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## 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: (ft3, m3)

• Vt: Total volume

• Vs: Volume of solid

• Vv: Volume of void

• Vw: volume of water

• Va: Volume of air

Weights: (lbs, kg, kN)

• Wt: total weight

• Ws: weight of solid

• Ww: weight of water

• Weight of air = 0

### Volume-volume relationship:

• Void ratio, e = Vv/Vs  (no unit)

• Porosity, n = Vv/Vt      (no unit)

• Degree of saturation, S (%) = Vw/Vv ´ 100 (%)

### Weight-weight relationship:

Water (Moisture) content: w (%) = Ww/Ws ´ 100 (%)

### Weight-Volume relationship:

(Unit weight or density, lbs/ft3, g/cm3, kN/m3)

• Moisture (total) unit weight, gt = Wt / Vt

• Dry unit weight, gd  = Ws / Vt

• Solid unit weight, gs = Ws / Vs

• Saturated unit weight, gsat = Wt / Vt     (when soil is completely saturated, S = 100%,Va=0)

• Submerged (buoyant) unit weight, gb =gsat -gw (when soil is below ground water table, S = 100%)

### Unit weight to unit weight relationship

• Specific gravity, Gs = gs / gw

• (Unit weight of water, gw = 62.4 lbs/ft3 = 1 g/cm3 = 9.8 kN/m3)

## Simple problems

### Example 1:Determine unit weights, water content, based on known volume and weight (English units)

Given: (English units)

• Volume of soil mass:  1 ft3.

• 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:

1. Moist unit weight gt = Wt / Vt  (Wt = 100 lbs, Vt=1 ft3,  are given)
2. Dry unit weight, gd  = Ws / Vt  (Weight of solid is weight of soil after dried in oven ,Ws = 80 lbs, Vt=1 ft3,  are given)
3. Water content, w (%) = Ww/Ws (Ws = 80 lbs , weight of water, Ww not known)
4. Find weight of water, from phase relationship diagram, Ww = Wt – Ws.

Solution:

1. Moist (total) unit weight, gt = Wt / Vt = 100/1 = 100 pcf (lbs/ft3)

2. Dry unit weight, gd  = Ws / Vt = 80/1= 80 pcf (lbs/ft3).

3. Weight of water = 100-80=20 lbs

4. Water (Moisture) content: w (%) = Ww/Ws ´ 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 m3.

• Weight of soil mass at moist condition: 45.5 kg

• Weight of soil after dry in oven: 36.4 kg

Problem solving technique:

1. Moist unit weight gt = Wt / Vt  (both value are given)
2. Dry unit weight, gd  = Ws / Vt  (both value are given)
3. Water content, w (%) = Ww/Ws (Weight of solid is weight of soil after dried in oven is given, weight of water not known)
4. Find weight of water, from phase relationship diagram, Ww = Wt – Ws.

Requirements:

Determine moist unit weight of soil, dry unit weight of soil, and water content.

Solution:

1. Moisture (total) unit weight, gt = Wt / Vt = 45.5/0.0283 = 1608 kg/m3 = 1.608 g/cm3

2. Dry unit weight, gd  = Ws / Vt = 36.4/0.0283= 1286 kg/m3=1.286 g/cm3

3. Weight of water = 45.5-36.4=9.1 lbs

4. Water (Moisture) content: w (%) = Ww/Ws ´ 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 ft3.

• 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:

1. Void ratio, e = Vv/Vs   (Vv, Vs, not given)
2. Find Vs = Ws/gs          (Ws = 100 lbs, gs is not given)
3. Find gs = Gsgw            (Gs is given, gw =62.4 lbs/ft3 is a know value)
4. Find Vv = 1-Vs            (e can be calculated)
5. Porosity, n = Vv/Vt      (Vv from step 4, Vs from step 2)
6. Degree of saturation, S = Vw/Vv          (Vv from step 4, need to find Vw)
7. Vw =Ww/gw               (Ww, not given, gw=62.4 lbs/ft3)
8. Find Ww = Wt – Ws    (Both Wt, Ww are given)

Solution:

1. Solid unit weight, gs = Gsgw=2.65*62.4=165.4 lbs/ft3

2. Volume of solid, Vs = Ws/gs = 100/165.4=0.6 ft3

3. Volume of void = Vt – Vs = 1 –0.6=0.4 ft3

4. Void ratio, e = Vv/Vs = 0.4/0.6=0.66

5. Porosity, n = Vv/Vt = 0.4/1 = 0.4

6. Weight of water = 125-100=25 lbs

7. Volume of water, Vw = Ww/gw = 25/62.4=0.4 ft3

8. Degree of saturation, S = Vw/Vv = 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 m3.

• 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:

1. Void ratio, e = Vv/Vs   (Vv, Vs, not given)
2. Find Vs = Ws/gs          (Ws = 45.5 kg, gs is not given)
3. Find gs = Gsgw            (Gs is given, gw =1 g/cm3 is a know value)
4. Find Vv = 1-Vs            (e can be calculated)
5. Porosity, n = Vv/Vt      (Vv from step 4, Vs from step 2)
6. Degree of saturation, S = Vw/Vv          (Vv from step 4, need to find Vw)
7. Vw =Ww/gw               (Ww, not given, gw=62.4 lbs/ft3)
8. Find Ww = Wt – Ws    (Wt = 56.6 kg, Ws = 45.5 kg are given)

Solution:

1. Solid unit weight, gs = Gsgw=2.65*1=2.65 g/cm3 = 2650 kg/m3

2. Volume of solid, Vs = Ws/gs = 45.5/2650=0.0171 m3

3. Volume of void = Vt – Vs = 0.0283 –0.0171=0.0112 m3

4. Void ratio, e = Vv/Vs = 0.0112/0.0171=0.65

5. Porosity, n = Vv/Vt = 0.0111/0.0283 = 0.39

6. Weight of water = 56.6-45.5=11.1 kg

7. Volume of water, Vw = Ww/gw = 11.1 kg/1 g/cm3= 11100 cm3= 0.0111m3

8. Degree of saturation, S = Vw/Vv = 0.0111/0.0111x100% = 100%. ### Related topics

Soil parameters

Lateral earth pressure