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 1 of 4.
Jayakanth said:
1 decade ago
How to calculate the safe load of column?
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.
Jaan said:
1 decade ago
Multiply the above answer by 4 because we have provided 4 number of bars.
Subhamay Ghosh said:
1 decade ago
Its strength of column as a whole and not for individual bars.
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.
Hari said:
1 decade ago
How answer is 41200 kg?
Gowri said:
1 decade ago
pu = 0.4*Fck*Ac+0.67*Fy*.
As use this formula and multiply 4 no.of columns.
As use this formula and multiply 4 no.of columns.
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.
Haftu said:
9 years ago
What is the rcc slab?
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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