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Design of Reinforced Concrete Shear wallDesign code: ACI 318-05
Design of shear reinforcement1. The shear strength, fVn must equal or be greater than the factored shear, Vu fVn ³ Vu ACI eq (11-1) where f = 0.75 is strength reduction factor, and Vn = Vc + Vs ACI eq (11-2) where Vc is shear strength of concrete, and Vs is strength of shear reinforcement. Shear strength of concrete2. Shear strength of concrete shear wall shall be calculated as a. Vc = 2 Öfc' h d Or the lesser of b. Vc = 3.3 Öfc' h d + Nu d/ (4 lw) ACI eq (11-29) c. Vc = { 0.6 Öfc' + lw ( 1.25 Öfc' + 0.2 [Nu/(lw* h)]) /( Mu/Vu - lw/2)} h d ACI eq (11-30) (It does not apply when Mu/Vu - lw/2 < 0) where fc' is compressive strength of concrete, h is thickness of the wall, d is effective depth, lw is overall length of wall. Nu is axial force, positive for compression, negative for tension, Mu is moment parallel to the direction of the wall. Shear reinforcementWhen Vu > f Vc,shear strength provided by horizontal shear reinforcement shall be calculated aS Vs = Av fy d /s ACI eq. (11-31) where Av is area of reinforcement, fy is yield strength of reinforcement, d is effective depth, and s is spacing of reinforcement. Maximum shear strength permittedfVn = 10 Öfc' ACI section 11.10.3 Effective depth1. d = 0.8 lw 2. Calculate based on strain compatibility analysis (same as in a beam section). Location of Critical Section1. One-half of the wall height. 2. lw /2 whichever is smaller. Minimum shear reinforcementWhen f Vc > Vu ³ 0.5 f Vc1. Minimum horizontal shear reinforcement ratio to gross section area, rt = 0.0025. 2. Minimum vertical shear reinforcement ratio to gross section area, rl = 0.0025 + 0.5 (2.5 - hw/ lw )( rt - 0.0025) ³ 0.0025 ACI eq. (11-32) and need not be greater than calculated based on ACI eq. 11-31. When Vu < 0.5 f Vc1. Minimum vertical wall reinforcement: (Section 14.3.2) a. 0.0012 for deformed bars smaller than #5 and fy > 6000 psi b. 0.0015 for deformed bars #5 or larger 2. Minimum horizontal wall reinforcement: (Section 14.3.3) a. 0.0020 for deformed bars smaller than #5 and fy > 6000 psi b. 0.0025 for deformed bars #5 or larger. Bar placement1. Maximum spacing of horizontal reinforcementa. lw /5. b. 3 h c. 18 inches. 2. Bar shall be placed uniformly across the length and height of walls Design of Flexural reinforcement1. Non-load bearing shear wallDesign as a cantilever beam: fMn ³ Mu 2. Load bearing shear wallDesign as a column subject to axial load and bending fPn ³ Pu fMn ³ Mu Location of flexural reinforcement:Flexural reinforcement shall be placed at each end of walls Design examples
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