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)
Discussion:
21 comments Page 2 of 3.
Sanjay said:
8 years ago
tan2(45+π/2).
Where π=angle of friction.
Where π=angle of friction.
Romy mehra said:
8 years ago
The value. Tan^2 (45 +φ/2) is for passive earth pressure.
Avinash said:
8 years ago
Y and h are constants.
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).
Yaseen said:
8 years ago
Right. Thanks for the given explanation.
Abhishikta said:
9 years ago
Ka = 1/tan^2(45 + φ/2).
Gopalakrishnan said:
9 years ago
@Tanupa. Passive earth pressure = Tan^2(45+ angle\2).
Tanupa said:
9 years ago
What is equation of passive earth pressure?
Zatak said:
1 decade ago
Active earth pressure = ka*y*h where ka = (1-sinφ)/(1+sinφ) = tan2(45° - φ/2)
Astik said:
1 decade ago
What is this process?
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