--- title: "Cut slope design: limestone N48E/52" subject: "Engineering Geology II" chapter: 6 key: "semester_4/engineering_geology_ii:cut-slope-limestone-n48e-52" source: https://www.bctengineer.com/q/semester_4/engineering_geology_ii/cut-slope-limestone-n48e-52 site: BCT Engineering --- # Cut slope design: limestone N48E/52 Engineering Geology II · Chapter 6: Measurement, analysis and interpretation of structural geological data - Asked in: 2079 Chaitra, 5 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/engineering_geology_ii/cut-slope-limestone-n48e-52 - Whole chapter: https://www.bctengineer.com/md/semester_4/engineering_geology_ii/chapter-6 ## Question An outcrop of limestone containing discontinuity has an orientation of N48°E/52° in the natural slope and has an attitude of N60°E/75°. The friction angle of the plane of discontinuity is 30°. Design a cut slope for a stable road with the possible mode of failure. ## Short answer Assuming the discontinuity is the 52° plane and the natural slope is 75°, the discontinuity (52/138) is nearly parallel to the slope, daylights and exceeds the 30° friction angle, so plane failure is possible. Design the cut no steeper than 52°, preferably 45–50°, striking parallel to the discontinuity. Alternatively support with bolts, shotcrete, drainage and benches. ## Answer **Method (lower-hemisphere stereonet, equal-angle):** plot the great circle of the slope face and of each discontinuity, draw the friction circle of radius $\phi$, and mark intersections of great circles. Then apply the kinematic tests below. #### Data and assumptions Read in the same way as similar exam questions: the discontinuity is N48°E/52° dipping SE = **52/138**, and the natural slope is N60°E/75° dipping SE = **75/150**; $\phi = 30^\circ$. #### Possible failure mode - The dip directions differ by $12^\circ$ (< 20°), so the plane is nearly parallel to the slope. - $\psi_p = 52^\circ < \psi_f = 75^\circ$: the plane **daylights** in the slope face. - $\psi_p = 52^\circ > \phi = 30^\circ$: the block can slide under gravity. All the conditions are met, so the possible mode of failure is **plane failure** along the discontinuity. #### Design of the cut slope The plane is not exposed in the cut face if the cut angle does not exceed the dip of the discontinuity: $$ \psi_{cut} \leq \psi_p = 52^\circ $$ **Answer: make the road cut at not more than 52° (say about 45° to 50° for a margin of safety), with the face striking parallel to the discontinuity (about N48°E to N60°E). If a steeper face is unavoidable, use rock bolts, shotcrete, drainage and benches.** (If the figures are read the other way round, with a 52° slope and a 75° joint, the joint does not daylight and the natural slope is stable; plane failure would then occur only if the cut were made steeper than 75°, so a cut of up to about 70° would be safe.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The slope (52/138) and the discontinuity (75/150) differ in dip direction by 12°. What does this tell us? A. The plane is within about 20° of the slope direction, so plane failure is kinematically possible B. The plane is perpendicular to the slope, so wedge failure occurs C. The plane dips into the hill, so toppling occurs D. The plane is too far from the slope direction for any sliding Answer: A. A dip-direction difference below about 20° means the plane is nearly parallel to the face, which is the first condition for plane failure. (Key point: Plane failure condition on dip direction) 2. Why is the natural slope (52°) stable against the 75° discontinuity? A. Because the plane dips less than the friction angle B. Because the plane dips more steeply than the slope, so it does not daylight C. Because the slope is steeper than the plane, so the plane daylights but is locked D. Because the strikes differ by 12° Answer: B. A plane steeper than the slope face does not emerge (daylight) in the face, so no block can slide out. (Key point: Daylighting in the natural slope) 3. What limit should the road cut angle satisfy? A. Cut angle greater than 75° B. Cut angle equal to the friction angle of 30° exactly C. Cut angle less than 75° (the dip of the discontinuity), face striking parallel to the joint D. Cut angle greater than 52°, the natural slope Answer: C. The cut must be flatter than the plane's dip so the plane is not exposed; the answer suggests about 70°. (Key point: Design rule for the cut angle) 4. If the cut were steeper than 75° so the plane daylights, why would it slide? A. Because 75° is smaller than the friction angle B. Because the dip direction difference exceeds 20° C. Because limestone is always weak D. Because the plane dip (75°) exceeds the friction angle (30°) Answer: D. Sliding along a daylighting plane needs its dip to exceed the friction angle; 75° > 30°. 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