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Introduction on Soil Phase Relationships :
Soil mass is generally a three-phase system. It consists of solid particles, liquid, and gas. For all practical purposes, the liquid may be considered to be water (although in some cases, the water may contain some dissolved salts) and the gas as air.The phase system may be expressed in SI units either in terms of mass-volume or weight-volume relationships. The interrelationships of the different phases are important since they help to define the condition or the physical make-up of the soil.
Concepts and Formulas of Soil Phase Relationships:
Phase relationship diagram:
In a mass of soil, there are three physical components: solid, water, and air. A phase relationship diagram is normally used to represent the relationship as follows:
Definitions:
Volume: (ft^{3} , m^{3} )
V_{t} : Total volume
V_{s} : Volume of solid
V_{v} : Volume of void
V_{w} : volume of water
V_{a} : Volume of air
Weights: (lbs, kg, kN)
W_{t} : total weight
W_{s} : weight of solid
W_{w} : weight of water
Weight of air = 0
Phase Relationships:
Volume-volume relationship:
Weight-weight relationship:
Water (Moisture) content (%):
Weight-Volume relationship:
(Unit weight or density, lbs/ft^{3} , g/cm^{3} , kN/m^{3} )
Moisture (total) unit weight:
Dry unit weight:
Solid unit weight:
Saturated unit weight (when soil is completely saturated, S = 100%,V_{a} =0):
Submerged (buoyant) unit weight (when soil is below ground water table, S = 100%):
Following relations are very handy in solving problems:
Unit weight to unit weight relationship
Specific gravity:
(Unit weight of water = 62.4 lbs/ft^{3} = 1 g/cm^{3} = 9.8 kN/m^{3} )
Average value of Gs for granular soils is 2.65, while the average value of Gs for cohesive soils is 2.80.
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Solved sample problems of Soil Phase Relationships:
Example 1: Determine unit weights, water content, based on known volume and weight (English units)
Given: (English units)
Volume of soil mass: 1 ft^{3} .
Weight of soil mass at moist condition: 100 lbs
Weight of soil after dried in oven: 80 lbs
Requirements:
Determine moist unit weight of soil, dry unit weight of soil, and water content.
Problem solving technique:
Moist unit weight g_{t} = W_{t} / V_{t} (W_{t} = 100 lbs, V_{t} =1 ft^{3} , are given)
Dry unit weight, g_{d} = W_{s} / V_{t} (Weight of solid is weight of soil after dried in oven ,Ws = 80 lbs, V_{t} =1 ft^{3} , are given)
Water content, w (%) = W_{w} /W_{s} (W_{s} = 80 lbs , weight of water, W_{w} not known)
Find weight of water, from phase relationship diagram, W_{w} = W_{t} – W_{s} .
Solution:
Moist (total) unit weight, g_{t} = W_{t} / V_{t} = 100/1 = 100 pcf (lbs/ft^{3} )
Dry unit weight, g_{d} = W_{s} / V_{t} = 80/1= 80 pcf (lbs/ft^{3} ).
Weight of water = 100-80=20 lbs
Water (Moisture) content: w (%) = W_{w} /W_{s} ´ 100 (%) = 20/80x100% = 25%
Example 2: Determine unit weights, water content, based on known volume and weight (SI units)
Given: (SI units)
Volume of soil mass: 0.0283 m^{3} .
Weight of soil mass at moist condition: 45.5 kg
Weight of soil after dry in oven: 36.4 kg
Problem solving technique:
Moist unit weight g_{t} = W_{t} / V_{t} (both value are given)
Dry unit weight, g_{d} = Ws / V_{t} (both value are given)
Water content, w (%) = W_{w} /W_{s} (Weight of solid is weight of soil after dried in oven is given, weight of water not known)
Find weight of water, from phase relationship diagram, W_{w} = W_{t} – W_{s} .
Requirements:
Determine moist unit weight of soil, dry unit weight of soil, and water content.
Solution:
Moisture (total) unit weight, g_{t} = W_{t} / V_{t} = 45.5/0.0283 = 1608 kg/m3 = 1.608 g/cm3
Dry unit weight, g_{d} = Ws / V_{t} = 36.4/0.0283= 1286 kg/m^{3} =1.286 g/cm^{3}
Weight of water = 45.5-36.4=9.1 lbs
Water (Moisture) content: w (%) = W_{w} /W_{s} ´ 100 (%) = 9.1/36.4x100% = 25%
Example 3: Determine void ratio, porosity, and degree of saturation based on known volume, weight, and specific gravity (English units)
Given: (English units)
Volume of soil mass: 1 ft^{3} .
Weight of soil mass at moist condition: 125 lbs
Weight of soil after dry in oven: 100 lbs
Specific gravity of solid = 2.65
Requirements:
Determine void ratio, porosity, and degree of saturation
Problem solving technique:
Void ratio, e = V_{v} /V_{s} (Vv, Vs, not given)
Find V_{s} = W_{s} /g_{s} (Ws = 100 lbs, gs is not given)
Find g_{s} = G_{s} g_{w} (Gs is given, g_{w} =62.4 lbs/ft^{3} is a know value)
Find V_{v} = 1-V_{s} (e can be calculated)
Porosity, n = V_{v} /V_{t} (V_{v} from step 4, Vs from step 2)
Degree of saturation, S = V_{w} /V_{v} (V_{v} from step 4, need to find V_{w} )
V_{w} =W_{w} /g_{w} (W_{w} , not given, g_{w} =62.4 lbs/ft3)
Find W_{w} = W_{t} – W_{s} (Both W_{t} , W_{w} are given)
Solution:
Solid unit weight, g_{s} = G_{s} g_{w} =2.65*62.4=165.4 lbs/ft^{3}
Volume of solid, V_{s} = W_{s} /g_{s} = 100/165.4=0.6 ft^{3}
Volume of void = V_{t} – V_{s} = 1 –0.6=0.4 ft^{3}
Void ratio, e = V_{v} /V_{s} = 0.4/0.6=0.66
Porosity, n = V_{v} /V_{t} = 0.4/1 = 0.4
Weight of water = 125-100=25 lbs
Volume of water, V_{w} = W_{w} /g_{w} = 25/62.4=0.4 ft^{3}
Degree of saturation, S = V_{w} /V_{v} = 0.4/0.4x100% = 100%.
Example 4: Determine void ratio, porosity, and degree of saturation based on known volume, weight, and specific gravity (English units)
Given: (metric units)
Volume of soil mass: 0.0283 m^{3} .
Weight of soil mass at moist condition: 56.6 kg
Weight of soil after dry in oven: 45.5 kg
Specific gravity of solid = 2.65
Requirements:
Determine void ratio, porosity, and degree of saturation
Problem solving technique:
Void ratio, e = V_{v} /V_{s} (V_{v} , V_{s} , not given)
Find V_{s} = W_{s} /g_{s} (W_{s} = 45.5 kg, g_{s} is not given)
Find g_{s} = G_{s} g_{w} (G_{s} is given, g_{w} =1 g/cm^{3} is a know value)
Find V_{v} = 1-V_{s} (e can be calculated)
Porosity, n = V_{v} /V_{t} (V_{v} from step 4, V_{s} from step 2)
Degree of saturation, S = V_{w} /V_{v} (V_{v} from step 4, need to find V_{w} )
V_{w} =W_{w} /g_{w} (W_{w} , not given, g_{w} =62.4 lbs/ft^{3} )
Find W_{w} = W_{t} – W_{s} (W_{t} = 56.6 kg, W_{s} = 45.5 kg are given)
Solution:
Solid unit weight, g_{s} = G_{s} g_{w} =2.65*1=2.65 g/cm^{3} = 2650 kg/m^{3}
Volume of solid, V_{s} = W_{s} /g_{s} = 45.5/2650=0.0171 m^{3}
Volume of void = Vt – Vs = 0.0283 –0.0171=0.0112 m3
Void ratio, e = V_{v} /V_{s} = 0.0112/0.0171=0.65
Porosity, n = V_{v} /V_{t} = 0.0111/0.0283 = 0.39
Weight of water = 56.6-45.5=11.1 kg
Volume of water, V_{w} = W_{w} /g_{w} = 11.1 kg/1 g/cm^{3} = 11100 cm^{3} = 0.0111m^{3}
Degree of saturation, S = V_{w} /V_{v} = 0.0111/0.0111x100% = 100%.
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