From bar bending schedule to steel order: kg and bars for each size
Your engineer hands you a bar bending schedule. The dealer wants kilograms of each size. This guide reads the schedule's columns as IS 2502 lays them out and turns each row into a cutting length and a weight. Then it works out how many full bars to buy so the fixer can cut them with little waste.
By Bulan Sarkar · Updated
Work out one total for each bar size. In every row of the schedule, multiply the total number of bars by the cutting length, then add up the metres for each diameter and multiply by the IS 1786 mass per metre: 0.395 kg for 8 mm, 0.617 for 10 mm, 0.888 for 12 mm, 1.58 for 16 mm, 2.47 for 20 mm and 3.85 for 25 mm. Hooks and bends add length. IS 2502:1963 gives the allowances, and its clause 5.1.1 rounds each cutting length up to the next 25 mm. Order kg for each size, never one tonnage figure. To count whole bars, divide the dealer's bar length by the cutting length, round down to get pieces per bar, and then divide the number of pieces you need by that.
Source: IS 2502:1963 Code of Practice for Bending and Fixing of Bars for Concrete Reinforcement: Table I, cl. 3.2.1 with Fig. 1 and Table II, Tables III and VIII, cl. 5.1 to 5.1.2 with Table XI, checked Oct 2026.
Reading the schedule: eight columns, one row per bar mark
A bar bending schedule lists every bar the structure needs, one row for each bar mark. IS 2502:1963, the BIS code for bending and fixing bars, sets out a standard form in Table I (cl. 2.1). It has eight columns:
| Column | What it holds | Example row |
|---|---|---|
| 1 Location | Where the bars go | Column |
| 2 Mark designation | The label tied to the bars | C4 4R 25N |
| 3 Size and type | Steel quality and bar size | MS round, 25 mm |
| 4 Number of sets | How many identical members | 5 |
| 5 Number per set | Bars of this mark in one member | 4 |
| 6 Total number | Column 4 × column 5 | 20 |
| 7 Length | The cutting length of one bar | 3,000 mm |
| 8 Shape | A sketch with its bending dimensions | Straight |
The mark in column 2 is what the bar bender writes on the label, and Note 2 to Table I explains how it is built. In the 1963 example, the 4 is the number of identical bars in that spot, R means a round bar, 25 is its size in mm and N stands for the bar's serial number. Modern schedules use their own short marks such as B1-T or S3, which is fine as long as the label on the bundle matches the schedule. Clause 9.1.2 asks for each bundle of bent bars to carry the bar mark, the schedule number and where the bars go.
Column 3 in the example says MS round because the code dates from 1963, when mild steel was common. On a house today, column 3 will usually read something like Fe 500D TMT 12. Check the grade as well as the size, because a dealer quoting for Fe 500 and a schedule asking for Fe 500D are not quoting for the same steel.
Before any arithmetic, check column 6 against columns 4 and 5 on every row. A typing slip there carries straight into the order.
Cutting length: what hooks and bends add
For a straight bar, the cutting length is just its length. A bent bar needs extra steel for each hook or bend, and IS 2502 tables the extra so nobody works out arcs on site. Table III gives the total length of the common shapes, where l is the straight length shown on the sketch, H is the hook allowance and B the bend allowance:
- Straight (shape A): l
- Hook at one end (B): l + H; hooks at both ends (C): l + 2H
- Right-angle bend at one end (D): l + B; bends at both ends (E): l + 2B
H and B come from Table II and depend on the bar size and on k, which sets how tightly the hook or bend is formed (Fig. 1, Note 1). The 1963 code gives k = 4 for bars to IS 1786, which were cold-twisted bars back then. Today's TMT bars are made to a later edition of IS 1786, so that column is the closest match, but the allowance your engineer used is the one that counts. If the schedule already gives cutting lengths, use them.
| Bar | Hook allowance H | Bend allowance B |
|---|---|---|
| 8 mm | 105 mm | 75 mm |
| 10 mm | 130 mm | 75 mm |
| 12 mm | 155 mm | 75 mm |
| 16 mm | 210 mm | 95 mm |
| 20 mm | 260 mm | 120 mm |
| 25 mm | 325 mm | 150 mm |
Stirrups and ties have their own table. Table VIII, shape A, a closed link with a hook at each end, takes 2(A + E) + 24d, where A and E are the two sides and d is the bar size. Note 2 to that table says the schedule should mark a side OD or ID if there could be any doubt about whether it is an outside or inside size. Our RCC calculator uses inside sizes. The code's clause on binders and stirrups (printed as 3.8, right after 3.2.1) also asks for the straight end beyond each hook of a stirrup to be at least eight times the bar size.
Last, clause 5.1.1 says the cutting length is the sum of the bending dimensions and allowances rounded up to the next whole 25 mm. A 10 mm bar with hooks at both ends and l = 3,000 mm adds up to 3,000 + 2 × 130 = 3,260 mm, so it is cut at 3,275 mm.
From metres to kilograms for each bar size
Once every row has a cutting length, the order works in three steps:
- Metres in each row = total number (column 6) × cutting length (column 7).
- Metres for each size: add up the rows with the same diameter.
- Kg for each size = metres × the nominal mass per metre in IS 1786:2008 Table 1.
| Bar | kg per metre |
|---|---|
| 8 mm | 0.395 |
| 10 mm | 0.617 |
| 12 mm | 0.888 |
| 16 mm | 1.58 |
| 20 mm | 2.47 |
| 25 mm | 3.85 |
Keep the sizes apart all the way to the order. A one-tonnage order such as "500 kg of steel" leaves the dealer to guess the split, and you end up with too much 12 mm and too little 8 mm on the day the stirrups are bent.
Take the standard's own row as a check. Twenty 25 mm bars at 3,000 mm make 60.0 m, and 60.0 × 3.85 = 231.0 kg.
From kilograms to bars to buy: cut from the dealer's bar length
Dealers sell whole bars. Kilograms alone don't tell you how many to buy, because each cut leaves an offcut. Ask the dealer what length they supply. The examples here assume 12 m bars.
Pieces per bar = bar length ÷ cutting length, rounded down. Bars to buy = pieces needed ÷ pieces per bar, rounded up.
A 4,150 mm bar comes out of a 12 m length twice, leaving a 3.70 m offcut. A 1,300 mm stirrup comes out nine times, leaving 0.30 m. A 3,000 mm bar comes out exactly four times with nothing left over. So the waste depends on how well each cutting length fits into the bar length. Before ordering, look for another row in the same size that could use the offcuts. A 3.70 m offcut of 12 mm can still be bent into chairs or used as a short bar.
A bar longer than the dealer's length has to be joined. The lap length and where the laps go come from the structural drawing, and the schedule should already show the bar in two pieces with the lap added. If it shows one 14 m bar, ask the engineer how to split it before you order.
IS 1786 clause 7.1 covers bars ordered cut to length: each piece may be up to 75 mm longer or 25 mm shorter than specified. IS 2502 Table XI allows ±25 mm when the fixer cuts a straight bar on site, and bent bars get tighter limits on each bent dimension (for example +3 / −5 mm up to 75 cm). That is why clause 5.1.2 leaves one dimension of every bent bar unspecified, to take up the cutting tolerance.
Worked examples
1. Steel order for three identical plinth beams
- Schedule: 3 beams (sets), each 230 × 450 mm, cover 25 mm; bottom bars B1: 3 per beam, 12 mm, bent up both ends, l = 4,000 mm; top bars B2: 2 per beam, 10 mm, same shape; stirrups B3: 20 per beam, 8 mm closed links
- Cutting lengths: B1 = 4,000 + 2 × 75 = 4,150 mm; B2 = 4,000 + 2 × 75 = 4,150 mm
- B3 inside sizes: A = 450 − 2 × 25 − 2 × 8 = 384 mm, E = 230 − 2 × 25 − 2 × 8 = 164 mm; 2 × (384 + 164) + 24 × 8 = 1,288 mm, cut at 1,300 mm
- Totals: B1 3 × 3 = 9 bars, B2 3 × 2 = 6 bars, B3 3 × 20 = 60 stirrups
- 12 mm: 9 × 4.150 = 37.35 m × 0.888 = 33.17 kg
- 10 mm: 6 × 4.150 = 24.90 m × 0.617 = 15.36 kg
- 8 mm: 60 × 1.300 = 78.00 m × 0.395 = 30.81 kg
Result: The schedule needs 79.34 kg: 33.17 kg of 12 mm, 15.36 kg of 10 mm and 30.81 kg of 8 mm.
2. How many 12 m bars to buy for the same beams
- 12 mm: 12,000 ÷ 4,150 = 2.89, so 2 pieces per bar; 9 ÷ 2 = 4.5, so 5 bars
- 10 mm: 2 pieces per bar; 6 ÷ 2 = 3 bars
- 8 mm: 12,000 ÷ 1,300 = 9.23, so 9 pieces per bar; 60 ÷ 9 = 6.67, so 7 bars
- Bought: 12 mm 5 × 12 × 0.888 = 53.28 kg; 10 mm 3 × 12 × 0.617 = 22.21 kg; 8 mm 7 × 12 × 0.395 = 33.18 kg
- Offcuts: four 3.70 m and one 7.85 m of 12 mm, three 3.70 m of 10 mm, six 0.30 m ends and one 4.20 m of 8 mm
Result: Buy 5 bars of 12 mm, 3 of 10 mm and 7 of 8 mm, 108.67 kg in all, and plan the 12 mm offcuts into chairs or another member before they turn into scrap.
3. Checking one row of a column schedule
- Row: Column, mark C4, 25 mm, 5 sets, 4 per set, total 20, length 3,000 mm, straight (the IS 2502 Table I example)
- Total number: 5 × 4 = 20, matching column 6
- Metres: 20 × 3.000 = 60.0 m; kg: 60.0 × 3.85 = 231.0 kg
- 12 m bars: 12,000 ÷ 3,000 = 4 pieces per bar, no offcut; 20 ÷ 4 = 5 bars
Result: 231.0 kg, or 5 bars of 25 mm with no offcut, because 3,000 mm divides the 12 m bar exactly.
Tips and tricks
- Ask for the schedule as a spreadsheet as well as a printout. Totals by size are then one sum, and you can check column 6 against columns 4 and 5 in seconds.
- Read the grade on each row, not just the size. Fe 500 and Fe 500D are different steels and are priced differently.
- Before you order, check the stirrup length the schedule gives against 2(A + E) + 24d. A stirrup that is too short shows up only when the fixer tries to close it around the cage.
- Give the bar bender the 12 mm and 16 mm rows first and plan cuts across rows, so the long offcuts from one member go into the next.
- Keep one copy of the schedule with the steel stack. The bender cuts from it, and the labels on the bundles should match its marks.
What we would do
We'd ask the engineer for the schedule early, check column 6 on every row, and recompute two or three cutting lengths with the Table II allowances before trusting the totals. Then we'd order kg for each size plus whole 12 m bars as worked out above, with the bill agreed on weight, and check the truck with the TMT delivery weight check. For a budget figure before the schedule exists, use the steel calculator. The material order for one floor fits the steel into the rest of the order.
Frequently asked questions
How do you calculate steel quantity from a bar bending schedule?
What is the cutting length formula for a stirrup?
What are the hook and bend allowances in IS 2502?
What does 'total number' mean in a bar bending schedule?
Sources and date checked
- IS 2502:1963 Code of Practice for Bending and Fixing of Bars for Concrete Reinforcement (BIS): cl. 2.1 and Table I with Notes 1 to 3 (standard schedule form, printed p. 5); cl. 3.1, 3.2.1, Fig. 1 Note 1 (k values) and the binders clause printed as 3.8 on p. 6 (8d straight beyond a stirrup hook); Table II hook and bend allowances (p. 7); Table III shapes A to E (p. 8); Table VIII shape A, 2(A + E) + 24d, with Notes 1 to 3 (p. 13); cl. 5.1, 5.1.1 (round up to the next whole 25 mm), 5.1.2 and Table XI tolerances (p. 15); cl. 9.1.2 (bundle labels, p. 26). Read 20 Oct 2026 from the page images.. Checked Oct 2026.
- IS 1786:2008 High Strength Deformed Steel Bars and Wires for Concrete Reinforcement (BIS): Table 1 nominal mass per metre, cl. 7.1 cut lengths +75 / −25 mm, Table 2 mass tolerances. Read 20 Oct 2026. The 12 m bar length, the beam schedule and the bar counts are our worked examples, not figures from either standard.. Checked Oct 2026.
This guide turns a schedule into an order and checks its arithmetic. Bar sizes, spacing, laps, anchorage and any change to the schedule are for your structural engineer to decide. Have the engineer inspect the steel before every pour.