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Conditions for plane failure

Short answer

Plane failure occurs when a discontinuity strikes nearly parallel to the slope face (±20°), dips out of the face at an angle less than the slope but greater than the friction angle, and is bounded laterally by release surfaces or a tension crack. On a stereonet, the plane's pole lies inside the daylight envelope and outside the friction circle.

What are the conditions for plane failure of a rock slope?

Plane failure is the sliding of a rock block along a single plane of weakness (joint, bedding or foliation) that daylights in the slope face. According to Hoek and Bray, it occurs when all these conditions are satisfied:

  1. Strike: the plane strikes parallel, or nearly parallel (within about ±20∘\pm 20^\circ), to the slope face. That is, the dip directions of the plane and the slope differ by no more than about 20∘20^\circ.
  2. Daylighting: the plane dips less steeply than the slope face, ψp<ψf\psi_p < \psi_f, so that the plane is exposed in the face.
  3. Friction: the dip of the plane is greater than its friction angle, ψp>ϕ\psi_p > \phi, so that the block can slide under gravity.
  4. Release surfaces: the sliding block must be free at its sides, either through lateral release joints of negligible resistance or because the plane passes through a convex nose of the slope; a tension crack usually bounds it at the top.
   crest   tension crack
      \     |
       \    |__
   face \     \ \   sliding plane
         \     \ \ (dip psi_p)
          \_____\_\___

Stereonet test: on a great-circle plot, the dip vector of the plane lies between the slope great circle and the friction circle, within about ±20∘\pm 20^\circ of the slope dip direction; equivalently, on a pole plot, the pole lies inside the daylight envelope and outside the friction circle. If ψp≤ϕ\psi_p \leq \phi, the slope is stable against plane sliding.

✓ Answer checked by Claude Opus 5.5 (high effort)

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