A slope is a structure nobody designed. Gravity pulls a wedge of soil down; the soil's own shear strength holds it back. The whole subject is one ratio — and understanding why rain collapses that ratio explains most of the landslides that kill people.
Every other structure in this course was designed by someone. A slope wasn't. It's just soil sitting at an angle, and the only question that matters is whether the shear strength along some potential sliding surface is bigger than the shear stress gravity is applying along it.
Slope stability is resisting shear strength ÷ driving shear stress along a failure surface. Everything else in this chapter is just different ways of computing those two numbers for different failure-surface shapes.
FS = 1 means the soil is using every bit of strength it has just to stand still — failure is imminent. FS = 1.5 means it's only using two-thirds of its strength. Design targets are typically 1.3–1.5 for permanent slopes.
The soil doesn't use all its strength unless it has to. The portion actually in play is the mobilized strength τd. You can define a separate factor for cohesion and for friction; a true factor of safety is the value where both are reduced equally.
When FSc' = FSφ', that common value is the factor of safety FSs.
"Infinite" means the slope is long and uniform, and the failure surface runs parallel to the ground at some shallow depth H. That sounds artificial, but it's an excellent model for real shallow slides — a thin mantle of weathered soil sliding on stronger rock beneath, which is exactly what most rainfall-triggered hillside failures are.
Because every slice of an infinite slope looks identical to every other, the forces on the sides of a slice cancel. That's what makes it the one case you can solve exactly, in one line, with no iteration.
Two independent contributions: a cohesion term that shrinks as the slide gets deeper, and a friction term that doesn't depend on depth at all.
The depth at which the slope fails. Note that if β ≤ φ', the denominator goes zero or negative — a cohesionless slope flatter than its friction angle is stable at any depth.
For sand (c' = 0), FS = tanφ'/tanβ — depth cancels completely. A dry sand slope is stable if and only if it's flatter than the friction angle. That's why a poured sand pile always settles to the same cone: its angle of repose is φ'.
Now saturate the same slope, with water seeping parallel to the surface. Two things change, and only one of them is obvious.
γ becomes γsat. More weight means more driving stress. But it also means more normal stress, so friction rises too — these largely cancel. This is not the main problem.
Friction depends on effective stress — grain-on-grain contact force. Pore water pressure pushes the grains apart and carries part of the load itself. Water has no shear strength, so every kilopascal it carries is a kilopascal not generating friction.
Set up a slope, then toggle seepage on and off. Watch the friction term collapse.
Real embankments and hillsides have a top and a bottom, so the failure surface curves. Observation and theory both say it's very close to a circular arc — which is convenient, because a circle lets you take moments about its centre and make every normal force on the arc disappear from the equation (they all point at the centre, so they have zero moment arm).
The arc exits exactly at the toe of the slope. The most common mode for steep slopes and reasonably uniform, frictional soil.
The arc exits partway up the slope face, taking a shallower bite. Typical where a weaker layer sits near the surface, or the slope is very steep.
The arc passes below the toe, scooping out the foundation as well. Happens when soft material extends beneath the slope — and it's the most destructive, because it takes ground well beyond the slope with it.
Before computers, searching hundreds of circles by hand was impossible. Taylor did the searching once, for every combination of slope angle and friction angle, and published the answer as a single dimensionless number.
m depends only on β and φ', read from Taylor's chart. It bundles the entire critical-circle search into one lookup.
c'/(γH) has no units — it compares the soil's cohesion against the stress its own weight generates. Two slopes with the same β, φ' and same c'/(γH) behave identically regardless of scale. That's what lets one chart cover every slope ever built.
Taylor's method answers "how tall can this slope be?" in one step. Its limits: it assumes a homogeneous soil, a circular arc, and no pore pressure. The moment groundwater matters — which, per Section 4, is when slopes actually fail — you need the method of slices.
Real slopes have layers, water tables, and irregular profiles, so the stress along the failure arc varies from one end to the other. The fix is the oldest trick in engineering: chop the problem into pieces small enough that conditions are constant within each piece.
Assume a circle, slice the mass above it, and sum. Then move the circle and do it again — the reported factor of safety is the lowest value found.
Slices push and pull on each other across their vertical faces. Those interslice forces are genuinely unknown, and how a method handles them is the only real difference between the two you'll use.
Resolve each slice perpendicular to its own base and assume neighbours don't interact. Direct, no iteration — and conservative, typically underestimating FS by 5–20%.
Assumes interslice forces are horizontal. Much more accurate — within a few percent of rigorous methods.
Notice FS appears on both sides of Bishop's equation, inside mα. There's no way to isolate it, so you guess a value (the OMS result is a good first guess), compute a new FS, and repeat until it stops changing. It typically converges in 3–4 rounds.
This runs the real analysis: it tries hundreds of trial circles, computes both OMS and Bishop for each, and shows you the critical one — the circle with the lowest factor of safety. The ru slider is the pore-pressure ratio u/(γh); slide it up to simulate a slope saturating during a storm.
Problem: A long uniform hillside has β = 22°, with a potential failure plane 4 m down on stronger rock. Soil: c' = 12 kPa, φ' = 28°, γ = 18 kN/m³. Find FS.
Problem: The slope of Example 1 becomes fully saturated with seepage parallel to the surface. γsat = 20 kN/m³. Nothing else changes. Find the new FS.
Problem: A cut is planned in saturated clay under undrained conditions (φ = 0), cu = 30 kPa, γ = 18 kN/m³, β = 45°. Taylor's chart gives m ≈ 0.17. Find the critical height, and FS for a 6 m cut.
Problem: A trial circle is divided into 5 slices, each b = 2.0 m wide (so ℓ = b/cosα). Soil: c' = 20 kPa, φ' = 20° (tanφ' = 0.364). Compute FS by the Ordinary Method, then refine with Bishop's.
| Slice | W (kN/m) | α (°) | ℓ (m) | u (kPa) | W sinα | OMS resist. |
|---|---|---|---|---|---|---|
| 1 | 60 | −12 | 2.045 | 10 | −12.47 | 54.82 |
| 2 | 140 | 2 | 2.001 | 22 | 4.89 | 74.93 |
| 3 | 180 | 16 | 2.081 | 28 | 49.61 | 83.39 |
| 4 | 150 | 31 | 2.333 | 24 | 77.25 | 73.08 |
| 5 | 70 | 47 | 2.933 | 12 | 51.20 | 63.22 |
| Σ | 170.48 | 349.44 | ||||
343 mm of rain fell in one day on Rangamati, Bandarban and Chattogram. Landslides killed 152 people, severed the Chittagong–Rangamati highway, and overturned valley-side retaining walls when their foundations lost support. Islam, Islam & Jeet (2021) collected undisturbed samples from six of the failed slopes and back-analysed them. Here is what the soil actually was:
| Site | H (m) | β | γ (kN/m³) | c (kPa) | φ (°) | USCS | FS | Failure mode |
|---|---|---|---|---|---|---|---|---|
| S-1 Manikchari | 15 | 70° | 18.54 | 1.3 | 39.7 | SM | 0.90 | Shallow wedge |
| S-2 Manikchari S. | 10 | 50° | 19.02 | 8.6 | 37.3 | SC | 1.30 | Toe general |
| S-3 Shapchari | 10 | 65° | 18.98 | 18.1 | 10.3 | CL | 0.54 | Base failure |
| S-4 Deppoyachari | 16 | 70° | 19.00 | 9.4 | 27.4 | CL-ML | 0.86 | Toe general |
| S-5 Moddhapara | 13 | 45° | 19.34 | 12.9 | 32.7 | CL | 1.28 | Shallow surface |
| S-6 Kaching | 12 | 50° | 16.80 | 1.8 | 30.2 | SM-SC | 0.82 | Toe surficial |
Slope heights, angles and properties from Tables 1–2; factors of safety from the PLAXIS 2D Mohr–Coulomb analyses in Figure 9 of the paper.
FS = 0.54, the lowest of the six, and the only base failure. Look at the numbers: φ = 10.3° is almost nothing, so the friction term is nearly absent — this slope stood on cohesion alone. Its natural moisture was 25.6% against a 27.7% saturation limit, so rain pushed it to saturation and softened the very cohesion holding it up.
Both have high φ (39.7° and 30.2°) but almost no cohesion (1.3 and 1.8 kPa). Section 3 told you a cohesionless slope is stable only while β < φ'. At β = 70° and 50°, both are far steeper than their friction angle — so with cohesion gone, failure is arithmetic.
The only two with FS > 1 are the two flattest slopes (50° and 45°), and both have meaningful cohesion and friction. S-2 also sat at 13.9% moisture against a 21.6% saturation limit — it had capacity to absorb rain before strength began dropping.
1. A dry sand slope (c' = 0) has φ' = 34° and β = 30°. What happens to FS if the potential failure surface is 2 m deeper?
2. Saturating a slope roughly halves the friction term. What is the physical reason?
3. Why must slope analysis search many trial circles rather than compute just one?
4. Why does Bishop's Simplified Method require iteration when the Ordinary Method doesn't?
5. Site S-3 at Chattogram had the highest cohesion of the six (18.1 kPa) yet the lowest FS (0.54). Why?