How to Calculate Rebar for a Slab

Step 1 — Establish the Grid Layout

A reinforced concrete slab typically contains a two-way orthogonal grid: bars running in the X-direction (often called the “bottom layer” in the span direction) and bars running in the Y-direction (the “distribution layer”). For a two-way spanning slab, both layers are structural; for a one-way slab, the perpendicular layer is distribution reinforcement only.

Define your slab dimensions: length (L) and width (W) in metres, plus the bar spacing (s) in mm specified on the structural drawing. The number of bars in one direction is:

n = floor((L × 1000) / s) + 1

Add edge bars as required by the drawing (commonly 2 × edge bars at each side).

Step 2 — Calculate Bar Lengths Including Laps

Stock bars are available in lengths of 6–18 m (commonly 12 m). Where the slab dimension exceeds the stock length, bars must be lapped. The minimum lap length (l₀) under EN 1992-1-1 for B500B in good bond conditions is typically 40–50 × bar diameter depending on concrete class and position (bottom vs. top, edge vs. centre). For a 12 mm bar, that gives a lap of approximately 480–600 mm.

Total bar length for one bar line across width W:

Bar length = W + 2 × cover + laps (if W > stock length)

Step 3 — Compute Total Linear Metres

Multiply the number of bars in each direction by their respective lengths:

• X-direction bars: n_X bars, each of length W + covers + laps
• Y-direction bars: n_Y bars, each of length L + covers + laps

Total linear metres = (n_X × length_X) + (n_Y × length_Y)

Step 4 — Convert to Kilograms Using the Weight Formula

The DIN 488 / EN 10080 linear weight formula is: kg/m = d² × 0.00617 (where d is diameter in mm).

Total mass (kg) = Total linear metres × kg/m for the chosen diameter.

Step 5 — Apply a Wastage Allowance

Site cutting generates offcuts, particularly at laps and around openings. A typical wastage allowance is 5–10 % for regular rectangular slabs with few openings, rising to 12–15 % for irregular shapes or heavily penetrated slabs. Apply the factor to your calculated mass to arrive at the order quantity.

Worked Example

Slab: 15 m × 10 m. Bar diameter: 12 mm B500B. Spacing: 200 mm c/c both ways. Cover: 35 mm. Stock length: 12 m. No laps required in the 10 m direction; one lap of 500 mm required in the 15 m direction.

• X-direction (across 10 m): n_X = floor(15 000 / 200) + 1 = 76 bars; each 10.07 m → 76 × 10.07 = 765 m
• Y-direction (across 15 m): n_Y = floor(10 000 / 200) + 1 = 51 bars; each 15.50 m (lap) → 51 × 15.50 = 791 m
• Total linear metres = 765 + 791 = 1,556 m
• Mass = 1,556 × 0.888 = 1,382 kg
• Add 7 % wastage: 1,382 × 1.07 ≈ 1,479 kg (≈ 1.5 tonnes)

Related Reading

• Reinforcing Steel Applications — structural use cases and specification guidance
• B500B Reinforcing Steel Bar — diameters 8–40 mm, cut lengths, mill certificates
• Reinforcing Steel Mesh — standard and bespoke panels to DIN 488-4
• How to Tie Rebar — wire gauges, patterns, and spacing guidance