Concrete Beam Design Calculator
Flexural strength of singly reinforced rectangular beams — stress block, φMₙ, and the minimum-steel check. Answer in five seconds; every step spelled out below.
Every equation verified against the ACI 318-19 registry · No account required
The full calculation, spelled out
Below is the live CalcPackage editor in public view — the same worksheet an engineer would stamp: bar areas from the ASTM tables, the equivalent stress block, the cited φ factor, and both pass/fail checks in sequence.
Change any input — every dependent block recomputes. Drop to two bars and watch minimum reinforcement take over from flexure. Click ✦ Explain on any block — including why the 3√f'c constant secretly carries √psi units and how the worksheet makes that explicit.
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How this calculator works
The calculator checks a simply supported, singly reinforced rectangular concrete beam using LRFD per ACI 318-19. The factored load is wᵤ = 1.2w_D + 1.6w_L (ASCE 7-22 load combination 2) with Mᵤ = wᵤL²/8 at midspan. From horizontal equilibrium the equivalent stress block depth is a = A_s f_y / (0.85 f'_c b), and the design strength is φMₙ = 0.90 · A_s f_y (d − a/2) for a tension-controlled section. Minimum flexural reinforcement is the larger of the 3√f'_c/f_y and 200/f_y bounds (Eq. 9.6.1.2) times b·d.
Assumptions and limitations
Simply supported single span, uniform load, singly reinforced rectangular section, normalweight concrete, and a tension-controlled section (εt ≥ 0.005, so φ = 0.90 — verified for the default geometry). Shear and stirrup design, deflection, crack control, development lengths, and T-beam behavior are covered in the full editor templates. This calculator is a design aid — engineering judgment and final responsibility rest with the licensed engineer of record.
Frequently asked questions
Is d the beam height?
No — d is the effective depth, from the compression face to the centroid of the tension steel. It already nets out cover, the stirrup, and half the bar diameter. For a 24-inch-deep beam with #8 bars and #4 stirrups, d ≈ 21.5 in.
Why does minimum steel govern with small bars?
Lightly reinforced beams can fail abruptly when the concrete cracks — ACI 9.6.1.2 requires enough steel that the reinforced section is at least as strong as the uncracked one. Drop to two #5 bars and watch that check, not flexure, become the story.
Is the calculation verified?
Yes. Every equation is matched against CalcPackage's ACI 318-19 equation registry — including the hidden-unit √f'c constant, which is registered with explicit psi^0.5 units — and dimension-checked on every recomputation.