BAR BENDING SCHEDULE
The schedule for bar bending is known as BBS. During this process, the reinforcing Steel was bent into the various shapes needed for RCC constructions. This procedure is frequently carried out there. The cut, bend, bundle, and location of the bars are easily determined by Bar bending schedules.
BBS is shown in tabular form and is a list of reinforcing bars. This table lists all the necessary information about bars, including their diameter, the type of bending they undergo, the length and width of each bend, the angles at which they turn, the total length of each Bar, and the number of each type of Bar.
Scheduling is creating a list of the type of Bar, its location, size, length, mark, bending, and all other provided details. The (IS: 5525) displays the various suggestions and ensures that the calculation for bar length uses the standard size. The length tolerance is +- 25mm.
Calculation For Footings bar bending schedule involves the following steps:
Remember that the quantity of bars or kilogrammes of Steel needed for a project is ordered. Each Bar is standardised to be 12m in length. The number of "12m" bars or kilogrammes is the final output of the BBS calculation.
The calculation is broken up into two parts, X bar calculations and Y bar calculations, to make it easier to understand. Estimation of Footing Reinforcement
Vertical bars projected in the Y direction are known as Y bars, while horizontal bars are known as X bars. Calculating Footing Reinforcement
- Deduct the concrete cover to find the dimensions of the bars.
- Find the Length of single X Bars & Y Bars
- Find the total length of X bars. & Y bars
- Calculate the weight of Steel required per 1m
- Calculate the total number of 12m bars required
- Find the total weight of Steel needed.
- Bar Bending Schedule Of Footing
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Let us prepare the bar bending schedule of the footing
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| BBS Footing |
Given data :
Footing Lengths are 2000 mm long, 1500 mm wide, and 300 mm deep.
Diameter of rebar is 12 mm. There is a 50 mm cover on all sides with a 150 mm spacing.
number of bars running along the x-axis
= [{(footing length) - (2 cover) / spacing}] + 1
= [(2000mm) - (2x50mm) /150mm] +1
= [{ 1900 mm } ÷ 150 mm.] +1
= 12.67 +1
= 14 nos.
by rounding off
Number of bars along the y-axis Number of bars along the y-axis
= [[(footing width) - (2 x cover) / spacing)] + 1
= [(1500mm) - (2x50mm) /150mm] +1
= [{ 1400 mm } ÷ 150 mm] +1
= 9.33 +1
= 10 nos.
Cutting the Bar's length in the x direction
= [(x-axis bar length + 2 no.s of x (L- bend length)] - 2nos x (2 times the bar's diameter for a 90° bend).
=[ { footing length – 2 × cover } + 2nos.×{ footing height – 2 × cover}] – 2× ( 2 × bar dia)
= [[2000 mm - 2 x 50mm] + 2 x [300 mm. - 2 x 50mm]] – 2 × ( 2 × 12 mm.)
= [1900 + 400] - 48 millimetres.
= 2300 mm – 48 mm
= 2252 mm i.e. 2.252 m
Bar length should be reduced in the y direction.
= ["bar length in y-axis" + { 2 Num of x (L- bend length)}] - 2nos. (2 times the Bar's diameter for a 90° bend).
= [ { footing width – 2 × cover } + 2nos.×{ footing height – 2 × cover}] – 2× ( 2 × bar dia. ) 12.67 +1ar bending schedule of the footing for the below-given drawing.
= [{1500 mm – 2 x 50 mm} + 2 x { 300mm – 2 x 50 mm] - 2 × (2 × 12 mm.)
= [1400 mm + 400 mm] - 48 millimetres.
= 1800 mm - 48 mm
= 1752 mm. i.e. 1.752 m.
Drawing of the column's bar-bending schedule
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| Drawing |
Bar dia. lateral ties. spacing = 250 mm, d1 = 8 mm.. cover =40 mm
Column size x = 300mm & y = 230 mm.
Development length Ld = 50d
Length of the longitudinal Bar
= up to ground level + GL to plinth level + plinth level to slab bottom + slab cover + Ld + L- bend in footing – distance from footing bottom.= {1200 mm+ 450 mm +3000 mm + 20 mm + 50d +300 mm} – 70 mm
={ 4670 +( 50 × 16mm )+ 300 mm } – 70 mm.
= 5770 mm – 70 mm
= 5700 mm i.e. 5.70 m
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Length of the lateral ties
= perimeter of the lateral relations + total hook length – no. of bends= 2sides × ( x – 2 × cover ) + 2 sides × ( y – 2 × cover ) +( 2nos × hook length) – (3 nos. × bend )
( Here, we have taken hook length = 10d1 for 135°∠ & bend = 2d1 for 90°∟)
={ [ 2 × (300mm – 2× 40mm.) ]+[ 2 × ( 230 mm – 2 × 40 mm.) ] } + { 2 × 10 × 8mm }- {3 × 2 × 8mm }
={ [ 2 × 220 mm ] + [2 × 150 mm ]} + 160 mm – 48 mm.
= {440 mm + 300 mm} + 112 mm
= 852 mm i.e. 0.852 m
Total number of lateral ties ( stirrups )
={ [length of the longitudinal bar – (Ld + L bend over footing)] ÷ stirrup spacing } + 1
Note: Ld + L bend is deducted from the length as no stirrups are provided over that length.
={[ 5700 mm – (50 × 16 mm + 300 mm.)] ÷ 250 mm.} + 1
= {[ 5700mm – 1100mm ] ÷ 250 mm.} + 1
= {4600 mm ÷ 250 mm.} + 1
= 18.4 + 1
= 19.4 nos.
Rounding off, the number of stirrups required = 20 nos
Let's now get the BBS table ready for the column.
BBS of Rectangular Footing
Given Data From Footing Drawing
Footing Size = 1 m x 1.2 m
Spacing of Bar = 100 mm C/c
Footing Cover = 50 mm
Bar Use in Footing = 12 mm
Step 1
Cutting Length of Y-AxisCutting Length = Length (Y-Axis) – (2x Cover) + (2x Height) – 4 Cover (For “L” Upside) – Bend
Cutting Length = {1200 – (2 x 50) }+ (2 x 300) – (4 x 50) – (2 x 2 x 12)
Cutting Length = 1100 + 600 – 200 – 48
Cutting Length = 1100 + 600 – 248
Cutting Length = 1700 – 248
Cutting Length = 1452
Cutting Length = 1.45 m
Step 2
Cutting Length of X-AxisCutting Length = Length (X-Axis) – (2x Cover) + (2x Height) – 4 Cover (For “L” Upside) – Bend
Cutting Length = {1000-(2 x 50)} + (2 x 300) – (4 x 50) – (2 x 2 x 12)
Cutting Length = 900 + 600 – 200 – 48
Cutting Length = 900 + 600 – 248
Cutting Length = 1500 – 248
Cutting Length = 1252
Cutting Length= 1.25 m
Step 3
No's of Bar Along With X-Axis ( Length – 1.45 m)No's of Bar X-Axis = [ Length – (2x Cover) ] /Spacing + 1
No's of Bar X-Axis = [ 1000 – (2x 50) ] /100 + 1
No's of Bar X-Axis = ( 900 / 100 ) + 1
No's of Bar X-Axis = 10 Nos
We Need 10 Numbers of a bar for the length of 1.45 m.
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No'sof Bar X-Axis = [ Length – (2x Cover) ] /Spacing + 1
No's of Bar X-Axis = [ 1200 – (2x 50) ] /100 + 1
No's of Bar X-Axis = ( 1100 / 100 ) + 1
No's of Bar X-Axis = 12 Nos
We Need 12 Numbers of a bar for the length of 1.25 m.