Design of Reinforced Concrete Shear wall

Design code: ACI 318-05

Design data:

Seismic shear force: (service load)

Roof: V_r = 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: l_w = 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.

Design reinforcement for shear wall

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,  f = 0.75

  fV_n = 10 Öf_c'  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

M_u = 1.4 [100 (60-7.5)+75(45-7.5)+50(30-7.5)+25 (25-7.5)] = 13130 ft-kips

N_u = (0.15)(10/12)(18)(60-7.5) = 118.1 kips

Calculate shear strength of concrete:

fV_c = 0.75 [3.3 Öf_c' h d + N_u d/ (4 l_w)] = 288.2 kips

M_u/V_u - l_w/2 = 28.5 ft

fV_c = 0.75 { 0.6 Öf_c' + l_w ( 1.25 Öf_c' + 0.2 [N_u/(l_w* h)]) /( M_u/V_u - l_w/2)} h d   = 163.8 kips

Or fV_c = 0.75 (2 Öf_c' h d) = 163.9 kips    Use

V_s = V_u - fV_c = 186.1 kips

Use #5 bar in one layer, area of reinforcement, Av = 0.3 in².

Spacing: S = fA_v f_y d /V_s = 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: r_t = 0.3 in² / (12x10) = 0.0025   O.K.

Design vertical reinforcement:

r_l = 0.0025 + 0.5 (2.5 - h_w/ l_w )( r_t - 0.0025)  =  0.0025

Use r_l = 0.0025 

Area of reinforcement: A_v = 0.0025 (10)(12) = 0.3 in²/ft

Use #5 bars at 12" O.C

M_u = 1.4 [(100)(60)+(75)(45)+(50)(30)+(25)(15)]=15750 ft-kip

Tension control section,  f = 0.9

Factor: R_n = (15750)(12000)/[0.9(10)(14.4x12)²] = 703 psi, and m = f_y/(0.85fc')=17.7

Reinforcement ratio, r = (1/m)[1-(1- 2 m R_n/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

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 / f h d² = 514.7 psi, m = 17.

Reinforcement ratio,  r = 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.

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: P_D = 40 kips

Live load at each floor and roof: P_L = 25 kips

Seismic shear force: (service load)

Roof: V_r = 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: l_w = 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: V_u = 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,  f = 0.75

  fV_n = 10 Öf_c'  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

M_u = 1.4 [100 (60-7.5)+75(45-7.5)+50(30-7.5)+25 (25-7.5)] = 13130 ft-kips

N_u = 1.2 [(0.15)(10/12)(18)(60-7.5)+4 P_D ]+1.0 (4 P_L ) = 462.1 kips

Calculate shear strength of concrete:

fV_c = 0.75 [3.3 Öf_c' h d + N_u d/ (4 l_w)] = 393.9 kips

M_u/V_u - l_w/2 = 28.5 ft

fV_c = 0.75 { 0.6 Öf_c' + l_w ( 1.25 Öf_c' + 0.2 [N_u/(l_w* h)]) /( M_u/V_u - l_w/2)} h d   = 228.2 kips    (Use)

Or fV_c = 0.75 (2 Öf_c' h d) = 196.7 kips 

V_s = V_u - fV_c = 112.1 kips

Use #4 bar in two layer, area of reinforcement, A_v = 0.4 in². (Code requires two layers for 12" wall)

Spacing: S = fA_v f_y d /V_s = 25.7 in

Check maximum spacing: (18x12)/5 = 43 in, 3 (10) = 30 in, or 18 in     Use 18" 

Check minimum reinforcement: r_t = 0.4 in² / (18x10) = 0.0019   < 0.0025

Use r_t =0.0025, spacing S = 0.4 in² / (0.0025)(h) = 13.3 in   Use 12 in

Design vertical reinforcemnt

r_l = 0.0025 + 0.5 (2.5 - h_w/ l_w )( r_t - 0.0025)  =  0.0025

Use r_l = 0.0025 

Use #4 bars in two layers at 12" O.C

M_u = 1.4 [(100)(60)+(75)(45)+(50)(30)+(25)(15)]=15750 ft-kip

N_u = 1.2 [(0.15)(10/12)(18)(60)+4 P_D ]+1.0 (4 P_L ) = 486.4 kips

Design as a column subjected to axial load and bending 

Gross area, A_g = (18)(12)(12) = 2592 in².

Assume tension control section, f = 0.9

fN_u/A_g = 0.141 ksi

fM_u/(A_g l_w) = 0.253 ksi

From ACI column design chart (See column design section), Area of reinforcement, r = 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 r >  0.01.  Use #4 ties at 12" O.C.