1.3 Soil Phase Relationships.pdf

1.3 Soil Phase Relationships
Mr. Shade S. Muluti
smuluti@unam.na
School of Engineering & The Built Environment
Department of Civil and Mining Engineering
Soil Mechanics
TCVG3682/2021

2
Learning Outcomes
TCVG3682/2021 – Mr. S.S. Muluti
✓ Determine the proportions of the main constituents in a
soil.
✓ Understand and differentiate between the two-phase
systems of soils.
✓ Determine and understand the relationships between
different variables of soils.
▪ Upon completion of this chapter, students should be able to
do the following:

3
Phase System of Soils
▪ Soil is not a coherent solid material like steel and concrete
but is a particulate material.
▪ Soils, as they exist in nature, consist of solid particles
(mineral grains, rock fragments) with water and air in the
voids between the particles.
▪ The water and air contents are readily changed by changes
in ambient conditions and location.
▪ As the relative proportions of the three phases vary in any
soil deposit, it is useful to consider a soil model which will
represent these phases distinctly and properly quantify the
amount of each phase.

4
Soil Model
Source: Adopted from Chappidi (2017)

5
▪ The compositions of natural soils may include diverse
components which may be classified into three large
groups:
❑ Solid Phase (minerals, cementation & organic material)
❑ Liquid Phase (water with dissolved salts)
❑ Gaseous Phase (air or other some gas)
▪ The spaces between the solids (solid particles) are called
voids.
▪ Water is often the predominant liquid and air is the
predominant gas.
▪ We will use the terms water and air instead of liquid and
gases.

6
▪ Soils are made up of either two-phase or three-phase
systems:
Two Phase System
1) Fully Saturated Soil - made up of two phases solid
particles and pore water (no air content).
2) Completely Dry Soil - made up of two phases solid
particles and pore air (no water content).
Three Phase System
3) Partially Saturated Soil - made up of three phases -
solid particles, pore water and pore air.
Partially Saturated soil is often attributed a fourth phase

7
Two Phase System
1. Fully Saturated Soil
Mineral Skeleton Fully Saturated

8
Two Phase System
2. Completely Dry Soil
Mineral Skeleton Dry Soil

9
Three Phase System
3. Partially Saturated Soil
Mineral Skeleton Partially Saturated

10
Phase Relations of Soils
▪ The soil model is given dimensional values for the solid,
water and air components:

11
▪ The volume of the various constituents of a soil is
quantified by following quantities:
o VT = Total Volume of a soil.
o Vv = Volume of the voids (pores).
o Vs = Volume of the soil solids.
o Va = Volume of the air in the voids.
o Vw = Volume of the water.
▪ Thus, for all soils:
VT = Vv + Vs = (Va + Vw) + Vs

12
▪ The weight of the various constituents of a soil is quantified
by following quantities:
o WT = Total Weight of a soil.
o Wa = Weight of the air in the voids (pores) ≈ 0.
o Ww = Weight of the water.
o Ws = Weight of the soil solids.
▪ Thus, for all soils:
WT = Ww + Ws
❖ If Vv = Vw (→Va = 0) and Ww ≠ 0, the soil is said to be
saturated; otherwise, it is unsaturated.
NOTE:

13
▪ For the purpose of engineering analysis and design, it is
necessary to express relations between the weights and
the volumes of the three phases.
▪ The various relations can be grouped into:
1) Weight relations
2) Volume relations
3) Weight-Volume relations

14
1. Weight Relationships
▪ The following are the basic weight relations:
a) Water content or Moisture content (w)
b) Specific gravity (Gs)
WT = Ww + Ws
Where:
o WT = Total Weight of a soil.
o Wa = Weight of the air ≈ 0.
o Ww = Weight of the water.
o Ws = Weight of the soil solids.

15
a) Water content or Moisture content (w)
▪ The ratio of the mass of water present to the mass of
solid particles is called the water content (w), or
sometimes the moisture content.
▪ The water content of a soil is found by weighing a
sample of the soil and then placing it in an oven at
110 ± 5°C until the weight of the sample remains
constant, that is, all the absorbed water is driven out.
𝑤 =
𝑊
𝑤
𝑊
𝑠
× 100%

16
b) Specific gravity (Gs)
▪ The specific gravity of a solid substance is the ratio of
the weight of a given volume of material to the weight
of an equal volume of water (at 20°C).
▪ Gs is useful because it enables the volume of solid
particles to be calculated from mass or weight.
▪ The specific gravity of soil solids is often needed for
various calculations in soil mechanics.
▪ For most inorganic soils, the value of Gs lies between
2.60 and 2.80.
𝐺𝑠 =
𝑊
𝑠
𝑊
𝑤
=
γ𝑠 ∙ 𝑉
𝑠
γ𝑤 ∙ 𝑉
𝑤
=
γ𝑠
γ𝑤
∴ 𝐺𝑠=
γ𝑠
γ𝑤

17
2. Volume Relationships
▪ The following are the basic volume relations:
a) Void ratio (e)
b) Porosity (n)
c) Degree of saturation (S)
d) Air content (a)
Where:
o VT = Total Volume of a soil.
o Vv = Volume of the voids (pores).
o Vs = Volume of the soil solids.
o Va = Volume of the air in the voids.
o Vw = Volume of the water.
VT = Vs + Vv + Vs
Vv = Va + Vw

18
a) Void ratio (e)
▪ Is defined as the ratio of the volume of voids (Vv) to the
volume of soil solids (Vs) and is expressed as a
decimal.
▪ The void ratio of real coarse-grained soils vary
between 0.3 and 1.0.
▪ Clay soils can have void ratio greater than one.
𝑒 =
𝑉
𝑣
𝑉
𝑠
𝑒 =
𝑉
𝑣
𝑉
𝑠
=
𝑉
𝑣
𝑉𝑇 − 𝑉
𝑣
=
𝑉
𝑣
𝑉𝑇(1 −
𝑉
𝑣
𝑉𝑇
)
=
𝑛
1 − 𝑛
∴ 𝑒 =
𝑛
1 − 𝑛

19
b) Porosity (n)
▪ Is defined as the ratio of the volume of voids (Vv) to the
total volume of soil (VT) and is expressed as a
percentage.
▪ Void ratio and porosity are inter-related to each other
as follows:
𝑛 =
𝑉
𝑣
𝑉𝑡
× 100%
𝑛 =
𝑉
𝑣
𝑉𝑡
=
𝑉
𝑣
𝑉
𝑠 + 𝑉
𝑣
=
𝑉
𝑣
𝑉
𝑠(1 +
𝑉
𝑣
𝑉
𝑠
)
=
𝑒
1 + 𝑒
∴ 𝑛 =
𝑒
1 + 𝑒

20
c) Degree of Saturation (S)
▪ Is defined as the ratio of the volume of water (Vw) to
the volume of voids (Vv) and is expressed as a
percentage.
▪ The degree of saturation tell us what percentage of
the volume of voids contains water.
o For fully saturated soil: S = 1 or 100% (Vv = Vw).
o For dry soil: S = 0.
o For partially saturated soil: 1 < S < 0
𝑆 =
𝑉
𝑤
𝑉
𝑣
× 100%

21
d) Air Content (a)
▪ Is defined as the ratio of air volume (Va) to total volume
(VT) and is expressed as a percentage.
▪ The air voids (Va) , is that part of the voids space not
occupied by water.
o For a completely dry soil: a = n
o For a saturated soil: a = 0
𝑎 =
𝑉
𝑎
𝑉𝑇
× 100%
𝑎 = 𝑛(1 − 𝑆)

22
3. Weight-Volume Relationships
▪ The following are the basic weight-volume relations:
a) Unit Weight (γ)
b) Dry Unit Weight (γd)
c) Saturated unit weight (γsat)
d) Effective unit Weight (γ')

23
a) Unit Weight (𝜸)
▪ The unit weight of a soil is the ratio of the weight of soil
to the total volume.
γ =
𝑊𝑇
𝑉𝑇
▪ Density is a measure of the quantity of mass in a unit
volume of material.
▪ Unit weight is a measure of the weight of a unit volume
of material.
▪ Both can be used interchangeably. The units of density
are ton/m³, kg/m³ or g/cm³.
▪ The unit of unit weight is kN/m³.

24
b) Dry Unit Weight (𝜸d)
▪ Is defined as the ratio of the weight of solids to the total
volume.
γ𝑑 =
𝑊
𝑠
𝑉𝑇
▪ The dry unit weight can also be determined as:
γ =
𝑊𝑇
𝑉𝑇
=
𝑊
𝑠 1 +
𝑊
𝑤
𝑊
𝑠
𝑉𝑇
= γ𝑑(1 + 𝑤) ∴ γ𝑑=
γ
1 + 𝑤
γ𝑑 =
𝑊
𝑠
𝑉
𝑠(1 + 𝑒)
=
γ𝑠
(1 + 𝑒)
=
γ𝑤 ∙ 𝐺𝑠
1 + 𝑒
∴ γ𝑑=
γ𝑤 ∙ 𝐺𝑠
1 + 𝑒

25
c) Saturated Unit weight (𝜸sat)
▪ For a saturated soil, the unit weight becomes:
γ𝑠𝑎𝑡 =
𝑊𝑇
𝑉𝑇
γ𝑠𝑎𝑡 =
𝑊
𝑠 1 +
𝑊
𝑤
𝑊
𝑠
𝑉
𝑠(1 + 𝑒)
=
γ𝑠 1 +
𝑒
𝐺𝑠
1 + 𝑒
=
γ𝑤𝐺𝑠 1 +
𝑒
𝐺𝑠
1 + 𝑒
=
γ𝑤 𝐺𝑠 + 𝑒
1 + 𝑒
▪ The saturated unit weight can also be determined as:
∴ γ𝑠𝑎𝑡=
γ𝑤 𝐺𝑠 + 𝑒
1 + 𝑒

26
d) Effective Unit Weight (𝜸')
▪ The effective or submerged unit weight of the soil is
given as:
γ𝑠𝑢𝑏 = γ′ = γ𝑠𝑎𝑡 − γ𝑤
Ground Surface
G.W.T
γ𝑠𝑎𝑡
γ𝑑𝑟𝑦
γ
γ𝑠𝑢𝑏
𝑺 = 𝟎
𝑺 = 𝟎 𝒕𝒐 𝟏
𝑺 = 𝟏
𝜸𝒘 = 𝟗. 𝟖𝟏 𝒌𝑵/𝒎𝟑

27
Density-Volume Relationships
▪ In the SI system, the unit used for unit weight is kilo
Newtons per cubic meter (kN/m3).
▪ Because the Newton is a derived unit, working with mass
densities (ρ) of soil may sometimes be convenient.
▪ The SI unit of mass density is kg/m3.
▪ We can write the density equations as follows:
𝜌 =
𝑀𝑇
𝑉𝑇
𝜌𝑑 =
𝑀𝑠
𝑉𝑇
Dry Density:
Buoyant Density:
Density:
𝜌′ = 𝜌 − 𝜌𝑤 𝝆𝒘 = 𝟏𝟎𝟎𝟎 𝒌𝒈/𝒎𝟑

28
▪ The density-volume relations may be derived by referring to
the soil element in which the volume of soil solids is equal
to 1 and the volume of voids is equal to e:
𝑀𝑤 = 𝑤𝐺𝑠𝜌𝑤
𝑀𝑠 = 𝐺𝑠𝜌𝑤 𝑉
𝑠 = 1
𝑉
𝑣 = 𝑒
Mass Volume

29
▪ As mentioned before, due to the convenience of working
with densities in the SI system, the following equations,
similar to unit–weight relationships will be useful:
𝜌 =
1 + 𝑤 𝐺𝑠𝜌𝑤
1 + 𝑒
𝜌𝑑 =
𝐺𝑠𝜌𝑤
1 + 𝑒
Dry Density:
Saturated Density:
Density:
𝜌𝑠𝑎𝑡 =
(𝐺𝑠+𝑒)𝜌𝑤
1 + 𝑒
Moisture Content: 𝑤 =
𝑆𝑒
𝐺𝑠

30
Relationships among γ, e, w, and Gs
▪ Consider the following three phase diagram where Vs = 1:
𝑊
𝑤 = 𝑤𝐺𝑠γ𝑤
𝑊𝑇
𝑊
𝑠 = 𝐺𝑠γ𝑤
𝑉𝑇 = 1 + 𝑒
𝑉
𝑤 = w𝐺𝑠
𝑉
𝑠 = 1
𝑉
𝑣 = 𝑒

31
Relationships among γ , e, w, and Gs
▪ The weights of soil solids and water can be given as:
𝑊
𝑠 = 𝐺𝑠γ𝑤
𝑊
𝑤 = 𝑤𝑊
𝑠 = 𝑤𝐺𝑠γ𝑤
▪ If Gs is expresses as:
𝐺𝑠 =
𝑊
𝑠
𝑉
𝑠γ𝑤
▪ Hence:
γ𝑑 =
𝑊
𝑠
𝑉𝑇
=
𝐺𝑠γ𝑤
1 + 𝑒
e =
𝐺𝑠γ𝑤
γ𝑑
− 1
OR
γ =
𝑊𝑇
𝑉𝑇
=
𝑊
𝑠 + 𝑊
𝑤
𝑉𝑇
=
𝐺𝑠γ𝑤 + 𝑤𝐺𝑠γ𝑤
1 + 𝑒
=
1 + 𝑤 𝐺𝑠γ𝑤
1 + 𝑒
Unit Weight:
Dry Unit Weight:

32
▪ If the weight and volume of water of the soil can be given
as: 𝑊
𝑤 = 𝑤𝑊
𝑠 = 𝑤𝐺𝑠γ𝑤
▪ Hence:
e = 𝑤𝐺𝑠
γ𝑠𝑎𝑡 =
𝑊𝑇
𝑉𝑇
=
𝑊
𝑠 + 𝑊
𝑤
𝑉𝑇
=
𝐺𝑠γ𝑤 + 𝑒γ𝑤
1 + 𝑒
=
𝐺𝑠 + 𝑒 γ𝑤
1 + 𝑒
Saturated Unit Weight:
Void ratio:
𝑉
𝑤 =
𝑊
𝑤
γ𝑤
=
𝑤𝐺𝑠γ𝑤
γ𝑤
= 𝑤𝐺𝑠
𝑆 =
𝑉
𝑤
𝑉
𝑣
=
𝑤𝐺𝑠
𝑒
𝑆𝑒 = 𝑤𝐺𝑠
OR
Degree of Saturation:

33
▪ If the soil sample is saturated—that is, the void spaces are completely
filled with water, the soil element becomes a two-phase diagram where
Vs = 1:
𝑉
𝑣 = 𝑉
𝑤 = 𝑒
𝑊
𝑤 = 𝑒γ𝑤
𝑊𝑇
𝑊
𝑠 = 𝐺𝑠γ𝑤 𝑉
𝑠 = 1
𝑉𝑇 = 1 + 𝑒

34
Relationships among γ, n and w
▪ Consider the following three phase diagram where VT = 1:
𝑊
𝑤 = 𝑤𝐺𝑠γ𝑤(1 − 𝑛)
𝑊
𝑠 = 𝐺𝑠γ𝑤(1 − 𝑛)
𝑉𝑇 = 1
𝑉
𝑠 = 1 − 𝑛
𝑉
𝑣 = 𝑛

35
Relationships among γ, n and w
▪ If VT is equal to 1, then Vv is equal to n, so Vs = 1− n.
▪ Therefore, the weight of soil solids (Ws) and the weight of
water (Ww) can then be expressed as follows:
𝑊
𝑊
𝑤 = 𝑤𝑊
𝑠 = 𝑤𝐺𝑠γ𝑤(1 − 𝑛)
𝑛 =
𝑉
𝑠
𝑉𝑇
▪ Recall:

36
Relationships among γ, n and w.
▪ Hence:
𝑤𝑠𝑎𝑡 =
𝑊
𝑤
𝑊
𝑠
=
𝑛γ𝑤
(1 − 𝑛)γ𝑤𝐺𝑠
=
𝑛
(1 − 𝑛) 𝐺𝑠
γ𝑠𝑎𝑡 =
𝑊𝑇
𝑉𝑇
=
𝑊
𝑠 + 𝑊
𝑤
𝑉𝑇
=
(1 − 𝑛)𝐺𝑠γ𝑤 + 𝑒γ𝑤
1 + 𝑒
= [(1 − 𝑛)𝐺𝑠+𝑛]γ𝑤
Saturated Unit Weight:
Moisture Content:
γ =
𝑊𝑇
𝑉𝑇
=
𝑊
𝑠 + 𝑊
𝑤
𝑉𝑇
= 𝐺𝑠γ𝑤(1 − 𝑛)(1 + 𝑤)
Unit Weight:
γ𝑑 =
𝑊
𝑠
𝑉𝑇
=
𝐺𝑠γ𝑤(1 − 𝑛)
1
= 𝐺𝑠γ𝑤(1 − 𝑛)
Dry Unit Weight:

37
Relationships among γ, n and w.
▪ If the soil sample is saturated—that is, the void spaces are completely
filled with water, the soil element becomes a two-phase diagram where
VT = 1:
𝑉
𝑣 = 𝑉
𝑤 = 𝑛
𝑉
𝑠 = 1 − 𝑛
𝑉𝑇 = 1
𝑊
𝑤 = 𝑛γ𝑤
𝑊

38
Various Unit Weight Relationships
Note: The 3rd column is a special case of the 1st column when S = 1.

39
Typical values of e, w, and γ
▪ Some typical values of void ratio, moisture content in a
saturated condition, and dry unit weight for soils in a natural
state are given in the Table below.

40
Summary - Phase Relations
▪ In summary, for the easy solution of phase relation
problems, you don’t have to memorize lots of complicated
formulas.
▪ Most of them can easily be derived from the phase
diagram.
▪ Just remember the following simple rules:
(a) Remember the basic definitions of properties.
(b) Draw a phase diagram.
(c) Assume either VS = 1 or VT = 1.
(d) Solve the Problem.

41
Example 1 - Phase Relations
▪ For a soil core sample, you are given the following data:
▪ Weight of soil sample, WT = 1013 g
▪ Vol. of soil sample, Vs = 585.0 cm3
▪ Specific Gravity, Gs= 2.65
▪ Dry weight of soil, Ws = 904.0 g
i. Moisture Content
ii. Void ratio
iii. Porosity
iv. Degree of Saturation
v. Dry unit Weight
▪ Determine:

42
𝑉
𝑠 =
𝑊
𝑠
𝐺𝑠𝛾𝑤
=
(904)
(2.65)(1.0)
= 341.1 𝑐𝑚3
𝑊𝑇 = 1013 g 𝑊
𝑠 = 904 g 𝑉𝑇 = 585 g
▪ Given:
𝑉
𝑣 = 𝑉𝑇 − 𝑉
𝑠 = 585 − 341.1 = 243.9 𝑐𝑚3
𝑉
𝑤 =
𝑊
𝑤
𝛾𝑤
=
(109)
(1.0)
= 109 𝑐𝑚3
𝑉
𝑎 = 𝑉
𝑣 − 𝑉
𝑤 = 243.9 − 109 = 134.9 𝑐𝑚3
𝑊
𝑤 = 𝑊𝑇 − 𝑊
𝑠 = 1013 − 904 = 109 g

43
▪ From the three-phase diagram, we can determine:
i. Moisture Content (w):
ii. Void ratio (e):
iii. Porosity (n):
iv. Degree of Saturation (S):
v. Dry unit Weight (γd):
𝑤 =
𝑊
𝑤
𝑊
𝑠
=
(109)
(904)
× 100 = 𝟏𝟐. 𝟏%
𝑒 =
𝑉
𝑣
𝑉
𝑠
=
(243.9)
(341.1)
= 𝟎. 𝟕𝟏𝟓
𝑛 =
𝑉
𝑣
𝑉𝑇
=
(243.9)
(585)
× 100 = 𝟒𝟏. 𝟕%
𝑆 =
𝑉
𝑤
𝑉
𝑣
=
(109)
(243.9)
× 100 = 𝟒𝟒. 𝟕%
𝛾𝑑 =
𝑊
𝑠
𝑉𝑇
=
(904)
(585)
= 𝟏. 𝟓𝟓 𝐠/𝒄𝒎𝟑

44
▪ For a saturated soil, show that:
𝛾𝑠𝑎𝑡 =
1 + 𝑤𝑠𝑎𝑡
1 + 𝑤𝑠𝑎𝑡𝐺𝑠
𝐺𝑠𝛾𝑤
Solution:
𝛾𝑠𝑎𝑡 =
𝑊𝑇
𝑉𝑇
=
𝑊
𝑤 + 𝑊
𝑠
𝑉𝑇
=
𝑊𝑠𝑎𝑡𝑊
𝑤 + 𝑊
𝑠
𝑉𝑇
= (1 + 𝑊
𝑤)
𝑊
𝑠
𝑉𝑇
𝑉
𝑣 = 𝑉
𝑤 = 𝑒
𝑊
𝑤 = 𝑒γ𝑤
𝑊𝑇
𝑊
𝑠 = 𝐺𝑠γ𝑤 𝑉
𝑠 = 1
𝑉𝑇 = 1 + 𝑒
Recall:

45
▪ Recall:
𝑊
𝑠 = 𝐺𝑠𝑉
𝑠𝛾𝑤
𝑒 + 1 =
𝑉
𝑣 + 𝑉
𝑠
𝑉
𝑠
=
𝑉𝑇
𝑉
𝑠
▪ Substituting we get:
𝛾𝑠𝑎𝑡 = (1 + 𝑊𝑠𝑎𝑡)
𝐺𝑠𝛾𝑤
1 + 𝑒
▪ Recall again:
𝑒 = 𝑊𝑠𝑎𝑡𝐺𝑠
▪ Finally, substituting we get :
∴ 𝜸𝒔𝒂𝒕=
𝟏 + 𝒘𝒔𝒂𝒕
𝟏 + 𝒘𝒔𝒂𝒕𝑮𝒔
𝑮𝒔𝜸𝒘

46
Homework - Phase Relations
▪ Prove the following relationships::
a)
b)
c)
d)
e)
f)
𝛾𝑑 = (1 − n)𝛾𝑤𝐺𝑠
𝛾𝑠𝑎𝑡 = [(𝐺𝑠 − n 𝐺𝑠 − 1 ]𝛾𝑤
𝑊(𝑠𝑎𝑡) =
𝑛𝛾𝑤
𝛾𝑠𝑎𝑡 − 𝑛𝛾𝑤
𝐺𝑠 =
𝛾𝑠𝑎𝑡
𝛾𝑤 − 𝑊
𝑐(𝛾𝑠𝑎𝑡 − 𝛾𝑤)
𝑒 =
𝛾𝑠𝑎𝑡 − 𝛾𝑑
𝛾𝑑 − 𝛾𝑠𝑎𝑡 + 𝛾𝑤)
𝛾𝑠𝑎𝑡 = 𝑛
1 + 𝑤𝑠𝑎𝑡
𝑤𝑠𝑎𝑡
𝛾𝑠𝑎𝑡

47
▪ Das, B., M. (2014), “Principles of Geotechnical
Engineering”, Eighth Edition, CENGAGE Learning, ISBN-
13: 978-1-133-10867-2
▪ Budhu, M. (2015), “Soil Mechanics Fundamentals”, First
Edition, John Wiley & Sons, ISBN: 978-1-119-01965-7.
▪ Chebet, F. (2020), Geotechnical Engineering 1 - Lecture
Notes, University of Cape Town (UCT).
▪ Orabi, A. (2016), 303322 Soil Mechanics – Lecture 1:
Introduction & Properties of Soil, International University for
Science & Technology (IUST).
▪ Ahmed, S.S. (n.d), CE 210 Soil Mechanics and Foundation
Engineering I - Lecture Notes: Chapter 2 - Soil Definition,
Classification and Properties, College of Engineering - MU.
References

1.3 Soil Phase Relationships.pdf

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