A bar bending schedule (BBS) for a beam is a table that lists every reinforcement bar in the beam with its bar mark, diameter, shape, number, cutting length and weight. To prepare it, you take each bar from the structural drawing, add hook lengths, subtract bend deductions to get the cutting length, then multiply the total length by the unit weight of steel (d²/162 kg per metre) to get the quantity in kilograms.
On a Nepali site the BBS is what the steel fixer (rod mistri) follows when cutting 12 m bars, and it is also what the contractor or owner uses to check the steel bill. A good BBS saves steel, reduces off-cuts and makes site checking easy. This post explains the cutting length formula, the bend and hook allowances used on most sites, and works out a full BBS for a common house beam.
What a bar bending schedule contains
A typical BBS has these columns:
- Bar mark: a short code such as B1-a or ST1 that links the bar to the drawing.
- Description and shape: bottom straight bar, top bar with L-bends, stirrup and so on, often with a small sketch.
- Diameter in mm.
- Number of bars in one member, and number of members.
- Cutting length of one bar in metres.
- Total length = number × cutting length.
- Unit weight in kg/m and total weight in kg.
Cutting length formula and bend deductions
When a bar is bent, the outside face stretches and the inside face is compressed. If you add up the outer dimensions of a bent bar, you get a length slightly more than the actual straight piece needed. That extra is removed using a bend deduction. Hooks, on the other hand, need extra length, so a hook allowance is added.
The general formula is:
Cutting length = total of straight dimensions + hook allowances − bend deductions
The values below are common site conventions used in Nepal and India, in the style of IS 2502 (Code of practice for bending and fixing of bars). They are rules of thumb for estimating, not exact code requirements. Some firms use slightly different numbers, so follow the convention your office or consultant uses and stay consistent.
| Item | Common allowance | For 8 mm bar | For 12 mm bar |
|---|---|---|---|
| 45° bend deduction | 1d | 8 mm | 12 mm |
| 90° bend deduction | 2d | 16 mm | 24 mm |
| 135° bend deduction | 3d | 24 mm | 36 mm |
| 135° stirrup hook allowance | 9d or 10d (10d used below) | 80 mm (10d) | 120 mm (10d) |
Here d is the bar diameter. For earthquake-resistant work, stirrups must end in 135° hooks with a straight extension into the core, as required by ductile detailing rules (IS 13920 and NBC 105 practice). The 10d hook allowance gives enough length for that extension.
Worked example: stirrup cutting length for a 230 x 350 beam
Data: beam size 230 mm x 350 mm, clear cover 25 mm, stirrup diameter 8 mm.
Step 1: Stirrup dimensions. On most sites the stirrup size is taken as the beam size minus cover on both sides.
- a = 230 − 2 × 25 = 180 mm
- b = 350 − 2 × 25 = 300 mm
Step 2: Count the bends and hooks. A closed rectangular stirrup has three 90° corners and one corner where both ends overlap with 135° hooks. So there are 3 bends of 90°, 2 bends of 135° and 2 hooks.
Step 3: Apply the formula.
Cutting length = 2(a + b) + 2 × 10d − 3 × 2d − 2 × 3d
- 2(180 + 300) = 960 mm
- Hooks: 2 × 10 × 8 = 160 mm
- 90° bends: 3 × 2 × 8 = 48 mm
- 135° bends: 2 × 3 × 8 = 48 mm
Cutting length = 960 + 160 − 48 − 48 = 1024 mm, say 1.024 m.
If your office uses a 9d hook, the length becomes 960 + 144 − 96 = 1008 mm. Some engineers also work with centre-line dimensions of the stirrup. All these methods give answers within a few centimetres; the key is to stick to one method for the whole project.
Number of stirrups
With a clear span of 4000 mm, stirrup spacing 150 mm and the first stirrup placed 50 mm from each column face:
Number = (4000 − 2 × 50) / 150 + 1 = 3900 / 150 + 1 = 26 + 1 = 27 stirrups.
In real earthquake-resistant beams, spacing is closer near the supports and wider at mid-span. In that case, count each zone separately using the same method.
Main bar cutting length
Assume the beam is an end beam framing into 300 mm columns, with the bars carried to the far face of the column (less 25 mm cover) and then bent with a 300 mm leg. The leg length here is only for illustration; in practice it comes from the development length shown on the structural drawing.
- Straight part = 4000 + 2 × (300 − 25) = 4550 mm
- Bottom bar, 16 mm: 4550 + 2 × 300 − 2 × (2 × 16) = 4550 + 600 − 64 = 5086 mm
- Top bar, 12 mm: 4550 + 2 × 300 − 2 × (2 × 12) = 4550 + 600 − 48 = 5102 mm
Sample BBS table for beam B1
Unit weights are from d²/162: 16 mm = 1.580 kg/m, 12 mm = 0.889 kg/m, 8 mm = 0.395 kg/m. You can check these with the steel weight calculator for rebar.
| Bar mark | Description / shape | Dia (mm) | Nos | Cutting length (m) | Total length (m) | Unit wt (kg/m) | Weight (kg) |
|---|---|---|---|---|---|---|---|
| B1-a | Bottom bar, L-bend both ends | 16 | 3 | 5.086 | 15.258 | 1.580 | 24.11 |
| B1-b | Top bar, L-bend both ends | 12 | 2 | 5.102 | 10.204 | 0.889 | 9.07 |
| B1-c | Stirrup 180 x 300, 135° hooks | 8 | 27 | 1.024 | 27.648 | 0.395 | 10.92 |
| Total for one beam | 44.10 | ||||||
Adding about 3% for off-cuts and cutting wastage gives roughly 45.4 kg for this beam. For the whole building, repeat the table for every beam, column, slab and footing, then sum the weights by diameter. That summary is what you use for ordering steel bundles.
Tips for preparing a BBS on site
- Plan cutting from 12 m bars. In Nepal TMT bars usually come in 12 m lengths. Combine cutting lengths so the off-cut is small. For example, two 5.086 m bars use 10.172 m and leave a 1.828 m piece that can become extra top bars or chairs.
- Show laps clearly. Where a bar is longer than 12 m, add the lap length from the drawing and place laps away from the zones the drawing marks as critical.
- Check cover before cutting. A stirrup cut for 25 mm cover will not fit correctly if the shuttering is out of size. Measure the formwork first.
- Use bar marks on site. Tag bundles of cut bars with their marks so the fixer does not mix 5.086 m and 5.102 m bars.
- Cross-check with the estimate. Compare BBS steel against your concrete volume using the concrete calculator. Typical house beams fall in a reasonable kg per cubic metre range; a very high or very low figure usually means a counting mistake.
For a rough idea of how steel fits into the full building budget, the house construction cost estimator gives an overall picture before you prepare detailed schedules.
Frequently asked questions
Why do we subtract length at bends?
Dimensions on a drawing are usually outer dimensions. A bent bar curves around the corner, so the real bar length needed is a little less than the sum of the outer sides. The bend deduction (1d, 2d or 3d by angle) corrects for that.
Is the hook length 9d or 10d?
Both are used on sites. 10d is common for 135° stirrup hooks in seismic areas because it leaves a comfortable straight extension inside the core. Use whatever your consultant specifies and stay consistent.
What is the formula for steel weight per metre?
Weight in kg/m = d²/162, where d is the bar diameter in mm. For example, 12 mm bar weighs 144/162 = 0.889 kg/m.
Does the BBS change for beams in different floors?
Only if the drawing changes. Beams with the same size, span and bars can share one BBS row, with the number of members multiplied in.
Comments
Post a Comment