Solution
Content
Solution 1
Date of assignment: 13/14
January
Date of solution: 16 January
E-Course on IS:1893-2002 (Part I)
1
Part II: True / False
1.1) Earthquake Magnitude and Intensity are two different
things.
True.
1.2)
Shaking intensity is expressed in Arabic
numerals (for
example, 6.5)
False.
Shaking intensity is expressed in Roman numerals (I, II, III, …XII).
Earthquake magnitude is expressed in Arabic numerals (e.g.,
6.6)
1.3) Indian seismic zone map gives probabilistic zone map.
False
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
True / False…
1.4) Peak Ground Acceleration (PGA) and Zero Period
Acceleration
(ZPA) are same.
True
1.5) A rigid structure experiences same motion as that of
the
ground.
True
1.6) Base isolation technique is very effective for flexible
structures.
False.
Base isolation is most effective for stiff structures.
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
Part III: Questions
1.1) What is the expected shaking intensity in
seismic zone IV?
Areas in zone IV are liable to shaking intensity of VIII on
the Modified Mercalli or MSK intensity scale.
Sudhir K. Jain,
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Solution 1 /
Questions…
1.2) What is the likely peak ground
acceleration in seismic zone IV?
Zone IV corresponds to intensity VIII. There can be a large
variation in the PGA value recorded in intensity VIII area
itself. Based on the recorded data from intensity VIII, an
average value can be estimated. As seen in the table on
slide 22 of Lecture 1, different empirical relationships
estimate the average value of PGA in the range of 0.15g to
0.30g. Of course, these are average values only and the
real value can be higher than 0.30g or lower than 0.15g.
In this context, notice that the new code provides a Z value
of 0.24g for zone IV which is consistent with the above
values.
Sudhir K. Jain,
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Solution 1 /
Questions…
1.3) How does the peak ground acceleration
increase with increase in shaking intensity?
Notice the values in table on slide 22 of Lecture 1. Every
time intensity increases by one, average value of PGA
almost doubles.
Compare this with the values of Z used in IS:1893-2002. As
we go one zone higher, the Z value increases by a factor of
1.5 or 1.6. Again, this is consistent with the above.
Sudhir K. Jain,
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Solution 1 /
Solution 2
7
Part II: True/ False
2.1) Response Spectrum gives the maximum response of a
single-degree-of-freedom system to the ground motion.
True
2.2) For each accelerogram, there is a unique response
spectrum for a given level of damping.
True
2.3) For 5% damping acceleration spectrum, the ordinates
in the period range 0.1-0.3sec are about 5 times the
peak ground acceleration
False.
The ordinates in the period range 0.1-0.3 sec are about 2.5
times the PGA
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Solution 1 /
True/False...
2.4) Modal Mass depends on how the mode shapes have been
scaled
False
Modal mass does not depend on how the mode shapes are
scaled. Modal participation factor depends on the modal
shape scaling.
2.5) Indian seismic zone map in IS:1893 is based on 2%
probability of its being exceeded in 50 years.
False
2.6) Soft storey and weak storey mean the same things.
False.
Soft storey relates to stiffness, weak storey relates to
strength.
Sudhir K. Jain,
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Solution 1 /
True/False...
2.7) The code restricts the inter-storey displacement to be
within 0.4% of the storey height
True.
Sudhir K. Jain,
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Solution 1 /
Part III: Solutions
Maximum
Acceleration, g
2.1) Consider the acceleration response spectrum in Fig. A1. What is
the peak ground acceleration (PGA) of this ground motion.
PGA =
0.33g
Undamped Natural Period T
(sec)
The peak ground acceleration (PGA) of this ground motion is 0.33g
Sudhir K. Jain,
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Solution 1 /
Solutions...
2.2) Consider the acceleration spectrum of Fig. A1. Obtain
maximum acceleration experienced by a single degree of freedom
system of natural period 0.5 sec and damping 5%.
Maximum
Acceleration, g
Ordinate for T = 0.5 sec is
1.03g
Undamped Natural Period T
(sec)
Maximum acceleration experienced by the system
of period 0.5 sec is 1.03g
Sudhir K. Jain,
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Solution 1 /
Solutions...
2.3) A ground motion is such that its PGA is 0.6g but it has the same
shape of response spectrum as of Fig. A1. Calculate maximum
acceleration experienced by a SDOF system of natural period 0.5
sec and damping 5%.
The spectrum of Fig. A1 pertains to PGA value of 0.33g (see
solution to problem 2.1). For PGA of 0.6g, we need to scale
up the ordinates in the same ratio.
Ordinate for T=0.5sec is 1.03g when PGA is 0.33g.
Hence, corresponding to PGA of 0.60g, the ordinate at
T=0.5sec is
= 1.03g X (0.60/0.33) = 1.87g
Sudhir K. Jain,
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Solutions...
2.4) Read the ordinates of the acceleration spectra of Fig. A2 for natural
period 0.4 sec, and for
a) Damping is 5% of critical
b) Damping is 2% of critical
What is the ratio of the maximum response for the two values of damping.
Maximum response for
5% damping, Sa = 1.5
Maximum response for
2% damping, Sa = 2.2
Hence, the ratio of the
maximum response for
the two values of
damping is
1.5
1
0.68
2.2 1.47
Sudhir K. Jain,
Period
(sec)
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Solutions...
2.5) A cycle stand can be modeled as a SDOF system (see figure). It is
to be analyzed for the ground motion of Fig. A1. Obtain the
maximum seismic base shear and base moment.
Weight = 60 kN
Natural period = 0.4sec
Mass = 60,000/9.81 kg = 6116 kg
2.5m
From Response Spectrum:
Spectral Acceleration (for T=0.4sec) =
0.9 g
Natural period = 0.4sec
Max. Base Shear = Mass x Spectral
Accln. =
(6116 kg) x (0.9x9.81m/sec2) = 54,000 N
= 54 kN
Max. Base Moment
Sudhir K. Jain,
kN) x (2.5m)
=Solution
135 kN- 1 /
E-Course=(54
on IS:1893
/
m
Solution
3
Date of assignment: 16 January
Date of solution: 18 January
E-Course on IS:1893-2002 (Part I)
16
Part II: True/ False
3.1) In the 2002 version of the code, vertical seismic
coefficient is taken as one half of the horizontal seismic
coefficient.
False
3.2) Horizontal acceleration is of more serious concern
than the vertical acceleration for seismic safety.
True
3.3) Consideration of vertical acceleration is important for
cantilevers.
True
Sudhir K. Jain,
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Solution 1 /
True / False (contd…)
3.4) Small earthquakes occur far more frequently than
the large earthquakes.
True
3.5) The concept of seismic design aims to ensure that
the structure will not be damaged in case of strong
earthquake shaking.
False
3.6) Design seismic force is much less than the
maximum expected seismic force.
True
Sudhir K. Jain,
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Solution 1 /
True / False (contd…)
3.7) From seismic safety view point, it is better to have
more redundancy in the structure.
True
3.8) A structure with higher ductility can be designed for
lower seismic load.
True
3.9) Precast structures are best for high seismic regions.
False
Sudhir K. Jain,
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Solution 1 /
True / False (contd…)
3.10) Soil structure interaction is more significant in case
of soft soils.
True
Sudhir K. Jain,
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Solution 1 /
Part III: Questions
3.1) The floor of a building does not have uniform distribution of mass (see
the floor plan below). Locate its centre of mass.
10
Let us divide the roof slab into
m
three rectangular parts as shown
I
in fig.
1200
4
kg/sq.m CM
Mass of Part I is 1200kg/sq.m, whilem
that of the other two parts is
1000
y
1000kg/sq.m
kg/sq.
Let origin be at point A, and the
A
m
coordinates of the centre of
x
20 m
mass be at (xcm,ycm)
(10 4 1200) 5 (10 4 1000) 15 (20 4 1000) 10
xcm
9.76m
(10 4 1200) (10 4 1000) ( 20 4 1000)
Ycm
II
III
8
m
(10 4 1200) 6 (10 4 1000) 6 (20 4 1000) 2
4.1m
(10 4 1200) (10 4 1000) (20 4 1000)
Hence, coordinates of centre of mass are (9.76,
4.1)
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Solution 1 /
Sudhir K. Jain,
Questions (contd…)
3.2) The plan of a simple one storey building is as shown below. All
columns and beams are same. Obtain its centre of stiffness.
In the X-direction, there are three identical
frames located at uniform spacing. Hence, the
y-coordinate of centre of stiffness is located
symetrically, 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.
5
m
5m
CS
y
x
5m
xcs
5m
10
m
k 0 k 5 k 10 k 20
8.75m
(k k k k)
Hence, coordinates of centre of stiffness are
(8.75, 5.0)
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Sudhir K. Jain,
Solution 1 /
Question 3.2 (contd…)
k
k
k
k
Unit
displaceme
nt
Another way to look at this problem is to deform the building such that
there is a unit translation in y-direction and no rotation. Each transverse
frame resists this movement by a force equal to its lateral stiffness. The
force required to cause this displacement must pass through the centre of
stiffness and should balance these forces of resistance from the frames.
Sudhir K. Jain,
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Solution 1 /
Solution
4
Date of assignment: 17 January
Date of solution: 20 January
E-Course on IS:1893-2002 (Part I)
24
Part II: True/ False
4.1) Soft soils may locally amplify earthquake ground motion.
True
4.2) One should consider the possibility of strong earthquake
shaking occurring simultaneously with strong wind.
False
4.3) Modulus of elasticity of concrete does not depend on the
stress level in the member.
False
Sudhir K. Jain,
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Solution 1 /
Part II: True/ False (contd…)
4.4) One should assume that design seismic loads in Xand Y-directions will occur simultaneously.
False
4.5) Load combination 0.9DL 1.5EL is to be used only
when stability of the structure is an issue.
False
4.6) Bottom-face reinforcement in beams near joints is
often governed by the load combination 0.9DL 1.5EL.
True
Sudhir K. Jain,
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Solution 1 /
Part II: True/ False (contd…)
4.7) Design of a RC frame building, circular in
plan, requires 100% + 30% rule for load
combinations.
True
Sudhir K. Jain,
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Solution 1 /
Part III: Problem 4.1
B
a) Earthquake load in X-direction
only.
50kN
A
MAB = 13.67 kN-m
C
D
MAB=13.67kNm
B
b) Earthquake load in Y-direction
only.
MAB = 13.67 kN-m
A
C
50k
N
D
Sudhir K. Jain,
MAB=13.67kN
m
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Solution 1 /
Problem 4.1 (contd…)
c).
c) Earthquake load acting parallel
to beam AB
B
A
C
MAB = 19.37 kN-m
50k
N
D
MAB=19.36kN
m
Notice that if we consider only loads a) and b),
the beam AB will be under-designed (for a
moment of 13.76 kN-m, against 19.36 kN-m it
experiences when shaking is in the direction
shown.
Sudhir K. Jain,
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Solution 1 /
Problem 4.1 (contd…)
B
d)
d) Earthquake load (full design
load) acting in X- and Ydirection simultaneously.
MAB = 27.34 kN-m
.
A
50k
N
C
50k
N
D
MAB=27.34kN
m
Notice that if we consider full EQ load in both
directions simultaneously, beam AB will be overdesigned (for a moment of 27.34 kN-m, against a
maximum of 19.36 kN-m it experiences for any
possible direction of shaking.)
Sudhir K. Jain,
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Solution 1 /
Problem 4.1 (contd…)
B
e).
e) 100% Design Load in X-direction,
and 30% Design Load in Y-direction
acting simultaneously.
A
50k
N
C
15k
N
MAB = 17.77 kN-m
D
MAB=17.77kN
m
e) 100% Design Load in Y-direction,
and 30% Design Load in X-direction
acting simultaneously.
B
A
MAB = 17.77 kN-m
15k
N
C
50k
N
D
MAB=17.77kN
m
Sudhir K. Jain,
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Solution 1 /
Problem 4.1 (contd…)
Note that 100% + 30% rule gives design
moment 17.77 kN-m against the
maximum of 19.37 kN-m.
This is close enough (within 10%).
If we use 100% + 40% rule (used in some
codes), we will get a design value of 19.14
kN-m, which is even better.
Sudhir K. Jain,
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Solution 1 /
Problem 4.1 (contd…)
As an alternate approach of Cl. 6.3.4.2,
the design moment =
(13.67) (13.67) 2 19.33kN m
This is also fairly close to the maximum
value of 19.37 kN-m
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
Solution 5
Date of assignment: 18
January
Date of solution: 21 January
E-Course on IS:1893-2002
34
Part II: True/ False
5.1) Design Spectrum is developed considering expected
ground motion, expected behaviour of structure, and
the assumptions in analysis and design.
True
5.2) In 2002 edition of the code, a flexible structure
located on soft soil will be designed for higher design
force than that on rock sites.
True
5.3) Response reduction factor should be higher for
structures more vulnerable to earthquake damage.
False
Sudhir K. Jain,
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Solution 1 /
Part II: True/ False (contd…)
5.4) Underground structures should be designed for higher
level of
design force than the over-ground structures.
False
5.5) 2002 edition of the code specifies vertical coefficient
to be two-thirds of the horizontal coefficient.
True
5.6) Spectrum ordinates for lower damping will be lower
than those for higher value of damping.
False
Sudhir K. Jain,
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Solution 1 /
Part II: True/ False (contd…)
5.7) Major projects, such as dams, major bridges, etc
require development of a site-specific design criteria.
True
Sudhir K. Jain,
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Solution 1 /
Question 5.1: RC Ductile Buildings
(T=0.3sec)
Old Code:
New Code:
ZI S a
2R g
Design seismic force Vb= oIKCW
Vb
Design seismic force
Value of o =0.02, 0.04, 0.05, and
0.08, for zones II, III, IV and V,
respectively.
Importance Factor I = 1.0
Soil-Foundation Factor = 1.0
Performance Factor for Ductile
System K = 1.0
Value of C (for T=0.3sec) = 1.0
Value of Z = 0.10, 0.16, 0.24,
0.36 for zones II, III, IV and
V, respectively.
Importance Factor I = 1.0
Response Reduction Factor for
Ductile Frame (SMRF) = 5
Value of S/g (For rock sites, and
T = 0.3 sec) = 2.5
See comparison of design force on next slide.
Sudhir K. Jain,
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Solution 1 /
Question 5.1: RC Ductile Buildings (T=0.3sec) (contd…)
Comparison of design seismic force:
Old Code
New Code
Zone II
0.02W
0.025W
Zone III
0.04W
0.04W
Zone IV
0.05W
0.06W
Zone V
0.08W
0.09W
Sudhir K. Jain,
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Solution 1 /
Question 5.2: RC Ductile Buildings
(T=1.0sec)
Old Code:
New Code:
ZI S a
2R g
Design seismic force Vb= oIKCW
Vb
Design seismic force
Value of o =0.02, 0.04, 0.05, and
0.08, for zones II, III, IV and V,
respectively.
Importance Factor I = 1.0
Soil-Foundation Factor = 1.0
Performance Factor for Ductile
System K = 1.0
Value of C (for T=1.0sec) = 0.53
Value of Z = 0.10, 0.16, 0.24,
0.36 for zones II, III, IV and
V, respectively.
Importance Factor I = 1.0
Response Reduction Factor for
Ductile Frame (SMRF) = 5
Value of S/g (For rock sites, and
T = 1.0 sec) = 1.0
See comparison of design force on next slide.
Sudhir K. Jain,
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Solution 1 /
Question 5.2: RC Ductile Buildings (T=1.0 sec) (contd…)
Comparison of design seismic force:
Old Code
New Code
Zone II
0.0106W
0.010W
Zone III
0.0212W
0.016W
Zone IV
0.0265W
0.024W
Zone V
0.0424W
0.036W
Sudhir K. Jain,
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Solution 1 /
Question 5.3: Ordinary RC Frame Building
(T=0.3sec)
Old Code:
New Code:
ZI S a
2R g
Design seismic force Vb= oIKCW
Vb
Design seismic force
Value of o =0.02, 0.04, 0.05, and
0.08, for zones II, III, IV and V,
respectively.
Importance Factor I = 1.0
Soil-Foundation Factor = 1.0
Performance Factor for Ordinary
Frame System K = 1.6
Value of C (for T=0.3sec) = 1.0
Value of Z = 0.10, 0.16, 0.24,
0.36 for zones II, III, IV and
V, respectively.
Importance Factor I = 1.0
Response Reduction Factor for
Ordinary Frame (OMRF) = 3
Value of S/g (For rock sites, and
T = 0.3 sec) = 2.5
See comparison of design force on next slide.
Sudhir K. Jain,
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Solution 1 /
Question 5.3: Ordinary RC Frame Building (T=0.3sec)
(contd…)
Comparison of design seismic force:
Old Code
New Code
Zone II
0.032W
0.042W
Zone III
0.064W
0.067W
Zone IV
0.080W
0.100W
Zone V
0.128W
0.150W
Sudhir K. Jain,
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Solution 1 /
Q.5.4: Unreinf. Masonry Building
(T=0.3sec)
Old Code:
Design seismic force Vb= oIKCW
Value of o =0.02, 0.04, 0.05, and
0.08, for zones II, III, IV and V,
respectively.
Importance Factor I = 1.0
Soil-Foundation Factor = 1.0
Performance Factor for K = 1.6
(separate value for masonry not
given; taken highest value)
Value of C (for T=0.3sec) = 1.0
New Code:
Vb
Design seismic force
ZI S a
2R g
Value of Z = 0.10, 0.16, 0.24,
0.36 for zones II, III, IV and
V, respectively.
Importance Factor I = 1.0
Response Reduction Factor for
Ordinary Masonry Bldg =
1.5
Value of S/g (For rock sites, and
T = 0.3 sec) = 2.5
See comparison of design force on next slide.
Sudhir K. Jain,
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Solution 1 /
Q.5.4: Unreinf. Masonry Building (T=0.3sec) (contd…)
Comparison of design seismic force:
Old Code
New Code
Zone II
0.02W
0.083W
Zone III
0.04W
0.133W
Zone IV
0.05W
0.20W
Zone V
0.08W
0.30W
Sudhir K. Jain,
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Solution 1 /
Q. 5: Observations
Note that the design seismic force for the
old and the new code are about the same for
Ductile Frame (T = 0.3 sec)
Ductile Frame (T = 1.0 sec)
Ordinary Frame (T = 0.3 sec)
However, there is significant increase in
design force for ordinary masonry buildings.
This is because the earlier edition of code did not
pay specific attention to design seismic force.
For example, as per NEHRP code, an ordinary
masonry building is to be designed for 5.33 times
that for a similar RC SMRF.
Sudhir K. Jain,
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Solution 1 /
Solution
6
Date of assignment: 20 January
Date of solution: 23 January
E-Course on IS:1893-2002 (Part I)
47
Part II: True/ False
6.1) A structural engineer can provide excellent earthquake
resistance in a building regardless of its configuration.
False
6.2) For buildings with better performance, the code
specifies lower value of R.
False
6.3) One is allowed to ordinary RC frame (OMRF) in zone III
as long as one is providing higher design force than that
for a special moment resisting frame (SMRF).
False
Sudhir K. Jain,
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Solution 1 /
Part II: True/ False (contd…)
6.4) If the share of gravity load on a 2-D frame is small, it
will also carry only a small share of design seismic load.
False
6.5) In general, one can accurately predict the natural
period of a building by carrying out dynamic analysis
using a sophisticated computer programme.
False
6.6) Rigid diaphragm implies that the beams are rigid and
hence the column stiffness can be taken as (12EI/L 3).
False
Sudhir K. Jain,
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Solution 1 /
Part III: Problem 6.1
(a) Floor diaphragm rigid in its own plane:
The seismic forces are distributed among the frames in
proportion to their lateral stiffness.
0.042
m
150
kN
100
kN
100
kN
50
kN
50
kN
Frames A and
C
Lateral stiffness of the frames:
kA, kC = [300/0.042] kN/m = 7500kN/m
kB
0.096m
150
kN
Frame
B
= [300/0.096] kN/m = 3125kN/m
kA/ kB = 2.4
Sudhir K. Jain,
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Solution 1 /
Problem 6.1 (contd…)
Table 1: Lateral forces for rigid floor diaphragm
(distributed in proportion to the stiffness)
Floor
Lateral loads (kN)
Frame A
Frame B
Frame C
Top
62.0
26.0
62.0
Middle
41.3
17.4
41.3
Ground
20.7
8.6
20.7
(b) Floor diaphragm extremely flexible in its own plane:
The seismic forces are distributed among the frames in
proportion to their tributary areas.
Tributary areas for frames A
and
C
Tributary
areas for frame B
Sudhir K. Jain,
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Solution 1 /
Problem 6.1 (contd…)
Tributary area ratio for frame A =
0.25
area ratio for frame B =
0.50
Table 2: Lateral forces for extremely flexible floor
area ratio for frame C =
diaphragm
Floor
Lateral loads (kN)
0.25
Frame A
Frame B
Frame C
Top
37.5
75.0
37.5
Middle
25.0
50.0
25.0
Ground
12.5
25.0
12.5
Notice the large difference in design force by the
two methods. By distributing loads in tributary area
ratio, the end frames are being underdesigned while
the middle frame is being over designed.
Sudhir K. Jain,
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Solution 1 /
Alternate method to compute lateral stiffness of 2-D
frames:
The lateral stiffness of the frames A, B and C are calculated by
applying point load at the roof:
100
kN
0.018
m
Frames A and
C
Lateral stiffness of the frames:
kA, kC = [100/0.018] kN/m = 5556kN/m
kB
0.042m
100
kN
Frame
B
= [100/0.042] kN/m = 2381kN/m
kA/ kB = 2.3
Thus it is observed that the ratio of the stiffness of the frames
A and B
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/
Solution
are almost
the same as obtained
in case
(a).
Sudhir
K. Jain,
1/
Problem 6.1: Some comments
In real buildings, most of the time we will be using 3-D
computer analysis. There, one can simply model the rigid
floor diaphragm action in the analysis. However, two
dimensional methods such as shown here are important for
preliminary analysis and for checking the computer
analysis results.
Sudhir K. Jain,
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Solution 1 /
Solution
7
Date of assignment: 21 January
Date of solution: 27 January
E-Course on IS:1893-2002 (Part I)
55
Part II: True/ False
7.1) It is quite difficult to accurately assess the natural
period of buildings.
True
7.2) Brick masonry infills do not contribute much lateral
stiffness to the building.
False
7.3) Empirical expressions for natural period are based on
data from real as-built buildings.
True
Sudhir K. Jain,
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Solution 1 /
Part II: True/ False (contd…)
7.4) Choosing a lower value of damping will give a lower
design force on the building.
False
7.5) As per 2002 edition of the code, it is sufficient to
consider first three modes of vibration of the building.
False
7.6) SRSS is a good method for modal combination if the
natural modes are well separated.
True
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
Part III: Problem 7.1
The seismic weight given in the problem includes the weight of
the infills.
(a) Ignoring stiffness contribution of infills, the time periods are
Txx= 0.826 secs.
(b) The stiffness contribution of infills is modelled by diagonal struts
(Holmes, 1961).
Strut width = one-third of diagonal length, strut thickness= wall
thickness
Assuming compressive strength of masonry as 5 MPa,
Modulus of elasticity (Drysdale et. al., 1993) = 550x5 = 2750
MPa
Cross sectional area along XX dirn.
= 0.54 sq.m.
Fundamental
periods,
Txx = 0.462 secs.
Strut
Model
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
Problem 7.1 (contd…)
(c) The stiffness contribution of infills is modelled by shell
elements.
Properties of elements: Modulus of elasticity = 2750 MPa
Thickness
= 230 mm
Fundamental periods,
Txx = 0.473 secs.
Shell
Model
Using the expression for buildings with infill panels, as
given in Cl.7.6.2 of the new code, Txx= 0.281 secs.
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
Problem 7.1 (contd…)
Observations:
Note that the natural period is much higher if
the stiffness contribution of infill panels is
ignored.
Considerable assumptions are needed for
masonry properties, for instance,
Modulus of elasticity of masonry
Width of strut: different researchers have given
different values.
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
Solution 8
Date of assignment: 22
January
Date of solution: 28 January
E-Course on IS:1893-2002
61
Part II: True/ False
8.1) Torsion provisions in the new code require
consideration of accidental torsion.
True
8.2) Centre of rigidity for a multistorey building can be
defined in more than one ways.
True
8.3) As per all floor definition, location of centre of rigidity
depends not only on the building properties, but also on
the lateral load profile.
True
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
Part II: True/ False
8.4) Soft storey and weak storey are one and the same
things.
False
8.5) Soft storey buildings behave well during the
earthquake because of the base isolation effect the soft
storey produces.
False
8.6) Since response spectrum shows higher seismic force
for stiffer buildings, it is best to have a very flexible
building from earthquake safety view point.
False
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
Part II: True/ False (contd…)
8.7) Gravity columns are safest during earthquake shaking
since they are not responsible for seismic loads.
False
8.8) Pounding effects are most detrimental when roof of a
building is liable to hit the adjacent building at the
middle of its column heights.
True
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
Part III: Problem 8.1
y
Grade of conc. M 25
2m
E = 5000√(25)=25,000 N/mm2
CR CM
A
Storey ht h=4500mm
Thickness of the walls =200mm
Length of the walls = 4000mm
4m
C
B
8m
x
D
16m
All walls are same and hence, 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 =
0.0m
Design eccentricity, ed = 0.0 ±0.05 x 8 = ± 0.4m
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
Part III: Problem 8.1 (contd…)
Lateral forces in the walls due to translation:
FCT
FDT
KC
F 50.0kN
KC KD
100
kN
KD
F 50.0kN
KC KD
C
50.0 kN
CR
A
D
CM
B
50.0 kN
Lateral forces in the walls due to torsional moment:
FiR
K i ri
K
i A , B ,C , D
i
ri
2
(F e d )
where ri is the distance of the shear wall from CR
As all walls have same stiffness, KA = KB = KC = KD = k
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
Part III: Problem 8.1 (contd…)
rA= -6.0m
rB = 6.0m
rC =4.0m
rD =-4.0m
and ed = 0.4m
FAR
rA k
2
2
2
2
(rA rB rC rD )k
(F e d ) -2.31kN
Similarly,
FBR = 2.31 kN
FCR = 1.54 kN
FDR = -1.54kN
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
Part III: Problem 8.1 (contd…)
51.54kN
Total lateral forces in the walls:
FA = -2.31 kN
FB = 2.31 kN
FC = 50 + 1.54 =51.54 kN
2.31kN
C
100
kN
CR
A
2.31kN
CM* B
48.46kN
D
FD = 50 - 1.54 = 48.46 kN
Similarly, when ed = -0.4m, then the total lateral forces in the walls will be,
FA = 2.31 kN
FB = -2.31 kN
FC = 50 - 1.54 = 48.46 kN
FD = 50 + 1.54 = 51.54 kN
Design lateral forces in the walls C and D
are
FC = FD = 51.54 kN
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
Part III: Problem 8.1 (contd…)
50kN
50kN
EQ Force in Y-direction
C
CR
Calculated eccentricity = 2.0m
A
Design eccentricity, ed = 2.0 x 1.5 +0.05 x 16
=3.8m
or
2.0due
- 0.05
x 16 = 1.2m
Lateral forces in the walls
to translation
:
FAT
KA
F 50.0kN
KA KB
FBT
KB
F 50.0kN
KA KB
Sudhir K. Jain,
CM
B
D
100
kN
E-Course on IS:1893 /
Solution 1 /
Part III: Problem 8.1 (contd…)
Lateral forces in the walls due to torsional moment:
When ed = 3.8m
FAR
rA k
2
2
2
2
(rA rB rC rD )k
(F e d ) -21.92kN
Similarly,
FBR = 21.92 kN
FCR = 14.62 kN
28.08kN
FDR = -14.62kN
C
Total lateral forces in the walls:
FA = (50.0 – 21.92)= 28.08 kN
14.62k 71.92kN
N
CR CM CM*
A
B
FB = (50.0 + 21.92) = 71.92 kN
FC = 14.62 kN
14.62kND
FD = -14.62 kN
Sudhir K. Jain,
E-Course on IS:1893 /
100 kN
Solution 1 /
Part III: Problem 8.1 (contd…)
Similarly, when ed = 1.2m, then the total lateral forces in the walls will be,
FA = (50.0 – 6.92)= 43.08 kN
FB = (50.0 + 6.92) = 56.92 kN
FC = 4.62 kN
FD = -4.62 kN
Hence, maximum lateral forces in walls A and B are
FA = 43.08 kN
FB = 71.92 kN
However, note the statement in Cl. 7.9.1 “negative torsional
shear shall be neglected. This implies that wall A should
still be designed for lateral force of 50kN. Thus, the design
forces are:
FA = 50.00 kN, FB = 71.92 kN, FC = FD = 51.54 kN
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
Solution 9
Date of assignment: 24 January
Date of solution : 28 January
E-Course on IS:1893-2002 (Part I)
72
Part II: True/ False
9.1) Code now requires ties to be provided for isolated
footings and pile caps (unless footing is supported on
rock directly).
True
9.2) Cantilever projections are particularly vulnerable to
damage during earthquakes.
True
9.3) It is better to concentrate shear walls near the centre
of the building rather than place them near the
perimeter.
False
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
Part II: True/ False (contd…)
9.4) The best treatment for non-parallel systems is to
perform dynamic analysis.
False
9.5) Irregular buildings should be designed for a higher
value of importance factor.
False
9.6) As a general rule, buildings with irregular
configuration perform poorly in earthquakes even when
good engineering has been carried out.
True
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
Date of assignment: 13/14
January
Date of solution: 16 January
E-Course on IS:1893-2002 (Part I)
1
Part II: True / False
1.1) Earthquake Magnitude and Intensity are two different
things.
True.
1.2)
Shaking intensity is expressed in Arabic
numerals (for
example, 6.5)
False.
Shaking intensity is expressed in Roman numerals (I, II, III, …XII).
Earthquake magnitude is expressed in Arabic numerals (e.g.,
6.6)
1.3) Indian seismic zone map gives probabilistic zone map.
False
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
True / False…
1.4) Peak Ground Acceleration (PGA) and Zero Period
Acceleration
(ZPA) are same.
True
1.5) A rigid structure experiences same motion as that of
the
ground.
True
1.6) Base isolation technique is very effective for flexible
structures.
False.
Base isolation is most effective for stiff structures.
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
Part III: Questions
1.1) What is the expected shaking intensity in
seismic zone IV?
Areas in zone IV are liable to shaking intensity of VIII on
the Modified Mercalli or MSK intensity scale.
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
Questions…
1.2) What is the likely peak ground
acceleration in seismic zone IV?
Zone IV corresponds to intensity VIII. There can be a large
variation in the PGA value recorded in intensity VIII area
itself. Based on the recorded data from intensity VIII, an
average value can be estimated. As seen in the table on
slide 22 of Lecture 1, different empirical relationships
estimate the average value of PGA in the range of 0.15g to
0.30g. Of course, these are average values only and the
real value can be higher than 0.30g or lower than 0.15g.
In this context, notice that the new code provides a Z value
of 0.24g for zone IV which is consistent with the above
values.
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
Questions…
1.3) How does the peak ground acceleration
increase with increase in shaking intensity?
Notice the values in table on slide 22 of Lecture 1. Every
time intensity increases by one, average value of PGA
almost doubles.
Compare this with the values of Z used in IS:1893-2002. As
we go one zone higher, the Z value increases by a factor of
1.5 or 1.6. Again, this is consistent with the above.
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
Solution 2
7
Part II: True/ False
2.1) Response Spectrum gives the maximum response of a
single-degree-of-freedom system to the ground motion.
True
2.2) For each accelerogram, there is a unique response
spectrum for a given level of damping.
True
2.3) For 5% damping acceleration spectrum, the ordinates
in the period range 0.1-0.3sec are about 5 times the
peak ground acceleration
False.
The ordinates in the period range 0.1-0.3 sec are about 2.5
times the PGA
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
True/False...
2.4) Modal Mass depends on how the mode shapes have been
scaled
False
Modal mass does not depend on how the mode shapes are
scaled. Modal participation factor depends on the modal
shape scaling.
2.5) Indian seismic zone map in IS:1893 is based on 2%
probability of its being exceeded in 50 years.
False
2.6) Soft storey and weak storey mean the same things.
False.
Soft storey relates to stiffness, weak storey relates to
strength.
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
True/False...
2.7) The code restricts the inter-storey displacement to be
within 0.4% of the storey height
True.
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
Part III: Solutions
Maximum
Acceleration, g
2.1) Consider the acceleration response spectrum in Fig. A1. What is
the peak ground acceleration (PGA) of this ground motion.
PGA =
0.33g
Undamped Natural Period T
(sec)
The peak ground acceleration (PGA) of this ground motion is 0.33g
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
Solutions...
2.2) Consider the acceleration spectrum of Fig. A1. Obtain
maximum acceleration experienced by a single degree of freedom
system of natural period 0.5 sec and damping 5%.
Maximum
Acceleration, g
Ordinate for T = 0.5 sec is
1.03g
Undamped Natural Period T
(sec)
Maximum acceleration experienced by the system
of period 0.5 sec is 1.03g
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
Solutions...
2.3) A ground motion is such that its PGA is 0.6g but it has the same
shape of response spectrum as of Fig. A1. Calculate maximum
acceleration experienced by a SDOF system of natural period 0.5
sec and damping 5%.
The spectrum of Fig. A1 pertains to PGA value of 0.33g (see
solution to problem 2.1). For PGA of 0.6g, we need to scale
up the ordinates in the same ratio.
Ordinate for T=0.5sec is 1.03g when PGA is 0.33g.
Hence, corresponding to PGA of 0.60g, the ordinate at
T=0.5sec is
= 1.03g X (0.60/0.33) = 1.87g
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
Solutions...
2.4) Read the ordinates of the acceleration spectra of Fig. A2 for natural
period 0.4 sec, and for
a) Damping is 5% of critical
b) Damping is 2% of critical
What is the ratio of the maximum response for the two values of damping.
Maximum response for
5% damping, Sa = 1.5
Maximum response for
2% damping, Sa = 2.2
Hence, the ratio of the
maximum response for
the two values of
damping is
1.5
1
0.68
2.2 1.47
Sudhir K. Jain,
Period
(sec)
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Solution 1 /
Solutions...
2.5) A cycle stand can be modeled as a SDOF system (see figure). It is
to be analyzed for the ground motion of Fig. A1. Obtain the
maximum seismic base shear and base moment.
Weight = 60 kN
Natural period = 0.4sec
Mass = 60,000/9.81 kg = 6116 kg
2.5m
From Response Spectrum:
Spectral Acceleration (for T=0.4sec) =
0.9 g
Natural period = 0.4sec
Max. Base Shear = Mass x Spectral
Accln. =
(6116 kg) x (0.9x9.81m/sec2) = 54,000 N
= 54 kN
Max. Base Moment
Sudhir K. Jain,
kN) x (2.5m)
=Solution
135 kN- 1 /
E-Course=(54
on IS:1893
/
m
Solution
3
Date of assignment: 16 January
Date of solution: 18 January
E-Course on IS:1893-2002 (Part I)
16
Part II: True/ False
3.1) In the 2002 version of the code, vertical seismic
coefficient is taken as one half of the horizontal seismic
coefficient.
False
3.2) Horizontal acceleration is of more serious concern
than the vertical acceleration for seismic safety.
True
3.3) Consideration of vertical acceleration is important for
cantilevers.
True
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
True / False (contd…)
3.4) Small earthquakes occur far more frequently than
the large earthquakes.
True
3.5) The concept of seismic design aims to ensure that
the structure will not be damaged in case of strong
earthquake shaking.
False
3.6) Design seismic force is much less than the
maximum expected seismic force.
True
Sudhir K. Jain,
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Solution 1 /
True / False (contd…)
3.7) From seismic safety view point, it is better to have
more redundancy in the structure.
True
3.8) A structure with higher ductility can be designed for
lower seismic load.
True
3.9) Precast structures are best for high seismic regions.
False
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
True / False (contd…)
3.10) Soil structure interaction is more significant in case
of soft soils.
True
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
Part III: Questions
3.1) The floor of a building does not have uniform distribution of mass (see
the floor plan below). Locate its centre of mass.
10
Let us divide the roof slab into
m
three rectangular parts as shown
I
in fig.
1200
4
kg/sq.m CM
Mass of Part I is 1200kg/sq.m, whilem
that of the other two parts is
1000
y
1000kg/sq.m
kg/sq.
Let origin be at point A, and the
A
m
coordinates of the centre of
x
20 m
mass be at (xcm,ycm)
(10 4 1200) 5 (10 4 1000) 15 (20 4 1000) 10
xcm
9.76m
(10 4 1200) (10 4 1000) ( 20 4 1000)
Ycm
II
III
8
m
(10 4 1200) 6 (10 4 1000) 6 (20 4 1000) 2
4.1m
(10 4 1200) (10 4 1000) (20 4 1000)
Hence, coordinates of centre of mass are (9.76,
4.1)
E-Course on IS:1893 /
Solution 1 /
Sudhir K. Jain,
Questions (contd…)
3.2) The plan of a simple one storey building is as shown below. All
columns and beams are same. Obtain its centre of stiffness.
In the X-direction, there are three identical
frames located at uniform spacing. Hence, the
y-coordinate of centre of stiffness is located
symetrically, 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.
5
m
5m
CS
y
x
5m
xcs
5m
10
m
k 0 k 5 k 10 k 20
8.75m
(k k k k)
Hence, coordinates of centre of stiffness are
(8.75, 5.0)
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Sudhir K. Jain,
Solution 1 /
Question 3.2 (contd…)
k
k
k
k
Unit
displaceme
nt
Another way to look at this problem is to deform the building such that
there is a unit translation in y-direction and no rotation. Each transverse
frame resists this movement by a force equal to its lateral stiffness. The
force required to cause this displacement must pass through the centre of
stiffness and should balance these forces of resistance from the frames.
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
Solution
4
Date of assignment: 17 January
Date of solution: 20 January
E-Course on IS:1893-2002 (Part I)
24
Part II: True/ False
4.1) Soft soils may locally amplify earthquake ground motion.
True
4.2) One should consider the possibility of strong earthquake
shaking occurring simultaneously with strong wind.
False
4.3) Modulus of elasticity of concrete does not depend on the
stress level in the member.
False
Sudhir K. Jain,
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Solution 1 /
Part II: True/ False (contd…)
4.4) One should assume that design seismic loads in Xand Y-directions will occur simultaneously.
False
4.5) Load combination 0.9DL 1.5EL is to be used only
when stability of the structure is an issue.
False
4.6) Bottom-face reinforcement in beams near joints is
often governed by the load combination 0.9DL 1.5EL.
True
Sudhir K. Jain,
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Solution 1 /
Part II: True/ False (contd…)
4.7) Design of a RC frame building, circular in
plan, requires 100% + 30% rule for load
combinations.
True
Sudhir K. Jain,
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Solution 1 /
Part III: Problem 4.1
B
a) Earthquake load in X-direction
only.
50kN
A
MAB = 13.67 kN-m
C
D
MAB=13.67kNm
B
b) Earthquake load in Y-direction
only.
MAB = 13.67 kN-m
A
C
50k
N
D
Sudhir K. Jain,
MAB=13.67kN
m
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Solution 1 /
Problem 4.1 (contd…)
c).
c) Earthquake load acting parallel
to beam AB
B
A
C
MAB = 19.37 kN-m
50k
N
D
MAB=19.36kN
m
Notice that if we consider only loads a) and b),
the beam AB will be under-designed (for a
moment of 13.76 kN-m, against 19.36 kN-m it
experiences when shaking is in the direction
shown.
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
Problem 4.1 (contd…)
B
d)
d) Earthquake load (full design
load) acting in X- and Ydirection simultaneously.
MAB = 27.34 kN-m
.
A
50k
N
C
50k
N
D
MAB=27.34kN
m
Notice that if we consider full EQ load in both
directions simultaneously, beam AB will be overdesigned (for a moment of 27.34 kN-m, against a
maximum of 19.36 kN-m it experiences for any
possible direction of shaking.)
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
Problem 4.1 (contd…)
B
e).
e) 100% Design Load in X-direction,
and 30% Design Load in Y-direction
acting simultaneously.
A
50k
N
C
15k
N
MAB = 17.77 kN-m
D
MAB=17.77kN
m
e) 100% Design Load in Y-direction,
and 30% Design Load in X-direction
acting simultaneously.
B
A
MAB = 17.77 kN-m
15k
N
C
50k
N
D
MAB=17.77kN
m
Sudhir K. Jain,
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Solution 1 /
Problem 4.1 (contd…)
Note that 100% + 30% rule gives design
moment 17.77 kN-m against the
maximum of 19.37 kN-m.
This is close enough (within 10%).
If we use 100% + 40% rule (used in some
codes), we will get a design value of 19.14
kN-m, which is even better.
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
Problem 4.1 (contd…)
As an alternate approach of Cl. 6.3.4.2,
the design moment =
(13.67) (13.67) 2 19.33kN m
This is also fairly close to the maximum
value of 19.37 kN-m
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
Solution 5
Date of assignment: 18
January
Date of solution: 21 January
E-Course on IS:1893-2002
34
Part II: True/ False
5.1) Design Spectrum is developed considering expected
ground motion, expected behaviour of structure, and
the assumptions in analysis and design.
True
5.2) In 2002 edition of the code, a flexible structure
located on soft soil will be designed for higher design
force than that on rock sites.
True
5.3) Response reduction factor should be higher for
structures more vulnerable to earthquake damage.
False
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
Part II: True/ False (contd…)
5.4) Underground structures should be designed for higher
level of
design force than the over-ground structures.
False
5.5) 2002 edition of the code specifies vertical coefficient
to be two-thirds of the horizontal coefficient.
True
5.6) Spectrum ordinates for lower damping will be lower
than those for higher value of damping.
False
Sudhir K. Jain,
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Solution 1 /
Part II: True/ False (contd…)
5.7) Major projects, such as dams, major bridges, etc
require development of a site-specific design criteria.
True
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
Question 5.1: RC Ductile Buildings
(T=0.3sec)
Old Code:
New Code:
ZI S a
2R g
Design seismic force Vb= oIKCW
Vb
Design seismic force
Value of o =0.02, 0.04, 0.05, and
0.08, for zones II, III, IV and V,
respectively.
Importance Factor I = 1.0
Soil-Foundation Factor = 1.0
Performance Factor for Ductile
System K = 1.0
Value of C (for T=0.3sec) = 1.0
Value of Z = 0.10, 0.16, 0.24,
0.36 for zones II, III, IV and
V, respectively.
Importance Factor I = 1.0
Response Reduction Factor for
Ductile Frame (SMRF) = 5
Value of S/g (For rock sites, and
T = 0.3 sec) = 2.5
See comparison of design force on next slide.
Sudhir K. Jain,
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Solution 1 /
Question 5.1: RC Ductile Buildings (T=0.3sec) (contd…)
Comparison of design seismic force:
Old Code
New Code
Zone II
0.02W
0.025W
Zone III
0.04W
0.04W
Zone IV
0.05W
0.06W
Zone V
0.08W
0.09W
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
Question 5.2: RC Ductile Buildings
(T=1.0sec)
Old Code:
New Code:
ZI S a
2R g
Design seismic force Vb= oIKCW
Vb
Design seismic force
Value of o =0.02, 0.04, 0.05, and
0.08, for zones II, III, IV and V,
respectively.
Importance Factor I = 1.0
Soil-Foundation Factor = 1.0
Performance Factor for Ductile
System K = 1.0
Value of C (for T=1.0sec) = 0.53
Value of Z = 0.10, 0.16, 0.24,
0.36 for zones II, III, IV and
V, respectively.
Importance Factor I = 1.0
Response Reduction Factor for
Ductile Frame (SMRF) = 5
Value of S/g (For rock sites, and
T = 1.0 sec) = 1.0
See comparison of design force on next slide.
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
Question 5.2: RC Ductile Buildings (T=1.0 sec) (contd…)
Comparison of design seismic force:
Old Code
New Code
Zone II
0.0106W
0.010W
Zone III
0.0212W
0.016W
Zone IV
0.0265W
0.024W
Zone V
0.0424W
0.036W
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
Question 5.3: Ordinary RC Frame Building
(T=0.3sec)
Old Code:
New Code:
ZI S a
2R g
Design seismic force Vb= oIKCW
Vb
Design seismic force
Value of o =0.02, 0.04, 0.05, and
0.08, for zones II, III, IV and V,
respectively.
Importance Factor I = 1.0
Soil-Foundation Factor = 1.0
Performance Factor for Ordinary
Frame System K = 1.6
Value of C (for T=0.3sec) = 1.0
Value of Z = 0.10, 0.16, 0.24,
0.36 for zones II, III, IV and
V, respectively.
Importance Factor I = 1.0
Response Reduction Factor for
Ordinary Frame (OMRF) = 3
Value of S/g (For rock sites, and
T = 0.3 sec) = 2.5
See comparison of design force on next slide.
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
Question 5.3: Ordinary RC Frame Building (T=0.3sec)
(contd…)
Comparison of design seismic force:
Old Code
New Code
Zone II
0.032W
0.042W
Zone III
0.064W
0.067W
Zone IV
0.080W
0.100W
Zone V
0.128W
0.150W
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
Q.5.4: Unreinf. Masonry Building
(T=0.3sec)
Old Code:
Design seismic force Vb= oIKCW
Value of o =0.02, 0.04, 0.05, and
0.08, for zones II, III, IV and V,
respectively.
Importance Factor I = 1.0
Soil-Foundation Factor = 1.0
Performance Factor for K = 1.6
(separate value for masonry not
given; taken highest value)
Value of C (for T=0.3sec) = 1.0
New Code:
Vb
Design seismic force
ZI S a
2R g
Value of Z = 0.10, 0.16, 0.24,
0.36 for zones II, III, IV and
V, respectively.
Importance Factor I = 1.0
Response Reduction Factor for
Ordinary Masonry Bldg =
1.5
Value of S/g (For rock sites, and
T = 0.3 sec) = 2.5
See comparison of design force on next slide.
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
Q.5.4: Unreinf. Masonry Building (T=0.3sec) (contd…)
Comparison of design seismic force:
Old Code
New Code
Zone II
0.02W
0.083W
Zone III
0.04W
0.133W
Zone IV
0.05W
0.20W
Zone V
0.08W
0.30W
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
Q. 5: Observations
Note that the design seismic force for the
old and the new code are about the same for
Ductile Frame (T = 0.3 sec)
Ductile Frame (T = 1.0 sec)
Ordinary Frame (T = 0.3 sec)
However, there is significant increase in
design force for ordinary masonry buildings.
This is because the earlier edition of code did not
pay specific attention to design seismic force.
For example, as per NEHRP code, an ordinary
masonry building is to be designed for 5.33 times
that for a similar RC SMRF.
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
Solution
6
Date of assignment: 20 January
Date of solution: 23 January
E-Course on IS:1893-2002 (Part I)
47
Part II: True/ False
6.1) A structural engineer can provide excellent earthquake
resistance in a building regardless of its configuration.
False
6.2) For buildings with better performance, the code
specifies lower value of R.
False
6.3) One is allowed to ordinary RC frame (OMRF) in zone III
as long as one is providing higher design force than that
for a special moment resisting frame (SMRF).
False
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
Part II: True/ False (contd…)
6.4) If the share of gravity load on a 2-D frame is small, it
will also carry only a small share of design seismic load.
False
6.5) In general, one can accurately predict the natural
period of a building by carrying out dynamic analysis
using a sophisticated computer programme.
False
6.6) Rigid diaphragm implies that the beams are rigid and
hence the column stiffness can be taken as (12EI/L 3).
False
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
Part III: Problem 6.1
(a) Floor diaphragm rigid in its own plane:
The seismic forces are distributed among the frames in
proportion to their lateral stiffness.
0.042
m
150
kN
100
kN
100
kN
50
kN
50
kN
Frames A and
C
Lateral stiffness of the frames:
kA, kC = [300/0.042] kN/m = 7500kN/m
kB
0.096m
150
kN
Frame
B
= [300/0.096] kN/m = 3125kN/m
kA/ kB = 2.4
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
Problem 6.1 (contd…)
Table 1: Lateral forces for rigid floor diaphragm
(distributed in proportion to the stiffness)
Floor
Lateral loads (kN)
Frame A
Frame B
Frame C
Top
62.0
26.0
62.0
Middle
41.3
17.4
41.3
Ground
20.7
8.6
20.7
(b) Floor diaphragm extremely flexible in its own plane:
The seismic forces are distributed among the frames in
proportion to their tributary areas.
Tributary areas for frames A
and
C
Tributary
areas for frame B
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
Problem 6.1 (contd…)
Tributary area ratio for frame A =
0.25
area ratio for frame B =
0.50
Table 2: Lateral forces for extremely flexible floor
area ratio for frame C =
diaphragm
Floor
Lateral loads (kN)
0.25
Frame A
Frame B
Frame C
Top
37.5
75.0
37.5
Middle
25.0
50.0
25.0
Ground
12.5
25.0
12.5
Notice the large difference in design force by the
two methods. By distributing loads in tributary area
ratio, the end frames are being underdesigned while
the middle frame is being over designed.
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
Alternate method to compute lateral stiffness of 2-D
frames:
The lateral stiffness of the frames A, B and C are calculated by
applying point load at the roof:
100
kN
0.018
m
Frames A and
C
Lateral stiffness of the frames:
kA, kC = [100/0.018] kN/m = 5556kN/m
kB
0.042m
100
kN
Frame
B
= [100/0.042] kN/m = 2381kN/m
kA/ kB = 2.3
Thus it is observed that the ratio of the stiffness of the frames
A and B
E-Course
on IS:1893
/
Solution
are almost
the same as obtained
in case
(a).
Sudhir
K. Jain,
1/
Problem 6.1: Some comments
In real buildings, most of the time we will be using 3-D
computer analysis. There, one can simply model the rigid
floor diaphragm action in the analysis. However, two
dimensional methods such as shown here are important for
preliminary analysis and for checking the computer
analysis results.
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
Solution
7
Date of assignment: 21 January
Date of solution: 27 January
E-Course on IS:1893-2002 (Part I)
55
Part II: True/ False
7.1) It is quite difficult to accurately assess the natural
period of buildings.
True
7.2) Brick masonry infills do not contribute much lateral
stiffness to the building.
False
7.3) Empirical expressions for natural period are based on
data from real as-built buildings.
True
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
Part II: True/ False (contd…)
7.4) Choosing a lower value of damping will give a lower
design force on the building.
False
7.5) As per 2002 edition of the code, it is sufficient to
consider first three modes of vibration of the building.
False
7.6) SRSS is a good method for modal combination if the
natural modes are well separated.
True
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
Part III: Problem 7.1
The seismic weight given in the problem includes the weight of
the infills.
(a) Ignoring stiffness contribution of infills, the time periods are
Txx= 0.826 secs.
(b) The stiffness contribution of infills is modelled by diagonal struts
(Holmes, 1961).
Strut width = one-third of diagonal length, strut thickness= wall
thickness
Assuming compressive strength of masonry as 5 MPa,
Modulus of elasticity (Drysdale et. al., 1993) = 550x5 = 2750
MPa
Cross sectional area along XX dirn.
= 0.54 sq.m.
Fundamental
periods,
Txx = 0.462 secs.
Strut
Model
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
Problem 7.1 (contd…)
(c) The stiffness contribution of infills is modelled by shell
elements.
Properties of elements: Modulus of elasticity = 2750 MPa
Thickness
= 230 mm
Fundamental periods,
Txx = 0.473 secs.
Shell
Model
Using the expression for buildings with infill panels, as
given in Cl.7.6.2 of the new code, Txx= 0.281 secs.
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
Problem 7.1 (contd…)
Observations:
Note that the natural period is much higher if
the stiffness contribution of infill panels is
ignored.
Considerable assumptions are needed for
masonry properties, for instance,
Modulus of elasticity of masonry
Width of strut: different researchers have given
different values.
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
Solution 8
Date of assignment: 22
January
Date of solution: 28 January
E-Course on IS:1893-2002
61
Part II: True/ False
8.1) Torsion provisions in the new code require
consideration of accidental torsion.
True
8.2) Centre of rigidity for a multistorey building can be
defined in more than one ways.
True
8.3) As per all floor definition, location of centre of rigidity
depends not only on the building properties, but also on
the lateral load profile.
True
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
Part II: True/ False
8.4) Soft storey and weak storey are one and the same
things.
False
8.5) Soft storey buildings behave well during the
earthquake because of the base isolation effect the soft
storey produces.
False
8.6) Since response spectrum shows higher seismic force
for stiffer buildings, it is best to have a very flexible
building from earthquake safety view point.
False
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
Part II: True/ False (contd…)
8.7) Gravity columns are safest during earthquake shaking
since they are not responsible for seismic loads.
False
8.8) Pounding effects are most detrimental when roof of a
building is liable to hit the adjacent building at the
middle of its column heights.
True
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
Part III: Problem 8.1
y
Grade of conc. M 25
2m
E = 5000√(25)=25,000 N/mm2
CR CM
A
Storey ht h=4500mm
Thickness of the walls =200mm
Length of the walls = 4000mm
4m
C
B
8m
x
D
16m
All walls are same and hence, 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 =
0.0m
Design eccentricity, ed = 0.0 ±0.05 x 8 = ± 0.4m
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
Part III: Problem 8.1 (contd…)
Lateral forces in the walls due to translation:
FCT
FDT
KC
F 50.0kN
KC KD
100
kN
KD
F 50.0kN
KC KD
C
50.0 kN
CR
A
D
CM
B
50.0 kN
Lateral forces in the walls due to torsional moment:
FiR
K i ri
K
i A , B ,C , D
i
ri
2
(F e d )
where ri is the distance of the shear wall from CR
As all walls have same stiffness, KA = KB = KC = KD = k
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
Part III: Problem 8.1 (contd…)
rA= -6.0m
rB = 6.0m
rC =4.0m
rD =-4.0m
and ed = 0.4m
FAR
rA k
2
2
2
2
(rA rB rC rD )k
(F e d ) -2.31kN
Similarly,
FBR = 2.31 kN
FCR = 1.54 kN
FDR = -1.54kN
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
Part III: Problem 8.1 (contd…)
51.54kN
Total lateral forces in the walls:
FA = -2.31 kN
FB = 2.31 kN
FC = 50 + 1.54 =51.54 kN
2.31kN
C
100
kN
CR
A
2.31kN
CM* B
48.46kN
D
FD = 50 - 1.54 = 48.46 kN
Similarly, when ed = -0.4m, then the total lateral forces in the walls will be,
FA = 2.31 kN
FB = -2.31 kN
FC = 50 - 1.54 = 48.46 kN
FD = 50 + 1.54 = 51.54 kN
Design lateral forces in the walls C and D
are
FC = FD = 51.54 kN
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
Part III: Problem 8.1 (contd…)
50kN
50kN
EQ Force in Y-direction
C
CR
Calculated eccentricity = 2.0m
A
Design eccentricity, ed = 2.0 x 1.5 +0.05 x 16
=3.8m
or
2.0due
- 0.05
x 16 = 1.2m
Lateral forces in the walls
to translation
:
FAT
KA
F 50.0kN
KA KB
FBT
KB
F 50.0kN
KA KB
Sudhir K. Jain,
CM
B
D
100
kN
E-Course on IS:1893 /
Solution 1 /
Part III: Problem 8.1 (contd…)
Lateral forces in the walls due to torsional moment:
When ed = 3.8m
FAR
rA k
2
2
2
2
(rA rB rC rD )k
(F e d ) -21.92kN
Similarly,
FBR = 21.92 kN
FCR = 14.62 kN
28.08kN
FDR = -14.62kN
C
Total lateral forces in the walls:
FA = (50.0 – 21.92)= 28.08 kN
14.62k 71.92kN
N
CR CM CM*
A
B
FB = (50.0 + 21.92) = 71.92 kN
FC = 14.62 kN
14.62kND
FD = -14.62 kN
Sudhir K. Jain,
E-Course on IS:1893 /
100 kN
Solution 1 /
Part III: Problem 8.1 (contd…)
Similarly, when ed = 1.2m, then the total lateral forces in the walls will be,
FA = (50.0 – 6.92)= 43.08 kN
FB = (50.0 + 6.92) = 56.92 kN
FC = 4.62 kN
FD = -4.62 kN
Hence, maximum lateral forces in walls A and B are
FA = 43.08 kN
FB = 71.92 kN
However, note the statement in Cl. 7.9.1 “negative torsional
shear shall be neglected. This implies that wall A should
still be designed for lateral force of 50kN. Thus, the design
forces are:
FA = 50.00 kN, FB = 71.92 kN, FC = FD = 51.54 kN
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
Solution 9
Date of assignment: 24 January
Date of solution : 28 January
E-Course on IS:1893-2002 (Part I)
72
Part II: True/ False
9.1) Code now requires ties to be provided for isolated
footings and pile caps (unless footing is supported on
rock directly).
True
9.2) Cantilever projections are particularly vulnerable to
damage during earthquakes.
True
9.3) It is better to concentrate shear walls near the centre
of the building rather than place them near the
perimeter.
False
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
Part II: True/ False (contd…)
9.4) The best treatment for non-parallel systems is to
perform dynamic analysis.
False
9.5) Irregular buildings should be designed for a higher
value of importance factor.
False
9.6) As a general rule, buildings with irregular
configuration perform poorly in earthquakes even when
good engineering has been carried out.
True
Sudhir K. Jain,
E-Course on IS:1893 /
Solution 1 /
