Response Spectrum Method of Analysis

8/12/2019 Response Spectrum Method of Analysis

1/71

T.K. Dat taDepartment Of Civil Engineerin g, IIT Delhi Response Spectrum Method Of Analysis

Chapters 5 & 6

Chap ter -5

RESPONSE SPECTRUM

METHOD OF ANALYSIS


2/71


Introduction

Response spectrum method is favoured by

earthquake engineering community because of:

It provides a technique for performing an

equivalent static lateral load analysis.

It allows a clear understanding of thecontributions 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


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Contd The concept of equivalent lateral forces for earth-

quake is a unique concept because it converts adynamic 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


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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

Theresponse spectrum method of analysis is

developed using the following steps.


5/71


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 applicationsshow that it provides mostly a good estimate of

peak responses.

Method is developed for single point, single

component excitation for classically dampedlinear systems. However, with additional

approximations it has been extended for multi

point-multi component excitations & for non-

classically damped systems.

Contd1/4


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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 = z

si i i if = Kx = K z

i

2

i i iK = M

2

si i i if = M z

2

simax i i imaxf = M z

1/6


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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 ithmode

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

isi i e

f M P

1 1

max max

i

i si e x K f K P

2

a dS S

Pe i

1/7

Pe i


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Since both response spectrum & mode shape

properties are required in obtaining , it is knownas 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


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Three different types of modal combination rulesare popular

ABSSUM

SRSS

CQC

ContdModal 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.maxi

x

2/1


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The combination rule gives an upper bound to the

computed values of the total response for tworeasons:

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.

Contd2/2


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0 5 10 15 20 25 30-0.4

-0.2

0

0.2

0.4

Topfloordisplacemen

t(m)

t=6.15

0 5 10 15 20 25 30-0.4

-0.2

0

0.2

0.4

Time (sec)

Firstgenera

lized

disp

lacemen

t(m)

t=6.1

(a) Top storey displacement

(b) First generalized displacement

2/3

Fig 5.1


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Contd

0 5 10 15 20 25 30-0.06

-0.04

-0.02

0

0.02

0.04

0.06

Time (sec)

Secondgen

eralizeddisplacement(m

)

t=2.5

(c) Second generalized displacement

Fig 5.1 (contd.)

2/3


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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

i

i

x x

2/4


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It is used for structures having closely spaced

frequencies:

Second term is valid for & includes the effect

of degree of correlation.

Due to the second term, the peak response may be

estimated less than that of SRSS.

Various expressions for are available; here

only two are given :

Contd

2

1 1 1

(5.13)m m m

i ij i j

i i j

x x x x

i j

2/5

i

j


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(Rosenblueth & Elordy) (5.14)

(Der Kiureghian) (5.15)

Both SRSS & CQC rules for combining peak modalresponses are best derived by assuming

earthquake as a stochastic process.

If the ground motion is assumed as a stationary

random process, then generalized coordinate ineach mode is also a random process & there

should exist a cross correlation between

generalized coordinates.

Contd

22

22

1

1 4

ij

ij

ij ij

32 2

2 22

8 1

1 4 1

ij ij

ij

ij ij ij

2/6


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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 randomvibration in deriving the combination rules, the

peak response would be better termed as mean

peak response.

Fig 5.2 shows the variation of with frquencyratio. rapidly decreases as frequency ratio

increases.

Contd2/7

ij

ijij


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Fig 5.2

Contd 2/8


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As both response spectrum & PSDF represent

frequency contents of ground motion, a relationshipexists between the two.

This relationship is investigated for the smoothed

curves of the two.

Here a relationship proposed by Kiureghian is

presented

Contd

0

2.8( ) 2ln (5.16 b)

2p

2/9

22

0

,2 4(5.16 a)

g

ff

x

DS

p


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Contd

0 10 20 30 40 50 6000.01

0.02

0.03

0.04

0.05

Frequency (rad/sec)

PSDFofac

celeration

(m

2sec-3/rad)

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)

PSDFsofacceleration(m

2sec-3/rad)

Eqn.5.16aFourier spectrum of El Centro

Unsmoothed

5 Point smoothedFig5.3

2/10

Example 5.1 : Compare between PSDFs obtained

from the smoothed displacement RSP and FFT ofElcentro record.


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Degree of freedom is sway degree of freedom.

Sway d.o.f are obtained using condensationprocedure; 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)

N

r

ir

r

i N

r

ir

r

W

W

2/11


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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 between1st& 2ndfloors.

Contd( 1... )

j ejR RP j r Rj

234

1 2

3

=5.06rad/s; =12.56rad/s;

=18.64rad/s; = .5rad/s

2/12

Solution :

;

;

T T

1 2

T T

3 4

= -1 -0.871 -0.520 -0.278 = -1 -0.210 0.911 0.752

= -1 0.738 -0.090 -0.347 = 1 -0.843 0.268 -0.145


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Approaches

Disp (m) Base shear in terms ofmass (m) Drift (m)

2 modes all modes 2 modes all modes 2 modes all modes

SRSS 0.9171 0.917 1006.558 1006.658 0.221 0.221

CQC 0.9121 0.905 991.172 991.564 0.214 0.214

ABSSUM 0.9621 0.971 1134.546 1152.872 0.228 0.223

Time history 0.8921 0.893 980.098 983.332 0.197 0.198

Table 5.1

Contd2/13


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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 findcondensed K matrix & R matrix as for 2D frames.

Repeat the same steps as described for 2D

frame

Application to 3D tall frames3/1


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3/2

C.G. of mass line

1CM

2CM

3CM

L

L

L

L g

x

x

(a)

C.G. of mass line

1CM

2

CM

3

CM

L

L

L

L Lg

x

x(b)

Figure 5.4:


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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 ofsection 5.3.4.

Results are shown in Table 5.2.

Contd

1 2 3

4 5 6

=13.516rad/s; =15.138rad/s; = 38.731rad/s;

= 39.633rad/s ; = 45.952rad/s; =119.187rad/s

X YV and V

3/3

Solution :


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T.K. Dat ta

Department Of Civil Engineerin g, IIT Delhi Response Spectrum Method Of Analysis

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


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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 excitation3/5


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Contd3/6

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


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Contd

2

1

2 ; 1.. (5.18)s

i i i i i ki k k

z z z u i m

(5.19)i kkii i

T

T

M R

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:


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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 ki

k i k

r t a u t z t

r t a u t z

r t

T T

a u t z t

3/8

k kiu is z

1

(5.20)s

i ki ki

k

z z

and z


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Assuming to be random

processes, PSDF of is given by:

Performing integration over the frequency rangeof interest & considering mean peak as peak

factor multiplied by standard deviation,

expected peak response may be written as:

Contd

T

1 11 1 21 1 31 2 12 2 22 2 32

T

11 21 31 12 22 32

= (5.24a

z = z z z z z z (5.24b)

tr t ,u t and z

( )r t

(5.25)rrS T T T T

uu zz uz zu a S a S a S S a

3/9


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Contd

....

1 2 3 S

1 2T T T T

uu uz D D zz D D zu

T

1 p 2 p 3 p S p

T

D 1 11 11 1 21 21 1 s1 s1 m 11 1m

ij i j j

E max r t = b b +b + + b (5.26

b = a u a u a u a u (5.27a)

= D D D ... D (5.27b

D =D , i =1,..,s ; j =1,..,m (5.27c)

and are the correlation matrices

whose elements are given by:

,uu u z l l z zl

i j i ji j

uu uu

u u -

1= S d (5.28

3/10


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Contd

i kj i ki kj

*

u z j uu

u z -

1= h S d (5.29

ki lj k lki lj

*

z z i j u u

z z -

1= h h S d (5.30

i k i k g

1 12 2

u u u u u2 2

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

k l k l g

1 1

2 2u u u u uS =S S coh k,l =coh k,l S (5.33)

i j i j g

1 1

2 2u u u u u4 4

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

3/11


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Contd

ij i j jD = D , For a single train of seismic wave,

that is displacement response spectrum for aspecified ; correlation matrices can be obtained

if is additionally provided; can be

determined from (Eqn 5.6).

If only relative peak displacement is required,thirdterm of Eqn.5.26is only retained.

Steps for developing the program in MATLAB is

given in the book.

coh(i, j )

j jD ,u gS

Example 5.4 Example 3.8 is solved for El centro

earthquake spectrum with time lag of 5s.

3/12


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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 -T

D

w w

T T

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) 0.011m0

5 10, 0 ; exp ; exp

2 20

D D D D

D D D D

coh i j

3/13


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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

Contd3/14


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Mean peak values determined are:

Contd

1 2

1 2

( ) 0.106 ; ( ) 0.099( ) 0.045 ; ( ) 0.022

tot tot

rel rel

u m u mu m u m

For perfectly correlated ground motion

1 0 0

0 1 0 null matrix0 0 1

1 1 1 0 0 0

1 1 1 0 0 0

1 1 1 0 0 00 0 0 1 1 1

0 0 0 1 1 1

0 0 0 1 1 1

u u u z

zz

3/15


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ContdMean peak values of relative displacement

1

2

RSA RHA

u =0.078m ; 0.081m

u = 0.039m ; 0.041m

It is seen thats the results of RHA & RSA matchwell.

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

4/1


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Cascaded analysisCascaded analysis is popular for seismic analysis

of secondary systems (Fig 5.5).

RSA cannot be directly used for the total systembecause of degrees of freedom become

prohibitively large ; entire system becomes

nonclasically damped.

4/1

Secondary System

xg..

..k

c

m

xa= xf+ xg.. .. ..

Secondary system mounted

on a floor of a building frame

SDOF is to be analyzed for

obtaining floor response spectrum

xfFig 5.5

4/2


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ContdIn 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

4/3


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Contd

Example 5.6 For example 3.18, find the mean peak

displacement of the oscillator for El Centro earthquake.for secondary system = 0.02 ; for the main

system = 0.05 ;floor displacement spectrum shown in

the Fig5.6 is used

Solution

4/3

0 5 10 15 20 25 30 35 400

0.5

1

1.5

Frequency (rad/sec)

Displacement(m)Using this spectrum,

peak displacement of the

secondary system with

T=0.811s is 0.8635m.

The time history analysis

for the entire system (with

C matrix for P-S system) is

found as 0.9163m.Floor displacement response

spectrum (Exmp. 5.6)

A i t d l RSA 4/4


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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

S i i ffi i t th d 4/5


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Seismic coefficient method

Seismic coefficient method uses also a set of

equivalent lateral loads for seismic analysis ofstructures & is recommended in all seismic codes

along with RSA & RHA.

For obtaining the equivalent lateral loads, it uses

some empirical formulae. The method consists ofthe 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

h

C

4/5

4/6


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ContdBase 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

( )i

f h

hC & ( )if h

4/6

4/7


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4/7

Distribution of lateral forces can be written as

j j j1

j j j1

j j1

j b

j j1

j j

j bj j

k

j j

j b k

j j

Sa1F = W (5.36)

1j j j1 g

F W = (5.37)

F W W

F = V (5.38)W

W h

F = V (5.39)W h

W hF = V (5.40)

W h

4/8


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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 = F =( W ) ( 5.41b ji j ji igi

SV = W (5.42)

g

V V (5.43

S W i = 1to n (5.44g

SV = W (5.45)

g

5/1


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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 forseismic analysis:

Approximate calculation of time period for

seismic coefficient method.

plot.

Effect of soil condition on

hC Vs T

a & hSA

or Cg g

5/1

5/2


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ContdSeismicity 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

NBCC1995

EURO CODE 1995

NZS 4203 1992

IS 1893 2002

5/2

5/3


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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

n

n

T T

A Tg

TT

1

1

1

1.0 0.4s

(5.46)0.40.4sh

T

CT

T

5/3

C td5/4


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Contd

T may be computed by

can have any reasonable distribution.

Distribution of lateral forces over the height

is given by

iF

1

(5.49)

k

j j

i b Nk

j j

j

W hF V

W h

2

1

1

1

2 (5.48)

N

i i

i

N

i i

i

W u

T

g Fu

5/4

5/5


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Contd

Distribution of lateral force for nine story frame is

shown in Fig5.8 by seismic coefficient method .

1 1 1 1k={1; 0.5 T +1.5 ; 2 for T 0.5s ; 0.5 T 2.5s; T 2.5s (5.50)

0 2 41

2

3

4

5

6

7

8

9

Storey force

Storey

T=2secT=1secT=0.4sec

W2

W

W

W

W

W

W

W

W

9@3m

Fig5.8

5/5

5/6


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Contd NBCC 1995

is given by

For U=0.4 ; I=F=1, variations of with T

are given in Fig 5.9.

hC

eh 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

44.5

Time period (sec)

Seism

icresponsefactorS

Fig5.9

5/6

5/7


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ContdFor 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 = 2 (5.53

g Fu

A

g

n

n

n

1.2 0.03 T 0.427sA

= (5.52)0.512T > 0.427sg

T

A/g

h vz = z

5/7

C5/8


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T.K. Dat ta


Contd

0 0.5 1 1.5 2 2.5 3 3.5 4

0

0.2

0.4

0.6

0.8

1

1.2

1.4

Time period (sec)

A/g

S

S

orA/g

Fig5.10

5/8

C d5/9


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T.K. Dat ta


Contd

Distribution of lateral forces is given by

1

t 1 b 1

b 1

0 T 0.7 s

F = 0.07TV 0.7 < T


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T.K. Dat ta


Contd

1 c

1e -3

c1 c

1

A0 T T

g

C = (5.57)TA

T Tg T

5/10

EURO CODE 8 1995

Base shear coefficient is given by

is given by

Pseudo acceleration in normalized form is given

by Eqn 5.58 in which values of Tb,Tc,Td

sC

eC

es

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

C td5/11


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T.K. Dat ta


Contd5/11

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

T2.5 T T T

A= (5.58)T

2.5 T T TuT

T T2.5 T T

T

5/12


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T.K. Dat ta


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

WF = V (5.59)

WWh

F = V (5.60)

Wh

/ & /e go goc u A u

C td5/13


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T.K. Dat ta


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

/ug0orA/ug0

..

..

....

..

Fig 5.11

C td6/1


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T.K. Dat ta


ContdNEW ZEALAND CODE ( NZ 4203: 1992)

Seismic coefficient & design response curvesare 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

C td6/2


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T.K. Dat ta


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

Rayleighs 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

C td6/3


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T.K. Dat ta


Contd

0 0.5 1 1.5 2 2.5 3 3.5 4

0

0.2

0.4

0.6

0.8

1

1.2

Time period (sec)

Category 1

Category 2

Category 3

Cb

Fig5.12

Contd 6/4


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T.K. Dat ta


Contd

IS CODE (1893-2002)

are the same; they are

given by:

ae

SC vs T & vs T

g

Time period is calculated by empiricalformula and distribution of force is given by:

2

j j

j b N2

j j

j=1

WhF = V (5.65)

Wh

a1+15T 0 T0.1sS

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

10.4 T 4. 0s

T

C td6/5


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T.K. Dat ta


Contd

a

a

1+15T 0 T 0 . 1 s

S = 2.5 0.1 T 0.55s for medium soil ( 5.63g

1.360.55 T 4 . 0 s

T

1+15T 0 T 0 . 1 sS= 2.5 0.1 T 0.67s for soft soil (5.64

g1.67

0.67 T 4 . 0 sT

For the three types of soil Sa/g are shown in Fig5.13

Sesmic zone coefficients decide about the PGA

values.

6/6

Contd


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T.K. Dat ta


0 0.5 1 1.5 2 2.5 3 3.5 4

0

0.5

1

1.5

2

2.5

3

Time period (sec)

Hard Soil

Medium Soil

Soft Soil

Spectralaccelerationcoefficient(Sa

/g)

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

Contd

Contd6/7


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T.K. Dat ta


ContdExample 5.7: Seven storey frame shown in Fig 5.14

is analyzed with

For mass: 25% for the top three & rest 50% of live

load are considered.

1 2 3T = 0.753s ; T = 0.229s ; T = 0.111s

R = 3; PGA = 0.4g ; for NBCC, PGA 0.65g

Solution:

First period of the structure falls in the falling

region of the response spectrum curve.

In this region, spectral ordinates are different

for different codes.

-3 7 -2

-1

Concrete density = 24kNm ; E = 2.510 kNm

Live load =1.4kNm

6/8Contd


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T.K. Dat ta


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

Contd6/9


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T.K. Dat ta


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

420337.18 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

Contd6/10


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T.K. Dat ta


Contd

0 2 4 6 8 10 12 14 161

2

3

4

5

6

7

Displacement (mm)

Nu

mbero

fstorey

IBCNBCC

NZ 4203

Euro 8

Indian

Comparison of displacements obtained by different codes

Fig 5.15

Lec-1/74


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Documents

Response Spectrum Method of Analysis