1.1 COMPONENTS OF SOILS In natural occurrence, soils are three-phase systems consisting of soil solids, water and air. It is important to know the void volume in a given soil and its moisture content to determine its unit weight in the field.

1.1 COMPONENTS OF SOILS A soil mass, consists of solid soil particles, containing void spaces between them. These voids may be filled either with air or water, or both. The soil will behave as a two phase system, when its void space is filled with either water or air alone. In other words, the soil mass will consists of only two phases, i.e. solid and water (liquid), or solid and air (gas), respectively. Such a condition is possible when either the soil is fully saturated (the voids are totally filled with water, and there is no air); or when the soil is fully dry (the voids are totally filled with air, and no water). But, as and when the soil mass is partially saturated ; its void space will be filled up by water as well as air, and hence it will behave as a three phase system.

1.1 COMPONENTS OF SOILS

1.1 COMPONENTS OF SOILS

1.2 Weight-Volume Relationships

1.2 Weight-Volume Relationships V = Vs + Vv = Vs + (Vw + Va) Where Vs = volume of soil solids Vv = volume of voids Vw = volume of water in the voids Va = volume of air in the voids WEIGHT W = Ws + Ww (Assume Wa = 0) where Ws = weight of soil solids Ww = weight of water Void ratio, e = Porosity, n = Degree of saturation, Sr = x 100%

1.2 Weight-Volume Relationships The relationship between void ratio (e) and porosity (n) n = Moisture content, Unit weight Dry Unit weight or

1.2 Weight-Volume Relationships In English units Unit weight,  in pounds per cubic foot (lb/ft3). In SI units Unit weight,  in kilo Newtons per cubic meter (kN/m3). REMEMBER THIS NOTE Gravity Acceleration = 9.81 m/sec2. Unit weight of water, w 9.81 kN/m3 or 62.4 lb/ft3 Density of water, w 1000 kg/m3. 1g/cm3 Density Dry Density where  = density of soil (kg/m3) d - dry density of soil (kg/m3) M = total mass of the soil sample (kg) Ms = mass of soil solids in the sample (kg) V = Total volume (m3) The unit weight in kN/m3 can be obtained from densities in kg/m3 as

Content, and Specific Gravity Relationships among Unit Weight, Void Ratio, Moisture Content, and Specific Gravity e = Vv/Vs e= Vv Gs = s/w Gs = (Ws/Vs)/w Gs = Ws/(w x Vs) If Vs =1, Ws = Gsw Ww = wWs = wGsw where Gs = specific gravity of soil solids w = moisture content w = unit weight of water w =Ww/Vw Vw = Ww/w = wGsw/w = wGs Assume Vs = 1

Content, and Specific Gravity Relationships among Unit Weight, Void Ratio, Moisture Content, and Specific Gravity The unit weight and dry unit weight can be calculated as below: Sre = wGs

Content, and Specific Gravity Relationships among Unit Weight, Void Ratio, Moisture Content, and Specific Gravity If the soil sample is saturated w = Ww/Vw w = Ww/e So, Ww = e w Sr = 1 when saturated e = wGs

Content, and Specific Gravity Relationships among Unit Weight, Void Ratio, Moisture Content, and Specific Gravity Sometimes, the density of the soil have to determine, so these equation is similar to the unit weight relationship except w need to be changed to w. w = density of water = 1000 kg/m3.

Relationships among Unit Weight, Porosity, and Moisture Content If V is equal to 1, then Vv is equal to n, so Vs = 1 - n

Relationships among Unit Weight, Porosity, and Moisture Content

Worked example 1 A specimen of clay was tested in the laboratory and the following data were collected: Mass of wet specimen M1 = 148.8 g Mass of dry specimen M2 = 106.2 g Volume of wet specimen V = 86.2 cm3 Specific gravity of particles Gs = 2.70 Determine: (a) the water content, (b) the bulk and dry densities, (c) the void ratio and porosity, and (d) the degree of saturation. Solution :

Worked example 1

Worked Example 2 Given : 1. The weight of a moist soil sample is 45.6 Ib. 2. The volume of the soil measured before drying is 0.40 ft3. 3. After the sample is dried out in an oven, its weight is 37.8 Ib. 4. The specific gravity of solids is 2.65. Required (a) Water content. (b) Unit weight of moist soil. (c) Void ratio. (d) Porosity (e) Degree of saturation. Solution:

Worked Example 2 Unit weight of moist soil () Void ratio (e) Porosity (n) Degree of saturation (s)

Worked Example 3 Given: 1. The moist mass of a soil specimen is 20.7 kg. 2. The specimen's volume measured before drying is 0.011 m3. 3. The specimen's dried mass is 16.3 kg. 4. The specific gravity of solids is 2.68. Required (a) Void ratio (b)Degree of saturation (c) Wet unit mass (d) Dry unit mass (e) Wet unit weight (f) Dry unit weight Solution:

Worked Example 3 Void ratio (e) Degree of saturation (s) Wet unit mass ()

Worked Example 3 Dry unit mass (d) Wet unit weight () Dry unit weight (d)

Worked Example 4 Given: An undisturbed soil sample has the following data: 1. Void ratio = 0.78. 2. Water content = 12%. 3. Specific gravity of solids = 2.68. Required (a) Wet unit weight (b) Dry unit weight (c) Degree of saturation. (d) Porosity. Solution: Assume

Worked Example 4 void ratio (e) =

Worked Example 4 Wet unit weight () Dry unit weight (d) Degree of saturation (s) Porosity (n)

Worked Example 5 Given: 1. A 100% saturated soil has a wet unit weight of 120 lb/ft3. 2. The water content of this saturated soil was determined to be 36%. Required (a) Void ratio (b) Specific gravity of solids. Solution:

Worked Example 5 Void ratio (e) Specific gravity of solids (Gs)

Worked Example 6 Given: A soil sample has the following data: 1. Void ratio = 0.94. 2. Degree of saturation = 35%. 3. Specific gravity of solids = 2.71. Required 1. Water content. 2. Unit weight. Solution :

Worked Example 6 From the given void ratio, Water content. Substitute Eq. (A) into Eq. (B) Unit weight From the given degree of saturation, S = VW/VV = 0.35