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Geomechanics & Ground Control

Mohr–Coulomb Failure

The shear-strength envelope τ = c + σ tanφ and the principal-stress failure criterion — the backbone of rock and soil strength.

PART 1

Topic Breakdown & Traps

The Engineering Principle

Rock fails in shear when the shear stress on a plane reaches the material's resistance, which has two parts: a cohesion (the strength at zero normal stress) and a frictional part that grows with the normal stress through the angle of internal friction . Plotted as , this is the Mohr–Coulomb envelope; a stress state is safe while its Mohr circle stays below the line and fails when the circle just touches it. Re-cast in principal stresses, the criterion predicts the major principal stress a sample can carry for a given confinement .

The Core Formula Matrix

Shear-strength envelope:

Principal-stress form (at failure):

Uniaxial compressive strength ():

= cohesion, = friction angle, = normal stress.
σ (normal)τσ₃=5σ₁=49.64φ=30°, c=10
Mohr circle for σ₃ = 5, σ₁ = 49.64 MPa just touching the c = 10 MPa, φ = 30° envelope.

The ‘IIT Traps’

  • ** uses , not .** The half-angle is added for the strength (active-failure) direction.
  • Cohesion is the intercept, friction is the slope. Confusing (a stress) with (dimensionless) breaks the units.
  • **The envelope is a straight line in space, not the Mohr circle.** Failure is the circle *tangent* to the line, not where the circle peaks.
PART 2

Progressive 3-Tier Question Suite

Q1BASIC1 Mark · MCQ
A rock has cohesion and friction angle . The shear strength on a plane carrying a normal stress of is:
Q2MEDIUM2 Marks · NAT
A rock has cohesion and friction angle . Its uniaxial compressive strength is ______ MPa. (Round off to two decimal places.)
MPa
Q3HARD2 Marks · NAT
For and , the major principal stress at failure under a confining stress is ______ MPa. (Round off to two decimal places.)
MPa