---
title: "Stability: HS 202/65, joints J1 to J4"
subject: "Engineering Geology II"
chapter: 6
key: "semester_4/engineering_geology_ii:stability-hs202-65-j4-52"
source: https://www.bctengineer.com/q/semester_4/engineering_geology_ii/stability-hs202-65-j4-52
site: BCT Engineering
---

# Stability: HS 202/65, joints J1 to J4

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

- Asked in: 2070 Magh, 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/stability-hs202-65-j4-52
- Whole chapter: https://www.bctengineer.com/md/semester_4/engineering_geology_ii/chapter-6

## Question

Discuss the stability analysis based on the following data.
HS = 202°/65°, F = 350°/27° [last digit unclear in the scan], J1 = 180°/81°, J2 = 78°/69°, J3 = 21°/73°, J4 = 293°/52°.
[Figure: stereonet showing HS, F, J1, J2, J3, J4]

## Short answer

With phi assumed 30°, lower-hemisphere stereonet analysis reveals toppling on J3 (dips 73° opposite face, >55° limit) and wedge failure along J1-J4 intersection (plunge 47.1° towards 260°, between phi and face apparent dip). Slope unstable, needs flattening/reinforcement.

## 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 65/202; F 27/350; J1 81/180; J2 69/078; J3 73/021; J4 52/293. The friction angle is not given, so $\phi = 30^\circ$ is assumed (typical for rock joints). (The last digit of F in the scan is unclear; $27^\circ$ is used. As F dips into the hill at less than $\phi$, it does not affect the result.)

#### Single planes
| Plane (dip/dip dir) | Diff. of dip dir from slope | Daylights ($\psi_p<\psi_f$) | $\psi_p>\phi$ | Result |
|---|---|---|---|---|
| F (27/350) | 148° | yes | no | Safe |
| J1 (81/180) | 22° | no | yes | Safe |
| J2 (69/078) | 124° | no | yes | Safe |
| J3 (73/021) | 179° | no | yes | **Toppling** |
| J4 (52/293) | 91° | yes | yes | Safe |

#### Intersections
| Pair | Intersection plunge/trend | Face apparent dip along trend | Result |
|---|---|---|---|
| F x J1 | 4.7° / 269° | 39.7° | Stable (plunge below $\phi$) |
| J1 x J4 | 47.1° / 260° | 48.5° | **Wedge failure** |

(The other 8 intersections plunge into the hill, away from the face, and cannot slide.)

#### Result
- **J3** (73/021) dips into the hill almost exactly opposite to the face (the face dips towards 202°, so the opposite direction is 022°). Its dip of $73^\circ$ is more than $90^\circ - 65^\circ + 30^\circ = 55^\circ$, so **toppling failure** is possible.
- **J1** strikes nearly parallel to the face but dips $81^\circ$, steeper than the face, so it does not daylight.
- F (27°) dips into the hill and is below $\phi$; J2 and J4 do not daylight individually.
- The intersection **J1 x J4** plunges about 47.1° towards 260°, between $\phi$ and the face apparent dip (48.5°), so **wedge failure** is also possible.

**Conclusion: the slope is unstable. The main modes are toppling along J3 and wedge sliding along the J1-J4 intersection.** Reduce the slope angle, remove the potential wedge, and use bolts or anchors and drainage.

## Check yourself

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

1. J3 (73/021) dips almost exactly opposite to the face. What makes toppling possible?
   A. Its dip 73° exceeds 90° − 65° + 30° = 55°
   B. Its dip is below 30°
   C. It daylights in the face
   D. Its dip direction equals the face dip direction
   Answer: A. For toppling, a steep plane dipping into the hill must have dip greater than 90° − psi_f + phi. (Key point: Toppling criterion)

2. J1 (81/180) is 22° from the face direction. Why does it not cause plane failure?
   A. Its dip is below phi
   B. It is steeper (81°) than the face (65°) so it does not daylight
   C. It dips into the hill
   D. It strikes perpendicular to the face
   Answer: B. A plane steeper than the face cannot daylight, so it cannot slide. (Key point: Why J1 is safe alone)

3. Which intersection gives wedge failure and why?
   A. F x J1 because its plunge is 4.7°
   B. J2 x J3 because both are steep
   C. J1 x J4, plunging between phi and the face apparent dip
   D. No intersection can slide
   Answer: C. J1 x J4 plunges about 47.1° towards 261°, above phi = 30° and below the face apparent dip (48.5°). (Key point: Wedge failure)

4. The friction angle is not given. What is assumed and why does it matter?
   A. phi = 0°, so everything slides
   B. phi = 90°
   C. No value is needed
   D. phi = 30°, typical for rock joints, used in all friction tests
   Answer: D. The stability tests (psi > phi, toppling limit, wedge plunge) all need phi, so a typical 30° is assumed. (Key point: Assumed friction angle)

## Related in Engineering Geology II

- [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