Seismic evaluation of masonry building a case study

Seismic Evaluation of Masonry Building A Case Study, Suhail Shafi, Dr. A R Dar, Danishzafar Wani,
Mohd Hanief Dar, Journal Impact Factor (2015): 9.1215 (Calculated by GISI) www.jifactor.com
www.iaeme.com/ijciet.asp 79 editor@iaeme.com
1
B Tech, Department of Civil Engineering, NIT, Srinagar, India
2
Professor and Head, Department of Civil Engineering, NIT, Srinagar, India
3
M Tech, Department of Civil Engineering, Kurukshetre University, India
4
M Tech, Department of Civil Engineering, NIT, Srinagar, India
ABSTRACT
Most of the construction in Kashmir Valley is still done in brick masonry, our research is
intended to check out the earthquake resistance of on-going masonry building construction (mostly
important buildings like schools) and to point out flaws in design and construction that result
in poor seismic performance of such structures.
Key Words: Earthquake Hazard, Masonry Buildings, Ignorance of Earthquake Hazards in Kashmir,
Ill Workmanship, Poor Design.
1. INTRODUCTION
The valley of Kashmir lies in the seismic zone V as per IS 1893 (part 1): 2002 annex E. As
such it is the hot spot for occurrence of earthquakes not only mild ones but also very severe ones.
However the people here are sleeping in a deep slumber and they keep on constructing buildings
which are no more than death traps considering their Earthquake susceptibility. Leaving alone proper
earthquake design of buildings, even the earthquake resistant guidelines or tips are not even followed
properly here. It is same for both private residential buildings with lesser Importance factor and
government constructed public buildings (like Hospitals, schools and colleges) with high value of
Importance factor.
Most of the loss of life in past earthquakes has occurred due to the collapse of
buildings, constructed in traditional materials like stone, brick, adobe and wood, which were not
initially engineered to be earthquake resistant. In view of the continued use of such buildings in
most countries of the world, it is essential to introduce earthquake resistance features in their
construction.
The main objective of this research is to deal with the basic concepts involved in
achieving appropriate earthquake resistance of masonry buildings; to include suitable
illustrations to explain the important points, and to present such data which could be used to
proportion the critical strengthening elements. Masonry building is defined as buildings which are
SEISMIC EVALUATION OF MASONRY BUILDING A
CASE STUDY
Suhail Shafi1
, Dr. A R Dar2
, DanishZafar Wani3
, Mohd Hanief Dar4
Volume 6, Issue 6, June (2015), Pp. 79-91
Article ID: 20320150606008
International Journal of Civil Engineering and Technology (IJCIET)
© IAEME: www.iaeme.com/Ijciet.asp
ISSN 0976 – 6308 (Print)
ISSN 0976 – 6316 (Online)
IJCIET
© I A E M E

Seismic Evaluation of Masonry Building A Case Study,
Mohd Hanief Dar, Journal Impact Factor (2015): 9.1215 (
www.iaeme.com/ijciet.asp
spontaneously and informally constructed in the traditional manner with bricks or stones with
intervention by qualified architects and engineers in their design.
Reinforced masonry, reinforced
types of structural systems, and
although some of the principles stated herein will
2. SEISMIC ANALYSIS AND DESIGN OF AMAR SINGH COLLEGE
Masonry buildings in India are generally designed for vertical loads based on IS 1905.
It is not confirmed whether the lateral load effects from wind or earthquake have been
considered in analysis or not, particularly when the buildings are considered in seismic prone
areas.
2.1 Building Data
The plan is shown in fig 1.
2.2 Material Strength
Permissible compressive strength of masonry (f
(Assuming unit strength = 35 MP
Permissible stress in steel in tension
(Use high strength deformed bars (Fe 415)
2.3 Live load data
Live load on roof = 1.0 KN/m2 (
for seismic calculations = 0)
Live load on first floor = 3.0 KN/m
f Masonry Building A Case Study, Suhail Shafi, Dr. A R Dar, Danishzafar Wani,
, Journal Impact Factor (2015): 9.1215 (Calculated by GISI) www.jifactor.com
80
chitects and engineers in their design.
reinforced concrete or steel frame buildings, tall buildings using various
and major industrial buildings, etc., are excluded from
stated herein will equally apply to these constructions.
2. SEISMIC ANALYSIS AND DESIGN OF AMAR SINGH COLLEGE
s not confirmed whether the lateral load effects from wind or earthquake have been
Figure 1 Plan of the Building
Permissible compressive strength of masonry (fm) = 2.5 N/mm2
(Assuming unit strength = 35 MPa and mortar H1 type)
Permissible stress in steel in tension = 0.55 fy
ngth deformed bars (Fe 415) i.e. fy= 230 MPa)
for seismic calculations = 0)
Live load on first floor = 3.0 KN/m2
Suhail Shafi, Dr. A R Dar, Danishzafar Wani,
www.jifactor.com
editor@iaeme.com
buildings using various
excluded from consideration
to these constructions.
s not confirmed whether the lateral load effects from wind or earthquake have been

2.4 Dead load data
Thickness of floor and roof slab = 120 mm
Weight of slab = 3 KN/m2
(Assuming weight density of concrete =25 KN/m3
)
Thickness of ground Storey wall = 350mm
Thickness of first Storey wall = 250mm
2.5 Seismic data
Seismic zone = v,
Zone factor = 0.36,
Importance factor (I) = 1.5
Response reduction factor (R) = 3 (as per IS 1893 (part1): 2002)
Direction of seismic force = E-W direction
3. DETERMINATION OF DESIGN LATERAL LOAD
3.1 Seismic weight calculations
Description Load calculations Total weight (KN)
DL and LL at roof level
(i) Weight of roof 15.7 × 11 × 3 518.1
(ii) Weight of walls ½ × {2 × (15.7 + 11) × 3.66× 5} 488.6
(Assume half weight of walls at
Second Storey lumped at roof)
(iii) Weight of live load (LL) 0 0
(Wr) Weight at roof level
(i) + (ii) + (iii) 1006.7
DL and LL at floor level
(i) Weight of floor 15.7 × 11 × 3 518.1
(ii) Weight of walls
(Assume half weight of walls at ½ × {2 × (15.7 + 11) × 3.66× 7} 1368.1
2nd
Storey and half weight of walls
At 1st Storey is lumped at roof)
(iii) Weight of live load (LL) 3 × 15.7 × 11 518.1
(Wf) weight at second level (i) +(ii) + (iii) 2404.3
Total seismic weight of building (Wr+ Wf) 3411 KN
3.2 Time period calculation
The approximate fundamental natural time period of a masonry building can be calculated
from the clause 7.6.2 of IS (part1): 2002 as,
Ta = 0.09 h/vd = 0.199
Where,
h = height of building in m, {i.e. 3.66 (first Storey) + 3.66 (second Storey) = 7.32m}
d = base dimension of building at the plinth level, in m, along the considered
Direction of lateral force (i.e. 11m, assuming earthquake in E-W direction)
Soil medium type, for which average response acceleration coefficient are as:
Sa/g = 2.5, for T = 0.199
Ah = ZISa/2Rg = (0.36/2) (1.5/3) (2.5) = 0.225

The total design base shear (Vb) along the direction of motion is given by,
Vb = AhW = .225 × 3411 = 767.475 KN
Lateral force at roof level = VB
= 767.475 × 1006.7 × 7.322
/ (1006.7 × 7.32
= 438.475 KN
Lateral force first floor level = V
= 767.475 × 1006.7 × 3.662
/ (1006.7 × 7.32
= 329 KN
Figure 2 (a) Elevation
4. DETERMINATION OF WALL RIGIDITIES
In the second step, we will calculate the relative stiffness of exterior shear walls. It is
assumed here that all the lateral force will be resisted by the exterior shear walls. Therefore,
the stiffness and masses of interior wall may
Rigidity of cantilever pier is given by R
Rigidity of fixed pier is given by R
Rigidity of North shear wall = 0.5Et
Rigidity of south shear wall = Rigidity of North sh
Rigidity of east shear wall = 0.69Et
Rigidity of west shear wall = 1.16Et
Relative stiffness of walls
North shear wall = 0.5 / (0.5 + 0.5)
South shear wall = 0.5
East shear wall = 0.69/ (0.69 + 1.16)
West shear wall = 1.16/ (0.69 + 1.16) = 0.63
5. DETERMINATION OF TORSIONAL FORCES
To calculate the shear forces due to torsion, first calculation of Center of mass and center of
rigidity are done as:
82
) along the direction of motion is given by,
= 767.475 KN
BWihi
2
/∑ Wihi
2
/ (1006.7 × 7.322
+ 3411×3.662
)
VBWihi
2
/∑ Wihi
2
/ (1006.7 × 7.322
+ 3411×3.662
)
Elevation of Building (b) Seismic Load (c) Storey
4. DETERMINATION OF WALL RIGIDITIES
the stiffness and masses of interior wall may be neglected in seismic analysis.
Rigidity of cantilever pier is given by Rc= Et/ {4(h/d) 3
+ 3(h/d)}
Rigidity of fixed pier is given by Rf = Et/ {(h/d) 3
+ 3(h/d)}
Rigidity of North shear wall = 0.5Et
Rigidity of south shear wall = Rigidity of North shear wall = 0.5Et
Rigidity of east shear wall = 0.69Et
Rigidity of west shear wall = 1.16Et
North shear wall = 0.5 / (0.5 + 0.5) = 0.5
East shear wall = 0.69/ (0.69 + 1.16) = 0.37
wall = 1.16/ (0.69 + 1.16) = 0.63
5. DETERMINATION OF TORSIONAL FORCES
www.jifactor.com
editor@iaeme.com
(c) Storey Shear
be neglected in seismic analysis.

5.1 Location of the center of mass
Centre of mass, XCM and YCM is calculated by taking statical moments about a point, say,
south west corner, using the respective weights of walls as forces in the moment summation. The
calculation of Centre of mass is shown in table 1.
Table 1 Calculation of Centre of Mass
Item Weight i (KN) X (m) Y (m) WX (KN-m) WY (KN-m)
Roof slab
518.1
5.5 7.85 2849.6 4067
North Wall 224.7 5.5 15.7 1235.85 3527.8
South Wall 224.7 5.5 0 1235.85 0
East Wall 302.3 5.5 7.85 3325.3 2373
West Wall 367 0 7.85 0 2881
Total ∑W = 1636.8
∑WX =
8646.6
∑WY =
12848.8
XCM = ∑WX/∑ W = 8646.6/1636.8 = 5.28
YCM = ∑WY/∑ W = 12848.8/1636.8 =7.85
5.2 Location of center of rigidity
The center of rigidity, XCR and YCR, is calculated by taking statical moments about apoint,
say, south-west corner, using the relative stiffness of the walls as the forces inthe moment
summation. The stiffness of slab is not considered in the determinationof center of rigidity. The
calculation for the center of rigidity is shown in table 2.
Table 2 Calculation For The Center Of Rigidity
ITEM RX RY X(m) Y(m) YRX XRY
NORTH
WALL
0.5 - - 15.7 7.85 -
SOUTH
WALL
0.5 - - 0 0 -
EAST
WALL
- .37 11 - - 4.04
WEST
WALL
- .63 0 - - 0
TOTAL ∑RX = 1 ∑RY = 1
∑YRX =
7.85
∑XRY =
4.07
XCR = ∑XRY/∑RY = 4.07
YCR = ∑YRX/∑RX = 7.85
5.3 Torsional eccentricity
Torsional eccentricity in y-direction
Eccentricity between center of mass and center of rigidity
eY = 7.85 -7.85 = 0m

Add 5% accidental eccentricity = .05 × 15.7 = 0.785m
Total eccentricity = 0 + 0.785 = 0.785m
Torsional eccentricity in x-direction
eX = 5.5 – 4.07 = 1.43m
Add 5% accidental eccentricity = .05 × 11 = 0.55m
Total eccentricity = 1.43 + 0.55 = 1.98m
5.4 Torsional moments
The torsional moment due to E
MTX = VXeY = 767.475 × 0.785 = 602.5
MTY = VYeX = 767.475 × 1.98 = 1519.6
(VY = VX, because SA/g is constant value of
5.5 Distribution of direct shear force and torsional shear force
Since, we are considering the seismic force only in E
direction will resist the forces and the walls in
the calculation of distribution of direct shear and torsional shear.
Table 3 calculation of distribution of direct shear and torsional shear
ITEM RX dY R
NORTH
WALL
.5 7.85 3.925
SOUTH
WALL
.5 7.85 3.925
dY = distance of considered wall from center of rigidity (15.7
Direct shear force = relative stiffness of wall × total base shear (0.5 × 767.475
Torsional force in North wall = R
Torsional force in South wall = R
Figure
84
entricity = .05 × 15.7 = 0.785m
Total eccentricity = 0 + 0.785 = 0.785m
direction
Add 5% accidental eccentricity = .05 × 11 = 0.55m
icity = 1.43 + 0.55 = 1.98m
The torsional moment due to E-W seismic force rotate the building in y
= 767.475 × 0.785 = 602.5
= 767.475 × 1.98 = 1519.6
/g is constant value of 2.5 for the time period 0.11 = T = 0.55)
Distribution of direct shear force and torsional shear force
Since, we are considering the seismic force only in E-W direction, the walls in N
direction will resist the forces and the walls in E-W direction may be ignored. Table 7.3 shows
the calculation of distribution of direct shear and torsional shear.
calculation of distribution of direct shear and torsional shear
RX dY RX dY
2 DIRECT SHEAR
FORCE(KN)
TORSIONAL
SHEAR
FORCE(KN)
3.925 30.8 383.7 38.4
3.925 30.8 383.7 -38.4
= distance of considered wall from center of rigidity (15.7 – 7.85 = 7.85)
lative stiffness of wall × total base shear (0.5 × 767.475
Torsional force in North wall = RXdy × VXeY/ RXdy
2
= 38.4 KN
Torsional force in South wall = RXdy × VXeY/ RXdy
2
= 38.4 KN
Figure 3 Torsional forces in building
www.jifactor.com
editor@iaeme.com
W seismic force rotate the building in y-direction, hence
2.5 for the time period 0.11 = T = 0.55)
W direction, the walls in N-S
W direction may be ignored. Table 7.3 shows
calculation of distribution of direct shear and torsional shear
TORSIONAL
SHEAR
FORCE(KN)
TOTAL
SHEAR
FORCE(KN)
422.1
345.3

The torsional forces are additive on the north wall and subtractive on the south wall as
shown, since the code directs that negative torsional shear shall be neglected. Hence the total shear
acting on the south wall is simply direct shear only.
5.6 Distribution of the total shear to individual piers within the wall
The shear carried by the north and south shear wall is now distributed to individual piers on
the basis of their respective stiffness. Since the relative stiffness of all the piers is same, hence, the
shear force will be distributed equally among all the individual piers.
Shear force in North shear wall piers = 422.1/5 =84.42 KN
Shear force in South shear wall piers = 345.3/5 = 69.06 KN
5.7 Increase in axial load due to overturning
Total overturning moment due to lateral force acting on building is,
MOVT = total shear (VX) × vertical distance between first floor level to critical plane ofweakness,
assuming at the level of sill + applied overturning moment at first floorlevel.,
Assume the stiffness of second Storey walls is the same e as first Storey, the
totaldirect shear in E-W direction of seismic load i.e. in X-direction is divided in North andSouth
shear wall in the proportion of their stiffness
Direct shear in north wall (VNX) = 383.7 KN = Direct shear in south wall (VSX)
Distribution of lateral force along the height of North and South wall is:
North shear wall
Lateral force at roof level = VNX × Wrhr
2
/ Wihi
2
= 383.7 × 0.57 = 218.7KN
Lateral force at first floor level = VNX × W1h1
2
/ Wihi
2
= 383.7 × 0.43 = 165KN
South shear wall
Lateral force at roof level = VSX × Wrhr
2
/ Wihi
2
= 383.7 × 0.57 = 218.7KN
Lateral force at first floor level = VSX × W1h1
2
/ Wihi
2
= 383.7 × 0.43 = 165KN
Increase in axial load in piers of North shear wall
Overturning moment in North wall (MOVT) is:
MOVT = 383.7 × 2.56 + 218.8 × 3.66 = 1783 KN
Increase in axial load due to overturning moment
Povt = Movtȴ iAi/In
Where,ȴiAi = centroid of net section of wall is calculated as shown in table 4
In = moment of inertia of net section of wall is calculated as shown in table 5
Table 4 centroid of net section of net section of wall
Pier Area (Ai) m2 ȴ (distance of left edge of wall to centroid of
piers) m
Aiȴ (m3
)
1 1.24 × 0.35 = 0.434 0.62 0.269
2 1.24 × 0.35 = 0.434 2.44 1.058
3 1.24 × 0.35 = 0.434 4.26 1.848
4 1.24 × 0.35 = 0.434 6.08 2.638
5 1.24 × 0.35 = 0.434 7.9 3.429
Σ = 2.17 Σ = 9.2416
Distance from left edge to centroid of net section of wall = 9.2416/2.12 = 4.26

Table 5 Calculation of Moment of Inertia of Net Section of Wall
Pier Area (Ai) m2 ȴi (m) Aiȴi (m3) Aiȴi2 (m4) I = td3/12 In = I + Aiȴi2
1 1.24 × 0.35 = 0.434 3.64 1.58 5.75 0.35 × 1.73/12 = 0.143 5.89
2 1.24 × 0.35 = 0.434 1.82 0.79 1.44 0.35 × 1.73/12 = 0.143 1.58
3 1.24 × 0.35 = 0.434 0.00 0.00 0.00 0.35 × 1.73/12 = 0.143 0.14
4 1.24 × 0.35 = 0.434 1.82 0.79 1.44 0.35 × 1.73/12 = 0.143 1.58
5 1.24 × 0.35 = 0.434 3.64 1.58 5.75 0.35 × 1.73/12 = 0.143 5.89
Σ = 2.17 Σ = 15.08
Table 6 Increase in Axial Load in the Pier on North Wall
Pier Aiȴi (m3
)
Povt = Movtȴ iAi/In
(KN)
1 1.58 186.85
2 0.79 93.42
3 0.00 0.00
4 0.79 93.42
5 1.58 186.85
Movt = 1783 KN-m
In = 15.08
The increase in axial load of south shear wall is the same as that for the North shear wall as
the two walls are symmetric.
6. DETERMINATION OF PIER LOADS, MOMENTS AND SHEAR
The total axial load (due to dead load, live load and overturning), shear and moments
in the individual piers of both the shear walls are calculated in table 7and 8 as below
Table 7 Axial load, moments, shear in piers of North shear wall
pier
Effective width of
pier (m)
Pd*
(KN)
PL#
(KN)
Povt
(KN)
Shear VE for
moment (KN)
Moment (KN-m) =
VE × h/2
1 1.84 152.54 57.77 186.85 84.42 71.8
2 2.44 202.276 76.62 93.42 84.42 71.8
3 2.44 202.276 76.62 0.00 84.42 71.8
4 2.44 202.276 76.62 93.42 84.42 71.8
5 1.84 152.54 57.77 186.85 84.42 71.8

Table 8 Axial load, moments, shear in piers of south shear wall
pier
Effective width of
pier (m)
Pd*
(KN)
PL#
(KN)
Povt
(KN)
Shear VE for
moment (KN)
Moment (KN-m) =
VE × h/2
1 1.84 152.54 57.77 186.85 69.1 58.8
2 2.44 202.276 76.62 93.42 69.1 58.8
3 2.44 202.276 76.62 0.00 69.1 58.8
4 2.44 202.276 76.62 93.42 69.1 58.8
5 1.84 152.54 57.77 186.85 69.1 58.8
Where,
* Pd = effective loading width of pier × dead load intensity in KN/m
Effective loading width of pier = width of pier + ½ of each adjacent opening of pier
h = height of pier = 1.7m
Dead load intensity is calculated as (per meter length of wall) below:
North wall first storey
1. Weight of first storey (from level of first floor to sill level) = 2.56 × 0.35 × 20 = 17.5 KN/m
2. Weight of second storey = 3.66 × 0.25 ×20 = 18.3 KN/m
3. Weight of floor at second storey level = ½ × (.12 × 15.7 ×25) = 23.55
(Assume north and south shear wall will take equal amount of load)
4. Weight of roof = ½ × (.12 × 15.7 ×25) =23.55
Total load = 82.8 KN/m
South wall first storey
1. Weight of first storey (from level of first floor to sill level) = 2.56 × 0.35 × 20 = 17.5 KN/m
2. Weight of second storey = 3.66 × 0.25 ×20 = 18.3 KN/m
3. Weight of floor at second storey level = ½ × (.12 × 15.7 ×25) = 23.55
4. Weight of roof = ½ × (.12 × 15.7 ×25) =23.55
# PL = effective loading width of pier × live load intensity in KN/m
Effective loading width of pier = width of pier + ½ of each adjacent opening of pier
Live load intensity is calculated as (per meter length of wall) below:
North wall: first storey
1. Live load on floor (3 KN/m2) = ½ × (3 × 15.7) = 23.55 KN/m
2. Live load on roof (1KN/m2) = ½ × (1 × 15.7) = 7.85 KN/m
South wall: first storey
1. Live load on floor (3 KN/m2) = ½ × (3 × 15.7) = 23.55 KN/m
2. Live load on roof (1KN/m2) = ½ × (1 × 15.7) = 7.85 KN/m

7. DESIGN OF SHEAR WALLS FOR AXIAL LOAD AND MOMENTS
Determination of jamb steel at the pier boundary
North shear wall
The design is tabulated in table9
Table 9 design of north shear wall
Pier
Moment
(KN-m)
Effective depth
(mm)
Area of jamb steel
As* (mm2)
No. of bars P (KN)
(total)
d
(mm)
t
(mm)
fa/Fa fb/Fb fa/Fa
+ fb/Fb
result
1 71.8 1140 304.3 2@16Ø 397.2 1.24 0.35 0.37 0.26 0.63 OK
2 71.8 1140 304.3 2@16Ø 372.3 1.24 0.35 0.34 0.26 0.60 OK
3 71.8 1140 304.3 2@16Ø 278.9 1.24 0.35 0.26 0.26 0.52 OK
4 71.8 1140 304.3 2@16Ø 372.3 1.24 0.35 0.34 0.26 0.60 OK
5 71.8 1140 304.3 2@16Ø 397.2 1.24 0.35 0.37 0.26 0.63 OK
South shear wall
The design is tabulated in table10.
Table 10 design of south shear wall
Pier
Moment
(KN-m)
Effective
depth
(mm)
Area of jamb
steel
As* (mm2)
No. of
bars
P
(KN)
(total)
d
(mm)
t
(mm)
fa/Fa fb/Fb
fa/Fa
+
fb/Fb
result
1 71.8 1140 304.3 2@16Ø 397.2 1.24 0.35 0.37 0.26 0.63 OK
2 71.8 1140 304.3 2@16Ø 372.3 1.24 0.35 0.34 0.26 0.60 OK
3 71.8 1140 304.3 2@16Ø 278.9 1.24 0.35 0.26 0.26 0.52 OK
4 71.8 1140 304.3 2@16Ø 372.3 1.24 0.35 0.34 0.26 0.60 OK
5 71.8 1140 304.3 2@16Ø 397.2 1.24 0.35 0.37 0.26 0.63 OK
* Jamb steel at the pier boundary is given by
As = M/{fs × 0.9 × defective}
Fs = 0.55 Fe = 0.55 × 415 = 230 N/mm2
deffective = dtotal – cover
Assume cover of 50mm on both sides
** Adequacy of individual pier under compression and moment is checked by interaction formula
i.e.
fa/Fa + fb/Fb ≤ 1
fa = Ptotal/{width of pier (d) × t } where, Ptotal = Pd + PL + Povt
fb = M/(td2/6)
Fa = permissible compressive stress = 2.5N/mm2 (as per IS: 1905)
Fb = permissible bending stress = 2.5 + 0.25 × 2.5 = 3.125 N/mm2 (as per IS: 1905)

8. DESIGN OF SHEAR WALLS FOR SHEAR
Shear in building may be resisted by providing the bands or bond beams. The bands represent
a horizontal framing system, which transfer the horizontal shear induced by the earthquakes from the
floors to shear (structural) walls. It als
action. In combination with vertical reinforcement, it improves the strength, ductility and energy
dissipation capacity of masonry walls. Depending upon its location in the building it may be term
as roof, lintel, and plinth band.in case of a flexible diaphragm, both roof and lintel band is required
however, in case of rigid diaphragm, a band at lintel level is sufficient. Plinth band is useful in
sustaining differential settlements, particularly
8.1 Design of bond beam
Total shear force in E-W direction (V) = 767.475 =768
Moment produced by this is given by
M = V × L/8 = 768 × 15.7/8 = 1507.2
Now,
T = M/d = 1507.2/11 = 137 KN
As = T/fs =137 × 1000/230 = 595.65 mm2
Provide 2 @ 16Ø and 2 @ 12Ø i.e. a total of four bars
A provided = 628mm2
Therefore, provide 150mm thick band and four bars of steel tied with stirrups 6mm dia at 150 c/c.
9. STRUCTURAL DETAILS OF THE BUILDING
Figure
89
AR WALLS FOR SHEAR
floors to shear (structural) walls. It also connects all the structural walls to improve the integral
dissipation capacity of masonry walls. Depending upon its location in the building it may be term
sustaining differential settlements, particularly, when foundation soil is soft or has uneven properties.
W direction (V) = 767.475 =768
Moment produced by this is given by
= 1507.2
/fs =137 × 1000/230 = 595.65 mm2
Provide 2 @ 16Ø and 2 @ 12Ø i.e. a total of four bars
9. STRUCTURAL DETAILS OF THE BUILDING
gure 4 structural details of the building
www.jifactor.com
editor@iaeme.com
o connects all the structural walls to improve the integral
dissipation capacity of masonry walls. Depending upon its location in the building it may be termed
, when foundation soil is soft or has uneven properties.

10. CHECKING ADEQUACY OF AMAR SINGH COLLEGE BUILDING FOR ABOVE
DESIGN REQUIREMENTS
The design details are now compared with the existing details of the building.
Table 11 Checking Adequacy of Amar Singh College Building for above Design Requirements
S.NO Data of Building under Assessment
Required as per design/IS
code
Whether
Complying?
1. Number of storeys, S =2 Equal to or less than 4 yes
2.
Wall building unit:
BURNED BRICK CONSTRUCTION
Compressive strength = 35 kg/cm2
Compressive strength ≥
35 kg/cm2 yes
3.
Thickness of load bearing walls, t
External wall =350mm
Internal wall =250mm
BB = 230 mm
CCB = 200 mm yes
4. Mortar used =1:6 C:S = 1:6 or richer yes
5.
Longest wall in room, L = 11m BB ≤ 8 m
CCB ≤ 7 m
No
6.
Height of wall, floor to ceiling
h = 3.66 m
BB = 3.45 m
CCB = 3.0 m
No
7.
Door, Window openings (See fig.3)
Overall (b1 + b2+….)/l,
=0.403
One storeyed 0.50
Two Storeyed 0.42
3 or 4 Storeyed 0.33
b4 min 560 mm
b5 min 450 mm
Yes
8.
Floor type
Reinforced Concrete slab
OK
With RC screed
With bracing
With ties & bracing
Yes
9.
Roof type
Sloping trussed,
With bracing
OK
With bracing
With ties &bracing
Yes
10. Seismic Bands
(i) at plinth = provided Required No
(ii) at lintel level = provided Required NO
(iii) at window = not provided
sill level
Required
No
(iv) at ceiling = not provided
eave level
Required
No
(v) at gable ends provided / not provided Required NO
(vi) at ridge top provided / not provided Required NO
11. Vertical bar
(i) at external corners = provided Required in all masonry
Buildings
NO
(ii) at external T-junctions = provided Required in all masonry
Buildings
NO
(iii) at internal corners = provided Required in all masonry
Buildings
NO
(iv) at internal T-junctions = provided Required in all masonry NO

CONCLUSION
Earthquakes in Kashmir have been causing extensive damage to buildings and loss of many
valuable lives since times immortal. Such devastations due to earthquake demand the need of making
structures that are earthquake resistant. Since most of the construction in Kashmir is still being done
in masonry, the aim of this project was to study various earthquake resistant measures in masonry
buildings in context with Kashmir then point out the various flaws in construction practices in the
current ongoing masonry constructions particularly important buildings like school buildings etc. and
compare them with the actual seismic requirements.
For the current building it was concluded that the design of the building is susceptible to
earthquake and can fail under design base earthquake because the earthquake resistant features were
either missing or not complying with the design requirements.
REFERENCES
1. Introduction to international disaster management by Damon P. Coppola.
2. Indian Standard Code of practices IS 1893:2002.
3. Indian Standard Code of Practices IS 13828:1993.
4. Mohd Hanief Dar, Zahid Ahmad Chat and Suhail Shafi, “Flaws In Construction Practices of
Masonry Buildings In Kashmir With Reference To Earthquakes (A Case Study)” International
Journal of Advanced Research in Engineering & Technology (IJARET), Volume 6, Issue 1,
2015, pp. 67 - 72, ISSN Print: 0976-6480, ISSN Online: 0976-6499.
5. Mohammed S. Al-Ansari,Qatar University, “Building Response To Blast and Earthquake
Loading” International journal of Computer Engineering & Technology (IJCET), Volume 3,
Issue 2, 2012, pp. 327 - 346, ISSN Print: 0976 – 6367, ISSN Online: 0976 – 6375.
6. Dharane Sidramappa Shivashaankar and Patil Raobahdur Yashwant, “Earthquake Resistant
High Rise Buildings –New Concept” International Journal of Advanced Research in
Engineering & Technology (IJARET), Volume 5, Issue 6, 2014, pp. 121 - 124, ISSN Print:
0976-6480, ISSN Online: 0976-6499.

Seismic evaluation of masonry building a case study

Recommended

Recommended

More Related Content

What's hot

What's hot (19)

Viewers also liked

Viewers also liked (17)

Similar to Seismic evaluation of masonry building a case study

Similar to Seismic evaluation of masonry building a case study (20)

More from IAEME Publication

More from IAEME Publication (20)

Recently uploaded

Recently uploaded (20)

Seismic evaluation of masonry building a case study