Inelastic Seismic Response of Structures Introduction Under relatively strong earthquakes structures undergo inelastic deformation due to current seismic design philosophy Therefore structures should have sufficient ID: 311122
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Slide1
Chapter - 6
Inelastic Seismic Response of StructuresSlide2
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.
1/1Slide3
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..
1/2Slide4
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.
1/3Slide5
Contd..
f
Loading
Loading
Unloading
Unloading
x
x
D
Fig.6.1
1/4Slide6
Contd..
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
x
y
f
y
x
f
x
y
f
y
f
f
x
y
x
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 curve
Fig.6.2
1/5Slide7
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..
1/6Slide8
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
Contd..
1/7Slide9
Contd..
1/8Slide10
10
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 k
t
.
Contd..
1/9Slide11
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
1/10Slide12
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
1/11Slide13
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:
Contd..
2/1
x
x
f
m
c
g
x
..
SDOF system with non-linear spring
0.15mg
0.0147m
x
x
f
Force-displacement behaviour of the spring
Fig . 6.3Slide14
14
2/2Slide15
Contd..
2/3Slide16
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
3
y
x
2
y
x
1
y
x
3
p
V
2
p
V
1
p
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.
2/4Slide17
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..
2/5Slide18
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..
2/6Slide19
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 g
y
f
=
0.01475m
y
x
=
3 storey frame
Force displacement
curve of the column
Contd..
Fig.6.5
2/7Slide20
Contd..
2/8Slide21
Contd..
2/9Slide22
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
3/1Slide23
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..
3/2Slide24
Contd..
x
ey
e
D
D
Colm. 1
Colm. 2
Colm. 3
Colm. 4
CR
Y
X
C.M.
Fig.6.6
3/3Slide25
Transient stiffness remaining constant over
is given by
The elements of the modification matrix are
Contd..
3/4Slide26
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
Contd..
3/5Slide27
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.
Contd..
3/6Slide28
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.
Contd..
3/7Slide29
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.
Contd..
3/8Slide30
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:Contd..
3/9Slide31
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
Slide32
32
3/11
Solution:
Forces in the columns are pulled back (Eq. 6.23) & displacements at the centreSlide33
Contd..
3/12Slide34
With the e values calculated as above, the forces in the columns are pulled back
Contd..
3/13Slide35
Contd..
3/14Slide36
36
Contd..
3/15Slide37
Contd..
3/16Slide38
Contd..
3/17
Because yield condition is practically satisfied, no further iteration is required.Slide39
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 ).
4/1Slide40
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. ( ).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.
4/2Slide41
4/3
k
k
k
k
3m
3m
3m
1.5k
A
1.5k
m
m
m
1
x
2
x
3
x
k = 23533 kN/m
m = 235.33
10
3
kg
i
K
d
K
0.1
di
KK
=
0.01471m
Displacement (m)
346.23kN
Force (kN)
Frame
Force-displacement
curve of column
Contd..
Fig.6.9 Slide42
4/4
Contd..
Fig.6.10 Fig.6.11 Slide43
4/5
Contd..
Fig.6.13 Fig.6.12 Slide44
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 byContd..
4/6Slide45
Non zero elements of K
p 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..4/7Slide46
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.
4/8Slide47
Elasto-plastic nature of the yield section is
shown in Fig 6.16.
Considering anti-symmetry : Contd..4/9
p
q
M
1
, M
2
M
p
1
= M
p
2
= M
p
3
q
Moment-rotation relationship
of
elasto
-plastic beam
fig. 6.16Slide48
4/10Slide49
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 4/11Slide50
& 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.)Contd..
4/12Slide51
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 6.14/13 k
k
k
k
3m
3m
3m
5m
1
θ
2
θ
3
θ
4
θ
5
θ
6
θ
k
k
1
D
2
D
3
D
E = 2.48
10
7
kN/m
2
Beam 30
40 cm
Column 30
50 cm
Frame
1
3
5
2
4
6
50KN-m
M
θ
Y
= 0.00109 rad
θ
Force-displacement curve
Fig. 6.17Slide52
52
4/14
JointTimeStepx
θ
M
sec
m
m/s
m/s
2
rad
rad/s
rad/s
2
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
Table shows that sections 1 & 2 undergo yielding;
recognising this, stiffness matrices are given below:
Table 6.1
Contd..Slide53
Contd..
4/15Slide54
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 analysis5/1Slide55
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).Contd..5/2Slide56
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
5/3Slide57
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..
5/4Slide58
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.Example 6.6
Carry out an equivalent static nonlinear analysis
for the frame shown in Fig 6.19
.
5/5Slide59
Cross section
Location
b (mm)d (mm)(kNm)(rad)(rad)C1G,1st, 2nd 400400168.9
9.025E-3
0.0271
C2
3
rd
,4
th
, 5
th
& 6
th
300
300
119.15
0.0133
0.0399
B1
G,1
st
, 2
nd 400
500
205.226.097E-30.0183B23rd ,4th, 5th & 6th
300
300
153.888.397E-3
0.0252
Contd..
3m
3m
3m
3m
3m
4m
4m
3m
3m
y
M
y
q
c
q
Frame
Moment rotation
curve for beams
Fig.6.19
Table 6.2
5/6Slide60
Contd..
D (m)
Base shear (KN)Plastic Hinges at section0.110891316.82510.118891317.8661,20.134891319.4571,2,3
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
0.518891342.546
1,2,3,4,5,6,7,8,9,10,11,12,13
0.534891343.207
1,2,3,4,5,6,7,8,9,10,11,12,13,140.622891
346.843
1,2,3,4,5,6,7,8,9,10,11,12,13,14,151.448699
307.822
1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,161.456699
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.
5/7Slide61
Contd..
0.9143
10.7548
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
Base shear (kN)
Roof displacement (m)
Fig.6.21
5/8Slide62
Contd..
2
1
3
3
3
4
5
16
17
5
6
7
8
8
12
11
13
10
9
14
15
15
Fig.6.22
5/9Slide63
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.6/1Slide64
Contd..
How much the structure will deform beyond
the yield limit depends upon its ductility; ductility factor is defined as For explaining ductility , two SDOFs are considered with elasto – plastic behavior & the other a corresponding elastic system shown in Fig 6.23.
f
x
y
f
y
x
o
x
o
f
Stiffness k
Elastic
Elasto-plastic
m
x
Fig. 6.23
6/2Slide65
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:Contd..
An associated factor, called yield reduction
factor, is defined as inverse of
:
6/3Slide66
Contd..
depends upon
6/4Slide67
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 .
Contd..
is known if for a & can be
calculated.
6/5Slide68
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).
Contd..
6/6Slide69
Contd..
0
1mgyxxf=0mgxx
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
T
a
=
0
.
0
3
5
T
f
=
1
5
T
b
=
0
.
1
2
5
T
c
=
0
.
5
T
d
=
3
T
e
=
1
0
Spectral Regions
x
0
/
x
g
0
o
r
x
m
/
x
g
0
T
(sec)
n
0.5
y
f
=
0.25
y
f
=
0.125
y
f
=
1
y
f
=
Disp.
sensitive
Vel.
sensitive
Acc.
sensitive
0.01
0.05
0.1
0.5
1
5
10
50
100
0.1
0.5
1
5
10
Spectral Regions
x
m
/
x
0
0.1
0.5
1
5
10
T
(sec)
n
T
a
=
0
.
0
3
5
T
f
=
1
5
T
b
=
0
.
1
2
5
T
c
=
0
.
5
T
d
=
3
T
e
=
1
0
Normalized peak
deformations for elasto-plastic
system and elastic system
Ratio of the
peak deformations
Fig. 6.24
6/7Slide70
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.
Inelastic response spectra
6/8Slide71
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 .
Contd..
6/9Slide72
Contd..
For different values of , the process is
repeated to obtain the ductility spectrum. 0
0.5
1
1.5
2
2.5
3
0
0.2
1
0.4
0.6
0.8
1
m
=
1.5
2
4
8
(sec)
n
T
f
y
/
w
=
A
y
/
g
Fig. 6.25
6/10Slide73
From the ductility spectrum, yield strength
to limit for a given set of & can be
obtained. Peak deformation .
Contd..
0.01
0.05
0.1
0.5
1
5
10
50
100
0.05
0.1
0.5
1
T
a
=
0
.
0
3
5
T
f
=
1
5
T
b
=
0
.
1
2
5
T
c
=
0
.
5
T
d
=
3
T
e
=
1
0
f
y
20
1
5
10
2
T
n
(sec)
R
y
0.12
0.195
0.37
8
m
=
4
m
=
2
m
=
1.5
m
=
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
6/11Slide74
Above plot for a number of earthquakes are
used to obtain idealized forms of &
. Contd..
0.01
0.05
0.1
0.5
1
5
10
50
100
0.05
0.1
0.5
1
T
a
=
1
/
3
3
T
f
=
3
3
T
b
=
=
1
1
/
/
8
2
T
c
T
e
=
1
0
f
y
T
n
(sec)
8
m
=
4
m
=
2
m
=
0.2
1
m
=
T
c'
1.5
m
=
Fig. 6.27
7/1Slide75
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 .
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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 .
Contd..
Natural vibration period
T
n
(sec) (log scale)
go
V
=
&
T
a
=1/33 sec
T
f
=33 sec
T
b
=1/8 sec
T
e
=10 sec
a
a'
b
b'
c
c'
d
d'
e
e'
f
f '
Elastic design
spectrum
Inelastic design
spectrum
Pseudo-velocity
V
or
V
y
(log scale)
/
V
m
A=
x
go
A=
A
x
go
D=
x
go
D/
µ
D/
µ
D=
x
go
A/
v
2µ-1
V
D
x
.
Illustration of the Method
Fig. 6.28
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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.Contd..Inelastic design spectrum for = 2 Fig. 6.28b
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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 .
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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.Contd..
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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.Contd..
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