Difficulty: Easy
Correct Answer: D = (q / gamma) * ((1 - sin phi) / (1 + sin phi))^2
Explanation:
Introduction / Context:
Determining a safe minimum depth for a shallow footing is a classic task in foundation engineering. The depth must be sufficient so that the overburden confinement provides the necessary shear resistance in the underlying soil and the bearing pressure at the base can be carried without shear failure. Empirical design aids often recast Rankine/Prandtl concepts into practical formulas that relate depth to soil strength and allowable bearing pressure.
Given Data / Assumptions:
Concept / Approach:
For granular soils, a convenient expression for minimum embedment is obtained by equating the available passive confinement from the overburden to the mobilized active tendency beneath the footing edge. Using Rankine relationships, the ratio (1 - sin phi)/(1 + sin phi) appears naturally, and squaring it reflects the conversion to a depth term when expressed against unit weight. This yields a compact depth estimate that increases if q is larger and decreases for stronger soils (higher phi) or denser soils (higher gamma).
Step-by-Step Solution:
Verification / Alternative check:
Check limits: if phi increases, the fraction (1 - sin phi)/(1 + sin phi) decreases, so D reduces, matching intuition. If q becomes larger, D increases to supply more confinement. Designers should still verify against bearing capacity (e.g., Terzaghi's equation) and settlement criteria.
Why Other Options Are Wrong:
tan^2(45 + phi/2) and cot^2(45 - phi/2) correspond to passive trends, not minimum embedment for active control. The inverted ratio ((1 + sin phi)/(1 - sin phi))^2 grows with phi, contradicting expected behavior. sec^2(phi) is not a recognized embedment relation.
Common Pitfalls:
Using total instead of effective unit weight when groundwater reduces gamma; ignoring surcharge q_s that should be combined with overburden; treating the result as a final design without checking settlement.
Final Answer:
D = (q / gamma) * ((1 - sin phi) / (1 + sin phi))^2
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