Bar Bending Schedule (BBS) Calculator
A complete BBS — shapes, cutting lengths, counts, laps and steel weight by diameter.
Schedule settings
Add bars for a member
Bar marks
No bar marks yet — add the bars for a member above, or a single bar mark.
Bar bending schedule
| Member | Mark | Dia | Shape | Dimensions (mm) | Members | Per member | Total no. | Cutting length (mm) | Total length (m) | kg/m | Weight (kg) |
|---|
Steel by diameter
| Diameter | Bars | Total length (m) | Stock bars (12 m) | Offcuts (m) | Weight (kg) | Share |
|---|
Results are estimates from the formulas shown, not a professional design or certification. Have a qualified engineer verify anything safety-critical (structures, electrical installations, gas or pressure systems).
About the Bar Bending Schedule (BBS) Calculator
Build a bar bending schedule for slabs, beams, columns, footings and stairs. Each bar mark has a shape — straight with hooks or bends, L, U, cranked (bent-up), closed or open link, circular tie or helix — a bar size and its dimensions; the number of bars comes from a count or from the spacing over a length. The calculator works out each cutting length with the IS 2502 formulas and hook and bend allowances, rounds it up to the next 25 mm, adds IS 456 laps where a bar is longer than a stock bar, and weighs everything with the IS 1786 masses.
Quick-start buttons fill the bar marks for a slab, beam, column, footing or stair flight from its size, cover and spacing; every row stays editable. You get the schedule with shape sketches, the totals by diameter and stock bars, and CSV and PDF downloads.
How to use it
- Set the steel (TMT/HYSD to IS 1786 or mild steel), its grade and the concrete grade — they set the bend radius, the hook and bend allowances and the lap length.
- Add bar marks one by one, or use Add bars for a member to fill them from a slab, beam, column, footing or stair size, cover and spacing.
- For each bar mark choose the shape and bar size, enter the dimensions shown on the drawing (outside dimensions, or the member size and cover for links), and give the number per member — directly or as a spacing over a length — and the number of members.
- Check the schedule, the notes and the totals by diameter, then copy it, or download the CSV or the PDF.
Examples
3,000 + 2 × 155 = 3,310 mm → cutting length 3,325 mm · 3.325 m × 0.888 kg/m = 2.95 kg
Inside A = 450 − 2 × 25 − 2 × 8 = 384 mm; E = 230 − 50 − 16 = 164 mm
2 × (384 + 164) + 24 × 8 = 1,288 mm → 1,300 mm · 4,000 ÷ 150 → 26 + 1 = 27 stirrups
D = 125 − 40 − 10 = 75 mm · 4,190 + 2 × 75 × tan 22.5° = 4,252 mm → 4,275 mm
Ld = 16 × 0.87 × 500 ÷ (4 × 1.4 × 1.6) = 777 mm (lap = Ld, more than 30φ) · one lap → 20,777 mm → 20,800 mm
D = 354 mm · 3,000 ÷ 50 + 2 × 1.5 = 63 turns · 63 × π × 362 + 8 × 8 = 71,711 mm → 71,725 mm
Cutting lengths (IS 2502 Tables III–IX)
Lengths are measured along the centre line of the bar. With outside dimensions, hook allowance H and bend allowance B (Table II) and internal bend radius R = k·d:
- Straight: l, + H for each standard hook, + B for each standard 90° bend
- L-bar: A + E − ½R − d; U-bar: A + E + C − 2(½R + d)
- Cranked bar: l + C − √(C² − D²) for each crank, which is l + D·tan(θ/2) — 0.414 D at 45°, the familiar "0.42 D"
- Closed link with 135° hooks: 2(A + E) + 24d; with 90° ends: 2(A + E) + 20d; open U-stirrup: 2A + E + 28d, with A and E the inside height and width
- Circular tie: π(D + d) + 2H; helix: N·π(D + d) + 8d when the pitch is not more than D/5 (N = turns, D = inside diameter)
IS 2502 cl. 5.1.1 then specifies cutting lengths to the next whole 25 mm. Clause 3.3 asks for a straight end of at least 8d on links: the 24d of a closed 135° link works out to about that with mild-steel corners (k = 2), but with k = 4 corners on TMT bars the same link with 8d ends is about 2.4d longer along its centre line — enter the allowance from your drawings in the row's override where it matters. For a helix the 8d end allowance can be overridden the same way.
Hook and bend allowances (IS 2502 Table II)
Fig. 1 bends hooks and 90° bends to an internal radius of k·d: k = 2 for mild steel and 4 for cold-worked or deformed (HYSD/TMT) steel — SP 34 Table 4.1 applies the same to all grades. For bars above 25 mm, k = 3 and 6 are recommended. H is about 9d (k = 2) or 13d (k = 4) and B about 5d or 6d, to the nearest 5 mm and at least 75 mm:
- TMT/HYSD (k = 4): 8 mm H 105, B 75 · 10 mm 130, 75 · 12 mm 155, 75 · 16 mm 210, 95 · 20 mm 260, 120 · 25 mm 325, 150 · 32 mm 415, 190
- Mild steel (k = 2): 8 mm H 75, B 75 · 10 mm 90, 75 · 12 mm 110, 75 · 16 mm 145, 80 · 20 mm 180, 100 · 25 mm 225, 125 · 32 mm 290, 160
The overrides in each row let you use the allowances on your drawings instead.
Number of bars, laps and weights
Bars spread at a spacing s over a length S number S ÷ s rounded down, plus one. The schedule multiplies by the number of identical members.
Where a bar is longer than the stock length (12 m by default), laps are added to IS 456 cl. 26.2.5.1: in flexural tension the larger of Ld and 30φ, and for column bars the compression lap — Ld in compression, at least 24φ. Ld = φ × 0.87 fy ÷ 4τbd with τbd = 1.2, 1.4, 1.5, 1.7 and 1.9 N/mm² for M20, M25, M30, M35 and M40 and above, plus 60 % for deformed bars and another 25 % in compression. Bars larger than 32 mm are not lapped; the schedule counts couplers or welded splices instead.
Weights use the IS 1786 Table 1 masses — 0.395, 0.617, 0.888, 1.58, 2.47 and 3.85 kg/m for 8 to 25 mm — or D²/162.
Stock bars to buy come from a cutting plan: the pieces of each size are taken longest first and each is cut from the first stock bar with enough length left (first-fit decreasing). Dividing the total length by the stock length undercounts whenever pieces do not fit neatly — in the example schedule, 22 ten-millimetre bars of about 4.2 m need 11 stock bars of 12 m, not the 8 that 93.2 m ÷ 12 m suggests.
Cover (IS 456 cl. 26.4)
Nominal cover to all bars including links: 20 mm in mild exposure, 30 mm moderate, 45 mm severe, 50 mm very severe and 75 mm extreme (Table 16); at least 40 mm for column bars (25 mm for columns of 200 mm or less with bars up to 12 mm) and 50 mm for footings. The member templates subtract the cover you give at each end and face.
Sources
- IS 2502:1963, Code of Practice for Bending and Fixing of Bars for Concrete Reinforcement: cl. 3.2, 3.3, 5.1.1, Fig. 1 and Tables II–IX — archive.org
- SP 34:1987, Handbook on Concrete Reinforcement and Detailing: Table 4.1 and Section 5 — archive.org
- IS 456:2000 (with Amendments 1–3): cl. 26.2.1, 26.2.5.1, 26.4 and 26.5.3.2 — archive.org
- IS 1786:2008 (with Amendment 1), Table 1 — archive.org
Limitations
- This is a scheduling aid, not a design: bar sizes, spacings, anchorage and lap positions come from the structural drawings and your engineer.
- IS 2502 gives approximate lengths for its standard shapes and bend radii; special shapes, other radii or seismic detailing to IS 13920 may need their own allowances — use the overrides.
- The templates make common assumptions (bars stop at the cover, legs run to the top cover, laps only at the stock length). Check each generated row against the drawing.
- Stock bars are planned for each size by first-fit decreasing, which is close to the fewest bars but not always exactly it; offcuts of one size are not reused for another, saw cuts are ignored, and a size with more than 20,000 pieces is counted by length.
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Frequently asked questions
How do I calculate the cutting length of a stirrup?
For a closed rectangular link with 135° hooks IS 2502 Table VIII gives 2(A + E) + 24d, with A and E the inside height and width. For a 230 × 450 mm beam with 25 mm cover and 8 mm links: A = 450 − 50 − 16 = 384 mm, E = 230 − 50 − 16 = 164 mm, so 2 × 548 + 192 = 1,288 mm, specified as 1,300 mm.
What is the hook length of a 12 mm TMT bar?
IS 2502 Table II gives a hook allowance of 155 mm for a 12 mm cold-worked (deformed) bar bent to a radius of 4d, and 110 mm for a 12 mm mild steel bar bent to 2d. The allowance is what the hook adds to the straight length measured to the outside of the hook.
Why is a crank 0.42 D?
A 45° crank of height D has an inclined length of D ÷ sin 45° = 1.414 D but covers only D horizontally, so it adds 0.414 D to the bar — usually rounded to 0.42 D. IS 2502 Table IV writes it as C − √(C² − D²). At 30° it adds 0.268 D.
How many bars are needed at 150 mm spacing over 3 m?
Divide the length by the spacing, round down and add one: 3,000 ÷ 150 = 20, so 21 bars. Use the length between the end covers, for example the slab width minus two covers.
Why are cutting lengths rounded to 25 mm?
IS 2502 cl. 5.1.1 specifies cutting lengths to the next greater whole 25 mm of the sum of the bending dimensions and allowances; any excess goes into the end anchorages. You can switch rounding off to see the exact lengths.
What is the lap length of a 16 mm Fe 500 bar in M25 concrete?
Ld = 16 × 0.87 × 500 ÷ (4 × 1.4 × 1.6) = 777 mm, which is more than 30φ = 480 mm, so a tension lap is about 777 mm (about 49φ). In compression τbd rises by 25 %, giving 621 mm, still more than 24φ.