The surface that has to be searched for

Everything else in this collection is checked somewhere. A beam is checked at mid-span and over its supports; a column at its base; a connection at the bolt group and at the net section. The engineer draws a line through the structure, and the calculation happens on that line.

A slope has no line to draw. The failure surface is whatever shape the ground finds cheapest, it is somewhere inside a homogeneous mass with no features to attach a check to, and it is not known until it has been looked for.

A ratio of moments #

Cut the mass above a trial surface into vertical slices. Each one supplies both halves of the answer.

The weight of a slice, resolved along its own base, drives the rotation: $W\sin\alpha$, with $\alpha$ the base inclination. Sum those over the surface and multiply by the radius, and that is the driving moment about the circle’s centre.

The same weight, resolved across its base, presses the surface together and generates friction: $N' \tan\phi'$, added to the cohesion $c'l$ that acts whether anything is pressing or not. Sum those and multiply by the same radius, and that is the resisting moment.

$$F = \frac{\sum \left(c'l + N'\tan\phi'\right)}{\sum W \sin\alpha}$$

with the radius cancelling from both. That is a ratio of moments, which nothing else on this site computes a factor of safety as. A steel beam’s utilisation is a ratio of stresses; a footing’s is a ratio of pressures; a slope’s is a ratio of two rotations that do not happen.

The sign of $\alpha$ matters more than anything else in the assembly, and getting it wrong is the classic error: the material behind the crest drives the rotation and the material at the toe resists it. Slices near the toe have negative $\alpha$ and contribute negatively to the driving sum. A spreadsheet with the sign convention reversed reports a slope holding itself up by its own weight.

The number that divides rather than multiplies #

$F$ appears in the numerator’s strength terms and nowhere else, which makes it a strength reduction factor: the slope is in equilibrium if the available $c'$ and $\tan\phi'$ are divided by $F$.

That is a different object from a load factor, and the difference shows up in what it protects against. Dividing the strength by 1.2 is 20 per cent on two quantities that behave differently: $\tan\phi'$ is reliable and will still be there in fifty years, while $c'$ is often an artefact of suction, cementation or root reinforcement and may not be. For the slope drawn, cohesion supplies 25 per cent of the resistance — so a slope at $F = 1.2$ whose cohesion turns out to be zero is at $F = 0.89$.

Fellenius and Bishop, and the gap that never changes sign #

The slices push on each other, and what they push with is unknown. Every method of slices is a different way of not knowing it.

Fellenius (1936) ignores the inter-slice forces entirely and resolves each slice’s weight normal to its own base. That is not statically consistent — the slices are not in equilibrium individually — but it is arithmetic anybody can do by hand, and it is conservative.

Bishop (1955) keeps vertical equilibrium of each slice and assumes the inter-slice forces are horizontal. The resulting expression has $F$ on both sides:

$$F = \frac{1}{\sum W\sin\alpha}\sum \frac{c'b + (W - ub)\tan\phi'}{\cos\alpha\left(1 + \dfrac{\tan\alpha\tan\phi'}{F}\right)},$$

so it is iterated — three or four passes from any sensible start.

For the critical circle here the two give 1.081 and 1.192, an 11 per cent gap. At the trial circle it is 22 per cent. The gap never changes sign: Fellenius is always the lower, because dropping the inter-slice forces always removes normal force from the bases where it was most needed.

That is a useful property and an uncomfortable one. Useful, because a hand calculation is safe. Uncomfortable, because a difference of 11 per cent between two methods of the same problem is larger than most of the margins being argued about — and neither of them is exact, since both are approximations to a problem whose statics is genuinely indeterminate.

The assumption nobody states #

The Bishop expression above has $F$ inside the sum, once per slice, and it is the same $F$ for every slice.

That is the assumption the whole method rests on, and there is no reason for it to be true. It says that at the moment of failure every point of the surface is the same fraction of the way to its own strength — that the base under the crest and the base at the toe are both, say, 84 per cent mobilised. In reality the toe is fully mobilised long before the back scarp is, which is why real slope failures are progressive and why a soil with a brittle strength — one that peaks and then falls — can fail at a load the method says it can carry.

A single factor of safety is a number about a surface, not about a point, and it is meaningful only if the surface is going to fail all at once.

It is the same shape of assumption as the one under a bolt group’s elastic analysis, where every bolt is assumed to be at the same fraction of its capacity and the real group redistributes until they are; and under the lower-bound theorem generally, where a structure is licensed to find any equilibrium it likes provided it is ductile enough to get there. In each case the licence is ductility, and in each case what invalidates it is a material that softens. Soil is the one that softens most.

The search is the calculation #

The critical circle for the slope drawn is centred at (13.8, 16.8) with a radius of 17.1 m, and it is neither the deepest circle nor the shallowest. It is deeper than a toe circle drawn by eye and shallower than the deepest one available, because the two effects that decide it run opposite ways: a deeper circle mobilises more surface — more cohesion — and also more mass — more driving weight — and the balance between them has a minimum somewhere in between.

That minimum is the answer, and finding it is not incidental to the calculation. It is the calculation. A modern program tries thousands of surfaces, circular and not, and the number it reports is the smallest one it found — which means the answer depends on the search having been wide enough, a dependence no other check in this collection has.

The dependence is not academic. Restricting the search to circles through the toe, as the eighty-one drawn are, is a restriction: a deep-seated circle passing below the toe is available in a soft foundation layer and is often the critical one, and a search that excludes it returns a number about a mechanism that will not occur. Every slope analysis is a statement about a search space, and the search space is chosen by whoever set the problem up rather than by the ground.

There is a family resemblance here to the upper-bound theorem’s mechanism search: try mechanisms, take the smallest collapse load, and know that the true answer is at or below whatever was found. Limit equilibrium is not a bound method, but it inherits that method’s discipline — the answer is only as good as the worst thing anyone thought to try.

What the water does #

Effective stress is the whole of soil strength, so the pore pressure enters the resisting sum and nothing else.

$$N' = W\cos\alpha - u\,l$$

and $u$ is subtracted before the friction is computed. Raising the pore pressure ratio $r_u = u/\gamma z$ from 0 to 0.25 to 0.40 takes the factor of safety from 2.14 to 1.65 to 1.36 — a 36 per cent loss, with no load applied by anybody and nothing about the soil changed.

This is why slopes fail after rain rather than during it, why drainage is the cheapest slope stabilisation there is, and why a factor of safety quoted without the piezometric assumption beside it is not a number.

The undrained case, which is a different problem #

Everything above is an effective stress analysis: the strength depends on how hard the surface is pressed together, so the pore pressure has to be known and the answer depends on it.

For a clay loaded faster than water can leave it, none of that applies. The undrained strength $c_u$ is a total-stress quantity with no friction term in it at all, so $\phi = 0$, the normal force drops out of the resisting sum entirely, and

$$F = \frac{c_u \, N_s}{\gamma H}$$

with $N_s$ a stability number that depends only on the geometry — 5.52 for a deep circle in a slope steeper than about 53°, and larger for flatter ones. The soil’s weight appears in the denominator and its strength in the numerator, and nothing else appears at all.

The 10 m slope drawn, if it were undrained clay of $c_u = 10$ kPa, would have $F = 0.29$ — it would not stand for a minute. It needs $c_u = 34$ kPa to reach unity, which is a firm clay. That comparison is the reason the two analyses are done on the same slope for different times in its life: undrained immediately after excavation, when the strength is high and the pore pressures have not equalised, and drained in the long term, when they have. For a cutting the long term is worse and for an embankment the short term is, and which one governs is the first decision anybody makes.

Which free body produced the number #

The free body is the entire mass above the surface, and the slices are a device for integrating over it rather than free bodies in their own right — which is exactly what Fellenius’s inconsistency is. Take the whole mass and it is in equilibrium under its weight, the normal and shear stresses on the base, and nothing else; the moment equation about the circle’s centre is the only one that can be written without knowing how the stresses are distributed, because on a circle every normal stress passes through the centre and drops out.

That is why the surface is a circle. Not because the ground prefers one, but because a circle is the shape for which one equation can be written without an assumption.

Where the model stops #

The soil is one material with one strength. Real ground is layered, and the critical surface in a layered profile is not circular at all — it runs along the weak layer and cuts up steeply at each end, which is why non-circular searches exist and why they usually find lower numbers.

The slices are vertical and there are twenty-four of them. Both are conventions. The base angle of a slice is taken as the angle at its midpoint, so a coarse division misrepresents the geometry near the toe where the surface is steepest — and the answer creeps as the count rises, by a per cent or so between twelve slices and fifty. It is the same dependence on how the thing was divided that every discretised calculation has, and here it is small and real.

The strength is a peak strength. For a clay that softens after failure, the relevant number is the residual, sometimes half the peak. A first-time slide is a peak-strength problem and a reactivated one is a residual-strength problem, and they are different calculations on the same slope.

Nothing here is a bound theorem. The upper and lower bound theorems close on the bearing capacity problem and give an exact answer; limit equilibrium does neither. It satisfies overall equilibrium of the mass and violates equilibrium of the parts, so its answer is not guaranteed on either side — it is close in practice and it is not a bound.

And the drawing shows a surface that has not happened. The circle is a hypothesis about a rotation nobody has observed. The mass above it is intact, the ground is standing, and every quantity on this page is about a motion that has been imagined and then insisted upon — which is the site’s founding move, applied to a body with no edges.

The ladder from here #

Later rungs on this anchor: non-circular surfaces and the wedge methods that handle a weak layer, where the search space is much larger and the minimum much harder to be sure of. Janbu, Morgenstern–Price and Spencer, which differ in what they assume about the inter-slice force direction and which bracket the answer to within a few per cent of each other and rather further from Fellenius. Progressive failure, where the single-$F$ assumption is dropped and the surface mobilises from the toe backwards. Reinforced slopes, where a nail or a geogrid adds a force to the resisting sum and the question becomes where to put it. Seismic slope stability, where a horizontal acceleration is added as a body force and the critical surface moves. And the reverse problem — back-analysis, in which a slope that has already failed is assumed to have had $F = 1$ and the equation is solved for the strength instead, which is the only way anybody ever measures the strength of a slope.