Seismic Analysis of Structures - III

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

Response Spectrum Method Of Analysis

Chapters – 5 & 6

Chapter -5RESPONSE SPECTRUM METHOD OF ANALYSIS



IntroductionResponse 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



Contd… The concept of equivalent lateral forces for earthquake is a unique concept because it converts a dynamic analysis partly to dynamic & partly to static analysis for finding maximum stresses.

For seismic design, these maximum stresses are of interest, not the time history of stress.

Equivalent lateral force for an earthquake is defined as a set of lateral force which will produce the same peak response as that obtained by dynamic analysis of structures .

The equivalence is restricted to a single mode of vibration.

1/2



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.



The first step is the dynamic analysis while , the second step is a static analysis.

The first two steps do not have approximations, while the third step has some approximations.

As a result, response spectrum analysis is called an approximate analysis; but applications show that it provides mostly a good estimate of peak responses.

Method is developed for single point, single component excitation for classically damped linear systems. However, with additional approximations it has been extended for multi point-multi component excitations & for nonclassically damped systems.

Contd…1/4



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

22 ; 1 (5.2)i i i i i i i gz z z x i m Ti

i Ti i

MI

M

1/5

iλi ξi



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

(5.7)

Contd…

i i ix =φ z

si i i if =Kx =Kφ z

i2

i i iKφ =ω Mφ

2si i i if =ω Mφ z

2simax i i imaxf =Mφ ω z

1/6



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 S

max ii aS is i i ef M P

1 1max max

ii s i ex K f K P

2a dS S

Pe i

1/7

Pe i



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…

Pe i

1/8



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

max1

m

ii

x x

(5.11)

is the absolute maximum value of response in the ith mode.

maxix

2/1



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



0 5 10 15 20 25 30-0.4

-0.2

0

0.2

0.4

Top

floor

dis

pla

cem

en

t (m

)

t=6.15

0 5 10 15 20 25 30-0.4

-0.2

0

0.2

0.4

Time (sec)

Fir

st g

ener

aliz

ed d

isp

lace

men

t (m

)

t=6.1

(a) Top storey displacement

(b) First generalized displacement

2/3

Fig 5.1



Contd…

0 5 10 15 20 25 30-0.06

-0.04

-0.02

0

0.02

0.04

0.06

Time (sec)

Secon

d g

en

era

lize

d d

isp

lacem

en

t (m

)

t=2.5

(c) Second generalized displacement

Fig 5.1 (contd.)

2/3



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…

2max

1

(5.12)m

ii

x x

2/4



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 ji i j

x x x x

i j

2/5

i j



(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

32 2

2 22

8 1

1 4 1

ij ij

ij

ij ij ij

2/6



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

i j

i j

i j



Fig 5.2

Contd… 2/8



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( ) 2 ln (5.16 b)

2p

2/9

22

0

,2 4(5.16 a)

gff

x

DS

p



Contd…

0 10 20 30 40 50 600

0.01

0.02

0.03

0.04

0.05

Frequency (rad/sec)

PS

DF

of

acce

lera

tio

n

(m

2 sec-3 /r

ad)

Unsmoothed PSDF from Eqn 5.16aRaw 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)

PS

DF

s of

acc

eler

atio

n (

m2 se

c-3 /rad

)

Eqn.5.16aFourier 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.



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

i

1

2

1

(5.17)

Nr

irr

i Nr

irr

W

W

2/11



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 T1 2

T T3 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



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

history0.8921 0.893 980.098 983.332 0.197 0.198

Table 5.1

Contd…2/13



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



3/2

C.G. of mass line

1CM

2CM

3CM

L

L

L

L

gx

x

(a)

C.G. of mass line

1CM

2CM

3CM

L

L

L

L L

gx

x

(b)

Figure 5.4:



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 :



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



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



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



Contd…

2

1

2 ; 1.. (5.18)s

i i i i i ki kk

z z z u i m

(5.19)i kki

i i

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:



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 ik i

s m s

k k i ki kik i k

r t a u t z t

r t a u t z

r t

T Ta u t z t

3/8

k kiu is z

1

(5.20)s

i ki kik

z z

βφ and z



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…

Tβ 1 11 1 21 1 31 2 12 2 22 2 32

T11 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 Tuu zz uz zua S a S a S S a

3/9



Contd…

....

1 2 3 S

12T T T Tuu uz βD βD zz βD βD zu

T1 p 2 p 3 p S p

TβD 1 11 11 1 21 21 1 s1 s1 m 11 1m

ij i j j

E max r t = b b+b φ +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 =D ω,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

i j i j

i j

α

uu uuu u -α

1= S ω dω ( 5.28)

σ σ

3/10



Contd…

i kj i k

i kj

α*

uz j uuu z -α

1= h S ω dω ( 5.29)

σ σ

ki lj k l

ki lj

α*

z z i j u uz z -α

1= hh S ω dω ( 5.30)

σ σ

i k i k g

1 12 2

uu u u u2 2

coh i,k1S = S S coh i,k = S ( 5.31)

ω ω

k l k l g

1 12 2

u u u u uS =S S coh k,l =coh k,l S ( 5.33)

i j i j g

1 12 2

uu u u u4 4

coh i, j1S = S S coh i, j = S ( 5.32)

ω ω

3/11



Contd… ij i j jD =D ω,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 jD ω,ξ

u gS

Example 5.4 Example 3.8 is solved for El centro earthquake spectrum with time lag of 5s.

3/12



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

12.24 rad/s ; 24.48rad/s

1 1 11;

1 1 13

0.0259 0.0259 0.0259 -TD

w w

T 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

05 10

, 0 ; exp ; exp2 2

0

D D D D

D D D D

coh i j

3/13



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

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



Mean peak values determined are:

Contd…

1 2

1 2

( ) 0.106 ; ( ) 0.099

( ) 0.045 ; ( ) 0.022tot tot

rel rel

u m u m

u m u m

For perfectly correlated ground motion1 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

uu uz

zz

3/15



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



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 mountedon a floor of a building frame

SDOF is to be analyzed forobtaining floor response spectrum

xf Fig 5.5



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



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 usedSolution

4/3

0 5 10 15 20 25 30 35 400

0.5

1

1.5

Frequency (rad/sec)

Dis

pla

cem

en

t (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)



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

CC

C

TC

4/4



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



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



4/7

Distribution of lateral forces can be written as

j j j1

j j j1

j j1j b

j j1

j jj b

j j

kj j

j b kj j

Sa1F =ρ ×W ×φ × ( 5.36)1j j j1 g

F W ×φ= ( 5.37)

∑F ΣW ×φ

W ×φF =V × ( 5.38)

ΣW ×φ

W ×hF =V ( 5.39)

ΣW ×h

W ×hF =V ( 5.40)

ΣW ×h



4/8

Computation of base shear is based on first mode.

Following basis for the formula can be put forward.

i

i

i

i

aeb i

b b

a ei

a1b

SaiV =ΣF =( ΣW ×φ × )×λ ( 5.41)b ji j ji igiS

V =W ( 5.42)g

V ≤Σ V ( 5.43)

S≤Σ W i=1ton ( 5.44)

g

SV = ×W ( 5.45)

g



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

SAor C

g g

5/1



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



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)

0.40.4s

n n

n

nn

T TA

Tg

TT

1

11

1.0 0.4s

(5.46)0.40.4sh

T

CT

T

5/3



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 ji b N

kj j

j

W hF V

W h

2

11

1

2 (5.48)

N

i ii

N

i ii

WuT

g Fu

5/4



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

Sto

rey

T=2secT=1secT=0.4sec

W2

W

W

W

W

W

W

W

W

9@3m

Fig5.8

5/5



Contd… NBCC – 1995

• is given by

• For U=0.4 ; I=F=1, variations of with T are given in Fig 5.9.

hCe

h e

C UC = ; C =USIF ( 5.51a);( 5.51b)

RA

S &g

0 0.5 1 1.51

1.5

2

2.5

3

3.5

4

4.5

Time period (sec)

Seis

mic

res

pons

e fa

ctor

S

Fig5.9

5/6



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)

1N 22

i i11 N

i i1

FuT =2π ( 5.53)

g Fu

Ag

n

nn

1.2 0.03≤T ≤0.427sA

= ( 5.52)0.512T >0.427sg

T

A/g

h vz =z

5/7



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

SS

or

A/g

Fig5.10

5/8



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 ii b t N

i ii=1

WhF = V -F ( 5.54)

Wh

5/9



Contd…

1 c

1e -3

c1 c

1

A0≤T ≤T

gC = ( 5.57)

TAT ≥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

eCe

s

CC = (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



Contd…

5/11

• Pseudo acceleration in normalized form ,

is given by

0

nn b

b

b n c

cc n dg

n

c dn d2

n

T1+1.5 0≤T ≤T

T

2.5 T ≤T ≤TA

= ( 5.58)T2.5 T ≤T ≤Tu

T

T T2.5 T ≥T

T



Rayleigh's method may be used for calculating T.

Distribution of lateral force is

Variation of are shown in Fig 5.11.

i i1i b N

i i1i=1

i ii b N

i ii=1

WφF =V ( 5.59)

Wφ

WhF =V ( 5.60)

Wh

/ & /e go goc u A u

5/12



Contd…

0 0.5 1 1.5 2 2.5 3 3.5 40

0.5

1

1.5

2

2.5

3

Time period (sec)

A/ug0

Ce/ug0

Ce /u

g0 o

r A

/ug

0

..

....

..

..

Fig 5.11

5/13



Contd… NEW ZEALAND CODE ( NZ 4203: 1992)

• Seismic coefficient & design response curves are the same.

• For serviceability limit,

is a limit factor.

• For acceleration spectrum, is replaced by T.

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



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

sl

6/2



Contd…

0 0.5 1 1.5 2 2.5 3 3.5 40

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



Contd… IS CODE (1893-2002)

• are the same; they are given by:

ae

SC vs T & vs T

g

• Time period is calculated by empirical formula and distribution of force is given by:

2j j

j b N2

j jj=1

WhF =V ( 5.65)

Wh

a

1+15T 0≤T ≤0.1sS

= 2.5 0.1≤T ≤0.4s for hard soil ( 5.62)g

10.4≤T ≤4.0s

T

6/4



Contd…

a

a

1+15T 0≤T ≤0.1sS

= 2.5 0.1≤T ≤0.55s for mediumsoil ( 5.63)g

1.360.55≤T ≤4.0s

T

1+15T 0≤T ≤0.1sS

= 2.5 0.1≤T ≤0.67s for soft soil ( 5.64)g

1.670.67≤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.



6/6

0 0.5 1 1.5 2 2.5 3 3.5 40

0.5

1

1.5

2

2.5

3

Time period (sec)

Hard Soil

Medium Soil

Soft Soil

Sp

ectr

al accele

rati

on

coeffi

cie

nt

(Sa/g

)

Variations of (Sa/g) with time period TFig 5.13

Contd…



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 ≈0.65gSolution: 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



6/8

A Seven storey-building frame for analysisFig 5.14

5m 5m 5m

7@3m

All beams:-23cm 50cm

Columns(1,2,3):-55cm 55cm

Columns(4-7):-:-45cm 45cm

Contd…



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



Contd…

0 2 4 6 8 10 12 14 161

2

3

4

5

6

7

Displacement (mm)

Nu

mb

er o

f st

orey

IBC

NBCC

NZ 4203

Euro 8

Indian

Comparison of displacements obtained by different codesFig 5.15

6/10



Lec-1/74



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

Loading

Unloading

Unloading

x

x

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

yf

f

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 x

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 &

t

Method

Contd..

k

22

k 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 Δt

δ 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 2β

x =x +Δx x =x +Δx x 1 k=x +Δx ( 6.9)

1/8


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



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



p e t( Δx) for( 1-a )Δpwithk =0

e 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 Δptk ≠ 0

aΔx=( Δx) +Factored Δx

x=0x

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=10140Nm

Δx =-0.00312gg

Δp=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 0.15mg

f = 0.95664 < 0.15mg and x>0k0.63432 0.15mg

10260 symδ 1

K=K + C + M= -124 10260t t 2βΔt β( Δt) 0 -124 10137

Δx =0.5913g

M δ M δΔp=-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



k/2

k/2

m

m

m k/2

k/2

k/2 k/2

3m

3m

3m

1x

2x

3x

yxx

yf

0.15m gyf

0.01475myx

3 storey frame

Force displacementcurve of the column

Contd..

Fig.6.5

2/7



Contd..

0.0032 0.16K

Δx= -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 3

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

e 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 0.02009

Δx = 0.0027 ; Δx=Δx +Δx = 0.0018 ; x =x +Δx= 0.0083312 2 k+1 k0.0026 0.00074 0.0114

0.-0.0509 0δ 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 1.47151Δf= 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 Interaction3/1



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 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



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

t

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 G

G =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 2

yixii

pxi 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..

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 +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 i

i F F

pU

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:

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

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 =1.5M

kM =2M ; k =k =k =k ;k =k =1.5k ;k =2k ; =50

min 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

For

ce (

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 Ay

V =154.24 V =19.56Bx By888 ; x =-0.08613g gkV =158.66 V =15.15Dx Dy773.51

V =211.54 V =26.08Cx CySolution:

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



Contd..

2

2

ex exi 0 ey eyi 0 θ 0

e

2 2

yixii

pxi pyi

A B C D

t e

t t t2

KDK = K =6k ; K = K =6k ;K = =3k 3.5

4186000 0 54250

K = 0 186000 54250

54250 54250 1139250

VV= +

V V

=0.462 ; =0.465; =0.491; =0.488

K =K

δ 1K=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

16282M δ M δ

Δp=-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 476.1

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 =183.891103.36

V =275.84F =F +ΔF= 81.09 ; ;

V =275.84954.75

V =367.79

Ay

By

Dy

Cy

2 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 V

1 1 1 1e = 0.824 e = =0.824 e = =0.824 e = =0.824

Contd.. 3/13



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;

xee x

x

i ix y

i

xA

xBxixi

pxi

Ue = etc

U

e

K xe e

K

h

hVh

V

U e

p p

2

0.00059

0.0003953; ;

0.000390.0053

0.000290.00398

yA

yByiyi

yDxD pyi

yCxC

h

hVh

hh V

hh

Contd.. 3/14



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 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 B

B BB K h B K h

B B

B By

G

GG K h K h

G

G

BB DK K K K K K

G G

xi yi

i

B B

G

Contd.. 3/15



Contd..

3

5

2

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

0.0026

0.0001 ;

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

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



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

pipxy ey py pyi i

AyAx

ByBx

DyDx

CyCx

yixii

pxi pyi

A C D

K K K U

K K K U

VV

VV

VV

VV

VV

V V

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 ).

t

t

tt

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. ( ).

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 6.9under 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 2.0

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

For

ce (

kN

)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)

Mo

men

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)

For

ce (

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)

Shear

Fo

rce(N

)-800000

-600000

-400000

-200000

0

200000

400000

600000

800000

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

t

t

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 0Δ2

K= ( 6.28a)kl kl kl kl θ

- α +0.672 2 2 6 θkl kl kl

- 0 α +1.332 6 2

-1Δ Δ Δθ θ θΔ

-1

1-1θ 2

2

-1

1

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

tK

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 i 12 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 6EI

M Mr = ; r =

θ θ

4/12

11 iM 12 iMiM1 iM 2 1 2 iM1

tKiM 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

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



4/14

JointTimeStep

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

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

Δ1.067

Δ-1.067 2.133 sym

Δ0 -1.067 2.133

θ0.8 -0.8 0 2.4

Κ = 4.83×10 × θ0.8 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

14

Δ 2

3

0.4451 sym Δ

Κ = 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

1

1

5/3

rr 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..1 p

iBn VBV 1 1n i

1BV

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

3m

4m 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)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

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

YRyf

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



yy

2yy n 0 y

f( x,x)f( x,x)= ; x( t)=μ( t)x ;

f

fa = =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

0.125y

f

0.25y

f

0.5y

f

Disp.sensitive

Vel.sensitive

Acc.sensitive

0.01 0.05 0.1 0.5 1 5 10 50 1000.001

0.005

0.01

0.05

0.1

0.51

5

10

0.001

0.005

0.01

0.05

0.1

0.51

5

10T a

=0.

035

T f=

15

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

0.5y

f

0.25y

f 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

24

8

(sec)nT

f y/w

=A y

/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

Tf=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

Pse

ud

o-ve

loci

ty V

or

Vy

(log

sca

le)

/V

A=xgo

A= A

xgo

D=xgo

D/µ

D/µ

D=x

go

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

Seismic Analysis of Structures - III

Education

Transcript of Seismic Analysis of Structures - III