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    Calculates locally in your browser · ACI 318 shear
    Concrete / shear

    Beam shear reinforcement calculator

    Select ACI 318-19 or ACI 318-19(22). The calculator selects the concrete shear equation from the provided Av compared with Av,min, computes the longitudinal reinforcement ratio and axial-force contribution, and checks strength, minimum stirrup demand and the applicable spacing limit.

    d440 mmSpacingOKAv / Av,minMetSection maximumOKVerdictPass
    Shear capacity
    197.9 kN
    φ(Vc + Vs)
    Concrete Vc
    118.7 kN
    minimum stirrups provided; Eq. (a) applied
    Stirrups Vs
    145.1 kN
    2-leg 10 mm at 200 mm
    Utilisation
    75.8%
    Demand 150 kN · ACI 318-19
    Shear check
    ACI 318-19: concrete contribution plus stirrups, reduced by φ = 0.75.
    197.9 kN
    Shear check
    TermValue
    Effective depth d440 mm
    Tension reinforcement As1257 mm² (4 × 20 mm bars)
    Longitudinal ratio ρw0.0095
    Axial term Nu/(6Ag)0.00 MPa (compression capped at 0.05f′c)
    ACI editionACI 318-19
    Stirrup area Av157 mm²
    Minimum stirrup area Av,min50 mm²
    Minimum shear demand threshold43.5 kN
    Minimum area required by demandRequired and met
    Vc equation branchACI 318-19 / 318-19(22) Eq. (a), Av ≥ Av,min
    Vc118.7 kN
    Vs145.1 kN
    φVn197.9 kN
    Nominal strength limit461.0 kN
    Maximum stirrup spacing220 mm
    Shear stress vu/(bw·d)1.14 MPa

    The Av,min branch is applied even when minimum stirrups are not required by demand. Exceptions for shallow beams, slab-integral beams, steel-fiber beams and joists are not evaluated.

    Inspect the model
    Choose a quick result view, a scaled comparison, or the complete diagnostic set.
    Capacity against stirrup spacing
    The maximum spacing tightens when the stirrup contribution exceeds the ACI threshold.
    Capacity against stirrup spacing0.00077.45154.89232.34309.79387.247575100100125125150150175175200200225225250250275275300300325325φVn kN
    Canonical equations
    The exact relationships implemented by this engine.

    Vc = (0.17 λ √f′c + Nu/(6Ag)) bw d · ACI 318-19 and 318-19(22) Table 22.5.5.1; branch selected from Av versus Av,min.

    Vs = Av fy d / s · Vertical stirrups, 45° cracking assumed.

    Nu/(6Ag) ≤ 0.05f′c · Compression positive; tension negative.

    Av,min = max(0.062√f′c, 0.35) bw s / fy · ACI 318-19 §9.6.3.4, SI units.

    Provide Av,min where Vu > 0.083 φ λ √f′c bw d · ACI 318-19 §9.6.3.1 in SI units; Table 9.6.3.1 exceptions are not evaluated.

    φVn = φ min(Vc + Vs, Vn,max) ≥ Vu · Checks shear strength, nominal maximum, minimum reinforcement trigger and spacing.

    engineering methodology

    A clear path from inputs to results.

    Select ACI 318-19 or 318-19(22), reapproved 2022. The calculator selects Vc from Av versus Av,min, uses the size-effect and longitudinal-steel ratio branch when Av is below minimum, includes the axial-force term, and checks strength, minimum reinforcement and stirrup spacing.

    Step-by-step methodWorked exampleAssumptions explained

    How to use

    1. Choose ACI 318-19 or 318-19(22), then enter web width, depth, clear cover to the stirrup outside, and tension-bar size/count.
    2. Set f′c, stirrup fy and factored axial load Nu (compression positive).
    3. Choose the number of legs, stirrup diameter and spacing.
    4. Enter the factored shear.

    The formula

    ϕVn=0.75(Vc+Vs),Vs=Avfyd/s;Vc={(0.17λfc′+Nu/6Ag)bwd,Av≥Av,min(0.66λsλρw1/3fc′+Nu/6Ag)bwd,Av<Av,min\phi V_n = 0.75(V_c+V_s),\quad V_s=A_v f_y d/s;\quad V_c=\begin{cases}(0.17\lambda\sqrt{f'_c}+N_u/6A_g)b_wd,&A_v\ge A_{v,min}\\(0.66\lambda_s\lambda\rho_w^{1/3}\sqrt{f'_c}+N_u/6A_g)b_wd,&A_v<A_{v,min}\end{cases}

    Reference: ACI 318-19 / 318-19(22) Table 22.5.5.1 and §§22.5.5, 22.5.10, 9.6.3 and 9.7.6; λs per Table 22.5.5.1.3.

    Worked example

    300 × 500 mm, four 20 mm tension bars, 2-leg 10 mm at 200 mm, Vu 150 kN

    Inputs
    • bw 300 mm, h 500 mm, clear cover to stirrup outside 40 mm, 20 mm main bars.
    • Four tension bars in one layer; no axial force; ACI 318-19.
    • f′c 28 MPa, stirrup fy 420 MPa.
    • 2 legs of 10 mm at 200 mm.
    • Vu 150 kN.
    Calculation path
    1. 1

      d = 500 − 40 cover − 10 stirrup − 10 = 440 mm.

    2. 2

      As = 4 × π × 20²/4 = 1,257 mm²; ρw = As/(bw·d) = 0.0095.

    3. 3

      Av = 157 mm².

    4. 4

      Av,min = max(0.062√28, 0.35) × 300 × 200 / 420 = 50 mm²; Av ≥ Av,min, so use Eq. (a).

    5. 5

      Vc = (0.17 × √28) × 300 × 440 = 118.7 kN.

    6. 6

      Vs = 157 × 420 × 440 / 200 = 145.1 kN.

    7. 7

      φVn = 0.75 × 263.8 = 197.9 kN; utilisation 75.8%.

    8. 8

      Vu > 0.083φλ√f′c bw d, so minimum stirrups are required. Spacing limit d/2 = 220 mm, so 200 mm is acceptable.

    Result

    φVn 197.9 kN · utilisation 75.8% · spacing OK.

    Interpreting the result

    Shear failures are brittle. For this ACI 318-19 / 318-19(22) model, the selected concrete equation depends on Av relative to Av,min; minimum stirrups are also checked against the demand trigger. If utilisation is close to 1.0, add reinforcement only after checking the section maximum and detailing limits.

    Assumptions & limitations

    • Vertical stirrups only.
    • Normal-weight concrete (λ = 1.0).
    • Rectangular, nonprestressed beam with one layer of tension steel; As is estimated from the entered bar count and diameter.
    • Effective depth is measured to the tension-bar centroid using clear cover to the outside of the stirrup; bar fit and clear spacing are not checked.
    • Axial force is uniform and compression is positive; Nu/(6Ag) is capped at 0.05f′c.
    • Torsion, prestress, shear demand along the span, strain compatibility and seismic detailing are not checked. Exceptions to minimum shear reinforcement are not evaluated.
    • ACI 318-25 updates one-way shear provisions and is not implemented; only 318-19 and 318-19(22) are offered.

    Frequently asked questions

    Why is nominal shear strength capped?

    ACI limits the combined nominal shear strength to prevent diagonal-compression failure of the web.

    Does this apply to slabs?

    No. Slabs and footings use different provisions.

    When is minimum shear reinforcement required?

    The model checks the ACI 318-19 §9.6.3.1 SI trigger Vu > 0.083φλ√f′c bw d, then compares the provided Av with Av,min. ACI requires the size-effect equation (c) when Av is below Av,min even if demand does not trigger minimum stirrups.

    Methodology last reviewed 2026-10-21. Results are informational and not financial or engineering advice — see our disclaimer.