10.2 Fundamental Soil Properties and Mass Balance
Key Takeaways
- Soil phases are linked by Se = wGs, where S is saturation, e is void ratio, w is moisture content, and Gs is specific gravity.
- Dry unit weight (gamma_d) relates to moist unit weight (gamma) via gamma_d = gamma / (1 + w).
- Darcy's Law is Q = k*i*A. Layered soil permeability is parallel: kh = sum(ki*Hi)/sum(Hi), perpendicular: kv = sum(Hi)/sum(Hi/ki).
- Shear strength is governed by Mohr-Coulomb: tau = c' + (sigma - u)tan(phi'), where u is pore water pressure.
- Seepage parallel to an infinite cohesionless slope surface reduces the Factor of Safety by approximately 50%.
Fundamental Soil Properties and Mass Balance
Understanding fundamental soil properties is essential for geotechnical and transportation engineering. Soil mechanics forms the foundation for earthworks, pavement design, and retaining structures. In civil engineering, soil is generally considered a three-phase system consisting of solid particles, water, and air. A deep comprehension of these volumetric and gravimetric relationships, alongside fluid flow and shear strength principles, is critical for PE Transportation candidates.
1. Three-Phase Soil System and Weight-Volume Relationships
Soil mechanics relies heavily on the three-phase diagram, which separates a soil mass into its constituent phases:
- Solids (s): The mineral or organic matter.
- Water (w): The liquid filling the void spaces.
- Air (a): The gas filling the remaining void spaces.
The total volume ($V_t$) is the sum of the volume of solids ($V_s$) and the volume of voids ($V_v$). The volume of voids consists of the volume of water ($V_w$) and the volume of air ($V_a$).
Similarly, the total weight ($W_t$) is the sum of the weight of solids ($W_s$) and the weight of water ($W_w$). The weight of air is assumed to be zero.
Key relationships include:
- Void Ratio ($e$): Ratio of the volume of voids to the volume of solids. $e = V_v / V_s$
- Porosity ($n$): Ratio of the volume of voids to the total volume. $n = V_v / V_t$. It relates to void ratio by $n = e / (1 + e)$.
- Degree of Saturation ($S$): Percentage of void volume occupied by water. $S = V_w / V_v$. For fully saturated soils, $S = 100%$.
- Moisture Content ($w$): Ratio of the weight of water to the weight of solids. $w = W_w / W_s$.
- Specific Gravity of Solids ($G_s$): Ratio of the unit weight of soil solids to the unit weight of water. $G_s = \gamma_s / \gamma_w$.
A fundamental identity linking these parameters is:
Detailed Example Calculation: Consider a soil sample with a total volume $V_t = 1.2 \text{ ft}^3$ and a total weight $W_t = 145 \text{ lb}$. After oven drying, the weight of the dry soil is $W_s = 128 \text{ lb}$. The specific gravity of solids $G_s = 2.65$. Calculate the void ratio, degree of saturation, and dry unit weight. Assume $\gamma_w = 62.4 \text{ lb/ft}^3$.
Step 1: Calculate moisture content ($w$) $W_w = W_t - W_s = 145 - 128 = 17 \text{ lb}$ $w = \frac{17}{128} = 0.1328 \text{ or } 13.28%$
Step 2: Calculate dry unit weight ($\gamma_d$) $\gamma_d = \frac{W_s}{V_t} = \frac{128}{1.2} = 106.67 \text{ lb/ft}^3$
Step 3: Calculate volume of solids ($V_s$) $V_s = \frac{W_s}{G_s \gamma_w} = \frac{128}{2.65 \times 62.4} = 0.774 \text{ ft}^3$
Step 4: Calculate void ratio ($e$) $V_v = V_t - V_s = 1.2 - 0.774 = 0.426 \text{ ft}^3$ $e = \frac{V_v}{V_s} = \frac{0.426}{0.774} = 0.550$
Step 5: Calculate degree of saturation ($S$) Using $S \cdot e = w \cdot G_s$: $S \times 0.550 = 0.1328 \times 2.65$ $S = \frac{0.3519}{0.550} = 0.64 \text{ or } 64%$
2. Darcy's Law and Hydraulic Gradient
Water flows through interconnected voids in soil due to differences in total head. The flow of groundwater is generally laminar and obeys Darcy's Law, which states that the discharge velocity ($v$) is proportional to the hydraulic gradient ($i$): Where:
- $k$ is the coefficient of permeability (or hydraulic conductivity).
- $i$ is the hydraulic gradient, defined as the head loss ($\Delta h$) over the flow length ($L$): $i = \frac{\Delta h}{L}$.
The total flow rate ($Q$) through a cross-sectional area ($A$) is given by:
Permeability is highly dependent on soil type. Coarse-grained soils like gravels and sands have high permeability ($k \approx 10^{-1} \text{ to } 10^{-3} \text{ cm/s}$), while fine-grained soils like clays have very low permeability ($k \approx 10^{-7} \text{ to } 10^{-9} \text{ cm/s}$). In stratified soils, horizontal equivalent permeability is typically dominated by the most permeable layer, while vertical equivalent permeability is dominated by the least permeable layer.
3. Boiling and Quicksand Condition
Upward seepage of water in cohesionless soils (like fine sands) can lead to a dangerous phenomenon known as a boiling or quicksand condition. This occurs when the upward seepage force equals or exceeds the downward buoyant weight of the soil, causing the effective stress to drop to zero.
Effective stress ($\sigma'$) is defined as total stress ($\sigma$) minus pore water pressure ($u$):
In an upward flow scenario, the pore water pressure increases by an amount equal to the seepage pressure ($i \cdot z \cdot \gamma_w$). The critical hydraulic gradient ($i_c$) at which the effective stress becomes zero is given by:
For most soils, $G_s \approx 2.65$ and $e \approx 0.65$, making the critical gradient $i_c$ typically around 1.0. If the actual hydraulic gradient $i$ reaches $i_c$, boiling occurs. This is critical in the design of cofferdams, excavation bracing, and retaining walls where groundwater is pumped out, creating upward flow conditions at the excavation base. A factor of safety (FS) against boiling is calculated as $FS = i_c / i_{actual}$, with a typical minimum requirement of 3.0.
4. Shear Strength of Soils and Testing Methods
The shear strength of soil represents its internal resistance to deformation and failure. The classic Mohr-Coulomb failure criterion defines shear strength ($\tau$) as: Where:
- $c'$ is the effective cohesion.
- $\sigma'$ is the effective normal stress on the failure plane.
- $\phi'$ is the effective angle of internal friction.
To determine these parameters, several laboratory tests are commonly used:
Direct Shear Test: The simplest method to determine shear strength parameters. A soil specimen is placed in a split shear box. A normal load is applied, and one half of the box is displaced horizontally at a constant rate until the soil fails in shear. While easy to perform, it forces the failure plane along the split in the box, which may not be the weakest plane.
Triaxial Tests: More versatile and reliable than the direct shear test. A cylindrical soil specimen is encased in a rubber membrane and subjected to an all-around confining pressure (cell pressure, $\sigma_3$). An axial load (deviator stress, $\Delta\sigma$) is then applied until failure. Triaxial tests are conducted in three main conditions:
- Consolidated-Drained (CD) Test: The sample is allowed to consolidate under the confining pressure (drainage open). Then, the deviator stress is applied very slowly, allowing full drainage so no excess pore pressure builds up. It yields effective stress parameters ($c', \phi'$). It takes a very long time for low-permeability soils.
- Consolidated-Undrained (CU) Test: The sample is consolidated under the cell pressure, but drainage is closed during the application of the deviator stress. Pore water pressure is measured, allowing calculation of both total and effective stress parameters. It is faster than the CD test and widely used.
- Unconsolidated-Undrained (UU) Test: No drainage is allowed during any stage. It yields total stress parameters ($c_u, \phi_u$). For saturated cohesive soils, $\phi_u$ is often 0, simplifying the strength to the undrained shear strength ($S_u$). This simulates rapid loading conditions on low-permeability soils (short-term stability).
5. Slope Stability and Countermeasures
Slope stability analysis involves evaluating the ratio of resisting forces (shear strength of the soil) to driving forces (gravity and seepage forces). The Factor of Safety (FS) is generally defined as:
Failure typically occurs along a slip surface, which can be planar (often in coarse-grained soils or along rock joints) or rotational/circular (typical in homogeneous clayey soils). Methods of slices (e.g., Bishop's, Spencer's) are common analytical techniques.
When a slope has an inadequate factor of safety, various countermeasures can be employed:
- Geometric Modifications: Flattening the slope angle, removing material from the top (crest), or adding a stabilizing berm at the toe (which provides a counter-balancing moment and increases the length of the critical slip surface).
- Drainage Control: Surface drainage (ditches) prevents water infiltration. Subsurface drainage (horizontal drains, trench drains) lowers the groundwater table, significantly reducing pore water pressures and thereby increasing effective stress and shear strength.
- Retaining Structures: Gravity walls, cantilever walls, or mechanically stabilized earth (MSE) walls provide lateral support at the toe.
- Soil Reinforcement: Ground anchors (tiebacks), soil nails, or installing piles/drilled shafts through the slip surface to provide shear resistance.
- Vegetation: Planting deep-rooted vegetation can help reduce surface erosion and provide some minor root reinforcement, although its structural benefit is limited compared to engineered solutions.
Understanding these foundational concepts ensures that transportation infrastructure, from embankments to bridge foundations, performs safely under long-term and short-term loading conditions.
A moist soil sample has a total volume of 0.05 ft³ and a total weight of 5.75 lbs. The moisture content is determined to be 14.8%, and the specific gravity of soil solids is 2.65. What is the degree of saturation (S) of this soil? (Assume the unit weight of water is 62.4 pcf).
An infinite sandy slope has an effective friction angle (phi') of 32° and a slope angle (beta) of 18°. If steady-state seepage parallel to the slope surface exists with the water table coincident with the slope surface, what is the Factor of Safety (FS) against sliding? Assume the saturated unit weight of the sand is 125 pcf, and the unit weight of water is 62.4 pcf.
A three-layered soil deposit consists of three horizontal layers (from top to bottom): Layer 1 (thickness = 4.0 ft, permeability k1 = 2.0 * 10^-4 cm/s), Layer 2 (thickness = 6.0 ft, permeability k2 = 5.0 * 10^-5 cm/s), and Layer 3 (thickness = 10.0 ft, permeability k3 = 1.0 * 10^-6 cm/s). What is the equivalent horizontal permeability (kh,eq) of this soil deposit for flow parallel to the layers?