---
title: "Stability: H 49/249, F 56/243, three joints"
subject: "Engineering Geology II"
chapter: 6
key: "semester_4/engineering_geology_ii:stability-h49-249-planes"
source: https://www.bctengineer.com/q/semester_4/engineering_geology_ii/stability-h49-249-planes
site: BCT Engineering
---

# Stability: H 49/249, F 56/243, three joints

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

- Asked in: 2077 Chaitra, 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/stability-h49-249-planes
- Whole chapter: https://www.bctengineer.com/md/semester_4/engineering_geology_ii/chapter-6

## Question

Discuss the stability analysis of the given planes.
H = 49/249, F = 56/243, J1 = 58/071, J2 = 30/175, J3 = 83/286 and ϕ = 25°.
[Figure: stereonet showing H, F, J1, J2, J3 and the friction circle]

## Short answer

Using phi=25°, stereonet checks find no plane or toppling failure; however, wedge failure along J2-J3 intersection is possible (plunge 27.7° towards 200°, between friction angle and face apparent dip 36.9°). Slope needs bolting or flattening to below 39° to prevent wedge sliding.

## 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)
H (hill slope) 49/249; F 56/243; J1 58/071; J2 30/175; J3 83/286; $\phi = 25^\circ$.

#### Single planes
| Plane (dip/dip dir) | Diff. of dip dir from slope | Daylights ($\psi_p<\psi_f$) | $\psi_p>\phi$ | Result |
|---|---|---|---|---|
| F (56/243) | 6° | no | yes | Safe (does not daylight) |
| J1 (58/071) | 178° | no | yes | Safe |
| J2 (30/175) | 74° | yes | yes | Safe |
| J3 (83/286) | 37° | no | yes | Safe |

#### Intersections
| Pair | Intersection plunge/trend | Face apparent dip along trend | Result |
|---|---|---|---|
| F x J2 | 30.0° / 176° | 18.5° | Safe (does not daylight) |
| F x J3 | 49.1° / 204° | 39.2° | Safe (does not daylight) |
| J2 x J3 | 27.7° / 200° | 36.9° | **Wedge failure** |

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

#### Result
- **F** has almost the same dip direction as the slope but is steeper ($56^\circ > 49^\circ$), so it does not daylight: no plane failure.
- **J1** dips into the hill (071, against 249 - 180 = 069). The toppling limit is $90^\circ - 49^\circ + 25^\circ = 66^\circ$, but $\psi = 58^\circ < 66^\circ$, so toppling is not expected.
- J3 is very steep, and J2 is gentle and dips almost at right angles to the face direction: neither slides alone.
- The intersection **J2 x J3** plunges $27.7^\circ$ towards $200^\circ$. This is greater than $\phi = 25^\circ$ and smaller than the apparent dip of the face in that direction ($36.9^\circ$), so it plots between the friction circle and the slope great circle.

**Conclusion: the slope is stable against plane and toppling failure, but wedge failure along the intersection of J2 and J3 is possible.** It can be treated by bolting, or by flattening the face to below about $39^\circ$, so that the apparent dip of the face towards 200° becomes less than the $27.7^\circ$ plunge of the wedge ($\tan\psi\cos 49.7^\circ = \tan 27.7^\circ \Rightarrow \psi \approx 39^\circ$).

## Check yourself

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

1. F (56/243) has nearly the same direction as the face (49/249). Why is there no plane failure?
   A. It is steeper (56°) than the face (49°), so it does not daylight
   B. It dips into the hill
   C. Its dip is below phi
   D. Its direction differs by 90°
   Answer: A. Plane failure needs psi_p < psi_f; here the plane is steeper. (Key point: Plane failure on F)

2. What is the toppling limit for J1 here and the result?
   A. 49° + 25° = 74°; toppling occurs
   B. 90° − 49° + 25° = 66°; J1 dips 58°, so no toppling
   C. 25°; toppling occurs
   D. 90°; no toppling
   Answer: B. Toppling needs dip above 90° − psi_f + phi = 66°; 58° is lower. (Key point: Toppling limit for J1)

3. Why is J2 x J3 a wedge failure?
   A. Its plunge 27.7° is below phi = 25°
   B. It plunges into the hill
   C. Its plunge 27.7° lies between phi = 25° and face apparent dip 36.9°
   D. Its plunge is greater than 36.9°
   Answer: C. The line plunges more than phi but less than the face apparent dip. (Key point: Wedge criterion)

4. How can the wedge be treated?
   A. Steepen the face
   B. Remove drainage
   C. Do nothing
   D. Flatten the face below about 27° in that direction or bolt it
   Answer: D. A face flatter than the wedge plunge keeps the line from daylighting. (Key point: Remedy)

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