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 one vertical bar is 4.640 m. we have total bar 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 25 nos of stirrups, which are going to use,
Total length
= 25 x 1.16
= 29 m long 8 mm bar.
Civil Notes APP
Quantity surveying, Concrete, Steel, Notes available in this Android App.
Click here to Download free from play store
This article first appeared on Engineering Discoveries
Discover more from Makao Bora
Subscribe to get the latest posts sent to your email.
