Response Spectrum Method Of Analysis
2/11
Application to 2D frames
 Degree of freedom is sway degree of freedom.
 Sway d.o.f are obtained using condensation
procedure; during the process, desired response
quantities of interest are determined and stored in
an array R for unit force applied at each sway
d.o.f.
 Frequencies & mode shapes are determined
using M matrix & condensed K matrix.
 For each mode find  (Eq. 5.2) & obtain Pei
i
(Eq. 5.9)
N
i 
r
W
 ir
r 1
N
 W r ir

(5.17)
2
r 1
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Department Of Civil Engineering, IIT Delhi
Response Spectrum Method Of Analysis
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Contd…
 Obtain R j  RPej ( j 1...r ) ; R j is the modal peak
response vector.
 Use either CQC or SRSS rule to find mean peak
response.
Example 5.2 : Find mean peak values of top displacement, base shear and inter storey drift between
1st & 2nd floors.
Solution :
ω1 =5.06rad/s; ω2 =12.56rad/s;
ω3 =18.64rad/s; ω 4 = 23.5rad/s
φ1T =  -1 -0.871 -0.520 -0.278  ; φ 2T = -1 -0.210 0.911 0.752
φ3T =  -1 0.738 -0.090 -0.347  ; φ T4 = 1 -0.843 0.268 -0.145
T.K. Datta
Department Of Civil Engineering, IIT Delhi
Response Spectrum Method Of Analysis
2/13
Contd…
Table 5.1
Disp (m)
Base shear in terms of
mass (m)
Drift (m)
Approaches
2 modes
all modes
2 modes
all modes
2 modes
all modes
SRSS
0.9171
0.917
1006.558
1006.658
0.221
0.221
CQC
0.9121
0.905
991.172
991.564
0.214
0.214
ABSSUM
0.9621
0.971
1134.546
1152.872
0.228
0.223
Time history
0.8921
0.893
980.098
983.332
0.197
0.198
T.K. Datta
Department Of Civil Engineering, IIT Delhi
Response Spectrum Method Of Analysis
Application to 3D tall frames
 Analysis is performed for ground motion applied to
each principal direction separately.
 Following steps are adopted:
 Assume the floors as rigid diaphragms & find
the centre of mass of each floor.
 DYN d.o.f are 2 translations & a rotation; centers
of mass may not lie in one vertical (Fig 5.4).
 Apply unit load to each dyn d.o.f. one at a
time & carry out static analysis to find
condensed K matrix & R matrix as for 2D frames.
 Repeat the same steps as described for 2D
frame
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Department Of Civil Engineering, IIT Delhi
Response Spectrum Method Of Analysis
3/1
3/2
C.G. of mass line
C.G. of mass line
CM1
CM1
L
L
CM2
CM2
L
L
CM3
CM3
L
L
L
(a)
L
L
xg


(b)
xg
x
x
Figure 5.4:
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Department Of Civil Engineering, IIT Delhi
Response Spectrum Method Of Analysis
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Contd…
Example 5.3 : Find mean peak values of top floor
displacements , torque at the first floor & VX and VY
at the base of column A for exercise for problem
3.21. Use digitized values of the response spectrum
of El centro earthquake ( Appendix 5A of the book).
Solution :
ω1 =13.516rad/s;
ω2 =15.138rad/s; ω3 = 38.731rad/s;
ω4 = 39.633rad/s ; ω5 = 45.952rad/s; ω6 =119.187rad/s
Results are obtained following the steps of
section 5.3.4.
Results are shown in Table 5.2.
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Department Of Civil Engineering, IIT Delhi
Response Spectrum Method Of Analysis
3/4
Contd…
TABLE 5.2
Torque
Approac
displacement (m)
(rad)
Vx(N)
Vy(N)
hes
(1)
(2)
(3)
SRSS
0.1431
0.0034
0.0020
214547
44081
CQC
0.1325
0.0031
0.0019
207332
43376
0.1216
0.0023
0.0016
198977
41205
Time
history
 Results obtained by CQC are closer to those of
time history analysis.
T.K. Datta
Department Of Civil Engineering, IIT Delhi
Response Spectrum Method Of Analysis
RSA for multi support excitation
 Response spectrum method is strictly valid for
single point excitation.
 For extending the method for multi support
excitation, some additional assumptions are
required.
 Moreover, the extension requires a derivation
through random vibration analysis. Therefore, it is
not described here; but some features are given
below for understanding the extension of the
method to multi support excitation.
 It is assumed that future earthquake is
represented by an averaged smooth response
spectrum & a PSDF obtained from an ensemble
of time histories.
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Department Of Civil Engineering, IIT Delhi
Response Spectrum Method Of Analysis
3/5
3/6
Contd…
 Lack of correlation between ground motions at
two points is represented by a coherence function.
 Peak factors in each mode of vibration and the
peak factor for the total response are assumed to
be the same.
 A relationship like Eqn. 5.16 is established
between S d and PSDF.
 Mean peak value of any response quantity r
consists of two parts:
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Department Of Civil Engineering, IIT Delhi
Response Spectrum Method Of Analysis
3/7
Contd…
• Pseudo static response due
displacements of the supports
to
the
• Dynamic response of the structure with
respect to supports.
Using normal mode theory, uncoupled
dynamic equation of motion is written as:
s
zi  2i zi  i2 zi    ki uk ; i  1..m (5.18)
k 1
i T MRk
 ki  T
i Mi
T.K. Datta
Department Of Civil Engineering, IIT Delhi
(5.19)
Response Spectrum Method Of Analysis
3/8
Contd…
 If the response of the SDOF oscillator to uk is zki
s
then
zi    ki z ki
(5.20)
k 1
 Total response is given by
s
m
r  t    ak uk  t    i zi  t 
k 1
i 1
s
m
s
r  t    ak uk  t    i   ki zki
(5.22)
r  t   a T u  t   T z  t 
(5.23)
k 1

(5.21)
i 1
k 1
φβ and z are vectors of size m x s (for s=3 &
m=2)
T.K. Datta
Department Of Civil Engineering, IIT Delhi
Response Spectrum Method Of Analysis
3/9
Contd…
φβT = 1 β11  1 β21  1 β31  2 β12 2β22  2β32  (5.24a)
z T = z11 z21 z31 z12
z22
z32 
 
(5.24b)

 Assuming r t ,u t and z t to be random
processes, PSDF of r (t ) is given by:
S rr  a S uua    S zz   a S uz    S zua
T
T
T
T
(5.25)
 Performing integration over the frequency range
of interest & considering mean peak as peak
factor multiplied by standard deviation,
expected peak response may be written as:
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Department Of Civil Engineering, IIT Delhi
Response Spectrum Method Of Analysis
3/10
Contd…
E max r  t   = b
T
b +b
uu
b T = a1up1 a 2up2 a3up3
T
uz
φβD + φ
T
βD zz
φβD + φ
aSupS 
(5.27a)
T
φβD
=  φ1β11D11 φ1β 21D21 .... φ1β s1Ds1 ...φ mβ11D1m 
Dij = Di  ω j ,ξ j 
i =1,..,s ; j =1,..,m
 luu , lu z and
lz z
12
b  (5.26)
T
βD zu
(5.27b)
(5.27c)
are the correlation matrices
whose elements are given by:
uiu j
1
=
σ ui σ u j
α
 S  ω  dω
-α
uiu j
T.K. Datta
Department Of Civil Engineering, IIT Delhi
(5.28)
Response Spectrum Method Of Analysis
3/11
Contd…
ui zkj
zki zlj
1
=
σui σ zkj
1
=
σ zki σ zlj
α
*
h
 j Suiuk  ω dω
(5.29)
-α
α
*
h
h
 i j Sukul  ω  dω
(5.30)
-α
Suiuk
coh  i,k 
1 21 21
= 2 Sui Suk coh i,k  =
Sug
2
ω
ω
(5.31)
Suiuj
coh  i, j 
1 21 21
= 4 Sui Suj coh  i, j  =
Sug
4
ω
ω
(5.32)
1
2
uk
1
2
ul
Sukul = S S coh k,l = coh k,l  Sug
T.K. Datta
Department Of Civil Engineering, IIT Delhi
(5.33)
Response Spectrum Method Of Analysis
3/12
Contd…
 For a single train of seismic wave, Dij = Di  ω j ,ξ j 
that is displacement response spectrum for a
specified ξ ; correlation matrices can be obtained
if coh(i, j )
is additionally provided; Su g can be
determined from D  ω j ,ξ j  (Eqn 5.6).
 If only relative peak displacement is required,third
term of Eqn.5.26 is only retained.
 Steps for developing the program in MATLAB is
given in the book.
Example 5.4 Example 3.8 is solved for El centro
earthquake spectrum with time lag of 5s.
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Department Of Civil Engineering, IIT Delhi
Response Spectrum Method Of Analysis
3/13
Contd…
Solution :The quantities required for calculating the
expected value are given below:
1
1
 1
 1
1 1 1 1
φ
 ; φ   0.5 1 ; r  3 1 1 1 ,
0.5

1






w1  12.24 rad/s ; w2  24.48 rad/s
 1111 11 21 11 31 12 12 12  22 12  32 
1 1 1 1
T
a  
;   

3 1 1 1
 2111  21 21  21 31  22 12  22  22  22  32 
 0.0259 0.0259 0.0259 -0.0015 -0.0015 -0.0015
 T D  

 0.0129 0.0129 0.0129 0.0015 0.0015 0.0015 
D11  D21  D31  D (1  12.24)  0.056m
T
D12  D22  D32  D (2  24.48)  0.011m
0
coh  i, j    1
  2
1
0
1
2 
 5 
 10 
1  ; 1  exp 
;


exp
2



 2 
 2 
0 
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Department Of Civil Engineering, IIT Delhi
Response Spectrum Method Of Analysis
3/14
Contd…
uu
uz
zz
 1
  0.873
 0.765
 0.0382
  0.0063

 0.0027
 1
 0.0008

 0.0001

 0.0142
 0.0007

 0.0001
0.873
1
0.873
0.765 
0.873
1 
0.0061 0.0027
0.0443 0.0062
0.0387 0.0063
0.0068 0.0447
0.0063 0.0387
0.0029 0.0068
0.0008
0.0001
0.0142 0.0007
1
0.0008
0.0007 0.0142
0.0008
1
0.0001 0.0007
0.0007
0.0007
0.0001 1
0.0142
0.0007
0.0007 1
0.0007
0.0142
0.0001 0.0007
T.K. Datta
Department Of Civil Engineering, IIT Delhi
0.0029 
0.0068 
0.0447 
0.0001 
0.0007 
0.0142 

0.0001 
0.0007 

1

Response Spectrum Method Of Analysis
3/15
Contd…
 Mean peak values determined are:
(u1 )tot  0.106m ; (u2 )tot  0.099m
(u1 ) rel  0.045m ; (u2 ) rel  0.022m
 For perfectly correlated ground motion
1
0

uu

 0
1
1

1


zz
0
0

 0
0
1
0
1
1
1
0
0
0
0
0 
uz  null matrix
1 
1 0 0 0
1 0 0 0 
1 0 0 0

0 1 1 1
0 1 1 1

0 1 1 1 
T.K. Datta
Department Of Civil Engineering, IIT Delhi
Response Spectrum Method Of Analysis
3/16
Contd…
 Mean peak values of relative displacement
RSA
RHA
u1 =0.078m
;
0.081m
u2 = 0.039m
;
0.041m
 It is seen that’s the results of RHA & RSA match
well.
 Another example (example 3.10) is solved for a time
lag of a 2.5 sec.
Solution is obtained in the same way and results
are given in the book. The calculation steps
are self evident.
T.K. Datta
Department Of Civil Engineering, IIT Delhi
Response Spectrum Method Of Analysis
4/1
Cascaded analysis
 Cascaded analysis is popular for seismic analysis
of secondary systems (Fig 5.5).
Secondary System
k
..
xf
m
Fig 5.5
..
xg
Secondary system mounted
on a floor of a building frame
c
.. .. ..
xa = xf + xg
SDOF is to be analyzed for
obtaining floor response spectrum
 RSA cannot be directly used for the total system
because of degrees of freedom become
prohibitively large ; entire system becomes
nonclasically damped.
T.K. Datta
Department Of Civil Engineering, IIT Delhi
Response Spectrum Method Of Analysis
4/2
Contd…
In the cascaded analysis two systems- primary
and secondary are analyzed separately; output of
the primary becomes the input for the secondary.
 In this context, floor response spectrum of the
primary system is a popular concept for
cascaded analysis.
The absolute acceleration of the floor in the figure
is xa
 Pseudo acceleration spectrum of an SDOF is
obtained for xa ; this spectrum is used for RSA of
secondary systems mounted on the floor.
T.K. Datta
Department Of Civil Engineering, IIT Delhi
Response Spectrum Method Of Analysis
4/3
Contd…
Using this spectrum,
peak displacement of the
secondary system with
T=0.811s is 0.8635m.
 The time history analysis
for the entire system (with
C matrix for P-S system) is
found as 0.9163m.
T.K. Datta
Department Of Civil Engineering, IIT Delhi
Displacement (m)
Example 5.6 For example 3.18, find the mean peak
displacement of the oscillator for El Centro earthquake.
 for secondary system = 0.02 ;  for the main
system = 0.05 ;floor displacement spectrum shown in
the Fig5.6 is used
Solution
1.5
1
0.5
0
0
5
10
15
20
25
30
35
40
Frequency (rad/sec)
Floor displacement response
spectrum (Exmp. 5.6)
Response Spectrum Method Of Analysis
Approximate modal RSA
 For nonclassically damped system, RSA cannot
be directly used.
 However, an approximate RSA can be performed.
 C matrix for the entire system can be obtained
(using Rayleigh damping for individual systems
& then combining them without coupling terms)


C1
C 
0
0
C2 
matrix is obtained considering all d.o.f. &
 T C  becomes non diagonal.
 Ignoring off diagonal terms, an approximate
modal damping is derived & is used for RSA.
T.K. Datta
Department Of Civil Engineering, IIT Delhi
Response Spectrum Method Of Analysis
4/4
4/5
Seismic coefficient method
 Seismic coefficient method uses also a set of
equivalent lateral loads for seismic analysis of
structures & is recommended in all seismic codes
along with RSA & RHA.
 For obtaining the equivalent lateral loads, it uses
some empirical formulae. The method consists of
the following steps:
• Using total weight of the structure, base
shear is obtained by
Vb  W  Ch
(5.34)
Ch is a period dependent seismic coefficient
T.K. Datta
Department Of Civil Engineering, IIT Delhi
Response Spectrum Method Of Analysis
4/6
Contd…
• Base shear is distributed as a set of lateral
forces along the height as
Fi  Vb  f (hi )
(5.35)
f (hi ) bears a resemblance with that for the
fundamental mode.
• Static analysis of the structure is carried out
with the force Fi (i = 1,2...... n) .
 Different codes provide different recommendations
for the values /expressions for Ch & f (hi ) .
T.K. Datta
Department Of Civil Engineering, IIT Delhi
Response Spectrum Method Of Analysis
4/7
 Distribution of lateral forces can be written as
S
F = ρ × W × φ × a1
1
j
j
j1
g
Fj
Wj × φ j1
=
∑ Fj ΣWj × φ j1
Fj = Vb ×
Fj = Vb
Fj = Vb
Wj × φ j1
ΣWj × φ j1
Wj ×h j
( 5.37)
(5.38)
(5.39)
ΣWj ×h j
Wj ×h jk
ΣWj ×h j
(5.36)
k
T.K. Datta
Department Of Civil Engineering, IIT Delhi
(5.40)
Response Spectrum Method Of Analysis
4/8
 Computation of base shear is based on first mode.
Following basis for the formula can be put forward.
Sa
i ×) λ
V = ΣF =(ΣW × φ ×
bi
ji
j
ji
i
g
Sa i
e
Vb i = Wi
g
 
Vb ≤ Σ Vb i
≤Σ
Sai
Wie
g
S
Vb = a1 × W
g
(5.41)
(5.42)
(5.43)
i = 1to n 
T.K. Datta
Department Of Civil Engineering, IIT Delhi
(5.44)
(5.45)
Response Spectrum Method Of Analysis
5/1
Seismic code provisions
 All countries have their own seismic codes.
 For seismic analysis, codes prescribe all three
methods i.e. RSA ,RHA & seismic coefficient
method.
Codes specify the following important factors for
seismic analysis:
• Approximate calculation of time period for
seismic coefficient method.
• Ch Vs T plot.
• Effect of soil condition on
T.K. Datta
Department Of Civil Engineering, IIT Delhi
S
A
or a
g
g
& Ch
Response Spectrum Method Of Analysis
5/2
Contd…
• Seismicity of the region by specifying PGA.
• Reduction factor for obtaining design forces
to include ductility in the design.
• Importance factor for structure.
 Provisions of a few codes regarding the first three
are given here for comparison. The codes include:
•
•
•
•
•
IBC – 2000
NBCC – 1995
EURO CODE – 1995
NZS 4203 – 1992
IS 1893 – 2002
T.K. Datta
Department Of Civil Engineering, IIT Delhi
Response Spectrum Method Of Analysis
5/3
Contd…
IBC – 2000
• Ch for class B site,
 1.0 T1  0.4s

Ch   0.4
T1  0.4s
T
 1
(5.46)
A
• for the same site,
is given by
g

0.4  7.5T
n
A 
  1.0
g 
0.4

Tn

0  Tn  0.08s
0.08  Tn  0.4s
T.K. Datta
Department Of Civil Engineering, IIT Delhi
(5.47)
Tn  0.4s
Response Spectrum Method Of Analysis
5/4
Contd…
 T may be computed by
 N
2
 Wi ui
T1  2  i 1N
g
Fi ui


 i 1







(5.48)
Fi can have any reasonable distribution.
 Distribution of lateral forces over the height
is given by
Fi  Vb
W j h kj
N
k
W
h
 j j
(5.49)
j 1
T.K. Datta
Department Of Civil Engineering, IIT Delhi
Response Spectrum Method Of Analysis
5/5
Contd…
k={1; 0.5 T +1.5
; 2 for T 1 ≤ 0.5s ; 0.5 ≤ T1 ≤ 2.5s; T1 ≥ 2.5s (5.50)
1
 Distribution of lateral force for nine story frame is
shown in Fig5.8 by seismic coefficient method .
W
2
9
W
8
W
7
Storey
W
W
9@3m
W
6
5
W
4
W
3
W
2
10
T=2sec
T=1sec
T=0.4sec
2
Storey force
4
Fig5.8
T.K. Datta
Department Of Civil Engineering, IIT Delhi
Response Spectrum Method Of Analysis
5/6
Contd…
 NBCC – 1995
CeU
; Ce = USIF (5.51a)
;( 5.51b)
• C is given by Ch =
R
h
A
• For U=0.4 ; I=F=1, variations of S & with T
g
are given in Fig 5.9.
Seismic response factor S
4.5
4
3.5
3
2.5
2
1.5
10
0.5
Time period (sec)
1
1.5
Fig5.9
T.K. Datta
Department Of Civil Engineering, IIT Delhi
Response Spectrum Method Of Analysis
5/7
Contd…
• For PGV = 0.4ms-1 , A
g
 1.2
A 
=  0.512
g 
 Tn
is given by
0.03 ≤ Tn ≤ 0.427s
Tn > 0.427s
(5.52)
• T may be obtained by
  Fu

i
1

T1 = 2π 
N
 g1 Fu

i i 
N
2
i
1
2
(5.53)
• S and A/g Vs T are compared in Fig 5.10 for
v = 0.4ms-1 , I = F = 1; zh = z v (acceleration and
velocity related zone)
T.K. Datta
Department Of Civil Engineering, IIT Delhi
Response Spectrum Method Of Analysis
5/8
Contd…
1.4
A/g
S
1.2
S or A/g
1
0.8
0.6
0.4
0.2
0
0
0.5
1
1.5
2
2.5
3
3.5
4
Time period (sec)
Fig5.10
T.K. Datta
Department Of Civil Engineering, IIT Delhi
Response Spectrum Method Of Analysis
5/9
Contd…
• Distribution of lateral forces is given by
Fi =  Vb - Ft 
Wh
i i
N
 Wh
i=1
0


Ft = 0.07T1Vb
 0.25V
b

i
(5.54)
i
T1 ≤ 0.7 s
0.7 < T1 < 3.6 s
T1 ≥ 3.6 s
T.K. Datta
Department Of Civil Engineering, IIT Delhi
(5.55)
Response Spectrum Method Of Analysis
5/10
Contd…
 EURO CODE 8 – 1995
• Base shear coefficient Cs is given by
C
Cs = e
• Ce is given by
q
A


g

1
Ce = 
 A  Tc  3
g  T 
  1
(5.56)
0 ≤ T1 ≤ Tc
(5.57)
T1 ≥ Tc
• Pseudo acceleration in normalized form is given
by Eqn 5.58 in which values of Tb,Tc,Td are
Tb
Tc
Td
hard
med
0.1
0.15
0.4
0.6
3.0
3.0
soft
0.2
0.8
3.0 (A is multiplied by 0.9)
T.K. Datta
Department Of Civil Engineering, IIT Delhi
Response Spectrum Method Of Analysis
5/11
Contd…
• Pseudo acceleration in normalized form
,
is given by
Tn

1+1.5 T
b

 2.5

A
=
 Tc 
ug0
 2.5  T 
 n


Tc Td
2.5

2
T

n
0 ≤ Tn ≤ Tb
Tb ≤ Tn ≤ Tc
Tc ≤ Tn ≤ Td
T.K. Datta
Department Of Civil Engineering, IIT Delhi
(5.58)
Tn ≥ Td
Response Spectrum Method Of Analysis
5/12
 Rayleigh's method may be used for calculating T.
 Distribution of lateral force is
Fi = Vb
Wφ
i
i1
 Wφ
i=1
Fi = Vb
(5.59)
N
i
i1
Wh
i i
(5.60)
N
 Wh
i=1
i
i
 Variation of ce / u go & A / u go
Fig 5.11.
T.K. Datta
Department Of Civil Engineering, IIT Delhi
are shown in
Response Spectrum Method Of Analysis
5/13
Contd…
3
..
A/ug0
..
2.5
Ce/ug0
..
Ce /ug0 or A
../ug0
2
..
1.5
1
0.5
0
0
0.5
1
1.5
2
2.5
3
3.5
4
Time period (sec)
Fig 5.11
T.K. Datta
Department Of Civil Engineering, IIT Delhi
Response Spectrum Method Of Analysis
6/1
Contd…
 NEW ZEALAND CODE ( NZ 4203: 1992)
• Seismic coefficient & design response curves
are the same.
• For serviceability limit,
C  T  = Cb  T1,1 RzL s
= Cb  0.4,1 RzL s
T1 ≥ 0.45
T1 ≤ 0.45
(5.61a)
(5.61b)
Ls is a limit factor.
• For acceleration spectrum,
T.K. Datta
Department Of Civil Engineering, IIT Delhi
T1 is replaced by T.
Response Spectrum Method Of Analysis
6/2
Contd…
• Lateral load is multiplied by 0.92.
• Fig5.12 shows the plot of cb vs T for   1
• Distribution of forces is the same as Eq.5.60
• Time period may be calculated by using
Rayleigh’s method.
• Categories 1,2,3 denote soft, medium and hard.
• R in Eq 5.61 is risk factor; Z is the zone factor;
ls is the limit state factor.
T.K. Datta
Department Of Civil Engineering, IIT Delhi
Response Spectrum Method Of Analysis
6/3
Contd…
1.2
Category 1
Category 2
1
Category 3
0.8
Cb
0.6
0.4
0.2
0
0
0.5
1
1.5
2
2.5
3
3.5
4
Time period (sec)
Fig5.12
T.K. Datta
Department Of Civil Engineering, IIT Delhi
Response Spectrum Method Of Analysis
Contd…
6/4
 IS CODE (1893-2002)
• Time period is calculated by empirical
formula and distribution of force is given by:
Fj = Vb
2
Wh
j j
N
2
Wh
 j j
(5.65)
j=1
• Ce vs T &
given by:
Sa
vs T are the same; they are
g

1+15T 0 ≤ T ≤ 0.1s
Sa 
=  2.5
0.1≤ T ≤ 0.4s
g 
1

0.4 ≤ T ≤ 4.0s
 T
T.K. Datta
Department Of Civil Engineering, IIT Delhi
for hard soil (5.62)
Response Spectrum Method Of Analysis
6/5
Contd…

1+15T
Sa 
=  2.5
g 
1.36

 T

1+15T
Sa 
=  2.5
g 
1.67

 T
0 ≤ T ≤ 0.1s
0.1≤ T ≤ 0.55s
for medium soil
(5.63)
for soft soil
(5.64)
0.55 ≤ T ≤ 4.0s
0 ≤ T ≤ 0.1s
0.1≤ T ≤ 0.67s
0.67 ≤ T ≤ 4.0s
For the three types of soil Sa/g are shown in Fig
5.13
Sesmic zone coefficients decide about the PGA
values.
T.K. Datta
Department Of Civil Engineering, IIT Delhi
Response Spectrum Method Of Analysis
6/6
Contd…
3
Spectral acceleration coefficient (Sa/g)
Hard Soil
2.5
Medium Soil
Soft Soil
2
1.5
1
0.5
0
0
0.5
1
1.5
2
2.5
3
3.5
4
Time period (sec)
Variations of (Sa/g) with time period T
Fig 5.13
T.K. Datta
Department Of Civil Engineering, IIT Delhi
Response Spectrum Method Of Analysis
6/7
Contd…
Example 5.7: Seven storey frame shown in Fig 5.14
is analyzed with
Concrete density = 24kNm-3 ; E = 2.5×107 kNm-2
Live load = 1.4kNm-1
For mass: 25% for the top three & rest 50% of live
load are considered.
T1 = 0.753s ; T2 = 0.229s ; T3 = 0.111s
R = 3; PGA = 0.4g ; for NBCC, PGA ≈ 0.65g
Solution:
 First period of the structure falls in the falling
region of the response spectrum curve.
 In this region, spectral ordinates are different
for different codes.
T.K. Datta
Department Of Civil Engineering, IIT Delhi
Response Spectrum Method Of Analysis
6/8
Contd…
All beams:-23cm  50cm
Columns(1,2,3):-55cm  55cm
7@3m
Columns(4-7):-:-45cm  45cm
5m
5m
5m
A Seven storey-building frame for analysis
Fig 5.14
T.K. Datta
Department Of Civil Engineering, IIT Delhi
Response Spectrum Method Of Analysis
6/9
Contd…
Table 5.3: Comparison of results obtained by different codes
1st Storey Displacement
(mm)
Base shear (KN)
Top Storey Displacement
(mm)
Codes
SRSS
CQC
SRSS
CQC
SRSS
3
all
3
all
3
all
3
all
IBC
33.51
33.66
33.52
33.68
0.74
0.74
0.74
0.74
NBCC
35.46
35.66
35.46
35.68
0.78
0.78
0.78
NZ
4203
37.18
37.26
37.2
37.29
0.83
0.83
Euro 8
48.34
48.41
48.35
48.42
1.09
Indian
44.19
44.28
44.21
44.29
0.99
T.K. Datta
Department Of Civil Engineering, IIT Delhi
3
all
CQC
3
all
10.64 10.64
10.64
10.64
0.78
11.35 11.35
11.35
11.35
0.83
0.83
12.00 12.00
12.00
12.00
1.09
1.09
1.09
15.94 15.94
15.94
15.94
0.99
0.99
0.99
14.45 14.45
14.45
14.45
Response Spectrum Method Of Analysis
6/10
Contd…
7
Number of storey
6
5
4
3
2
1
0
2
4
6
8
10
Displacement (mm)
12
IBC
NBCC
NZ 4203
Euro 8
Indian
14
16
Comparison of displacements obtained by different codes
Fig 5.15
T.K. Datta
Department Of Civil Engineering, IIT Delhi
Response Spectrum Method Of Analysis
Lec-1/74
T.K. Datta
Department Of Civil Engineering, IIT Delhi
Response Spectrum Method Of Analysis