Analysis and Design of Earth Quake Resistant Masonry Building

Analysis and Design of Earth Quake Resistant Masonry Building

A. Thulasi (1), Dr. N.Ramesh (2) , Dr. S. Gunasekar (3)

(1) Research Scholar, Civil Engineering Department, K.S.Rangasamy College of Technology, Tiruchengode, TamilNadu, India

(2) Professor, Civil Engineering Department, K.S.Rangasamy College of Technology, Tiruchengode, Tamil Nadu, India.

(3) Assistant Professor, Civil Engineering Department, K.S.Rangasamy College of Technology, Tiruchengode, Tamil Nadu, India

Abstract – Masonry buildings are among the most widely used construction systems for housing worldwide. Brick and stone masonry structures are preferred due to theirdurability, fire resistance, thermal performance, and architectural flexibility. The easy availability of materials, cost-effectiveness, and adaptability to various environmental conditions make masonry construction common in rural, urban, and hilly regions. Despite these advantages, masonry buildings exhibit poor seismic performance. Post-earthquake investigations have consistently shown that masonry structures are highly vulnerable and have suffered extensive damage during major earthquakes such as Bhuj (2001), Chamoli (1999), Jabalpur (1997), Killari (1993), Uttarkashi (1991), and BiharNepal (1988). A significant proportion of casualties in these events resulted from the collapse of low-strength, non-engineered masonry buildings.

Although earthquake engineering knowledge has advanced over the past three decades, a comprehensive and standardized procedure for the seismic analysis and design of masonry buildings remains inadequately developed and insufficiently addressed in the Indian academic curriculum, despite nearly 90% of the population residing in masonry structures. This study presents a systematic procedure for the seismic analysis and design of a two-storeyed residential masonry building. The methodology is structured into distinct steps to enhance clarity and promote confidence in designing masonry buildings as engineered structures. The analysis and design are carried out using STAAD.Pro software.

Key Words: Two Story Masonry Building, Seismic Performance Evaluation, Shear Walls, Reinforced Masonry Building, STAAD Pro V8i.

1. INTRODUCTION

   Recent seismic events in India and worldwide have demonstrated the significant vulnerability of masonry buildings to earthquake-induced lateral forces. Although masonry construction remains the most prevalent housing system due to its durability, thermal efficiency, fire

   resistance, cost-effectiveness, and availability of local materials, its inherent characteristicssuch as low tensile strength, brittle behavior, and limited ductilityresult in poor seismic performance. Post-earthquake damage assessments from Bhuj (2001), Chamoli (1999), Jabalpur (1997), Killari (1993), Uttarkashi (1991), and BiharNepal (1988) indicate that the majority of structural failures and associated casualties were attributed tonon-engineered, low-strength unreinforced masonry (URM) buildings. In India, masonry structures are generally designed in accordance with IS 1905, while seismic actions are evaluated as per IS 1893 (Part 1). However, comprehensive analytical methodologies for seismic evaluation and design of masonry buildings are still inadequately addressed in practice and academia, despite the high percentage of the population residing in such structures.

   The seismic response of masonry buildings is governed by parameters such as material strength, stiffness characteristics, mass distribution, geometric regularity, boundary conditions, and foundation system. Improved seismic resistance can be achieved through the use of higher-grade cement-sand mortar, regular and symmetric plan configurations to minimize torsional irregularities, controlled distribution of openings to maintain shear capacity, and appropriate height-to-thickness ratios to prevent out-of-plane instability. Ensuring a continuous and reliable load path through diaphragms, shear-resisting walls, and foundation elements is critical for effective force transfer. Reinforced masonry shear walls function as vertical cantilever elements resisting in-plane shear, bending, and overturning moments, with boundary reinforcement enhancing ductility and energy dissipation capacity. This study presents a systematic procedure for the seismic analysis and design of a multi-storeyed masonry residential building, demonstrating the feasibility of treating masonry construction as an engineered structural system through analytical modeling and code-based design approaches.

2. LITERATURE REVIEW

   Hongwei Qi, Xiaoning Huang (2011): Seismic Reinforcement of Existing Masonry Structure on Conceptual Design:

   He Summarized that Earthquake-resistance behaviors of existing masonry structures are evaluated based on a certain masonry building. The existing masonry structure is evaluated in its defects of primary design and construction measurements, seismic bearing capacity and durability. The masonry building is reinforced for avoiding collapse due to the weak positions of its bottom and top layers, stair halls, short walls and wall corners. Evaluation results and the corresponding seismic reinforcement design are significant for safety of both existing and newly designed masonry structures in earth-quake areas. The strengthen building prolong its life, decrease the damage and collapse in severe and strong earthquakes, so as to protect human life and property.

   Matthew J. DeJong(2009): Seismic Assessment Strategies for Masonry Structures:

   He summarized that Masonry structures are vulnerable to earthquakes, but their seismic assessment remains a challenge. This dissertation develops and improves several strategies to better understand the behavior of masonry structures under seismic loading, and to determine their safety. The primary focus is on historic arched or vaulted structures, but more modern unreinforced masonry structures are also considered. Assessment strategies which employ simplified quasi-static loading to simulate seismic effects are initially addressed. New analysis methods which focus on stability or strength are presented, and the merits of these strategies are clarified. First, a new parametric graphical equilibrium method is developedwhich allows real-time analysis and illuminates the complex stability of vaulted masonry structures.

   Willis, C. R.1 Griffith, M. C.2 Lawrence, S. J.3(2009): Earthquake Design of Unreinforced Masonry Residential Buildings up to 15 meters in Height:

   This report presents deemed to satisfy solutions for earthquake loading of unreinforced masonry structures upto 15 m in height. The structures under analysis are Importance Level 2 buildings and are hence subject to Earthquake Design Category II (EDC II). This study considers the wall forces and associated actions due to earthquake loads. Out-of-plane bending tends to govern as wall span (L), site sub-soil class, hazard factor (Z) and the number of levels increase. This applies for both the office building and the home unit building. This finding is based on the failure criterion of the strength of walls in two-way bending being exceeded. Out-of-plane shear governs in relatively few cases and, when it occurs, it is in conjunction with out-of-plane bending and/or in-plane shear failure. There is no difference in this respect between the office building and the home unit building. This study sets out to identify which seismically induced actions are critical to the life-safety design objective embodied in the earthquake loading code. Out of-plane wall actions are related to the earthquake induced accelrations and in plane wall actions are related to the earthquake induced in-plane shear force. The 15 m height limit for the study means that typical structures up to and including five stories in height are considered. This also covers all domestic construction

   (Building Class 1a and 1b), whether less than or greater than

   8.5 m tall.

   C.V.R.Murthy(2012):IITK-BMTPC Earthquake Tipsonthe Behavior of Reinforced Concrete (RC) and Masonry Buildings

   He summarized that Masonry buildingsare brittle structures and one of the most vulnerable of the entire building stock under strong earthquake shaking. The large number of human fatalities in such constructions during the past earthquakes in India corroborates this. Thus, it is very important to improve the seismic behaviour of masonry buildings. A number of earthquake-resistant features can be introduced to achieve this objective. Ground vibrations during earthquakes cause inertia forces at locations of mass in the building. These forces travel through the roofandwalls to the foundation. The main emphasis is on ensuring that these forces reach the ground without causing majordamage or collapse. Of the three components of a masonry building (roof, wall and foundation) , the walls are most vulnerable to damage caused by horizontal forces due toearthquake. Awall topples down easily if pushed horizontally at the top in a direction perpendicular to its plane (termed weak direction), but offers much greater resistance if pushed along its length (termed strong direction). The ground shakessimultaneously in the vertical and two horizontal directions during earthquakes (IITK-BMTPC Earthquake Tip 5). However, the horizontal vibrations are the most damaging to normal masonry buildings. Horizontal inertia force developed at the roof transfers to the walls acting either in the weak or in the strong direction. If all the walls are not tied together like a box, the walls loaded in their weak direction tend to topple. To ensure good seismic performance, allwalls must bejoined properly to the adjacent walls. In this way, walls loaded in their weak direction can take advantage of the good lateral resistance offered by walls loaded in their strong direction. Further, walls also need to be tied to the roof and foundation to preserve their overall integrity.

   LITERATURE

   SURVEY

   ANALYSIS

   PACKAGE

   MANUAL

   CALCULATION

   RESULT AND

   DISCUSSION

   ANALYSIS AND

   DESIGN

3. METHODOLOGY

4. WORK PROCESS

   1. BASIC DATA

      1. Type of Building Residential Building

      2. Type of Structure Two Storey Masonry Building

      3. Floor to Floor Height 3.5 m

         1. PLAN OF RESIDENTIAL BUILDING

            Fig1. Building Plan

            Fig2(a). North Wall Elevation

            Fig2(b): South Wall Elevation

         2. MATERIAL STRENGTH

            1. Permissible compressive strength (fm) = 2.5 N/mm2 (Assuming unit strength = 35 MPa and mortar H1type)

               In code IS 4326 : 1993 specifies that well burnt bricks and solid concrete bricks possessing a compressive / crushing strength < 35 MPa shall be used.

            2. Permissible stresses in steel in tension = 0.05 fy.

            ( Use high strength deformed bar ( Fe 415)

            i.e. fy = 230 N/mm2.)

         3. LOAD

         4.4 (a) LIVE LOAD DATA

         Live load on roof = 1.0 kN/m2 ( for seismic calculation = 0). Live load on floor = 1.0 kN/m2

         4.4 (b) 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 wall = 250 mm. Weight of wall = 5 kN/m2. (Assuming weight density of masonry 20 kN/m3.)

         (Wr) Weight at roof level	360+460+0	820 kN
         DL and LL at floor level
         Weight of floor	3 x 8 x 15	360 kN
         Weight of walls (Assume half weight of walls at second storey and half weight of walls at first storey lumped at roof).	2 x 1/2 { 2(8+15) x 4 x5}	920 kN
         Weight of live load	1 x 8 x 15	120 kN
         ( Wf) Weight at second level	360+920+120	1400 kN
         Total seismic weight of building (Wr) + ( Wf)	820 + 1400	2220 kN

         1. (c) SEISMIC DATA

            1. Seismic zone = V

            2. Zone factor ( Z) = 0.36 , Zonefactorgiven is for maximum considered earthquake (MCE) and service life of structure in a zone. The factor 2 is used so as reduce the maximum consider earthquake (MCE) zone factor to the factor for design basis earthquake

            3. Importance factor (I) = 1

            4. Response reduction factor ( R) = 3 ( as per IS 1893 ( Part 1) : 2002 )

            5. Soil medium type, for which average responseacceleration co-efficient are as

            6. Direction of seismic force = E-W direction.

         2. LOAD CALCULATION

         STEP: 1 DETERMINATION OF DESIGN LATERAL LOAD

         For determination of lateral load earthquake Equivalent Static lateral forces procedure is adopted.

         1. SEISMIC WEIGHT CALCULATION

            The seismic weight of each floor is its full dead load plus appropriate amount of imposed load. While computing the seismic weight of each floor, the weight of walls in any storey shall be equally distributed to the floors above and below the storey. The weight of live load for seismic calculation is taken as zero.

         2. TIME PERIOD CALCULATION

            The approximate fundamental natural period of a masonry building can be calculated from the clause 7.6.2 of IS 1893 (Part 1) : 2002 as,

            where ,

            h = height of building in m, {i.e. 4.0 (first storey) + 4.0 (second storey) = 8.0 m}

            d = Base dimension of building at the plinth level in m, along the considered

            direction of lateral force (i.e. 8m, assuming earthquake in E-W direction.

            Table -1: Seismic Weight Calculation

            Description	Load calculation	Total
            DL and LL load at roof level
            Weight of roof	3 x 8 x 15	360 kN
            Weight of walls (Assume half weight of walls at second storey is lumped at roof)	1/2{2(8+15) x 4 x 5}	460 kN
            Weight of live load (For seismic calculation, LL on roof is zero)	0 x 8 x 15	0 kN

            The total design lateral base shear (VB) along the direction of motion is given by

            VB = AhW = 0.15 x 2220 = 333 kN.

         3. DISTRIBUTION OF BASE SHEAR TO DIFFERENT FLOOR LEVELS

         The design lateral base shear (VB) is distributed along the height of building as per the following expression.

         Where ,

         Qi = Design lateral force at floor I, Wi = Seismic weight of floor I,

         hi = Height of floor I measured from base.

         n = Number of storeys in the buildings is the number of levels at which mass are located.

         Lateral Force at Roof Level

         = (333x 820 x 82) / (820 x 82 + 1400 x 42)

         = 233.35 kN

         Lateral Force at Roof Level

         = (333 x 1400 x 42) / (820 x 82 + 1400 x 42)

         = 9.65 kN

         (a) (b)

         Fig:4.5 (a) Elevation of Building (b) Seismic load or Storey shear

         STEP: 2 DETERMINATIONS OF WALL RIGIDITIES

         Rigidity of North Shear Wall

         wall = solid wall(c) – strip A(c) + 1,2,3,9,4(f)

         1,2,3,9,4(f) = 1/ R1,2,3,9,4(f)

         R1,2,3,9,4(f) = R1,2,3,9(f) + R4(f)

         R1,2,3,9(f) = 1/1,2,3,9(f)

         1,2,3,9(f) = solid1,2,3,9(c) – stripB(f) + 1,2,3(f)

         1,2,3(f) = 1/(R1(f) + R2(f) + R3(f) )

         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)}

         Rsolid c) = Et / {4(4/8) 3 + 3(4/8)}

         = 0.5 Et

         solid(c) = 2.0/ Et

         RstripA(c) = Et / {4(2.5/8)3 + 3(2.5/8)}

         = 0.944 Et

         stripA(c) = 1.06/Et

         Rsoild1,2,3,9(f) = Et / {(2.5/6)3 + 3(2.5/6)}

         = 0.756 Et

         soild1,2,3,9(f) = 1.322/ Et

         RstripB(f) = Et / {(1/6)3 + 3(1/6)}

         = 1.98 Et

         stripB(f) = 0.546/Et

         (R1(f) + R2(f) + R3(f) ) = Et / {(1/1) 3 + 3(1/1)}

         = 0.25 Et

         1,2,3(f) = 1.33 / Et

         1,2,3,9(f) = 1.322/ Et 0.5046/ Et + 1.33/ Et

         = 2.15/ Et R1,2,3,9(f) = Et/2.15 = 0.465 Et

         R4(f) = Et / {(2.5/1)3 + 3(2.5/1)}

         = 0.043 Et

         1,2,3,9,4(f) = 1.968/ Et

         wall = 2.0/ Et 1.06/ Et + 1.96/ Et

         = 2.908/ Et

         Rwall = 0.343 Et

         Rigidity of South Shear Wall

         wall = solid wall(c) – strip A2(c) + 5,6,7(f)

         5,6,7(f) = 1/ R5,6,7(f)

         R5,6,7(f) = R5(f) + R6(f) + R7(f)

         R5(f) = R7(f) = Et/{(1/1) 3 + 3(1/1)}

         = 0.615 Et

         R5,6,7(f) = 2 x 0.25 Et + 0.615 = 1.115 Et

         5,6,7(f)

         Rsolid(c)

         = 1/ R5,6,7(f) = 0.896/ Et

         = Et/{4(4/8) 3 + 3(4/8)}

         = 0.5 Et

         YCM = WY/W = 7.5m from east wall

         Location of the Centre of Rigidity

         The centre of rigidity, XCR and YCR, is calculated by taking

         solid(c) = 2.0/ Et

         RstripA2(c) = Et/{4(1/8)3 + 3(1/8)}

         = 2.612 Et

         stripA2(c) = 0.382/ Et

         wall = solid wall(c) – strip A2(c) + 5,6,7(f)

         = 2/ Et 0.382/ Et + 0.896/ Et

         = 2.513/ Et

         Rwall = 0.398 Et

         Relative Stiffness of Walls

         North shear wall = 0.343/(0.343 + 0.398) = 0.462 South shear wall = 0.398/(0.343 + 0.398) = 0.538

         STEP: 3 DETERMINATIONS OF TORSIONAL FORCES

         Location of the center of Mass

         Centre of mass, XCM and YCM is calculated by taking statical moments about a point, say southwest corner, using the respective weights of walls as forces in the moment summation. Because of symmetrical layout of building, the centre of the mass will occur near the centre of building i.e. XCM = 4.0m, YCM = 7.5m.However for methodology purpose the calculations for the centre of mass is shown in Table

         Table 2 : Calculation of centre of mass

         statical moments about a point, say south-west corner using the relative stiffness of the walls as forces in the moment summation. The stiffness of slab is not considered in the determination of centre of rigidity. The calculation for the rigidity is shown in Table

         Table 3 : Calculation of Centre of Rigidity

         Item	RX	RY	X(m)	Y(m)	Y RX	X RY
         N – Wall	0.462	–	–	15	6.93	–
         S – Wall	0.538	–	–	0.0	0	–
         E – Wall	–	0.5	8.0	–	–	4.0
         W – wall	–	0.5	0.0	–	–	0.0
         	RX = 1.0	RY =1.0			Y RX = 6.93	XRY = 4.0

         XCR = XRY/RY = 4.0m from W- Wall YCR = YRX/RX = 6.93m from S Wall

         Torsional Eccentricity

         Torsional Eccentricity in Y direction

         Eccentricity between centre of mass and centre of rigidity

         	Weight i kN	X(m)	Y(m)	WX(kNm)	WY(kNm)
         Roof slab	8 x 15 x 3 = 360 kN	4.0	7.5	1440	2700
         N – Wall	8 x 4 x 5 = 160 kN	4.0	15	640	2400
         S – Wall	8 x 4 x 5 = 160 kN	4.0	0.0	640	0
         E – Wall	15 x 4 x 5 = 300 kN	8.0	7.5	2400	2250
         W – wall	15 x 4 x 5 = 300 kN	0.0	7.5	0	2250
         	W = 1280			WX = 5120	WY = 9600

         ey = 7.50 6.72 = 0.78m

         XCM = WX/W = 4.0m from west wall

         Add minimum 5% accidental eccentricity

         = 0.05 x 15 = 0.75m

         Total eccentricity = 0.78 + 0.75 = 1.53m Torsional Eccentricity in X direction

         Eccentricity between centre of mass and centre of rigidity ex = 4.0 4.0 = 0.0 m

         Add minimum 5% accidental eccentricity

         = 0.05 x 8 = 0.40m

         Total eccentricity = 0.0 + 0.40 = 0.40 m

         Torsional Moment

         The torsional moment due to E W seismic force rotate the building in Y direction, hence

         MTX = Vx ey = 333 x 1.53 = 509.50 kNm

         Similarly, if considered seismic force in N S direction MTY = Vy ex = 333 x 0.40 = 133.2 kNm

         (Vy = Vx,because Sa/g is considered value of 2.5 for the ime 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 in N S direction will resist the forces and the walls in E W direction may be ignored. Table shows the calculation of distribution of direct shear and torsionalshear.

         Table.4 – Distribution of forces in North and South shear walls

         Ite m	Rx	dy (m)	Rx dy	Rx dy2	Direct shear force (kN)	Torsiona l shear for(kN)	Total shear (kN)
         N – Wall	0.46 2	8.0 7	3.72 8	31.67	153.8 5	+33.94	187.8 0
         S – Wall	0.53 8	6.9 3	3.72 8	24.30	179.1 5	-33.94	179.1 5
         				= 55.9 6

         Distance considered wall from centre of rigidity ( 15 6.93 = 8.07 m)

         Torsional forces in N wall =

         = (3.728 x 509.44)/55.96

         = 33.94 kN

         Torsional forces in S wall =

         = (3.728 x 509.44)/55.96

         = 33.94 kN

         Fig 3 : Torsional Forces

         The torsional forces additive on the north wall and subtractive on the south wall as shown. Since thecodedirects that negative torsional shear shall be neglected. Hence the total shear acing on the south wall is simply direct shearonly.

         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.

         Table.5 North Shear Wall

         Pier Group	Stiffness	Relative stiffness	Shear Force
         1,2,3,9	0.465	0.915	171.80 kN
         Pier 4	0.043	0.085	15.96 kN

         Shear171.8kN in pier group 1,2,3,9 is further divided in vertical piers 1, 2 and 3 in proportion to their stiffness. The stiffness of pier 1,2 and 3 are 0.25 each so the shear force carried by each pier is

         Table.5 North Shear Wall

         Pier	Stiffness	Relative stiffness	Shear Force
         1	0.25	0.33	56.70 kN
         2	0.25	0.33	56.70 kN
         3	0.25	0.33	56.70 kN
         4	0.043	0.085	15.96 kN
         5	0.25	0.225	40.30 kN
         6	0.615	0.55	98.53 kN
         7	0.25	0.225	40.30 kN

         STEP: 4 DETERMINATION INCREASE IN AXIAL LOAD DUE TO OVERTURNING

         Total overturning moment due to lateral force acting on the building is,

         Movt = Total shear (Vx) x vertical distance between second floor level to critical

         plane of weakness, assuming at the level of sill + applied overturning

         Where ,

         Movt = ( 153.85 x 2.5 ) + (107.78 x 4 ) = 815.75

         kN.m

         Increase in axial load due to overturning moment

         moment at second floor level.

         Assume the stiffness of second storey walls is the same as first storey, the total direct shear in E W direction of seismic load i.e. in x direction is divided in North and South shear wall in the proportion to their stiffness

         Direct Shear in North wall (VNX ) = 153.85 kN Direct Shear in South wall (VSX) = 179.15 kN

         Distribution of lateral force along the height of North and South wall is

         North Shear Wall

         Lateral force at roof level =

         = 153.85(820 x 82)/{(820 x 82) + (1400 x 42)}

         = 107.78 kN

         Lateral force at second floor level =

         =153.85(1400x 42)/{(820 x 82) + (1400 x 42)}

         = 46.07 kN

         South Shear Wall

         Lateral force at roof level =

         = 179.15(820 x 82)/{(820 x 82) + (1400 x 42)}

         = 125.56 kN

         Lateral force at second floor level =

         = 179.15(1400x 42)/{(820 x 82) + (1400 x 42)}

         = 53.59 kN.

         Increase in axial load in piers of north shear wall :

         Overturning moment in north wall ( Movt ) is

         Movt = (Total shear at second floor x Critical height ) + ( lateral load at roof level x Storey height )

         liAi = Centroid of net section of wall.

         In = Moment of inertia of net section of wall.

         Table.6 Calculation of centroid of net section of wall

         Pier	Area (Ai) m2	L m	Ail ( m3 )
         1	1 x 0.25 = 0.25	0.5 m	0.125
         2	1 x 0.25 = 0.25	3.0 m	0.750
         3	1 x 0.25 = 0.25	5.5 m	1.375
         4	1 x 0.25 = 0.25	7.5 m	1.875
         	= 1.0		=4.125

         Distance from left edge to centriod of net section of wall =

         4.125 / 1.0 = 4.125 m.

         Table.7 Calculation of moment of inertia of net section of wall

         Pie r	Ai m2	li m	Ai li m3	Ai li2 m4	I = td3/1 2	In=I+ Ai li2
         1	0.25	3.62 5	0.906	3.285	0.02	3.305
         2	0.25	1.12 5	0.281	0.316	0.02	0.326
         3	0.25	1.37 5	0.344	0.472	0.02	0.492
         4	0.25	3.37 5	0.844	2.848	0.02	2.868
         	= 1.0					=6.99=7m 4

         Increase in axial load in individual piers of north wall is determined in table

         Table.7 Increase in axial load in individual piers of north wall

         Pier	Ai li m3	Povt= Movt x (Ai li / In ) kN
         1	0.906	105.58
         2	0.281	32.75
         3	0.344	40.09
         4	0.844	98.36

         Fig -4: Over turning moment in pier of north wall

         Movt = 815.75 kN.m In = 7.0 m4

         Increase in axial load in piers of south shear wall :

         Overturning moment in south wall (Movt) is,

         Movt = ( Total shear at second floor x Critical height ) + ( lateral load at roof level x storey height )

         Movt = ( 179.15 x 2.5 ) + ( 125.56 x 4.0 ) = 950.12 kN.m

         Increase in axial load due to overturning moment

         Table.8 Calculation of centroid of net section of wall

         Pier	Ai m2	l m	Ail m2
         5	0.25	0.5	0.125
         6	0.50	4.0	2.0
         7	0.25	7.5	1.875
         	= 1.0		= 4.0

         Distance from left edge to centroid = 4.0 / 1.0 =4.0 m

         Table.9 Calculation of moment of inertia of net section of wall

         Pier	Ai m2	li m	Ai li m3	Ai li2 m4	I = td3/12	In=I+ Ai li2
         5	0.25	3.50	0.875	3.06	0.02	3.08
         6	0.50	0	0	0	0.04	0.04
         7	0.25	3.50	0.875	3.06	0.02	3.08
         	= 1.0					=6.20 m4

         Increase in axial load in individual piers of south shear wall is determined as in table.

         Table 10 Increase in axial load in individual piers of south shear wall

         Pier	Ai li m3	Povt= Movt x (Ai li / In ) kN
         5	0.875	134.08
         6	0	0
         7	0.875	134.08

         Movt = 950.12 kN.m, ln = 6.20 m4

         Where,

         Aili = centriod of net section of wall.

         ln = moment of inertia of ne section of wall.

         Fig 5 : Over turning moment in pier of south wall

         STEP: 5 DETERMINATIONS OF PIER LOADS, MOMENTS AND SHEAR

         The total axial load (due to dead load, live load and overturning), shear and moment in the individual piers of both the sear walls are calculated in tables

         Table 11 Axial load, moment, shear in piers of North shear wall

         North Wall: First Storey

         Effective loading width of pier = width of pier + ½ of each adjacent opening of pier dead load intensity is calculated as (per meter length of wall)

         North Wall : First Storey

         Weight of first storey(from level of IInd to sill level)

         = 2.5 x 0.25 x 20 = 12.5 kNm

         Weight of second storey= 4 x 0.25 x 20 = 20 kNm Weight of floor at IInd storey level(Assume North and South shear wall will equal amount of load)

         Pie r	Effecti ve width of pier	Pd1 (kN)	PL2 (kN )	Povt(k N)	Shear VE for mome nt	Moment(kN m) = VE x h/2
         1	1.75	135.6 2	26.2 5	105.58	56.70	56.70 x ½ =28.35
         2	2.5	193.7 5	37.5 0	32.75	56.70	56.70 x ½ =28.35
         3	2.25	174.3 7	33.7 5	40.09	56.70	56.70 x ½ =28.35
         4	1.5	116.2 5	22.5 0	98.36	15.96	15.96 x 2.5/2 =28.35

         = ½ (0.12 x 15 x 25) = 22.5 kNm

         Weight of roof = ½ (0.12 x 15 x 25) = 22.5 kNm

         Total Load = 77.5 kNm

         Table 12 Axial load, moment, shear in piers of South shear wall

         South Wall: First Storey

         South Wall: First Storey

         Weight of first Storey (from level of IIndto sill level)

         = 2.5 x 0.25 x 20 = 12.5 kNm

         Weight of second Storey = 4 x 0.25 x 20 = 20 kNm Weight of floor at IInd storey level(Assume North and South shear wall will equal amount of load)

         = ½ (0.12 x 15 x 25) = 22.5 kNm

         Weight of roof = ½ (0.12 x 15 x 25) = 22.5 kNm

         Total Load = 77.5 kNm

         II. PL = effective loading width of pier x live load intensity in kN/m

         Effective loading width of pier = width of pier + ½ of each adjacent opening pier live load intensity (per meter length of wall) calculated as

         North Wall: First Storey

         Pie r	Effecti ve width of pier	Pd1 (k N)	PL2(k N)	Povt(k N)	Shear VE for mome nt	Moment(k Nm) = VE x h/2
         5	2	155	30	134.0 8	40.30	40.30 x ½ =20.15
         6	4	310	60	0	98.53	98.53x ½ =49.27
         7	2	155	30	134.0 8	40.30	40.30x ½ =20.15

         Live load on floor (1 kN/m2)(Assume North and South shear wall take equal amount of load) = ½ (1 x 15) = 7.5 kN/m

         Live load on roof (1 kN/m2)(Assume North and South shear wall take equal amount of load) = ½ (1 x 15) = 7.5 kN/m

         Total Load = 15 kNm

         South Wall: First Storey

         Live load on floor (1 kN/m2)(Assume North and South shear wall take equal amount of load) = ½ (1 x 15) = 7.5 kN/m

         Live load on roof (1 kN/m2)(Assume North and South shear wall take equal amount of load) = ½ (1 x 15) = 7.5 kN/m

         Total Load = 15 kNm

         I. Pd = effective loading width of pier x dead load intensity in kNm

         STEP: 6 DESIGN OF SHEAR WALLS FOR AXIAL LOADS AND MOMENTS

         North Shear Wall

         Table 13 Determination on jamb steel at the pier boundary

         Pie r	Moment (kNm)	Effective depth (mm)	Area of jamb steel AS*(mm2)	No of bar s	P(kN)
         1	28.35	900	152.17	2 @ 10	267.45
         2	28.35	900	152.17	2 @ 10	264
         3	28.35	900	152.17	2 @ 10	248.21
         4	19.95	900	107.08	2 @ 10	237.11

         Table 14 **Check for Adequacy for piers

         Pier	P(kN)	d(m)	t(m)	Fa/Fa	Fb/Fb	Fa/Fa+ fb/Fb
         1	267.45	1	0.25	0.427	0.217	0.644	OK
         2	264	1	0.25	0.422	0.217	0.639	OK
         3	248.21	1	0.25	0.397	0.217	0.614	OK
         4	237.11	1	0.25	0.379	0.153	0.532	OK

         South Shear Wall

         Pie r	Moment(kNm )	Effectiv e depth (mm)	Area of jamb steel AS*(mm2 )	No of bar s	P(kN)
         5	20.15	900	108.15	2 @ 10	309.0 8
         6	49.27	1800	132.23	2 @ 10	370.0 0
         7	20.15	900	108.15	2 @ 10	319.0 8

         Table 15 Determination on jamb steel at the pier boundary

         Table 16 **Check for Adequacy for piers

         Pier	P(kN)	d(m)	t(m)	Fa/Fa	Fb/Fb	Fa/Fa+ fb/Fb
         5	309.08	1	0.25	0.494	0.154	0.648	OK
         6	370.00	2	0.25	0.296	0.094	0.390	OK
         7	319.08	1	0.25	0.456	0.154	0.610	OK

         * Jamb steel at the pier boundary is given by, AS = M / (fs x 0.9 x deffective)

         fs = 0.55 Fe = 0.55 x 415 = 230 N/mm2

         deffective = dtotal Cover

         ** Adequacy of individual piers under compression and moment is checked by interaction formula i.e.

         (fa/Fa) + (fb/Fb) 1.33

         fa = [Ptotal i.e.(Pd+PL+Povt)] / [width of pier (d) x t] fb = M/(td2/6)

         Fa = Permissible compressive stress = 2.5 N/mm2(as per IS:1905)

         Fb = Permissible bending stress = 2.5 + 0.25 x 2.5

         = 3.125 N/mm2 (as per ISs: 1905)

         STEP 7: 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.

         Design of Bond Beam:

         Total seismic shear in E W direction = 333kN Moment prodced (M) = V x L/8 = 333 x 15/8

         = 624.37 kNm T = M/d = 624.37/8 = 78.04kN

         A = T/fs = 78.04 x 1000/230= 339.33 mm2

         Use 2 @ 16 (= 402 ssmm2)

         Fig 6 : Design of Bond Beam

5. STRUCTURAL MODELLING AND DESIGN OF A TWO STOREYED MASONRY BUILDING

In this design, analysis of the two stoyered masonry building is done. Based on the analysis, all the Surfaces are designed for strength and serviceability. STAAD-PRO is used for the analysis.

Fig 7 : Masonry Building 3D view

1. Load Combinations

   Loads on any structure may be arranged in design so that the maximum force or moment is achieved at the point in the structure being considered. Hence all loadcombinations must be considered. Following are the different load combinations used in this design of masonry building.

   1. 1.5 DL+LL

      1. Dead Load + Live Load

   2. 1.2DL+1.2LL+1.2ELX

      1.2Dead Load + 1.2 Live Load + 1.2 Earthquake load at X direction

   3. 1.2DL+1.2LL-1.2ELX

      1.2Dead Load + 1.2 Live Load – 1.2 Earthquake load at X direction

   4. 1.2DL+1.2LL+1.2ELZ

      1.2Dead Load + 1.2 Live Load + 1.2 Earthquake load at Z direction

   5. 1.2DL+1.2LL-1.2ELZ

      1.2Dead Load + 1.2 Live Load – 1.2 Earthquake load at Z direction

   6. 1.5DL + 1.5 ELX

      1.5Dead Load + 1.5 Earthquake load at X direction

   7. 1.5DL- 1.5 ELX

      1.5Dead Load – 1.5 Earthquake load at X direction

   8. 1.5DL+1.5 ELZ

      1.5Dead Load + 1.5 Earthquake load at Z direction

   9. 1.5DL – 1.5 ELZ

      1.5Dead Load – 1.5 Earthquake load at Z direction

   10. 0.9DL + 1.5 ELX

       0.9 Dead Load + 1.5 Earthquake load at X direction

   11. 0.9 DL 1.5 ELX

       0.9 Dead Load 1.5 Earthquake load at X direction

   12. 0.9 DL + 1.5 ELZ

       0.9 Dead Load + 1.5 Earthquake load at Z direction

   13. 0.9DL 1.5 ELZ

   0.9 Dead Load 1.5 Earthquake load at Z direction

   Fig 8 : DL and LL acting on slab

   Fig 9 : Seismic motion acting on the building in X direction

   Fig 10 : Seismic motion acting on the building in Z direction

   Fig 11 : Displacement occurred due to load combinations

   Fig 12 : Displacement occurred on the slab

   Fig 13 : Displacement Deatils

   Fig 14 : Effect of Maximum Moment at X direction

   Fig 15: Effect of Maximum Moment at Y direction

   Fig 16: Effect of Maximum Moment at XY direction

   Fig 17: Effect of Maximum Stress at Top

   3	107.13	1	0.25	0.428	1.43	0.374	OK
   4	60.2	1	0.25	0.2408	1.14	0.461	OK

   Fig 18: Effect of Maximum Stress at bottom

2. DESIGN OF SHEAR WALLS FOR AXIAL LOADAND MOMENTS

North Shear Wall

Table 17- Determination of jamb steel at the pier boundary

Table 18- ** Check for Adequacy of piers

South Shear Wall

Table 19 Determination of jamb steel at the pier boundary

Table 20 **Check for adequacy of piers

* Jamb steel at the pier boundary is given by, AS = M / (fs x 0.9 x deffective)

fs = 0.55 Fe = 0.55 x 415 = 230 N/mm2

deffective = dtotal Cover

** Adequacy of individual piers under compression and moment is checked by interaction formula i.e.

(fa/Fa) + (fb/Fb) 1.33

fa = [Ptotal i.e.(Pd+PL+Povt)] / [width of pier (d) x t] fb = M/(td2/6)

Fa = Permissible compressive stress = 2.5 N/mm2(as per IS:1905)

Fb = Permissible bending stress = 2.5 + 0.25 x 2.5

= 3.125 N/mm2 (as per ISs: 1905)

CONCLUSION

Masonry buildings are the most common type of construction used for all housing around the world. But the post-earthquake survey has proved that the masonry buildings are most vulnerable to and have suffered maximum damages in the past earthquakes. A survey of the affected areas in past earthquakes demonstrated that the major losses of lives were due to collapse of low-strength masonry buildings. Due to the brittleness of the masonry material, lack of ductility, strength and locally used traditional material in a traditional manner without the earthquake-resistant features are the main causes of collapse of building during earthquake.

The present work is a step towards with regard to illustrate a procedure for seismic analysis and design of masonry building. The procedure has been presented by considering each clause as mentioned in IS 1905 and IS 4326:1993 with the help of an example of a three-storeyed residential masonry building.

REFERENCES

1. Seismic Reinforcement of Existing Masonry Structure On Conceptual Design Hongwei Qi1,*, Xiaoning Huang2 The Open Construction and Building Technology Journal, 2011.

2. Behavior of Masonry Walls uder Lateral Loads R.Meli.

3. Seismic Assessment Strategies for Masonry Structures by Matthew J. DeJong

4. Design Of Masonry Structures For Earthquake In Australia Rod Johnston

5. Earthquake Loads & Earthquake Resistant Design of Buildings by Andrew King

6. Earthquake Performance of Unreinforced Masonry Residential Buildings up to 15 m in Heigh WILLIS, C.

   R.1 GRIFFITH, M. C.2 LAWRENCE, S. J.3

7. How to Brick Masonry Structure Behave During Earthquake? C.V.R.Murthy

8. IS 4326, 1993, Indian Standard Code Of Practice for Earthquake Resistant Design And Construction Of Building (2nd Revision).

9. IS 1893 (PART I), 2002, Indian Standard Criteria For Earthquake Resistant Design Of Structures (5th Revision).

10. IS 1905: Code Of

Structural Use Of Unreinforced Masonry (3rRevision)