Difficulty: Medium
Correct Answer: z_tc = 2 c_u / γ
Explanation:
Introduction / Context:Tension cracks can develop near the crest of slopes or excavations in purely cohesive soils (φu = 0) because the soil has finite tensile capacity related to its undrained shear strength. Estimating the crack depth informs safe excavation and stability checks, including the possibility of block toppling or piping paths for water.
Given Data / Assumptions:
Concept / Approach:For φu = 0 soils, the maximum tensile stress that can be mobilized at the ground surface is related to c_u. Equilibrium of a thin vertical slice leads to a limiting crack depth proportional to c_u/γ. Classical analysis gives z_tc = 2 c_u / γ. Beyond this depth, vertical total stress exceeds the tensile capacity and crack propagation ceases.
Step-by-Step Solution:
Assume a potential vertical crack of depth z at the crest.Relate vertical total stress at depth z: σ_v = γ * z.Equate available tensile resistance ≈ 2 * c_u (cohesion acting on both crack faces) to σ_v at the limit.Solve: γ * z_tc = 2 * c_u ⇒ z_tc = 2 c_u / γ.Verification / Alternative check:
Check dimensions: c_u (kPa) / γ (kN/m^3) yields metres; result is dimensionally consistent.Why Other Options Are Wrong:
c_u/γ: underestimates depth by factor of 2; 4 c_u/γ overestimates.(c_u * γ)/2 and √(c_u/γ): dimensionally incorrect for a length.Common Pitfalls:
Mixing drained (φ′) and undrained (φu) parameters; forgetting that the derivation uses total stresses and undrained cohesion.Final Answer:
z_tc = 2 c_u / γ
Discussion & Comments