Example
suppose we have a column. The height of the column is 4 m and having a cross-sectional area is 300 x 400 mm and having a 40 mm of clear cover. Six bars are going to use having a diameter of 16 mm. The diameter of the stirrup is 8 mm and having a spacing @ 150 mm and @200 mm at L/3 respectively.
Given Data
Height = 4 meter
Cross section = 300 x 400 mm
Clear cover = 40 mm.
No of vertical bars = 6 no’s.
The diameter of the vertical bar = 16 mm.
Diameter of stirrup = 8 mm.
stirrups center to center spacing = @150 or @ 200 mm.
BBS of Column =?
Solution
The calculation was to proceed into two steps.
- vertical bars calculation
- Cutting length of stirrups
Step 1:Vertical Bar Calculation
Length of 1 bar = H + Ld
Where
Ld = development length
H = Height of column
Length of 1 bar
= 4000 mm + 40d <where d is dia of bar>
= 4000 + 40 x 16
= 4000 + 640
= 4640 mm or 4.640 m Ans.
The length of one vertical bar is 4.640 m. we have a total bar of six bars,
Total length
= 6 x 4.640
= 27.84 m long vertical bar is required.
Step 2:Cutting The Length Of Stirrups
The cross-sectional area of the column is 300 mm x 400 mm
A: is the vertical cross-section area of the stirrup
B: is a horizontal cross-section area of the stirrup
Calculation of length A
A = Horizontal Distance – 2 Side clear cover
A = 300 – 2 x clear cover
A = 300 – 2 x 40
A = 300 – 80
A = 220 mm
The length of B
B = Vertical distance – 2 x Top, Bottom cover
B = 400 – 2 x clear cover
B = 400 – 2 x 40
B = 400 – 80
B = 320 mm
No of stirrups
= 4000/3
=1333.3 mm or 1.33 m
Formula = L/3 / spacing + 1
(no of stirrups in end zone)
=1333.3 / 150
= 8.8 nos say 9 nos
There are total two zones of 150 mm spacing and one zone of 200 mm spacing.
= 2 x 9
= 18 nos (at end zones)
At Mid Zone
=1333.3 / 200
= 6.6 nos say 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 x10d – 5 x 2d
where
hook = 10d
bend = 5 x 2d (we have 5 bends in one stirrup)
d = is diameter of bar
= (2 x 220) + ( 2 x 320) + 2 x 10 x 8 – 2 x 5 x 8
= 440 + 640 + 160 – 80
= 1160 mm or 1.16 m
We have a total of 25 nos of stirrups, which are going to use,
Total length
= 25 x 1.16
= 29 m long 8 mm bar.
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