--- title: "Cut slope design: HS 110/40, B 130/20" subject: "Engineering Geology II" chapter: 6 key: "semester_4/engineering_geology_ii:cut-slope-hs110-40" source: https://www.bctengineer.com/q/semester_4/engineering_geology_ii/cut-slope-hs110-40 site: BCT Engineering --- # Cut slope design: HS 110/40, B 130/20 Engineering Geology II · Chapter 6: Measurement, analysis and interpretation of structural geological data - Asked in: 2072 Magh, 4 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-hs110-40 - Whole chapter: https://www.bctengineer.com/md/semester_4/engineering_geology_ii/chapter-6 ## Question The attitudes of different planes are given below. HS = 110°/40°; B = 130°/20°; J1 = 100°/40°; J2 = 200°/50°, ϕ = 32°. Design the cut slope inclination to be stable for the given discontinuities from different types of failure. [Figure: stereonet showing HS, B, J1, J2] ## Short answer Stereonet analysis with phi=32° shows J1 plane failure if cut >40° and J1-J2 wedge (plunge 32.1° towards 142°) if face apparent dip exceeds 32.1°. Keeping cut slope ≤35° avoids both failures, as face angle is below J1 dip and wedge limit. Design cut at 35° for stability. ## 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 (dip direction/dip) HS 110/40; B (bedding) 130/20; J1 100/40; J2 200/50; $\phi = 32^\circ$. In dip/dip direction: HS 40/110, B 20/130, J1 40/100, J2 50/200. #### Failure checks for the existing 40° face - **B**: its dip direction is within 20° of the slope, but $20^\circ < \phi = 32^\circ$, so it is stable. - **J1**: its dip direction differs by $10^\circ$ and its dip of $40^\circ$ equals the face angle. A steeper cut would daylight J1, and as $40^\circ > \phi$ it would slide (**plane failure**). - **J2**: dips at $90^\circ$ to the face direction and cannot slide on its own. - **J1 x J2** intersection plunges $32.1^\circ$ towards 142°, just above $\phi$. It daylights in the face if the apparent dip of the face towards 142° is greater than $32.1^\circ$, i.e. if the cut angle is greater than $36.4^\circ$ (from $\tan\psi\cos 31.7^\circ = \tan 32.1^\circ$). - **B x J1** plunges $15.4^\circ$ and **B x J2** $20.0^\circ$ (both below $\phi$): safe. #### Design | Cut angle $\psi$ | Apparent dip of face along 142° | Wedge ($32.1^\circ<$ app. dip)? | Plane on J1 ($\psi>40^\circ$)? | |---|---|---|---| | 40° | 35.5° | yes | no (equal) | | 37° | 32.7° | yes (marginal) | no | | 36° | 31.7° | no | no | | 35° | 30.8° | no | no | **Answer: the cut slope should be limited to about 35° (maximum 36°, below both the J1 dip of 40° and the wedge limit of 36.4°). Then plane, wedge and toppling failure are all kinematically avoided.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is bedding B (20/130) stable although its direction is close to the face? A. Its dip 20° is below phi = 32° B. It is steeper than the face C. It dips into the hill D. Its dip is 90° Answer: A. A plane dipping less than the friction angle cannot slide. (Key point: Bedding stability) 2. Why would a cut steeper than 40° cause plane failure on J1? A. J1 dips less than phi B. J1 would daylight and 40° > phi C. J1 dips into the hill D. J1 is perpendicular to the face Answer: B. A face steeper than J1's dip exposes J1, and J1 dip exceeds phi. (Key point: Plane failure on J1) 3. Which cut angle limit comes from the J1 x J2 wedge? A. Exactly 50° B. 20° C. About 36.4°, where the face apparent dip equals the 32.1° plunge D. The J2 dip of 50° Answer: C. The wedge daylights when apparent dip along 142° exceeds 32.1°, i.e. cut angle above about 36.4°. (Key point: Wedge limit) 4. What cut inclination is recommended? A. 40° B. 45° C. 60° D. About 35° (maximum 36°) Answer: D. It must be below both the J1 dip (40°) and the wedge limit (36.4°). 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