# 1. Analytical Calculation of Factor of Safety (FoS) ## **Problem Restatement** - **Slope height, \( h \):** 15 m - **Slope angle, \( \beta \):** \( 68^\circ \) - **Discontinuity dip, \( \theta \):** \( 45^\circ \) (strike parallel to slope face) - **Friction angle, \( \phi \):** \( 32^\circ \) - **Cohesion, \( c \):** 15 kN/m\(^2\) - **Rock unit weight, \( \gamma \):** 25 kN/m\(^3\) - **Slope width (assume unit thickness):** 1 m Assume no water, no seismic forces. ## **Step 1: Find the Geometry of the Sliding Block** - **Length of failure plane, \( L \):** \[ L = \frac{h}{\sin \theta} = \frac{15}{\sin 45^\circ} = \frac{15}{.7071} \approx 21.21\,\text{m} \] - **Area of failure surface (per unit thickness):** \[ A = L \times 1 = 21.21\,\text{m}^2 \] ## **Step 2: Calculate the Weight of the Sliding Block** - **Block volume (per unit thickness):** \[ V = h \times \text{base length} \times 1\,\text{m} \] The "base length" is the horizontal projection: \[ \text{Base length} = h / \tan \beta = 15 / \tan 68^\circ = 15 / 2.475 = 6.06\,\text{m} \] So, \[ V = 15 \times 6.06 \times 1 = 90.9\,\text{m}^3 \] But for planar failure, the sliding mass is a prism above the failure plane: \[ \text{Area of block} = L \times 1 = 21.21\,\text{m}^2 \] The thickness perpendicular to the slope is 1 m (as per the problem statement). So, \[ \text{Weight},\, W = \gamma \cdot L \cdot 1 \cdot \sin(\theta) \cdot 1 \] But more precisely, \[ W = \gamma \cdot h \cdot 1 = 25 \times 15 = 375\,\text{kN} \] (This is the weight per meter width of the slope. For planar failure, use the block sitting above the failure plane. Since the discontinuity extends from toe to crest, the block is a prism of height \( h \), thickness 1 m, and width along the slope face: \( h / \sin \theta \).) The correct method is to use the area of the failure plane and the height perpendicular to it: \[ W = \gamma \cdot L \cdot 1 \cdot h_{\perp} \] But if we consider a unit width, the weight per meter width is: \[ W = \gamma \cdot h \cdot 1 = 375\,\text{kN} \] (For planar failures, it's common to use weight per unit width.) ## **Step 3: Forces Acting on the Block** - **Shear strength along the plane (\( S \)):** \[ S = c \cdot L + (N) \cdot \tan \phi \] Where \( N \) is the normal force on the plane. - **Driving force (\( D \)):** \[ D = W \cdot \sin \theta \] **Normal force (\( N \)):** \[ N = W \cdot \cos \theta \] ## **Step 4: Factor of Safety Formula** \[ \text{FoS} = \frac{c L + (W \cos\theta) \tan\phi}{W \sin\theta} \] ## **Step 5: Plug in the Numbers** - \( c = 15 \) kN/m\(^2\) - \( L = 21.21 \) m - \( W = 375 \) kN - \( \theta = 45^\circ \) - \( \phi = 32^\circ \) - \( \sin 45^\circ = .7071 \) - \( \cos 45^\circ = .7071 \) - \( \tan 32^\circ = .6249 \) ### Calculate Shear Resistance: \[ c L = 15 \times 21.21 = 318.15\,\text{kN} \] \[ W \cos\theta = 375 \times .7071 = 265.16\,\text{kN} \] \[ (W \cos\theta) \tan\phi = 265.16 \times .6249 = 165.74\,\text{kN} \] \[ \text{Numerator} = 318.15 + 165.74 = 483.89\,\text{kN} \] ### Calculate Driving Force: \[ W \sin\theta = 375 \times .7071 = 265.16\,\text{kN} \] ### Compute FoS: \[ \text{FoS} = \frac{483.89}{265.16} = 1.83 \] --- ## **Summary Table** | Parameter | Value | |------------------------|------------------------| | Slope height | \( 15\,\text{m} \) | | Slope angle | \( 68^\circ \) | | Discontinuity dip | \( 45^\circ \) | | Friction angle | \( 32^\circ \) | | Cohesion | \( 15\,\text{kN/m}^2 \)| | Unit weight | \( 25\,\text{kN/m}^3 \)| | FoS | \( 1.83 \) | --- # 2. Evaluation and Reinforcement (if required) ## **Acceptable FoS Criteria** - For permanent slopes: **minimum FoS = 1.3** - For temporary slopes: minimum FoS can be lower. ### **Conclusion:** - **Calculated FoS = 1.83 > 1.3** - The slope is **stable** as per conventional design guidelines. ## **If FoS < 1.3: Example Reinforcement Scheme** *Since the FoS is above 1.3, reinforcement is not strictly required. However, if it were below 1.3:* ### **Reinforcement Options** **Rock Bolts:** - **Assumptions:** - Bolt length: .7–1. × slope height, say 10 m - Bolt spacing: 2 m × 2 m - Bolt capacity: 150 kN per bolt **Design Rationale:** - Bolts are installed perpendicular to the failure plane to increase normal stress and shear resistance. - They prevent sliding by increasing the effective normal force on the discontinuity. --- # 3. Numerical Verification Using RocPlane **(No actual screenshot can be provided in text, but here are the steps and expected entries.)** ## **RocPlane Input Parameters** - Slope height: 15 m - Slope angle: 68° - Discontinuity dip: 45° - Friction angle: 32° - Cohesion: 15 kN/m² - Unit weight: 25 kN/m³ - Unit thickness: 1 m ## **Expected Output** - **FoS (RocPlane):** Should be *very close* to 1.83 **Discussion:** If RocPlane gives a slightly different value, possible reasons: - RocPlane may consider tension cracks, water, or partial plane length. - Analytical solution assumes a simple planar prism, unit thickness. --- # 4. Re-analysis After Reinforcement **(If reinforcement was added, e.g., bolts as above):** - Enter bolt parameters in RocPlane - Re-run analysis **Expected Outcome:** - **FoS > 1.3** - Compare to design target. ## **Technical & Economic Justification** - **Technical:** Reinforcement increases normal force, raising FoS above target. - **Economic:** Minimize number and length of bolts to achieve FoS > 1.3. --- # **Summary Table** | Step | Result/Design | |-----------|--------------| | Analytical FoS | 1.83 | | Design Target | 1.3 | | Reinforcement (if needed) | Rock bolts, 10 m long, 2x2 m spacing, 150 kN capacity | | RocPlane FoS (before reinforcement) | ≈1.83 | | RocPlane FoS (after reinforcement) | >1.3 | --- # **Conclusion** - The slope is stable with FoS ≈ 1.83 (well above 1.3). - No reinforcement is required. - Analytical and numerical (RocPlane) results are expected to match closely. - If FoS were below 1.3, reinforcement (e.g., rock bolts) could be designed as above to achieve an acceptable FoS. --- **If you need a worked example with FoS < 1.3 (for reinforcement), or screenshots from RocPlane, please specify.**