---
title: "Slope stability: N40W/55 slope, two joints"
subject: "Engineering Geology II"
chapter: 6
key: "semester_4/engineering_geology_ii:slope-stability-n40w-55"
source: https://www.bctengineer.com/q/semester_4/engineering_geology_ii/slope-stability-n40w-55
site: BCT Engineering
---

# Slope stability: N40W/55 slope, two joints

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

- Asked in: 2081 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/slope-stability-n40w-55
- Whole chapter: https://www.bctengineer.com/md/semester_4/engineering_geology_ii/chapter-6

## Question

A rock slope has orientation of N40°W/55° exposed with two sets of joints with attitudes of N50°W/37° and N70°E/67°. The friction angle of discontinuity is 28°. Describe the stability condition of the rock slope by stereographic projection of the discontinuities.

## Short answer

Plot the great circles and friction circle on a lower-hemisphere equal-angle stereonet. J1 (37/220) daylights and dips more than 28°, so plane failure is possible; J1-J2 intersection plunges 36.4° toward 232°, within the face, so wedge failure is also possible. The slope is unstable.

## 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
Slope face N40°W/55° dips south-west, so it is taken as **55/230** (dip/dip direction). Joint set 1 N50°W/37° dips SW: **37/220**. Joint set 2 N70°E/67° dips SE: **67/160**. $\phi = 28^\circ$.

#### Individual planes
| Plane (dip/dip dir) | Diff. of dip dir from slope | Daylights ($\psi_p<\psi_f$) | $\psi_p>\phi$ | Result |
|---|---|---|---|---|
| J1 (37/220) | 10° | yes | yes | **Plane failure** |
| J2 (67/160) | 70° | no | yes | Safe |

#### Intersection of the joint sets
| Pair | Intersection plunge/trend | Face apparent dip along trend | Result |
|---|---|---|---|
| J1 x J2 | 36.4° / 232° | 55.0° | **Wedge failure** |

#### Result
- **J1** strikes almost parallel to the face (10° difference), daylights ($37^\circ < 55^\circ$) and dips more than $\phi$ ($37^\circ > 28^\circ$), so **plane failure along J1 is possible**.
- J2 is steeper than the face ($67^\circ > 55^\circ$), so it alone cannot slide.
- J1 and J2 form a wedge whose line plunges $36.4^\circ$ towards $232^\circ$, within the face (apparent dip of face $55.0^\circ$) and above $\phi$, so **wedge failure is also possible**.

**Conclusion: the slope is unstable (plane failure along J1 and wedge failure along J1-J2).** To stabilise it, flatten the cut to below about $35^\circ$ (less than the intersection plunge $36.4^\circ$ and the J1 dip $37^\circ$), or use rock bolts, drainage and benching.

## Check yourself

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

1. Why is plane failure possible along J1 (37/220) in slope 55/230?
   A. J1 dips opposite the slope
   B. J2 is steeper
   C. Dip directions differ by 10°, it daylights (37<55) and 37 exceeds phi 28
   D. J1 dips less than phi
   Answer: C. All plane failure conditions are met. (Key point: Plane failure on J1)

2. Why is J2 (67/160) safe alone?
   A. Its dip direction differs by 70° and it is steeper than the face, so it does not daylight
   B. It dips less than phi
   C. It is parallel to the slope
   D. It has no friction
   Answer: A. It is steeper than the slope and not near the dip direction. (Key point: J2)

3. The J1 x J2 line plunges 36.4° towards 232°; face apparent dip is 55°. Wedge failure?
   A. Not possible, 36.4 < phi
   B. Not possible, plunge > face dip
   C. Possible, since phi (28) < 36.4 < 55
   D. Possible only if dry
   Answer: C. Condition psi_f > psi_i > phi is satisfied. (Key point: Wedge)

4. What is a suggested cut angle for stabilisation?
   A. Steeper than 55°
   B. Flatter than about 35°, below the plunge 36.4° and J1 dip 37°
   C. Exactly 60°
   D. No change
   Answer: B. Flattening below both gives non-daylighting geometry. (Key point: Remedy)

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