Two slips are a shear test

A slope that has slipped is the one full-scale strength test the ground ever gives, and read backwards — its factor of safety set to one and the equation solved for the soil — it gives not a soil but a line of them: every pair of cohesion $c'$ and friction angle $\phi'$ on the line fails the slope exactly. That essay found the second equation in the scar itself. A cohesive soil fails along a deep circle and a frictional one along a shallow surface near the face, so the depth of the slip picks one point off the line, and a metre of error in the depth spreads the answer from 16° to 27°.

It ended on a second route. A slope that has failed twice — at two points along one cutting, or once before a regrade and once after — gives two lines, and where they cross is a soil that fits both failures with no appeal to the depth. Railway and highway authorities hold inventories of exactly this: a century of slips in cuttings through the same geology, each one back-analysed alone. The question is what a pair of them is worth together.

A soil, and two slopes that fail in it #

To have an answer to check against, start from the soil. Take a soil of 19 kN/m³ with an effective cohesion of 8 kPa, a friction angle of 24° and a pore pressure that is a quarter of the overburden everywhere, $r_u = 0.25$. Cut a face at 35° into it and it stands until it is 8.2 m high; cut one at 28° and it stands to 14.1 m. Those are the two slopes. Each is the height at which its face, searched over every circle by Bishop’s method, reaches a factor of safety of exactly one, with a firm stratum half the slope’s height below its toe.

Now forget the soil and back-analyse each slope as an engineer would, knowing only its geometry, its pore pressure and the fact that it moved. Each gives a line.

They cross at 24.1° and 8.0 kPa, which is the soil. Two failures do carry two equations, and with nothing else known the pair has found what one slip could not.

The patch around the crossing #

The trouble is visible in the same figure: the two lines cross at a shallow angle. Near the crossing the 35° slope’s line falls by about half a kilopascal for every degree of friction and the 28° slope’s by about 0.9, so a small movement of either line slides the crossing a long way along them.

And each line does move, because neither slope is known to have failed at a factor of exactly one. A slip is recorded when the ground has moved, by which time it has already lost part of its strength along the surface — friction supplies what equilibrium asks right up to the moment it cannot, and nobody measures that moment; the slope was three-dimensional and the analysis is not; the pore pressure on the day was not the design value. A doubt of five per cent in either factor is modest.

That doubt costs nothing to compute, for a reason worth knowing. Bishop’s method contains the strength only as $c'/F$ and $\tan\phi'/F$, so the soil that gives a slope a factor of $1.05$ is the soil that gives it one, with both its cohesion and the tangent of its friction angle multiplied by $1.05$. A line at $F = 1.05$ is the $F = 1$ line stretched outward from the origin — no new search is needed. With each slope allowed anywhere between 0.95 and 1.05, the four extreme crossings put the friction angle anywhere from 18° to 30° and the cohesion from 4.4 to 12.4 kPa.

That is no better than the single slip with its depth measured to a metre. The second failure has replaced one uncertain equation with another.

At one angle, the lines never cross #

It can be worse, and the commonest case is the worst.

A cutting is usually dug to one standard angle along its length and varies in height with the ground. Back-analyse slips at three heights on one angle and the three lines are the same line stretched: at every friction angle the cohesion each requires is in proportion to the slope’s height — 6.1, 10.1 and 15.1 kPa at 20° for slopes of 4.9, 8.2 and 12.3 m. That is exact, and it follows from dimensions. For a fixed angle, pore pressure ratio and shape, the factor of safety depends on the cohesion only through the stability number $c'/\gamma H$, so doubling the height and doubling the cohesion leaves the slope exactly as safe.

Proportional lines meet only where all of them reach zero cohesion, at the friction angle on which an infinitely long slope of that angle just stands — here 48.1°. So slips at several heights on one angle can be consistent with one soil only if that soil has no cohesion at all. If they are all genuinely at failure, that is what they say; if the soil has cohesion, they cannot all have been at failure, and the inventory is telling the analyst which slips to distrust rather than what the soil is.

This is not only a statement about models. Skempton’s back-analyses of first-time slips in the cuttings of the London Clay, slopes of similar angle and many heights, found a cohesion of about a kilopascal — near enough nothing that the softened clay is now designed as if it had none. The geometry of the inventory pointed there before any interpretation did.

Each slip is a shear test #

The way to see what makes two failures informative is to stop looking at the lines and look at what each failure did to the ground.

Along a slip at the moment of failure every slice of soil is on its strength line, $\tau = c' + \sigma'\tan\phi'$. Average over the slip’s length and the average shear stress equals the cohesion plus the average effective normal stress times $\tan\phi'$ — exactly, because the law is linear. A failure is therefore one point on the soil’s Mohr–Coulomb line, at the slip’s mean normal stress, which is exactly what one shear-box test or one triaxial test gives — the same line that the circle nobody draws is tangent to, sampled at one place.

And the slopes that fail in one soil test it at very different stresses. A face at 55° fails when it is only 3.8 m high, along a short slip at a mean effective normal stress of 8 kPa, where the 8 kPa of cohesion is half of the strength. A face at 22° stands until it is 41.5 m high, and its slip runs at 109 kPa, where cohesion is a seventh. Steep, short slopes measure the cohesion; flat, tall ones measure the friction.

That is the whole of the conditioning. A strength line found from two shear tests is well determined when the two normal stresses are far apart and badly when they are close — no laboratory would fit a line to two tests at 23 and 42 kPa and believe its slope to a few degrees, and a strength that belongs to the test programme is exactly the kind a pair of nearby tests produces. The 35° and 28° slopes are those two tests. Two slopes at the same angle are the same test run twice at different scales, which is why their lines never cross: the stability number makes them the same normal stress, relative to the cohesion, and a line cannot be fitted to one point.

How different the second slope has to be #

Hold the 35° slope and vary the second. Close to it on either side the crossing is useless: a 31° or a 40° slope admits anything from 11° to 37°. Moving the second slope further away improves things on both sides, but unevenly. Steeper second slopes improve slowly — at 55° the band is still ten degrees wide — because the steep slope’s slip runs at 8 kPa against the first slope’s 23, and both are at the low-stress end. Flatter second slopes improve fast, because their failure height grows rapidly as the face approaches the angle at which the soil’s friction alone would hold it: at 25° the second slope is 21 m high and the band is seven degrees, and at 22° it is 41.5 m high and the band is four degrees, 22° to 26°.

So the second failure worth having is not just a different one but a much flatter, much taller one. In a cutting inventory that is rare — the tall slips of a line are usually in its deepest cuttings, which were dug at the same standard angle — and it is the reason natural landslides on long, gentle valley sides, which fail at heights of tens of metres and normal stresses of a hundred kilopascals, are the field evidence for a clay’s friction angle, and short, steep cuttings for its cohesion.

Both scars, to one scale #

The crossing used no depth, so the scars become a third equation, and with three equations for two unknowns the problem is over-determined: it can now be wrong. At the crossing soil the 35° slope should have slipped 3.6 m deep and the 28° one 5.6 m. If the two scars measured in the field were both much deeper than that, no single homogeneous soil fits both slips and their depths together, and the likeliest explanation is that the soil is not homogeneous — a softer layer at depth, or a pre-existing shear surface — which is a finding about the site that neither slip could have produced alone.

What the crossing is wanted for #

The reason anyone back-analyses a slip is to design the repair, and here the patch around the crossing turns out to matter much less than it looks. Regrade the 35° slope to 25°. On the true soil it has a factor of safety of 1.32. On the four corner soils of the patch — from 18° with 12.4 kPa to 30° with 4.4 — it has 1.26, 1.32, 1.34 and 1.39.

The two corners where the slopes err in opposite directions, which are the ones that throw the friction angle furthest, give 1.32 and 1.34: almost exactly the true answer. The spread comes only from the two corners where both slopes err the same way, and those are simply the whole line scaled by five per cent, which scales the regraded slope’s factor by five per cent with it. The soil is badly determined and the repair is not, because a regraded slope at 25° tests the soil at normal stresses between those of the two slopes that failed, and anywhere along the patch the soils agree about that range.

Compare the single slip. Along the 35° slope’s line alone, the same regrade has a factor anywhere from 1.10 to 1.59 depending on which soil is picked — the problem the earlier essay ended on. The second failure has cut that to a band set by the doubt in the failures themselves. The crossing is worth much more as a prediction than as a soil, and the prediction is good exactly where it interpolates between the two failures: a repair that took the ground somewhere neither slip tested — a much taller cutting, a much flatter face — would be an extrapolation along the badly determined direction.

A wetter winter #

Every number above assumes both slopes failed at the pore pressure they are analysed with, and that is the assumption most likely to be wrong. Suppose the 28° slope failed in a wet winter at $r_u = 0.35$ — so it gave way at 9.9 m rather than 14.1 — and is back-analysed at 0.25 like its neighbour. Its line moves down, because a slope that failed while wet looks weaker than its soil when analysed dry, and the crossing moves to 11.1° and 15.7 kPa: the friction halved and the cohesion doubled.

The error is large because it is systematic in one failure and not the other, which is precisely the kind of error the crossing is most sensitive to. A five per cent doubt in each factor spread the answer over twelve degrees; a pore pressure misread by a tenth of the overburden in one slope moves it thirteen degrees in one direction. Two slips from one season, at one pore pressure, are worth more than two slips that bracket a wet winter, unless the pore pressures of both are known.

By hand: the stretch, and the shear test #

Two pieces of arithmetic carry most of this, and both can be checked without a computer.

The stretch: Bishop’s factor for a slope in a soil $(c', \phi')$ is the $F$ that solves $F = \sum [c' b + (W - ub)\tan\phi'] / [m_\alpha \sum W\sin\alpha]$ with $m_\alpha = \cos\alpha + \sin\alpha\tan\phi'/F$. Replace $c'$ by $f c'$ and $\tan\phi'$ by $f\tan\phi'$ and every term is unchanged if $F$ is replaced by $fF$. So the soil with $c' = 1.05 \times 10.1$ kPa and $\phi' = \tan^{-1}(1.05 \tan 20^\circ)$ — 10.6 kPa and 20.9° — gives the 35° slope a factor of 1.05, which a direct search confirms to three figures.

The shear test: the 28° slope’s slip at the crossing soil averages 42 kPa of effective normal stress, so its mean shear stress at failure is $8 + 42 \tan 24° = 26.7$ kPa; the 35° slope’s averages 23 kPa and $8 + 23 \tan 24° = 18.2$. The line through those two points has a slope of $8.5/19 = 0.45$, which is $\tan 24°$, and an intercept of 8 — the soil. A five per cent error in each mean shear stress, in opposite directions, changes the rise between those two points from 8.5 kPa to 6.3 or 10.7 — a line through them at 18° or at 29° — which is the patch, recovered from a shear-box calculation.

A homogeneous soil, circles, and one pore pressure #

The soil is the same everywhere in both slopes, with one cohesion and one friction angle. Real cuttings cross layers, and a back-analysis of a layered slope has a line for every layer’s parameters, which is exactly the dimension the crossing cannot see.

The slips are circles. Bishop’s method searches circles, as the mechanism under a footing is searched for a wedge; many real slips are non-circular, following a weak layer or the base of a weathered zone, and their back-analysis needs a non-circular method whose lines can differ by a few per cent from the circular ones — the same order as the doubt allowed here.

The pore pressure is a fixed fraction of the overburden. A ratio $r_u$ is a convenience; in a real slope the pore pressure follows a phreatic surface, and in a cutting it rises slowly over decades as the clay swells after excavation, which is why first-time slips in cuttings often come thirty or fifty years after the digging.

Progressive failure, softening and the ends of a slip #

They cannot show progressive failure. A stiff clay slope does not mobilise its peak strength all along its slip at once; the toe softens first, and by the time the slip moves the mean strength along it is somewhere between peak and residual. A back-analysis measures that mean, and two slips of different lengths average it differently — another reason their lines need not cross where any laboratory test would place the soil.

They cannot show time. The cohesion of an overconsolidated clay falls as it softens after unloading — the reverse of the ground that gains strength under a load — so two slips decades apart in the same cutting may be testing two different soils. The crossing then mixes them.

And they cannot show the three-dimensional slip. A real slip has ends, where it shears through soil on its sides, and its resistance is a few per cent larger than a two-dimensional section assumes. That bias is in the same direction for every slip, so it scales both lines together — which, by the stretch above, moves the crossing along the direction the regrade does not care about.

Two tests, far apart #

Two failures in one soil cross at that soil. A 35° slope 8.2 m high and a 28° slope 14.1 m high, back-analysed, cross at 24.1° and 8.0 kPa.

Each slip is one shear test, at the mean effective normal stress on its surface, and two tests fix a line only if their stresses are far apart. Slopes at one angle are one test at different scales; they cross only at zero cohesion.

The doubt is large unless the slopes are very different. Five per cent in each factor puts the 35° and 28° crossing anywhere from 18° to 30°; a 22° second slope, 41.5 m high, narrows it to four degrees.

The repair is better determined than the soil. A 25° regrade has a factor of 1.26 to 1.39 across the whole patch, against 1.10 to 1.59 along a single slip’s line. And a second slip that failed in a wetter winter, analysed as if it had not, moves the crossing to 11° and 15.7 kPa.

Still open: a hundred slips, and which of them failed #

Two slips give a crossing; an inventory gives a hundred lines, and a hundred lines do not meet at a point. Fitting a soil to all of them is a regression — and a design strength taken from it is a property of the scatter, not of any one slip — and a regression needs to know which slips were genuinely at a factor of one and which were helped over the edge by a burst main, a blocked drain or a heavy train — errors that are not five per cent and not random. Whether a cutting inventory can be read as a scatter of shear tests, with the outliers found by the same geometry that made two slips at one angle useless, and whether the friction angle that survives is closer to the peak or to the residual strength of the clay, is the question a century of recorded slips puts to any single back-analysis.