---
title: "Failure mode: HS 170/50, BP 160/42"
subject: "Engineering Geology II"
chapter: 6
key: "semester_4/engineering_geology_ii:failure-mode-hs170-50"
source: https://www.bctengineer.com/q/semester_4/engineering_geology_ii/failure-mode-hs170-50
site: BCT Engineering
---

# Failure mode: HS 170/50, BP 160/42

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

- Asked in: 2079 Asoj, 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-hs170-50
- Whole chapter: https://www.bctengineer.com/md/semester_4/engineering_geology_ii/chapter-6

## Question

Analyze the failure mode by using stereographic projection from the following data. H.S.: 170°/50°, B.P.: 160°/42°, Friction angle: 26°.

## Short answer

On the stereonet, the bedding plane (42/160) lies in the daylight envelope: its dip direction is only 10° from the slope, it daylights in the 50° face, and 42° exceeds the 26° friction angle. Failure mode is plane failure along the bedding plane. Reduce slope below 42° or use bolts and drainage.

## 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)
Hill slope HS = 170°/50°, bedding plane BP = 160°/42°, $\phi = 26^\circ$. In dip/dip direction form: HS = 50/170, BP = 42/160.

| Plane (dip/dip dir) | Diff. of dip dir from slope | Daylights ($\psi_p<\psi_f$) | $\psi_p>\phi$ | Result |
|---|---|---|---|---|
| BP (42/160) | 10° | yes | yes | **Plane failure** |

#### Result
- The dip directions differ by only $10^\circ$ (within $\pm 20^\circ$).
- $\psi_p = 42^\circ < \psi_f = 50^\circ$, so the bedding plane daylights in the face.
- $\psi_p = 42^\circ > \phi = 26^\circ$.

On the stereonet, the dip vector of BP lies between the slope great circle and the friction circle (equivalently, its pole falls inside the daylight envelope). **Possible mode of failure: plane failure (sliding along the bedding plane).** It can be controlled by reducing the slope angle below $42^\circ$ or by bolting and drainage.

## Check yourself

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

1. HS is 170/50 and BP is 160/42 (dip direction/dip). Why is the dip-direction check satisfied?
   A. The directions differ by 10°, within the ±20° limit
   B. The directions differ by 90°
   C. The directions are opposite
   D. The directions differ by 42°
   Answer: A. 170° − 160° = 10°, inside the ±20° range needed for plane failure. (Key point: Dip direction criterion)

2. Which pair of conditions makes the bedding plane liable to slide?
   A. psi_p > psi_f and psi_p < phi
   B. psi_p < psi_f and psi_p > phi (42° < 50° and 42° > 26°)
   C. psi_p = 90° and phi = 26°
   D. psi_p < phi only
   Answer: B. The plane must daylight (dip below the face dip) and dip more than the friction angle. (Key point: Daylight and friction)

3. What is the possible mode of failure and a control measure?
   A. Wedge failure; add more joints
   B. Toppling; steepen the slope
   C. Plane failure along bedding; reduce slope below 42° or bolt and drain
   D. No failure; do nothing
   Answer: C. BP daylights and exceeds phi, so plane failure occurs; flattening below the bedding dip, bolting and drainage control it. (Key point: Result and control)

4. Where does the pole/great circle of BP plot in the stereonet for plane failure?
   A. Outside the slope great circle, beyond the friction circle's opposite side
   B. At the centre of the net
   C. On the primitive circle only
   D. Between the slope great circle and the friction circle (daylight envelope)
   Answer: D. The great circle of a sliding plane lies between the slope great circle and the friction circle. (Key point: Stereonet position)

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