# Step-by-Step Solution: Factor of Safety for Surface Mine Bench Slope ## **Given Data** - **Planned slope angle (\(\beta\))**: \(65^\circ\) - **Angle of discontinuity from horizontal (\(\theta\))**: \(41^\circ\) - **Angle of internal friction (\(\phi\))**: \(30^\circ\) - **Cohesion (\(c\))**: \(1.6\, \mathrm{kN/m^2}\) - **Unit weight (\(\gamma\))**: \(1875\, \mathrm{kg/m^3}\) - **Bench height (\(h\))**: \(50\, \mathrm{m}\) - **Thickness of block (\(t\))**: \(1\, \mathrm{m}\) - **Assume drained conditions** --- ## **Step 1: Convert Unit Weight to kN/m³** \[ \gamma = 1875\, \mathrm{kg/m^3} \times 9.81\, \mathrm{m/s^2} = 18378.75\, \mathrm{N/m^3} = 18.38\, \mathrm{kN/m^3} \] --- ## **Step 2: Geometry of the Potential Sliding Block** Assume a **block failure** along the discontinuity plane. - **Length of base along slope (\(L\))**: For unit thickness, \[ L = \frac{h}{\sin{\theta}} = \frac{50}{\sin{41^\circ}} \approx \frac{50}{.6561} \approx 76.21\, \mathrm{m} \] --- ## **Step 3: Calculate Forces** ### **Weight of Block (\(W\)):** \[ W = \text{Volume} \times \gamma = L \times t \times 1\,\mathrm{m} \times \gamma = L \times 1 \times 1 \times \gamma \] \[ W = 76.21 \times 1 \times 1 \times 18.38 = 1401.3\, \mathrm{kN} \] --- ## **Step 4: Resolve Forces Along and Perpendicular to the Discontinuity** ### **a. Normal Force (\(N\)):** \[ N = W \cos{\theta} \] \[ N = 1401.3 \times \cos{41^\circ} = 1401.3 \times .7547 = 1058.5\, \mathrm{kN} \] ### **b. Downslope Force (\(S\)):** \[ S = W \sin{\theta} \] \[ S = 1401.3 \times \sin{41^\circ} = 1401.3 \times .6561 = 919.3\, \mathrm{kN} \] --- ## **Step 5: Shear Strength Along Discontinuity** \[ \tau_f = c + N \tan{\phi} \] Where: - \(c\): Cohesion (kN/m²) - \(N\): Normal force per meter block (\(kN\)) - \(\phi\): Angle of internal friction ### **Calculate Shear Strength (\(R\)):** \[ R = c \times L + N \times \tan{\phi} \] - \(c \times L = 1.6 \times 76.21 = 121.94\, \mathrm{kN}\) - \(\tan{30^\circ} = .5774\) - \(N \times \tan{\phi} = 1058.5 \times .5774 = 611.3\, \mathrm{kN}\) \[ R = 121.94 + 611.3 = 733.2\, \mathrm{kN} \] --- ## **Step 6: Factor of Safety (FoS) Calculation** \[ \text{FoS} = \frac{\text{Shear Strength (R)}}{\text{Driving Force (S)}} \] \[ \text{FoS} = \frac{733.2}{919.3} = .80 \] --- ## **Step 7: Final Answer** ### **Factor of Safety** \[ \boxed{\text{FoS} = .80} \] --- ## **Conclusion** - **The factor of safety for the given slope is \(.80\).** - **Interpretation:** Since FoS < 1, the slope is **unstable** under these conditions and would likely fail along the discontinuity. --- ### **Summary Table** | Parameter | Value | |----------------------|-----------------| | Slope angle (\(\beta\)) | \(65^\circ\) | | Discontinuity angle (\(\theta\)) | \(41^\circ\) | | Internal friction (\(\phi\)) | \(30^\circ\) | | Cohesion (\(c\)) | \(1.6\, \mathrm{kN/m^2}\) | | Unit weight (\(\gamma\)) | \(18.38\, \mathrm{kN/m^3}\) | | Bench height (\(h\)) | \(50\, \mathrm{m}\) | | Length along discontinuity (\(L\)) | \(76.21\, \mathrm{m}\) | | FoS | \(.80\) | --- **Note:** - If you need graphical representation or more detail, please specify. - Always verify units and assumptions for your specific case.