The ultimate compressive load for an axially loaded short column is determined on the following assumption

a) The maximum compressive strain in concrete is 0.002
b) Strain in concrete is equal to strain in steel
c) Stress-strain relation for steel is the same in compression or tension

For an absolutely axially loaded short column, at ultimate stage, the ultimate compressive load is resisted partly by concrete and partly by steel. Thus, at ultimate stage,

Ultimate load = Pu = Puc + Pus

Where,

• Puc = Ultimate load on concrete = 0.45 fck Ac
• Pus = Ultimate load on steel = 0.75 fy Asc
• Ac = Area of concrete
• Asc = Area of longitudinal steel

After putting the value of Puc and Pus in the above equation

Pu = 0.45 fck Ac + 0.75 fy Asc

This relation is applicable for the ideal condition of axial loading. In the practical conditions, the loading is never absolutely axial and there will always be some eccentricity which cannot be avoided. Hence we may consider the possibility of a minimum eccentricity of 0.05 times the lateral dimension and assume an 11% reduction in the ultimate strength of the column.

After 11% reduction we can write as,

Pu = 0.40 fck Ac + 0.67 fy Asc

Assume, Ag = Gross sectional area of the column

Therefore, Ag = Ac + Asc

Now,

Pu = 0.40 fck (Ag – Asc) + 0.67 fyAsc [ Note, Ag = Ac + Asc So, Ac = Ag – Asc].

Or, Pu = 0.40 fck Ag – 0.40 fck Asc + 0.67 fy Asc ……………….(a)

Or, Pu = 0.40 fck Ag + (0.67 fy – 0.40 fck) Asc

If ‘p’ = percentage of steel provided = $\frac{A_{sc}}{A_{g}}\times 100$

Or, Asc = $\frac{p}{100}\times A_{g}$

After putting the value of Asc in the equation (a)

Pu = 0.40 fck Ag – 0.40 fck $\frac{p}{100}\times A_{g}$ + 0.67 fy $\frac{p}{100}\times A_{g}$

Or, Pu = 0.40 fck (Ag $\frac{p}{100}\times A_{g}$) + 0.67 fy $\frac{p}{100}\times A_{g}$

Or, Pu/Ag = 0.40 fck + $\frac{p}{100}$ (0.67 fy – 0.40 fck)