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Soil Mechanics / Shear strength of soils

Drained and undrained strength

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

Lesson review: expert verification remains open

This lesson has received a full-context editorial review and a check of its technical explanations. Selected arithmetic has been checked separately. This is not independent engineering sign-off.

Checks needed for project use

  • A geotechnical reviewer must check laboratory interpretation, classification conventions and the limits of drained, undrained and one-dimensional idealisations.
  • Stress-path and critical-state illustrations simplify stress anisotropy, total vs excess pressure, loading history, drainage, structured soils, strain softening and cyclic behaviour. Full specialist review remains necessary; selected corrections do not certify the models.

These are educational models and exercises. Check their assumptions before using the results. The full technical content and every diagram have not yet been independently verified.

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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: τf=c′+σ′tan⁡φ′\tau_f = c' + \sigma'\tan\varphi'. 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′c' and φ′\varphi'. 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 sus_u, with an apparent friction angle φu=0\varphi_u = 0. This is the short-term condition, immediately after construction.
Undrained shear strength
su=σ1−σ32∣failure,φu=0s_u = \dfrac{\sigma_1 - \sigma_3}{2}\Big|_{\text{failure}}, \qquad \varphi_u = 0

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

su
undrained shear strength (radius of the total-stress failure circle) (kPa)
σ1, σ3\sigma_1,\ \sigma_3
total principal stresses at failure (kPa)

In plain terms: su 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 su. It is a convenience for the short term, not a fundamental parameter like φ′.

The reason sus_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 (φu=0\varphi_u = 0) total-stress envelope. Change the consolidation stress and you change the water content, and sus_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 su, 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 su, 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 su with φu = 0; su depends on the consolidation state, not a constant.
    • Short-term loading and long-term unloading often govern clay stability, but check both conditions. Establish drainage for sands from the loading rate and drainage conditions.