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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.
Principal-stress form (at failure):
Uniaxial compressive strength ():
= cohesion, = friction angle, = normal stress.
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