Module 9 · Lesson 9.1

Representing a stress state

The Mohr circle, the s–t and p–q invariants, and the total–effective split.

Why this matters

A soil element rarely fails the instant a load arrives — it is carried there along a path of changing stress. To follow that path we first need a compact way to represent a stress state as a single point, so that loading traces a line we can plot against the failure line. That is what the stress invariants give us.

What you should already know

Effective stress σ′ = σ − u (Module 8)
Principal stresses and the idea of shear on a plane
The Mohr circle (met again here for soil)

A two-dimensional stress state is captured by the two principal stresses $σ_{1}$ and $σ_{3}$ , and drawn as a Mohr circle with:

centre $s = \frac{σ _{1} + σ _{3}}{2}$ — a measure of the mean (normal) stress, and
radius $t = \frac{σ _{1} - σ _{3}}{2}$ — a measure of the shear (deviatoric) stress.

Instead of drawing the whole circle every time, we can represent the state by the single point $(s, t)$ — its top. As the load changes, that point moves, and its track is the stress path.

Stress invariants (MIT s–t and Cambridge p–q)

s = \frac{σ _{1} + σ _{3}}{2}, t = \frac{σ _{1} - σ _{3}}{2}; p = \frac{σ _{1} + 2 σ _{3}}{3}, q = σ_{1} - σ_{3}

What it calculates: compact mean and deviatoric measures of a stress state

$s, p$: mean-stress measures (circle centre; triaxial mean) (kPa)
$t, q$: deviatoric-stress measures (circle radius; deviator) (kPa)

In plain terms: The s–t pair matches the Mohr circle directly; the p–q pair is standard for triaxial work. Both split a stress state into a mean part and a shearing part — the split that matters for soil.

Every invariant comes in a total and an effective version. Because effective stress subtracts the porewater pressure equally from every normal stress, $σ_{1}^{'} = σ_{1} - u$ and $σ_{3}^{'} = σ_{3} - u$ . The mean measures shift by $u$ ; the deviatoric measures do not — as the next lesson derives. So the total and effective points sit at the same height on a stress-path plot, offset horizontally by the pore pressure.

Try it

Total and effective Mohr circle

The pore pressure slides the effective circle left of the total circle by exactly u. Watch it approach the failure line.

Major total stress σ₁200 kPa

Minor total stress σ₃80 kPa

Pore pressure u40 kPa

Friction angle φ′ (envelope)28 °

Total s, t: 140, 60 kPa
Effective s′, t′: 100, 60 kPa
Cambridge p, q: 120, 120 kPa
Gap TSP→ESP = u: 40 kPa

The effective circle is touching the failure line — the soil is at (or past) failure. Reduce the pore pressure or the deviator stress.

Worked example

Total and effective invariants of a stress state

Given

Total principal stresses σ1 = 200 kPa, σ3 = 80 kPa
Pore pressure u = 40 kPa

Find

The s–t and p–q invariants in both total and effective stress.

Assumptions

Saturated soil; effective stress σ′ = σ − u applies to both principal stresses.

Summary

A stress state is a Mohr circle: centre s = (σ1+σ3)/2, radius t = (σ1−σ3)/2.
s–t (MIT) and p–q (Cambridge) invariants split stress into mean and deviatoric parts.
Deviatoric measures (t, q) are the same in total and effective stress; mean measures (s, p) differ by u.
The effective stress point is the total point shifted left by the pore pressure.

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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