How to Compute Rebar Quantity for Slabs, Beams, and Columns (Step-by-Step with Worked Examples)

Rebar take-off turns a structural drawing into a bill of materials and a cutting list a fabricator can use. A wrong cutting-length or bar-count assumption can waste hundreds of kilograms of steel. This guide covers the unit-weight formula, cutting length with bend/hook allowances, lap splice length, and spacing-to-bar-count conversion — with three worked examples: a slab strip, a beam, and a column. Values follow NSCP 2015 and ACI 318-19 by family; verify project figures against the structural drawings.

Why Rebar Take-off Accuracy Matters

Steel is priced by weight but placed by length and count. A take-off based only on "kg per m³ of concrete" can price a bid but not generate a usable cutting list. A proper take-off yields total tonnage for procurement, individual cutting lengths for the bending yard, and a bar count the placing crew can verify.

Step 1 — Unit Weight of Reinforcing Steel

Unit weight comes from a bar's cross-sectional area and steel density, about 7,850 kg/m³. For diameter d in millimeters:

W = d² ÷ 162  (kg per linear meter)

This rounds the exact relationship W = (π/4) × d² × 7,850 ÷ 10⁶ = d²/162.19. Using 162 is standard regional practice, accurate to about 0.1–0.2%. Common bar sizes and unit weights:

Our free Reinforcement Calculator automates this unit-weight lookup (plus bar-area and spacing conversions) so you can move straight to the cutting-length and total-weight math shown below.

Step 2 — Building the Cutting Length

Cutting length is clear length plus whatever each end needs — a hook, a bend, or an embedment into an adjoining member:

L_cut = L_clear + Σ(hook or bend allowances) − Σ(bend deductions, if used)

Per ACI 318-19 Chapter 25 (mirrored by NSCP 2015): a 180° hook needs about 4d_b extension (not less than ~65 mm); a 90° main-bar hook needs about 12d_b. For stirrup/tie hooks (135° seismic hooks on closed stirrups and ties), the minimum extension is 6d_b, not less than ~75 mm — so 75 mm governs for 10 mm bars, as used in the examples below. A small per-corner bend deduction is sometimes applied, but is under 1–2% for these diameters and is ignored here.

Step 3 — Lap Splice Length

A lap splice is needed when a required length exceeds available stock (6, 9, or 12 m bars) or where drawings call for a construction joint, such as column verticals spliced at every floor. The common rule of thumb:

L_lap ≈ 40 × d_b

This 40d figure is a planning number for ordinary tension splices of Grade 40/60 bars in normal-strength concrete — not a substitute for a real development-length check, which depends on bar size, f′c, steel grade, spacing/cover, and splice class (Class B ≈ 1.3×l_d). Verify against NSCP 2015 / ACI 318-19 provisions before finalizing a BBS.

Step 4 — Converting Spacing to Bar Count

For a specified spacing (e.g., 12 mm ϕ @ 150 mm o.c.), bar count to cover length L:

N = ceiling(L ÷ s) + 1

Rounding the space count up before adding 1 keeps the actual spacing at or tighter than specified — never wider.

Worked Example 1 — One-Way Slab Strip

Slab panel 4.0×3.0 m, 150 mm thick, spanning the 3.0 m direction. Main bars: 12 mm ϕ @ 150 mm o.c. across the span, spaced along the 4.0 m length. Temp bars: 10 mm ϕ @ 200 mm o.c., spaced along the 3.0 m width. Embedment: 100 mm/end.

Main bars (12 mm): L_cut = 3,000 + 100 + 100 = 3,200 mm = 3.2 m.
N = ceiling(4,000 ÷ 150) + 1 = 27 + 1 = 28 bars.
Total length = 28 × 3.2 = 89.6 m. Weight = 89.6 × 0.889 = 79.65 kg.

Temperature bars (10 mm): L_cut = 4,000 + 100 + 100 = 4,200 mm = 4.2 m.
N = ceiling(3,000 ÷ 200) + 1 = 15 + 1 = 16 bars.
Total length = 16 × 4.2 = 67.2 m. Weight = 67.2 × 0.617 = 41.46 kg.

Slab subtotal = 79.65 + 41.46 = 121.11 kg before wastage (≈10.1 kg/m² for the 12 m² panel). With 7% wastage: 121.11 × 1.07 = 129.59 kg.

Worked Example 2 — Rectangular Beam (Main Bars + Stirrups)

Beam 300×500 mm, 5.0 m clear span, 40 mm cover. Bars: 4–20 mm ϕ bottom, 2–16 mm ϕ top, 300 mm embedment/support. Stirrups: 10 mm ϕ closed @ 150 mm o.c. (uniform for clarity).

Bottom bars (20 mm): L_cut = 5,000 + 300 + 300 = 5,600 mm = 5.6 m.
Total length = 4 × 5.6 = 22.4 m. Weight = 22.4 × 2.469 = 55.31 kg.

Top bars (16 mm): L_cut = 5.6 m (same embedment).
Total length = 2 × 5.6 = 11.2 m. Weight = 11.2 × 1.580 = 17.70 kg.

Stirrups (10 mm): outer dimensions = (300 − 80) × (500 − 80) = 220 × 420 mm.
Perimeter = 2 × (220 + 420) = 1,280 mm. + two 75 mm hooks: L_cut = 1,280 + 150 = 1,430 mm = 1.43 m.
N = ceiling(5,000 ÷ 150) + 1 = 34 + 1 = 35 stirrups.
Total length = 35 × 1.43 = 50.05 m. Weight = 50.05 × 0.617 = 30.88 kg.

Beam subtotal = 55.31 + 17.70 + 30.88 = 103.89 kg before wastage. With 7% wastage: 103.89 × 1.07 = 111.16 kg.

Worked Example 3 — Tied Column (Verticals + Ties)

Column 450×450 mm, 3.0 m story height, 40 mm cover. Verticals: 8–20 mm ϕ, spliced every floor. Max tie spacing = least of 16×d_b=320 mm, 48×d_tie=480 mm, or 450 mm — 320 mm governs (ACI 318-19 / NSCP 2015 provisions); this example uses 150 mm o.c.

Vertical bars (20 mm): lap = 40 × 20 = 800 mm. L_cut = 3,000 + 800 = 3,800 mm = 3.8 m.
Total length = 8 × 3.8 = 30.4 m. Weight = 30.4 × 2.469 = 75.06 kg.

Ties (10 mm): outer dimensions = 450 − 80 = 370 mm each side.
Perimeter = 2 × (370 + 370) = 1,480 mm. + two 75 mm hooks: L_cut = 1,480 + 150 = 1,630 mm = 1.63 m.
N = ceiling(3,000 ÷ 150) + 1 = 20 + 1 = 21 ties.
Total length = 21 × 1.63 = 34.23 m. Weight = 34.23 × 0.617 = 21.12 kg.

Column subtotal = 75.06 + 21.12 = 96.18 kg before wastage. With 7% wastage: 96.18 × 1.07 = 102.91 kg.

Summary Table

Wastage Allowance (5–10%)

Even a correct cutting list loses material on site — offcuts, bending errors, damaged bars, unusable remnants. A 5–10% wastage allowance on computed steel weight is standard practice: 5% for simple elements like straight slab bars, closer to 10% for congested joints or heavily bent members. The 7% used above is a reasonable mid-range figure — confirm the value your project uses.

Assumptions & Limitations

• W = d²/162 is a rounded approximation (true ≈ d²/162.19); mill certificates may differ slightly.
• Hook allowances (75 mm min. for stirrup/tie hooks, 4d_b/12d_b for main-bar hooks) are typical; governing figures come from the drawings and applicable NSCP 2015 / ACI 318-19 clauses.
• The 40d lap rule is for quick take-off only; verify against calculated l_d for the actual f′c, f_y, spacing/cover, and splice class.
• Embedment allowances (100 mm slab, 300 mm beam) are illustrative, not calculated development lengths.
• N = ceiling(L/s) + 1 assumes bars start near the edge; half-spacing detailing shifts the count slightly.
• Examples ignore congestion, splice staggering, and joint clashes — check the placing drawings.
• This is an estimating guide, not a substitute for a signed and sealed structural design.

Frequently Asked Questions

Why does my computed steel weight differ from the supplier's receipt?

Mill-rolled bars vary slightly from nominal mass, and suppliers bill by actual weighed tonnage, not the theoretical d²/162 figure. A 1–3% gap is normal — one reason wastage is built into procurement.

Is 40d always the correct lap length?

No — it's a planning shortcut. Actual length depends on f′c, f_y, bar size, spacing/cover, casting position, and splice class per NSCP 2015 / ACI 318-19. Confirm against the design calculations for critical or congested members.

How much wastage should I allow when ordering rebar?

5% for simple, repetitive elements like straight slab bars; 10% for heavily bent members or congested columns. Many Philippine contractors use 7–8% as a practical average.

For a full project-wide take-off with auto-generated column, beam, and slab schedules, see RHCES's Bar Bending Schedule calculator — part of 150+ free structural and civil engineering calculators.