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Cut slope design: quartzite N55W/50

Short answer

Plane failure along the discontinuity (50/215) is possible because it daylights in the 70° slope and dips above the 28° friction angle. Design the cut at not more than 50°, say 45°, striking parallel to the joint (N55°W). The existing 70° face should be cut back or reinforced with bolts and drainage.

An outcrop of quartzite containing discontinuity has orientation N55°W/50° in the rock slope, which has attitude N45°W/70°. If the planes of discontinuities have an average friction angle of 28°, design a cut slope for a stable road with possible mode of failure.

Method (lower-hemisphere stereonet, equal-angle): plot the great circle of the slope face and of each discontinuity, draw the friction circle of radius ϕ\phi, and mark intersections of great circles. Then apply the kinematic tests below.

Data and assumptions

Rock slope N45°W/70° dipping SW = 70/225; discontinuity N55°W/50° dipping SW = 50/215; ϕ=28∘\phi = 28^\circ.

Possible failure mode

  • Dip directions differ by 10∘10^\circ (< 20°).
  • ψp=50∘<ψf=70∘\psi_p = 50^\circ < \psi_f = 70^\circ: the plane daylights.
  • ψp=50∘>ϕ=28∘\psi_p = 50^\circ > \phi = 28^\circ: sliding is possible.

So the existing slope is liable to plane failure along the discontinuity.

Design

The plane is not exposed in the cut face if the cut angle does not exceed the dip of the plane:

ψcut≤50∘\psi_{cut} \leq 50^\circ

Answer: design the road cut at not more than 50° (say 45° for a margin of safety), striking parallel to the joint plane (N55°W). The existing 70° face should be cut back by about 20° or reinforced with bolts and drainage.

✓ Answer checked by Claude Opus 5.5 (high effort)

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