Soil is commonly represented as three phases: solid particles, water and air. Phase relationships connect the mass and volume of those phases and provide the basis for routine calculations involving moisture content, density, void ratio, porosity, saturation and unit weight.
These relationships are useful for checking laboratory results and preparing geotechnical records. They do not determine soil strength, compressibility or design suitability on their own.
Three-phase soil model
The total volume and total mass can be written as:
$$ V = V_s + V_w + V_a $$ $$ M = M_s + M_w $$The mass of air is treated as negligible.
| Symbol | Meaning | Common unit |
|---|---|---|
| \(V\) | Total volume | m3 or cm3 |
| \(V_s\) | Volume of solids | m3 or cm3 |
| \(V_w\) | Volume of water | m3 or cm3 |
| \(V_a\) | Volume of air | m3 or cm3 |
| \(M\) | Total or wet mass | kg or g |
| \(M_s\) | Dry mass of solids | kg or g |
| \(M_w\) | Mass of water | kg or g |
All masses in a calculation must use the same unit. All volumes must also use the same unit.
Moisture content
Gravimetric moisture content is the mass of water divided by the dry mass of solids:
$$ w = \frac{M_w}{M_s} $$Because \(M_w=M_{wet}-M_{dry}\):
$$ w = \frac{M_{wet}-M_{dry}}{M_{dry}} $$Multiply the decimal result by 100 to report moisture content as a percentage.
Example
A specimen has a wet mass of 575 g and a dry mass of 500 g:
$$ M_w = 575-500 = 75\ \mathrm{g} $$ $$ w = \frac{75}{500}=0.15=15.0\% $$The dry mass must not exceed the wet mass. If it does, the sample identification, balance readings and data entry should be checked.
Bulk density and dry density
Bulk density uses the total mass and total specimen volume:
$$ \rho = \frac{M}{V} $$Dry density uses the dry mass:
$$ \rho_d = \frac{M_s}{V} $$Where bulk density and moisture content are known:
$$ \rho_d = \frac{\rho}{1+w} $$The moisture content in this equation is a decimal. Entering 15 instead of 0.15 would produce an invalid result.
Using the 575 g specimen and a volume of 300 cm3:
$$ \rho = \frac{575}{300}=1.9167\ \mathrm{g/cm^3} $$ $$ \rho_d = \frac{1.9167}{1.15}=1.6667\ \mathrm{g/cm^3} $$This is equivalent to approximately 1917 kg/m3 bulk density and 1667 kg/m3 dry density.
Specific gravity of soil solids
The particle specific gravity is:
$$ G_s = \frac{\rho_s}{\rho_w} $$It is dimensionless. A measured project value should be used where the result is sensitive to \(G_s\). An assumed value must be identified as an assumption.
Void ratio and porosity
Void ratio compares void volume with solid volume:
$$ e = \frac{V_v}{V_s} $$Porosity compares void volume with total volume:
$$ n = \frac{V_v}{V} $$The two quantities are related:
$$ n = \frac{e}{1+e} $$ $$ e = \frac{n}{1-n} $$Porosity may be shown as a decimal or percentage, while void ratio is normally shown as a decimal. The labels must make that distinction clear.
Where dry density and specific gravity are known:
$$ e = \frac{G_s\rho_w}{\rho_d}-1 $$Degree of saturation
Degree of saturation is the proportion of void volume occupied by water:
$$ S_r = \frac{V_w}{V_v} $$Using moisture content, specific gravity and void ratio:
$$ S_r = \frac{wG_s}{e} $$For \(w=0.15\), \(G_s=2.65\) and \(e=0.65\):
$$ S_r = \frac{0.15\times2.65}{0.65}=0.6115 $$The calculated degree of saturation is approximately 61.2 percent.
A result materially above 100 percent indicates inconsistent inputs, assumptions, measurement uncertainty or data-entry error. It should be investigated rather than automatically capped at 100 percent.
Dry, saturated and submerged unit weight
Using void ratio and specific gravity:
$$ \gamma_d = \frac{G_s\gamma_w}{1+e} $$ $$ \gamma_{sat} = \frac{G_s+e}{1+e}\gamma_w $$ $$ \gamma' = \gamma_{sat}-\gamma_w $$Where:
| Symbol | Meaning | Common unit |
|---|---|---|
| \(\gamma_d\) | Dry unit weight | kN/m3 |
| \(\gamma_{sat}\) | Saturated unit weight | kN/m3 |
| \(\gamma'\) | Submerged or buoyant unit weight | kN/m3 |
| \(\gamma_w\) | Unit weight of water used in the calculation | kN/m3 |
For \(G_s=2.65\), \(e=0.65\) and \(\gamma_w=9.81\) kN/m3:
- porosity = 0.3939
- dry unit weight = 15.755 kN/m3
- saturated unit weight = 19.620 kN/m3
- submerged unit weight = 9.810 kN/m3
Calculation checks
Before using the result, confirm that:
- wet mass is not less than dry mass
- mass and volume units are consistent
- percentages have been converted to decimals inside equations
- specific gravity is measured or clearly identified as assumed
- void ratio is not negative
- porosity lies between 0 and 1
- degree of saturation is physically consistent
- density has not been confused with unit weight
What these formulas do not establish
Phase relationships organise physical measurements. They do not directly provide:
- allowable bearing pressure
- settlement
- shear strength
- permeability
- compaction acceptance
- soil classification
- pavement design parameters
Those assessments require appropriate investigation, testing, applicable standards and engineering judgement.
Related resources
- Relative Compaction
- Atterberg Limits
- Soil Classification
- Effective Stress in Soil
- Darcy Flow and Hydraulic Gradient
Authoritative references
This guide is an educational and checking aid. Results depend on the quality, units and applicability of the inputs. It does not replace project-specific testing, applicable standards or review by a suitably qualified geotechnical professional.