A slip measures the slope it fell from

A failed slope gives a line, not a soil took one slip and found that it fixes a line of possible soils rather than a soil, every point on the line agreeing that the slope stood at a factor of exactly one. Two slips are a shear test took two and found that their lines cross at the soil that made them, provided the slopes are different enough; and each slip, it said, is really a single shear test at the mean normal stress on its surface. Then it looked ahead to the scale at which a real owner has the data. A railway company’s records hold hundreds of slipped cuttings. Surely a hundred lines, fitted with due respect for scatter and with the helped-along cases set aside, would give what no borehole can: the operational strength of the clay at the scale of the slope.

It is a good idea, and it does not work the way it sounds as though it should. A hundred slips are not a hundred shear tests of the clay. What goes wrong can be shown in one line of statics and it is not a matter of having too few of them.

The ray each slip lies on #

Take the simplest reading that keeps the point: every slip a plane parallel to the face of the cutting, at some depth $z$. The infinite-slope equilibrium then gives its stresses in closed form, the driving shear and the effective normal stress,

$$\tau = \gamma z \sin\beta\cos\beta, \qquad \sigma' = \gamma z \cos^2\!\beta\,(1 - r_u).$$

The back-analysis knows the geometry of the slip and writes down $\tau$. It rarely knows the pore pressure at the moment of failure, so it assumes a ratio $r_u$, and writes down $\sigma'$. That is a point on a shear-strength plot, and the point is said to lie on the clay’s envelope, since the slope was at the limit. Now divide one coordinate by the other:

$$\frac{\tau}{\sigma'} = \frac{\tan\beta}{1 - r_u}.$$

Neither $\gamma$, nor $z$, nor the clay appears. Each slip’s point lies on a ray from the origin, and the ray’s slope is the cutting’s own angle and the analyst’s own assumption. A slip’s depth moves it along its ray. Nothing about its strength can move it off. Whatever clay this cutting is cut in, if it slipped, its point is on this ray.

So where is the clay? Only in the decision about which points are on the plot at all. A cutting of a given angle and height either slipped or it did not, and that selection is what carries the strength. The plot of slips alone has discarded the half of the information that contained it.

An inventory, built so the answer is known #

To test this, build the site. Six thousand cuttings in one clay: heights uniform from 3 to 14 m, angles uniform from 10° to 20°, each with a slip plane at three tenths of its height. The clay has $\phi' = 22°$ and $c' = 4$ kPa, each cutting with its own scatter of 1.5° and 1.5 kPa about those values. Each has its own worst pore pressure in its life, $r_u$ scattered around 0.3, and a tenth of them have also had a burst main or a blocked drain worth another 0.2 of $r_u$. A cutting slips if its clay cannot carry its shear at its worst pore pressure. Of the six thousand, 655 do, or 11 per cent.

The analyst has the geometry of all of them and assumes $r_u = 0.3$, which is exactly the mean. That is generous. And the analyst fits a straight line through the 655 points.

The fit is tight, and its answer is $\phi' = 24.1°$, $c' = 0.7$ kPa. The clay’s cohesion has all but vanished, and two degrees have appeared in its friction. Both effects follow from the rays. Lines from the origin fitted by a line will pass near the origin, whatever is true of the clay, and so a small cohesion is what a slip plot says with nothing behind it. The gradient is close to the mean of $\tan\beta/(1 - r_u)$ for the cuttings that fell. Those have a mean angle of 18°, $\tan 18° = 0.325$, and over 0.7 that is 0.464, or 24.9°. The fit has found the angle of the slopes that fell, divided by 0.7, and very little else.

A small cohesion intercept is a celebrated result on real clays: Skempton’s back-analyses of first-time slides in London Clay gave about 1 kPa, and it has been widely repeated. Those slides came with measured pore pressures and careful reconstruction, so they are not what this inventory imitates, and nothing here says that result is wrong. What this shows is that a small cohesion from a line through slips with an assumed ratio is not evidence of it: this clay has 4 kPa and reads 0.7.

The assumed pore pressure comes straight back out #

It takes only one change to see how much of the friction angle was never measured. Run the same 655 slips with a different assumption about pore pressure, and the fitted friction angle goes from 17.3° at no pore pressure to 38° at 0.6. This is a simple rescaling of the $x$ coordinate by $1 - r_u$, so the fit’s gradient rescales by the reciprocal, and $(1 - r_u)\tan\phi'$ stays exactly 0.31 throughout. That product is what a slip inventory measures, and it does not come apart without a piezometer. Two slips are a shear test found the same trade when a second, wetter failure had an unknown ratio; here the trade is wholesale. Any friction angle anyone would like to report can be obtained, with a plausible $r_u$ and the same data.

The line between fell and stood has the same trouble, since it too has been drawn on the assumed $x$. Knowing the pore pressure is the one thing nothing else can stand in for. A strength for the regrade design can still be had, however, because the regrade is designed with the same pore pressure, and the product is all it needs.

Same clay, different slopes #

If the fit were measuring the clay, then building the cuttings steeper would change the number of slips and not the result. Build the same site with the angles between 8° and 16° and a mere one per cent of the cuttings slip, those on the weakest clay and the highest pore pressure; the line through them says $\phi' = 20.0°$. Build it between 18° and 30° and 73 per cent slip, including all sorts of ordinary clay well above the limit; the line says 29.5°. At the 30 kPa of a typical slip plane that is a strength of 11.4 kPa from one railway and 19.9 kPa from another, of one clay. A pair of neighbouring lines built by different engineers to different specifications would read as two clays.

The line between fell and stood is a logistic regression of the outcome, slipped or not, on the same two coordinates, and its contour of even odds is the place where a cutting of those coordinates is as likely to go as to stay. It uses every cutting, and it is a real attempt to find the limit rather than to fit what lies past it. At 30 kPa it reads 15.9 to 16.7 kPa across all seven inventories, against the clay’s 16.1. Its own friction angle and cohesion are not good: 18.8° and 6.3 kPa on the base inventory, against 22° and 4. The scatter tilts the line, and the two parameters trade against each other along a narrow valley. It has the strength in the range it was asked about, though, and that is the number a design uses. The strength no specimen had made a related point about a fractile drawn from scatter: a characteristic value can be well known when the distribution’s parameters separately are not.

The practical trouble is obvious. An inventory is a record of failures. Slips are reported, photographed and given an incident number, while the cuttings that stood have only a drawing in a filing cabinet, if that. Yet the ones that stood are half of the shear test. Asset registers that hold the geometry of everything are now the thing that lets this be done, and it is a reason to keep them.

More slips, and the wrong answer more precisely #

The usual remedy for a poor estimate is more data, and here it does what such a remedy does when the bias is structural. Five slips give a median friction angle of 25.5°, and the middle four fifths of inventories span from 19.2° to 29.6°, so the clay is well inside the band. A hundred give 24.1°, with a narrower band. By three hundred slips the band runs 23.4° to 24.9°, and the clay at 22° is now outside it.

A confidence interval on a regression would be reported from the same arithmetic, and would look excellent. With a hundred points and a well-fitting line an analyst would quote $24° \pm 1°$ and be, by a clear statistical test, wrong. The method is not giving a poor estimate of the clay; it is giving a good estimate of something else: the angle at which the site’s cuttings tend to fall, divided by $1 - r_u$.

Setting aside the helped ones #

The obvious scruple is that some of these slips were helped over the edge by a burst main or a heavy train. They are not at a factor of one under the conditions recorded, and they will look weak. The earlier essay’s proposal was that the same geometry that made two slips at one angle useless might find them, since they should lie below the line. A standard robust fit trims points well below the rest and refits.

It does not help, and the reason is the same. A cutting that was helped has its $\tau$ and its assumed $\sigma'$ set by its geometry, and so its point lies on the ray of its slope, as any slip’s does. Only if it is low for its $\sigma'$ can it be picked out, which is to say only if it has an unusually flat angle, and that is not the same thing as having been helped. Of the 231 helped slips in this inventory, 176 stay in after trimming, and the 76 removed include 21 slips that were helped by nothing. More importantly, the flat-angled low points are the ones that had been pulling the line towards the clay. Without them it rises to 26.1°, two degrees further away.

This is an outlier test doing what outlier tests do, and removing the points least like the rest. In a setting where the rest are biased, those points are the ones nearest the truth. A burst main is a fact that is found in a maintenance log and not on a graph. An asset owner who wants these slips out must go and look.

Peak or residual, or neither #

The earlier essay’s other question was whether the friction angle that survives is nearer the peak or the residual strength of the clay. It turns out that a real inventory mixes two populations. First-time slides, in a clay softened over decades, have something close to a fully softened strength. Reactivated slides, on a surface that has moved before, have the residual strength, far lower. Give some share of the cuttings an old surface, at 13° with no cohesion, and those slip at much flatter angles and much more often.

Through the slips, the strength at 30 kPa falls modestly from 14.1 to 11.9 kPa at two fifths of the cuttings, nowhere near either population. It cannot say which it is reading, since a slope on a residual surface is just another ray. Between fell and stood, the reading holds up to a fifth of the cuttings, at about 17 kPa, the softened clay’s, because the cuttings on old surfaces are slipping regardless of angle, and a slip that has nothing to do with the angle cannot shift the boundary; once two fifths are on old surfaces it falls away to 12.6.

Neither line says “residual”. Neither line says what fraction is. The way to separate them is to know from the ground which slips were reactivated, and a shear surface with polished, slickensided faces is a thing that can be seen in a trial pit. As in the surface that has to be searched for, what statics provides is a family of candidates, and it is the ground that picks one.

What it does to a regrade #

The reading matters because somebody will use it. A regrade is designed to a limit angle at a slope’s height, and that needs only the strength at its working stress, with the same $r_u$ as the analysis. For a cutting 8.5 m high, the plane is at 2.55 m and the clay itself stands at about 20.4°. A line through the slips of the base inventory says 18.1°, which is safe, and costs about an eighth more land on each side. A line through the slips of the steep inventory, $\phi' = 29.5°$ and $c' = 2.9$ kPa, says 24.8°. That is a design angle more than four degrees steeper than the clay can hold, reached from a tight fit with a good correlation.

The fit errs on the safe side where the site’s cuttings are flat and on the unsafe side where they are steep, so the owner of a railway built steep, on a clay already failing, with a lot of slips to fit, gets the most reassuring number and the least reason to believe it. The cores that agree for the wrong reason made the same warning about a tight agreement among concrete cores: a small scatter says the sample is alike, and it may be alike in the wrong respect. And since a design strength comes with its own statistical apparatus, the strength that belongs to the test programme is a reminder that the number comes from the programme that produced it and not only from the material; here the programme is the slopes that were built. The strength three cores revise goes the other way, showing what a few direct measurements can add to a prior belief, and a handful of piezometers would have that effect here on the one number the slips cannot give.

By hand #

The base inventory’s line, again. A mean slipped angle of 18° gives $\tan\beta/(1 - r_u) = 0.325/0.7 = 0.464$, or 24.9°, against a fitted 24.1°: the fit’s gradient falls a little short of the mean ray because the points’ spread along $x$ is largest at the steeper slopes, and a small intercept takes up the difference. $(1 - 0.3)\tan 24.1° = 0.312$, and $(1 - 0.6)\tan 38.0° = 0.312$.

And the boundary, directly. A cutting at the middle of the site’s heights, 8.5 m, has its plane at 2.55 m, so $\gamma z = 51$ kPa. Its clay is at the limit at $r_u = 0.3$ when $\tan\beta\cos^2\!\beta \cdot 51 = 4 + 51\cos^2\!\beta \times 0.7 \times 0.404$, which holds at about $\beta = 20°$: $\sin\beta\cos\beta = 0.321$ gives 16.4 kPa, and the right side is $4 + 45.0 \times 0.283 = 16.7$ kPa. The strength 4 + 0.404 × 31.5 at its normal stress of 31.5 kPa is 16.7 kPa, and the fell-or-stood line at that stress reads 17.0. Its neighbours’ cuttings at 18° have stood, as they should.

What the plane leaves out #

An infinite slope. Real slips are curved and the back-analysis would be Bishop’s, and the essays before this one use it. A circular slip’s $\tau/\sigma'$ does not reduce to a single ray quite as neatly, since the angle at the base varies around the circle and the depth is chosen by the search. The effect is a matter of degree: a circle’s mean stresses still follow mainly from its slope’s angle and from the same assumed $r_u$. This essay uses the plane because it makes the mechanism visible, and it should not be taken for a calculation of a real site.

A stationary population. A real inventory has cuttings of different ages and with different histories of softening, so that a first-time slide at 100 years is in weaker clay than one at 20, and the true strength drifts with age. The fell-or-stood reading could carry age as another coordinate, and the slip plot could not.

A known geometry. Records of a slipped cutting’s depth are usually poor, and a slope’s angle before it fell often worse, and the depth share here is taken as known. Noise in the angle moves a point along neither of its axes independently, so the rays smear.

A logistic curve. The line between fell and stood is an assumption as well. The scatter here is well behaved because it was built that way, and the 50 per cent contour at the working stress benefits from that. On a real site, standing cuttings are not chosen at random; many were flattened after trouble, or are on better ground than their neighbours for reasons not recorded.

Still open: a cutting that has not slipped yet #

Everything here reads a site backwards, from what is known to have happened. An owner’s real interest is forwards. Which of the cuttings still standing is nearest to going, in the next wet winter? A fell-or-stood line gives every standing cutting a probability, and it is a statement about the site’s scatter rather than about that cutting, whose own clay and pore pressure are in neither coordinate. Whether a piezometer in the few cuttings just below the line would tell an owner more than the same money spent on regrading the worst of them, and whether the line’s own uncertainty is small enough for that distinction to matter, is what a hundred slips would be asked once somebody had paid to read them in both directions.