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.
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.
Dayanand Pathak said:
1 decade ago
As per IS-456 strength of the RCC columm is 0.4*Fck*Ac + 0.67*Fy*Asc.
As per above Quation strength will be 0.4*40*(20*20-20)+ 0.67*300*20 = 10,100 KG.
please clearify the same.
As per above Quation strength will be 0.4*40*(20*20-20)+ 0.67*300*20 = 10,100 KG.
please clearify the same.
Anbarasan muthaiah said:
1 decade ago
Pu = 0.4fck Ac+ .67fy As.
Ac = 20 * 20 - 4(20).
= 400-80 = 320.
As = 4 * 20 = 80.
fck = 40.
fy = 300.
Pu = 0.4 * 40 * 320 + 0.67 * 300 * 80 = 21200kg.
Ac = 20 * 20 - 4(20).
= 400-80 = 320.
As = 4 * 20 = 80.
fck = 40.
fy = 300.
Pu = 0.4 * 40 * 320 + 0.67 * 300 * 80 = 21200kg.
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.
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.
Asha said:
10 years ago
This not ultimate load it is require only safe load.
So the equation shall be p = Ac*FC+As*FS.
p = 20*20*40+(4*20)*300 = 40000 kg.
So the equation shall be p = Ac*FC+As*FS.
p = 20*20*40+(4*20)*300 = 40000 kg.
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.
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.
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.
Durgesh said:
9 years ago
Put the formula, P = sigma sc * Asc + sigma cc * Ac.
All data are given you just find P value.
All data are given you just find P value.
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