For a soil sample tested under uniaxial compression, the failure occurs at an angle of 45° + φ/2, where φ is the internal angle of friction. This failure corresponds to
- AVon Mises criterion.
- BTresca criterion.
- CMohr-Coulomb failure criterion.Correct
- DDrucker-Prager criterion.
Solution
Concept
In Mohr-Coulomb theory, the failure plane makes an angle of 45° + φ/2 with the major principal plane.
Formula
θf = 45° + φ/2
Step-by-step solution
1. The failure angle depends on φ, the angle of internal friction. 2. In Mohr-Coulomb theory, the failure envelope is a straight line at angle φ to the σ axis. 3. The Mohr circle touches this line. From geometry, the failure plane angle with the major principal plane is 45° + φ/2. 4. So the criterion is Mohr-Coulomb.
Explanation
The Mohr-Coulomb criterion uses φ and gives the failure plane at 45° + φ/2 with the major principal plane. This angle comes from the geometry of the Mohr circle touching the failure envelope. So option C is correct.
Why the other options are wrong
A: Von Mises is for ductile metals and has no φ term. B: Tresca gives a failure plane at 45° (φ = 0). D: Drucker-Prager is a smooth cone criterion and does not give this simple angle.
Shortcut
If the failure angle has φ in it, choose Mohr-Coulomb.
Common mistake
Choosing Tresca because it also involves shear. Tresca gives 45° only, as it has no friction angle.
Correct answer: C) Mohr-Coulomb failure criterion.