1. Inelastic Seismic Response of Structures
Chapter - 6
Inelastic Seismic Response of Structures

2. Introduction Under relatively strong earthquakes, structures
1/1
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.

3. Contd.. For this, nonlinear time history analysis of SDOF
1/2
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.

4. Non linear dynamic analysis
1/3
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.

5. Contd..
1/4 f
Loading
Unloading x D
Fig.6.1

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

7. Contd.. Equation of motion for non linear analysis takes the form
1/6
Contd..
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.

8. Contd.. and are taken as that at the beginning of
1/7
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

9. Contd..
1/8

10. Contd.. For more accurate value of acceleration, it is
1/9
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.

11. Elasto-plastic non linearity
1/10
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

12. Contd.. Use Eq. 6.7, find Plastic to plastic state
1/11
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

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

14. 2/2

15. Contd..
2/3

16. Solution for MDOF System
2/4
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
0.5m m 3 y x 2 1 p V
Fig.6.4
For the solution of Eqn. 6.1, state of the yield
section is examined at each time step.

17. Contd.. Depending upon the states of yield
2/5
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.

18. Contd.. Example 6.2: Refer to Fig 6.5; K/m = 100; m = 1 kg;
2/6
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:

19. Force displacement curve of the column
Contd..
2/7
k/2 m 3m 1 x 2 3 y f
0.15m g = m
3 storey frame
Force displacement curve of the column
Fig.6.5

20. Contd..
2/8

21. Contd..
2/9

22. Bidirectional Interaction
3/1
Bidirectional Interaction
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.

23. Contd.. If the interaction is ignored, yielding in two
3/2
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

24. Contd.. 3/3 Fig.6.6 x e y D Colm. 1 Colm. 2 Colm. 3 Colm. 4 CR Y X
C.M.
Fig.6.6

25. Contd.. Transient stiffness remaining constant over is given by
3/4
Transient stiffness remaining constant over
is given by
The elements of the modification matrix are

26. Contd.. When any of the column is in the full plastic state
3/5
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

27. Contd.. For , curve is circular ; ,
3/6
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.

28. Contd.. If any element violates the yield condition at the
3/7
Contd..
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.

29. Contd.. After convergence , forces are calculated &
3/8
Contd..
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.

30. 3/9
Contd..
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:

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

32. 3/11
Solution:
Forces in the columns are pulled back (Eq. 6.23) & displacements at the centre

33. Contd..
3/12

34. Contd..
3/13
With the e values calculated as above, the forces in the columns are pulled back

35. Contd..
3/14

36. Contd..
3/15

37. Contd..
3/16

38. Contd..
3/17
Because yield condition is practically satisfied, no further iteration is required.

39. Multi Storey Building frames
4/1
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 ).

40. Contd.. At the end of each , velocity is calculated at
4/2
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. ( ).
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 & 6.11 are for the result of elasto
-plastic case Figs & 6.13 are for the result
of bilinear case
Moment in Fig 6.12 does not remain constant
over time unlike elasto-plastic case.

41. Force-displacement curve of column
Contd..
4/3 k 3m
1.5k A m 1 x 2 3
k = kN/m
m =  103 kg i K d
0.1 di KK = m
Displacement (m)
346.23kN
Force (kN)
Frame
Force-displacement curve of column
Fig.6.9

42. Contd..
4/4
Fig.6.10
Fig.6.11

43. Contd..
4/5
Fig.6.12
Fig.6.13

44. Contd.. For nonlinear moment rotation relationship,
4/6
Contd..
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

45. Contd.. Non zero elements of Kp are computed using
4/7
Non zero elements of Kp are computed using
Eqns & 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.

46. Contd.. For 2D & 3D frames having weak beam strong
4/8
Contd..
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).
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.

47. Moment-rotation relationship of elasto-plastic beam
Contd..
4/9
Elasto-plastic nature of the yield section is
shown in Fig 6.16.
Considering anti-symmetry : p q
M1, M2
Mp1 = Mp2 = Mp3
Moment-rotation relationship
of elasto-plastic beam
fig. 6.16

48. 4/10

49. Contd.. Equation of motion for the frame is given by:
4/11
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

50. Contd.. & are obtained using Eqn. 6.29b
4/12
Contd..
& 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.)

51. Force-displacement curve
Contd..
4/13
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 k 3m 5m 1 θ 2 3 4 5 6 D
E = 2.48  107 kN/m2
Beam 30  40 cm
Column 30  50 cm
Frame
50KN-m M
θY = rad θ
Force-displacement curve
Fig. 6.17

52. Contd.. Table shows that sections 1 & 2 undergo yielding;
4/14
Table 6.1
Joint
Time
Step x θ M
sec m
m/s
m/s2
rad
rad/s
rad/s2
kNm 1
1.36
0.0341
0.013
-0.452 50 3
0.0883
0.014
-0.297
-23.18 5
0.1339
-0.098
42.89 2
-50 4
23.18 6
-42.89
Table shows that sections 1 & 2 undergo yielding;
recognising this, stiffness matrices are given below:

53. Contd..
4/15

54. Push over analysis Push over analysis is a good nonlinear static
5/1
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.

55. Contd.. It can be force or displacement control depending
5/2
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).

56. Contd.. Assumed collapse mechanism
5/3
Assumed collapse mechanism
Moment rotation relationship of yielding
section.
Limiting displacement.
Rotational capacity of plastic hinge.
Displacement controlled pushover analysis is
carried out in following steps:
Choose suitable
Corresponding to , find

57. Contd.. Obtain ; obtain At nth increment,
5/4
Contd..
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.

58. Contd.. Rotations at the hinges are calculated at each
5/5
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.
Example 6.6
Carry out an equivalent static nonlinear analysis
for the frame shown in Fig 6.19.

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

60. Plastic Hinges at section
Contd..
5/7
Solution is obtained by SAP2000.
D (m)
Base shear (KN)
Plastic Hinges at section 1
1,2
1,2,3
1,2,3,4
1,2,3,4,5
1,2,3,4,5,6
1,2,3,4,5,6,7
1,2,3,4,5,6,7,8
1,2,3,4,5,6,7,8,9
1,2,3,4,5,6,7,8,9,10
333.11
1,2,3,4,5,6,7,8,9,10,11
1,2,3,4,5,6,7,8,9,10,11,12
1,2,3,4,5,6,7,8,9,10,11,12,13
1,2,3,4,5,6,7,8,9,10,11,12,13,14
1,2,3,4,5,6,7,8,9,10,11,12,13,14,15
1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16
1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17
Table 6.3

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

62. Contd..
5/9 2 1 3 4 5 16 17 6 7 8 12 11 13 10 9 14 15
Fig.6.22

63. Ductility & Inelastic spectrum
6/1
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.

64. Contd.. How much the structure will deform beyond
6/2
How much the structure will deform beyond
the yield limit depends upon its ductility;
ductility factor is defined as f x y o
Stiffness k
Elastic
Elasto-plastic m
For explaining ductility , two SDOFs are considered with elasto – plastic behavior & the other a corresponding elastic system shown in Fig 6.23.
Fig. 6.23

65. Contd.. An associated factor, called yield reduction
6/3
An associated factor, called yield reduction
factor, is defined as inverse of :
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:

66. Contd..
6/4
depends upon

67. Contd.. Time history analysis shows the following :
6/5
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 .
is known if for a & can be
calculated.

68. Contd.. Effect of time period on are illustrated in Fig 6.24.
6/6
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).

69. Ratio of the peak deformations
Contd..
6/7 1 m g y x f = xx
0.125
0.25
0.5
Disp.
sensitive
Vel.
Acc.
0.01
0.05
0.1 5 10 50
100
0.001
0.005 T a . 3 b 2 c d e
Spectral Regions / o r
(sec) n
0.5 y f =
0.25
0.125 1
Disp.
sensitive
Vel.
Acc.
0.01
0.05
0.1 5 10 50
100
Spectral Regions x m / T
(sec) n a . 3 b 2 c d e
Normalized peak deformations for elasto-plastic system and elastic system
Ratio of the peak deformations
Fig. 6.24

70. Inelastic response spectra
6/8
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.

71. Contd.. Using value, find for the E-P system.
6/9
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 .

72. Contd.. For different values of , the process is
6/10
For different values of , the process is
repeated to obtain the ductility spectrum.
0.5 1
1.5 2
2.5 3
0.2
0.4
0.6
0.8 m = 4 8
(sec) n T f y / w A g
Fig. 6.25

73. Contd.. From the ductility spectrum, yield strength
6/11
From the ductility spectrum, yield strength
to limit for a given set of & can be
obtained.
Peak deformation
0.01
0.05
0.1
0.5 1 5 10 50
100 T a = . 3 f b 2 c d e y 20 n
(sec) R
0.12
0.195
0.37 8 m 4
1.5
0.0
If spectrum for
is known ,it is possible
to plot vs for
different values of .
The plot is shown in
Fig
Fig. 6.26

74. Contd.. Above plot for a number of earthquakes are
7/1
Above plot for a number of earthquakes are
used to obtain idealized forms of &
0.01
0.05
0.1
0.5 1 5 10 50
100 T a = / 3 f b 8 2 c e y n
(sec) m 4
0.2 c'
1.5
Fig. 6.27

75. Construction of the spectra
7/2
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

76. Contd.. Similarly, divide V ord inates of segments by ; to
7/3
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
Natural vibration period Tn (sec) (log scale) go V =
& T a
=1/33 sec f
=33 sec b
=1/8 sec e
=10 sec a' b' c c' d d' e'
f '
Elastic design
spectrum
Inelastic design
Pseudo-velocity or y
(log scale) / m A= x A D= D/ A/ v
2µ-1 V D .
Illustration of the Method
Fig. 6.28

77. Inelastic design spectrum for  = 2
Contd..
7/4
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.
Inelastic design spectrum for  = 2
Fig. 6.28b

78. Ductility in multi-storey frames
7/5
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 .

79. Contd.. Shear frames are designed following seismic
7/6
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.

80. Contd.. Results show that
7/7
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.