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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Example

You enter

Concrete strength f'c (psi) 4000
Stirrup yield fyt (psi) 60000
Web width bw (in) 12
Effective depth d (in) 21.5
Stirrup area Av, both legs (in²) 0.22
Factored shear Vu (kip, 0 = capacity only) 40
Lightweight factor lambda (1.0 normalweight) 1

You get

Concrete shear Vc 32.6 kip
Design concrete shear phi Vc (phi = 0.75) 24.5 kip
Stirrup shear demand Vs,req 20.7 kip
Required stirrup spacing 13.7 in
Code max spacing d/2 10.75 in (d/4 if Vs is high)

Details, formula, and sources

The 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 d/2 code maximum. 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; the section upper limit and minimum-reinforcement triggers are separate checks. A design aid, not a substitute for a licensed engineer's design.

Vc = 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 = d/2.

ACI 318-19 (Building Code Requirements for Structural Concrete), by name: §22.5.5.1 (simplified Vc), §22.5.10.5.3 (vertical-stirrup Vs), §21.2.1 (phi = 0.75 shear), §9.7.6.2.2 (d/2 max spacing).

ACI 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.

Estimate. AHJ and licensed professional govern.

Field names used by the API: fc, fyt, bw, d, av_in2, vu, lambda, vc_kip, phi_vc, vs_req_kip, s_req_in, s_max_in

Simplified Vc the traditional 2 x lambda x sqrt(f'c) x bw x d concrete term for a member without axial load, not the detailed 318-19 expression with rho_w and size-effect termsACI 318-19 §22.5.5.1
Vertical stirrups Vs = Av x fyt x d / s assumes stirrups perpendicular to the axis; Av is the total area crossing the crack (both legs)ACI 318-19 §22.5.10.5.3
Limits separate the §22.5.1.2 section upper limit, §9.6.3 minimum shear reinforcement, and the d/4 tightening when Vs is high are separate checks the detailer appliesACI 318-19