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:
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:
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)
Discussion & Comments