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 = 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 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,
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 = [latex] \frac{A_{sc}}{A_{g}}\times 100 [/latex]
Or, Asc = [latex] \frac{p}{100}\times A_{g} [/latex]
After putting the value of Asc in equation (a)
Pu = 0.40 fck Ag – 0.40 fck [latex] \frac{p}{100}\times A_{g} [/latex] + 0.67 fy [latex] \frac{p}{100}\times A_{g} [/latex]
Or, Pu = 0.40 fck (Ag – [latex] \frac{p}{100}\times A_{g} [/latex]) + 0.67 fy [latex] \frac{p}{100}\times A_{g} [/latex]
Or, Pu/Ag = 0.40 fck + [latex] \frac{p}{100} [/latex] (0.67 fy – 0.40 fck)
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