4) Soil Composition
Prepared From The Coduto’s Text Book
by
Instr. Nurullah AKBULUT
.in engineering practice difficulties with
soils are almost exclusively do not to the
soils themselves but to the water contained
in their voids. On a planet without any
water there would be NO need for soil
mechanics.
Karl Terzaghi, 1939
Once the soil and rock samples have been brought to the
laboratory, we need to conduct appropriate tests to develop data
for our analyses. Some of these tests measure familiar
engineering properties, such as shear strength, while others focus
on the sample's composition and structure.
The composition of soil and rock is quite different from that of
other civil engineering materials, such as steel, concrete, or
wood. These differences include:
 Soil and rock are natural materials, not manufactured products
 Soil is a particulate material that consists of individual particles
 Soil can contain all three phases of matter (solid, liquid, and
gas) simultaneously, and these three phases can be present in
varying proportions. Rock also can contain all three phases,
although the liquid and gas phases may be confined to the
fissures .
The most Important Properties of Soils
•
•
•
•
•
Elastoplastic material
Particulate structure
Heterogenity
Non-Isotropi
Three phases of matter
This chapter discusses the methods we use
to assess the composition of soils and the
parameters we use to describe this
composition
SOIL AS A PARTICULATE MATERIAL
Most civil engineering materials consist of a continuous mass held
together with molecular bonds, and the mechanical properties of such
materials depend on their chemical makeup and on the nature of these
bonds. For example, the shear strength of steel depends on the
strength of the molecular bonds, and shear failure requires breaking
them.
In contrast, soil is a particulate material that consists of individual
particles assembled together as shown in Figure 4.1.
Its engineering properties depend largely on the interaction between
these particles, and only secondarily on their internal properties. This
is especially true in gravels, sands, and silts. For example, when soils
fail in shear, they do so because the particles begin to roll and slide
past each other, not because the particles break internally. Breakage of
individual particles is typically minimal.
Thus, the shear strength depends on factors such as the coefficient of
friction between the particles, the tightness of packing, and so on,
rather than the chemical bonds inside the particles.
Clays also have a particulate structure, but the nature of the particles is
quite different, as discussed later in this chapter. In clays, there is
much more interaction between the particles and the pore water,
so their behavior is more complex than that of other soils.
THE THREE PHASES
• Soil also is different from most civil engineering
materials in that it can simultaneously contain solid,
liquid, and gas phases.
• The liquid and gas phases are contained in the voids
or pores between the solid particles.
• The three phases often interact, and these
interactions have important effects on the soil's
behavior.
The Mineral Skeleton-Three Phase
Solid Particles
Volume
Voids
• The solid phase: is always present in soil, and usually
consists of particles derived from rocks. It also can
include organic material.
• The liquid phase: is usually present, and most often
consists of water. However, it also can include other
materials, such as:gasoline,leachate, seawater,
petroleum seeps.
• The gas phase: If the liquid phase does not completely
fill the voids, then the remaining space is occupied by
the gas phase. It is usually air, but can include other
gasses, such as: methane, carbon diokside, hydrogen
sulfide, e.t.c.
Clearly, components other than water and air
are very important. However, they generally
represent only a small portion of the soil
weight and volume.
Therefore, for purposes of this
chapter, we will simply refer
to the liquid and gas phases as
"pore water" or “water” and
"pore air“ or “air”.
WEIGHT-VOLUME
RELATIONSHIPS
It is helpful to identify the relative proportions
of solids, water, and air in a soil, because
these proportions have a significant effect on
its behavior.
Therefore, geotechnical engineers have
developed quantitative methods of assessing
these components.
Phase Diagrams
Three Phase System
Va
Air
Vw
Water
Wa=0
Vv
V
Ww
W
Vs
Volume
Solid
Ws
Weight,
or Mass
Definitions of Weight-Volume Parameters
Moisture Content or Water Content
A small w indicates a dry soil, while a large w indicates a wet one.
Values in the field are usually between 3 and 70%.
But values greater than 100% are sometimes found in soft soils
below the groundwater table, which simply means such soils have
more water than solids.
Degree of Saturation
 This is similar to moisture content in that both are equal to zero when
there is no water. However, S has a maximum value of 100%, which
occurs when all of the voids are filled with water. We use the term
saturated to describe this condition.
 Soils below the groundwater table are generally saturated.
 Values of S above the groundwater table are usually between 5 and
100%, although values approaching zero can be found in very arid areas.
 Capillary effects can draw water upward from the groundwater table,
often producing soils with S =100% well above the groundwater table.
Therefore, do not use saturation computations to determine the location
of the groundwater table. Instead, use observation wells.
Unit Weight and Density
In all calculations,
w = 9.81 kN/ m3= 62.4 Ib/ft3, rw= 1000 kg/m3 for fresh water.
 The unit weight of undisturbed soil samples can easily
be determined in the laboratory by measuring their
physical dimensions and weighing them. This method
produces reliable assessments of  for many soils.
 However, it is affected by sample disturbance,
especially in sandy and gravelly soils.
 Sometimes unit weight measurements are made on
supposedly "undisturbed" samples that in reality have
significant disturbance.
 Such measurements are very misleading, so it is best to
not even attempt unit weight measurements on poor
quality samples
 For most geotechnical computations, unit weight is
more useful than density because we use it to compute
stresses due to the weight of the soil.
Specific Gravity of Solids
 The specific gravity of any material is the ratio of its density to that of
water. In the case of soils, we compute it for the solid phase only, and
express the results as the specific-gravity of solids;
 This is quite different from the specific gravity of the entire soil mass,
which would include solid, water, and air. Therefore, do not make the
common mistake of computing Gs as  / w.
 Nearly all real soils have 2.60 <GS< 2.80, which is a very narrow
range. The additional precision obtained by performing a test is
generally not worth the expense.
For most practical problems, it is sufficient to estimate ,from the following list:
• Clean, light colored sand of quartz and feldspar
2.65
• Dark colored sand
2.72
• Sand-silt-clay mixtures
2.72
• Clay
2.65
Nevertheless, some unusual soils have olivene values well
outside these limits. For example, the olivene sands in
Hawaii have Gs values as high as 4.50, while organic soils
have low Gs values, sometimes less than 2.0.
Void Ratio
 The relative volumes of voids and solids may be
expressed as the void ratio
 Densely packed soils have a low void ratio.
 Typical values in the field range from 0.1 to 2.5.
Porosity
 It is a similar parameter and related with e
 It typically is between 9 and 70%
 Sometimes geotechnical engineers divide the porosity into
two parts: the water porosity, nw, and the air porosity, na:
Relative Density
 It is a special weight-volume parameter used in sandy and
gravelly soils.
 The values of emax and emin represent the soil in very dense and
very loose conditions, respectively. Thus, loose soils have low
values of Dr while dense soils have high values.
 In theory, the lowest possible value of Dr is 0% and the highest
possible value Dr is 100%.
 Thus, Dr is often more useful than e because we can easily
compare the field value to the lowest and highest possible
values.
These are not intended to be used in lieu of laboratory or in-situ tests, but
could be used to check test results or for preliminary analyses.
Derived Equations
with Weight Volume Relationships by Phase Diagrams
For all of these equations, parameters normally expressed as a percentage
must be inserted in decimal form.
For example, a degree of saturation of 45% in Equation would be expressed
as 0.45 not 45
 From the definition of void ratio, if the volume of solids is 1 unit then the volume of
voids is e units.
 The mass of solids is then Gsrw and, from the definition of water content, the mass
of water is Gsrw .
 The volume of water is thus wGs. These volumes and masses are represented in
figure.
 Then a number of relationships can also be derived by using the phase diagram.
Solving Weight-Volume Problems
We often encounter problems where one or more of the weight-volume parameters is known
and others need to be determined.
For example, some parameters, such as moisture content, can be measured in the laboratory,
while others, such as void ratio, cannot. Therefore, we need to have a means of
computing them.
Often we can perform these computations using the derived equations presented last.
However, if this method does not work, we go back to fundamentals and solve the
problem using a phase diagram as follows:
1. Draw a phase diagram and annotate all of the dimensions presented in the problem
statement. Remember to set Wa = 0. If the soil is saturated, we also can set Va = 0.
2. Sometimes no weights or volumes are given in the problem statement, or the only
dimensions given are equal to zero (i.e., no air or no water). If this is the case, then the
given data is applicable to the entire soil strata, regardless of the sample size. However,
the phase diagram analysis method requires that a certain quantity of soil be specified,
and will not work unless we do so. Therefore, we must assume a quantity. Any one
dimension may be assumed (but only one!). Usually we assume V= 1 m3 or V= 1 ft3.
3. Using basic equations and obvious addition and subtraction from the phase diagram (i.e.,
Va= V - Vs - Vw), determine all of the remaining dimensions.
4. Compute the required parameters using basic equations from these dimensions in phase
diagram
Example 1
A 27.50 Ib soil sample has a volume of 0.220 ft3,
a moisture content of 10.2%, and a specific
gravity of solids of 2.65. Compute the unit
weight, dry unit weight, degree of saturation,
void ratio, and porosity.
Solution using fundamental and derived equations
Solution using a phase diagram
• Although the fundamental and derived equations
were sufficient to solve this problem, and would
be the easiest method, we also will illustrate a
solution using a phase diagram.
• IN EXAMINATIONS and QUIZES, YOU
MUST SOLVE RELEVANT PROBLEMS BY
USING ONLY THIS METHOD
Step 1: Draw and annotate a phase diagram
Step 2: Assume dimension
Not applicable to this problem, because weights and
volumes have been given
Step 3: Determine dimensions in phase diagram.
Step 4: Compute parameters.
Example 2
A certain soil has the following properties:
Gs= 2.71
n= 41.9%
w = 21.3%
Find the degree of saturation, S and the unit
weight, by using phase relations?
Solution
Step 1: Draw and annotate phase diagram
Step 2: Assume dimension.
No weights or volumes were stated, so all of the given data applies to
the entire soil strata. Therefore, we will develop a phase diagram
for an assumed volume of 1 m3
Step 3: Determine dimensions in phase diagram.
Step 4: Compute parameters.
Example 3
The standard method of measuring the specific gravity of solids (ASTM D854) uses
a calibrated glass flask known as a pycnometer, as shown in Figure 4.6. The
pycnometer is first filled with water and set on a balance to find its mass. Then, it is
refilled with a known mass of dry soil plus water so the total volume is the same as
before. Again, its mass is determined. From this data, we can compute Gs
Using this technique on a certain soil sample, we have obtained the following data:
•
•
•
•
•
Mass of soil = 81.8 g
Moisture content of soil = 11.2%
Mass of pycnometer+water = 327.12 g
Mass of pycnometer+soil+water = 373.18 g
Volume of pycnometer = 250.00 ml
Compute Gs for this soil.
Solution:
PARTICLE SIZE AND SHAPE
• The individual solid particles in a soil can
have different sizes and shapes.
• These characteristics also have a significant
effect on its engineering behavior.
• Therefore, geotechnical engineers often
assess particle size and shape.
Particle Size Classification
• Several systems have been
developed to classify soil particles
based on their size.
• We will now examine only one:
the ASTM system.
• According to ASTM, particles
larger than 3 in (75 mm) in
diameter are known as rock
fragments. Smaller particles are
defined as soil, and are classified
according their ability to pass
through certain size sieves.
• A sieve is a carefully manufactured
mesh of wires with a specified
opening size, as shown in Figure
4.8.
Photographs of samples from each category
Natural deposits often include a mixture of both rock fragments and soil.
Laboratory Tests
Although the distribution of particle sizes can
often he estimated by eye, two laboratory
tests are commonly used to provide more
precise assessments:
1) sieve analysis
2) hydrometer analysis.
Sieve Analysis
• A sieve analysis is a laboratory test that measures the grain-size
distribution of a soil by passing it through a series of sieves.
• The larger sieves are identified by their opening size. For
example, a 3/4-inch sieve will barely pass a 3/4-inch diameter
sphere.
• Smaller sieves are numbered, with the number indicating the
openings per inch. For example, a #8 sieve has 8 openings per
inch or 64 per square inch.
• However, the size of these openings is less than 1/8 inch because
of the width of the wire. Table 4.7 presents opening sizes for
standard sieves used in North America.
Most people can just barely see 0.1 mm
diameter objects without using a
magnifying glass, and this nearly
corresponds to the #200 sieve. This
diameter also represents the border
between gritty and smooth textures. Thus,
referring back to Table 4.6, sands and silts
could be visually distinguished by looking
at the particles (if you can see them, it's a
sand) and by feeling for grittiness (if it
feels gritty, it's a sand). Both are best done
with wetted soil samples.
The test (ASTM D422) consists of preparing a
soil sample with known weight of solids,
Ws and passing it through the sieves. The
sieves are arranged in order with the
coarsest one on top. A pan is located below
the finest sieve. The weight retained on
each sieve is then expressed as a
percentage of the total weight.
The percentage passing the #200 sieve is
especially noteworthy. Soils that have less
than about 5 percent passing the #200 sieve
are called "clean," while those that have
more are called "dirty." For example, clean
sand is primarily sand, with less than 5
percent silt or clay.
Hydrometer Analysis
• Although sieve analyses work very well for
particles larger than the #200 sieve (sands and
gravels) and they determine the total amount of
fines, they do not give any insight on the
distribution of finer particles (silts and clays).
• The smallest clay particles are only about 1xl0-4
mm in diameter, which is about the same size as
a smoke particle. It is impossible to manufacture
sieves this small, so we need to use another
technique:
The hydrometer analysis
This procedure (ASTM D422)
consists of placing a soil
sample with a known Ws into
a 1000 ml graduated cylinder
and filling it with water.
The laboratory technician
vigorously shakes the cylinder
to place the soil in suspension,
and then places it upright on
a table as shown in Figure
4.12.
Once the cylinder has been set
upright, the soil particles
begin to settle to the bottom.
 The velocity is proportional to the square of the particle
diameter, so large particles settle much more quickly than
small ones.
 In addition, we can determine the mass of solids still in
suspension by measuring the specific gravity of the soilwater mixture, which is done using the hydrometer
 Therefore, by making a series of specific gravity
measurements, usually over a period of 24 hours, and
employing Stoke's law, we can determine the distribution
of particles sizes in the soil sample.
• The hydrometer analysis is unsuitable for
particles larger than about a #100 sieve
because they settle more quickly than we
can measure the specific gravity.
• However, by performing a sieve analysis,
hydrometer analysis, or both, we can
determine the distribution of particle sizes
for virtually any soil.
Grain Size Distribution Curves
 Real soils rarely fall completely within only one of the
categories listed in particle size classification. They almost
always contain a variety of particles sizes mixed together.
Therefore, we need to have an effective means of presenting
the distribution of particle sizes in a soil.
 That method is the grain-size distribution curve.
 These are plots of the grain diameter (on a logarithmic
scale) vs. the percentage of the solids by weight smaller
than that diameter. We call the latter "percent finer" or
"percent passing," which generates mental images of that
portion of the soil passing through a sieve.
• Soil A, indicates primarily fine-grained soils (silts and clays)
• Soil B, C, E indicate primarily coarse-grained soils (sands and
gravels)
Acording to Grain Size Disstribution Shape:
• Poorly-graded soils (or uniformly-graded soils). If Steep grainsize distribution curves, such as soil C, reflect soils with a
narrow range of particle sizes.
• Well-graded soils. If soils with flat curves, such as soil D,
containing a wide range of particle sizes
• Gap-graded soils. If soils have missing particles in a certain size
range in their grain-size distribution curve, such as soil E. Gapgraded soils are sometimes considered a type of poorly
graded soil. Aggregates used to make concrete are typically
gap graded.
We can determine the percentage of each type of soil
by weight by comparing the percents passing the
appropriate sieve sizes as listed in particle size
classification.
For example, to determine the amount of sand in a
soil, subtract the percentage passing the #200 sieve
from the percentage passing the #4 sieve.
The particle diameters that correspond to certain
percent-passing values for a given soil are known as
the D-sizes. For example, D10, is the grain size that
corresponds to 10 percent passing. In other words,
10 percent of the soil is finer then D10.
Some Engineering Parameters for Grain Size Distribution
Curves
•Coefficient of Uniformity
•Coefficient of curvature
Steep curves, which reflect poorly graded soils, have
low values of Cu , while flat curves (well-graded
soils) have high values.
Soils with smooth curves have Cc values between
about 1 and 3, while irregular curves have higher or
lower values. For example, most gap-graded soils
have a Cc outside this range.
Particle Shape
 The shape of silt, sand, and gravel particles varies from very
angular to well rounded. Angular particles are most often found
near the rock from which they were formed, while rounded
particles are most often found farther away where the soil has
experienced more abrasion.
 Angular particles have a greater shear strength than smooth ones
because it is more difficult to make them slide past one another.
This is why aggregate base material used beneath highway
pavements is often made of rocks that have been passed through a
rock crusher to create very angular gravel. Clay particles have an
entirely different shape, and are discussed later in the next
section.
 Some non-clay particles are much flatter than any of the samples.
One example is mica, which is plate-shaped. Although mica never
represents a large portion of the total weight, even a small amount
can affect a soil's behavior. Sands that include mica are known as
micaceous sands.
Example 4
Determine the following for soil D
in Figure 4.13 given the next:
• Percent gravel, sand, and fines
• Cu and Cc
Solution
Percent finer data from grain size curve:
#200 — 40%
#40 — 88%
3 in — 100%
Percentage of each type of soil:
Gravel = 3 in - #4 = 100% - 88% = 12%
Sand = #4 - #200 = 88% - 40% = 48%
Fines = #200 = 40%
D-sizes from grain-size curve
D10 = 0.0019 mm
D30 = 0.030 mm
D60 = 0.49 mm
Cu and Cc :
-Answer
-Answer
-Answer
Note:
This is an exceptionally large value of Cu
which reflects the very flat grain-size curve.
Most Cu values are less than 20.
CLAY SOILS
• Soils that consist of silt, sand, or gravel are primarily
the result of physical and mild chemical weathering
processes and retain much of the chemical structure
of their parent rocks.
• However, this is not the case with clay soils because
they have experienced extensive chemical
weathering and have been changed into a new
material quite different from the parent rocks.
• As a result, the engineering properties and behavior
of clays also are quite different from other soils.
Formation and Structure of Clay Minerals
Several different chemical weathering processes form clay
minerals which are the materials from which clays are made. We
will examine one of these processes as an example. This process
changes orthoclase feldspar, a mineral found in granite and many
other rocks, into a clay mineral called kaolinite. The process
begins when water acquires carbon dioxide as it falls through the
atmosphere and seeps through soil, thus forming a weak
carbonic acid (H2CO3) solution. This acid reacts with orthoclase
feldspar according to the following chemical formula:
4KAlSi3O8 + 2H2CO3 + 2H2O -> 2K2CO3 + Al4(OH)8Si4O10 +8SiO2
orthoclase + carbonic acid + water - potassium carbonate + kaolinite + silica
Finally, the potassium carbonate and silica are carried off in
solution by groundwater, ultimately to be deposited elsewhere,
leaving kaolinite clay where orthoclase feldspar once existed.
 In these various chemical weathering processes, it is formed sheetlike chemical structures.
There are two types of sheets:
1) Tetrahedral or silica sheets, consist of silicon and oxygen
atoms;
2) Octahedral or alumina sheets, have aluminum atoms
and hydroxyls (OH).
 Sometimes octahedral sheets have magnesium atoms instead of
aluminum, thus forming magnesia sheets.
 These sheets then combine in various ways to form dozens of
different clay minerals, each with its own chemistry and
structure.
The three most common ones are:
• Kaolinite
• Montmorillonite
(also called smectite)
• Illite
Kaolinite
Consists of alternating silica and alumina sheets. These
sheets are held together with strong chemical bonds,
So kaolinite is very stable clay.
Unlike most other clay minerals, kaolinite
does not expand appreciably when wetted.
So it is used to make pottery.
It also is an important ingredient in paper, paint, and other
products, including Pharmaceuticals (i.e., kaopectate).
Since kaolinite does not expand
appreciably when wetted,
It is more prefered for construction of
core section with clay in
embankmant dams especailly if
easily obtained around
Montmorillonite
Has layers made of two silica sheets and one
alumina sheet. The bonding between these layers is
very weak, so large quantities of water can easily
enter and separate them,
thus causing the clay to swell. This
property can be very troublesome or very useful,
depending on the situation
Problems with soil expansion include
extensive distortions in structures, highways,
and other civil engineering projects .
However, this expansive behavior and the low
permeability of montmorillonite can be useful
for sealing borings (especially in deep
excavation by slury trench) or providing
groundwater barriers.
Bentonite, a type of montmorillonite, is commercially
mined and sold for such purposes.
Illite
Has layers similar to those in montmorillonite, but
contains potassium ions between each layer.
The chemical bonds in this structure are
stronger than those in montmorillonite but
weaker than those in kaolinite, so illite expands
slightly when wetted. Glacial clays in the Great
Lakes region are primarily illite.
Other clay minerals include vermiculite, attapulgite,
and chlorite.
Properties of Clays
Because of the small particle diameter and plate-like
shape of clays, the surface area to mass ratio is
much greater than in other soils. This ratio is
known as the specific surface. For example:
Montmorillonite has a specific surface of about
800 m2/g, which means 3.5 g of this clay has a
surface area equal to that of a football field
Illite=> (80-100 m2/g)
Kaolinite => (10-20) m2/g
The large specific surface of clays provides more
contact area between particles, and thus more
opportunity for various interparticle forces to
develop.
It also provides more places for water molecules
to attach, thus giving clays a much greater
affinity for absorbing water. Some clay can
easily absorb several times their dry weight in
water.
Montmorillonite clays have the greatest specific
surface, so it is no surprise that they have the
greatest affinity for water.
The interactions between this water and the
clay minerals are quite complex, but the
net effect is that:
The engineering properties vary as the
moisture content (water content)
varies. For example, the shear strength
of given clay at a moisture content of
50% will be less than at a moisture
content of 10%.
This behavior is quite different from that in
sands, because their specific surface is
much smaller and the particles are more
inert. Other than changes in pressure
within the pore water, variations in
moisture content have very little effect on
the behavior of sands.
Clay Minerals compared with each other
• Strength: K>I>M
• Compressibilty: M>I>K
• Hydraulic Conductivity: K>I>M
Formation of Clay Soils
• On a slightly larger but still microscopic scale, clay
minerals are assembled in various ways to form clay
soils. These microscopic configurations are called
the soil fabric, and depend largely on the history of
formation and deposition.
• For example, residual clay, which has weathered inplace and is still at its original location, will have a
fabric much different from marine clay, which has
been transported and deposited by sedimentation.
These differences are part of the reason such soils
behave differently.
• Although we sometimes encounter soil strata that
consist of nearly pure clay, most clays are mixed
with silts and/or sands.
• Nevertheless, even a small percentage of clay
significantly impacts the behavior of a soil.
• When the clay content exceeds about 50 percent,
the sand and silt particles are essentially floating in
the clay, and have very little effect on the
engineering properties of the soil.
PLASTICITY AND THE
ATTERBERG LIMITS
• Silts and clays are two very different kinds of soils. Some
classification systems draw the line between them based on
particle size as determined from a hydrometer test, typically
at 0.001 to 0.005 mm.
• Although classification systems can be useful, they also can
be misleading because the biggest difference between silt
and clay is not their particle sizes, but their physical and
chemical structures is important.
• In addition, It would be impractical to use electron
microscopes or other sophisticated equipment to distinguish
between clays and silts on a routine works.
Therefore, in geotechnical engineering
Plasticity
is used for assesing engineering properties of clay and
silt
Plasticity, is described as the response
of a soil to changes in moisture
content.
 When adding water to a soil changes its consistency
from hard and rigid to soft and pliable, the soil is said to
be exhibiting plasticity.
 Clays can be very plastic and silts only slightly plastic,
whereas clean sands and gravels do not exhibit any
plasticity at all.
 This assessment can be made using visual-manual
procedures, and with experience one can distinguish
between clays and silt simply with the hands and a
water bottle.
More formal assessments of plasticity are
performed in the laboratory using the
Atterberg limits tests.
The Atterberg Limits
• In 1911, the Swedish soil scientist Albert Atterberg (18461916) developed a series of tests to evaluate the relationship
between moisture content and soil consistency. Then, in the
1930s, Karl Terzaghi and Arthur Casagrande adapted these
tests for civil engineering purposes, and they soon became a
routine part of geotechnical engineering.
• This series includes three separate tests: the liquid limit test,
the plastic limit test, and the shrinkage limit test. Together
they are known as the Atterberg limits tests
(ASTMD427andD4318).
• The liquid limit and plastic limit tests are routinely
performed in many soil mechanics laboratories. However,
the shrinkage limit test is less useful, and is rarely performed
by civil engineers.
Liquid Limit Test: Liquid Limit
• The liquid limit test uses the liquid limit device, a standard laboratory
apparatus. A soil sample is placed in the device, and a groove is cut
using a standard tool. The cup is then repeatedly dropped, and the
number of drops required for the groove to close for a distance of onehalf inch (13 mm) is recorded. The soil is then removed and its
moisture content is determined. This test is then repeated at various
moisture contents, producing a plot of number of drops vs. moisture
content.
• By definition, the soil is said to be at its liquid limit when exactly
25 drops are required to close the groove for a distance of one-half
inch (13 mm). The corresponding moisture content is determined
from the test data.
•
The liquid limit, LL or wL. is this moisture content expressed
without a percentage sign. For example, if the moisture content is
45%, the liquid limit is 45.
Liquid Limit Determination
The Plastic Limit Test:Plastic Limit
The plastic limit test procedure involves
carefully rolling the soil sample into threads. As
this rolling process continues, the thread
becomes thinner and eventually breaks. If the
soil is dry, it breaks at a large diameter. If it is
wet, it breaks at a much smaller diameter.
By definition, the soil is at the plastic limit when
it breaks at a diameter of one-eighth of an inch
(3 mm). The plastic limit, PL or wp , is this
moisture content with the percent sign dropped.
Consistency and Plasticity Assessments
Based on Atterberg Limits
The Atterberg limits test
results help engineers assess
the plasticity of a soil and its
consistency at various
moisture contents.
 Consistency of fine-grained soil varies in
proportion to the water content
liquid
Liquid limit
plastic
Plastic limit
semi-solid
Shrinkage limit
solid
Plasticity
Index
• Plasticity Index:
When the soil is at a moisture content between the liquid limit
and the plastic limit, it is said to be in a plastic state. It can
be easily molded without cracking or breaking. The
children's toy play-dough has a similar consistency. This
property relates to the amount and type of clay in the soil.
 The plasticity index, PI or IP ,measure of the range of
moisture contents that encompass the plastic state:
Soils with a large clay content retain this
plastic state over a wide range of moisture
contents, and thus have a large plasticity
index. The opposite is true of silty soils.
Table 4.8 describes soil characteristics at
various ranges of plasticity index. Clean
sands and gravels are considered to be no
plastic (NP).
Liquidity index:
The liquidity index, IL
compares the current
moisture content of a
soil, w, to its Atterberg
limits:
Thus, a liquidity index of 0
means the soil is
currently at the plastic
limit, and 1 means it is at
the liquid limit.
STRUCTURED VS. UNSTRUCTURED SOILS
Many soils contain additional physical features beyond a "simple"
particulate assemblage. These are known as structured soils and
include the following:
 • Cemented soils contain cementing agents that bind the
particles together. The most common cementing agents are
calcium carbonate (CaCO3) and iron oxides (Fe2O3). Both are
usually transmitted into the soil in solution within the
groundwater.
 • Fissured soils contain discontinuities similar to fissures in rock.
Stiff clays are especially likely to contain fissures.
 • Sensitive clays are those with a flocculated structure of clay
particles that resemble a house of cards. These soils are very
sensitive to disturbance, which destroys this delicate structure.
Unstructured soils are those that do not contain such special
features. Most geotechnical analyses are based on unstructured
soils, and thus often need to be modified when working with
structured soils.
ORGANIC SOILS
 An organic soil is one that contains a significant amount
of organic material recently derived from plants or
animals. It needs to be fresh enough to still be in the
process of decomposition, and thus retains a distinctive
texture, color, and odor.
 The identification of organic soils is very important,
because they are much weaker and more compressible
than inorganic soils, and thus do not provide suitable
support for most engineering projects. If such soils are
present, we usually avoid them, excavate them, or drive
piles through them to reach more suitable deposits.
SUMMARY
 1. Soil is a particulate material, so its engineering properties
depend primarily on the interaction between these particles. This
is especially true of gravels, sands, and silts. Clays also are
particulates, but their behavior is much more complex because of
the interaction between the particles and the pore water.
 2. Soil can include all three phases of matter simultaneously, and
their relative proportions are important. Geotechnical engineers
have developed a series of weight-volume parameters to describe
these proportions.
 3. The distribution of particle sizes in a soil also is important,
and this distribution can be determined by performing a sieve
analysis and/or a hydrometer analysis. The results are presented
as a grain-size distribution curve.
 4. The solid particles also have various shapes, which impact
their behavior.
 5. Clays are formed by chemical weathering processes. The
individual particles are much smaller than sands or silts, and their
engineering behavior is much more dependent on the moisture
content.
 6. Clays and silts are often distinguished from each other by assessing
their plasticity, which reflects their affinity for water. The Atterberg
limits tests, especially the plastic limit and liquid limit, help us do this.
 7. Structured soils are those with special features, such a
cementation, fissures, or flocculated structures. They behave
differently from unstructured soils, which do not contain these
features.
 8. Organic soils are those with a significant quantity of organic
matter. Their engineering properties are much worse than those of
inorganic soils.
THE END