A failed slope gives a line, not a soil

The surface that has to be searched for computed a slope’s factor of safety the way it is normally computed: take the soil’s strength, cut the slope into slices on a trial circle, compare the moment the strength can resist with the moment the weight drives, and search over circles for the smallest ratio. It ended on the reverse problem, “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.”

That sentence is right about the value of the exercise and hides its difficulty. A laboratory specimen is a few centimetres of soil, taken from one borehole, disturbed by the taking, and sheared at a rate and a stress path chosen for the machine. A slip is the whole mass of soil failing along the surface it chose, at its own pore pressures, over its own time. Nothing measures field strength better. But the slip gives one number — that the factor of safety was one — and a soil whose strength is written as a cohesion and a friction angle has two.

One equation, two unknowns #

The slope here is 10 m high with a face at 35°, in soil of 19 kN/m³, with pore water pressures that are a quarter of the weight of soil above any point ($r_u = 0.25$), over a firm stratum 5 m below the toe that no slip surface can pass through. It has slipped. Every factor of safety below is Bishop’s simplified method on a searched critical circle — the minimum over centres and depths, as the earlier essay insisted — so each number is the slope’s factor, not one circle’s.

For each friction angle there is exactly one cohesion that makes the critical circle’s factor one, because the factor rises steadily with cohesion. Stepping the friction angle from nothing to 40° and solving for the cohesion each time traces the whole of what the failure knows.

Every point on that line failed that slope. A soil with 32 kPa of cohesion and no friction, one with 12.4 kPa and 20°, and one with 2 kPa and 40° are, as far as the fact of the failure can tell, indistinguishable. The line is not straight — it bends because the friction a slip mobilises depends on the normal force on its base, which depends on where the base is — and it does not stop at 40°. It runs on to the axis at 48.1°, the friction angle that fails the slope with no cohesion at all.

That last point is worth a moment, because it is the one most often chosen. Setting the cohesion to zero is a common convention in back-analysis, on the reasoning that cohesion is unreliable and friction is not. On this slope the convention yields a friction angle of 48°, which no soil that fails at 35° has, and it yields it for a specific reason: with no cohesion the critical surface shrinks to a skin sliding parallel to the face, and the factor is the infinite-slope expression

$$F = \left(1 - \frac{r_u}{\cos^2\beta}\right)\frac{\tan\phi'}{\tan\beta}$$

which is one at $\tan\phi' = \tan 35°/0.627$. The convention has chosen, without saying so, a failure a few centimetres deep.

Each soil fails along its own surface #

That is the clue to the second equation. The pairs on the line agree about the factor and disagree about where the slip is.

The three surfaces are three different landslides. The nearly cohesive soil fails deep, on a long, flat-bottomed arc that passes under the toe and breaks out six and a half metres behind the crest. The frictional soil fails shallow, on a tight arc through the toe that barely reaches behind the crest. The middle soil is between.

The reason is in what each kind of strength rewards. Cohesion resists in proportion to the length of the surface, whatever its depth — the opposite of friction, which ignores the area in contact and counts only the force pressing on it — so a cohesive soil can be failed most cheaply by a surface that carries a great deal of weight for its length — a deep one, which takes a large driving mass in exchange for a surface not much longer than a shallow one. Friction resists in proportion to the normal force on the surface, which grows with depth as fast as the driving weight does, so going deep buys nothing and a frictional soil fails where the geometry is worst: close to the face. The balance between the two is carried by a single dimensionless number, $\lambda = \gamma H \tan\phi'/c'$, which is 0.7 for the soil at 5°, 5.6 for the soil at 20° and 33 for the soil at 35°. Slopes with the same $\lambda$ fail along geometrically similar surfaces, whatever their size.

The scar is the second equation #

So a slip whose depth has been measured — by an inclinometer that bent, a borehole that found the slickensided surface, or simply the shape of the scar once the debris is moved — carries a second piece of information that the factor of safety does not.

A slip measured 5 m deep fixes the soil at about 21° and 12 kPa. The curve is steep at the frictional end and shallow at the cohesive end, so the depth is a sharp instrument for telling a frictional soil from a moderately cohesive one and a blunt one for telling two cohesive soils apart. It also jumps: between no friction and 5° the critical surface stops running down to the firm stratum and pulls up under the toe, and the depth falls from 13.6 m to 8.8 m across a few degrees. A slip that reached a hard layer says the soil was nearly cohesive and says little more.

The measurement has an error, and the line turns the error into a spread of soils. A metre either side of 5 m admits every soil from 16° with 15 kPa to 27° with 8 kPa. That is not a small range in soil mechanics, but it is a range: without the depth, the slope admits every soil from nothing to 48°.

There is a caution in the curve’s shape, too. The depth plotted is the depth of the critical circle — the surface Bishop’s method says is weakest. A real slip is not obliged to be circular, and in a soil with any layering it will follow the weakest layer instead. The depth is an equation only where the model’s surface is the kind of surface the slope actually used; where a slip follows a clay band, the band is the answer and the back-analysis should be done on a surface through it.

Why it matters: the repair #

None of this would matter if every soil on the line predicted the same thing about any other slope. They do not, and the slope that matters is the repaired one.

The commonest repair is to flatten the face. Here it is regraded to 25° by cutting the crest back 7.2 m, which in homogeneous ground is the same geometry as building the toe out and gives the same factor.

The regrade is excellent for the frictional soil and nearly useless for the cohesive one, and the picture shows why. The frictional soil’s slip lived near the face; flattening the face moves the whole of the weak geometry away and its factor rises to 1.45, nearly the ratio of the two faces’ tangents, as the infinite-slope formula would say. The cohesive soil’s slip lived deep, under the toe, driven by the weight of the whole block above it. Flattening the face removes a wedge from the top of that block — some of the driving weight — but leaves a long, deep surface that is not much shorter than before, and the factor rises only to 1.11.

Swept along the whole line, the same regrade is worth anything from 1.07 to 1.51 on soils that all reproduce the failure. Drainage is worse still for the cohesive end: lowering the pore pressure ratio from 0.25 to 0.10 raises the normal effective stress on the slip surface, which is worth something only through friction, so it gives a purely cohesive soil nothing at all — exactly 1.00 — and a soil at 40° a factor of 1.28. A designer who back-analysed with cohesion set to zero, and then drained the slope, has specified a remedy whose whole value comes from the assumption.

The measured depth narrows it. With the slip known to within a metre of 5 m the regrade is worth 1.24 to 1.35 — a spread that still matters against a target of 1.3, but one a designer can bracket rather than guess.

How far to flatten #

The design question is usually asked the other way: what face angle gives the repaired slope a factor of 1.3?

The three curves start together at 35°, where every soil gives one by construction, and fan out as the face flattens. For the frictional soil a regrade to 27.7° is enough; for the middle soil, 24.5°; for the nearly cohesive one, 16.6° — a face less than half as steep as the one that failed, and a cut that removes about four times as much ground. The difference between 27.7° and 16.6° on a 10 m slope is 15 m of land at the crest, which on a road or a railway cutting is often the difference between a regrade and a retaining wall.

The assumption about the soil decides the size of the repair more than the target factor does. Moving the target from 1.3 to 1.2 on the middle soil changes the angle by three degrees; moving along the line from one soil that fits the failure to another changes it by eleven.

Two soils that fit, by hand #

Two of the points on the line can be checked without a computer, and they bracket the rest.

At the frictional end, with no cohesion, the infinite-slope formula above gives $\tan\phi' = \tan 35°/(1 - 0.25/\cos^2 35°) = 0.700/0.627 = 1.117$, so $\phi' = 48.1°$, which is where the line meets the axis.

At the cohesive end, with no friction, the pore pressure does not enter and the slope is Taylor’s problem: the factor is $c'N/\gamma H$ for a stability number that depends on the face angle and the depth to a firm layer. The line gives 32.1 kPa at $F = 1$, a stability coefficient of $32.1/(19 \times 10) = 0.169$ — close to the 0.18 Taylor’s chart gives for a deep failure in a slope this flat, and a little below it because the firm stratum 5 m under the toe stops the circle going deeper. Taylor’s chart also carries the regrade’s result: below about 53°, a purely cohesive slope on a deep stratum has a stability number that barely depends on its face angle, because the critical circle is a base failure whose weight and length both scale with the block. That is the 1.07.

What a back-analysis needs before it starts #

The argument is not that back-analysis is unreliable. It is that it needs two observations, and the second is usually available and usually unused.

The factor of safety at failure. One, by definition, and itself an assumption: a slope that moved slowly for months was at one; a slope that slid in a storm was at one under the pore pressures of the storm, which were probably higher than the $r_u$ of 0.25 assumed here. Every point on the line moves if the pore pressure at failure is wrong, and it is the least well known number in the calculation.

The surface. Its depth at least, and ideally its breakout behind the crest and its toe, which between them fix the circle’s centre and radius. Two soils on the line that give the same depth — which cannot happen here, because the curve is monotonic beyond 5° — would need the breakout to separate them.

An independent estimate of one strength. Laboratory tests give a friction angle far more reliably than a cohesion, because friction is a property of the grains and cohesion is largely an artefact of fitting a straight line to curved test data. A friction angle from tests, placed on the line, gives the cohesion; if the slip’s depth then agrees, the two observations corroborate each other, and if it does not, one of them — the surface, the pore pressure or the test — is wrong.

This is the same logic as the strength no specimen had: a design strength is a statement about a population inferred from a sample, and a slope that failed is a sample of one, whose single measurement has to be read together with everything else known about the ground.

Where the model stops #

The surface is a circle and the soil is uniform. Real slips are often non-circular, and in layered ground they follow the weakest layer — a clay band, an old shear surface, the contact between fill and natural ground. On such a slope the back-analysis should be done on the surface that was observed, with a method that allows non-circular surfaces, and the line of pairs is a property of that layer rather than of the slope.

The strength is linear in normal stress. A cohesion and a friction angle describe a straight line on a plot of shear strength against effective normal stress, and soils have curved envelopes: a soil tested at low stress shows a high friction angle and little cohesion, and at high stress the reverse. A shallow slip samples the envelope at low stress and a deep slip at high stress, so the line of pairs is partly an artefact of forcing a straight envelope onto one stress level — which is another reason the slip’s depth matters, since it says at what stress the strength was measured.

The pore pressure is a ratio. Real pore pressures come from a water table and a flow net, and a ratio of 0.25 is a convenient average that no point in the slope actually has. A basement is a boat makes the general point that water is the load most often wrong; a back-analysis that gets the water wrong recovers a wrong soil exactly, and every repair designed on it inherits the error.

Bishop’s method assumes the factor is the same on every slice, which is what makes a single number meaningful and which a slope failing progressively from the toe does not obey. Nor is a limit-equilibrium factor either of the two bounds plasticity offers: it satisfies neither the mechanism’s compatibility nor equilibrium everywhere, so its error has no guaranteed sign. A progressive failure has a factor above one on part of its surface and below one elsewhere at the moment of failure, and the back-analysed strength is an average over a surface that was not all at its strength at once.

What the pictures cannot show #

That the soil after the slip is not the soil before it. A slip surface that has moved has been sheared to large strains, and in a clay the strength along it falls towards a residual value with a friction angle perhaps half the peak one and no cohesion at all. The back-analysis recovers the strength that failed the slope; the repaired slope has to stand on the surface the failure left behind, which is weaker. A regrade designed on back-analysed peak strengths for a slope that will reuse the old surface is designed on a soil that no longer exists — which is the case for being on the frictional, pessimistic end of the line only if the residual friction angle is actually there.

Nor can they show time. The ground that arrives after the weight found clay gaining strength as it consolidates under load; a cutting does the opposite, unloading the clay, which swells and softens over years. A cutting that stood for twenty years and then slipped failed in drained conditions at the end of that softening, and back-analysis of it measures the softened soil — the right one for the next twenty years, and a different one from the soil that was dug.

The same line elsewhere #

The argument is not special to slopes. Any structure whose collapse is computed by a mechanism with two strength parameters gives, on failure, a line rather than a point: a footing that has failed gives a line of cohesions and friction angles through its bearing-capacity factors, and a retaining wall that has moved gives a line through its active pressure. In each case the geometry of the mechanism — the depth of the wedge, the extent of the heave — is the second equation, and in each case it is usually photographed and rarely used.

And it is a general statement about inverse problems in statics. The force that is whatever it needs to be observed that friction supplies what equilibrium asks for until it cannot. A failure is the moment it cannot, and it tells the observer that the capacity equalled the demand — one equation. How the capacity was made up is a separate question, and the answer is in the shape of the failure rather than in the fact of it.

Still open: the slope that has failed twice #

A single slip gives one line. A slope that has failed twice — once in a wet winter and again after a regrade, or at two points along a cutting with different heights — gives two lines, and where they cross is a soil that fits both failures with no appeal to the depth at all. Whether two failures of slightly different geometry cross at a well-conditioned angle, or whether their lines are so nearly parallel that the crossing is as uncertain as a single line, is a question about how different two slopes have to be before their failures carry independent information — and it is the question behind every railway cutting inventory that holds a century of slips and uses each one alone.