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

Answer: Option C

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?

Post your comments here:

Name *:

Email   : (optional)

» Your comments will be displayed only after manual approval.