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IJSRD - International Journal for Scientific Research & Development| Vol. 7, Issue 02, 2019 | ISSN (online): 2321-0613
Analysis, Design and Estimation of RC Shear Walls G Plus 13 Multi-
Storied Residential Building
1 2 3 4 5
L Ravi Kumar K. V. Ganesh T Bhanu Prakash P Geetha Afjal Basha
1Associate Professor 2,3,4,5BE Student
1,2,3,4,5Department of Civil Engineering
1,2,3,4,5Kuppam Engineering, Ananatapur, A.P, India
Abstract— Computer Aided “Analysis, design and estimation we require formwork of greater strength, which means the
of RC Shear Walls G PLUS 13 Multi-Storied Residential conventional formwork, is not suitable for the construction.
Building” involves analysis of building frames by using Hence we require an aluminum formwork which is called
STAAD Pro. Conventional method of analysis involves lot of mivan shuttering.
complications and tedious calculations such analysis is a time The proposed building is G+13 stories R.C.C framed
consuming task. Analysis can be made quickly by using structure. The building consists of four flats in each floor and
software’s. STAAD Pro is the leading design software in the the total flats in the building are 56 numbers. This project
market. Many design companies use this software for their which is going on in Bangalore comprises of development of
project design purposes. Hence this project mainly deals with residential Building along with other necessary utilities. This
the analysis of the building by using STAAD Pro , drafting document pertains to the structural designs carried out for a
by AutoCAD, Architecture design by REVIT architecture part of above said residential township project for various
and Estimation done by the MS excel. structures. The development is in the seismic Zone – II. The
Keywords: STAAD Pro, REVIT Architecture, RC Wall Load basic wind speed at location of the development is 33 m/s.
SBC of soil according to soil investigations is 250 KN/m2.
I. INTRODUCTION The design parameters considered are as per Indian Standard
Day to day increase in the industrial growth and population Code of practice.
established the problem of constructing number of buildings Block 3 Block 2 Block 1
either for the residence or for office or for industry. On Living room 5.1x3.20 6.5x4.7 5.5x3.3
account of the high cost involved in acquiring land especially Bed room 1 4.1x3.2 4.5x3.7 3.9x3.3
in densely populated areas, there is invariably a need for the Bed room 2 4.4x3.2 3.7x3.2 4.0x3.2
construction of multi-storied buildings. So, the advanced Bed room 3 4.5x4.2
knowledge in technology especially knowledge in reinforced Dining/kitchen 4.1x3.2 4.2x2.7 5.6x3.2
cement concrete has come to the rescue of the engineer for Bathroom 1 1.8x1.4 2.4x1.2 2.1x1.5
planning and designing multi-storied buildings. The stage has Bathroom 2 2.3x1.2 2.5x1.2 2.3x1.5
come in which multi-storied construction is essential and Shaft 1 1.2x0.7 3.1x1.5 1.5x0.7
inevitable. Shaft 2 3.2x0.9
In this project, the concept of monolithic Utility 3.2x1.0 2.0x1.2 3.2x1.2
construction technology is adapted. That means the whole
structure along with the slab is casted at a single pour [at a Balcony 2.0x1.2 3.3x1.2
time]. In order to construct a monolithic structure obviously Table 1: Dimensions of the rooms
Fig. 1: Building layout
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Analysis, Design and Estimation of RC Shear Walls G Plus 13 Multi-Storied Residential Building
(IJSRD/Vol. 7/Issue 02/2019/427)
II. LITERATURE REVIEW E. A Study on Construction of RC Shear Walls for Multi-
Storied Residential Building-by A. Shiva Shankar, S. Sunil
A. IS: 875 (Part 1) for Dead Loads: Prathap Reddy
Indian Standard Code of Practice For Design Loads (Other A study has been carried out to determine the strength of RC
Than Earthquake) For Buildings and Structures. The dead shear wall of a multistoried building by changing shear wall
load comprises of the weights of walls, partitions floor location. Three different cases of shear wall position for a
finishes, false ceilings, false floors and the other permanent multi-story building have been analyzed. Incorporation of
constructions in the buildings. The dead loads may be shear wall has become inevitable in multi-store building to
calculated from the dimensions of various members and their resist lateral forces
unit weights. The unit weights of plain concrete and III. ANALYSIS AND DESIGN
reinforced concrete made with sand and gravel or crushed
natural stone aggregate may be taken as 24 KN/m3 and 25 A G+13 floor residential building is considered whose
3
KN/m respectively. architectural plan and structural framing plans were prepared
B. IS: 875 (Part 2) for Imposed Loads: as shown in figure below, before it is modeled in STAAD Pro.
Indian Standard Code Of Practice For Design Loads (Other The entire analysis of building has been done in one stage
Than Earthquake), For Buildings And Structures, Imposed keeping the IS code provision in view wherever necessary.
load is produced by the intended use or occupancy of a The whole building has been split into its structural
building including the weight of movable partitions, components viz., slab, beams, columns and footings.
distributed and concentrated loads, load due to impact and A. Statement of the Project
vibration and dust loads. Imposed loads do not include loads The design data shall be as follows:
due to wind, seismic activity, snow, and loads imposed due 2
Live load : 3.0 kN/m
to temperature changes to which the structure will be Floor finish : 1.0 kN/m2
subjected to, creep and shrinkage of the structure, the Location : Bangalore (Zone -II)
differential settlements to which the structure may undergo. Depth of foundation below ground: 3.25 m
C. IS: 875 (Part 3) for Wind Loads: Safe bearing capacity (SBC) of the soil: 250 kN/m2
Indian Standard Code Of Practice For Design Loads (Other Ground floor and first floor height : 3.8 m
Than Earthquake) For Buildings And Structures, This First floor to fourteenth Floor Height: 2.825m
standard gives wind forces and their effects (static and Floors : G + 13 floors.
dynamic ) that should that taken into account when designing Inner and exterior Wall: 150 mm thick reinforced concrete
buildings, structures and components thereof. Wind is air in walls
motion relative to the surface of the earth. 1) Material Properties Concrete:
All components unless specified in design: M25 grade all
The primary cause of wind is traced to earth’s 2) Material Properties Steel:
rotation and differences in terrestrial radiation. The effects are HYSD reinforcement of grade Fe 415 confirming to IS: 1786
primarily responsible for convection either upwards or is used throughout.
downwards. The wind generally blows horizontal to the
ground at high wind speeds. Since vertical components of B. Load Combinations:
atmospheric motion are relatively small the term ‘Wind’ Dead load + Live load
denotes almost exclusively the horizontal wind, vertical Dead load + live load + seismic load
winds are always identified as such. The wind speeds are Dead load + live load + wind load
assessed with the aid of anemometers or anemographs which Dead load + wind load
are installed at meteorological observatories at heights Dead load + seismic load
generally varying from 10 to 30 meters above ground. Live load + wind load
D. IS: 1893 (Part 1) for Earthquake Resistant Design of Partial safety factor for loads: 1.5
Structures: Partial safety factor for steel: 1.15
Indian Standard Criteria for Earthquake Resistant Design of Partial safety Factor for Concrete: 1.5
Structures, (Part 1-General Provisions and Buildings), It deals 1) Dead Load
with assessment of seismic loads on various structures and Dead load consists of the permanent construction material
earthquake resistant design of buildings. Its basic provisions load compressing the roof, floor, wall, and foundation
are applicable to buildings; elevated structures; industrial and systems, including claddings, finishes and fixed equipment.
stack like structures; bridges; concrete masonry and earth Dead load is the total load of all the components of the
dams; embankments and retaining walls and other structures. building (or) structure.
Temporary elements such as scaffolding, temporary Weight = volume x density
Slab weight = 0.15x25 = 3.75 kN/m2
excavations need not be designed for earthquake forces. 2
Self-weight of Floor finishing = 0.01x25 = 0.25 kN/m
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Analysis, Design and Estimation of RC Shear Walls G Plus 13 Multi-Storied Residential Building
(IJSRD/Vol. 7/Issue 02/2019/427)
k1 = risk coefficient
k2 = terrain roughness and height factor
k3 = topography factor
k4 = Importance factor for cyclonic region.
Basic wind speed in Bangalore 33 m/s
Fig. 2: Dead load
2) Live load
Live loads are produced by the use and occupancy of a
building. Load includes those from human occupants,
furnishings, no fixed equipment, storage, and construction
and maintenance activities. In STAAD we assign live load in Fig. 4: wind load
terms of UDL. We have to create a load case at live load and 4) RC Wall Load
select all the beams to carry such load. After the assignment Floor load is calculated based on the load on the RC WALL.
of the live load the structure appears as shown below. Assignment of floor load is done by creating load case for RC
Live load is taken as 3 kN/m2 WALL load. After the assignment of RC WALL load of a
structure look as shown in below.
The intensity of RC WALL load is taken as 3 .75 kN/m2
Fig. 3: live load
3) Wind Load Fig. 5: shear wall load
In the list of loads we can see wind load is present both in
vertical and horizontals loads. This is because wind causes IV. MANUAL DESIGN OF REINFORCED STRUCTURES
uplift of the roof by creating a negative pressure on the top of
the roof. Wind produces non static loads on a structure at A. Design of Beam:
highly variable magnitudes. The variation in pressure at Beam is the horizontal member of a structure, carrying
different locations on a building is complex to the point that transverse loads. Beams are rectangular in cross-section.
pressure may become too analytically intensive for precise Beam carries the floor slab or the roof slab. Beam transfer all
consideration in design. the loads including its self –weight to the columns or walls.
a) Design Wind Pressure: Beam is subjected to bending moments and shear forces.
It can be mathematically expressed as follows It is typically used for resisting vertical loads, shear forces
Vz = Vb x k1x k2 x k3 x k4 {Pg.no 5 clause no 6.3 of IS 875- and bending moments.
2015} Types of Beams
pz = 0.6Vz2 {Pg.no 9 clause no 7.2 of IS 875-2015} 1) Simply supported beam
pd = Kd x Ka x Kc x pz {Pg.no 9 clause no 7.2 of IS 875- 2) Fixed beam
2015} 3) Cantilever beam
0.7 pz {Pg.no 9 clause no 7.2 of IS 875-2015} 4) Continuous beam
Design Wind Pressure is Maximum of ( pd , 0.7pz) 5) Overhanging beam
Where Vb = design wind speed at height Z in m/s {Pg.no 51 6) We are considering the fixed beam.
Annexure A of IS 875 Part3-2015}
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1553
Analysis, Design and Estimation of RC Shear Walls G Plus 13 Multi-Storied Residential Building
(IJSRD/Vol. 7/Issue 02/2019/427)
1) Details of Beam Materials: Length available beyond the actual cutoff point from center
For M25 Concrete, Fck = 25 N/mm2 of supports
2 l 4
For Fe 415 Steel, fy = 415 N/mm eff 1.85 1.850.15m
2 2
Clear span l = 7.4 m
Effective length leff = 7.26+0.23 = 7.49m Provision of shear reinforcement
D = 600 mm w l 73.24
Vu = u eff
Mu = 361.73 KN-m 2 2
Moment of Resistance
X max = 146.4KN
Xumax u 2 Design shear force at distance d = 412 face to support at a
Mu,lim 0.36 10.42 bd fck distance of
d d
l support width
= eff dCentre of beam
2 2
d = 520mm
4 0.23
D = 600mm = 0.412
2 2
Assuming ϕ 20 for main reinforcement and ϕ8for shear = 1.47m
reinforcement From similarity of triangle
146.42
Vu = 1.873= 137.10 KN
4
D 2D
Breath of the beam between and V 137.10103
u
3 3 c bd 230412
B = 450 mm 2
Hence provide b = 450 mm = 0.426N/mm
Tensile reinforcement (Ast) % Ps = 1002314.1= 0.66%
A f 230412
st y
Mu 0.87fy Ast d 1
bdfck = 0.42
c
D = 412mm = (412+20+8) = 440mm
Ast = 2004.79 mm2 At the upper lever of bars does not continue up to the critical
Curtailment and detailing of reinforcement section for shear force.
V=0.87fyAsvd=0.87415100412
16 ϕ of 8 bars S S 1.78103
A f v
st y
Mu 0.87fy Astd 1 =8356.269mm
bdf
ck Sv, min = 0.75×d = 0.75×412 = 309mm
(or)
Sv, min = 300mm
Mu = 308.98106N-mm Sv, min given by the following formula
Asv 0.4
Let theoretical cut off section from support=x
bS 0.87f
vmin y
100 0.4
Mu =
230S 0.87415
X = 1.00 vmin
Sv min 1000.87415
2300.4
= Sv min =300mm
from center of the beam Provide 2-Legged stirrups ϕ8 @ 300mm c/c
Actually upper bars will be curtailed at 2.745+ (12 ϕ or d (or) Check shear strength at cut off point as per Cl 26.2.3.2(a)
whichever greater) from Centre of the beam Shear at cut off point =
= 2.745+0.412 W l l
= u ef ef
= 3.157 w 1.85
2 u 2
The curtailment of upper bars will be done at 3.157 m from
Centre of the beam
=73.24 4
73.2 1.85
φσ
LD = st
2 2
4τ
bd = 135.4KN
200.87415
= Shear strength of the section with ϕ8 @300mmc/c stirrups
41.61.2 2
=
τ bdV
= 0.940 < 2.02m 3 c us
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