9.1 Soil Composition, Weight-Volume Phase Relationships

9.1 Soil Composition, Weight-Volume Phase Relationships

Unlike continuous manufactured structural materials such as structural steel or reinforced concrete, soil is a natural, particulate, multi-phase system. An undisturbed mass of soil consists of solid mineral particles forming an interlocking skeleton, interspersed with void spaces filled with liquid (water), gas (air), or both. The mechanical and hydraulic behavior of a soil deposit—its compressibility, shear strength, permeability, and load-bearing capacity—is directly governed by the volumetric and gravimetric proportions of these three constituent phases.

In geotechnical engineering and on the Civil Engineering Licensure Examination (CELE), phase relationships form the computational bedrock of the Hydraulics and Geotechnical Engineering (HGE) examination cluster. Mastery of three-phase transformations is essential for solving earthwork volume changes, compaction controls, settlement predictions, and effective stress profiles.

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The Three-Phase Soil Model

To analyze soil phase proportions systematically, the physical soil mass is idealized into a separated three-phase diagram (or block diagram), dividing the total volume $V$ and total weight $W$ into their discrete constituents:

1. Solid Phase (Mineral grains): Volume $V_s$, Weight $W_s$, Mass $M_s$.
2. Liquid Phase (Pore water): Volume $V_w$, Weight $W_w$, Mass $M_w$.
3. Gas Phase (Pore air): Volume $V_a$, Weight $W_a \approx 0$, Mass $M_a \approx 0$.

+------------------+ --- Top of Soil Profile
|     Air (Va)     | Wa ≈ 0
+------------------+ --- Vv (Total Void Volume)
|    Water (Vw)    | Ww
+------------------+ --- Separates Voids from Solids
|    Solids (Vs)   | Ws
+------------------+ --- Base of Specimen

The fundamental volume and weight summation equations are:

$V = V_s + V_v = V_s + V_w + V_a$

$W = W_s + W_w + W_a = W_s + W_w$

where $V_v = V_w + V_a$ represents the volume of voids, and the weight of the air phase ($W_a$) is neglected in terrestrial engineering calculations.

Limiting Two-Phase States

Soil deposits frequently transition between three-phase and two-phase states based on groundwater conditions and environmental exposure:

• Completely Dry Soil ($S = 0\%$): The void spaces are occupied entirely by air ($V_w = 0, V_v = V_a, W_w = 0$). The total weight equals the dry solids weight ($W = W_s$).
• Fully Saturated Soil ($S = 100\%$): The void spaces are completely filled with water ($V_a = 0, V_v = V_w$). All air is expelled, and the total weight is $W = W_s + W_w$.
• Partially Saturated (Moist) Soil ($0\% < S < 100\%$): The typical vadose zone state where voids contain both pore water and pore air.

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Volumetric Phase Ratios

Volumetric ratios define the relative spacing and distribution of voids and solids within the soil matrix.

1. Void Ratio ($e$)

The void ratio ($e$) is defined as the ratio of the volume of voids to the volume of solid soil grains:

$e = \frac{V_v}{V_s}$

• The void ratio is expressed as a pure decimal. Unlike porosity, $e$ can theoretically exceed $1.0$.
• In clean dense sands, $e$ typically ranges from $0.40$ to $0.65$; in loose sands, $e$ ranges from $0.70$ to $0.90$.
• In soft clays and organic soils, void ratios commonly exceed $1.5$ and can exceed $3.0$; highly plastic sodium bentonite can reach still higher values.

2. Porosity ($n$)

The porosity ($n$) is the ratio of the volume of voids to the total soil volume:

$n = \frac{V_v}{V} = \frac{V_v}{V_s + V_v}$

Porosity is commonly expressed as a percentage ($0\% < n < 100\%$) or decimal ($0 < n < 1.0$). By dividing the numerator and denominator by $V_s$, the rigorous interconversion formulas between $e$ and $n$ are derived:

$n = \frac{V_v / V_s}{(V_s + V_v) / V_s} = \frac{e}{1+e}$

$e = \frac{n}{1-n}$

3. Degree of Saturation ($S$ or $S_r$)

The degree of saturation ($S$) measures the volumetric fraction of void space occupied by liquid water:

$S = \frac{V_w}{V_v} \times 100\% = \frac{V_w}{V_w + V_a} \times 100\%$

4. Air Content ($a_c$) and Percent Air Voids ($n_a$)

• Air Content ($a_c$): The fraction of void volume occupied by air: $a_c = \frac{V_a}{V_v} = 1 - S$
• Percent Air Voids ($n_a$): The ratio of air volume to the total soil volume: $n_a = \frac{V_a}{V} = \frac{V_v(1-S)}{V} = n(1 - S)$

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Gravimetric Phase Ratios & The Master Identity

Gravimetric ratios compare the weights (or masses) of the constituent phases.

1. Moisture Content / Water Content ($w$)

The gravimetric moisture content ($w$) is the ratio of the weight (or mass) of pore water to the weight (or mass) of dry solid soil grains:

$w = \frac{W_w}{W_s} \times 100\% = \frac{M_w}{M_s} \times 100\% = \frac{W - W_s}{W_s} \times 100\%$

2. Specific Gravity of Soil Solids ($G_s$)

The specific gravity of solid grains ($G_s$) is the ratio of the unit weight of the solid soil skeleton ($\gamma_s$) to the unit weight of distilled water ($\gamma_w$) at a standard reference temperature of $4^\circ\text{C}$:

$G_s = \frac{\gamma_s}{\gamma_w} = \frac{\rho_s}{\rho_w} = \frac{W_s}{V_s \gamma_w} = \frac{M_s}{V_s \rho_w}$

Standard unit weight values for pure water:

• SI Metric: $\gamma_w = 9.81\text{ kN/m}^3$ (or mass density $\rho_w = 1.00\text{ g/cm}^3 = 1000\text{ kg/m}^3$).
• US Customary: $\gamma_w = 62.4\text{ lb/ft}^3$.

Typical values of $G_s$ across geological materials:

• Standard Quartz Sands: $2.65$
• Inorganic Silts: $2.66 - 2.70$
• Lean and Fat Clays: $2.70 - 2.82$
• Organic soils and peats: often well below $2.60$, because organic matter is light
• Iron-rich lateritic soils: can exceed $2.80$

3. Derivation of the Fundamental Master Identity: $S \cdot e = w \cdot G_s$

Consider an idealized soil element whose solid skeleton has a unit volume ($V_s = 1.0$):

1. From the definition of void ratio, $V_v = e \cdot V_s = e$.
2. From specific gravity, the weight of solids is $W_s = G_s \gamma_w V_s = G_s \gamma_w$.
3. From moisture content, the weight of water is $W_w = w W_s = w G_s \gamma_w$.
4. The volume of water is $V_w = \frac{W_w}{\gamma_w} = \frac{w G_s \gamma_w}{\gamma_w} = w G_s$.
5. From the definition of saturation: $S = \frac{V_w}{V_v} = \frac{w G_s}{e}$.

Rearranging produces the Master Phase Identity:

$S \cdot e = w \cdot G_s$

For a fully saturated soil where $S = 1.0$ (or 100%), this simplifies to the indispensable relationship:

$e = w_{sat} \cdot G_s$

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Formulations for Soil Unit Weights

The unit weight (or density) of a soil mass describes its gravitational weight per unit volume under varying degrees of saturation and drainage.

1. Moist / Bulk Unit Weight ($\gamma$)

The bulk unit weight (also called moist, total, or wet unit weight) represents the in-situ weight of all phases divided by the total volume:

$\gamma = \frac{W}{V} = \frac{W_s + W_w}{V_s + V_v}$

Substituting $W_s = G_s \gamma_w V_s$, $W_w = S e \gamma_w V_s$, and $V = V_s(1+e)$:

$\gamma = \frac{G_s \gamma_w V_s + S e \gamma_w V_s}{V_s(1+e)} = \frac{(G_s + S e)\gamma_w}{1+e}$

Using the identity $S e = w G_s$, this can alternatively be formulated in terms of gravimetric moisture content:

$\gamma = \frac{(G_s + w G_s)\gamma_w}{1+e} = \frac{G_s \gamma_w(1+w)}{1+e}$

2. Dry Unit Weight ($\gamma_d$)

The dry unit weight reflects the packing density of the solid mineral framework in the absence of water weight ($W_w = 0$ or $S = 0$):

$\gamma_d = \frac{W_s}{V} = \frac{G_s \gamma_w}{1+e}$

Dividing the moist unit weight formula by $(1+w)$ yields the primary field compaction relationship:

$\gamma_d = \frac{\gamma}{1+w}$

3. Saturated Unit Weight ($\gamma_{sat}$)

When every void space is filled with water ($S = 1.0$), the soil reaches its saturated unit weight:

$\gamma_{sat} = \frac{(G_s + e)\gamma_w}{1+e}$

4. Submerged / Effective / Buoyant Unit Weight ($\gamma'$)

When soil lies below the groundwater table, Archimedes' buoyant force acts upward on the soil particles. The submerged unit weight represents the net effective gravitational force transferred to underlying strata:

$\gamma' = \gamma_{sat} - \gamma_w = \frac{(G_s + e)\gamma_w}{1+e} - \frac{(1+e)\gamma_w}{1+e} = \frac{(G_s - 1)\gamma_w}{1+e}$

5. Zero-Air-Voids Unit Weight ($\gamma_{zav}$)

In soil compaction, the zero-air-voids dry unit weight is the theoretical maximum dry density achievable at a given moisture content if all air could be expelled ($S = 100\%$):

$\gamma_{zav} = \frac{G_s \gamma_w}{1 + w G_s}$

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Relative Density of Granular Soils ($D_r$)

For coarse-grained, cohesionless deposits (sands and gravels), void ratio alone does not convey whether a soil is structurally dense or loose because the particle size distribution and grain angularity dictate the minimum and maximum possible void volumes. The state of packing is therefore quantified through Relative Density ($D_r$ or $I_D$):

$D_r = \frac{e_{\max} - e}{e_{\max} - e_{\min}} \times 100\%$

where:

• $e_{\max}$ = maximum void ratio of the soil in its loosest laboratory state (ASTM D4254).
• $e_{\min}$ = minimum void ratio of the soil in its densest vibrated state (ASTM D4253).
• $e$ = in-situ natural void ratio.

Formulation in Terms of Dry Unit Weights

Because field quality control evaluates unit weight rather than void ratio directly, substituting $e = \frac{G_s \gamma_w}{\gamma_d} - 1$ produces:

$D_r = \frac{\left(\frac{G_s \gamma_w}{\gamma_{d,\min}} - 1\right) - \left(\frac{G_s \gamma_w}{\gamma_d} - 1\right)}{\left(\frac{G_s \gamma_w}{\gamma_{d,\min}} - 1\right) - \left(\frac{G_s \gamma_w}{\gamma_{d,\max}} - 1\right)} = \frac{\frac{1}{\gamma_{d,\min}} - \frac{1}{\gamma_d}}{\frac{1}{\gamma_{d,\min}} - \frac{1}{\gamma_{d,\max}}} \times 100\%$

Algebraic simplification yields the widely used reciprocal product formula:

$D_r = \left[\frac{\gamma_d - \gamma_{d,\min}}{\gamma_{d,\max} - \gamma_{d,\min}}\right] \left[\frac{\gamma_{d,\max}}{\gamma_d}\right] \times 100\%$

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CELE Board-Exam Worked Situational Problem

Problem Statement

An undisturbed soil sample recovered using a thin-walled Shelby tube from a bridge pier exploration site in Batangas has a cylindrical volume $V = 0.00100\text{ m}^3$ ($1.00\text{ liter}$ or $1000\text{ cm}^3$). The total moist mass of the specimen is $M = 1.950\text{ kg}$. After drying in an oven at $105^\circ\text{C}$ for 24 hours, the dry mass is $M_s = 1.625\text{ kg}$. Independent pycnometer laboratory testing establishes that the specific gravity of the solid soil grains is $G_s = 2.70$. Take $\gamma_w = 9.81\text{ kN/m}^3$ and $\rho_w = 1.000\text{ g/cm}^3 = 1000\text{ kg/m}^3$.

Calculate:

1. The gravimetric moisture content ($w$), void ratio ($e$), and porosity ($n$).
2. The degree of saturation ($S$) and moist (bulk) unit weight ($\gamma$).
3. The saturated unit weight ($\gamma_{sat}$), submerged buoyant unit weight ($\gamma'$), and the additional mass of water (in kilograms) required to bring the specimen to full 100% saturation.

Step-by-Step Solution

Part 1: Moisture Content, Void Ratio, and Porosity

1. Moisture Content ($w$): $M_w = M - M_s = 1.950\text{ kg} - 1.625\text{ kg} = 0.325\text{ kg}$ $w = \frac{M_w}{M_s} \times 100\% = \frac{0.325\text{ kg}}{1.625\text{ kg}} \times 100\% = 20.00\%$

2. Volume of Solid Grains ($V_s$): $V_s = \frac{M_s}{G_s \rho_w} = \frac{1.625\text{ kg}}{2.70 \times 1000\text{ kg/m}^3} = 6.0185 \times 10^{-4}\text{ m}^3$

3. Volume of Voids ($V_v$): $V_v = V - V_s = 0.00100\text{ m}^3 - 6.0185 \times 10^{-4}\text{ m}^3 = 3.9815 \times 10^{-4}\text{ m}^3$

4. Void Ratio ($e$): $e = \frac{V_v}{V_s} = \frac{3.9815 \times 10^{-4}\text{ m}^3}{6.0185 \times 10^{-4}\text{ m}^3} = 0.6615 \approx 0.662$

5. Porosity ($n$): $n = \frac{e}{1+e} = \frac{0.6615}{1 + 0.6615} = \frac{0.6615}{1.6615} = 0.3981 = 39.81\%$

Part 2: Degree of Saturation and Moist Unit Weight

1. Degree of Saturation ($S$): $V_w = \frac{M_w}{\rho_w} = \frac{0.325\text{ kg}}{1000\text{ kg/m}^3} = 3.250 \times 10^{-4}\text{ m}^3$ $S = \frac{V_w}{V_v} \times 100\% = \frac{3.250 \times 10^{-4}\text{ m}^3}{3.9815 \times 10^{-4}\text{ m}^3} \times 100\% = 81.63\%$ Cross-check using master identity: $S = \frac{w G_s}{e} = \frac{0.2000 \times 2.70}{0.6615} = \frac{0.5400}{0.6615} = 81.63\%$. (Verified).

2. Moist (Bulk) Unit Weight ($\gamma$): $W = M \times g = 1.950\text{ kg} \times 9.81\text{ m/s}^2 = 19.130\text{ N} = 0.019130\text{ kN}$ $\gamma = \frac{W}{V} = \frac{0.019130\text{ kN}}{0.00100\text{ m}^3} = 19.13\text{ kN/m}^3$ Alternative formula: $\gamma = \frac{G_s \gamma_w(1+w)}{1+e} = \frac{2.70 \times 9.81 \times 1.200}{1.6615} = \frac{31.784}{1.6615} = 19.13\text{ kN/m}^3$.

Part 3: Saturated Unit Weight, Submerged Unit Weight, and Water to Satiate

1. Saturated Unit Weight ($\gamma_{sat}$): $\gamma_{sat} = \frac{(G_s + e)\gamma_w}{1+e} = \frac{(2.70 + 0.6615) \times 9.81}{1 + 0.6615} = \frac{3.3615 \times 9.81}{1.6615} = 19.85\text{ kN/m}^3$

2. Submerged Unit Weight ($\gamma'$): $\gamma' = \gamma_{sat} - \gamma_w = 19.85 - 9.81 = 10.04\text{ kN/m}^3$ Alternative check: $\gamma' = \frac{(G_s - 1)\gamma_w}{1+e} = \frac{1.70 \times 9.81}{1.6615} = \frac{16.677}{1.6615} = 10.04\text{ kN/m}^3$.

3. Additional Water Mass for 100% Saturation ($\Delta M_w$): At saturation, the volume of water equals total void volume ($V_{w,sat} = V_v = 3.9815 \times 10^{-4}\text{ m}^3$): $M_{w,sat} = V_{w,sat} \times \rho_w = 3.9815 \times 10^{-4}\text{ m}^3 \times 1000\text{ kg/m}^3 = 0.39815\text{ kg}$ $\Delta M_w = M_{w,sat} - M_w = 0.39815\text{ kg} - 0.32500\text{ kg} = 0.07315\text{ kg} = 73.15\text{ grams}$

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CELE Board Examination Traps & Critical Pitfalls