Civil Engineering - Soil Mechanics and Foundation Engineering - Discussion

Discussion Forum : Soil Mechanics and Foundation Engineering - Section 1 (Q.No. 2)
2.
The active earth pressure of a soil is proportional to (where φ is the angle of friction of the soil)
tan (45° - φ)
tan2 (45° + φ/2)
tan2 (45° - φ/2)
tan (45° + φ)
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
21 comments Page 1 of 3.

ViVEK said:   1 year ago
Ka = (1-sin¢)/(1+sin ¢)
= Cot^2 (Alpha) = Cot^2 (45+¢/2)
=Cot^2 [90-(45-¢/2)]
={Cot [90-(45-¢/2)]}^2
= {tan(45-¢/2)}^2
= tan^2 (45-¢/2)

Kp = (1+sinĀ¢) / (1-sin¢)
= tan^2 (Alpha) = tan^2 (45+¢/2)

=>In active earth pressure failure plane make angle of 45+¢/2 with horizontal .

=>In the Passive earth pressure, failure plane makes an angle of 45-¢/2 with horizontal.

=> Remember that angles are not indicated of Alpha.
In the above formulas, Alpha is the Critical Angle.
(8)

Ravi said:   2 years ago
I think this term is also called flow value.
(1)

Joseph Nigus said:   2 years ago
The correct answer is option B.

Harsh Shah said:   3 years ago
What will be for passive earth pressure of soil? Please explain.

Manoj said:   3 years ago
According to me, Pa is proportional to cot^2(45+π/2).

Mahesh said:   5 years ago
How to calculate the angle? Please, anyone, tell me.

Staffana aishwarya said:   6 years ago
Soil at rest K0 = 1 - sinΠ.

Paul said:   6 years ago
What will be for soil at rest?

Sudipta said:   6 years ago
ka=1/tan^2(45°+Φ/2).
=tan^2(45°+Φ/2).
=flow value.

Akhil said:   7 years ago
Active earth pressure = 1/(tan2 (45° + π/2)) = tan2 (45° + π/2)
ka = 1/ kp.
(2)


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