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 2 of 4.
Vicky and himalaya said:
7 years ago
For concrete-(400-20)*40 = 15200kg.
For steel-20*1300 = 26000kg so total 41200kg answer.
For steel-20*1300 = 26000kg so total 41200kg answer.
Rahul said:
7 years ago
Here, it should be 1300kg/cm2 in the place of 300kg/cm2.
Anurag Singh said:
6 years ago
P = Asc*σsc + Acc * σcc.
Acc = 20*20. σcc = 40,
Asc = 4*20=80. σsc = 300,
Slove = 40000kg.
So, I think it's None of these.
Acc = 20*20. σcc = 40,
Asc = 4*20=80. σsc = 300,
Slove = 40000kg.
So, I think it's None of these.
Aswathy said:
6 years ago
@Vicky.
Can you explain why you take. 400- 20 for concrete. And why you take 1300 instead of 300 (permissible compressive stress in steel)? Please explain.
Can you explain why you take. 400- 20 for concrete. And why you take 1300 instead of 300 (permissible compressive stress in steel)? Please explain.
Santraj said:
6 years ago
I am not getting it, please anyone explain in detail.
Suraj said:
5 years ago
P=sigma cc *(total area-area of steel)+sigma sc*area of steel.
Total area=20*20=400sq cm ..hence p=36800.
Hence nearly option B.
Total area=20*20=400sq cm ..hence p=36800.
Hence nearly option B.
Sanjeet kushwaha said:
5 years ago
You are correct, Thanks @Suraj.
Gautom Kumar Mondal said:
5 years ago
P=0.4FckXAc+0.67XFyXAst [ Hear-Fck=40 kg/sm2,Fy=300Kg/sm2, Ac= 20x20 sqcm,Ast= 4x20=80 sqcm]
P= 0.4x40x(20x20-4x20)+0.67x300x4x20=21200 Kg This factored load.
So the safe load which column can carry-= 21200/1.5=14133.33 kg. == 14133 kg.
P= 0.4x40x(20x20-4x20)+0.67x300x4x20=21200 Kg This factored load.
So the safe load which column can carry-= 21200/1.5=14133.33 kg. == 14133 kg.
Uddhav said:
4 years ago
Compressive stress in concrete = 40kg/cm^2.
Compressive stress in steel = 300kg/cm^2,
Area of concrete = 400cm^2,
Area of steel=4*20 = 80cm^2 , Pu=40*400+80*300=40,000kg so answer (B) is correct.
Compressive stress in steel = 300kg/cm^2,
Area of concrete = 400cm^2,
Area of steel=4*20 = 80cm^2 , Pu=40*400+80*300=40,000kg so answer (B) is correct.
Sudhakar Singh said:
3 years ago
Pu = (0.45fck x Ac) + (0.67fy Asc) -----> (1 eqn.)
0.45 fck = 40Kg/cm^2.
0.67fy = 300Kg/cm^2.
Put values in equation 1.
= 40 x (400-20) + (300 x 20),
= 15200 + 6000,
Pu = 21200 Kg.
0.45 fck = 40Kg/cm^2.
0.67fy = 300Kg/cm^2.
Put values in equation 1.
= 40 x (400-20) + (300 x 20),
= 15200 + 6000,
Pu = 21200 Kg.
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