--- title: "Embankment 36°, method of slices FoS" subject: "Soil Mechanics" chapter: 11 key: "semester_4/soil_mechanics:embankment-method-of-slices-fos" source: https://www.bctengineer.com/q/semester_4/soil_mechanics/embankment-method-of-slices-fos site: BCT Engineering --- # Embankment 36°, method of slices FoS Soil Mechanics · Chapter 11: Stability of Slopes - Asked in: 2078 Poush - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/soil_mechanics/embankment-method-of-slices-fos - Whole chapter: https://www.bctengineer.com/md/semester_4/soil_mechanics/chapter-11 ## Question An embankment 10 m high is inclined at an angle of 36° to the horizontal. A stability analysis by the method of slices gives the following forces per running meter: Σ Shearing forces = 450 kN; Σ Normal forces = 900 kN; Σ Neutral forces = 216 kN. The length of the failure arc is 27 m. Laboratory tests on the soil indicate the effective values of c' and $\phi'$ as 20 kN/m² and 18° respectively. Determine the factor of safety of the slope with respect to (a) shearing strength and (b) cohesion. ## Short answer Factor of safety with respect to shear strength: F_s = (c′L + ΣN′ tanφ′)/ΣT = (20×27 + 684×tan18°)/450 = 1.69. Factor of safety with respect to cohesion: c_m = (ΣT − ΣN′ tanφ′)/L = 8.44 kN/m², giving F_c = c′/c_m = 2.37. ## Answer **Data (per metre run):** $\Sigma T = 450$ kN, $\Sigma N = 900$ kN, $\Sigma U = 216$ kN (neutral force, pore pressure force), arc length $L = 27$ m, $c' = 20$ kN/m², $\phi' = 18^\circ$. **Effective normal force:** $\Sigma N' = \Sigma N - \Sigma U = 900 - 216 = 684$ kN. #### (a) Factor of safety with respect to shear strength The factor of safety is the ratio of the available shear strength to the shear force: $$ \begin{aligned} F_s &= \frac{c'L + (\Sigma N-\Sigma U)\tan\phi'}{\Sigma T} \\ &= \frac{20(27) + 684\tan18^\circ}{450} \\ &= \frac{540 + 222.2}{450} = 1.69 \end{aligned} $$ #### (b) Factor of safety with respect to cohesion Here the full friction is mobilised ($F_\phi = 1$) and only the cohesion is reduced to the value needed for equilibrium: $$ \Sigma T = c_m L + (\Sigma N - \Sigma U)\tan\phi' \;\Rightarrow\; c_m = \frac{450 - 222.2}{27} = 8.44\ \text{kN/m}^2 $$ $$ F_c = \frac{c'}{c_m} = \frac{20}{8.44} = 2.37 $$ **Answer: (a) $F_s \approx 1.69$; (b) $F_c \approx 2.37$.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the effective normal force ΣN' from ΣN = 900 kN and ΣU = 216 kN? A. 1116 kN B. 684 kN C. 450 kN D. 216 kN Answer: B. ΣN' = ΣN - ΣU = 900 - 216 = 684 kN. (Key point: Effective normal force) 2. Which expression gives Fs in the method of slices? A. [c'L + (ΣN - ΣU) tanφ']/ΣT B. [c'L + ΣN tanφ']/ΣT C. ΣT/[c'L + (ΣN - ΣU) tanφ'] D. (ΣN - ΣU) tanφ'/(c'L) Answer: A. Fs is available shear strength (cohesion plus frictional resistance from effective normal force) over the shearing force. (Key point: Factor of safety wrt shear strength) 3. What is Fs for c' = 20, L = 27 m, φ' = 18°, ΣT = 450 kN? A. 0.59 B. 1.20 C. 2.37 D. 1.69 Answer: D. (540 + 222.2)/450 = 1.69. (Key point: Value of Fs) 4. How is the factor of safety with respect to cohesion found? A. Keep full friction (F_phi = 1), find mobilised cohesion cm for equilibrium, then Fc = c'/cm B. Divide both c' and tanφ' by the same factor C. Fc = Fs D. Fc = cm/c' Answer: A. With friction fully mobilised, cm = (450 - 222.2)/27 = 8.44 kN/m2 and Fc = 20/8.44 = 2.37. 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