2.1 Weight-Volume Phase Relationships & Index Properties

Key Takeaways

  • Soil is a three-phase system comprising solid mineral grains, water, and air; phase relationships establish fundamental volumetric (e, n, S) and gravimetric (w, γ, γd, γsat, γ') definitions.
  • The foundational phase identity S · e = w · Gs couples saturation degree, void ratio, gravimetric water content, and specific gravity of soil solids.
  • Atterberg limits (LL, PL, SL) define consistency boundaries for fine-grained soils; Plasticity Index (PI = LL - PL) and Liquidity Index (LI = (w - PL)/PI) quantify soil state and stress history.
  • Relative density (Dr = (emax - e)/(emax - emin) × 100%) quantifies the degree of compaction for coarse-grained soils.
  • Non-HAZMAT chemical, electrical, and thermal soil properties—organic content, pH, sulfate/chloride, resistivity, and thermal resistivity—control corrosion protection and buried-utility design even when strength parameters look adequate.
Last updated: July 2026

2.1 Weight-Volume Phase Relationships & Index Properties

1. The Three-Phase Soil Model

Natural soil deposits are non-homogeneous, multi-phase systems composed of three distinct constituents:

  1. Solid Phase ($V_s, W_s$): Mineral grains or organic matter forming the structural soil skeleton.
  2. Liquid Phase ($V_w, W_w$): Pore water occupying a fraction or all of the void space.
  3. Gas Phase ($V_a, W_a \approx 0$): Air or soil vapor filling the remaining void space.

The total volume of a soil mass $V$ and total weight $W$ are defined as: V=Vs+Vv=Vs+Vw+VaV = V_s + V_v = V_s + V_w + V_a W=Ws+Ww+Wa=Ws+Ww(since Wa0)W = W_s + W_w + W_a = W_s + W_w \quad (\text{since } W_a \approx 0)

Where:

  • $V_v = V_w + V_a$ is the total volume of voids.
  • $W_s$ is the dry weight of solid grains.
  • $W_w$ is the weight of pore water.

The density of water is standardized as $\rho_w = 1.00 \text{ g/cm}^3 = 1000 \text{ kg/m}^3$, corresponding to a unit weight of $\gamma_w = 62.4 \text{ lb/ft}^3 = 9.81 \text{ kN/m}^3$.


2. Volumetric Relationships

Void Ratio ($e$)

The ratio of void volume to solid volume: e=VvVse = \frac{V_v}{V_s}

  • Loose sands/soft clays: $e > 0.8$
  • Dense sands/stiff clays: $e < 0.5$

Porosity ($n$)

The ratio of void volume to total volume, expressed as a fraction or percentage: n=VvV×100%n = \frac{V_v}{V} \times 100\%

Interconversion between void ratio and porosity is derived directly from $V = V_s + V_v$: n=e1+ee=n1nn = \frac{e}{1 + e} \quad \Longleftrightarrow \quad e = \frac{n}{1 - n}

Degree of Saturation ($S$)

The ratio of water volume to void volume: S=VwVv×100%S = \frac{V_w}{V_v} \times 100\%

  • Dry soil: $S = 0%$
  • Saturated soil: $S = 100%$ ($1.0$)
  • Unsaturated/Moist soil: $0% < S < 100%$

3. Gravimetric Relationships & Specific Gravity

Moisture Content ($w$)

The ratio of water weight to dry solid weight: w=WwWs×100%w = \frac{W_w}{W_s} \times 100\%

Specific Gravity of Soil Solids ($G_s$)

The dimensionless ratio of the unit weight of solid grains ($\gamma_s$) to the unit weight of water ($\gamma_w$): Gs=γsγw=WsVsγwG_s = \frac{\gamma_s}{\gamma_w} = \frac{W_s}{V_s \gamma_w} Typical values for common minerals:

  • Quartz sand: $G_s = 2.65$
  • Clay minerals (illite, kaolinite): $G_s = 2.68 - 2.75$
  • Organic soils: $G_s < 2.40$
  • Heavy iron-bearing soils (hematite): $G_s > 3.00$

4. Fundamental Phase Identity: $S \cdot e = w \cdot G_s$

The identity connecting saturation, void ratio, moisture content, and specific gravity is fundamental to geotechnical analysis.

Derivation: Se=(VwVv)(VvVs)=VwVsS \cdot e = \left(\frac{V_w}{V_v}\right) \cdot \left(\frac{V_v}{V_s}\right) = \frac{V_w}{V_s} From weight definitions: Ww=Vwγw    Vw=WwγwW_w = V_w \gamma_w \implies V_w = \frac{W_w}{\gamma_w} Ws=VsGsγw    Vs=WsGsγwW_s = V_s G_s \gamma_w \implies V_s = \frac{W_s}{G_s \gamma_w} Substituting $V_w$ and $V_s$: VwVs=Ww/γwWs/(Gsγw)=(WwWs)Gs=wGs\frac{V_w}{V_s} = \frac{W_w / \gamma_w}{W_s / (G_s \gamma_w)} = \left(\frac{W_w}{W_s}\right) \cdot G_s = w \cdot G_s Thus: Se=wGsS \cdot e = w \cdot G_s

For fully saturated soil ($S = 1.0$), $e = w \cdot G_s$.


5. Unit Weight Definitions

Moist (Total) Unit Weight ($\gamma$)

γ=WV=Gs+Se1+eγw=Gs(1+w)1+eγw=γd(1+w)\gamma = \frac{W}{V} = \frac{G_s + S e}{1 + e} \gamma_w = \frac{G_s (1 + w)}{1 + e} \gamma_w = \gamma_d (1 + w)

Dry Unit Weight ($\gamma_d$)

γd=WsV=Gsγw1+e=γ1+w\gamma_d = \frac{W_s}{V} = \frac{G_s \gamma_w}{1 + e} = \frac{\gamma}{1 + w}

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

When $S = 1.0$: γsat=Gs+e1+eγw\gamma_{sat} = \frac{G_s + e}{1 + e} \gamma_w

Submerged (Buoyant) Unit Weight ($\gamma'$)

The effective weight per unit total volume underwater: γ=γsatγw=Gs+e1+eγwγw=Gs11+eγw\gamma' = \gamma_{sat} - \gamma_w = \frac{G_s + e}{1 + e} \gamma_w - \gamma_w = \frac{G_s - 1}{1 + e} \gamma_w


6. Fine-Grained Index Properties: Atterberg Limits

Atterberg limits quantify the moisture contents at which fine-grained soils transition between physical states:

Boundary / IndexNotationDefinition / FormulaEngineering Significance
Liquid Limit$LL$Water content transitioning liquid to plastic state (Casagrande cup or fall cone test).Higher $LL$ indicates greater compressibility and swell potential.
Plastic Limit$PL$Water content at which soil crumbles when rolled into a 3.2 mm (1/8 in) thread.Defines lower limit of plastic behavior.
Shrinkage Limit$SL$Water content below which further moisture loss produces no volume reduction.Used for expansive soil evaluation.
Plasticity Index$PI$$PI = LL - PL$Range of water content over which soil remains plastic.
Liquidity Index$LI$$LI = \frac{w - PL}{PI}$Relative in-situ soil consistency and stress history indicator.
Activity$A$$A = \frac{PI}{% < 2,\mu\text{m}}$Mineralogical sensitivity (Kaolinite ~0.4, Illite ~0.9, Montmorillonite >1.5).

Interpretation of Liquidity Index ($LI$):

  • $LI < 0$: Dry, hard, brittle soil; highly overconsolidated.
  • $0 \le LI \le 1$: Plastic behavior; normal to light overconsolidation.
  • $LI > 1$: Water content exceeds liquid limit; sensitive or quick clay vulnerable to complete strength loss upon shearing.

7. Coarse-Grained Index Properties

Relative Density ($D_r$)

Used to describe the compactness of cohesionless soils (sands and gravels): Dr=emaxeemaxemin×100%D_r = \frac{e_{max} - e}{e_{max} - e_{min}} \times 100\% Alternatively expressed in terms of dry unit weight: Dr=[γdγd,minγd,maxγd,min]γd,maxγd×100%D_r = \left[\frac{\gamma_d - \gamma_{d,min}}{\gamma_{d,max} - \gamma_{d,min}}\right] \cdot \frac{\gamma_{d,max}}{\gamma_d} \times 100\%

Relative Density ($D_r%$)Descriptive ClassificationEngineering Behavior
$0 - 15%$Very LooseHigh liquefaction susceptibility, large settlement
$15 - 35%$LooseContractive behavior during shear
$35 - 65%$Medium DenseModerate shear strength, acceptable bearing capacity
$65 - 85%$DenseDilative shear response, high friction angle
$85 - 100%$Very DenseExcellent bearing capacity, minimal settlement

Grain Size Distribution Parameters

From the sieve analysis particle size distribution curve:

  • $D_{10}$ (Effective size): Diameter corresponding to 10% passing.
  • $D_{30}$: Diameter corresponding to 30% passing.
  • $D_{60}$: Diameter corresponding to 60% passing.

Coefficient of Uniformity ($C_u$): Cu=D60D10C_u = \frac{D_{60}}{D_{10}}

Coefficient of Curvature ($C_c$): Cc=(D30)2D10D60C_c = \frac{(D_{30})^2}{D_{10} \cdot D_{60}}

USCS Grading Criteria:

  • Well-Graded Gravel (GW): $C_u \ge 4$ and $1 \le C_c \le 3$.
  • Well-Graded Sand (SW): $C_u \ge 6$ and $1 \le C_c \le 3$.
  • If criteria are not met, classified as Poorly-Graded (GP or SP).

8. Detailed Worked Example

Problem Statement: A moist soil sample retrieved from a bore pit has a total volume of $0.012 \text{ m}^3$ and a total mass of $22.8 \text{ kg}$. After oven-drying at $110^\circ\text{C}$ to constant mass, the dry mass is $19.5 \text{ kg}$. Laboratory testing yields $G_s = 2.68$. Calculate:

  1. Moisture content ($w$)
  2. Moist unit weight ($\gamma$) and dry unit weight ($\gamma_d$)
  3. Void ratio ($e$) and porosity ($n$)
  4. Degree of saturation ($S$)
  5. Saturated unit weight ($\gamma_{sat}$) and submerged unit weight ($\gamma'$)

Solution Steps:

  1. Calculate Moisture Content ($w$): Ww=(MMs)g=(22.819.5)9.81=32.37 NW_w = (M - M_s) \cdot g = (22.8 - 19.5) \cdot 9.81 = 32.37 \text{ N} Ws=19.59.81=191.30 NW_s = 19.5 \cdot 9.81 = 191.30 \text{ N} w=MwMs=22.819.519.5=3.319.5=0.1692    16.92%w = \frac{M_w}{M_s} = \frac{22.8 - 19.5}{19.5} = \frac{3.3}{19.5} = 0.1692 \implies 16.92\%

  2. Calculate Moist and Dry Unit Weights: γ=WV=22.8 kg×9.81 m/s20.012 m3=223.67 N0.012 m3=18,639 N/m3=18.64 kN/m3\gamma = \frac{W}{V} = \frac{22.8 \text{ kg} \times 9.81 \text{ m/s}^2}{0.012 \text{ m}^3} = \frac{223.67 \text{ N}}{0.012 \text{ m}^3} = 18,639 \text{ N/m}^3 = 18.64 \text{ kN/m}^3 γd=γ1+w=18.641+0.1692=15.94 kN/m3\gamma_d = \frac{\gamma}{1 + w} = \frac{18.64}{1 + 0.1692} = 15.94 \text{ kN/m}^3

  3. Calculate Void Ratio ($e$) and Porosity ($n$): Using $\gamma_d = \frac{G_s \gamma_w}{1 + e}$: 1+e=Gsγwγd=2.68×9.8115.94=26.2915.94=1.6491 + e = \frac{G_s \gamma_w}{\gamma_d} = \frac{2.68 \times 9.81}{15.94} = \frac{26.29}{15.94} = 1.649 e=1.6491=0.649e = 1.649 - 1 = 0.649 n=e1+e=0.6491.649=0.3936    39.36%n = \frac{e}{1 + e} = \frac{0.649}{1.649} = 0.3936 \implies 39.36\%

  4. Calculate Degree of Saturation ($S$): Using $S \cdot e = w \cdot G_s$: S=wGse=0.1692×2.680.649=0.45350.649=0.6987    69.87%S = \frac{w \cdot G_s}{e} = \frac{0.1692 \times 2.68}{0.649} = \frac{0.4535}{0.649} = 0.6987 \implies 69.87\%

  5. Calculate $\gamma_{sat}$ and $\gamma'$: γsat=Gs+e1+eγw=2.68+0.6491.649×9.81=3.3291.649×9.81=19.81 kN/m3\gamma_{sat} = \frac{G_s + e}{1 + e} \gamma_w = \frac{2.68 + 0.649}{1.649} \times 9.81 = \frac{3.329}{1.649} \times 9.81 = 19.81 \text{ kN/m}^3 γ=γsatγw=19.819.81=10.00 kN/m3\gamma' = \gamma_{sat} - \gamma_w = 19.81 - 9.81 = 10.00 \text{ kN/m}^3


Soil Chemical, Electrical, and Thermal Properties (Non-HAZMAT)

Beyond index and strength properties, several non-hazardous chemical, electrical, and thermal soil characteristics govern durability of buried infrastructure, concrete mix design, and utility performance. These properties are typically screened from disturbed bulk samples and do not require environmental (HAZMAT) sampling protocols.

Organic Content, pH, Sulfate, and Chloride

Organic content (ASTM D2974, loss-on-ignition) quantifies the percentage of organic matter by mass. Soils exceeding roughly $5%$ organic content show increased compressibility and reduced strength relative to inorganic soils of similar index properties; soils exceeding $20$–$75%$ (agency-dependent threshold) are classified as peat (Pt) under USCS. pH (ASTM G51) below $5.5$ indicates acidic conditions that accelerate corrosion of buried ferrous metals and can aggressively attack certain concrete admixtures. Water-soluble sulfate content controls ACI 318 concrete exposure class (S0–S3) selection and cement type (Type II vs. Type V) for foundations in contact with soil. Chloride content independently drives reinforcing-steel corrosion risk and is tested alongside sulfate for "aggressive soil" classification on transportation and buried-structure projects.

Cation Exchange Capacity (CEC)

CEC measures a clay's capacity to hold and exchange cations on particle surfaces and is a strong proxy for clay mineralogy and expansive potential. High-CEC smectite/montmorillonite clays ($80$–$150\text{ meq}/100\text{g}$) are associated with high swell potential, while low-CEC kaolinite clays ($3$–$15\text{ meq}/100\text{g}$) are relatively inert. CEC results are cross-checked against Atterberg limits and activity ($A = PI/%\text{clay}$) when screening for expansive-soil risk.

Electrical Resistivity and Corrosion Classification

Soil electrical resistivity, measured in the field by the Wenner four-pin method (ASTM G57), is the primary screening parameter for corrosivity of buried metallic infrastructure (steel pipe, ductile iron, corrugated metal culverts). Lower resistivity indicates a more electrically conductive, more corrosive soil environment.

Resistivity Range (ohm-cm)Typical Corrosivity Classification
$> 20{,}000$Essentially noncorrosive
$10{,}000\text{–}20{,}000$Mildly corrosive
$5{,}000\text{–}10{,}000$Moderately corrosive
$3{,}000\text{–}5{,}000$Corrosive
$1{,}000\text{–}3{,}000$Highly corrosive
$< 1{,}000$Severely corrosive

(Bands are representative; agency-specific tables such as AASHTO/state DOT criteria may shift thresholds slightly.)

Thermal Conductivity and Resistivity

Soil thermal properties govern buried electrical duct ampacity (heat dissipation from cables) and frost-depth prediction (via the modified Berggren equation). Thermal resistivity (the inverse of conductivity, in $^\circ\text{C}\cdot\text{cm}/\text{W}$) is lowest for saturated, dense, coarse-grained soils and highest for dry, low-density soils — moisture and density, not mineralogy alone, dominate the result.

ConditionApprox. Thermal Resistivity ($^\circ\text{C}\cdot\text{cm}/\text{W}$)
Dry sand$\approx 120$
Moist clay$\approx 60$
Saturated sand/clay$40\text{–}60$

Worked Example: Resistivity-Based Corrosivity Classification

A four-pin Wenner resistivity survey along a proposed steel water-transmission main alignment returns a native-moisture reading of $\rho = 2{,}400\ \Omega\text{-cm}$. Comparing against the classification table, this value falls in the $1{,}000$–$3{,}000\ \Omega\text{-cm}$ band, classifying the soil as Highly Corrosive. Per typical utility corrosion-control practice, this classification triggers a recommendation for cathodic protection (sacrificial anode or impressed current) and/or a dielectric polyethylene encasement on the buried steel main, rather than relying on standard mill-applied coatings alone.

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Three-Phase Soil Element Representation
Test Your Knowledge

A soil specimen has a moist unit weight of 122.0 pcf and a water content of 18.5%. What is its dry unit weight?

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Test Your Knowledge

A fine-grained soil has a specific gravity Gs = 2.68, moisture content w = 22.0%, and degree of saturation S = 90.0%. What is the void ratio e of the soil?

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Test Your Knowledge

A soil test indicates a natural moisture content of w = 45%, Liquid Limit LL = 40%, and Plastic Limit PL = 20%. Which statement correctly characterizes the soil's consistency state and Liquidity Index (LI)?

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Test Your Knowledge

A clean sand has maximum and minimum void ratios of e_max = 0.85 and e_min = 0.45. If the field void ratio is e = 0.57, what is the relative density Dr of the sand?

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