Slope stability condition: Failure of an earth slope will occur only when the total driving shear force acting along the potential slip surface is how related to the total available shearing strength?

Difficulty: Easy

Correct Answer: Greater than total shearing strength

Explanation:


Introduction / Context:
Slope stability in soil mechanics is evaluated by comparing the total mobilized (driving) shear along a potential failure surface with the total available shearing strength (resisting) along that surface. The safety margin is commonly summarized by the factor of safety. This item tests the fundamental failure criterion.


Given Data / Assumptions:

  • Potential slip surface (often circular for simple analyses).
  • Shear strength follows Mohr–Coulomb: τ_f = c + σ′ tan φ.
  • Driving shear arises from self-weight, surcharge, and seepage forces.


Concept / Approach:

Failure occurs when the mobilized shear equals or exceeds the available strength. In practice, failure is identified when the driving shear slightly exceeds strength and progressive failure ensues. The factor of safety FS is defined as Strength / Mobilized shear. Stability requires FS > 1; incipient failure corresponds to FS ≈ 1; failure corresponds to FS < 1 (i.e., driving > strength).


Step-by-Step Solution:

Set criterion: Driving shear ≥ Shear strength.Translate to factor of safety: FS = Strength / Driving < 1 at failure.Therefore, failure when driving shear is greater than total shearing strength.


Verification / Alternative check:

Limit equilibrium methods (Fellenius, Bishop, Janbu) all pivot on this comparison. Numerical analyses also use a “shear strength reduction” method to find the state where mobilized shear just exceeds strength.


Why Other Options Are Wrong:

(a) Equality indicates incipient failure (limit state), not general stability. (c) If driving is less than strength, slope is stable. (d) and (e) ignore the full Mohr–Coulomb strength.


Common Pitfalls:

Confusing equality with actual observed failure; overlooking pore pressure effects that reduce σ′ and therefore strength.


Final Answer:

Greater than total shearing strength

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