Ultimate Compressive Load for an Axially Loaded Short Column

Ultimate Compressive Load for an Axially Loaded Short Column

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 = 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 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,

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_yA_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 =

Or, A_sc =

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

P_u = 0.40 f_ck A_g – 0.40 f_ck + 0.67 f_y

Or, P_u = 0.40 f_ck (A_g – ) + 0.67 f_y

Or, P_u/A_g = 0.40 f_ck + (0.67 f_y – 0.40 f_ck)

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