t.k. datta department of civil engineering, iit delhi response spectrum method of analysis chapters...

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