Concrete beam design calculator
Reinforced concrete is designed by making the steel carry the tension the concrete cannot. This calculator solves the ACI 318 rectangular-section equation for the steel area a factored moment requires, applies the minimum-reinforcement floor, and checks that the section stays tension-controlled.
| Step | Formula | Value |
|---|---|---|
| β1 | 0.85 − 0.05(f′c − 28)/7 | 0.850 |
| Required moment | Mu / φ | 166.7 kN·m |
| As required | solve φMn = Mu | 940 mm² |
| As minimum | max(0.25√f′c/fy, 1.4/fy)·b·d | 450 mm² |
| Stress block a | As·fy/(0.85·f′c·b) | 55.3 mm |
| φMn | φ·As·fy·(d − a/2) | 150.0 kN·m |
Singly reinforced, rectangular section, tension steel only. Shear and detailing are separate checks.
a = As·fy/(0.85·f′c·b) · Equivalent rectangular block, ACI 22.2.
φMn = φ·As·fy·(d − a/2) · φ = 0.90 for tension-controlled sections.
max(0.25√f′c/fy, 1.4/fy)·b·d · §9.6.1.2.
ρ ≤ 0.319·β1·f′c/fy · Ensures εt ≥ 0.005.
A clear path from inputs to results.
Reinforced concrete is designed by making the steel carry the tension the concrete cannot. This calculator solves the ACI 318 rectangular-section equation for the steel area a factored moment requires, applies the minimum-reinforcement floor, and checks that the section stays tension-controlled.
How to use
- Enter the section width, overall depth, cover and bar size.
- Set f′c and fy.
- Enter the factored moment Mu.
- Read the required steel, the capacity and the tension-control check.
The formula
Reference: ACI 318-19 §22.2 (equivalent rectangular stress block), §9.6.1.2 (minimum steel), §21.2 (tension control).
Worked example
300 × 500 mm beam, 28 MPa concrete, 150 kN·m
- b = 300 mm, h = 500 mm, cover 40 mm, 20 mm bars.
- f′c = 28 MPa, fy = 420 MPa.
- Mu = 150 kN·m.
- 1
Effective depth: 500 − 40 − 10 = 450 mm.
- 2
β1 = 0.85 at 28 MPa.
- 3
Required steel from φMn = Mu: about 940 mm².
- 4
Minimum steel: max(0.25√28/420, 1.4/420) × 300 × 450 = 450 mm², so strength governs.
- 5
Stress block a = 940 × 420/(0.85 × 28 × 300) = 55.3 mm.
- 6
φMn = 0.9 × 940 × 420 × (450 − 27.6) = 150.0 kN·m.
As = 940 mm² (3.1 mm² per mm of width) · φMn 150.0 kN·m · tension-controlled.
Interpreting the result
The reinforcement ratio is the number to watch. Below the tension-control limit the section fails ductilely, with the steel yielding before the concrete crushes. Above it, φ drops and the failure mode becomes brittle — which is why the check matters as much as the steel area.
Assumptions & limitations
- Singly reinforced rectangular section in flexure only.
- Tension-controlled φ = 0.90; the engine flags sections above the limit.
- Shear, torsion, development length, crack control and fire detailing are excluded.
Frequently asked questions
Why is the minimum steel so large?
To prevent brittle failure immediately after cracking. The 1.4/fy term usually governs for normal-strength steel.
Does this design shear reinforcement?
No. Shear, torsion and detailing are separate checks and usually govern the stirrup spacing.
What if the section is too small?
The required steel becomes imaginary. Increase the depth, the width, or both — depth is far more efficient.
Methodology last reviewed 2026-10-21. Results are informational and not financial or engineering advice — see our disclaimer.