Module 11 · Lesson 11.3

Drained and undrained strength

Why the same clay has two strengths, and when each governs.

Why this matters

A saturated clay has, in effect, two strengths — one if it is loaded slowly enough to drain, another if it is loaded too fast to drain. Which one governs depends on how quickly the load is applied relative to how quickly the clay can shed the excess pore pressure. Getting this choice right is one of the most consequential judgements in geotechnical engineering.

The true, fundamental strength is always frictional on effective stress: $τ_{f} = c^{'} + σ^{'} tan φ^{'}$ . What differs is what happens to the porewater when the clay is sheared.

Drained loading is slow enough that no excess pore pressure builds up (or it has fully dissipated). Effective stresses are known throughout, and strength is expressed by $c^{'}$ and $φ^{'}$ . This is the long-term condition.
Undrained loading is fast enough that the water cannot leave in time. Shearing generates an excess pore pressure, the effective stress changes even though the total stress may not, and it is simplest to express strength in total stress as a single undrained shear strength $s_{u}$ , with an apparent friction angle $φ_{u} = 0$ . This is the short-term condition, immediately after construction.

Undrained shear strength

s_{u} = \frac{σ _{1} - σ _{3}}{2}_{failure}, φ_{u} = 0

What it calculates: the total-stress strength of a saturated clay sheared without drainage

s_u: undrained shear strength (radius of the total-stress failure circle) (kPa)
$σ_{1}, σ_{3}$: total principal stresses at failure (kPa)

In plain terms: s_u is not a soil constant: it depends on the water content and the effective stress the clay consolidated under. A clay compressed more (denser, lower void ratio) has a higher s_u. It is a convenience for the short term, not a fundamental parameter like φ′.

The reason $s_{u}$ works is that a saturated clay shears at essentially constant volume when undrained: the void ratio, and therefore the effective stress at failure, is fixed by whatever the clay consolidated under before the fast load. So every undrained test at that state fails at the same deviator stress regardless of the total confining pressure — a horizontal ( $φ_{u} = 0$ ) total-stress envelope. Change the consolidation stress and you change the water content, and $s_{u}$ moves with it.

Worked example

Undrained strength from an unconsolidated–undrained test

Given

Saturated clay, unconsolidated–undrained (UU) triaxial test
At failure: cell (total confining) pressure σ3 = 100 kPa
Axial total stress at failure σ1 = 220 kPa

Find

The undrained shear strength s_u, and a check that a second cell pressure would give the same value.

Assumptions

Saturated clay sheared without drainage; total-stress interpretation with φu = 0.

Practice

A saturated clay in an undrained triaxial test fails at a cell pressure σ3 = 120 kPa and an axial stress σ1 = 300 kPa. What is its undrained shear strength s_u, in kPa?

Check yourself

A permanent cutting is to be made in stiff clay. Which condition is critical for its long-term stability?

Summary

The fundamental strength is frictional on effective stress (c′, φ′); drained and undrained differ in the pore-pressure response.
Drained = slow/long-term: effective stresses known, strength from c′ and φ′.
Undrained = fast/short-term: express strength as s_u with φu = 0; s_u depends on the consolidation state, not a constant.
Loading on clay: short-term critical. Unloading (cuttings): long-term critical. Sands: always drained.

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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