Physical properties of soil

Contents:

• Phase relationship diagram  

  Volume-volume relationship

  Weight-weight relationship

  Weight-Volume relationship

  Unit weight to unit weight relationship

• 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)

• Example 3:Determine void ratio, porosity, and degree of saturation based on known volume, weight, and specific gravity (English units)

• Example 4:Determine void ratio, porosity, and degree of saturation based on known volume, weight, and specific gravity (English units)

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)

• 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

Volume-volume relationship:

• 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 (%)

Weight-weight relationship:

Water (Moisture) content: w (%) = W_w/W_s ´ 100 (%)

Weight-Volume relationship: 

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

• 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%)

Unit weight to unit weight relationship

• Specific gravity, G_s = g_s / g_w   

• (Unit weight of water, g_w = 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 g_t = W_t / V_t  (W_t = 100 lbs, V_t=1 ft³,  are given)
2. 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 ft³,  are given)
3. Water content, w (%) = W_w/W_s (W_s = 80 lbs , weight of water, W_w not known)
4. Find weight of water, from phase relationship diagram, W_w = W_t – W_s.

Solution:

1. Moist (total) unit weight, g_t = W_t / V_t = 100/1 = 100 pcf (lbs/ft³)

2. Dry unit weight, g_d  = W_s / V_t = 80/1= 80 pcf (lbs/ft³).

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

4. Water (Moisture) content: w (%) = W_w/W_s ´ 100 (%) = 20/80*100% = 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 g_t = W_t / V_t  (both value are given)
2. Dry unit weight, g_d  = Ws / V_t  (both value are given)
3. 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)
4. 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:

1. Moisture (total) unit weight, g_t = W_t / V_t = 45.5/0.0283 = 1608 kg/m3 = 1.608 g/cm3

2. Dry unit weight, g_d  = Ws / V_t = 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 (%) = W_w/W_s ´ 100 (%) = 9.1/36.4*100% = 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 = V_v/V_s   (Vv, Vs, not given)
2. Find V_s = W_s/g_s          (Ws = 100 lbs, gs is not given)
3. Find g_s = G_sg_w            (Gs is given, g_w =62.4 lbs/ft³ is a know value)
4. Find V_v = 1-V_s            (e can be calculated)
5. Porosity, n = V_v/V_t      (V_v from step 4, Vs from step 2)
6. Degree of saturation, S = V_w/V_v          (V_v from step 4, need to find V_w)
7. V_w =W_w/g_w               (W_w, not given, g_w=62.4 lbs/ft3)
8. Find W_w = W_t – W_s    (Both W_t, W_w are given)

Solution:

1. Solid unit weight, g_s = G_sg_w=2.65*62.4=165.4 lbs/ft³

2. Volume of solid, V_s = W_s/g_s = 100/165.4=0.6 ft³

3. Volume of void = V_t – V_s = 1 –0.6=0.4 ft³

4. Void ratio, e = V_v/V_s = 0.4/0.6=0.66

5. Porosity, n = V_v/V_t = 0.4/1 = 0.4

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

7. Volume of water, V_w = W_w/g_w = 25/62.4=0.4 ft³

8. Degree of saturation, S = V_w/V_v = 0.4/0.4*100% = 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 = V_v/V_s   (V_v, V_s, not given)
2. Find V_s = W_s/g_s          (W_s = 45.5 kg, g_s is not given)
3. Find g_s = G_sg_w            (G_s is given, g_w =1 g/cm³ is a know value)
4. Find V_v = 1-V_s            (e can be calculated)
5. Porosity, n = V_v/V_t      (V_v from step 4, V_s from step 2)
6. Degree of saturation, S = V_w/V_v          (V_v from step 4, need to find V_w)
7. V_w =W_w/g_w               (W_w, not given, g_w=62.4 lbs/ft³)
8. Find W_w = W_t – W_s    (W_t = 56.6 kg, W_s = 45.5 kg are given)

Solution:

1. Solid unit weight, g_s = G_sg_w=2.65*1=2.65 g/cm³ = 2650 kg/m³

2. Volume of solid, V_s = W_s/g_s = 45.5/2650=0.0171 m³

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

4. Void ratio, e = V_v/V_s = 0.0112/0.0171=0.65

5. Porosity, n = V_v/V_t = 0.0111/0.0283 = 0.39

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

7. Volume of water, V_w = W_w/g_w = 11.1 kg/1 g/cm³= 11100 cm³= 0.0111m³

8. Degree of saturation, S = V_w/V_v = 0.0111/0.0111*100% = 100%.