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Slope Factor of Safety

FS = \frac{resisting}{driving}, FS_{c = 0}^{dry} = \frac{tan ϕ}{tan β}

FS is resisting shear (or moment) divided by driving shear (or moment). For an infinite cohesionless dry slope, FS = tanφ / tanβ, independent of depth; failure impends when β → φ. Seepage parallel to the slope reduces FS to (γ'/γ_sat) tanφ / tanβ. Cohesion adds c/(γ z cosβ sinβ). Method of slices (Bishop, Ordinary) is used for circular arcs.

FS

Factor of safety

ϕ

(deg)

Effective friction angle

β

(deg)

Slope angle

c

(kPa)

Cohesion

Step-by-step solved example

Infinite dry sand slope, c = 0, φ = 34°, β = 20°. Find FS.

Infinite slope (c = 0, dry): FS = tanφ / tanβ.

1. Tangents
tan 34° = 0.6745, tan 20° = 0.3640.
$tan 3 4^{\circ} = 0.6745$
2. FS
FS = 0.6745/0.3640 = 1.85.
$FS = 1.85$

Answer: 1.85

10 practice questions

0/10 correct

1.Dry cohesionless infinite slope with φ = 30°, β = 18°. FS is nearest

2.For c = 0 dry sand, FS = 1 when

3.Seepage parallel to a cohesionless slope multiplies the dry FS by about

4.Adding a cohesion intercept, all else equal, makes infinite-slope FS

5.Ordinary method of slices takes FS as

6.A rapid drawdown is critical because

7.Typical minimum static FS for a permanent slope is about

8.Infinite-slope analysis assumes the failure surface is

9.φ = 32°, β = 32°, c = 0, dry. FS =

10.Tension cracks at the crest of a clay slope