Infinite Slope Factor of Safety with Cohesion and Pore Pressure

This is the general infinite-slope equation, and both of the simpler pages are special cases of it. Mohr-Coulomb says the shear strength available on a plane is $\tau_f = c' + \sigma'\tan\phi'$. On a slope-parallel plane at depth $z$, the total normal stress is $\gamma z\cos^{2}\beta$ and the pore pressure is $u$, so the effective normal stress is $\gamma z\cos^{2}\beta - u$. The shear stress the plane actually has to carry is $\gamma z\sin\beta\cos\beta$. Divide strength by demand and you have the whole page.

The important structural fact is that $z$ no longer cancels. In the cohesionless case, both the resisting and driving terms grew in proportion to depth and the ratio was depth-free. Here the cohesion $c'$ is a constant that does not grow with depth at all, while the driving stress and the friction term both do. So the factor of safety falls as the failure surface goes deeper, and the critical plane is the deepest one the soil actually extends to — normally a soil-rock contact, a relict shear surface, or the base of a weathered mantle. Run the calculation at that depth. Choosing a depth for convenience, or because it is where the borehole sample came from, produces a number for a plane that is not the one at risk.

Set $FS = 1$ and solve for depth and you get the classical critical depth of a cohesive infinite slope, which is why a shallow soil mantle can stand on a slope that a thicker mantle of the same material could not. Forestry and highway engineers meet this constantly: the same hillside is stable where the till is thin and unstable in the hollows where it collects.

Be suspicious of a large $c'$. Effective cohesion is small in real soils — a few kPa in a stiff clay and genuinely zero in a clean sand — and a good deal of what a laboratory reports as $c'$ is curvature in the strength envelope being forced through a straight line over a stress range that does not include your slope. Some of it is apparent cohesion from partial saturation, which vanishes when the soil wets. Some of it is root reinforcement, which is real but temporary: root cohesion in a clear-felled block decays over roughly three to fifteen years as the old roots rot before new ones establish, and that decay window is exactly when the shallow slides arrive. A great many landslides on logged ground happen a decade after the trees left rather than the winter after, and this equation is where you can see why.

The pore pressure is the other soft input, and the useful thing to do is invert for it. Solve for $u$ at $FS = 1.0$ and you get the triggering pore pressure — the value at which this plane moves. Then compare it with what the ground can physically deliver. Under steady slope-parallel seepage with the water table at the surface the ceiling is $u = \gamma_w z\cos^{2}\beta$, a little under half the total normal stress. If the triggering value sits above that ceiling, rainfall alone will not do it and something else — a toe excavation, a fill surcharge, an earthquake — has to be involved. If it sits below, the slope is rainfall-triggerable, and the question stops being geotechnical arithmetic and becomes hydrology: how often does that pore pressure occur, and can you measure it?

Skempton and DeLory's 1957 work on the London Clay slopes is the classic demonstration that the answer to a slope problem is very often the pore pressure and not the strength. FS = 1.0 remains the definition of impending failure and not a target. This page is a teaching aid; a real slope is designed to the governing code, on a site investigation, by a qualified practitioner.