Active earth pressure – numerical application: Find the active lateral earth pressure intensity at a depth of 10 m in dry, cohesionless sand where φ = 30° and γ = 1.8 t/m^3. Assume a vertical wall with horizontal backfill.

Difficulty: Easy

Correct Answer: 6 t/m^2

Explanation:


Introduction / Context:
Active earth pressure represents the minimum lateral pressure mobilized when a retaining wall yields sufficiently away from the backfill. For cohesionless backfill with level ground, Rankine’s active coefficient provides a quick way to compute lateral stress with depth.


Given Data / Assumptions:

  • Internal friction angle φ = 30 degrees.
  • Unit weight γ = 1.8 t/m^3 (tonnes per cubic metre).
  • Depth z = 10 m; dry, c = 0; vertical wall, horizontal backfill; no surcharge.


Concept / Approach:
For cohesionless soil with horizontal backfill, Rankine active earth pressure coefficient is Ka = (1 − sin φ) / (1 + sin φ). The vertical overburden stress is σv = γ * z. The active horizontal pressure is σh = Ka * σv. Units must be consistent (t/m^3 * m → t/m^2).


Step-by-Step Solution:

Compute sin φ: sin 30° = 0.5.Ka = (1 − 0.5) / (1 + 0.5) = 0.5 / 1.5 = 1/3 ≈ 0.333.σv = γ * z = 1.8 * 10 = 18 t/m^2.σh = Ka * σv = (1/3) * 18 = 6 t/m^2.


Verification / Alternative check:
Using Coulomb with zero wall friction and vertical wall reduces to Rankine for level backfill; results match. A simple proportionality check: σh increases linearly with depth; at z = 0, σh = 0; at 10 m, the computed 6 t/m^2 is reasonable.


Why Other Options Are Wrong:

  • 4, 5, 7, 8 t/m^2 correspond to incorrect Ka or arithmetic; only 6 t/m^2 satisfies Ka = 1/3 with given γ and z.


Common Pitfalls:
Using degrees vs radians incorrectly; confusing kN/m^3 with t/m^3; forgetting that cohesionless c = 0 so no additive term appears.


Final Answer:
6 t/m^2

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