Civil Engineering - UPSC Civil Service Exam Questions - Discussion
Discussion Forum : UPSC Civil Service Exam Questions - Section 16 (Q.No. 3)
3.
A rectangular footing 1 m x 2 m is placed at a depth of 2 m in a saturated clay having an unconfined compressive strength of 100 kN/m2. According to Skempton, the net ultimate bearing capacity is
Discussion:
7 comments Page 1 of 1.
Naik yawar said:
3 years ago
Clay is saturated that means we have to take (1+0.2B/L).
Rajkumar said:
4 years ago
Qnu = (1+0.3B/L)C Nc.
So the answer is Qnu = 442.75 kN/m2.
So the answer is Qnu = 442.75 kN/m2.
Monisha said:
5 years ago
D=2 is given, then how will you substitute For D=1?
Koyel Jana said:
7 years ago
@Prit.
100 is divided because of cohesion C = UCS/2.
100 is divided because of cohesion C = UCS/2.
Prit said:
8 years ago
Why is 100 divided?
Arul raj said:
8 years ago
As the soil is saturated then angle of internal friction 0,
Then bearing capacity factor(as per skemton) Nc=(1+0.2*2/1)*(1+0.2*1/2)*5=7.7
AND Nq=1.
N(GAMA)=0 then net ulmitate bearing capacity=7.7*100/2=385 kn/m2.
Then bearing capacity factor(as per skemton) Nc=(1+0.2*2/1)*(1+0.2*1/2)*5=7.7
AND Nq=1.
N(GAMA)=0 then net ulmitate bearing capacity=7.7*100/2=385 kn/m2.
ABHISHEK ANAND said:
9 years ago
c = 100/2 = 50.
Nc = 5(1 + 0.2(B/L)(1 + 0.2(D/B) = 7.7.
Qnu = 50*7.7 = 385.
Nc = 5(1 + 0.2(B/L)(1 + 0.2(D/B) = 7.7.
Qnu = 50*7.7 = 385.
(1)
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