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.

Prasanta naskar said:   1 decade ago
Active earth pressure is directly proportional to cot2 (45° + φ/2) or tan2 (45° - φ/2).
(1)

Astik said:   1 decade ago
What is this process?

Zatak said:   10 years ago
Active earth pressure = ka*y*h where ka = (1-sinφ)/(1+sinφ) = tan2(45° - φ/2)

Tanupa said:   9 years ago
What is equation of passive earth pressure?

Gopalakrishnan said:   8 years ago
@Tanupa. Passive earth pressure = Tan^2(45+ angle\2).

Abhishikta said:   8 years ago
Ka = 1/tan^2(45 + φ/2).

Yaseen said:   8 years ago
Right. Thanks for the given explanation.

Harshdeep said:   8 years ago
Hers, cot^2(45 +B÷2 ) or by converting cot into tan^2 (90 -45 - B÷2) = tan^2 (45 - B÷2).

Avinash said:   8 years ago
Y and h are constants.

Romy mehra said:   8 years ago
The value. Tan^2 (45 +φ/2) is for passive earth pressure.


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