Civil Engineering - Soil Mechanics and Foundation Engineering - Discussion
Discussion Forum : Soil Mechanics and Foundation Engineering - Section 3 (Q.No. 27)
27.
The passive earth pressure of a soil, is proportional to (where φ is the angle of internal friction of the soil.)
Discussion:
10 comments Page 1 of 1.
Jitendra Ahirwar said:
4 years ago
For active case;
Ka =tan^2(45-π/2)
And for the passive case,
it is Kp=tan^2(45+π/2).
Ka =tan^2(45-π/2)
And for the passive case,
it is Kp=tan^2(45+π/2).
(3)
Krishan said:
8 years ago
It is asking about proposal, not for equal.
(1)
Nikhil Dongre said:
5 years ago
Earth pressure at rest proportional to =(1-sinθ).
Active earth pressure proportional to = cotθ (45°+ φ/2) or tanθ (45°- φ/2).
Passive earth pressure proportional to = tanθ (45°+ φ/2).
Active earth pressure proportional to = cotθ (45°+ φ/2) or tanθ (45°- φ/2).
Passive earth pressure proportional to = tanθ (45°+ φ/2).
(1)
Yogesh said:
9 years ago
Isn't it tan2 (45°+ φ/2)?
Souvik said:
8 years ago
Yes, you are right @ Yogesh.
Hari meena said:
8 years ago
Yes, Right.
Melkie said:
8 years ago
It shoud be tan2 (45+q/2).
Tushar said:
7 years ago
Tan2 (45-π/2) is the right answer.
Akhil Sharma said:
6 years ago
Please explain the correct answer.
Dheeraj said:
6 years ago
The Right answer is tan^2(45π/2).
For Active earth pressure: tan^2(45+π/2).
Earth pressure at rest :( 1-sinθ).
For Active earth pressure: tan^2(45+π/2).
Earth pressure at rest :( 1-sinθ).
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