2. Va
Solid
Vw
Water
Air Vs
• Soil is generally a three phase material
• Contains solid particles and voids
• Voids can contain liquid and gas phases
3. Va
Solid
Vw
Water
Air Vs
• Soil is generally a three phase material
• Contains solid particles and voids
• Voids can contain liquid and gas phases
4. Va
Solid
Vw
Water
Air Vs
• Soil is generally a three phase material
• Contains solid particles and voids
• Voids can contain liquid and gas phases
Phase Volume Mass Weight
Air Va 0 0
Water Vw Mw Ww
Solid Vs Ms Ws
5. Units
• Length metres
• Mass tonnes (1 tonne = 103 kg)
• Density t/m3
• Weight kilonewtons (kN)
• Stress kilopascals (kPa) 1 kPa= 1 kN/m2
• Unit weight kN/m3
• Accuracy Density of water, w = 1 t/m3
Stress/Strength to 0.1 kPa
6. Weight and Unit weight
• Force due to mass (weight) more important than mass
• W = Mg
• Unit weight
7. Weight and Unit weight
• Force due to mass (weight) more important than mass
• W = Mg
W
• Unit weight
V
Mg
V
= g
8. Weight and Unit weight
• Force due to mass (weight) more important than mass
• W = Mg
W
• Unit weight
V
Mg
V
= g
z v = gz
v
v = z
9. Specific Gravity
This is defined by
Density of Material
G
Density of Water w
Unit Weight of Material
G
Unit Weight of Water w
• Gs 2.65 for most soils
• Gs is useful because it enables the volume of solid particles
to be calculated from mass or weight
10. Specific Gravity
This is defined by
Density of Material
G
Density of Water w
Unit Weight of Material
G
Unit Weight of Water w
• Gs 2.65 for most soils
• Gs is useful because it enables the volume of solid particles
to be calculated from mass or weight
Ms Ms Ws Ws
Vs
s Gs w s Gs w
11. Voids ratio
• It is not the actual volumes that are important but rather the
ratios between the volumes of the different phases. This is
described by the voids ratio, e, or porosity, n, and the degree of
saturation, S.
12. Voids ratio
• It is not the actual volumes that are important but rather the
ratios between the volumes of the different phases. This is
described by the voids ratio, e, or porosity, n, and the degree of
saturation, S.
Vv
• The voids ratio is defined as e
Vs
13. Voids ratio
• It is not the actual volumes that are important but rather the
ratios between the volumes of the different phases. This is
described by the voids ratio, e, or porosity, n, and the degree of
saturation, S.
Vv
• The voids ratio is defined as e
Vs
Vv
• and the porosity as n
V
14. Voids ratio
• It is not the actual volumes that are important but rather the
ratios between the volumes of the different phases. This is
described by the voids ratio, e, or porosity, n, and the degree of
saturation, S.
Vv
• The voids ratio is defined as e
Vs
Vv
• and the porosity as n
V
The relation between these quantities can be simply determined
as follows
Vs = V - Vv = (1 - n) V
15. Voids ratio
• It is not the actual volumes that are important but rather the
ratios between the volumes of the different phases. This is
described by the voids ratio, e, or porosity, n, and the degree of
saturation, S.
Vv
• The voids ratio is defined as e
Vs
Vv
• and the porosity as n
V
The relation between these quantities can be simply determined
as follows
Vs = V - Vv = (1 - n) V
Hence
Vv Vv n
e
Vs (1 n )V 1 n
16. Degree of Saturation
• The degree of saturation, S, has an important influence on soil
behaviour
Vw Vw
• It is defined as S
Va Vw Vv
17. Degree of Saturation
• The degree of saturation, S, has an important influence on soil
behaviour
Vw Vw
• It is defined as S
Va Vw Vv
• The phase volumes may now be expressed in terms of e, S and Vs
• Vw = e S Vs Va = Vv - Vw = e Vs (1 - S)
18. Degree of Saturation
• The degree of saturation, S, has an important influence on soil
behaviour
• It is defined as Vw Vw
S
Va Vw Vv
• The phase volumes may now be expressed in terms of e, S and Vs
• Vw = e S Vs Va = Vv - Vw = e Vs (1 - S)
Assuming Vs = 1 m3, the following table can be produced
Phase Volume Mass Weight
Air e (1 - S) 0 0
Water eS eS w eS w
Solid 1 Gs w Gs w
19. Unit Weights
• The bulk unit weight
W w Gs Vs w e S Vs w ( Gs e S)
bulk
V Vs eVs 1 e
20. Unit Weights
• The bulk unit weight
W w Gs Vs w e S Vs w ( Gs e S)
bulk
V Vs eVs 1 e
• The saturated unit weight (S = 1)
w ( Gs e)
sat
1 e
21. Unit Weights
• The bulk unit weight
W w Gs Vs w e S Vs w ( Gs e S)
bulk
V Vs eVs 1 e
• The saturated unit weight (S = 1)
w ( Gs e)
sat
1 e
• The dry unit weight (S = 0)
w Gs
dry
1 e
22. Unit Weights
• The bulk unit weight
W w Gs Vs w e S Vs w ( Gs e S)
bulk
V Vs eVs 1 e
• The saturated unit weight (S = 1)
w ( Gs e)
sat
1 e
• The dry unit weight (S = 0)
w Gs
dry
1 e
• The submerged unit weight
' sat w
23. Moisture Content
• The moisture content, m, is defined as
Weight of Water Ww
m
Weight of Solids Ws
24. Moisture Content
• The moisture content, m, is defined as
Weight of Water Ww
m
Weight of Solids Ws
In terms of e, S, Gs and w
Ww = w Vw = w e S Vs
Ws = s Vs = w Gs Vs
25. Moisture Content
• The moisture content, m, is defined as
Weight of Water Ww
m
Weight of Solids Ws
In terms of e, S, Gs and w
Ww = w Vw = w e S Vs
Ws = s Vs = w Gs Vs
hence eS
m
Gs
26. Example 1
• Distribution by mass and weight
Phase Trimmings Mass Sample Mass, M Sample Weight, Mg
(g) (g) (kN)
-6
Total 55 290 2845 10
-6
Solid 45 237.3 2327.9 10
-6
Water 10 52.7 517 10
27. Example 1
• Distribution by mass and weight
Phase Trimmings Mass Sample Mass, M Sample Weight, Mg
(g) (g) (kN)
-6
Total 55 290 2845 10
-6
Solid 45 237.3 2327.9 10
-6
Water 10 52.7 517 10
• Distribution by volume (assume Gs = 2.65)
Total Volume V = r2 l
28. Example 1
• Distribution by mass and weight
Phase Trimmings Mass Sample Mass, M Sample Weight, Mg
(g) (g) (kN)
-6
Total 55 290 2845 10
-6
Solid 45 237.3 2327.9 10
-6
Water 10 52.7 517 10
• Distribution by volume (assume Gs = 2.65)
Total Volume V = r2 l
Ww
Water Volume Vw
w
29. Example 1
• Distribution by mass and weight
Phase Trimmings Mass Sample Mass, M Sample Weight, Mg
(g) (g) (kN)
-6
Total 55 290 2845 10
-6
Solid 45 237.3 2327.9 10
-6
Water 10 52.7 517 10
• Distribution by volume (assume Gs = 2.65)
Total Volume V = r2 l
Ww
Water Volume Vw
w
Ws
Solids Volume Vs
w Gs
30. Example 1
• Distribution by mass and weight
Phase Trimmings Mass Sample Mass, M Sample Weight, Mg
(g) (g) (kN)
-6
Total 55 290 2845 10
-6
Solid 45 237.3 2327.9 10
-6
Water 10 52.7 517 10
• Distribution by volume (assume Gs = 2.65)
Total Volume V = r2 l
Ww
Water Volume Vw
w
Ws
Solids Volume Vs
w Gs
Air Volume Va = V - Vs - Vw
32. Ww 10
Moisture content m 0.222 22.2 %
Ws 45
Vv Va Vw
Voids ratio e 0.755
Vs Vs
33. Ww 10
Moisture content m 0.222 22.2 %
Ws 45
Vv Va Vw
Voids ratio e 0.755
Vs Vs
Vw Vw
Degree of Saturation S 0.780 78.0 %
Vv Va Vw
34. Ww 10
Moisture content m 0.222 22.2 %
Ws 45
Vv Va Vw
Voids ratio e 0.755
Vs Vs
Vw Vw
Degree of Saturation S 0.780 78.0 %
Vv Va Vw
W
Bulk unit weight bulk 181 kN / m3
.
V
35. Ww 10
Moisture content m 0.222 22.2 %
Ws 45
Vv Va Vw
Voids ratio e 0.755
Vs Vs
Vw Vw
Degree of Saturation S 0.780 78.0 %
Vv Va Vw
W
Bulk unit weight bulk 181 kN / m3
.
V
Ws
Dry unit weight dry 14.8 kN / m3
V
36. Ww 10
Moisture content m 0.222 22.2 %
Ws 45
Vv Va Vw
Voids ratio e 0.755
Vs Vs
Vw Vw
Degree of Saturation S 0.780 78.0 %
Vv Va Vw
W
Bulk unit weight bulk 181 kN / m3
.
V
Ws
Dry unit weight dry 14.8 kN / m3
V
6
(W 14.9 10 9.81)
Saturated unit weight sat 19.04 kN / m3
V
Note that dry < bulk < sat
