Rankine active earth pressure – total thrust on a dry sand backfill A vertical retaining wall of height H = 3 m retains a horizontal backfill of dry sand with unit weight γt = 20 kN/m³ and angle of shearing resistance φ' = 30°. The groundwater table is at the base (i.e., the backfill above is dry). Compute the total active thrust per metre length on the wall.

Difficulty: Medium

Correct Answer: 30 kN/m

Explanation:


Introduction:
Retaining wall design requires the active earth pressure resultant acting on the wall. For dry, level backfills the Rankine theory provides closed-form expressions using the active pressure coefficient Ka. This question checks application of Ka and correct integration to find total thrust.


Given Data / Assumptions:

  • Wall height H = 3 m; backfill horizontal and dry.
  • Unit weight γt = 20 kN/m³; friction angle φ' = 30°.
  • Rankine active condition; no wall friction; water table at base (no pore pressure above).


Concept / Approach:

For Rankine active state with horizontal backfill, Ka = (1 − sinφ) / (1 + sinφ). The active pressure at depth z is pa(z) = Ka * γt * z (triangular distribution). The total thrust is the area of the triangle: Pa = (1/2) * Ka * γt * H², acting at H/3 from the base.


Step-by-Step Solution:

1) Compute Ka: sin 30° = 0.5 ⇒ Ka = (1 − 0.5) / (1 + 0.5) = 0.5 / 1.5 = 1/3 ≈ 0.3333.2) Total active thrust Pa = 0.5 * Ka * γt * H² = 0.5 * (1/3) * 20 * (3²).3) Evaluate: 0.5 * (1/3) * 20 * 9 = 10 * 3 * (1/3) = 30 kN/m.


Verification / Alternative check:

Location of resultant at H/3 above base (not asked) is consistent with triangular pressure; magnitude matches standard tables for φ = 30°.


Why Other Options Are Wrong:

35 and 45 kN/m overestimate; 24 and 4 kN/m underestimate drastically for given γ and H.


Common Pitfalls:

Using γ_sub instead of γt for dry fill; forgetting the 1/2 factor; mixing Coulomb and Rankine expressions.


Final Answer:

30 kN/m

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