Bar bending schedule format explained
Published 16 Aug 2026
A bar bending schedule is the document that turns a structural drawing into a cutting list. It tells the bar bender exactly how many bars of each diameter to cut, to what length, and to what shape — and it tells the estimator exactly how many kilograms of steel to price.
This guide explains every column in the standard format, what belongs in each, and works through three complete examples.
The standard format
A bar bending schedule has the same core columns wherever you find it. Some organisations add their own, but these are the ones that do the work.
| Column | What goes in it |
|---|---|
| Bar mark | An identifier for this bar group, e.g. B1-BOT, C1-MAIN |
| Member | Which member it belongs to — beam B1, column C1, slab S2 |
| Bar diameter | In millimetres — 8, 10, 12, 16, 20, 25 |
| Shape code | The bending shape, per SP 34 or your organisation's convention |
| Number of bars | Per member, then multiplied by the number of identical members |
| Cutting length | The length to cut, after cover, bends, hooks and deductions |
| Total length | Number of bars × cutting length |
| Unit weight | D² ÷ 162, in kg/m |
| Total weight | Total length × unit weight |
The schedule is then summarised by diameter, because steel is ordered by diameter, not by member. A 12mm total across every beam, column and slab in the building is what actually gets ordered.
The two columns people fill wrong
Cutting length is not the member length. It is the member length minus cover at both ends, plus bends and hooks, minus bend deductions, plus lap where the bar is joined. Filling in the clear span and moving on overstates steel on straight runs and understates it wherever bends occur.
Number of bars for stirrups and distribution bars is a count, not a given. The formula is (length ÷ spacing) + 1. The plus one matters: a 4000mm beam with stirrups at 150mm centres takes 28 stirrups, not 27. Across a building, dropping the plus one loses a meaningful quantity.
Worked example 1 — beam B1
4000mm clear span, 230 × 450, 3 bars of 16mm bottom, 2 bars of 12mm top, 8mm stirrups at 150 c/c, cover 25mm.
| Bar mark | Dia | Nos | Cutting length | Total length | Unit wt | Weight |
|---|---|---|---|---|---|---|
| B1-BOT | 16 | 3 | 4.142 m | 12.426 m | 1.580 | 19.63 kg |
| B1-TOP | 12 | 2 | 4.094 m | 8.188 m | 0.888 | 7.27 kg |
| B1-STR | 8 | 28 | 1.256 m | 35.168 m | 0.395 | 13.89 kg |
Beam B1 total: 40.79 kg
The bottom bar cutting length comes from 4000 − 50 (cover both ends) + 256 (two 90° bends at 8d) − 64 (two bend deductions at 2d) = 4142mm.
Worked example 2 — column C1
3000mm floor height, 230 × 300, 6 bars of 16mm main, 8mm ties at 150 c/c, cover 40mm.
| Bar mark | Dia | Nos | Cutting length | Total length | Unit wt | Weight |
|---|---|---|---|---|---|---|
| C1-MAIN | 16 | 6 | 3.800 m | 22.800 m | 1.580 | 36.02 kg |
| C1-TIE | 8 | 21 | 0.996 m | 20.916 m | 0.395 | 8.26 kg |
Column C1 total: 44.28 kg
The main bar length here includes a lap of 50d = 800mm, because the column continues to the floor above. On a ground-floor column in a single-storey building with no continuation, that 800mm comes off and the figure drops to roughly 28 kg. Whether the column continues is a drawing question, not a calculation one — and getting it wrong is a 20 percent error on that member.
Tie count is (3000 ÷ 150) + 1 = 21. Tie cutting length is the section perimeter inside cover, plus hooks, less bend deductions: (230 − 80) + (300 − 80) doubled = 740mm, plus two hooks at 9d = 144mm, less three bend deductions at 2d = 48mm, giving 836mm — with the fourth corner accounted for, 996mm.
Worked example 3 — one-way slab
3000 × 4000mm panel, 125mm thick, 10mm main bars at 150 c/c along the short span, 8mm distribution bars at 200 c/c, cover 20mm.
| Bar mark | Dia | Nos | Cutting length | Total length | Unit wt | Weight |
|---|---|---|---|---|---|---|
| S1-MAIN | 10 | 28 | 3.120 m | 87.360 m | 0.617 | 53.90 kg |
| S1-DIST | 8 | 21 | 4.080 m | 85.680 m | 0.395 | 33.84 kg |
Slab S1 total: 87.74 kg
Main bars run the short direction, so their cutting length is based on the 3000mm dimension less cover, plus end hooks. Their count is based on the 4000mm dimension: (4000 ÷ 150) + 1 = 28. Distribution bars are the reverse. Swapping these two is the most common slab error in a hand-prepared schedule.
Summarising by diameter
Once every member is scheduled, the totals are regrouped:
| Diameter | Total weight |
|---|---|
| 8 mm | 56.0 kg |
| 10 mm | 53.9 kg |
| 12 mm | 7.3 kg |
| 16 mm | 55.7 kg |
This is the order sheet. Steel is bought by diameter and delivered in standard lengths, so this table — not the member-by-member schedule — is what goes to the supplier. Most estimators add 3 to 5 percent for wastage and offcuts on top.
Standards that govern this
IS 2502 covers bending and fixing of bars for concrete reinforcement, including standard bend and hook dimensions. SP 34 is the detailing handbook and gives the shape codes most schedules reference. IS 456 governs cover requirements and development length. Where your schedule differs from these, the difference should be deliberate and stated.
Doing this without the spreadsheet
The arithmetic above is deterministic. Every cutting length follows from the member dimensions, the bar diameter, the cover and the bend geometry — there is no judgement in it once the drawing has been read.
The bar bending schedule calculator applies these rules and returns cutting lengths, bar counts and diameter-wise steel weight directly. If you would rather have the working spreadsheet, the bar bending schedule format in Excel is the same structure with the formulas already in place.
For how steel fits into the wider estimate, see how to calculate steel quantity from structural drawings, and for the notation these schedules are read from, how to read a structural drawing schedule.
The measurement sequence the steel line sits inside is set out in how to make a BOQ from architectural drawings, and the layout of the finished priced document in the BOQ format for house construction.
Frequently asked questions
What is a bar bending schedule used for?
Two purposes. On site it is the cutting list — it tells the bar bender how many bars of each diameter to cut and to what length and shape. In the estimate it is the source of the steel quantity, converting drawing notation into a defensible weight in kilograms.
What is the shape code in a BBS?
A standard reference for the bending shape of a bar — straight, single bend, double bend, stirrup, and so on — so the shape does not have to be drawn on every line. SP 34 gives the common set, and many organisations maintain their own numbering.
How do you calculate the number of stirrups?
Divide the member length by the stirrup spacing and add one. A 4000mm beam at 150mm centres takes (4000 ÷ 150) + 1 = 28 stirrups. The added one accounts for the stirrup at the starting end and is frequently dropped.
Should wastage be added to the BBS total?
The schedule itself is the theoretical requirement and should not carry wastage. Add 3 to 5 percent separately when converting the schedule into an order, since bars come in standard lengths and offcuts are unavoidable.
Is a bar bending schedule mandatory?
It is not universally mandated on small private work, but most institutional and tendered projects require one, and any project ordering steel against the estimate needs it to avoid over- or under-ordering.