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    Calculates locally in your browser · ACI 318-19 §11.5.3
    Concrete / Shear walls

    Shear wall calculator

    The in-plane shear strength of a reinforced concrete wall: the concrete contribution including the axial term, the horizontal steel contribution, and the design strength against the demand — with the minimum reinforcement ratio checked.

    Wall3.0 mAspect h/l1.00VerdictPass
    Design shear
    788 kN
    φVn = 0.75 × 1051
    Concrete Vc
    547 kN
    Includes the axial term
    Steel Vs
    504 kN
    ρh 0.25%
    Utilisation
    51%
    Demand 400 kN
    Shear strength
    ACI 318-19 §11.5.3, with the axial contribution separated.
    Shear strength
    ComponentValueUnit
    Concrete Vc546.9kN
    Horizontal steel Vs504.0kN
    Nominal Vn1050.9kN
    Design φVn788.2kN
    Demand400kN
    Minimum ρh0.20%

    ACI 318-19 §11.5.3: Vc = 0.17·λ·√f'c·h·d + Nu·d/(4·lw), φ = 0.75.

    Inspect the model
    Choose a quick result view, a scaled comparison, or the complete diagnostic set.
    Design shear against horizontal steel
    The concrete carries a fixed share; the steel adds linearly.
    Design shear against horizontal steel0.000295.01590.02885.021180.031475.040%0%0.1%0.1%0.2%0.2%0.25%0.25%0.3%0.3%0.4%0.4%0.6%0.6%φVn (kN)
    Canonical equations
    The exact relationships implemented by this engine.

    Vc = 0.17·λ·√f'c·h·d + Nu·d/(4·lw) · Nu positive in compression.

    Vs = ρh · h · d · fy · Horizontal reinforcement only.

    φVn = 0.75 (Vc + Vs) · ACI 318-19 §11.5.3.

    engineering methodology

    A clear path from inputs to results.

    A shear wall resists lateral load in its own plane, and its strength comes from two sources: the concrete, which gains capacity from axial compression, and the horizontal reinforcement. This separates the two and checks the minimum steel ratio.

    Step-by-step methodWorked exampleAssumptions explained

    How to use

    1. Enter the wall length, thickness and effective depth.
    2. Set the concrete and steel strengths and the axial compression.
    3. Compare the design shear with the demand, and check the reinforcement ratio.

    The formula

    Vc=0.17λfc′ h d+Nud4lw,ϕVn=0.75(Vc+Vs)V_c = 0.17\lambda\sqrt{f'_c}\,h\,d + \frac{N_u d}{4 l_w}, \qquad \phi V_n = 0.75(V_c + V_s)

    Reference: ACI 318-19 §11.5.3; minimum horizontal steel §11.6.2.

    Worked example

    3 m × 200 mm wall, 500 kN axial, 0.25% steel

    Inputs
    • f'c 30 MPa, fy 420 MPa, d = 2,400 mm, 400 kN shear demand.
    Calculation path
    1. 1

      Vc = 0.17 × 5.477 × 200 × 2400 + 500,000 × 2400/(4 × 3000) = 447 + 100 = 547 kN. Vs = 0.0025 × 200 × 2400 × 420 = 504 kN. φVn = 0.75 × 1,051 = 788 kN.

    Result

    788 kN capacity against 400 kN demand — 51% utilised.

    Interpreting the result

    Axial compression is worth real shear capacity, which is why a wall in a tall building performs better than the same wall tested alone. Net tension removes the benefit entirely.

    Assumptions & limitations

    • In-plane shear only.
    • Axial force is compression; tension needs the tension provisions.
    • No boundary-element or stability check.
    • Slender walls also need Ch. 11 and Ch. 18 checks.

    Frequently asked questions

    Why does compression increase shear strength?

    It delays diagonal cracking; the Nu·d/(4lw) term is the code’s linearisation of that effect.

    Does this cover seismic walls?

    No — special structural walls need the Ch. 18 provisions, including boundary elements and the shear amplification.

    What is the minimum horizontal steel?

    ρ = 0.0020 for bars up to No. 16 and 0.0015 for larger, in each direction.

    What depth should I use?

    d = 0.8·lw is common for a wall with a concentrated end zone; use the actual distance to the centroid of the tension steel where it is known.

    Methodology last reviewed 2026-10-22. Results are informational and not financial or engineering advice — see our disclaimer.