### BBS OF COLUMN

BAR BENDING SCHEDULE OF COLUMN

• The height of the column is 3 meter and having a cross sectional dimensions are 450 x 600 mm and having a 40 mm of clear cover. There are 12 bars used of 16 mm as a main vertical bars. The diameter of stirrups are used at 100 mm and 150 mm c/c at L/4. So calculate the cutting length of bars and stirrups?
Given data:-
Height - 3 meter
Cross sectional dimensions - 450x600 mm
Clear cover - 40 mm
No.of vertical bars - 12 nos
Dia. Of vertical - 16 mm
Dia. Of stirrups - 10 mm
C/C distance of stirrups - 100 & 150 mm

1. Vertical bar calculation
Length of 1 bar - H + Ld
Where, H - Height of column
LD - development length

Length of 1 bar = 3000 mm + 47d
= 3000 + 47*16
= 3000 + 752
= 3752 mm

We have total 12 bars so total length of all bars are,
3.752 m X 12 nos = 45.024 m

2. Stirrup calculation
Cross sectional dimensions - 450 X 600 mm
Let assume the 450 mm as A side and
600 mm as B side of column

Calculation of A = 450 - 2 X clear cover
= 450 - 2 X 40 mm
= 450 - 80 mm
= 370 mm

Calculation of B = 600 - 2 X clear cover
= 600 - 2 X 40 mm
= 600 - 80 mm
= 520 mm

No of stirrups = 3000 / 4 = 750 mm
There are two zones of 100 mm
Spacing and two zones of 150 mm.

At end zones,
1500 / 100 = 15 Rings
At mid zones,
1500 / 150 = 10 Rings
Total no. Of stirrups = 15 + 10 = 25 Rings

Cutting length of one stirrup

Formula- (2xA) + (2xB) + hook - bend.
Where,
Hook = 10d
Bend = 5x 2d (because we have 5 bend)
d = Dia of bar
= (2x370) + (2x520) + (10x10) - (2x5x10)
= 740 + 1040 + 100 -100
= 1780 mm
We have total 25 stirrups so total length is,
= 25x 1.78 m
= 44.5 m long 10 mm bar.