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BASIC DEFINITIONS AND TERMINOLOGY OF SOILS
Soil is a three phase material which consists of solid particles which make up the soil skeleton and voids which may be full of water if the soil is saturated, may be full of air if the soil is dry, or may be partially saturated as shown in Figure 1.
Solid
Water
Air
Figure 1: Air, Water and Solid phases in a typical soil
It is useful to consider each phase individually as shown in Table 1.
Phase Volume Mass Weight Air VA 0 0 Water VW M (^) W WW Solid VS M (^) S WS
Table 1 Distribution by Volume, Mass, and Weight
2.1 Units
For most engineering applications the following units are used:
Length metres Mass tonnes (1 tonne = 10 3 kg) Density (mass/unit volume) t/m^3 Weight kilonewtons (kN) Stress kilopascals (kPa) 1 kPa = 1 kN/m^2 Unit Weight kN/m^3
To sufficient accuracy the density of water ρw is given by
ρw = 1 tonne/m^3 = 1 g/cm^3
In most applications it is not the mass that is important, but the force due to the mass, and the weight, W, is related to the mass, M, by the relation
W = M g
where g is the acceleration due to gravity. If M is measured in tonnes and W in kN, g = 9.8 m/s 2
Because the force is usually required it is often convenient in calculations to use the unit weight, γ (weight per unit volume).
W
V
M g V
= ρ g
Hence the unit weight of water, γw = 9.8 kN/m^3
2.2 Specific Gravity
Another frequently used quantity is the Specific Gravity, G, which is defined by
G
Density of Material = (^) Density of Water = w
G
Unit Weight of Material = (^) Unit Weight of Water = w
It is often found that the specific gravity of the materials making up the soil particles are close to the value for quartz, that is
Gs ≈ 2.
For all the common soil forming minerals 2.5 < Gs < 2.
We can use Gs to calculate the density or unit weight of the solid particles
ρs = Gs ρw γs = Gs γw
and hence the volume of the solid particles if the mass or weight is known
V
M
G
W
s G
s s w
s s w
2.3 Voids Ratio and Porosity
Va = Vv - Vw = e Vs (1 - S)
Phase Volume Mass Weight Air e (1 - S) 0 0 Water e S e S ρw e S γw Solid (^1) Gs ρw Gs γw
Table 2 Distribution by Volume, Mass and Weight in Soil
Note that Table 2 assumes a solid volume Vs = 1 m^3 , All terms in the table should be multiplied by Vs if this is not the case.
2.5 Unit Weights
Several unit weights are used in Soil Mechanics. These are the bulk, saturated, dry, and submerged unit weights.
The bulk unit weight is simply defined as the weight per unit volume
γ bulk W
V
When all the voids are filled with water the bulk unit weight is identical to the saturated unit weight, γsat , and when all the voids are filled with air the bulk unit weight is identical with the dry unit weight, γdry. From Table 2 it follows that
γ bulk W γ w^ s^ γ w^ γ w^ s
V
G e S e
G e S e
γ sat γ w^ Gs^ e
e
S = 1
γ dry γ w^ Gs
e
S = 0
Note that in discussing soils that are saturated it is common to discuss their dry unit weight. This is done because the dry unit weight is simply related to the voids ratio, it is a way of describing the amount of voids.
The submerged unit weight, γ´, is sometimes useful when the soil is saturated, and is given by
γ´ = γsat - γw
2.6 Moisture content
The moisture content, m , is a very useful quantity because it is simple to measure. It is defined as the ratio of the weight of water to the weight of solid material
m W W
w s
If we express the weights in terms of e, S, Gs and γw as before we obtain
Ww = γw Vw = γw e S Vs
Ws = γs Vs = γw Gs Vs
and hence
m e S G (^) s
Note that if the soil is saturated (S=1) the voids ratio can be simply determined from the moisture content.
Example – Mass and Volume fractions
A sample of soil is taken using a thin walled sampling tube into a soil deposit. After the soil is extruded from the sampling tube a sample of diameter 50 mm and length 80 mm is cut and is found to have a mass of 290 g. Soil trimmings created during the cutting process are weighed and found to have a mass of 55 g. These trimmings are then oven dried and found to have a mass of 45 g. Determine the phase distributions, void ratio, degree of saturation and relevant unit weights.
- Distribution by mass and weight
Phase Trimmings Mass (g)
Sample Mass, M (g)
Sample Weight, Mg (kN)
Total (^55 290 2845) × 10 - Solid 45 237.3 (^) 2327.9 × 10 - Water 10 52.7 (^517) × 10 -
- Distribution by Volume
Sample Volume , V = π (0.025) 2 (0.08) = 157.1 × 10 -6^ m^3
WaterVolume V (^) w W^ w m w
= = ×^ =. ×
− −
(^66 )
Solid Volume V W G s s m s w
= = ×.
×
= ×
− −
(^66 )
- Dry unit weight γ (^) dry kN m
= 26 0 = kN m 17
- Saturated unit weight γ (^) sat = (^.^ +^.^ ) = kN m .
If the soil were fully saturated the moisture content would be
- Moisture content m = 6 87 = = 26 0
Alternatively the unit weights may be calculated from the expressions given earlier which are on p. 5 of the Data Sheets
γ dry Gs^ γ w
e
γ sat Gs^ e γ w
e
See also pages 4 and 5 of the Data Sheets for most of the definitions and equations given in this section.
