---
title: "Failure mode: HS 75/230, BP 68/045"
subject: "Engineering Geology II"
chapter: 6
key: "semester_4/engineering_geology_ii:failure-mode-hs75-230-figure"
source: https://www.bctengineer.com/q/semester_4/engineering_geology_ii/failure-mode-hs75-230-figure
site: BCT Engineering
---

# Failure mode: HS 75/230, BP 68/045

Engineering Geology II · Chapter 6: Measurement, analysis and interpretation of structural geological data

- Asked in: 2073 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/failure-mode-hs75-230-figure
- Whole chapter: https://www.bctengineer.com/md/semester_4/engineering_geology_ii/chapter-6

## Question

Suggest the possible mode of failure from the following figure.
Hill slope (HS) 75°/230°; Bedding plane (BP) 68°/045°; Joint (J1) 49°/215°; Joint (J2) 55°/265°; Joint (J3) 65°/199°; Internal friction angle (ϕ) 32°.
[Figure: stereonet showing HS, BP, J1, J2, J3]

## Short answer

With phi=32°, kinematic tests reveal plane failure on J1 (dip 49° toward 215°, within 20° of face, >phi, <face angle), toppling on BP (steep, opposite direction, dip 68° >47° limit), and multiple wedge failures (J1-J2, J2-J3, J1-J3). Slope highly unstable; flatten to <45° and reinforce.

## 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/dip direction)
HS 75/230; BP 68/045; J1 49/215; J2 55/265; J3 65/199; $\phi = 32^\circ$.

#### Single planes
| Plane (dip/dip dir) | Diff. of dip dir from slope | Daylights ($\psi_p<\psi_f$) | $\psi_p>\phi$ | Result |
|---|---|---|---|---|
| BP (68/045) | 175° | yes | yes | **Toppling** |
| J1 (49/215) | 15° | yes | yes | **Plane failure** |
| J2 (55/265) | 35° | yes | yes | Safe |
| J3 (65/199) | 31° | yes | yes | Safe |

#### Intersections
| Pair | Intersection plunge/trend | Face apparent dip along trend | Result |
|---|---|---|---|
| J1 x J2 | 48.4° / 227° | 75.0° | **Wedge failure** |
| J1 x J3 | 32.1° / 272° | 70.2° | **Wedge failure** |
| J2 x J3 | 54.0° / 249° | 74.2° | **Wedge failure** |

(The other 3 intersections plunge into the hill, away from the face, and cannot slide.)

#### Result
- **J1** dips $49^\circ$ at 215°, only $15^\circ$ from the face direction. $32^\circ < 49^\circ < 75^\circ$: **plane failure** along J1.
- **BP** dips $68^\circ$ at 045°, opposite to the face (230° - 180° = 050°) and steeper than $90^\circ - 75^\circ + 32^\circ = 47^\circ$: **toppling failure**.
- J1 x J2, J2 x J3 (and marginally J1 x J3) intersect with plunges above $32^\circ$ but below the face apparent dip: **wedge failure** is also possible because the face is very steep.

**Conclusion: the main modes are plane failure along J1 and toppling of bedding slabs (BP), with secondary wedge sliding.** The slope needs to be flattened to less than about $45^\circ$ (J1 dip is $49^\circ$) and reinforced.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Why does J1 (49/215) undergo plane failure?
   A. It dips only 15° from the face direction, with 32° < 49° < 75°
   B. It dips into the hill
   C. Its dip is below phi
   D. Its dip is steeper than the face
   Answer: A. Direction within 20°, daylights and dip exceeds phi = 32°. (Key point: Plane failure on J1)

2. Why is BP (68/045) toppling?
   A. It dips in the same direction as the face
   B. It dips into the hill, opposite the face, steeper than 90° − 75° + 32° = 47°
   C. Its dip is less than phi
   D. It is horizontal
   Answer: B. A steep plane dipping into the hill with dip above 47° gives slab toppling. (Key point: Toppling of BP)

3. Why is wedge failure also possible here?
   A. The face is flat
   B. All joints are vertical
   C. The face is very steep (75°), so intersections plunging between phi and the apparent dip exist
   D. phi is zero
   Answer: C. A steep face gives high apparent dips so several intersections plunge between phi and the face apparent dip. (Key point: Wedge failures)

4. What flattening is needed?
   A. Above 75°
   B. Exactly 90°
   C. Remove bolting
   D. Less than about 45° (below the J1 dip of 49°) with reinforcement
   Answer: D. Cutting below J1 dip stops J1 from daylighting. (Key point: Remedy)

## Related in Engineering Geology II

- [Failure mode: HS N45E/70, bedding N55E/45](https://www.bctengineer.com/question.md?key=semester_4%2Fengineering_geology_ii%3Afailure-mode-hs-n45e-70) — Engineering Geology II, chapter 6
- [Failure mode from stereonet: HS 40/050](https://www.bctengineer.com/question.md?key=semester_4%2Fengineering_geology_ii%3Afailure-mode-hs40-050-figure) — Engineering Geology II, chapter 6
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- [Stability: HS 310/43, BP 306/31](https://www.bctengineer.com/question.md?key=semester_4%2Fengineering_geology_ii%3Astability-hs310-43-bp306-31) — Engineering Geology II, chapter 6
- [Failure mode: HS 170/50, BP 160/42](https://www.bctengineer.com/question.md?key=semester_4%2Fengineering_geology_ii%3Afailure-mode-hs170-50) — Engineering Geology II, chapter 6
- [Stability: HS 202/65 with four joints](https://www.bctengineer.com/question.md?key=semester_4%2Fengineering_geology_ii%3Astability-hs202-65-j4-50) — Engineering Geology II, chapter 6

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