Bar Bending Schedule for RCC Slab
Bar Bending Schedule for RCC Slab.
This article is about How to prepare BBS for Slab
To preparing BBS for a Slab. there are some criteria and assumptions that require to be met Let’s get started.
Forces on Slab
U should know the normal forces action on a Slab.
If you see the shear force and bending moment diagram, the slab has high moments at center and less at ends. The shear force is maximum at the ends while at the center shear force is zero or minimum.
 Use less cutting of steel and provide desirable lapping.
 Lap another steels at same time.
 Don’t Overlap at the same position/ place.
 Between the laps end to end spacing should be provided.
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BBS Used by
BBS is used by different kind of peoples and officials on construction site which is given below:
 Quantity surveyor
 Steel fixer
 Contractor
 Project manager
 Inspectors
 Clerks of construction
In Construction site all estimation and costing operation is the responsibilities of Quantity Surveyor. The B.B.S gives the details to Quantity Surveyor for finding out the number of steel its type, size and shape. Contractor are using B.B.S for ordering the steel bar for proceeding the construction. Steel fixer using B.B.S for doing their work or installing the bar in Column, Beam or Slab according to the B.B.S that how much cut length or bending of bar will be needed and which size of steel should be use. The inspector & clerks of construction refers to the B.B.S to make are that the reinforcement work in the field is in proper with the design reinforcement as per drawing.
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Bar Bending Schedule for Slab
Details from the diagram:
 Thickness of slab is150 mm
 Development Length is 50d
 Bottom bar of slab will be provided 12 mm dia of steel is at 200 mm center to center spacing
 Top bar of slab will be provided 12 mm dia of steel is at 200 mm center to center spacing
 Extra bar of slab at x axis will be provided 12 mm dia of steel is at 125 mm center to center spacing
 Extra bar of slab at y axis will be provided 12 mm dia of steel is at 125 mm center to center spacing
 Stirrup spacing is 8mm @ 150 mm C/C
 Concrete Cover of slab is 25 mm
Steps for Bar Bending Schedule of Slab
Generally, there are five steps for preparing the B.B.S of column these five is given below:
 Number of steel
 Cut length of steel
 Find out lapping
 B.B.S
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Number of Steels
Numbers of steel in Main bar
Number of steels = (Total length of slab ÷ spacing of bar) +1
Number of stirrups = 5000 ÷ 200 + 1
Number of stirrups = 26 numbers
Numbers of steel in Distribution bar
Number of steels = (Total length of slab ÷ spacing of bar) +1
Number of stirrups = 5000 ÷ 200 + 1
Number of stirrups = 26 numbers
Numbers of Extra bar at xaxis
Number of steels = (Total length of slab ÷ spacing of bar) +1
Number of stirrups = 5000 ÷ 125 + 1
Number of stirrups = 41 numbers
Numbers of Extra bar at yaxis
Number of steels = (Total length of slab ÷ spacing of bar) +1
Number of stirrups = 5000 ÷ 125 + 1
Number of stirrups = 41 numbers
Cut length of steel
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Cut Length of Main Bar
Length of Main bar = slab length – 2(concrete cover) + Development length (Ld) + Overlap Length
= 5000 – 2(25) + (12×12) + (50×12)
= 5694 mm or 5.69 m
Now we find the length of bar
The total length of Main bar = 5694 x number of steels in slab
The total length of bar = 5694 x 26
The total length of bar = 148044 mm or 148 meter
Cut Length of Distribution Bar
Length of Distribution bar = slab length – 2(concrete cover) + Development length (Ld) + Overlap Length
= 5000 – 2(25) + (12×12) + (50×12)
= 5694 mm or 5.69 m
Now we find the length of bar
The total length of Distribution bar = 5694 x number of steels in slab
The total length of Distribution bar = 5694 x 26
The total length of bar = 148044 mm or 148 meter
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Cut Length of Extra Bar at xaxis
Total Length of extra bar = Length of extra bar – concrete cover
= 1500 – 2(25)
= 1450 mm or 1.450 meter
Now we find the length of bar
The total length of extra bar at x axis = 1450 x number of extra bar in slab
The total length of bar = 1450 x 41
The total length of bar = 59450 mm or 59.45 meter
Cut Length of Extra Bar at yaxis
Length of Extra Bar at yaxis = slab length – 2(concrete cover)
= 1500 – 2(25)
= 1450 mm or 1.40 m
Now we find the length of bar
The total length of Extra bar = 1450 x number of extra bar in slab
The total length of bar = 1450 x 41
The total length of bar = 59450 mm or 59.45 meter
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Find out lapping
Main Bar
length of lapping is 50d
Lapping length = 50 x Dia of steel
Length of lapping = 50×12 = 600 mm.
We know that the length of typical steel is 12.25 m or 40 feet length
Total Length of Main Bar = 148044 m that is more than 12.25 m or 40 feet so lapping should be provided.
Length of lapping is 600 mm the result will be 148044 + 600 = 148644 m
It is mentioned already in this article, consider the points while lapping the bar. There is no one right process we can use. Just remember the lines.
 Don’t tie a column at bottom and also at top
 Lap at another bars every time
 Overlap top bars with bottom bars always
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Distribution Bar
length of lapping is 50d
Lapping length = 50 x Dia of steel
Length of lapping = 50×12 = 600 mm.
We know that the length of typical steel is 12.25 m or 40 feet length
Total Length of Main Bar = 148044 m that is more than 12.25 m or 40 feet so lapping should be provided.
Length of lapping is 600 mm the result will be 148044 + 600 = 148644 m
It is mentioned already in this article, consider the points while lapping the bar. There is no one right process we can use. Just remember the lines.
 Don’t tie a column at bottom and also at top
 Lap at another bars every time
 Overlap top bars with bottom bars always
Number of Steels
Numbers of steel in Main bar
Number of steels = (Total length of slab ÷ spacing of bar) +1
Number of stirrups = 5000 ÷ 200 + 1
Number of stirrups = 26 numbers
Numbers of steel in Distribution bar
Number of steels = (Total length of slab ÷ spacing of bar) +1
Number of stirrups = 5000 ÷ 200 + 1
Number of stirrups = 26 numbers
Numbers of Extra bar at xaxis
Number of steels = (Total length of slab ÷ spacing of bar) +1
Number of stirrups = 5000 ÷ 125 + 1
Number of stirrups = 41 numbers
Numbers of Extra bar at yaxis
Number of steels = (Total length of slab ÷ spacing of bar) +1
Number of stirrups = 5000 ÷ 125 + 1
Number of stirrups = 41 numbers
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Bar Bending Schedule
The B.B.S of column is shown below in table
S. NO 
Bar Mark 
Bar dia 
No of bars 
Length (m) 
unit weight(m/kg) 
weight of steel (Kg) 
1 
Main Bar 
12 
26 
148 
0.888 
3417 
2 
Distribution Bar 
12 
26 
148 
0.888 
3417 
3 
Extra Bar xaxis 
12 
41 
59.45 
0.888 
2164.45 
4 
Extra Bar yaxis 
12 
41 
59.45 
0.888 
2164.45 
Total Weight of Reinforced Concrete Slab 
11163 
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