Fatih Küçükkarakurt

Dilatancy angle: what the choice of psi actually costs

4 min read
Abstract

The dilatancy angle controls how much a soil expands while it shears, and setting it equal to the friction angle — associated flow — is a mathematical convenience with no physical basis for soil. Using Davis's reduced parameters on a strip footing with a friction angle of 30 degrees and a cohesion of 10 kPa, associated flow gives a collapse pressure of 301.4 kPa while non-dilatant flow gives 200.8 kPa, a reduction of 33.4%. In undrained analysis the same parameter is worse than inaccurate: any positive dilatancy angle generates suction without limit and the computed strength never converges.

A soil sheared at constant stress changes volume. Dense sand expands, because grains riding over one another need room; loose sand contracts. The dilatancy angle ψ\psi is the parameter that governs it, through the plastic potential:

g=σ1−σ32−σ1+σ32sin⁡ψg = \frac{\sigma_1 - \sigma_3}{2} - \frac{\sigma_1 + \sigma_3}{2}\sin\psi

and in plane strain the plastic volumetric strain rate follows directly:

ε˙vpγ˙p=sin⁡ψ\frac{\dot\varepsilon^p_v}{\dot\gamma^p} = \sin\psi

Set ψ=φ\psi = \varphi and the plastic potential becomes the yield function; the flow is associated and the plastic strain increment is normal to the yield surface. This is mathematically desirable — it makes the stiffness matrix symmetric, guarantees uniqueness, and puts limit analysis theorems at your disposal. It is also, for soil, false.

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How large psi really is

Associated flow at φ=30°\varphi = 30° implies sin⁡ψ=0.5\sin\psi = 0.5: for every unit of shear strain, half a unit of volumetric expansion, continuing indefinitely. No soil does this. Dilation is a transient of the dense state — the material expands until it reaches its critical void ratio and then shears at constant volume.

Bolton’s correlation for plane strain ties the excess of peak friction over critical state to the dilatancy:

φpeak′−φcv′=0.8 ψ\varphi'_{\text{peak}} - \varphi'_{\text{cv}} = 0.8\,\psi

A dense sand with a peak friction angle of 40° and a critical-state angle of 32° therefore has ψ=10.0°\psi = 10.0°, so sin⁡ψ=0.1736\sin\psi = 0.1736. Associated flow would have used sin⁡ψ=0.5000\sin\psi = 0.5000 and overstated the dilation rate by a factor of 2.88 — and would have kept it up to arbitrary strain.

For a loose sand or a normally consolidated clay, ψ\psi is zero or negative, and the associated assumption is not an overestimate but a sign error.

What it costs in a collapse load

Non-associated flow makes the problem harder to solve and the answer smaller, and the second part is what matters. Davis’s substitution gives a way to quantify it without a nonlinear analysis: replace the real parameters with reduced ones,

β=cos⁡ψ cos⁡φ1−sin⁡ψ sin⁡φ,tan⁡φ∗=βtan⁡φ,c∗=β c\beta = \frac{\cos\psi\,\cos\varphi}{1 - \sin\psi\,\sin\varphi}, \qquad \tan\varphi^{*} = \beta\tan\varphi, \qquad c^{*} = \beta\,c

and solve the associated problem with those. For φ=30°\varphi = 30° and c=10c = 10 kPa:

ψ\psiφ∗\varphi^{*}c∗c^{*} (kPa)NcN_cc∗Ncc^{*}N_c (kPa)
30° (associated)30.00°10.0030.14301.4
10° (Bolton)28.33°9.3426.47247.2
0° (non-dilatant)26.57°8.6623.19200.8

with Nc=(Nq−1)cot⁡φN_c = (N_q - 1)\cot\varphi and Nq=eπtan⁡φtan⁡2(45°+φ/2)N_q = e^{\pi\tan\varphi}\tan^2(45° + \varphi/2) as usual.

The associated assumption inflates the collapse pressure from 200.8 to 301.4 kPa. That is 33.4% — larger than the margin most partial factor systems apply, arrived at by a choice that is usually made by leaving a field at its default.

Where it is not merely inaccurate

In an undrained analysis with a positive dilatancy angle, the consequences are qualitative rather than quantitative.

Undrained means the volume cannot change, so any plastic volumetric expansion the flow rule demands must be cancelled by an elastic volumetric contraction, which requires a drop in pore pressure. The pore pressure falls, the effective stress rises, the available shear strength rises with it, and the material shears further — demanding more dilation, more suction, more strength.

There is no limit in the model. The computed strength grows without bound, the analysis either fails to converge or converges to a collapse load that is a function of how far it was pushed, and no mesh refinement helps because the mechanism is in the constitutive model, not the discretisation.

Real soil escapes this because water cavitates, because the dilation stops at the critical state, and because drainage is never perfect. A Mohr-Coulomb model with constant ψ\psi knows about none of those. The practical rule follows: in undrained effective stress analysis, set ψ=0\psi = 0 unless the model has an explicit dilatancy cut-off, and if it has one, know what void ratio it is set to.

The rule that is worth following

  • ψ=0\psi = 0 for undrained analysis, always, unless a cut-off is active.
  • ψ=φ′−30°\psi = \varphi' - 30° for dense sands, floored at zero, is the common workable rule and is consistent with Bolton to within the accuracy of either.
  • ψ=0\psi = 0 for loose sands, normally consolidated clays, and anything at or near critical state.
  • ψ=φ\psi = \varphi only when you are deliberately computing an upper bound and will say so.

And, separately: report it. A model description that gives φ′\varphi' and c′c' without ψ\psi is under-specified by a third of the collapse load, which is the same class of omission as leaving out the initial stress state or the integration rule. The dilatancy angle also sets the return direction in the plasticity algorithm, so it is not only the answer that changes but which region of the yield surface the return lands in.

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