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 © 2020, IRJET | Impact Factor value: 7.529 | ISO 9001:2008 Certified Journal | Page 1396 Structural Design of Reinforced Concrete Shear wall Darshitkumar R. Gohel ---------------------------------------------------------------------***---------------------------------------------------------------------- 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 have shown very good performance in past earthquakes. Shear walls in high seismic regions require special detailing. However, in past earthquakes, even buildings with sufficient amount of walls that were not specially detailed for seismic performance (but had enough well- distributed reinforcement) were saved from collapse. Shear wall buildings are a popular choice in many earthquake prone countries, like Chile, New Zealand and USA. Shear walls are easy to construct, because reinforcement detailing of walls is relatively straight- forward and therefore easily implemented at site. Shear walls are efficient, both in terms of construction cost and effectiveness in minimizing earthquake damage in structural and non-structural elements (like glass windows and building contents). 2. Problem Statement 18-storey building has plan dimensions as shown in Fig. 1. Two shear walls are to be provided in each direction to resist the seismic forces. The axial load on the each shear wall is 15000kN due to both dead and live loads. The height between floors is 3.0m. dead load per unit area of the floor, which consists of floor slab, finishes. etc., is 4.5kN/m 2 and the weight of partitions on floor is 3kN/m 2 . The intensity of live load on each floor is3.5kN/m 2 and on roof is 1.5kN/m 2 . The soil below the foundation is hard and the building is located in Roorkee. Figure 1. Plan 3. Solution 3.1 Seismic weight of the building As per the code provisions, the percentage of design live load to be considered for the calculation of earthquake forces is 25% for the floors and live load for the roof is not to be accounted for. Hence, the effective weight at each floor will be = 4.5 + 3.0 + 0.25 × 3.5 = 8.375kN/m 2 and that at the roof = 4.5kN/m 2 Weight of 144 beams, each of 3.5 m span, at each floor and roof = 0.3×0.6× (3.5×144) ×25 = 2268kN Weight of 81 columns at each floor = 0.3×0.6×2.4×61×25 = 658.8kN Weight of columns at roof = ×658.8 = 329.4kN
Structural Design of Reinforced Concrete Shear wall
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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
© 2020, IRJET | Impact Factor value: 7.529 | ISO 9001:2008 Certified Journal | Page 1396
Structural Design of Reinforced Concrete Shear wall
Darshitkumar R. Gohel
---------------------------------------------------------------------***----------------------------------------------------------------------
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
have shown very good performance in past earthquakes.
Shear walls in high seismic regions require special
detailing. However, in past earthquakes, even buildings
with sufficient amount of walls that were not specially
detailed for seismic performance (but had enough well-
distributed reinforcement) were saved from collapse.
Shear wall buildings are a popular choice in many
earthquake prone countries, like Chile, New Zealand and
USA. Shear walls are easy to construct, because
reinforcement detailing of walls is relatively straight-
forward and therefore easily implemented at site. Shear
walls are efficient, both in terms of construction cost and
effectiveness in minimizing earthquake damage in
structural and non-structural elements (like glass windows
and building contents).
2. Problem Statement
18-storey building has plan dimensions as shown in Fig. 1.
Two shear walls are to be provided in each direction to
resist the seismic forces. The axial load on the each shear
wall is 15000kN due to both dead and live loads. The
height between floors is 3.0m. dead load per unit area of
the floor, which consists of floor slab, finishes. etc., is
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
roof is 1.5kN/m2. The soil below the foundation is hard and
the building is located in Roorkee.
Figure 1. Plan
As per the code provisions, the percentage of design live
load to be considered for the calculation of earthquake
forces is 25% for the floors and live load for the roof is not
to be accounted for.
= 4.5 + 3.0 + 0.25 × 3.5 = 8.375kN/m2
and that at the roof = 4.5kN/m2
Weight of 144 beams, each of 3.5 m span, at each floor and
roof
= 2268kN
= 0.3×0.6×2.4×61×25
= 658.8kN
= 329.4kN
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
© 2020, IRJET | Impact Factor value: 7.529 | ISO 9001:2008 Certified Journal | Page 1397
Plan area of building is 28 m × 28 m = 784 m2
Equivalent load at roof level = 4.5×784 + 2268 + 329.4
= 6125.4kN
= 9492.8kN
= 167503kN
3.2 Base shear The fundamental natural period of vibration T for the
buildings having shear walls is given by T =
= (d, the plan dimension = 28m)
Building is situated in Roorkee, i.e., in Zone IV. Zone factor
Z = 0.24, importance factor I = 1, response reduction factor
R = 4.0 for 5% damping and type I soil, average response
acceleration coefficient = 1.81 Design horizontal seismic
coefficient Ah = × ×
Lateral loads and shear forces at different floor level are
given in Table 1.
Mass No. Wi(kN) hi(m) Wihi 2
Qi(kN) Vi(kN)
∑Wihi2 = 170363498.4
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
© 2020, IRJET | Impact Factor value: 7.529 | ISO 9001:2008 Certified Journal | Page 1398
3.3 Bending moment and shear force
Two shear walls are provided in each direction to resist the seismic forces. Therefore, the lateral forces acting on one shear wall will be half the calculated shears and is as shown in Fig. 2.
Figure 2. Lateral Forces on shear wall The shear wall will be designed as a cantilever fixed at the
base and free at the top.
Maximum shear force at base V = 4547.705kN
Maximum bending moment at base,
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) +
(184.73×27) + (228.06×30) + (275.96×33) + (328.41×36) +
(385.42×39) + (447×42) + (513.14×45) + (583.84×48) +
(659.1×51) + (476.8×54)
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
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
τv = (Clause 9.2.1, IS: 13920-1993)
Where, dw= effective depth of the wall section For rectangular wall sections, dw= 0.9 × Lw (Clause 9.2.1, IS: 13920-1993) ∴ dw = 0.9 × 14000 = 12600mm
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
© 2020, IRJET | Impact Factor value: 7.529 | ISO 9001:2008 Certified Journal | Page 1399
∴ τv= = 1.8046MPa
For M30 grade concrete τcmax= 3.5MPa [Table-20, IS 456:2000] 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 For pt = 0.25%, τc= 0.37MPa [Table 19, IS 456:2000] Since τv>τc,
Shear reinforcement is required.
4.2 Horizontal shear reinforcement
Provide horizontal shear reinforcement as per 9.2.5 of IS 13920:1993 Assume 2-legged 12 mm diameter horizontally aligned closed Stirrups along the height of the shear wall, Spacing of stirrups along the height of shear wall
∴ Spacing of stirrups = Sv =
∴ Sv=
∴ Sv = 189.59 mm c/c Minimum horizontal reinforcement = (Ash)min = 0.0025 of gross sectional area of the wall in elevation (Clause 9.1.4, IS 13920:1993) ∴ (Ash)min = 0.0025 ×300 × 1000 = 750 mm2 Hence, provide 2-legged 12 mm diameter horizontally aligned closed stirrups at 250mm c/c along the entire height of the shear wall.
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 stirrups.
Spacing of stirrups = Sv=
∴ Sv =
∴ Sv = 189.59 mm c/c Minimum vertical reinforcement = 0.0025 of gross sectional area of the wall in plan (Clause 9.1.4, IS 13920:1993) ∴ (Asv) = 0.0025 × 300 × 14000 = 10500 mm2 Hence, provide 2-legged 12 mm diameter vertically aligned closed stirrups at 250mm c/c along the entire length of the shear wall. Asv>Ash, Hence ok.
4.4 Check for flexural strength
With reference to Annex 'A' and Clause 9.3.1 of IS 13920:1993
xu /Lw =
xu’/Lw =
xu /Lw = = 0.4965
xu’/Lw = = 0.6597
Since xu’/Lw>xu /Lw, Eq.(a) of Annex “A” (Clause 9.3.1) IS 13920:1993, is applicable.
4.5 Moment of resistance of Rectangular Shear wall section
∴ =
= 107127.0892kNm
Balance moment to be resisted by the edge reinforcement in each shear wall = (278862.64 – 107127.08) = 171735.56kNm
4.6 Check on boundary elements
To check the necessity of providing boundary elements in the shear wall (Figure 3). Gross area of wall section, Ag = 14000×300 = 4200000mm2
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
© 2020, IRJET | Impact Factor value: 7.529 | ISO 9001:2008 Certified Journal | Page 1400
Ixx = = 68.6×1012 mm4
∴ σ =
∴ σ =
∴ σ = 5.3571±28.4553 Maximum stress = σmax = 33.8124MPa As per Clause 9.4.1 of IS 13920:1993, if the extreme fibre compressive stress in the wall due to factored gravity loads plus factored earthquake loads exceeds 0.2fck = 0.2×30 = 6 < 33.8124MPa Hence, boundary elements are required in the shear wall. Provide a boundary element of length 1400mm each and width 500mm at each edge of the shear wall, Figure 1.
4.7 Design of boundary elements
Boundary element is essentially treated as a column. The vertical reinforcement in boundary elements shall not be less than 0.80% nor greater than 6%. (Clause 9.4.4, IS 13920 1993) Adopt 3% vertical reinforcement in the boundary elements. Asc provided = 0.03×1400×500 = 21000mm2 Axial compression load on the boundary element due seismic forces = (Mu-Muv)/Cw (Clause 9.4.2, IS 13920:1993) Mu = Factored design moment on the entire shear wall section = 278862.64kNm Muv = Moment of resistance provided by the distributed Vertical reinforcement across the wall section = 171735.56kNm Cw = center to center distance between the boundary elements along the two vertical edges of the shear wall = 12.6 m
Axial compressive load =
= 8502.15kN Required axial load capacity of boundary element = axial load due to gravity effects + axial load due to seismic forces = 2250 + 8502.15 = 10752.15kN As per clause 9.4.2, IS 13920:1993, the boundary element shall be assumed to behave as an axially loaded short column. If the design strength of an axially loaded short column = PuD, then 0.4fck Ac + 0.674fyAsc PuD = 0.447fck Ag + (fc– 0.447 fck )Asc fc= 0.790 fy(for Fe415 steel) PuD = 0.447×30×1400×500+ (0.790×415 – 0.447×30) ×21000 = 15990.24kN> 10752.15kN, Hence ok Area of steel for each boundary element = 21000mm2 Provide 10 nos 40Φ + 12 nos 32Φ in each element Asc provided = 22217.34 mm2>21000 mm2 Hence, Ok
As per (Clause 9.4.5, IS13920:1993), boundary elements shall be provided with special confining reinforcement throughout their height. Area of special confining reinforcement
Ash = 0.18 Sh
(Clause 7.4 8, IS 13920:1993) Assume 12 mm diameter closed hoops as confining reinforcement with a clear cover of 40 mm The gross area of the boundary element section Ag =1400×500 = 700000 mm2 The size of the core = Ak= (1400-40 - 40) × (500 -40-40) = 1320×420 mm Since both dimension of the core is greater than 300mm, a cross tie will have to be used. If the cross-tie is placed at mid-length of the longer dimension of the core then, h = 1320/2 = 660 mm and for shorter dimension, h = 420/2 = 210 mm. The spacing of the confining hoops, S, shall not exceed the smaller of
1. Minimum member dimension, i.e. =
125mm 2. 100mm.
The spacing however, need not be less than 100mm. (Clause 7.4.6 of IS 13920:1993) Hence adopt 100mm spacing of the confining hoops S = 100mm
Ash = = 225.55 mm2>
113.09 mm2 (Area of 12 mm diameter bar) Hence, 12 mm diameter confining reinforcement is insufficient. Adopt 20 mm diameter confining hoops at a spacing of 100 mm c/c along the entire height of the boundary element Ash provided = 314 mm2 > 225.55 mm2. Hence, ok.
5. Detailing of reinforcement
The detailing of reinforcement in the shear wall is shown in Figure 4.
REFERENCES [1] IS 13920:1993 Ductile detailing of reinforced concrete
structures subjected to seismic forces - Code of practice
[CED 39: Earthquake Engineering].
[2] IS 456 (2000): Plain and Reinforced Concrete - Code of Practice [CED 2: Cement and Concrete]
[3] IS 1893-1 (2002): Criteria for Earthquake Resistant Design of Structures, Part 1: General Provisions and Buildings [CED 39: Earthquake Engineering]
[4] IS:4326-1993: Earthquake Resistant Design and Construction of Buildings
[5] SP:22(S&T)-1982 : Indian Standard Explanatory Handbook On codes for Earthquake Engineering - IS:1893-1975 & IS:4326-1976
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
© 2020, IRJET | Impact Factor value: 7.529 | ISO 9001:2008 Certified Journal | Page 1401
[6] IITK-BMTPC Earthquake Tips: Learning Seismic Design and Construction, IIT Kanpur, India.
BIOGRAPHIES
Volume: 07 Issue: 05 | May 2020 www.irjet.net p-ISSN: 2395-0072
© 2020, IRJET | Impact Factor value: 7.529 | ISO 9001:2008 Certified Journal | Page 1396
Structural Design of Reinforced Concrete Shear wall
Darshitkumar R. Gohel
---------------------------------------------------------------------***----------------------------------------------------------------------
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
have shown very good performance in past earthquakes.
Shear walls in high seismic regions require special
detailing. However, in past earthquakes, even buildings
with sufficient amount of walls that were not specially
detailed for seismic performance (but had enough well-
distributed reinforcement) were saved from collapse.
Shear wall buildings are a popular choice in many
earthquake prone countries, like Chile, New Zealand and
USA. Shear walls are easy to construct, because
reinforcement detailing of walls is relatively straight-
forward and therefore easily implemented at site. Shear
walls are efficient, both in terms of construction cost and
effectiveness in minimizing earthquake damage in
structural and non-structural elements (like glass windows
and building contents).
2. Problem Statement
18-storey building has plan dimensions as shown in Fig. 1.
Two shear walls are to be provided in each direction to
resist the seismic forces. The axial load on the each shear
wall is 15000kN due to both dead and live loads. The
height between floors is 3.0m. dead load per unit area of
the floor, which consists of floor slab, finishes. etc., is
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
roof is 1.5kN/m2. The soil below the foundation is hard and
the building is located in Roorkee.
Figure 1. Plan
As per the code provisions, the percentage of design live
load to be considered for the calculation of earthquake
forces is 25% for the floors and live load for the roof is not
to be accounted for.
= 4.5 + 3.0 + 0.25 × 3.5 = 8.375kN/m2
and that at the roof = 4.5kN/m2
Weight of 144 beams, each of 3.5 m span, at each floor and
roof
= 2268kN
= 0.3×0.6×2.4×61×25
= 658.8kN
= 329.4kN
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
© 2020, IRJET | Impact Factor value: 7.529 | ISO 9001:2008 Certified Journal | Page 1397
Plan area of building is 28 m × 28 m = 784 m2
Equivalent load at roof level = 4.5×784 + 2268 + 329.4
= 6125.4kN
= 9492.8kN
= 167503kN
3.2 Base shear The fundamental natural period of vibration T for the
buildings having shear walls is given by T =
= (d, the plan dimension = 28m)
Building is situated in Roorkee, i.e., in Zone IV. Zone factor
Z = 0.24, importance factor I = 1, response reduction factor
R = 4.0 for 5% damping and type I soil, average response
acceleration coefficient = 1.81 Design horizontal seismic
coefficient Ah = × ×
Lateral loads and shear forces at different floor level are
given in Table 1.
Mass No. Wi(kN) hi(m) Wihi 2
Qi(kN) Vi(kN)
∑Wihi2 = 170363498.4
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
© 2020, IRJET | Impact Factor value: 7.529 | ISO 9001:2008 Certified Journal | Page 1398
3.3 Bending moment and shear force
Two shear walls are provided in each direction to resist the seismic forces. Therefore, the lateral forces acting on one shear wall will be half the calculated shears and is as shown in Fig. 2.
Figure 2. Lateral Forces on shear wall The shear wall will be designed as a cantilever fixed at the
base and free at the top.
Maximum shear force at base V = 4547.705kN
Maximum bending moment at base,
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) +
(184.73×27) + (228.06×30) + (275.96×33) + (328.41×36) +
(385.42×39) + (447×42) + (513.14×45) + (583.84×48) +
(659.1×51) + (476.8×54)
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
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
τv = (Clause 9.2.1, IS: 13920-1993)
Where, dw= effective depth of the wall section For rectangular wall sections, dw= 0.9 × Lw (Clause 9.2.1, IS: 13920-1993) ∴ dw = 0.9 × 14000 = 12600mm
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
© 2020, IRJET | Impact Factor value: 7.529 | ISO 9001:2008 Certified Journal | Page 1399
∴ τv= = 1.8046MPa
For M30 grade concrete τcmax= 3.5MPa [Table-20, IS 456:2000] 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 For pt = 0.25%, τc= 0.37MPa [Table 19, IS 456:2000] Since τv>τc,
Shear reinforcement is required.
4.2 Horizontal shear reinforcement
Provide horizontal shear reinforcement as per 9.2.5 of IS 13920:1993 Assume 2-legged 12 mm diameter horizontally aligned closed Stirrups along the height of the shear wall, Spacing of stirrups along the height of shear wall
∴ Spacing of stirrups = Sv =
∴ Sv=
∴ Sv = 189.59 mm c/c Minimum horizontal reinforcement = (Ash)min = 0.0025 of gross sectional area of the wall in elevation (Clause 9.1.4, IS 13920:1993) ∴ (Ash)min = 0.0025 ×300 × 1000 = 750 mm2 Hence, provide 2-legged 12 mm diameter horizontally aligned closed stirrups at 250mm c/c along the entire height of the shear wall.
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 stirrups.
Spacing of stirrups = Sv=
∴ Sv =
∴ Sv = 189.59 mm c/c Minimum vertical reinforcement = 0.0025 of gross sectional area of the wall in plan (Clause 9.1.4, IS 13920:1993) ∴ (Asv) = 0.0025 × 300 × 14000 = 10500 mm2 Hence, provide 2-legged 12 mm diameter vertically aligned closed stirrups at 250mm c/c along the entire length of the shear wall. Asv>Ash, Hence ok.
4.4 Check for flexural strength
With reference to Annex 'A' and Clause 9.3.1 of IS 13920:1993
xu /Lw =
xu’/Lw =
xu /Lw = = 0.4965
xu’/Lw = = 0.6597
Since xu’/Lw>xu /Lw, Eq.(a) of Annex “A” (Clause 9.3.1) IS 13920:1993, is applicable.
4.5 Moment of resistance of Rectangular Shear wall section
∴ =
= 107127.0892kNm
Balance moment to be resisted by the edge reinforcement in each shear wall = (278862.64 – 107127.08) = 171735.56kNm
4.6 Check on boundary elements
To check the necessity of providing boundary elements in the shear wall (Figure 3). Gross area of wall section, Ag = 14000×300 = 4200000mm2
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
© 2020, IRJET | Impact Factor value: 7.529 | ISO 9001:2008 Certified Journal | Page 1400
Ixx = = 68.6×1012 mm4
∴ σ =
∴ σ =
∴ σ = 5.3571±28.4553 Maximum stress = σmax = 33.8124MPa As per Clause 9.4.1 of IS 13920:1993, if the extreme fibre compressive stress in the wall due to factored gravity loads plus factored earthquake loads exceeds 0.2fck = 0.2×30 = 6 < 33.8124MPa Hence, boundary elements are required in the shear wall. Provide a boundary element of length 1400mm each and width 500mm at each edge of the shear wall, Figure 1.
4.7 Design of boundary elements
Boundary element is essentially treated as a column. The vertical reinforcement in boundary elements shall not be less than 0.80% nor greater than 6%. (Clause 9.4.4, IS 13920 1993) Adopt 3% vertical reinforcement in the boundary elements. Asc provided = 0.03×1400×500 = 21000mm2 Axial compression load on the boundary element due seismic forces = (Mu-Muv)/Cw (Clause 9.4.2, IS 13920:1993) Mu = Factored design moment on the entire shear wall section = 278862.64kNm Muv = Moment of resistance provided by the distributed Vertical reinforcement across the wall section = 171735.56kNm Cw = center to center distance between the boundary elements along the two vertical edges of the shear wall = 12.6 m
Axial compressive load =
= 8502.15kN Required axial load capacity of boundary element = axial load due to gravity effects + axial load due to seismic forces = 2250 + 8502.15 = 10752.15kN As per clause 9.4.2, IS 13920:1993, the boundary element shall be assumed to behave as an axially loaded short column. If the design strength of an axially loaded short column = PuD, then 0.4fck Ac + 0.674fyAsc PuD = 0.447fck Ag + (fc– 0.447 fck )Asc fc= 0.790 fy(for Fe415 steel) PuD = 0.447×30×1400×500+ (0.790×415 – 0.447×30) ×21000 = 15990.24kN> 10752.15kN, Hence ok Area of steel for each boundary element = 21000mm2 Provide 10 nos 40Φ + 12 nos 32Φ in each element Asc provided = 22217.34 mm2>21000 mm2 Hence, Ok
As per (Clause 9.4.5, IS13920:1993), boundary elements shall be provided with special confining reinforcement throughout their height. Area of special confining reinforcement
Ash = 0.18 Sh
(Clause 7.4 8, IS 13920:1993) Assume 12 mm diameter closed hoops as confining reinforcement with a clear cover of 40 mm The gross area of the boundary element section Ag =1400×500 = 700000 mm2 The size of the core = Ak= (1400-40 - 40) × (500 -40-40) = 1320×420 mm Since both dimension of the core is greater than 300mm, a cross tie will have to be used. If the cross-tie is placed at mid-length of the longer dimension of the core then, h = 1320/2 = 660 mm and for shorter dimension, h = 420/2 = 210 mm. The spacing of the confining hoops, S, shall not exceed the smaller of
1. Minimum member dimension, i.e. =
125mm 2. 100mm.
The spacing however, need not be less than 100mm. (Clause 7.4.6 of IS 13920:1993) Hence adopt 100mm spacing of the confining hoops S = 100mm
Ash = = 225.55 mm2>
113.09 mm2 (Area of 12 mm diameter bar) Hence, 12 mm diameter confining reinforcement is insufficient. Adopt 20 mm diameter confining hoops at a spacing of 100 mm c/c along the entire height of the boundary element Ash provided = 314 mm2 > 225.55 mm2. Hence, ok.
5. Detailing of reinforcement
The detailing of reinforcement in the shear wall is shown in Figure 4.
REFERENCES [1] IS 13920:1993 Ductile detailing of reinforced concrete
structures subjected to seismic forces - Code of practice
[CED 39: Earthquake Engineering].
[2] IS 456 (2000): Plain and Reinforced Concrete - Code of Practice [CED 2: Cement and Concrete]
[3] IS 1893-1 (2002): Criteria for Earthquake Resistant Design of Structures, Part 1: General Provisions and Buildings [CED 39: Earthquake Engineering]
[4] IS:4326-1993: Earthquake Resistant Design and Construction of Buildings
[5] SP:22(S&T)-1982 : Indian Standard Explanatory Handbook On codes for Earthquake Engineering - IS:1893-1975 & IS:4326-1976
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
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[6] IITK-BMTPC Earthquake Tips: Learning Seismic Design and Construction, IIT Kanpur, India.
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