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Steel & Rebar

How to Prepare a Bar Bending Schedule (BBS): Step-by-Step, with Formulas

A BBS turns structural drawings into an exact steel order. Here's the full method — cutting length, bend deductions, lap length and unit weights — with a worked beam example.

Key takeaways

  • A bar bending schedule lists every bar in a structural member with its shape, cutting length, count and weight.
  • Cutting length = the sum of the bar's leg dimensions + hooks − bend deductions (1d per 45°, 2d per 90°, 3d per 135°).
  • Weight comes from the d²/162 rule, which gives kilograms per metre directly from the bar diameter in millimetres.
  • A schedule prepared from drawings typically brings steel wastage down from around 5% to under 2%.

Walk onto any well-run site and ask how much steel is in the third-floor slab — the site engineer will pull out one document. The bar bending schedule is the bridge between the structural drawing and the steel yard: it converts every bar shown on the drawing into a shape, a cutting length, a count and a weight you can order, cut and bill against.

This guide follows IS 2502 (bending and fixing of bars) and IS 456 (plain and reinforced concrete), which is what most projects in India and much of the Gulf specify. The method is identical under ACI 315 or BS 8666 — only the shape codes and the standard bend allowances change.

What a bar bending schedule actually contains

Every row of a BBS describes one bar mark — one shape, one diameter, repeated a known number of times. The columns rarely change from one company to the next: bar mark, member, shape code, diameter, spacing, number of bars, cutting length, total length and weight.

The shape code is a small sketch of the bent bar — a stirrup, an L-bend, a crank — with each leg dimensioned. It exists so the bar bender never has to interpret the structural drawing. If your schedule needs a phone call to be understood, it is not finished.

Bar mark Dia (mm) Shape Nos. Cutting length Weight (kg)
B1 — bottom main16Straight + L47.44 m47.02
B2 — top main12Straight + L27.08 m12.59
S1 — stirrups8Rectangular, 135° hooks411.348 m21.83

The schedule built in the worked example below — a 6 m simply supported beam, 300 × 450 mm.

Step 1: Work out the cutting length

Cutting length is the straight length of bar you actually cut, before bending. Take the sum of the bar's leg dimensions measured along its centre line, add any hooks, then subtract a deduction for every bend.

Cutting length = Σ (leg dimensions) + Σ (hooks) − Σ (bend deductions)

For a straight bar anchored at both ends: Cutting length = clear span + (2 × Ld) − (2 × 2d)

Bend deductions trip up more freshers than any other part of the schedule. Steel stretches on the outside of every bend, so a bent bar measures longer than the sum of its legs. The standard corrections are:

  • 45° bend — deduct 1d
  • 90° bend — deduct 2d
  • 135° hook — deduct 3d

Hooks work the other way — they add length. A 135° hook on a stirrup adds 9d per end under IS 2502, subject to a 75 mm minimum. Miss the hooks and every stirrup arrives short; miss the deductions and every bar arrives a few centimetres long. On a thousand-bar order, either is real money.

Step 2: Add lap length where bars are joined

Steel is rolled in 12 m lengths. Anything longer than that — a continuous beam, a column running up several floors — has to be joined, and the joint is made by overlapping two bars so that force transfers from one to the other through the surrounding concrete. That overlap is the lap length.

Lap length is derived from development length, Ld, which is the length of embedment needed to develop the bar's full design strength:

Ld = (φ × σs) / (4 × τbd)

φ = bar diameter · σs = 0.87 fy · τbd = design bond stress (IS 456 cl. 26.2.1.1), increased 60% for deformed bars

For Fe415 bars in M20 concrete this works out at almost exactly 47d — which is where the familiar site rule of thumb comes from. IS 456 cl. 26.2.5.1 then requires the lap in flexural tension to be at least Ld or 30d, whichever is greater, and in compression at least Ld in compression or 24d.

In practice most Indian project specifications simplify this to 50d in tension and 40d in compression, and that is what you will usually see called up on the drawing. Two rules that matter more than the number itself: never lap all the bars in a section at the same level — stagger the joints — and never lap in a zone of maximum bending moment if the drawing gives you any choice.

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Step 3: Convert length to weight

Steel is bought by weight, not length, so every schedule ends with the same conversion. The unit weight of a round bar in kilograms per metre is its diameter in millimetres, squared, divided by 162:

Unit weight (kg/m) = d² / 162

d = bar diameter in mm. For US bar sizes the parallel rule is n² / 24 lb/ft, where n is the bar number in eighths of an inch.

A 16 mm bar weighs 1.580 kg/m; an 8 mm stirrup, 0.395 kg/m. Multiply by total length per bar mark, sum the column, add 2–3% for wastage, and that is your order quantity. The full derivation of the d²/162 rule is worth reading once — it makes the number impossible to forget.

Worked example: a 6 m beam

Take a simply supported beam with a clear span of 6.0 m and a section of 300 × 450 mm. Clear cover is 25 mm. Reinforcement is 4 nos 16 mm bottom, 2 nos 12 mm top, and 8 mm two-legged stirrups at 150 mm centres. Concrete is M20, steel is Fe415, so Ld = 47d.

  1. Bottom bars, 4 nos 16 mm. Ld = 47 × 16 = 752 mm at each end, with one 90° bend at each end.
    Cutting length = 6000 + (2 × 752) − (2 × 2 × 16) = 6000 + 1504 − 64 = 7440 mm = 7.44 m
  2. Top bars, 2 nos 12 mm. Ld = 47 × 12 = 564 mm.
    Cutting length = 6000 + (2 × 564) − (2 × 2 × 12) = 6000 + 1128 − 48 = 7080 mm = 7.08 m
  3. Stirrups, 8 mm. Bent dimensions inside the cover are (300 − 2 × 25) = 250 mm and (450 − 2 × 25) = 400 mm.
    Perimeter = 2 × (250 + 400) = 1300 mm. Add two 135° hooks at 9d: 2 × 9 × 8 = 144 mm.
    Deduct three 90° bends and two 135° hooks: (3 × 2 × 8) + (2 × 3 × 8) = 48 + 48 = 96 mm.
    Cutting length = 1300 + 144 − 96 = 1348 mm = 1.348 m
  4. Number of stirrups. (6000 ÷ 150) + 1 = 41 nos
  5. Weights. 16 mm → 1.580 kg/m; 12 mm → 0.889 kg/m; 8 mm → 0.395 kg/m.
    Bottom: 4 × 7.44 × 1.580 = 47.02 kg
    Top: 2 × 7.08 × 0.889 = 12.59 kg
    Stirrups: 41 × 1.348 × 0.395 = 21.83 kg
  6. Total. 47.02 + 12.59 + 21.83 = 81.44 kg. Add 3% wastage → order 84 kg for this beam.

One note on step 1: the example anchors the full development length into the support because that is the general case. Where the drawing details a specific anchorage — a standard hook, a shorter embedment into a wide support — schedule what the drawing says, not what the formula gives.

Common BBS mistakes

  • Measuring to the outside of the bar. Leg dimensions are taken along the centre line, inside the cover. Working to outside faces makes every stirrup too big.
  • Forgetting the extra stirrup. Spacing gives you the gaps; bars are gaps + 1. Dropping the +1 loses a stirrup on every member.
  • Laps that never got scheduled. Any bar longer than 12 m needs a lap, and that lap is extra steel. Schedules that ignore it under-order on every long beam.
  • Applying deductions to unbent bars. A straight bar with no bends gets no deduction. Blanket-applying 2d shortens bars that were fine.
  • Mixing standards. IS 2502 hook allowances with a BS 8666 shape code is a schedule nobody can fabricate reliably. Pick the standard the specification names and stay in it.

Frequently asked questions

What is the difference between cutting length and total length?

Cutting length is the straight length of one bar before it is bent, already accounting for hooks and bend deductions. Total length is cutting length multiplied by the number of bars of that mark. Steel is ordered against total length converted to weight.

Why do you subtract bend deductions at all?

Steel stretches on the outside of every bend, so the finished bar measures longer than the sum of its leg dimensions. Deducting 1d per 45° bend, 2d per 90° bend and 3d per 135° hook corrects for that stretch so the bar fits inside the formwork at the specified cover.

How much wastage should I add?

Two to three per cent is normal for a schedule prepared from drawings, because bars are supplied in fixed 12 m lengths and off-cuts are unavoidable. Sites ordering from thumb rules rather than a schedule typically waste around five per cent.

Is a BBS the same everywhere in the world?

The columns and the method are effectively universal. The standard values are not. This article follows IS 2502 and IS 456; ACI 315 and BS 8666 use their own shape codes and bend allowances. Check which standard your project specification names before fixing hook lengths and development lengths.