1. Chapter III
PHYSICAL AND INDEX PROPERTIES
Specific objectives
To identify the basic components of soil and their effect on total soil mass.
To identify various basic terms associated with soil.
To identify inter relationship among these parameters.
To be able to calculate some parameters given the required other parameters.
Soil composition
Soil deposits comprise the accumulated solid particles of soil or other materials plus the void spaces that exist
between those particles. The void spaces are partially or completely filled with water or other liquid. Void
spaces not occupied by the fluid are filled with air or other gas. Since the volume occupied by a soil bulk may
generally be expected to include material in three states of matter – solid, liquid and gas – soil deposits are
referred to as three phase systems.
Significant engineering properties of a soil deposit, such as strength and compressibility, are directly related to
or at least affected by basic factors such as how much volume or weight of a bulk soil is solid particles or water
or air. Some terms involving these basic compositions are defined to understand the state or condition of a soil.
Calculations to determine bearing capacity of soil for foundations, to estimate foundation settlement, to
determine stability of slopes require these parameters. For this reason, an understanding of the terminology
and definitions relating to soil composition is fundamental to the study of soil mechanics.
Three Phase System
We know soil bulk consists of solid particles, water and air in random arrangement, and for convenience of
analysis let us represent it as three separate quantities as shown in fig.no.1 which is known as three-phase
model of soil. It helps us to define and understand several basic properties of soil. Now we can conveniently
define several terms related to soil.
water Volume Weight
Va air Wa=0
Vv
Vw water Ww
V
W
Vs solid Ws
air solids
Fig.No.1. Three phase model for soil
Volume relationships
Specific volume (Sv): It is the total volume of soil bulk containing unit volume of solid particles. It can be
denoted by ‘Sv’. Specific volume for a soil is the sum of unity and volume of voids. For a hypothetical spherical
rhombic packing Sv =1.35 and for cubic packing Sv=1.92.Generally the value of specific volume ranges
between 1.43 and 1.67, for well graded sands and between 1.51 and 1.85, for uniform sands.

2. HAWASSA UNIVERSITY FACULTY OF TECHNOLOGY CIVIL ENGINEERING DEPARTMENT
Void ratio (e): It is the ratio of the volume of voids to the volume of soil solids in a given soil mass.
e = Vv /Vs
Then e = Vv -------- if we take Vs = 1
Therefore V = 1 + e
The volumes of various components of soil in the phase diagram are worked out or are considered w.r.t above
two definitions. Volume of solids is taken equal to unity and volume of voids equal to e.
Porosity (n): It is the ratio of the volume of voids to the total volume of soil mass.
n = Vv / V
V = total volume of soil specimen
Also n = Vv / (Vs + Vv)
Dividing both numerator and denominator by Vs we get
= (Vv / Vs) / [(Vs /Vs) + (Vv / Vs)]
= e / (1 + e)
Degree of saturation (S): It is the ratio of the volume of water in the void spaces to the volume of voids,
generally expressed as a percentage.
S (%) = (Vw / Vv) 100
For saturated soils, the degree of saturation is 100%.
Weight relationships
Specific gravity of soil solids (Gs): It is the ratio of weight of some volume of soil solids to weight of same
volume of water, both being measured at zero degree temperature.
Specific gravities of some soils
Soil type quartz sand silt clay chalk loess peat
Gs 2.64 – 2.66 2.67 – 2.73 2.7 –2.9 2.6 –2.75 2.65 – 2.73 1.3 –1.9
Moisture content (w): It is the ratio of weight of water to the weight of solid particles usually expressed in
percentage.
Moisture content w% = (Ww / Ws) 100
Moist unit weight (γ): It is the ratio of total weight of soil mass to total volume of soil mass.
Moist unit weight = γ = W / V
Dry unit weight (γd): It is the ratio of weight of solid particles to total volume of soil mass. It could also be
defined as the ratio of total weight of soil mass to its total volume in dry state (absence of moisture).
γd = Ws / V
Saturated unit weight (γsat): It is the ratio of total weight of soil mass to its total volume when soil is completely
saturated. In other words when a soil mass is completely saturated, the moist unit weight of soil becomes equal
to the saturated unit weight.
γ = γsat if Vv = Vw.
Submerged unit weight: (γ' ) In many soil mechanics problems it is necessary to determine the net inter
granular weight or effective weight of a soil when it is below the ground water table. In this state soil particles
are buoyed up by the pressure of the surrounding body of water. The effective soil weight then becomes the
weight of the soil material when it is weighed under water. The weight of the soil solids is reduced by the weight
of the volume of water they displace.
Therefore Wsub = VsGsγw - VsGwγw
= Vsγw ( Gs – Gw )
= Vsγw (Gs – 1)
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γsub = Wsub / V
= [ Vsγw (Gs – 1 ) ] / [ Vs (1 + e ) ]
∴ γsub = ( Gs – 1 ) γw / (1 + e ) ------------------------------------- ------------------------(1)
Precise determination of submerged unit weight of a soil requires accurate determination of specific gravity of
soil solids and void ratio. As such for many practical problems it is usually taken half of the bulk unit weight.
In-situ unit weight: It is nothing but moist unit weight of a soil as it is in nature. Care has to be taken not to
disturb the natural arrangement of solids, water and air during determination of in situ unit weight.
Relative density (Dr): The actual packing of sand and gravel soils is conveniently expressed in terms of the
loosest and the densest states through the parameter relative density Dr.
Dr = {( emax – e0 ) / ( emax – emin )}100
= {γd max / γd }{(γd - γd min ) / (γd max - γd min )}100
Where e0 and γd are the actual values of the void ratio and the dry unit weight of soil. emax and emin are the
values of void ratios of the soil in its densest and loosest states. The values of emax and emin depend primarily
on the particle roundness and the uniformity co efficient, they increase with an increase of the angularity and
decrease for larger values of the uniformity co efficient.
Denseness of granular soils
Relative density Dr,% 0 –20 20 – 40 40 – 60 60 – 80 80 -100
Description very loose loose medium dense very dense
Typical values of void ratio, moisture content and dry unit weight of soils
natural moisture content Dry unit weight in
Type of soil void ratio
saturated condition, % KN/M3
Loose uniform sand 0.80 30 14.5
Dense uniform sand 0.45 16 18
Loose angular grained silty sand 0.65 25 16
Dense angular grained silty sand 0.40 15 19
Stiff clay 0.60 21 17
Soft clay 0.9 – 1.4 30 – 50 11.5 – 14.5
Loess 0.9 25 13.5
Soft organic clay 2.5 – 3.2 90 – 120 6–8
Glacial till 0.3 10 21
Relationships between parameters
Va air Wa=0
Vw=wGs=e Ww=wGsγw=eγw Vv =e
Vv=e Vw=wGs water Ww=wGsγw
Vs=1 Ws=Gsγw Vs =1 solid Ws=Gsγw
Saturated soil Unsaturated soil
Fig.No.2. Weight - Volume relationship
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Assuming volume of solids equal to unity some useful relationships among the earlier defined parameters can
be obtained as follows.
Since Vs = 1, Vv = e and Ws = Vs Gs γw
= Gs γw -------------------- (as Vs = 1)
Weight of water Ww = w Ws
∴ Ww = w Gs γw
Then bulk unit weight γ = W / V
= (Ws + Ww) / (Vs + Vv)
∴ γ = Gs γw (1 + w) / (1 + e) -------------------------------------------------(2)
Similarly for dry unit weight γd = Ws / V
= Ws / (Vs + Vv)
= Gs γw / (1 + e)
∴γd = Gs γw / (1 + e) --------------------------------------------------(3)
From equations (2) and (3)
γ = γd (1 + w)
∴γd = γ / (1 + w ) --------------------------------------------------(4)
If soil specimen is completely saturated Vv =e
Also Vv = Ww / γw
= w Gs γw / γw
= wGs
∴ e = wGs (for saturated soils only) -------------------------------(5)
Then γsat = (Ws + Ww ) / ( Vs + Vv )
=( Gsγw + eγw ) / (1 + e )
Therefore γsat =( Gs + e )γw / (1 + e ) -----------------------------------------------(6)
Then from equations (1) and (6)
γsub = γsat - γw ------------------------------------------------(7)
wkt S = Vw / Vv
= (Ww / γw ) / Vv
=[ ( w Gs γw) / γw ] / Vv
= ( w Gs ) / e
∴ S = ( w Gs ) / e -------------------------------------------------(8)
From equations (6) and (7)
γsub = γsat - γw
= ( Gs + e )γw / (1 + e ) - γw
= [( Gs + e ) / (1 + e ) –1] γw
= [ (Gs + e - 1 - e ) / ( 1 + e )] γw
= [ (Gs - 1 ) / ( 1 + e )] γw ----------------------------------------------(9)
Similar relationships in terms of porosity ‘n’ can be arrived at by taking unit total mass of soil and
correspondingly redefining the volumes of other components as given below.
γ = Gsγw ( 1- n )(1+w) ------------------------------------------------(10)
γd = (1-n) Gsγw ------------------------------------------------(11)
γsat = [(1-n)Gs + n ]γw ------------------------------------------------(12)
γ′ = (1-n)( Gs – 1) γw -------------------------------------------------(13)
S = wγ / [n(1 +w)γw ] -------------------------------------------------(14)
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Particle size and shape
The individual solid particles in a soil can have different sizes and shapes, and these characteristics have a
significant effect on its engineering behavior. Also fine grained soils, because of the small particle diameter and
plate like shape posses surface area to mass ratio much greater than in other soils. Hence they behave
different from soil of large size particles. As such particle size and its shape have great influence on the
properties of soil. It is essential for geotechnical engineers to know particle size, their shape to understand soil.
Particle size distribution
Natural soil does not comprise particles of single diameter; rather it is a mixture of particles of different
diameters (sizes) in various proportions. As such it is necessary to find out the proportions of different sized
particles present in a soil to get an idea about the properties of the soil.
Although, the distribution of particle sizes in a soil can often be estimated by eye, two laboratory tests are
commonly used to provide more precise and uniform assessments from person to person. They are sieve
analysis and hydrometer analysis.
A sieve analysis is a lab test that measures the grain size distribution of a soil by passing it through a series of
sieves. The test consists of preparing a soil sample with known weight of soil solids, Ws, and passing it through
a set of 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 #200 sieve (size 0.075 mm) is noteworthy. In many classifications of soils they are
termed fine fraction of soils. Fraction retained over # 200 sieve are termed coarse fraction.
Sieve analysis is very well for particles larger than # 200 sieves (sands and gravels) but the distribution of finer
particles (< 0.075 mm) cannot be obtained by this analysis, as it is impossible to manufacture sieves of these
sizes. We use another technique, the hydrometer analysis for this purpose. In this test we place a soil sample
with known Ws in to a 1000ml graduated cylinder and fill it with water. After shaking it vigorously it is kept
upright on a table. The particles begin to settle to the bottom. We describe this downward motion using Stoke’s
law.
v = D2γw (Gs – Gl ) / (18η)
Where v = velocity of settling particle
D = particle diameter
γw = unit weight of water
Gs = Specific gravity of soil solids
Gl = specific gravity of soil water mixture
and η = dynamic viscosity of soil water mixture.
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 soil mass of solids still in suspension by measuring the
specific gravity measurements, usually over a period of 24 hrs, and employing Stoke’s law together with
specific gravity measurements, we can determine the distribution of particles sizes in the soil sample.
However, by performing sieve analysis, hydrometer analysis or both, we can determine the distribution of
particles sizes for virtually any soil.
Natural soils always contain a variety of particles sizes mixed together, which is effectively presented by a 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 ‘percent finer’ or ‘percent passing’.
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Fig.No.3. Grain size distribution curves for different soils
Curve ‘A’, (refer fig. no.3) indicate primarily fine grained soils (silts & clays) while these on the right side, such
as ‘B’, indicates coarse grained soils. Steep grain size distribution, such as soil ‘C’, reflects soils with a narrow
range of particle sizes. These are known as poorly graded soils (or uniformly graded soils). Conversely soils
with flat curves such as soil ‘D’ contain a wide range of particle sizes and are known as well graded soils.
Some soils have a nearly flat zone in their grain-size distribution curve, such as soil ‘E’. These are called gap-
graded because they are missing particles in a certain size range. Gap graded soils are sometimes considered
a type of poorly graded soil.
For further simple presentation of grain size distribution of a soil two additional parameters, the co-efficient of
uniformity(Cu) and the co-efficient of curvature(Cc) are defined as below.
2
D D30
C u = 60 and C c =
D10 D60 xD10
Where D10, D30, and D60 are sizes whose percent finer is 10, 30, and 60 respectively. 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 1 and 3, while irregular curves have higher or lower values.
The shape of silt and gravel particles varies from very angular to well round. Angular particles are most often
found near the rock from which they are formed, while rounded particles are most often found farther where the
soil has experienced more abrasion.
Angular particles have greater shear strength than smooth ones because it is more difficult to make them slide
past one another. Some non-clay particles are also much flatter. 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 soils
behavior.
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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 rock. However, this is not the case with clay
soils because they experience extensive chemical weathering and have been changed in to a new material
quite different from parent rocks.
Several different chemical weathering processes form clay minerals, which are the materials from which clays
are made. For example
4KalSi3O8 + 2H2CO3 + 2H2O → 2K2CO3 + Al4(OH)8Si4O10 + 8SiO2
(Orthoclase feldspar) (Carbonic acid) (Water) (Pot. Carbonate) (Kaolinite) (Silica)
Potassium carbonate and silica are carried off in solution by ground water. Ultimately to be deposited
elsewhere, leaving kaolinite clay where orthoclase feldspar once existed.
These various chemical processes form sheet like chemical structures. There are two types of sheets;
tetrahedral or silica sheets consist of silicon and oxygen atoms; octahedral or alumina sheets have aluminium
atoms and hydroxyls (OH), Fig.No.4. Sometimes octahedral sheets have magnesium atoms instead of
aluminium, thus forming magnesia sheets. These sheets then combine in various ways to form dozen of
different clay minerals, each with its own chemistry and structure.
Single silica tetrahedron Diagramatic sketch showing sheet structure of silica
tetrahedrons arranged in a hexagonal network
Single Octahedral Unit Diagramatic sketch showing sheet structure of octahedral units
Silicons
Oxygens
Aluminiums,magnesiums etc.
Fig.No.4. Atomic structure of tetrahedral and octahedral building blocks of clay minerals
Kaolinite--consists of alternating silica and alumina sheets. The name of this group of mineral is derived
from the name of a place in China, “Kauling” meaning high ridge of a hill near Jauchan Fu, China, where this
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mineral was available in large quantity. Each mineral block is approximately 7A0 in thickness. These sheets
are held together with strong chemical bonds, so Kaolinite is very stable clay.
Montmorillonite -- (Smectite) contains layers made of two silica sheets and one alumina sheet. The bonding
between these layers is very weak as it is only Vander walls forces hold them together, so large quantities of
water can easily enter and separate them, thus causing the clay to swell.
Illite-- has layers similar to those in Montmorillonite and contains potassium ions between each layer. The
chemical bonds in this structure are stronger than those in Montmorillonite but are weaker than those in
kaolinite. So Illite expands slightly when wetted. (Fig.No.5)
Fig.No.5. Minerological structure of clay minerals
F
Properties of clays
Individual clay particles are extremely small (less than 2 μm dia) and plate like in shape, the surface to mass
ratio is much greater. This ratio known as specific surface is 800 M2 /g for montmorillonite. The large specific
surface of clays provides more contact area between particles, and thus more opportunity for various inter
particle forces to develop. It also provides much greater affinity for absorbing water. The interaction between
this water and clay minerals are quite complex, but the net effect is that the engineering properties vary as the
moisture content varies. This behavior is quite different from that in sands, because their specific surface is
much smaller and the particles are more inert.
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. Residual soils will have a fabric
much different from marine clay, which has been transported and deposited by sedimentation.
Although we sometimes encounter soil strata that consist of rarely pure clay, most clay 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%, the sand and silt particles are essentially floating in clay and have very little
effect on the engineering properties of soil.
Consistency of soils
Consistency refers to the texture and firmness of soil and usually related to the strength. Consistency of soil is
conventionally described as soft, medium stiff, stiff and hard. As the terms are relative and mean different
states to different people, there is necessity to evolve more convenient and clear definitions for these states of
soil. The consistency of coarse grained soils is not much important. Consistency of clays is very important as
they exhibit drastic variation of strength with change in its consistency. Consistency of field clays is associated
with cohesion and unconfined compressive strength of soils.
Consistency of disturbed soils or remoulded soils varies with water content. At high water contents soils exhibit
the properties of liquid; at lesser water contents, the volume of the mixture is decreased and the material
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exhibits the plastic properties; at still less water contents, the mixture behaves as semisolid and finally as a
solid.
The water content indicating the division between the liquid and plastic state has been designated as liquid
limit. The water content at the division between the plastic and semi-solid state is the plastic limit. The water
content at the division between the semisolid and solid state is the shrinkage limit.
At water contents above the shrinkage limit, the total volume of the soil-water mixture changes in proportion to
change in the water content. Below the shrinkage limit, there is little or no change in the volume as water
content varies.
Standardized methods to determine these limits have been developed which are studied in laboratory.
Several other indices are also evolved to conveniently understand the consistency and properties of the soil.
One such parameter is plasticity index. Plasticity index is the numerical difference between liquid and plastic
limits for a soil. Another index known as Liquidity index gives an idea about the consistency of soil in field,
which is mathematically represented as below.
w% − w p w% − w p
LI = =
wl − w p PI %
Where w is the water content of the soil, wl is the liquid limit and wp is the plastic of the soil. A value of greater
than 1 for the above index indicates that the soil is in liquid state. A value less than 1 indicates the soil is in
plastic state and value near to zero indicates the water content is near plastic limit wher soils exhibits relatively
high strength. Negative values of LI are also possible and indicate a desiccated (hard), hard soil.
Fig.No.6. relationship between volume water and volume of soil
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Problems
1. A representative soil specimen collected in the field weighs 1.8KN and has a volume of 0.1 M3. The
moisture content as determined in laboratory is 12.6%. For Gs=2.71, determine the e,n,S,γ and γd.
2. A 72 cm3 sample of moist soil weighs 141.5 g. When it is dried out in an oven, it weighs 122.7 g. The
sp. gr. of solids is found to be 2.66. Find w,e,n,S,γ and γd.
3. An undisturbed soil sample has a void ratio of 0.56, water content of 15% and sp. gr. solids of 2.64.
Find n,S,γ and γd.
4. A fine sand has an in-place unit wt. of 18.85KN/M3 and a water content of 5.2%. The sp. gr. of solids is
2.66. Void ratios at densest and loosest conditions are 0.38 and 0.92, respectively. Find the relative
density.
5. Calculate the dry unit wt, the sat. unit wt and the buoyant unit wt. of a soil having a void ratio of 0.7 and
a value of Gs of 2.72. Calculate also the unit wt. and water content at a degree of saturation of 75%.
6. A soil specimen is 38mm in dia. and 76mm long in its natural condition and weighs 168.0 g. when dried
completely in an oven the specimen weighs 130 5g. The value of Gs is 2.73. What is the degree of
saturation of the specimen?
7. The in situ dry density of a sand is 1.72 mg/m3. The max. and min. dry densities, determined by
standard lab. tests, are 1.81Mg/m3 and 1.54Mg/m3 respectively. Determine the density index of the
sand.
8. A large soil sample obtained from a borrow pit has a wet mass of 26.5 kg. The in-place volume
occupied by the sample is 0.013 m3. A small portion of the sample is used to determine the water
content; the wet mass is 135g, and after drying in an oven, the mass is 117g.
a) Determine the soil’s water content.
b) Determine the soil wet and dry density for conditions at borrow pit. (David F Mc Carthy)
9. An undisturbed soil sample has a dry mass of 59 kg and an in situ (in-ground) volume of 0.035 m3. The
specific gravity of the soil particles is 2.65. Determine the void ratio, e. (David F Mc Carthy)
10. A volume of undisturbed soil, 0.015 m 3, obtained from a construction site has a wet mass of 31 kg and
a dry mass of 27.5 kg. Gs of soil solids is 2.71. Determine the void ratio, water content, and degree of
saturation for the in-place condition. (David F Mc Carthy)
11. For a sand, the maximum and minimum possible void ratios were determined in the laboratory to be
0.94 and 0.33, respectively. Find the most unit weight of sand in kN/m3 compacted in the field at a
relative density of 60% and moisture content of 10%. Given: Gs =2.65. Also calculate the maximum
and minimum possible dry unit weights that the sand can have. (Braja M Das)
12. For a saturated soil, given γd = 15.29 kN/m 3 and w=21%, determine:
a) γsat b) e c) Gs d) γmoist when the degree of saturation is 50% (Braja M Das)
13. A soil deposit has a void ratio of 10.0 and the specific gravity of its solids is 2.0.
a) if S=100% determine its γb b) if S=90% determine its γb (Gulhati S K and Datta M)
14. A clay soil is found to have a liquid limit of 75 %, a plastic limit of 45%, and a shrinkage limit of 25% if a
soil sample of this soil has a total volume of 30 cm3 at the liquid limit and a volume of 16.7 cm3 at the
shrinkage limit, what is the specific gravity of the soil solids. (David F Mc Carthy)
15. A soil when tested for consistency limits in laboratory revealed the following values. wl= 70%, wp=40%
and ws= 20%. The value of liquidity index was reported as 0.22. What is the water content of the soil in
field and in which consistency it is in the field?
16. The laboratory test results of a sand are emax= 0.91,emin=0.48 and Gs=2.67. What would be the dry and
moist unit weights of this sand when compacted at a moisture content of 10% to a relative density of
65%?
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