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Dr. KY Sambath
1
Soil Mechanic
Dr. KY Sambath
Lecturer and researcher, Institute of Technology of Cambodia
Ph.D of Civil Engineering, Master CE (INSA Rennes, France)
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1. Classification and Physical Properties of Soils
 In the field of civil engineering, nearly all projects are built on to, or
into, the ground. Whether the project is a structure, a roadway, a
tunnel, or a bridge, the nature of the soil at that location is of great
importance to the civil engineer.
Civil engineering
Structural engineering
 Soil mechanics is the subject within a branch of Geotechnical
engineering that looks at the behavior of soils in civil engineering.
 Geotechnical engineering is the term given to the branch of engineering
that is concerned with aspects pertaining to the ground.
Soil mechanics
Geotechnical engineering
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1.1 Agricultural and engineering soil
 Topsoil : A layer of organic soil, usually not more than 500 mm thick,
in which humus (highly organic partly decomposed vegetable matter)
is often found.
 Hardpan : In humid climates humid acid can be formed by
rainwater causing decomposition of humus. This acid leaches out
iron and alumina oxides down into the lower layers where they act
as cementation agents to form a hard, rock-like material. Hardpan
is difficult to excavate and, as it does not soften when wet, has a
high resistance to normal soil drilling methods. A hardpan layer is
sometimes found at the junction of the topsoil and the subsoil.
 Subsoil : The portion of the Earth’s crust affected by current
weathering, and lying between the topsoil and the unweathered soil
below.
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1.1 Agricultural and engineering soil
 Soil : The soft geological deposits extending from the subsoil to
bedrock. In some soils there is a certain amount of cementation
between the grains which affects the physical properties of the soil. If
this cementation is such that a rock-hard material has been produced,
then the material must be described as rock. A rough rule is that if the
material can be excavated by hand or hand tools, then it is a soil.
 Groundwater : A reservoir of underground water. The upper surface
of this water may occur at any depth and is known as the water table
or groundwater level (GWL).
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1.2 Engineering definitions
Rock are made from various types of minerals. Minerals are substances
of crystalline form made up from a particular chemical combination. The
main minerals found in rocks include quartz, feldspar, calcite and mica.
Geologists classify all rocks into three basic groups:
1. Igneous rocks
2. Sedimentary rocks
3. Metamorphic rocks
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1.2 Engineering definitions
Soil : The actions of frost, temperature, gravity, wind, rain and chemical
weathering are continually forming rock particles that eventually become
soils. There are 3 types of soil when considering modes of formation.
1. Transported soil (gravels, sands, silts and clays)
2. Residual soil (topsoil, laterites)
3. Organic soil
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1.2 Engineering definitions
 Geotechnical engineers classify soils as either granular or cohesive.
• Granular soils (sometimes referred to as cohesionless soils) are
formed from loose particles without strong inter-particle forces, e.g.
sands and gravels.
• Cohesive soils (e.g. clays, clayey silts) are made from particles
bound together with clay minerals. The particles are flaky and sheetlike and retain a significant amount of adsorbed water on their
surfaces. The ability of the sheet-like particles to slide relative to one
another, gives a cohesive soil the property known as plasticity.
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1.3 Clay soils
 It is generally believed that rock fragments can be reduced by
mechanical means to a limiting size of about 0.002 mm, so that a soil
containing particles above this size has a mineral content similar to the
parent rock from which it was created.
 For the production of particles smaller than 0.002 mm some form of
The three main groups of clay minerals are:
1. Kaolinite
2. Illite
3. Montmorillonite
chemical action is generally necessary before breakdown can be
achieved. Such particles, although having a chemical content similar to
the parent rock, have a different crystalline structure and are known as
clay particles.
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1.3 Clay soils
The three main groups of clay minerals are:
1. Kaolinite
Kaolinite
2. Illite
3. Montmorillonite
Illite
Montmorillonite
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1.4 Field identification of soils
 Gravels, sands and peats are easily recognizable, but difficulty arises
in deciding when a soil is a fine sand or a coarse silt or when it is a
fine silt or a clay.
 Organic silts and clays are invariably dark grey to blue-black in color
and give off a characteristic odor, particularly with fresh samples.
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1.5 Laboratory classification of soils, water content
 Drying soils: Soils can be either oven or air dried. It has become
standard practice to oven dry soils at a temperature of 105°C but it
should be remembered that some soils can be damaged by such a
temperature.
 The water content, also called the moisture content, is given the
symbol w and is the ratio of the amount of water to the amount of dry
soil.
Example: A sample of soil was placed in a water
content tin of mass 19.52 g. The combined mass
of the soil and the tin was 48.27 g. After oven
drying the soil and the tin had a mass of 42.31 g.
Determine the water content of the soil.
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1.5 Laboratory classification of soils, Particle size distribution
 Granular soils- Particle size distribution
The usual method in the classification of soils is based on the
determination of the particle size distribution by shaking an oven dried
sample of the soil (usually after washing the sample over a 63μm sieve)
through a set of sieves and recording the mass retained on each sieve.
 The classification system of each country are differences!
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1.5 Laboratory classification of soils, Particle size distribution
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1.5 Laboratory classification of soils, Particle size distribution
 Four systems for describing oils based on particle size
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1.5 Laboratory classification of soils, Particle size distribution
 Classification des Grains Solides: Système Français
Argiles
Limons (ou silts)
0.002 mm
Sables fins
0.02 mm
Sables grossiers
0.2 mm
Graviers
2 mm
Cailloux
20 mm
Blocs rocheux
200 mm
les sols pulvérulents
Forces capillaires
1
est proportionnel à
pesanteur
D
=> D est relativement grand dans le cas des sols pulvérulents
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1.5 Laboratory classification of soils, Particle size distribution
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Example : The results of a sieve analysis on a
soil sample were:
 The results of the sieve analysis are plotted with the particle sizes
horizontal and the summation percentages vertical. As soil particles
vary in size from molecular to boulder it is necessary to use a log
scale for the horizontal plot so that the full range can be shown:
mass retained cumulative
×100%
total mass
Percentage passing = 100% - Percentage retained
Percentage retained =
at the bottom 2.3 g passed through the 63 μm sieve.
Solution:
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1.5 Laboratory classification of soils, Particle size distribution
1. Effective size of a distribution (D10): This parameter is the
diameter in the particle-size distribution curve corresponding to 10%
Example : The sieve analysis on a soil sample
above:
finer. The effective size of a granular soil is a good measure to
estimate the hydraulic conductivity and drainage through soil.
2. Uniformity coefficient (Cu) (le coefficient d’uniformité): This
parameter is defined as
Cu 
D60
D10
where D60 diameter corresponding to 60% finer.
If Cu < 4.0 then the soil is uniformly graded.
If Cu > 4.0 then the soil is either well graded or gap and a glance
at the grading curve should be sufficient for the reader to decide
which is the correct description.
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1.5 Laboratory classification of soils, Particle size distribution
3. Coefficient of gradation (CC) (le facteur de courbure): This
parameter is defined as
 From the particle-size distribution curve, the type of
distribution of various-size particles can be seem:
D302
Cc 
D60  D10
Ex. Different types of particle-size distribution curves
4. Sorting coefficient (S0): This parameter is defined as
S0 
D75
D 25
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1.5 Laboratory classification of soils, Particle size distribution
1. granulométrie étalée et discontinue (alluvions
de sables et graviers),
2. granulométrie étalée et continue (arène
granitique),
3. granulométrie serrée (sable de Fontainebleau),
4. limon argileux,
5. argile limoneuse,
6. argile pratiquement pure (bentonite),
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1.5 Laboratory classification of soils, Particle size distribution
ASTM D6913: Standard Test Methods for Particle-Size
Distribution (Gradation) of Soils Using Sieve Analysis.
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ASTM D7928 : Standard Test Method for ParticleSize Distribution (Gradation) of Fine-Grained Soils
Using the Sedimentation (Hydrometer) Analysis
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1.5 Laboratory classification of soils, Consistency
 Cohesive soils: Consistency limit (or index) test
Albert Mauritz Atterberg,
Sweden (1846 – 1916)
Clay minerals can be remolded with some moisture. This cohesive
nature is caused by the adsorbed water surrounding the clay particles.
 A Swedish scientist named Atterberg developed a method to describe
the consistency of fine-grained soils with varying moisture contents.
 The behavior of soil can be divided into
four basic states
solid, semisolid, plastic, and liquid
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1.5 Laboratory classification of soils, Consistency
 Liquid limit (wL or LL) (la limite de liquidité) is the water content
when the soil stops acting as a liquid.
 It is obtained by experimental : by a brass cup and a hard rubber
base.
place a soil paste in the cup, use
a groove cut the center of the
soil pat with the standard
grooving tool.
The cup is lifted and dropped with h = 10 mm.
The Liquid limit is defined at 25 blows the
distance of 12.5 mm of the 2 sides soil pat is
closed.
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1.5 Laboratory classification of soils, Consistency
 It is hard to make a test and obtain the Liquid limit (wL or LL). Thus,
ASTM required 5 test from 15 to 35 blows.
IF 
w1  w2
N
log( 2 )
N1
where: IF is flow index; w1 moisture content of
soil, in percent, corresponding to N1 blows and w2
moisture content corresponding to N2 blows.
w   I F log N  C
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1.5 Laboratory classification of soils, Consistency
 Another method of determining liquid limit by the fall cone method.
 The LL is defined when penetrate a distance d = 20 mm in 5 seconds.
IF 
w2  w1
d
log( 2 )
d1
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1.5 Laboratory classification of soils, Consistency
 Plastic Limit (la limite de plasticité) : the limit at which plastic failure
changes to brittle failure (when the soil crumbles) is known as the
 The fall cone method can be used to find the
PL of the soil by changing the cone weight.
plastic limit (wP or PL).
The plastic limit test is simple and is performed by repeated rolling of
an ellipsoidal-sized soil mass by hand on a ground glass plate. (Check
ASTM D-4318)
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1.5 Laboratory classification of soils, Consistency
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1.5 Laboratory classification of soils, Consistency
 The plasticity index (PI) (Indice de plasticité) is defined as:
PI  LL  PL
 Liquidity index of soil (LI) (l’indice de
liquidité) is defined as:
LI 
w  PL
LL  PL
where w is in situ moisture content of soil
• classified by Burmister (1949)
 Consistency index of soil (CI) (l’indice de
consistance) is defined as:
CI 
LL  w
LL  PL
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1.5 Laboratory classification of soils, Consistency
 The Shrinkage Limit (SL) is water content that give the minimum
volume of the soil
 Shrinkage limit test, ASTM D-4943, describes a
method where volume Vi is determined by
filling the wax.
V

 w (%) 
i
SL 
 V f  w
M2
 100
 M 1  M 2   100  Vi  V f   w  100
M2
M2
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1.5 Laboratory classification of soils, Consistency
 The Activity (A) (l’activíté d’une argile) is a quantity use as an
index for identifying the swelling potential of clay soils, it is
Sir Alec Westley
Skempton
English, (19142001)
defined as:
A
PI
(% of clay-size fraction, by w eight)
A=
Indice de plasticité
(% éléments inférieurs à 2  m )
Mineral
A
Montmorillonite
7.2
Illite
0.9
Kaolinite
0.38
Activity (Based on
Skempton, 1953)
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1.5 Laboratory classification of soils, Consistency
 Plasticity Chart:
 Liquid and plastic limits are determined by relatively simple laboratory
tests that provide information about the nature of cohesive soils.
 Casagrande (1932) studied the relationship of the plasticity index to the
liquid limit of a wide variety of natural soils, and he proposed a Chart
which is called Plasticity Chart.
 The plasticity chart is of great value and is the basis for the
classification of fine-grained soils in the Unified Soil Classification
System.
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1.5 Laboratory classification of soils, Consistency
Désignation
Symbole
international
Symbole
français
Limons peu
plastiques
ML
Lp
Argiles peu
plastiques
CL
Ap
Limons très
plastiques
MH
Lt
Argiles très
plastiques
CH
At
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1.5 Laboratory classification of soils, Consistency
 Use the plasticity Chart to estimate the shrinkage limit of a soil
(Adapted from Holtz and Kovacs, 1981)
1. Plot the plasticity index against the liquid limit
of a given soil such as point A in the Figure.
2. Project the A-line and the U-line downward to
meet at point B. Point B will have the
coordinates of LL = -43.5 and PI = -46.4.
3. Join points B and A with a straight line. This
will intersect the liquid limit axis at point C.
The abscissa of point C is the estimated
shrinkage limit.
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1.6 Soil classification and description
 the American Association of State Highway and
 Soil classification systems have been in use for a very long time with
the first recorded use being in China over 4000 years ago.
 United States Department of Agriculture firstly proposed in 1896,
which the classification system were purely on particle size.
 There are several classification schemes available for soil
classification with the purpose in;
1. used to specify a certain soil type that is best suited for a given
application.
2. used to establish a soil profile along a desired cross section of a soil
mass.
Transportation Officials (AASHTO) developed
one scheme that classifies soils according to
their usefulness in roads and highways,
 Unified Soil Classification System (USCS) was
originally developed for use in airfield
construction but was later modified for general
use.
 American Society for Testing and Materials
(ASTM) Classification System
 The British Soil Classification System (BSCS)
 The textural classification systems developed by
the U.S. Department of Agriculture (USDA).
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1.6 Soil classification and description
 AASHTO soil classification system (1929):
 used to determine the suitability of soils for earthworks, embankments,
and road bed materials (subgrade—natural material below a constructed
pavement; subbase—a layer of soil above the subgrade; and base—a
layer of soil above the subbase that offers high stability to distribute
wheel loads).
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 The AASHTO system
classifies soils into 7
major groups;
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1.6 Soil classification and description
 To evaluate the quality of a soil as a highway subgrade material, the
group index (GI) is used. The group index is given as:
GI  (F200  35) 0.2  0.005  LL  40    0.01 F200  15  PI  10 
 The liquid limit and the plasticity index for
soils that fall into groups A-2, A-4, A-5, A-6,
and A-7.
where F200 is percent passing No. 200 sieve,
1. A negative value for GI, it is taken as 0.
2. The value of group index have to rounded off to the nearest whole
number.
3. There is no upper limit for the group index.
4. The group index of soils belonging to groups A-1-a, A-1-b, A-2-4, A2-5, and A-3 is always 0.
5. When calculating the group index for soils that belong to groups A2-6 and A-2-7, use the partial group index for PI, or
GI  0.01 F200  15  PI  10 
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1.6 Soil classification and description
 USCS: The original form of this system was proposed by Casagrande
in 1942.
 The USCS uses symbols for the particle size groups. These
symbols and their representations are
G—gravel,
• W for well graded
S—sand,
• P for poorly graded
M—silt,
C—clay.
Arthur Casagrande
(1902-1981)
 H for high plasticity (LL > 50)
 L for low plasticity (LL < 50)
 a symbol, O, indicating the presence of organic material.
ex. CL means a clay soil with low plasticity, and SP means a poorly
graded sand.
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 ASTM is nearly
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identical to the
USCS.
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1.6 Soil classification and description
 BSCS: The system divides soil into two main
categories. If at least 35% of a soil can pass
through a 63 μm sieve then it is a fine soil.
Conversely, if the amount of soil that can pass
through the 63 μm sieve is less than 35% then it
W: Well graded
L: Low plasticity (LL < 35%)
P: Poorly graded
I: Intermediate (35 ≤ LL ≤ 70)
Pu: Uniform
H: High plasticity (50 ≤ LL ≤ 70)
Pg: Gap graded
V: Very high (70 ≤ LL ≤ 90)
O: Organic
E: Extremely high (LL > 90%)
is a coarse soil.
The main soil types are designated by capital letters:
G: Gravel
M: Silt, M-soil
S: Sand
C: Clay
F: Fine soil, Fines
Pt: Peat
Example:
 ‘M-soil’ is to classify soils that plot below the
A-line that particle size distributions not wholly
in the range of silt sizes.
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1.6 Soil classification and description
 The plasticity chart is the relationship
of the soil LL to its PI.
 A-line in the plasticity chart, is an
empirical boundary between inorganic
clays, whose points lie above the line,
and organic silts and clays whose points
lie below. The A-line is defined as:
PI  0.73( LL  20)
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1.6 Soil classification and description
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1.6 Soil classification and description
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1.6 Soil classification and description
 The textural classification systems (USDA): by its
surface appearance.
 The soil are named after their principal components,
sandy clay, silty clay etc.
 The names are classify into a chart, and the
classification method is based on the particle-size limits
of each organization. For the chart on the right is based
on USDA.
Example: the particle size distribution of soil A in
the figure has; 30% sand, 40% silt, and 30% clay.
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2. Soil Compaction (Le compactage des sols)
• Compaction of soil is densification the soil by removing its void.
• To improving a soil, compaction is normally the least expensive
method.
Proctor Compaction Test—ASTM D 1140 and ASTM D 1557
Essai Proctor (NF P 94-093)
1933, PROCTOR l’Ingénieur américain mit en évidence que le poids spécifique
sec d’un sol est influencée de la teneur en eau et de l’énergie de compactage;
grâce à l’essai qui porte son nom : Essai proctor.
Compaction apparatus
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2. Soil Compaction
•
For most projects, the standard Proctor (Proctor Normal) test is satisfactory. For
projects involving heavy loads, such as runways to support heavy aircraft loads,
a modified Proctor (Proctor Modifié) test was developed.
1.
Mix dry soil thoroughly with water to prepare a uniform specimen
with designed water content.
2.
Pour the loose specimen at a little over one-third depth into a
standard-size mold 101.6 mm (4 in.) in diameter and 116.43 mm
(4.584 in.) high without an extension collar, with 944 cm3 (1/30 ft3)
volume.
3.
Compact the specimen by 25 free drops of a rammer (W = 24.5 N
(2.5 kilogram force [kgf] or 5.5 pound force [lbf]) from 304.8 mm
(12 in.) high. The mold should be placed on hard ground to avoid
possible compaction energy loss.
4.
Repeat steps 1 to 3 for the second and third layers to fill the mold
with soil slightly above the top level of the mold. For the third-layer
compaction, an extension collar is attached.
5.
Remove the extension collar and trim the specimen surface by a
straight edge to get exactly 944 cm3 (1/30 ft3) volume of the
specimen.
6.
Weigh the whole mold with soil in it to obtain the wet weight of the
specimen.
7.
Extrude the specimen from the mold and obtain a representative soil
specimen in a container for water content determination.
8.
Repeat steps 1 to 7 for several different water contents. In general,
soils from the previous experiment could be reused for the next test
by breaking them down to particles and remixing with additional
water.
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2. Soil Compaction
Basic Concept

 Gs

 G 
 d   s  w 

1  w  1  wG s
1 e 
S%

The test is identified when additional
water causes the bulk unit weight of
the soil to decrease?
e


 w


wG s
Sr
if S r  100%
 e  emin  wG s
 G 
  d   d ,max   s   w
1 e 
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2. Soil Compaction
Energy imparted by the hammer
• The S = 100% curve is called the zero air void (ZAV) curve.
E comp  m h g
hd
Nb Nl
V
hd height of fall hammer,
Nb number of blows,
Nl number of layers.
• When the compaction energy is increased, γd,max increases.
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2. Soil Compaction
Energy imparted by the hammer
E comp  m h g
hd
Nb Nl
V
hd height of fall hammer,
Nb number of blows,
Nl number of layers.
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2. Soil Compaction
• Facteurs d’influence
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2. Soil Compaction
 Field Compaction:
Relative Compaction (R.C.) is defined as:
 d , field
R .C . 
 d ,max

,
is the specified dry unit weight,
that shall be achieved in the field
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2. Soil Compaction
 Field Compaction Equipment:
Pneumatic rubber tire rollers used for both sandy
soils and clayey soils. Soils are compacted with
both tire pressure and kneading action.
Sheep’s-foot rollers have unique wheel surfaces
that can effectively compact the clayey soils and
the deeper part of soils in earlier passes.
Smooth-wheel (drum) rollers used for sandy and
clayey soils for the finishing and smoothing
process. It is not used to compact thicker layers.
Vibratory wheels are usually part of all the
foregoing rollers; they are especially
effective in compacting granular soils.
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2. Soil Compaction
 Field Compaction:
Number of passes: In general construction several
or more passes of rollers are required to obtain a
specified dry unit weight. The more passes applied,
the higher the dry unit weight is obtained.
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2. Soil Compaction
 Field Compaction:
 One more simple and effective compaction technique is
dynamic compaction.
 Dropping a heavy weight repeatedly on the ground, the
weight is about 80kN to 360kN with the drop height from
10m to 30m.
 This method is good for sandy soil but not silt and clay soils.
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2. Soil Compaction
 Field Density test, Sand Cone Method (ASTM D 1556 and AASHTO T 191):
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2. Soil Compaction
 Méthode du densitometer à membrane (NF P 94-061-2):
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2. Soil Compaction
 California Bearing Ratio (CBR), Standard ASTM (D 1883) or NF P 94-078:
 This is a penetration test which was developed by the California
Department of Transportation and became standards.
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2. Soil Compaction
 California Bearing Ratio (CBR), Standard ASTM (D 1883) or NF P 94-078:
• CBR = maximum of the two formulars
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2. Soil Compaction
 Proctor - CBR:
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3. Seepage (l’écoulement sous terrain)
 The water can flow through soil due to the void in the soil.
Daniel Bernoulli,
 It is important in the construction of earth dams, levees,
1700 – 1782, Swiss.
embankments, excavation and underground structures, etc.
 The basic energy flow equation (Bernoulli’s equation), which
defined the head as 3 parts:
v2
ht  hz  h p  hv  z 

 w 2g
u
ℎ:
ℎ :
ℎ
;ℎ :
ℎ
;ℎ :
ℎ
;
ℎ
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3. Seepage (l’écoulement sous terrain)
 The velocity of water flow through soil is small, thus the velocity
head is neglected.
ht  hz  h p  z 
u
w
Henry Darcy, 1803 –
1858, French.
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3. Seepage (l’écoulement sous terrain)
 When water flows in soils, it creates friction and lead to head loss.
 Darcy’s Law:
:
:ℎ
v  k i
ℎ
;
; :
ℎ
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3. Seepage (l’écoulement sous terrain)
Ex. Shows water flow though the soil specimen in a cylinder (D = 6mm). The specimen’s k value is 3.4 × 10−4 cm/s.
a) Calculate pressure heads hp at Points A, B, C, and D and draw the levels of water height in standpipes.
b) Compute the amount of water flow q through the specimen.
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3. Seepage (l’écoulement sous terrain)
Ex. Shows water flow though the soil specimen in a cylinder (D = 6mm). The specimen’s k value is 3.4 × 10−4 cm/s.
a) Calculate pressure heads hp at Points A, B, C, and D and draw the levels of water height in standpipes.
b) Compute the amount of water flow q through the specimen.
Solution:
q = k (Δh/L)A = 3.4 × 10−4 × (8/18) × π(6/2)2
= 4.27 × 10−3 cm3/s ←
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3.1 Coefficient of hydraulic conductivity (Permeability)
 Hazen’s (1911) empirical formula:
k  C ( D10 ) 2
k has the unit in [cm/s] and D10 has the unit in [mm]
C: Hazen’s empirical coefficient, between 0.4 and 10.0 (mostly
 Chapuis (2004) empirical formula:
3


e
2
k  2.4622 ( D10 )

1

e


0.7825
e: void ratio of soil
0.4 to 1.5)
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3.2 Laboratory Test
1. Constant head permeability test:
k
QL
A h t
where
Q: collected amount of water for a time period t
L: length of soil specimen in the flow direction
A: cross-sectional area of a soil specimen
Δh: hydraulic head loss in constant head test setup
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3.2 Laboratory Test
2. Falling head permeability test:
aL
h1
k
ln
A (t2  t1 ) h2
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3.3 Field Test
1. Unconfined aquifer under impervious layer:
k
q
r2
ln
 (h22  h12 ) r1
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3.3 Field Test
2. Confined aquifer:
k
q
r
ln 2
2 H (h2  h1 ) r1
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3.4 Flow Net
1. One- dimensional flow net:
 The cylinder is equally divided into Nf (3) called flow channels.
 The specimen length L is equally divided into Nd (4) called equipotential lines.
h
N
1 a
qa  k i a  k d a  k h
b
Nd b
Nf a
q A  qa N f  k h
Nd b
a=b
q A  k h
Nf
Nd
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3.4 Flow Net
2. Two- dimensional flow net:
Note:
1. Flow lines and equipotential lines are perpendicular.
2. Each opening of the net should be a square shape or close to it (i.e., a = b).
Acceptable near-squares in flow net construction
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3.4 Flow Net
2. Two- dimensional flow net:
1. Draw the geometry of structure 2. Select a proper Nf
value (3 or 4)\the first trial. 3. Identify the boundary flow
lines and boundary equipotential lines (EL). 4. Draw trial
flow lines with selected Nf for the entire earth structure.
There are equal amounts of water flow through all flow
channels. 5. Starting from the upstream, draw the first EL
to have all net openings squares or near-squares. 6.
Continue the foregoing step for the second and third EL. 7.
At the downstream exit point, it may not get to full squares
with the last EL. draw an imaginary EL to have full nearsquare sections. Based on partial squares for the last
elements, obtain an average fractional number for the last
EL, such as 5.4 in Figure.
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3.4 Flow Net
2. Two- dimensional flow net:
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3.5 Pressure Heads in Flow Net
 Total heads at any points on the same equipotential line should be the same.
ht  hz  h p  z 
u
w
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3.6 Boundary Water Pressure
 The water under a dam creates uplift pressure on its base and may lead to instability of the dam.
 The unbalanced water pressure is one of the key parameters for sheet pile design.
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3.6 Boundary Water Pressure
 Ex:
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3.6 Boundary Water Pressure
 Ex:
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3.6 Boundary Water Pressure
Ex: A flow net under a concrete dam is drawn in Figure.
(a) Calculate and plot the water pressure distribution along the base of the dam.
(b) Compute the resultant uplift force against the base of the dam.
(c) Calculate the point of application of the resultant uplift force.
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3.6 Boundary Water Pressure
(a) Calculate and plot the water pressure distribution along the base of the dam.
Solution:
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3.6 Boundary Water Pressure
(b) Compute the resultant uplift force against the base of the dam.
(c) Calculate the point of application of the resultant uplift force.
Solution:
(c) Point of application of P = Σ(moment)/P =
6478.2/926.2 = 6.99 m from Point “a.”
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4. Soil Total and Effective Stress
 A major problem in geotechnical analysis is the estimation of the
state of stress at a point at a particular depth in a soil mass.
 The total vertical stress acting at a point in the soil is due to the
weight of everything that lies above that point including soil, water
and any load applied to the soil surface.
z  z
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4. Soil Total and Effective Stress
 The pore pressure at a point in the soil depends on the depth below
the water table and the flow conditions. For a horizontal ground
water table, the pore pressure at a point beneath the ground is
84
 Terzaghi, 1925 developed the effective stress that
controls the changes in the volume and strength of
a soil which is known as the effective stress.
' u
considered as hydrostatic pressure.
 Above the water table the soil is saturated with capillary water in a
state of suction, and the pore pressures is negative.
u   w zw
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4. Soil Total and Effective Stress
 Soil Layers with steady water table:
 The total vertical stress can be calculated by cumulated weight from the upper layers of the soil:
 The effective vertical stress is computed by the Terzaghi equation:
 H
i
i
' u
u   w zw
 The effective stress σ′ under the
water table can be directly by γ ′
(submerged unit weight).
 '   w
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4. Soil Total and Effective Stress
Ex. 1, A 3 m layer of sand, of saturated unit weight 18 kN/m3, overlies a
4 m layer of clay, of saturated unit weight 20 kN/m3. If the groundwater
level occurs within the sand at 2 m below the ground surface, determine
the total and effective vertical stresses acting at the center of the clay
layer. The sand above groundwater level may be assumed to be
saturated.
Ex. 2, A 5 m deep deposit of silty sand lies above a 4 m deep deposit of
gravel. The gravel is underlain by a deep layer of stiff clay. The ground
water table is found 2 m below the ground surface. The soil properties
are:
ρb sand above GWT = 1.7 Mg/m3
ρsat sand below GWT = 1.95 Mg/m3
ρsat gravel = 2.05 Mg/m3
Draw the distributions of vertical total stress, pore water pressure, and
vertical effective stress with depth down to the clay layer.
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4. Soil Total and Effective Stress
Ex. 3, Details of the subsoil conditions at a site are shown in Figure
below together with details of the soil properties. The ground surface is
subjected to a uniform loading of 60 kPa and the groundwater level is 1.2
m below the upper surface of the silt. It can be assumed that the gravel
has a degree of saturation of 50% and that the silt layer is fully saturated.
NB γw = 9.81 kN/m3.
Determine the vertical effective stress acting at a point 1 m above the
silt/rock interface.
h  w
G s  eS r
1 e
 '  140.6kPa
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4. Soil Total and Effective Stress
 Capillary rise:
 Above water table are not completely dry, this is due to suction created by surface tension. It is called capillary rise.
 Hazen, 1930 gave an empirical approximate:
hcapillary ( mm ) 
C
e D10
C is a constant with a range of 10 to 50.
 Pore pressure in the capillary zone is defined as:
u   S r  w hcapillary
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4. Soil Total and Effective Stress
 Effective stress with water flow:
 There is an extra water pressure at the bottom of soil specimen
γwΔh, it is called upward seepage pressure
uz   w (H1  z)   w
w
h
  wi
H
h
z
H
Is the hydraulic gradient unit forces
 'z   z  u z   ' z   w
h
z
H
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4. Soil Total and Effective Stress
 Effective stress with water flow:
 'z   z  u z   ' z   w
h
z
H
If the σ′z = 0
h
 'z   ' z   w
z0
H
'
 ic
iB  C 
hB  hC
BC
w
ic is the critical hydraulic gradient.
 Based on experimental observation, Terzaghi (1922)
If i > ic the condition cause quick sand or sand boiling is
suggested the section of d × d/2 (area BCED) for evaluating
occurred.
the factor of safety against quicksand.
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4. Soil Total and Effective Stress
 Excavation:
 '    u   H clay  H ex   clay  hw  w
 '    u  hex  w   H clay  H ex   clay  hw  w
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5. Consolidation of low permeability soil
 The delay in the deformation of the soil mass
 When additional stresses applied to soil mass, it causes the soil to
deform. ie. The footing load or pile load.
 The response instantaneously of the soil mass (instantaneous
deformation) is the elastic response of the material (soil grain,
water, and air).
 The water will continuously drain out from the void of the soil mass
due to the additional stresses, this phenomena will cause the soil to
deform continuously during a period of time; it is defined as
consolidation phenomena.
 The theory of consolidation was originally developed by Karl
Karl von Terzaghi (October 2, 1883 – October 25,
1963), Austrian, the "father of soil
Terzaghi (1925); the study of soil deformation under compressive
mechanics and geotechnical engineering"
load with slow drain rate of water through soil void spaces.
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5. Consolidation of low permeability soil
 One dimensional consolidation
 When the stress increment is applied in one direction only, typically
vertical direction. The lateral strain are constrain, t = 0.
 Consolidation settlement is the settlement corresponding to the
consolidation process.
 Under negative excess pore pressure the soil swell, it is the reverse
process of the consolidation.
 To predict the magnitude and rate of consolidation settlement in 1-D
conditions, the oedometer test is proposed.
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5.1 Consolidation Oedometer test
 Oedometer test
 The specimen is in the form of a disc of soil, held inside a metal ring
and lying between two porous stones.
 The upper porous stone is pushed by a steel cap, which can move
inside the ring with a small clearance. The whole assembly is put
inside an open cell of water, thus, the pore water in the specimen can
flow freely.
 To reduce side friction,
inside surface of the ring
should be smoothed.
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5.1 Consolidation Oedometer test
 Oedometer test
 When the load is applied at the top cap, the excess pore pressure will
increase
 The water drained at top and bottom of the specimen easily, thus, the
excess pore pressure at that places is drop to zero rapidly.
 At the center of the specimen, the excess pore pressure initially (t =
0) equal to the applied stress u(t=0) = .
 At the center of the specimen, the pore pressure is equal to 0, when t
go to infinity. t  
  '(t  0)    (t  0)   u (t  0)
  '(t  0)  0
  '(t   )    (t   )  0
 ' u
 '    u
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5.1 Consolidation Oedometer test
 The initial void ratio can be
calculated from the specific
gravity Gs and unit weight.
e0 
e0 
Gs   w
d
Vv

Vs
1
Vv
Md
Gs   w
 Gs   w
 Md
Vv  
 1
 d
 Gs   w
V
Md
hs  s 
A A  Gs   w
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5.1 Consolidation Oedometer test
 The void ratio of each increment period can be calculated from the
displacement readings.
h1  h0   h1
h2  h0   h2
h
h  hs
e1  1
hs
h  hs
e2  2
hs
 Typical plots of void ratio (e) after
consolidation against effective stress (σ′)
la contrainte effective σ′ et l’indice des vides
« e » pour chaque palier de chargement.
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5.1 Consolidation Oedometer test
 The plots show that the shapes of the curves are related to the
stress history of the soil.
 A normally consolidated soil is linear and is called the virgin (onedimensional) compression line (1DCL).
 Permanent (irreversible) changes in soil structure continuously
take place and the soil does not revert to the original structure
during expansion.
 A soil is overconsolidated, its state will be represented by a point
on the expansion or recompression parts,
 The recompression curve ultimately rejoins the virgin compression
line. The changes in soil structure along this line are almost
recoverable.
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5.1 Consolidation Oedometer test
 The compression index Cc (l’indice de compression) is the
slope of 1 DCL,
 The swelling index Cs (l’indice de gonflement) is the slope
of the unloading portion.
 The preconsolidation pressure, σ′c, (La contrainte de
préconsolidation σ ‘p) is the maximum past effective
overburden pressure to which the soil specimen has been
subjected.
 Over consolidation Ratio OCR (rapport de
surconsolidation); the ratio of the σ′c, to the present
effective pressure, σ′0.
 c'
OCR  '
0
 The constrained modulus or modulus of oedometer,
(Le module oedométrique) (also called one-dimensional
elastic modulus) E′oed is the reciprocal of the coefficient
of volume compressibility mv.
1
1

m

v
'
E oed
1  e0
 e0  e1  1  h0  h1 

 '
 '
' 
' 



h



0 
0 
1
0 
 1
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5.2 Consolidation settlement
 To estimate the consolidation settlement,
 V e0  e1

V0
1  e0
dsoed
e0  e1
e0  e1  1'   0'

dz  '
dz
'
1  e0
 1   0 1  e0
dsoed  m v   ' dz
soed 

H
0
m v   ' dz
soed  m v   ' H
soed
  1' 
C c log  ' 
0 


H
1  e0
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5.2 Consolidation settlement
 For overconsolidated clay:
 c'
OCR  '  1
0
'
1
 
soed
 1'   c'   0'
'
c
  1' 
C s log  ' 
0  H

1  e0
soed
  c' 
  1' 
C s log  ' 
C c log  ' 

c 
0 



H
H
1  e0
1  e0
Terzaghi and peck, 1967
C c  0.009  LL  0.090
Mayne, 1980 and Biarrez, 1994
C c  0.0092  LL  0.119
C c  0.009   LL  13 
Rendon-Herrero, 1983
 1  e0 
C c  0.141G s1.2  

G
s


2.38
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 A soil profile is shown in Figure. If a uniformly distributed load,
σ, is applied at the ground surface, what is the settlement of the
clay layer caused by primary consolidation if
a. The clay is normally consolidated,
b. The preconsolidation pressure σ’c = 200kPa,
c. σ’c = 150kPa ; Use Cs = 1/5 Cc.
a / . S c  191mm
b / . S c  38 mm
c / . S c  67.5 mm
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103
5.2 Consolidation settlement
• Cas Particuliers; Multi-couche Compressible:
 Dans la situation rarement, les sols multi-couche, E. Absi, 1964 à donné les résultasts
approchée comme:
C va 
Hypothèses pour cette approchée:
H2
 n
 
 1
hi 

C vi 
2
 Il repose sur une couche sous-jacente qui peut être, soit parfaitement
impermeable, soil parfaitement permeable.
n
Où
H   hi
1
 Le sol est chargé uniformément.
L’épaisseur totale
 Le sol compressible est compose d’une superposition de n couches ayant
chacune des caractéristiques Cv, k, et une épaisseur h définies.
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5.3 Degree of consolidation
 Degree of consolidation Uv:
e0  e  '   0'
Uv 
 '
e0  e1  1   0'
Uv = 0 ; means consolidation not yet begun,
Uv = 1 ; means consolidation is complete,
Uv 
ui  ue
u
 1 e
ui
ui
ui : initial excess pore pressure due to applied load
ue : excess pore pressure at time t
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5.4 Theory of one-dimensional consolidation
 Terzaghi (1943) developed an analytical model for determining
the degree of consolidation. The theory was made according to
the assumptions that:
1. The soil is homogeneous and fully saturated.
2. The solid particles and water are incompressible.
3. Compression and flow are one-dimensional (vertical).
4. Strains are small.
5. Darcy’s law is valid at all hydraulic gradients.
6. The coefficient of permeability and the coefficient of volume
v z  kiz   k
h
k u e

z
 w z
compressibility remain constant throughout the process.
7. There is a unique relationship, independent of time, between
void ratio and effective stress.
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5.4 Theory of one-dimensional consolidation
• The condition of continuity equation:
k  2ue
dV

dxdydz

 w z 2
dt
• The rate of volume change:
dV
 '
 mv
dxdydz
dt
t
u
dV
  m v e dxdydz
dt
t
u e
k  2ue

m
v
 w z 2
t
 2u e u e
cv

z 2
t
k
cv 
mv w
v z  kiz   k
h
k u e

z
 w z
• cv being defined as the coefficient
of consolidation
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5.4 Theory of one-dimensional consolidation
• The solution for the excess pore water pressure ue:
 2ue u e
cv

2
z
t
u e ( t  0)  u i ; u e ( t   )  0
m 
1
u e (t )   
m0  d
m 
u e (t ) 

m0
m
ue 

m0
Tv 
M 

2d
0
2ui
n z

1  cos  n    sin
n
2d

2ui
M
2
 n 2 2 c v t 


 exp  
2
4
d



 n 2 2 c v t 


 exp  
2
4
d



Mz 

2
sin

 exp   M Tv 
d 

cv t
d2

n z
n z

u i sin
dz   sin
2d
2d

v z  kiz   k
h
k u e

z
 w z
• Tv is called the time factor
 2 m  1
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5.4 Theory of one-dimensional consolidation
• The degree of consolidation at depth z and time t can be obtained
by substituting the value of ue.
m
ue
2 
Mz 
2
Uv  1
 1 
sin

 exp   M Tv 
ui
d 
m 0 M 
• The average degree of consolidation at time t for constant ui can
be obtained by substituting the value of ue as:
 2 d u dz 
 e 1
Uv  1  0
  2 d   ui


m 
2
2
Uv  1 
exp

M
Tv 

2
m0 M
v z  kiz   k
h
k u e

z
 w z
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5.4 Theory of one-dimensional consolidation
 The variation of U with Tv can be
calculated from:
m 
Uv  1

2
2
m0
4
 2 m  1
2
 2

2
exp  
 2 m  1 Tv 
 4

 Can also be approximated by:
for U v  0 to 60%
Tv 

U v 
2
4
for U v  60%
Tv  0.085  0.933 log(1  U v )
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Variation Uv and Tv
Dr. KY Sambath
110
Uv (%)
Tv
Uv (%)
Tv
Uv (%)
Tv
Uv (%)
Tv
Uv (%)
Tv
Uv (%)
Tv
Uv (%)
Tv
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
5.5
6
6.5
7
7.5
8
8.5
9
9.5
10
10.5
11
11.5
12
12.5
13
13.5
14
14.5
0.000019
0.000079
0.000177
0.000314
0.000491
0.000707
0.000962
0.001257
0.001590
0.001963
0.002376
0.002827
0.003318
0.003848
0.004418
0.005026
0.005674
0.006362
0.007088
0.007854
0.008659
0.009503
0.010387
0.011310
0.012272
0.013273
0.014314
0.015394
0.016513
15
15.5
16
16.5
17
17.5
18
18.5
19
19.5
20
20.5
21
21.5
22
22.5
23
23.5
24
24.5
25
25.5
26
26.5
27
27.5
28
28.5
29
0.017671
0.018869
0.020106
0.021382
0.022698
0.024053
0.025447
0.026880
0.028353
0.029864
0.031416
0.033006
0.034636
0.036305
0.038013
0.039760
0.041547
0.043373
0.045239
0.047143
0.049087
0.051070
0.053093
0.055154
0.057255
0.059395
0.061575
0.063794
0.066052
29.5
30
30.5
31
31.5
32
32.5
33
33.5
34
34.5
35
35.5
36
36.5
37
37.5
38
38.5
39
39.5
40
40.5
41
41.5
42
42.5
43
43.5
0.068349
0.070685
0.073061
0.075476
0.077931
0.080424
0.082957
0.085529
0.088141
0.090792
0.093482
0.096211
0.098980
0.101788
0.104635
0.107522
0.110449
0.113414
0.116419
0.119464
0.122548
0.125673
0.128836
0.132040
0.135284
0.138567
0.141891
0.145256
0.148660
44
44.5
45
45.5
46
46.5
47
47.5
48
48.5
49
49.5
50
50.5
51
51.5
52
52.5
53
53.5
54
54.5
55
55.5
56
56.5
57
57.5
58
0.152106
0.155593
0.159120
0.162689
0.166300
0.169953
0.173648
0.177386
0.181167
0.184991
0.188860
0.192772
0.196730
0.200733
0.204782
0.208877
0.213020
0.217211
0.221450
0.225738
0.230077
0.234466
0.238908
0.243402
0.247950
0.252552
0.257211
0.261926
0.266699
58.5
59
59.5
60
60.5
61
61.5
62
62.5
63
63.5
64
64.5
65
65.5
66
66.5
67
67.5
68
68.5
69
69.5
70
70.5
71
71.5
72
72.5
0.271532
0.276425
0.281380
0.286398
0.291481
0.296630
0.301847
0.307134
0.312491
0.317922
0.323427
0.329009
0.334671
0.340413
0.346239
0.352150
0.358149
0.364240
0.370423
0.376703
0.383083
0.389565
0.396152
0.402849
0.409659
0.416585
0.423632
0.430804
0.438105
73
73.5
74
74.5
75
75.5
76
76.5
77
77.5
78
78.5
79
79.5
80
80.5
81
81.5
82
82.5
83
83.5
84
84.5
85
85.5
86
86.5
87
0.445541
0.453116
0.460835
0.468704
0.476729
0.484916
0.493272
0.501804
0.510520
0.519428
0.528535
0.537852
0.547389
0.557155
0.567162
0.577423
0.587950
0.598758
0.609862
0.621280
0.633028
0.645127
0.657598
0.670465
0.683754
0.697494
0.711715
0.726455
0.741750
87.5
88
88.5
89
89.5
90
90.5
91
91.5
92
92.5
93
93.5
94
94.5
95
95.5
96
96.5
97
97.5
98
98.5
99
99.5
100
0.757646
0.774190
0.791439
0.809454
0.828308
0.848081
0.868870
0.890782
0.913947
0.938517
0.964673
0.992635
1.022669
1.055108
1.090372
1.128999
1.171699
1.219434
1.273551
1.336024
1.409913
1.500346
1.616932
1.781247
2.062129
inf
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5.4 Coefficient of consolidation
• Determination of cv:
•
Log time method
 Obtained by the oedometer test;
plotting the dial gauge readings
against the time in minutes,
plotted on a logarithmic axis.
cv t
d2
Tv d 2
cv 
t
0.19673 d 2
cv 
t50
Tv 
Uv (%)
50
Tv
0.196730
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5.4 Coefficient of consolidation
• Determination of cv:
•
Root time method
 Obtained by the oedometer test; plotting the dial
gauge readings against the square root of time in
minutes.
cv t
d2
Tv d 2
cv 
t
0.84808 d 2
cv 
t90
Tv 
Uv (%)
90
Tv
0.848082
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5.5 Compression ratios
 There are three parts of the compression are observed from the test; the initial compression, the primary
consolidation compression, and the secondary compression.
a0 
initial a s 
primary a100 
secondary a f
 The compression ratios can be defined as:
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Ex. The following compression readings were taken during an oedometer test on a saturated clay specimen
(Gs = 2.73) when the applied pressure was increased from 214 to 429 kPa:
Time (min)
0
1_4
1_2
1
2 1_4
4
9
16
25
Gauge (mm)
5.00
4.67
4.62
4.53
4.41
4.28
4.01
3.75
3.49
Time (min)
36
49
64
81
100
200
400
1440
Gauge (mm)
3.28
3.15
3.06
3.00
2.96
2.84
2.76
2.61
• After 1440 min, the thickness of the specimen was 13.60 mm and the water content was 35.9%. Determine
the coefficient of consolidation from both the log time and the root time methods and the values of the three
compression ratios. Determine also the value of the coefficient of permeability.
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Solution
Total change in thickness during
increment = 5.00 – 2.61 = 2.39 mm
Average thickness during increment
= 13.60 + 2.39/2 = 14.80 mm
Length of drainage path d = 14.80/2
= 7.40 mm
The log time plot from the test as
shown at the left
t50  12.5 min
0.1967 d 2
cv 
 0.45 m 2 / year
t50
r0  0.088
rp  0.757
rs  0.155
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Solution
Total change in thickness during
increment = 5.00 – 2.61 = 2.39 mm
Average thickness during increment
= 13.60 + 2.39/2 = 14.80 mm
Length of drainage path d = 14.80/2
= 7.40 mm
The root time plot from the test as
shown at the left
t50  53.3 min
0.8481d 2
cv 
 0.46 m 2 / year
t90
r0  0.080
rp  0.785
rs  0.135
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Solution
In order to determine the permeability, the value of mv must be calculated.
Final void ratio e1  w1G s  0.359  2.73  0.98
Final void ratio e0  e1   e
 e 1  e0

H
H0
e
1.98

2.39 15.99
 e  0.35 ; e0  1.33
•
mv 
1 e0  e1
 7.0  10 4 m 2 / kN
'
'
1  e0  1   0
2
m v  0.70 m / MN
Coefficient of permeability:
k  cv m v  w
k  1.0  1010 m / s
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5.6 Secondary compression
 In the Terzaghi theory, due to a change in effective stress, the pore pressure increase and continuously
dissipate. This phenomena is governed by the permeability of the process. However, experimental results show
that when the excess pore pressure has dissipated to zero the compression does not stop but continues at a
gradually decreasing rate under constant effective stress. This is defined as secondary compression.
 It take place by the readjustment of fine-grained particles into a more stable configuration after the disturbance
caused by the decrease in void ratio, especially if the soil is laterally confined. An additional factor is the
gradual lateral displacements which take place in thick layers subjected to shear stresses.
 Secondary compression’s rate is controlled by the highly viscous film of adsorbed water surrounding clay
mineral particles in the soil. The viscosity of the film increases as the particles move closer, resulting in a
decrease in the rate of compression of the soil.
 Secondary compression’s rate in oedometer test is defined by the slope Cα
 Mitchell and Soga, 2005, normalizing Cα = (0.01 to 0.1) Cc
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5.7 Correction for construction period
 In practice, the applied loads are continuously over a period of
time, not instantaneously as shown in Figure below:
 Terzaghi (1943) proposed an empirical method of correcting
the instantaneous time–settlement curve to allow for the
Pt1
construction period.
 The net load P’ applied uniformly over the time tc .
 The degree of consolidation at time tc is the same as if the load
P′ had been acting as a constant load for the period 1/2tc.
 Thus the settlement at any time during the construction period
t
sc ( 1 )
2
is equal to that occurring for instantaneous loading at half that
t
Pt
t
time. sc ,corrected (t1 )  1 sc ( 1 )
sc , corrected (t c )  sc ( c )
P'
2
2
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6. Shear Strength of Soils
 Stress Transformation and Mohr circle diagram:
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6. Shear Strength of Soils
 Stress Transformation and Mohr circle diagram:
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6. Shear Strength of Soils
 Stress Transformation the concept of the pole:
 The known stresses σA and τA on a known A–A plane are drawn, in
addition the Mohr’s circle of this element is also known with a pole A
(point A).
 On the Mohr’s circle draw a parallel line to A-A plane connected to A.
 Find the “pole”, an intersection of the line AA on the circle.
 On the Mohr’s circle draw any parallel line to
any plane which connected to the pole, at the
other intersection between the line and the
Mohr’s circle, is the stresses value that act on
that plane.
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6. Shear Strength of Soils
 Stress Transformation and Mohr circle diagram:
 
1   3
2
sin 2
 n   3  ( 1   3 ) cos 2 
 n  OE  OA  AE   3  AB cos 2 
  3  AB cos 2 
  3  ( 1   3 ) cos 2 
  DE  DC sin(180   2 )
 DC sin(2 )
 3
 1
sin 2
2
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6. Shear Strength of Soils
The Mohr circle:
Ex. On a failure plane in a purely frictional mass of dry sand the total
stresses at failure were: shear = 3.5 kPa; normal = 10.0 kPa.
Determine (a) by calculation and (b) graphically the resultant stress on
the plane of failure, the angle of shearing resistance of the soil, and the
angle of inclination of the failure plane to the major principal plane.
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6. Shear Strength of Soils
 Coulomb’s law of soil shear strength:
Charles Coulomb (1736-1806) a
physicist, using the laws of friction and
cohesion to determine the true sliding
surface.
 The shear resistance offered by a particular soil is made up of the two
components of friction and cohesion.
 In 1776, Coulomb suggested that the equation of the strength envelope
of a soil could be expressed by the straight line equation:
  c   tan 
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6. Shear Strength of Soils
 The equation gave satisfactory predictions for sands and gravels, for
which it was originally intended, but it was not so successful when
' u
applied to silts and clays. The reasons is the lack of understanding
of the drainage conditions that later proposed by Terzaghi, 1925,
Effective stress, σ′.
Modified Coulomb’s law:
  c '  ' tan  '
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6. Shear Strength of Soils
 The Mohr Coulomb’s theory assumes that the difference between the
major and minor principal stresses is a function of their sum,
(σ1 − σ3) = f(σ1 + σ3). Any effect due to σ2 is ignored.
( '1   '3 )  2 c ' cos  ' ( '1   '3 ) sin  '
q '  2 c ' cos  ' p sin  '
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6. Shear Strength of Soils
 Determination of the shear strength parameters:
 The direct shear box test
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6. Shear Strength of Soils
 Determination of the shear strength parameters:
 The direct shear box test
 The usual plan size of the sample is 60 × 60 mm2, but for
testing granular materials such as gravel or stony clay it is
necessary to use a larger box, generally 300 × 300 mm2,
although even greater dimensions are sometimes used.
h 
Movement of box
Length of sample
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6. Shear Strength of Soils
 Determination of the shear strength parameters:
Example : Shear box test (i)
Drained shear box tests were carried out on a series of soil samples
with the following results:
Test No
Total normal Stress
(kPa)
Total shear stress at
failure (kPa)
1
100
98
2
200
139
3
300
180
4
400
222
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6. Shear Strength of Soils
Example : Shear box test (ii)
A drained shear box test with the cross-sectional area of the box was 3600
mm2 and the test was carried out in a fully instrumented shear box
apparatus. Determine the strength parameters of the soil in terms of
effective stress.
φ′ = 25°;
c′ = 13 kPa
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6. Shear Strength of Soils
 The initial density of a soil tested in a shear
 The effect of density on shear strength
box determines the way in which the soil will
behave during shearing.
 The initially dense soil expands in volume
(dilates) – indicted by the increase in vertical
displacement of the sample.
This phenomenon was demonstrated by Osborne
Reynolds, 1842-1912.
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6. Shear Strength of Soils
 Unconfined Compression Test
 The specimen height-to-diameter ratio should be at least 2.0 or
more to avoid the end boundary effect during the shear.
 The test is completed about 10 to 20 minutes.
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6. Shear Strength of Soils
 The Triaxial Compression Test
UU: Unconsolidated undrained test
CU: Consolidated undrained test
CD: Consolidated drained test
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6. Shear Strength of Soils
 The Triaxial test
 During the application of this load, the sample experiences shortening
in the vertical direction with a corresponding expansion in the
horizontal direction.
A
Volume of sample
H 0  H
Types of failure in the triaxial test
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6. Shear Strength of Soils
Ex. A sample of clay was subjected to an
 The Triaxial test: the undrained shear test
 A sample can be tested in 15 minutes or less, so that there is no time
undrained triaxial test with a cell pressure of 100
kPa and the additional axial stress necessary to
for any pore pressures developed to dissipate or to distribute
cause failure was found to be 188 kPa. Assuming
themselves evenly throughout the sample. Thus, the results of the test
that φu = 0°, determine the value of additional
is the total stress.
axial stress that would be required to cause failure
UU: Unconsolidated undrained test
of a further sample of the soil if it was tested
undrained with a cell pressure of 200 kPa.
cu 
1   3
2
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6. Shear Strength of Soils
 The Triaxial test: the effective stress parameters c’ and ’
CD: Consolidated drained test
Ex. A series of drained triaxial tests were performed
on a soil. Each test was continued until
failure and the effective principal stresses for the
tests were:
 A porous disc is placed on the pedestal before the test sample is placed
in position so that water can drain out from the soil.
 The cell pressure creates a pore water pressure within the soil sample
and the apparatus is left until the sample has consolidated, when the
pore water pressure reaches zero the sample is consolidated.
 The sample is sheared by a low rate of strain which the pore pressure
can dissipate through the porous discs.
 An average test time for a clay sample is about three days but with
some soils a test may last as long as two weeks.
φ′ = 29°
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6. Shear Strength of Soils
 The Triaxial test: the effective stress parameters c’ and ’
CU: Consolidated undrained test
 The sample is consolidated exactly as for the drained test, but at this
Ex. (i) The following results were obtained from a
series of consolidated undrained triaxial tests
carried out on undisturbed samples of a compacted
soil, each sample, originally 76 mm long and 38
mm in diameter, experienced a vertical
deformation of 5.1 mm.
stage the drainage connection is shut off and the sample is sheared
under undrained conditions.
 For most soils a strain rate of 0.05 mm/min is satisfactory, which
means that the majority of samples can be sheared in under three
hours.
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6. Shear Strength of Soils
 The Triaxial test: the effective stress parameters c’ and ’
Ex. (ii) A series of undisturbed samples from a normally consolidated
clay was subjected to consolidated undrained tests.
sin  ' 
 '1   '3
 '1   '3
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6. Shear Strength of Soils
 Vane Shear Test
 The vane shear device, a rigid, cross-shaped vane used in the field or lab.
 Inserted into the soil twisted applied torque T.
 D3 D2 
T f   Cu 

H
2
 8

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Reference
1. Smith L., Elements of Soil Mechanics, 9th Edit., Wiley Blackwell, 2014.
2. Budhu M. Soil mechanics and foundations, 3rd Edit. John Wiley & Son, 2011.
3. Philipponnat G., Hubert B., Fondations et ouvrages en terre, 1er Edit. Eyrooles, 1992.
4. Ishibashi I., Hazarika H., Soil Mechanics Fundamentals and Applications, 2nd Edit., CRC Press, Taylor & Francis
Group, 2015.
5. J.A. Knappett and R. F. Craig, Craig’s Soil Mechanics, 8th Edit., Spon Press, London and New York, 2012.
6. DAS B. M., Principles of Geotechnical Engineering, 8th Edit., CENGAGE Learning, 2012.
7. M. Khemakhem and Z. Bouarada, Elements de Géotechnique.
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