Index Properties of Soil — Void Ratio, Porosity, Degree of Saturation

Three-phase diagram, all volume–mass relationships, unit weights, specific gravity, and relative density — with derivations and GATE-level worked examples

Last Updated: March 2026

Key Takeaways

  • Soil is a three-phase material: solid particles, water, and air. All index property relationships come from the three-phase diagram.
  • Void ratio e = Vv/Vs; porosity n = Vv/V; relationship: e = n/(1−n) and n = e/(1+e).
  • Degree of saturation S = Vw/Vv × 100 %; for fully saturated soil S = 100 %, for dry soil S = 0 %.
  • Water content w = Mw/Ms × 100 % (mass of water / mass of dry solids); key relationship: Se = wGs.
  • Specific gravity of soil solids Gs = 2.60–2.80 for most soils; Gs = 2.68 is the standard assumed value for sandy soils.
  • Unit weights: γd (dry) = γsat / (1+e); γsat = (Gs+e)γw/(1+e); γ′ (submerged) = γsat − γw.
  • Relative density Dr = (emax−e)/(emax−emin) × 100 % classifies the density state of coarse-grained soils.

1. Three-Phase Diagram

Unlike steel or concrete, soil is not a continuous solid. It is a particulate material consisting of solid mineral grains with voids between them. The voids are filled partly or wholly with water. This three-component nature — solid, water, air — is captured in the three-phase diagram (also called the phase diagram or block diagram), which is the starting point for all index property relationships.

1.1 Notation

SymbolMeaningUnit
VTotal volume of soil samplem³ or cm³
VsVolume of solid particlesm³ or cm³
VvVolume of voids (= Vw + Va)m³ or cm³
VwVolume of waterm³ or cm³
VaVolume of airm³ or cm³
MTotal masskg or g
MsMass of solid particles (= dry mass)kg or g
MwMass of waterkg or g
MaMass of air ≈ 0

1.2 Phase Relationships

V = Vs + Vv = Vs + Vw + Va

M = Ms + Mw   (Ma ≈ 0)

For a fully saturated soil: Va = 0, so Vv = Vw.
For a completely dry soil: Vw = 0, so Vv = Va.

2. Volume Ratios — Void Ratio & Porosity

2.1 Void Ratio (e)

e = Vv / Vs

Range: e > 0 (no upper limit; loose soils can have e > 1)

Typical values: Dense sand 0.4–0.6 | Loose sand 0.6–0.9 | Soft clay 0.8–1.5 | Peat > 2.0

Void ratio is preferred over porosity in geotechnical calculations because when the volume of soil changes (due to consolidation or compaction), the volume of solids Vs remains constant — making e easier to track mathematically.

2.2 Porosity (n)

n = Vv / V

Range: 0 < n < 1 (expressed as a fraction or percentage)

Typical values: Sand 25–50 % | Clay 40–70 % | Gravel 25–40 %

2.3 Relationship Between e and n

Since V = Vs + Vv:

e = n / (1 − n)

n = e / (1 + e)

Derivation: n = Vv/V = Vv/(Vs+Vv). Dividing numerator and denominator by Vs: n = (Vv/Vs) / (1 + Vv/Vs) = e/(1+e). ✓

2.4 Air Voids Ratio (av) and Air Content (Ac)

av = Va / V = n (1 − S)    (fraction of total volume occupied by air)

Air content Ac = Va / Vv = 1 − S

3. Water Content & Degree of Saturation

3.1 Water Content (w)

w = Mw / Ms × 100 %

Also written as a decimal: w = Mw / Ms

Measured by IS 2720 Part 2: oven-drying at 105–110 °C for 24 hours

Typical values: Sand 10–40 % | Stiff clay 20–40 % | Soft clay 40–120 % | Peat 200–1000 %

Water content is always expressed relative to the dry mass of solids, not total mass. This is because Ms is constant for a given soil sample regardless of wetting or drying, whereas total mass M changes.

3.2 Degree of Saturation (S)

S = Vw / Vv × 100 %

S = 0 % → completely dry soil

S = 100 % → fully saturated soil (all voids filled with water)

0 % < S < 100 % → partially saturated (three-phase soil)

4. Specific Gravity of Soil Solids (Gs)

Gs = ρs / ρw = Ms / (Vs × ρw)

ρw = 1000 kg/m³ (density of water at 4 °C)

Soil TypeGs RangeTypical Value
Gravel & sand2.65–2.682.67
Silt2.65–2.702.68
Clay (inorganic)2.68–2.802.72
Organic soil2.40–2.602.50
Peat1.30–1.901.60

Gs is determined by the pycnometer method (IS 2720 Part 3). Unless stated otherwise in problems, use Gs = 2.67 for sand and Gs = 2.70 for clay.

5. The Fundamental Relation: Se = wGs

This is the single most useful equation in the index properties topic — it connects four key parameters.

S e = w Gs

Derivation

S = Vw/Vv   →   Vw = S × Vv = S × e × Vs

Mw = Vw × ρw = S e Vs ρw

Ms = Vs × ρs = Vs × Gs × ρw

w = Mw/Ms = (S e Vs ρw) / (Vs Gs ρw) = Se/Gs

S e = w Gs

Special Cases

ConditionS valueSimplification
Fully saturatedS = 1e = wGs
Completely dryS = 0w = 0 (no water)

6. Unit Weights of Soil

Unit weight γ = W/V (weight per unit volume, kN/m³). All unit weight formulas are derived from the three-phase diagram using Gs, e, S, and γw = 9.81 kN/m³ ≈ 10 kN/m³.

6.1 Summary of Unit Weight Formulas

TypeSymbolFormulaTypical Range (kN/m³)
Bulk (moist)γ(Gs + Se) γw / (1+e)16–22
DryγdGs γw / (1+e)14–20
Saturatedγsat(Gs+e) γw / (1+e)18–23
Submerged (buoyant)γ′γsat − γw = (Gs−1) γw / (1+e)8–13

6.2 Relationship Between Bulk and Dry Unit Weight

γd = γ / (1 + w)

This is the most-used conversion in compaction and field density problems.

6.3 Derivation of γsat

Take Vs = 1 unit (reference volume for phase diagram):

Vv = e, V = 1 + e

For saturation: Vw = e, Mw = e ρw

Ms = Gs ρw

Wsat = (Gs + e) ρw g = (Gs + e) γw

γsat = Wsat/V = (Gs+e) γw / (1+e) ✓

6.4 Zero Air Voids Line (ZAV)

For S = 100 % (fully saturated): γd = Gs γw / (1 + wGs)

This is the theoretical maximum dry density at a given water content — the ZAV curve on a compaction plot. No point on a compaction curve can lie to the right of the ZAV line.

7. Relative Density (Dr)

Relative density (also called density index ID) describes how dense a coarse-grained soil (sand or gravel) is relative to its loosest and densest possible states.

Dr = (emax − e) / (emax − emin) × 100 %

Equivalent form in terms of dry density:

Dr = [(γd − γd,min) / (γd,max − γd,min)] × (γd,max / γd) × 100 %

Dr (%)StateDescription
0–15Very looseLiable to liquefaction; low bearing capacity
15–35LoosePoor foundation material without improvement
35–65Medium denseAcceptable for moderate loads
65–85DenseGood foundation material
85–100Very denseExcellent bearing capacity

emax is measured by pouring dry sand loosely into a mould (minimum density test, IS 2720 Part 14). emin is measured by vibrating saturated sand to maximum density (IS 2720 Part 14).

8. IS 2720 Test Methods

PropertyIS 2720 PartMethod
Water contentPart 2Oven drying at 105–110 °C for 24 h
Specific gravityPart 3Pycnometer method
Grain size (coarse)Part 4Dry sieve analysis
Grain size (fine)Part 4Hydrometer analysis (Stokes’ law)
Atterberg limitsParts 5 & 6Casagrande cup (LL), thread rolling (PL)
Field densityPart 28Core cutter method
Field densityPart 29Sand replacement method
Max/min density (sand)Part 14Pouring & vibration methods

9. Worked Examples

Example 1 — Complete Phase Diagram (GATE CE type)

Problem: A soil sample has a bulk unit weight of 19.2 kN/m³, water content of 12 %, and Gs = 2.68. Find: (a) dry unit weight, (b) void ratio, (c) porosity, (d) degree of saturation.

(a) Dry Unit Weight

γd = γ / (1+w) = 19.2 / (1 + 0.12) = 19.2 / 1.12 = 17.14 kN/m³

(b) Void Ratio

γd = Gs γw / (1+e)
17.14 = 2.68 × 9.81 / (1+e)
1+e = 2.68 × 9.81 / 17.14 = 26.29 / 17.14 = 1.534
e = 0.534

(c) Porosity

n = e/(1+e) = 0.534/1.534 = 0.348 = 34.8 %

(d) Degree of Saturation

Se = wGs
S = wGs/e = (0.12 × 2.68) / 0.534 = 0.3216 / 0.534 = 0.602 = 60.2 %

Example 2 — Saturated Unit Weight and Submerged Unit Weight

Problem: A clay sample has void ratio e = 0.80 and Gs = 2.72. Compute the saturated unit weight and submerged unit weight. Take γw = 9.81 kN/m³.

γsat = (Gs+e) γw / (1+e) = (2.72+0.80) × 9.81 / (1+0.80)
= 3.52 × 9.81 / 1.80 = 34.53 / 1.80 = 19.18 kN/m³

γ′ = γsat − γw = 19.18 − 9.81 = 9.37 kN/m³

Alternatively: γ′ = (Gs−1)γw/(1+e) = (2.72−1)×9.81/1.80 = 1.72×9.81/1.80 = 9.37 kN/m³ ✓


Example 3 — GATE-Style: Find Missing Parameters

Problem (GATE CE 2018 type): A saturated soil sample (S = 100 %) has water content w = 40 % and Gs = 2.70. Find: (a) void ratio, (b) porosity, (c) saturated unit weight.

(a) Void Ratio

For saturated soil: Se = wGs with S = 1
e = wGs = 0.40 × 2.70 = 1.08

(b) Porosity

n = e/(1+e) = 1.08/2.08 = 0.519 = 51.9 %

(c) Saturated Unit Weight

γsat = (Gs+e)γw/(1+e) = (2.70+1.08)×9.81/2.08
= 3.78 × 9.81 / 2.08 = 37.08 / 2.08 = 17.83 kN/m³

Note: High void ratio (e = 1.08) is typical for soft saturated clay — the result γsat ≈ 17.8 kN/m³ is consistent with soft clay behaviour.

10. Common Mistakes

Mistake 1: Expressing Water Content as Mw/M (Total Mass) Instead of Mw/Ms

What happens: Water content is computed as water mass divided by total mass, giving a value that is always less than 100 % regardless of how wet the soil is. This is sometimes called “moisture ratio” but is NOT what IS 2720 defines as water content.

Fix: Water content w = Mw/Ms × 100 %. In very soft clays this can exceed 100 % — that is perfectly valid and means the mass of water exceeds the mass of dry solids.

Mistake 2: Confusing Void Ratio e with Porosity n

What happens: Both measure void space but relative to different references. Substituting e into a formula that requires n (or vice versa) gives wrong answers. For example, using e = 0.4 where n = 0.4 are very different states (n = 0.4 corresponds to e = 0.667).

Fix: e = Vv/Vs (always ≥ 0, can exceed 1); n = Vv/V (always between 0 and 1). Convert using e = n/(1−n) or n = e/(1+e) before substituting.

Mistake 3: Forgetting that γd = γ / (1+w) Requires w as a Decimal

What happens: Writing γd = γ / (1 + 12) instead of γ / (1 + 0.12) when w = 12 %. This gives an answer roughly 12 times too small.

Fix: In all formulas, w must be a decimal fraction (e.g., w = 12 % → w = 0.12). The × 100 % is only for reporting the final value of w.

Mistake 4: Using γw = 10 kN/m³ Without Checking Whether the Problem Uses 9.81

What happens: Many GATE problems specify γw = 9.81 kN/m³ in the data. Using 10 kN/m³ introduces a ~2 % error which can shift answers between options. Some problems explicitly use 10 kN/m³ — always check.

Fix: Use whatever γw the problem gives. When not stated, use 9.81 kN/m³ for precision or 10 kN/m³ for quick mental arithmetic — but be consistent throughout.

Mistake 5: Applying Se = wGs With S as a Percentage Instead of a Decimal

What happens: Writing Se = wGs with S = 60 and w = 0.12 gives 60 × e = 0.12 × 2.68, yielding e = 0.00536 — clearly wrong. Both S and w must be decimals in this formula.

Fix: In Se = wGs, S is a decimal (e.g., 60 % → 0.60) and w is a decimal (e.g., 12 % → 0.12).

11. Frequently Asked Questions

Q1. Can void ratio e exceed 1.0? What does that mean physically?

Yes — void ratio can and does exceed 1.0 for many natural soils, particularly soft clays, silts, and organic soils. A void ratio of e = 1.0 simply means the volume of voids equals the volume of solid particles; e = 2.0 means the voids are twice the volume of solids. There is no theoretical upper limit for void ratio (peat has e values up to 10 or more). High void ratio indicates a loose, compressible soil that will undergo large settlements under load. Only for porosity n is the range physically restricted to 0–1, because n = Vv/V — you cannot have more void volume than total volume.

Q2. Why is void ratio e preferred over porosity n in geotechnical formulas?

When soil compresses (consolidates) or expands (swells), the volume of solid particles Vs remains essentially constant — only Vv changes. Because e = Vv/Vs and Vs is constant, tracking changes in e directly tracks changes in void volume: Δe = ΔVv/Vs. In contrast, porosity n = Vv/V, and V itself changes during consolidation, making the mathematics more complex. For this reason, consolidation theory, void ratio–pressure relationships (e-log p curves), and all settlement calculations use e as the primary variable.

Q3. What is the practical significance of submerged (buoyant) unit weight?

When soil is submerged below the water table, each soil element experiences an upward buoyant force equal to the weight of water it displaces. The net downward unit weight is therefore γ′ = γsat − γw. This submerged unit weight is used whenever computing vertical effective stress below the water table: σ′v = γ′ × z (for soil below water table). Using γsat in place of γ′ for submerged soil would compute total stress, not effective stress — a fundamental error that appears repeatedly in GATE problems.

Q4. How is relative density different from degree of compaction used in pavement design?

Relative density Dr is used exclusively for coarse-grained soils (sands and gravels) and describes how compact a deposit is relative to its possible loosest and densest states — it is based on void ratio. Degree of compaction (or relative compaction), used in pavement subgrade and embankment engineering, is the ratio of the field dry density to the maximum dry density from the Proctor test: RC = (γd,fieldd,max) × 100 %. It is applicable to any soil being compacted in a fill operation. The two measures are conceptually similar (both describe how compact the soil is relative to a reference state) but they use different reference states and different test procedures.

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