Soil Mechanics-I.pptx gggggggghhhhhhhhbbbbbb

By
Dr. Amanullah Marri
Professor
THE MECHANICS OF SOIL
I
NED UNIVERSITY OF ENGINEERING &TECHNOLOGY, KARACHI-75270

SYLLABUS
1. INTRODUCTION TO SOIL MECHANICS
2. ORIGIN AND FORMATION OF SOIL
3. CLAY MINERALOGY
4. COMPOSITION & PHYSICAL PROPERTIES OF
SOIL
5. SOIL CLASSIFICATION
3

SYLLABUS
6. SOIL COMPACTION
7. SOIL HYDRAULICS
8. SEEPAGE THEORY
9. THE CONCEPT OF EFFECTIVE STRESS IN SOIL
10.STRESS DISTRIBUTION IN SOIL FROM
SURFACE LOADS
11.CONSOLIDATION THEORY
12. SHEAR STRENGTH OF SOIL
4

LIST OF BOOKS
1. Soil Mechanics: Basic Concepts and
Engineering Applications by A. Aysen
2. Soil mechanics: concepts and applications by
William Powrie
3. Advanced soil mechanics by Braja M. Das
4. Soil mechanics by T. William Lambe, Robert V.
Whitman
5. Soil mechanics in engineering practice by Karl
Terzaghi
5

LIST OF BOOKS
6. elements of soil mechanics by Ian Smith
7. Soil Mechanics and Foundations by Dr. B.C.
Punmia
8. Soil mechanics by Robert F. Craig
9. Soil Mechanics & Foundation Engineering by
Purushothama Raj
10. Soil Mechanics and Foundations by Muniram
Budhu
6

LIST OF BOOKS
11. Fundamentals of geotechnical engineering by
Braja M. Das
12. Principles of Geotechnical Engineering by
Braja M. Das
13. Geotechnical engineering of dams by Robin
Fell, Patrick MacGregor, David Stapledon
14. Geotechnical engineering: principles and
practices of soil by V. N. S. Murthy
7

PREREQUISITE KNOWLEDGE AND/OR SKILLS
 Students should have a basic understanding
of the following:
 basic tools for stress, strain and strength
analysis
 to design structures, predict failures and
understand the physical properties of
materials.
8

 analyze and design structural members
subjected to tension, compression, torsion
and bending using fundamental concepts of
stress, strain, elastic behavior and inelastic
behavior.
 conduct themselves professionally and with
regard to their responsibilities to society,
especially with respect to designing structures
to prevent failure.
9

 a fundamental knowledge and understanding
of the mechanics of fluid at rest and in motion
by describing and observing fluid phenomena
and by developing and using the principles
and laws for analyzing fluid interactions with
natural and constructed systems.
10

COURSE OBJECTIVES
 Develop technical competence in basic
principles of soil mechanics and fundamentals
of application in engineering practice.
(Outcomes b, e, k)
 Ability to list the salient engineering properties
of soils and their characteristics and describe
the factors which control these properties.
(Outcomes c)
11

COURSE OBJECTIVES
 Ability to apply laboratory methods of
determining the properties of soils. (Outcomes
a, c, d, k)
 Ability to identify common situations when the
soil becomes a factor in an engineering or
environmental problem. (Outcomes a, c, i, k)
 Capable of performing basic analytical
procedures in these situations to obtain the
engineering quantity desired given the formuli,
tables, and the soil properties and understand
their limitations. (Outcomes a, c, d, k) 12

RELATIONSHIP OF COURSE TO UNDERGRADUATE DEGREE
PROGRAM OBJECTIVES AND OUTCOMES
This course serves students in a variety of
engineering majors. The information below
describes how the course contributes to the
college's educational objectives.
13

 Soil Mechanics refers to the application of the
mechanics of materials and fluids to describe
the behavior of soils. The course covers soil
mechanics along with GeoEngineering in a less
rigorous and more qualitative manner. Upon
completion of the course, students should have
an understanding of the properties of soils, the
standard analyses performed by a
GeoEngineer, and the qualitative aspects of soil
behavior which are used in the analysis of
problems in GeoEngineering. 14

 Within the Geological Engineering Program, this course
helps provide key educational outcomes as listed below:
 a. an ability to apply knowledge and principles of
mathematics, science, and engineering to geological
engineering problems. This includes differential
equations, calculus-based physics, chemistry, and
geological science topics that emphasize geologic
processes, the identification of minerals and rocks,
geophysics, and field methods. This also includes
engineering science topics such as statics,
properties/strength of materials, and geomechanics.
15

 b. an ability to design and conduct experiments, as
well as to analyze and interpret data
 c. an ability to design a system, component, or
process to meet desired needs within realistic
constraints such as economic, environmental,
social, political, ethical, health and safety,
constructability, and sustainability. This requires
exposure to topics such as surface and near-surface
natural processes, the impacts of construction
projects, disposal of wastes, and site remediation.
16

 d. an ability to function on multi-disciplinary teams
 e. an ability to identify, formulate, and solve
geological engineering problems in space and
time. This includes the knowledge of the physical
and chemical properties of earth materials,
surface water, ground water and their distribution.
 f. an understanding of professional and ethical
responsibility.
17

 g. an ability to communicate effectively
 h. the broad education necessary to understand the
impact of engineering solutions in a global,
economic, environmental, and societal context.
 i. a recognition of the need for, and an ability to
engage in life-long learning
 j. a knowledge of contemporary issues
 k. an ability to use the techniques, skills, and
modern engineering tools necessary for engineering
practice.
18

ENGINEERING NOMENCLATURE
Science: is the study of the physical and natural
world using theoretical models and data from
experiments and observations. OR Knowledge
gained through experiments and observations.
Technology: technology is the practical
application of science.
Engineering is the application of scientific
methods to solve discrete problems.
19

INTRODUCTION TO SOIL MECHANICS
21

WHAT IS SOIL AND WHAT IS ROCK…?
 To an engineer, Soil is the un-aggregated and
un-cemented deposits of minerals and organic
particles covering the earth crust.
 To an engineer, ROCK is the aggregated and
cemented deposits of minerals and organic
particles covering the earth crust.
22

WHAT IS SOIL AND ROCK
MECHANICS…?
 Soil Mechanics is a branch of engineering
mechanics that describes the properties and
behavior of soils to the force field of their
physical environment.
 Rock Mechanics is a branch of engineering
mechanics that describes the properties and
behavior of Rocks to the force field of their
physical environment.
23

WHAT IS SOIL AND MECHANICS…?
24

WHAT IS FOUNDATION
ENGINEERING…?
Foundation engineering is the application of soil
mechanics and rock mechanics in the design of
foundation elements of structures.
25

GEOMECHANICS
 Geomechanics involves the geologic study of
the behavior of soil and rock. The two main
disciplines of Geomechanics are soil
mechanics and rock mechanics.
26

GEOTECHNICAL ENGINEERING…??
Geotechnical engineering is the branch of civil
engineering concerned with the engineering behavior of
earth materials. Geotechnical engineering uses
principles of soil mechanics and rock mechanics to
investigate subsurface conditions and materials;
determine the relevant physical/mechanical and
chemical properties of these materials; evaluate stability
of natural slopes and man-made soil deposits; assess
risks posed by site conditions; design earthworks and
structure foundations; and monitor site conditions,
earthwork and foundation construction.
27

GEOTECHNICAL ENGINEERING…??
28

ground
SOIL FOUNDATION INTERACTION

TYPICAL GEOTECHNICAL PROJECT
construction site
Geo-Laboratory
for testing
Design Office
for design & analysis
s
o
i
l
s
a
m
p
l
e
s
soil properties
design details

bed rock
firm
ground
SHALLOW FOUNDATIONS
for transferring building loads to underlying ground
mostly for firm soils or light loads

bed rock
weak soil
DEEP FOUNDATIONS
P
I
L
E
for transferring building loads to underlying ground
mostly for weak soils or heavy loads

34
DEEP FOUNDATIONS
Driven timber piles, Pacific Highway

RETAINING WALLS
for retaining soils from spreading laterally
Road
Train
`
retaining
wall

EARTH DAMS
for impounding water
soil
reservoir
clay
core shell

CONCRETE DAMS
reservoir
soil
concrete dam

CONCRETE DAMS
Three Gorges Dam, Hong Kong

EARTHWORKS
Roadwork, Pacific Highway
preparing the ground prior to construction

CONSTRUCTION HAZARD
an unwelcome visitor at an earthwork site
A dead Anaconda python (courtesy: J. Brunskill)
What does it have
to do with Geo?#!

42
GEOFABRICS
used for reinforcement, separation, filtration and
drainage in roads, retaining walls, embankments…
Geofabrics used on Pacific Highway

REINFORCED EARTH WALLS
using geofabrics to strengthen the soil

44
SOIL NAILING
steel rods placed into holes drilled into the walls and
grouted

45
SHEET PILES
Resist lateral earth pressures
Used in excavations, waterfront structures, ..

46
SHEET PILES
used in temporary works

47
SHEET PILES
interlocking sections

COFFERDAM
sheet pile walls enclosing an area, to prevent
water seeping in

SHEET PILES
sheets of interlocking steel or timber driven into the
ground, forming a continuous sheet
ship
warehouse
sheet pile

SHORING
propping and supporting the exposed walls to resist
lateral earth pressures

BLASTING
For ore recovery in mines

GROUND IMPROVEMENT
Impact Roller to Compact the Ground

GROUND IMPROVEMENT
Sheepsfoot Roller to Compact Clay Soils

GROUND IMPROVEMENT
Smooth-wheeled Roller

GROUND IMPROVEMENT
Big weights dropped
from 25 m, compacting
the ground.
Craters formed in compaction

ENVIRONMENTAL GEOMECHANICS
Waste Disposal in Landfills

INSTRUMENTATION
to monitor the performances of earth and earth
supported structures.
to measure loads, pressures, deformations, strains,…

SOIL TESTING
Cone Penetration Test Truck – Lavarach Barracks, Townsville

SOIL TESTING
More Field Tests
Standard Penetration Test
Vane Shear Test

SOIL TESTING
Triaxial Test on Soil Sample in Laboratory

SOIL TESTING
Variety of Field Testing Devices

TYPICAL SAFETY FACTORS
Type of Design
Safety
Factor
Probability of
Failure
Earthworks 1.3-1.5 1/500
Retaining
structures
1.5-2.0 1/1500
Foundations 2.0-3.0 1/5000

SOME UNSUNG HEROES OF CIVIL ENGINEERING…
… buried right under your feet.
foundations soil
exploration
tunneling

GREAT CONTRIBUTORS TO THE DEVELOPMENTS
IN GEOTECHNICAL ENGINEERING
HALL OF FAME

Karl Terzaghi
1883-1963
C.A.Coulomb
1736-1806
WJM Rankine
1820-1872 A.Casagrande
1902-1981
L. Bjerrum
1918-1973
A.W.Skempton
1914-
G.F.Sowers
1921-1996
G.A. Leonards
1921-1997

GEOTECHNICAL ENGINEERING
LANDMARKS
CHALLENGES

LEANING TOWER OF PISA
Our blunders become monuments!

HOOVER DAM, USA
Tallest (221 m) concrete dam

PETRONAS TOWER, MALAYSIA
Tallest building in the world

SOME SUGGESTIONS
Attend the lectures
Develop a good feel for the subject
It takes longer to understand from the
lecture notes
It is practical, interesting and makes lot of
sense.

SOME SUGGESTIONS
Work in groups

SOME SUGGESTIONS
Thou shall not wait till the last
minute

EXAMS
My mama always said, “Exam is like a box of
chocolates; you never know what you are
gonna get”

To an agriculturist, soil is the substance existing
on the earth’s surface, which grows and
develops plants.
To a geologist, soil is the material in a relatively
thin surface zone within which roots occur, and
rest of the crust is termed as rock irrespective of
hardness.
To an engineer, soil is the un-aggregated and un-
cemented deposits of minerals and organic
particles covering the earth crust. 79
ORIGIN AND FORMATION OF SOIL

80
SOIL FORMATION
Parent Rock
Residual soil Transported soil
in situ weathering (by
physical & chemical
agents) of parent rock
weathered and
transported far away
by wind, water and ice.

81
RESIDUAL SOILS
Formed by in situ weathering of parent rock

83
TRANSPORTED SOILS
Transported by: Special name:
 wind “Aeolian”
sea (salt water) “Marine”
lake (fresh water) “Lacustrine”
river “Alluvial”
ice “Glacial”

GEOLOGICAL CYCLE OF SOIL
FORMATION
84
Soil
Formation
Weathering
Upheaval
Transportation
Deposition

PARENT ROCK
Formed by one of these three different processes
Igneous Sedimentary Metamorphic
formed by cooling of
molten magma (lava)
formed by gradual
deposition, and in layers
formed by alteration of
igneous & sedimentary
rocks by
pressure/temperature
e.g., limestone, shale
e.g., marble
e.g., granite

ROCK TYPES
There are three major type of rocks:
90

IGNEOUS ROCKS FORMATION
 Igneous rocks form when magma (molten matter) such
as that produced by exploding volcanoes cools
sufficiently to solidify.
 Igneous rocks can be coarse-grained or fine-grained
depending on weather cooling occurred slowly or rapidly.
 Relatively slow cooling occurs when magma is trapped in
the crust below the earth surface (such as at the core of
mountain range), while more rapid cooling occurs if the
magma reaches the surface while molten lava flow.
92

 In coarse-grained igneous rocks the most common is
granite, a hard rock rich in quartz, widely used as a
construction material and for monuments.
 Most common in fine-grained igneous rocks is basalt,
a hard dark coloured rock rich in ferromagnesian
minerals and often used in road construction.
 Being generally hard, dense, and durable, igneous
rocks often make good construction materials.
 Also, they typically have high bearing capacities and
therefore make good foundation material.
93

There are over 700 types of igneous rocks
identified and most of them are formed below the
surface of the earth. Based on their mode of
occurrence, igneous rocks are either intrusive or
plutonic rocks and extrusive or volcanic rocks.
97

SEDIMENTARY ROCKS FORMATION
 Sedimentary rocks are formed when mineral particles,
fragmented rock particles and remains of certain
organisms are transported by wind, water, and ice and
deposited, typically in layers, to forms sediments.
Over a period of time as layers accumulate at a site;
pressure on lower layers resulting from the weight of
overlying strata hardens the deposits, forming
sedimentary rocks.
 Sedimentary rocks comprise the great majority of
rocks found on the earth’s surface.
98

 Additionally deposits may be solidified and cemented
by certain minerals (e.g., silica, iron oxides, calcium
carbonate).
 Sedimentary rocks can be identified easily when their
layered appearance is observable. The most common
sedimentary rocks are shale, sandstone, limestone,
and dolomite.
 The strength and hardness of sedimentary rocks are
variable, and engineering usage of such rocks varies
accordingly.
99

 Relatively hard shale makes a good foundation
martial. Sandstones are generally good construction
materials. Limestone and dolomite, if strong, can be
both good foundation and construction materials.
Clastic sedimentary rocks
 The fragments of pre-existing rocks or minerals that
make up a sedimentary rock are called clasts.
Sedimentary rocks made up of clasts are called clastic
( clastic indicates that particles have been broken and
transported). Clastic sedimentary rocks are primarily
classified on the size of their clasts.
100

101

102

103

METAMORPHIC ROCKS FORMATION
When sedimentary or igneous rocks literally
change their texture and structure as well as
mineral and chemical composition, as a result of
heat, pressure, and shear; metamorphic rocks
are formed.
 Metamorphic rocks are much less common at
the earth’s surface than sedimentary rocks
 Metamorphic rocks can be hard and strong if
un-weathered.
104

METAMORPHIC ROCKS
 They can be a good construction material like
marble is often used for buildings and
monuments
 but foliated metamorphic rocks often contain
planes of weakness that diminish strength.
 Metamorphic rocks sometimes contain weak
layers between very hard ones.
 The most common metamorphic rocks are
gneiss, schist, slat, quartzite and marble.
105

IDENTIFICATION OF ROCK TYPES
 Igneous: A tough, frozen melt with little texture or
layering; mostly black, white and/or gray
minerals; may look like granite or like lava.
 Sedimentary: Hardened sediment with layers
(strata) of sandy or clayey stone; mostly brown to
gray; may have fossils and water or wind marks.
 Metamorphic: Tough rock with layers (foliation) of
light and dark minerals, often curved; various
colours; often glittery from mica.
109

110
1. Examine the specimen closely. Begin by looking for
differences in texture like solid, porous, crystallized or grainy
elements.
2. Pour vinegar over the porous or smooth rock. If it sizzles, you
have a common sedimentary rock known as limestone or
possibly a metamorphic rock known as marble. If not, you
will need to further examine the rock for answers.
3. Note the arrangement of crystals, the colour of its holes or the
size of any particles. Small particles are usually sedimentary
rock like sandstone, while large particles are another
sedimentary rock called conglomerate. and schist.

111
4. Pay special attention to rocks with holes, as they can
fall into either of two categories. Light-coloured holes
represent a sedimentary rock called tufa, while dark
holes are surprisingly found to be igneous rock such as
pumice.
5. Check the surface of the rock. Is it glassy, shiny or
dull?
6. Observe the layers within a layered rock. Although
sedimentary rock is known for its layered
characteristics, several types of metamorphic rock also
reveal layers such as slate and schist.

IGNEOUS ROCKS
112
Andesite Basalt Diorite
Gabbro
Granite

SEDIMENTARY ROCKS
113
Seventy percent of all the rocks on earth are sedimentary rocks
Limestone
Peat
Gypsum
Rock salt Sandstone
Shale

SEDIMENTARY ROCKS
114
Siltstone
Coal
Conglomerate
Sandstone
Conglomerate Limestone

METAMORPHIC ROCKS
115
Gneiss Marble Quartzite
Schist Slate Phyllite

126
ELEMENTS OF EARTH
12500 km dia
8-35 km crust % by weight in crust
O = 49.2
Si = 25.7
Al = 7.5
Fe = 4.7
Ca = 3.4
Na = 2.6
K = 2.4
Mg = 1.9
other = 2.6
82.4%

ELEMENTS OF EARTH
Geotechnical engineers are interested mainly in
the top 100 metres of the earth crust. As you can
see from the table, 82% of the elements are
oxygen, silicon and aluminium.
127

CLAY MINERALOGY
Clay minerals are very tinny crystalline
substances evolved primarily from chemical
weathering of certain rock forming minerals. All
clay minerals are very small, colloidal sized
crystals and they can only be seen with an
electron microscope. Most of the clay minerals
have sheet or layered structures. A few have
elongated tubular or fiber structures.
128

CLAY MINERALOGY
Soil masses generally contain mixture of several
clay minerals named for the predominating clay
mineral with varying amounts of non clay
minerals. Clay minerals are complex aluminium
silicates composed of two basic units:
 Silica tetrahedron, and
 Aluminium octahedron
129

130
BASIC STRUCTURAL UNITS
0.26 nm
oxygen
silicon
0.29 nm
aluminium or
magnesium
hydroxyl or
oxygen
Clay minerals are made of two distinct structural
units.
Silicon tetrahedron Aluminium Octahedron

CLAY MINERALOGY
The particular way, in which these are stacked,
together with different bonding and different
metallic ions in the crystal lattice, constitute the
different clay minerals.
131

SILICA TETRAHEDRON
Each tetrahedron unit consists of four oxygen
atoms surrounding a silica atom. Together oxygen
and silicon account for 80 % of the earth crust.
This must mean that crustal minerals are mainly
combinations of these two elements.
132

1.Silicate Clays
SILICA TETRAHEDRON

OCTAHEDRON
The octahedral unit consists of six hydroxyl units
surrounding an aluminium (or Magnesium) atom.
The octahedral layer is composed of cations in
octahedral coordination with oxygen. The
combination of the aluminium octahedral units
forms a Gibbsite. If the main metallic atoms in
the octahedral units are Magnesium these
sheets are referred to as Bruce sheets.
139

140
B or G
OCTAHEDRON
http://www.luc.edu/faculty/spavko1/minerals/

GIBBSITE
The octahedral unit is called gibbsite if the
metallic atom is mainly aluminium. Chemical
Formula for gibbsite is Al (OH)3, Aluminium
Hydroxide. Gibbsite is an important ore of
aluminium and is one of three minerals that make
up the rock Bauxite. Bauxite is often thought of as
a mineral but is really a rock composed of
aluminium oxide and hydroxide minerals such as
gibbsite, boehmite, AlO (OH) and diaspore, HAlO2,
as well as clays, silt and iron oxides and
hydroxides. 141

BRUCITE
The octahedral unit is called brucite if the
metallic atom is mainly magnesium. Chemical
Formula of brucite is Mg (OH) 2, Magnesium
Hydroxide. Brucite is a mineral that is not often
used as a mineral specimen but does have some
important industrial uses. It is a minor ore of
magnesium metal and a source of magnesia. It is
also used as an additive in certain refractory.
143

TETRAHEDRAL SHEET
The tetrahedral sheet is composed of tetrahedron
units. Each unit consists of a central four-
coordinated atom (e.g. Si, Al, or Fe) surrounded
by four oxygen atoms that, in turn, are linked with
other nearby atoms (e.g. Si, Al, or Fe), thereby
serving as inter-unit linkages to hold the sheet
together. Sheet of horizontally linked, tetrahedral-
shaped units; serve as one of the basic structural
components of silicate clay minerals.
144

TETRAHEDRAL SHEET
148
Several tetrahedrons joined together form a tetrahedral sheet.

CLAY MINERALS
 Kaolinite
 Illite
 Montmorillonite
 Vermiculite
 Smectite
152

KAOLINITE
The chemistry of Kaolinite is Al2Si2O5 (OH) 4,
Aluminium Silicate Hydroxide and the skeletal is
shown in the figure. Kaolinite's structure is
composed of silicate sheets (Si2O5) bonded to
aluminium oxide/hydroxide layers (Al2 (OH) 4)
called gibbsite layers. The silicate and gibbsite
layers are tightly bonded together with only weak
bonding existing between these silicate/gibbsite-
paired layers (called s-g layers).
153

KAOLINITE
The weak bonds between these s-g layers cause
the cleavage and softness of this mineral. The
Kaolinite structural unit consists of alternating
layers of silica tetrahedra with tips embedded in
an aluminum (gibbsite) octahedral unit. This
alternating of silica and gibbsite layers produces
what is some times called a 1:1 basic unit. The
resulting flat sheet unit is about 7 Å thick and
extends infinitely in other two dimensions.
154

KAOLINITE
155
7 Å
(a) Schematic of basic unit
(b) Kaolinite cluster
H bond
H bond
H bond
Kaolinite is used in the production of ceramics, as
filler for paint, rubber and plastics and the largest
use is in the paper industry to produce a glossy
paper such as is used in most magazines.

156
KAOLINITE
Different combinations of tetrahedral and
octahedral sheets form different clay minerals:
1:1 Clay Mineral (e.g., kaolinite, halloysite):

KAOLINITE
Si
Al
Si
Al
Si
Al
Si
Al
joined by strong H-bond
no easy separation
0.72 nm
Typically
70-100
layers
joined by oxygen
sharing
Kaolinite is used for making paper, paint and in pharmaceutical
industry.

Click here, for 3D view of Kaolinite
KAOLINITE

ILLITE
The chemistry of Illite is KAl2 [(OH) 2|AlSi3O10],
Potassium aluminium silicate hydroxide hydrate
mineral of the clay group. The Illite clay mineral
consists of an octahedral layer of gibbsite
sandwiched between two layers of silica
tetrahedral. This produces a 1:2 mineral with the
additional difference that some of the silica
positions are filled with aluminium atoms and
potassium ions are attached between layers to
make up the charge deficiency.
160

ILLITE
This bonding results in a less stable condition
than for Kaolinite, and thus the activity of Illite is
greater. Illite is found in a wide variety of
environments, the materials of this series form by
either weathering or hydrothermal alteration of
muscovite. Illite can be formed from kaolin, if
potassium is added.
161

ILLITE
162
k k k
k k k
G
G
G
G
Si
Si
(a) Schematic of basic unit (b) Illite cluster

163
ILLITE
Si
Al
Si
Si
Al
Si
Si
Al
Si
0.96 nm
joined by K+
ions
fit into the hexagonal
holes in Si-sheet

MONTMORILLONITE
The chemistry of Montmorillonite is (Na, Ca) (Al,
Mg)6 (Si4O10)3 (OH)6 - nH2O, Hydrated Sodium
Calcium Aluminium Magnesium Silicate
Hydroxide. Montmorillonite is a member of the
general mineral groups the clays. It swells in
water and posses high cation-exchange
capacities. It typically forms microscopic or at
least very small platy micaceous crystals. The
water content is variable, and in fact when the
crystals absorb water they tend to swell to several
times their original volume. 165

MONTMORILLONITE
This makes montmorillonite a useful mineral for
several purposes. It is the main constituent in a
volcanic ash called bentonite, which is used in
drilling muds. The bentonite gives the water
greater viscosity ("thickness" of flow), which is
very important in keeping a drill head cool during
drilling and facilitating removal of rock and dirt
from within a drill hole.
166

MONTMORILLONITE
Another important use of Montmorillonite is as
an additive to soils and rocks. The effect of the
Montmorillonite is to slow the progress of water
through the soil or rocks. This is important to
farmers with extended dry periods, engineers of
earthen dams or levees or perhaps to plug up old
drill holes to prevent leakage of toxic fluids from
bottom levels to higher aquifers used for drinking
water.
167

MONTMORILLONITE
As a mineral specimen, Montmorillonite does not
get much consideration. Usually, pure samples of
Montmorillonite are massive, dull and not very
attractive. However, as with all minerals, there
are those exceptional specimens that defy the
norm. Montmorillonite has been found as
attractive pink inclusions in quartz crystals, and
these make for interesting specimens.
168

MONTMORILLONITE
169
G or B
Si
Si
(a) Schematic of basic unit (b) Montmorillonite cluster
G
G
G
Na, Ca, -nH2O
bond
Na, Ca, -nH2O
bond

170
MONTMORILLONITE
Different combinations of tetrahedral and
octahedral sheets form different clay minerals:
2:1 Clay Mineral (e.g., Montmorillonite, Illite)

171
MONTMORILLONITE
Si
Al
Si
Si
Al
Si
Si
Al
Si
0.96 nm
joined by weak
van der Waal’s bond
easily separated
by water
 also called smectite; expands on contact with water

173
SUMMARY - MONTMORILLONITE
 Montmorillonite have very high specific surface,
cation exchange capacity, and affinity to water. They
form reactive clays.
 Bentonite (a form of Montmorillonite) is frequently
used as drilling mud.
 Montmorillonite have very high liquid limit (100+),
plasticity index and activity (1-7).

VERMICULITE
Vermiculite is a 2:1 clay, meaning it has 2
tetrahedral sheets for every one octahedral
sheet. It is a limited expansion clay with a
medium shrink-swell capacity. Vermiculite has a
high cation exchange capacity at 100-150 meq
/100 g. Vermiculite clays are weathered micas in
which the potassium ions between the molecular
sheets are replaced by magnesium and iron ions.
174

SOIL FABRIC
The arrangement and organisation of particles
and other features within a soil mass is termed
its structure or fabric. This includes bedding
orientation, stratification, layer thickness, the
occurrence of joints and fissures, the occurrence
of voids, artefacts, tree roots and nodules, the
presence of cementing or bonding agents
between grains.
176

SOIL FABRIC
The soil fabric is the brain it retains the memory
of the birth of the soil and subsequent changes
that occur. Two common types of soil fabric
flocculated and dispersed are formed.
flocculated structure: If a flocculated structure,
formed under a salt environment, many particles
tend to orient parallel to each other. When a
flocculated structure formed under fresh water,
many particles tend to orient perpendicular to
each other. 177

SOIL FABRIC
Dispersed structure: A dispersed structure is the
result when a majority of the particles orient
parallel to each other.
178

179
CLAY FABRIC
Flocculated Dispersed
edge-to-face contact
face-to-face contact

180
CLAY FABRIC
Electrochemical environment (i.e., pH, acidity,
temperature, cations present in the water) during
the time of sedimentation influence clay fabric
significantly.
Clay particles tend to align perpendicular to the
load applied on them.

DEFUSE DOUBLE LAYER
Diffuse double layer (DDL) is an ionic structure
that describes the variation of electric potential
near a charged surface, such as clay, and
behaves as a capacitor. The surface charges on
fine-grained soils are negative. These negative
surface charges attract cations and the positively
charged side of water molecules from
surrounding water.
182

DEFUSE DOUBLE LAYER
Consequently, a thin film or layer of water, called
adsorbed water, is bonded to the mineral
surfaces. The thin film or layer of water is known
as the diffuse double layer. The largest
concentration of cations occurs at the mineral
surface and decreases exponentially with
distance away from the surface.
183

186
CATION CONCENTRATION IN WATER
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ +
+
+
+ +
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ + +
+
+ +
+
+
+
+
+
+
+
+ +
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ +
+
cations
 cation concentration drops with distance from clay particle
- -
- -
- -
- -
- -
- -
- -
clay particle
double layer free water

187
ADSORBED WATER
- -
- -
- -
- -
- -
- -
- -
 A thin layer of water tightly held to particle; like a skin
 1-4 molecules of water (1 nm) thick
 more viscous than free water
adsorbed water

188
CLAY PARTICLE IN WATER
- -
- -
- -
- -
- -
- -
- -
free water
double layer
water
adsorbed water
50 nm
1nm

189
A CLAY PARTICLE
Plate-like or Flaky Shape

190
ISOMORPHOUS SUBSTITUTION
 substitution of Si4+
and Al3+
by other lower valence
(e.g., Mg2+
) cations
 results in charge imbalance (net negative)
+
+
+ + +
+
+
_
_ _
_ _
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
positively charged edges
negatively charged faces
Clay Particle with Net negative Charge

CATION EXCHANGE CAPACITY
191
Cation-exchange capacity is defined as the
degree to which a soil can adsorb and
exchange cations.
Cation-a positively charged ion (NH4
+
, K+
,
Ca2+
, Fe2+
, etc...)
Anion-a negatively charged ion (NO3
-
, PO4
2-
,
SO4
2-
, etc...)

192
CATION EXCHANGE CAPACITY (C.E.C)
 capacity to attract cations from the water (i.e., measure of
the net negative charge of the clay particle)
 measured in meq/100g (net negative charge per 100 g of clay)
milliequivalents
known as exchangeable cations
 The replacement power is greater for higher valence and
larger cations.
Al3+
> Ca2+
> Mg2+
>> NH4
+
> K+
> H+
> Na+
> Li+

SPECIFIC SURFACE
193
Specific surface area "SSA" is a property of solids
which is the total surface area of a material per
unit of mass, solid or bulk volume, or cross-
sectional area.
The specific surface area of natural particles is
an important parameter to quantify processes
such as mineral dissolution and sorptive
interactions in soils and sediments.

194
SPECIFIC SURFACE
 surface area per unit mass (m2
/g)
 smaller the grain, higher the specific surface
e.g., soil grain with specific gravity of 2.7
10 mm cube
1 mm cube
spec. surface = 222.2 mm2
/g spec. surface = 2222.2 mm2
/g

195
A COMPARISON
Mineral Specific surface
(m2
/g)
C.E.C
(meq/100g)
Kaolinite 10-20 3-10
Illite 80-100 20-30
Montmorillonite 800 80-120
Chlorite 80 20-30

197
clay mineral indentification.pdf

198
OTHERS…
X-Ray Diffraction (XRD)
Differential Thermal Analysis (DTA)
 to identify the molecular structure and minerals present
 to identify the minerals present

199
SCANNING ELECTRON MICROSCOPE
 common technique to see
clay particles
 qualitative
plate-like
structure

200
CASAGRANDE’S PI-LL CHART
0 10 20 30 40 50 60 70 80 90 100
0
10
20
30
40
50
60
Liquid Limit
Plasticity
Index
A-line
U-line
montmorillonite illite
kaolinite
chlorite
halloysite

SOIL TYPES
Bentonite is a clay formed by the decomposition
of volcanic ash with a high content of
montmorillonite. It exhibits the properties of clay
to an extreme degree.
Varved Clays consist of thin alternating layers of
silt and fat clays of glacial origin. They possess
the undesirable properties of both silt and clay.
The constituents of varved clays were
transported into fresh water lakes by the melted
ice at the close of the ice age. 201

SOIL TYPES
Kaolin, China Clay are very pure forms of white
clay used in the ceramic industry.
Boulder Clay is a mixture of an unstratified
sedimented deposit of glacial clay, containing
unsorted rock fragments of all sizes ranging from
boulders, cobbles, and gravel to finely pulverized
clay material.
202

SOIL TYPES
Calcareous Soil is a soil containing calcium
carbonate. Such soil effervesces when tested
with weak hydrochloric acid.
Marl consists of a mixture of calcareous sands,
clays, or loam.
Hardpan is a relatively hard, densely cemented
soil layer, like rock which does not soften when
wet. Boulder clays or glacial till is also sometimes
named as hardpan..
203

SOIL TYPES
Caliche is an admixture of clay, sand, and gravel
cemented by calcium carbonate deposited from
ground water.
Peat is a fibrous aggregate of finer fragments of
decayed vegetable matter. Peat is very
compressible and one should be cautious when
using it for supporting foundations of structures.
Loam is a mixture of sand, silt and clay.
204

SOIL TYPES
Loess is a fine-grained, air-borne deposit
characterized by a very uniform grain size, and
high void ratio. The size of particles ranges
between about 0.01 to 0.05 mm. The soil can
stand deep vertical cuts because of slight
cementation between particles. It is formed in dry
continental regions and its color is yellowish light
brown.
205

SOIL TYPES
Shale is a material in the state of transition from
clay to slate. Shale itself is sometimes
considered a rock but, when it is exposed to the
air or has a chance to take in water it may rapidly
decompose.
206

207
SUMMARY - CLAYS
 Clay particles are like plates or needles. They are
negatively charged.
 Clays are plastic; Silts, sands and gravels are non-plastic.
 Clays exhibit high dry strength and slow dilatancy.

FUTURE TASKS
Organic clay
Black cotton soil
humic acid
208

ORGANIC CLAY
ORGANIC clay is defined as a clay with sufficient
organic content to influence the soil properties
[ASTM, D2487]. Organic clays are normally
characterized by high plasticity, high water
content, high compressibility, low permeability
and low shear strength. These characteristics
make organic clays unsuitable for engineering
construction unless a suitable improvement
method is applied.
209

HUMIC ACID
Humic acid is a principal component of humic
substances, which are the major organic
constituents of soil. Humic acid has the average
chemical formula C187H186O89N9S1 and is insoluble
in strong acid (pH = 1).
210

LIME TREATED CLAY
Lime can be used to treat soils in order to
improve their workability and load-bearing
characteristics in a number of situations.
215

COMPOSITION & PHYSICAL PROPERTIES OF
SOIL

SOIL COMPOSITION
While a nearly infinite variety of substances may
be found in soils, they are categorized into four
basic components: minerals, organic matter, air
and water.
a) Mineral matter obtained by the disintegration
and decomposition of rocks;
b) Organic matter, obtained by the decay of plant
residues, animal remains and microbial
tissues;
217

SOIL COMPOSITION
c) Water, obtained from the atmosphere and the
reactions in soil (chemical, physical and
microbial);
d) Air or gases, from atmosphere, reactions of
roots, microbes and chemicals in the soil
218

SOIL PHASE RELATIONSHIPS
Soil mass is generally a three phase system. It
consists of solid particles (i.e., minerals and
organic matters), liquid and gas. The phase
system may be expressed in terms of mass-
volume or weight-volume relationships.
219

VOLUMETRIC RATIOS
There are three volumetric ratios that are very
useful in geotechnical engineering and these can
be determined directly from the phase diagram:
222

VOLUMETRIC RATIOS
Voids ratio
Degree of saturation
223
s
v
V
V
e 
100


v
w
V
V
Sr

VOLUMETRIC RATIOS
Porosity
224
V
Vv



VOLUMETRIC RATIOS
Percentage air voids:
225

VOLUMETRIC RATIOS
Air content:
Air content, ac is the ratio of the volume of air
voids to the volume of voids.
226

MASS-VOLUME RELATIONSHIP
Unit weight
Unit weight of solid
227
V
W


s
s
s
V
W



SOIL COMPOSITION
Saturated unit weight
Submerged unit weight
228
V
Wsat
sat 

w
sat
sub 

 


INDEX PROPERTIES OF SOIL
Index properties are the properties of soil that
help in identification and classification of soil.
Water content, Specific gravity, Particle size
distribution, In situ density (Bulk Unit weight of
soil), Consistency Limits and relative density are
the index properties of soil. These properties are
generally determined in the laboratory. In situ
density and relative density require undisturbed
sample extraction while other quantities can be
determined from disturbed soil sampling.
229

The following are the Index Properties of soil.
1. Water content
2. Specific Gravity
3. In-situ density
4. Particle size distribution
5. Consistency limits
6. Relative Density
230

Water content
231
100


s
w
W
W


Specific Gravity: The ratio of the specific weight of
a substance to the specific weight of water is
called specific gravity.
 This number indicates how much heavier or
lighter a material is than water.
 In soils, SG refers to the mass of solid matter of
a given soil sample as compared to an equal
volume of water.
232
w
s
s
G




RELATIVE DENSITY BOTTLE (RD-
BOTTLE)
233

SPECIFIC GRAVITY
1 ft3
Soil Solids Water
W = 187.2 lbs W= 62.4 lbs
SG= 187.2/62.4 = 3.0
1 ft3

GENERAL RANGES OF SG FOR SOILS
 Sand 2.63 – 2.67
 Silt 2.65-2.7
 Clay & Silty Clay2.67-2.9
 Organic Soils <2.0

SPECIFIC GRAVITY
 For example: A material with SG of 2 is twice as
heavy as water (2x 62.4 lbs/ft3
) = 124.8 lbs/ft3

IN-SITU DENSITY
𝛾=
𝑊
𝑉
237

DENSITY OF WATER
238
-30 -20 -10 0 10 20 30 40 50 60 70 80 90 100
950
960
970
980
990
1000
1010
Temperature ( C)

Density
of
water
(kg/m3)

IN-SITU DENSITY TEST METHODS
 Sand Cone Method
 Rubber Balloon Method
 Nuclear Density Method
239

IN-SITU DENSITY BY SAND CONE APPARATUS
241

242

243

NUCLEAR DENSITY METHOD
A nuclear density gauge is a tool used in civil
construction and the petroleum industry, as well
as for mining and archaeology purposes. It
consists of a radiation source that emits a
directed beam of particles and a sensor that
counts the received particles that are either
reflected by the test material or pass through it.
By calculating the percentage of particles that
return to the sensor, the gauge can be calibrated
to measure the density and inner structure of the
test material. 245

CONSISTENCY LIMITS
Liquid limit
Plastic limit
Shrinkage limit
248

EXAMPLE
No, of
blows
Can no.
mass of
can, M1
(g)
Mass of
the
can+wet
soil, M2
(g)
mass of
the can
+dry soil,
M3 (g)
moisture
content,
w(%)
30 25 15.4 32.36 30.35 13.44
16 20 15.6 40.2 36.96 15.17
9 10 15.39 42.65 38.75 16.70
5 33 15.85 52.9 47.25 17.99
251

SOLUTION
252
1 10 100
0
10
20
30
No. of blows
Moisture
Content
(%)

FREE SWELL INDEX OF SOIL
Free Swell
Index
Degree of
expansiveness
LL PL SL
<20 Low 0.50 0-35% >17%
20-35 Moderate 40-60% 25-50% 8-18%
35-50 High 50-75% 35-65% 6-12%
>50 Very high >60% >45% <10%
260

RELATIVE DENSITY
in which
emax= void ratio of sand in its loosest state having
a dry density of pdmin
emin void ratio in its densest state having a dry
density of pdMax
e = void ratio under in-situ condition having a dry
density of pd
261

MAXIMUM VOID RATIO
262
e
1
.
G
γ w
s
d




MINIMUM VOID RATIO
263
 
e
1
e
G
γsat


 w


INTERRELATIONSHIPS
Void ratio (e) in terms of porosity () i.e. e = f
()
264

INTERRELATIONSHIPS
Void ratio (e) in terms of porosity () i.e. e = f ()
265

INTERRELATIONSHIPS
Porosity in terms of void ratio
266
1
1
v
v
v s
v s
v
v s
v v
V
V
V
V V
V V
V
V V
V V









 

INTERRELATIONSHIPS
267
1
1
1
1
1
1






e
e
e
e
e




EXERCISE
Determine the void ratio of a soil whose porosity
is 32%. A laboratory test reveals that the void
ratio of a certain soil is 1.234. What is the
porosity of this soil?
[Ans: e = 0.471, =55.3%]
268

INTERRELATIONSHIPS
Prove that
269

INTERRELATIONSHIPS
Dry unit weight in terms of specific gravity, void ratio and
unit weight of water. i.e,
271
e
1
.
G
γ w
s
d




INTERRELATIONSHIPS
272
e
1
.
G
γ
e
1
γ
γ
e
1
γ
Vv
Vs
Vs
γ
γ
V
W
γ
w
s
d
s
d
s
s
d
s
d











INTERRELATIONSHIPS
Dry unit weight as a function of in-situ unit weight and water
content :
273
1
d






INTERRELATIONSHIPS
274















1
γ
γ
)
(1
γ
γ
γ
ω.γ
γ
V
W
V
ω.W
γ
V
W
V
W
γ
V
W
W
V
W
γ
d
d
d
d
s
s
s
w
s
w

INTERRELATIONSHIPS
Dry unit weight in terms of specific gravity, water
content, degree of saturation and specific weight
of water i.e.,
275
)
γ
ω,
,
G
,
(S
γ w
s
r
d f


INTERRELATIONSHIPS
276
e
1
.
G
γ w
s
d



r
s
S
ωG

e
r
s
w
s
d
S
ωG
1
.γ
G
γ



INTERRELATIONSHIPS
Specific weight in terms of specific gravity, void
ratio, and degree of saturation
277
)
S
e,
,
(G r
s
f



INTERRELATIONSHIPS
278
e
S
e
G
e
e
S
G
V
V
V
S
V
G
V
V
V
V
W
W
V
W
r
w
r
w
w
v
s
v
r
w
s
w
w
w
s
s
w
s














1
)
.
(
1
.
.
1
.
.
.
.
.
.













INTERRELATIONSHIPS
Saturated unit weight in terms of specific gravity,
voids ratio and specific weight of water i.e.,
279
)
γ
e,
(G,
γ w
sat
f


INTERRELATIONSHIPS
280
v
s
v
r
w
s
s
sat
w
w
s
s
sat
w
s
sat
sat
V
V
.V
.S
γ
.V
γ
γ
V
.V
γ
.V
γ
γ
V
W
W
V
W
γ








 
e
1
e
G
γ
e
1
e
γ
G.γ
γ
e
1
.1.e
γ
.1
G.γ
γ
sat
w
w
sat
w
w
sat









w


INTERRELATIONSHIPS
Submerged unit weight in terms of specific
gravity, voids ratio and unit weight of water i.e.
281
)
γ
e,
,
(G
γ` w
s
f


INTERRELATIONSHIPS
282
w
sat 

 

`  
e
1
e
G
γsat


 w

 
 
e
1
1)
(G
`
e
1
e)
1
e
(G
`
e
1
)
e
1
(
e
G
`
e
1
e
G
`

















w
w
w
w
w
w











INTERRELATIONSHIPS
Prove that
283
(1 )
1
s w
G w
e






INTERRELATIONSHIPS
284
v
s
s
s
v
s
s
v
s
s
s
v
s
w
s
V
V
w
V
V
V
w
W
V
V
wW
W
V
V
W
W
V
W













)
1
(
.
)
1
(







INTERRELATIONSHIPS
285
e
w
G
e
w
w
s
s






1
)
1
(
1
)
1
(





EXAMPLE
The weight of a chunk of moist soil sample is
45.6 lb. the volume of the soil chunk measured
before drying is 0.40 ft3
. After being dried out in
an oven, the weight of dry soil is 37.8 lb. the
specific gravity of soil is 2.65. Determine water
content, unit weight of moist soil void ratio,
porosity, and degree of saturation.
286

EXAMPLE
The dry density of a soil is 1.78 g/cm3
. It is
determined that the void ratio is 0.55. What is
the moist unit weight if S = 50 %? If S = 100 %?
Assume G = 2.76
287

EXAMPLE
For a soil in natural state, given e = 0.8, w = 24%,
and Gs = 2.68, Determine the moist unit weight,
dry unit weight, and degree of saturation.
288

EXERCISE -I
An oven tin containing a sample of moist soil was
weighed and had a mass of 37.82 g; the empty
tin had a mass of 16.15 g. After drying, the tin
and soil were weighed again and had a mass of
34.68 g. Determine the void ratio of the soil if the
air void content is (a) zero (b) 5 % take Gs = 2.70.
[Ans: (a) e = 0.456, (b) e = 0.533]
289

EXERCISE-II
The moist mass of a soil specimen is 20.7 kg.
The specimen’s volume measured before drying
is 0.011 m3
. The specimens dried mass is 16.3
kg. The specific gravity of solids is 2.68.
Determine void ratio, degree of saturation, wet
unit mass, dry unit mass, wet unit weight, and dry
unit weight.
290

EXERCISE-III
An undisturbed soil sample has e = 0.78, w =
12%, and Gs = 2.68. Determine wet unit weight,
dry unit weight, degree of saturation, and
porosity.
Ans: =16.49 kN/m3
, =14.72 kN/m3
, Sr = 40.9 %,
= 44.0 %]
291

EXERCISE-IV
A sample of dry sand having a unit weight of 16.5
kN/m3
and specific gravity of 2.70 is placed in
the rain. During the rain the volume of the
sample remains constant but the degree of
saturation increases to 40 %. Determine the unit
weight and water content of the soil after being in
the rain.
292

EXERCISE-V
For a given soil, the in situ void ratio is 0.72 and
Gs = 2.61. Calculate the porosity, dry unit weight
(lb/ft3
, kN/m3
), and the saturated unit weight.
What would the moist unit weight be when the
soil is 60% saturated?
293

EXERCISE-VI
A 100 % saturated soil has a wet unit weight of
120 lb/ft3
. The water content of this saturated
soil was determined to be 36 %. Determine void
ratio, and specific gravity of solids.
[Ans: e =1.04, Gs =2.88]
294

EXERCISE-VII
A proposed earth dam will contain 50,000,000
m3
of earth. Soil to be taken from a borrow pit will
be compacted to a void ratio of 0.78. The void
ratio pf soil in the borrow pit is 1.12. Estimate the
volume of soil that must be excavated from the
borrow pit.
295

EXERCISE-VIII
A fibre reinforced cemented sandy soil sample of
50 mm diameter and 100 mm height is to be
prepared for the subsequent testing of its
engineering properties. The sample is to be
compacted to a targeted dry density of (15
+RN/10) kN/m3
at an optimum moisture content
of 10%. If the soil is added with 5% cement and
0.25% fibre.
296

EXERCISE-VIII
 Total volume, Volume of solid, Volume of voids,
Volume of water, Volume of air voids, Void ratio,
Porosity.
 Determine the amount of sand, cement, fibre
and water to be mixed for sample preparation.
297

SOIL GRADATION
Soil gradation is very important to geotechnical
engineering. It is an indicator of other
engineering properties such as compressibility,
shear strength, and hydraulic conductivity.
In a design, the gradation of the in situ or on site
soil often controls the design and ground water
drainage of the site. A poorly graded soil will have
better drainage than a well graded soil because
there are more void spaces in a poorly graded
soil.
300

SOIL GRADATION
When a fill material is being selected for a project
such as a highway embankment or earthen dam,
the soil gradation is considered. A well graded
soil is able to be compacted more than a poorly
graded soil. These types of projects may also
have gradation requirements that must be met
before the soil to be used is accepted.
301

SOIL GRADATION
 Grain size distribution
 Hydrometer Analysis
302

303
GRAIN SIZE DISTRIBUTION
In coarse grain soils …... By sieve analysis
Determination of GSD:
In fine grain soils …... By hydrometer analysis
Sieve Analysis Hydrometer Analysis
hydrometer
stack of sieves
sieve shaker

SIEVE ANALYSIS
304
Sieve Stack

SIEVE ANALYSIS
305
Sieve Shaker

306
Soil Dispersion Mixer
HYDROMETER ANALYSIS

WHAT IS A HYDROMETER?
 Device used to determine directly the
specific gravity of a liquid
 Consists of a thin glass tube closed at both
ends
 Large bulb contains lead shot to cause the
instrument to float upright in liquid.
 Scale is calibrated to indicate the specific
gravity of the liquid.

MEASURE SAMPLE
Collect 50 g of fine from mechanical sieving procedure
1 ft3

REQUIRED TEST EQUIPMENT
Mixer Scale Hydrometer Jar Scale Deflocculating
Solution

DEFLOCCULATING STAGE
 Add sample to 125 ml of 40g/L sodium
hexometaphosphate (deflocculating solution)
 Allow to soak for at least 12 hours

Mix sample with spatula to dislodge settled particles

SAMPLE PREPARATION
 Pour sample into mixing cup
 Use distilled water to rinse beaker

 Add distilled water to the soil in mixer cup to
make it about two-thirds full.
 Mix using mixer for two minutes
SAMPLE PREPARATION

 Seal the jar or simply place hand over end
and rotate about 60 times.
 Set the jar on the bench and record the time.
This is time (t=0) on your data sheet.
 Take a hydrometer reading and temperature
reading at prescribed intervals.
 After each reading the hydrometer is put into
the transparent cylinder
STARTING THE TEST

316
DEFINITION OF GRAIN SIZE
Boulders Cobbles
Gravel Sand Silt and
Clay
Coarse Fine Coarse Fine
Medium
300 mm 75 mm
19 mm
No.4
4.75 mm
No.10
2.0 mm
No.40
0.425 mm
No.200
0.075
mm
No specific
grain size-use
Atterberg limits

317
GENERAL GUIDANCE
Coarse-grained soils:
Gravel Sand
Fine-grained soils:
Silt Clay
NO.200
0.075 mm
• Grain size
distribution
• Cu
• Cc
• PL, LL
• Plasticity chart
50 %
NO. 4
4.75 mm
Required tests: Sieve analysis
Atterberg limit
LL>50
LL <50
50%

EXAMPLE
Plot the gradation curve and determine the PSD
parameters.
318
Sieve No. 4 10 20 50 80 100 120 200 Pan
Sieve opening (mm) 4.3 2.0 0.9 0.3 0.2 0.2 0.1 0.1 0.0
Mass retained (grams) 2.0 2.5 18.4 124.0 476.6 192.2 10.1 72.2 68.0
percent retained 0.2 0.3 1.9 12.8 49.3 19.9 1.0 7.5 7.0
cumulative % retained 0.2 0.5 2.4 15.2 64.5 84.4 85.5 93.0 100.0
Cumulative % passing 99.8 99.5 97.6 84.8 35.5 15.6 14.5 7.0 0.0

EXAMPLE
319
0.01 0.1 1 10
0
10
20
30
40
50
60
70
80
90
100
Grain Size in mm
Probability
Finer
by
Weight
(%)
Cumulative Probability Curve
0.20131
Fine Sand Gravel
0.20131
Fine Sand Gravel

EXAMPLE
320
< 1%
93%
7%
Gravel
Sand
Fine

WELL GRADED SOIL
A well graded soil is a soil that contains particles
of a wide range of sizes and has a good
representation of all sizes from the No. 4 to No.
200 sieves. A well graded gravel is classified as
GW while a well graded sand is classified as SW.
322

POORLY GRADED SOIL
A poorly graded soil is a soil that does not have a
good representation of all sizes of particles from
the No. 4 to No. 200 sieve. Poorly graded soils
are either uniformly graded or gap-graded. A
poorly graded gravel is classified as GP while a
poorly graded sand is classified as SP. Poorly
graded soils are more susceptible to
soil liquefaction than well graded soils.
323

PSD PARAMETERS
the Coefficients of Uniformity: Calculating the
coefficients of uniformity and curvature requires
grain diameters. The grain diameter can be found
for each percent of the soil passing a particular
sieve. This means that if 40% of the sample is
retained on the No. 20 sieve then there is 60%
passing the No. 20 sieve.
324

GAP-GRADED SOIL
A gap-graded soil is a soil that has an excess or
deficiency of certain particle sizes or a soil that
has at least one particle size missing. An example
of a gap-graded soil is one in which sand of the
No. 10 and No. 40 sizes are missing, and all the
other sizes are present.
325

PSD PARAMETERS
the Coefficients of Uniformity : The coefficient of
uniformity, Cu is a crude shape parameter and is
calculated using the following equation:
where D60 is the grain diameter at 60% passing,
and D10 is the grain diameter at 10% passing.
327

PSD PARAMETERS
The Coefficients of Curvature : The coefficient of
curvature, Cc is a shape parameter and is
calculated using the following equation:
328

CRITERIA FOR GRADING SOILS
The following criteria are in accordance with the
Unified Soil Classification System:
For a gravel:
Cu > 4 & 1 < Cc < 3
If both of these criteria are not met, the gravel is
classified as poorly graded or GP. If both of these
criteria are met, the gravel is classified as well
graded or GW.
330

CRITERIA FOR GRADING SOILS
For a sand :
Cu ≥ 6 & 1 < Cc < 3
If both of these criteria are not met, the sand is
classified as poorly graded or SP. If both of these
criteria are met, the sand is classified as well
graded or SW.
331

PSD PARAMETERS
Fineness modulus (FM): An empirical factor
obtained by adding the total percentages of a
sample of the aggregate retained on each of a
specified series of sieves, and dividing the sum
by 100.
332

FINENESS MODULUS
The fineness modulus is an empirical factor that
gives a relative measure of the proportions of
fine and coarse particles in an aggregate. It is a
value widely used to indicate the relative fineness
or coarseness of an aggregate.
333

FINENESS MODULUS
Soil Type Fineness Modulus
Fine sand 2.20 to 2.60
Medium sand 2.60 to 2.90
Coarse sand 2.90 to 3.20
Fine aggregate
2.70 to 3.0
2.0 to 4.0
Coarse aggregate 6.50 to 8.0
334

GROUP INDEX
he group index was devised to provide a basis for
approximating within-group evaluations. Group
indexes range from 0 for the best subgrade soils
to 20 for the poorest.
The higher the value of GI the weaker will be the
soil and vice versa. Thus, quality of performance
of a soil as a subgrade material is inversely
proportional to GI.
335

336
GROUP INDEX
 
)
10
PI
)(
15
F
(
01
.
0
)
40
LL
(
005
.
0
2
.
0
)
35
F
(
GI
200
200







)
10
PI
)(
15
F
(
01
.
0
GI 200 


For Group A-2-6 and A-2-7
The first term is determined by the LL
The second term is determined by the PI
In general, the rating for a pavement subgrade is
inversely proportional to the group index, GI.
use the second term only
F200: percentage passing through the No.200 sieve

EXAMPLE
U.S. sieve size Size opening (mm) Weight retained (g)
¾ in. 19.0 0
3/8 in. 9.50 158
No.4 4.75 308
No. 10 2.00 608
No. 40 0.425 652
No. 100 0.150 224
No. 200 0.075 42
Pan -- 8
337
An air-dry soil sample weighing 2000g is brought to the
soils laboratory for mechanical grain size analysis. The
laboratory data are as follows: Plot a grain size distribution
curve for this soil sample. And with the help of distribution
curve determine Cc and Cu. as well.

SOLUTION
U.S. sieve
size
Size
opening
(mm)
Weight
retained (g)
Percenta
ge
retained
Cumulati
ve
percenta
ge
retained
Percenta
ge
passing
¾ in. 19.0 0 0 0 100
3/8 in. 9.50 158 7.9 7.9 92.1
No.4 4.75 308 15.4 23.3 76.7
No. 10 2.00 608 30.4 53.7 46.3
No. 40 0.425 652 32.6 86.3 13.7
No. 100 0.150 224 11.2 97.5 2.5
No. 200 0.075 42 2.1 99.6 0.4
Pan -- 8 0.4 100 --
338

SOLUTION
339
0.001 0.01 0.1 1 10 100 1000
0
20
40
60
80
100
Grain Diameter (mm)
%
Passing

IDENTIFICATION AND CLASSIFICATION OF SOILS

IDENTIFICATION AND CLASSIFICATION OF SOILS
 Introduction
 Field identification of soils
 Engineering classification of soils
 Purpose of soil classification
 Unified soil classification system
 AASHTO soil classification system
 Textural soil classification system
341

INTRODUCTION
 It is necessary to have a standard language for
a careful description and classification of a soil.
 Soil classification is the arrangement of soils
into various groups or sub groups so as to
express briefly the primary material
characteristics without detailed descriptions.
342

343
INTRODUCTION
Classifying soils into groups with similar behavior, in
terms of simple indices, can provide geotechnical
engineers a general guidance about engineering
properties of the soils through the accumulated
experience.
Simple indices
GSD, LL, PI
Classification
system
(Language)
Estimate
engineering
properties
Achieve
engineering
purposes
Use the accumulated
experience
Communicate
between engineers

FIELD IDENTIFICATION OF SOILS
Field identification of soil is of great importance for
civil engineering. Sometimes the lack of time and
facilities makes laboratory soil testing impossible in
construction. Even when laboratory tests are to
follow, field identification tests must be made during
the soil exploration.
Experience is the greatest asset in field
identification; learning the technique from an
experienced technician is the best method of
acquiring the skill.
344

FIELD IDENTIFICATION OF SOILS
Advantages of field identification:-
•It is very economical.
•It can be carried out in short duration of time.
•No pre-setting is required
Disadvantages of field identification:-
•It is just an approximation.
•You cannot completely rely on it.
•Experienced people are required.
345

VISUAL EXAMINATION
Colour of soil:-Visual examination should
establish the colour, grain size, grain shapes (of
the coarse-grained portion), some idea of the
gradation, and some properties of the
undisturbed soil. It helps in Unified classification
of soil. Colour is often helpful in distinguishing
between soil types, and with experience, one may
find it useful in identifying the particular soil type.
346

VISUAL EXAMINATION
Compressive strength:- Knowing the consistency
index (consistency) of the soil we can have the
value of compressive strength by field
determination. However, this value is an
approximate value and can be used as a
guideline. It can be tabulated as follows:-
347

TYPES OF SOIL
Sand, silt, and clay are the basic types of soil.
Most soils are made up of a combination of the
three. The texture of the soil, how it looks and
feels, depends upon the amount of each one in
that particular soil. The type of soil varies from
place to place on our planet and can even vary
from one place to another in your own backyard.
349

TYPES OF SOIL
Loam is a rich soil consisting of a mixture of sand
and clay and decaying organic materials. Loam is
composed of sand, silt, and clay in relatively even
concentration (about 40-40-20% concentration
respectively).
350

351
ORGANIC SOILS
• Highly organic soils- Peat (Group symbol PT)
- A sample composed primarily of vegetable tissue in various stages
of decomposition and has a fibrous to amorphous texture, a dark-
brown to black color, and an organic odor should be designated
as a highly organic soil and shall be classified as peat, PT.
• Organic clay or silt( group symbol OL or OH):
- “The soil’s liquid limit (LL) after oven drying is less than 75 % of its
liquid limit before oven drying.” If the above statement is true,
then the first symbol is O.
- The second symbol is obtained by locating the values of PI and LL
(not oven dried) in the plasticity chart.

PURPOSE OF SOIL CLASSIFICATION
1. Provides a concise and systematic method for
designating various types of soil.
2. Enables useful engineering conclusions to be
made about soil properties.
3. Provides a common language for the
transmission of information.
4. Permits the precise presentation of boring
records and test results.
352

353
CLASSIFICATION SYSTEMS
Two commonly used systems:
Textural Classification of soil
Unified Soil Classification System (USCS).
American Association of State Highway and
Transportation Officials (AASHTO) System

EXAMPLES
Example 1. What is the classification of a soil
sample with 18% sand, 58% silt, and 24% clay?
Entering the left axis at 18%, the bottom axis at
58%, and the right axis at 24%, and moving to
the intersection point, the soil's classification is
Silty Clay Loam.
355

EXAMPLES
sample with 47% sand, 32% silt, and 21% clay?
Entering the left axis at 47%, the bottom axis at
32%, and the right axis at 21%, and moving to
the intersection point, the soil's classification is
Clay Loam.
sample with 32% gravel, 38% sand, 22% silt, and
8% clay?
356

General Classification
Granular Materials
(35% or less passing No. 200 sieve)
Silt Clay Materials
(More than 35% passing No. 200
sieve)
Group Classification
A-1 A-2 A-7
A-1-a A-1-b A-3 A-2-4 A-2-5 A-2-6 A-2-7 A-4 A-5 A-6
A-7-5
A-7-6
Sieve Analysis %
2.00 mm (No .10) 50 max - - - - - - - - - -
0.425 mm (No. 40) 30 max
50
max
51 min - - - - - - - -
0.075 mm (No. 200) 15 max 25 max 10 max 35 max 35 max 35 max 35 max 36 min 36 min 36 min 36 min
Characteristics of
fraction passing
Liquid Limit - - 40 max 41 min 40 max 41 min 40 max 41 min 40 max 41 min
Plasticity Index 6 max N.P 10 max 10 max 11 min 11 min 10 max 10 max 11 min
11
min1
Usual types of
significant constituent
materials
Stone fragments,
gravel, and sand
Fine
sand
Silty or clayey gravel and sand Silty soils Clayey soils
General ratings as
subgrade
Excellent to good Fair to poor
357

EXAMPLE
For the given sieve analysis data classify the soil
according to the AASHTO Classification System.
358
Sieve No. 4 10 20 50 80 100 120 200 Pan
Sieve opening (mm) 4.3 2.0 0.9 0.3 0.2 0.2 0.1 0.1 0.0
Mass retained (grams) 2.0 2.5 18.4 124.0 476.6 192.2 10.1 72.2 68.0
percent retained 0.2 0.3 1.9 12.8 49.3 19.9 1.0 7.5 7.0
cumulative % retained 0.2 0.5 2.4 15.2 64.5 84.4 85.5 93.0 100.0
Cumulative % passing 99.8 99.5 97.6 84.8 35.5 15.6 14.5 7.0 0.0

359
0.01 0.1 1 10
0
10
20
30
40
50
60
70
80
90
100
Grain Size in mm
Probability
Finer
by
Weight
(%)
Cumulative Probability Curve
0.20131
Fine Sand Gravel
0.20131
Fine Sand Gravel

360
< 1%
93%
7%
Gravel
Sand
Fine

362
Major Division Group Symbols Laboratory Tests Typical names
Fine
grained
soils
(More
than
half
of
material
is
smaller
than
No.200
sieve)
Silts
and
Clays
(Liquid
limit
less
than
50
%)
ML
PI < 4 or
Plots below A line
Inorganic silts and very
fine sands, rock flour,
Silty or clayey fine sands,
or clayey silts with slight
plasticity
CL
PI >7 and Plots on or
above a line
Inorganic clays of low to
medium plasticity,
gravelly clays, sandy
clays, Silty clays, lean
clays
OL
Liquid limit oven dried /
liquid limit not oven
dried < 0.75
Organic silts and organic
silty clays of low
plasticity
Silts
and
Clays
(Liquid
limit
more
than
50
%)
MH PI plots below A line
Inorganic silts, micaceous
or diatomaceous fine
sandy or silty soils,
elastic silts
CH
PI plots on or above A
line
Inorganic clays of high
plasticity, fat clays
OH
Liquid limit oven dried /
liquid limit not oven
dried < 0.75
Organic clays of medium
to high plasticity, organic
silts
Highly
organic
soils
Pt
Organic odor and dark
color
Peat and other highly

363
Major Division Group Symbols Typical names Laboratory classification criteria
Coarse
grained
soils
(More
than
half
material
is
larger
than
No.200
Sieve)
Gravels
(More
than
half
of
coarse
fraction
is
larger
than
No.4
sieve)
Clean
gravels
Less
than
5
%
fines
GW
Well graded
gravels
Gravel-sand
mixtures, little or
no fines
3
and
1
/
4
w
b
C
C
c
u 
GP
Poorly graded
gravels, gravel-
sand mixtures
little or no fines
Not meeting Cu or Cc requirements for
GW
Cu < 4 and 1 > Cc >3
Gravel
with
fines
More
than
12
%
fines
GM
Silty gravels,
gravel-sand-silt
mixtures
Atterberg limits
below A line or Ip
less than 4
Fines classify as
ML or MH
Limits
plotting
in
hatched
zone
with
Ip
b/w
4
and
7
are
border
line
cases
requiring
use
of
dual
symbols
GC
Clayey gravels,
gravel-sand-clay
mixtures
Atterberg limits
above A line with Ip
greater than 7
Fines classify as
CL or CH
Sands
(More
than
half
of
coarse
fraction
is
smaller
than
No.4
sieve
Clean
sands
Less
than
5%
fine
SW
Well graded
sands, gravelly
sands, little or no
fines
3
and
1
/
6
w
b
C
C
c
u 
SP
Poorly graded
sands, gravelly
sands, little or no
fines
Not meeting Cu or Cc requirements for
SW
Sands
with
fines
More
than
12%
fines
SM
Silty sands, sand-
silt mixtures
Atterberg limits
below A line or Ip
less than 4 fines
classify as
ML or MH
Limits
plotting
in
hatched
zone
with
Ip
b/w
4
and
7
are
border
line
cases
requiring
use
of
dual
symbols
SC
Clayey sands,
sand-clay
mixtures
Atterberg limits
above A line with Ip
greater than 7
Fines classify as
CL or OH

364
PLASTICITY CHART
(Holtz and Kovacs, 1981)
LL
PI
H
L
•The A-line generally
separates the more
claylike materials
from silty materials,
and the organics
from the inorganics.
•The U-line indicates
the upper bound for
general soils.
Note: If the measured
limits of soils are on
the left of U-line,
they should be
rechecked.

365
BORDERLINE CASES (DUAL SYMBOLS)
 For the following three conditions, a dual symbol
should be used.
 Coarse-grained soils with 5% - 12% fines.
 About 7 % fines can change the hydraulic
conductivity of the coarse-grained media by
orders of magnitude.
 The first symbol indicates whether the coarse
fraction is well or poorly graded. The second
symbol describe the contained fines. For
example: SP-SM, poorly graded sand with silt.

366
BORDERLINE CASES (DUAL SYMBOLS)
 Fine-grained soils with limits within the shaded
zone. (PI between 4 and 7 and LL between
about 12 and 25).
 It is hard to distinguish between the silty and
more claylike materials.
 CL-ML: Silty clay, SC-SM: Silty, clayed sand.
 Soil contain similar fines and coarse-grained
fractions.
 possible dual symbols GM-ML

367
BORDERLINE CASES (SUMMARY)
(Holtz and Kovacs, 1981)

368
DEFINITION OF GRAIN SIZE
Boulders Gravel Sand Silt-Clay
Coarse Fine
75 mm No.4
4.75 mm
No.40
0.425 mm
No.200
0.075
mm
No specific
grain size-use
Atterberg
limits

369
GENERAL GUIDANCE
8 major groups: A1~ A7 (with several subgroups) and organic soils A8
The required tests are sieve analysis and Atterberg limits.
The group index, an empirical formula, is used to further evaluate soils
within a group (subgroups).
 The original purpose of this classification system is used for road
A4 ~ A7
A1 ~ A3
Granular Materials
 35% pass No. 200 sieve
Silt-clay Materials
 36% pass No. 200 sieve
Using LL and PI separates silty materials
from clayey materials
Using LL and PI separates silty materials
from clayey materials (only for A2 group)

CLASSIFICATION OF SOIL BASED ON SENSITIVITY
The degree of disturbance of undisturbed clay
sample due to remoulding can be expressed as:
370
Classification of soil
on the basis of St
(After Skempton and
Northey)

EXERCISE-I
An air-dry soil sample is brought to the soils
laboratory for mechanical grain size analysis. The
laboratory data is given in the table. Plot a grain
size distribution curve for this soil sample. And
with the help of distribution curve determine Cc
and Cu. as well. Also determine the percentage of
gravel, sand, silt, and clay particles present in the
sample, by using USC system.
371

EXERCISE-I
U.S. sieve No. Size opening (mm) Mass retained (g)
No.4 4.75 0.0
No. 10 2.00 40.0
No. 20 0.085 60.0
No. 40 0.425 89.0
No. 60 0.250 140.0
No. 80 0.177 122.0
No. 100 0.150 210.0
No. 200 0.075 56.0
Pan -- 12.0
372

EXERCISE-II
A borrow pit site soil is to be checked for its
suitability for using as a subgrade and
embankment material for an earthen dam. The
results of a sieving analysis are tabulated herein
in Table 1. The liquid and plastic limit values
were found to be (30+RN/10) % and (20+RN/10)
% respectively.
Classify the soil according to AASHTO and USCS
Classification system.
373

EXERCISE-II
Using these data, check the suitability of the soil
to be used as subgrade or Earthfill dam
embankment material.
Suggest suitable type of soil improvement
method if required for both purposes.
374

377
REFERENCES
Main References:
Das, B.M. (1998). Principles of Geotechnical Engineering, 4th edition,
PWS Publishing Company. (Chapter 3)
Holtz, R.D. and Kovacs, W.D. (1981). An Introduction to Geotechnical
Engineering, Prentice Hall. (Chapter 3)
Others:
Santamarina, J.C., Klein, K.A., and Fam, M.A. (2001). Soils and Waves,
John Wiley & Sons, LTD.

SOIL COMPACTION
 Introduction
 Compaction test
 Saturation (zero air void) line
 Laboratory compaction tests
 Field compaction
379

INTRODUCTION
Compaction of soil may be defined as the
process by which the soil particles are artificially
rearranged and packed together into a state of
closer contact by mechanical means in order to
decrease its porosity and thereby increase its dry
density. This is usually achieved by dynamic
means such as tamping, rolling, or vibration. The
process of compaction involves the expulsion of
air only.
380

INTRODUCTION
The following are the important effects of
compaction :
I. Compaction increases the dry density of the
soil, thus increasing its shear strength and
bearing capacity through an increase in
frictional characteristics;
II. Compaction decreases the tendency for
settlement of soil ; and,
III. Compaction brings about a low permeability
of the soil. 381

Solids
Water
Air
Solids
Water
Air
Compressed
soil
Load
Soil
Matrix
gsoil (1) =
WT1
VT1
gsoil (2) =
WT1
VT2
gsoil (2) > gsoil (1)
INTRODUCTION

Factor Affecting Soil Compaction:
1- Soil Type
2- Water Content (wc)
3- Compaction Effort Required (Energy)
Why Soil Compaction:
1- Increase Soil Strength
2- Reduce Soil Settlement
3- Reduce Soil Permeability
4- Reduce Frost Damage
5- Reduce Erosion Damage
Types of Compaction : (Static or Dynamic)
1- Vibration
2- Impact
3- Kneading
4- Pressure
Water is added to
lubricate the contact
surfaces of soil
particles and improve
the compressibility of
the soil matrix

SOIL COMPACTION
The response of a soil to compaction depends on
both the method of compaction and the moisture
content. All other things being equal, the amount of
compaction is greatest at an “optimum moisture
content”; the terminology comes from engineers. In
general, soils compacted “wet-of-optimum” are more
liable to either swell or to collapse when wetted,
having higher permeability, a lower air entry value,
greater ultimate strength at the moulding moisture
content, and greater rigidity both at the moulding
moisture content and when swollen.
384

SOIL COMPACTION
Moulding moisture content, however, seems to
have little effect on strength when swollen;
judging from tests on foundry sands, dry strength
increases with the moulding moisture content.
There are reports of failure by piping erosion of
earth dams that had been compacted dry-of-
optimum.
385

COMPACTION TEST
To determine the soil moisture-density
relationship and to evaluate a soil as to its
suitability for making fills for a specific purpose,
the soil is subjected to a compaction test.
Proctor (1933) showed that there exists a definite
relationship between the soil moisture content
and the dry density on compaction and that, for a
specific amount of compaction energy used.
386

COMPACTION TEST
There is a particular moisture content at which a
particular soil attains its maximum dry density.
Such a relationship provides a satisfactory
practical approach for quality control of fill
construction in the field.
387

DRY AND WET OF OPTIMUM
The water content at compaction is also
sometimes specified because of its effect on soil
fabric, especially for clays. Clays compacted dry
of optimum have a flocculated fabric, which
generally corresponds to higher permeability,
greater strength and stiffness, and increased
brittleness. Conversely, clays compacted wet of
optimum to the same equivalent dry density tend
to have a more oriented or dispersed fabric,
which typically corresponds to lower permeability,
lower strength and stiffness, but more ductility. 389

DRY AND WET OF OPTIMUM
390
Effect of compacted water content on soil fabric for clays (Coduto,
1999).

ZERO AIR VOID LINE
A line showing the relation between water
content and dry density at a constant degree of
saturation S may be established from the
equation:
391

ZERO AIR VOID LINE
If one substitutes S = 100% and plots the
corresponding line, one obtains the theoretical
saturation line, relating dry density with water
content for a soil containing no air voids. It is said
to be ‘theoretical’ because it can never be
reached in practice as it is impossible to expel
the pore air completely by compaction.
392

Moisture
Content
Dry Density
gd max
Compaction Curve
for Standard Proctor
(OMC)
gd max
(OMC)
Zero Air Void Curve
Sr < 100%
Zero Air Void Curve
Sr =100%
Zero Air Void Curve
Sr = 60%
Compaction
Curve for
Modified
Proctor
ZERO AIR VOID LINE

COMPACTION EFFORTS
The amount of compaction ( energy applied per
unit of volume) is called compaction efforts.
Increase in the compaction efforts results in an
increase in the maximum dry density and
decrease the OMC.
In laboratory compaction efforts are applied
through Standard Proctor Test or Modified Proctor
Test. In both the cases the compaction energy is
given as:
394
mould
of
Volume
hammer
of
drop
of
Height
hammer
of
Weight
layers
of
Number
layer
per
blows
of
Number 



E

COMPACTION EFFORTS
The degree of compaction is not directly
proportional to compaction efforts and dry
density doesn’t increase indefinitely. When the
soil is initially loose, the compaction increases
the dry density, but further compaction beyond
certain point doesn’t increase the density.
395

Water Content
Dry
Density
Effect of Energy on Soil Compaction
Higher
Energy
Z
A
V
Increasing compaction energy Lower OWC and higher dry density
In the field
increasing compaction
energy = increasing
number of passes or
reducing lift depth
In the lab
increasing
compaction energy
= increasing
number of blows
EFFECT OF COMPACTION EFFORTS

Soil Compaction in the
Lab:
1- Standard Proctor Test
2- Modified Proctor Test
3- Gyratory Compaction Standard Proctor Test Modified Proctor Test
Gyratory Compaction

Soil Compaction in the
Lab:
1- Standard Proctor Test
wc1 wc2 wc3
wc4
wc5
gd
1
gd
2
gd
3
gd
4
gd
5
Optimum
Water
Content
Water
Content
Dry Density
gd max
Zero Air Void Curve
Sr =100%
Compaction
Curve
1
2
3
4
5
(OWC)
4 inch diameter compaction mold.
(V = 1/30 of a cubic foot)
5.5 pound hammer
25 blows
per layer
H = 12 in
Wet to
Optimum
Dry to
Optimum
Increasing Water Content
e
G w
s
dry


1


gdr
y
=
gwet
Wc
100
%
1+
gZAV =
Gs gw
Wc Gs
1+
Sr

COMPACTION OF UNSATURATED SOILS
399
Dry of optimum Optimum moisture
content
Wet of optimum
liable to either swell
or to collapse when
wetted, having
higher permeability

COMPACTION CHARACTERISTICS
400
Some cohesionless soils
exhibit two peaks in the
compaction curve; one at
very dry conditions,
where there are no
capillary tensions to resist
the compaction effort, and
the other at the optimum
moisture content, where
optimum lubrication
between particles occurs.

401
15
16
17
18
19
20
0 2 4 6 8 10
Dry
density,
g
(kN/m
3
)
Moisture content, w (%)
0.25 % Fibre
0% Fibre
0.5 % Fibre

402
15
16
17
18
19
20
0 0.25 0.5 0.75 1
Dry
density,
g(kN/m
3
)
Fiber Content, FC (%)
Water content = 0%

1- Rammers
2- Vibratory Plates
3- Smooth Rollers
4- Rubber-Tire
5- Sheep foot Roller
6- Dynamic Compaction
FIELD COMPACTION

Field Soil Compaction
Because of the differences between lab and field compaction
methods, the maximum dry density in the field may reach 90% to
95%.
Moisture
Content
Dry Density
gd max
(OMC)
ZAV
95% gd max

Example:
The laboratory test for a standard proctor is shown below. Determine the optimum water content and maximum dry
density. If the Gs of the soil is 2.70, draw the ZAV curve.
Solution:
gdr
y
=
gwet
Wc
100
%
1+
Volume of
Proctor Mold
(ft3
)
1/30
1/30
1/30
1/30
1/30
1/30
Weight of wet
soil in the
mold (lb)
3.88
4.09
4.23
4.28
4.24
4.19
Water Content
(%)
12
14
16
18
20
22
Volume of
Mold
(ft3
)
1/30
1/30
1/30
1/30
1/30
1/30
Weight of wet
soil in the
mold (lb)
3.88
4.09
4.23
4.28
4.24
4.19
Water Content
(%)
12
14
16
18
20
22
Wet Unit
Weight
(lb/ft3
)
116.4
122.7
126.9
128.4
127.2
125.7
Dry Unit
Weight
(lb/ft3
)
103.9
107.6
109.4
108.8
106.0
103.0
10 11 12 13 14 15 16 17 18 19 20 21 22 23
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
gZAV =
Gs gw
Wc
Gs
1+
Sr

10 11 12 13 14 15 16 17 18 19 20 21 22 23
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
gdry max
Optimum
Water
Content

EXAMPLE
A proctor compaction test was conducted on a
soil sample, and the following observations were
made:
If the volume of the mold used was 950 cm3
and
the specific gravity of soils grains was 2.65.
410
Water content,
percent
7.7 11.5 14.6 17.5 19.7 21.2
Mass of wet soil, g 1739 1919 2081 2033 1986 1948

EXAMPLE
Make necessary calculations and draw, (i)
compaction curve and (ii) 80% and 100%
saturation lines.
411

Checking Soil Density in the Field:
1- Sand Cone (ASTM D1556-90)
2- Balloon Dens meter
The same as the sand cone, except a rubber
balloon is used to determine the volume of the hole
3- Nuclear Density (ASTM D2292-91)
Nuclear Density meters are a quick and fairly accurate way of determining density and moisture content. The meter uses a radioactive
isotope source (Cesium 137) at the soil surface (backscatter) or from a probe placed into the soil (direct transmission). The isotope
source gives off photons (usually Gamma rays) which radiate back to the mater's detectors on the bottom of the unit. Dense soil
absorbs more radiation than loose soil and the readings reflect overall density. Water content (ASTM D3017) can also be read, all
within a few minutes.
A small hole (6" x 6" deep) is dug in the compacted material to be tested. The soil is removed
and weighed, then dried and weighed again to determine its moisture content. A soil's
moisture is figured as a percentage. The specific volume of the hole is determined by filling it
with calibrated dry sand from a jar and cone device. The dry weight of the soil removed is
divided by the volume of sand needed to fill the hole. This gives us the density of the
compacted soil in lbs per cubic foot. This density is compared to the maximum Proctor
density obtained earlier, which gives us the relative density of the soil that was just compacted.

Compaction Specifications:
Compaction performance parameters are given on a construction
project in one of two ways:
1- Method Specification
detailed instructions specify machine type, lift depths, number of
passes, machine speed and moisture content. A "recipe" is given
as part of the job specifications to accomplish the compaction
needed.
2- End-result Specification
Only final compaction requirements are specified (95% modified
or standard Proctor). This method, gives the contractor much more
flexibility in determining the best, most economical method of
meeting the required specs.

EXERCISE-I
An embankment for a highway is to be
constructed from a soil compacted to a dry unit
weight of 18 kN/m3
. The clay to be trucked to the
site from a borrow pit. The bulk unit weight of the
soil in the borrow pit is 17 kN/m3
and its natural
moisture content is 5%. Calculate the volume of
clay from the borrow pit required for 1 cubic
meter of embankment. Assume Gs =2.7.
416

EXERCISE-I (CONTINUE)
Estimate the amount of water to be added per
cubic meter of the soil, if the optimum moisture
content to achieve the targeted dry density of 18
kN/m3
is 7%.
417

EXERCISE-II
Soil from a borrow pit to be used for construction
of an embankment as shown in Figure gave the
following laboratory results when subjected to the
ASTM D 698 Standard Proctor test:
Maximum dry unit weight = 118.5 lb/ft3
,
Optimum moisture content = 12.5%
The soil on the borrow pit site is found to have:
Maximum dry unit weight = (110 +RN/10) lb/ft3
,
Moisture content = (5 +RN/10) %.
418

EXERCISE-II
Determine the following for 1 Km Road section:
 Amount of soil to be excavated from borrow pit.
 Amount of water to be added in the
embankment.
419

EXERCISE-III
The contractor, during construction of the soil
embankment, conducted a sand-cone in-place
density test to determine whether the required
compaction was achieved. The following data
were obtained during the test:
Weight of sand used to fill test hole and funnel of
sand-cone device = 845 g.
Weight of sand to fill funnel = 323g.
420

EXERCISE-II
Unit weigh of sand = 100 lb/ft3
.
Weigh of wet soil from test hole = 600g.
Moisture content of soil from test hole = 17%.
Based on the contract, the contractor is
supposed to attain the 95% compaction. Will you
approve the contractor’s work?
421

EXERCISE-III
Condition
 The combined weight of a mold and the specimen
of compacted soil it contains is 9.0 lb.
 The mold’s volume is 1/35ft3
 The mold’s weight is 4.35 lb.
 The specimen’s water content is 12%.
What is dry unit weight of the specimen?
422

SOIL HYDRAULICS
425
Baluchistan 2012

PERMEABILITY
Measurement of soil/rock sample permeability is
of significant importance. For instance,
 Pumping water, oil or gas into or out of a
porous formation
 Disposing of brine wastes in porous formations
 Storing fluids in mined caverns for energy
conversions
 Predicting water flow into a tunnel
427

GEOLOGICAL FORMATIONS OF SOIL
Aquifer-An aquifer is an underground layer of
water-bearing permeable rock or unconsolidated
materials (gravel, sand, or silt) from which
groundwater can be usefully extracted using a
water well.
429

430
AQUIFER

Aquitard-Geological formation that may contain
groundwater but is not capable of transmitting
significant quantities of it under normal hydraulic
gradients. May function as confining bed.
431

432
AQUITARD

Aquifuge - a formation which has no
interconnected openings and hence cannot
absorb or transmit water.
433

Aquiclude-a formation which, although porous
and capable of absorbing water, does not permit
its movement at rates sufficient to furnish an
appreciable supply for a well or spring.
434

435

436

aqueduct -An aqueduct is a water supply or
navigable channel constructed to convey water. In
modern engineering, the term is used for any
system of pipes, ditches, canals, tunnels, and
other structures used for this purpose.
437

438

DARCY’S LAW
439
Sand filter
h
u/s
d/s
q q

440
v







A
q
ki
or
iA
q
k
kiA
q
iA
q
A
q
i
q
A
dx
dh
k
q 

COEFFICIENT OF PERMEABILITY
441

442
Coefficient of permeability versus particle size (Novac 1982)

FLOW VELOCITY
443
This velocity is an average velocity, since it
represents flow rate divided by gross cross-sectional
area of the rock mass/ soil. This area however
includes both solid soil material and voids. Since
water or oil moves only through the voids, the actual
(interstitial) velocity is,
v
ki
A
q

 Av
q 

s
vv
A
q 

FLOW VELOCITY
444
v
v
vs
.
.
. .
 

 

v s
s
v
s
v v
s
Q A v Av
Av
v
A
LAv V v
v
LA V
v
v
 1
e
e
 

(1 )
s
v e
v
e


q

EXAMPLE
In a rock permeability test, it took 160.0 minutes
for 156 cm3
of water to flow through a rock
sample, the cross-sectional area of which was
50.3 cm2
. The voids ratio of the rock sample was
0.45. Determine the average and actual velocity
of the water through the rock sample.
445

PERMEABILITY TESTS
446
 Field permeability tests
 Laboratory Permeability tests
 Falling head permeability test
 Constant head permeability test

AQUIFER RECUPERATION (CONFINED)
448
Aquifer H
r2
r
r1
h1
h h2
Impermeable
layer

449
From Darcy’s law
kiA
q 
kHdh
r
dr
q
rH
dr
dh
k
q
2
)
2
(




AQUIFER RECUPERATION

450
Integrating for whole of the cross-section
)
(
2
)
ln(
2
ln
2
1
1
2
1
2
2
1
2
1
2
1
2
1
h
h
kH
r
r
q
kHh
r
q
dh
kH
dr
r
q
h
h
r
r
h
h
r
r









451
)
r
/
r
ln(
)
h
h
(
kH
2
q
1
2
1
2 


2 1
2 1
ln( / )
2 H(h -h )
q r r
k



452
(UNCONFINED)

UNCONFINED AQUIFER
Consider a pumping well is located in an
unconfined, homogeneous aquifer; in this case
the piezometric surface lies within the aquifer
and water can be drawn from the hole thickness
h. thus From Darcy’s law
For very small portion with distance dr and height
dh the discharge will be
453
kiA
q 

UNCONFINED AQUIFER
454
krhdh
r
dr
q
rh
dr
dh
k
q
2
)
2
(




2
1
2
1
2
1
2
1
2
2
ln
2
1
2 h
h
r
r
h
h
r
r
h
k
r
q
dh
kh
dr
r
q






UNCONFINED AQUIFER
455
)
h
h
(
k
)
r
r
ln(
q 1
2
2
2
1
2


)
r
/
r
ln(
)
h
h
(
k
q
1
2
1
2
2
2



2 1
2 2
2 1
ln( / )
(h -h )
q r r
k



SCAVENGER WELL
A well located between a good well (or group of
wells) and a source of potential contamination,
which is pumped (or allowed to flow) as waste to
prevent the contaminated water from reaching
the good well.
456

INJECTION WELLS
An Injection well is a device that places fluid deep
underground into porous rock formations, such
as sandstone or limestone, or into or below the
shallow soil layer. These fluids may be water,
wastewater, brine (salt water), or water mixed
with chemicals.
Injection wells have a range of uses that include
long term (CO2) storage, waste disposal,
enhancing oil production, mining, and preventing
salt water intrusion. 458

CONSTANT HEAD PERMEABILITY TEST
460

461
From Darcy’s law
kiA
q 
CONSTANT HEAD PERMEABILITY TEST
A
L
h
k
q 

EXAMPLE-1
A constant head permeability test was carried out
on a cylindrical sample of sand 4 in. in diameter
and 6 in. in height. 10 in3
of water was collected
in 1.75 min, under a head of 12 in. Compute the
hydraulic conductivity in ft/year and the velocity
of flow in ft/sec.
462

FALLING HEAD PERMEABILITY TEST
463

464
From Darcy’s law
kiA
q 
dt
dh
v 

With negative sign used to indicate a falling head. The flow of water
into the specimen is therefore
-a(dh/dt)
q in
The flow of water through and out of the specimen is
A
L
h
k
qout 

465
2 2
1 1
h t
h t
h
-a(dh/dt) k A
L
dh A
a k dt
h L
1 A
a dh k dt
h L

 
 
 

466
   
 
 
2
2
1
1
t
h
h
t
1 2
1 2
A
a ln h k
L
A
a ln h / h k t
L
aL
k ln h / h
At
t
 



Where
a = area of cross section of burette
A = area of cross section of sample
L = length of sample
t = time elapsed (t1-t2)
h1 = initial head of burette at time t1
h2 = final head of burette at time t2
467

EXAMPLE
The hydraulic conductivity of a soil sample was
determined in a soil mechanics laboratory by
making use of a falling head permeameter. The
data used and the test results obtained were as
follows: diameter of sample = 2.36 in, height of
sample = 5.91 in, diameter of stand pipe = 0.79
in, initial head hQ = 17.72 in. final head hl =
11.81 in. Time elapsed = 1 min 45 sec.
Determine the hydraulic conductivity in ft/day.
468

471
Project: Upper and Lower Nara Canal Date
Client: Sindh Irrigation & Drainage Authority Location
Consultants: Mott MacDonald BH No.
Main Contractor: China Intl. Water & Electric Corp Test
1 1 0.01 1.39 1.00
2 2 0.02 1.38 0.99 8.67E-06
3 3 0.02 1.38 0.99 0.00E+00
4 4 0.02 1.38 0.99 0.00E+00
5 5 0.03 1.37 0.99 8.73E-06
6 10 0.03 1.37 0.99 0.00E+00
7 15 0.04 1.36 0.98 1.76E-06
8 20 0.04 1.36 0.98 0.00E+00
9 25 0.05 1.36 0.97 8.85E-07
10 30 0.05 1.35 0.97 7.10E-07
11 35 0.05 1.35 0.97 7.12E-07
12 40 0.06 1.34 0.97 7.14E-07
13 45 0.06 1.34 0.96 7.16E-07
14 50 0.06 1.34 0.96 0.00E+00
15 55 0.07 1.33 0.96 1.62E-06
16 60 0.07 1.33 0.96 0.00E+00
17 65 0.07 1.33 0.96 0.00E+00
18 70 0.08 1.32 0.95 1.81E-06
19 75 0.08 1.32 0.95 0.00E+00
20 80 0.08 1.32 0.95 0.00E+00
k
(m/min)
S.No.
Time
Elapsed
(min)
Water Level
(m)
h
(m)
ht/h
Height of Casing Top above
GWT (m)
1.4
Bore hole Depth(m) 22.86
Casing Depth of (m) 11
Depth of Test Section L (m) 11.86
Intake Factor F
(BS 5930:1999) fig.d
14.717
Basic Tim
e
Lag
=
(T)
t1 (min) 1
19/12/2011
Old Jamrao Canal
3
1
Dia of Bore hole (m) 0.15
Dia of Casing (m) 0.076
Area of Bore hole (m
2
) 0.018
G. Water Table(m) 0.8
Casing Top above GL (m) 0.6
21 85 0.08 1.32 0.95 0.00E+00
22 90 0.08 1.32 0.95 0.00E+00
23 95 0.08 1.32 0.95 0.00E+00
Basic Tim
e
Lag
=
(T)
t1 (min) 1
t2 (min) 85
h1 (m) 1.000
h2 (m) 0.950
t: Upper and Lower Nara Canal Date
: Sindh Irrigation & Drainage Authority Location
ltants: Mott MacDonald BH No.
Contractor: China Intl. Water & Electric Corp Test
kavg = 1.20E-06
Coefficient of permeability (k) m/min
k = 7.39E-07

EXAMPLE-2
A sand sample of 35 cm2
cross sectional area
and 20 cm long was tested in a constant head
permeameter. Under a head of 60 cm, the
discharge was 120 ml in 6 min. The dry weight of
sand used for the test was 1 120 g, and Gs =
2.68. Determine (a) the hydraulic conductivity in
cm/sec, (b) the discharge velocity, and (c) the
seepage velocity.
472

PINHOLE TEST
Pinhole tests differentiates between dispersive
and non dispersive clays. For dispersive clay the
water flowing through the specimen carries a
cloudy colored suspension of colloids, whereas
water running through ordinary, erosion resistant
clays is crystal clear.
This test is used for evaluating clay soils for
erodibility by flowing water through a small hole
that is drilled through the compacted specimen.
473

PINHOLE TEST
under hydraulic heads (H) ranging between 50
and 1020 cm. Dispersibility is assessed by
observing effluent colour and flow discharge
through the hole, by visual inspection of the hole
after the completion of the test.
Prior to each pinhole experiment the soil is air-
dried at room temperature (20°C) for 3-4 days
and is sieved at a mesh size of 1.25 cm
diameter.
474

DISPERSIVE SOIL
A dispersive soil is structurally unstable. In
dispersive soils the soil aggregates – small clods
– collapse when the soil gets wet because the
individual clay particles disperse into solution.
Using dispersive clay soils in hydraulic structures,
embankment dams, or other structures such as
roadway embankments can cause serious
engineering problems if these soils are not
identified and used appropriately.
475

EQUATION OF CONTINUITY
dz
dx
dy
q
q +q
x
z
y

dxdy
z
h
k
A
i
k
q
dxdz
y
h
k
A
i
k
q
dydz
x
h
k
A
i
k
q
kiA
q
z
z
z
z
z
y
y
y
y
y
x
x
x
x
x














 A
di
i
k
dq
q 


dydz
dx
x
h
x
h
k
dq
q x
x
x 













 2
2
dxdy
dz
z
h
z
h
k
dq
q
dxdz
dy
y
h
y
h
k
dq
q
z
z
z
y
y
y






























2
2
2
2

     
z
z
y
y
x
x
z
y
x dq
q
dq
q
dq
q
q
q
q 
















dxdy
z
h
k
dxdz
y
h
k
dydz
x
h
k z
y
x
dxdy
dz
z
h
z
h
k
dxdz
dy
y
h
y
h
k
dydz
dx
x
h
x
h
k z
y
x 








































2
2
2
2
2
2

0
0
2
2
2
2
2
2
2
2
2
2
2
2


















z
h
k
y
h
k
x
h
k
dzdxdy
z
h
k
dydxdz
y
h
k
dxdydz
x
h
k
z
y
x
z
y
x
0
2
2
2
2
2
2









z
h
y
h
x
h

0
2
2
2
2
2
2









z
h
y
h
x
h
0
2
2
2
2






y
h
x
h
Three dimensional equation of continuity for
isotropic and homogeneous soil.
Two dimensional equation of continuity for
isotropic and homogeneous soil.

THE CONCEPT OF EFFECTIVE STRESS

INTRODUCTION
488
The pressure transmitted
through grain to grain at the
contact points through a soil
mass is termed as
intergranular or effective
pressure. It is known as
effective pressure since this
pressure is responsible for
the decrease in the void ratio
or increase in the frictional
resistance of a soil mass.

INTRODUCTION
The importance of the forces transmitted through
the soil skeleton from particle to particle was
recognized in 1923 when Terzaghi presented the
principle of effective stress, an intuitive
relationship based on experimental data. The
principle applies only to fully saturated soils and
relates the following three stresses:
489

INTRODUCTION
I. The total normal stress () on a plane within the soil
mass, being the force per unit area transmitted in a
normal direction across the plane, imagining the soil
to be a solid (single-phase) material;
II. The pore water pressure (u), being the pressure of the
water filling the void space between the solid
particles;
III. The effective normal stress (`) on the plane,
representing the stress transmitted through the soil
skeleton only.
490

INTRODUCTION
Effective stress (σ') acting on a soil is calculated
from two parameters, total stress (σ) and pore
water pressure (u) according to:
The principle of effective stress is the most
important principle in soil mechanics.
Deformations of soils are a function of effective
stresses, not total stresses.
The principle of effective stresses applies only to
normal stresses and not to shear stresses.
491

SPRING ANALOGY
Valve
Springs
The concept of effective stress in soil is analogue
to the spring analogy which is expressed in the
following figure

DURING CONSOLIDATION
494
 remains the same (=q) during consolidation. u
decreases (due to drainage) while ’ increases transferring
the load from water to the soil.
GL
saturated clay
q kPa
A

u
’

u
’
q

DURING CONSOLIDATION
495
 remains the same (=q) during consolidation. u
decreases (due to drainage) while ’ increases transferring
the load from water to the soil.
GL
saturated clay
q kPa
A


u

’


u
’
q

EXAMPLE-I
Calculate the effective stress for a soil element at
depth 5 m in a uniform deposit of soil, as shown
in Figure. Assume that the pore air pressure is
zero.
500

EXAMPLE-II
A borehole at a site reveals the soil profile shown
in Figure. Plot the distribution of vertical total and
effective stresses with depth. Assume pore air
pressure is zero.
503

EXAMPLE-III
The soil profile shown in figure, determine the
present effective overburden pressure at the
midheight of compressible clay layer.
509
Elev. 710 ft
Elev. 732 ft
Elev. 752 ft
Elev. 760 ft
Water table
Sand and gravel
Unit weight = 132.0 lb/ft
3
Sand and gravel
3
Clay
3

EXAMPLE-IV
The depth of water in a well is 3 m. Below the
bottom of the well lies a layer of sand 5 meters
thick overlying a clay deposit. The specific gravity
of the solids of sand and clay are respectively
2.64 and 2.70. Their water contents are
respectively 25 and 20 percent. Compute the
total, inter-granular and pore water pressures at
points A and B shown in Figure.
510

EXAMPLE-V
A trench is excavated in fine sand for a building
foundation, up to a depth of 13 ft. The excavation
was carried out by providing the necessary side
supports for pumping water. The water levels at
the sides and the bottom of the trench are as
given Fig. Examine whether the bottom of the
trench is subjected to a quick condition if Gs =
2.64 and e = 0.7. If so, what is the remedy?
514

EXERCISE-I
A uniform layer of sand 10 m deep overlays
bedrock. The water table is located 2 m below
the
surface of the sand which is found to have a
voids ratio e = 0.7. Assuming that the soil
particles have a specific gravity Gs = 2.7
calculate the effective stress at a depth 5 m
below the surface.
516

EXERCISE-II
A stratum of sand 2.5 m thick overlies a stratum
of saturated clay 3 m thick. The water table is 1
m below the surface. For the sand, Gs = 2.65, e
= 0.50 and for the clay G = 2.72, e = 1.1.
Calculate the total and effective vertical stresses
at depths of 1 m, 2.5 m and 5.5 m below the
surface assuming that the sand above the water
table is completely dry.
517

EXERCISE-III
In a deep deposit of clay the water table lies 3 m
below the soil surface. Calculate the effective
stresses at depths of 1 m, 3 m and 5 m below
the surface. Assume that the soil remains fully
saturated above the water table.
518

EXERCISE-IV
Plot the distribution of total stress, effective
stress, and pore-water pressure with depth for
the soil profile shown in Figure. Neglect capillary
action and pore air pressure.
519

EXERCISE-V
A clay layer 3.66 m thick rests beneath a deposit
of submerged sand 7.92 m thick. The top of the
sand is located 3.05 m below the surface of a
lake. The saturated unit weight of the sand is
19.62 kN/m3
and of the clay is 18.36 kN/m3
.
Compute (a) the total vertical pressure, (b) the
pore water pressure, and (c) the effective vertical
pressure at mid height of the clay layer.
520

EXERCISE-VI
A large excavation is made in a stiff clay whose
saturated unit weight is 109.8 lb/ft3
. When the
depth of excavation reaches 24.6 ft, cracks
appear and water begins to flow upward to bring
sand to the surface. Subsequent borings indicate
that the clay is underlain by sand at a depth of
36.1 ft below the original ground surface.
What is the depth of the water table outside the
excavation below the original ground level?
522

INTRODUCTION
Consolidation may be defined as the” Time rate
of compression of soil under static loading
resulting to the dissipation of pore water
pressure”.
It is the gradual reduction in volume of a fully
saturated soil of low permeability due to drainage
of some of the pore water, the process continuing
until the excess pore water pressure set up by an
increase in total stress has completely
dissipated; 526

INTRODUCTION
the simplest case is that of one-dimensional
consolidation, in which a condition of zero lateral
strain is implicit. The process of swelling, the
reverse of consolidation, is the gradual increase
in volume of a soil under negative excess pore
water pressure.
527

CONSOLIDATION THEORY
A general theory for consolidation, incorporating
three-dimensional flow vectors is complicated
and only applicable to a very limited range of
problems in geotechnical engineering. For the
vast majority of practical settlement problems, it
is sufficient to consider that both seepage and
strains take place in one direction only; this
usually being vertical.
528

CONSOLIDATION THEORY
One-dimensional consolidation specifically
occurs when there is no lateral strain, e.g. in the
Oedometer test. One-dimensional consolidation
can be assumed to be occurring under wide
foundations.
529

TERZAGHI 1-D CONSOLIDATION EQUATION
531
x
z
y
dx
dy
dz
Qi
Qou
t

TERZAGHI ONE-DIMENSIONAL
EQUATION
To derive the equation for time rate of settlement
using an element of the soil sample of thickness dz
and cross-sectional area of dA = dxdy, we will
assume the following:
 The soil is saturated, isotropic and homogeneous
 Darcy’s law is valid
 Flow only occurs vertically
 The strains are very small
532

EQUATION
533
dxdydt
v
dt
Av
dt
q
Q z
z
v
in 



  dxdydt
dz
z
v
v
dt
q
q
Q z
z
z
z
out 













dzdxdydt
z
v
dt
v
A
dt
q
dt
t
V
Q
Q
dt
t
V
V
z
z
z
in
out
















EQUATION
Since the change in volume of the soil (V) is
equal to the change in volume of pore water
expelled (Vw), which is equal to the change in
volume of the voids (Vv) therefore,
534

EQUATION
535
dt
t
V
e
dt
t
e
V
dt
t
V
dt
t
eV
dt
t
V
dt
t
V
dt
t
V
s
s
s
v


















)
(

EQUATION
536
dxdydzdt
t
e
e
dt
t
e
e
dxdydz
dt
t
V










1
1
1
t
e
e
z
v
dxdydzdt
t
e
e
dzdxdydt
z
v
z
z












1
1
1
1

EQUATION
537
From Darcy’s law
w
w
w
z
z
z
z
z
z
u
h
h
u
z
u
k
v
z
h
i
z
h
k
v
i
k
v


















EQUATION
538
By Partial Differentiation with respect to depth z gives
 
2
2
z
u
k
z
v
z
u
k
z
v
z
w
z
z
w
z
z























ONE-DIMENSIONAL EQUATION
539
As there is no change in the volume of solid
therefore,
dt
t
e
V
dt
t
V
s





e
dxdydz
e
V
Vs
e
V
Vs







1
1
1
1
As we know that

EQUATION
540
As we know that the change in total vertical pressure
is equal to the change in pore water pressure. i.e., u
=v we can write
v
v
v
v
e
a
t
u
a
t
e
t
u
e
t
e



















EQUATION
541
t
u
e
a
z
v v
z






1 v
v
m
e
a


1
where
t
u
m
z
v
v
z





Therefore,

542
2
2
z
u
C
t
u
v





2
2
2
2
z
u
m
k
t
u
z
u
k
t
u
m
w
v
z
w
z
v












v
w
v
z
C
m
k




543
2
2
z
u
C
t
u
v






SOLUTION TO TERZAGHI EQUATION
544
 
v
m
m
av T
M
M
U 2
0
2
exp
2
1 

 



2
/
)
1
2
( 

 m
M
Where
2
2
z
u
C
t
u
v






545
 
%)
100
log(
933
.
0
781
.
1
T
:
%
100
53
100
%
4
T
:
%
53
0
v
av
2
v
av
av
av
U
U
For
U
U
For
















DEGREE OF CONSOLIDATION
The ratio, expressed as a percentage, of the
amount of consolidation at a given time within a
soil mass, to the total amount of Consolidation
obtainable under a given stress condition.
It is the ratio of the settlement occurred at a
particular time and depth to the total expected
settlement. This parameter can be expressed as
546

547
t
v
t
t
v
s
s
U
s
s
s
U




1

548
0
0
0
1
u
u
U
u
u
u
U
v
v




0
0
0
1
e
e
U
e
e
e
U
v
v





549

550
Tv Tv U
0.00 0.00 0
0.00 0.04 5
0.01 0.09 10
0.02 0.13 15
0.03 0.18 20
0.05 0.22 25
0.07 0.27 30
0.10 0.31 35
0.13 0.35 40
0.16 0.40 45
0.20 0.44 50
0.24 0.49 55
0.28 0.53 60
0.34 0.58 65
0.40 0.63 70
0.48 0.69 75
0.57 0.75 80
0.68 0.83 85
0.85 0.92 90
1.13 1.06 95
  100

U-Tv RELATION
551
0 0.25 0.5 0.75 1 1.25 1.5 1.75 2
0
20
40
60
80
100
Tv
U
(%)

U-Tv RELATION
552
0 0.25 0.5 0.75 1 1.25 1.5
0
20
40
60
80
100
ÖTv
U
(%)

U-Tv RELATION
553
0 0.25 0.5 0.75 1 1.25 1.5
0
20
40
60
80
100
ÖTv
U
(%)

U-Tv RELATION
554
0 0.25 0.5 0.75 1 1.25 1.5
0
20
40
60
80
100
ÖTv
U
(%)

U-Tv RELATION
555
0 0.25 0.5 0.75 1 1.25 1.5
0
20
40
60
80
100
ÖTv
U
(%)

U-Tv RELATION
556
0 0.25 0.5 0.75 1 1.25 1.5
0
20
40
60
80
100
ÖTv
U
(%)

U-Tv RELATION
557
0 0.25 0.5 0.75 1 1.25 1.5
0
20
40
60
80
100
ÖTv
U
(%)
a
b

U-Tv RELATION
558
0 0.25 0.5 0.75 1 1.25 1.5
0
20
40
60
80
100
ÖTv
U
(%)
a
b = 0.15 a
b/a =0.15

U-Tv RELATION
559
0 0.25 0.5 0.75 1 1.25 1.5
0
20
40
60
80
100
ÖTv
U
(%)
50
90
10
0

CONSOLIDATION TEST
The consolidation or Oedometer test is used to
determine the compressibility characteristics of
saturated undisturbed or remoulded soil.
560

COMPRESSIBILITY CHARACTERISTICS
Compressibility is the degree to which a soil mass
decreases in volume when supporting a load.
A soil is a particulate material, consisting of solid
grains and void spaces enclosed by the grains.
The voids may be filled with air or other gas, with
water or other liquid, or with a combination of
these. The volume decrease of a soil under
stress might be conceivably attributed to:
563

1. Compression of the solid grains;
2. Compression of pore water or pore air;
3. Expulsion of pore water or pore air from the
voids, thus decreasing the void ratio or
porosity.
564

Compressibility characteristics of soils forms one
of the important soil parameters required in
design considerations. Compression index, Cc,
which is the slope of the linear portion of void
ratio, e vs. logarithm of effective pressure p(log p)
relationship, is extensively used for settlement
determination. The e–log p is most often
assumed to be linear at higher pressure range
and hence Cc is taken as a constant.
565

SWELLING CHARACTERISTICS
The swelling behavior of expansive soils often
causes unfavorable problems, such as
differential settlement and ground heaving.
566

OBTAINING T90 FROM THE EXPERIMENTAL RESULTS

CONSOLIDATION PARAMETERS
r
S
wG
e 
0
90
2
90
2
90 848
.
0
t
t
T
C H
H d
d
v 

)
1
( 0
0
e
H
H
e 


 k = cvmvw
)
1
(
)
1
`(
0
0
e
a
m
e
e
m
v
v
v
v







`
v
v
e
a



modulus
Constraint
1


v
m
E

EXAMPLE-1
For the following given information of an
Oedometer test calculate the consolidation
parameters of the soil.
Moisture content, % w= 32
Specific gravity, G = 2.7
Diameter, mm D = 74.94
Height, mm H =19.22

EXAMPLE-1
t[min] t After stress-1: 50 kPa After stress-2: 100 kPa After stress-3: 200 kPa
0 0.00 0.0 0.0 0.0
0.25 0.50 0.5 0.177 0.213
1 1.00 0.58 0.254 0.314
2 1.41 0.689 0.316 0.383
4 2.00 0.81 0.399 0.468
6 2.45 0.887 0.454 0.51
9 3.00 0.977 0.506 0.543
16 4.00 1.095 0.553 0.569
25 5.00
36 6.00
1.095 0.553 0.569
Elasped time Vertical displacment /Dial Gauge reading (mm)
Total Displacement

EXAMPLE-1
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5
0.0
0.2
0.4
0.6
0.8
1.0
1.2
Square root of time
Vertical
displacement
3.4

TIME FOR 90% CONSOLIDATION T90
From the graph
90
90
3.4
11.56 minutes
t
t



0 0
0 0.32 2.7 0.864
s
e w G
e

  
0.41mm
 
0
0
`
19.22 0.41 18.81
H H
H mm

 
  
From the graph

Total vertical displacement after each stage
1
2
3
1.095 0 1.095 (for 50 kPa)
0.553 0 0.553 (for 100 kPa)
0.569 0 0.569 (for 200 kPa)
H mm
H mm
H mm
   
   
   
0
0
1
18.81
10.09
1 0.864
s
s
H
H
e
H mm


 


1
1
2
2
3
3
1.095
0.108 (for 50 kPa)
10.09
0.553
0.055 (for 100 kPa)
10.09
0.569
0.056 (for 200 kPa)
10.09
s
s
s
H
e
H
H
e
H
H
e
H

   

   

   

0
1 0 1
2 0 2
3 0 3
0.864 0.108 0.756
0.756 0.055 0.701
0.701 0.056 0.645
f
f
f
f
e e e
e e e
e e e
e e e
  
     
     
     
Final void ratio

COMPRESSION CURVE
10 100 1000
0.56
0.6
0.64
0.68
0.72
0.76
0.8
Effective pressure, kPa
Void
ratio
,e

COEFFICIENT OF CONSOLIDATION CV
Coefficient of consolidation Cv
2 2 3 2
90 90
6 2
0.848 0.848(18.81 10 / 2)
11.56
6.49 10 / min
v
v
v
T d d
C
t t
C m



  
 
Coefficient of volume compressibility mv
6 2
3
0
0.108
10 1.158 m /MN
(1 ) 50 10 (1 0.864)
v
e
m
e


   
   

COEFFICIENT OF PERMEABILITY, K
6 6 3
10
3
(6.49 10 / 60) (1.158 10 ) 9.8 10
12.275 10 /
4.4 10 /
v v w
k c m
k
k m s
k mm hour

 


  
     
 
 

ANALYSIS OF THE LABORATORY MEASUREMENTS FOR
EACH STAGE
Loading stage 1 2 3
Pressure `z[kPa]
50 100 200
Initial void ratio, e0
0.864 0.756 0.701
Initial height, H0 [mm]
18.81 17.715 17.162
Height change, H
1.095 0.553 0.569
Void ratio change, e
0.108 0.055 0.056
Final void ratio, ef
0.756 0.701 0.645
t90 [min] 11.56
Cv [m2
/in]
6.4910-6

An 8 ft clay layer beneath a building is overlain by
stratum of permeable sand and gravel and is underlain
by impermeable bedrock. The total expected total
settlement for the clay layer due to the footing load is
2.5 in. the coefficient of consolidation (cv) is 2.68 10-3
in2
/min. How many years it will take for 90 % of total
expected consolidation settlement to take place.
Compute the amount of consolidation settlement that
will occur in one year? How many years will it take for
consolidation settlement of one inch to take place?
EXAMPLE-2

EXAMPLE-2
Sand and gravel
Impermeable bedrock
Clay 8.0 ft
St = 2.5 in
cv = 2.68 10-3
in2
/min
tyears = ? for U = 90%
S 1 year = ? for t = 1 year
t = ? for S = 1 in

v
dr
v
dr
v
v
C
T
t
t
C
T
H
H
2
2


years
55
.
5
min
10
9168
.
2
10
68
.
2
)
96
)(
848
.
0
(
6
3
2




 
t
t
t
15
.
0
)
96
(
)
60
24
365
1
(
10
68
.
2
2
3
2








v
v
dr
v
v
T
T
t
C
T
H
For TV = 0.15, U = 43 %
in
075
.
1
)
5
.
2
(
43
.
0
)
(








s
s
s
U
s
s
s
U
in
075
.
1
)
5
.
2
(
43
.
0
)
(








s
s
s
U
s
s
s
U
EXAMPLE-2

EXAMPLE-2
%
40
4
.
0
5
.
2
1





U
U
s
s
U
For U = 40 %, TV = 0.126 years
82
.
0
min
10
33
.
2
10
68
.
2
)
96
(
126
.
0
5
3
2
2







t
t
t
C
T
t
v
dr
v H

EXAMPLE-3
A foundation is to be constructed on a site
where the soil profile is as shown in the figure,
the coefficient of consolidation CV = 4.9610-6
m2
/min. How long it will take for half the
expected consolidation settlement to take place
if the clay layer is underlain by (a) permeable
sand and gravel? (b) Impermeable bedrock?

EXAMPLE-3
Elev. 185.6 m
Elev. 192 m
Elev. 198 m
Elev. 200 m
Water table
Sand and gravel
Unit weight = 19.83.0 kN/m3
Clay
Unit weight = 17.10 kN/m3
Water table
Elev. 195.5 m

EXAMPLE
Elev. 185 m
Elev. 190 m
Elev. 198 m
Elev. 200 m
Water table
Sand and gravel
Unit weight = 20 kN/m3
Clay
Unit weight = 17 kN/m3
Water table
Elev. 195 m

EXAMPLE
CV = 4.9610-6
m2
/min
Hdr = H/2 = 6.4/2 = 3.2 m (two way drainage)
t =?
U = 50 %
For U = 50 %, Tv = 0.196
years
77
.
0
min
10
046
.
4
10
96
.
4
)
2
.
3
(
196
.
0
5
6
2
2







t
t
t
C
T
t
v
dr
v H

FIELD CONSOLIDATION LINE
 The modified curve of the logarithm of vertical effective stress
versus void ratio (e- log`) is called the field consolidation
line. There are two methods for determining the field
consolidation line, one for normally consolidated clay, and the
other for over consolidated clay.
 In the case of normally consolidated clay, determination of the
field consolidation line is fairly simple. However for over
consolidated clay, finding the field consolidation line is
somewhat difficult.

In the case of normally consolidated clay, with the
given (e- log`) curve develop from the laboratory
test, the point on the (e- log`) curve
corresponding to 0.4 eo is determined let point f. a
straight line connecting points a (point a is the
point designated by a pressure of `0 and void
ratio of eo) and f gives the field consolidation line
for normally consolidated clay.
NORMALLY CONSOLIDATED CLAY

Normally consolidated clay
e
log
`

a
e
eo
`o
log
`

a
e
eo
`o
0.4 eo
log
`
f

a
e
eo
`o
0.4 eo
log
`
f
af is called field
consolidation line

COMPRESSION INDEX
a
e
e1
log `1
e2
log
`
f
log `2
The slope of the field consolidation line is called compression index
1
2
1
2
`
log
`
log 
 


e
e
Cc

COMPRESSION INDEX







 











 
















0
0
0
0
1
2
1
2
1
2
)
(
log
)
(
log
log
log
log










c
c
C
e
e
e
e
e
C

EXAMPLE-4
When the total pressure acting at midheight of a
consolidating clay layer is 200 kN/m2
, the
corresponding void ratio of the clay is 0.98. When the
total pressure acting at the same location is 500
kN/m2
, the corresponding void ratio decreases to 0.81.
Determine the void ratio of the clay if the total pressure
acting at midheight of the consolidating clay layer is
1000 kN/m2
.

MAXIMUM PAST PRESSURE
e
` (log scale)

g
e
` (log scale)
 Point of maximum curvature

g
h
i
e
` (log scale)
Horizontal line
Tangent line

MAXIMUM OVERBURDEN PRESSURE
g
h
j
e
` (log scale)
i
Bisector line

MAXIMUM OVERBURDEN PRESSURE
k
g
h
j
e
` (log scale)
i

MAXIMUM PAST OVERBURDEN
PRESSURE
g
h
j
e
` (log scale)
`m
k
i

EXAMPLE-5
611
Elev. 710 ft
Elev. 732 ft
Elev. 752 ft
Elev. 760 ft
Water table
Sand and gravel
3
Sand and gravel
3
Clay
3
The soil profile shown in figure, determine the present
effective overburden pressure at the midheight of
compressible clay layer.

NORMAL AND OVER CONSOLIDATED
CLAY
e e
log ` log `

CLAY
e
eo
e
eo
`o
`o
log ` log `
a
b a

CLAY
a
e
eo
e
eo
`o
`o
log ` log `
b a

CLAY
b a a b
e
eo
e
eo
`o
`o
log ` log `

CLAY
a b
e
eo
e
eo
`o
`o
log ` log `
Normally consolidated clay Over consolidated clay
b a

NORMAL CONSOLIDATION LINE
'
p
ln
N
v 


'
p
ln
1
N
e 




EXAMPLE-6
During a consolidation test, a sample of fully
saturated clay 3 cm thick is consolidated under a
pressure increment of 200 kN/m2
. When
equilibrium is reached, the sample thickness is
reduced to 2.60 cm. The pressure is then
removed and the sample is allowed to expand
and absorb water. The final thickness is observed
as 2.8 cm (ft,) and the final moisture content is
determined as 24.9%. If the specific gravity of the
soil solids is 2.70, find the void ratio of the
sample before and after consolidation. 618

EXAMPLE-8
A soil sample has a compression index of 0.3. If
the void ratio e at a stress of 2940 Ib/ft2
is 0.5,
compute (i) the void ratio if the stress is
increased to 4200 Ib/ft2
, and (ii) the settlement
of a soil stratum 13 ft thick.
620

EXAMPLE-9
A 2.5 cm thick sample of clay was taken from the
field for predicting the time of settlement for a
proposed building which exerts a uniform
pressure of 100 kN/m2 over the clay stratum.
The sample was loaded to 100 kN/m2 and
proper drainage was allowed from top and
bottom. It was seen that 50 percent of the total
settlement occurred in 3 minutes. Find the time
required for 50 percent of the
621

EXAMPLE-9
total settlement of the building, if it is to be
constructed on a 6 m thick layer of clay which
extends from the ground surface and is underlain
by sand.
622

EXAMPLE-10
A laboratory sample of clay 2 cm thick took 15
min to attain 60 percent consolidation under a
double drainage condition. What time will be
required to attain the same degree of
consolidation for a clay layer 3 m thick under the
foundation of a building for a similar loading and
drainage condition?
623

EXERCISE-1
The soil profile at a site for a proposed office
building consists of a layer of fine sand 10.4 m
thick above a layer of soft normally consolidated
clay 2.0 m thick. Below the soft clay there is a
deposit of coarse sand. The ground water table
was observed at 1.0 m below the ground level.
The void ratio of the sand is 0.75 and the water
content of the clay is 45 %.
624

EXERCISE-1 (CONTINUE)
The building will impose a vertical stress increase
of 150 kPa at the mid height of the clay layer.
Estimate the primary consolidation settlement of
the clay. Assume the soil above the water table to
be saturated. Take Cc = 0.45 and Gs = 2.67.
625

STRESS DISTRIBUTION IN SOIL
627

STRESS DISTRIBUTION IN SOIL
628

WESTERGAARD EQUATION
629
2
/
3
2
2
)
/
(
)
2
2
(
)
2
1
(
2
)
2
2
(
)
2
1
(













z
r
z
Q







630
where
 = Vertical stress increment at depth z
Q = Concentrated load
 = Poisson’s ratio (ratio of lateral strain to axial
stress in a material)
z = depth

If Poisson’s ratio is zero
631
 
  2
/
3
2
2
2
/
3
2
2
/
3
2
2
/
3
2
2
)
/
(
2
1
)
/
(
2
1
2
1
2
2
1
)
/
(
2
1
2
2
1
z
r
z
Q
z
r
z
Q
z
r
z
Q


























BOUSSINESQ EQUATION
632
  2
/
5
2
2
)
/
(
1
2
3
z
r
z
Q
q



This equation also gives stress q as a function of both the
vertical distance z and horizontal distance r. for low r/z
ratio; the Boussinesq equation gives higher values of q
than the Westergaard equation. The Boussinesq equation is
more widely used. Boussinesq’s equation considers a point
load on a semi-infinite, homogeneous, isotropic,
weightless, elastic half-space.

EXAMPLE
A concentrated load of 1000 kN is applied at the
ground surface. Compute the vertical pressure (i)
at a depth of 4 m below the load, (ii) at a
distance of 3 m at the same depth. Use
Boussinesq's equation.
633

EXAMPLE
634
  2
/
5
2
2
)
/
(
1
2
3
z
r
z
Q





Q = 1000 kN
z = 4 m
(a)
r = 0
 =?
(b)
r = 3 m
 =?

EXAMPLE
A concentrated load of 250 tons is applied to the
ground surface. Determine the vertical stress
increment due to this load at a depth of 20 ft:
(a) directly below the ground surface and
(b) 16 ft from the line of the concentrated load
using Westergaard and Boussinesq equations.
635

EXAMPLE
Data
Q = 250 tons
z = 20 ft
(a)
r = 0
 =?
(b)
r = 16 ft
 =?
636
Westergaard method
  2
/
3
2
2
)
/
(
2
1 z
r
z
Q





Boussinesq method
  2
/
5
2
2
)
/
(
1
2
3
z
r
z
Q
q




VERTICAL PRESSURE BELOW A UNIFORM LOAD
Analysis of stress distribution resulting from
uniform loaded surface area is generally more
complicated than those resulting from
concentrated loads. Two methods are discussed
here
1) Approximate method
2) Method based on Elastic theory
637

APPROXIMATE METHOD
The approximate method based upon the
assumption that the area of stress below a
concentrated load increases with depth with the
slope 2:1. Accordingly, stress at a depth z is given
by :
638
1
2 2
1

Vertical pressure below a loaded surface area (uniform load)

APPROXIMATE METHOD
639
)
)(
( z
L
z
B
Q





EXAMPLE-1
A 10-ft by 15-ft rectangular area carrying a
uniform load of 5000 lb/ft2
is applied to the
increment due to this load at a depth of 20 ft
below the ground surface by the approximate
method.
640

EXAMPLE-1
Data
Q = 5000  (10 15) = 750,000 lb
B = 10 ft
L = 15 ft
z =20 ft
 = q = ?
641
)
)(
( z
L
z
B
Q





EXAMPLE-2
Determine vertical soil pressure using 2:1
method:
Given:-
Footing: 8 feet x 4 feet rectangular footing
Column load = 25 kips
Requirement:-
Determine vertical soil pressure at 6’ below
bottom of footing
642

SOLUTION
644
2
kips/ft
178
.
0
)
6
4
)(
6
8
(
25
)
)(
(













z
L
z
B
Q

EXERCISE-3
A concentrated load of 45000 Ib acts at
foundation level at a depth of 6.56 ft below
ground surface.
Find the vertical stress along the axis of the load
at a depth of 32.8 ft and at a radial distance of
16.4 ft at the same depth by (a) Boussinesq, and
(b) Westergaard formulae for  = 0. Neglect the
depth of the foundation.
645

EXERCISE-4
A rectangular raft of size 30 x 12 m founded at a
depth of 2.5 m below the ground surface is subjected
to a uniform pressure of 150 kPa. Assume the center
of the area is the origin of coordinates (0, 0). and the
corners have coordinates (6, 15). Calculate stresses
at a depth of 20 m below the foundation level by the
methods of (a) Boussinesq, and (b) Westergaard at
coordinates of (0, 0), (0, 15), (6, 0) (6, 15) and (10,
25). Also determine the ratios of the stresses as
obtained by the two methods. Neglect the effect of
foundation depth on the stresses.
646

STRIP LOADS
The state of stress encountered in this case also is
that of a plane strain condition. Such conditions are
found for structures extended very much in one
direction, such as strip and wall foundations,
foundations of retaining walls, embankments, dams
and the like. For such structures the distribution of
stresses in any section (except for the end portions
of 2 to 3 times the widths of the structures from its
end) will be the same as in the neighbouring
sections, provided that the load does not change in
directions perpendicular to the plane considered.
650

METHOD BASED ON ELASTIC THEORY
651

653
I
z
Q
R
Qz
z
r
z
Q
z
2
5
3
2
/
5
2
2 2
3
)
/
(
1
1
2
3



































 2
/
1
2
2
2
2
2
/
5
2
2
2
)
(
2
1
)
(
3
2 z
r
z
z
r
z
r
z
r
Q
r



  















 2
/
1
2
2
2
2
2
/
3
2
2
)
(
1
)
(
2
1
2 z
r
z
z
r
z
r
z
Q











 2
/
5
2
2
2
)
(
2
3
z
r
rz
Q
rz



654
 
2
/
5
2
/
1
1
2
3










z
r
I

)
0
(
2
3
)
/
(
1
1
2
3
2
2
/
5
2
2












 r
z
Q
z
r
z
Q
z




EXAMPLE
A 20-ft by 30-ft rectangular foundation carrying a
uniform load of 6000 lb/ft2
is applied to the
increment due to this uniform load at a depth of
20 ft below the center of the loaded area.
656

EXAMPLE
657
A
10 ft
10 ft
15 ft 15 ft

INFLUENCE FACTORS
658
Influence factors for vertical stress increase due to a point
load (Craig, 1997)

r/z I r/z I r/z I
0.0 0.478 0.8 0.139 1.6 0.020
0.1 0.466 0.9 0.108 1.7 0.016
0.2 0.433 1.0 0.084 1.8 0.013
0.3 0.385 1.1 0.066 1.9 0.011
0.4 0.329 1.2 0.051 2.0 0.009
0.5 0.273 1.3 0.040 2.2 0.006
0.6 0.221 1.4 0.032 2.4 0.004
0.7 0.176 1.5 0.025 2.6 0.003
659

INFLUENCE FACTORS
660
Influence coefficients for points under uniformly loaded circular area

INFLUENCE FACTORS
661
Influence coefficients for points under uniformly loaded rectangular
areas

EXAMPLE-1
A rectangular footing as shown in Figure exerts a
uniform pressure of 500 kN/m2
. Determine the
vertical stress at the center of the footing for a
depth of 2 m.
663

EXAMPLE-2
A rectangular L-shape footing as shown in Figure,
exerts a uniform pressure of 500 kN/m2
.
Determine the vertical stress at point A for a
depth of 2 m. Use table for influence factor.
665

EXAMPLE-3
A water tank is supported by a ring foundation
having outer diameter of 8m and inner diameter
of 6m. The uniform load intensity on the
foundation is 200 kN/m2
. compute the vertical
stress caused by the water tank at a depth 4 m
below the centre of the foundation
667

EXAMPLE-7
A column of a building carries a load of 1000
kips. The load is transferred to sub soil through a
square footing of size 16 x 16 ft founded at a
depth of 6.5 ft below ground level. The soil below
the footing is fine sand up to a depth of 16.5 ft
and below this is a soft compressible clay of
thickness 16 ft. The water table is found at a
depth of 6.5 ft below the base of the footing.
668

EXAMPLE-7
The specific gravities of the solid particles of
sand and clay are 2.64 and 2.72 and their
natural moisture contents are 25 and 40 percent
respectively. The sand above the water table may
be assumed to remain saturated. If the plastic
limit and the plasticity index of the clay are 30
and 40 percent respectively, estimate the
probable settlement of the footing.
669

1

1

3

3

3

3
STRESSES IN SOIL

STRESSES IN SOIL
pressure 1
pressure 1
zero
pressure
zero
pressure
pressure 1
pressure 1

STRESSES IN SOIL
1

1
direct stress
1
shear
stress
 = 0

1



STRESSES IN SOIL
1

1

1





STRESSES IN SOIL
1

1


1





STRESSES IN SOIL
1

1



1



STRESSES IN SOIL
1

1
 = 0
 = 0

1


1

STRESSES IN SOIL
1

1



1





STRESSES IN SOIL
1

1

1





STRESSES IN SOIL
1

1
1


1



STRESSES IN SOIL
1

1
1
 = 0

1



STRESSES IN SOIL
1

1


Mohr’s
Circle of
Stress
1
0
Question: What
is the stress at
this point?
Answer: This
circle

STRESSES IN SOIL
pressure 1
pressure 1
pressure
3
pressure
3
1
1
3
3

STRESSES IN SOIL
1

1
direct stress
 = 1
shear
stress
 = 0

1


3
3

3
3

STRESSES IN SOIL
1

1

1





3
3
3

3

STRESSES IN SOIL
1

1


1





3
3
3

3

STRESSES IN SOIL
1

1



1


3

3
3

3

STRESSES IN SOIL
1

1

1




3

3
3

3

STRESSES IN SOIL
1

1
3
 = 0

1

1


3 
3
3

3

STRESSES IN SOIL
1

1



1





3
3
3

3

STRESSES IN SOIL
1

1



1





3

3
3

3

STRESSES IN SOIL
1

1

1





3
3
3

3

STRESSES IN SOIL
1

1
3


1



3
3
3

3

STRESSES IN SOIL
1

1
3


1



1

3

3
3

3 
3

STRESSES IN SOIL
1

1

Mohr’s
Circle of
Stress


1
3

3 
3
Question: What
is the stress at
this point?
Answer: This
circle

STRESSES IN SOIL
1

1

Mohr’s
Circle of
Stress


1
1

1

3

3

3

3
3

3 
3

SOIL WITH COHESION AND FRICTION

 = c + 
tan 
Mohr’s Circle of
Stress
c
c

Soil fails when
Mohr’s circle
touches these
lines

1
3

1

c

3



(C,) AND (3, 1) RELATIONSHIP

c

1

A E
F
3



B C D

c

1

A E
F
3



B C D
ccot (

1
-

3
)
/
2
(1-3)





















cos
sin
2
2
sin
cot
sin
2
2
2
/
)
(
cot
2
/
)
(
sin
sin
3
1
3
1
3
1
3
1
3
1
3
1
c
c
c
BD
AB
CD



















































sin
1
sin
1
sin
1
sin
1
2
sin
1
sin
1
sin
1
sin
1
2
sin
1
sin
1
sin
1
cos
2
)
sin
1
(
cos
2
)
sin
1
(
cos
2
)
sin
1
(
)
sin
1
(
cos
2
sin
sin
3
1
3
2
1
3
1
3
1
3
1
3
1
3
1





























c
c
c
c
c
c





N





)
2
/
45
(
tan
sin
1
sin
1 0
2






sin
1
sin
1
sin
1
sin
1
2 3
1





 c

 
 N
N
C 3
1 2 


COEFFICIENT OF LATERAL EARTH PRESSURE,
K0






sin
1
sin
1
sin
1
sin
1
2 3
1





 c
For cohesionless soils, c = 0











N
K
1
sin
1
sin
1
sin
1
sin
1
sin
1
sin
1
0
1
3
3
1










And
3
1
3
1
sin









O
O
C
A (x, xy)
B(y, yx)
3

1
2

1
1
3
3
3
n



n
c D
O
E
F
G
O
f
F`
H
max

RELATION B/W  AND 
G
F
C
 2
90
0
(180
0
-2)
 + 900
+ 1800
- 2 = 1800
2 = 900
+ 
q = 450
+ /2
q = (/4 + /2)

Maximum shear stress max
2
2
2
2
3
1
max
3
1












R
OE
OD
ED
R

SHEAR STRESS AT FAILURE F
F`
F
C
1 3
`
sin `
sin(180 2 )
sin 2
sin 2
sin 2
2
f
f
f
f
FF
FCF
CF
R
R
R




 
 
 
 
 




2

NORMAL STRESS N
1 3 1 3
` `
cos(180 2 )
cos2
2 2
n
n
n
OF OC F C
OC FC

 
   
 
  
  
 
 
 c
F(n, f)
max

POLE POINT
The pole on Mohr’s circle identifies a point
through which any plane passing will
intersect the Mohr’s circle at a point that
represents the stresses on that plane. The
pole point is determined from drawing a line
parallel to the face of the element, which
the stresses are acting.

POLE POINT
O
C
A(y ,y
x)
B
(x ,xy)

3

1


D
E

POLE POINT
O
C
A(y ,y
x)
B
(x ,xy)

3

1


D
E
Pole

POLE POINT
O
C
A(y ,y
x)
B
(x ,xy)

3

1


D
E
ma
x Pole

POLE POINT
O
C
A(y ,y
x)
B
(x ,xy)

3

1


D
E
ma
x Pole
A`
B`

RADIUS OF CIRCLE
From the geometry of the figure
` `
`
2 2
y x
A B
CA
 

 
AA` = yx
The radius of the circle can be determined as:
2 2
2
2
` `
2
y x
xy
R AC
R CA AA
R
 


 

 
 
 
 

CENTRE OF CIRCLE
The location of the centre of Mohr circle in terms
of stress components can be located as:
` `
2
2
y x
x
y x
OC OB B C
OC
OC
 

 
 

 



MAXIMUM SHEAR
Therefore the maximum shear stress can also be
expressed in terms of stress components as:
max
2
2
max
2
y x
xy
R

 
 


 
 
 
 

MAJOR AND MINOR PRINCIPAL
STRESSES
1
2
2
1
2 2
y x y x
xy
OD OC CD

   
 
  
 
 
  
 
 
3
2
2
3
2 2
y x y x
xy
OE OC EC

   
 
  
 
 
  
 
 

731
100 150 200
-50
0
50
A (100, -45)
B (200, 45)

max

max

1

2
 (MPa)

(MPa)

SHEAR STRENGTH OF SOIL
Shear strength is a term used in soil mechanics
to describe the magnitude of the shear stress
that a soil can sustain. The shear resistance of
soil is a result of friction and interlocking of
particles, and possibly cementation or bonding at
particle contacts.
733

STRENGTH THEORIES FOR SOILS
A number of theories have been proposed for
explaining the shearing strength of soils. Of all
such theories, the Mohr’s strength theory and the
Mohr-Coulomb theory, a generalisation and
modification of the Coulomb’s equation, meet the
requirements for application to a soil in an
admirable manner.
736

LABORATORY TESTS FOR DETERMINATION OF SHEAR
STRENGTH PARAMETERS
1) Direct shear test
2) Triaxial test
3) Simple shear test
4) Plane strain triaxial test
5) Torsional ring shear test
6) Unconfined Compression Test
7) Laboratory Vane Shear Test
737

Motor
drive
Load cell to
measure
Shear Force
Normal load
Rollers
Soil
Porous plates
Top platen
Measure relative horizontal displacement, dx
vertical displacement of top platen, dy
SHEAR BOX TEST

The limiting shear stress (soil strength) is given by
t = c + sn tan f
where c = cohesion (apparent)
f = friction angle
t
sn
MOHR-COULOMB FAILURE
CRITERION

NORMAL AND SHEAR STRESS
741
n

1 1
2 2
3 3
4 4
5 5

742
t
sn
SHEAR PARAMETERS (C, )

743
t
sn

744
t
sn

745
t
sn

746
t
c
  
 
c tan
f
sn

MAXIMUM SHEAR AND SHEAR AT FAILURE (MAX , F)

748
t   
 
c tan
f
sn

749
t   
 
c tan
f
sn

750
t   
 
c tan
f
sn
m
ax
f

n

EXAMPLE
The stresses at failure on the failure plane in a
cohesionless soil mass were:
Shear stress = 4 kN/m2
; normal stress = 10
kN/m2
. Determine the resultant stress on the
failure plane, the angle of internal friction of the
soil and the angle of inclination of the failure
plane to the major principal plane.
751

GRAPHICAL SOLUTION
The procedure is first to draw the σ-and τ-axes
from an origin O and then, to a suitable scale,
set-off point D with coordinates (10,4), Joining O
to D, the strength envelope is got. The Mohr
Circle should be tangential to OD to D. DC is
drawn perpendicular to OD to cut OX in C, which
is the centre of the circle. With C as the centre
and CD as radius, the circle is completed to cut
OX in A and B.
752

EXERCISE
The following results were obtained in a shear
box text. Determine the angle of shearing
resistance and cohesion intercept:
Normal stress (kN/m2
) 100 200 300
Shear stress (kN/m2
) 130 185 240
754

EXAMPLE
Clean and dry sand samples were tested in a
large shear box, 25 cm × 25 cm and the following
results were obtained :
Normal load (kN) 5 10 15
Peak shear load (kN) 5 10 15
Ultimate shear load (kN) 2.9 5.8 8.7
Determine the angle of shearing resistance of the
sand in the dense and loose states.
756

SOLUTION
The value of  obtained from the peak stress
represents the angle of shearing resistance of the
sand in its initial compacted state; that from the
ultimate stress corresponds to the sand when
loosened by the shearing action.
The area of the shear box = 25 × 25 = 625 cm2
.
= 0.0625 m2
.
Normal stress in the first test = 5/0.0625 kN/m2
=
80 kN/m2
757

SOLUTION
Similarly the other normal stresses and shear
stresses are obtained by dividing by the area of
the box and are as follows in kN/m2
:
Normal stress, σ 80 160 240
Peak shear stress, max 80 160
240
Ultimate shear stress, f 46.4 92.8 139.2
758

761
y = 0.5706x
R² = 0.8253
0
10
20
30
40
50
0 10 20 30 40 50
Shear
Stress

(kN/m
2
)
Normal Stress n (kN/m2)
BH7-SPT04
 = 29.70
C = 0
PRACTICAL WORKED EXAMPLE

762
Borehole
No.
Sample
No.
Depth
(m)
Strength
parameters
Bulk
Density
(
)
(kN/m³)
Moisture
Content
(

)
(%
)
Dry
Density
(

d)
(kN/m
3
)
c
(kPa)

(degree)
BH14 SPT2 2.0 0.0 17.3 19.2 13.1 17.0
BH14 SPT3 3.0 1.7 22.9 19.2 34.3 14.3
BH14 SPT6 6.0 3.9 26.4 19.2 37.0 14.0
PRACTICAL WORKED EXAMPLE

EXAMPLE
The following data were obtained in a direct
shear test. Normal pressure = 20 kN/m2
,
tangential pressure = 16 kN/m2
. Angle of internal
friction = 20°, cohesion = 8 kN/m2
.
Represent the data by Mohr’s Circle and compute
the principal stresses and the direction of the
principal planes.
763

UNCONFINED COMPRESSION TEST
765

766

767
0
1
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
Stress
(N/mm2)
0 10
1 2 3 4 5 6 7 8 9
Stroke Strain (%)
Max

TRIAXIAL TEST
Cell
pressure
Pore pressure
and volume
change
Rubber membrane
Cell water
O-ring seals
Porous filter disc
Confining cylinder
Deviator load
Soil

sr sr = Radial stress (cell
pressure)
sa = Axial stress
F = Deviator load
sr
STRESSES IN TRIAXIAL SPECIMEN

sr sr = Radial stress (cell
pressure)
sa = Axial stress
F = Deviator load
sr
 
a r
F
A
 
From equilibrium we have
STRESSES IN TRIAXIAL SPECIMEN

EXAMPLE-1
A series of shear tests were performed on a soil.
Each test was carried out until the sample
sheared and the principal stresses for each test
were :
Test No. (kN/m2
) (kN/m2
)
1 200 600
2 300 900
3 400 1200
773

EXAMPLE-2
From a direct shear test on an undisturbed soil
sample, the following data have been obtained.
Evaluate the undrained shear strength
parameters. Determine shear strength, major
and minor principal stresses and their planes in
the case of specimen of same soil sample
subjected to a normal stress of 100 kN/m2
. :
sn (kN/m2
) 70 96 114
t (kN/m2
) 138 156 170
775

EXAMPLE-3
A cylindrical sample of saturated clay 4 cm in
diameter and 8 cm high was tested in an
unconfined compression apparatus. Find the
unconfined compression strength, if the
specimen failed at an axial load of 360 N, when
the axial deformation was 8 mm. Find the shear
strength parameters if the angle made by the
failure plane with the horizontal plane was
recorded as 50°.
777

TRIAXIAL TESTING PROCEDURE
 Flushing
 Saturation ramp
 B-check
 Consolidation
 Shearing

TYPE OF TRIAXIAL TESTS
Depending on the nature of loading and drainage
condition, triaxial tests are conducted in three
different ways
 Unconsolidated Undrained test (UU-test)
 Consolidated Undrained test (CU-test)
 Consolidated Drained test (CD-test)

EXAMPLE
A cylindrical sample of soil having a cohesion of
80 kN/m2
and an angle of internal friction of 20°
is subjected to a cell pressure of 100 kN/m2
.
Determine: (i) the maximum deviator stress
(s1- s3) at which the sample will fail, and (ii) the
angle made by the failure plane with the axis of
the sample.
788

EXAMPLE
A normally consolidated clay was consolidated
under a stress of 3150 lb/ft2
, then sheared
undrained in axial compression. The principal
stress difference at failure was 2100 lb/ft2
, and
the induced pore pressure at failure was 1848
lb/ft2
. Determine (a) the Mohr-Coulomb strength
parameters, in terms of both total and effective
stresses analytically, (b) compute (s1/s3), and
(s’1/s’3), and (c) determine the theoretical angle
of the failure plane in the specimen.
790

EXAMPLE
The following results were obtained at failure in a
series of consolidated-undrained tests, with pore
pressure measurement, on specimens of
saturated clay. Determine the values of the
effective stress parameters c’ and ’ by drawing
Mohr circles.
791

STRESS-STRAIN BEHAVIOUR
A graph that describes the relationship between
stress and strain. Stress-strain graphs indicate
the elastic and plastic regions for a given
material.
Stress- stress in soil/rock is the body's reaction to
a change due to external agencies that requires a
physical adjustment or response. The intensity of
stress is expressed in units of force divided by
units of area.
793
A
F


 
Area
Force
Stress

 Deviator stress- is the difference between the
major and minor principal stresses in a triaxial
test.
 Mean effective stress
794
3
1
q 
 

3
2
p 3
1 
 


795
Strain- is the deformation of a physical body
under the action of applied forces.
is defined as the ratio of the change in dimension
of the soil/rock mass to the original dimension.

796
V
V
A
F
v
r
a



















 ,
,
r
r
,
H
H
dimension
Original
dimension
in
Change
Strain
Area
Force
Stress

LDT
External
a (LDT)
E sec
q
=

1
-
3
q
=

1
-
3
E tan
qmax
a (external)

DILATANCY
804
Vo
lum
etric
strain,
v
(%
)
Dilation
Contraction
Elastic contraction
Pre-peak onset of dilation
Increased dilation rate
Post-peak dilation
Axial strain
Axial strain at peak
strength

YIELDING DURING ISOTROPIC
COMPRESSION

Brittle Ductile
0 10 20 30
0
10000
20000
30000
CD-15C6M
ea (%)
q
(MPa)
0 10 20 30
0
10000
20000
30000
CD-15C8M
ea (%)
q
(MPa)
0 10 20 30
0
10000
20000
30000
CD-10C12M
ea (%)
q
(MPa)
Transitional
FAILURE MODE
807

Brittle Ductile
Transitional
FAILURE MODE
808

Global behaviour Local behaviour
STRAIN LOCALIZATION
809

SHEAR BAND PATTERN
Weakly developed Diagonally crossing
Fully developed
810

1
1
3
3
n
f
s
SHEAR BAND PATTERN

SHEAR BAND PATTERN
Weakly developed Diagonally crossing
Fully developed
813

ILLUSTRATION OF PROGRESSIVE
FAILURE
814
(Klein et al., 2001)

BRITTLENESS INDEX (IB)
0 5 10 15 20
0
1
2
3 C = 0%
Logarithmic
(C = 0%)
C = 5%
s`3 (MPa)
IB
=
(qf/qu-1)
815
Transitional
Brittle
Ductile

STRAIN LOCALIZATION
Portaway
sand
SHEAR BANDING
816
F
A
I
L
U
R
E
g=17.4kN/m3, s’3 = 50 kPa
SUMMARY

FACTORS AFFECTING THE MODE OF FAILURE
1) Initial relative density
2) Confining pressure
3) Type and amount of reinforcement
817

EFFECT OF INITIAL RELATIVE DENSITY
818
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.01 0.1 1 10 100
e
p' (MPa)
e = 0.657
e = 0.584
e = 0.523
Cement = 10%
Isotropic compression of cemented Portaway sand

819

820
-2
-1
0
1
2
0 2 4 6 8 10
e
v
(%)
ea(%)
C = 0%, Dr = 80%
C = 0%, Dr = 55%
C = 0%, Dr = 21%
s3 = 300 kPa
Dilation
Compression

821
0.0
0.4
0.8
1.2
1.6
0 10 20 30
q
(MPa
)
ea (%)
Loose sand
500 kPa
1
2
3
-0.2
0
0.2
0.4
0.6
0 10 20 30
Du
(MPa)
ea (%)
Loose sand
500 kPa
1
2
3
0
0.4
0.8
1.2
1.6
0.0 0.5 1.0 1.5
q
(MPa)
p(kPa)
Loose sand
500 kPa
1
2
3
0
0.4
0.8
1.2
1.6
0 10 20 30
q
(MPa)
ea (%)
Dense sand
500 kPa
1
2
3
4
-0.2
0
0.2
0.4
0.6
0 10 20 30
Du
(MPa)
ea (%)
Dense sand
500 kPa
1
2
3
4
0
0.4
0.8
1.2
1.6
0.0 0.5 1.0 1.5
q
(MPa)
p(kPa)
Dense sand
500 kPa
1
2
3 4

EFFECT OF CONFINING PRESSURE
822
1 MPa 4 MPa
8 MPa 12 MPa
Cement = 5%
1 MPa 4 MPa
8 MPa 12 MPa
Cement = 5%
0 10 20 30
ea(%)
1 MPa 4 MPa
8 MPa 12 MPa
Cement = 5%
1 MPa 4 MPa
8 MPa 12 MPa
Cement = 10%
1 MPa 4 MPa
8 MPa 12 MPa
Cement = 10%
0 10 20 30
ea (%)
1 MPa 4 MPa
8 MPa 12 MPa
Cement = 10%
1 MPa 4 MPa
8 MPa 12 MPa
Cement = 15%
1 MPa 4 MPa
8 MPa 12 MPa
Cement = 15%
0 10 20 30
ea (%)
1 MPa 4 MPa
8 MPa 12 MPa
Cement = 15%
0
10
20
30
40
q
(MPa)
1 MPa 4 MPa
8 MPa 12 MPa
Cement = 0%
0.0
0.5
1.0
1.5
2.0
2.5
q/p'
1 MPa 4 MPa
8 MPa 12 MPa
Cement = 0%
-8.0
-4.0
0.0
4.0
8.0
12.0
0 10 20 30
e
v
(%)
ea (%)
1 MPa 4 MPa
8 MPa 12 MPa
Cement = 0%

EFFECT OF CONFINING PRESSURE
823
-15
-10
-5
0
5
10
15
0 10 20 30 40
e
v
(%)
ea (%)
0.05MPa
0.1MPa
0.3MPa
0.5MPa
1MPa
4MPa
8MPa
12MPa
20MPa
Cement = 0%
gd = 17.4 kN/m3
-15
-10
-5
0
5
10
15
0 10 20 30 40
e
v
(%)
ea (%)
0.05MPa
0.1MPa
0.3MPa
0.5MPa
1MPa
4MPa
8MPa
12MPa
20MPa
Cement = 5%
gd = 17.4 kN/m3
-15
-10
-5
0
5
10
15
0 10 20 30 40
e
v
(%)
ea (%)
0.05MPa
0.5MPa
1MPa
4MPa
8MPa
12MPa
20MPa
Cement = 10%
gd = 17.4 kN/m3
-15
-10
-5
0
5
10
15
0 10 20 30 40
e
v
(%)
ea (%)
1MPa
4MPa
8MPa
12MPa
20MPa
Cement = 15%
gd = 17.4 kN/m3

EFFECT OF CEMENT CONTENT
824
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.01 0.1 1 10 100
e
p' (MPa)
C = 0%
C = 5%
C = 10%
C = 15%
e 0.49

825
0
10
20
30
40
q
(MPa)
C = 0% C = 5%
C = 10% C = 15%
`3 = 1MPa
0.0
0.5
1.0
1.5
2.0
2.5
q/p`
C = 0% C = 5%
C = 10% C = 15%
`3 = 1MPa
-8.0
-4.0
0.0
4.0
8.0
12.0
0 10 20 30
e
v
(%)
ea (%)
C = 0% C = 5%
C = 10% C = 15%
`3 = 1MPa
C = 0% C = 5%
C = 10% C = 15%
`3 = 4MPa
C = 0% C = 5%
C = 10% C = 15%
`3 = 4MPa
0 10 20 30
a (%)
C = 0% C = 5%
C = 10% C = 15%
`3 = 4MPa
C = 0% C = 5%
C = 10% C = 15%
`3 = 8MPa
C = 0% C = 5%
C = 10% C = 15%
`3 = 8MPa
0 10 20 30
ea (%)
C = 0% C = 5%
C = 10% C = 15%
`3 = 8MPa
C = 0% C = 5%
C = 10% C = 15%
`3 = 12MPa
C = 0% C = 5%
C = 10% C = 15%
`3 = 12MPa
0 10 20 30
ea (%)
C = 0% C = 5%
C = 10% C = 15%
`3 = 12MPa

826
0
5
10
15
20
25
30
35
40
45
0 5 10 15 20 25 30 35 40 45
q
f
(MPa)
pf (MPa)
Cement = 0%
Cement = 5%
Cement =10%
Cement =15%
Uncemented
Portawaysand

827
0
1
2
3
4
5
6
7
8
9
10
0 1 2 3 4 5 6 7 8 9 10
q
f
(MPa)
pf (MPa)
Cement content 0%
5%
10%
15%
Uncemented sand

828
0.3
1.1
3.2
7.4
0
2
4
6
8
10
0 5 10 15
C
o
h
e
s
i
o
n
,
c
(
M
P
a
)
Cement content (%)
50.0 50.9 49.5
46.7
30
50
70
90
0 5 10 15
F
r
i
c
t
i
o
n
a
l
a
a
n
g
l
e
,


()
Cement content (%)
0
2
4
6
8
10
0 5 10 15
30
50
70
90
0 5 10 15
0
5
10
15
0 5 10 15
q
f
(
M
P
a
)
p
f (MPa)
a
b
c
d
a- Cement = 0%, c = 268 kPa, = 50
b- Cement = 5%, c = 1080kPa, = 50
c- Cement = 10%, c = 3186kPa, = 50
d- Cement = 15%, c = 7431kPa, = 50

829
0.3
1.1
3.2
7.4
0
2
4
6
8
10
0 5 10 15
C
o
h
e
s
i
o
n
,
c
(
M
P
a
)
Cement content (%)
50.0 50.9 49.5
46.7
30
50
70
90
0 5 10 15
F
r
i
c
t
i
o
n
a
l
a
a
n
g
l
e
,


()
Cement content (%)
0
2
4
6
8
10
0 5 10 15
30
50
70
90
0 5 10 15
0
5
10
15
0 5 10 15
q
f
(
M
P
a
)
p
f (MPa)
a
b
c
d
a- Cement = 0%, c = 268 kPa, = 50
b- Cement = 5%, c = 1080kPa, = 50
c- Cement = 10%, c = 3186kPa, = 50
d- Cement = 15%, c = 7431kPa, = 50

SAND CEMENT PROPORTION
The following is the procedure for the
determination of the void ratio of a
cemented specimen, Determine
 The specific gravity of sand G sand
 The specific gravity of cement G cement
 The dry mass of specimen M dry i.e., the
mass of solid M solid
 Specimen dimensions i.e., height H and
diameter D

AVERAGE SPECIFIC GRAVITY
 The average specific gravity of the specimen G (taking
cement content = C %)
cement
soil G
C
G
C
G 




100
100
100

INITIAL VOID RATIO
The volume of solid (i.e., sand + cement) V solid
The total volume of the specimen V total
The volume of voids V voids
w
solid
solid
G
M
V



2
4
D
VTotal


solid
total
voids V
V
V 


INITIAL VOID RATIO
The initial void ratio of the specimen e
solid
voids
V
V
e 

CONSTANT DRY DENSITY WITH INCREASING CEMENT
CONTENT
The mass of sand M sand and mass of cement M
cement can be calculated as follow:
total
sand M
C
M 







100
100
total
cement M
C
C
M 







100

SAND CEMENT MIXING BUCKET
Once dry sand and cement
mixed thoroughly, water of
required percent must be
added and mixed to get a
uniform and consistent sand-
cement paste.

SAND CEMENT MIXING
The targeted dry unit weight of the material for a
standard size specimen can be maintained by
varying the weight of the moist sample using the
following equation.
The required weight of mixture should be taken for
specimen preparation





1
d

MEMBRANES
Rubber membranes: (a) Latex; (b) Neoprene.

Concrete sand
Gs = 2.65
D50 = 0.35
Cu= 2.2
emax = 0.79
emin = 0.46
Polypropylene fibre
Length= 22mm
Diameter = 0.023mm
Gs = 0.9
Elastic modulus = 8 GPa
Ordinary Portland cement
TS EN 197-1-CEM 1
Gs = 3.15
MATERIALS

SPECIMEN PREPARATION ACCESSORIES

CEMENTED SPECIMEN PREPARATION
Grine, K. and Glendinning, S. (2007) Geotech Geol Eng 25, 441–448
Grine & Glendinning (2007)

Ismail, M. A., Joer, H. A., and Randolph, M. F., (2000) Sample Preparation Technique for Artificially
Cemented Soils,” Geotechnical Testing Journal, 23( 2),, pp. 171–177.
Ismail, et al (2000)

Ladd, R. S. (1978) Preparing test specimens using undercompaction," Geotechnical Testing Journal, 1(1),
16-23.
Ladd, (1978)

Bradshaw, A. S. and Baxter, C. D. P. (2007) Sample preparation of silts for liquefaction testing,
Geotechnical Testing Journal, 30(4), 1-9.
Bradshaw and Baxter (2007)

UNCEMENTED SPECIMEN
PREPARATION

Pore pressure
control
Back pressure /
volume control
Porous stones
Top cap
Top drainage
valve
Bottom drainage
valve
Cell chamber
Load cell
Pore pressure valve
Spacers
Computer
Cell pressure /
volume control
Displacement
control
Specimen
Cross beam

FINAL YEAR PROJECT LIST
1. Development of Soil Map for Karachi city.
2. Use of Geosynthetics as reinforced subgrade
materials.
3. Use of Geosynthetics as drainage and filter
materials.
4. Groundwater and Seepage modelling using
DrainMod
5. Mechanical behaviour of reinforced soil using
Matlab
862

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