---
title: "Short note: Factor of safety"
subject: "Engineering Geology I"
chapter: 6
key: "semester_3/engineering_geology_i:short-note-factor-of-safety"
source: https://www.bctengineer.com/q/semester_3/engineering_geology_i/short-note-factor-of-safety
site: BCT Engineering
---

# Short note: Factor of safety

Engineering Geology I · Chapter 6: Physical Geology

- Asked in: 2063 Baisakh (old course), 2 marks; 2062 Baisakh (old course), 2 marks (asked 2 times)
- Video: [Slope stability: failure definition and factor of safety](https://www.youtube.com/watch?v=lfrRzlPlv-I) — GEOTECH VIDEOS
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/engineering_geology_i/short-note-factor-of-safety
- Whole chapter: https://www.bctengineer.com/md/semester_3/engineering_geology_i/chapter-6

## Question

Write a short note on factor of safety.

## Short answer

Factor of safety (FS) is the ratio of resisting forces (shear strength) to driving forces (shear stress) on a slope. FS>1 indicates stability; typical design values are 1.3–1.5 for permanent slopes and 1.1–1.2 for temporary cuts. For a planar rock slide, FS = (cA+W cosψ tanφ)/(W sinψ) gave 1.61 for a 30° slope.

## Answer

**Factor of safety (FS)** is the ratio of the forces (or shear strength) resisting failure of a slope to the forces (or shear stress) driving it.

$$
FS = \frac{\text{Resisting force (shear strength)}}{\text{Driving force (shear stress)}}
$$

- $FS > 1$: stable; $FS = 1$: limiting equilibrium; $FS < 1$: failure.
- Usual design values are 1.3 to 1.5 for permanent slopes, and 1.1 to 1.2 for temporary cuts (higher when failure would be disastrous).
- For a planar rock slide on a plane dipping at $\psi_p$, with block weight $W$, cohesion $c$, area $A$ of the sliding plane and friction angle $\phi$:

$$
FS = \frac{cA + W\cos\psi_p\tan\phi}{W\sin\psi_p}
$$

**Example:** $W = 1000$ kN, $\psi_p = 30^\circ$, $\phi = 35^\circ$, $c = 10$ kPa, $A = 20$ m$^2$. Resisting $= 200 + 606.4 = 806.4$ kN; driving $= 500$ kN. **Answer: $FS = 1.61$ (stable).**

## Check yourself

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

1. Factor of safety is:
   A. Resisting force (shear strength) divided by driving force (shear stress)
   B. Driving force divided by resisting force
   C. Sum of the two forces
   D. Weight divided by area
   Answer: A. FS = resisting/driving. (Key point: Definition)

2. What does FS < 1 mean?
   A. Stable
   B. Failure
   C. Limiting equilibrium
   D. Very safe
   Answer: B. FS = 1 is limiting equilibrium and FS > 1 is stable. (Key point: Interpretation)

3. In the planar slide formula, what is the driving force?
   A. W cos(psi_p) tan(phi)
   B. cA
   C. W sin(psi_p)
   D. W tan(phi)
   Answer: C. Weight component along the plane drives; cA plus normal friction resist. (Key point: Planar slide)

4. For the example (W=1000 kN, 30°, phi 35°, c=10 kPa, A=20 m^2), FS is about:
   A. 0.62
   B. 1.00
   C. 2.5
   D. 1.61
   Answer: D. Resisting 200+606.4 = 806.4 kN over driving 500 kN gives 1.61. (Key point: Numerical)

## Related in other subjects

- [Factor of safety and partial safety factor](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Afactor-safety-partial-safety) — Design of Steel and Timber Structure, chapter 3, asked 2×
- [Factor of safety](https://www.bctengineer.com/question.md?key=semester_4%2Fengineering_geology_ii%3Afactor-of-safety) — Engineering Geology II, chapter 6
- [Gumbel design flood, safety factor, confidence limits](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Agumbel-20yr-life35-rel87) — Engineering Hydrology, chapter 6
- [Gumbel design flood, factor of safety 1.3, 45-year life](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Agumbel-25yr-life45-fos1-3) — Engineering Hydrology, chapter 6