Module 13 · Lesson 13.1

The critical state

The unique end-point of shearing, and the critical-state line.

Why this matters

So far the course has treated compression (Module 10) and strength (Module 11) as separate topics. Critical-state soil mechanics shows they are two faces of one framework. Its central observation is simple and powerful: shear any soil far enough and it forgets its starting state, arriving at a unique condition that depends only on the stress it is under.

What you should already know

Void ratio and the e–log σ′ compression line (Module 10)
Shear strength, φ′ and the q–p′ invariants (Modules 9, 11)
Dilation and the peak/critical-state distinction (Module 11)

At the critical state a soil shears continuously at constant volume and constant stress — it has stopped changing. Dense soils reach it after dilating; loose soils after contracting; but both arrive at the same state for a given stress. Because the void ratio no longer changes there, the critical state links a soil's volume to the stress at which it can flow — the bridge between compression and strength.

Plotted with the invariants of Module 9, the critical states of a soil fall on a single critical-state line (CSL):

in q–p′ space (deviatoric vs mean effective stress) the CSL is a straight line through the origin, $q = M p′$;
in e–ln p′ space (void ratio vs mean effective stress) it is a line parallel to the normal compression line.

The slope $M$ in q–p′ is the strength at the critical state, and it is fixed by the critical-state friction angle.

Critical-state line and stress ratio

q = M p^{'}, M = \frac{6 sin φ _{cs}^{'}}{3 - sin φ _{cs}^{'}}

What it calculates: the deviatoric stress at the critical state, and its link to the friction angle

q: deviatoric stress at critical state (kPa)
p′: mean effective stress (kPa)
M: critical-state stress ratio (slope of the CSL)
$φ_{cs}^{'}$: critical-state friction angle (degrees)

In plain terms: M and φ′cs carry the same information (triaxial compression). φ′cs = 30° gives M = 1.2; a stronger soil has a larger M and a steeper CSL.

Try it

Critical-state explorer

The critical-state line q = Mp′, and a drained test path climbing to meet it. Failure is where the path reaches the line.

Critical-state friction φ′cs30 °

Confining stress σ′₃ (= p′₀)100 kPa

Stress ratio M: 1.200
p′ at failure: 167 kPa
q at failure: 200 kPa
Axial σ′₁f = σ′₃ + q: 300 kPa

What the plot shows

Red line = critical-state line q = Mp′ (slope M from φ′cs).
Orange path = drained triaxial compression (slope 3 in q–p′).
Failure is the intersection — the same stress Mohr–Coulomb gives.

φ′cs = 30° gives M = 1.20. The drained path meets the critical-state line at p′ = 167, q = 200 kPa, i.e. σ′₁ = 300 kPa — identical to the Mohr–Coulomb failure stress.

Worked example

Failure state of a drained triaxial test in q–p′ space

Given

Critical-state friction angle φ′cs = 30° (so M = 1.2)
Drained triaxial compression at cell pressure σ′3 = 100 kPa

Find

The mean and deviatoric stress at failure, and the axial stress σ′1f.

Assumptions

Drained test: σ′3 constant; the drained path in q–p′ has slope 3.

Summary

At the critical state a soil shears at constant volume and constant stress, independent of its start.
The critical states lie on the critical-state line: q = Mp′ in stress space, parallel to the NCL in e–ln p′.
M = 6 sinφ′cs/(3 − sinφ′cs) links the CSL slope to the friction angle (M = 1.2 at φ′cs = 30°).
Critical-state and Mohr–Coulomb give the same failure strength; critical state adds the volume dimension.

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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