“BBS” The word BBS Plays a significant role in any construction of High rise buildings. BBS refers to Bar Bending Schedule. Well, What’s the use of BBS? Why we use BBS? What is BBS? Bar Bending Schedule is termed as “Calculation of the total Steel required for the construction of a building” We use steel to make concrete to be reinforced and for tension requirements. But how much steel required for constructing 4 floors building? How much Steel I have to order? All these questions are answered in BBS.
Before calculating the steel read carefully given a drawing of the footing and given all important points like the following:
1. Footing. (Length, Width, Thickness)
2. Diameter of Footing reinforcement
3. Grade of reinforcement is going to use
4. Spacing of the reinforcement (c/c)
5. Hook’s length (If needed)
6. Concrete Covers of the Footing (Top and Bottom)
Let’s solve this example to understand well
EXAMPLE:
Suppose we have a Column Footing having a Length 2 m, Width 2 m and having a Thickness 0.250 m. The main bars is 12 mm @ 150 c/c and Distribution bar is also 12 mm @ 150 c/c. The footing clear cover is 50 mm from top and 75 mm from the bottom. Calculate the number of the steel is going to use in this Column Footing.
GIVEN DATA:
Length of the Footing = 2 m
Width of the Footing = 2 m
Thickness = 0.250 m
Main bar = 12 mm @ 150 c/c
Distribution bar = 12 mm @ 150 c/c
Clear cover = 50 mm from 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 second 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 distance – 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
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 /162.2 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)
= 2 m - 2 (0.112) + 2 (0.119)
= 2 - 0.224 + 0.283
= 2.059 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 /162.2 x 28.826 m
= 25.32 kg
So that I can share through this short article, although there are still many shortcomings and there may be errors apologize. Hopefully this article useful and rewind your work.
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