---
title: "Failure mode from stereonet: HS 40/050"
subject: "Engineering Geology II"
chapter: 6
key: "semester_4/engineering_geology_ii:failure-mode-hs40-050-figure"
source: https://www.bctengineer.com/q/semester_4/engineering_geology_ii/failure-mode-hs40-050-figure
site: BCT Engineering
---

# Failure mode from stereonet: HS 40/050

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

- Asked in: 2078 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/failure-mode-hs40-050-figure
- 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.
[Figure: equal-angle, lower-hemisphere stereonet with planes HS, J1, J2, J3, J4 plotted]
Orientations (Dip/Direction): HS 40/050; J1 46/040; J2 28/200; J3 70/103; J4 39/348. Friction angle: 25°.

## Short answer

Only wedge failure is kinematically possible, formed by joints J3 and J4; their intersection plunges 32.4° toward 026°, between the 25° friction angle and the 37.5° face apparent dip. Single planes do not daylight in a sliding sense, and the other intersections are stable or plunge into the hill.

## 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 40/050; J1 46/040; J2 28/200; J3 70/103; J4 39/348; $\phi = 25^\circ$.

#### Single planes
| Plane (dip/dip dir) | Diff. of dip dir from slope | Daylights ($\psi_p<\psi_f$) | $\psi_p>\phi$ | Result |
|---|---|---|---|---|
| J1 (46/040) | 10° | no | yes | Safe (does not daylight) |
| J2 (28/200) | 150° | yes | yes | Safe |
| J3 (70/103) | 53° | no | yes | Safe |
| J4 (39/348) | 62° | yes | yes | Safe |

#### Intersections
| Pair | Intersection plunge/trend | Face apparent dip along trend | Result |
|---|---|---|---|
| J1 x J2 | 6.9° / 123° | 13.6° | Stable (plunge below $\phi$) |
| J1 x J3 | 45.9° / 035° | 39.0° | Safe (does not daylight) |
| J1 x J4 | 38.4° / 360° | 28.3° | Safe (does not daylight) |
| J3 x J4 | 32.4° / 026° | 37.5° | **Wedge failure** |

(The other 2 intersections plunge into the hill, away from the face, and cannot slide.)

#### Result
- **J1** dips almost with the slope but is steeper ($46^\circ > 40^\circ$), so it does not daylight.
- J2 dips back into the hill at $200^\circ$ (far from the face) and is not toppling-type since it is gentle.
- J3 and J4 individually do not daylight in a sliding sense.
- The intersection **J3 x J4** plunges $32.4^\circ$ towards $026^\circ$, which is between $\phi = 25^\circ$ and the face apparent dip $37.5^\circ$ and points out of the face.

**Possible mode of failure: wedge failure formed by joints J3 and J4, sliding along their line of intersection.** (Plane and toppling failure are not kinematically possible.)

## Check yourself

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

1. J1 (46/040) has almost the same direction as the face (40/050). Why does it not slide as a plane?
   A. It does not daylight because it is steeper (46°) than the face (40°)
   B. Its dip 46° is below the friction angle
   C. It dips into the hill
   D. Its dip direction differs by 90°
   Answer: A. A plane steeper than the face does not daylight, so no plane failure. (Key point: Why J1 is safe alone)

2. What must a line of intersection satisfy for wedge failure?
   A. Plunge less than the friction angle
   B. Plunge between phi and the face apparent dip, pointing out of the face
   C. Plunge greater than the face apparent dip
   D. Plunge equal to 90°
   Answer: B. Wedge sliding needs phi < plunge < apparent dip of face in that direction, with the line plunging out of the face. (Key point: Wedge criterion)

3. Which intersection gives the possible failure?
   A. J1 x J2
   B. J1 x J3
   C. J3 x J4 (plunge 32.4° / 026°)
   D. J1 x J4
   Answer: C. 32.4° lies between phi = 25° and the face apparent dip of 37.5°, so J3 x J4 forms a wedge. (Key point: Critical intersection)

4. Why is the J1 x J2 intersection (6.9° plunge) stable?
   A. It plunges into the hill
   B. Its plunge exceeds the face apparent dip
   C. J2 is vertical
   D. Its plunge is below phi = 25°
   Answer: D. A plunge below the friction angle cannot slide. (Key point: J1 x J2 stability)

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