Reinforced Concrete Beam Shear and Stirrup Spacing (ACI 318-19)

Whether a rectangular concrete beam carries its shear and what stirrup spacing it needs.

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Checked against published worked examples from: StructurePoint.

Details, formula, and sources

ScopeThe simplified concrete contribution Vc = 2 x lambda x sqrt(f'c) x bw x d, the design phi Vc at phi = 0.75, the stirrup demand Vs = Vu/phi - Vc when the demand exceeds phi Vc, the required spacing s = Av x fyt x d / Vs, and the governing maximum spacing: d/2 (d/4 when Vs is high) or the spacing where the stirrups still meet the Av,min minimum. It flags where Av,min is required (Vu above half of phi Vc) and a section too small for its shear. A 12 x 21.5 in beam on 4,000 psi concrete carries 24.5 kip on concrete alone; a 40 kip demand needs #3 stirrups at 10 in (the d/2 cap governs over the computed 13.7 in). Uses the simplified Vc for a member without axial load. A design aid, not a substitute for a licensed engineer's design.

Formula or tableVc = 2 x lambda x sqrt(f'c) x bw x d; phi Vc = 0.75 x Vc; Vs,req = Vu/0.75 - Vc when Vu > phi Vc; s = Av x fyt x d / Vs; s_max = min(d/2 <= 24 in, or d/4 <= 12 in when Vs > 4 sqrt(f'c) bw d; Av fyt / (max(0.75 sqrt(f'c), 50) bw)); Av,min required when Vu > 0.75 lambda sqrt(f'c) bw d; section OK when Vs <= 8 sqrt(f'c) bw d.

Source and editionACI 318-19 (Building Code Requirements for Structural Concrete), by name: §22.5.5.1 and Table 22.5.5.1 (simplified Vc, valid with Av >= Av,min), §22.5.10.5.3 (vertical-stirrup Vs), §21.2.1 (phi = 0.75 shear), §9.7.6.2.2 (max spacing), §9.6.3.1 and §9.6.3.4 (Av,min), §22.5.1.2 (section limit).

Free accessACI 318-19 is viewable through ACI's online public-access reader; the 2 x sqrt(f'c) x bw x d and Av x fyt x d / s arithmetic is public.

What governsEstimate. AHJ and licensed professional govern.

API field namesfc, fyt, bw, d, av_in2, vu, lambda, vc_kip, phi_vc, vs_req_kip, s_req_in, s_max_in

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