Sh Wall Example

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Design of Reinforced Concrete Shear wall

Example 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

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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/V

u- l

w/2 = 28.5 ft

Vc

= 0.75 { 0.6 fc' + l

w( 1.25 f

c' + 0.2 [N

u/(l

w* h)]) /( M

u/V

u- l

w/2)} h d = 163.8 kips

OrVc = 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.




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Check minimum reinforcement: t= 0.3 in2 / (12x10) = 0.0025 O.K.

Design vertical reinforcement:

l= 0.0025 + 0.5 (2.5 - h

w/ l

w)(

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




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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




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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 + N

ud/ (4 l

w)] = 393.9 kips

Mu/V

u- l

w/2 = 28.5 ft

Vc

= 0.75 { 0.6 fc' + l

w( 1.25 f

c' + 0.2 [N

u/(l

w* h)]) /( M

u/V

u- l

w/2)} h d = 228.2 kips (Use)

OrVc

= 0.75 (2 fc' h d) = 196.7 kips

Design horizontal shear reinforcement:

Vs

= Vu

- Vc

= 112.1 kips




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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 - h

w/ l

w)(

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/(A

glw

) = 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.




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