International Research Journal of Engineering and Technology (IRJET) e-ISSN: 2395-0056
Volume: 07 Issue: 05 | May 2020 www.irjet.net p-ISSN: 2395-0072
Structural Design of Reinforced Concrete Shear wall
Darshitkumar R. Gohel
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Abstract - Shear walls are specially designed structural
walls incorporated in buildings to resist lateral forces that
are produced the plane of the wall due to wind, earthquake
and other forces. The term 'shear wall' is rather misleading
as such walls behave more like flexural members. They are
usually provided in tall buildings and have been found to be
of immense use to avoid total collapse of buildings under
seismic forces. It is always advisable to incorporate them
built in regions likely to experience earthquake of large
intensity or high winds. Shear walls for wind are designed as
simple concrete walls.
Key Words: Reinforced concrete shear wall, Axial force,
Bending moment, Shear force, Base Moment, Base Shear,
Shear strength.
1. INTRODUCTION
Properly designed and detailed buildings with shear walls Figure 1. Plan
have shown very good performance in past earthquakes.
Shear walls in high seismic regions require special 3. Solution
detailing. However, in past earthquakes, even buildings
with sufficient amount of walls that were not specially 3.1 Seismic weight of the building
detailed for seismic performance (but had enough well-
distributed reinforcement) were saved from collapse. As per the code provisions, the percentage of design live
Shear wall buildings are a popular choice in many load to be considered for the calculation of earthquake
earthquake prone countries, like Chile, New Zealand and forces is 25% for the floors and live load for the roof is not
USA. Shear walls are easy to construct, because to be accounted for.
reinforcement detailing of walls is relatively straight-
forward and therefore easily implemented at site. Shear Hence, the effective weight at each floor will be
walls are efficient, both in terms of construction cost and
= 4.5 + 3.0 + 0.25 × 3.5 = 8.375kN/m2
effectiveness in minimizing earthquake damage in
structural and non-structural elements (like glass windows and that at the roof = 4.5kN/m2
and building contents).
Weight of 144 beams, each of 3.5 m span, at each floor and
2. Problem Statement roof
18-storey building has plan dimensions as shown in Fig. 1. = 0.3×0.6× (3.5×144) ×25
Two shear walls are to be provided in each direction to
= 2268kN
resist the seismic forces. The axial load on the each shear
wall is 15000kN due to both dead and live loads. The Weight of 81 columns at each floor
height between floors is 3.0m. dead load per unit area of
the floor, which consists of floor slab, finishes. etc., is = 0.3×0.6×2.4×61×25
4.5kN/m2 and the weight of partitions on floor is 3kN/m2.
The intensity of live load on each floor is3.5kN/m2 and on = 658.8kN
roof is 1.5kN/m2. The soil below the foundation is hard and Weight of columns at roof = ×658.8
the building is located in Roorkee.
= 329.4kN
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Volume: 07 Issue: 05 | May 2020 www.irjet.net p-ISSN: 2395-0072
Plan area of building is 28 m × 28 m = 784 m2 = (d, the plan dimension = 28m)
Equivalent load at roof level = 4.5×784 + 2268 + 329.4 Building is situated in Roorkee, i.e., in Zone IV. Zone factor
= 6125.4kN Z = 0.24, importance factor I = 1, response reduction factor
Equivalent load at each floor = 8.375×784 + 2268 + 658.8 R = 4.0 for 5% damping and type I soil, average response
= 9492.8kN acceleration coefficient = 1.81 Design horizontal seismic
Seismic weight of the building, W = 6125.4 + 9492.8×17
= 167503kN coefficient Ah = × ×
3.2 Base shear
= = 0.0543
The fundamental natural period of vibration T for the
Base shear Vb = Ah×W = 0.0543×167503 = 9095.41kN
buildings having shear walls is given by T =
Lateral loads and shear forces at different floor level are
given in Table 1.
Table 1 Calculation of lateral loads and shear
Mass No. Wi(kN) hi(m) Wihi2 Qi(kN) Vi(kN)
1 6125.40 54 17861666.40 0.104844 953.60 953.60
2 9492.8 51 24690772.80 0.144930 1318.20 2271.80
3 9492.8 48 21871411.20 0.128381 1167.68 3439.47
4 9492.8 45 19222920.00 0.112835 1026.28 4465.75
5 9492.8 42 16745299.20 0.098292 894.00 5359.75
6 9492.8 39 14438548.80 0.084751 770.85 6130.60
7 9492.8 36 12302668.80 0.072214 656.82 6787.42
8 9492.8 33 10337659.20 0.060680 551.91 7339.33
9 9492.8 30 8543520.00 0.050149 456.12 7795.45
10 9492.8 27 6920251.20 0.040621 369.46 8164.91
11 9492.8 24 5467852.80 0.032095 291.92 8456.83
12 9492.8 21 4186324.80 0.024573 223.50 8680.33
13 9492.8 18 3075667.20 0.018054 164.20 8844.54
14 9492.8 15 2135880.00 0.012537 114.03 8958.57
15 9492.8 12 1366963.20 0.008024 72.98 9031.55
16 9492.8 9 768916.80 0.004513 41.05 9072.60
17 9492.8 6 341740.80 0.002006 18.24 9090.85
18 9492.8 3 85435.20 0.000501 4.56 9095.41
∑Wihi2 = 170363498.4
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3.3 Bending moment and shear force M = (2.28×3) + (9.12×6) + (20.52×9) +(36.49×12) +
(57.01×15) + (82.1×18) + (111.75×21) + (145.96×24) +
Two shear walls are provided in each direction to resist the
(184.73×27) + (228.06×30) + (275.96×33) + (328.41×36) +
seismic forces. Therefore, the lateral forces acting on one
shear wall will be half the calculated shears and is as shown (385.42×39) + (447×42) + (513.14×45) + (583.84×48) +
in Fig. 2. (659.1×51) + (476.8×54)
M = 185908.425kNm
3.4 Loads and material properties
Typical stress resultants for a shear wall obtained From
above analysis part and are as follows:
Taking partial safety factor 1.5
Factored shear force, Vu = 1.5 × 4547.705 = 6821.56kN
Factored bending moment, Mu = 1.5 × 185908.425
= 278862.64kNm
Factored axial load, Pu = 1.5 × 15000 = 22500kN
Axial force on boundary element = 2250kN
The material properties for reinforced concrete shear wall
and reinforcing steel are assumed as follows: Concrete grade
M30, fck = 30MPa, Reinforcement grade HYSD Fe 415
4. General requirements for a shear wall
Thickness of Wall, tw = 300mm (Minimum thickness, as per
Clause 9.1.2, IS13920:1993 shall be 150 mm)
Length of wall, Lw =14000mm
Since wall thickness > 200 mm. vertical as well as horizontal
reinforcement shall be provided in two layers or in two
curtains in the shear wall (Clause 9.1.5, IS 13920:1993)
Maximum diameter or reinforcement < (tw)/10
(Clause 9.1.6, IS 13920:1993)
Hence, maximum reinforcement diameter shall be 30 mm.
Since 30 mm size bars are not available in the market, the
maximum practical reinforcement bar diameter shall be 28
mm.
Maximum spacing of reinforcement shall not exceed the
smaller of the following:
Lw/5 =14000/5 = 2800mm
3tw = 3 × 300 = 900 mm
450 mm
(Clause 9.1.7, IS 13920:1993)
Hence, maximum spacing of any reinforcement ≤450mm
Minimum in-plane reinforcement in the longitudinal and
transverse directions in the shear wall shall be 0.25% of the
respective gross sectional area of the wall.
(Clause 9.1.4, IS 13920:1993)
4.1 Shear strength requirements
Figure 2. Lateral Forces on shear wall Let the nominal shear stress in the wall = τv
The shear wall will be designed as a cantilever fixed at the τv = (Clause 9.2.1, IS: 13920-1993)
base and free at the top. Where,
Maximum shear force at base V = 4547.705kN dw= effective depth of the wall section
Maximum bending moment at base, For rectangular wall sections,
dw= 0.9 × Lw (Clause 9.2.1, IS: 13920-1993)
∴ dw = 0.9 × 14000 = 12600mm
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∴ τv= = 1.8046MPa xu /Lw =
For M30 grade concrete τcmax= 3.5MPa
[Table-20, IS 456:2000] xu’/Lw =
Since τv<τcmax, wall section is adequate forshear.
Assume the minimum 0.25% steel in the wall in the vertical
as well as in the horizontal direction Φ=
For pt = 0.25%, τc= 0.37MPa [Table 19, IS 456:2000]
Since τv>τc,
λ=
Shear reinforcement is required.
4.2 Horizontal shear reinforcement ρ = vertical reinforcement ratio =
Provide horizontal shear reinforcement as per 9.2.5 of = = 0.0030133
IS 13920:1993
Assume 2-legged 12 mm diameter horizontally aligned Φ= = 0.0362654
closed Stirrups along the height of the shear wall,
Spacing of stirrups along the height of shear wall λ= = 0.1785
∴ Spacing of stirrups = Sv = xu /Lw = = 0.4965
∴ Sv= xu’/Lw = = 0.6597
∴ Sv = 189.59 mm c/c
Since xu’/Lw>xu /Lw, Eq.(a) of Annex “A” (Clause 9.3.1)
Minimum horizontal reinforcement = (Ash)min
IS 13920:1993, is applicable.
= 0.0025 of gross sectional area of the wall in elevation
(Clause 9.1.4, IS 13920:1993)
4.5 Moment of resistance of Rectangular Shear wall
∴ (Ash)min = 0.0025 ×300 × 1000 = 750 mm2
section
Hence, provide 2-legged 12 mm diameter horizontally
aligned closed stirrups at 250mm c/c along the entire height A-1 the moment of resistance of a slender rectangular shear
of the shear wall. wall section with uniformly vertical reinforcement may be
estimated as follows:
4.3 Vertical shear reinforcement
∴ =
AS per Clause 9.2.6 of IS 13920:1993, the vertical
reinforcement, which shall be uniformly distributed in the
wall section, shall not be less than the horizontal
reinforcement.
Assume 2-legged 12 mm diameter vertically oriented β= = 0.5158 & = 0.088678
stirrups.
Spacing of stirrups = Sv= ∴ = {0.0362654 × [(5.9220469) × (0.293456) –
∴ Sv = 0.063271127]}
∴ Sv = 189.59 mm c/c
∴ = 0.06072964239
Minimum vertical reinforcement = 0.0025 of gross sectional
area of the wall in plan ∴ Muv = 0.06072964239 × 30 × 300 × 140002
(Clause 9.1.4, IS 13920:1993) = 107127.0892kNm
∴ (Asv) = 0.0025 × 300 × 14000 = 10500 mm2 Balance moment to be resisted by the edge reinforcement in
Hence, provide 2-legged 12 mm diameter vertically aligned each shear wall
closed stirrups at 250mm c/c along the entire length of the = (278862.64 – 107127.08)
shear wall. = 171735.56kNm
Asv>Ash, Hence ok.
4.6 Check on boundary elements
4.4 Check for flexural strength
To check the necessity of providing boundary elements in
With reference to Annex 'A' and Clause 9.3.1 of the shear wall (Figure 3).
IS 13920:1993 Gross area of wall section, Ag = 14000×300
= 4200000mm2
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As per (Clause 9.4.5, IS13920:1993), boundary elements
Ixx = = 68.6×1012 mm4
shall be provided with special confining reinforcement
Combined stress at edge of wall: throughout their height.
∴σ= Area of special confining reinforcement
Ash = 0.18 Sh
∴σ=
(Clause 7.4 8, IS 13920:1993)
∴ σ = 5.3571±28.4553 Assume 12 mm diameter closed hoops as confining
Maximum stress = σmax = 33.8124MPa reinforcement with a clear cover of 40 mm
As per Clause 9.4.1 of IS 13920:1993, if the extreme fibre The gross area of the boundary element section
compressive stress in the wall due to factored gravity loads Ag =1400×500 = 700000 mm2
plus factored earthquake loads exceeds 0.2fck The size of the core = Ak= (1400-40 - 40) × (500 -40-40) =
= 0.2×30 1320×420 mm
= 6 < 33.8124MPa Since both dimension of the core is greater than 300mm, a
Hence, boundary elements are required in the shear wall. cross tie will have to be used.
Provide a boundary element of length 1400mm each and If the cross-tie is placed at mid-length of the longer
width 500mm at each edge of the shear wall, Figure 1. dimension of the core then, h = 1320/2 = 660 mm and for
shorter dimension, h = 420/2 = 210 mm.
4.7 Design of boundary elements The spacing of the confining hoops, S, shall not exceed the
smaller of
Boundary element is essentially treated as a column. The 1. Minimum member dimension, i.e. =
vertical reinforcement in boundary elements shall not be
less than 0.80% nor greater than 6%. (Clause 9.4.4, IS 13920 125mm
1993) 2. 100mm.
Adopt 3% vertical reinforcement in the boundary elements. The spacing however, need not be less than 100mm.
Asc provided = 0.03×1400×500 = 21000mm2 (Clause 7.4.6 of IS 13920:1993)
Axial compression load on the boundary element due Hence adopt 100mm spacing of the confining hoops
seismic forces S = 100mm
= (Mu-Muv)/Cw (Clause 9.4.2, IS 13920:1993) Ash = = 225.55 mm2>
Mu = Factored design moment on the entire shear wall 113.09 mm2 (Area of 12 mm diameter bar)
section = 278862.64kNm Hence, 12 mm diameter confining reinforcement is
Muv = Moment of resistance provided by the distributed insufficient.
Vertical reinforcement across the wall section Adopt 20 mm diameter confining hoops at a spacing of 100
= 171735.56kNm mm c/c along the entire height of the boundary element
Cw = center to center distance between the boundary Ash provided = 314 mm2 > 225.55 mm2.
elements along the two vertical edges of the shear wall Hence, ok.
= 12.6 m
Axial compressive load = 5. Detailing of reinforcement
= 8502.15kN The detailing of reinforcement in the shear wall is shown in
Required axial load capacity of boundary element = axial Figure 4.
load due to gravity effects + axial load due to seismic forces =
2250 + 8502.15 = 10752.15kN REFERENCES
As per clause 9.4.2, IS 13920:1993, the boundary element
shall be assumed to behave as an axially loaded short
[1] IS 13920:1993 Ductile detailing of reinforced concrete
column.
If the design strength of an axially loaded short column = PuD, structures subjected to seismic forces - Code of practice
then [CED 39: Earthquake Engineering].
0.4fck Ac + 0.674fyAsc [2] IS 456 (2000): Plain and Reinforced Concrete - Code of
PuD = 0.447fck Ag + (fc– 0.447 fck )Asc Practice [CED 2: Cement and Concrete]
fc= 0.790 fy(for Fe415 steel) [3] IS 1893-1 (2002): Criteria for Earthquake Resistant
PuD = 0.447×30×1400×500+ (0.790×415 – Design of Structures, Part 1: General Provisions and
0.447×30) ×21000 Buildings [CED 39: Earthquake Engineering]
= 15990.24kN> 10752.15kN, Hence ok [4] IS:4326-1993: Earthquake Resistant Design and
Construction of Buildings
Area of steel for each boundary element = 21000mm2
[5] SP:22(S&T)-1982 : Indian Standard Explanatory
Provide 10 nos 40Φ + 12 nos 32Φ in each element
Handbook On codes for Earthquake Engineering -
Asc provided = 22217.34 mm2>21000 mm2 Hence, Ok IS:1893-1975 & IS:4326-1976
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International Research Journal of Engineering and Technology (IRJET) e-ISSN: 2395-0056
Volume: 07 Issue: 05 | May 2020 www.irjet.net p-ISSN: 2395-0072
[6] IITK-BMTPC Earthquake Tips: Learning Seismic Design
and Construction, IIT Kanpur, India.
BIOGRAPHIES
Darshitkumar Rameshbhai
Gohel,
He has completed his Bachelor of
civil engineering with first-class
with distinction from R. H. Sapat
College of Engineering, Nasik,
Maharashtra. (Savitribai Phule
Pune University). He is skilled in
managing Project Control activities
inclusive of Procurement,
Tendering & Estimation, Contracts,
Quantity Surveying, Billing &
Estimation and Project Cost
Control.
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