CE-REF.COM
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Example 1:Determine unit weights, water content, based on known volume and weight (English units)
Example 2:Determine unit weights, water content, based on known volume and weight (SI units)
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:
Volume: (ft^{3}, m^{3})
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
Void ratio, e = V_{v}/V_{s} (no unit)
Porosity, n = V_{v}/V_{t} (no unit)
Degree of saturation, S (%) = V_{w}/V_{v} ´ 100 (%)
Water (Moisture) content: w (%)
= W_{w}/W_{s} ´ 100
(%)
(Unit weight or density, lbs/ft^{3}, g/cm^{3}, kN/m^{3})
Moisture (total) unit weight, g_{t} = W_{t} / V_{t}
Dry unit weight, g_{d} = W_{s} / V_{t}
Solid unit weight, g_{s} = W_{s} / V_{s}
Saturated unit weight, g_{sat} = W_{t} / V_{t} (when soil is completely saturated, S = 100%,V_{a}=0)
Submerged (buoyant) unit weight, g_{b} =g_{sat} -g_{w} (when soil is below ground water table, S = 100%)
Specific gravity, G_{s} = g_{s} / g_{w}
(Unit weight of water, g_{w} = 62.4 lbs/ft^{3} = 1 g/cm^{3} = 9.8 kN/m^{3})
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:
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%
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:
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%
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:
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%.
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:
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%.