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Bar bending schedule and beam reinforcement details

There are two columns of 600 x 600. We’ve taken beam size as 300 x 600 mm and clear cover = 300 mm. Beam or column face to rod outer face or part is clear cover. In the beam, there are 4 main rods. Bottom is divided into three parts, Left, Mid and Right.

In bottom, left, mid and right side, there is two 25 ϕ rod plus two 20 ϕ rod. In top, left, mid and right, there is two 20 ϕ. For stirrup, specification is 8 ϕ @ 100 (left), 150 (mid), 100 (right) centre to centre.

Bottom Bar

1st bottom bar layer

Development length = Ld = 50d (depending on grade of concrete and steel)
25ϕ = beam length + development length – 90 degree Bend
= 5000 + (50 x 25 x 2) – (2 x 25 x 1 x 2) = 7.4 m or 7400 mm
Total number = 2 and therefore for 2 numbers = 7.4 x 2 = 14.8 m

2nd bottom bar layer

Applying Same formula
20ϕ = 5000 + (50 x 20 x 2) – (2 x 20 x 1 x 2) = 6.92 m or 6920 mm
Total = 2 and therefore for 2 numbers = 6.92 x 2 = 13.84 m

Top Bar

20ϕ = similar to 2nd bottom bar layer

Stirrups

L/3 (For Zone 1)
L/3 = 5000/3 = 1666.67 mm
Quantity of stirrups = Length (L/3)/Spacing + 1 = 1666.67/100 + 1 = 17.7 = approximately 18 numbers
Whole quantity of stirrups for two Zone 1 = 18 x 2 = 36 numbers

Now For Zone 2

Quantity of stirrups = Length (L/3)/Spacing – 1 = 1666.67/150 – 1 = 10 numbers
Quantity (Total) = 36 + 10 = 46 numbers

Stirrups Cutting Length

a = 300 – (2 x clear cover) – (2 x bar half)
a = 300 – (2 x 30) – (2 x 4) = 232 mm
b = 600 – (2 x clear cover) – (2 x bar half)
b = 600 – (2 x 30) – (2 x 4) = 532 mm

Cutting length of stirrup = (2 x a) + (2 x b) + (2 x 10d) [Hook] – (3 x d x 2) – (2d x 3) [2 and 3 are number of band]
Cutting length = (2 x 232) + (2 x 532) + (2 x 10 x 8) – (3 x 8 x 2) – (2 x 8 x 3)
Cutting length = 1.592 m = 1592 mm
So, for 46 stirrups = 46 x 1.592 = 73.232 m

For more information, go through the following tutorial.

Lecturer: L & T - Learning Technology