Civil Engineering - RCC Structures Design - Discussion

Yes, you correct @Ramesh.

Does anyone here ever heard about Working Stress Method? In WSM nighter fck or fy is used and does not apply = 0.4 x fck x Asc + 0.67 x fy x Ast.
It is based on WSM only not LSM.
Load bearing capacity should be calculated based on
= Sigma cc * Ac + sigma sc * Asc. this formula only.

Hence,
P = ((20*20)-20)*40 + 1300*20 = 41200 kg.

Note. Here compressive stress in steel is 1300kg/cm^2 instade of 300kg/cm^2 and total area of steel is 20 cm^2 (Including all 4 no. bars).
P.S. - 4x20=80 cm^2 Means 20% of Steel of Column section 20 x 20 which is far beyond Max limit of 6%.

Thank you @Ramesh.

This question solve by the working state method because of the given data is permissible stress not a grade of concrete.

Ans - 36800 kg.

Safe load of column = 0.4*Fck*Asc + 0.67*Fy*Ast.

P= 0.4fck*Asc+0.67fy*Ast,
= 0.4 x 40 x 380 + 0.67 x 300 x 20,
=10,100 Kgs.

Actually, permissible compressive stress in concrete = .4*FCK=40kg/cm^2.
Permissible compressive stress in steel = 0.67*fy=1300 kg/cm^2.
From safe load of column,

P= 0.4fck*Asc+0.67fy*Ast.
40*380+1300*20.
= 41200kg.

0.4 Fck Ac + 0.67 Fy Asc.

This formula is used when given parameters are:-
Fck i.e Characterstic strength of concrete
Fy i.e Yield strength of Bar

But
Given papameters are:-
Permissible compressive stress in concrete i.e sigma cc
Permissible compressive stress in steel i.e sigma sc

So formula used will be:-
Sigma cc * Ac + sigma sc * Asc.
Where Ac = cross sectional area of concrete.
Asc = cross sectional area of steel in compression.
I hope it's clear.

Yes, 21200 kg is the correct answer.

Agree @Anbarasan.

[0.4X40X(20X20)] + [0.67X300X20] = 41,680.

Put the formula, P = sigma sc * Asc + sigma cc * Ac.
All data are given you just find P value.