## Ultimate Compressive Load for Axially Loaded Short Column

The ultimate compressive load for axially loaded short column is determined by 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 the ultimate stage, the ultimate compressive load is resisted partly by concrete and partly by steel. Thus, at the ultimate stage,

Ultimate load = **P _{u} = P_{uc} + P_{us} **

Where,

- P
_{uc}= Ultimate load on concrete = 0.45 f_{ck}A_{c} - Pus = Ultimate load on steel = 0.75 f
_{y}A_{sc} - Ac = Area of concrete
- Asc = Area of longitudinal steel

After putting the value of P_{uc} and P_{us} in the above equation

P_{u} = 0.45 f_{ck} A_{c} + 0.75 f_{y} A_{sc}

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 that 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 an 11% reduction, we can write as,

P_{u }= 0.40 f_{ck} Ac + 0.67 f_{y} A_{sc}

Assume, A_{g} = Gross sectional area of the column

Therefore, A_{g} = A_{c} + A_{sc}

Now,

P_{u} = 0.40 f_{ck} (A_{g} – A_{sc}) + 0.67 f_{y}A_{sc} [** Note**, A_{g} = A_{c} + A_{sc} So, A_{c} = A_{g} – A_{sc}].

Or, P_{u} = 0.40 f_{ck} A_{g} – 0.40 f_{ck} A_{sc} + 0.67 f_{y} A_{sc} ……………….(a)

Or, P_{u} = 0.40 f_{ck} A_{g} + (0.67 f_{y} – 0.40 f_{ck}) A_{sc}

If ‘p’ = percentage of steel provided = [latex] \frac{A_{sc}}{A_{g}}\times 100 [/latex]

Or, A_{sc} = [latex] \frac{p}{100}\times A_{g} [/latex]

After putting the value of A_{sc} in equation (a)

P_{u} = 0.40 f_{ck} A_{g} – 0.40 f_{ck} [latex] \frac{p}{100}\times A_{g} [/latex] + 0.67 f_{y} [latex] \frac{p}{100}\times A_{g} [/latex]

Or, P_{u} = 0.40 f_{ck} (A_{g} – [latex] \frac{p}{100}\times A_{g} [/latex]) + 0.67 f_{y} [latex] \frac{p}{100}\times A_{g} [/latex]

Or, P_{u}/A_{g} = 0.40 f_{ck} + [latex] \frac{p}{100} [/latex] (0.67 f_{y} – 0.40 f_{ck})

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