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Design of Reinforced Concrete Shear wall. analysis and design of shearwall.
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Design of Reinforced Concrete Shear wallExample 1: Design of reinforced concrete non-load bearing shear wall.
Example 2: Design of Reinforced Concrete load bearing shear wall
Design examples
Example 1: Design of reinforced concrete non-load bearing shear wall.
Design code: ACI 318-05
Design data:
Seismic shear force: (service load)
Roof: Vr = 100 kips
4th floor: V4 = 75 kips, ,
3rd floor: V3 = 50 kips
2nd floor: V2 = 25 kips
Floor height: H = 15 ft
Length of wall: lw = 18 ft
Width of wall: h = 10 in
Concrete strength: fc' = 4000 psi
Yield strength of steel: fy = 60 kis
Assumption:
1.out-of-plan moment is neglectable.
2.The wall is an exterior wall.
Requirement:
Design reinforcement for shear wall
Solution:
Maximum shear occurs at load combination: 1.2D+1.4E+1.0L
Calculate maximum vertical and shear force at first floor
Maximum factored shear: Vu = 1.4 (100+75+50+25) = 350 kips
Check maximum shear strength permitted
Assume effective depth, d = 0.8 (18) = 14.4 ft
Strength reduction factor, = 0.75
Vn = 10 fc' h d = 819 kips > 350 kips O.K.
Critical section for shear at smaller of 18 ft/2 = 9 ft , H/2 = 7.5 ft
Calculate factored overturning moment and weight of wall at critical section
Mu = 1.4 [100 (60-7.5)+75(45-7.5)+50(30-7.5)+25 (25-7.5)] = 13130 ft-kips
Nu = (0.15)(10/12)(18)(60-7.5) = 118.1 kips
Calculate shear strength of concrete:
Vc = 0.75 [3.3 fc' h d + Nu d/ (4 lw)] = 288.2 kips
Mu/Vu - lw/2 = 28.5 ft
Vc = 0.75 { 0.6 fc' + lw ( 1.25 fc' + 0.2 [Nu/(lw* h)]) /( Mu/Vu - lw/2)} h d = 163.8 kips
Or Vc = 0.75 (2 fc' h d) = 163.9 kips Use
Design horizontal shear reinforcement:
Vs = Vu - Vc = 186.1 kips
Use #5 bar in one layer, area of reinforcement, Av = 0.3 in2.
Spacing: S = Av fy d /Vs = 12.6 in, Use 12" O.C.
Check maximum spacing: (18x12)/5 = 43 in, 3 (10) = 30 in, or 18 in O.K.
Check minimum reinforcement: t = 0.3 in2 / (12x10) = 0.0025 O.K.
Design vertical reinforcement:
l = 0.0025 + 0.5 (2.5 - hw/ lw )( t - 0.0025) = 0.0025
Use l = 0.0025
Area of reinforcement: Av = 0.0025 (10)(12) = 0.3 in2/ft
Use #5 bars at 12" O.C
Design flexural reinforcement:
Calculate factored moment at base:
Mu = 1.4 [(100)(60)+(75)(45)+(50)(30)+(25)(15)]=15750 ft-kip
Tension control section, = 0.9
Factor: Rn = (15750)(12000)/[0.9(10)(14.4x12)2] = 703 psi, and m = fy/(0.85fc')=17.7
Reinforcement ratio, = (1/m)[1-(1- 2 m Rn/fy)] = 0.013
Area of reinforcement, As = 0.013 h d = 22.9 in2.
Use #10 bars, number of bars, n = 22.9/1.27 = 18
Check effective depth
Concrete cover = 2" for exterior wall. Use 3" spacing between #10 bars in two layers
Effective depth, d = (18)(12) - 2-(3)(8)/2 =202 in
Recalculate reinforcement, Factored Rn = Mu / h d2 = 514.7 psi, m = 17.
Reinforcement ratio, = 0.0094
Area of reinforcement, As = 18.9 in2.
Use #10 bars, number of bars, n = 18.9 /1.27 = 15, Use 16 # 10
Use #4 closed shape ties to enclose tension reinforcement,
Area of reinforcement for shear As = 0.4 in2.
Check clear spacing between bars, S = 10-(2)(2)-(0.5)(2)-1.27 = 3.73 in O.K.
Reinforcement detail
Example 2: Design of Reinforced Concrete load bearing shear wall
Situation: A reinforced concrete load bearing shear wall supporting for a four story building
Design code: ACI 318-05
Design data:
Vertical load: (service load)
Dead load at each floor and roof: PD = 40 kips
Live load at each floor and roof: PL = 25 kips
Seismic shear force: (service load)
Roof: Vr = 100 kips
4th floor: V4 = 75 kips, ,
3rd floor: V3 = 50 kips
2nd floor: V2 = 25 kips
Floor height: H = 15 ft
Length of wall: lw = 18 ft
Width of wall: h = 12 in
Concrete strength: fc' = 4000 psi
Yield strength of steel: fy = 60 kis
Assumptions:
1. out-of-plan moment is neglectable.
2. The wall is an exterior wall.
Requirement:
Design reinforcement for shear wall
Solution
Maximum shear occurs at load combination: 1.2D+1.4E+1.0L
Calculate maximum vertical and shear force at first floor
Maximum factored shear: Vu = 1.4 (100+75+50+25) = 350 kips
Check maximum shear strength permitted
Assume effective depth, d = 0.8 (18) = 14.4 ft
Strength reduction factor, = 0.75
Vn = 10 fc' h d = 819 kips > 350 kips O.K.
Critical section for shear at smaller of 18 ft/2 = 9 ft , H/2 = 7.5 ft
Calculate factored overturning moment and weight of wall at critical section
Mu = 1.4 [100 (60-7.5)+75(45-7.5)+50(30-7.5)+25 (25-7.5)] = 13130 ft-kips
Nu = 1.2 [(0.15)(10/12)(18)(60-7.5)+4 PD ]+1.0 (4 PL ) = 462.1 kips
Calculate shear strength of concrete:
Vc = 0.75 [3.3 fc' h d + Nu d/ (4 lw)] = 393.9 kips
Mu/Vu - lw/2 = 28.5 ft
Vc = 0.75 { 0.6 fc' + lw ( 1.25 fc' + 0.2 [Nu/(lw* h)]) /( Mu/Vu - lw/2)} h d = 228.2 kips (Use)
Or Vc = 0.75 (2 fc' h d) = 196.7 kips
Design horizontal shear reinforcement:
Vs = Vu - Vc = 112.1 kips
Use #4 bar in two layer, area of reinforcement, Av = 0.4 in2. (Code requires two layers for 12" wall)
Spacing: S = Av fy d /Vs = 25.7 in
Check maximum spacing: (18x12)/5 = 43 in, 3 (10) = 30 in, or 18 in Use 18"
Check minimum reinforcement: t = 0.4 in2 / (18x10) = 0.0019 < 0.0025
Use t =0.0025, spacing S = 0.4 in2 / (0.0025)(h) = 13.3 in Use 12 in
Design vertical reinforcemnt
l = 0.0025 + 0.5 (2.5 - hw/ lw )( t - 0.0025) = 0.0025
Use l = 0.0025
Use #4 bars in two layers at 12" O.C
Calculate factored moment and axial load at base:
Mu = 1.4 [(100)(60)+(75)(45)+(50)(30)+(25)(15)]=15750 ft-kip
Nu = 1.2 [(0.15)(10/12)(18)(60)+4 PD ]+1.0 (4 PL ) = 486.4 kips
Design as a column subjected to axial load and bending
Gross area, Ag = (18)(12)(12) = 2592 in2.
Assume tension control section, = 0.9
Nu/Ag = 0.141 ksi
Mu/(Ag lw) = 0.253 ksi
From ACI column design chart (See column design section), Area of reinforcement, = 0.011
Area of reinforcement, As = (0.01)(18x12)(12) = 22.8 in2.
Use #10 bars, number of bar, n = 22.8/1.27 = 18
Use 10#10 bars at each end of shear wall, column ties is required since > 0.01. Use #4 ties at 12" O.C.