3 phase models

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Soil composition - phaserelationships

Bahareh Kaveh

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Soil composition - phaserelationshipsSoil consists of a skeleton formed from solid particleswith voids between the particles filled with air and/orwater.Soil is, therefore, a multiphase material with a solid

phase and a fluid phase. As soil is composed of solids, liquid and gas, it canbe regarded as a three-phase material:Saturated: Where the pores are entirely filled withwaterPartially saturated: where the pores contain both airand water (the term unsaturated is also used).The water may contain dissolved minerals.

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Composition of soil

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Three phase models

Unit solid volume: 1 volume unitUnit solid mass: 1 mass unitUnit total volume: 1 volume unit of all 3

phasesUnit solid volume is the most convenient.

A given soil is depicted as a fixed volume ofsolid material together with varyingamounts of water and air.

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Three phase models

a) Unit solid volume b) Unit solid mass c) Unit total volume

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Basic properties

Void volume: Amount of volume not occupied bysolids.

Void Ratio, e is ratio of void volume to solid volume.

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Basic properties Porosity, n is ratio of void

volume to the total volume.

n = e / (1+e) Specific Volume, v is the

total volume of the soil

model. v = 1 + e

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Basic properties Degree of Saturation, S T is quantity

of water expressed as a fraction of

the void volume.

S T = V w / Vv Vw = S T Vv [Transpose for V w ]

Vw = S T e For a perfectly dry soil, S T = 0

For a saturated soil, S T = 1

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Basic properties

Air-void volume: It is that

part of the void volume which

is not occupied by water. Va = V v Vw V

a = e S

Te

Va = e (1 S T)

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Basic properties

Air-void Ratio A V:

It is ratio of the air-void volume

to the specific volume Av = {e (1 S T)} / (1+e)

Av = n (1

S T)

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Basic properties Example 1

For a soil having a void ratio of 0.750 and

percentage saturation of 85%,

determine the porosity and air-void ratio.

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Grain specific gravity (G S )

The ratio of the mass of a given volume of material to

the mass of the same volume of water.

The mass of one unit of solid volume in the soil

model will therefore be:

Ms = G s w

w is the density of water which can be taken as 1000

kg / m 3 or 1 Mg/ m 3 (mega gram)

The mass of water in the soil model will be:

Mw = S T e w (Vw= S T e)12


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Moisture content, m

The ratio of mass of water to the mass of solids.

m = Mass of water / Mass of solids

m = M w / M s

m = (S T e w) / (G s w )m = (S T e) / (G s) m G s = S T e

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Air-void Ratio A V re-visited

Air-void Ratio A V: It is ratio of the air-void

volume to the specific volume Av = {e (1 S T)} / (1+e)

Av = n (1

S T) m G s = S T e S T = m G s / e Av = (e m G s ) / (1+e)

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Basic properties Example 2

An oven tin containing a sample of moist soilwas weighed and had a mass of 37.82 g;the empty tin had a mass of 16.15 g. Afterdrying, the tin and soil were weighed againand had a mass of 34.68 g. (G S = 2.70)

Determine the void ratio of the soil if the air-void ratio is:

Zero 5%

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Soil Densities

Density relates to quantity of a materialrelated to the amount of space it occupies.

Units: Mg / m 3, kg / m 3 or g / ml.

Dry density = Mass of solids/Total volumeWe know that:

Ms = G s wv = 1 + eTherefore d = G s w / 1 + e

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Soil Densities

Bulk density = Total mass / Total volume b = (M S + M W) / Total volumeWe know that: Ms = G s w Mw = S T e w v = 1 + e

b = (G s w + S T e w) / (1 + e)

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Soil densities

Saturated density ( sat ) is bulk

density when S T=1

Submerged density ( sub ) is the

difference between saturated density

and density of water.

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Unit weight

Unit weight is obtained by multiplyingdensity with g.

d = d g b = b g sat = sat g

Unit: KN / m3

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Soil Densities- example 3

Determine the relationship between DryDensity and Bulk Density as a ratio ofthe two OR

b /

d

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Soil densities-example 4

In its natural condition, a soil sample has amass of 2290g and a volume of 1.15x10 -3m 3.

After being completely dried in an oven, the

mass of the sample is 2035g. The volume ofGs for the soil is 2.68. Determine the bulkdensity, unit weight, water content, void ratio,

porosity, degree of saturation and air content.

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In a sample of moist clay soil, the voidratio is 0.788 and the degree ofsaturation is 0.93. Assuming G

S =

2.70, determine the dry density, thebulk density and the moisture content.

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The bulk density of sand in drainedcondition above the water table was foundto be 2.06 Mg/m 3 and its water content was

18%. Assume G S = 2.70 and calculate:The drained unit weight

The submerged unit weight and moisturecontent of the same sand below the watertable.

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A specimen of clay was tested in the laboratory andthe following data collected:Mass of wet specimen = 148.8 g,Mass of dry specimen = 106.2 g

Volume of wet specimen = 86.2 cm3, Specific gravity

of particles = 2.70Determine: The moisture content The bulk and dry densities The void ratio The degree of saturation

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3 phase models

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