CIVIL Engineer

Bar Bending Schedule for beam concept

We are taking the below example of single reinforcement – bent up bar to explain
Given

Diameter of the Bar = 12 mm
Clear Cover = 25 mm
Clear Span (i.e “L”) = 6000 (Length of the slab)
Slab Thickness = 150 mm
Ld = 40d (Development Length). We know the difference between lap length and development length.
Calculation
Therefore,

Cutting Length = Clear Span of Slab + (2 X Developement Length) +(2 x inclined length) – (45° bend x 4) – (90° bend x 2)
Cutting Length = Clear Span of Slab + (2 X Ld) +(2 x 0.42D) – (1d x 4) – (2d x 2) [BBS Shape Codes]
In the above formula, we have all the value, (except D)
Ld – Development Length which is equal to 40d
d – Diameter of Bar
So D = height of the bend bar (refer the image)
“D” = Slab Thickness – (2 x clear cover) – (diameter of bar)
= 150 – (2 x 25) – 12 = 88 mm
“D” = 88 mm
Now we know the “D” value which is the clear height of the bar (refer the image).
In order to find the inclined bar length using (Trignometry Function)
Inclined length = d/(sin 45°) – d / (tan 45°) = (d/0.7071) – (d/1)= (1d – 0.7071d)/0.7071= 0.42 D
we are providing four 45°bends at inner side (1,2,3 & 4) and two 90° bends ( a,b ).
45° = 1d ; 90° = 2d
Coming back to the formula,
Cutting Length = Clear Span of Slab + (2 X Ld) +(2 x 0.42D) – (1d x 4) – (2d x 2)
= 6000 mm + ( 2 x 40 x 12) + (2 x 0.42 x 88) – (1x12x4) – (2x12x2)
Cutting Length = 6938 mm or 6.94 m
So far the above dimensions, you need to cut the main bars in 6.94 m length.