---
title: "Canal slope stability: HS 138/45"
subject: "Engineering Geology II"
chapter: 6
key: "semester_4/engineering_geology_ii:canal-slope-stability-hs138-45"
source: https://www.bctengineer.com/q/semester_4/engineering_geology_ii/canal-slope-stability-hs138-45
site: BCT Engineering
---

# Canal slope stability: HS 138/45

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

- Asked in: 2074 Bhadra, 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/canal-slope-stability-hs138-45
- Whole chapter: https://www.bctengineer.com/md/semester_4/engineering_geology_ii/chapter-6

## Question

Interpret the stability condition of the rock slope where a canal alignment has to pass. The orientations of the discontinuities, hill slope and the internal friction angle are as follows: HS = 138°/45°, J1 = 234°/38°, J2 = 098°/58°, J3 = 315°/60° and ϕ = 25°.
[Figure: stereonet]

## Short answer

Stereonet analysis with phi=25° shows no planes daylight in a dangerous orientation; J1-J2 intersection plunges 21.3° (<phi) so no wedge failure; toppling condition not met. The rock slope is kinematically stable; provide drainage and lining.

## 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 138°/45°; J1 234°/38°; J2 098°/58°; J3 315°/60°; $\phi = 25^\circ$. In dip/dip direction: HS 45/138, J1 38/234, J2 58/098, J3 60/315.

#### Single planes
| Plane (dip/dip dir) | Diff. of dip dir from slope | Daylights ($\psi_p<\psi_f$) | $\psi_p>\phi$ | Result |
|---|---|---|---|---|
| J1 (38/234) | 96° | yes | yes | Safe |
| J2 (58/098) | 40° | no | yes | Safe |
| J3 (60/315) | 177° | no | yes | Safe |

#### Intersections
| Pair | Intersection plunge/trend | Face apparent dip along trend | Result |
|---|---|---|---|
| J1 x J2 | 21.3° / 174° | 39.0° | Stable (plunge below $\phi$) |

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

#### Interpretation
- **J1** dips toward 234°, $96^\circ$ away from the face direction: it runs across the slope and cannot slide out.
- **J2** is steeper ($58^\circ$) than the face ($45^\circ$): it does not daylight.
- **J3** dips into the hill (315° against 138° + 180° = 318°) at $60^\circ$; toppling needs $\psi \geq 90^\circ - 45^\circ + 25^\circ = 70^\circ$, so it is below the limit.
- The only intersection that points out of the face is **J1 x J2**, plunging $21.3^\circ$ towards 174°. Its plunge is **less than $\phi = 25^\circ$**, so the wedge cannot slide.

**Conclusion: the rock slope is kinematically stable against plane, wedge and toppling failure** and the canal alignment is acceptable. Provide drainage and lining to prevent seepage water from raising pore pressure, and check local loosening.

## Check yourself

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

1. J2 (58/098) differs by 40° from the face. Why is it safe?
   A. It dips steeper (58°) than the face (45°) and is outside ±20°
   B. It dips less than phi
   C. It strikes parallel to canal
   D. It is horizontal
   Answer: A. It is steeper than the face, so it does not daylight. (Key point: J2 does not daylight)

2. What toppling limit applies to J3 and the outcome?
   A. 45° + 25° = 70°; toppling occurs
   B. 90° − 45° + 25° = 70°; J3 dips 60°, so no toppling
   C. 25°; toppling occurs
   D. 90°; no toppling possible
   Answer: B. Toppling needs dip ≥ 90° − psi_f + phi = 70°; J3 is 60°. (Key point: Toppling limit J3)

3. Why can the J1 x J2 wedge not slide?
   A. Its plunge exceeds the face apparent dip
   B. It plunges into the hill
   C. Its plunge 21.3° is less than phi = 25°
   D. J1 is vertical
   Answer: C. If the line plunges less than phi, friction prevents sliding. (Key point: Wedge test)

4. What is the overall conclusion?
   A. Unstable against plane failure
   B. Unstable against toppling
   C. Needs complete excavation
   D. Kinematically stable; provide drainage and lining
   Answer: D. No plane, wedge or toppling mode is possible; drainage controls seepage pressure. (Key point: Conclusion and remedy)

## Related in Engineering Geology II

- [Stability: HS 202/65, joints J1 to J4](https://www.bctengineer.com/question.md?key=semester_4%2Fengineering_geology_ii%3Astability-hs202-65-j4-52) — Engineering Geology II, chapter 6
- [Stability: HS 175/52, BP 150/48](https://www.bctengineer.com/question.md?key=semester_4%2Fengineering_geology_ii%3Astability-hs175-52-bp150-48) — Engineering Geology II, chapter 6
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- [Failure mode: HS 75/230, BP 68/045](https://www.bctengineer.com/question.md?key=semester_4%2Fengineering_geology_ii%3Afailure-mode-hs75-230-figure) — Engineering Geology II, chapter 6
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