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Bar bending schedule

Intermediate Estimation and Quantity Surveying Reviewed 24 Oct 2026 2 min read

A bar bending schedule (BBS) is a table listing every reinforcement bar in a member or structure: its mark, location, diameter, shape, dimensions, cutting length, number and total weight. It turns reinforcement drawings into instructions for the bar bending yard and into quantities for ordering, billing and checking.

Each bar is identified by a mark that matches the drawing. Its shape is sketched with the dimensions of each segment. The cutting length is the total length of straight bar needed before bending: the sum of the segment lengths, plus hooks and anchorage, minus allowances for elongation at bends, because bars stretch slightly when bent. Typical bend deductions depend on angle and bar diameter. Laps are added where bars are longer than stock lengths. The number of bars comes from the member size and spacing, for example span divided by spacing plus one.

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Diagram: Indiacon original, not to scale.

Weights are found from length times unit weight, d²/162 kg per metre for bar diameter d in mm. Totals are summarised by diameter, which feeds material orders, and by member, which feeds billing. A good BBS cuts waste by planning cutting from standard 12 m bars, avoids confusion on site and speeds inspection before concreting. It must follow the bending rules in the code and the detailing standard, such as minimum bend diameters.

Key formulas
  • Cutting length = Σ segment lengths + hook allowances − bend deductions
  • Stirrup (rectangular): 2(a + b) + 2 hooks − 3 × 90° bends − 2 × 135° bends
  • Number of bars = (length / spacing) + 1
  • Weight = cutting length × number × d² / 162
Solved example

Problem. Find the cutting length and weight of 2-legged 8 mm stirrups for a 230 × 450 mm beam with 25 mm clear cover, 135° hooks of 10d each, and bend deductions of 2d per 90° bend and 3d per 135° bend. Beam length 5.0 m, spacing 150 mm.

  1. Inner dimensions: (230 − 50) = 180 mm by (450 − 50) = 400 mm. Perimeter = 2 × (180 + 400) = 1 160 mm.
  2. Hooks: 2 × 10 × 8 = 160 mm. Deductions: 3 × 2 × 8 + 2 × 3 × 8 = 48 + 48 = 96 mm.
  3. Cutting length = 1 160 + 160 − 96 = 1 224 mm.
  4. Number = 5 000 / 150 + 1 = 34.3, say 35. Weight = 1.224 × 35 × 64 / 162 = 16.9 kg.

Answer. Cutting length ≈ 1.22 m; 35 stirrups ≈ 17 kg

Quick quiz

1What is a bar bending schedule?

A table of every bar with shape, cutting length, number and weight.

2Why deduct at bends?

Bars elongate slightly when bent.

3How is bar weight found?

Length × d²/162 kg per metre.

Code note
  • IS 2502 (India): Code of practice for bending and fixing of bars for concrete reinforcement.
  • IS 456 (India): Plain and reinforced concrete: general rules for materials, durability, design and construction.

Clauses and documents to read

  • IS 2502:1963 (bending and fixing of bars for concrete reinforcement)
  • IS 5525:1969 (detailing of reinforcement)
  • SP 34:1987 (handbook on concrete reinforcement and detailing, cited by number)

Cited by number and clause only. We do not reproduce clause text, tables or figures (BIS Act 2016, s.10(5) and s.11). Check the current edition and amendments before use.

References and further reading

Standard texts, listed in no order of merit. We do not copy text, figures, tables or solved problems from any book; our explanations and examples are original.

  • B. N. Dutta, Estimating and Costing in Civil Engineering, UBS Publishers, latest edition.

Free and official sources

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