Bar bending schedule is a table form data in which various diameters of steel, the shape of bending, length of each bent and straight portion, angles of bending, the total length of each bar, and the number of each type of bar.

This information is very useful for site engineer who is working on-site and while preparing estimate and bills of work.

**Bar Bending Schedule**

Let’s take the example of RCC slab reinforcement to calculate and prepare a Bar Bending Schedule.

**RCC Slab Reinforcement Details**

- Main Bent Up Bars = 12 mm Dia. @ 150 c/c
- Distribution Steel Bars = 8 mm Dia. @ 180 c/c
- Side Cover = 50 mm
- Top & Bottom Cover = 20 mm
- All Hook’s = 90 Degree
- Bent Up = 45 Degree

SR. No. | Descriptions | No | Length | Breadth | Height | Quantity |

1 | 12 mm dia main Steel bars @ 150 mm c/c Alternate bent up. | |||||

L = Span Length + Wall Width(Right) + Wall Width (Left) + 2 x 9D (Hooks) – 2xCover L = 3 + 0.23 + 0.23 + (2 x 9 x 0.012) – (2 x 0.05) | ||||||

L = 3.58 m (Straight length without bent) | ||||||

Span = 4 + 0.23 + 0.23 – (2 x 0.05)= 4.36 m | ||||||

No. of Bars = (4.36/0.15) + 1 = 30 nos. | ||||||

Extra Length of Bent up: | ||||||

X – Value | ||||||

Extra Length = 0.45 X | ||||||

X = 0.12(Slab Thickness) – 2 x 0.02 (Top & Bottom Cover) – 0.012/2(Harf dia top) – 0.012/2( Half dia bottom) | ||||||

L = 3.58 + 0.45X L = 3.58 + 0.45 x 0.068 L = 3.61 m | 30 | 3.61 | @ | 0.89 | 96.38kg | |

(Note: In case of bent up bars are provided on both sides, we have to add 0.45 X twice in length) | ||||||

8 mm dia. Distribution steel @ 180 mm c/c | ||||||

Bars at Bottom: | ||||||

Hook Length = 9d = 9 x 0.008= 0.072 < 0.075 ( Mini. Hook length = 0.075) | ||||||

L = 4 + 0.23 + 0.23 + 2 x 0.075(Both side hook(9d) | ||||||

L = 4.51 m | ||||||

Width of Slab = 3 + 0.23 + 0.23 – 2 x 0.05 (Both Side Cover) = 3.36 m | ||||||

No. of Bars = (3.36 / 0.18) + 1 = 20 nos. | ||||||

Bars at Top: | ||||||

Width of Slab at one end for bent up at Top | ||||||

L= 0.23 + 0.45 – 0.068 – 0.05(cover) | ||||||

L = 0.562 m | ||||||

Nos. of Bars at one End:= (0.562/0.18) + 1 = 5 nos. | ||||||

Nos. of bars at both side = 2 x 5 = 10 | ||||||

Total Bars = 20 + 10 =30 | 30 | 4.51 | @ | 0.40 | 54.12 kg | |

Read More: How to Calculate Steel Quantity from Drawing | BBS of Slab | Steel Quantity Calculation

**How Is Bend Deducted In Bar Bending Schedules?**

The Bend deductions are very much important while calculating the Bar Bending Schedule. Basically, we will be having a thought that while bending a bar how much length of the bar should be cut.

Many of us think that we require more than the required length so that the bar will come to its perfect length after Bending. But this is not true.

The Reinforcement bars which come to the market are of Length 12 m. So as the steel has the deformation property, due to the elasticity the length of the bar increases while we bend them. So the Deductions will be different for the different bents based on the angles.

The Deductions are as follows:

d is the diameter of the bar.

1. For a 45-degree bend, the deduction is d.

For example, if the length of the bar is 12 m (assume), the diameter is 25 mm. So for one bend, it is deducted by d.

i.e.

Cutting Length = 12- d= 12- 0.025= 11.975 m.

Therefore only 11.975 m bar is necessary to get a bar of Length 12 m after bending.

If we bend a bar of 12 m then the length of the bar after bent will be

= 12+ d= 12+0.025= 12.025 m which will be more than required.

2. Similarly for a 90-degree bend, the deduction is 2d.

For example, if the length of the bar is 12 m (assume), the diameter is 25 mm. So for one bend, it is deducted by 2d as it is 90 degrees bend.

i.e.

Cutting Length = 12- 2d= 12- 2*0.025= 11.95 m.

3. Similarly for 135-degree bend, the deduction is 3d.

For example, if the length of the bar is 12 m (assume), the diameter is 25 mm. So for one bend, it is deducted by 3d as it is 135 degrees bend.

i.e.

Cutting Length = 12- 3d= 12- 3*0.025= 11.925 m.

4. Similarly for a 180-degree bend, the deduction is 4d.

For example, if the length of the bar is 12 m (assume), the diameter is 25 mm. So for one bend, it is deducted by 4d as it is 180 degrees bend

**i.e.**

Cutting Length = 12- 3d= 12- 4*0.025= 11.900 m.

Note: For n number of bends there should be n deductions in the original length to get the cutting length.

For Example,

Length=12m, Diameter=25mm, bent angle= 90 degree & no of bents are 2.

Therefore Cutting Length= 12- 2*(2d)= 12- 2(2*0.025)= 11.9 m.

Note: Practically in the construction site we will be deducting only 2d for a single bend irrespective of the angle (i.e. even though the angles are 45 degrees, 90 degrees,135 degrees, 180 degrees we will deduct only 2d for each bent).

Refer to “IS:2502 Code of Practice for Bending and Fixing of bars” for further information.

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