Sample Calculations

Design of Reinforced Concrete Shear wall

Project: 5255 Collins Ave, Miami Beach, Florida 33140

By: Chaiban Engineering Consultants, Inc,.

Design of Reinforced Concrete load bearing shear wall

Situation: A reinforced concrete load bearing  shear wall supporting for a 15 story building 

Florida Building Code: 2010

Design code: ACI 318-05

Design data:

Vertical load: (service load)

Uniform factored Load ( DL +LL) = 210 PSF

Dead load at each floor and roof: PD = 217 kips

Live load at each floor and roof: PL = 133 kips

Wind shear force: (service load)

15th floor:  V15 = 40 kips, , 

14 floor: V14 = 40 kips

13 floor: V13 = 40 kips

12 floor: V12 = 40 kips

11 floor: V11 = 39 kips

10 floor: V10 = 39 kips

9 floor: V9 = 38 kips

8 floor: V8 = 38 kips

7 floor: V7 = 37 kips

6 floor: V6 = 37 kips

5 floor: V5 = 37 kips

4 floor: V4 = 36 kips

3 floor: V3 = 34 kips

2 floor: V2 = 33 kips

1 floor: V1= 31 kips

Total Shear: 559 kips =Vu

Factored Shear ( 1.6 x 559 k= 894.4 kips)

Vu= 894.4 kips

Floor height: H = 10 ft

Length of wall: lw = 18 ft

Width of wall: h = 12 in

Concrete strength: fc' = 4000 psi

Yield strength of steel: fy = 60 ksi

Assumptions: 

1. out-of-plan moment is negligible  

2. The wall is an interior wall.

Requirement: 

Design reinforcement for shear wall

Solution

Maximum shear occurs at load combination: 1.3D+1.6W+1.6L

Calculate maximum shear force at first floor;

Maximum factored shear: Vu = 40 kips

Check maximum shear strength permitted

Assume effective depth, d = 0.8 (18) = 14.4 ft ( 173 in)

Strength reduction factor,  f = 0.75

  fVc = 2 Ø Öfc'  b d = 197 kips     > 40 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 =         = 59,473 ft-kips

Nu =         = 350 kips

Calculate shear strength of concrete:

fVc = 0.75 [3.3 Öfc' b d + Nu d/ (4 lw)] = 314.2 kips

Mu/Vu - lw/2 = 57.5 ft

fVc = 0.75 { 0.6 Öfc' + lw ( 1.25 Öfc' + 0.2 [Nu/(lw* h)]) /( Mu/Vu - lw/2)} h d   = 239.9 kips    (Use)

Or fVc = 0.75 (2 Öfc' h d) = 197  kips 

Design horizontal shear reinforcement:

Vs = Vu - fVc = 894.4- 239.9 = 654.5 kips

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

Spacing: S = fAv fy d /Vs = 4.50 in

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

Check minimum reinforcement: rt = 0.4 in2 / (18x10) = 0.0019   < 0.003

Use rt =0.003, spacing S = 0.4 in2 / (0.003)(h) = 11.11 in   Use 11 in

But As Built wall horizontal reinforcement found based on X Ray tests is #4 bars ( double) @ 4 inches o.c.

OK

Design vertical reinforcement

rl = 0.003 + 0.5 (2.5 - hw/ lw )( rt - 0.003)  =  0.003

Use rl = 0.003 

Use #4 bars in two layers at 4" O.C ( As Built)

Calculate factored moment and axial load at base:

Mu = 59,473 ft-kip

Nu =  350.0 kips

Design as a column subjected to axial load and bending 

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

Assume tension control section, f = 0.9

fNu/Ag = 0.122 ksi

fMu/(Ag lw) = 0.095 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 each

As Built: #11 bars @ 6 “ on center as checked by GC based on test holes. ( 18 each vertical bars per 4.50 ft section x 2 = 9 ft.

Use 18#11 bars at each end of shear wall, column ties is required since r >  0.01.  Use #4 ties at 4" O.C.

Check Deflection at top of Shear Wall:

Deflection: wL4/8EI

W= 3,200 Lbs/Ft or 266.67 Lbs/In

L= 150 LF or 1800 Inch

I= 2,985,984 in4

E= 57,000 Sqrt (4000 psi)= 3,604,996

Δ= 32 inches at top of shear wall. ( worst case)

See Wall section attached.

See hand calculations attached for worst load combination analysis ( Dead and Wind Loads).

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