Chapter - 6 Inelastic Seismic Response of Structures

T.K. Datta Department Of Civil Engineering, IIT Delhi

Response Spectrum Method Of Analysis

Chapter - 6

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 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



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



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



Contd..f

Loading

LoadingUnloading

Unloading

x

xD

Fig.6.1

1/4



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

yx

f

x

yf

yff

xyx

yf

yx

f

x

Variation of force with displacement under cyclic loading

Idealized model of forcedisplacement curve

Idealized model of forcedisplacement curveFig.6.2

1/5



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 xD D D D

KtCt

Ct Kt



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 DD &

tD

Method

Contd..

k2

2k k

Δx=Δtx +δΔtΔx ( 6.3)ΔtΔx=Δtx + x +β Δt Δx ( 6.4)2

1/7



Contd..

k k2

k k

t t 2

g t k t k

k+1 k k+1 k k+

1 1 1Δx= Δx- x - x ( 6.5)βΔt 2ββ Δtδ δ δΔx= Δx- x +Δt 1- x ( 6.6)βΔt β 2β

kΔx=Δp ( 6.7)δ 1k=k + c + m ( 6.8a)βΔt β Δt

m δ m δΔp=-mΔx + + c x + +Δt -1 c x ( 6.8b)βΔt β 2β 2βx =x +Δx x =x +Δx x 1 k=x +Δx ( 6.9)

1/8


Response Spectrum Method Of Analysis10

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



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 xD D

1/10



p e t( Δx) for( 1-a )Δpwithk =0e pΔx=( Δx) +( Δx)

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( Δx) 1-e Δp tk ≠ 0aΔx=( Δx) +Factored Δx

x=0xD

1/11



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

Contd.. 2/1

xxf

m

c

gx..

SDOF system with non-linear spring

0.15mg

0.0147m x

xf

Force-displacement behaviour of the spring

Fig . 6.3



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 =100Nmt

-1k=10140NmΔx =-0.00312ggΔp=37.55N

-1Δx=0.0037m; Δx=-0.01ms ; Δf=k Δxt( f ) =1.7243N>0.15mgx t+Δt( f ) +eΔxk =0.15mg( e=0.3176)x t t

2/2



Contd..

X +ΔX =0t 1

-1kΔx = 1-e Δp; k =0; Δx=0.00373m; Δx=-0.00749mst2-1x =x +Δx +Δx =0.01725m x =0.1827mst 1t+Δt 2 t+Δt

P -c x - ftk+1 k+1 x( k+1) -1x = = 0.279msk+1 mAt 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=-0.033t tt+Δt t+Δtx from Eqn =3.28; f =f +k Δx =1.4435Nxt tt+Δt t+Δt 2

2/3



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

3yx

2yx

1yx

3pV

2pV

1pV

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



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 1K =K + C + M ( 6.12a)βΔt β Δt

M δ M δΔp=-M rΔx + + C x + +Δt -1 C x ( 6.12b)βΔt β 2β 2β

2/5



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 0.15mgf = 0.95664 < 0.15mg and x>0k

0.63432 0.15mg10260 sym

δ 1K=K + C + M= -124 10260t t 2βΔt β( Δt) 0 -124 10137Δx =0.5913g

M δ M δΔp=-MIΔx + + C x + +Δt -1 C xg t tk kβΔt β 2β 2β

32.6224= 18.0256

8.43760.0032

-1Δx=K Δp= 0.00180.0009

2/6



k/2

k/2

m

m

m k/2

k/2

k/2 k/2

3m

3m

3m

1x

2x

3x

yx x

yf

0.15m gyf

0.01475myx

3 storey frame

Force displacementcurve of the column

Contd..

Fig.6.5

2/7



Contd..

0.0032 0.16KΔx= -0.0014 ; Δf= Δx= -0.072-0.0009 -0.045

e( 0.16)1.60977 0.15mg1f =f +Δf = 0.88664 f +eΔf=f + e ( -0.07) = ≤0.15mgk+1 k k k 2

0.58932 ≤0.15mge( -0.045)3e =0.136;1

e =1; e =12 3e Δx e Δx1 1 1 1

Δx =Δx = e Δx +e ( Δx -Δx ) = e Δxe 1 1 1 12 2 2 2e Δx +e ( Δx -Δx )+e( Δx -Δx ) e Δx1 1 12 2 3 3 2 3 3

e0.000435 0.13581= -0.000965 e = 0.68932

-0.001865 3.07e3

2/8



0.0028 0.00324 0.02009Δx = 0.0027 ; Δx=Δx +Δx = 0.0018 ; x =x +Δx= 0.0083312 2 k+1 k

0.0026 0.00074 0.01140.-0.0509 0

δ δ δΔx= Δx- x +Δt 1- x = -0.0406 ; x =x +Δx=k k k+1 kβΔt β 2β -0.0524

.13610.07

0.0165

e( 0.16) 0.00 0.0218 1.47151Δf= e ( -0.07) + -0.005 = -0.075 ; f =f +Δf= 0.8822 k+1 k

-0.005 -0.05 0.584e( -0.045)3-2.289

-1x =M P -C x -F = -1.7tk+1 k+1 k+1 k+1

018-2.2825

Contd.. 2/9



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



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 x θ

ex ex ey ey θ ex y ey x

K 0 K eK = 0 K K e ( 6.13a)

K e K e KK = K ; K = K ; K = K e + K e ( 6.13b)

3/2



Contd..

xe

ye

D

D

Colm. 1

Colm. 2

Colm. 3

Colm. 4

CR

Y

XC.M.

Fig.6.6

3/3



Transient stiffness remaining constant over is given by

The elements of the modification matrix are

tD

pK

Contd..

t e pK =K -K ( 6.14)

22yi xi yixi

pxi pyi pxyi pyxii i i

2 2i exi xi eyi yi

xi exi xi yi eyi yi

yixixi yi2 2

pxi pyi

B B BBK = ; K = ; K =K = ( 6.15)G G GG =K h +K h ( 6.16a)B =K h ; B =K h ( 6.16b)

VVh = ; h = ( 6.16c)V V

3/4

tK



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..

2 2yixi

ipxi pyi

VV= + ( 6.19)V V

3/5



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 1i 1i

1i

1i

Contd.. 3/6



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.

1i

Contd..

1UD taK

1 1

1

1

(6.21)

(6.22 )

(6.22 )

ta

i i

i i

force tolerance a

displacement tolerance b

D D

D D

D D

F K U

F F

U U

3/7

1 'K = K +Kta tt02



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 ii

F F

pUD

3/8



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:

tD

Contd..

1

(6.25)

(6.26)pi i pi

pi i ei i

Fw U

U U K F

D D

D D D

px py p 0 p p oBA D

op 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 =1.5MkM =2M ; k =k =k =k ;k =k =1.5k ;k =2k ; =50m

in which m =m =m=620kg and V =152.05

3/9



find Initial stiffness & stiffness at t = 1.38s, given that t = 1.36s

Contd.. 3/10

2k

k

y

x3.5m

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

Forc

e (N

)

Force-displacementcurve of column A



3/11

U UU 0.00336 0.13675 -0.16679x xxU = 0.00037 U = 0.00345 U = -0.11434y y y

0.00003 0.00311 -0.06153θ θ θk k k

V 627.27xF = V = 70.yk

Vθ k

V =102.83 V =10.10Ax AyV =154.24 V =19.56Bx By888 ; x =-0.08613g gkV =158.66 V =15.15Dx Dy773.51V =211.54 V =26.08Cx Cy

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



Contd..

2

2ex exi 0 ey eyi 0 θ 0

e

2 2yixi

ipxi pyi

A B C D

t e

t t t2

KDK = K =6k ; K = K =6k ;K = =3k 3.54186000 0 54250

K = 0 186000 5425054250 54250 1139250

VV= +V V=0.462 ; =0.465; =0.491; =0.488

K =K

δ 1K=K + C + M=K +1βΔt β( Δt)

4

g t k t k

638.6 sym0 638.6

5.425 5.425 1379.760000M= ×10

16282M δ M δΔp=-MΔx + + C U + +Δt -1 C U = 286βΔt β 2β 2β 612

3/12



With the e values calculated as above, the forces in the columns are pulled back

-1t

k+1 k

Ax

Bxk+1 k

Dx

Cx

0.0025 476.1ΔU=K Δp= 0.000001 and ΔF=K ΔU= 10.2

0.000001 181.20.0059

U =U +ΔU= 0.00040.0001

VV =183.891103.36 V =275.84F =F +ΔF= 81.09 ; ;V =275.84954.75 V =367.79

Ay

By

Dy

Cy2 2

yixii A B C D

pxi pyi

A B D CA B D C

=13.52V =20.27V =20.27V =27.03

VV= + & =1.47 ; =1.47 ; =1.47 ; =1.47 V V1 1 1 1e = 0.824 e = =0.824 e = =0.824 e = =0.824

Contd.. 3/13



5 3

2

2

0.254 0.830.00122 10 ; is calculed as 0.83 10 in which .0.00124 0.83

16268.57(1 ) 285.76

631.48

0.29167

0.007950.00

;

xee x

x

i ix y

i

xA

xBxixi

pxi

Ue = etc

U

e

K xe e

K

hhVh

V

D D D

D D

U e

p p

2

0.000590.0003953

; ;0.000390.00530.000290.00398

yA

yByiyi

yDxD pyi

yCxC

hhV

hhh Vhh

Contd.. 3/14



2 2

22 2

246.56 18.12246.56 18.12

; ; ;246.56 18.12246.56 18.12

1.9721.314

;1.3140.98

; ; ;2

xA yA

xB yBxi exi xi yi eyi yi

xD yD

xC xC

A

Bi exi xi eyi yi

D

C

yixipxi pyi p p pxyi pyxi

i i

B BB B

B K h B K hB BB By

GG

G K h K hGG

BB DK K K K K KG G

xi yi

i

B BG

Contd.. 3/15



Contd..

3

52

1

1.0

16.06 185 10

0.29 53.96 0

621

10000 10 0.16 63.85( )

0 0.54 126.6

0.2917; 0.2917

0.00260.0001 ;0.00

t

t t t

Pyi i Pxi ixp yp

Pyi Pxi

p

sym

sym

t t

K x K ye e

K K

D D

D D D

2 2

K

K K C M = K M =

U K p U

0.002598 0.00009830.002598 0.0001018

;0.002602 0.00009830.002602 0.0001018

x py

D

U

3/16



Contd..

2 2

9.78151.91311.54228.043

;11.37227.74510.98303.471

1.002; 1.002; 1.0; 1.00

ex px pxy pxpi

pxy ey py pyi i

AyAx

ByBx

DyDx

CyCx

yixii

pxi pyi

A C D

K K K UK K K U

VVVVVVVV

VVV V

D D D

B

V

3/17

Because yield condition is practically satisfied, no further iteration is required.



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 ).

tD

tD

tt D

4/1



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. ( ).

tD

tD

tD~t smallD

Contd..

Example 6.4 Find the time history of moment at A & the force-displacement plot for the frame shown in Fig 6.9under 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 2.0D

4/2



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

iK

dK

0.1diKK

0.01471m Displacement (m)

346.23kN

Forc

e (k

N)Frame

Force-displacementcurve of column

Contd..

Fig.6.9



4/4

-600000

-400000

-200000

0

200000

400000

600000

0 5 10 15 20 25 30

Time (sec)M

omen

t (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)

Forc

e (N

)

Contd..

Fig.6.10

Fig.6.11



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)Shea

r For

ce(N

)

-800000-600000-400000-200000

0200000400000600000800000

0 5 10 15 20 25 30

Time (sec)

Mom

ent (

N-m

)

Contd..

Fig.6.13

Fig.6.12



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

tD

tD

Contd.. 4/6

tK

(6.27)t e p K K K



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



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



Elasto-plastic nature of the yield section is shown in Fig 6.16.

Considering anti-symmetry :

Contd.. 4/9

p

M1, M2

Mp1 = Mp2 = Mp3

Moment-rotation relationship

of elasto-plastic beamfig. 6.16



4/10

1

22 2

11

22 2

2

kl klk -k - -2 2Δkl-k 2k 0 Δ2K= ( 6.28a)kl kl kl kl θ- α +0.672 2 2 6 θ

kl kl kl- 0 α +1.332 6 2

-1Δ Δ Δθ θ θΔ

-11-1

θ 22

-11

2-1

1Δ

2

K =K -K K K ( 6.28b)3 α +0.67 16K = ( 6.29a)1 3 α +1.33kl

3 α +0.67 11 -13θ= Δ ( 6.29b)1 3 α +1.331 0l

3 α +0.67 11 -1 -1 -1 1 -13kK =k - ( 6.30)1 3 α +1.33-1 2 1 0 1 0l



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

tKD

21 &

x =x +Δx ; x =x +Δx ( 6.33a)i i-1 i-1 i i-1 i-1M =M +ΔM ; M =M +ΔM ( 6.33b)1i 1i-1 1i-1 2i 2i-1 2i-1



& 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 D i 12 D i

Contd..

1i-1 2i-11 2

c c

1i-1 2i-11i-1 2i-1

1i-1 2i-1

r l r lα = & α =6EI 6EIM Mr = ; r =θ θ

4/12

11 D iM 12 D iMiM1 iM 2 1 2 iM1

tKDiM 2

021 P 21

21 &



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

2θ

3θ

4θ

5θ

6θ

k k

1D

2D

3D

E = 2.48 107 kN/m2

Beam 30 40 cmColumn 30 50 cm

Frame

1

3

5

2

4

6

50KN-m

M

θY = 0.00109 rad θ

Force-displacement curve Fig. 6.17



4/14

Joint TimeStep x θ M

sec m m/s m/s2 rad rad/s rad/s2 kNm1 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

xx

Table shows that sections 1 & 2 undergo yielding; recognising this, stiffness matrices are given below:

x

x x θ θ

Table 6.1

Contd..



1

2

3

14

2

3

4

5

6

Δ1.067Δ-1.067 2.133 symΔ0 -1.067 2.133θ0.8 -0.8 0 2.4

Κ = 4.83×10 × θ0.8 0 -0.8 0.8 4θ0 0.8 0 0 0.8 3.2θ0.8 -0.8 0 0.4 0 0 2.4θ0.8 0 -0.8 0 0.4 0 0.8 4θ0 0.8 0 0 0 0 0 0.8 3.2

14

Δ 2

3

0.4451 sym ΔΚ = 4.83×10 × -0.6177 1.276 Δ

0.2302 -1.0552 1.811 Δ

Contd.. 4/15



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



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



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

1D

1D

5/3

rr DD 11



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..1DD p

D iBn VBV 1 1n iD D

1BVD

n

PMM ||

5/4



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 Vs Δ

Example 6.6

Carry out an equivalent static nonlinear analysis for the frame shown in Fig 6.19.

5/5



Cross section

Location b (mm) d (mm) (kNm) (rad) (rad)

C1 G,1st, 2nd 400 400 168.9 9.025E-3 0.0271C2 3rd ,4th, 5th & 6th 300 300 119.15 0.0133 0.0399B1 G,1st, 2nd 400 500 205.22 6.097E-3 0.0183B2 3rd ,4th, 5th & 6th 300 300 153.88 8.397E-3 0.0252

yMymax

Contd..

3m

3m

3m

3m

3m4m 4m

3m

3m

yM

y c

Frame Moment rotationcurve for beams

Fig.6.19

Table 6.2

5/6



Contd..

D (m)Base shear

(KN) Plastic Hinges at section0.110891 316.825 10.118891 317.866 1,20.134891 319.457 1,2,30.142891 320.006 1,2,3,40.150891 320.555 1,2,3,4,50.174891 322.201 1,2,3,4,5,60.190891 323.299 1,2,3,4,5,6,70.206891 324.397 1,2,3,4,5,6,7,80.310891 331.498 1,2,3,4,5,6,7,8,90.318891 332.035 1,2,3,4,5,6,7,8,9,100.334891 333.11 1,2,3,4,5,6,7,8,9,10,110.350891 334.185 1,2,3,4,5,6,7,8,9,10,11,120.518891 342.546 1,2,3,4,5,6,7,8,9,10,11,12,130.534891 343.207 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.

5/7



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 20

100

200

300

400

Bas

e sh

ear

(kN

)

Roof displacement (m)

Fig.6.21

5/8



Contd..

2 1

3 3 3

4

516 17

5

67

8 8

1211

1310 9

14 1515

Fig.6.22

5/9



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



Contd.. How much the structure will deform beyond the yield limit depends upon its ductility; ductility factor is defined as

xmμ = ( 6.35a)xyFor 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

mxFig. 6.23

6/2



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:

2YR

YR yf

Contd.. An associated factor, called yield reduction factor, is defined as inverse of :

Y YY

0 0

f xf = = ( 6.35b)f x

ym my

0 y 0 y

xx x μ= × =μf = ( 6.36)x x x R

6/3



y

y

2yy n 0 y

f( x,x)f( x,x)= ; x( t)=μ( t)x ;ffa = =ω x fm

Contd..

2n n y y g

g2 2n n y n

y

x+2ξω x+ω x f( x,x)=-x ( 6.37)xμ+2ξω x+ω x f( μ,μ)=-ω ( 6.38)a

depends upon .,, yn f

6/4



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 .

1Yf1Yf

1Yf

Yf

Contd..

is known if for a & can be calculated.

mx Yf 0x

6/5



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 f

goom xxx

Yf

YR

mxox

YR

Contd..

om xx TfY & YR

6/6



Contd..

0 1m g yx x f

0m gxx

0.125yf

0.25yf

0.5y

f

Disp.sensitive

Vel.sensitive

Acc.sensitive

0.01 0.05 0.1 0.5 1 5 10 50 1000.001

0.0050.01

0.050.1

0.51

510

0.001

0.0050.01

0.050.1

0.51

510

T a= 0

. 035

T f=1

5

T b= 0

.12 5

T c= 0

.5

T d=3

T e=1

0

Spectral Regions

x 0/x

g 0or

x m/x

g0

T (sec)n

0.5yf

0.25yf 0.125yf

1yf

Disp.sensitive

Vel.sensitive

Acc.sensitive

0.01 0.050.1 0.5 1 5 10 50 1000.1

0.5

1

5

10

Spectral Regions

x m/x 0

0.1

0.5

1

5

10

T (sec)n

T a=0

.035

T f=1

5

T b=0

.125

T c=0

.5

T d=3

T e= 1

0

Normalized peakdeformations for elasto-plastic

system and elastic system Ratio of thepeak deformations Fig. 6.24

6/7



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.

YYY AVD ,,nT

Inelastic response spectra

2y y y n y y n yD =x V =ω x A =ω x ( 6.39)

y yf =mA ( 6.40)

6/8



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 Yf

oo xKf ox

f Yf

Contd..

Yf

Yf

6/9



Contd.. For different values of , the process is repeated to obtain the ductility spectrum.

nT

0 0.5 1 1.5 2 2.5 30

0.2

1

0.4

0.6

0.8

1

1.5

248

(sec)nT

f y/w

=Ay

/g

Fig. 6.25

6/10



From the ductility spectrum, yield strength to limit for a given set of & can be obtained.

Peak deformation .

nT

2m Y Y nx =μx =μA ω

1

f nT

Contd..

0.01 0.050.1 0.5 1 5 10 50 1000.05

0.1

0.5

1

Ta=0

.035

T f=1

5

T b=0

.125

Tc=0

.5

T d=3

Te=1

0

fy

20

1

5

10

2

Tn (sec)

Ry

0.12

0.195

0.37

8

4

2

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. 6.26.

Fig. 6.26

6/11



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 1000.05

0.1

0.5

1

Ta=1

/33

T f=3

3

T b= =1 1/ /8 2

T c Te=1

0

fy

Tn (sec)

8

4

2

0.2

1

Tc'

1.5

Fig. 6.27

7/1



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

cb 12 YR '' cb

7/2



Similarly, divide V ord inates of segments by ; to get ; D ordinates of segments by to get ; ordinate by to get .

Join & ; draw for ; take as the same ; join .

Draw for .

)( dc YR'' dc

)( ed YR '' ed f

'f

'f

'egoY xD sTn 33'a

a ''&bagoY xA sTn 33

1

Contd..

Natural vibration period Tn (sec) (log

scale)

goV &

Ta =1/33 sec Tf =33 secTb =1/8 sec Te =10 sec

aa'

b

b'

c

c'

d

d' e

e'f

f '

Elastic designspectrum

Inelastic designspectrum

Pseu

do-v

eloc

ity V

or

Vy

(log

scal

e)

/V

A=xgo

A= A

xgo

D=xgoD/µ

D/µ

D=xgo

A/v2µ

-1

V

D

x.

Illustration of the Method

Fig. 6.28

7/3



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

Contd..

Inelastic design spectrum for = 2 Fig. 6.28b

0.01 0.02 0.05 0.1 0.2 0.3 0.5 0.7 1 2 3 4 5 6 7 10 20 30 50 70 100

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

aTbT eT fTcT dT

Time period (sec)

Pseu

do v

eloc

ity(m

/sec)

Elastic design spectrum

Inelastic design spectrum

(b)

7/4



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 .

yx

7/5



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



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..

7/7

Chapter - 6 Inelastic Seismic Response of Structures

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Transcript of Chapter - 6 Inelastic Seismic Response of Structures