A load of 300kN is applied on a short concrete column 250 mm x 250 mm....
Given Data:
Load applied on the column, P = 300kN
Dimensions of column, b = 250mm and h = 250mm
Total area of steel reinforcement, As = 5600mm2
Modulus of elasticity of steel, Es = 15Ec
Maximum permissible stress in concrete, σc = 4N/mm2
Load to be supported, P' = 600kN
Stress in Concrete and Steel:
Given that the column is short, the load is carried by both concrete and steel. Let σc be the stress in concrete and σs be the stress in steel.
Total area of cross-section of column, A = b x h = 250 x 250 = 62500mm2
From equilibrium, P = σc x Ac + σs x As
Where, Ac = A - As (area of concrete)
Therefore, σc = (P - σs x As) / Ac
Also, Es = 15Ec
Therefore, σs = σc x Es / Ec
Substituting the values, σc = 1.68N/mm2 and σs = 25.2N/mm2
Area of Steel Required:
To find the area of steel required to support a load of 600kN, let σc' be the stress in concrete.
From equilibrium, P' = σc' x Ac + σs x As
Also, σs = σc' x Es / Ec
Substituting the values, σc' = 2.52N/mm2
Let As' be the area of steel required.
Therefore, As' = (P' - σc' x Ac) / σs
Substituting the values, As' = 14000mm2
Therefore, additional steel of area 14000 - 5600 = 8400mm2 is required.
Thus, the total area of steel required to support a load of 600kN is 14000mm2.