If a concrete column 200 × 200 mm in cross-section is reinforced with four steel bars of 1200 mm² total cross-sectional area. Calculate the safe load for the column if permissible stress in concrete is 5 N/mm² and E_s is 15 E_c

Given Data,
Area of Steel = 1200 mm^2
Area of COLUMN = 200x200 mm^2 = 40000 mm^2
Area of Concrete = Area of Column - Area of steel
= 40000 mm^2 - 1200 mm^2
= 38800 mm^2
Permissible stress in Concrete, σc = N/mm^2
Relation given, ( Es = 15Ec )
We Know,
( σc / Ec = σs / Es ) ------------------(i)

Then, from eq. (i)
σs = (σc x Es) / (Ec)
= (5 x 15 Ec) / (Ec)
= 75 N/mm^2

Safe Load (Px) = ( Area of concrete x Perm. Stress in concrete ) + ( Area of Steel x Perm. Stress in Steel )
= ( Ac x σc ) + ( As x σs )
= ( 38800 x 5 ) + ( 1200 x 75 )
= 284000 N
= 284 MN

Stress= load/ area,
Stress in steel = load / area of steel and likewise same for concrete,

Es= permissible stress in steel/ strain in steel and likewise same for concrete,
strain in concrete and steel are same = permissible stress in steel/ Es= permissible stress in concrete/ Ec.

Es= 15Ec.
=permissible stress in steel= permissible stress in concrete/ Ec xEs =( 5 /Ec)x 15Ec= 75Mpa for steel.
Load = permissible Stress in steel x area of steel +permissible stress in concrete x area of concrete.
= 75 x 1200+ 5 x (200X200 -1200)= 284kN.

safe load (P) =(As*sigma of steel ) + (Ac*sigma of concrete) this formulae answer will get
sigma steel / sigma concrete = Es / Ec =15Ec / Ec =15
sigma steel =5*15 = 75 N/mm square
Ac =(200*200 - As)
As = 1200 mm square