One problem, taken completely apart: a 12.7 m two-layer slope with seepage, and a slip circle that cuts through both soils. We'll build the geometry, cut the mass into slices, do one slice entirely by hand, then let the equation chase its own tail until it converges.
Compute the long-term factor of safety for the given failure surface using the Modified Bishop's Method.
That phrase is not decoration — it decides which strength parameters you are allowed to use.
Right after excavation, water has not had time to move. Use total stresses with φ = 0 and cu. Pore pressures are unknown, so they never appear in the equation.
Years later, pore pressures have reached steady seepage. Use effective stress with c' and φ', and you must subtract u explicitly. The problem gives c' and φ' for both layers, which is the giveaway.
"Long term" means drained, effective-stress analysis. For a cut slope this is usually the critical case: excavation unloads the ground, pore pressures start low and rise over time toward equilibrium, so the slope gets weaker as the years pass. A cut that stands fine on the day it is dug can fail a decade later.
Most students lose this problem here, before any soil mechanics happens. Put down coordinates and everything becomes arithmetic.
Toe at the origin, crest 19.05 m back and 12.7 m up, centre O floating 22.7 m above the toe line. The arc leaves the toe, dips below it, crosses into the upper layer at x = 21.85 m, and re-emerges 9.6 m behind the crest.
Mid-point x, height h, weight W, base angle α, base length ℓ, and pore pressure u. Plus which soil the base sits in.
A slice straddling the layer boundary is part 17.9 and part 18.2 kN/m³. Compute each piece separately — \( W = b(\gamma_1 h_1 + \gamma_2 h_2) \).
c' and φ' come from whichever soil the base passes through — not the soil above it. Shearing happens on the base.
Weight is about everything stacked above the base; strength is about what the base is made of. Mixing these up is the single most common slicing error. A slice can be 90% stiff clay by volume and still shear through weak sand because that is where its base lies.
\( \sin\alpha = (x_m - x_O)/R \). Slices uphill of the centre get positive α and drive the failure; slices near the toe, downhill of the centre, get negative α and resist it. In the table below slice 1 has α = −13.7° and contributes −16.6 kN/m — it is holding the slope back, and the arithmetic says so on its own.
Ten slices done ten times is not ten times the learning. Here is slice 7 from end to end; every other slice is the same six lines with different numbers.
The denominator is the easy half: it needs no iteration and no strength parameters at all. Just add up W sinα.
Multiply by R and you have the driving moment about O. Since R divides out of both sides of the factor-of-safety equation, we never actually need it.
FS appears on both sides. It is in the answer and it is buried inside mα in every single term. There is no algebra that isolates it — the equation is implicit, so it must be solved by iteration.
mα is where Bishop earns its accuracy. The Ordinary Method simply ignores the forces between slices; Bishop keeps horizontal equilibrium, and mα is exactly the correction that bookkeeping introduces. That single term is the entire difference between the two methods — and it is worth roughly 7% here.
Start from the Ordinary Method answer — it is free, and always a little low, which makes it an excellent first guess.
The whole solution, rebuilt on every keystroke. Click any slice in the drawing to see its numbers. Change the slice count and watch the answer converge; switch the water off and watch it jump.
Modified Bishop, converged, using 20+ slices. With the 10 slices tabulated above you get 1.887; refining to 20 gives 1.906. Quote it as 1.9 — the geometry is read off a drawing and the phreatic surface is approximate, so three decimals would be false precision.
Never report a factor of safety without testing it. Three checks, each taking seconds:
The Ordinary Method gives 1.79 — about 6–7% lower. Bishop should always come out higher, by roughly 5–20%. If yours came out lower, the iteration or a sign is wrong.
1.887, 1.906, 1.901. Changes in the second decimal only, so the answer has converged. If it were still marching in one direction you would need more slices.
Dry, the same slope gives 2.14. Water costs 11% of the factor of safety. It must move in that direction, and by a plausible amount.
Cut the sliding mass into strips, work out what gravity asks of each strip and what its base can supply, add both columns up, and — because the strength term secretly depends on the answer — guess, recompute, and repeat until the number stops moving.