Civil Engineering - UPSC Civil Service Exam Questions - Discussion

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

 [A]. 420 kN/m2 [B]. 412.5 kN/m2 [C]. 385 kN/m2 [D]. 350 kN/m2

Explanation:

No answer description available for this question.

 Abhishek Anand said: (May 6, 2016) c = 100/2 = 50. Nc = 5(1 + 0.2(B/L)(1 + 0.2(D/B) = 7.7. Qnu = 50*7.7 = 385.

 Arul Raj said: (Jun 15, 2017) 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.

 Prit said: (Aug 23, 2017) Why is 100 divided?

 Koyel Jana said: (Aug 22, 2018) @Prit. 100 is divided because of cohesion C = UCS/2.

 Monisha said: (Apr 6, 2020) D=2 is given, then how will you substitute For D=1?

 Rajkumar said: (Nov 8, 2021) Qnu = (1+0.3B/L)C Nc. So the answer is Qnu = 442.75 kN/m2.