Infinite Slope Factor of Safety, Dry Cohesionless Soil

This is the cleanest result in soil mechanics, and it is worth pausing on how much falls out of it. Take a long slope of dry cohesionless soil — sand, gravel, dry colluvium — and imagine a failure surface running parallel to the ground at some depth $z$. Cut out a slice of that sliding layer. Gravity pulls the slice down with a weight $W = \gamma z$ per unit of plan area. Resolve that weight onto the failure plane: the driving component along the slope is $\gamma z\sin\beta\cos\beta$ and the normal component pressing the slice onto the plane is $\gamma z\cos^{2}\beta$. Friction resists with $\gamma z\cos^{2}\beta\tan\phi'$. Divide resistance by demand and everything except two tangents cancels:

$FS = \dfrac{\gamma z\cos^{2}\beta\tan\phi'}{\gamma z\sin\beta\cos\beta} = \dfrac{\tan\phi'}{\tan\beta}$

Say the consequence out loud, because students routinely fail to believe it. Neither the depth of the failure surface nor the unit weight of the soil appears in the answer. They are both in the numerator and both in the denominator and they cancel exactly. A dry sand slope therefore stands at its friction angle regardless of how big the pile is — ankle-high or the side of a quarry, the stable angle is the same. Making the pile bigger adds weight to the sliding block, but that same extra weight presses the block harder onto the surface it would slide on, and friction grows in exact step. This is why a sandpile always has the same shape, why a conveyor stockpile of one material always makes the same cone, and why the angle of repose is a material property rather than a size effect.

Set $FS = 1$ and the equation says $\beta = \phi'$. That is the angle of repose, arrived at from a completely different direction than the block-on-an-incline problem in first-year physics — and the same answer, because it is the same physics with the block made of the surface itself.

Now the warnings, which matter more than the algebra. $FS = 1.0$ is not "safe". It is the definition of impending failure: the resisting force exactly equals the driving force, and the slope is on the point of moving. Design targets for permanent slopes commonly sit at 1.3 to 1.5, and the reason for the margin is not that the division is doubtful — it is a division — but that $\phi'$ is uncertain. A friction angle from a correlation is easily two or three degrees off, and at ordinary slope angles two degrees moves the factor of safety by roughly a tenth. The margin is carried against the inputs, not against the arithmetic.

And the mechanism is a real restriction. The infinite-slope analysis assumes the failure surface is planar, parallel to the ground, and long compared with its depth, so that the forces on the two ends of the slice cancel and drop out. That describes a shallow translational slide in a weathered mantle over rock extremely well. It describes a deep rotational failure in a homogeneous clay not at all — for that you need Bishop, Janbu or Morgenstern-Price, none of which is a closed form. Applying this page to a rotational problem produces a confident number about the wrong mechanism, which is the most dangerous kind of wrong.

This site is a teaching and checking aid. Any slope that will be built, cut, or lived below is designed to the governing geotechnical code, on strengths from a site investigation, by a qualified practitioner who has seen the ground.