Codes referenced
Bend shape codes, BBS column layout, and standard bending-practice conventions.
Bend diameter (4d for ≤ Ø16, 5d for Ø20–Ø25, 6d or 8d for larger), hook extension (12d).
Spacing, cover, anchorage and lap-length requirements that the BBS must reflect.
Illustrated shape codes (00 to 99) and standard bend schedules that BBS in India still follows.
Why the BBS is the spine of rebar work
The bar bending schedule is the single document that converts a structural drawing into tonnage, length, and yard work. Three downstream artefacts depend on it:
- · Steel billing — total length per bar diameter × unit weight (0.617 kg/m for Ø12, 0.785 kg/m for Ø16, 1.210 kg/m for Ø20, 1.580 kg/m for Ø25) drives the procurement and wastage calculation.
- · Cutting and bending yard — cutting length per bar drives machine setup. A 10 mm error on every bar accumulates into a site-wide wastage that does not show on the BBS until the take-off reconciliation.
- · Placing and lap scheduling — bar marks drive placement sequences. Without a clean BBS, the site crew orders rebar by intuition and ends up with leftover off-cuts and missing bars.
IS 2502:1963 is short — barely 12 pages — and 60 years old. But the column layout it defines (mark · nos · dia · length/bar · total · shape) is the same one every BBS in India uses, and the same one rebar suppliers quote against.
BBS notation — the columns, and what each means
A working BBS has the columns shown below. Total length and weight are derived; the others come from the design drawings.
| Column | Meaning | Source |
|---|---|---|
| Member | Structural element: beam B1, column C3, slab S2, footing F1 | Drawing GA / RC |
| Mark | Unique bar identifier: A1, A2 (bottom main); S1, S2 (stirrups); C1 (crank); L1 (lap zone) | Designer |
| Type | Main / stirrup / crank / lap | Designer |
| Nos | Number of bars of this mark in the member × number of members | Drawing |
| Dia (mm) | Bar diameter | Drawing |
| Length / bar (mm) | Cutting length — straight + hooks − bend deductions − cover adjustments | BBS calc |
| Total length (m) | Nos × Length / bar ÷ 1000 | Derived |
| Weight (kg) | Total length × unit weight (per IS 1786: 0.617 kg/m Ø12, 0.785 kg/m Ø16, 1.210 kg/m Ø20, 1.580 kg/m Ø25, 2.470 kg/m Ø32) | Derived |
| Shape | Standard IS 2502 / SP 34 shape code with dimensions marked | Designer + standard |
Reference: IS 2502:1963 standardises the BBS column layout; SP 34:1987 illustrates the standard shape codes used in practice.
Shape codes — the visual shorthand
SP 34:1987 and IS 2502 share the same numbered shape-code convention (00 = straight bar, 11 = 90° hook both ends, 21 = stirrup, 33 = crank, etc.). The shape code number is sketched with all dimensions marked in the BBS — clear cover, total length, hook extension, bend diameter. A BBS without shape sketches is not a BBS; it is a length table.
Hooks and bends per IS 1786
IS 1786:2008 specifies two things: the internal bend diameter (which controls bend ductility without cracking the bar) and the straight extension beyond the bend (which gives the anchorage).
| Bar diameter d (mm) | Internal bend diameter (mandatory) | 90° / 135° straight extension |
|---|---|---|
| ≤ 10 | 3d | 12d |
| 12 – 16 | 4d | 12d |
| 20 – 25 | 5d | 12d |
| 32 | 6d | 12d |
| 40 | 8d | 12d |
Reference: IS 1786:2008 + IS 2502:1963. The bend diameter is the diameter of the pin around which the bar is bent; the straight extension is the part of the bar beyond the bend.
Bend deduction — why the bar is shorter than its dimensions
When a bar is bent through an angle θ around an internal bend diameter D, the bar's centreline traces an arc of length (π/180) × θ × (D + d). The straight-line distance between the two ends of the bend is shorter than the arc by the bend deduction.
| Bend angle | Arc length consumed | Straight chord length | Bend deduction (working rule) |
|---|---|---|---|
| 45° | 0.785 × (D + d) | 0.765 × (D + d) | ≈ 1d at D = 4d |
| 90° | 1.571 × (D + d) | 1.414 × (D + d) | ≈ 2d at D = 4d |
| 135° | 2.356 × (D + d) | 2.121 × (D + d) | ≈ 3d at D = 4d |
| 180° | 3.142 × (D + d) | 2.000 × (D + d) | ≈ 4d at D = 4d |
Working rule (D = 4d is the most common case for ≤ Ø16 bars). For Ø20–Ø25 bars use D = 5d; the deduction still works out to ≈ 2d / 3d / 1d / 4d within ±0.1d precision.
Cutting length — the BBS number that drives the yard
Cutting length is the straight-bar length that comes off the cutting machine, before bending. Three shapes account for 90% of BBS rows: straight with hook, stirrup, and cranked slab bar.
1 · Straight with single 90° hook
Cutting length = clear span + cover at both ends + 1 × hook extension (12d) − bend deduction (2d)
For a 20 mm bar, 6 m clear span, 25 mm cover at each end:
Cutting length = 6000 + 25 + 25 + 12 × 20 − 2 × 20 = 6050 + 240 − 40 = 6250 mm
2 · Rectangular stirrup (most common)
Cutting length = 2 × (B + D) + 2 × 12d (for two 135° hooks) − bend deductions
Where B = clear width (b − 2 × cover), D = clear depth (h − 2 × cover). Bend deductions: 2 × 3d for the two 135° hooks + 2 × 2d for the two 90° corners = 10d total.
Cutting length = 2 × (B + D) + 24d − 10d = 2 × (B + D) + 14d
3 · Cranked slab bar (45° crank)
Cutting length = clear span + 2 × crank length − bend deduction (1d at the crank)
Crank length = √2 × crank height ≈ 1.414 × crank height. Crank height = slab depth − 2 × cover − bar Ø.
Worked example — beam B1 BBS
Scenario. RCC beam B1, clear span 6000 mm, cross-section 300 × 600 mm, cover 25 mm, Fe500 reinforcement.
Mark A1 — 4 bottom bars Ø20 (continuous, with 135° hook at both ends)
Cut length per bar = 6000 + 25 + 25 + 12 × 20 + 12 × 20 − 2 × 3 × 20 = 6100 + 240 + 240 − 120 = 6460 mm. Total = 4 × 6460 = 25 840 mm = 25.84 m. Weight = 25.84 × 1.210 = 31.3 kg.
Mark A2 — 2 top bars Ø16 (continuous, with 135° hook at both ends)
Cut length per bar = 6100 + 2 × 12 × 16 − 6 × 16 = 6100 + 384 − 96 = 6388 mm. Total = 2 × 6388 = 12 776 mm = 12.78 m. Weight = 12.78 × 0.785 = 10.0 kg.
Mark S1 — stirrups Ø8 @ 150 c/c
Clear dimensions B = 300 − 2 × 25 = 250 mm; D = 600 − 2 × 25 = 550 mm. Number = (6000 / 150) + 1 = 41 stirrups per beam (c/c spacing + 1 at each end).
Cutting length = 2 × (250 + 550) + 14 × 8 = 1600 + 112 = 1712 mm. Total = 41 × 1712 = 70 192 mm = 70.2 m. Weight = 70.2 × 0.395 = 27.7 kg.
| Mark | Type | Nos | Dia (mm) | Length/bar (mm) | Total (m) | Weight (kg) | |
|---|---|---|---|---|---|---|---|
| A1 | Bottom main | 4 | 20 | 6460 | 25.84 | 31.3 | |
| A2 | Top main | 2 | 16 | 6388 | 12.78 | 10.0 | |
| S1 | Stirrup Ø8 @ 150 | 41 | 8 | 1712 | 70.20 | 27.7 | |
| Total per beam | 108.82 | 69.0 | |||||
Per-beam total steel ≈ 69 kg — useful as a sanity check against the structural designer's estimate (typical for M30, Fe500, 0.8–1.0% steel content by volume). For a row of 10 such beams, total ≈ 690 kg, which is one full truckload of Ø16–Ø20.
Common deviations seen on Indian sites
BBS shows straight length, no deduction for bends. Site orders 5% extra steel as "wastage." Apply 1d/2d/3d deductions at BBS stage, not at the yard.
Stirrups bent with 90° hooks cannot seat into the core under shear and tend to open out. IS 1786 calls for 135° at both stirrup ends. This is the most-cited rebar detailing defect in India.
A 9d hook saves 3d × bar area × density × number of hooks in steel weight but loses anchorage. IS 1786 mandates 12d. Inspect at receipt, not just at placement.
Bending Ø20 Fe 500 around a 60 mm pin is inside the 4ϕ = 80 mm that Table 4 permits for that grade and size — it will not crack the bar. The pin to worry about is the one bigger than the table allows. IS 1786 Table 4 sets the maximum mandrel for the grade: for Fe 500 it is 4ϕ up to and including 20 mm and 5ϕ above it. Confirm the yard's bending pins.
Crew installs stirrups at "roughly" 150 mm c/c. Actual spacing varies 130–200 mm. Measure five consecutive stirrups with a tape before accepting the placement.
A stirrup cut to clear dimensions only (B × D) ignores cover. Cutting length = 2 × (B − 2 × cover) + 2 × (D − 2 × cover) + 14d accounts for cover at all four edges.
Field-checklist blocks
A short questionnaire, an inspector's checklist, and three worked examples that show how a cutting length, a hook angle and a bend deduction resolve into a bar the yard can bend and the engineer can bill.
Questionnaire — ask before the BBS is released for cutting
- Does the schedule carry the full IS 2502 column set (mark, type, nos, dia, length per bar, total length, weight) plus a shape sketch with clear cover, total length, hook extension and bend diameter marked? A length table without the sketch is not a BBS.
- Are the bend deductions already in the cutting length at 1d for the 45° crank, 2d for the 90° hook and 3d for the 135° hook, or is the yard expected to work them out from the shape?
- Is the hook extension on the schedule 12d for both the 90° and the 135° hook, and will it be measured on the hooks at receipt rather than trusted at placement?
- Is every stirrup, tie and crosstie scheduled 135° at both ends, the 90° hook used only where a longitudinal bar terminates at a support that is not a continuation, and the 45° crank used only in a slab at the column face where negative moment peaks?
- Is the stirrup cutting length written as 2 × (B + D) + 14d, with B and D taken as clear dimensions (b − 2 × cover and h − 2 × cover) rather than overall section dimensions?
- Does the weight column use the right unit weight for each diameter (0.617 kg/m for Ø12, 0.785 for Ø16, 1.210 for Ø20, 1.580 for Ø25, 2.470 for Ø32)?
- Is the bend pin at the yard sized for the bar being bent, 4d for Ø12 to Ø16 and 5d for Ø20 to Ø25 per IS 1786?
- Does the nos column multiply bars per member by the number of members, so the tonnage covers the whole pour and not one bay?
Inspector's checklist — what to verify in the yard and on the placing floor
- Bend deduction applied on the schedule at 1d/2d/3d/4d, before the sheet reaches the yard, not discovered at the bending machine.
- Hook extension measured on a sample of hooks at receipt, 12d on both 90° and 135° hooks, checked before the bars go into the pour.
- Stirrup, tie and crosstie hook angle confirmed as 135° at both ends, seated into the core so the stirrup cannot open under shear.
- Bend pin diameter checked against bar size, 4d for Ø12–Ø16 and 5d for Ø20–Ø25 per IS 1786.
- Stirrup spacing measured with a tape across five consecutive stirrups before the placement is accepted, since spacing on the floor drifts from the schedule.
- Cover carried into the stirrup cut length, so the closing hook lands inside the cover instead of standing proud of the formwork.
- Weight column re-derived from the unit weights and the member total compared with the designer's estimate.
- Shape sketch issued with the drawing for every mark, with the crank location marked on the slab bars.
What happens if… three worked examples
What happens. Stirrup S1 (Ø8 at 150 mm c/c, B = 250 mm, D = 550 mm) is scheduled at 2 × (250 + 550) + 2 × 12 × 8 = 1,792 mm instead of 2 × (250 + 550) + 14 × 8 = 1,712 mm. The 80 mm difference is exactly the 10d of bend deduction on the four bends. At (6000 / 150) + 1 = 41 stirrups per beam that is 3,280 mm of over-cut bar per beam, and on a row of 10 beams, 32.8 m of Ø8 or about 13 kg.
Likely outcome. Every stirrup comes 80 mm long, so the closing hook lands outside the cover line and the stirrup stands proud of the formwork. The site then papers over it by ordering 5% extra steel as wastage, which on 108.82 m per beam is another 5.44 m of bar per beam and 54.4 m on the row. Two payments for one error, and the cover is still wrong. The fix is the deduction on the BBS.
What happens. On beam B1 the saving is trivial. Mark A1 has 4 Ø20 bars with two hooks each, so 8 hooks per beam; going from 12d to 9d saves 3 × 20 = 60 mm per hook, or 480 mm per beam = 0.48 m of Ø20 = 0.58 kg. The 41 stirrups with two Ø8 hooks each save 2 × 41 × 3 × 8 = 1,968 mm = 1.97 m = 0.78 kg per beam. Across 10 beams that is 5.8 kg on the bottom bars and 7.8 kg on the stirrups, about 13.6 kg out of the 690 kg in that row, roughly 2% of the steel.
Likely outcome. You buy 2% of the weight and give up the anchorage the 12d extension exists to provide. On a 135° stirrup hook the short extension also stops the leg seating into the core, so the stirrup is weaker in shear at the same time as the bar is weaker in bond. IS 1786 mandates 12d, and this is the one saving that is never worth taking.
What happens. The BBS bills 41 stirrups per 6,000 mm beam, (6000 / 150) + 1 = 41. At a uniform 200 mm the floor fits (6000 / 200) + 1 = 31 stirrups, so 10 per beam are missing, or 10 × 1,712 = 17,120 mm = 17.12 m of Ø8 and about 6.8 kg a beam. Over 10 beams that is 171.2 m of stirrup steel, roughly 68 kg, simply not in the concrete.
Likely outcome. The drawing still shows Ø8 at 150 mm c/c, the BBS still bills 41, and the beam quietly gets 10 of its 41 stirrups missing, with no drawing change to flag it. Five stirrups measured with a tape is the whole test.
Frequently Asked Questions
What is the standard hook length for an IS 1786 bar?
What is the bend-deduction formula for a bar?
How is the cutting length of a stirrup calculated?
What does the IS 2502 BBS notation look like?
Related articles
Reinforcement Steel — IS 1786 Walkthrough
Fe500, Fe550, ductility, weldability and bend-test acceptance for HYSD bars.
BBS Cutting-Length Calculator — Significance
Companion to the BBS cutting-length calculator — what the numbers mean.
IS 456:2000 — Annotated Walkthrough
Cl 26 — Detailing of reinforcement: spacing, cover, anchorage, laps.
Lap Length Calculator — Significance
Tension/compression lap lengths per IS 456 for Fe500 — companion to BBS.
Amit Haridas
Founder & Proprietor, ConcreteInfo. 25+ years experience in construction QA/QC, concrete technology, RMC plant operations, consulting and technical training across India. For a project-specific BBS review or bar-bending audit, contact amit@concreteinfo.in.
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