About Shear Wall Design Calculator
The shear wall calculator performs the in-plane (horizontal) shear strength design of a reinforced-concrete structural wall using the ACI 318 provisions in SI form. It evaluates the concrete contribution Vc, the distributed horizontal reinforcement contribution Vs, the nominal shear strength Vn (subject to an upper limit), and the design strength phi*Vn, then compares it against the factored shear demand Vu.
Enter the wall length lw, web thickness t, height hw, concrete strength f'c, reinforcement yield strength fy, the factored shear Vu, and the provided horizontal reinforcement ratio rho_t. The tool reports the gross web area Acv, each strength component, the demand/capacity utilization, whether the wall passes, the minimum reinforcement check, and the horizontal reinforcement ratio required to carry the demand.
How It Works
- Compute the gross web shear area Acv = lw * t and the wall aspect ratio hw / lw.
- Evaluate the concrete shear strength Vc = 0.17 * lambda * sqrt(f'c) * Acv (SI, MPa), with lambda for lightweight concrete.
- Add the horizontal reinforcement contribution Vs = rho_t * fy * Acv, where rho_t = Av / (t * s).
- Cap the nominal strength at Vn = min(Vc + Vs, 0.66 * sqrt(f'c) * Acv), apply phi = 0.75, and check phi*Vn against Vu while verifying rho_t >= 0.0025.
Worked Example
A wall with lw = 3000 mm, t = 250 mm gives Acv = 750,000 mm^2. With f'c = 28 MPa (sqrt = 5.2915) and normal-weight concrete (lambda = 1), Vc = 0.17 * 1 * 5.2915 * 750000 = 674,667 N. With rho_t = 0.0025 and fy = 420 MPa, Vs = 0.0025 * 420 * 750000 = 787,500 N. The uncapped Vn = 674,667 + 787,500 = 1,462,167 N, below the cap 0.66 * 5.2915 * 750000 = 2,619,294 N, so Vn = 1,462,167 N. The design strength phi*Vn = 0.75 * 1,462,167 = 1,096,625 N. Against Vu = 900,000 N the utilization is 900000 / 1096625 = 0.82, so the wall is adequate and rho_t = 0.0025 meets the 0.0025 minimum.
Formulas
- Gross web shear area
Acv = lw * t- Concrete shear strength (SI)
Vc = 0.17 * lambda * sqrt(fc) * Acv- Horizontal reinforcement strength
Vs = rho_t * fy * Acv- Nominal strength with upper limit
Vn = min(Vc + Vs, 0.66 * sqrt(fc) * Acv)- Design strength and check
phi*Vn >= Vu, phi = 0.75, rho_t >= 0.0025
Standards & References
- ACI 318 Building Code Requirements for Structural Concrete (Chapter 11/18 shear walls)
- ACI 318 §11.5.4 wall shear strength
- ACI 318 minimum distributed reinforcement (rho_t, rho_l >= 0.0025)
Frequently Asked Questions
What units does the shear wall calculator use?
It uses a consistent SI set: lengths in mm, concrete strength f'c and yield strength fy in MPa, and forces in N. The Vc coefficient 0.17 and the 0.66 cap are the SI (MPa) forms of the ACI 318 expressions.
Why is the nominal shear strength capped at 0.66*sqrt(fc)*Acv?
ACI 318 limits the maximum nominal shear strength of a wall to prevent web crushing of the concrete, regardless of how much horizontal reinforcement is provided. Once Vc + Vs reaches 0.66*sqrt(f'c)*Acv (8.3*sqrt(f'c) in psi units) the section is governed by diagonal compression, so adding more steel does not increase strength.
What is the minimum horizontal reinforcement ratio?
ACI 318 requires a minimum distributed reinforcement ratio of 0.0025 in both the horizontal (rho_t) and vertical (rho_l) directions for shear walls. The calculator flags whether the provided rho_t meets this minimum.
How is the required reinforcement ratio computed?
If the concrete alone cannot carry the demand, the required steel strength is Vs,req = Vu/phi - Vc, and the required ratio is rho_t = Vs,req / (fy * Acv). When phi*Vc already exceeds Vu the required ratio is reported as zero, though the minimum 0.0025 still applies.