Bar Bending Schedule of Column Footing
Bar Bending Schedule of Column Footing
Here we teach how to calculate Bar Bending Schedule of Column Footing give below at clear explanation.
Before calculating the steel read carefully given footing drawing and note all the important points like.
1.Footing(Length,Width,Thinkness)
2.Diameter of footing reinforcement
3.Grate of reinforcement
4.Spacing of the reinforcement(c/c)
5.Hook’s length(If needed)
6.Concrete covers of the footing(Top and Bottom)
Example:
Suppose we have a column footing having a length 2m,width 2m and having a thickness 0.250m.The main bars is 12mm @ 150c/c and distribution bar is also 12 mm @150 c/c.The footing clear cover is 50 mm from top and 75mm from the bottom.Calculate the number of the steel is going to use in this column footing.
Given Data:
Length of the Footing = 2m
Width of the footing = 2m
Thickness = 0.250m
Main bar = 12mm @ 150c/c
Distribution bar = 12 mm@150 c/c
Clear cover =50 mm form top and sides and 75 mm from bottom
Solution:
The first step is to calculate the number of the bars which are going to use in footing.
In seconds step we will calculate the cutting length of the bars and at the last we calculate the weight of the reinforcement bars.
Footing Main Bar’s
= (Total didtance – clear cover) / c/c +1
= (2 m – (0.05+0.05) / 0.15 + 1
= 14 Bars
Cutting length of one main bar in footing:
Formula:
= {Total length – 2(Half diameter of the bar + clear cover) + 2(Thickness of the slab – Bottom and top clear cover – Half diameter of the bar)}
Length of 1 Main Bar:
= 2 m – 2 (0.006 + 0.05) + 2 (0.250 – 0.075 – 0.05 – 0.006)
= 2 m -2 (0.112) + 2 (0.119)
= 2 – 0.224 + 0.283
= 2.059 m
Weight of the bar:
= D ²/ 162.2 x Length
= D ²/ 162.2 x 28.826 m
= 25.32 kg
Footing Distribution Bar:
No’s of Distribution bar
= Total distance – clear cover) / c/c + 1
= (2 m – (0.05 + 0.05) / 0.15 + 1
= 14 Bars
Cutting Length
Formula:
= {Total length – 2 (Half diameter of the bar + Clear cover) + 2(Thickness of the slab – Bottom and Top clear cover – Half diameter of the bar)}
Length of 1 Distribution bar:
= 2 m – 2(0.006 + 0.05) +2 (0.250 – 0.075 – 0.05 -0.006)
= 2m – 2(0.112) + 2(0.119)
= 2 – 0.224 + 0.283
= 2.05 m
Total Length:
= Length of one bar x No’s of bars
= 2.059 x 14
= 28.826 m
Weight of main bar:
= D² / 162.2 x Length
= D² / 162.2x 28.826 m
= 25.32 kg
Conclusion:
1.Footing Main Bar
= 12 mm @ 150 c/c
= 14 No’s
= 25.32 kg
2.Footing Distribution Bar
= 12 mm @150 c/c
= 14 No’s
= 25.32 kg
