Page 1
Phase Relations
2.1 INTRODUCTION
Soils generally contain soil grains, water and air, which are known as the three phases. The
relative proportions of these three phases play an important role in the engineering behaviour of
the soils. Two extreme cases are dry soils and saturated soils, both having only two phases. Dry
soils do not have water, and all the voids are filled by air. Saturated soils do not have air, and the
voids are filled by water only. Very often, in geotechnical problems (e.g., earth works) and
laboratory tests, it is necessary to compute the masses (or weights) and volumes of these three
phases.
In this chapter, you will learn how to compute masses (or weights) and volumes of the soil
grains, water and air in the soils. Let’s define some simple terms and develop expressions
relating them. These would be very useful in the phase relation calculations. It is important to
understand (They are quite logical. You don’t have to memorise them.) these definitions. They
will appear in almost every chapter in this subject.
2.2 DEFINITIONS
Let’s consider a soil mass shown in Fig. 2.1 (a), where all three phases are present. The soil
grains, water and air are separated in Fig. 2.1 (b), which is known as the phase diagram. In the
phase diagram, volumes are shown on the left and weights (or masses) are shown on the right.
voids filled by
air and water
(a) Soil
M
a
M
M
w
s
soil
water
air
(b) Phase Diagram
M
t
soil grain
V
t
w
V
s
a
V
V
v
V
Figure 2.1 Phase Relations of Soils
Water content (w) is a measure of the amount of water present in the soil. It is defined as:
Page 2
Phase Relations
2.1 INTRODUCTION
Soils generally contain soil grains, water and air, which are known as the three phases. The
relative proportions of these three phases play an important role in the engineering behaviour of
the soils. Two extreme cases are dry soils and saturated soils, both having only two phases. Dry
soils do not have water, and all the voids are filled by air. Saturated soils do not have air, and the
voids are filled by water only. Very often, in geotechnical problems (e.g., earth works) and
laboratory tests, it is necessary to compute the masses (or weights) and volumes of these three
phases.
In this chapter, you will learn how to compute masses (or weights) and volumes of the soil
grains, water and air in the soils. Let’s define some simple terms and develop expressions
relating them. These would be very useful in the phase relation calculations. It is important to
understand (They are quite logical. You don’t have to memorise them.) these definitions. They
will appear in almost every chapter in this subject.
2.2 DEFINITIONS
Let’s consider a soil mass shown in Fig. 2.1 (a), where all three phases are present. The soil
grains, water and air are separated in Fig. 2.1 (b), which is known as the phase diagram. In the
phase diagram, volumes are shown on the left and weights (or masses) are shown on the right.
voids filled by
air and water
(a) Soil
M
a
M
M
w
s
soil
water
air
(b) Phase Diagram
M
t
soil grain
V
t
w
V
s
a
V
V
v
V
Figure 2.1 Phase Relations of Soils
Water content (w) is a measure of the amount of water present in the soil. It is defined as:
The natural water content for most soils would be well below 100%, but organic soils and some
marine clays can have water contents greater than 100%.
w
M
M
w
s
=× 100 (%)
The void content of a soil is expressed through two simple terms, void ratio (e) and porosity(n),
defined as:
e
V
V
v
s
=
and
n
V
V
v
t
=× 100 (%)
It should be noted that while e is expressed as a decimal number, n is traditionally expressed as a
percentage ranging from 0 to 100%. Void ratios of sands may range from 0.4 to 1 and for clays
these can vary from 0.3 to 1.5. For soft clays and organic soils, e can be even more.
It is necessary to understand the definitions of the specific terms introduced in
this chapter. The expressions relating these terms, including the ones in boxes, can
be derived from the first principles within a few minutes. When carrying out phase
computations, it is a good practice to go from the first principles.
The degree of saturation (S) is a measure of the void volume that is filled by water, expressed as
a percentage ranging from 0 to 100. It is defined as:
S
V
V
w
v
=× 100 (%)
For a completely dry soil S = 0%, and for a soil where the voids are completely filled with water
(saturated soil) S = 100%. Soils below the water table are often saturated.
Unit weight ( ?) of a soil is simply the weight per unit volume. However, because of the different
phases present in the soil, several forms of unit weights are used in geotechnical engineering.
The most common one is the bulk unit weight ( ?
m
), which is also known as total, wet or moist
unit weight. It is the total weight divided by the total volume, and is written as:
?
m
t
t
M
V
=
Dry unit weight ( ?
d
) is the unit weight of the soil when dry. Therefore, it can be written as:
Page 3
Phase Relations
2.1 INTRODUCTION
Soils generally contain soil grains, water and air, which are known as the three phases. The
relative proportions of these three phases play an important role in the engineering behaviour of
the soils. Two extreme cases are dry soils and saturated soils, both having only two phases. Dry
soils do not have water, and all the voids are filled by air. Saturated soils do not have air, and the
voids are filled by water only. Very often, in geotechnical problems (e.g., earth works) and
laboratory tests, it is necessary to compute the masses (or weights) and volumes of these three
phases.
In this chapter, you will learn how to compute masses (or weights) and volumes of the soil
grains, water and air in the soils. Let’s define some simple terms and develop expressions
relating them. These would be very useful in the phase relation calculations. It is important to
understand (They are quite logical. You don’t have to memorise them.) these definitions. They
will appear in almost every chapter in this subject.
2.2 DEFINITIONS
Let’s consider a soil mass shown in Fig. 2.1 (a), where all three phases are present. The soil
grains, water and air are separated in Fig. 2.1 (b), which is known as the phase diagram. In the
phase diagram, volumes are shown on the left and weights (or masses) are shown on the right.
voids filled by
air and water
(a) Soil
M
a
M
M
w
s
soil
water
air
(b) Phase Diagram
M
t
soil grain
V
t
w
V
s
a
V
V
v
V
Figure 2.1 Phase Relations of Soils
Water content (w) is a measure of the amount of water present in the soil. It is defined as:
The natural water content for most soils would be well below 100%, but organic soils and some
marine clays can have water contents greater than 100%.
w
M
M
w
s
=× 100 (%)
The void content of a soil is expressed through two simple terms, void ratio (e) and porosity(n),
defined as:
e
V
V
v
s
=
and
n
V
V
v
t
=× 100 (%)
It should be noted that while e is expressed as a decimal number, n is traditionally expressed as a
percentage ranging from 0 to 100%. Void ratios of sands may range from 0.4 to 1 and for clays
these can vary from 0.3 to 1.5. For soft clays and organic soils, e can be even more.
It is necessary to understand the definitions of the specific terms introduced in
this chapter. The expressions relating these terms, including the ones in boxes, can
be derived from the first principles within a few minutes. When carrying out phase
computations, it is a good practice to go from the first principles.
The degree of saturation (S) is a measure of the void volume that is filled by water, expressed as
a percentage ranging from 0 to 100. It is defined as:
S
V
V
w
v
=× 100 (%)
For a completely dry soil S = 0%, and for a soil where the voids are completely filled with water
(saturated soil) S = 100%. Soils below the water table are often saturated.
Unit weight ( ?) of a soil is simply the weight per unit volume. However, because of the different
phases present in the soil, several forms of unit weights are used in geotechnical engineering.
The most common one is the bulk unit weight ( ?
m
), which is also known as total, wet or moist
unit weight. It is the total weight divided by the total volume, and is written as:
?
m
t
t
M
V
=
Dry unit weight ( ?
d
) is the unit weight of the soil when dry. Therefore, it can be written as:
?
d
s
t
M
V
=
Saturated unit weight ( ?
sat
) is the bulk unit weight of a soil when it is saturated. Submerged unit
weight ( ?') is the effective unit weight of a submerged soil, and is given by:
? ? ? ' = 
sat w
where, ?
w
is the unit weight of water, which is 9.81 kN/m
3
.
Densities ( ?) are similar to unit weights, except that mass, instead of weight, is used in the
computations. Thus, bulk density ( ?
m
), dry density ( ?
d
), saturated density ( ?
sat
) and submerged
density ( ?') can be defined in a similar manner. Density of water ( ?
w
) is 1 g/cc, 1 t/m
3
or 1000
kg/m
3
. You may remember that ? = ?g.
Specific gravity of a soil grain (G
s
) is the ratio of the density of soil grain to the density of water.
It tells us how many times the soil grain is heavier than water. For most soils, G
s
varies in a very
narrow range of 2.6 to 2.8. Nevertheless, there are exceptions where for mine tailings rich in
minerals, we have measured specific gravity values as high as 3.8. For organic soils it can be as
low as 2. In phase computations, if G
s
is not given, it is reasonable to assume a value in this
range.
2.3 PHASE RELATIONSHIPS
All the terms introduced above (e.g., w,
?
m
) are ratios and thus do not depend on
the amount of soil under consideration. In
a homogeneous soil mass they should be
the same anywhere. Let us consider a
portion of the soil shown in Fig. 2.1a,
where the volume of the soil grains is 1
unit volume. Using the terms defined
above, the volumes and weights of the
three phases can be determined as shown
in Fig. 2.2.
soil
water
air
G
Se
s
w
w
Se
1
e
Figure 2.2 Phase Relations when V
s
= 1
For V
s
=1, V
v
and V
w
are e and Se
respectively. The weights are obtained
multiplying the volumes by the appropriate
unit weights.
Now let us develop some simple and very
useful expressions for w, n, ?
m
, etc. using Fig. 2.2.
Water content, when expressed as a decimal number, is:
Page 4
Phase Relations
2.1 INTRODUCTION
Soils generally contain soil grains, water and air, which are known as the three phases. The
relative proportions of these three phases play an important role in the engineering behaviour of
the soils. Two extreme cases are dry soils and saturated soils, both having only two phases. Dry
soils do not have water, and all the voids are filled by air. Saturated soils do not have air, and the
voids are filled by water only. Very often, in geotechnical problems (e.g., earth works) and
laboratory tests, it is necessary to compute the masses (or weights) and volumes of these three
phases.
In this chapter, you will learn how to compute masses (or weights) and volumes of the soil
grains, water and air in the soils. Let’s define some simple terms and develop expressions
relating them. These would be very useful in the phase relation calculations. It is important to
understand (They are quite logical. You don’t have to memorise them.) these definitions. They
will appear in almost every chapter in this subject.
2.2 DEFINITIONS
Let’s consider a soil mass shown in Fig. 2.1 (a), where all three phases are present. The soil
grains, water and air are separated in Fig. 2.1 (b), which is known as the phase diagram. In the
phase diagram, volumes are shown on the left and weights (or masses) are shown on the right.
voids filled by
air and water
(a) Soil
M
a
M
M
w
s
soil
water
air
(b) Phase Diagram
M
t
soil grain
V
t
w
V
s
a
V
V
v
V
Figure 2.1 Phase Relations of Soils
Water content (w) is a measure of the amount of water present in the soil. It is defined as:
The natural water content for most soils would be well below 100%, but organic soils and some
marine clays can have water contents greater than 100%.
w
M
M
w
s
=× 100 (%)
The void content of a soil is expressed through two simple terms, void ratio (e) and porosity(n),
defined as:
e
V
V
v
s
=
and
n
V
V
v
t
=× 100 (%)
It should be noted that while e is expressed as a decimal number, n is traditionally expressed as a
percentage ranging from 0 to 100%. Void ratios of sands may range from 0.4 to 1 and for clays
these can vary from 0.3 to 1.5. For soft clays and organic soils, e can be even more.
It is necessary to understand the definitions of the specific terms introduced in
this chapter. The expressions relating these terms, including the ones in boxes, can
be derived from the first principles within a few minutes. When carrying out phase
computations, it is a good practice to go from the first principles.
The degree of saturation (S) is a measure of the void volume that is filled by water, expressed as
a percentage ranging from 0 to 100. It is defined as:
S
V
V
w
v
=× 100 (%)
For a completely dry soil S = 0%, and for a soil where the voids are completely filled with water
(saturated soil) S = 100%. Soils below the water table are often saturated.
Unit weight ( ?) of a soil is simply the weight per unit volume. However, because of the different
phases present in the soil, several forms of unit weights are used in geotechnical engineering.
The most common one is the bulk unit weight ( ?
m
), which is also known as total, wet or moist
unit weight. It is the total weight divided by the total volume, and is written as:
?
m
t
t
M
V
=
Dry unit weight ( ?
d
) is the unit weight of the soil when dry. Therefore, it can be written as:
?
d
s
t
M
V
=
Saturated unit weight ( ?
sat
) is the bulk unit weight of a soil when it is saturated. Submerged unit
weight ( ?') is the effective unit weight of a submerged soil, and is given by:
? ? ? ' = 
sat w
where, ?
w
is the unit weight of water, which is 9.81 kN/m
3
.
Densities ( ?) are similar to unit weights, except that mass, instead of weight, is used in the
computations. Thus, bulk density ( ?
m
), dry density ( ?
d
), saturated density ( ?
sat
) and submerged
density ( ?') can be defined in a similar manner. Density of water ( ?
w
) is 1 g/cc, 1 t/m
3
or 1000
kg/m
3
. You may remember that ? = ?g.
Specific gravity of a soil grain (G
s
) is the ratio of the density of soil grain to the density of water.
It tells us how many times the soil grain is heavier than water. For most soils, G
s
varies in a very
narrow range of 2.6 to 2.8. Nevertheless, there are exceptions where for mine tailings rich in
minerals, we have measured specific gravity values as high as 3.8. For organic soils it can be as
low as 2. In phase computations, if G
s
is not given, it is reasonable to assume a value in this
range.
2.3 PHASE RELATIONSHIPS
All the terms introduced above (e.g., w,
?
m
) are ratios and thus do not depend on
the amount of soil under consideration. In
a homogeneous soil mass they should be
the same anywhere. Let us consider a
portion of the soil shown in Fig. 2.1a,
where the volume of the soil grains is 1
unit volume. Using the terms defined
above, the volumes and weights of the
three phases can be determined as shown
in Fig. 2.2.
soil
water
air
G
Se
s
w
w
Se
1
e
Figure 2.2 Phase Relations when V
s
= 1
For V
s
=1, V
v
and V
w
are e and Se
respectively. The weights are obtained
multiplying the volumes by the appropriate
unit weights.
Now let us develop some simple and very
useful expressions for w, n, ?
m
, etc. using Fig. 2.2.
Water content, when expressed as a decimal number, is:
w
M
M
Se
G
w
s
w
sw
==
?
?
w
Se
G
s
=
Porosity, when expressed as a decimal number, is:
n
V
V
e
e
v
t
==
+ 1
Bulk unit weight is:
??
m
t
t
s
w
M
V
GSe
e
==
+
+ 1
An expression for saturated unit weight can be obtained by substituting S = 1 in the above
expression (Note that w and S are expressed as decimal numbers in these expressions).i.e.,
??
sat
s
w
Ge
e
=
+
+ 1
EXAMPLES
1. A cylindrical specimen of moist clay has a diameter of 38 mm, height of 76 mm and mass
of 174.2 grams. After drying in the oven at 105
o
C for about 24 hours, the mass is reduced
to 148.4 grams. Find the dry density, bulk density and water content of the clay.
Assuming the specific gravity of the soil grains as 2.71, find the degree of saturation.
Solution:
Volume of the specimen = V
t
= p (1.9)
2
(7.6) = 86.2 cm
3
M
t
= 174.2 g
M
s
= 148.4 g
??
d
= 148.4/86.2 = 1.722 g/cm
3
?
m
= 174.2/86.2 = 2.021 g/cm
3
w = M
w
/M
s
= (174.2148.4)/148.4 = 0.174 or 17.4%
?
?
d
s w
G
e
=
+ 1
Page 5
Phase Relations
2.1 INTRODUCTION
Soils generally contain soil grains, water and air, which are known as the three phases. The
relative proportions of these three phases play an important role in the engineering behaviour of
the soils. Two extreme cases are dry soils and saturated soils, both having only two phases. Dry
soils do not have water, and all the voids are filled by air. Saturated soils do not have air, and the
voids are filled by water only. Very often, in geotechnical problems (e.g., earth works) and
laboratory tests, it is necessary to compute the masses (or weights) and volumes of these three
phases.
In this chapter, you will learn how to compute masses (or weights) and volumes of the soil
grains, water and air in the soils. Let’s define some simple terms and develop expressions
relating them. These would be very useful in the phase relation calculations. It is important to
understand (They are quite logical. You don’t have to memorise them.) these definitions. They
will appear in almost every chapter in this subject.
2.2 DEFINITIONS
Let’s consider a soil mass shown in Fig. 2.1 (a), where all three phases are present. The soil
grains, water and air are separated in Fig. 2.1 (b), which is known as the phase diagram. In the
phase diagram, volumes are shown on the left and weights (or masses) are shown on the right.
voids filled by
air and water
(a) Soil
M
a
M
M
w
s
soil
water
air
(b) Phase Diagram
M
t
soil grain
V
t
w
V
s
a
V
V
v
V
Figure 2.1 Phase Relations of Soils
Water content (w) is a measure of the amount of water present in the soil. It is defined as:
The natural water content for most soils would be well below 100%, but organic soils and some
marine clays can have water contents greater than 100%.
w
M
M
w
s
=× 100 (%)
The void content of a soil is expressed through two simple terms, void ratio (e) and porosity(n),
defined as:
e
V
V
v
s
=
and
n
V
V
v
t
=× 100 (%)
It should be noted that while e is expressed as a decimal number, n is traditionally expressed as a
percentage ranging from 0 to 100%. Void ratios of sands may range from 0.4 to 1 and for clays
these can vary from 0.3 to 1.5. For soft clays and organic soils, e can be even more.
It is necessary to understand the definitions of the specific terms introduced in
this chapter. The expressions relating these terms, including the ones in boxes, can
be derived from the first principles within a few minutes. When carrying out phase
computations, it is a good practice to go from the first principles.
The degree of saturation (S) is a measure of the void volume that is filled by water, expressed as
a percentage ranging from 0 to 100. It is defined as:
S
V
V
w
v
=× 100 (%)
For a completely dry soil S = 0%, and for a soil where the voids are completely filled with water
(saturated soil) S = 100%. Soils below the water table are often saturated.
Unit weight ( ?) of a soil is simply the weight per unit volume. However, because of the different
phases present in the soil, several forms of unit weights are used in geotechnical engineering.
The most common one is the bulk unit weight ( ?
m
), which is also known as total, wet or moist
unit weight. It is the total weight divided by the total volume, and is written as:
?
m
t
t
M
V
=
Dry unit weight ( ?
d
) is the unit weight of the soil when dry. Therefore, it can be written as:
?
d
s
t
M
V
=
Saturated unit weight ( ?
sat
) is the bulk unit weight of a soil when it is saturated. Submerged unit
weight ( ?') is the effective unit weight of a submerged soil, and is given by:
? ? ? ' = 
sat w
where, ?
w
is the unit weight of water, which is 9.81 kN/m
3
.
Densities ( ?) are similar to unit weights, except that mass, instead of weight, is used in the
computations. Thus, bulk density ( ?
m
), dry density ( ?
d
), saturated density ( ?
sat
) and submerged
density ( ?') can be defined in a similar manner. Density of water ( ?
w
) is 1 g/cc, 1 t/m
3
or 1000
kg/m
3
. You may remember that ? = ?g.
Specific gravity of a soil grain (G
s
) is the ratio of the density of soil grain to the density of water.
It tells us how many times the soil grain is heavier than water. For most soils, G
s
varies in a very
narrow range of 2.6 to 2.8. Nevertheless, there are exceptions where for mine tailings rich in
minerals, we have measured specific gravity values as high as 3.8. For organic soils it can be as
low as 2. In phase computations, if G
s
is not given, it is reasonable to assume a value in this
range.
2.3 PHASE RELATIONSHIPS
All the terms introduced above (e.g., w,
?
m
) are ratios and thus do not depend on
the amount of soil under consideration. In
a homogeneous soil mass they should be
the same anywhere. Let us consider a
portion of the soil shown in Fig. 2.1a,
where the volume of the soil grains is 1
unit volume. Using the terms defined
above, the volumes and weights of the
three phases can be determined as shown
in Fig. 2.2.
soil
water
air
G
Se
s
w
w
Se
1
e
Figure 2.2 Phase Relations when V
s
= 1
For V
s
=1, V
v
and V
w
are e and Se
respectively. The weights are obtained
multiplying the volumes by the appropriate
unit weights.
Now let us develop some simple and very
useful expressions for w, n, ?
m
, etc. using Fig. 2.2.
Water content, when expressed as a decimal number, is:
w
M
M
Se
G
w
s
w
sw
==
?
?
w
Se
G
s
=
Porosity, when expressed as a decimal number, is:
n
V
V
e
e
v
t
==
+ 1
Bulk unit weight is:
??
m
t
t
s
w
M
V
GSe
e
==
+
+ 1
An expression for saturated unit weight can be obtained by substituting S = 1 in the above
expression (Note that w and S are expressed as decimal numbers in these expressions).i.e.,
??
sat
s
w
Ge
e
=
+
+ 1
EXAMPLES
1. A cylindrical specimen of moist clay has a diameter of 38 mm, height of 76 mm and mass
of 174.2 grams. After drying in the oven at 105
o
C for about 24 hours, the mass is reduced
to 148.4 grams. Find the dry density, bulk density and water content of the clay.
Assuming the specific gravity of the soil grains as 2.71, find the degree of saturation.
Solution:
Volume of the specimen = V
t
= p (1.9)
2
(7.6) = 86.2 cm
3
M
t
= 174.2 g
M
s
= 148.4 g
??
d
= 148.4/86.2 = 1.722 g/cm
3
?
m
= 174.2/86.2 = 2.021 g/cm
3
w = M
w
/M
s
= (174.2148.4)/148.4 = 0.174 or 17.4%
?
?
d
s w
G
e
=
+ 1
e==
(. )( )
.
. .
271 1
1722
10 0574
?
w
Se
G
s
=
? S= =
(. )( . )
.
.
0174 2 71
0574
0821 or 82.1%
2. Field density testing (e.g., sand replacement method) has shown bulk density of a
compacted road base to be 2.06 t/m
3
with water content of 11.6%. Specific gravity of the
soil grains is 2.69. Calculate the dry density, porosity, void ratio and degree of saturation.
Solution:
w
Se
G
s
=
?Se = (0.116)(2.69) = 0.312
??
m
s
w
GSe
e
=
+
+ 1
?206
269 0312
1
10 .
..
. =
+
+
×
e
?e = 0.457
3. 5 kg of soil, at natural water content of 3%, is to be mixed with water to achieve water
content of 12%. How much water would you add to the above soil?
Solution:
Let the mass of the dry soil be x kg.
?x = 4.854 kg
x
x
w

= =
5
03 .0
At w = 12%, M
w
= (4.854)(0.12) = 0.582 kg
At w = 3%, M
w
= (4.854)(0.03) = 0.146 kg
? amount of water to add = 0.582  0.146 kg
= 436 g = 436 ml
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