The K0 procedure: initial stresses before anything moves

Before a finite element model can be loaded it has to be given a stress state, and that state is not zero. Soil at depth is already carrying its own weight, and it has been doing so for long enough that the horizontal stress has settled to whatever value its deposition and unloading history produced.

The vertical component is easy: it is the weight of what is above, which is the arithmetic half of effective stress. The horizontal component is not deducible from it. It is an independent piece of information about the soil’s history, and in almost every analysis it is supplied by assumption.

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What K0 isPermalink to “What K0 is”#

The coefficient of earth pressure at rest is defined by

$$K_0 = \frac{\sigma'_h}{\sigma'_v}$$

under the condition that no horizontal strain has ever occurred — one-dimensional deposition, no lateral yielding. It is a stress ratio, not a material property, and its value records what the ground has been through.

For a normally consolidated soil, Jaky’s expression is the standard:

$$K_0 = 1 - \sin\varphi'$$

At $\varphi' = 32°$, $\sin\varphi' = 0.5299$ and

$$K_0 = 1 - 0.5299 = 0.470$$

Jaky’s formula is empirical, fitted to granular materials, and it is accurate to about ±0.05 across the soils it was fitted to. That is a wide band, and it is the good case.

The first way to get it wrong: let the model generate itPermalink to “The first way to get it wrong: let the model generate it”#

The obvious way to establish initial stresses in a finite element model is to switch gravity on and solve. It is obvious, it is easy, and for a horizontal ground surface with a linear elastic material it produces

$$K_0 = \frac{\nu}{1 - \nu}$$

because the lateral confinement of a laterally restrained elastic body is set entirely by Poisson’s ratio. At $\nu = 0.30$ that is 0.429 — 8.83% below Jaky’s 0.470, from a parameter that was chosen for its effect on the deformation analysis and has nothing to do with the soil’s stress history.

The mismatch is not fixable by tuning $\nu$ unless you are willing to distort the deformation response, because the value that would reproduce Jaky here is

$$\nu = \frac{K_0}{1 + K_0} = 0.3198$$

and there is no reason for the soil to have that Poisson’s ratio.

This is why programs have a K0 procedure: a separate initialisation step that writes $\sigma'_h = K_0\,\sigma'_v$ directly, with $K_0$ supplied by the engineer, instead of deriving it from elastic constants. Gravity loading is still the right choice where the ground surface is not horizontal, or where layers are inclined, because there the assumption of one-dimensional deposition is false and a direct $K_0$ assignment produces a state that is not in equilibrium.

The second way: forget the soil was unloadedPermalink to “The second way: forget the soil was unloaded”#

An overconsolidated soil has been under more vertical stress than it is under now. Vertical unloading is not matched by horizontal unloading — the horizontal stress is partly locked in — so $K_0$ rises, and can exceed one. The standard expression is

$$K_0^{\text{OC}} = (1 - \sin\varphi')\,\mathrm{OCR}^{\sin\varphi'}$$

At an overconsolidation ratio of 4 the horizontal effective stress equals the vertical one. At 8 it exceeds it, and the major principal stress in the ground is horizontal — which inverts the failure mechanism that most intuition is built on.

What it costs in a real profilePermalink to “What it costs in a real profile”#

Take a uniform deposit, $\gamma = 19.5$ kN/m³, water table at 2 m, and look at 8 m depth:

$$\sigma_v = 156.0\ \text{kPa}, \qquad u = 58.86\ \text{kPa}, \qquad \sigma'_v = 97.14\ \text{kPa}$$

The elastic route is 4.03 kPa low, which is 8.83% and is arguably tolerable. Missing the overconsolidation is 49.53 kPa, which is 108.5% and is not. Integrated over an 8 m wall, including the water, the total horizontal thrust goes from 386.9 kN/m to 615.0 kN/m — a factor of 1.59 on the load the structure is designed for.

None of that difference is visible in the model geometry, the mesh, the element type or the material stiffness. It is one number, entered once, at the start.

Why it propagatesPermalink to “Why it propagates”#

An initial stress error does not stay an initial stress error. In a Mohr-Coulomb or Hardening Soil analysis the initial state determines how far each point is from the yield surface, so it decides where plasticity begins, in what order zones yield, and therefore the shape of the failure mechanism. The Mohr circle picture makes this concrete: changing $K_0$ changes the radius and the centre of the initial circle, and a circle that starts closer to the envelope reaches it under less load.

Two consequences follow for practice.

A K0 state can be inadmissible. If the assumed $K_0$ places the initial stress point outside the yield surface, the program has been handed a state the material cannot sustain. Well-behaved codes detect this and either refuse or return the stresses to the surface — but the second option quietly changes the state you asked for, and the run continues. For $\varphi' = 32°$ with no cohesion, the passive limit is $K_p = 3.25$ and the active limit $K_a = 0.307$, so any $K_0$ outside that range is inadmissible before the first load step.

The initialisation must be reported. “Mohr-Coulomb, $\varphi' = 32°$, $c' = 5$ kPa” does not specify the analysis. The same parameters with $K_0 = 0.47$ and with $K_0 = 0.98$ are different problems with different answers, and only one of them is the site. A model description that omits the initial stress procedure is not reproducible, which is the same failure as omitting the mesh and rather more consequential.