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

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1B Tech, Department of Civil Engineering, NIT, Srinagar, India 2Professor and Head, Department of Civil Engineering, NIT, Srinagar, India

3M Tech, Department of Civil Engineering, Kurukshetre University, India 4M 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 Dar

2, DanishZafar Wani

3, Mohd Hanief Dar

4

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 MPPermissible 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

spontaneously and informally constructed in the traditional manner with bricks or stones with chitects and engineers in their design.

reinforced concrete or steel frame buildings, tall buildings using variousand major industrial buildings, etc., are excluded from

stated herein will equally apply to these constructions.

2. SEISMIC ANALYSIS AND DESIGN OF AMAR SINGH COLLEGE

Masonry buildings in India are generally designed for vertical loads based on IS 1905. s 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

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,

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[email protected]

spontaneously and informally constructed in the traditional manner with bricks or stones with

buildings using various excluded from consideration

to these constructions.

Masonry buildings in India are generally designed for vertical loads based on IS 1905. s 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




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 2ndStorey 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 KNLateral 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 mayRigidity of cantilever pier is given by RRigidity of fixed pier is given by RRigidity of North shear wall = 0.5EtRigidity of south shear wall = Rigidity of North shRigidity of east shear wall = 0.69EtRigidity 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

BWihi2/∑ Wihi

2

/ (1006.7 × 7.322+ 3411×3.662)

VBWihi2/∑ Wihi

2

/ (1006.7 × 7.322+ 3411×3.662)

Elevation of Building (b) Seismic Load (c) Storey

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 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



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(c) Storey Shear

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,

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 = 0.785mTotal eccentricity = 0 + 0.785 = 0.785mTorsional eccentricity in x-directionEccentricity between center of mass and center of rigidityeX = 5.5 – 4.07 = 1.43m Add 5% accidental eccentricity = .05 × 11 = 0.55mTotal eccentricity = 1.43 + 0.55 = 1.98m

5.4 Torsional moments The torsional moment due to E

MTX = VXeY = 767.475 × 0.785 = 602.5MTY = 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 Edirection will resist the forces and the walls inthe 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 × 767.475Torsional force in North wall = RTorsional force in South wall = R

Figure



84

entricity = .05 × 15.7 = 0.785m Total eccentricity = 0 + 0.785 = 0.785m

direction Eccentricity between center of mass and center of rigidity

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/ RXdy2 = 38.4 KN

Torsional force in South wall = RXdy × VXeY/ RXdy2 = 38.4 KN

Figure 3 Torsional forces in building


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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 × Wrhr2 / Wihi

2 = 383.7 × 0.57 = 218.7KN Lateral force at first floor level = VNX × W1h1

2 / Wihi2 = 383.7 × 0.43 = 165KN

South shear wall

Lateral force at roof level = VSX × Wrhr2 / Wihi

2 = 383.7 × 0.57 = 218.7KN Lateral force at first floor level = VSX × W1h1

2 / Wihi2 = 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 (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 # 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 (Assume north and south shear wall will take equal amount of load) 2. Live load on roof (1KN/m2) = ½ × (1 × 15.7) = 7.85 KN/m (Assume north and south shear wall will take equal amount of load) Total load = 31.4 KN/m South wall: first storey 1. Live load on floor (3 KN/m2) = ½ × (3 × 15.7) = 23.55 KN/m (Assume north and south shear wall will take equal amount of load) 2. Live load on roof (1KN/m2) = ½ × (1 × 15.7) = 7.85 KN/m (Assume north and south shear wall will take equal amount of load) Total load = 31.4 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 alsaction. 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 termas 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 barsA 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

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 also connects all the structural walls to improve the integral 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 termas 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, 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

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

gure 4 structural details of the building


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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

o connects all the structural walls to improve the integral 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 termed 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

, when foundation soil is soft or has uneven properties.

Therefore, provide 150mm thick band and four bars of steel tied with stirrups 6mm dia at 150 c/c.




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.