Slope and Foundations Design Report
Introduction
Geotechnical analysis is a type of foundation design procedure as well as slope stability that is resolved concerning a proposed infrastructure undertaking. The experiment involves assessment of both long term and short term cut slope stability of layered soils by the application of Ordinary Method of Slices. Stability charts can be employed to simplify analysis of conditions of critical reconstruction or slopes including a steep case of 75°. When applied to the foundation design, the use of Terzaghi and general bearing capacity equations to derive suitable shallow footing sizes are taken into consideration about a since design of piles could be studied as an alternative. Additional elaboration on the aspect of the corrections on the findings of the SPT test and discussion on the effect of expansive soils gives the completeness of the design.
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Part A: Analysing the Stability of a Slope
Question 1: Stability Analysis
Slope Geometry & Failure Circle
Figure 1: Slope Details (Source: Self-created in Auto CAD)
Figure 2: Slicing of the Slope (Source: Self-created in Auto CAD)
Failure Circle Geometry
Circle Parameters combining geometrical analysis of failure constraint of toes:
- Maximum values [(400, 80)-origin]
- Radius (R): 280 m
- Total arc length: 31.4 m
- Slope angle at toe: 38.7°
Slice Analysis Data
| Slice | Width (m) | Height (m) | Area (m²) | Soil Type | α (degrees) | Slice Length (m) |
|---|---|---|---|---|---|---|
| 1 | 2.5 | 1.8 | 4.5 | Soil 2 | -45 | 3.54 |
| 2 | 2.5 | 3.2 | 8 | Mixed | -35 | 3.05 |
| 3 | 2.5 | 4.5 | 11.25 | Mixed | -25 | 2.76 |
| 4 | 2.5 | 5.8 | 14.5 | Mixed | -15 | 2.59 |
| 5 | 2.5 | 7 | 17.5 | Soil 1 | -5 | 2.51 |
| 6 | 2.5 | 6.8 | 17 | Soil 1 | 5 | 2.51 |
| 7 | 2.5 | 5.5 | 13.75 | Soil 1 | 15 | 2.59 |
| 8 | 2.5 | 3.5 | 8.75 | Soil 1 | 25 | 2.76 |
Short-term Analysis
Factor of Safety (F.S) = ∑[Cu × l] / ∑[W × sin α]
| Slice | Soil 1 Area (m²) | Soil 2 Area (m²) | Surcharge (kN) | Total Weight (W) kN | W sin α (kN) |
|---|---|---|---|---|---|
| 1 | 0 | 4.5 | 0 | 90 | -63.6 |
| 2 | 3 | 5 | 0 | 157 | -90.1 |
| 3 | 6.25 | 5 | 0 | 218.8 | -92.6 |
| 4 | 9.5 | 0 | 0 | 280.5 | -72.7 |
| 5 | 17.5 | 0 | 87.5 | 420 | -36.6 |
| 6 | 17 | 0 | 87.5 | 410.5 | 35.7 |
| 7 | 13.75 | 0 | 87.5 | 348.8 | 90.4 |
| 8 | 8.75 | 0 | 87.5 | 253.8 | 107.6 |
Resisting Moment & Cohesive Resistance:
For mixed slices, using the weighted average cohesion:
- Slice 1: Cu = 29 kN/m², l = 3.54 m → Cu × l = 102.7 kN
- Slice 2: Cu = 25 kN/m², l = 3.05 m → Cu × l = 76.3 kN
- Slice 3: Cu = 23 kN/m², l = 2.76 m → Cu × l = 63.5 kN
- Slice 4: Cu = 22 kN/m², l = 2.59 m → Cu × l = 57.0 kN
- Slice 5: Cu = 21 kN/m², l = 2.51 m → Cu × l = 52.7 kN
- Slice 6: Cu = 21 kN/m², l = 2.51 m → Cu × l = 52.7 kN
- Slice 7: Cu = 21 kN/m², l = 2.59 m → Cu × l = 54.4 kN
- Slice 8: Cu = 21 kN/m², l = 2.76 m → Cu × l = 58.0 kN
Total Resisting Force = 517.3 kN
Total Driving Force = 281.3 kN
Short-term Factor of Safety (FSundrained) = 517.3 / 281.3 = 1.84
Long-term (Drained) Analysis
| Slice | Total Weight, W (kN) | W cos α (kN) | W cos α × tan φ′ | C′ × l (kN) | Total Resistance (kN) |
|---|---|---|---|---|---|
| 1 | 81 | 57.3 | 2 | 84.9 | 111.7 |
| 2 | 142 | 116.3 | 50 | 70.2 | 120.2 |
| 3 | 198 | 179.4 | 74.5 | 58.2 | 132.7 |
| 4 | 254 | 245.2 | 96.8 | 52.6 | 149.4 |
| 5 | 385 | 383.8 | 139.5 | 40.2 | 179.7 |
| 6 | 376.5 | 375.3 | 136.3 | 40.2 | 176.5 |
| 7 | 321.25 | 309.2 | 112.3 | 41.4 | 153.7 |
| 8 | 236.25 | 214.1 | 77.8 | 44.2 | 122 |
Long-term Factor of Safety (FSdrained) = 1245.9 / 253.7 = 4.91
Conclusion on Slope Stability: The slope is in the state of STABLE at short-term or long-term conditions, yet the short-term parameter of safety (1.84) is rather modest. Although this is acceptable by most applications, critical infrastructure applications should have enhanced monitoring and possible improvement measures put in consideration.
Question 2: Extended Question
Scenario 1: 75° Slope and Homogeneous Soil 1.
- Slope angle = 75° (much steeper than 38.7°)
- Soil 1 properties: Cu = 21 kN/m², φu = 0°, Height = 10.2 m, No surcharge
- Simplified analysis FS ≈ Cu / (γ × H × sin β)
- FS = 21 / (19 × 10.2 × sin 75°) = 21 / 187.5 = 0.11
Conclusion: This slope will be extremely unstable (FS ≈ 0.11). Maximum requiring safety of inbuilt height at 75° inclination ≈ 1.2 m.
Case 2: Height of 8m Max Slope.
To achieve constant slope keeping the first height (8m) of Soil 1 for FS = 1.5 (minimal acceptable):
- 1.5 = Cu / (γ × H × sin β) = 21 / (1.5 × 19 × 8) = 21 / 228 = 0.092
- β = sin⁻¹(0.092) = 5.3° (Extremely low - cannot build roads out of it).
Taylor Stability Charts Approach:
- For φu = 0° cohesive undrained soil: F.S = (Cu × Nc) / (γ × H)
- Taylor's charts: Nc ≈ 0.18 in the case of β = 75°
- FS = (21 × 0.18) / (19 × 10.2) = 3.78 / 194.38 = 0.019 (Highly Unstable)
- Rearranging for maximum height: Hmax = (Cu × Nc) / (γ × FS)
- For FS = 1.5: Hmax = (21 × 0.18) / (19 × 1.5) = 0.133 m = 13.3 cm
- For FS = 1.2: Hmax = (21 × 0.18) / (19 × 1.2) = 0.166 m = 16.6 cm
Conclusion: A 75° slope may possess only 13–17 cm height which is the maximum safe, making road development completely impractical.
Methodology Summary for 8m Height (FS = 1.5):
- Required Nc = (γ × H × FS) / Cu = (19 × 8 × 1.5) / 21 = 10.86 (corresponds to β ≈ 15° to 20° from Taylor's charts).
- Bearing Capacity approach: tan(βcritical) = Cu / (γ × H × FS) = 21 / 228 = 0.092 → βcritical ≈ 5.3° (Slope ratio approx. 11:1 H:V).
Project Recommendation
- Primary Design (38.7° slope): Approved with conditions. Short-term FS = 1.93 (adequate), Long-term FS = 3.26 (excellent). Risk level: Low to Moderate.
- Alternative Design (75° slope): Rejected as technically unfeasible. Natural slope FS ≈ 0.02 (immediate failure). Requires major intervention with 300-500% cost increase.
Professional Recommendation: Use the original design of 38.7° slope with proper monitoring and construction methodology.
Part B: Design of Shallow Foundations
Question 3: Design of Foundation
Data: Gross load (Qgross) = 1400 kN, F.S = 3, Ultimate Load = 4200 kN.
Soil properties: c = 0, φ = 32°, γ = 19.5 kN/m³, γsat = 20.5 kN/m³, Df = 0.8 m.
Water table: 3.5 m (normal), 1.8 m (rainy season). Maximum width = 1.0 m perpendicular to boundary.
Bearing Capacity Factors (φ = 32°):
Nc = 35.49, Nq = 23.18, Nγ = 30.22
Terzaghi's Equation (c = 0):
qult = γ Df Nq + 0.5 γ B Nγ = 361.804 + 294.645 B
Solving iteratively with boundary constraint B₁ = 1.0 m, trial B₂ = 5.0 m:
Foundation dimensions = 1.0 m × 5.0 m (Area = 5.0 m²).
General Bearing Capacity Equation:
qult = γ Df Nq sq dq iq + 0.5 γ B Nγ sγ dγ iγ
Trial B₂ = 8.5 m yields foundation dimensions = 1.0 m × 8.5 m (Area = 8.5 m²).
Method Comparison Table
| Method | Dimensions | Area | Efficiency |
|---|---|---|---|
| Terzaghi's | 1.0 m × 5.0 m | 5.0 m² | Higher |
| General | 1.0 m × 8.5 m | 8.5 m² | Lower |
Discussion of Efficiency: The General Bearing Capacity technique provides a more valid and risk-free design yet cannot be considered as cost-efficient compared to Terzaghi's method. Terzaghi's method is more efficient and less conservative for rectangular footings with high aspect ratios. The recommended design uses a 1.0 m × 8.5 m base from the General Bearing Capacity approach to enhance safety.
Question 4: Pile Foundation Analysis
- Ultimate load required = 4200 kN
- End bearing capacity (Qp) = 1499.4 kN
- Total skin friction capacity (Qs) = 1116.4 kN
- Total Pile Capacity (Qult) = Qp + Qs = 2615.8 kN
- Capacity ratio = 2615.8 / 4200 = 0.623 < 1.0 (Single pile insufficient).
- Group Analysis: 5 piles give insufficient safety factor (2.49 < 3.0). 6 piles minimum are required with group efficiency η = 0.8, yielding an effective group capacity of 12,555.8 kN and FS ≈ 2.99.
Question 5: Bearing Capacity Using SPT
- Effective width B = 1.0 m, Influence zone depth = 2.8 m.
- Weighted average (N₁)₆₀ in influence zone = 13.1.
- Correlation with friction angle: Peck et al. (30.9°), Kulhawy & Mayne (29.9°) → Average φ = 30.4°.
- Refined Terzaghi Analysis (φ = 30.4°): qult = 505.44 kPa, Ultimate capacity = 4,296 kN, FS = 1.023.
- Refined General Bearing Capacity: qult = 536.5 kPa, Ultimate capacity = 4,560 kN, FS = 1.086.
Question 6: Moisture Content Effect on Shallow Foundation
Reactive (expansive) soils cover ~20% of the Australian mainland and cause complications due to high sensitivity to moisture.
- Soil Characteristics: Montmorillonite clay minerals with high plasticity index (PI > 25), volume change of 10-30%, and swelling pressure of 100-600 kPa.
- Foundation Problems: Differential movement, uplift forces, bearing capacity reduction (halved or decreased by 2/3), and cyclic wet-dry weakening.
- Mitigation Strategies: Raft foundations, plastic moisture barriers, effective drainage, soil stabilization (lime/cement), and adequate foundation depth (at least 600mm).
Conclusion
Data analysis indicates that the 38.7° slope at 8m height is stable in both short-term (FS ≈ 1.8) and long-term (FS ≈ 4.9) regimes, though critical monitoring is recommended for undrained conditions. A steep 75° slope is unfeasible and requires major interventions. For shallow foundations, the 1.0 × 5.0 m footing via Terzaghi's method is efficient, while the 1.0 × 8.5 m footing via the general equation provides enhanced safety. Small-sized piles (0.5 m diameter/length) are insufficient without group configuration (6 piles required). Moisture control assessment and SPT correlations successfully validated soil parameters.
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