Structural Engineering & Statics
where the load goes
Series

Slope stability — the series

3 essays on one idea, from the one that introduces it to the one that assumes the rest.
  1. The circle is searched for, and the first guess is 39 per cent optimistic. The same slope with 81 trial circles evaluated, each one through the toe and each one giving its own factor of safety. There is no equation whose solution is the answer: the slip surface is a shape the ground chooses, so the calculation is a search over shapes and the answer is the smallest number found — 1.191 against 1.650 for the circle a first guess puts through the toe from above the middle of the slope, which is 39 per cent optimistic. A slope analysis that reports one circle has reported nothing.

    The surface that has to be searched for

    Every other check in this collection is made at a section somebody drew. A slope has no section — the failure surface is a shape the ground chooses, so the calculation is a search over shapes, and the answer is the smallest number found rather than the solution of anything.

    part 1 · equilibrium
  2. Every one of these soils failed that slope. For a 10 m slope at 35° in soil of 19 kN/m³ with a pore pressure ratio of 0.25, over a firm stratum 5 m below its toe, which has slipped: the pairs of effective cohesion and friction angle that put its critical circle at a factor of safety of exactly one, by Bishop's method. At no friction the cohesion is 32.1 kPa; at 20° it is 12.4; at 40°, 2.0; and the line reaches no cohesion at 48.1°, where a skin of soil slides parallel to the face. Labelled, the depth of each pair's slip below the original ground: 13.6 m at 0°, 8.8 m at 5°, 7.4 m at 10°, 5.2 m at 20°, 3.6 m at 30°, 2.3 m at 40°.

    A failed slope gives a line, not a soil

    A slope that has slipped is the one full-scale strength test geotechnics ever gets, and the standard way to read it is to set its factor of safety to one and solve for the soil. One equation cannot give two strengths. What it gives is a line of cohesions and friction angles, every one of which fails the slope exactly, and the repair designed on one point of that line can be worth a factor of 1.07 or of 1.51. The slip itself says which point: each soil on the line fails along a different surface, and the depth of the scar is the second equation.

    part 2 · equilibrium
  3. Each slip is a shear test at one normal stress. The soil's strength, τ = 8 + σ′·tan 24° (line), and the mean effective normal stress and shear stress along the critical slip of each of five slopes that fail in it: 55° and 3.8 m high, 8 kPa; 45° and 5.2 m high, 13 kPa; 35° and 8.2 m high, 23 kPa; 28° and 14.1 m high, 42 kPa; 22° and 41.5 m high, 109 kPa. Every slip is one point on the line. Two slips fix the line as two shear-box tests do, and fix it well only if their normal stresses are far apart: the steep, short slopes test the soil where cohesion is most of its strength, the flat, tall ones where friction is.

    Two slips are a shear test

    One failed slope gives a line of soils, every one of which fits the failure. Two failed slopes give two lines, and where they cross is a soil that fits both. Whether the crossing means anything depends on how differently the two slopes loaded the ground: two slips at one angle cross only at zero cohesion, two at nearby angles cross anywhere in a twenty-degree band, and a flat, tall slope beside a steep, short one pins the soil to within two degrees — because each slip is a shear test at one normal stress, and a strength line needs two tests far apart.

    part 3 · equilibrium

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