Slab reinforcement estimator

The tool lays out an orthogonal two-way bar grid. In each direction the number of bars is the span divided by the on-centre spacing plus one (the “fence-post” rule — a run of length D at spacing s needs D/s gaps but D/s+1 bars): n_x = floor(W/s)+1 and n_y = floor(L/s)+1. Total steel length = (n_x·L + n_y·W) × mats.

Bar weight = total length × the standard unit mass. Metric bars follow the exact rule w = D²/162 ≈ 0.006165·D² kg/m (the mass of a round steel section at 7,850 kg/m³): 8 mm 0.395, 10 0.617, 12 0.888, 16 1.578, 20 2.466, 25 3.853 kg/m. Imperial bars use the ASTM A615 nominal mass (#3 0.560, #4 0.994, #5 1.552, #6 2.235 kg/m).

Tie wire = number of bar crossings (n_x·n_y·mats) × ~0.2 m of 16-gauge wire per tie (16 ga = 1.588 mm = 0.01555 kg/m), converted to 3.5 lb (≈1.6 kg) coils. Chairs = (L/grid+1)×(W/grid+1), about one per m² at a 1 m grid. The mesh alternative = slab area ÷ ~12 m² effective coverage per 6.0×2.4 m sheet (14.4 m² gross minus one 300 mm lap each way).

Bars ship in fixed stock lengths and any long span needs lap splices (~40–50× bar diameter), so a waste/lap allowance (default 10%) is applied to both the steel and the tie-wire figures — this is the single factor most take-offs omit and the only place an estimate can under-order.

A 6×4 m slab at 200 mm both ways, 12 mm bar, single mat: n_x = floor(4/0.2)+1 = 21, n_y = floor(6/0.2)+1 = 31; length = 21×6 + 31×4 = 250 m; weight = 250×0.888 = 222 kg, ×1.10 waste ≈ 244 kg. Crossings = 21×31 = 651, tie wire ≈ 651×0.2×0.01555×1.10 ≈ 2.2 kg (≈ 2 coils).

Bars are assumed to run at the nominal spacing and every intersection tied (an upper bound — crews often tie a ~50–60% checkerboard). This is a material take-off, not a structural design: bar size, spacing, cover, laps and development length must come from the engineer’s design to ACI 318 / Eurocode 2.