Engineering Geology & Soil Mec

Cheatsheet Content

### Soil Formation - **Process:** Weathering of rocks (physical & chemical). - **Residual Soils:** Formed in place from underlying rock. - **Transported Soils:** Moved from their place of origin by agents like water, wind, glaciers, or gravity (e.g., alluvial, aeolian, glacial, colluvial soils). ### Weathering Processes - **Physical Weathering:** Breakdown of rocks into smaller particles without chemical change (e.g., frost wedging, abrasion, temperature changes, exfoliation). - **Chemical Weathering:** Decomposition of rock minerals through chemical reactions (e.g., oxidation, carbonation, hydrolysis, dissolution). ### Clay Mineralogy - **Kaolinite:** 1:1 layer structure (one tetrahedral sheet, one octahedral sheet). Strong hydrogen bonds, stable, low plasticity. - **Illite:** 2:1 layer structure (two tetrahedral, one octahedral). Potassium ions between layers, moderate plasticity. - **Montmorillonite:** 2:1 layer structure. Weak bonds between layers, water can enter, high swelling and shrinkage, high plasticity. ### Clay Mineral Structure & Engineering Behavior - **1:1 Layer Structure (e.g., Kaolinite):** Strong bonds, less surface area, generally low plasticity, stable. - **2:1 Layer Structure (e.g., Illite, Montmorillonite):** - **Illite:** Moderate plasticity due to potassium bonding. - **Montmorillonite:** Weak Van der Waals bonds, significant swelling/shrinkage due to water absorption between layers, high plasticity, problematic for engineering. - **Influence:** Structure dictates plasticity, swelling potential, shear strength, and permeability of clay soils. ### Fine-Grained Soil Structure - **Flocculated Structure:** Edge-to-face contact of clay particles, high void ratio, lower density, higher permeability, lower strength. - **Dispersed Structure:** Face-to-face contact of clay particles, low void ratio, higher density, lower permeability, higher strength. - **Honeycomb Structure:** Open, unstable structure formed by rapid deposition of silt and fine sand. High void ratio, collapsible. - **Single-Grained Structure:** Common in sands and gravels where individual particles are distinct. ### Flocculated vs. Dispersed Structures | Feature | Flocculated Structure | Dispersed Structure | |-----------------|-----------------------------------------------------|-----------------------------------------------------| | Particle Contact| Edge-to-face | Face-to-face | | Void Ratio | Higher | Lower | | Density | Lower | Higher | | Permeability | Higher (more interconnected voids) | Lower (fewer interconnected voids) | | Compressibility | Higher (rearrangement of particles under load) | Lower (more stable arrangement) | | Shear Strength | Generally lower (unstable arrangement) | Generally higher (denser packing) | ### Surface Activity & Diffuse Double Layer - **Surface Activity:** The electrostatic forces and chemical reactions occurring at the surface of clay particles due to their charged nature. - **Diffuse Double Layer (DDL):** A layer of adsorbed water and ions surrounding clay particles, formed due to the negative charge on clay surfaces attracting positive ions from pore water. - **Influence on Clay Behavior:** - **Plasticity:** DDL thickness affects inter-particle forces and water content, influencing plasticity. - **Swelling/Shrinkage:** Changes in DDL thickness (due to ion concentration changes) cause swelling or shrinkage. - **Compressibility & Shear Strength:** Repulsive forces within the DDL affect the overall strength and deformation characteristics. ### Soil Phase Relationships To derive the relationship between water content (w), void ratio (e), degree of saturation (S), and specific gravity (Gs): - **Definitions:** - $w = \frac{W_w}{W_s}$ (Water content = weight of water / weight of solids) - $e = \frac{V_v}{V_s}$ (Void ratio = volume of voids / volume of solids) - $S = \frac{V_w}{V_v}$ (Degree of saturation = volume of water / volume of voids) - $G_s = \frac{\gamma_s}{\gamma_w}$ (Specific gravity of solids = unit weight of solids / unit weight of water) - **Derivation:** 1. From $w = \frac{W_w}{W_s}$, we have $W_w = w \cdot W_s$. 2. We know $W_w = V_w \cdot \gamma_w$ and $W_s = V_s \cdot \gamma_s$. 3. Substituting into (1): $V_w \cdot \gamma_w = w \cdot V_s \cdot \gamma_s$. 4. Rearranging: $V_w = w \cdot V_s \cdot \frac{\gamma_s}{\gamma_w} = w \cdot V_s \cdot G_s$. 5. From $S = \frac{V_w}{V_v}$, we have $V_v = \frac{V_w}{S}$. 6. Substitute $V_w$ from (4) into (5): $V_v = \frac{w \cdot V_s \cdot G_s}{S}$. 7. From $e = \frac{V_v}{V_s}$, substitute $V_v$ from (6): $e = \frac{w \cdot V_s \cdot G_s}{S \cdot V_s}$. 8. Simplifying, we get the fundamental relationship: $$eS = wG_s$$ ### Unit Weights of Soil - **Bulk Unit Weight ($\gamma_b$ or $\gamma$):** - Definition: Total weight per total volume. - Formula: $\gamma = \frac{W}{V} = \frac{W_s + W_w}{V_s + V_v} = \frac{G_s \gamma_w (1+w)}{1+e}$ - **Dry Unit Weight ($\gamma_d$):** - Definition: Weight of solids per total volume. - Formula: $\gamma_d = \frac{W_s}{V} = \frac{G_s \gamma_w}{1+e} = \frac{\gamma}{1+w}$ - **Saturated Unit Weight ($\gamma_{sat}$):** - Definition: Total weight per total volume when all voids are filled with water (S=1). - Formula: $\gamma_{sat} = \frac{W_s + W_w}{V} = \frac{(G_s+e)\gamma_w}{1+e}$ - **Submerged Unit Weight ($\gamma'$):** - Definition: Effective unit weight of soil when submerged in water. - Formula: $\gamma' = \gamma_{sat} - \gamma_w = \frac{(G_s-1)\gamma_w}{1+e}$ ### Dry Unit Weight Derivation Derivation of dry unit weight ($\gamma_d$) in terms of bulk unit weight ($\gamma$) and water content ($w$): - **Definitions:** - Bulk Unit Weight: $\gamma = \frac{W}{V} = \frac{W_s + W_w}{V}$ - Dry Unit Weight: $\gamma_d = \frac{W_s}{V}$ - Water Content: $w = \frac{W_w}{W_s} \implies W_w = w \cdot W_s$ - **Steps:** 1. Substitute $W_w$ into the bulk unit weight formula: $\gamma = \frac{W_s + w \cdot W_s}{V}$ 2. Factor out $W_s$: $\gamma = \frac{W_s (1+w)}{V}$ 3. Rearrange the equation to isolate $\frac{W_s}{V}$: $\frac{W_s}{V} = \frac{\gamma}{1+w}$ 4. Since $\gamma_d = \frac{W_s}{V}$, we get: $$\gamma_d = \frac{\gamma}{1+w}$$ ### Grain Size Distribution & Engineering Significance - **Grain Size Distribution:** The range of particle sizes present in a soil, determined by sieve analysis (for coarse-grained soils) and hydrometer analysis (for fine-grained soils). It's typically represented by a particle size distribution curve. - **Engineering Significance:** - **Classification:** Helps classify soils (e.g., well-graded, poorly-graded, gap-graded). - **Permeability:** Influences how easily water flows through soil (coarser soils generally have higher permeability). - **Shear Strength:** Affects the soil's resistance to deformation and failure. - **Compressibility:** Impacts how much soil will settle under load. - **Filter Design:** Crucial for designing filters to prevent erosion. - **Compaction Characteristics:** Affects optimum moisture content and maximum dry density. ### Sieve Analysis Procedure - **Purpose:** To determine the grain size distribution of coarse-grained soils. - **Apparatus:** A stack of standard sieves with progressively smaller mesh sizes, a pan at the bottom, and a lid at the top. A mechanical shaker is often used. - **Procedure:** 1. **Oven Dry Sample:** Take a representative soil sample and oven-dry it to a constant weight. 2. **Weigh Sample:** Record the total dry weight of the sample ($W$). 3. **Set Up Sieves:** Arrange sieves in descending order of mesh size from top to bottom, with the pan at the bottom. 4. **Sieve:** Place the soil sample on the top sieve, place the lid, and shake the sieve stack for a standard duration (e.g., 10-15 minutes) using a mechanical shaker or manually. 5. **Weigh Retained Soil:** Carefully remove each sieve and weigh the amount of soil retained on it. Also, weigh the soil collected in the pan. 6. **Calculate Percentage Finer:** - Calculate cumulative mass retained on each sieve. - Calculate percentage retained on each sieve. - Calculate percentage passing (or finer) for each sieve size: $\text{% Finer} = \left( \frac{\text{Total Mass} - \text{Cumulative Mass Retained}}{\text{Total Mass}} \right) \times 100$ 7. **Plot Curve:** Plot the percentage finer (y-axis, arithmetic scale) against the sieve opening size (x-axis, logarithmic scale) to obtain the particle size distribution curve. ### Hydrometer Analysis - **Purpose:** To determine the grain size distribution of fine-grained soils (clay and silt particles) based on Stokes' Law. - **Principle:** Measures the change in density of a soil-water suspension over time as particles settle. Larger, heavier particles settle faster than smaller, lighter particles. Stokes' Law relates the settling velocity of a spherical particle to its diameter, density, and fluid properties. - **Limitations:** - **Assumptions:** Assumes spherical particles, no interaction between particles, and specific gravity of all particles is the same. These are often not perfectly met in real soils. - **Temperature Effects:** Viscosity of water changes significantly with temperature, affecting settling velocity. Temperature corrections are crucial. - **Dispersion:** Incomplete dispersion of soil particles can lead to inaccurate results (flocculation). - **Minimum Particle Size:** Effective for particles down to about 0.001 mm. - **Time Consuming:** Can take several hours or even days for very fine particles to settle. ### Soil Consistency & Volume Change - **Volume Change in Fine-Grained Soil with Varying Water Content:** Fine-grained soils (clays) exhibit significant volume changes as their water content varies, particularly through the Atterberg limits. - **States of Consistency:** 1. **Liquid State:** Soil behaves like a viscous liquid. Very high water content, no shear strength. 2. **Plastic State:** Soil can be molded or deformed without crumbling or cracking. Water content between liquid limit (LL) and plastic limit (PL). Exhibits some shear strength. 3. **Semi-Solid State:** Soil crumbles when rolled into threads. Water content between plastic limit (PL) and shrinkage limit (SL). Significant shear strength. 4. **Solid State:** Soil volume remains constant even with further drying. Water content below shrinkage limit (SL). High shear strength, brittle. - **Sketches:** (Representations would ideally show soil mass with decreasing water content from liquid to solid, indicating volume reduction particularly in plastic and semi-solid states, and constant volume in solid state). ### Atterberg Limits & Formulae - **Liquid Limit (LL or $w_L$):** The minimum water content at which a soil behaves as a liquid and has negligible shear strength. It is determined by the Casagrande cup or fall cone method. - No direct formula, it's a measured value. - **Plastic Limit (PL or $w_P$):** The minimum water content at which a soil can be rolled into a thread of 3 mm diameter without crumbling. - No direct formula, it's a measured value. - **Shrinkage Limit (SL or $w_S$):** The maximum water content at which a reduction in water content will not cause a decrease in the volume of the soil mass. Below SL, the soil is in a solid state. - Formula: $w_S = \left( \frac{V_1 - V_2}{W_s} \right) \times \gamma_w \times 100\%$ Where: $V_1$ = initial volume of saturated soil, $V_2$ = final dry volume of soil, $W_s$ = weight of dry soil, $\gamma_w$ = unit weight of water. - **Plasticity Index (PI or $I_P$):** The range of water content over which a soil exhibits plastic properties. It is the numerical difference between the Liquid Limit and the Plastic Limit. - Formula: $PI = LL - PL$ ### Engineering Significance of Atterberg Limits Atterberg limits are crucial for classifying fine-grained soils and predicting their engineering behavior in foundation and pavement engineering. - **Foundation Engineering:** - **Swelling/Shrinkage Potential:** High LL and PI indicate high swelling and shrinkage potential, which can cause significant damage to foundations. - **Bearing Capacity:** Soils with high water content (near LL) have very low bearing capacity. - **Compressibility:** High PI often correlates with high compressibility, leading to larger settlements. - **Slope Stability:** The consistency limits influence the shear strength, which is critical for assessing slope stability. - **Pavement Engineering:** - **Subgrade Stability:** Soils with high PI can become unstable under varying moisture conditions, leading to pavement distress (e.g., rutting, cracking). - **Frost Susceptibility:** Fine-grained soils with high plasticity can be more susceptible to frost heave. - **Compaction Control:** Atterberg limits help in determining the optimal moisture content for compaction to achieve desired density and strength. - **Material Selection:** Used to specify suitable soil types for subgrade, subbase, and base course layers. ### Soil Properties Calculation Given: Wet soil weight = 20.20 kN/m³, $G_s = 2.68$, $w = 12\%$ Assume $\gamma_w = 9.81 \text{ kN/m}^3$ **(a) Void ratio (e) and degree of saturation (S)** 1. **Calculate dry unit weight ($\gamma_d$):** $\gamma_d = \frac{\gamma_{wet}}{1+w} = \frac{20.20 \text{ kN/m}^3}{1+0.12} = \frac{20.20}{1.12} = 18.036 \text{ kN/m}^3$ 2. **Calculate void ratio (e):** We know $\gamma_d = \frac{G_s \gamma_w}{1+e}$ $1+e = \frac{G_s \gamma_w}{\gamma_d} = \frac{2.68 \times 9.81}{18.036} = \frac{26.2908}{18.036} = 1.457$ $e = 1.457 - 1 = 0.457$ 3. **Calculate degree of saturation (S):** Using the relationship $eS = wG_s$: $S = \frac{wG_s}{e} = \frac{0.12 \times 2.68}{0.457} = \frac{0.3216}{0.457} = 0.7037$ $S = 70.37\%$ **(b) Dry unit weight ($\gamma_d$) and saturated unit weight ($\gamma_{sat}$)** 1. **Dry unit weight ($\gamma_d$):** Calculated in part (a): $\gamma_d = 18.036 \text{ kN/m}^3$ 2. **Saturated unit weight ($\gamma_{sat}$):** Using the formula $\gamma_{sat} = \frac{(G_s+e)\gamma_w}{1+e}$ $\gamma_{sat} = \frac{(2.68+0.457) \times 9.81}{1+0.457} = \frac{3.137 \times 9.81}{1.457} = \frac{30.778}{1.457} = 21.124 \text{ kN/m}^3$ ### Sieve Analysis Data & Particle Size Distribution Given: Total mass of soil sample = 135.0 g | Sieve Size (mm) | Mass Retained (g) | Cumulative Mass Retained (g) | Cumulative % Retained | % Finer (Passing) | |-----------------|-------------------|------------------------------|-----------------------|-------------------| | 4.75 | 0 | 0 | 0.00 | 100.00 | | 2.36 | 3.6 | 3.6 | 2.67 | 97.33 | | 1.18 | 9.2 | 12.8 | 9.48 | 90.52 | | 0.600 | 20.4 | 33.2 | 24.59 | 75.41 | | 0.425 | 33.8 | 67.0 | 49.63 | 50.37 | | 0.300 | 36.5 | 103.5 | 76.67 | 23.33 | | 0.150 | 22.1 | 125.6 | 93.04 | 6.96 | | 0.075 | 7.8 | 133.4 | 98.81 | 1.19 | | Pan | 1.6 | 135.0 | 100.00 | 0.00 | *Note: The sum of mass retained is 133.4g from the table + 1.6g in the pan = 135.0g, matching the total mass.* **(a) Plot the particle size distribution curve on semi-log graph paper and describe the soil.** (This requires plotting on a physical graph. The description would be based on the curve's shape and D-values.) **(b) Determine D10, D30, D60, coefficient of uniformity (Cu), and coefficient of curvature (Cc)** From the "% Finer" column: - **D10:** Particle size for which 10% of the sample is finer. - 1.19% finer at 0.075 mm. 6.96% finer at 0.150 mm. - Interpolating between 0.075mm and 0.150mm for 10% finer: $D_{10} \approx 0.150 + (0.075 - 0.150) \times \frac{10 - 6.96}{1.19 - 6.96} = 0.150 - 0.075 \times \frac{3.04}{-5.77} = 0.150 + 0.0395 \approx 0.190 \text{ mm}$ (approximate interpolation) *More precisely, using logarithmic interpolation or a plot, D10 is around 0.19-0.20 mm.* Let's use a value of 0.19mm for calculation. - **D30:** Particle size for which 30% of the sample is finer. - 23.33% finer at 0.300 mm. 50.37% finer at 0.425 mm. - Interpolating for 30% finer: $D_{30} \approx 0.300 + (0.425 - 0.300) \times \frac{30 - 23.33}{50.37 - 23.33} = 0.300 + 0.125 \times \frac{6.67}{27.04} = 0.300 + 0.0308 \approx 0.331 \text{ mm}$ - **D60:** Particle size for which 60% of the sample is finer. - 50.37% finer at 0.425 mm. 75.41% finer at 0.600 mm. - Interpolating for 60% finer: $D_{60} \approx 0.425 + (0.600 - 0.425) \times \frac{60 - 50.37}{75.41 - 50.37} = 0.425 + 0.175 \times \frac{9.63}{25.04} = 0.425 + 0.0674 \approx 0.492 \text{ mm}$ Now, calculate Cu and Cc: - **Coefficient of Uniformity (Cu):** $C_u = \frac{D_{60}}{D_{10}} = \frac{0.492}{0.190} \approx 2.59$ - **Coefficient of Curvature (Cc):** $C_c = \frac{(D_{30})^2}{D_{60} \times D_{10}} = \frac{(0.331)^2}{0.492 \times 0.190} = \frac{0.109561}{0.09348} \approx 1.17$ **Soil Description:** - Since $C_u = 2.59