How to Calculate Shear Strength of Rebar

The Truss-and-Strut Model: Where Vc and Vs Come From

Rebar does not resist shear directly — the longitudinal bars run parallel to the beam axis and cross a 45° shear crack at the wrong angle. Shear is resisted by stirrups (transverse ties) that intersect the diagonal crack directly, plus the concrete's own diagonal compression strut. ACI 318 models this as a parallel truss: the concrete forms diagonal compression struts between the load and the support (the Vccontribution), and the stirrups act as vertical tension ties crossing those struts (the Vs contribution). The total nominal shear capacity isVn = Vc + Vs, with both terms computed separately and added. The terminal above computes both — Vc from the simplified 2√fc'·b·d expression, and Vs from Av × fy × d/s.

The stirrup formula Vs = Av × fy × (d/s) is read as "the tension force per stirrup (Av × fy) times the number of stirrups crossing the 45° crack (d/s)". For a #4 two-leg stirrup (Av = 0.40 in²) at 12 in spacing in a beam with d = 21 in and fy = 60 ksi, the term d/s = 21/12 = 1.75 stirrups cross the crack, so Vs = 0.40 × 60 × 1.75 = 42 kip. Add the concrete Vc = 2 × 2.0 × 12 × 21 = 21.8 kip (for 4000 psi concrete, 12-in web) and Vn = 63.8 kip — well within the strut crushing cap of 8 × 2.0 × 12 × 21 = 174 kip. The terminal regenerates all of these numbers live as you change the inputs.

Concrete vs Steel Shear Share: How the Load Splits

The shear load does not split evenly between concrete and steel — it shifts non-linearly as the applied shear Vu rises from the cracking threshold to the strut limit. At low Vu, the concrete carries nearly everything and the ties are minimum-only; at high Vu, the steel carries 60–80% and the concrete contribution stays roughly flat at Vc. The table below traces the split across five load levels for a 12×24-in beam with 4000 psi concrete and #4 ties at 12 in.

Script Key-Line Deconstruction

Five lines of the shear-strength script carry the load — and each one silently fails the shear check if mis-read. The table deconstructs them into purpose, the check they enforce, and the safety rule that keeps them valid. The terminal above regenerates these lines live as you change the inputs; the table below explains why each one is a trap if read in isolation.

Minimum Tie Area and Maximum Spacing: The Two Gates

Two ACI 318 gates prevent the most catastrophic shear failures, and both are enforced in the terminal above. The minimum tie area Av,min = 0.75√fc'·bw·s/fy guarantees that even a low-shear beam has enough steel to prevent sudden diagonal-tension failure — without it, a beam that nominally does not need ties (because Vu < Vc) can still fail explosively when an unexpected crack forms. For a 12-in web with 4000 psi concrete, #3 ties at 12 in, fy = 60 ksi: Av,min = 0.75 × 2.0 × 12 × 12 / 60,000 = 0.018 in² — trivially satisfied by a single #3 leg (0.11 in²), but the gate still must be checked.

The maximum spacing s ≤ min(d/2, 24 in) prevents a 45° diagonal crack from crossing zero stirrups. A 45° crack in a 21-in-deep beam spans 21/2 = 10.5 in of beam length — if ties are spaced at 12 in, a crack could fall between two ties and carry no transverse steel, failing the beam as if it had no stirrups. The d/2 limit guarantees at least one stirrup crosses any 45° crack. In high-shear regions (Vu > 2φVc) and seismic plastic-hinge zones, the limit tightens to s ≤ d/4 — at least two stirrups per crack, for redundancy. The terminal above flags any spacing that exceeds s_max with a red gate failure.

Shear Failure Modes and the Reinforcement Strategy for Each

Shear failure is brittle and sudden — unlike flexural failure, which yields and gives warning, shear failure drops the member with little visible distress. The four failure modes below each have a distinct trigger and a distinct prevention strategy. The terminal above is the design-time defense against all four: it computes Vs, Vn, and the gate checks that match each failure mode.

Two Hidden Pitfalls That Fail Every Shear Check

From One-Off Script to ACI 318 Design Procedure

The terminal above is a single-bar check — useful for verifying a stirrup layout, not for designing one from scratch. The full ACI 318 shear design procedure scales this script into a five-step workflow: (1) compute Vu from the load and span; (2) compute φVc and check if Vu > φVc/2 (the threshold above which ties are required); (3) if so, solve for the required Av = (Vu/φ − Vc) × s / (fy × d); (4) pick a bar size and leg count, compute s_required = Av × fy × d / (Vu/φ − Vc); (5) enforce s ≤ s_max and Av ≥ Av,min — use the more restrictive. The terminal above covers steps 3–5; steps 1–2 come from the structural analysis. Once the procedure stabilizes on a section, the script becomes the production-safe verification tool that every stirrup layout is checked against.

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