# Guideline, Basics and Formulas of Bar Bending Schedule [BBS]

Bar bending schedule is the ultimate list of reinforcement bars for any type of structural element. This schedule include shape, size, mark, length, location, and bending detail of reinforcement. BBS is the tabular view representation of every reinforcement bar used in different structural element.

Bar bending is the special process to bend, cut, and to fix the reinforcement bars according to the particular drawing.

### The Guideline

Each and every RCC structural elements must have separate BBS. There is no need of grouping them. They must need to follow IS guideline for bending and length of hook.

It is very easy to memorize. Users must need to keep the shape codes of bar bending with themselves always. Users must use the thumb rule calculation of steel reinforcement to assure the estimation.

### BBS Basics & Formulas to be remembered

1. Diameter of bars (in mm) – 6, 8, 10, 12, 16, 20, 25, 28, 32, 36, 40 mm

2. The standard length of reinforcement bar – 12 meter or 40 feet

3. Weight of bar (Kg) per meter formula – D2/162

### Extension Length Formulas

Footing lap length formula – 40d

Column Lap length formula – 50d

Development length for dowel bars – 16d

Hook Length – 9d

Concrete Cover

Footing – 75 mm

Column & Beam – 25-50 mm

### Bend Deduction

45° – 1d

90° – 2d

135° – 3d

### Crank Length Formula

45° – 0.42d

30° – 0.27d

60° – 0.58d

Extra Bars Length – L/4 or L/5

### Example of Bar Bending Schedule

Step 1 – Find Cutting Length of Bars

Cutting Length of Bottom Bar = Lclear – (2x concrete cover) + (2×500)

= 3000 – (2×40) + (2×500) = 3920 mm

Cutting Length of top bar = Lclear – (2x concrete cover) = 3000 – (2×40) = 2920 mm

Step 2 – Number of Stirrups

Number of Stirrups = (Length of Beam / Spacing) + 1 = (3000/200) + 1 = 16 numbers

Step 3 – Cutting Length of Stirrups

Cutting Length of stirrup = Perimeter of stirrup + (3 x 90° bend) + (2 x hook length)

= 2 (520+220) + (3 x 2d) + (2x1d) = 1480+(3x2x10) +(2×10) = 1560 mm

Bar Bending Schedule Format |
||||||||

Bar Shape | Dia of Bar | Cutting Length | Number of Bar | Length of Bar (mm) | ||||

8 | 10 | 12 | 16 | 20 | ||||

10 | 3920 | 2 | - | 7840 | – | – | – | |

10 | 2920 | 2 | - | 5840 | – | – | – | |

8 | 1560 | 16 | 24960 | - | – | – | – |

Total Length (m) | 24.96 | 13.68 | - | - | - | |||

Weight Per Metre (Kg/m) | 0.4 | 0.62 | 0.89 | 1.58 | 2.47 | |||

Total Weight in Kg | 9.984 | 8.482 | – | – | – |

### The Preparation

1. At first this schedule suggests about the amount of steel to be procured.

2. Then This also instruct the contractor for planning of steel purchase in a week.

3. After that it also instruct the bar bender for fabrication the reinforcement into the desired shape according to the structural drawing.

4. To estimate the material, it can also instruct the quantity surveyor.

### Making Process of Bar Bending Schedule

There are different type of bar bending schedule. These are mentioned below.

1. Bar bending schedule for slab

2. Bar bending schedule for beam

3. Bar bending schedule for column.

There are five steps of it. The steps are

(a) User has to find the amount of reinforcement bars for using

(b) User must find the length of cutting of each and every bar.

(c) Users have to find number of distribution bars’ calculation in one condition that is if the material is slab.

(d) Users must find the cutting length of stirrups.

(e) At last user just need to list down all the bars in the table and search the amount of steel.

### Wrap up

In this article we discussed about bar bending schedule, the guideline, formula, example, format, preparation, and making process of bar bending schedule, and many formulas regarding bar bending schedule. If readers like this article then please let us know about this in below the article. We are very hopeful to know about your opinion.