The apparent density of a sample is defined as the quotient between its mass and the unaltered volume, which includes all the spaces or pores it contains. If there is air in these spaces, the apparent density ρb, or bulk density it is:
ρb = Mass / Volume = Mass particles + Mass air /Volume particles+ Volume air
When the apparent density of a soil sample is calculated, it must be previously dried in an oven at 105 ºC until the mass is constant, indicating that all the air has evaporated..
According to this definition, the apparent density of soils or dry density, it is calculated in this way:
ρs = Weight of solid elements / Volume solid + Volume pores
Denoting as Ms to dry weight or mass and Vt = Vs + Vp as the total volume, the formula is:
ρs = Ms / Vt
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The units of bulk density in the International System of Units are kg / m3. However, other units such as g / cm3 and megagrams / cubic meter: Mg / m3 are also widely used.
The concept of apparent density is very useful when it comes to heterogeneous and porous materials such as soils, as it is indicative of their drainage and aeration capacity, among other qualities..
For example, low-porous soils have high bulk densities, are compact, and tend to water easily, unlike porous soils..
When there is water or another fluid in the pores of the sample, the volume after drying decreases, therefore, when making the calculations, it is necessary to know the original water proportion (see resolved example).
The apparent density of materials in general, including the soil, is highly variable, since there are factors such as the degree of compaction, the presence of organic matter, its texture, structure, depth and others, which affect the shape and the amount of pore spaces.
Soils are defined as a heterogeneous mixture of inorganic substances, organic substances, air and water. To the touch they can be texture fine, medium, or coarse, while the component particles can be arranged in various ways, a parameter known as structure.
Fine, well-structured soils with a high percentage of organic matter tend to have low values of apparent density. On the contrary, thick soils, with less organic matter and little structure, tend to have higher values.
According to its texture, the apparent density has the following values:
| Texture | Apparent Density (g / cm3) |
|---|---|
| Fine | 1.00 - 1.30 |
| Median | 1.30 - 1.50 |
| Gross | 1.50 - 1.70 |
These values serve as a general reference. In peaty soils, abundant in plant residues, the apparent density can be as low as 0.25 g / cm3, if it is a volcanic mineral soil it is around 0.85 g / cm3, while in very compacted soils it reaches 1.90 g / cm3.
The apparent density value also increases with depth, since the soil is generally more compacted and has a lower percentage of organic matter..
The interior of the terrain is composed of horizontal layers or strata, called horizons. Horizons have different textures, composition, and compaction. Therefore they present variation in terms of the apparent density.
A soil study is based on its profile, which consists of various horizons that occur in an orderly vertical way.
Since the variability in bulk density is very large, it often has to be measured directly by various procedures..
The simplest method is to extract a sample from the soil, inserting into it a drill with a space metal cylinder of known volume and making sure not to compact the soil. The extracted sample is sealed, to prevent loss of moisture or alteration of the characteristics.
Then in the laboratory the sample is extracted, weighed and then placed in an oven at 105ºC to dry for 24 hours..
Although it is the simplest way to find the dry density of the soil, it is not the most recommended for soils with very loose textures or full of stones..
For these, the method of digging a hole and saving the extracted earth, which will be the sample to dry, is preferable. The volume of the sample is determined by pouring dry sand or water into the dug hole.
In any case, from the sample it is possible to determine very interesting properties of the soil to characterize it. The following solved exercise describes how to do it.
A clay sample of length 100 mm is drawn from the sample cylinder, the internal diameter of which is also 100 mm. When weighed, a mass of 1531 g was obtained, which once dry was reduced to 1178 g. The specific gravity of the particles is 2.75. It is asked to calculate:
a) The bulk density of the sample
b) Moisture content
c) The void ratio
d) Dry density
e) The degree of saturation
f) Air content
The volume without altering Vt is the original volume of the sample. For a cylinder of diameter D and height h, the volume is:
V cylinder = Vt = Area of base x height = πDtwo/ 4 = π x (100 x 10-3 m)two x 100 x 10 -3 m / 4 = 0.000785 m3
The statement states that the mass of the sample is Ms = 1531 g, therefore according to the equation given at the beginning:
ρb = Ms / Vt = 1531 g / 0.000785 m3 = 1950 319 g / m3 = 1.95 Mg / m3
Since we have the original mass and the dry mass, the mass of the water contained in the sample is the difference of these two:
M Water = 1531 g - 1178 g = 353 g
The percentage of moisture in the sample is calculated as follows:
% Humidity = (Mass Water / Ms) x 100% = (353 g / 1178 g) = 29. 97%
To find the void ratio, it is necessary to break down the total volume of the sample Vt on:
V t = V particles + Volume pores
The volume occupied by the particles is obtained from the dry mass and the specific gravity, data obtained from the statement. The specific gravity sg is the quotient between the density of the material and the density of water under standard conditions, therefore the density of the material is:
ρ = sg x ρWater = 2.75 x 1 g / cm3 = 2.75 g / cm3
ρ = Ms / Vs → Vs = 1.178 g / 2.75 g / cm3 = 0.428 cm3 = 0.000428 m3
The volume of voids in the sample is Vv = Vt - Vs = 0.000785 m3 - 0.000428 m3 = 0.000357 m3.
The void ratio and it is:
e = Vv / Vs = 0.000357 m3 / 0.000428 m3 = 0.83
The dry density of the sample is calculated as indicated in the introduction:
ρs = Weight of solid elements / Volume solid + Volume pores= 1178 g / 0.000785 m3 = 1.5 Mg / m3
The degree of saturation is S = (VWater / Vv ) x 100%. Since we know the mass of water in the sample, calculated in item b) and its density, the calculation of its volume is immediate:
ρWater = MWater / V Water → VWater = 353 g / 1 g / cm3 = 353 cm3 = 0.000353 m3
On the other hand, the volume of voids was calculated in item c)
S = (0.000353 m3 / 0.000357 m3) x 100% = 98.9%
Finally the percentage air content is A = (Vair / Vt) x 100%. The air volume corresponds to:
Vv - VWater = 0.000357 m3 - 0.000353 m3 = 0.000004 m3
A = (Vair / Vt) x 100% = (0.000004 m3/ 0.000785 m3) x100% = 0.51%
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