--- title: "Infinite dry slope, c = 10 kPa, φ = 25°" subject: "Soil Mechanics" chapter: 11 key: "semester_4/soil_mechanics:infinite-dry-slope-c10-phi25" source: https://www.bctengineer.com/q/semester_4/soil_mechanics/infinite-dry-slope-c10-phi25 site: BCT Engineering --- # Infinite dry slope, c = 10 kPa, φ = 25° Soil Mechanics · Chapter 11: Stability of Slopes - Asked in: 2077 Chaitra, 6 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/soil_mechanics/infinite-dry-slope-c10-phi25 - Whole chapter: https://www.bctengineer.com/md/semester_4/soil_mechanics/chapter-11 ## Question Carry out the stability analysis for an infinite dry slope with strength properties of c = 10 kPa and $\phi$ = 25°. Assume the plane failure surface lies at a depth of 5 m from the slope surface. Take the unit weight of soil above the failure plane as 16 kN/m³ and the inclination of the slope as 10°. What happens if the cohesion of the soil reduces to zero? ## Short answer For the given dry slope (c=10 kPa, φ=25°, z=5 m, γ=16 kN/m³, β=10°), F = c/(γ z sinβ cosβ) + tanφ/tanβ = 3.38. If cohesion reduces to zero, F = tanφ/tanβ = 2.64; the slope remains stable as β < φ. ## Answer **Data:** $c = 10$ kPa, $\phi = 25^\circ$, $z = 5$ m (depth of the failure plane, as given), $\gamma = 16$ kN/m³, $\beta = 10^\circ$, no water. For a dry infinite slope: $$ F = \frac{c}{\gamma z\sin\beta\cos\beta} + \frac{\tan\phi}{\tan\beta} $$ **Step 1: Values** $$ \sin10^\circ\cos10^\circ = 0.1710, \quad \tan25^\circ = 0.4663, \quad \tan10^\circ = 0.1763 $$ **Step 2: Cohesion term** $$ \frac{10}{16\times5\times0.1710} = \frac{10}{13.68} = 0.731 $$ **Step 3: Friction term** $$ \frac{0.4663}{0.1763} = 2.645 $$ $$ F = 0.731 + 2.645 = 3.38 $$ $F = 3.38 > 1$, so the slope is **stable** with a large margin. **If cohesion becomes zero** $$ F = \frac{\tan\phi}{\tan\beta} = 2.645 $$ The factor of safety drops by about 22% but remains well above 1, so the slope is still stable. The stability then depends only on friction ($\beta=10^\circ<\phi=25^\circ$). **Answer: $F = 3.38$ (stable); with $c = 0$, $F = 2.64$ (still stable).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which formula applies to a dry infinite slope with c and φ? A. F = c/(γ z sinβ cosβ) + tanφ/tanβ B. F = c/(γ z) + tanφ/tanβ C. F = c tanφ/(γ z tanβ) D. F = tanφ/tanβ only Answer: A. Strength is c + σ tanφ divided by τ = γ z sinβ cosβ, giving the cohesion term plus the friction term. (Key point: Formula with cohesion) 2. With c = 10 kPa, γ = 16 kN/m3, z = 5 m, β = 10°, what is the cohesion term (sin10°cos10° = 0.1710)? A. 0.073 B. 0.731 C. 1.37 D. 2.645 Answer: B. 10/(16 x 5 x 0.1710) = 10/13.68 = 0.731. (Key point: Cohesion term) 3. What is the factor of safety and the conclusion? A. F = 0.73, unsafe B. F = 2.64, stable C. F = 3.38, stable D. F = 1.0, limiting Answer: C. F = 0.731 + 2.645 = 3.38, which is greater than 1, so the slope is stable. (Key point: Final factor of safety) 4. If cohesion drops to zero, what happens to this slope? A. It fails because F < 1 B. F = 2.64 and it remains stable since β < φ C. F becomes 1.0 D. F doubles Answer: B. F = tanφ/tanβ = 0.4663/0.1763 = 2.645, still well above 1 because β = 10° is less than φ = 25°. 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