Earth pressure on a retaining wall: The total horizontal thrust per metre length (active condition) on a wall of height h retaining cohesionless soil (unit weight w, angle of repose φ) is

Difficulty: Medium

Correct Answer: (w * h^2 / 2) * tan^2(45° − φ/2)

Explanation:

Introduction / Context:Retaining structures must resist lateral earth pressures. The Rankine active earth pressure theory provides a convenient closed-form for cohesionless backfill, frequently used in preliminary design and checks.

Given Data / Assumptions:

  • Backfill is dry, cohesionless, with unit weight w and angle of internal friction φ (equal to angle of repose).
  • Vertical, smooth wall; horizontal backfill surface (Rankine conditions).
  • Active state of earth pressure (wall yields sufficiently).

Concept / Approach:Rankine active pressure coefficient K_a = tan^2(45° − φ/2) = (1 − sin φ)/(1 + sin φ). Resultant active thrust P_a per metre length isP_a = (1/2) * w * h^2 * K_aacting at h/3 above the base. Substituting K_a gives the expression in terms of φ.

Step-by-Step Solution:

Compute K_a = tan^2(45° − φ/2).Total thrust: P_a = (w * h^2 / 2) * K_a.Hence P_a = (w * h^2 / 2) * tan^2(45° − φ/2).

Verification / Alternative check:Using identity K_a = (1 − sin φ)/(1 + sin φ), P_a = (w h^2 / 2) * (1 − sin φ)/(1 + sin φ), consistent with Rankine theory.

Why Other Options Are Wrong:

  • tan^2(45° + φ/2) corresponds to K_p (passive), not active.
  • w h^2 tan φ lacks the correct coefficient and dimensions.
  • (w h^2 / 3) times the ratio mixes location of resultant with magnitude.
  • (w h^2 / 2) ignores soil friction angle entirely.

Common Pitfalls:Confusing K_a and K_p signs; forgetting that the resultant acts at h/3; applying Coulomb theory coefficients to Rankine conditions without accounting for wall friction or backface inclination.

Final Answer:(w * h^2 / 2) * tan^2(45° − φ/2)

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