Punching Shear in Flat Slabs and Footings: ACI 318-25 Guide

Punching shear failure is characterised by a diagonal crack propagating from the column face outward and downward through the slab at approximately 45°. The failure surface is a truncated cone or pyramid. The concrete within the critical perimeter is "punched" downward relative to the surrounding slab.

This failure mode is brittle — there is little warning and sudden loss of load-carrying capacity. The catastrophic progressive collapse at the L'Ambiance Plaza in 1987 and numerous parking garage failures have been attributed to punching shear. Good detailing — specifically continuous bottom reinforcement passing through the column cage — can prevent progressive collapse even after a punching failure (ACI §8.7.4.2).

Key variables:

• Slab effective depth d (average of x and y directions)
• Column size (c₁ × c₂) or circular diameter
• f'_c — concrete compressive strength (tensile cracking mechanism)
• Column location — interior, edge, corner (affects perimeter and moment transfer)

ACI 318-25 §22.6.4.1: The critical section for two-way (punching) shear is located at d/2 from the column face (or concentrated load area). For a square column c×c:

For a rectangular column c₁×c₂:

When φV_c < V_u, shear reinforcement is required. Options:

Shear studs (headed stud reinforcement — HSR) are the dominant solution in modern flat plate construction. They are pre-welded to flat steel rails and placed on rebar before concrete casting. Their higher V_n ceiling (0.66√f'_c vs 0.50√f'_c for stirrups) and ease of placement make them preferred.

Minimum Slab Thickness Without Shear Reinforcement (Rule of Thumb)

For interior columns carrying ~100 kN/m² total factored load, and column size c≈400–600 mm, a slab without shear reinforcement typically requires d ≥ 200 mm (h ≥ 250 mm). Use the Foundation Design Calculator to check specific cases.

Given: Interior column 500×500 mm, flat plate h=250 mm, f'_c=30 MPa, normal weight, factored column load V_u=1,400 kN. Flexural reinforcement ρ=0.006 in each direction.

Step 1 — Effective depth: d = 250−40−16/2 = 202 mm (average for two orthogonal bar layers, Ø16 bars)

Step 2 — Critical perimeter: b_o = 4×(500+202) = 4×702 = 2,808 mm

Step 3 — Three V_c formulas (interior, β_c=1.0, α_s=40, ρ_w=0.006, λ=1.0):

V_c1 = [0.33×1.0×(0.006)^1/3×√30]×2,808×202×10⁻³ = [0.33×0.1817×5.477]×567,216×10⁻³ = 0.3280×567.2 = 186.0 kN

V_c2 = [0.17×(1+2/1.0)×0.1817×5.477]×567.2 = [0.17×3×0.995]×567.2 = 0.5073×567.2 = 287.7 kN

V_c3 = [0.083×(2+40×202/2808)×0.1817×5.477]×567.2 = [0.083×(2+2.88)×0.995]×567.2 = [0.083×4.88×0.995]×567.2 = 0.4031×567.2 = 228.5 kN

Governing: V_c = V_c1 = 186.0 kN → φV_c = 0.75×186.0 = 139.5 kN

Wait — V_u=1,400 kN >> φV_c=139.5 kN. Note these V_c formulas give unit stress × area:

V_c1,total = 0.3280 × 2,808 × 202 / 1,000 = 186.0 kN (already includes b_o·d) → correct total.