Seismic Analysis of Structures - III

T.K. Datta
Department Of Civil Engineering, IIT Delhi
Response Spectrum Method Of Analysis
Chapters – 5 & 6
Chapter -5
RESPONSE SPECTRUM
METHOD OF ANALYSIS

T.K. Datta
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.
1/1

T.K. Datta
Contd…
 The concept of equivalent lateral forces for earth-
quake 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.
1/2

T.K. Datta
Contd…
 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.
1/3
 The response spectrum method of analysis is
developed using the following steps.

T.K. Datta
 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 non-
classically damped systems.
Contd…
1/4

T.K. Datta
 Equation of motion for MDOF system under
single point excitation
(5.1)
 Using modal transformation, uncoupled sets of
equations take the form
 is the mode shape; ωi is the natural frequency
is the more participation factor; is the
modal damping ratio.
Development of the method
gx+ + = −&& & &&Mx Cx Kx MI
2
2 ; 1 (5.2)i i i i i i i gz z z x i mξ ω ω λ+ + = − =&&&& LL
T
i
i T
i i
φ
λ
φ φ
=
MI
M
1/5
iφ
λ
i ξ
i

T.K. Datta
 Response of the system in the ith mode is
(5.3)
 Elastic force on the system in the ith mode
(5.4)
 As the undamped mode shape satisfies
(5.5)
 Eq 5.4 can be written as
(5.6)
 The maximum elastic force developed in the ith
mode
Contd…
i i ix =φ z
si i i if = Kx = Kφ z
iφ
2
i i iKφ = ω Mφ
2
si i i if =ω Mφ z
2
simax i i imaxf = Mφ ω z
1/6

T.K. Datta
 Referring to the development of displacement
response spectrum
(5.8)
 Using , Eqn 5.7 may be written as
(5.9)
 Eq 5.4 can be written as
(5.10)
 is the equivalent static load for the ith mode
of vibration.
 is the static load which produces structural
displacements same as the maximum modal
displacement.
Contd…
( )max ,ii i d i iz Sλ ω ξ=
max ii aSλ= = isi i ef M Pφ
1 1
max max
− −
= = ii si ex K f K P
2
a dS Sω=
Pei
1/7
Pei

T.K. Datta
 Since both response spectrum & mode shape
properties are required in obtaining , 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.
Contd…
Pei
1/8

T.K. Datta
 Three different types of modal combination rules
are popular
 ABSSUM
 SRSS
 CQC
Contd…
Modal combination rules
 ABSSUM stands for absolute sum of maximum
values of responses; If is the response quantity
of interest
x
max
1
m
i
i
x x
=
=∑ (5.11)
is the absolute maximum value of
response in the ith mode.
maxix
2/1

T.K. Datta
 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.
Contd…
2/2

T.K. Datta
0 5 10 15 20 25 30
-0.4
-0.2
0
0.2
0.4
Topfloordisplacement(m)
t=6.15
0 5 10 15 20 25 30
-0.4
-0.2
0
0.2
0.4
Time (sec)
Firstgeneralizeddisplacement(m)
t=6.1
(a) Top storey displacement
(b) First generalized displacement
2/3
Fig 5.1

T.K. Datta
Contd…
0 5 10 15 20 25 30
-0.06
-0.04
-0.02
0
0.02
0.04
0.06
Time (sec)
Secondgeneralizeddisplacement(m)
t=2.5
(c) Second generalized displacement
Fig 5.1 (contd.)
2/3

T.K. Datta
 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.
 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.
Contd…
2
max
1
(5.12)
m
i
i
x x
=
= ∑
2/4

T.K. Datta
 It is used for structures having closely spaced
frequencies:
 Second term is valid for & includes the effect
of degree of correlation.
 Due to the second term, the peak response may be
estimated less than that of SRSS.
 Various expressions for are available; here
only two are given :
Contd…
2
1 1 1
(5.13)
m m m
i ij i j
i i j
x x x xρ
= = =
= +∑ ∑∑
i j≠
2/5
iρ j

T.K. Datta
(Rosenblueth & Elordy) (5.14)
(Der Kiureghian) (5.15)
 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.
Contd…
( )
( )
22
2 2
1
1 4
ij
ij
ij ij
ξ β
ρ
β ξ β
+
=
− +
( )
( ) ( )
3
2 2
2 22
8 1
1 4 1
ij ij
ij
ij ij ij
ξ β β
ρ
β ξ β β
+
=
− + +
2/6

T.K. Datta
 Because of this, 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 with frquency
ratio. rapidly decreases as frequency ratio
increases.
Contd…
2/7
iρ j
iρ j
iρ j

T.K. Datta
Fig 5.2
Contd… 2/8

T.K. Datta
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
Contd…
0
2.8
( ) 2ln (5.16 b)
2
p
ωτ
ω
π
 
=  ÷
 
2/9
( )
( )
( )
2
2
0
,2 4
(5.16 a)g
ff
x
D
S
p
θ
θ θ
ω ξω ξω
ω
ω ω π πτ ω
+
  
= +   +    
&&

T.K. Datta
Contd…
0 10 20 30 40 50 600
0.01
0.02
0.03
0.04
0.05
Frequency (rad/sec)
PSDFof
acceleration(m
2
sec-3
/rad)
Unsmoothed PSDF from Eqn 5.16a
Raw PSDF from fourier spectrum
0 10 20 30 40 50 60 70 80 90 1000
0.005
0.01
0.015
0.02
0.025
Frequency (rad/sec)
PSDFsofacceleration(m
2
sec
-3
/rad)
Eqn.5.16a
Fourier spectrum of El Centro
Unsmoothed
5 Point smoothed
Fig5.3
2/10
Example 5.1 : Compare between PSDFs
obtained from the smoothed displacement RSP
and FFT of Elcentro record.

T.K. Datta
 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
(Eq. 5.9)
Application to 2D frames
iλ
( )
1
2
1
(5.17)
N
r
ir
r
i N
r
ir
r
W
W
φ
λ
φ
=
=
=
∑
∑
2/11

T.K. Datta
 Obtain ; 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 dis-
placement, base shear and inter storey drift between
1st
& 2nd
floors.
Contd…
( 1... )j ejR RP j r= = R j
234
1 2
3
ω =5.06rad/s; ω =12.56rad/s;
ω =18.64rad/s; ω = .5rad/s
2/12
Solution :
[ ] [ ]
[ ] [ ]
;
;
T T
1 2
T T
3 4
φ = -1 -0.871 -0.520 -0.278 φ = -1 -0.210 0.911 0.752
φ = -1 0.738 -0.090 -0.347 φ = 1 -0.843 0.268 -0.145

T.K. Datta
Approaches
Disp (m)
Base shear in terms
of mass (m)
Drift (m)
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
Table 5.1
Contd…
2/13

T.K. Datta
 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
Application to 3D tall frames
3/1

T.K. Datta
3/2
C.G. of mass line
1CM
2CM
3CM
L
L
L
L
gx&&
θ
x
(a)
C.G. of mass line
1CM
2CM
3CM
L
L
L
L L
g
x&&
θ
x
(b)
Figure 5.4:

T.K. Datta
Example 5.3 : Find mean peak values of top floor
displacements , torque at the first floor &
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).
Results are obtained following the steps of
section 5.3.4.
Results are shown in Table 5.2.
Contd…
1 2 3
4 5 6
ω =13.516rad/s; ω =15.138rad/s; ω = 38.731rad/s;
ω = 39.633rad/s ; ω = 45.952rad/s; ω =119.187rad/s
X YV and V
3/3
Solution :

T.K. Datta
Approac
hes
displacement (m)
Torque
(rad)
Vx(N) Vy(N)
(1) (2) (3)
SRSS 0.1431 0.0034 0.0020 214547 44081
CQC 0.1325 0.0031 0.0019 207332 43376
Time
history
0.1216 0.0023 0.0016 198977 41205
TABLE 5.2
Contd…
 Results obtained by CQC are closer to those of
time history analysis.
3/4

T.K. Datta
 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.
RSA for multi support excitation
3/5

T.K. Datta
Contd…
3/6
 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 and PSDF.
 Mean peak value of any response quantity r
consists of two parts:
dS

T.K. Datta
Contd…
2
1
2 ; 1.. (5.18)
s
i i i i i ki k
k
z z z u i mξω ω β
=
+ + = =∑ &&&& &
(5.19)i k
ki
i i
β =
φ
φ φ
T
T
MR
M
3/7
• Pseudo static response due to the
displacements of the supports
• Dynamic response of the structure with
respect to supports.
Using normal mode theory, uncoupled
dynamic equation of motion is written as:

T.K. Datta
 If the response of the SDOF oscillator to
then
 Total response is given by
 are vectors of size m x s (for s=3 &
m=2)
Contd…
( ) ( ) ( )
( ) ( )
( ) ( ) ( )
1 1
1 1 1
(5.21)
(5.22)
(5.23)
s m
k k i i
k i
s m s
k k i ki ki
k i k
r t a u t z t
r t a u t z
r t
φ
φ β
= =
= = =
= +
= +
= +
∑ ∑
∑ ∑ ∑
βφT T
a u t z t
3/8
k kiu is z&&
1
(5.20)
s
i ki ki
k
z zβ
=
=∑
βφ and z

T.K. Datta
 Assuming to be random
processes, PSDF of is given by:
 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:
Contd…
{ }
ϕ ϕ ϕ ϕ ϕ ϕ  
T
β 1 11 1 21 1 31 2 12 2 22 2 32
T
11 21 31 12 22 32
φ = β β β β β β ( 5.24a)
z = z z z z z z ( 5.24b)
( ) ( ) ( )tr t ,u t and z
( )r t
(5.25)rrS = + + +T T T T
uu zz uz zua S a S a S S aβ β β βφ φ φ φ
3/9

T.K. Datta
Contd…
( )
[
( )
....
     
  
  
l l l l
L1 2 3 S
1 2T T T T
uu uzβD βD zz βD βD zu
T
1 p 2 p 3 p S p
T
βD 1 11 11 1 21 21 1 s1 s1 m 11 1m
ij i j j
E max r t = b b +bφ +φ φ + φ b ( 5.26)
b = a u a u a u a u ( 5.27a)
φ = φ β D φ β D φ β D ...φ β D ( 5.27b)
D =Dω ,ξ i=1,..,s ; j=1,..,m ( 5.27c)
 and are the correlation matrices
whose elements are given by:
,uu u zl l z zl
( )∫l i j i j
i j
α
u u u u
u u -α
1
= Sω dω ( 5.28)
σ σ
3/10

T.K. Datta
Contd…
( )∫ &&l i kj i k
i kj
α
*
u z j u u
u z -α
1
= h Sω dω ( 5.29)
σ σ
( )∫ && &&l ki lj k l
ki lj
α
*
z z i j u u
z z -α
1
= hh Sω dω ( 5.30)
σ σ
( )
( )
&& && && &&i k i k g
1 1
2 2
uu u u u2 2
coh i,k1
S = S S coh i,k = S ( 5.31)
ω ω
( ) ( )&& && && && &&k l k l g
1 1
2 2
u u u u uS = S S coh k,l = coh k,l S ( 5.33)
( )
( )
&& && &&i j i j g
1 1
2 2
u u u u u4 4
coh i, j1
S = S S coh i, j = S ( 5.32)
ω ω
3/11

T.K. Datta
Contd…
( )ij i j jD =Dω ,ξ For a single train of seismic wave,
that is displacement response spectrum for a
specified ξ ; correlation matrices can be
obtained
if is additionally provided; can be
determined from (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.
coh( i,j )
( )j jDω ,ξ
u gS&&
Example 5.4 Example 3.8 is solved for El centro
earthquake spectrum with time lag of 5s.
3/12

T.K. Datta
Contd…
Solution :The quantities required for calculating the
expected value are given below:
1 2
11 11 11 21 11 31 12 12 12 22 12 32
21 11 21 21 21 31 22 12 22 22 22 32
1 1 1 1 1 1 11
; ; ,
0.5 1 0.5 1 1 1 13
12.24rad/s ; 24.48rad/s
1 1 11
;
1 1 13
0.0259 0.0259 0.0259 -T
D
w w
β
ϕ β ϕ β ϕ β ϕ β ϕ β ϕ β
ϕ β ϕ β ϕ β ϕ β ϕ β ϕ β
ϕ
     
= = =     − −     
= =
  
= =   
   
=
βφT T
φ φ r
a
( )
11 21 31 1
12 22 32 2
1 2
1 1 1 2
2 1
0.0015 -0.0015 -0.0015
0.0129 0.0129 0.0129 0.0015 0.0015 0.0015
( 12.24) 0.056m
( 24.48) 0.011m
0
5 10
, 0 ; exp ; exp
2 2
0
D D D D
D D D D
coh i j
ω
ω
ρ ρ
ω ω
ρ ρ ρ ρ
π π
ρ ρ
 
 
 
= = = = =
= = = = =
 
− −    = = = ÷  ÷     
  
3/13

T.K. Datta
1 0.873 0.765
0.873 1 0.873
0.765 0.873 1
0.0382 0.0061 0.0027 0.0443 0.0062 0.0029
0.0063 0.0387 0.0063 0.0068 0.0447 0.0068
0.0027 0.0063 0.0387 0.0029 0.0068 0.0447
1 0.0008 0.0001 0.0142
0.0008 1 0
 
 =  
  
 
 =
 
  
=
l
l
l
uu
uz
zz
0.0007 0.0001
.0008 0.0007 0.0142 0.0007
0.0001 0.0008 1 0.0001 0.0007 0.0142
0.0142 0.0007 0.0001 1 0.0007 0.0001
0.0007 0.0142 0.0007 0.0007 1 0.0007
0.0001 0.0007 0.0142 0.0001 0.0007 1
 
 
 
 
 
 
 
 
 
Contd…
3/14

T.K. Datta
 Mean peak values determined are:
Contd…
1 2
1 2
( ) 0.106 ; ( ) 0.099
( ) 0.045 ; ( ) 0.022
tot tot
rel rel
u m u m
u m u m
= =
= =
 For perfectly correlated ground motion
1 0 0
0 1 0 null matrix
0 0 1
1 1 1 0 0 0
1 1 1 0 0 0
1 1 1 0 0 0
0 0 0 1 1 1
0 0 0 1 1 1
0 0 0 1 1 1
 
 = = 
  
 
 
 
 
= 
 
 
 
  
l l
l
uu uz
zz
3/15

T.K. Datta
Contd…
 Mean peak values of relative displacement
1
2
RSA RHA
u =0.078m ; 0.081m
u = 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.
3/16

T.K. Datta
Cascaded analysis
 Cascaded analysis is popular for seismic analysis
of secondary systems (Fig 5.5).
 RSA cannot be directly used for the total system
because of degrees of freedom become
prohibitively large ; entire system becomes
nonclasically damped.
4/1
Secondary System
xg
..
..
k
c
m
xa = xf + xg
.. .. ..
Secondary system mounted
on a floor of a building frame
SDOF is to be analyzed for
obtaining floor response spectrum
xf
Fig 5.5

T.K. Datta
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
 Pseudo acceleration spectrum of an SDOF is
obtained for ; this spectrum is used for RSA of
secondary systems mounted on the floor.
ax&&
ax&&
4/2

T.K. Datta
Contd…
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
4/3
ξ ξ
0 5 10 15 20 25 30 35 40
0
0.5
1
1.5
Frequency (rad/sec)
Displacement(m)
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.
Floor displacement response
spectrum (Exmp. 5.6)

T.K. Datta
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)
 matrix is obtained considering all d.o.f. &
becomes non diagonal.
 Ignoring off diagonal terms, an approximate
modal damping is derived & is used for RSA.
1
2
0
0
C
C
C
 
=  
 
φ
T
Cφ φ
4/4

T.K. Datta
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
is a period dependent seismic coefficient
(5.34)b hV W C= ×
hC
4/5

T.K. Datta
Contd…
• Base shear is distributed as a set of lateral
forces along the height as
bears a resemblance with that for the
fundamental mode.
• Static analysis of the structure is carried out
with the force .
 Different codes provide different recommendations
for the values /expressions for .
( ) (5.35)i b iF V f h= ×
(i = 1,2...... n)iF
( )if h
hC & ( )if h
4/6

T.K. Datta
4/7
 Distribution of lateral forces can be written as
j j j1
j j j1
j j1
j b
j j1
j j
j b
j j
k
j j
j b k
j j
S
a1F =ρ×W ×φ× ( 5.36)
1j j j1 g
F W ×φ
= ( 5.37)
∑F ΣW ×φ
W ×φ
F = V × ( 5.38)
ΣW×φ
W ×h
F = V ( 5.39)
ΣW×h
W ×h
F = V ( 5.40)
ΣW×h

T.K. Datta
4/8
 Computation of base shear is based on first mode.
Following basis for the formula can be put forward.
( )
( )
i
i
i
i
ae
b i
b b
a e
i
a1
b
SaiV =ΣF =( ΣW×φ× )×λ ( 5.41)
b ji j ji igi
S
V = W ( 5.42)
g
V≤ Σ V ( 5.43)
S
≤ Σ W i =1ton ( 5.44)
g
S
V = ×W ( 5.45)
g

T.K. Datta
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.
• plot.
• Effect of soil condition on
hC Vs T
a
& h
SA
or C
g g
5/1

T.K. Datta
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
5/2

T.K. Datta
Contd…
IBC – 2000
• for class B site,
• for the same site, is given by
hC
A
g
0.4 7.5 0 0.08s
1.0 0.08 0.4s (5.47)
0.4
0.4s
n n
n
n
n
T T
A
T
g
T
T

 + ≤ ≤

= ≤ ≤

 >

1
1
1
1.0 0.4s
(5.46)0.4
0.4sh
T
C
T
T
≤

= 
≥

5/3

T.K. Datta
Contd…
 T may be computed by
 can have any reasonable distribution.
 Distribution of lateral forces over the height
is given by
iF
1
(5.49)
k
j j
i b N
k
j j
j
W h
F V
W h
=
=
∑
2
1
1
1
2 (5.48)
N
i i
i
N
i i
i
W u
T
g Fu
π =
=
 
 
 =
 
  
∑
∑
5/4

T.K. Datta
Contd…
 Distribution of lateral force for nine story frame is
shown in Fig5.8 by seismic coefficient method .
( )1 1 1 1k={ 1; 0.5 T +1.5 ; 2 for T≤ 0.5s ; 0.5 ≤ T ≤ 2.5s; T ≥ 2. 5s ( 5.50)
0 2 41
2
3
4
5
6
7
8
9
Storey force
Storey
T=2sec
T=1sec
T=0.4sec
W
2
W
W
W
W
W
W
W
W
9@3m
Fig5.8
5/5

T.K. Datta
Contd…
 NBCC – 1995
• is given by
• For U=0.4 ; I=F=1, variations of with T
are given in Fig 5.9.
hC
e
h e
C U
C = ; C =USIF ( 5.51a);( 5.51b)
R
A
S &
g
0 0.5 1 1.51
1.5
2
2.5
3
3.5
4
4.5
Time period (sec)
SeismicresponsefactorS
Fig5.9
5/6

T.K. Datta
Contd…
• For PGV = 0.4ms-1
, is given by
• T may be obtained by
• S and Vs T are compared in Fig 5.10 for
v = 0.4ms-1
, I = F = 1; (acceleration and
velocity related zone)
 
 
  
∑
∑
1
N 22
i i1
1 N
i i1
Fu
T = 2π ( 5.53)
g Fu
A
g





n
n
n
1.2 0.03≤ T ≤ 0.427s
A
= ( 5.52)0.512
T > 0.427sg
T
A/g
h vz = z
5/7

T.K. Datta
Contd…
0 0.5 1 1.5 2 2.5 3 3.5 4
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Time period (sec)
A/g
S
SorA/g
Fig5.10
5/8

T.K. Datta
Contd…
• Distribution of lateral forces is given by





1
t 1 b 1
b 1
0 T≤ 0.7s
F = 0.07T V 0.7 < T < 3.6s ( 5.55)
0.25V T≥ 3.6s
( )
∑
i i
i b t N
i i
i=1
Wh
F = V -F ( 5.54)
Wh
5/9

T.K. Datta
Contd…




 
 ÷
 
1 c
1e -
3
c
1 c
1
A
0≤ T ≤ T
g
C = ( 5.57)
TA
T≥ T
g T
5/10
 EURO CODE 8 – 1995
• Base shear coefficient is given by
• is given by
• Pseudo acceleration in normalized form is given
by Eqn 5.58 in which values of Tb,Tc,Td
sC
eC
e
s
C
C = (5.56)
q′
b c dT T T
hard 0.1 0.4 3.0
med 0.15 0.6 3.0
soft 0.2 0.8 3.0 (A is multiplied by 0.9)
are

T.K. Datta
Contd…
5/11
• Pseudo acceleration in normalized form
,
is given by





 
 ÷
 



&&
0
n
n b
b
b n c
c
c n dg
n
c d
n d2
n
T
1+1.5 0≤ T ≤ T
T
2.5 T≤ T ≤ T
A
= ( 5.58)T
2.5 T≤ T ≤ Tu
T
T T
2.5 T≥ T
T

T.K. Datta
 Rayleigh's method may be used for calculating T.
 Distribution of lateral force is
 Variation of are shown in
Fig 5.11.
∑
∑
i i1
i b N
i i1
i=1
i i
i b N
i i
i=1
Wφ
F = V ( 5.59)
Wφ
Wh
F = V ( 5.60)
Wh
/ & /e go goc u A u&& &&
5/12

T.K. Datta
Contd…
0 0.5 1 1.5 2 2.5 3 3.5 4
0
0.5
1
1.5
2
2.5
3
Time period (sec)
A/ug0
Ce/ug0
Ce/ug0orA/ug0
..
....
..
..
Fig 5.11
5/13

T.K. Datta
Contd…
 NEW ZEALAND CODE ( NZ 4203: 1992)
• Seismic coefficient & design response curves
are the same.
• For serviceability limit,
is a limit factor.
( ) ( )
( )
b 1 s 1
b s 1
C T = C T ,1 RzL T≥ 0.45 ( 5.61a)
= C 0.4,1 RzL T≤ 0.45 ( 5.61b)
sL
1T
6/1

T.K. Datta
Contd…
• Lateral load is multiplied by 0.92.
• Fig5.12 shows the plot of
• 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;
is the limit state factor.
1bc vs T for µ =
sl
6/2

T.K. Datta
Contd…
0 0.5 1 1.5 2 2.5 3 3.5 4
0
0.2
0.4
0.6
0.8
1
1.2
Time period (sec)
Category 1
Category 2
Category 3
Cb
Fig5.12
6/3

T.K. Datta
Contd…
 IS CODE (1893-2002)
• are the same; they are
given by:
a
e
S
C vs T & vs T
g
• Time period is calculated by empirical
formula and distribution of force is given by:
∑
2
j j
j b N
2
j j
j=1
Wh
F = V ( 5.65)
Wh







a
1+15T 0≤ T ≤ 0.1s
S
= 2.5 0.1≤ T ≤ 0.4s for hard soil ( 5.62)
g
1
0.4≤ T ≤ 4.0s
T
6/4

T.K. Datta
Contd…














a
a
1+15T 0≤ T ≤ 0.1s
S
= 2.5 0.1≤ T ≤ 0.55s for medium soil ( 5.63)
g
1.36
0.55≤ T ≤ 4.0s
T
1+15T 0≤ T ≤ 0.1s
S
= 2.5 0.1≤ T ≤ 0.67s for soft soil ( 5.64)
g
1.67
0.67≤ T ≤ 4.0s
T
6/5
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
6/6
0 0.5 1 1.5 2 2.5 3 3.5 4
0
0.5
1
1.5
2
2.5
3
Time period (sec)
Hard Soil
Medium Soil
Soft Soil
Spectralaccelerationcoefficient(Sa/g)
Variations of (Sa/g) with time period T
Fig 5.13
Contd…

T.K. Datta
Contd…
Example 5.7: Seven storey frame shown in Fig 5.14
is analyzed with
For mass: 25% for the top three & rest 50% of live
load are considered.
1 2 3T = 0.753s ; T = 0.229s ; T = 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.
-3 7 -2
-1
Concrete density = 24kNm ; E = 2.5×10 kNm
Live load = 1.4kNm
6/7

T.K. Datta
6/8
A Seven storey-building frame for analysis
Fig 5.14
5m 5m 5m
7@3m
All beams:-23cm × 50cm
Columns(1,2,3):-55cm × 55cm
Columns(4-7):-:-45cm × 45cm
Contd…

T.K. Datta
Contd…
Table 5.3: Comparison of results obtained by different codes
Codes
Base shear (KN)
1st Storey Displacement
(mm)
Top Storey Displacement
(mm)
SRSS CQC SRSS CQC SRSS CQC
3 all 3 all 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 10.64 10.64 10.64 10.64
NBCC 35.46 35.66 35.46 35.68 0.78 0.78 0.78 0.78 11.35 11.35 11.35 11.35
NZ
4203
37.18 37.26 37.2 37.29 0.83 0.83 0.83 0.83 12.00 12.00 12.00 12.00
Euro 8 48.34 48.41 48.35 48.42 1.09 1.09 1.09 1.09 15.94 15.94 15.94 15.94
Indian 44.19 44.28 44.21 44.29 0.99 0.99 0.99 0.99 14.45 14.45 14.45 14.45
6/9

T.K. Datta
Contd…
0 2 4 6 8 10 12 14 16
1
2
3
4
5
6
7
Displacement (mm)
Numberofstorey
IBC
NBCC
NZ 4203
Euro 8
Indian
Comparison of displacements obtained by different codes
Fig 5.15
6/10

T.K. Datta
Lec-1/74

T.K. Datta
Chapter - 6
Inelastic Seismic
Response of Structures

T.K. Datta
Introduction
 Under relatively strong earthquakes, structures
undergo inelastic deformation due to current
seismic design philosophy.
 Therefore, structures should have sufficient
ductility to deform beyond the yield limit.
 For understanding the ductility demand imposed
by the earthquake, a study of an SDOF
system in inelastic range is of great help.
The inelastic excursion takes place when the
restoring force in the spring exceeds or equal to
the yield limit of the spring.
1/1

T.K. Datta
 For this, nonlinear time history analysis of SDOF
system under earthquake is required; similarly,
nonlinear analysis of MDOF system is useful for
understanding non-linear behaviour of MDOF
system under earthquakes.
 Nonlinear analysis is required for other reasons
as well such as determination of collapse state,
seismic risk analysis and so on.
 Finally, for complete understanding of the
inelastic behavior of structures, concepts of
ductility and inelastic response spectrum are
required.
The above topics are discussed here.
Contd.. 1/2

T.K. Datta
Non linear dynamic analysis
 If structure have nonlinear terms either in inertia
or in damping or in stiffness or in any form of
combination of them, then the equation of motion
becomes nonlinear.
 More common nonlinearities are stiffness and
damping nonlinearities.
 In stiffness non linearity, two types of non
linearity
are encountered :
• Geometric
• Material (hysteretic type)
 Figure 6.1 shows non hysteric type non linearity;
loading & unloading path are the same.
1/3

T.K. Datta
Contd..
f
Loading
Loading
Unloading
Unloading
x
x∆
Fig.6.1
1/4

T.K. Datta
Contd..
 Figure 6.2 shows hysteric type nonlinearity;
experimental curves are often idealised as
(i) elasto plastic; (ii) bilinear hysteretic ;
(iii) general strain hardening
yx
f
x
yf
y
x
f
x
y
f
yf
f
xyx
y
f
y
x
f
x
Variation of force with displacement under cyclic loading
Idealized model of force
displacement curve
Idealized model of force
displacement curveFig.6.2
1/5

T.K. Datta
 Equation of motion for non linear analysis takes
the form
and matrices are constructed for the
current time interval.
 Equation of motion for SDOF follows as
 Solution of Eqn. 6.2 is performed in incremental
form; the procedure is then extended for MDOF
system with additional complexity.
 and should have instantaneous values.
Contd..
mΔx + c Δx + k Δx = -mΔx (6.2)gt t&& & &&
1/6
(6.1)K gt t+ + = −&& & &&M x C x x Mr x∆ ∆ ∆ ∆
KtCt
Ct Kt

T.K. Datta
 and are taken as that at the beginning of
the time step; they should be taken as average
values.
 Since are not known, It requires an
iteration.
 For sufficiently small , iteration may be
avoided.
 NewMark’s in incremental form is
used for the solution
tc tk
xx &∆∆ &
t∆
Method−β
Contd..
( )
( )
& && &&
& && &&
k
2
2
k k
Δx = Δt x + δ Δt Δx ( 6.3)
Δt
Δx = Δt x + x +β Δt Δx ( 6.4)
2
1/7

T.K. Datta
Contd..
( )
( )
 
 ÷
 
    
  ÷  ÷
    
&& & &&
& & &&
&& & &&
& & & &&
k k2
k k
t t 2
g t k t k
k+1 k k+1 k k+
1 1 1
Δx = Δx - x - x ( 6.5)
βΔt 2ββ Δt
δ δ δ
Δx = Δx - x + Δt 1- x ( 6.6)
βΔt β 2β
kΔx = Δp ( 6.7)
δ 1
k = k + c + m ( 6.8a)
βΔt β Δt
mδ m δ
Δp = -mΔx + + c x + + Δt -1 c x ( 6.8b)
βΔt β 2β 2β
x = x +Δx x = x +Δx x && &&1 k= x +Δx ( 6.9)
1/8

T.K. Datta
Response Spectrum Method Of Analysis81
 For more accurate value of acceleration, it is
calculated from Eq. 6.2 at k+1th step.
 The solution is valid for non hysteretic non
linearity.
 For hysteretic type, solution procedure is
modified & is first explained for elasto - plastic
system.
 Solution becomes more involved because
loading and unloading paths are different.
 As a result, responses are tracked at every time
step of the solution in order to determine loading
and unloading of the system and accordingly,
modify the value of kt.
Contd.. 1/9

T.K. Datta
Elasto-plastic non linearity
 For material elasto plastic behaviour, is taken
to be constant.
 is taken as k or zero depending upon
whether the state is in elastic & plastic state
(loading & unloading).
 State transition is taken care of by iteration
procedure to minimize the unbalanced force;
iteration involves the following steps.
Elastic to plastic state
tc
tk
( ) ( )0
(6.10)ee
x a x∆ = ∆
1/10

T.K. Datta
p e t(Δ x) for( 1- a )Δp withk =0
e pΔ x =( Δ x ) +( Δ x )
Contd..
 Use Eq. 6.7, find
Plastic to plastic state
Eq. 6.7 with Kt=0 is used ; transition takes place if
at the end of the step; computation is then
restarted.
Plastic to elastic state
Transition is defined by
is factored (factor e) such that
is obtained for with
&x < 0
&x = 0
a(Δ x ) ( )1- eΔp tk≠ 0
aΔ x =( Δ x ) +Factored Δ x
&x = 0x∆
1/11

T.K. Datta
Example 6.1 Refer fig. 6.3 ; ; find
responses at t=1.52 s & 1.64s given responses at
t= 1.5s & 1.62s ; m=1kg
Solution:
sradn /10=ω
Contd.. 2/1
x
xf
m
c
gx
..
SDOF system with non-linear
spring
0.15mg
0.0147m x
x
f
Force-displacement behaviour
of the spring
Fig . 6.3

T.K. Datta
&
&&
t = 1.5s; x = 0.01315m; x = 0.1902m/s;
2 -1x = 0.46964 m/s ; f = 1.354 ; c =0.4Nsm ;x t
-1k =100Nm
t
&&
-1k = 10140Nm
Δx = - 0.00312gg
Δp = 37.55N
&& -1Δx = 0.0037m; Δx = - 0.01ms ; Δf =k Δx
t
( f ) =1.7243N> 0.15mgx t+Δt
( f ) + eΔxk = 0.15mg( e = 0.3176)x t t
2/2

T.K. Datta
Contd..
( )
& &
&
&
&
&&
&
& X +ΔX = 0
t 1
-1kΔx = 1-e Δp; k =0; Δx =0.00373m; Δx =-0.00749ms
t2
-1x = x +Δx +Δx = 0.01725m x = 0.1827ms
t 1t+Δt 2 t+Δt
P -c x -f
tk+1 k+1 x( k+1) -1x = = 0.279ms
k+1 m
At t =1.625s ; x >0 ; k = 10040;Δp =-0.4173; Δx =0 .000042;
Δx = -0.061;
( )
( )
&
& & & &
&&
; e =-6.8;Δx = eΔx =-0.000283;
1
-5kΔx = 1-e Δp; k =100; Δx = Δx +Δx =-4.44×10 ; Δx =-0.061;
t 12 2
x = x +Δx =0.0298; x = x +Δx= -0.033
t tt+Δt t+Δt
x from Eqn =3.28; f = f +kΔx = 1.4435N
xt tt+Δt t+Δt 2
2/3

T.K. Datta
Solution for MDOF System
 Sections undergoing yielding are predefined and
their force- deformation behaviour are specified
as shown in Fig 6.4.
0.5k
k
1.5k
k
0.5m
m
m
m
3y
x
2y
x
1y
x
3p
V
2p
V
1p
V
0.5k
k
1.5k
k
x
x
x
Fig.6.4
 For the solution of Eqn. 6.1, state of the yield
section is examined at each time step.
2/4

T.K. Datta
 Depending upon the states of yield
sections, stiffness of the members are changed &
the stiffness matrix for the incremental equation is
formed.
 If required, iteration is carried out as explained for
SDOF.
 Solution for MDOF is an extension of that of SDOF.
Contd..
( )
    
  ÷ 
    
&& & &&
t t 2
g t k t k
KΔx = Δp ( 6.11)
δ 1
K =K + C + M ( 6.12a)
βΔt β Δt
Mδ M δ
Δp =-M r Δx + + C x + +Δt -1 C x ( 6.12b)
βΔt β 2β 2β
2/5

T.K. Datta
Example 6.2: Refer to Fig 6.5; K/m = 100; m = 1 kg;
find responses at 3.54s. given those at 3.52s.
Solution:
Contd..
  
   
   
   
  
 
 
 
  
    
  ÷  ÷
    
&
&&
&& & &&
1.44977 0.15mg
f = 0.95664 < 0.15mg and x > 0
k
0.63432 0.15mg
10260 sym
δ 1
K = K + C + M = -124 10260
t t 2βΔt β( Δt) 0 -124 10137
Δx = 0.5913g
Mδ M δ
Δp = -MIΔx + + C x + + Δt -1 C xg t tk kβΔt β 2β 2β
 
 
 
  
 
 
 
  
32.6224
= 18.0256
8.4376
0.0032
-1Δx = K Δp = 0.0018
0.0009
2/6

T.K. Datta
k/2
k/2
m
m
m
k/2
k/2
k/2 k/2
3m
3m
3m
1
x
2
x
3
x
y
x x
y
f
0.15m gy
f =
0.01475myx =
3 storey frame
Force displacement
curve of the column
Contd..
Fig.6.5
2/7

T.K. Datta
Contd..
   
   
   
      
 
        
    
         
0.0032 0.16
K
Δx = -0.0014 ; Δf = Δx = -0.07
2
-0.0009 -0.045
e( 0.16)1.60977 0.15mg1
f = f +Δf = 0.88664 f + eΔf = f + e ( -0.07) =≤ 0.15mg
k+1 k k k 2
0.58932≤ 0.15mg e ( -0.045)
3
e =0.136;
1
  
     
   
   
     
 
        
     
     
    
e =1; e =1
2 3
eΔx e Δx
1 1 1 1
Δx = Δx = e Δx + e ( Δx - Δx ) = e Δxe 1 1 1 12 2 2 2
eΔx + e ( Δx - Δx )+ e ( Δx - Δx ) e Δx
1 1 12 2 3 3 2 3 3
e0.000435 0.13581
= -0.000965 e = 0.6893
2
-0.001865 3.07e
3
2/8

T.K. Datta
    
     
     
     
    
 
   
 ÷      
& & && & & &
0.0028 0.00324 0.02009
Δx = 0.0027 ; Δx = Δx + Δx = 0.0018 ; x = x + Δx = 0.00833
12 2 k+1 k
0.0026 0.00074 0.0114
0.-0.0509 0
δ δ δ
Δx = Δx - x + Δt 1- x = -0.0406 ; x = x +Δx =
k k k+1 kβΔt β 2β
-0.0524
( )
 
 
 
  
 
            
       
       
       
&& &
.1361
0.07
0.0165
e( 0.16) 0.00 0.0218 1.47151
Δf = e ( -0.07) + -0.005 = -0.075 ; f = f +Δf = 0.882
2 k+1 k
-0.005 -0.05 0.584e ( -0.045)
3
-2.289
-1x = M P -C x -F = -1.7
tk+1 k+1 k+1 k+1
 
 
 
  
018
-2.2825
Contd.. 2/9

T.K. Datta
 Bidirectional interaction assumes importance
under:
• Analysis for two component earthquake
• Torsionally Coupled System
 For such cases, elements undergo yielding
depending upon the yield criterion used.
 When bidirectional interaction of forces on
yielding is considered, yielding of a cross
section depends on two forces.
 None of them individually reaches yield value;
but the section may yield.
Bidirectional Interaction
3/1

T.K. Datta
If the interaction is ignored, yielding in two
directions takes place independently.
 In incremental analysis, the interaction effect is
included in the following way.
 Refer Fig 6.6; columns translate in X and Y
directions with stiffness and .
Contd..
eyikexik
 
 
 
 
 
∑ ∑ ∑ ∑i i i i
ex ex y
e ey ey x
ex y ey xθ
ex ex ey eyθ ex y ey x
K 0 K e
K = 0 K K e ( 6.13a)
K e K e K
K = K ; K = K ; K = K e + K e ( 6.13b)
3/2

T.K. Datta
Contd..
x
e
y
e
D
D
Colm. 1
Colm. 2
Colm. 3
Colm. 4
CR
Y
XC.M.
Fig.6.6
3/3

T.K. Datta
Transient stiffness remaining constant over
is given by
The elements of the modification matrix are
t∆
pK
Contd..
t e pK =K -K ( 6.14)
22
yi xi yixi
pxi pyi pxyi pyxi
i i i
2 2
i exi xi eyi yi
xi exi xi yi eyi yi
yixi
xi yi2 2
pxi pyi
B B BB
K = ; K = ; K =K = ( 6.15)
G G G
G =K h +K h ( 6.16a)
B =K h ; B =K h ( 6.16b)
VV
h = ; h = ( 6.16c)
V V
3/4
tK

T.K. Datta
 When any of the column is in the full plastic state
satisfying yield criterion, .
 During incremental solution changes as the
elements pass from E-P, P-P, P-E; the change
follows E-P properties of the element & yield
criterion.
 Yield criterion could be of different form; most
popular yield curve is
tk =0
tk
Contd..
φ
   
 ÷  ÷ ÷  ÷
   
2 2
yixi
i
pxi pyi
VV
= + ( 6.19)
V V
3/5

T.K. Datta
 For , curve is circular ; ,
curve is ellipse; shows plastic state,
shows elastic state, is inadmissible.
 If , internal forces of the elements are
pulled back to satisfy yield criterion; equilibrium
is disturbed, corrected by iteration.
 The solution procedure is similar to that for
SDOF.
 At the beginning of time , check the states of
the
elements & accordingly the transient stiffness
matrix is formed.
pyipxi VV = pyipxi VV ≠
1=iφ 1<iφ
1>iφ
1>iφ
Contd.. 3/6

T.K. Datta
If any element violates the yield condition at the
end of time or passes from E-P, then an
iteration scheme is used.
If it is P-P & for any element, then an
average stiffness predictor- corrector scheme
is employed.
The scheme consists of :
 is obtained with for the time internal Δt
&
incremental restoring force vector is obtained.
1>iφ
Contd..
1U∆ taK
1 1
1
1
(6.21)
(6.22 )
(6.22 )
ta
i i
i i
force tolerance a
displacement tolerance b
+
+
∆ = ∆
∆ − ∆ ≤
∆ − ∆ ≤
F K U
F F
U U
3/7
( )1 'K = K +K
ta tt02

T.K. Datta
 After convergence , forces are calculated &
yield criterion is checked ; element forces are
pulled back if criterion is violated.
 With new force vector is calculated &
iteration
is continued.
 For E-P, extension of SDOF to MDOF is done.
 For calculating , the procedure as given in
SDOF is adopted.
taK
Contd..
1
(6.23)i i
iϕ
′ =F F
pU∆
3/8

T.K. Datta
 If one or more elements are unloaded from
plastic to elastic state, then plastic work
increments for the elements are negative
 When unloaded, stiffness within , is taken as
elastic.
Example 6.3: Consider the 3D frame in Fig 6.8;
assume:
t∆
Contd..
1
(6.25)
(6.26)
pi i pi
pi i ei i
Fw U
U U K F−
∆ = ∆
∆ = ∆ − ∆
( ) ( ) ( )
( ) ( )
( )
px py p 0 p p oBA D
o
p o x y o B C o C oC A
x y P A
D =3.5m; h = 3.5m; M = M = M = M ; M = M =1.5M
k
M = 2M ; k = k = k = k ; k = k =1.5k ;k = 2k ; = 50
m
in which m = m = m = 620kg and V =152.05
3/9

T.K. Datta
find Initial stiffness & stiffness at t = 1.38s, given
that t = 1.36s
Contd.. 3/10
2k
k
y
x
3.5m
1.5k
1.5k
3.5m
3.5m
A
B C
D
3 D frame
For column A
Displacement (m)0.00467m
152.05 N
Force(N)
Force-displacement
curve of column A

T.K. Datta
3/11
                    
           
           
           


 
 
 
  
& &&
& &&
& &&
U UU 0.00336 0.13675 -0.16679x xx
U = 0.00037 U = 0.00345 U = -0.11434y y y
0.00003 0.00311 -0.06153θ θ θk k k
V 627.27x
F = V = 70.yk
V
θ k

 
 
 
 
&&
V =102.83 V =10.10
A x Ay
V =154.24 V =19.56
Bx By
888 ; x =-0.08613g
gkV =158.66 V =15.15
Dx Dy773.51
V = 211.54 V = 26.08
Cx Cy
Solution:
Forces in the columns are pulled back (Eq. 6.23)
& displacements at the centre

T.K. Datta
Contd..
( )
φ
φ φ φ φ
 
 
 
  
   
 ÷  ÷ ÷  ÷
   
∑ ∑ ∑
2
2
ex exi 0 ey eyi 0θ 0
e
2 2
yixi
i
pxi pyi
A B C D
t e
t t t2
KD
K = K = 6k ; K = K = 6k ; K = = 3k 3.5
4
186000 0 54250
K = 0 186000 54250
54250 54250 1139250
VV
= +
V V
=0.462 ; =0.465; =0.491; = 0.488
K = K
δ 1
K = K + C + M= K +1
βΔt β( Δt)
 
 
 
  
 
      
  ÷  ÷         
& &&&&
4
g t k t k
638.6 sym
0 638.6
5.425 5.425 1379.76
0000M= ×10
16282
Mδ M δ
Δp = -MΔx + + C U + + Δt -1 C U = 286
βΔt β 2β 2β
612
3/12

T.K. Datta
 With the e values calculated as above, the
forces in the columns are pulled back
   
   
   
      

 
 
 
 

 
 
 
 
-1
t
k+1 k
Ax
Bx
k+1 k
Dx
Cx
0.0025 476.1
ΔU = K Δp = 0.000001 and ΔF = K ΔU= 10.2
0.000001 181.2
0.0059
U = U +ΔU = 0.0004
0.0001
VV =183.89
1103.36
V = 275.84
F = F +ΔF = 81.09 ; ;
V = 275.84
954.75
V = 367.79
φ φ φ φ φ
φ φ φ φ
   
 ÷  ÷ ÷  ÷
   
=
Ay
By
Dy
Cy
2 2
yixi
i A B C D
pxi pyi
A B D C
A B D C
= 13.52
V = 20.27
V = 20.27
V = 27.03
VV
= + & = 1.47 ; = 1.47 ; = 1.47 ; =1.47
V V
1 1 1 1
e = 0.824 e = = 0.824 e = =0.824 e = = 0.824
Contd.. 3/13

T.K. Datta
5 3
2
2
0.254 0.83
0.00122 10 ; is calculed as 0.83 10 in which .
0.00124 0.83
16268.57
(1 ) 285.76
631.48
0.29167
0.00795
0.00
;
xe
e x
x
i i
x y
i
xA
xBxi
xi
pxi
U
e = etc
U
e
K x
e e
K
h
hV
h
V
− −
   
∆   ∆ = × ×    ∆
      
 
 ∆ = − ∆ =  
  
= = =
=
=
=
∑
∑
;U e
p p
2
0.00059
0.0003953
; ;
0.000390.0053
0.000290.00398
yA
yByi
yi
yDxD pyi
yCxC
h
hV
h
hh V
hh
=
=
=
==
==
Contd.. 3/14

T.K. Datta
2 2
22 2
246.56 18.12
246.56 18.12
; ; ;
246.56 18.12
246.56 18.12
1.972
1.314
;
1.314
0.98
; ; ;
2
xA yA
xB yB
xi exi xi yi eyi yi
xD yD
xC xC
A
B
i exi xi eyi yi
D
C
yixi
pxi pyi p p pxyi pyxi
i i
B B
B B
B K h B K h
B B
B By
G
G
G K h K h
G
G
BB D
K K K K K K
G G
θ
= =
= =
= =
= =
= =
=
=
= +
=
=
= = = =∑ xi yi
i
B B
G
=
Contd.. 3/15

T.K. Datta
Contd..
3
5
2
1
1.0
16.06 185 10
0.29 53.96 0
62
1
10000 10 0.16 63.85
( )
0 0.54 126.6
0.2917; 0.2917
0.0026
0.0001 ;
0.00
t
t t t
Pyi i Pxi i
xp yp
Pyi Pxi
p
sym
sym
t t
K x K y
e e
K K
δ
β β
−
= − ×
= + + + × −
∆ ∆
= = = =
 
 
 
  
 
 
 
  
 
 ∆ = ∆ = ∆ 
  
∑ ∑
∑ ∑
2 2
K
K K C M = K M =
U K p U
0.002598 0.0000983
0.002598 0.0001018
;
0.002602 0.0000983
0.002602 0.0001018
x py
   
   
   
= ∆ =   
   
      
U
3/16

T.K. Datta
Contd..
2 2
9.78151.913
11.54228.043
;
11.37227.745
10.98303.471
1.002; 1.002; 1.0; 1.00
ex px pxy px
pi
pxy ey py pyi i
AyAx
ByBx
DyDx
CyCx
yixi
i
pxi pyi
A C D
K K K U
K K K U
VV
VV
VV
VV
VV
V V
ϕ
ϕ ϕ ϕ ϕ
− − ∆   
∆ =   − − ∆   
==
==
==
==
   
= + ÷  ÷ ÷  ÷
   
= = = =B
V
3/17
Because yield condition is practically satisfied,
no further iteration is required.

T.K. Datta
Multi Storey Building frames
 For 2D frames, inelastic analysis can be done
without much complexity.
 Potential sections of yielding are identified &
elasto–plastic properties of the sections are
given.
 When IMI = Mp for any cross section, a hinge is
considered for subsequent & stiffness matrix
of the structure is generated.
 If IMI > Mp for any cross section at the end of
IMI is set to Mp, the response is evaluated with
average of stiffness at t and (IMI = Mp ).
t∆
t∆
tt ∆+
4/1

T.K. Datta
 At the end of each , velocity is calculated at
each potential hinge; if unloading takes place at
the end of , then for next , the section
behaves elastically. ( ).
t∆
t∆
t∆
~t small∆
Contd..
Example 6.4
Find the time history of moment at A & the force-
displacement plot for the frame shown in Fig 6.9
under El centro earthquake; ; compare the
results for elasto plastic & bilinear back bone curves.
• Figs. 6.10 & 6.11 are for the result of elasto
-plastic case Figs 6.12 & 6.13 are for the result
of bilinear case
• Moment in Fig 6.12 does not remain constant
over time unlike elasto-plastic case.
st 2.0=∆
4/2

T.K. Datta
4/3
k
k
k
k3m
3m
3m 1.5k
A
1.5k
m
m
m
1x
2x
3x
k = 23533 kN/m
m = 235.33 × 103
kg
i
K
dK
0.1diKK =
0.01471m Displacement (m)
346.23kN
Force(kN)
Frame
Force-displacement
curve of column
Contd..
Fig.6.9

T.K. Datta
4/4
-600000
-400000
-200000
0
200000
400000
600000
0 5 10 15 20 25 30
Time (sec)
Moment(N-m)
-400000
-300000
-200000
-100000
0
100000
200000
300000
400000
0 0.005 0.01 0.015 0.02 0.025 0.03 0.035
Displacement (m)
Force(N)
Contd..
Fig.6.10
Fig.6.11

T.K. Datta
4/5
-500000
-400000
-300000
-200000
-100000
0
100000
200000
300000
400000
500000
-0.005 0 0.005 0.01 0.015 0.02 0.025 0.03
Displacement(m)
ShearForce(N)
-800000
-600000
-400000
-200000
0
200000
400000
600000
800000
0 5 10 15 20 25 30
Time (sec)
Moment(N-m)
Contd..
Fig.6.13
Fig.6.12

T.K. Datta
 For nonlinear moment rotation relationship,
tangent stiffness matrix for each obtained
by considering slope of the curve at the
beginning of
 If unloading takes place, initial stiffness is
considered.
 Slopes of backbone curve may be interpolated ;
interpolation is used for finding initial stiffness.
 If columns are weaker than the beams, then top
& bottom sections of the column become
potential sections for plastic hinge.
 During integration of equation of motion is
given by
t∆
t∆
Contd..
4/6
tK
(6.27)t e p= −K K K

T.K. Datta
 Non zero elements of Kp are computed using
Eqns. 6.15 & 6.16 and are arranged so that they
correspond to the degrees of freedom affected by
plastification.
 The solution procedure remains the same as
described before.
 If 3D frame is weak beam-strong column system,
then problem becomes simple as the beams
undergo only one way bending.
The analysis procedure remains the same as
that of 2D frame.
Contd.. 4/7

T.K. Datta
 For 2D & 3D frames having weak beam strong
column systems, rotational d.o.f are condensed
out; this involves some extra computational
effort.
 The procedure is illustrated with a frame as
shown in the figure (with 2 storey).
Contd..
• Incremental rotations at the member ends
are calculated from incremental
displacements.
• Rotational stiffness of member is modified
if plastification/ unloading takes place.
• The full stiffness matrix is assembled &
rotational d.o.f. are condensed out.
4/8

T.K. Datta
 Elasto-plastic nature of the yield section is
shown in Fig 6.16.
 Considering anti-symmetry :
Contd.. 4/9
pθ
M1, M2
Mp1 = Mp2 = Mp3
θ
Moment-rotation relationship
of elasto-plastic beam
fig. 6.16

T.K. Datta
4/10
( )
( )
 
 
 
  
    
   
   
    
 
 
  
1
2
2 2
1
1
2
2 2
2
kl kl
k -k - -
2 2
Δkl
-k 2k 0
Δ2
K = ( 6.28a)
kl kl kl kl θ
-α + 0.67
2 2 2 6 θ
kl kl kl
- 0α +1.33
2 6 2
( )
( )
( )
( )
( )
( )
 
 
 
  
  
   
      
      
      
-1
Δ Δ Δθ θ θΔ
-1
1-1
θ 2
2
-1
1
2
-1
1
Δ
2
K = K -K K K ( 6.28b)
3α +0.67 16
K = ( 6.29a)
1 3α +1.33kl
3α +0.67 11 -13
θ = Δ ( 6.29b)
1 3α +1.331 0l
3α +0.67 11 -1 -1 -1 1 -13k
K = k - ( 6.30)
1 3α +1.33-1 2 1 0 1 0l

T.K. Datta
Contd..
 Equation of motion for the frame is given by:
The solution requires to be computed at time
t; this requires to be calculated.
 Following steps are used for the calculation
&& & &&Δt g
Δ0
MΔx +CΔx +K Δx = -MIΔx ( 6.31)
C =αK +βM ( 6.32)
4/11
tK∆
21 & αα
& & &x = x +Δx ; x = x + Δx ( 6.33a)
i i-1 i-1 i i-1 i-1
M =M +ΔM ; M =M + ΔM ( 6.33b)
1i 1i-1 1i-1 2i 2i-1 2i-1

T.K. Datta
 & are obtained using Eqn. 6.29b
in which values are calculated as:
 & are then obtained; and hence
& & are calculated from
and , is obtained using ( Eq. 6.30).
 If Elasto-plastic state is assumed, then
for at the beginning of the time
interval; for unloading are obtained
by (Eq.6.28a.)
11 −∆ iθ 12 −∆ iθ
α
Contd..
1i-1 2i-1
1 2
c c
1i-1 2i-1
1i-1 2i-1
1i-1 2i-1
r l r l
α = & α =
6EI 6EI
M M
r = ; r =
θ θ
4/12
11 −∆ iM 12 −∆ iM
iM1 iM2 1α 2α iM1
tK∆iM2
021 == αα
PΜ=Μ=Μ 21
21 & αα

T.K. Datta
Contd..
Example 6.5: For the frame shown in Fig 6.17, find
the stiffness matrix at t = 1.36 s given the response
quantities in Table 6.1
4/13
k
k
k
k3m
3m
3m
5m
1θ
2
θ
3θ
4
θ
5
θ
6
θ
k k
1
∆
2
∆
3
∆
E = 2.48 × 107
kN/m2
Beam 30 × 40 cm
Column 30 × 50 cm
Frame
1
3
5
2
4
6
50KN-m
M
θY = 0.00109 rad θ
Force-displacement curve
Fig. 6.17

T.K. Datta
4/14
Joint
Time
Step
x θ M
sec m m/s m/s2
rad rad/s rad/s2
kNm
1 1.36 0.00293 0.0341 -1.2945 0.00109 0.013 -0.452 50
3 1.36 0.00701 0.0883 -2.8586 0.00095 0.014 -0.297 -23.18
5 1.36 0.00978 0.1339 -3.4814 0.00053 0.009127 -0.098 42.89
2 1.36 0.00293 0.0341 -1.2945 0.00109 0.013 -0.452 -50
4 1.36 0.00701 0.0883 -2.8586 0.00095 0.014 -0.297 23.18
6 1.36 0.00978 0.1339 -3.4814 0.00053 0.009127 -0.098 -42.89
x&x&&θθ&θ&&
Table shows that sections 1 & 2 undergo yielding;
recognising this, stiffness matrices are given below:
x
&x &&x &θ &&θ
Table 6.1
Contd..

T.K. Datta
 
 
 
 
 
 
 
 
 
 
 
 
 
 
1
2
3
1
4
2
3
4
5
6
Δ1.067
Δ-1.067 2.133 sym
Δ0 -1.067 2.133
θ0.8 -0.8 0 2.4
Κ = 4.83×10 ×θ 0.8 0 -0.8 0.8 4
θ0 0.8 0 0 0.8 3.2
θ0.8 -0.8 0 0.4 0 0 2.4
θ0.8 0 -0.8 0 0.4 0 0.8 4
θ0 0.8 0 0 0 0 0 0.8 3.2
 
 
 
  
1
4
Δ 2
3
0.4451 symΔ
Κ = 4.83×10 × -0.6177 1.276Δ
0.2302 -1.0552 1.811Δ
Contd..
4/15

T.K. Datta
 Push over analysis is a good nonlinear static
(substitute) analysis for the inelastic dynamic
analysis.
 It provides load Vs deflection curve from rest to
ultimate failure.
 Load is representative of equivalent static load
taken as a mode of the structure & total load
is conveniently the base shear.
 Deflection may represent any deflection & can
be conveniently taken as the top deflection.
Push over analysis 5/1

T.K. Datta
 It can be force or displacement control depending
upon whether force or displacement is given an
increment.
 For both , incremental nonlinear static analysis is
‘performed by finding matrix at the beginning
of each increment.
 Displacement controlled pushover analysis is
preferred because, the analysis can be carried
out up to a desired displacement level.
 Following input data are required in addition to
the fundamental mode shape(if used).
tK
Contd.. 5/2

T.K. Datta
 Assumed collapse mechanism
 Moment rotation relationship of yielding
section.
 Limiting displacement.
 Rotational capacity of plastic hinge.
Contd..
 Displacement controlled pushover analysis is
carried out in following steps:
 Choose suitable
 Corresponding to , find
1δ∆
1∆δ
5/3
rr
φδδ ×∆=∆ 11

T.K. Datta
 Obtain ; obtain
 At nth increment,
 At the end of each increment , moments are
checked at all potential locations of plastic
hinge.
 For this, is calculated from condensation
relationship.
 If , then ordinary hinge is assumed
at that section to find K for subsequent
increment.
Contd..
1∆Κ=∆ δp
∑∆= iBn VBV 1 1n iδ∆ = ∆∑
1BV∆
nθ
PMM =||
5/4

T.K. Datta
Contd..
 Rotations at the hinges are calculated at each
step after they are formed.
 If rotational capacity is exceeded in a plastic
hinge, rotational hinge failure precedes the
mechanism of failure.
 is traced up to the desired displacement
level.
iB 1iV VsΔ
Example 6.6
 Carry out an equivalent static nonlinear analysis
for the frame shown in Fig 6.19.
5/5

T.K. Datta
Cross
section
Location b (mm) d (mm) (kNm) (rad) (rad)
C1 G,1st
, 2nd
400 400 168.9 9.025E-3 0.0271
C2 3rd
,4th
, 5th
& 6th
300 300 119.15 0.0133 0.0399
B1 G,1st
, 2nd
400 500 205.22 6.097E-3 0.0183
B2 3rd
,4th
, 5th
& 6th
300 300 153.88 8.397E-3 0.0252
yMyθmaxθ
Contd..
3m
3m
3m
3m
3m
4m 4m
3m
3m
yM
yθ cθ
Frame
Moment rotation
curve for beams
Fig.6.19
Table 6.2
5/6

T.K. Datta
Contd..
D (m)
Base shear
(KN) Plastic Hinges at section
0.110891 316.825 1
0.118891 317.866 1,2
0.134891 319.457 1,2,3
0.142891 320.006 1,2,3,4
0.150891 320.555 1,2,3,4,5
0.174891 322.201 1,2,3,4,5,6
0.190891 323.299 1,2,3,4,5,6,7
0.206891 324.397 1,2,3,4,5,6,7,8
0.310891 331.498 1,2,3,4,5,6,7,8,9
0.318891 332.035 1,2,3,4,5,6,7,8,9,10
0.334891 333.11 1,2,3,4,5,6,7,8,9,10,11
0.350891 334.185 1,2,3,4,5,6,7,8,9,10,11,12
0.518891 342.546 1,2,3,4,5,6,7,8,9,10,11,12,13
0.534891 343.207 1,2,3,4,5,6,7,8,9,10,11,12,13,14
0.622891 346.843 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15
1.448699 307.822 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16
1.456699 308.225 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17
Table 6.3
 Solution is obtained by SAP2000.
5/7

T.K. Datta
Contd..
0.9143
1
0.7548
0.5345
0.3120
0.1988
0.0833
Fig.6.20
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
0
100
200
300
400
Baseshear(kN)
Roof displacement (m)
Fig.6.21
5/8

T.K. Datta
Contd..
2 1
3 3 3
4
5
16 17
5
67
8 8
12
11
13
10 9
14 1515
Fig.6.22
5/9

T.K. Datta
Ductility & Inelastic spectrum
 A structure is designed for a load less than
that obtained from seismic coefficient method
or RSA (say, for
 The structure will undergo yielding, if it is
subjected to the expected design earthquake.
 The behavior will depend upon the force
deformation characteristics of the sections.
 The maximum displacements & deformations
of the structure are expected to be greater than
the yield displacements.
)43(./ −≈RRVB
6/1

T.K. Datta
Contd..
 How much the structure will deform beyond
the yield limit depends upon its ductility;
ductility factor is defined as
xmμ = ( 6.35a )
xy
For explaining
ductility , two SDOFs
are considered with
elasto – plastic
behavior & the other a
corresponding
elastic system shown
in Fig 6.23.
f
x
yf
yx ox
of
Stiffness k
Elastic
Elasto-plastic
mx
Fig. 6.23
6/2

T.K. Datta
 means that the strength of the SDOF
system is halved compared to the elastic system.
 With the above definitions, equation of motion
of SDOF system becomes:
2=YR
YR yf
Contd..
 An associated factor, called yield reduction
factor, is defined as inverse of :
Y Y
Y
0 0
f x
f = = ( 6.35b)
f x
ym m
y
0 y 0 y
xx x μ
= × =μ f = ( 6.36)
x x x R
6/3

T.K. Datta
&
& y
y
2y
y n 0 y
f( x,x)
f( x,x)= ; x( t)=μ( t)x ;
f
f
a = =ω x f
m
Contd..
&& & & &&
&&
&& & &
2
n n y y g
g2 2
n n y n
y
x +2ξω x +ω x f( x,x)=- x ( 6.37)
x
μ+2ξω x +ω x f( μ,μ)=- ω ( 6.38)
a
 depends upon .,, yn fξωµ
6/4

T.K. Datta
 Time history analysis shows the following :
 For , responses remain within elastic
limit & may be more than that for .
 For , two counteracting effects take
place (i) decrease of response due to
dissipation of energy (ii) increase of response
due to decreased equivalent stiffness.
 Less the value of , more is the permanent
deformation at the end .
1=Yf
1<Yf
1<Yf
Yf
Contd..
 is known if for a & can be
calculated.
µ mx Yf 0x
6/5

T.K. Datta
 Effect of time period on are
illustrated in Fig 6.24.
 For long periods, &
independent of ; .
 In velocity sensitive region, may be
smaller or greater than ; not significantly
affected by ; may be smaller or larger
than .
 In acceleration sensitive region, ;
increases with decreasing ; ;
for shorter period, can be very high
(strength not very less).
Yf
, , ,m o Yx x fµ
goom xxx ≈≈
Yf
YR=µ
mx
ox
µ
YR
Contd..
om xx >
TfY & YR>µ
µ
6/6

T.K. Datta
Contd..
0
1
m g y
x x f =
0m g
xx
0.125
y
f =
0.25
y
f =
0.5
y
f =
Disp.
sensitive
Vel.
sensitive
Acc.
sensitive
0.01 0.05 0.1 0.5 1 5 10 50 100
0.001
0.005
0.01
0.05
0.1
0.5
1
5
10
0.001
0.005
0.01
0.05
0.1
0.5
1
5
10
Ta=0.035
Tf=15
T
b=0.125
Tc=0.5
Td=3
Te=10
Spectral Regions
x0/xg0orxm/xg0
T (sec)
n
0.5
y
f =
0.25
y
f =
0.125yf =
1
y
f =
Disp.
sensitive
Vel.
sensitive
Acc.
sensitive
0.01 0.050.1 0.5 1 5 10 50 100
0.1
0.5
1
5
10
Spectral Regions
xm/x0 0.1
0.5
1
5
10
T (sec)
n
Ta=0.035
Tf=15
Tb=0.125
Tc=0.5
Td=3
Te=10
Normalized peak
deformations for elasto-plastic
system and elastic system
Ratio of the
peak deformationsFig. 6.24
6/7

T.K. Datta
 Inelastic response spectrum is plotted for :
 For a fixed value of , and plots of
against are the inelastic spectra or ductility
spectra & they can be plotted in tripartite plot.
 Yield strength of the E-P System.
 Yield strength for a specified is difficult to
obtain; but reverse is possible by interpolation
technique.
µ ξ YYY AVD ,,
nT
Inelastic response spectra
2
y y y n y y n yD = x V =ω x A = ω x ( 6.39)
y yf = mA ( 6.40)
µ
6/8

T.K. Datta
 For a given set of & , obtain response for
E-P system for a number of .
 Each solution will give a ; , is
maximum displacement of elastic system.
 From the set of & , find the desired &
corresponding .
 Using value, find for the E-P system.
 Through iterative process the desired
and are obtained .
nT ξ
Yf
µ oo xKf = ox
f µ µ
Yf
Contd..
Yf µ
µ
Yf
6/9

T.K. Datta
Contd..
 For different values of , the process is
repeated to obtain the ductility spectrum.
nT
0 0.5 1 1.5 2 2.5 3
0
0.2
1
0.4
0.6
0.8
1µ =
1.5
2
4
8
(sec)nT
fy/w=Ay/g
Fig. 6.25
6/10

T.K. Datta
 From the ductility spectrum, yield strength
to limit for a given set of & can be
obtained.
 Peak deformation .
µ nT ξ
2
m Y Y nx =μ x =μ A ω
1=µ
f nT
µ
Contd..
0.01 0.050.1 0.5 1 5 10 50 100
0.05
0.1
0.5
1
Ta=0.035
Tf=15
Tb=0.125
Tc=0.5
Td=3
Te=10
fy
20
1
5
10
2
Tn(sec)
Ry
0.12
0.195
0.37
8µ=
4µ=
2µ=
1.5µ=
0.0
 If spectrum for
is known ,it is possible
to plot vs. for
different values of .
 The plot is shown in
Fig. 6.26.
Fig. 6.26
6/11

T.K. Datta
 Above plot for a number of earthquakes are
used to obtain idealized forms of & .f nT
Contd..





1 T < Tn a
-1/2f = ( 2μ -1) T < T < T ( 6.41)y n cb
-1μ T > Tn c
0.01 0.050.1 0.5 1 5 10 50 100
0.05
0.1
0.5
1
Ta=1/33
Tf=33
Tb=
=
1
1
/
/
8
2Tc
Te=10
fy
Tn (sec)
8µ=
4µ=
2µ=
0.2
1µ=
Tc'
1.5µ=
Fig. 6.27
7/1

T.K. Datta
Construction of the spectra
 As , idealized inelastic design
spectrum for a particular can be constructed
from elastic design spectrum.
 Inelastic spectra of many earthquakes when
smoothed compare well with that obtained as
above.
 Construction of the spectrum follows the steps
below :
 Divide constant A-ordinates of segment
by to obtain .
YfYR 1=
µ
cb − 12 −= µYR '' cb −
7/2

T.K. Datta
 Similarly, divide V ord
inates of segments
by ; to
get ; D ordina-
tes of segments
by to get ;
ordinate by to
get .
 Join & ; draw
for
; take as the same
; join .
 Draw for
)( dc − µ=YR
'' dc −
)( ed −
µ=YR '' ed −
f
'f
µ
'f
'e
µgoY xD = sTn 33>
'a
a ''&ba
goY xA &&= sTn
33
1
<
Contd..
Natural vibration period Tn (sec) (log
scale)
goV= &
Ta=1/33 sec Tf=33 sec
Tb=1/8 sec Te=10 sec
a
a'
b
b'
c
c'
d
d'
e
e'
f
f '
Elastic design
spectrum
Inelastic design
spectrum
Pseudo-velocityVorVy(logscale)
/V µ
A=xgo
A=α
Axgo
D=xgo
D/µ
D/µ
D=
xgo
A/v2µ-1
αV
α
D
x
.
Illustration of the
Method
Fig. 6.28
7/3

T.K. Datta
Example 6.7 :
Construct inelastic design spectrum from
the elastic spectrum given in Fig 2.22.
The inelastic design spectrum is drawn & shown
in Fig 6.28b.
2=µ
Contd..
Inelastic design spectrum for µ = 2
Fig. 6.28b
0.01 0.02 0.05 0.1 0.2 0.3 0.50.7 1 2 3 4 567 10 20 30 50 70100
0.001
0.002
0.003
0.004
0.005
0.007
0.01
0.02
0.03
0.04
0.05
0.07
0.1
0.2
0.3
0.4
0.5
0.7
1
2
3
4
5
7
10
7/4

T.K. Datta
Ductility in multi-storey frames
 For an SDOF, inelastic spectrum can provide
design yield strength for a given ; maximum
displacement under earthquake is found as
 For multi-storey building , it is not possible
because
 It is difficult to obtain design yield strength of
all members for a uniform .
 Ductility demands imposed by earthquake on
members widely differ.
 Some studies on multi - storey frames
are summarized here to show how ductility
demands vary from member to member
when designed using elastic spectrum for
uniform .
µ
yxµ
µ
µ
7/5

T.K. Datta
 Shear frames are designed following seismic
coefficient method ; is obtained using
inelastic spectrum of El centro earthquake
for a specified ductility & storey shears
are distributed as per code.
 Frames are analysed assuming E-P behaviour
of columns for El centro earthquake.
 The storey stiffness is determined using
seismic coefficient method by assuming
storey drifts to be equal.
BYV
Contd.. 7/6

T.K. Datta
 Results show that
For taller frames, are larger in upper & lower
stories; decrease in middle storeys.
Deviation of storey ductility demands from the
design one increases for taller frames.
In general demand is maximum at the first
storey & could be 2-3 times the design
Study shows that increase of base shear by
some percentage tends to keep the demand
within a stipulated limit.
µ
µ
µ
Contd..
µ
7/7

T.K. Datta

Seismic Analysis of Structures - III

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Seismic Analysis of Structures - III