Module 11 · Lesson 11.2

The Mohr–Coulomb criterion

The failure line, the Mohr circle tangent to it, and the principal-stress form.

Why this matters

To use strength in a calculation we need a failure criterion: a rule that says, for a given stress state, whether the soil has failed. The Mohr–Coulomb criterion is that rule — a straight failure line on a plot of shear stress against effective normal stress, and the condition that the Mohr circle of stress just touches it.

Mohr–Coulomb failure criterion

τ_{f} = c^{'} + σ_{n}^{'} tan φ^{'}

What it calculates: the shear stress a plane can carry at failure, for a given effective normal stress

$τ_{f}$: shear strength on the plane (kPa)
c′: effective cohesion intercept (kPa)
$σ_{n}^{'}$: effective normal stress on the plane (kPa)
$φ^{'}$: effective angle of shearing resistance (degrees)

This assumes

Strength is expressed in effective stress (the meaningful frame for drained strength).
c′ and φ′ are the intercept and slope of the failure line over the stress range of interest.

In plain terms: For most soils the true cohesion c′ is zero or small; a fitted c′ > 0 is usually a curvature or an overconsolidation effect over a limited stress range, not a real strength at zero stress.

A stress state is drawn as a Mohr circle: with principal effective stresses $σ_{1}^{'}$ and $σ_{3}^{'}$ , the circle has centre $s^{'} = (σ_{1}^{'} + σ_{3}^{'}) /2$ and radius $t = (σ_{1}^{'} - σ_{3}^{'}) /2$ . Every point on the circle is the $(σ_{n}^{'}, τ)$ on some plane. The soil fails when the circle grows until it is tangent to the failure line: at that point one plane has reached its strength.

From first principles

Principal-stress form of Mohr–Coulomb

We want to show: Turn the failure line into a direct relationship between the principal stresses at failure.

Failure is the moment the Mohr circle just touches the failure line. That single geometric condition — radius equals the perpendicular distance from the centre to the line — gives an algebraic link between σ′1 and σ′3.

The tangent point also fixes the orientation of the failure plane. Because the tangent point subtends an angle $2 θ_{f} = 90 + φ^{'}$ at the circle centre, the failure plane makes an angle $θ_{f} = 45 + φ^{'} /2$ with the major principal plane. Sands ( $φ^{'} \approx 3 3^{\circ}$ ) fail on planes about $6 0^{\circ}$ to the horizontal in a triaxial sample — the inclined shear band you can see in a failed specimen.

Try it

Mohr–Coulomb envelope explorer

The failure Mohr circle grows until it is tangent to the strength line. Read the axial stress at failure and the failure-plane angle.

Soil

Confining stress σ′₃60 kPa

c′, φ′: 0 kPa, 40°
Kp = tan²(45+φ′/2): 4.60
Axial at failure σ′₁f: 276 kPa
Deviator (σ′₁−σ′₃)f: 216 kPa
Failure plane θf: 65°

At σ′₃ = 60 kPa this soil fails at σ′₁ = 276 kPa — a deviator of 216 kPa. The circle is tangent to the envelope; the tangent point marks a plane at 65° to the major principal plane, where the soil actually slips.

Worked example

Axial stress at failure in a drained triaxial test

Given

Drained test on sand
Confining effective stress σ′3 = 50 kPa
c′ = 0, φ′ = 32°

Find

The major principal effective stress at failure, the deviator stress, and the failure-plane angle.

Assumptions

Fully drained, so measured stresses are effective; c′ = 0 for the clean sand.

Practice

A drained triaxial test on a soil with c′ = 5 kPa and φ′ = 28° is run at a confining effective stress σ′3 = 80 kPa. What is the major principal effective stress at failure, in kPa?

Summary

Mohr–Coulomb: τ_f = c′ + σ′_n·tanφ′; failure is the Mohr circle tangent to this line.
Principal-stress form: σ′1f = σ′3·Kp + 2c′√Kp with Kp = tan²(45 + φ′/2).
The failure plane lies at 45 + φ′/2 to the major principal plane.
Be sceptical of a fitted c′ > 0 — most soils have little or no true cohesion.

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This is educational material. It uses simplified examples to teach principles, and must not be relied on for real design or safety-critical decisions. Module overview and checkpoint