How to use this calculator
- Enter or choose units.
- Enter or choose width to cover (across the bars).
- Enter or choose end cover each side.
- Enter or choose maximum centre-to-centre spacing.
- Enter or choose bar diameter.
- The number of bars updates as you type. Use Copy result or Share to keep it.
Method
Gaps = ⌈(W − 2c) ÷ s_max⌉; Bars = Gaps + 1; s = (W − 2c) ÷ Gaps; As per metre = (π d² ÷ 4) × 1000 ÷ s
- W
- width the bars are spread across
- c
- end cover to the first and last bar
- s_max
- maximum spacing
- d
- bar diameter
Worked example
- Clear width = 3000 − 2 × 50 = 2900 mm; 2900 ÷ 150 = 19.3 → 20 gaps.
- Bars = 20 + 1 = 21 at 2900 ÷ 20 = 145 mm.
- One 12 mm bar = 113.1 mm²; × 1000 ÷ 145 = 780 mm²/m. Length = 21 × 4 = 84 m.
A 3 m width at no more than 150 mm centres needs 21 bars at 145 mm, giving 780 mm² of steel per metre.
What the result means
Compare the steel area with the area your design requires. Repeat the calculation in the other direction for a two-way mat.
Assumptions and limits
- Bars are evenly spaced with the same cover at both ends.
- Does not check code minimum or maximum spacing, cover or lap requirements; those come from the design.
Questions
Why add one bar?
Bars sit at both ends of every gap, so there is always one more bar than gaps, like fence posts.
Why is the spacing smaller than the maximum?
The gaps are rounded up to a whole number and then spread evenly, so the real spacing never exceeds the maximum.
Please note: Quantities are estimates. Confirm requirements with your engineer or supplier before ordering materials.