Example:
Load = 400 kN
Column size = 300mm×300mm
No of steel bar used in column 4 nos
Diameter of bar used in column = 16mm
M20 grade concrete used
Grade of steel = Fe 415
Formula:
Pu = (0.4fck.Ac)+(0.67fy Asc)
( From IS-456 2000 code page no 71)
Where .
Pu = axial load on the member
Ac = area of concrete
fck = characteristic compressive
strength of the concrete
fy = characteristic strength of the
compression reinforcement
Asc = area of longitudinal reinforceme
-nt for column
fck = 20
fy = 415
Ac = Ag-Asc
Ag = gross area of column
Ag = 300×300 mm square
= 90000 mm square
Asc = 4×((3.14×16×16)÷4) mm square
= 804.24 mm square
Ac = (90000-804.24) mm square
= 89196 mm square
Pu = (0.4×20×89196)+(0.67×415×804)
= 937120.1 N
= 937.1 kN ans.
According to the IS-456 2000
Safety of factor = load /1.5
= 624.74 kN
Note: for more information watch video link below.
https://youtu.be/pFDmAt0EWvE