Mohr–Coulomb Failure Criterion

τ_{f} = c + σ^{'} tan ϕ

Shear strength on a failure plane is cohesion c plus effective normal stress times tan φ. Use effective-stress parameters c', φ' with σ' for drained or long-term conditions; total-stress cu, φu = 0 for undrained saturated clay. The Mohr circle touches the envelope at failure.

$τ_{f}$ (kPa): Shear strength at failure
$c$ (kPa): Cohesion (c' or c_u)
$σ^{'}$ (kPa): Effective normal stress
$ϕ$ (deg): Friction angle

Step-by-step solved example

c = 15 kPa, σ' = 80 kPa, φ = 28°. Shear strength on that plane?

Failure envelope τ = c + σ' tanφ on Mohr coordinates.

1. Friction term
tan 28° = 0.5317. σ' tanφ = 80×0.5317 = 42.54 kPa.
$σ^{'} tan ϕ = 42.5 kPa$
2. Strength
τ_f = 15 + 42.54 = 57.5 kPa.
$τ_{f} = 57.5 kPa$

Answer: 57.5 kPa

10 practice questions

0/10 correct

1.For c = 0, φ = 30°, σ' = 100 kPa, τ_f =

2.Undrained saturated clay is often modeled with

3.The Mohr circle for a soil element plots

4.Increasing pore pressure u while total σ is fixed

5.c = 20 kPa, φ = 0, σ = 150 kPa. τ_f =

6.Direct shear tests measure strength on

7.φ' of clean dense sand is typically

8.At failure the Mohr circle

9.c' = 10 kPa, φ' = 20°, σ' = 50 kPa. τ_f is nearest

10.Drained strength of NC clay is governed mainly by