Calculate the load carrying capacity:
We can get stress in concrete = (342.046 × 1000) /153566 = 2.227355 N/mm².
Therefore, Stress in steel = 18 × 2.227355 = 40.0924 N/mm².
To calculate the load carrying capacity, we ought to know the strength (permit able stress) of all of the components. For instance, let the permissible stress in concrete be 4 N/mm² and the allowable stress in steel be 120 N/mm². There is another limits, you might notice, that even though the strength of steel is 120 N/mm², we might not stress it to the full, because if we stress steel to its full strength, i.e. 120 N/mm², then concrete shall be stressed to 6.667 N/mm², that is 120/18, which is not permissible. Therefore, the maximum stress that might be induced in steel is just 72 N/mm² that is 4 × 18.
We can now calculate the maximum load the column might support or the load carrying capacity of the column as follows:
P = σs × As + σc × Ac
= 72 × 6434 + 4 × 153566
= 1077512 N or 1077.512 kN.
Another kind of problem that a designer faces is to choose the amount of steel needed if the column is to support a magnitude of load. For example, if the column is needed to carry an axial load of 960 kN, let us compute the steel requirement.
Maximum stress in steel = 72 N/mm²
Maximum stress in concrete = 4 N/mm²
Assume the area of steel be As.
Area of concrete = 160000 - As.
∴ 72 × As + 4 (160000 - As) = 960 × 1000
(72 - 4) As + 640000 = 960000
∴ As =(960000 - 640000)/ (72 - 4) = 4706 mm²
This might be provided suitably (say, 8 nos of 28 mm dia bars, which shall be a little more than adequate).