In this post, we are going to see How to prepare bar bending schedule for Beam alias BBS for Beam.
Note – There is no one single version which is an absolute right. so please use the below calculations as a guide.
We hope that you remember about steel weight calculation post. if you have missed that post, please go ahead and read it.
Now we are going to prepare BBS for a beam in two types.
- First one is BBS for the simple beam which we used normally at the site.
- The second one contains many details as shown in the respective diagram
Let’s get started
Bar Bending Schedule for Simple Beam
As you can see in the figure, the beam has clear span of 3metre consists of 2 numbers of 16 mm dia at bottom and 2 numbers of 12mm dia bars at top with 8mm dia stirrups at 150mm Clear Cover
Assuming Clear Cover of 25 mm at both ends and sides of the beam
Observations
- Clear Span of Beam = 3000 mm
- Development Length Ld = 50d (assumption)
- Clear Cover on any ends = 25 m
- Bottom – 2 numbers of 16⌀
- Top – 2 numbers of 12⌀
- Stirrups – 8⌀ @ 150mm clear cover
Step 1 – Find cutting length of top bar
Cutting length of top bar = Clear Span of Beam + Development length (Anchorage) Ld on 2 sides – Clear Cover on 2 ends
= 3000+(2 x 50d) – 2 x 25 = 3000 + (2 x 50 x 12) – 50
= 4150 mm
Step 2 – Find cutting length of bottom bar
Cutting length of bottom bar = Clear Span of Beam + Development length (Anchorage) Ld on 2 sides – Clear Cover on 2 ends
= 3000+(2 x 50d) – 225 mm = 3000 + (2 x 50 x 16) – 50
= 4550 mm
Step 3 – Find out Number of Stirrups
Number of Stirrups required = (Clear Span of Beam/Spacing Stirrups)+ 1
= (3000 / 150) +1 = 21 No.s
Step 4 – Find out cutting length of Each Stirrup
Let’s split the sides as (a, b) for easy calculation.
Length of One Hook = 9d (As per Estimating and Costing in Civil Engineering P.No.214)
Cutting length of Stirrup = Perimeter of stirrup + Number of Bends + Number of Hooks
= 2(a+b) + 3 numbers of 90 degree bends + 2 numbers of hooks
= 2(300+500) + (3 x 2d)+(2 x 9d) = 1600 + (3 x 2 x 8) + (2 x 9 x 8) = 1792 mm
| Diameter of Bar | Numbers | Cutting Length | Total Length | |
| Top Bar | 12 | 2 | 4150 mm | 8300 mm |
| Bottom Bar | 16 | 2 | 4550 mm | 9100 mm |
| Stirrups | 8 | 21 | 1792 mm | 37632 mm |
Bar Bending Schedule for Detailed Beam
As you can see that this diagram has more detailed and technical implementation of design aspects. Let’s get started
The above beam has the clear span of 6metre consists of two layers of the bottom (2 numbers of 20mm dia) and one layer of the top (2 numbers of 12mm dia).
It consists of 3 Zones where Zone 1, 3 has stirrups of 150 mm spacing & Zone 2 has stirrups of 200 mm spacing.
Observations
- Clear Span of Beam = 6000 mm
- Development Length Ld = 50d (assumption)
- Clear Cover on any ends = 25 mm
- Bottom – 4 numbers of 25⌀
- Top – 2 numbers of 12⌀
- Stirrups
- zone 1,3 = 8⌀ @ 150mm clear cover
- zone 2 = 8⌀ @ 200mm clear cover
Step 1 – Find cutting length of top bar
Cutting length of top bar = Clear Span of Beam + Development length (Anchorage) Ld on 2 sides – Clear Cover on 2 ends
= 6000+(2 x 50d) – 2 x 25 = 6000 + (2 x 50 x 12) – 50
= 7150 mm
Step 2 – Find cutting length of bottom bar
Cutting length of bottom bar = Clear Span of Beam + Development length (Anchorage) Ld on 2 sides – Clear Cover on 2 ends
= 6000+92 x 50d) – 2 x 25 = 6000 + (2 x 50 x 25) – 50
= 8450 mm
Step 3 – Find out Number of Stirrups
As we said this beam has three types of zones where Zone 1 & 3 has stirrups has 150 mm spacing and Zone 3 has 200 mm spacing.
First calculate the length of zones = L/3 = 6/3 = 2 metre.
Number of stirrups required for Zone 1 = (Clear Span of beam/ spacing Stirrups)+ 1
= (2000/150)+1 = 14.3 ♎ 14 no.s
Number of stirrups required for Zone 3 = same as Zone 1 = 14 no.s
Number of stirrups required for Zone 2 = (Clear Span of beam/ spacing Stirrups) – 1 = (2000/200)-1 = 10 no.s
The reason for the “minus 1 “ is already zone 1 has end stirrup so we don’t need to again start from new one. we just continue the stirrups already existing with extra spacing.
Step 4 – Find out cutting length of Each Stirrup
Length of One Hook = 9d (As per Estimating and Costing in Civil Engineering P.No.214)
Cutting length of Stirrup = Perimeter of stirrup + Number of Bends + Number of Hooks
= 2(a+b) + 3 numbers of 90 degree bends + 2 numbers of hooks
= 2(300+500) + (3 x 2d)+(2 x 9d) = 1600 + (3 x 2 x 8) + (2 x 9 x 8) = 1792 mm
| Diameter of Bar | Numbers | Cutting Length | Total Length | |
| Top Bar | 20 | 2 | 7150 mm | 14300 mm |
| Bottom Bar | 25 | 4 | 8450 mm | 33800 mm |
| Stirrups | 8 | 21 | 1792 mm | 37632 mm |
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