BECM 3241 (2).pdf

Geotechnical Engineering-I
BECM 3241
Md. Ashraful Islam Lecturer, Dept. of BECM, RUET
1

Syllabus
Course content:
Introduction to Geotechnical Engineering, formation. type and identification of
soils, soil composition, soil structure and fabric, index properties of soils,
Engineering classification of soils, soil compaction, principles of total and
effective stresses, permeability and seepage, capillarity and flow net, shear-
strength characteristics of soils, compressibility and settlement behavior of
soils.
2

Reference Books
1. Principles of Geotechnical Engineering(7th ,8th, 9th Edition) by Braja M. Das
2. Soil Mechanics and Foundations by B. C. Punmia
3. Principles of Foundation Engineering(7th ,8th, 9th Edition) by Braja M. Das
4. Soil Mechanics And Foundation Engineering by Dr. K. R. Arora
5. Foundation Analysis and Design by Joseph E. Bowles
3

Course Outcomes
At the end of the course, the student will be able to:
 Appreciate the necessity/scope/importance of Geotechnical Engineering for BECM
graduate.
 Characterization and classification of soils
 Understands soil’s physical and plasticity characteristics.
 Understand the profound impact of presence of water in soil on its behavior.
 Understand the principles of total and effective stresses in soil
 Calculate total and effective stresses in a soil mass.
 Understand and calculate permeability and seepage of soil
 Identify shear strength parameters for field conditions
 Compute and analyze the consolidation settlements
 Understand the principles of soil compaction 4

5
 Soil
 Soil is defined as the uncemented aggregate of mineral grains
and decayed organic matter (solid particles) with liquid and
gas in the empty spaces between the solid particles.
 ASTM D 653 defines soils as “(earth), sediments or other
unconsolidated accumulations of solid particles produced by
the physical and chemical disintegration of rocks, and which
may or may not contain organic matter.’’
 Each soil, like human fingerprints, is unique.
Introduction
Fig.:01: Soil Element

6
Introduction
 Soil is used as a construction material in various engineering projects, and it
supports structural foundations.
 Thus, Building engineers must study the properties of soil, such as its origin,
grain-size distribution, ability to drain water, compressibility, shear strength,
and load-bearing capacity.

7
Introduction
 Soil Mechanics
Soil mechanics is the branch of science that deals with the study of the physical
properties of soil and the behavior of soil masses subjected to various types of forces.
 Geotechnical Engineering
Geotechnical Engineering deals with the application of soil mechanics to Civil
engineering/Building engineering field problems( design of foundations, retaining
structures and earth structures.

8
Origin, Formation and types of soil
 Origin of soil and Rock cycle:
In general, soils are formed by weathering of rocks. The major rock types are
categorized as igneous, sedimentary and metamorphic.
 Igneous rocks: formed by the solidification of molten magma.
 Sedimentary rocks: formed from layers of cemented sediments.
 Metamorphic rocks: formed by the alteration of existing rocks(without melting)
due to heat and pressure.

9
Soils
Fig.:02: Rock cycle

10
 Formation of soil:
Soils are formed by weathering of rocks due to mechanical disintegration or
chemical decomposition. When a rock surface gets exposed to atmosphere for
an appreciable time, it disintegrates or decomposes into small particles and
thus the soils are formed.

11
Formation of soil:
 Physical disintegration or mechanical weathering
 Thermal expansion and contraction
 Crystal growth, including frost action
 Alternate wetting and drying
 Unloading( e.g. uplift, erosion, or change in fluid pressure)
 Organic activity (e.g. the growth of plant roots)

12
Formation of soil:
 Chemical decomposition of rocks:
When chemical decomposition or chemical weathering of rocks takes place,
original rock minerals are transformed into new minerals by chemical
reactions. The soils formed do not have the properties of the parent rock.
 Hydrolysis
 Chelation
 Cation exchange
 Oxidation and reduction
 Carbonation

13
 Types of soil
The products of weathering(soils) may stay in the same place or may be moved to
other places by ice, water, wind and gravity. Soils as they are found in different
regions can be classified into two broad categories:
i. Residual soils –to remain at the original place.
ii. Transported soils -to be moved and deposited to other places.
The transported soils may be classified into several groups, depending on their
mode of transportation and deposition:

14
 Types of soil
ii. Transported soils
 Glacial soils—formed by transportation and deposition of glaciers
 Alluvial soils—transported by running water and deposited along streams
 Lacustrine soils—formed by deposition in quiet lakes
 Marine soils—formed by deposition in the seas
 Aeolian soils—transported and deposited by wind
 Colluvial soils—formed by movement of soil from its original place by
gravity, such as during landslides

15
Particle-size and Particle-size distribution of soil
 Soil Particle-size
 The sizes of particles that make up soil vary over a wide range.
 Soils generally are called gravel, sand, silt, or clay, depending on the pre-
dominant size of particles within the soil.
 To describe soils by their particle size, several organizations have developed
particle-size classifications.

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 Particle-size Classifications
Name of organization Grain size (mm)
Gravel Sand Silt Clay
Massachusetts Institute of Technology(MIT) > 2 2 to 0.06 0.06 to 0.002 < 0.002
U.S. Department of Agriculture(USDA) > 2 2 to 0.05 0.05 to 0.002 < 0.002
American Association of State Highway and
Transportation Officials (AASHTO)
76.2 to 2 2 to 0.075 0.075 to 0.002 < 0.002
Unified Soil Classification System 76.2 to 4.75 4.75 to 0.075 Fines(i.e., silts and clays)
< 0.075

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 Soil-separate-size limits by various systems

 Particle-size distribution of soil
Generally two methods are used to find the particle-
size distribution of soil:
 Sieve analysis—for particle sizes larger than
0.075 mm in diameter, and
 Hydrometer analysis—for particle sizes
smaller than 0.075 mm in diameter.
Sieve analysis
Hydrometer analysis

19
 Sieve Analysis

20
 Particle-size Distribution Curve

21
 Particle-size Distribution Curve
𝑪𝒖=
𝑫𝟔𝟎
𝑫𝟏𝟎
𝑪𝒄=
𝑫𝟑𝟎
𝟐
𝑫𝟔𝟎 𝑿 𝑫𝟏𝟎
𝑺𝟎=
𝑫𝟕𝟓
𝑫𝟐𝟓
Where,
𝑪𝒖 = Uniformity coefficient
𝑪𝒄 = Coefficient of gradation
𝑺𝟎 = Sorting coefficient
𝑫𝟏𝟎= diameter corresponding to 10%
finer(Effective size)
𝑫𝟔𝟎= diameter corresponding to 60% finer

22
 Different types of Particle-size distribution curves

23
 Example-1: For a soil with 𝑫𝟏𝟎= 0.08 mm, 𝑫𝟑𝟎= 0.22 mm, and 𝑫𝟔𝟎= 0.41 mm.
Calculate the uniformity coefficient and the coefficient of gradation.
𝑺𝒐𝒍𝒏:
𝑪𝒖=
𝑫𝟔𝟎
𝑫𝟏𝟎
=
𝟎.𝟒𝟏
𝟎.𝟎𝟖
= 5.13
𝑪𝒄=
𝑫𝟑𝟎
𝟐
𝑫𝟔𝟎 𝑿 𝑫𝟏𝟎
=
𝟎.𝟐𝟐𝟐
𝟎.𝟒𝟏 𝑿 𝟎.𝟎𝟖
= 1.48

24
Plasticity and Structure of Soil

25
Plasticity of Soil
The plasticity of a soil is its ability to undergo deformation without
cracking. It is an important index property of fine grained soil, especially
for clayey soils. The adsorbed water in clayey soils is leads to the plasticity
of soil. The soil becomes plastic only when it has clay minerals.

26
Consistency of soil - Atterberg limits
 Consistency is meant the relative ease with which soil can be deformed.
 It is mostly used for fine grained soils for which the consistency is related to a large
extent to water content.
 Consistency denotes the degree of firmness of the soil which may be termed as soft,
firm, stiff or hard.
 In 1911, the Swedish agriculturist Atterberg developed a method to describe the
consistency of fine-grained soils with varying moisture contents.

27
 On an arbitrary basis, depending on the
moisture content, the behavior of soil can be
divided into four basic states—solid,
semisolid, plastic, and liquid—as shown in
figure.
 He sets arbitrary limits, known as consistency
limits or Atterberg limits, for these divisions
in terms of water content.
 Thus, the water content at which the soil
changes from one state to other are known
as Atterberg limits or consistency limits.
 Most useful Atterberg limits are liquid limit,
plastic limit and shrinkage limit.
Consistency of soil - Atterberg limits
Atterberg limits

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Liquid Limit (LL)
 The moisture content, in percent, at which the soil changes from plastic to liquid
state is defined as the liquid limit.
 The minimum water content at which the soil is still in the liquid state but has a
small shearing strength against flowing.
 It is defined as the minimum water content at which a part of soil cut by a groove
of standard dimensions will flow together for a distance of 12.50 mm(0.5 inch)
under an impact of 25 blows in the device.
 The procedure for the liquid limit test is given by ASTM in Test Designation D-
4318.

29
Liquid Limit test device
Liquid limit test device and grooving
tools
Soil pat in the liquid limit device: before test
and after test

30
Liquid Limit test device
Liquid limit test device and grooving tools
Soil pat in the liquid limit device: before test
and after test

31
Plastic Limit (PL)
 The moisture content at which soil changes from semisolid to plastic state is the
plastic limit.
 The plastic limit is defined as the moisture content in percent, at which the soil
crumbles, when rolled into threads of 3.2 mm(
1
8
inch) in diameter. The plastic limit is
the lower limit of the plastic stage of soil.
 The plastic limit test is simple and is performed by repeated rollings of an ellipsoidal-
sized soil mass by hand on a ground glass plate (Figure 4.8).
 The procedure for the plastic limit test is given by ASTM in Test Designation D-4318

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Plastic Limit Test Equipment
Plastic Limit Test

33
Plasticity Index (PI)
The range of consistency (water content) within which a soil exhibits plastic
properties is called plastic range and it is indicated by plasticity index. The plasticity
index (PI) is defined as the difference between the liquid limit and the plastic limit
of a soil, that is
PI = LL - PL

34
Liquidity Index(LI)
 Liquidity index indicates the nearness of its water content to its liquid limit.
 When the soil is at its liquid limit, its liquidity index is equal to unity and it
behaves as a liquid.
 When the soil is at the plastic limit, its liquidity index is zero.
 The liquidity index is also known as Water-Plasticity ratio.
 The ratio of the natural water content of a soil minus its plastic limit, to its
plasticity index is called Liquidity index , that is
LI =
𝒘 −𝑷𝑳
𝑷𝑰
=
𝒘 −𝑷𝑳
𝑳𝑳−𝑷𝑳
Where,
w = natural water content of the soil .

35
Consistency Index(CI)
 It is defined as the ratio of the liquid limit minus the natural water content to
the plasticity index of a soil, that is
CI =
𝑳𝑳− 𝒘
𝑷𝑰
=
𝑳𝑳− 𝒘
𝑳𝑳−𝑷𝑳
 It shows the nearness of the water content of the soil to its plastic limit.
 If the CI of a soil is equal to unity, it is at the plastic limit.
 A soil with a consistency index of zero is at the liquid limit.
 If CI exceeds unity, the soil is in a semi-solid state and will be stiff.
 A negative value of CI indicates that the soil has natural water content greater
than the liquid limit and behaves just like a liquid.

36
Shrinkage Limit (SL)
 The moisture content, in percent, at which the soil changes from solid to
semisolid state is defined as the shrinkage limit.
 The moisture content, in percent, at which the volume of the soil mass ceases to
change is known as the shrinkage limit.
 The maximum water content at which a reduction in water content will not cause
a decrease in the volume of a soil mass.
 It is lowest water content at which a soil can still be completely saturated.
 The procedure for the shrinkage limit test is given by ASTM (2010), ASTM Test
Designation D-427/ ASTM (2010) Test Designation D-4943.

37
Figure 4.11 Definition of shrinkage limit
From Figure 4.11, the shrinkage limit can be
determined as
SL = 𝑤𝑖(%) - ∆w(%) ……………….(i)
Where,
𝑤𝑖 = initial moisture content when the soil
is placed in the shrinkage limit dish
∆w = change in moisture content
However,
𝑤𝑖(%) =
𝑀1 − 𝑀2
𝑀2
x 100 ………………..(ii)
Where,
𝑀1 = mass of the wet soil pat in the dish at
the beginning of the test (g)
𝑀2 = mass of the dry soil pat (g)

38
Also,
∆w(%) =
(𝑉𝑖 − 𝑉𝑓)ρ𝑤
𝑀2
x 100 …………………….(iii)
Where,
𝑉𝑖 = initial volume of the wet soil pat (𝑐𝑚3)
𝑉𝑓 = volume of the oven-dried soil pat (𝑐𝑚3)
ρ𝑤 = density of water (g/𝑐𝑚3 )
Finally, combining Eqs. (i), (ii) and (iii) gives
SL =
𝑴𝟏 − 𝑴𝟐
𝑴𝟐
x 100 -
(𝑽𝒊 − 𝑽𝒇)𝝆𝒘
𝑴𝟐
x 100
 SL = [
𝑴𝟏 − 𝑴𝟐
𝑴𝟐
-
𝑴𝟐
] x 100 Shrinkage limit test: (a) soil pat before
drying; (b) soil pat after drying

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Shrinkage Ratio (SR)
The ratio of the volume change of soil as a percentage of the dry volume to the
corresponding change in moisture content that is
SR =
(
∆𝑽
𝑽𝒇
)
(
∆𝑴
𝑴𝟐
)
=
(
∆𝑽
𝑽𝒇
)
(
∆𝑽𝝆𝒘
𝑴𝟐
)
=
𝑴𝟐
𝑽𝒇𝝆𝒘
=
𝝆𝒅
𝝆𝒘
=
𝜸𝒅
𝜸𝒅
where ∆𝑉 = change in volume
∆𝑀 = corresponding change in the mass of moisture
Thus, the shrinkage ratio of a soil is equal to the mass specific gravity of the soil in its dry
state.
It can also be shown that
𝑮𝒔=
𝟏
𝟏
𝑺𝑹
−(
𝑺𝑳
𝟏𝟎𝟎
)
, where 𝐺𝑠= specific gravity of soil solids.

40
Volumetric Shrinkage(VS)
The volumetric shrinkage or volumetric change is defined as the decrease in
the volume of a soil mass, expressed as a percentage of the dry volume of the
soil mass, when the water content is reduced from a given percentage to the
shrinkage limit.
VS =
𝑽𝒊 − 𝑽𝒅
𝑽𝒅
x 100
It can also be shown that
VS(%) = SR [ω(%) - SL]

41
Linear Shrinkage (LS)
It is defined as the decrease in one dimension of a soil mass expressed as a
percentage of the original dimension, when the water content is reduced from a
given value to the shrinkage limit. It is calculated from the following formula:
LS(%) = 100[ 1 -
𝟏𝟎𝟎
𝑽𝑺 % +𝟏𝟎𝟎
𝟏
𝟑
]

42
Activity of Soils
 Skempton (1953) observed that the plasticity index of a
soil increases linearly with the percentage of clay-size
fraction (% finer than 2μm by weight) present .
 On the basis of these results, Skempton defined a quantity
called activity, which is the slope of the line correlating PI
and % finer than 2μm. This activity may be expressed as
A =
𝑷𝑰
(%of clay−size fraction, by weight)
,
where A = Activity.
 Thus, the Activity of soil is the ratio of the plasticity index
and the percentage of clay fraction (finer than 2μm).
 Activity is used as an index for identifying the swelling
potential of clay soils. Activity (Based on Skempton)

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Plasticity Chart
 Casagrande (1932) studied the relationship of the
plasticity index to the liquid limit of a wide variety of
natural soils.
 On the basis of the test results, he proposed a
plasticity chart as shown in Figure .
 The important feature of this chart is the empirical
A-line that is given by the equation PI =0.73(LL - 20).
 An A-line separates the inorganic clays from the
inorganic silts.
 Inorganic clay values lie above the A-line, and values
for inorganic silts lie below the A-line.
 The U-line[PI = 0.9(LL- 8)] lies above the A-line.
 The information provided in the plasticity chart is of
great value and is the basis for the classification of
fine-grained soils in the Unified Soil Classification
System.

44
Example -1: Following are the results of a shrinkage limit test:
• Initial volume of soil in a saturated state = 24.6 𝑐𝑚3
• Final volume of soil in a dry state = 15.9 𝑐𝑚3
• Initial mass in a saturated state = 44.0 g
• Final mass in a dry state = 30.1 g
Determinate the shrinkage limit, shrinkage ratio, specific gravity, volumetric shrinkage
and the linear shrinkage(at a moisture content of 28%).
𝑺𝒐𝒍𝒏
:
SL = [
𝑴𝟏 − 𝑴𝟐
𝑴𝟐
-
𝑴𝟐
] x 100
= [
𝟒𝟒 −𝟑𝟎.𝟏
𝟑𝟎.𝟏
-
𝟐𝟒.𝟔 −𝟏𝟓.𝟗 ∗𝟏
𝟑𝟎.𝟏
] x 100 = 17.28%
SR =
𝑴𝟐
𝑽𝒇𝝆𝒘
=
𝟑𝟎.𝟏
𝟏𝟓.𝟗 ∗𝟏
=1.89
𝑮𝒔=
𝟏
𝟏
𝑺𝑹
−(
𝑺𝑳
𝟏𝟎𝟎
)
=
𝟏
𝟏
𝟏.𝟖𝟗
−(
𝟏𝟕.𝟐𝟖
𝟏𝟎𝟎
)
= 2.81
Given,
𝑉𝑖 = 24.6 𝑐𝑚3
𝑉𝑓 = 15.9 𝑐𝑚3
𝑀1 = 44.0 g
𝑀2 = 30.1 g
𝝆𝒘= 1 g/𝑐𝑚3
ω = 28%
SL=?, SR =?, 𝑮𝒔= ?, VS= ?
and LS = ?

45
Example -1:
𝑺𝒐𝒍𝒏
:
VS(%) = SR [ω(%) - SL] = 1.89 [28 - 17.28]= 20.26%
LS(%) = 100[ 1 -
𝟏𝟎𝟎
𝑽𝑺 % +𝟏𝟎𝟎
𝟏
𝟑
] = 100[ 1 -
100
20.26+100
1
3
]= 5.96%

46
Example -2: An undisturbed saturated specimen of clay has a volume of 18.9 𝑐𝑚3and a
mass of 30.2 g. On oven drying, the mass reduces to 18.0g. The volume of dry specimen as
determined by displacement of mercury is 9.9 𝑐𝑚3.
Determine shrinkage limit, specific gravity , shrinkage ratio and volumetric shrinkage.
𝑺𝒐𝒍𝒏:
SL = [
𝑴𝟏 − 𝑴𝟐
𝑴𝟐
-
𝑴𝟐
] x 100
= [
30.2 −18.0
18.0
-
18.9 −9.9 ∗1
18.0
] x 100 = 17.78%
SR =
𝑴𝟐
𝑽𝒇𝝆𝒘
=
18.0
9.9 ∗1
=1.82
𝑮𝒔=
𝟏
𝟏
𝑺𝑹
−(
𝑺𝑳
𝟏𝟎𝟎
)
=
1
1
1.82
−(
17.78
100
)
= 2.69
VS =
𝑽𝒊 − 𝑽𝒅
𝑽𝒅
x 100 =
18.9 −9.9
9.9
x 100 = 90.91%
Given,
𝑉𝑖 = 18.9 𝑐𝑚3
𝑉𝑓 = 9.9 𝑐𝑚3
𝑀1 = 30.2 g
𝑀2 = 18.0 g
𝝆𝒘= 1 g/𝑐𝑚3
SL=?, SR =?, 𝑮𝒔= ?, VS= ?

47
Assignment
1. The mass specific gravity of a fully saturated specimen of clay having a water content of
36% is 1.86. On oven drying, the mass specific gravity drops to 1.72. Calculate the specific
gravity of clay and its shrinkage limit.[B.C. Punmia_Example- 3.7_p-70]
2. The Atterberg limits of a clay soil are: liquid limit 52%, plastic limit 30% and shrinkage limit
18%. If the specimen of this soil shrinks from a volume of 39.5 𝑐𝑚3
at the liquid limit to a
volume of 24.2 𝑐𝑚3
at the shrinkage limit, calculate the true specific gravity.
[B.C. Punmia_Example- 3.8_p-70-71]
3. A saturated soil sample has a volume of 25 𝑐𝑚3 at the liquid limit. If the soil has liquid limit
and shrinkage limit of 42% and 20% respectively, determine the maximum volume which
can be attained by the soil specimen. Take G = 2.72.[B.C. Punmia_Example- 3.10_p-70]

48
Soil Structure
 Soil structure is defined as the geometric arrangement of soil particles with
respect to one another.
 Among the many factors that affect the structure are the shape, size, and
mineralogical composition of soil particles, and the nature and composition of
soil water.
 In general, soils can be placed into two groups:
(i) Cohesionless soil and
(ii) Cohesive soil.

49
Structures in Cohesionless Soils
The structures generally encountered in
cohesionless soils can be divided into two major
categories:
i. Single grained structure and
ii. Honeycombed structure
Single Grained Structure
Honeycombed Structure

50
Single Grained Structure
 In single-grained structures, soil particles are in stable positions, with each
particle in contact with the surrounding ones.
 The shape and size distribution of the soil particles and their relative positions
influence the denseness of packing.
Single-grained structure: (a) loose; (b) dense

51
Honeycombed Structure
 In the honeycombed structure, relatively fine sand
and silt form small arches with chains of particles.
 Soils that exhibit a honeycombed structure have
large void ratios, and they can carry an ordinary
static load.
 However, under a heavy load or when subjected to
shock loading, the structure breaks down, which
results in a large amount of settlement.
Honeycombed structure

52
Structures in Cohesive Soils
 The structure of cohesive soils is highly complex.
 The macrostructure of clay soils can be broadly divided into
categories such as-
 Dispersed structures
 Flocculent structures
 Domains
 Clusters and
 Peds
 Dispersed structures- Formed by settlement of
individual clay particles; more or less parallel orientation
Dispersed structures

53
 Flocculent structures- Formed by settlement of flocs of clay particles.
Flocculent structures Nonsalt flocculation Salt flocculation

54
Arrangement of domains and clusters with silt-sized particles
 Domains- Aggregated or flocculated submicroscopic units of clay particles.
 Clusters- Domains group to form clusters; can be seen under light microscope.

55
Arrangement of peds and
macropore spaces
 Peds- Clusters group to form peds; can be seen without microscope.

56
Clay Minerals
There are two fundamental building blocks for the clay mineral structures-
(1)Silica tetrahedron and (2) Alumina octahedron.
= Oxygen
= Silicon
= Hydroxyl/Oxygen
= Aluminum/ Magnesium

57
 Each tetrahedron unit consists of four oxygen atoms surrounding a silicon atom.
 The combination of tetrahedral silica units gives a silica sheet.
 Three oxygen atoms at the base of each tetrahedron are shared by neighboring
tetrahedra.
 The silica sheet is represented by
Tetrahedron Unit

58
Octahedral Units
 The octahedral units consist of six hydroxyls surrounding an aluminum atom and the
combination of the octahedral aluminum hydroxyl units gives an octahedral sheet.
 This also is called a gibbsite sheet.
 Sometimes magnesium replaces the aluminum atoms in the octahedral units; in this
case, the octahedral sheet is called a brucite sheet.
 This unit is symbolished by

59
Kaolinite
 Kaolinite consists of repeating layers of elemental
silica-gibbsite sheets in a 1:1 lattice, as shown in
figures.
 Each layer is about 7.2 Å thick.(1Å =one angstrom=
10−7 mm =10−10m)
 The layers are held together by hydrogen bonding.
 Kaolinite occurs as platelets, each with a lateral
dimension of 1000 to 20,000 å and a thickness of 100
to 1000 å.
 The surface area of the kaolinite particles per unit
mass is about 15 𝑚2/g. The surface area per unit mass
is defined as specific surface. Kaolinite

60
Illite
 Illite consists of a gibbsite sheet bonded to two silica
sheets—one at the top and another at the bottom.
 It is sometimes called clay mica.
 The illite layers are bonded by potassium ions.
 Illite particles generally have lateral dimensions
ranging from 1000 to 5000 Å and thicknesses from 50
to 500 Å.
 The specific surface of the particles is about 80 𝑚2/g.
Illite

61
Montmorillonite
 Montmorillonite has a structure similar to that of
illite—that is, one gibbsite sheet sandwiched
between two silica sheets.
 Potassium ions are not present as in illite, and a
large amount of water is attracted into the space
between the layers.
 Particles of montmorillonite have lateral
dimensions of 1000 to 5000 Å and thicknesses of
10 to 50 Å.
 The specific surface is about 800 𝑚2/g.
Montmorillonite

63
Introduction
 Different soils with similar properties may be classified into groups and subgroups
according to their engineering behavior.
 Most of the soil classification systems that have been developed for engineering
purposes are based on simple index properties such as particle-size distribution and
plasticity.
 For general engineering purposes, soils may be classified by the following systems:
 Textural Classification(based on the particle-size distribution)
 The textural classification system developed by the U.S. Department of
agriculture.
 Classification by Engineering Behavior[based on the particle-size distribution and
the plasticity (i.e., liquid limit and plasticity index)].
 The AASHTO classification system, and
 The Unified classification system.

64
Textural Classification
 Texture of soil refers to its surface appearance.
 Soil texture is influenced by the size of the individual particles present in it.
 In the textural classification system, the soils are named after their principal
components, such as sandy clay, silty clay, and so forth.
 This classification method is based on the particle-size distribution of soil.
 U.S. Department of Agriculture System (USDA)
 By making use of the grain size limits, a triangular classification chart has been
developed for classifying mixed soils.
 With the given relative percentages of the sand, silt and clay, a point is located on the
triangular chart.
 Because of its simplicity, it is widely used by workers in the field of agriculture.

65
USDA Textural Classification
Sand = 30%,
Silt = 40%
Clay =30%
This soil falls into the zone of clay loam.

66
USDA Textural Classification
Soil Compositions
Gravel = 20%
Sand = 10%,
Silt = 30%
Clay = 40%
Modified Textural Compositions
Sand =
10 𝑥 100
100 −20
= 12.5%
Silt =
30 𝑥 100
100 −20
= 37.5%
Clay =
40 𝑥 100
100 −20
= 50.0%
The USDA textural classification is clay. However,
because of the large percentage of gravel, it may be
called gravelly clay.
 This chart is based on only the fraction of soil that passes through the
No. 10 sieve. Hence, if the particle-size distribution of a soil is such
that a certain percentage of the soil particles is larger than 2 mm in
diameter, a correction will be necessary.

67
Example- 1:Classify the following soils according to the USDA textural classification
system.
Particle-size
distribution (%)
Soil
A B C D
Sand 28.4 37.8 15 25
Silt 36.4 36.6 30 29.5
Clay 35.2 25.6 55 45.5
Solution
Step- 1: Calculate the modified percentages of sand, silt and clay as follows:
 Modified % sand =
% 𝑠𝑎𝑛𝑑
100 −% 𝑔𝑟𝑎𝑣𝑒𝑙
x 100
 Modified % silt =
% silt
x 100
 Modified % clay =
% clay
x 100
Particle-size
distribution (%)
Soil
A B C D
Gravel 12 18 0 12
Sand 25 31 15 22
Silt 32 30 30 26
Clay 31 21 55 40

68
Modified Textural Compositions
Particle-size
distribution (%)
Soil
A B C D
Sand 28.4 37.8 15 25
Silt 36.4 36.6 30 29.5
Clay 35.2 25.6 55 45.5
Classification of soil
A B C D
Gravelly clay
loam
Gravelly loam Clay Gravelly clay

69
Classification by Engineering Behavior
 Currently, two more elaborate classification systems are commonly used by soils
engineers.
i. American Association of State Highway and Transportation Officials
(AASHTO) classification system and
ii. Unified Soil Classification System.
 Both systems take into consideration the particle-size distribution and Atterberg
limits.
 The AASHTO classification system is used mostly by state and county highway
departments.
 Geotechnical engineers generally prefer the Unified system.

70
AASHTO Classification System
 The AASHTO system of soil classification was developed in 1929 as the Public Road
Administration classification system.
 According to this system, soil is classified into seven major groups: A-1 through A-7.
 Soils classified under groups A-1, A-2, and A-3 are granular materials of which 35% or
less of the particles pass through the No. 200 sieve.
 Soils of which more than 35% pass through the No. 200 sieve are classified under
groups A-4, A-5, A-6, and A-7. These soils are mostly silt and clay-type materials.
 This classification system is based on the following criteria:
1. Grain size
a. Gravel: fraction passing the 75-mm sieve and retained on the No. 10 (2-mm)
U.S. sieve
b. Sand: fraction passing the No. 10 (2-mm) U.S. sieve and retained on the No.
200 (0.075-mm) U.S. sieve
c. Silt and clay: fraction passing the No. 200 U.S. sieve

71
Table : Classification of Highway Subgrade Materials

72
Table : Classification of Highway Subgrade Materials

73
AASHTO Classification System
2. Plasticity: The term silty is applied when the fine fractions of the soil have a plasticity
index of 10 or less. The term clayey is applied when the fine fractions have a plasticity
index of 11 or more.
3. If cobbles and boulders (size larger than 75 mm) are encountered, they are excluded
from the portion of the soil sample from which classification is made. However, the
percentage of such material is recorded.
 To evaluate the quality of a soil as a highway subgrade material, one must also
incorporate a number called the group index (GI) with the groups and subgroups of
the soil.
 In general, the quality of performance of a soil as a subgrade material is inversely
proportional to the group index.

74
Group Index (GI)
 The group index is given by the equation
Partial group index determined
from the liquid limit
Partial group index
determined from the
plasticity index
GI = (𝑭𝟐𝟎𝟎 – 35) [0.2 +0.005(LL – 40)] + 0.01(𝑭𝟐𝟎𝟎 – 15)(PI – 10)
Where
𝑭𝟐𝟎𝟎= Percentage passing through the no. 200 sieve
LL = Liquid Limit
PI = Plasticity Index

75
Rules for Determining the Group Index
Following are some rules for determining the group index:
1. If GI yields a negative value , it is taken as 0.
2. The calculated group index is rounded off to the nearest whole number (for example,
GI = 3.4 is rounded off to 3; GI = 3.5 is rounded off to 4.
3. There is no upper limit for the group index.
4. The group index of soils belonging to groups A-1-a, A-1-b, A-2-4, A-2-5, and A-3 is
always 0.
5. When calculating the group index for soils that belong to groups A-2-6 and A-2-7,
use the partial group index for PI, or GI = 0.01(𝑭𝟐𝟎𝟎 – 15)(PI – 10)

76
Example- 2: Classify the following soil by the AASHTO Classification System:
Percentage passing No. 10 sieve = 100
Solution:
Since 58% of the soil is passing through the no. 200 sieve, it falls under silt-clay classifications—that is,
it falls under group A-4, A-5, A-6, or A-7. Starting from the left of the table, the soil falls under A-4(see
the table below).
Parameter Specifications in Table Parameters of the given soil
Percent passing sieve
No. 10
No. 40
No. 200
------
------
36 min. 58
Liquid limit
Plasticity index
40 max.
10 max.
30
10
GI = (𝐹200 – 35) [0.2 +0.005(LL – 40)] + 0.01(𝐹200 – 15)(PI – 10)
= (58– 35) [0.2 +0.005(30 – 40)] + 0.01(58– 15)(10 – 10) = 3.45 ≈ 3
So, the classification is A-4(3)
Liquid limit (–No. 40 fraction) = 30
Plasticity index (–No. 40 fraction) = 10

77
Example- 3: Classify the following soil by the AASHTO Classification System:
Solution:
The percentage passing through the No. 200 sieve is less than 35, so the soil is a granular material. From
Table , we see that it is type A-2-6.
Parameter Specifications in Table Parameters of the given soil
Percent passing sieve
No. 10
No. 40
No. 200
------
------
35max. 34
Liquid limit
Plasticity index
40 max.
11 min.
37
12
GI = 0.01(𝑭𝟐𝟎𝟎 – 15)(PI – 10)= 0.01(34– 15)(12 – 10)= 0.38 ≈0
Thus, the soil is type A-2-6(0)
Liquid limit (–No. 40 fraction) = 37
Plasticity index (–No. 40 fraction) = 12

78
Unified Soil Classification System
 The original form of this system was proposed by Casagrande in 1942 for use in the
airfield construction works during World War II.
 In cooperation with the U.S. Bureau of Reclamation and Army Corps of Engineers
this system was modified in 1952 to enable the system to be applicable to other
constructions like foundations, earth dams, earth canals and earth slopes etc.
 This system classifies soils into two broad categories:
a) Coarse-grained soils that are gravelly and sandy in nature with less than 50%
passing through the No. 200 sieve.
b) Fine-grained soils are with 50% or more passing through the No. 200 sieve.

79
Symbols Used in USCS
Soil type Prefix Sub group Suffix
Gravel G Well graded W
Sand S Poorly-graded P
Silt M Silty M
Clay C Clayey C
Organic O low plasticity (LL < 50) L
Peat Pt high plasticity (LL> 50) H
GW = well graded gravels
GP = Poorly graded gravels
SW = Well-graded sands
SP = Poorly graded sands
GM = Silty gravels
GC = Clayey gravels
SC = Clayey sands
SM = Silty sand
ML = Inorganic silts of low plasticity
CL = Inorganic clays of low to medium
plasticity
OL = organic silts of low plasticity
MH = Inorganic silts of high plasticity
CH = Inorganic clays of high plasticity
OH = Organic clays of high plasticity

80

81
Unified
Soil
Classification
System

82

84
Comparison between the AASHTO and Unified Systems
 Both soil classification systems, AASHTO and Unified, are based on the texture and
plasticity of soil.
 Also, both systems divide the soils into two major categories, coarse grained and
fine grained, as separated by the No. 200 sieve.
 According to AASHTO system, a soil is considered fine grained when more than
35% passes through the No. 200 sieve whereas According to the Unified system, a
soil is considered fine grained when more than 50% passes through the No. 200
sieve.
 In the AASHTO system, the No. 10 sieve is used to separate gravel from sand; in the
Unified system, the No. 4 sieve is used.

85
In the Unified system, the gravelly and sandy soils clearly are separated; in the
AASHTO system, they are not.
The A-2 group, in particular, contains a large variety of soils. Symbols like GW, SM, CH,
and others that are used in the Unified system are more descriptive of the soil
properties than the A symbols used in the AASHTO system.
The classification of organic soils, such as OL, OH, and Pt, is provided in the Unified
system. Under the AASHTO system, there is no place for organic soils.
Comparison between the AASHTO and Unified Systems

87
Permeability
 Soils are assemblages of solid particles with interconnected
voids where water can flow from a point of high energy to a
point of low energy.
 Permeability is the measure of the soil’s ability to permit water
to flow through its pores or voids.
 Permeability is defined as the property of a porous material
which permits the passage or seepage of water through its
interconnecting voids.
 Coarse- grained soils high permeability
 Fine- grained soils low permeability
Loose soil
Easy to flow high permeability
Dense soil
Difficult to flow low permeability

88
Importance of Permeability
 Influence the rate of settlement of a saturated soil under load
 Design of earth dams is very much based upon the permeability of the soils
used.
 Stability of slopes and retaining structures can be greatly affected by the
permeability of the soils involved
 Filters made of soils are design based upon their permeability.
 Ground water flow towards wells and drainage of soil.

89
Bernoulli’s Equation
The total head at a point in water under motion can be given by the sum of the
pressure, velocity, and elevation heads.
h
u
z

90
Bernoulli’s Equation
 If Bernoulli’s equation is applied to the flow of water through
a porous soil medium, the term containing the velocity head
can be neglected because the seepage velocity is small, and
the total head at any point can be adequately represented by
h =
𝒖
𝜸𝒘
+ Z
 The loss of head between two points, A and B, can be given by
∆h = 𝒉𝑨 - 𝒉𝑩 = (
𝒖𝑨
𝜸𝒘
+ 𝒁𝑨) - (
𝒖𝑩
𝜸𝒘
+ 𝒁𝑩)
 The head loss, ∆h , can be expressed in a nondimensional form as
i =
∆𝐡
𝐋
where i = hydraulic gradient
L = distance between points A and B—that is, the length of flow over which the loss of
head occurred

91
Factors Affecting Permeability
1. Grain size
2. Properties of the pore fluid
3. Void ratio of the soil
4. Structural arrangement of the soil particles
5. Entrapped air and foreign-matter
6. Adsorbed water in clayey soils

92
Darcy’s Law
In 1856, Darcy derived an empirical formula for laminar flow conditions in a saturated
soils, “The velocity of flow through soil is proportional to hydraulic gradient.”
v ꝏ i
 v = ki …………..………….(i)
 v =
𝑞
𝐴
= ki
Then the quantity of water flowing through the soil per unit time is q = kiA = k.
∆𝒉
𝑳
. A
Where,
v = velocity of flow or discharge velocity
k = hydraulic conductivity /coefficient of permeability
q = discharge per unit time
A = total cross-sectional area of soil mass, perpendicular to the direction of flow
i = hydraulic gradient =
∆𝒉
𝑳

93
Darcy’s Law
 If a soil sample of length L and cross-sectional
area A, is subjected to differential head of water,
ℎ1 - ℎ2, the hydraulic gradient i will be equal to
ℎ1 − ℎ2
𝑳
and, we have
q = k.
𝒉𝟏 − 𝒉𝟐
𝑳
. A
Flow of water through soil
 When hydraulic gradient is unity(Eq.(i)), k = v. Thus the coefficient of permeability is
defined as the average velocity of flow that will occur through the total cross-
sectional area of soil under unit hydraulic gradient.

94
Relationship Between the Discharge Velocity and the Seepage
Velocity
Discharge Velocity:
 It is defined as the quantity of water flowing in unit time through a unit gross cross-
sectional area of soil at right angles to the direction of flow.
 If A is the gross cross-sectional area and q is the quantity of water flowing through
the soil per unit time, then discharge velocity, v =
𝑞
𝐴
.
Seepage Velocity:
 It is the actual velocity of water through the void spaces and denoted by 𝑣𝑠.
 𝑣𝑠 is greater than v.

95
Velocity
Figure shows a soil of length L with a
gross cross-sectional area A.
If the quantity of water flowing
through the soil in unit time is q, then
q = vA = 𝐴𝑣𝑣𝑠 ………………………….(i)
Where,
𝑣𝑠= seepage velocity
𝐴𝑣=area of void in the cross section of the specimen
However,
A = 𝐴𝑣 + 𝐴𝑠 ………………………………(ii)
Where,
𝐴𝑠= area of soil solids in the cross section of the specimen

96
Combining Eqs. (i) and (ii) gives
 q = v(𝐴𝑣 + 𝐴𝑠) = 𝐴𝑣𝑣𝑠
 𝑣𝑠 =
v(𝐴𝑣 + 𝐴𝑠)
𝐴𝑣
=
v(𝐴𝑣 + 𝐴𝑠)L
𝐴𝑣𝐿
=
v(𝑉𝑣 + 𝑉𝑠)
𝑉𝑣
=
𝑣𝑉𝑠(1+
𝑉𝑣
𝑉𝑠
)
𝑉𝑣
𝑉𝑠
=
𝑣(1+𝑒)
𝑒
=
𝑣
𝑒
1+𝑒
=
𝑣
𝑛
 𝒗𝒔 =
𝒗
𝒏
or, v = 𝒏𝒗𝒔
Velocity
Where,
𝑉
𝑣=volume of voids in the specimen
𝑉
𝑠= volume of soil solids in the specimen
e =
𝑉𝑣
𝑉𝑠
= void ratio
𝑛 =
𝑒
1+𝑒
= porosity

97
Hydraulic Conductivity
 The coefficient of permeability(k) is also known as hydraulic conductivity, is a
measure of soil permeability. It is generally expressed in cm/sec or m/sec in SI units.
 The hydraulic conductivity of soils depends on several factors:
 Fluid viscosity
 Pore size distribution
 Grain-size distribution
 Void ratio
 Roughness of mineral particles
 Degree of soil saturation
 Two standard laboratory tests are used to determine the hydraulic conductivity of soil.
i. The constant-head test
ii. The falling-head test.

98
Constant-Head Test
 The constant-head test is used primarily for coarse-grained soils.
 A typical arrangement of the constant-head
permeability test is shown in Figure .
 In this type of laboratory setup, the water supply at
the inlet is adjusted in such a way that the difference
of head between the inlet and the outlet remains
constant during the test period.
 After a constant flow rate is established, water is
collected in a graduated flask for a known duration. Constant-head permeability
test

99
Constant-Head Test
Constant-head hydraulic conductivity test arrangement for
a granular soil in the laboratory

100
Constant-Head Test
The total volume of water collected may be expressed as
Q = Avt = A(ki)t = A(k
ℎ
𝐿
)t where v = ki and i =
𝒉
𝑳
Therefore, k =
𝑸𝑳
𝑨𝒉𝒕
where
Q = volume of water collected
A = area of cross section of the soil specimen
t = duration of water collection
L = length of the specimen

101
Falling-Head Test
 The falling head test is used for both coarse-grained soils as
well as fine-grained soils.
 A typical arrangement of the falling-head permeability test is
shown in Figure.
 Water from a standpipe flows through the soil.
 The initial head difference ℎ1 at time t = 0 is recorded, and
water is allowed to flow through the soil specimen such that
the final head difference at time t = 𝑡2 is ℎ2.
 Hence, from Darcy’s law, the rate of flow q is given by
q =
− 𝒅𝒉
𝒅𝒕
a = kiA
 k
𝒉
𝑳
A = -
𝒅𝒉
𝒅𝒕
a ….(i)
Where
q = flow rate
a = cross-sectional area of the standpipe
A = cross-sectional area of the soil specimen

102
Falling-Head Test
Rearrangement of Eq. (i) gives
dt =
𝑎𝐿
𝐴𝑘
(-
𝑑ℎ
ℎ
)
Integrating with limits of time from 0 to t and head difference from ℎ1 to ℎ2
‫׬‬
0
𝑡
𝑑𝑡 = -
𝑎𝐿
𝐴𝑘
‫׬‬
ℎ1
ℎ2 𝑑ℎ
ℎ
 t = -
𝑎𝐿
𝐴𝑘
[𝑙𝑛ℎ2 - 𝑙𝑛ℎ1] =
𝑎𝐿
𝐴𝑘
ln
ℎ1
ℎ2
Therefore,
k =
𝒂𝑳
𝑨𝒕
ln
𝒉𝟏
𝒉𝟐
or, k = 2.303
𝒂𝑳
𝑨𝒕
𝒍𝒐𝒈𝟏𝟎
𝒉𝟏
𝒉𝟐

103
Example-1: A constant-head permeability test gives the following values:
• L = 30 cm
• A = area of the specimen = 177 𝑐𝑚2
• Constant-head difference, h = 50 cm
• Water collected in a period of 5 min = 350 𝑐𝑚3
Calculate the hydraulic conductivity in cm/sec.
𝑺𝒐𝒍𝒏:
We have,
k =
𝑸𝑳
𝑨𝒉𝒕
 k =
𝟑𝟓𝟎∗𝟑𝟎
𝟏𝟕𝟕∗𝟓𝟎∗(𝟓∗𝟔𝟎)
= 3.95 x 𝟏𝟎𝟑 cm/sec
Given,
Q = 350 𝑐𝑚3
A = 177 𝑐𝑚2
t = 5 min= (5*60) sec
L = 30cm
h= 50 cm
k = ???

104
Example-2: For a falling-head permeability test, the following values are given:
• Length of specimen = 200 mm.
• Area of soil specimen = 1000 𝑚𝑚2.
• Area of standpipe = 40 𝑚𝑚2.
• Head difference at time t =0 = 500 mm.
• Head difference at time t =180 sec = 300 mm.
Determine the hydraulic conductivity of the soil in cm/sec.
𝑺𝒐𝒍𝒏:
We have,
k = 2.303
𝒂𝑳
𝑨𝒕
𝒍𝒐𝒈𝟏𝟎(
𝒉𝟏
𝒉𝟐
)
 k = 2.303
𝟒𝟎∗𝟐𝟎𝟎
𝟏𝟎𝟎𝟎∗𝟏𝟖𝟎
𝒍𝒐𝒈𝟏𝟎(
𝟓𝟎𝟎
𝟑𝟎𝟎
)
 k = 2.27 x 𝟏𝟎−𝟐𝐦m/sec or, k = 2.27 x 𝟏𝟎−𝟒 cm/sec
Given,
a = 40 𝑚𝑚2
L = 200 mm
A = 1000 𝑚𝑚2
t = 180 sec
ℎ1 = 500 mm
ℎ2 = 300 mm
k = ???

105
Example-3: A permeable soil layer is underlain by an impervious layer, as shown in
Figure below. With k = 5.3 x 10−5
m/sec for the permeable layer, calculate the rate of
seepage through it in 𝑚3 /hr/m width if H = 3 m and a = 8°.
𝑺𝒐𝒍𝒏:
From Figure,
i =
ℎ𝑒𝑎𝑑 𝑙𝑜𝑠𝑠
𝑙𝑒𝑛𝑔𝑡ℎ
=
𝐿 tan 𝛼
(
𝐿
cos 𝛼
)
= sin 𝛼
We have,
q = kiA = k*(𝑠𝑖𝑛 𝛼)*(3𝑐𝑜𝑠 𝛼)*(1)
 q = 5.3 x 10−5 * (60x60) *(𝑠𝑖𝑛 8°)*(3𝑐𝑜𝑠 8°)*(1) = 0.0789 𝒎𝟑 /hr/m
Given,
k = 5.3 x 10−5
m/sec
 k = 5.3 x 10−5
* (60x60) m/hr

106
Example-4: Find the flow rate in 𝑚2/sec/m length (at right angles to the cross
section shown) through the permeable soil layer shown in Figure below. given H = 8 m,
𝐻1 = 3 m, h = 4 m, L = 50 m, α = 8°, and k = 0.08 cm/sec.
𝑺𝒐𝒍𝒏:
Hydraulic gradient(i) =
ℎ
𝐿
cos 𝛼
We have,
q = kiA = k(
ℎ cos 𝛼
𝐿
)(𝐻1𝑐𝑜𝑠 𝛼)*(1)
q = 0.08 x 10−2 (
4 cos 8°
50
)(3𝑐𝑜𝑠 8°)*(1)
q = 0.19 x 𝟏𝟎−𝟑 𝒎𝟐/sec/m.
Given,
H = 8 m, 𝐻1 = 3 m,
h = 4 m, L = 50 m,
α = 8°
k = 0.08 cm/sec or,
k = 0.08 x
10−2m/sec

107
Equivalent Hydraulic Conductivity in Stratified Soil
 In a stratified soil deposit where the hydraulic conductivity for flow in a given
direction changes from layer to layer, an equivalent hydraulic conductivity can be
computed to simplify calculations.
 The equivalent hydraulic conductivity of the whole deposit will depend upon the
direction of flow with relation to the bedding planes. We shall consider both the
cases of flow:
i. Horizontal flow in stratified soil
ii. Vertical flow in stratified soil.

108
Equivalent Hydraulic Conductivity – Horizontal Flow in
Stratified Soil(Parallel to the bedding planes)
 Figure shows n layers of soil with flow in the horizontal
direction.
 Let us consider a cross section of unit length passing
through the n layer and perpendicular to the direction of
flow.
 In this case, the hydraulic gradient i will be the same for all
the layers(𝑖𝑒𝑞 = 𝑖1 = 𝑖2 = 𝑖3 = …… = 𝑖𝑛)
 The total flow through the cross section in unit time can be
written as
q = v.1.H= 𝑣1.1. 𝐻1 + 𝑣2.1. 𝐻2 + 𝑣3.1. 𝐻3 + ……..+ 𝑣𝑛.1. 𝐻𝑛
 vH= 𝑣1 𝐻1 + 𝑣2 𝐻2 + 𝑣3 𝐻3 + ……..+ 𝑣𝑛 𝐻𝑛 ………………….(i)
Where v = average discharge velocity
𝑣1, 𝑣2, 𝑣3,……., 𝑣𝑛 = discharge velocities of flow in layers denoted by the subscripts

109
If 𝑘𝐻1
, 𝑘𝐻2
, 𝑘𝐻3
,….., 𝑘𝐻𝑛
are the hydraulic conductivities of the individual layers in the
horizontal direction and 𝑘𝐻(𝑒𝑞)
is the equivalent hydraulic conductivity in the
horizontal direction, then, from Darcy’s law,
v = 𝒌𝑯(𝒆𝒒)
𝒊𝒆𝒒; 𝒗𝟏= 𝒌𝑯𝟏
𝒊𝟏; 𝒗𝟏= 𝒌𝑯𝟏
𝒊𝟏; 𝒗𝟐= 𝒌𝑯𝟐
𝒊𝟐; . . . 𝒗𝒏= 𝒌𝑯𝒏
𝒊𝒏
Substituting the preceding relations for velocities into Eq. (i), we get
Equivalent Hydraulic Conductivity – Horizontal Flow in
Stratified Soil(Parallel to the bedding planes)
𝒌𝑯(𝒆𝒒)
=
𝟏
𝑯
(𝒌𝑯𝟏
𝑯𝟏 + 𝒌𝑯𝟐
𝑯𝟐+ 𝒌𝑯𝟑
𝑯𝟑+ . . . . + 𝒌𝑯𝒏
𝑯𝒏)

110
Equivalent Hydraulic Conductivity – Vertical Flow in Stratified
Soil(Perpendicular to the bedding planes)
Figure shows n layers of soil with flow in the vertical
direction. In this case, the velocity of flow through all
the layers is the same. However, the total head loss, h, is
equal to the sum of the head losses in all layers. Thus,
v = 𝒗𝟏= 𝒗𝟐= 𝒗𝟑= . . . = 𝒗𝒏
h = 𝒉𝟏+ 𝒉𝟐+ 𝒉𝟑+ . . .+ 𝒉𝒏
 h = 𝑯𝟏𝒊𝟏 +𝑯𝟐𝒊𝟐+𝑯𝟑𝒊𝟑+ . . .+𝑯𝒏𝒊𝒏 . . . . . . . . . .. .(i)

111
If 𝑘𝑉1
, 𝑘𝑉2
, 𝑘𝑉3
,….., 𝑘𝑉𝑛
are the hydraulic conductivities of the individual layers in the
vertical direction and 𝑘𝑉(𝑒𝑞)
is the equivalent hydraulic conductivity, then, from darcy’s law,
v = 𝑘𝑉(𝑒𝑞)
.i = 𝑘𝑉(𝑒𝑞)
ℎ
𝐻
; Also, 𝑖1=
𝑣
𝑘𝑉1
; 𝑖2=
𝑣
𝑘𝑉2
, . . ., 𝑖𝑛=
𝑣
𝑘𝑉𝑛
;
 h =
𝑣𝐻
𝑘𝑉(𝑒𝑞)
. . . . . . . . . . . . . . . . . . . .(ii)
From Eq.(i) & Eq.(ii), we get,
vH
kV(eq)
= H1
v
kV1
+H2
v
kV2
+H3
v
kV3
+ . . .+Hn
v
kVn
Equivalent Hydraulic Conductivity – Vertical Flow in Stratified
Soil(Perpendicular to the bedding planes)
 𝒌𝑽(𝒆𝒒)
=
𝑯
𝑯𝟏
𝒌𝑽𝟏
+
𝑯𝟐
𝒌𝑽𝟐
+
𝑯𝟑
𝒌𝑽𝟑
+ .....+
𝑯𝒏
𝒌𝑽𝒏

112
Example-5: Show that for any stratified soil mass 𝒌𝑯(𝒆𝒒)
is always greater than𝒌𝑽(𝒆𝒒)
.
𝑺𝒐𝒍𝒏:
Consider a three-layer system, having 𝑘1= 2, 𝑘2= 1, 𝑘3= 4 units and 𝐻1= 4, 𝐻2= 1, 𝐻3= 2
units.
Therefore, H = 𝐻1+𝐻2+𝐻3 = 4+1+2 = 7 units
𝒌𝑯(𝒆𝒒)
=
(𝒌𝑯𝟏
𝑯𝟏 + 𝒌𝑯𝟐
𝑯𝟐+ 𝒌𝑯𝟑
𝑯𝟑)
𝑯
=
17
7
= 2.43
𝒌𝑽(𝒆𝒒)
=
𝑯
(
𝑯𝟏
𝒌𝑽𝟏
+
𝑯𝟐
𝒌𝑽𝟐
+
𝑯𝟑
𝒌𝑽𝟑
)
= 2
Thus, 𝒌𝑯(𝒆𝒒)
> 𝒌𝑽(𝒆𝒒)
[Showed]

113
Permeability Test in the Field by Pumping from Wells
 In the field, the average hydraulic conductivity of a soil
deposit in the direction of flow can be determined by
performing pumping tests from wells.
 Figure shows a case where the top permeable layer,
whose hydraulic conductivity has to be determined, is
unconfined and underlain by an impermeable layer.
 During the test, water is pumped out at a constant
rate from a test well that has a perforated casing.
Several observation wells at various radial distances
are made around the test well.
 Continuous observations of the water level in the test
well and in the observation wells are made after the
start of pumping, until a steady state is reached.
Pumping test from a well in an unconfined
permeable layer
 The steady state is established when the water level in the test and observation wells becomes
constant.

114
 k =
𝟐.𝟑𝟎𝟑𝒒𝒍𝒐𝒈𝟏𝟎(
𝒓𝟏
𝒓𝟐
)
π(𝒉𝟏
𝟐
− 𝒉𝟐
𝟐
)
. . . . . . . . . . . . .(i)
From field measurements, if q, 𝒓𝟏, 𝒓𝟐, 𝒉𝟏, and 𝒉𝟐 are known, the hydraulic conductivity
can be calculated from the simple relationship presented in Eq. (i).
 The expression for the rate of flow of groundwater into the well, which is equal to
the rate of discharge from pumping, can be written as
q = k(
𝑑ℎ
𝑑𝑟
)2πrh

𝑑𝑟
𝑟
= (
2π𝑘
𝑞
)hdh
 ‫׬‬
𝑟2
𝑟1 𝑑𝑟
𝑟
= (
2π𝑘
𝑞
)‫׬‬
ℎ2
ℎ1
hdh
 ln
𝑟1
𝑟2
=
2π𝑘
2𝑞
(ℎ1
2
- ℎ2
2
)

115
 The average hydraulic conductivity for a confined aquifer can also
be determined by conducting a pumping test from a well with a
perforated casing.
 Pumping is continued at a uniform rate q until a steady state is
reached.
 The steady state of discharge is q = k(
𝑑ℎ
𝑑𝑟
)2πrH
 ‫׬‬
𝑟2
𝑟1 𝑑𝑟
𝑟
= (
2π𝑘H
𝑞
)‫׬‬
ℎ2
ℎ1
dh
Pumping test from a well penetrating the full
depth in a confined aquifer

𝑑𝑟
𝑟
= (
2π𝑘H
𝑞
)dh
 k =
𝟐.𝟑𝟎𝟑𝒒 𝒍𝒐𝒈𝟏𝟎(
𝒓𝟏
𝒓𝟐
)
2π H(ℎ1−ℎ2)
or, k =
𝒒 𝒍𝒐𝒈𝟏𝟎(
𝒓𝟏
𝒓𝟐
)
2.727 H(ℎ1−ℎ2)
 ln
𝑟1
𝑟2
=
2π𝑘
𝑞
(ℎ1-ℎ2)

117
Laplace’s Equation of Continuity
 To derive the Laplace differential equation of continuity, let us
consider a single row of sheet piles that have been driven into a
permeable soil layer, as shown in Figure.
 For flow at a point A, we consider an elemental soil block. The block
has dimensions dx, dy, and dz.
 Let 𝑣𝑥 and 𝑣𝑧 be the entry velocity components in the horizontal
and vertical directions, respectively. Then (𝑣𝑥 +
𝜕𝑣𝑥
𝜕𝑥
𝑑𝑥) and (𝑣𝑧 +
𝜕𝑣𝑧
𝜕𝑧
𝑑𝑧) will be the corresponding velocity components at the exit of
the element.
 The rate of flow of water into the elemental block in the horizontal
and vertical directions respectively are, 𝒗𝒙 dzdy and 𝒗𝒛 dxdy.
 The rates of outflow from the block in the horizontal and vertical
directions respectively are,
(𝒗𝒙 +
𝝏𝒗𝒙
𝝏𝒙
𝒅𝒙)dzdy and (𝒗𝒛 +
𝝏𝒗𝒛
𝝏𝒛
𝒅𝒛))dxdy

118
 Assuming that water is incompressible and that no volume change in the soil mass occurs then,
Total rate of inflow = Total rate of outflow
 𝑣𝑥 dzdy + 𝑣𝑧 dxdy = [(𝑣𝑥 +
𝜕𝑣𝑥
𝜕𝑥
𝑑𝑥)dzdy + (𝑣𝑧 +
𝜕𝑣𝑧
𝜕𝑧
𝑑𝑧)dxdy ]
 𝑣𝑥 dzdy + 𝑣𝑧 dxdy = 𝑣𝑥 dzdy +
𝜕𝑣𝑥
𝜕𝑥
𝑑𝑥dzdy + 𝑣𝑧 dxdy +
𝜕𝑣𝑧
𝜕𝑧
𝑑𝑧dxdy

𝝏𝒗𝒙
𝝏𝒙
+
𝝏𝒗𝒛
𝝏𝒛
= 0 . . . . . . . . . . . . . . . . . . . . . . . . . . .(i)
 From Darcy’s law,
𝑣𝑥 = 𝑘𝑥𝑖𝑥= 𝑘𝑥
𝜕ℎ
𝜕𝑥
and 𝑣𝑧 = 𝑘𝑧𝑖𝑧= 𝑘𝑧
𝜕ℎ
𝜕𝑧
. . . . . . . . . . . . . . . . . . .(ii)
Where, 𝑘𝑥and 𝑘𝑧 are the hydraulic conductivities in the horizontal and vertical directions, respectively.
 From Eqs. (i) and (ii), we can write 𝑘𝑥
𝝏𝟐𝒉
𝝏𝒙𝟐 + 𝑘𝑧
𝝏𝟐𝒉
𝝏𝒛𝟐 = 0
 If the soil is isotropic with respect to the hydraulic conductivity—that is, 𝑘𝑥 = 𝑘𝑧.
 Then, the continuity equation for two-dimensional flow simplifies to
𝝏𝟐𝒉
𝝏𝒙𝟐 +
𝝏𝟐𝒉
𝝏𝒛𝟐 = 0
Laplace’s Equation of Continuity

𝜕𝑣𝑥
𝜕𝑥
𝑑𝑥dzdy +
𝜕𝑣𝑧
𝜕𝑧
𝑑𝑧dxdy = 0

119
Equipotential Line:
An equipotential line is a line along which the potential head at all points is equal. Thus, if
piezometers are placed at different points along an equipotential line, the water level will
rise to the same elevation in all of them.
Flow Line and Equipotential Line
Flow Line:
A flow line is a line along which a water particle will
travel from upstream to the downstream side in the
permeable soil medium.

120
Flow Nets
 To complete the graphic construction of a
flow net, one must draw the flow and
equipotential lines in such a way that
i. The equipotential lines intersect the
flow lines at right angles.
ii. The flow elements formed are
approximate squares Completed flow net
 A combination of a number of flow lines and equipotential lines is called a flow net.
 In these figures,Nf = 4 = the number of flow channels in the flow net, and Nd= 6 =
the number of potential drops

121
Properties of Flow Net
a) The flow lines and equipotential lines meet at right angles to one another.
b) The fields are approximately squares, so that a circle can be drawn touching all the
four sides of the square.
c) The quantity of water flowing through each flow channel is the same. Similarly, the
same potential drop occurs between two successive equipotential lines.
d) Smaller the dimensions of the field, greater will be the hydraulic gradient and
velocity of flow through it.
e) In a homogeneous soil, every transition in the shape of the curves is smooth, being
either elliptical or parabolic in shape.

122
A flow net can be utilized for the following purposes:
i. Determination of seepage
ii. Determination of hydrostatic pressure
iii. Determination of seepage pressure
iv. Determination of exit gradient.
Applications of Flow Net

123
Seepage Calculation from a Flow Net
Let,
b and l be the width and length of the field
∆h = head drop through the field
∆q = discharge passing through the flow channel
H = head difference between the upstream and downstream
sides
From Darcy’s law of flow through soils,
∆q = k.
∆h
𝑙
.(b x 1); [ Considering unit thickness]
Fig. Shows a portion of flow net. The portion between any two successive flow lines is
known as at flow channel. The portion enclosed between two successive equipotential
lines and successive flow lines is known as field such as that shown hatched in fig.
Portion of flow net

124
If 𝑁𝑑 = total number of potential drops in the complete flow net then, ∆h =
𝐻
𝑁𝑑
Hence, ∆q = k.
H
𝑁𝑑
.(
𝑏
𝑙
)
The total discharge through the complete flow net is given by
q = σ ∆q= k.
H
𝑁𝑑
.(
𝑏
𝑙
). 𝑁𝑓 = kH
𝑁𝑓
𝑁𝑑
.(
𝑏
𝑙
)
Where, 𝑁𝑓 = total number of flow channels in the net.
If the field is square(b = l)
Thus, q = kH
𝑵𝒇
𝑵𝒅
Seepage Calculation from a Flow Net
If the field is not square(b ≠ l) and
𝑏
𝑙
=
𝑏1
𝑙1
=
𝑏2
𝑙2
=
𝑏3
𝑙3
= . . . = n
Thus, q = kH
𝑵𝒇
𝑵𝒅
. n

125
Example-1: For homogeneous earth dam 52 m high and 2 m free board, a flow net was
constructed and following results were obtained:
Number of equipotential drops = 25
Number of flow channels = 4
The dam has a horizontal filter of 40 m length at its downstream end. Calculate the discharge
per metre length of the dam if the coefficient of permeability of the dam material is 3 x
10−3
cm/sec.
𝑺𝒐𝒍𝒏:
The discharge per unit length is given by q = kH
𝑁𝑓
𝑁𝑑
q = 3 x 10−5
* 50 *
4
25
=2.4 x 10−4
𝑚3
/sec/m
Here,
H = 52 -2 = 50 m
𝑁𝑓 = 4
𝑁𝑑 = 25
k = 3 x 10−3cm/sec.
= 3 x 10−5m/sec.

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