Civil Engineering - UPSC Civil Service Exam Questions - Discussion
Discussion Forum : UPSC Civil Service Exam Questions - Section 26 (Q.No. 19)
19.
A dry sand specimen is put through a triaxial test. The cell pressure is 50 kPa and the deviator stress at failure is 100 Pa. The angle of internal friction for the sand specimen is
Discussion:
4 comments Page 1 of 1.
Link said:
4 years ago
Sin(fi) = sigma 1 -sigma 3/sigma 1+ sigma 3.
Where;
Sigma 1 = 100+50=150.
Sigma 3= 50.
Sinθ= 150-50/150+50.
Sinθ=1/2.
Then,
θ = 30°.
Where;
Sigma 1 = 100+50=150.
Sigma 3= 50.
Sinθ= 150-50/150+50.
Sinθ=1/2.
Then,
θ = 30°.
Phani said:
7 years ago
Let say angle is x.
Sinx = (50div;100),
x = sin^-1(50div;100),
x = 30°.
Sinx = (50div;100),
x = sin^-1(50div;100),
x = 30°.
Priya said:
8 years ago
Σ 1= σ3*tan2(45+fai/2).
150=50*tan2(45+15).
30°.
150=50*tan2(45+15).
30°.
Karenina.c said:
8 years ago
I think given question in deviator stress=100kpa.
Normal stress= effective stress+deviator stress.
So normal stress=50+100=150kpa.
Normal stress= cell pressure [45+internal friction/2].
Then, get the answer 30.
Normal stress= effective stress+deviator stress.
So normal stress=50+100=150kpa.
Normal stress= cell pressure [45+internal friction/2].
Then, get the answer 30.
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