Minimum depth of foundation as per Rankine’s concept: According to Rankine’s formulation for cohesionless backfill with unit weight γ and friction angle φ, the minimum embedment depth D_f (to prevent shear failure at ground level under net foundation pressure q) is given by which expression?

Civil Engineering Soil Mechanics and Foundation Engineering Difficulty: Easy
Choose an option
  • A
    D_f = (q / γ) * [ (1 − sin φ) / (1 + sin φ) ]^2
  • B
    D_f = (q / γ) * [ (1 + sin φ) / (1 − sin φ) ]^2
  • C
    D_f = (q / γ) * tan^2(45° + φ/2)
  • D
    D_f = (q / γ) * tan^2(45° − φ/2)
  • E
    D_f = (γ / q) * [ (1 − sin φ) / (1 + sin φ) ]^2

Answer

Correct Answer: D_f = (q / γ) * [ (1 − sin φ) / (1 + sin φ) ]^2

Explanation

Introduction / Context:Designing a shallow foundation requires a sufficient embedment to prevent shear failure and to develop adequate passive resistance near the ground surface. Rankine’s earth pressure theory leads to a practical expression for the minimum depth of foundation in cohesionless soil in terms of soil unit weight γ, friction angle φ, and the net contact pressure q at the base.

Given Data / Assumptions:

  • Cohesionless soil (c = 0), homogeneous and horizontal ground surface.
  • Foundation base loaded with net pressure q at depth D_f.
  • Rankine active/passive states with no wall friction, planar ground.
  • Unit weight of soil = γ, friction angle = φ.

Concept / Approach:

At minimum embedment, the available passive resistance near the surface must balance the driving tendency from foundation pressure. Using Rankine’s coefficients, K_a = (1 − sin φ)/(1 + sin φ) and K_p = (1 + sin φ)/(1 − sin φ), and equating forces, the limiting embedment is expressed in terms of K_a (or K_p). Rearrangement gives the standard Rankine depth formula.

Step-by-Step Solution:

Use Rankine: K_a = tan^2(45° − φ/2) = (1 − sin φ)/(1 + sin φ).Minimum depth arises when overburden γ D_f produces enough lateral confining effect.Solving for D_f yields D_f = (q/γ) * [ (1 − sin φ) / (1 + sin φ) ]^2.

Verification / Alternative check:

Dimensional check: q/γ has units of length. The bracketed ratio is dimensionless; squaring preserves units. As φ increases (denser sand), the bracket decreases, giving smaller D_f, which is physically consistent.

Why Other Options Are Wrong:

(b) inverts the ratio, predicting larger D_f for stronger sands—unrealistic. (c) and (d) use tan half-angle forms unrelated to D_f here. (e) inverts q/γ, giving wrong dimensions.

Common Pitfalls:

Confusing gross vs. net pressure q; applying the expression to cohesive or sloping backfills without modification.

Final Answer:

D_f = (q / γ) * [ (1 − sin φ) / (1 + sin φ) ]^2

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