Civil Engineering - UPSC Civil Service Exam Questions - Discussion
Discussion Forum : UPSC Civil Service Exam Questions - Section 7 (Q.No. 43)
43.
The effective depth of a singly reinforced rectangular beam is 30 cm. The section is over-rein-forced and the neutral axis is 12 cm below the top. If the maximum stress attained by concrete is 50 kg/cm2 and the modular ratio is 18, then the stress developed in steel would be
Discussion:
13 comments Page 1 of 2.
Shiva said:
8 years ago
How to solve this?
Chikzz said:
8 years ago
Nc ={ m Qcb / (m Qcb + Qst) }. D
Gaja said:
8 years ago
mc/t = n/(d-n) substitute & find t.
Answer is 1350.
Answer is 1350.
Phani said:
7 years ago
C is the correct option.
Vpil said:
7 years ago
Explain clearly. Please.
Jack said:
7 years ago
Here m=18 I.e modular ratio and c=maximum stress attained =50 kg/cm^2 and d=overall depth of the beam =30cm and n=depth of N. A below the top=12cm.
Then by formula m.c/t=n/d-n.
t=18*50*18/12=1350 kg/cm^2 is the answer.
Then by formula m.c/t=n/d-n.
t=18*50*18/12=1350 kg/cm^2 is the answer.
Rakesh said:
6 years ago
D is correct balance is1350 but over reinforce me take less value.
Jitu said:
5 years ago
Here;
Ca/Xa=t/(d-x) m
=> 50/12=t/(30-12) * 18.
Ca/Xa=t/(d-x) m
=> 50/12=t/(30-12) * 18.
Susheel kumar said:
5 years ago
σ s= m σ c.
m modular ratio.
s stress in steel.
c stress in concrete.
m modular ratio.
s stress in steel.
c stress in concrete.
Ajay Sharma said:
5 years ago
(18*50)/t = (12/(30-12)) = >t = 1350kg/cm2.
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