Soil Phase Relationships and Classification

Key Takeaways

  • A soil phase diagram separates solids, water, and air volumes while soil weight is normally the sum of solids and water weights
  • Water content is based on dry-solids weight, void ratio on solids volume, porosity on total volume, and saturation on void volume
  • Total, dry, saturated, and submerged unit weights describe different water and buoyancy conditions and cannot be substituted silently
  • Percentages must be converted to decimals before using identities such as Se = wGs, and calculated saturation cannot exceed 100 percent
  • Gradation and Atterberg-limit classification requires the applicable 2026 handbook or AASHTO procedure rather than a soil name guessed from appearance
Last updated: July 2026

Soil Phase Relationships and Classification

Draw the boxes: Most phase-relation errors disappear when volumes and weights are placed in separate columns for solids, water, and air before a formula is selected.

The Three-Phase Model

Total volume contains solids and voids, while voids contain water and air:

V = V_s + V_v and V_v = V_w + V_a.

With air weight neglected, total weight is W = W_s + W_w. The standard dimensionless relationships are:

QuantityDefinitionDenominator to remember
Water content, wW_w/W_sDry-solids weight
Void ratio, eV_v/V_sSolids volume
Porosity, nV_v/VTotal volume
Saturation, SV_w/V_vVoid volume
Solids specific gravity, G_sγ_s/γ_wWater unit weight reference

Porosity and void ratio are linked by n = e/(1 + e) and e = n/(1 - n). For a consistent phase system, Se = wG_s, with w and S entered as decimals. Thus 12% water content is 0.12, not 12. A computed S > 1 signals inconsistent data, a percent conversion error, or an invalid assumption.

Moist-versus-dry measurement check

A specimen weighs 1,000 g before oven drying and 890 g after drying. The water mass is 1,000 - 890 = 110 g, but water content is

w = 110/890 = 0.1236 = 12.4%.

Dividing by the moist 1,000-g mass would give 11.0% and use the wrong denominator. The dry measurement represents solids mass; the lost mass represents water under the stated test assumptions.

Four Unit Weights

Unit weight always means weight divided by a stated volume, but the water condition changes the numerator:

  • dry: γ_d = G_sγ_w/(1 + e); water weight is omitted;
  • total or moist: γ = G_s(1 + w)γ_w/(1 + e) = γ_d(1+w);
  • saturated: γ_sat = (G_s + e)γ_w/(1 + e); every void is water-filled;
  • submerged or buoyant: γ' = γ_sat - γ_w = (G_s - 1)γ_w/(1 + e).

Submerged unit weight is useful for effective overburden below a water table. It is not the soil's total saturated unit weight and should not be used to calculate total gravity load. Use γ_sat for total soil weight and subtract pore-water pressure when effective stress is needed.

Worked Phase-Relationship Example

A soil has G_s = 2.70, e = 0.65, and w = 12% = 0.12. Use γ_w = 62.4 pcf.

First compute porosity:

n = 0.65/(1 + 0.65) = 0.3939 = 39.4%.

Use Se = wG_s to find saturation:

S = 0.12(2.70)/0.65 = 0.498 = 49.8%.

The dry unit weight is

γ_d = 2.70(62.4)/(1.65) = 102.1 pcf.

The actual moist unit weight is

γ = 102.1(1 + 0.12) = 114.4 pcf.

For the same solids and void ratio at complete saturation,

γ_sat = (2.70 + 0.65)(62.4)/1.65 = 126.7 pcf.

and

γ' = 126.7 - 62.4 = 64.3 pcf.

The ordering check is sensible: γ_d < γ < γ_sat, and the submerged value is saturated unit weight minus water unit weight. The actual sample is only 49.8% saturated, so γ_sat describes a hypothetical saturated state rather than its current total unit weight.

Gradation of Coarse-Grained Soil

A sieve analysis reports percent passing versus particle size. Determine the fractions retained above and passing the classification sieve specified by the applicable system; do not assume all sand- or gravel-sized material behaves as fines. On a semilog gradation curve, D_10, D_30, and D_60 are particle sizes at 10%, 30%, and 60% passing. Common shape measures are

C_u = D_60/D_10

and

C_c = D_30²/(D_10D_60).

For example, if D_10 = 0.15 mm, D_30 = 0.35 mm, and D_60 = 0.75 mm, then C_u = 5.0 and C_c = 0.35²/[0.15(0.75)] = 1.09. These numbers still do not determine a symbol until fines content and the classification limits are checked.

These coefficients describe spread and curve shape, but classification also depends on gravel/sand proportions, fines content, and the system's boundary criteria. Use the PE Civil handbook active for the test date or the specified AASHTO LRFD 8th-edition procedure, including its identified May 2018 errata, rather than memorizing a symbol without its limits.

Atterberg Limits for Fine-Grained Soil

The liquid limit LL marks the boundary between liquid and plastic behavior; the plastic limit PL marks the boundary between plastic and semisolid behavior. Plasticity index is

PI = LL - PL.

If LL = 48 and PL = 22, then PI = 26. Plot the stated LL and PI on the applicable plasticity chart to distinguish silt-like and clay-like behavior and low versus high plasticity. Boundary points may require dual symbols or additional information. Organic behavior, gradation, and visual description must be handled according to the requested classification system.

Classification Workflow

  1. Confirm whether data are masses, weights, or volumes and convert every percent to a decimal for equations.
  2. Draw the phase diagram and calculate independent quantities before using derived identities.
  3. Check 0 ≤ n < 1, e ≥ 0, and 0 ≤ S ≤ 1.
  4. For classification, determine coarse versus fine fraction from the required sieve.
  5. For coarse soil, evaluate gravel/sand split, fines, C_u, and C_c; for fine soil, calculate PI and use the chart.
  6. Report the classification symbol and descriptive name from the same system.

Do not mix a USCS symbol, an AASHTO group threshold, and a casual field description into one unsupported classification.

Test Your Knowledge

For Gs = 2.70, e = 0.65, and water content w = 12%, what degree of saturation does the phase identity give?

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Test Your Knowledge

Which relationship between saturated and submerged unit weight is correct for a soil below the water table?

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B
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D
Test Your Knowledge

A fine-grained soil has liquid limit LL = 48 and plastic limit PL = 22. What is its plasticity index?

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D