Civil Engineering - UPSC Civil Service Exam Questions - Discussion
Discussion Forum : UPSC Civil Service Exam Questions - Section 1 (Q.No. 32)
32.
For designing end bearing piles of square cross-section in clays having average unconfmed compressive strength of 6 t/m2, the net ultimate bearing capacity may be taken as
Discussion:
13 comments Page 1 of 2.
Zorox said:
9 years ago
According to terzaghi = 5.7*c .
Mayerhoff = 5.14c.
C = ucs/2.
C = 3.
For square footing;
Net bearing capacity = 1.3CNc.
Mayerhoff = 5.14c.
C = ucs/2.
C = 3.
For square footing;
Net bearing capacity = 1.3CNc.
RAHUL said:
8 years ago
I think the answer should be 27 because it is asking about piles that too end bearing only so formula is 9C. Where c=3.
Anuj said:
6 years ago
In the pile foundation according to Meyerhof theory the value of end bearing capacity is 9c then Qup=9*3=27.
Dharshit said:
8 years ago
End bearing resistance;
Q=9Cu.
Cu=Unconfined compression strength/Fos.
Cu=6/3 =2 t/m2.
Q=9*2 =18t/m2.
Q=9Cu.
Cu=Unconfined compression strength/Fos.
Cu=6/3 =2 t/m2.
Q=9*2 =18t/m2.
Yadvendra said:
5 years ago
Use Skempkons analysis for Nc.
take df/B =2.5 and b/l =1.
=> Nc=9.
Hence Q(be)=9*3,
= 27.
take df/B =2.5 and b/l =1.
=> Nc=9.
Hence Q(be)=9*3,
= 27.
(2)
Naresh Sreelam said:
7 years ago
Uc = 2c
C = 6/2 = 3.
In end bearing piles ultimate strength is 9c.
9*3=27.
C = 6/2 = 3.
In end bearing piles ultimate strength is 9c.
9*3=27.
(1)
Maxp50 said:
8 years ago
According to me, the exact answer is 22.23. So the nearest option is C.
Gopinath said:
9 years ago
I think compressive strength = net sbc, so the factor of safety 3.
Darwin said:
8 years ago
For pile design the factor of safety must be 3. I think so.
(1)
Sanjeev kr rai said:
8 years ago
Yes @Maxp50.
1.3cN = 1.3*6/2*5.72=22.3.
1.3cN = 1.3*6/2*5.72=22.3.
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