Civil Engineering - RCC Structures Design - Discussion
Discussion Forum : RCC Structures Design - Section 1 (Q.No. 27)
27.
A short column 20 cm x 20 cm in section is reinforced with 4 bars whose area of cross section is 20 sq. cm. If permissible compressive stresses in concrete and steel are 40 kg/cm2 and 300 kg/cm2, the Safe load on the column, should not exceed
Discussion:
39 comments Page 4 of 4.
Ashish said:
7 years ago
Thank you. You are extremely right, @Sankha.
Ujjwal said:
8 years ago
Yes, 21200 kg is the correct answer.
Agree @Anbarasan.
Agree @Anbarasan.
Ramesh said:
8 years ago
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.
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.
Sanjib said:
8 years ago
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.
P= 0.4fck*Asc+0.67fy*Ast,
= 0.4 x 40 x 380 + 0.67 x 300 x 20,
=10,100 Kgs.
Vishal said:
8 years ago
This question solve by the working state method because of the given data is permissible stress not a grade of concrete.
Ans - 36800 kg.
Ans - 36800 kg.
Saravanaraja said:
8 years ago
Thank you @Ramesh.
Sankha said:
8 years ago
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%.
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%.
Sandeep said:
8 years ago
Yes, you correct @Ramesh.
Jayakanth said:
1 decade ago
How to calculate the safe load of column?
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers