Bar Bending Schedule for Column [BBS]
How do you prepare Bar Bending Schedule for column?
okay Let's go, we have a column design 2d drawing with Column data below,
Height = 4 meter
Cross section = 300 x 400 mm
Clear cover = 40 mm
No of vertical bars = 6 no's
Diameter of vertical bar = 16 mm
Diameter of stirrup = 8 mm
Stirrups center to center spacing = @150 or @ 200mm
Bar Bending Schedule for One Way Slab
Step 1 : Vertical Reinforcement Bar calculation
Formula:-
Length of 1 bar = H + Ld
Where
- Ld = development length.
- h = Height of column Length of 1 bar.
- d = diameter of reinforcement bar.
= 4000 + 40d
= 4000+ 40 ×16
= 4640mm
So length of 1 vertical reinforcement bar is 4.640m and we have a total 6 nos of bars,
Total length = 6 x 4.640m
= 27.84 m
long vertical reinforcement bar required.
Bar Bending Schedule For Two Way Slab
Step 2: Cutting length of stirrups
Cross sectional area of column is 300mm*400mm
Calculation of 'A' is vertical cross section area of stirrup
= 300 — 2 x clear cover
= 300 - 2 x 40
= 300 - 80
= 220mm
Calculation of 'B' is horizontal cross section area of stirrup
= 400 - (2 x clear cover)
= 400- (2 x 40)
= 400 - 80
= 320mm
No of stirrups = 4000/3
=1333.3mm.
Formula = L/3 / spacing + 1
(No of stirrups in end zone)
= 1333.3 / 150
= 8.8 nos ~ 9 nos
There are total two zones of 150mm spacing and one zone of 200mm spacing.
=2x9
= 18 nos (at end zones)
At Mid Zone
=1333.3 / 200
= 6.6 nos ~ 7 nos
Total no of stirrups
= 18+7
= 25 nos
Cutting length of one stirrup
Formula: = (2 x A) + (2 x B) + hook - bend
Cutting length
= (2 x A) + (2 x B) +2 × 10 d 5×2d
where
Hook = I0 d
Bend = 5 x 2d (because we have 5 bends in stirrup) d = is diameter of bar
= (2 × 220) + (2 × 320) +2 ×10×8 -2×5×8
= 440 + 640 + 160 - 80
= 1160 mm.
So we have a total 25 nos of stirrups are going to use,
Total length = 25 x 1.16 = 29m long 8mm reinforcement bar.
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