---
title: "Failure mode: HS N45E/70, bedding N55E/45"
subject: "Engineering Geology II"
chapter: 6
key: "semester_4/engineering_geology_ii:failure-mode-hs-n45e-70"
source: https://www.bctengineer.com/q/semester_4/engineering_geology_ii/failure-mode-hs-n45e-70
site: BCT Engineering
---

# Failure mode: HS N45E/70, bedding N55E/45

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

- Asked in: 2069 Bhadra, 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/failure-mode-hs-n45e-70
- 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: N45°E/70°; Bedding: N55°E/45°; Joint 1: N47°W/31°; Joint 2: N77°E/36°; Joint 3: N20°E/10°.
[Figure: stereonet]

## Short answer

Using assumed 30° friction angle, bedding (45/145) daylights and exceeds friction, so plane failure is main. Bedding-J2 intersection plunges 32.6° toward 195° above friction and daylights, so wedge failure also possible. Reduce cut below 45° and support.

## 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
All planes strike as given and are taken to dip on the side that appears in the figure: Hill slope N45°E/70° SE = **70/135**; Bedding N55°E/45° SE = **45/145**; J1 N47°W/31° SW = **31/223**; J2 N77°E/36° SE = **36/167**; J3 N20°E/10° SE = **10/110**. The friction angle is not stated, so $\phi = 30^\circ$ is assumed.

#### Single planes
| Plane (dip/dip dir) | Diff. of dip dir from slope | Daylights ($\psi_p<\psi_f$) | $\psi_p>\phi$ | Result |
|---|---|---|---|---|
| Bedding (45/145) | 10° | yes | yes | **Plane failure** |
| J1 (31/223) | 88° | yes | yes | Safe |
| J2 (36/167) | 32° | yes | yes | Safe |
| J3 (10/110) | 25° | yes | no | Safe |

#### Intersections
| Pair | Intersection plunge/trend | Face apparent dip along trend | Result |
|---|---|---|---|
| Bedding x J1 | 29.1° / 201° | 48.0° | Stable (plunge below $\phi$) |
| Bedding x J2 | 32.6° / 195° | 53.8° | **Wedge failure** |
| Bedding x J3 | 6.7° / 062° | 38.4° | Stable (plunge below $\phi$) |
| J1 x J2 | 29.8° / 205° | 43.1° | Stable (plunge below $\phi$) |
| J1 x J3 | 8.1° / 147° | 69.6° | Stable (plunge below $\phi$) |
| J2 x J3 | 9.4° / 090° | 62.8° | Stable (plunge below $\phi$) |

#### Result
- **Bedding** dips $10^\circ$ off the face direction at $45^\circ$: $30^\circ < 45^\circ < 70^\circ$, so it daylights and can slide: **plane failure**.
- J3 dips only $10^\circ$, below $\phi$: stable. J1 dips across the slope (88° from the face direction).
- The **Bedding x J2** intersection plunges $32.6^\circ$ towards 195°, in the face (apparent dip $53.8^\circ$) and above $\phi$: **wedge failure** is also possible.

**Possible mode of failure: plane failure along the bedding planes (main), with wedge failure formed by bedding and J2.** Reduce the cut angle below $45^\circ$ and provide bolting and drainage.

## Check yourself

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

1. Why is bedding (45/145) a plane failure?
   A. 10° from the face direction, daylights and 30° < 45° < 70°
   B. It dips into the hill
   C. Its dip is below phi
   D. It is steeper than the face
   Answer: A. All three plane failure conditions are met. (Key point: Plane failure on bedding)

2. What friction angle was used and why?
   A. 0°
   B. 30°, since it is not stated
   C. 60°
   D. 45° from bedding
   Answer: B. No phi was given, so a typical 30° is assumed. (Key point: Assumed friction)

3. Why is J3 (10/110) stable?
   A. It is steeper than the face
   B. It dips into the hill
   C. Its dip 10° is below phi
   D. It is vertical
   Answer: C. It dips less than the friction angle. (Key point: J3 stability)

4. Which intersection gives the wedge?
   A. J1 x J3
   B. J2 x J3
   C. Bedding x J3
   D. Bedding x J2 (32.6° plunge towards 195°, above phi, in the face)
   Answer: D. Its plunge exceeds phi and the face apparent dip is 53.8°. (Key point: Wedge)

## Related in Engineering Geology II

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- [Failure mode: HS 75/230, BP 68/045](https://www.bctengineer.com/question.md?key=semester_4%2Fengineering_geology_ii%3Afailure-mode-hs75-230-figure) — Engineering Geology II, chapter 6
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