Culmann Planar Wedge Factor of Safety

Karl Culmann published this in 1866, which makes it the oldest slope stability calculation still in routine use, and it is the right tool for a case the infinite-slope equations cannot touch: a steep cut, where the face is short relative to its height and the failure surface cannot possibly be parallel to the ground.

The picture is a triangle. A cut of height $H$ stands at face angle $\beta$. A straight failure plane runs from the toe up to the crest at some flatter angle $\theta$, isolating a rigid wedge. Weigh the wedge, resolve its weight onto the plane, put Mohr-Coulomb strength on the plane, and divide. The wedge weight per unit width is $W = \tfrac{1}{2}\gamma H^{2}\sin(\beta-\theta)/(\sin\beta\sin\theta)$ and the plane is $H/\sin\theta$ long, and after the algebra settles the whole thing collapses to two terms:

$FS = \dfrac{2c'\sin\beta}{\gamma H\sin(\beta-\theta)\sin\theta} + \dfrac{\tan\phi'}{\tan\theta}$

Look at the second term. It is $\tan\phi'/\tan\theta$ — the dry infinite-slope answer, with the plane angle standing in for the ground angle. That is not a coincidence: friction never cared about the size of the block, so the entire height and unit weight of the cut sit in the cohesion term and nowhere else. Set $c' = 0$ and the height cancels completely, which is the same result as before said in a new setting: a cohesionless cut has no critical height, only a critical angle. Cohesion is the only thing that makes a critical height exist at all, and that is exactly why a clay bank stands vertically to some depth and a sand bank does not stand vertically at any depth.

The answer is for one trial plane and is not yet the factor of safety of the slope. This is the step most often skipped. Culmann's method requires you to minimise over $\theta$, and the governing value is the lowest one you find. The classical critical plane for the $FS = 1$ case sits at $\theta_{cr} = (\beta + \phi')/2$, neatly bisecting face angle and friction angle, and that is the place to start searching — but a single trial at a convenient angle gives an upper bound on the true factor of safety, which is the dangerous direction to be wrong in. Run a range of angles either side and take the worst.

Minimised, and with $FS = 1$, the same relation rearranges into Culmann's critical height $H_{cr} = (4c'/\gamma)\left[\sin\beta\cos\phi'/(1-\cos(\beta-\phi'))\right]$, which is the form most textbooks print.

What it assumes, and how real cuts actually fail. One straight plane through the toe, a perfectly rigid wedge, homogeneous soil, no water, and no tension crack. Real steep cuts fail on curved surfaces, which is why the planar assumption is known to be slightly unconservative for shallow slopes and reasonable for steep ones. A tension crack at the crest changes things twice over: it removes the cohesion along its length, and if it fills with water after a storm it adds a hydrostatic thrust pushing the wedge out. Both effects are absent here and both make matters worse.

The other absence is time. Temporary cuts in stiff clay routinely stand for weeks on undrained strength and then fail months later as negative pore pressures dissipate and the effective cohesion drains away — the classical delayed failure of a cutting. A cut standing on back-figured cohesion is standing on the least durable parameter in soil mechanics. And excavation depth limits in occupational health and safety regulations are law; this arithmetic is not.