Question
Download Solution PDFThe intensity of active earth pressure at a depth of 10 meters in dry cohesionless sand with an internal angle of friction of 30° and with a weight of 1.8 t/m3, is:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
Active earth pressure (Pa)
\(P_a=k_a\gamma H\)
Where ka = Coefficient of active earth pressure
\(\gamma=Unit\, weight\, of\, soil\)
H = Depth at which pressure calculate
Calculation:
Given
Depth of dry sand (H) = 10.0 m
Angle of shearing resistance (ϕ) = 30∘
Unit weight (\(\gamma\)) = 1.8 t/m3
Active earth pressure at depth H (Pa)
= \(k_a×\gamma× H\)
Where \(k_a={1-sin\phi \over 1+sin\phi}={1-sin30^\circ\over 1+sin30^\circ}\)
ka = 0.333
Active earth pressure at depth 10.0 m (Pa) = 0.333 × 1.8 × 10
Active earth pressure (Pa) = 6 t/m2
Last updated on Jun 4, 2025
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