Rankine active condition — the active earth pressure of a cohesionless soil is proportional to which expression in terms of the friction angle phi?

Introduction / Context:
For retaining walls with granular backfill and negligible wall friction, Rankine’s theory provides a simple coefficient relating horizontal and vertical stresses at failure. The active earth pressure coefficient Ka determines the magnitude of lateral pressure when the wall yields sufficiently away from the soil to mobilize active conditions.

Given Data / Assumptions:

• Backfill is level and cohesionless (c = 0).
• Wall friction and wall batter are neglected (pure Rankine state).
• phi is the soil’s angle of shearing resistance.

Concept / Approach:

Rankine’s expression for the active coefficient is Ka = tan^2(45 - phi/2) which is algebraically equal to (1 - sin phi)/(1 + sin phi). As phi increases, Ka decreases, reflecting the greater shear resistance and lower lateral thrust exerted by denser, more interlocked grains. The active pressure at depth z is then sigma_h = Ka * gamma * z (plus any surcharge term Ka * q).

Step-by-Step Solution:

Verification / Alternative check:

Check limiting cases: phi = 0 gives Ka = 1 (fluid-like); higher phi gives progressively smaller Ka, consistent with observed behavior of dense sands.

Why Other Options Are Wrong:

tan^2(45 + phi/2) corresponds to Kp (passive). Linear tan forms lack the square and are dimensionally inconsistent for the Rankine coefficients. cot^2(45 - phi/2) is the inverse, not Ka.

Common Pitfalls:

Confusing active and passive states; forgetting that surcharge contributes Ka * q uniformly; applying Rankine where wall friction or sloping backfills require Coulomb theory.

Final Answer:

tan^2(45 - phi/2)