 Open Access
 Total Downloads : 596
 Authors : Siva Kiran Kollimarla, Chadalawada Jagan Mohan, Chelli Jagadeesh Babu, M Venkateswara Rao
 Paper ID : IJERTV5IS040068
 Volume & Issue : Volume 05, Issue 04 (April 2016)
 DOI : http://dx.doi.org/10.17577/IJERTV5IS040068
 Published (First Online): 31032016
 ISSN (Online) : 22780181
 Publisher Name : IJERT
 License: This work is licensed under a Creative Commons Attribution 4.0 International License
Seismic Analysis of a Multi Storey Plane Frame using Static & Dynamic Methods
Comparison of Equivalent Static Force & Response Spectrum on Frame with & without Infills
Siva Kiran Kollimarla Department of Civil Engineering R.V.R&J.C college of engineering GunturA.P, INDIA
Chelli Jagadeesh babu R.V.R&J.C College of Engineering GunturA.P, INDIA
Chadalawada Jagan Mohan Department of Civil Engineering R.V.R&J.C college of engineering GunturA.P, INDIA
M Venkateswara Rao

V. R & J. C college of engineering
GunturA.P, INDIA
AbstractIn the context of seismic analysis and design of structures, in earthquake engineering, a verity of methods are available. Standard codes provide provision for certain methods for the analysis of wide range of structures of engineering interest. In this paper an attempt is made to present the provisions of IS 1893:2002, part1 for the analysis of structures and its suitability.
Keywords Lumped mass; infills; CQC; seismic analysis;

INTRODUCTION
In the context of seismic analysis and design of structures, a variety of methods are available for seismic analysis of structures. In this paper an attempt has made to present static and dynamic methods of seismic analysis as per Indian standard code IS 1893:2002 (Part 1). The basic concepts and codal provisions of Equivalent static force method and Response spectrum method as per IS 1893:2002 (Part 1) are explained in detail.

Equivalent static analysis (ESA)
This approach defines a series of forces acting on a building to represent the effect of earthquake ground motion, typically defined by a seismic design response spectrum. It assumes that the building responds in its fundamental mode. For this to be true, the building must be low rise and must not twist significantly when the ground moves. The response is read from design response spectrum, given the natural frequency of the building. The applicability of this method is extended in many building codes by applying factors to account for higher buildings with some higher modes and for low levels of twisting. To account for effects due to yielding of the structure, many codes apply modification factors that reduce the design forces.decomposing as a means of reducing degrees of freedom in the analysis.

Response spectrum analysis (RSM)
This approach permits the multiple modes of response of a building to be taken in to account (in the
frequency domain). This is required in many building codes for all except for very simple or very complex structures. The response of a structure can be defined as a combination of many special shapes (modes) that in a vibrating string correspond to the harmonics. Computer analysis can be used to determine these modes for a structure. For each mode, a response is read from the design spectrum, based on modal mass, and they are then combined to provide an estimate of the total response of the structure. In this we have to calculate the magnitude of forces in all directions i.e. X, Y & Z and then see the effects on the building. Combination methods include the following.

Absolute values are added together (AVS)

Square root of sum of the squares (SRSS)

Complete Quadratic Combinations (CQC).

A method that is an improvement on SRSS for closely shaped modes.
The results of a response spectrum analysis using the response spectrum from a ground motion is typically different from that which would be calculated from linear dynamic analysis using that ground motion directly, since phase information is lost in the process of generating the response spectrum.
In cases where structures are either too irregular, too tall are of significance to a community in disaster response, the response spectrum approach is no longer appropriate and more complex analysis is often required such as nonlinear static analysis or dynamic analysis.


STEP BY STEP PROCEDURE FOR STATIC AND DYNAMIC METHODS

Equivalent static force method

Calculation of Lumped masses to various floor levels The earthquake forces shall be calculated for the full dead load plus the percentage of imposed load as given in Table8 of IS
1893 (Part 1):2002. The imposed load on roof is assumed to be zero. The lumped mass of each floor are worked out as follows:

Mass of each floor is calculated as
Mass of columns +Mass of beams in longitudinal and transverse direction of that floor + Mass of slab+ Imposed load of that floor if permissible.

Roof
Mass of infill + Mass of columns + Mass of beams in longitudinal and transverse direction of that floor + Mass of slab + Imposed load of that floor if permissible
5) Calculation of moment due to design lateral forces
n
n
M Fihi i1



Response Spectrum Method

Determination of Eigen values and Eigen vectors
The mass and stiffness matrices at each floor level are calculated in S.I units and then modeled in the form of matrix as

50% of imposed load, if imposed load is greater than 3 kN/m2
M1 0
0 M
M 2
0 0
0
0
0
0 0
0
0
0
M3 0
M
M
0 4

Seismic weight of building



The seismic weight of each floor is its full dead load plus appropriate amount of imposed load, as specified in Clause
k1 k 2
k
K 2

k 2
k 2 k 3
0 0
3
3
k 0
7.3.1 and 7.3.2 of IS 1893 (Part 1):2002. Any weight supported in between stories shall be distributed to the floors above and below in inverse proportion to its distance from the floors.
0 k 3 k 3 k 4
0
0
0
0
4
4
k
12EI
k 4
k
k
4

Determination of Fundamental Natural Period
Column stiffness of storey,
k c
L3
The approximate fundamental natural period of a vibration (Ta), in seconds, of a moment resisting frame building without brick infill panels may be estimated by the empirical expression (Clause 7.6.1)
The stiffness of infill is determined by modeling the infill as an equivalent diagonal strut
1
E I h 4
h
f c
Ta = 0.075xh0.75
For RC frame building
2 2E m tsin2
1
Ta = 0.085xh0.75 For Steel frame building
E I l 4
l f b
Emtsin2
Where h is the height of the building, in meters.
If the structure considered with infills (Clause 7.6.2)
Ta = 0.09H/D
Width of the strut Where
W= 1 2 2
2 h l

Determination of Design Base Shear
Design seismic base shear, (Clause 7.5.3)
VB = AhW

Vertical Distribution of Base Shear
The design base shear (VB) computed shall be distributed along the height of the building as per the expression, (Clause 7.7.1)
Wihi 2
Ef = Youngs modulus of the concrete
Em = Youngs modulus of the masonry infill h = Height of infill wall
l = length of wall
t = Thickness of wall Ic=Moment of inertia of columns Ib =Moment of inertia of beam
A=cross sectional area of diagonal stiffness
<>Qi VB
n
n
Wih i 2 i1
ld =Diagonal length of strut = p + l2
Therefore, stiffness of infill is
Where
Qi = Design lateral forces at floor i Wi= Seismic weights of the floor i
hi = Height of the floor i measured from base and n = Number of stories
k d
AEm
ld
cos 2
2
2
Taking k/m = n
The design horizontal seismic coefficient Ah for various modes are,
A Z I Sa1
2K 2m
k2 0 0
p 2 R g
K 2M


k2
2K 2m
k3 0
0 k3
2K 2m
k4
For rocky, or hard soil sites
0
0 k4
k 0.575 2m
1 15T;
S a 2.5;
g
0.00
0.10
T 0.10
T 0.40
The solution for the above equation is given as
1.00/T;
0.40
T 4.0
Eigen values
2
For medium soil sites
2
2
1 0 0 0
2
2
2 0 2 0 0
1 15T;
S
0.00 T 0.10
4
4
0 0 3 0
a 2.5;
0.10 T 0.55
Eigen vectors
0 0
11
0
12
2
13
14
g
For soft soil sites
1.36/T;
0.55 T 4.0
1 15T;
S
0.00 T 0.10
1 2 3 4 21 22
23 24 .
a 2.5;
0.10 T 0.67
31
32
33
34
g 1.67/T;
0.67 T 4.0
41
42
43
44

Determination of storey shear in each mode
Natural frequency in various modes
The peak shear force will be obtained by
n
1 0 0 0
0
0
0
0
2 0
Vik
Qik ji1
Natural time period
0 0
0
0
0
T1 0
0
0
T T2
0 0
0
0
0
3 0
0 4
0 0
0
0
0
T3 0
T
T
0 4

Determination of storey shear force due to all modes
The peak storey shear force (Vi) in storey i due to all modes considering is obtained by combining those due to each mode in accordance with modal combination SRSS (Square Root of Sum of Squares) or CQC (Complete Quadratic Combination) methods.

Maximum Absolute Value Sum (AVS)
2) Determination of Modal Participation factor
4
4
Wiik
4
4
P i1
n
n
Vi vik
k1
3.15
2
2
Wi (ik )
i1

Determination of model mass
4 2


Square Root of Sum of Squares (SRSS)
If the building does not have closely spaced modes, the peak response quantity () due to all modes considered shall be obtained as,
r
Wii1
( )2
k
k
i
i
M i1
k
k
k1
g4 W ( )2
Where
i
i1
i1
k = Absolute value of quantity in mode k, and r is the

Determination of lateral force on each floor
The design lateral force (Qik) at floor i in mode k is given by,
Qik (Ak Pk ik Wk)
numbers of modes being considered.


Complete Quadratic Combination (CQC)
r r 8 2 1 ij 1.5
Seismic weight of building
= Seismic weight of all floors = M13+.2M52+M3+M4
Where
iij j i1 j1
ij =
1 ij
2 4 2ij 1 ij2
= 64.46+64.46+64.46+37.09 = 2303.4.276

Determination of Fundamental Natural Period
r = Number of modes being considered
ij = Cross modal coefficient
i = Response quantity in mode i (including sign)
j = Response quantity in mode j (including sign)
= Modal damping ration (in fraction)
The approximate fundamental natural period of a vibration (Ta), in seconds, of a moment resisting frame building without brick infill panels may be estimated by the empirical expression
T = 0.075xh0.75
= Frequency ratio / a
ij j i
i = Circular frequency in ith mode, and j = Circular frequency in jth mode.
7) Determination of lateral forces at each storey
The design lateral forces Froof and Fi, at roof and at ith floor, are calculated as,
= 0.075×140.75 = 0.5423 s
Where h is the height of the building, in meters.

Determination of Design Base Shear
Design seismic base shear, VB = AhW
Froof = Vroof, and Fi = Vi Vi+1
Z I S
0.24 1
Ah = a
= 1.842 0.04343.27
H: Calculation of moment due to design lateral forces
2 R g 2 5
n
n
M Fihi
For Ta
Sa 1
= 0.5423 => = = 1.842, for rock site from Fig 2
i1
g Ta


ANALYSIS OF MULTI STOREY PLANE FRAME


Equivalent static force method

Frame Without infills

Calculation of Lumped masses to various floor
levels
of IS 1893 (part 1): 2002
Design seismic base shear, VB = 0.0443 x (230.43×9.81) = 99.933 kN

Vertical Distribution of Base Shear
Qi VB
Qi VB
W h
W h
2
1 1
W1h 2 W2p2 W3p2 W4h42
Roof
= Mass of infill + Mass of columns + Mass of beams in longitudinal and transverse direction of that floor + Mass of
=99.93
1
632.25×3.52
2 2 2 2
slab + Imposed load of that floor if permissible
={((0.25x10x(3.5/2)+0.15x15x(3.5/2))20}+{(0.25x10x0.40+0. 25x15x0.35)25}+{0.10x5x10x25}+{(0.25×0.45x(0.35/2)x3)x
25}
632.25×3.5
=4.307kN
Similarly,
632.25×7
632.25×10.5
363×14
= 363.82 kN (weight) = 37.087 ton (mass) 3rd, 2nd, 1st Floors
={((0.25x10x3.5)+(0.15x15x3.5))20}+{(0.25x10x0.40+0.25x 15×0.35)25}+{0.10x5x10x25}+
{0.25×0.45×3.5x3x25}+{5x10x3.5×0.5}
= 632.43 kN (weight) = 64.45 ton (mass)

50% of imposed load, if imposed load is greater than 3 kN/m2
Q2 = 0.1724 x 99.933 = 17.23 kN Q3 = 0.3872 x 99.933 = 38.768kN Q4 = 0.3967 x 99.933 = 39.654kN
Maximum overturning moment=1097.9kNm

Vertical Distribution of Base Shear
Qi VB
W1h 2
1
1
W1h 2 W2p2 W3p2 W4h42
1
=135.6
632.25×3.52
632.25×3.52 632.25×72 632.25×10.52 363×142
Fig.1. Distribution of lateral forces and shear by Equivalent static force method on frame without infill




Frame With infills

Calculation of Lumped masses to various floor levels
Roof
Mass of infill + Mass of columns + Mass of beams in longitudinal and transverse direction of that floor + Mass of slab + Imposed load of that floor if permissible
= 363.82 kN (weight) = 37.087 ton (mass) 3rd, 2nd, 1st Floors
={((0.25x10x3.5)+(0.15x15x3.5))20}+{(0.25x10x0.40+0.25x 15×0.35)25}+{0.10x5x10x25}+
{0.25×0.45×3.5x3x25}+{5x10x3.5×0.5}
= 632.43 kN (weight) = 64.45 ton (mass)

50% of imposed load, if imposed load is greater than 3 kN/m2
Seismic weight of building
= Seismic weight of all loors = M1+M2+M3+M4
= 64.46+64.46+64.46+37.09 = 230.47


Determination of Fundamental Natural Period
Ta = 0.09H/DD
= 0.09×14/1010= 0.3984 sec

Determination of Design Bash Shear

= 5.8kN
Similarly,
Q2 = 0.1724 x 135.63 = 23.215kN Q3 = 0.3872 x 135.63 = 52.234kN Q4 = 0.3967 x 135.63 = 53.429kN
Maximum overturning moment=1479.3kNm
Fig.2. Distributions of lateral forces and shear by Equivalent static force method on frame with in fills


Response Spectrum Method

Frame without stiffness of infills

Determination of Eigen values and Eigen vectors
The mass and stiffness matrices at each floor level are calculated in S.I units and then modeled in the form of matrix as
Ah =
Z I Sa
= 0.24 1 x2.5 0.06
M1 0
0 0
2 R g 2 5
Sa 1
0
0
M
0
M2 0
0
0
0 M3 0
For Ta = 0.3984 => g
= = 2.5, for rock site from Figure 2
Ta
0
0
M
M
0
0
0
0
4
of IS 1893 (part 1): 2002
64.46 0 0 0
0
0
0
0
0
0
Design seismic base shear, VB = 0.06 x (230.43×9.81) = 134.68kN
M 103 0
0
64.46
0
0
64.46
0 kg
0
37.09
k1 k 2
k2
K

k 2
k 2 k3
0 0
k3 0


Modal participation factor
4
14.40
0 k3
k3 k4
k 4
Wiik
0 0 k k
i1
(W11k W22k ….. W44k) 4.30
4 4
P
4 W1(1k)2 W2(2k)2 ….. W4(4k)2 1.95
2
2
Column stiffness of storey,
Wi(ik)
i1
0.68
3 0.25×0.453
12EI
12x22360x10
12

Determination of model mass
k L3
11880.78 kN/m 3.53
[9.81(64.45(0.0328) … 37.08(0.0872))]2Total lateral stiffness of each storey,
k1= k2= k3= k4= 3×11880.78 = 35642.36 kN/m
M1
9.81[9.81(64.45(0.0328)2 … 37.08(0.0872)2 )]
207.60
Similarly, M2 = 18.54, M3 = 3.82, M4 = 0.47
71284.72 3564.36 0 0
3564.36 71284.72 3564.36
K
0
0
kN/m
Modal contributions of various modes
0
0
0
0
0 3564.36 71284.72 35642.36
M1 207.60
35642.36 35642.36
For mode 1,
For mode 2,
M
M 2
230.43
18.54
0.90 90%
0.0804 8.04%
Eigen values
81
0
0
0
0
657
0
0
0
0
1475
0
0
0
0
206
81
0
0
0
0
657
0
0
0
0
1475
0
0
0
0
206
2
For mode 3,
M
M 3
M
M
230.43
3.82
230.43
0.47
0.0165 1.65%
5
For mode 4, 4
M
230.43
0.0020 0.20%
Eigen vectors
0.0328
0.0795
0.0808
0.0397

Determination of lateral force on each floor
The design lateral force (Qik) at floor i in mode k is given by,
0.0608 0.0644 0.0540 0.0690
Q A P W
1 2 3 4 .
ik k ik k i
0.0798 0.0273 0.0448 0.0799
0.0872 0.0865 0.0839 0.0696
The design horizontal seismic coefficient Ah for various modes are,
Natural frequency in various modes
A Z I Sa1 0.24 11.433 0.0343
9 0 0 0
p 2 R g 2 5
0 25.63 0 0
Z I S
0.24 1
A p
a2
2.5 0.060
0 0
38.41 0
2 R g 2 5
0
0
0
Natural time period
0 45.44
Similarly,
Ap 0.060, Ah4 0.060.
0.6977 0 0 0
T 0 0
0.2450
0
0 0
0.1636 0
For rocky, or hard soil sites
0
0
0
0
0
0
0.1383
1 15T;
S
0.00 T 0.10
a 2.5;
0.10 T 0.40
g 1.00/T;
0.40 T 4.0
For
T 0.6978 Sa1 1.433

Determination of storey shear force due to all modes
1 g
S Maximum Absolute Value Sum (AVS)
For
T2 0.2450 a2 2.5
g
V1 70.056 36.51416.572 5.824
128.986
For
T 0.1636 Sa3 2.5
n V2 59.781 6.927 27.63215.925
110.282
3 g Vi vik
k1
V3 40.738 42.1271.867 21.796
106.544
S
For
T4 0.1382 a4 2.5
g
V415.720 27.207 26.385 21.878
91.207
Design lateral force
Square Root of Sum of Squares (SRSS)
V1 = [(V11)2 + (V12)2 + (V13)2 + (V14)2]1/2
(Ap
P1 11
W1 )
= [(70.056)2 + (36.514)2 + (16.572)2 + (5.824)2]1/2
Qi1
Qi1
(Ap P1 21 W2 )
(Ap P1 31 W3 )
= 80.930kN
Similarly,
V2 = 68.110kN
(A
(A
h4
P1 41
)
)
W4
V3 = 62.553kN V4 = 46.507kN
((0.0343)(14.40)(0.0328)(64.45 9.81))
((0.0343)(14.40)(0.0608)(64.45 9.81))
(10.275)
(19.043)
Complete Quadratic Combination (CQC)
kN
V1 =
((0.0343)(14.40)(0.0798)(64.45 9.81)) (25.018)
((0.0343)(14.40)(0.0872)(37.08 9.81)) (15.720)
1 0.0073
0.0031
0.0023 70.056
0.0073 1
0.0559
0.0278 36.51
[70.056 – 36.514 – 16.572 + 5.824]0.0031
0.0559
1 0.2597 16.57
Similarly,
43.44
35.199
44.204
29.499
21.749
37.722
V1 = [80.70] V2 =
0.0023
0.0278
0.2597
1 5.82
Qi2
, Qi3
, Qi2
14.920
24.517
43.675
1 0.0073
0.0031
0.0023 59.78
27.207
26.385
21.878
0.0073 1
0.0559
0.0278 6.92
[59.781+ 6.927 27.632 15.925]0.0031
0.0559 1
0.2597 27.63

Determination of storey shear in each mode
0.0023
0.0278
0.2597
1 15.92
The storey shear forces for the first mode is,
V2 = [66.61]
(Q11 Q21 Q31 Q41)
70.056
V3 =
n (Q21 Q31 Q41)
59.925
Vi1 Qi1
(Q Q )
kN
.7
1 0.0073
0.0031
0.003 40.738
ji1
31 41
40
38
(Q )
.7
[40.738 + 42.127 1.867 + 21.796]0.0073 10.0559
0.027842.127
41
15
20
0.0031
0.0559 1
0.25971.867
V12
36.514
V13
16.572
0.0023
0.0278
0.2597
1 21.796
V22
6.927
V
27.632
V3 = [62.95]
Vi2
, Vi3
23 ,
V32 42.127 V33 1.867
V4 =
V42
27.207
V43
26.385
1 0.0073
0.0031
0.0023 15.720
0.0073 1
0.0559
0.0278 27.207
V
5.824
[15.720 27.207 – 26.385 – 21.878]14 0.0031
0.0559 1
0.2597 26.385
V V24 15.925
0.0023
0.0278
0.2597
1 21.878
i4 V34
21.796
V = [48.48]
4
V44
21.878

Determination of Lateral Forces at Each Storey
Maximum Absolute Value Sum (AVS) Froof = V4 = 91.207 kN
Ffloor 3 = F3 = V3 V4 = 106.544 91.207 = 15.337 kN
Ffloor 2 = F2 = V2 V3 =110.2815 106.544 = 3.738 kN Ffloor 1 = F1 = V1 V2 = 128.986 110.282 = 18.705 kN
Square Root of Sum of Squares (SRSS) Froof = F4 = V4 = 46.508kN
Ffloor 3 = F3 = V3 V4 = 62.562 46.508 = 16.055kN
Ffloor 2 = F2 = V2 V3 = 68.119 62.562 = 5.56kN Ffloor 1 = F1 = V1 V2 = 80.941 68.119 = 12.823kN
Complete Quadratic Combination (CQC) Froof = F4 = V4 = 48.492kN
Ffloor 3 = F3 = V3 V4 = 62.957 48.492 = 14.465kN Ffloor 2 = F2 = V2 V3 = 66.625 62.957 = 3.67kN Ffloor 1 = F1 = V1 V2 = 80.714 66.625 = 14.089kN

Calculation of moment due to design lateral forces


Absolute Value Sum (AVS) Overturning moment = 1529.6 kNm
Square Root of Sum of Squares (SRSS) Overturning moment = 903.456 kNm
Complete Quadratic Combination (CQC) Overturning moment = 905.758 kNm
Fig.4. Distributions of lateral forces and shear by SRSS method on frame without infills
Fig.5. Distribution of lateral forces and shear by CQC method on frame without infills



Frame with the stiffness of Infills

Determination of Eigen values and Eigen vectors

M1 0
0
0
M M 2
0 0
0
0
0
0 0
0 0
M 3 0
0 M
0 M
4
0 
0 
64.46 
0 

0 
0 
0 
37.0 
9 
0 
0 
64.46 
0 

0 
0 
0 
37.0 
9 
64.46 0 0 0
Fig.3. Distribution of lateral forces and shear by Absolute value sum method on frame without infills
M 103 0
64.46 0
0 kg
Column stiffness of storey, Eigen values
12EI
12x22360x103 0.001893
1443 0 0 0
k
L3
3.53
11846.758 kN/m
2 0
0
11707
0
0 0
26247 0
Ef = Youngs modulus of the concrete = 22360 N/m2
Em = Youngs modulus of the masonry infill = 13800 N/m2 h = Height of infill wall = 3.5 m
Eigen vectors
0
0
0
0
0.0010
0
0.0025
36746
0.0025
0.0013
l = length of wall = 5 m
t = Thickness of wall = 250 mm
1 2 3 4
0.0019
0.0025
0.0028
0.0020
0.0009
0.0027
0.0017
0.0014
0.0027
0.0022
.
0.0025 0.0022
c
c
I =Moment of inertia of columns = 1 x (0.25 x 0.453)
12
Natural frequency in various modes
0 
0 
0 
108.198 
0 
0 
0 
162.008 
0 
0 
0 191.6 
0 
0 
0 
108.198 
0 
0 
0 
162.008 
0 
0 
0 191.6 
= 0.001893 m4
I =Moment of inertia of beam = 1 x (0.25 x 0.403)
37.99
b 12
= 0.001333 m4
W = 1x J2 + 2 = 0.7885
0
0
9
9
0 3
2 h 1
A=cross sectional area of diagonal stiffness
= W*t = 0.7885 x 0.25 = 0.1972 m2
Natural time period
0.1654 0 0 0
ld =Diagonal length of strut = p + l2 = 6.103 m 
T 
0 
0.0581 
0 
0 
0 
0.0388 
ld =Diagonal length of strut = p + l2 = 6.103 m 
T 
0 
0.0581 
0 
0 
0 
0.0388 
0
0
0
0
0
0
0
0
h
22360 x 0.001893 x 3.5
1/4
0.611m
0.03
28
2 2 x 13800 x 0.25 x sin 2 x 35
22360 x 0.001333 x 5.0 1/4

Determination of model participation factor
4
l 1.45 m
Wiik
13800 x 0.25 x sin 2 x 35
Therefore, stiffness of infill is
P
i1
4
4
Wi (ik ) 2 i1
14.40
AEm cos 2
0.1972 x 13800 x 106
0.8192
(W11k W2 2k ….. W4 4k )
4.30
ld 6.103
W1 (1k ) 2 W2 ( 2k ) 2 ….. W4
( 4k
) 2
1.95
299086.078 N/m

Determination of model mass
4 2
0.68
For the frame with two bays there are two struts participating
Wii1
in one direction, total lateral stiffness of each storey
M1
i1
4
k = k = k = k = 3×11846.758 + 2×299086078
gWi (i1 ) 2
1 2 3 4
i1
= 633712.430 kN/m
1.2686
9
9
0.6343
K 10
0.6343
1.2686
0
0.6343
0
0 N/m
[9.81(64.45(0.0328) … 37.08(0.0872))]2M1
9.81[9.81(64.45(0.0328) 2 … 37.08(0.0872) 2 )]
0 0.6343
0
0
0
0
1.2686
0.6343
0.6343 0.6343
Similarly, M2 = 18.54, M3 = 3.82, M4 = 0.47.
Modal contributions of various modes
For mode 1,
M1 207.60 0.90 90%
M 230.43
For mode 2,
M 2
M
M3
18.54
230.43
<>3.82
0.0804 8.04%
Similarly,
32.512
26.343
27.972
18.667
12.980
22.512
For mode 3,
0.0165 1.65%
M 230.43
Qi2
, Qi3
, Qi2
11.166
20.361
15.514
16.696
26.065
13.057
For mode 4,
M 4 0.47 0.0020 0.20%
M 230.43

Determination of storey shear in each mode


Determination of lateral force on each floor
The design lateral force (Qik) at floor i in mode k is given by
The peak shear force
n
n
Vik Qik
Q A P W
ji1
ik k ik k i
The design horizontal seismic coefficients Ah for various modes are,
The storey shear forces for the first mode is,
A Z I Sa1 0.24 1 2.5 0.060
(Q11 Q21 Q31 Q41) 70.056
p 2 R g 2 5
n (Q Q Q ) 59.925
Vi1 Qi1
21 31
41
kN
A Z I Sa2 0.24 1 1.871 0.045
ji1
(Q31 Q41)
40.738
p 2 R g 2 5
(Q )
0
Similarly,
Ap 0.037, Ah4 0.035
Similarly,
41
15.72
For rocky, or hard soil sites
V12
27.327
V13
10.487
V22
5.184
V23
17.485
Vi2
, Vi3
,
1 15T;
0.00 T 0.10
V32
31.528
V33
1.182
Sa
V
20.361
V
16.696
2.5;
0.10 T 0.40
42
43
g 1.00/T;
0.40 T 4.0
V14
3.476
V
9.5042
Vi2
24
For
T 0.1655 Sa1 2.5
V34
13.008
1 g V
13.057
For
T 0.0581 Sa2 1 15T 1.871
44
For For
2 g
T 0.0388 Sa3 1 15T 1.582
3 g
4
4
T 0.0382 Sa4 1 15T 1.492

Determination of storey shear force due to all modes
Maximum Absolute Value Sum (AVS)
V1 122.194 27.327 10.487 3.476
g n
Design lateral force
Vi
vik
V2 104.272 5.184 17.485 9.5042
V 71.057 31.528 1.182 13.008
k1
3
(Ap P1 11 W1 )
(Ap P1 21 W2 )
V4 27.419 20.36116.696 13.057
163.508
Qi1
136.463
(Ap P1 31 W3 )
kN
(A P W )
116.791
h4 1 41 4
((0.060)(14.40)(0.0328)(64.45 9.81))
((0.060)(14.40)(0.0608)(64.45 9.81))
(17.922)
(33.215)
77.544
Square Root of Sum of Squares (SRSS)
kN
((0.060)(14.40)(0.0798)(64.45 9.81)) (43.637) r
((0.060)(14.40)(0.0872)(64.45
(k )2
9.81)) (27.419)
k1
V1 = [(V11)2 + (V12)2 + (V13)2 + (V14)2]1/2
= [(122.194)2+(27.327)2+(10.487)2+(3.475)2]1/2
= 125.699 kN
Similarly,
V2 = 106.281kN V3 = 78.827kN V4 = 40.196kN
Complete Quadratic Combination (CQC)
Square Root of Sum of Squares (SRSS)
Froof = F4 = V4 = 40.202kN
Ffloor 3 = F3 = V3 V4 = 78.840 40.202 = 38.638kN Ffloor 2 = F2 = V2 V3 = 106.297 78.840 = 27.457kN Ffloor 1 = F1 = V1 V2 = 125.716 106.297 = 19.422kN
Complete Quadratic Combination (CQC) Froof = F4 = V4 = 44.990 – 0 = 44.990kN
r r Ffloor 3 = F3 = V3 V4 = 79.136 44.990 = 38.146kN
iij j i1 j1
Ffloor 2 = F2 = V2 V3 = 105.995 79.136 = 26.859kN Ffloor 1 = F1 = V1 V2 = 125.533 105.995 = 19.538kN
1 0.0073
0.0031
0.0023 122.272
0.0073 1
0.0559
0.0278 27.327
h) Calculation of moment due to design lateral forces
V1 =
[122.194 27.327 10.487 3.475]0.0031 0.0559 1 0.2597 10.487 n
i i
V1 = [125.512]
0.0023 0.0278 0.2597 1 3.475
M
Absolute Value Sum (AVS)
F h
i1
1 0.0073
0.0073 1
0.0031
0.0559
0.0023104.272
0.0278 5.184
Overturning moment=1730.1kNm
Square Root of Sum of Squares (SRSS)
[104.272 5.184 17.485 9.504]V = 0.0031
0.0559 1
0.2597 17.485
Overturning moment=1228.7kNm
2 0.0023
0.0278 0.2597
1 9.504
V2 = [105.977]
Complete Quadratic Combination (CQC) Overturning moment=1230.8kNm
1 0.0073
0.0031
0.0023 71.057
0.0073 1
0.0559
0.0278 31.528
[71.057 31.528 1.182 13.008]V3 =
0.0031
0.0559 1
0.2597 1.182
0.0023 0.0278 0.2597 1 13.008
V3 = [79.125]
1 0.0073
0.0031
0.0023 27.419
0.0073 1
0.0559
0.0278 20.362
[27.419 20.362 16.696 13.057]V4 =
0.0031
0.0559 1
0.259716.696
0.0023
0.0278 0.2597
1 13.057
V4 = [40.964]
Fig.6. Distribution of lateral forces and shear by AVS method on frame with infills

Determination of Lateral Forces at Each Storey
Maximum Absolute Value Sum (AVS) Froof = F4 = V4 = 77.544 kN
Ffloor 3 = F3 = V3 V4 =116.791 77.544 = 39.247 kN
Ffloor 2 = F2 = V2 V3 =136.463 116.791 = 19.672kN Ffloor 1 = F1 = V1 V2 =163.508 136.463 = 27.045kN
Fig.7. Distribution of lateral forces and shear by SRSS method on frame with infills
Fig.8. Distribution of lateral forces and shear by CQC method on frame with infills
CONCLUSION
In case of Equivalent Static Force method, the design base shear value is found maximum for frame with infills over the frame without infills with the seismic weight being unchanged. The increase in base shear is due to natural period of the structure according to IS 1890:2002. This method is suitable for structures with low to medium range heights.
In case of Response spectrum method, Complete Quadratic Combination (CQC) provides, reasonable accuracy over AVS and SRSS methods due to provision for modal contribution. This method is suitable for almost all kinds of engineering interest.
REFERENCES
IS 1893 (part1): 2002, Criteria for earthquake Resistant Design of StructuresPart1: general Provisions and Buildings
, Bureau of Indian Standards, New Delhi.
Pankaj Agarwal, Manish Shrikhande (2007),Earthquake Resistant Design Of Structures PrenticeHall of India Limited Private Limited, New Delhi.
T.K.Data(2010), Seismic Analysis of Structures , John Wiley &Sons (Asia) Pte Ltd.