Civil Engineering - UPSC Civil Service Exam Questions - Discussion
Discussion Forum : UPSC Civil Service Exam Questions - Section 24 (Q.No. 3)
3.
A saturated stiff clay has unit weight 2 gm/cm3 and unconfined compressive strength 2 kg/cm2. The depth of tension crack that would develop in this clay is
Discussion:
8 comments Page 1 of 1.
Priyanka said:
4 years ago
10 m is the correct answer.
Umesh JIT said:
7 years ago
C= q/2 = 1.
D = 2 c/ GAMMA.
= 2*1/ 0.002.
= 1000 cm.
= 10 m.
D = 2 c/ GAMMA.
= 2*1/ 0.002.
= 1000 cm.
= 10 m.
CHANDRU said:
8 years ago
Yes, the answer is 10m.
Mrinmoy Nath said:
8 years ago
Yes, correct @Abhijeet.
(1)
Abhijeet said:
8 years ago
Depth of tension crack= 2c/Y.
Where c= cohesion.
And from unconfined compression shear test cohesion = q/2 , q=unconfined compressive strenght.
So answer is 10 m.
@Deepak C = cohesion.=q/2.
Where c= cohesion.
And from unconfined compression shear test cohesion = q/2 , q=unconfined compressive strenght.
So answer is 10 m.
@Deepak C = cohesion.=q/2.
Karthika.c said:
8 years ago
The answer should be 20m. Not 2m as per unit conversion. Please explain it.
Chhaya said:
8 years ago
Thanks @Deepak.
Deepak said:
10 years ago
Depth of tension crack, d = (2xC)/Y = 2.0 m.
C = Unconfined compressive strength.
Y = Saturated unit weight.
And, maximum unsupported depth of cutting = 2d.
C = Unconfined compressive strength.
Y = Saturated unit weight.
And, maximum unsupported depth of cutting = 2d.
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