Explanatory Examples on Indian Seismic Code IS 1893 (Part I)

Document No. IITGN-WB-EQ3 V3.0
Final Report: IS1893 (Part 1) Solved Examples
IITGN – World Bank Project on Seismic Codes
Explanatory Examples on Indian Seismic
Code IS 1893 (Part I)
by
Sudhir K. Jain1
O R Jaiswal2
R K Ingle2
Debasis Roy3
with assistance from
Pothuraju Deepak2
1) Indian Institute of Technology Gandhinagar
2) Visvesvaraya National Intitute of Technology, Nagpur
3) Indian Institute of Technology Kharagpur
November 2019

 This document has been developed under the project on Seismic Codes
sponsored by the World Bank at Indian Institute of Technology
Gandhinagar.
 The solved examples presented in this document have been developed
to illustrate the provisions of IS:1893 Part 1 (2016) or the provisions that
are proposed under the current project as the case may be. Some of the
examples were developed and presented earlier in IITK-GSDMA
Project on Building Codes at IIT Kanpur (https://nicee.org/IITK-
GSDMA_Codes.php)and have been suitably modified wherever
appropriate.
 The views and opinions expressed are those of the authors and not
necessarily of the World Bank, IIT Gandhinagar, VNIT Nagpur or the
Bureau of Indian Standards.
 Comments and feedbacks may please be forwarded to: Prof. Sudhir K
Jain, email: skjain@iitgn.ac.in, skjain.iitk@gmail.com

Examples on IS 1893(Part 1)
IITGN-WB-EQ3 V3.0
CONTENTS
Sl.
No
Title Page No.
1. Calculation of Design Seismic Force by Static Analysis Method 4
2. Calculation of Design Seismic Force by Dynamic Analysis Method 7
3. Location of Centre of Mass 10
4 Location of Centre of Stiffness (a)in a Single-Storey Building, and (b) in a
Multi-Storey Building
11
5. Lateral Force Distribution as per Torsion Provisions of IS 1893-2016 (Part I) 16
6. Seismic Analysis Using Gross (Ig) and Effective (Ie) Moment of Inertia 18
7. Seismic Analysis Including Stiffness Effect of Infill Walls 24
8. Calculation of Time Period for a Building with Structural Walls 29
9. Seismic Loading for a Dual System 32
10. Design for Anchorage of an Equipment 33
11. Anchorage Design for an Equipment Supported on Vibration Isolator 35
12. Design of a Large Sign Board on a Building 37
13. Liquefaction Analysis Using SPT Data 38
14. Liquefaction Analysis Using CPT Data 40
15. Liquefaction Analysis with SPT data for a site with a structure imposing a
surcharge at ground surface
44

IITGN-WB-EQ3 V3.0 Example 1/Page 4
Example 1 – Calculation of Design Seismic Force by Static Analysis
Method
Problem Statement:
Consider a four-storey reinforced concrete office building shown in Figure 1.1. The building is located in
Shillong (seismic zone V). The soil conditions are medium stiff and the entire building is supported on a raft
foundation. The R. C. frames are infilled with brick-masonry. The lumped weight due to dead loads is 12
kN/m2
on floors and 10 kN/m2
on the roof. The floors are to cater for a live load of 4 kN/m2
on floors and
1.5kN/m2
on the roof. Determine design seismic load on the structure as per new code.
[Problem adopted from Jain S.K, “A Proposed Draft for IS:1893 Provisions on Seismic Design of Buildings;
Part II: Commentary and Examples”, Journal of Structural Engineering, Vol.22, No.2, July 1995, pp.73-90 ]
Figure 1.1 – Building configuration
3
@
5000
PLAN
3200
3200
3200
4200
ELEVATION
(4)
(1)
x
(D)
(C)
(A)
(B)
(3)
(2)
4 @ 5000
(5)
y
(4)

Solution:
Design Parameters:
For seismic zone V, the zone factor Z is 0.36
(Table 3 of IS: 1893 Part 1). Being an office
building, the importance factor, I, is 1.0 (Table
8of IS: 1893 Part 1). Building is required to be
provided with moment resisting frames detailed
as per IS: 13920-2016. Hence, the response
reduction factor, R, is 5 (Table 9of IS: 1893 Part
1).
Seismic Weights:
The floor area is 1520=300 sq. m. Since the live
load class is 4kN/sq.m, only 50% of the live load
is lumped at the floors. At roof, no live load is to
be lumped. Hence, the total seismic weight on the
floors and the roof is:
Floors:
𝑊1 = 𝑊2 = 𝑊3 = 300(12 + 0.5(4))
= 4200 kN
Roof:
𝑊4 = 300(10) = 3000 kN
(clause7.3.2, Table 10of IS: 1893 Part 1)
Total Seismic weight of the structure,
𝑊 = ∑ 𝑊𝑖 = 3(4200) + 3000 = 15600 kN
Fundamental Period:
Lateral load resistance is provided by moment
resisting frames infilled with brick masonry
panels. Hence, approximate fundamental natural
period:
(Clause 7.6.2.cof IS: 1893 Part 1)
EL in X-Direction:
𝑇 =
0.09ℎ
√𝑑
=
0.09(13.8)
√20
= 0.28 sec
The building is located on Type II (medium soil).
From Fig. 2 of IS: 1893, for T=0.28 sec,
𝑆𝑎
𝑔
= 2.5
𝐴ℎ =
𝑍
2
𝐼
𝑅
𝑆𝑎
𝑔
= (
0.36
2
)(
1
5
) (2.5) = 0.09
(Clause 6.4.2 of IS: 1893 Part 1)
Design base shear
𝑉𝐵 = 𝐴ℎ𝑊 = 0.09(15600) = 1404 kN
(Clause 7.6.1of IS: 1893 Part 1)
Force Distribution with Building Height:
The design base shear is to be distributed with
height as per clause 7.6.3a. Table 1.1 gives the
calculations. Fig. 1.2(a) shows the design seismic
force in X-direction for the entire building.
EL in Y-Direction:
𝑇 =
0.09ℎ
√𝑑
=
0.09(13.8)
√15
= 0.32 sec
𝑆𝑎
𝑔
= 2.5
𝐴ℎ =
𝑍
2
𝐼
𝑅
𝑆𝑎
𝑔
= (
0.36
2
)(
1
5
) (2.5) = 0.09
Therefore, for this building the design seismic
force in Y-direction is same as that in the X-
direction. Fig. 1.2(b) shows the design seismic
force on the building in the Y-direction.
Table 1.1 – Lateral Load Distribution with Height by the Static Method
Storey
Level
𝑊𝑖 (kN) ℎ𝑖 (m)
𝑊𝑖ℎ𝑖
2
1000
𝑊𝑖ℎ𝑖
2
∑ 𝑊𝑖ℎ𝑖
2
Lateral Force at ith
Level for EL in
direction (kN)
X Y
4 3,000 13.8 571.3 0.424 595 595
3 4,200 10.6 471.9 0.350 492 492
2 4,200 7.4 230.0 0.171 240 240
1 4,200 4.2 74.1 0.055 77 77
 1,347.3 1.000 1,404 1,404

(a) (b)
Figure 1.2–Design seismic force on the building for (a) X-direction, and (b) Y-direction.

Example 2 – Calculation of Design Seismic Force by Dynamic
Analysis Method
Problem Statement:
For the building of Example 1, the dynamic properties (natural periods, and mode shapes) for vibration in
the X-direction have been obtained by carrying out a free vibration analysis (Table 2.1). Obtain the design
seismic force in the X-direction by the dynamic analysis method outlined in cl. 7.7.5.4 and distribute it with
building height.
Table 2.1 – Free Vibration Properties of the building for vibration in the X-Direction
Natural Period (sec)
Mode 1 Mode 2 Mode 3
0.860 0.265 0.145
Mode Shape
Roof 1.000 1.000 1.000
3rd
Floor 0.904 0.216 -0.831
2nd
Floor 0.716 -0.701 -0.574
1st
Floor 0.441 -0.921 1.016
[Problem adopted from, Jain S.K, “A Proposed Draft for IS: 1893 Provisions on Seismic Design of
Buildings; Part II: Commentary and Examples”, Journal of Structural Engineering, Vol.22, No.2, July 1995,
pp.73-90]
Note – Natural periods given in Table 2.1 are obtained through dynamic analysis using structural analysis
software. For the same building, fundamental period (T = 0.28 sec) reported in Example 1 is obtained using
empirical formula, which is based on experimental data collected on various buildings. There can be a large
variation in computed fundamental period depending on how the building is modeled. For instance, in the
presence instance, the dynamic analysis did not consider stiffness contributions of infill walls and other non-
structural elements, making the analysis model of the building more flexible than the real building and hence
the computed fundamental period of 0.86 sec is much higher than that from the empirical equation.
Solution:
Table 2.2– Calculation of modal mass and modal participation factor (clause 7.7.5.4)
Storey
Level i
Weight
𝑊𝑖 (kN)
𝜙𝑖1 𝑊𝑖𝜙𝑖1 𝑊𝑖𝜙𝑖1
2
2
2
4 3,000 1.000 3,000 3,000 1.000 3,000 3,000 1.000 3,000 3,000
3 4,200 0.904 3,797 3,432 0.216 907 196 -0.831 -3,490 2,900
2 4,200 0.716 3,007 2,153 -0.701 -2,944 2,064 -0.574 -2,411 1,384
1 4,200 0.441 1,852 817 -0.921 -3,868 3,563 1.016 4,267 4,335
 15,600 11,656 9,402 -2,905 8,822 1,366 11,620
𝑀𝑘 =
[∑ 𝑊𝑖𝜙𝑖𝑘]2
𝑔 ∑ 𝑊𝑖𝜙𝑖𝑘
2
(11656)2
9402𝑔
=
14450
𝑔
kN
= 14,45,000 kg
(−2905)2
8822𝑔
=
957
𝑔
kN
=95,700 kg
(1366)2
11620𝑔
=
161
𝑔
kN
= 16,100 kg
% of Total weight 92.6% 6.1% 1.0%
𝑃𝑘 =
∑ 𝑊𝑖𝜙𝑖𝑘
∑ 𝑊𝑖𝜙𝑖𝑘
2 240
.
1
402
,
9
656
,
11
 329
.
0
822
,
8
905
,
2



118
.
0
620
,
11
366
,
1


It is seen that the first mode excites 92.6% of the
total mass. Hence, in this case, codal requirements
on number of modes to be considered such that at
least 90% of the total mass is excited, will be
satisfied by considering the first mode of
vibration only. However, for illustration, solution
to this example considers the first three modes of
vibration.
The lateral load Qikacting at ith
floor in the kth
mode is
𝑄𝑖𝑘 = 𝐴ℎ𝑘𝜙𝑖𝑘𝑃𝑘𝑊𝑖
(clause 7.7.5.4c of IS: 1893 Part 1)
The value of Ahk for different modes is obtained
from clause 6.4.2.
Mode 1:
T1 = 0.860sec
As the building located on Type 1 (rocky soil),
from the Clause 6.4.2b of IS 1893 Part 1, the
design acceleration coefficient is
(
𝑆𝑎
𝑔
) = (
1
0.86
) = 1.16
𝐴ℎ1 =
𝑍
2
𝐼
𝑅
𝑆𝑎
𝑔
= (
0.36
2
)(
1
5
) (1.16) = 0.0418
𝑄𝑖1 = 𝐴ℎ1𝜙𝑖1𝑃1𝑊𝑖 = (0.0418)(1.240)𝜙𝑖1𝑊𝑖
Mode 2:
T2 = 0.265sec;
(
𝑆𝑎
𝑔
) = 2.5;
𝐴ℎ2 =
𝑍
2
𝐼
𝑅
𝑆𝑎
𝑔
= (
0.36
2
)(
1
5
) (2.5) = 0.09
𝑄𝑖2 = 𝐴ℎ2𝜙𝑖2𝑃2𝑊𝑖 = (0.09)(−0.329)𝜙𝑖2𝑊𝑖
Mode 3:
T3 = 0.145 sec;
(
𝑆𝑎
𝑔
) = 2.5;
𝐴ℎ3 =
𝑍
2
𝐼
𝑅
𝑆𝑎
𝑔
= (
0.36
2
)(
1
5
) (2.5) = 0.09
𝑄𝑖3 = 𝐴ℎ3𝜙𝑖3𝑃3𝑊𝑖 = (0.09)(0.118)𝜙𝑖3𝑊𝑖
Table 2.3 summarizes the calculation of lateral
load at different floors in each mode.
Since all of the modes are well separated (clause
3.1), the contribution of different modes is
combined by the SRSS (square root of the sum of
the square) method
V4 = [(155.5)2
+ (88.8)2
+ (31.9)2
]1/2
= 182 kN
V3 = [(352.3)2
+ (115.6)2
+ (5.2)2
]1/2
= 371 kN
V2 = [(508.2)2
+ (28.4)2
+ (30.8)2
]1/2
= 510 kN
V1 = [(604.2)2
+ (86.2)2
+ (14.6)2
]1/2
=610 kN
(Clause 7.7.5.3b of IS: 1893 Part 1)
The externally applied design loads are then
obtained as:
Q4 = V4 = 182 kN
Q3 = V3 – V4 = 371 – 182 = 189 kN
Q2 = V2 – V3 = 510 – 371 = 139 kN
Q1 = V1 – V2 = 610 – 510 = 100 kN
(Clause 7.7.5.4f of IS: 1893 Part 1)
Clause 7.7.3 requires that the base shear obtained
by dynamic analysis (VB = 610 kN) be compared
with that obtained from empirical fundamental
period as per Clause 7.6.2c. If VB is less than that
from empirical value, the response quantities are
to be scaled up.
Table 2.3– Lateral load calculation by modal analysis method (earthquake in X-direction)
Floor
Level
(i)
Weight
𝑊𝑖(kN)
𝜙𝑖1 𝑄𝑖1 𝑉𝑖1 𝜙𝑖2 𝑄𝑖2 𝑉𝑖2 𝜙𝑖3 𝑄𝑖3 𝑉𝑖3
4 3,000 1.000 155.5 155.5 1.000 -88.8 -88.8 1.000 31.9 31.9
3 4,200 0.904 196.8 352.3 0.216 -26.8 -115.6 -0.831 -37.1 -5.2
2 4,200 0.716 155.9 508.2 -0.701 87.2 -28.4 -0.574 -25.6 -30.8
1 4,200 0.441 96.0 604.2 -0.921 114.6 86.2 1.016 45.4 14.6

We may interpret “base shear calculated using a
fundamental period as per 7.6.2” in two ways:
1. We calculate base shear as per Cl. 7.6.1. This
was done in the previous example for the same
building and we found the base shear as 1,404 kN.
Now, dynamic analysis gives us base shear of 610
kN which is lower. Hence, all the response
quantities are to be scaled up (Cl. 7.7.3) in the
ratio (1,404/610 = 2.30). Thus, the seismic forces
obtained above by dynamic analysis should be
scaled up as follows:
Q4 = 182  2.30 = 419 kN
Q3 = 189  2.30 = 435 kN
Q2 = 139  2.30 = 320 kN
Q1 = 100  2.30 = 230 kN
2. We may also interpret this clause to mean that
we redo the dynamic analysis but replace the
fundamental time period value by Ta(= 0.28 sec).
In that case, for mode 1:
T1= 0.28 sec;
(
𝑆𝑎
𝑔
) = 2.5;
𝐴ℎ1 =
𝑍
2
𝐼
𝑅
𝑆𝑎
𝑔
= (
0.36
2
)(
1
5
) (2.5) = 0.09
Modal mass times Ah1
= 14,450 (0.09) = 1,300 kN
Base shear in modes 2 and 3 is as calculated
earlier: Now, base shear in first mode of vibration
=1300 kN, 86.2 kN and 14.6 kN, respectively.
Total base shear by SRSS
=
= 1,303 kN
Notice that most of the base shear is contributed
by first mode only. In this interpretation of Cl
7.7.3, we need to scale up the values of response
quantities in the ratio (1,303/610 = 2.14). For
instance, the external seismic forces at floor levels
will now be:
Q4 = 182  2.14 = 389 kN
Q3 = 189  2.14 = 404 kN
Q2 = 139  2.14 = 297 kN
Q1 = 100  2.14 = 214 kN
Clearly, the second interpretation gives about
10% lower forces. We could make either
interpretation. Herein we will proceed with the
values from the second interpretation and
compare the design values with those obtained in
Example 1 as per static analysis:
Table 2.4 – Base shear at different storeys
Floor
Level
i
Q (static)
Q (dynamic,
scaled)
Storey Shear V
(static)
Storey ShearV
(dynamic,
scaled)
Storey Moment,
M (Static)
Storey
Moment, M
(Dynamic)
4 595 kN 389 kN 595 kN 389 kN 1,904 kNm 1,245kNm
3 492 kN 404 kN 1,087 kN 793 kN 5,382 kNm 3,782kNm
2 240 kN 297 kN 1,327 kN 1,090 kN 9,629 kNm 7,270kNm
1 77 kN 214 kN 1,404 kN 1,304 kN 15,526 kNm 12,750kNm
Note - Even though the base shear by the static and the dynamic analyses are comparable (because of
scaling as per Cl. 7.7.3), there is considerable difference in the lateral load distribution with building
height, and therein lies the advantage of dynamic analysis. For instance, the storey moments are
significantly affected by change in load distribution.
2
2
2
6
.
14
2
.
86
1300 


IITGN-WB-EQ3 V3.0 Example 3 /Page10
Example 3 – Location of Centre of Mass
Problem Statement:
Locate centre of mass of a building having non-uniform distribution of mass as shown in the Figure 3.1
Figure 3.1 – Plan
Solution:
Let us divide the roof slab into three rectangular
parts as shown in Figure 3.2
Figure 3.2 – Mass distribution
Mass of part I is 1200 kg/m2
, while that of the
other two parts is 1000 kg/m2.
.
Let origin be at point A, and the coordinates of
the centre of mass be at (X, Y)
𝑌 =
[(10×4×1200)×6+(10×4×1000)×6 +(20×4×1000)×2]
[(10×4×1200)+(10×4×1000)+(20×4×1000)]
𝑌 = 4.1 m
𝑋 =
[(10×4×1200)×5+(10×4×1000)×15+(20×4×1000)×10]
[(10×4×1200)+(10×4×1000)+(20×4×1000)]
𝑋 = 9.76 m
Hence, coordinates of centre of mass are
(9.76, 4.1) m.
1200 kg/m2
1000 kg/m2
20 m
4 m
10 m
A
1200 kg/m2
1000 kg/m2
I II
III
1000 kg/m
2
8 m
10 m
4 m
20 m
8 m

IITGN-WB-EQ3 V3.0 Example 4 /Page11
Example 4 – Location of Centre of Stiffness
Problem (a):
The plan of a simple one-storey building is shown in Figure 4.1. All columns and beams are same. Obtain its
centre of stiffness.
Figure 4.1 –Plan of Building
Solution:
In the X-direction there are three identical frames
located at uniform spacing. Hence, the y-
coordinate of centre of stiffness is located
symmetrically, i.e., at 5.0 m from the left bottom
corner.
In the Y-direction, there are four identical frames
having equal lateral stiffness. However, the
spacing is not uniform. Let the lateral stiffness of
each transverse frame be k, and coordinating of
center of stiffness be (X, Y).
k
k
k
k
k
k
k
k
X











20
10
5
0
= 8.75 m
Hence, coordinates of centre of stiffness are
(8.75, 5.0) m.
5 m
5 m
5 m 5 m 10 m
(1) (2) (3) (4)
(A)
(B)
(C)

Problem (b)
Figure presents plan view of a four-storey building. Columns are 400 mm × 400 mm each. Beams (250 mm
× 600 mm) connect columns in x-direction and beams (250 mm × 400 mm) connect column in y-direction.
All storeys are identical, except that the topmost storey is partially built (see Figure 4.2). Height of each
storey is 3 m. The design seismic loads are 15.72 kN, 62.89 kN, 139.06 kN and 162.45 kN for floors 1 to 4,
respectively. Determine the center of resistance (CR) following the (i) Single-floor definition, and (ii) All-
floor definition.
(a) Floor plan of floor 1, 2 and 3
(b) Floor plan of floor 4
5m
5m
5m 5m 10m
Y
X
Origin
5m
5m
5m 5m
Y
X
Origin

(c) 3D View of the building
Figure 4.2– Floor Plan and 3D view of the building
Solution:
The building has been modelled using the
commercially available software program
ETABS. Slab is modelled as a membrane, while a
rigid diaphragm constraint has been assigned to
the slab at every floor level. The columns and
beams are modelled as frame members.
The center of mass is assumed at the geometric
center of the floor plan (10.0 m, 5.0 m) for floor
1, 2 and 3, and (5.0 m, 5.0 m) for the fourth floor.
The building is symmetrical about x axis. The
center of resistance in y direction is at 5 m from
origin.
a) Single floor definition:
The steps involved in calculation of center of
resistance are as follows:
1. Apply lateral force (say F0) at the center
of mass of the floor of interest which will
cause (ϴ0) rotation to the floor.
2. Apply a moment (say M) at the center of
mass of the floor of interest which will
cause (ϴ) rotation to the floor.
3. In order for the floor to have purely
translational motion (with zero rotation),
moment (say M0) should be applied such
that it produces (- ϴ0) rotation.
0
0
M M


 
4. The center of resistance is given as
0
0
CR CM
M
x x
F
 
Table 4.5: Location of center of resistance (CR) following single-floor definition
Storey
F0
(kN)
ϴ0
(rad.)
M
(kNm)
ϴ
(rad.)
M0
(kNm)
x-coordinate
of CR
1 1000000 0.102945 1000000 0.082327 -1250440.32 8.75
2 1000000 0.356877 1000000 0.284235 -1255570.21 8.74
3 1000000 0.708285 1000000 0.530543 -1335019.03 8.66
4 1000000 -2.455199 1000000 1.000116 2454914.23 7.45

IITGn-World Bank -EQ2-V1.0 Example 4 /Page 14
X
CR
CM
Y
P
ex
F
P
Origin
L
b) All floor definition:
For calculation of center of resistance about y
axis, the lateral force profile was applied in y
direction at the center of mass of each floor.
The rotation of each floor was restricted by
applying the joint restraints at corners of each
floor which resulted in reaction force (P) at the
restraints (see Figure 4.3).
(a) Reaction force (P) at the restraints on floors 1, 2 and 3
(b) Reaction force (P) at the restraints on floor 4
Figure 4.3– Reaction force (P) at the restraints on each floor
The calculation of center of resistance of each
floor is calculated as follows,
CR CM
PL
x x
F
 
X
CR CM
Y
P
ex
F
P
Origin
L

Table 4.6: Location of center of resistance (CR) following all-floor definition
Storey
P (kN) F (kN)
x-coordinate of CR
(m)
1 -0.85 15.72 10.54
2 24.48 62.89 6.11
3 -57.78 139.06 14.16
4 -0.06 162.45 5.00
Summary
The center of resistance (CR) for each floor
following both the definitions are summarised in
Table 4.7
Table 4.7: Location of center of resistance (CR)
Storey
x-coordinate of CR (m)
Single-floor
definition
All-floor
definition
1 8.75 10.54
2 8.74 6.11
3 8.66 14.16
4 7.45 5.00

Example 5 –Lateral Force Distribution as per Torsion Provisions
of IS 1893-2016 (Part 1)
Problem Statement:
Consider a simple one-storey building having two shear walls in each direction. It has some gravity
columns that are not shown. All four walls are in M25 grade concrete, 200 thick and 4 m long. Storey
height is 4.5 m. Floor consists of cast-in-situ reinforced concrete. Design shear force on the building is
100 kN in either direction.
Compute design lateral forces on different shear walls using the torsion provisions of 2016 edition of
IS 1893 (Part 1).
Figure 5.1 – Plan
Solution:
Grade of concrete: M25
𝐸 = 5000√25 = 25000 N/mm2
Storey height h= 4500 m
Thickness of wall t= 200 mm
Length of walls L= 4000 mm
All walls are same, and hence, spaces have
same lateral stiffness, k.
Centre of mass (CM) will be the geometric
centre of the floor slab, i.e., (8.0, 4.0).
Centre of rigidity (CR) will be at (6.0, 4.0).
EQ Force in X-direction:
Because of symmetry in this direction,
calculated eccentricity = 0.0 m
Design eccentricity:
𝑒𝑑 = (1.5 × 0.0) + (0.05 × 8) = 0.4 m,
and
𝑒𝑑 = (0.0) − (0.05 × 8) = −0.4 m
(Clause 7.8.2 of IS 1893:2002)
Lateral forces in the walls due to translation:
𝐹𝐶𝑇 =
𝐾𝑐
𝐾𝑐 + 𝐾𝐷
𝐹 = 50 kN
𝐹𝐷𝑇 =
𝐾𝐷
𝐾𝑐 + 𝐾𝐷
𝐹 = 50 kN
Lateral forces in the walls due to torsional
moment:
𝐹𝑖𝑅 = (𝐾𝑖𝑟𝑖 ∑ 𝐾𝑖𝑟𝑖
2
𝑖=𝐴,𝐵,𝐶,𝐷
⁄ ) (𝐹𝑒𝑑)
where ri is the distance of the shear wall from
CR.
All the walls have same stiffness, KA= KB = KC
= KD = k, and
𝑟𝐴 = −6.0 m
𝑟𝐵 = +6.0 m
𝑟𝐶 = +4.0 m
𝑟𝐷 = −4.0 m
X
4m
2m
8m
16m
A B
C
D
4m
4m
Y

and
𝑒𝑑 = ±0.4 m
Therefore,
𝐹𝐴𝑅 =
𝑘𝑟𝐴
(𝑟𝐴
2 + 𝑟𝐵
2 + 𝑟𝐶
2 + 𝑟𝐷
2)𝑘
(𝐹𝑒𝑑)
𝐹𝐴𝑅 = ±2.31 kN
Similarly,
FBR= 31
.
2
 kN
FCR= 54
.
1
 kN
FDR= 54
.
1
 kN
Total lateral forces in the walls due to seismic
load in X direction:
FA= 2.31 kN
FB= 2.31 kN
FC= Max (50 54
.
1
 ) = 51.54 kN
FD= Max (50 54
.
1
 ) = 51.54 kN
EQ Force in Y-direction:
Calculated eccentricity= 2.0 m
Design eccentricity:
𝑒𝑑 = (1.5 × 2.0) + (0.05 × 16) = 3.8 m
or
𝑒𝑑 = (2.0) − (0.05 × 16) = 1.2 m
Lateral forces in the walls due to translation:
𝐹𝐴𝑇 =
𝐾𝐴
𝐾𝐴 + 𝐾𝐵
𝐹 = 50 kN
𝐹𝐵𝑇 =
𝐾𝐵
𝐾𝐴 + 𝐾𝐵
𝐹 = 50 kN
Lateral force in the walls due to torsional
moment: when ed = 3.8 m
𝐹𝐴𝑅 =
𝑘𝑟𝐴
(𝑟𝐴
2 + 𝑟𝐵
2 + 𝑟𝐶
2 + 𝑟𝐷
2)𝑘
(𝐹𝑒𝑑)
𝐹𝐴𝑅 = − 21.92 kN
Similarly,
𝐹𝐵𝑅 = + 21.92 kN
𝐹𝐶𝑅 = − 14.62 kN
𝐹𝐷𝑅 = + 14.62 kN
Total lateral forces in the walls:
FA= 50 - 21.92= 28.08 kN
FB= 50 +21.92 = 71.92 kN
FC= -14.62 kN
FD= 14.62 kN
Similarly, when ed = 1.2 m, then the total
lateral forces in the walls will be,
FA= 50 – 6.93 = 43.07 kN
FB= 50 + 6.93 = 56.93 kN
FC= - 4.62 kN
FD= 4.62 kN
Maximum forces in walls due to seismic load
in Y direction:
FA= Max (28.08, 43.07) = 43.07 kN;
FB= Max (71.92, 56.93) = 71.92 kN;
FC = Max (14.62, 4.62) = 14.62 kN;
FD = Max (14.62, 4.62) = 14.62 kN;
Combining the forces obtained from seismic
loading in X and Y directions:
FA= 43.07 kN
FB= 71.92 kN
FC= 51.54 kN
FD= 51.54 kN.
It is to be noted that in 2002 version of IS
1893 Part 1, Clause 7.9.1 states that
“However, negative torsional shear shall be
neglected”. Hence, wall A would have been
designed for not less than 50 kN. However, in
the 2016 version of IS 1893 there is no such
provision. Hence, wall A will now be
designed for 43.07 kN.

IITGN-WB-EQ3 V3.0 Example 6 /Page 18
Example 6 – Seismic Analysis Using Gross (Ig) and Effective (Ie)
Moment of Inertia
Problem Statement:
Consider a four-storey reinforced concrete moment resisting frame building shown in Figure 6.1. The
building is located in seismic zone V with medium soil. The R.C. frames are infilled with brick-
masonry. Slab thickness is 150 mm, beams are 300 × 400 mm and columns are 350 ×350 mm. The
objective of this example is to perform seismic analysis using gross moment of inertia (Ig) and
effective moment of inertia (Ie) and compare various response quantities.
Figure 6.1– Building configuration
3
@
5
m
PLAN
3.2 m
3.2 m
3.2 m
4.2 m
ELEVATION
(4)
(1)
(D)
(C)
(A)
(B)
(3)
(2)
4 @ 5 m
(5)
(4)
Y
X

Solution:
Other data considered in this example are:
Grade of concrete = M25, live load on floor =
4 kN/m2
, live load on roof = 1.5 kN/m2
, floor
finish = 1 kN/m2
and roof finish = 2.5 kN/m2
.
Seismic weights of various floors are :
1st
Floor, W1= 4,333 kN
2nd
Floor,W2= 4,300 kN
3rd
Floor, W3= 4,300 kN
4th
Floor, W4= 2,920 kN
Total seismic weight = 15,850 kN
Seismic analysis is performed for X-direction.
The approximate fundamental time period
along X direction is 0.28 sec (clause 7.6.2 cof
IS 1893 Part 1).
Here, zone factor, Z = 0.36, importance factor,
I = 1, response reduction factor, R = 5, soil
type is medium soil. Hence Design horizontal
acceleration coefficient, Ah = 0.09 (clause
6.4.2 of IS 1893 Part 1).
As per equivalent static analysis method,
design base shear, 𝑉𝐵
̅̅̅ = 0.09 × 15,850 =
1,427 kN. The lateral load distribution in X
direction by equivalent static analysis method
(clause 7.6.3 a of IS 1893 Part 1) is shown in
Table 6.1.
Table 6.1– Lateral load distribution with height by the equivalent Static method
Storey
Level
Wi (kN) hi(m) Wi hi
2 𝑊𝑖ℎ𝑖
2
∑ 𝑊𝑖ℎ𝑖
2
Lateral Force at ith
Level
for EL in X direction (kN)
4 2,920 13.8 5,56,085 0.412 587
3 4,300 10.6 4,83,148 0.358 510
2 4,300 7.4 2,35,468 0.174 249
1 4,330 4.2 76,381 0.057 81
 15,850 13,51,082 1.000 1,427
Dynamic analysis using Ig:
For this building, considering Young’s
modulus as E = 5000 fck, and using gross
moment of inertia of beam and column cross-
section, dynamic properties in X-direction are
obtained using SAP2000 software. Seismic
mass of each floor is lumped at floor level, and
rigid diaphragm is applied at every floor.
Beams and columns are modelled using frame
element and gross moment of inertia (i.e.,
bd3
/12 for a rectangular section) is considered.
Since seismic mass is lumped at floor level,
mass density of beam and column members
are neglected. Thus, only stiffness effect of
beam, column is considered. The dynamic
properties are shown in Table 6.2.
Table 6.2–Dynamic properties of building in the X- direction using Ig
Natural Period (sec) 1.271 0.397 0.222
Mode Shape coefficient
Roof 1.000 1.000 1.000
3rd
Floor 0.904 0.221 -0.786
2nd
Floor 0.709 -0.702 -0.547
1st
Floor 0.424 -0.898 0.990
Modal mass Mk(%) 92.0 6.7 1.2
Modal participation factor, Pk 1.247 -0.347 0.133

Using response spectrum analysis (7.7.5.4 of
IS 1893 Part 1), lateral forces at each floor are
obtained as:
Q4 = 171 kN
Q3 = 178 kN
Q2 = 127 kN
Q1 = 94 kN
The total base shear from this dynamic
analysis, VBG= 171+178+127+94 = 570 kN.
Since dynamic base shear is less than the base
shear obtained from equivalent static analysis,
the dynamic response quantities are scaled up
in the ratio of 𝑉𝐵
̅̅̅ 𝑉𝐵𝐺 (1,427 570
⁄ = 2.50
⁄ ).
(clause 7.7.3 of IS 1893 Part 1)
The scaled up lateral design seismic forces at
each floor are:
Q4 = 171  2.50 = 428 kN
Q3 = 178  2.50 = 445 kN
Q2 = 127  2.50 = 317 kN
Q1= 94 2.50 = 237 kN
Dynamic analysis using Ie:
Now, effective moment of inertia (Ie) is used.
As per clause 6.4.3.1 of IS 1893 Part 1
effective moment of inertia are as below:
Ie =0.35 Ig for beams
Ie =0.7 Ig for columns
Dynamic properties using effective moment of
Inertia are shown in Table 6.3
Table 6.3 –Dynamic properties of building
in the X- direction using Ie
Mode
1
Mode
2
Mode 3
Natural Period
(sec)
1.773 0.541 0.289
Roof 1.000 1.000 1.000
3rd
Floor 0.886 0.141 -0.872
2nd
Floor 0.669 -0.744 -0.442
1st
Floor 0.367 -0.836 1.059
Modal mass
Mk(%)
89.8 8.1 1.8
Modal
participation
factor, Pk
1.272 -0.391 0.156
obtained as:
Q4 = 151 kN
Q3 = 116 kN
Q2 = 73 kN
Q1 = 71 kN
analysis VBE= 151+116+73+71 = 411 kN.
Since dynamic base shear is less than base
̅̅̅ 𝑉𝐵𝐸 (1,427 411
⁄ = 3.47
⁄ ).
(clause 7.7.3 of IS 1893 Part 1)
each floor are:
Q4 = 151  3.47 = 524 kN
Q3 = 116  3.47 = 404 kN
Q2= 73 3.47 = 252 kN
Q1= 71 3.47 = 247 kN
Dynamic analysis using Ie as per proposed
modifications to IS 1893 Part 1:
As per proposed modifications, effective
moment of inertia are given as:
Ie =0.5 Ig for beams
Ie =0.7 Ig for columns
Analysis using these properties is also
performed and comparison of various response
quantities obtained from dynamic analysis of
model with Ig , Ie of IS 1893 and of proposed
modifications is given in Table 6.4.
Table 6.4 – Comparison of responses
Ig Ie Ie(proposed)
1st
modal period
(sec)
1.271 1.773 1.626
1st
modal mass
(%)
92.0 89.8 91.0
𝑉𝐵
̅̅̅/VB 2.50 3.47 3.18
Scaled up design forces (kN)
Q4 428 524 483
Q3 445 404 425
Q2 317 252 277
Q1 237 247 242

Observations:
Following observations are noted:
1. For building with Ie time period is more
since the analysis model of the building
becomes more flexible.
2. The mode shape and modal mass have
changed due to change in moment of
inertia.
3. After scaling with static base shear, total
base shear in both the cases is same, but
distribution of seismic forces on various
floors is different.
Drift calculation:
As per clause 7.11.1 of IS 1893 Part 1,
following points are to be noted:
1) Deformation of RC building shall be based
on analysis using effective moment of
inertia.
2) Storey Drift is to be obtained for lateral
earthquake loads without any load factors,
i.e., with partial safety of 1.0. (clause
7.11.1.1)
3) Displacement estimates obtained from
dynamic analysis methods shall not be
scaled up by the ratio 𝑉𝐵
̅̅̅ 𝑉𝐵
⁄ . (clause
7.11.1.2)
Here, in order to demonstrate the effect of Ig
and Ie, drift calculations are done for both the
cases. Unscaled dynamic forces of model with
Ig and Ie of IS 1893 and proposed
modifications are given in Table 6.5. The
displacement at each floor due to these lateral
earthquake forces is shown in Figure 6.2, 6.3
and 6.4 respectively for Ig , Ie of IS 1893and Ie
of proposed modifications.
Table 6.5– Lateral earthquake forces with Ig
and Ie
Unscaled lateral earthquake forces
model
with Ig
model with
Ie (IS 1893)
model with
Ie(proposed)
Q4 171 kN 151 kN 152 kN
Q3 178 kN 116 kN 134 kN
Q2 127 kN 73 kN 87 kN
Q1 94 kN 71 kN 76 kN
Base
shear
570 kN 411 kN 449 kN
Figure 6.2 Storey displacement and drift for
building model with Ig
building model with Ie (IS 1893)
building model with Ie (proposed
modifications)
D4 = 20.1 mm
D3 = 17.9 mm
D2 = 13.9 mm
D1 = 8.3 mm
3.2
m
3.2
m
3.2
m
4.2
m
Storey Drift d1
=(8.3 − 0) 4200
⁄
= 0.00197(0.197%)
Storey Drift d2
=(13.9 − 8.3) 3200
⁄
= 0.00175 (0.175%)
Storey Drift d3
=(17.9 − 13.9) 3200
⁄
= 0.00125 (0.125%)
Storey Drift d4
=(20.1 − 17.9) 3200
⁄
= 0.00069(0.069%)
D4 = 29.9 mm
D3 = 25.8 mm
D2 = 19.1 mm
D1 = 10.5 mm
3.2
m
3.2
m
3.2
m
4.2
m
Storey Drift d1
=(10.5 − 0) 4200
⁄
= 0.00250 (0.250%)
Storey Drift d2
=(19.1 − 10.5) 3200
⁄
= 0.00268 (0.268%)
Storey Drift d3
=(25.8 − 19.1) 3200
⁄
= 0.00209 (0.209*%)
Storey Drift d4
=(29.9 − 25.8) 3200
⁄
= 0.00128(0.128%)
D4 = 26.5 mm
D3 = 23.3 mm
D2 = 17.7
mm
D1 = 10.2 mm
3.
2
m
3.
2
m
3.
2
m
4.
2
m
Storey Drift d1
=(10.2 − 0) 4200
⁄
= 0.00242(0.242%)
Storey Drift d2
=(17.7 − 10.2) 3200
⁄
= 0.00235 (0.235%)
Storey Drift d3
=(23.3 − 17.7) 3200
⁄
= 0.00176 (0.176%)
Storey Drift d4
=(26.5 − 23.3) 3200
⁄
= 0.00100(0.100%)

A comparison of storey displacement and drift
obtained from the analysis of building model
with Ig and Ie and using respective lateral
earthquake forces is given in Table 6.6.
From these results it is noted that lateral
earthquake forces are less for the case of
building model with Ie, but drift and
deformation are higher for this model.
However, the drift are less than 0.4% as per
clause 7.11.1.1 of IS 1893 Part 1.
Note –It is reiterated here that in this
example, drift calculations with Ig and Ieare
done for the purpose of comparison only. In
practice drift is to be calculated using Ie
model.
Member forces:
Forces in various beams and columns are to be
obtained using scaled up forces, i.e., after
multiplying by, 𝑉𝐵
̅̅̅ 𝑉𝐵
⁄ . The member forces
are obtained using both the building model
with Ig and Ie, by applying respective
earthquake forces. For some of the members
these forces are shown in Table 6.7 and Table
6.8. In these Tables, the percentage difference
in maximum forces is also listed.
Table 6.6– Storey deformation and drift
Floor
Level
Ig Ie (IS 1893) Ie (proposed modifications)
Storey
Displac
ement
Di(mm)
Inter
Storey
Displace-
ment
Di – Di-1
(mm)
Inter Storey
Drift
(
𝐷𝑖 − 𝐷𝑖−1
ℎ𝑖
)
× 100
Storey
Displacem
ent
Di(mm)
Inter
Storey
Displace
ment
Di – Di-1
(mm)
Inter Storey
Drift
(
ℎ𝑖
)
× 100
Storey
Displace
ment
Di(mm)
Inter
Storey
Displacem
ent
Di – Di-1
(mm)
Inter Storey
Drift
(
ℎ𝑖
)
× 100
4 20.1 2.2 0.069 29.9 4.1 0.128 26.5 3.2 0.100
3 17.9 4.0 0.125 25.8 6.7 0.209 23.3 5.6 0.175
2 13.9 5.6 0.175 19.1 8.6 0.269 17.7 7.5 0.234
1 8.3 8.3 0.198 10.5 10.5 0.250 10.2 10.2 0.243
Table 6.7 – Comparison of Beam Forces by Considering Ig and Ie for EQ load case
Member force
using
Left end Centre Right end
Shear
(kN)
Moment
(kN-m)
Shear
(kN)
Moment
(kN-m)
Shear
(kN)
Moment
(kN-m)
Ground Floor Beam D1D2
Ig 58.7 157 58.7 10.4 58.7 -136.3
Ie (IS 1893) 51.7 134.6 51.7 5.3 51.7 -124
% difference*
11.9 14.3
Ie (Proposed) 55.5 146.3 55.5 7.6 55.5 -131.2
% difference*
5.5 6.8
Ig 48.8 120.8 48.8 -1.2 48.8 -123.2
Ie(IS 1893) 46.2 115 46.2 -0.4 46.2 -115.8
% difference*
5.4 6
Ie (Proposed) 47.8 118.8 47.8 -0.7 47.8 -120.3
% difference*
2.0 2.4
Roof Beam D1D2
Ig 11.2 30.5 11.2 2.4 11.2 -25.7
Ie(IS 1893) 15 39.5 15 2 15 -35.6
% difference*
-33.7 -29.7
Ie (Proposed) 13.3 35.4 13.3 2.2 13.3 -30.9
% difference*
-18.8 -16.1
Roof Beam D2D3
Ig 10 24.9 10 0 10 -25
Ie(IS 1893) 13.4 33.2 13.4 -0.1 13.4 -33.5
% difference*
-33.7 -34.1
Ie (Proposed) 11.6 28.8 11.6 -0.2 11.6 -29.1
% difference*
-16.0 -16.4
*
Here % difference is with respective to Ig model

Table 6.8 – Comparison of Column Forces by Considering Ig and Ie for EQ load Case
Member force
using
Bottom end Top end
Axial
(kN)
Shear
(kN)
Moment
(kN-m)
Axial
(kN)
Shear
(kN)
Moment
(kN-m)
Ground Floor Column D1
Ig 140.7 63.6 161.8 140.7 63.6 -105.2
Ie (IS 1893) 138.9 63.2 178.2 138.9 63.2 -87.2
% difference*
1.3 0.6 -10.2
Ie (Proposed) 141 63.3 168.4 141 63.3 -97.3
% difference*
-0.2 0.5 -4.1
Ig -18.7 77.1 180.5 -18.7 77.1 -143.2
Ie(IS 1893) -12.2 77.1 197.6 -12.2 77.1 -126.4
% difference*
34.6 -0.1 -9.5
Ie (Proposed) -15.9 77.2 187.8 -15.9 77.2 -136.4
% difference*
15.0 -0.1 -4.0
Roof Beam D1D2
Ig 11.2 13.3 12.2 11.2 13.3 -30.5
Ie(IS 1893) 15 15.3 9.4 15 15.3 -39.5
% difference*
-33.7 -14.6 -29.7
Ie (Proposed) 13.3 14.7 11.8 13.3 14.7 -35.4
% difference*
-18.8 -10.5 -16.1
Roof Beam D2D3
Ig -1.2 26.9 35.6 -1.2 26.9 -50.6
Ie(IS 1893) -1.7 33.8 39.4 -1.7 33.8 -68.8
% difference*
-34.2 -25.5 -35.9
Ie (Proposed) -1.7 30.7 38.6 -1.7 30.7 -59.8
% difference*
-41.7 -14.1 -18.2
*
Here % difference is with respective to Ig model
Comparison of member forces from building
with Ig and Ie indicates that roof beams and
columns have large difference in their forces.
Further, column forces are consistently more
in the building model with Ie. These
differences are due to change in dynamic force
distribution in the two models (Table 6.4).

Example 7 – Seismic Analysis Including Stiffness Effect Infill
Walls
Problem Statement:
For the building considered in Example 6 include the stiffness effect of external wall of 230 mm thick
(Figure 7.1). Compressive Strength of brickis 10.5 MPa and Compressive Strength of mortar is 16.67
MPa. Compare dynamic characteristics with those obtained without including the wall stiffness effect
in the X-direction.
Figure 7.1 – Plan of building in with infill walls
Solution:
Plan of the building in Example 6 is
considered with the infill wall on the periphery
along the X-direction is shown in Figure 7.1.
The building is considered to have
Unreinforced Masonry (URM) infill wall.
Compressive Strength of brick is 10.5 MPa
and Compressive Strength of mortar is 16.67
MPa.The thickness of infill is taken as 230
mm. Infill is modelled as equivalent strut as
per Clause 7.9.2.2 of IS 1893 Part 1 2016.
Diagonal strutsare considered to be pin-jointed
with the RC frame.
It is to be noted that analysis is being done in
X-direction only.
Equivalent Diagonal Strut for URM Infill
Wall:
There are various approaches for modelling
the infill wall as strut. Here, three cases are
considered:
A) Strut model as per IS 1893 Part 1
B) Strut model as per Paulay and Priestly,
1992
C) Strut model as per Paulay and
Priestly(1992) with 20% opening. Effect of
opening is taken as per Mondal and Jain
(2008).
A) Strut as per IS 1893 Part 1:
The modulus of elasticity Em (in MPa) of
masonry infill is calculated as per Clause
7.9.2.1 of IS 1893 Part 12016:
𝑓𝑚 = 0.433𝑓𝑏
0.64
𝑓
𝑚𝑜
0.36
where, fband fmo arethe Compressive Strengths
of brick and mortar respectively.
3
@
5
m
(1)
(A)
(3)
(2)
(B)
(C)
(D)
(5)
Y
X
4 @ 5 m
PLAN
(4)

𝑓𝑚 = 0.433(10.5)0.64(16.67)0.36
𝑓𝑚 = 5.4 MPa
𝐸𝑚 = 550𝑓𝑚 = 2970 MPa
Where, fmis Compressive Strength of masonry
prism, in MPa.
Diagonal strut are considered to be pin-joined
to RC frame at the junction of beam and
column (Figure 7.2). Width wds, of the
equivalent diagonal strut shall be taken as per
clause 7.9.2.2 of IS 1893 Part 1:
𝑤𝑑𝑠 = 0.175 × 𝛼ℎ
−0.4
× 𝐿𝑑𝑠
Where
𝛼ℎ = ℎ (√
𝐸𝑚 𝑡 𝑠𝑖𝑛(2𝜃)
4 𝐸𝑓 𝐼𝑐 ℎ
4
)
Efis the moduli of elasticity of RC frame
Icis the moment of inertia of adjoining column
tis thickness of the infill wall
θis the angle of the diagonal strut with
horizontal
his the clear height of URM infill wall
Lds is the length of the diagonal strut
At ground storey clear height (h) is 3.8 m and
𝜃 is 400
with the horizontal, clear length is of
4.65 m and length of the diagonal strut (𝐿𝑑𝑠)
of 6 m (√(3.8)2 + (4.65)2), 𝐸𝑓 of 25000
MPa(M25 grade of concrete) and Ic= 1.25 ×
10−3
m4
((0.35)x(0.35)3
/12)
𝛼ℎ and𝑤𝑑𝑠are:
𝛼ℎ
= 3.8 (√
2970 × 0.23 × sin(2 × 40)
4 × 25000 × 1.25 × 10−3 × 3.8
4
)
= 4.2
𝑤𝑑𝑠 = 0.175 × (4.2)−0.4
× 6 = 0.592 m
For the remaining storeys of higher floors,
height h = 2.8m and 𝜃 is 310
, 𝐿𝑑𝑠 =
5.42 m, 𝛼ℎ and𝑤𝑑𝑠 are:
𝛼ℎ
= 2.8 (√
2970 × 0.23 × sin(2 × 31)
4 × 25000 × 1.25 × 10−3 × 2.8
4
)
= 3.2
𝑤𝑑𝑠 = 0.175 × (3.2)−0.4
× 5.42 = 0.596 m
Figure 7.2 – Equivalent diagonal strut of
URM infill wall
Now, including the stiffness effect of external
wall in X-direction, dynamic analysis of
building is performed and dynamic properties
are shown in Table 7.1.
IS 1893 Part 1), lateral forces at each floor
are obtained as:
Q4 = 330 kN
Q3 = 417 kN
Q2 = 320 kN
Q1 = 204 kN
analysis VBG= 330+417+320+204 = 1271 kN.
̅̅̅ 𝑉𝐵𝐺 (1427 1271
⁄ = 1.12
⁄ ).
(Clause 7.7.3 of IS 1893 Part 1)
F
Pin Joint
of strut

Table 7.1–Dynamic properties of building in the X- direction using strut as per IS 1893 Part 1
Roof 1.000 1.000 1.000
3rd
Floor 0.904 0.251 -0.744
2nd
Floor 0.706 -0.712 -0.622
1st
Floor 0.428 -0.953 0.999
Modal mass Mk(%) 92.1 6.9 1.1
each floor are:
Q4 = 330  1.12 = 371 kN
Q3 = 417  1.12 = 468 kN
Q2 = 320  1.12 = 359 kN
Q1 = 204  1.12 = 229 kN
B) Strut as per Paulay and Priestley 1992:
The width of the equivalent diagonal strut
𝑤𝑑𝑠 = 0.25 × 𝐿𝑑𝑠, At ground storey clear
height (h) = 3.8 m, 𝜃 = 400
and clear
horizontal length = 4.65 m. Thus, one gets,
length of the diagonal strut (𝐿𝑑𝑠) = 6 m, and
𝑤𝑑𝑠 = 0.25 × 6 = 1.5 m. For remaining
storeys h = 2.8 m, 𝜃 = 310
, 𝐿𝑑𝑠 = 5.42 m,
and 𝑤𝑑𝑠 = 1.36 m.
Again dynamic analysis is performed and
dynamic properties in X-direction are shown
in Table 7.2.
obtained as:
Q4 = 348 kN
Q3 = 431 kN
Q2 = 323 kN
Q1 = 200 kN
analysis VBG= 348+431+323+200 = 1302 kN.
̅̅̅ 𝑉𝐵𝐺 (1427 1302
⁄ = 1.10
⁄ ).
each floor are:
Q4 = 348  1.10 = 382 kN
Q3 = 431  1.10 = 472 kN
Q2 = 323  1.10 = 354 kN
Q1 = 200  1.10 = 219 kN
C) Strut as per Paulay and Priestley 1992
with 20% openings:
For the case of opening in the wall, the
reduced width (wdo) of strut is given by
(Mondal and Jain 2008):
wdo = 𝜌𝑤wds
𝜌𝑤 = 1 − 2.5 𝐴𝑟
Here, Ar is ratio of face area of opening to face
area of infill. Considering 20% opening
𝜌𝑤 = 1 − 2.5 (0.2) = 0.5
Hence, at the ground floor, 𝑤𝑑𝑠 = 1.5 ×
0.5 = 0.75 m, and at remaining storey𝑤𝑑𝑠 =
1.36 × 0.5 = 0.68 m. It is interesting to note
that due to 20% opening, the width of strut has
reduced by 50%.
Dynamic properties of this model in X-
direction are shown in Table 7.3.
obtained as:
Q4 = 345 kN
Q3 = 432 kN
Q2 = 327 kN
Q1 = 203 kN
analysis VBG= 345+432+327+203 = 1307 kN.
̅̅̅ 𝑉𝐵𝐺 (1427 1307
⁄ = 1.09
⁄ ).

Table 7.2–Dynamic properties of building in the X- direction using strut model of Paulay and
Priestley 1992
Roof 1.000 1.000 1.000
3rd
Floor 0.889 0.207 -0.980
2nd
Floor 0.678 -0.750 -0.547
1st
Floor 0.395 -0.948 1.268
Modal mass Mk(%) 90.9 8.3 1.4
Table 7.3–Dynamic properties of building in the X- direction using strut as per Paulay and
Priestley 1992 with 20% opening
Roof 1.000 1.000 1.000
3rd
Floor 0.898 0.232 -0.588
2nd
Floor 0.692 -0.737 -0.743
1st
Floor 0.406 -0.945 0.973
Modal mass Mk(%) 91.3 7.5 1.2
each floor are:
Q4 = 345  1.09 = 377 kN
Q3 = 432  1.09 = 472 kN
Q2 = 327  1.09 = 357 kN
Q1 = 203  1.09 = 221 kN
Observations:
1) Due to inclusion of infill wall stiffness,
building lateral stiffness increases, and
hence, time period has reduced.
2) Width of strut as per IS 1893 Part 1 is
quite low as compared to that given by
Paulay and Priestley, 1992.
3) Due to opening, effective strut width
reduces considerably. Here, for 20%
opening, the width of strut gets reduced by
50%.
4) Distribution of forces on the floor changes
with the inclusion of strut.
A comparison of major dynamic
characteristics from various strut models is
shown in Table 7.4.

Table 7.4 – Comparison of dynamic proprieties using different strut models
Bare Frame
Strut Model
IS 1893
Part 1
Paulay and
Priestley
Paulay and Priestley
with 20% opening
1st
modal period (sec) 1.271 0.564 0.339 0.525
1st
modal mass (%) 92.0 92.1 90.9 91.3
Width of
strut (m)
Ground Floor 0.592 1.5 0.75
Other Floors 0.596 1.36 0.68
𝑉𝐵
̅̅̅/VB 2.5 1.12 1.10 1.09
Scaled up design forces (kN)
Q4 (kN) 428 371 382 345
Q3(kN) 445 468 472 432
Q2(kN) 317 359 354 327
Q1(kN) 237 229 219 203
References:
Paulay, T., and Priestley, M. J. N., 1992.
Seismic Design of Reinforced Concrete and
Masonry Buildings, John Wiley & Sons, Inc.,
New York, NY.
Mondal, G., and Jain, S. K., 2008. Lateral
stiffness of masonry infilled reinforced
concrete (RC) frames with central
opening. Earthquake spectra, 24(3), 701-723.
Changes as per proposed
modifications of IS 1893 Part 1:
As per the proposed modifications,
compressive strength of masonry
prism is given by
𝑓𝑚 = 0.63𝑓𝑏
0.49
𝑓𝑚0
0.32
Hence, in the present example, one gets
𝑓𝑚 = 0.63(10.5)0.49(16.67)0.32
= 4.9 MPa
and
𝐸𝑚 = 550𝑓𝑚 = 550(4.9) = 2695 MPa
Thus, as per proposed modifications,
value of E will get changed to 2693 MPa
as against 2970 MPa.

Example 8– Calculation of Time Period for a Building with
Structural Walls
Problem Statement:
Consider a four storey RC building from Example 6, where now shear walls are added as shown in
Figure 8.1. These shear walls extend up to the roof slab. Consider grade of concrete as M25. The total
height of the building is 13.8 m and the thickness of the structural wall is 0.2 m. Calculate time period
of building using clause 7.6.2 b of IS 1893 Part 1.
Figure 8.1 – Plan of building in example 6 with structural wall
(A) (B)
Figure 8.2 – Elevation of building A) XZ Plane B) YZ Plane
Solution:
For a building with a structural wall, an
approximate time period is obtained as per
clause 7.6.2 b of IS 1893 part 1.
𝑇𝑎 =
0.075ℎ0.75
√𝐴𝑤
≥
0.09ℎ
√𝑑
Where 𝐴𝑤 is total eeffective area (m2
) of walls
in the first storey of the building given by:
𝐴𝑤 = ∑ [𝐴𝑤𝑖 {0.2 + (
𝐿𝑤𝑖
ℎ
)
2
}]
𝑁𝑤
𝑖=1
Where
ℎ = height of the building
𝐴𝑤𝑖 = effective cross-sectional area of wall i in
first storey of the building, in m2
(2) (3) (4) (5)
(A)
(B)
(C)
(D)
3
@
5000
PLAN
(1)
x
4 @ 5000
y
x
z
5 m 5 m 5 m 5 m
4.
2
m
3.2
m
3.2
m
3.2
m
y
z
5 m 5 m 5 m
4.
2
m
3.2
m
3.2
m
3.2
m

𝐿𝑤𝑖 = length of structural wall i in first storey
in the considered direction of lateral forces, in
m
𝑑 = base dimension of the building at the
plinth level along the considered direction of
the earthquake shaking, in m
𝑁𝑤 = number of walls in the considered
direction of the earthquake shaking
For Earthquake in X Direction:
There are two shear walls are present with a
length of 5 m each.
𝐿𝑤1 = 𝐿𝑤2 = 5 m
𝐴𝑤1 = 𝐴𝑤2 = 5 × 0.2 = 1 m2
𝐴𝑤 = [1 {0.2 + (
5
13.8
)
2
}] × 2 = 0.66 m2
𝑇𝑎 =
0.075 × 13.80.75
√0.66
= 0.66 sec
which is greater than
0.09 × 13.8
√20
= 0.27 sec
For Earthquake in Y Direction:
There are two shear walls are present with
length of 5 m each.
𝐿𝑤1 = 𝐿𝑤2 = 5 m
𝐴𝑤1 = 𝐴𝑤2 = 5 × 0.2 = 1 𝑚2
𝐴𝑤 = [1 {0.2 + (
5
13.8
)
2
}] × 2 = 0.66 m2
𝑇𝑎 =
0.075 × 13.80.75
√0.66
= 0.66 sec
which is greater than
0.09 × 13.8
√15
= 0.32 sec
Proposed Modifications to IS 1893 (Part1):
Approximate time period as per suggested
modifications to IS 1893:2016 Part 1 is
𝑇𝑎 =
0.0058ℎ
√𝐴𝑤
Where 𝐴𝑤 is effective area factor of walls in
the first storey of the building given by:
𝐴𝑤 =
100
𝐴𝐵
∑ [
𝐴𝑤𝑖
{1 + 0.83 (
ℎ
𝐿𝑤𝑖
)
2
}
]
𝑁𝑤
𝑖=1
where,
ℎ = Height of the building as defined in 7.6.2a
of IS 1893 Part 1
𝐴𝑤𝑖 = effective cross-sectional area of wall i in
first storey of the building, in m2
𝐴𝐵 =plan area at the base of structure
𝐿𝑤𝑖 = length of structural wall i in first storey
in the considered direction of lateral forces, in
m
𝑁𝑤 = number of walls in the considered
direction of the earthquake shaking
For Earthquake in X Direction:
There are two shear walls are present with a
length of 5 m each.
𝐿𝑤1 = 𝐿𝑤2 = 5 m
𝐴𝑤1 = 𝐴𝑤2 = 5 × 0.2 = 1 m2
ℎ = 13.8 m
𝐴𝐵 = 20 × 15 = 300 m2
𝐴𝑤 =
100
𝐴𝐵
∑ [
𝐴𝑤𝑖
{1 + 0.83 (
ℎ
𝐿𝑤𝑖
)
2
}
]
𝑁𝑤
𝑖=1
=
100
300
[
1
{1 + 0.83 (
13.8
5
)
2
}
× 2]
𝐴𝑤 =0.091m2

𝑇𝑎 =
0.0058 × 13.8
√0.091
= 0.265 sec
For Earthquake in Y Direction:
There are two shear walls are present with
length of 5 m each.
𝐿𝑤1 = 𝐿𝑤2 = 5 m
ℎ𝑖 = 13.8 m
𝐴𝐵 = 20 × 15 = 300 m2
𝐴𝑤1 = 𝐴𝑤2= 5 × 0.2 = 1 m2
𝐴𝑤 =
100
𝐴𝐵
∑ [
𝐴𝑤𝑖
{1 + 0.83 (
ℎ
𝐿𝑤𝑖
)
2
}
]
𝑁𝑤
𝑖=1
=
100
300
[
1
{1 + 0.83 (
13.8
5
)
2
}
× 2]
𝐴𝑤 = 0.091m2
𝑇𝑎 =
0.0058 × 13.8
√0.091
= 0.265 sec
Time period of the building as per IS
1893:2016 Part 1 equation is 0.66 sec while it
is 0.265 sec with the proposed method which
is more reasonable.
Note:
As per Amendment No.1 dated September
2017 of IS1893:2016 Part 1, the formula for
Aw is changed as follows:
𝐴𝑤 = ∑ [𝐴𝑤𝑖 {0.2 + (
𝐿𝑤𝑖
ℎ
)}
2
]
𝑁𝑤
𝑖=1
However, the changed formula given in
amendment is not correct and one shall use
the original formula given in the code.

Example 9– Seismic Loading for a Dual System
Problem Statement:
Consider the building with shear walls given in Example 8. This dual system comprises of two shear
walls each in X- and Y-direction (Figure 9.1). Consider grade of concrete as M25, thickness of shear
wall as 200 mm. Obtain design base shear for shear wall and frame system as per Clause 7.2.7 of IS
1893 Part 1.
Figure 9.1 – Plan of Building
Solution:
For this building, design seismic loads on
various floors are obtained using static method
in Example 6. These seismic loads in X-
direction on each floor are given below:
Q4 = 595 kN
Q3 = 492 kN
Q2 = 240 kN
Q1 = 77 kN
Thus, total base shear is 1404 kN
This dual system is modelled and analyzed for
these lateral forces. For beams, columns, and
shear walls gross section properties, i.e., Ig are
used. Effect of infill wall stiffness is not
included in the analysis. Analysis results show
that the base shear resisted by two shear walls
in X-direction is 1329kN (i.e 94.7%) of the
total base shear) and base shear taken by frame
comprising of beams and columns is 75kN
(i.e 5.3%) of total base shear.
Since, base shear resisted by frame is less than
25%, as per Clause 7.2.7 of IS 1893 Part 1, in
the dual system, frame has to be designed for
at least 25% of total base shear.
Therefore design forces in frame, (i.e in
columns and beams) are to be scaled up to
25% of the design base shear.
Hence, for columns and beams of frame,
design forces are to be scaled by a factor of
25/5.3 = 4.7, so that, the base shear for
moment resisting frame is 75 × 4.7 =
354 𝑘𝑁.
Thus, shear wall will be designed for base
shear of 1329kN (94.7%) and frame will be
designed for 354kN (25%) base shear.
(2) (3) (4) (5)
(A)
(B)
(C)
(D)
3
@
5000
PLAN
(1)
x
4 @ 5000
y

Example 10 – Design for Anchorage of an Equipment
Problem Statement:
A 100 kN equipment (Figure 10.1) is to be installed on the roof of a five storey building in
Simla (seismic zone IV). It is attached by four anchored bolts, one at each corner of the
equipment, embedded in a concrete slab. Floor to floor height of the building is 3.0 m. except
the ground storey which is 4.2 m. Determine the shear and tension demands on the anchored
bolts during earthquake shaking.
Figure 10.1– Equipment installed at roof
Solution:
Zone factor, Z = 0.24 (for zone IV, Table 3 of
IS 1893 Part 1),
Height of point of attachment of the equipment
above the foundation of the building,
x = (4.2 +3.0 × 4) m = 16.2 m,
Height of the building, h = 16.2 m,
Amplification factor of the equipment, 𝑎𝑃 =
1(rigid component, Table 13),
Response modification factorRp = 2.5 (Table
13),
Importance factor Ip = 1 (not life safety
component, Table 14),
Weight of the equipment, Wp = 100 kN
The design seismic force (Clause 7.13.3.1 of
draft IS 1893)
𝐹𝑃 = (
𝑍
2
) (1 +
𝑥
ℎ
) (
𝑎𝑃
𝑅𝑃
) 𝐼𝑃𝑊𝑃
= (
0.24
2
) (1 +
16.2
16.2
) (
1.0
2.5
) (1 × 100)
𝐹𝑃 = 9.6 kN < 0.1𝑊𝑃 = 10.0 kN
Hence, design seismic force, for the equipment
𝐹𝑃 = 10.0 kN.
The anchorage of equipment with the building
must be designed for twice of this force
(C7.13.3.4 of draft IS 1893)
Shear per anchor bolt,
𝑉 =
2𝐹𝑃
4
=
2(10.0)
4
= 5.0 kN
The overturning moment is
𝑀𝑜𝑡 = 2.0 × 10.0 × 1.5 = 30.0 kNm
p
F
1.0 m
1.5 m
Anchor bolt
CG
p
W
Anchor
bolt

The overturning moment is resisted by two
anchor bolts on either side. Hence, tension per
anchor bolt from overturning is
𝐹𝑡 =
(30.0)
(1.0 × 2)
= 15.0 kN
If overturning moment is considered in
diagonal direction then tension per anchor bolt
from overturning is
𝐹𝑡 =
(30.0)
(1.0 × √2)
= 21.3 kN

Example 11 – Anchorage Design for an Equipment
Supported on Vibration Isolator
Problem Statement:
A 100 kN electrical generator of a emergency power supply system is to be installed on the
fourth floor of a 6-storey hospital building in Guwahati (zone V). It is to be mounted on four
flexible vibration isolators, one at each corner of the unit, to damp the vibrations generated
during the operation. Floor to floor height of the building is 3.0 m. except the ground storey
which is 4.2 m. Determine the shear and tension demands on the isolators during earthquake
shaking.
Figure 11.1 –Electrical generator installed on the floor
Solution:
Zone factor, Z = 0.36 (for zone V, Table 3
of IS 1893 Part 1),
Height of point of attachment of the
generator above the foundation of the
building,
𝑥 = 4.2 + (3.0 × 3) = 13.2 m
Height of the building,
ℎ = 4.2 + (3.0 × 5) = 19.2 m
Amplification factor of the generator,
𝑎𝑃 = 2.5 (flexible component, Table 13),
Response modification factorRp = 2.5
(vibration isolator, Table 13),
Importance factor Ip = 1.5 (life safety
component, Table 14),
Weight of the generator, Wp = 100 kN
The design lateral force on the generator,
(clause 7.13.3.1 of draft IS 1893)
𝐹𝑃 = (
𝑍
2
) (1 +
𝑥
ℎ
) (
𝑎𝑃
𝑅𝑃
) 𝐼𝑃𝑊𝑃
=
0.36
2
(1 +
13.2
19.2
) (
2.5
2.5
) (1.5 × 100)
= 45.6 kN > 0.1𝑊𝑃 = 10.0 kN
Hence, design force is taken as 45.6 kN
Since the generator is mounted on flexible
vibration isolator, the design force is
doubled i.e.,
𝐹𝑃 = 2 × 45.6 = 91.2 kN
Shear force resisted by each isolator,
𝑉 =
𝐹𝑃
4
= 22.8 kN
The overturning moment,
𝑀𝑜𝑡 = 91.2 × 0.8 = 73.0 kNm
Vibration
Isolator
CG

The overturning moment (Mot) is resisted
by two vibration isolators on either side.
Therefore, tension or compression on each
isolator,
𝐹𝑡 =
(73.0)
(1.2 × 2)
= 30.4 kN
If overturning moment is considered in
diagonal direction then tension per anchor
bolt from overturning is
𝐹𝑡 =
(73.0)
(1.2 × √2)
= 43.0 kN

Example 12 – Design of a Large Sign Board on a Building
Problem Statement:
A neon sign board is attached to a 5-storey building in Ahmedabad (seismic zone III). It is
attached by two anchors at a height 12.0 m and 8.0 m. From the elastic analysis under design
seismic load, it is found that the deflections of upper and lower attachments of the sign board
are 35.0 mm and 25.0 mm, respectively. Find the design relative displacement.
Solution:
Since sign board is a displacement
sensitive nonstructural element, it should
be designed for seismic relative
displacement.
(All references made as per clause 7.13.4
of IS 1893 (modified)
Height of level x to which upper
connection point is attached, hx= 12.0 m
Height of level y to which lower
connection point is attached, hy = 8.0 m
Deflection at building level x of structure
A due to design seismic load determined
by elastic analysis = 35.0 mm
Deflection at building level y of structure
A due to design seismic load determined
by elastic analysis = 25.0 mm
Response reduction factor of the building
R = 5 (special RC moment resisting frame,
Table 9 of IS 1893 of Part 1)
𝛿𝑥𝐴 = 5 × 35 = 175.0 mm
𝛿𝑦𝐴 = 5 × 25 = 125.0 mm
(i)𝐷𝑃 = (𝛿𝑥𝐴 − 𝛿𝑦𝐴) = 50.0 mm
Design the connections of neon board to
accommodate a relative motion of 50 mm.
(ii) Alternatively, assuming that the
analysis of building is not possible to
assess deflections under seismic loads, one
may use the drift limits (this presumes that
the building complies with seismic code).
Maximum interstorey drift allowance as
per clause 7.11.1.1 of IS 1893 Part 1 is
0.004 times the storey height, i.e.,
Δ𝑎𝐴
ℎ𝑠𝑥
= 0.004
𝐷𝑃 = 𝑅(ℎ𝑥 − ℎ𝑦) (
Δ𝑎𝐴
ℎ𝑠𝑥
)
= 5(12000 − 8000)(0.004)
= 80.0 mm
The neon board will be designed to
accommodate a relative motion of 80 mm.

Example 13 –Liquefaction Analysis using SPT data
Problem Statement:
This involves a hypothetical site in seismic zone IV with design earthquake magnitude of 7.5. No
structure exists at this site, ground surface is horizontal and groundwater table found at 6 m depth.
The site is underlain by horizontal layers of poorly graded sand and silty sand with 18 kN/m3
unit
weight. Liquefaction susceptibility of a layer of sand at 12.75 m depth with 8 % non plastic fines
(with a raw SPT blow count of 17 measured inside a 100-mm diameter borehole using a donut
hammer handled manually with a rope and pulley, and a split-spoon sampler without liner) at this
site was assessed proceeding as follows:
Solution:
Site Characterization:
The site is located in zone IV. The peak
horizontal ground acceleration value for the
site will be taken as 0.24g corresponding to
zone factor Z = 0.24
Liquefaction Potential of Underlying Soil
Step by step calculation for the depth of
12.75m is given below.
For Zone IV(𝑎𝑚𝑎𝑥 𝑔
⁄ ) = 0.24 and assuming
for the design earthquake𝑀𝑤 = 7.5, and
3
kN/m
5
.
18

sat
 and
3
kN/m
8
.
9

w

Initial stresses:
𝜎𝑣0 = 12.75 × 18.5 = 235.9 kPa
𝑢0 = (12.75 − 6.00) × 9.8 = 66.2 kPa
𝜎𝑣0
′
= (𝜎𝑣0 − 𝑢0) = 169.7 kPa
Assuming the location of water table at the
time of CPT and that under design condition
to be identical, 𝜎𝑣𝑑 = 𝜎𝑣0 and 𝜎𝑣𝑑
′
= 𝜎𝑣0
′
.
Stress reduction factor:
𝑟𝑑 = 1.174 − 0.0267𝑧
= 1.174 − (0.0267 × 12.75)
= 0.83
Cyclic stress ratio due to earthquake:
𝐶𝑆𝑅 = 0.65 (
𝑎𝑚𝑎𝑥
𝑔
) (
𝜎𝑣𝑑
𝜎𝑣𝑑
′ ) 𝑟𝑑
= 0.65(0.24)(235.9 169.7
⁄ )(0.83)
= 0.18
Correction for raw SPT blow count (N) for
overburden pressure:
(𝑁1)60 = 𝐶𝑁𝜂1𝜂2𝜂3𝜂4𝑁
𝐶𝑁 = (𝑃𝑎 𝜎𝑣0
′
⁄ )1/2
= (100 𝜎𝑣0
′
⁄ )1/2
= 0.77
(From Annex F of proposed modifications of
IS 1893 Part 1)
75
.
0
1 
 for donut hammer, rope and pulley
00
.
1
2 
 since depth is more than 10 m
00
.
1
3 
 since the sampler was without a
liner
00
.
1
4 
 since borehole was uncased with
diameter less than 120 mm
(𝑁1)60 = 0.77 × 0.75 × 1 × 1 × 1 × 17
≈ 10
Cyclic Resistance Ratio:
For(𝑁1)60 of 10and fines content of %
8
𝐶𝑅𝑅7.5 = 0.11(Figure 9 of Annex G of
proposed modifications of IS 1893 Part 1)
Corrected Critical Stress Ratio Resisting
Liquefaction:
𝐶𝑅𝑅 = (𝐶𝑅𝑅7.5)(𝑀𝑆𝐹)(𝐾𝜎)(𝐾𝛼)
The magnitude scaling factor, MSF, is 1 since
the design earthquake is of magnitude 7.5.
Correction factor, K, depends on the relative
density. Since the relative density in percent,
Dr, can be estimated from (Skempton 1986):

(𝑁1)60 = 10and𝜎𝑣0
′
= 169.7 kPa
relative density is about 37 %. Accordingly,
𝐾𝜎 = (𝜎𝑣0
′
𝑃𝑎
⁄ )𝑓−1
= (169.7 100
⁄ )0.8−1
≈ 0.90
Since initial static shear is zero (assuming
free field condition and the ground surface is
horizontal) 𝐾𝛼 = 1
𝐶𝑅𝑅 = (𝐶𝑅𝑅7.5)(𝑀𝑆𝐹)(𝐾𝜎)(𝐾𝛼)
= (0.11)(1)(0.90)(1)
≈ 0.10
Factor of safety against liquefaction:
𝐹𝑆 = (𝐶𝑅𝑅 𝐶𝑆𝑅
⁄ ) = (0.10/0.18) ≈ 0.56
Skempton, AW. 1986. Standard penetration
test procedures and the effects in sands of
overburden pressure, relative density, particle
size, ageing and over consolidation.
Géotechnique, 36(3), 425-447

Example 14 –Liquefaction Analysis using CPT data
Problem Statement:
Prepare a plot of factors of safety against liquefaction versus depth for the cone penetration test
(CPT) data provided in Table 14.1. For this site located in Zone V, the ground surface is horizontal
and water table was at 2.35 m below ground surface at the time of the CPT. The design earthquake
was of magnitude 7.5 and there is not structure to be constructed at this site. Assume the unit
weight of the soil to be18 kN/m3
.
Table 14.1 – Result of the Cone penetration Test
Depth (m) qc (bar) fs (bar) Depth (m) qc (bar) fs (bar) Depth (m) qc (bar) fs (bar)
0.50 144.31 0.652 7.50 45.46 0.132 14.50 46.60 0.161
1.00 95.49 0.602 8.00 39.39 0.135 15.00 46.77 0.155
1.50 39.28 0.281 8.50 36.68 0.099 15.50 47.58 0.184
2.00 20.62 0.219 9.00 45.30 0.129 16.00 41.99 0.130
2.50 150.93 1.027 9.50 51.05 0.185 16.50 48.94 0.329
3.00 55.50 0.595 10.00 46.39 0.193 17.00 56.69 0.184
3.50 10.74 0.359 10.50 58.05 0.248 17.50 112.90 0.392
4.00 9.11 0.144 11.00 48.94 0.159 18.00 104.49 0.346
4.50 33.69 0.297 11.50 63.75 0.218 18.50 77.75 0.256
5.00 70.69 0.357 12.00 53.93 0.193 19.00 91.58 0.282
5.50 49.70 0.235 12.50 53.60 0.231 19.50 74.16 0.217
6.00 51.43 0.233 13.00 62.39 0.275 20.00 115.02 0.375
6.50 64.94 0.291 13.50 54.58 0.208
7.00 57.24 0.181 14.00 52.08 0.173
Solution:
Liquefaction Potential of Underlying Soil:
Step by step calculation for the depth of 4.5m
is given below. Detailed calculations are
given in Table 14.2. This table provides the
factor of safety against liquefaction (FS).
The site is located in zone V. The peak
horizontal ground acceleration value for the
site will be taken as 0.36g corresponding to
zone factor Z = 0.36
(𝑎𝑚𝑎𝑥 𝑔
⁄ ) = 0.36,𝑀𝑤 = 7.5
𝛾𝑠𝑎𝑡 = 18 kN/m3
, 𝛾𝑤 = 9.8 kN/m3
Depth of water level below G.L. = 2.35m
Depth at which liquefaction potential is to be
evaluated = 4.5m
Initial stresses:
𝜎𝑣0 = 4.5 × 18 = 81.00 kPa
𝑢0 = (4.5 − 2.35) × 9.8 = 21.07 kPa
𝜎𝑣0
′
= (𝜎𝑣𝑜 − 𝑢0) = 59.93 kPa
Assuming the location of water table at the
time of CPT and that under design condition
to be identical, 𝜎𝑣𝑑 = 𝜎𝑣0 and 𝜎𝑣𝑑
′
= 𝜎𝑣0
′
.
Stress reduction factor:
𝑟𝑑 = 1 − 0.00765 𝑧
= 1 − (0.00765 × 4.5)
= 0.966
Critical stress ratio induced by
earthquake:
𝐶𝑆𝑅 = 0.65 (
𝑎𝑚𝑎𝑥
𝑔
) (
𝜎𝑣𝑑
𝜎𝑣𝑑
′ ) 𝑟𝑑
= 0.65(0.36)(81 59.93
⁄ )(0.966)
= 0.306
Factor of safety against liquefaction (FS):
These sample calculations are for the layer at
4.5 m for the stratigraphy of Table 14.1.
Correction factor for grain characteristics:

𝑘𝑐 = 1.0for𝐼𝑐 ≤ 1.64 and
𝑘𝑐 = {−0.403𝐼𝑐
4
+ 5.581𝐼𝑐
3
− 21.63𝐼𝑐
2
+33.75𝐼𝑐 − 17.88}for𝐼𝑐 > 1.64
The soil behavior type index,𝐼𝑐, is given by
𝐼𝑐 = √(3.47 − log 𝑄)2 + (1.22 + log 𝐹)2
= √(3.47 − log 42.19)2 + (1.22 + log 0.903)2
𝐼𝑐 = 2.19
where,
𝐹 = (
𝑓
(𝑞𝑐 − 𝜎𝑣0)
) 100
= (
29.7
(3369 − 81)
) 100
= 0.903
and
𝑄 = (
𝑞𝑐 − 𝜎𝑣0
𝑃𝑎
) (
𝑃𝑎
𝜎𝑣0
′ )
𝑛
= (
3369 − 81
101.35
) (
101.35
59.93
)
0.5
= 42.19
Since, 𝐼𝑐 = 2.19 > 1.64, thus
𝑘𝑐 = {−0.403(2.19)4
+ 5.581(2.19)3
−21.63(2.19)2
+ 33.75(2.19)
−17.88}
= 1.64
Normalized Cone Tip Resistance:
(𝑞𝑐1𝑁)𝑐𝑠 = 𝑘𝑐(𝑃𝑎 𝜎𝑣0
′
⁄ )𝑛(𝑞𝑐 𝑃𝑎
⁄ )
= 1.64(101.35 59.93
⁄ )0.5(3369 101.35
⁄ )
= 70.77
Factor of safety against liquefaction:
For(𝑞𝑐1𝑁)𝑐𝑠 = 70.77, CRR7.5 =0.11 (Figure
10of Annex G of proposed modifications of IS
1893 Part 1)
Since the design earthquake is of magnitude
7.5, MSF = 1. Since the ground surface is
horizontal, K = 1 and since v 100 kPa for
the soil layer being dealt with, K = 1. It
should be noted that estimation of K for layers
within which v> 100 will require estimates of
relative density, Dr. For cohesionless, normally
consolidated soils Dr can be approximated by
(Jamiolkowski et al. 1986)
𝐷𝑟 = −98 + 66 log (0.316
𝑞𝑐
√𝜎𝑣0
′
)
with𝐷𝑟expressed in% and𝑞𝑐and𝜎𝑣0
′
taken in
kPa units. Thus
𝐶𝑅𝑅 = (𝐶𝑅𝑅7.5)(𝑀𝑆𝐹)(𝐾𝜎)(𝐾𝛼) = 0.11
and
𝐹𝑆 = (𝐶𝑅𝑅 𝐶𝑆𝑅
⁄ ) = (0.11/0.306) = 0.36
Summary:
This assessment shows that except for the
soil layers between2.5 m and 20 m depth at
this site are liquefiable in earthquake with
peak ground acceleration of 0.36g, while the
shallower layers are nonliquefiable. The plot
for depth versus factor of safety is shown in
Figure 14.1.
Jamiolkowski M, Ladd CC, Germaine JT,
and Lancellotta R. 1985. New developments
in field and laboratory testing of soils. 11 Int
Conf on Soil Mech and Found Eng, San
Francisco, 1, 57-153.

Table 14.2– Liquefaction Assessment: Water Level 2.35 m below GL
Depth vd 'vd rd CSR qc fs F Q Ic kc (qc1N)cs Dr K CRR FS
m kPa kPa kPa kPa %
0.50 9 9.00 1.00 0.23 14431 65.20 0.45 241.91 1.40 1.00 242.06 100.00 1.00 100.00 Above
water
table:
non
lique-
fiable
1.00 18 18.00 0.99 0.23 9549 60.20 0.63 159.87 1.63 1.00 160.17 90.23 1.00 100.00
1.50 27 27.00 0.99 0.23 3928 28.10 0.72 65.43 1.97 1.27 83.53 58.96 1.00 0.13
2.00 36 36.00 0.98 0.23 2062 21.90 1.08 33.54 2.31 1.99 68.04 36.36 1.00 0.11
2.50 45 43.53 0.98 0.24 15093 102.70 0.68 226.55 1.53 1.00 227.23 90.70 1.00 100.00 424.79
3.00 54 47.63 0.98 0.25 5550 59.50 1.08 79.10 2.01 1.31 105.02 60.73 1.00 0.19 0.75
3.50 63 51.73 0.97 0.27 1074 35.90 3.55 13.96 2.92 5.92 87.81 12.47 1.00 0.14 0.51
4.00 72 55.83 0.97 0.29 911 14.40 1.72 11.15 2.83 5.01 60.64 6.66 1.00 0.10 0.34
4.50 81 59.93 0.97 0.31 3369 29.70 0.90 42.19 2.19 1.64 70.77 43.13 1.00 0.11 0.36
5.00 90 64.03 0.96 0.32 7069 35.70 0.51 86.63 1.79 1.10 96.60 63.43 1.00 0.16 0.50
5.50 99 68.13 0.96 0.33 4970 23.50 0.48 58.62 1.93 1.22 72.68 52.44 1.00 0.12 0.37
6.00 108 72.23 0.95 0.33 5143 23.30 0.46 58.85 1.92 1.21 72.45 52.58 1.00 0.12 0.36
6.50 117 76.33 0.95 0.34 6494 29.10 0.46 72.50 1.83 1.13 83.61 58.48 1.00 0.13 0.38
7.00 126 80.43 0.95 0.34 5724 18.10 0.32 62.00 1.83 1.13 71.56 54.11 1.00 0.11 0.32
7.50 135 84.53 0.94 0.35 4546 13.20 0.30 47.66 1.92 1.21 59.46 46.79 1.00 0.10 0.28
8.00 144 88.63 0.94 0.36 3939 13.50 0.36 40.04 2.02 1.33 55.18 42.00 1.00 0.10 0.28
8.50 153 92.73 0.93 0.36 3668 9.90 0.28 36.26 2.02 1.33 50.45 39.31 1.00 0.09 0.25
9.00 162 96.83 0.93 0.37 4530 12.90 0.30 44.09 1.95 1.24 56.79 44.74 1.00 0.10 0.27
9.50 171 100.93 0.92 0.37 5105 18.50 0.37 48.78 1.95 1.24 62.62 47.57 1.00 0.10 0.27
10.00 180 105.03 0.91 0.36 4639 19.30 0.43 43.22 2.02 1.33 59.94 44.26 0.99 0.10 0.27
10.50 189 109.13 0.89 0.36 5805 24.80 0.44 53.40 1.95 1.23 68.16 50.14 0.98 0.11 0.30
11.00 198 113.23 0.88 0.36 4894 15.90 0.34 43.84 1.98 1.27 58.01 44.72 0.97 0.10 0.27
11.50 207 117.33 0.87 0.36 6375 21.80 0.35 56.56 1.88 1.17 68.51 51.78 0.96 0.11 0.29
12.00 216 121.43 0.85 0.35 5393 19.30 0.37 46.67 1.97 1.26 61.23 46.50 0.96 0.10 0.27
12.50 225 125.53 0.84 0.35 5360 23.10 0.45 45.53 2.01 1.31 62.48 45.85 0.95 0.09 0.27
13.00 234 129.63 0.83 0.36 6239 27.50 0.46 52.39 1.96 1.25 68.09 49.74 0.94 0.10 0.29
13.50 243 133.73 0.81 0.34 5458 20.80 0.40 44.79 2.00 1.29 60.67 45.46 0.94 0.09 0.27
14.00 252 137.83 0.80 0.34 5208 17.30 0.35 41.93 2.00 1.30 57.21 43.68 0.93 0.09 0.27
14.50 261 141.93 0.79 0.34 4660 16.10 0.37 36.68 2.06 1.39 53.90 40.07 0.93 0.08 0.25
15.00 270 146.03 0.77 0.34 4677 15.50 0.35 36.23 2.06 1.38 53.24 39.77 0.93 0.08 0.25
15.50 279 150.13 0.76 0.32 4758 18.40 0.41 36.31 2.08 1.43 55.02 39.87 0.92 0.09 0.29
16.00 288 154.23 0.75 0.33 4199 13.00 0.33 31.28 2.11 1.47 49.44 35.90 0.93 0.08 0.25
16.50 297 158.33 0.73 0.33 4894 32.90 0.72 36.29 2.19 1.65 63.63 39.91 0.91 0.09 0.28
17.00 306 162.43 0.72 0.31 5669 18.40 0.34 41.80 2.00 1.30 57.28 43.76 0.90 0.09 0.29
17.50 315 166.53 0.71 0.32 11290 39.20 0.36 84.48 1.73 1.06 91.71 63.15 0.85 0.13 0.40
18.00 324 170.63 0.69 0.31 10449 34.60 0.34 76.99 1.75 1.07 85.35 60.58 0.85 0.12 0.39
18.50 333 174.73 0.68 0.31 7775 25.60 0.34 55.92 1.88 1.17 68.46 51.77 0.87 0.10 0.31
19.00 342 178.83 0.67 0.30 9158 28.20 0.32 65.48 1.81 1.11 75.57 56.13 0.85 0.10 0.34
19.50 351 182.93 0.65 0.30 7416 21.70 0.31 51.89 1.89 1.18 64.35 49.76 0.86 0.09 0.29
20.00 360 187.03 0.64 0.29 11502 37.50 0.34 80.93 1.73 1.06 88.47 62.02 0.82 0.12 0.40

Figure 14. 1– Factor of Safety against Liquefaction
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
Depth
(m)
Factor of Safety

Example 15–Liquefaction Analysis with SPT data for a site
with a structure imposing a surcharge at ground surface
Problem Statement:
A circular water tank is to be constructed at a horizontally layered site underlain by layers of
sand and cohesionless silt. The SPT data from the site are presented in Table 15.1. The
borehole was uncased and was of 150 mm diameter. Manually operated pulley and a donut
hammer were used for the SPT. No liner was used. The depth of water table during the
drilling was 8.65 m. The ground surface was horizontal. The design earthquake is of
magnitude 6.5, and the expected peak horizontal ground acceleration at the site is 0.24gat
ground surface. Structural loads imposed by the tank was approximated as 180 kPa uniform
surcharge over a circular are of 37 m diameter and the bulk (total) unit weight of all soil
layers was assumed to be 18 kN/m3
.
Table 15.1– SPT data
Depth
m
FC
%
SPT
N
Depth
m
FC
%
SPT
N
1.5 0 12 9.0 35 29
3.0 0 8 12.0 20 14
4.5 0 13 15.0 20 37
6.0 0 23 18.0 35 33
7.5 35 34 21.0 35 57
Solution:
Calculations and results for this
assessment, summarized in Table 15.2, are
procedurally similar to that illustrated in
Example 13 except that in this case the
structural load needs to be considered.
Recalling CSR is essentially the ratio of
estimated shear stress during the design
earthquake within a particular soil layer to
the effective normal stress within that
layer, the expression for CSR in the
absence of a structure,
𝐶𝑆𝑅𝑒𝑞 = 0.65 (𝑎𝑚𝑎𝑥 𝑔
⁄ )𝑟𝑑(𝜎𝑣𝑑/𝜎𝑣𝑑
′ ),
was modified to
𝐶𝑆𝑅𝑒𝑞 =
{0.65 (𝑎𝑚𝑎𝑥 𝑔
⁄ )𝑟𝑑𝜎𝑣𝑑 + ∆𝜏𝑣ℎ} (𝜎𝑣𝑑
′
+ ∆𝜎𝑣),
⁄
where vh

 is the shear stress increment
within the soil layer of interest due to the
structure and v

 is the corresponding
normal stress increment due to the
structure. The stress increments are
typically estimated from an appropriate
elastic solution. In this example, the
elastic solution of Foster and Ahlvin
(1954). Area was used for a point at the
edge of the circular surface surcharge. As
seen from Table 15.2, soil layers to a
depth of 12 m were found to be
susceptible to liquefaction.
It should be noted that under static
structural load the soil layers are expected
to settle and become stronger. Since
constrained modulus for sands is likely to
be at least a few thousand MPa, settlement
due to structural load and consequent
increase in cyclic shear strength is
expected to be small. The increase was
therefore neglected in this assessment.
Summary:
This assessment shows that except for the
soil layers between1.5 m and 7.5 m depth
and that between 9 m and 15 m depth at
this site are liquefiable in the magnitude
6.5 design earthquake with peak ground
acceleration of 0.24g. Other layers are
nonliquefiable. The plot for depth versus
factor of safety is shown in Figure 15.1.
Foster, CR and Ahlvin,
RG. 1954. Stresses and deflections
induced by a uniform circular
load. Proceedings of Highway Research
Board, 33, 467-470

Table 15.2–SPT based liquefaction assessment for the tank site
Depth
m
(N1)60
vd
kPa
'vd
kPa
v
kPa
vh
kPa
rd CSR Dr K CRR FS
1.5 13.25 27.00 27 179.59 59.34 0.99 0.20 0.62 1 0.14
Above water table:
non liquefiable
3.0 8.02 54.00 44.19 83.40 58.62 0.98 0.36 0.44 1 0.10 0.43
4.5 10.64 81.00 56.475 80.76 57.32 0.97 0.35 0.46 1 0.12 0.54
6.0 18.21 108.00 68.76 78.11 56.01 0.95 0.35 0.56 1 0.19 0.90
7.5 24.08 135.00 81.045 76.49 53.92 0.94 0.34 0.61 1 0.27 1.29
9.0 19.95 162.00 93.33 74.86 51.82 0.93 0.33 0.54 1 0.22 1.10
12.0 8.96 216.00 117.9 68.66 46.93 0.85 0.29 0.39 0.97 0.13 0.64
15.0 22.24 270.00 142.47 65.26 41.57 0.77 0.26 0.53 0.91 0.27 1.39
18.0 18.76 324.00 167.04 60.38 37.38 0.69 0.23 0.46 0.89 0.21 1.15
21.0 30.81 378.00 191.61 55.96 32.76 0.61 0.20 0.54 0.84 0.42 2.42
Figure 15. 1– Factor of Safety against Liquefaction
0
1.5
3
4.5
6
7.5
9
10.5
12
13.5
15
16.5
18
19.5
21
0 0.5 1 1.5 2 2.5
Depth
(m)
Factor of Safety

Explanatory Examples on Indian Seismic Code IS 1893 (Part I)

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