Active earth pressure relation (Rankine): If p1 is the vertical pressure intensity at depth h in a soil mass of unit weight w (or any given vertical stress) and φ is the angle of repose (internal friction), what is the lateral active pressure intensity p2?

Introduction / Context:Earth-retaining structures rely on earth pressure theories to estimate lateral pressures. Rankine's active earth pressure coefficient K_a links vertical stress to lateral stress in cohesionless backfills under active conditions.

Given Data / Assumptions:

• Backfill is dry, cohesionless, and extends to the surface.
• Wall yields sufficiently to mobilize active conditions.
• Plane strain and no wall friction (Rankine assumptions).

Concept / Approach:The active earth pressure coefficient K_a equals (1 − sin φ) / (1 + sin φ). An equivalent trigonometric form is K_a = tan^2(45° − φ/2). Therefore, p2 = K_a * p1 for the lateral stress intensity at the depth considered.

Step-by-Step Solution:Write K_a = (1 − sin φ) / (1 + sin φ).Use the identity tan^2(45° − φ/2) = (1 − sin φ) / (1 + sin φ).Hence p2 = p1 * K_a = p1 * tan^2(45° − φ/2).

Verification / Alternative check:For φ = 0°, K_a = 1, so p2 = p1; for φ = 30°, K_a ≈ 0.333, consistent with typical retaining wall design numbers.

Why Other Options Are Wrong:tan^2(45° + φ/2) and (1 + sin φ)/(1 − sin φ): These correspond to the passive coefficient, not active.(1 − sin φ)/(1 + sin φ) alone is correct for K_a but the option combining it directly with p1 is not the selected canonical trigonometric form here.p1 * cos φ: Not an earth pressure coefficient expression.

Common Pitfalls:

• Mixing up active and passive coefficients.
• Applying Rankine where Coulomb (with wall friction) would be more appropriate.

Final Answer:p2 = p1 * tan^2(45° − φ/2)