Bar bending schedule for Beam | BBS | Bar bending Schedule Farmula

 The bar arrangement according to the bar's length and diameter is known as the bar bending schedule for the Beam.

The structural drawing of a reinforced beam is prepared according to the design load of the structure and should adhere to the standard code of practice in a particular country, such as IS 456 in Asia.

We need the section plan of the Beam or the longitudinal section drawing of the RCC beam to prepare the bar bending schedule of the Beam.





Beam bbs Here is an example of a beam in which the steel bar is denoted by the number of bars and the diameter of the bars used in an RCC beam.

As you can see, the length of the Beam is not specified because it is located in a different location according to the design plan.

The clear span of the Beam is 5.50 meters, and the section is 600450 mm. On both sides, the Beam is supported by a 600600 mm column. You can select based on your Beam.

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Total beam reinforcement length

The length of the Beam is equal to the exact span of the Beam plus both column dimensions.

L = 0.60+5.50+0.60 = 6.70 m

Length of reinforcement bar = Total length minus cover on both sides

Length of reinforcement bar = 6.70-0.03=0.03 = 6.64 m

We also consider the developing length of the bar for extra length on both sides.

1. Provide Development length

The development length of the Beam is calculated as per IS 456







Here, in our Beam, the main bar is of dia 16mm and Fe415 (415 N/mm2) with M20 Grade concrete.

So the Development length for 16mm deformed bar= 16*415/(4*1.2*1.6) = 865 mm = 0.865 m.

Development length of 20mm bar = 1080 mm = 1.08 m

Length of Bar = total length – cover + 2x development length

2. Provide Hooks

We can also provide the hook of a U shape, which is x16 times the diameter of the bar. Here hook length for 16mm bar = 16×16 = 256 mm

Hook length for 20mm bar = 20×16 = 320 mm

So length of bar with hooks = 5.50+0.256 = 5.756 m

The ets bar bending schedule of a beam with development length without hooks

 Stirrups/Ring of 10mm size = 0.54×0.39 m with bend hook Length of Ring

= 2x(0.54+0.39+0.10) = 2.06 8 number of the Main bar of 16 mm

 with 8.37m (0.865+6.64+0.865) with u shape = 8*8.37 =

 66.96mso 16mm weight = 66.96*1.58 = 105.80 KG

3 number of the bottom bar of 20mm with 3.56m with straight length = 3*3.56 = 10.68m

6 number of the top extra bar of 20mm with 3.45m (1.08+2.37) with L shape = 6*3.45 = 20.70m

so 20mm weight = (10.68+20.70)*2.47 = 77.50 KG

35 number of Stirrups/ring = 35×2.06 = 72.10 m 

so 10mm weight = 72.10*0.617 = 44.48 KG

Total Weight of Steel in beam = 105.80+77.50+44.48 = 227.78 KG