--- title: "10 m cut in cohesive soil, bedrock 15 m" subject: "Soil Mechanics" chapter: 11 key: "semester_4/soil_mechanics:cut-10m-cohesive-phi-zero-bedrock" source: https://www.bctengineer.com/q/semester_4/soil_mechanics/cut-10m-cohesive-phi-zero-bedrock site: BCT Engineering --- # 10 m cut in cohesive soil, bedrock 15 m Soil Mechanics · Chapter 11: Stability of Slopes - Asked in: 2078 Baisakh - Page: https://www.bctengineer.com/q/semester_4/soil_mechanics/cut-10m-cohesive-phi-zero-bedrock - Whole chapter: https://www.bctengineer.com/md/semester_4/soil_mechanics/chapter-11 ## Question A cut 10 m deep is to be made in a stratum of cohesive soil (c = 35 kN/m², $\gamma$ = 18.5 kN/m³ and $\phi$ = 0). The bed rock is located 15 m below the original ground surface. [The remainder of the question is cut off in the scan, presumably asking for the factor of safety or stability.] ## Short answer Assuming a vertical cut in cohesive soil (φ=0, c=35 kN/m², γ=18.5 kN/m³, H=10 m, depth factor 1.5), Taylor's stability number N_s ≈ 0.261 yields F = c/(N_s γ H) = 0.72. The cut is unsafe; critical height is about 7.25 m. ## Answer **Note:** the rest of the question and the cut slope are missing. The usual form asks for the factor of safety (or safe height) of the cut. The solution below assumes a **vertical cut** (slope angle $\beta = 90^\circ$) and uses Taylor's stability number method; the same steps apply for any given slope angle. **Data:** $H = 10$ m, $c = c_u = 35$ kN/m², $\gamma = 18.5$ kN/m³, $\phi = 0$. Bedrock at 15 m below the original ground, so the depth factor is $$ n_d = \frac{D}{H} = \frac{15}{10} = 1.5 $$ **Taylor's stability number (for $\phi = 0$)** $$ N_s = \frac{c_m}{\gamma H} $$ For steep slopes ($\beta > 53^\circ$) the critical circle is a toe circle regardless of the depth factor, and Taylor's chart gives, for $\beta = 90^\circ$, $N_s \approx 0.261$. **Critical height** $$ H_c = \frac{c}{\gamma N_s} = \frac{35}{18.5\times0.261} = 7.25\ \text{m} $$ **Factor of safety** $$ F = \frac{H_c}{H} = \frac{7.25}{10} = 0.72 \quad \left(\text{or } F = \frac{c}{N_s\gamma H} = \frac{35}{0.261\times185} = 0.72\right) $$ **Result:** $F < 1$, so a 10 m vertical cut in this soil is **unsafe** and would fail. The maximum safe height of a vertical cut is about 7.25 m (with $F = 1$). For $F = 1.5$, the cut must be flattened (read $N_s$ for a smaller $\beta$ from Taylor's chart) or supported by a retaining system. **Answer: $F \approx 0.72$ for a vertical cut (unsafe).** The depth factor $n_d = 1.5$ matters only for flatter slopes ($\beta<53^\circ$). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With bedrock 15 m below the original ground and a 10 m cut, what is the depth factor nd? A. 0.67 B. 1.0 C. 1.5 D. 2.5 Answer: C. nd = D/H = 15/10 = 1.5. (Key point: Depth factor) 2. What is Taylor's stability number for a φ = 0 soil? A. Ns = γH/cm B. Ns = cm/(γH) C. Ns = cm tanφ/(γH) D. Ns = H/c Answer: B. Ns = cm/(γH), so Hc = c/(γ Ns). (Key point: Stability number) 3. For a vertical cut (Ns ≈ 0.261), c = 35 kN/m2 and γ = 18.5 kN/m3, what is Hc? A. About 7.25 m B. About 10 m C. About 13.8 m D. About 2.4 m Answer: A. Hc = 35/(18.5 x 0.261) = 7.25 m. (Key point: Critical height) 4. What is the result for the 10 m vertical cut? A. F = 1.38, safe B. F = 0.72, unsafe C. F = 1.5, safe D. F = 0.26, unsafe Answer: B. F = Hc/H = 7.25/10 = 0.72 < 1, so the cut would fail unless flattened or supported. (Key point: Conclusion on safety) 5. For which slopes does the depth factor matter in Taylor's φ = 0 analysis? A. Only steep slopes B. Only slopes flatter than about 53° C. All slopes equally D. Only vertical cuts Answer: B. For β > 53° the critical circle is a toe circle regardless of depth factor, so nd matters only for flatter slopes. 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