B.

TECH – SEMESTER V

CEB4303
GEO TECHNICAL
ENGINEERING

J. Karthick

A.P (S.G)

Department of Civil Engineering

OUTCOMES
At the end the course, the student will be able to
CO ‘ s OUTCOMES

CO1 Classify various types of soil. BTL2, BTL3

CO2 Analyze permeability, seepage and flow nets.
BTL2, BTL 3
CO3 Determine the stresses in soil through various
methods BTL2, BTL3
CO4 Describe compaction and consolidation and
determine the factors
affecting them. BTL3
CO5 Determine the strength of different types of soil
under various conditions. BTL 2, BTL 3
 MODULE 1

MODULE 1: INDEX PROPERTIES AND CLASSIFICATION OF

SOIL
 MODULE 1 Syllabus

• Nature of Soil - Soil - phase relation - sieve analysis - sedimentation

analysis - Atterberg limits - classification for engineering purposes - BIS
Classification system
 THE TERM “SOIL”

• The term "soil" can have different meanings, depending upon the field in which it is
considered

• To a geologist, it is the material in the relative thin zone of the Earth's surface
within which roots occur, and which are formed as the products of past surface
processes. The rest of the crust is grouped under the term "rock".

• To a pedologist, it is the substance existing on the surface, which supports plant

life.

• To an engineer, it is a material that can be construct foundations, buildings,

bridges, embankments, roads, dams, retaining walls etc.
 Soil Mechanics
• Soil Mechanics is a discipline of Civil Engineering involving the study of
soil, its behaviour and application as an engineering material.

• Soil Mechanics Concerns the application of various fields which

includes Mechanics, Hydraulics, chemistry to engineering problems
associated with Soils.

• The study of the science of Soil Mechanics will make the Civil Engineer
to understand the soil behaviour under various conditions.

• Geo technical Engineering a new term of studies which includes Soil

Mechanics, Rock Mechanics, Geology, Hydrology and Structural
Engineering.
 Formation of Soils
• Physical Weathering

This reduces the size of the parent rock material, without any change in the original
composition of the parent rock. Physical or mechanical processes taking place on the
earth's surface include the actions of water, frost, temperature changes, wind and ice.
They cause disintegration and the products are mainly coarse soils. The main process
involved in physical weathering are exfoliation, unloading, erosion, freezing, and
thawing. In exfoliation, the outer shell separates from the main rock. Heavy rain and wind
cause erosion of the rock surface.

• Chemical weathering

This process breaks up the material into smaller particles and also alters the nature of
the original parent rock itself. The main processes responsible are hydration, oxidation,
and carbonation. New compounds are formed due to the chemical alterations.
Classification of Soils (Based on Formation)

Based of the mode of formation, soils are broadly classified in to two broad
categories namely

• Residual soils

• Transported soils
 Residual Soils

• Residual soils are found at the same location where they have been formed.

Generally, the depth of residual soils varies from 5 to 20 m.

• Accumulation of residual soils takes place as the rate of rock decomposition

exceeds the rate of erosion or transportation of the weathered material. In humid

regions, the presence of surface vegetation reduces the possibility of soil

transportation.

• Residual soils comprise of a wide range of particle sizes, shapes and composition .
 Transported Soils

Weathered rock materials can be moved from their original site to new locations by one
or more of the transportation agencies to form transported soils. Transported soils are
classified based on the mode of transportation and the final deposition environment.
(a) Soils that are carried and deposited by rivers are called alluvial deposits.
(b) Soils that are deposited by flowing water or surface runoff while entering a lake are
called lacustrine deposits.
(c) If the deposits are made by rivers in sea water, they are called marine deposits. Marine
deposits contain both particulate material brought from the shore as well as organic
remnants of marine life forms.
(d) Melting of a glacier causes the deposition of all the materials scoured by it leading to
formation of glacial deposits.

(e) Soil particles carried by wind and subsequently deposited are known as aeolian

deposits.
 Types of Soil (Geological Consideration)
Following are the types of soil based on the geological agents or the
processes through which the soil deposits have been developed.

Glacial Soil: This type of soil is developed, transported and

deposited by the actions of glaciers. These deposits consists of
rocks fragments, boulders, gravels, sand, silt and clay in various
proportions (i.e., a heterogeneous mixture of all sizes of particles).

Residual Soil: This type of soil is found on nearly flat rock surfaces
were the weathering action has produced a soil with a little or no
tendency to move. Residual soil also occurs when the rate of
weathering is higher than the rate of removal.
• Alluvial Soil: The soil transported and deposited by water is called alluvial soil.
As flowing water (stream or river) looses velocity, it tends to deposit some of
particles that it was carrying in suspension or by rolling, sliding or skipping along
the river bed. Coarser or heavier particles are dropped first.

• Wind blown Soil or Aeolian Soil: The soil transported by the geological agent
‘wind’ and subsequently deposited is known as wind blown soil or Aeolian Soil.
Aeolian soil as two main types namely Dune sand (Desert soil) and Loess (Loose
soil without moisture).
 Types of Soil (Engineering Considerations)
• Clay: ( < .002mm): It is composed of very fine particles, less than .002
mm in size. They are flaky in shape and therefore have considerable
surface area. These surfaces carry electrical charge, which helps in
understanding the engineering properties of clay soils. In moist condition,
clay becomes sticky and can be rolled into threads.
• Silt: (.002mm < Size < .06mm): It is composed of very fine particles
ranging in size between .002 and .06mm. It has high capillarity, no
plasticity and very low dry strength. It possesses properties of both clay
and sand, i.e. it shows slight cohesion and also friction. The colour of silty
soil is mostly brown.
• Sand: (.06mm< Size < 2mm) It consists of particles ranging in size
from .06 and 2mm. It has a grey colour. These particles may be rounded to
angular in shape. It shows no plasticity, high strength in a confined state
and has considerable frictional resistance. It has high permeability and low
capillarity. Sand is the most wanted construction material. Abundant
quantities of sand are available in deserts and riverbeds.
 Types of Soil (Engineering Considerations)
• Gravels: (2mm< Size < 60mm):They consists of particles varying in size
from 2mm to 60mm. They form a good foundation material. They show
high frictional resistance. The frictional resistance depends upon the
particle size and shape. The gravels produced by crushing of rocks are
angular in shape, while those taken from river beds are sub-rounded to
rounded. They show very high permeability. When sand and silt are mixed
with gravels their bearing capacity is further increased but permeability
may be decreased.

• Cobbles and Boulders: Particles larger than gravels are commonly

known as cobbles and boulders. Cobbles generally range in size from
60mm t0 200mm. The material larger than 200mm is designated as
boulders.
• Density = ρ = = , in geo technical engineering all the units in terms of density is
expressed in terms of

• Unit weight = γ = = , in geo technical engineering all the units in terms of Unit
Weight of the soil is expressed in terms of .
 PHASE RULE
• The soil mass, in general three phase consist of Solid, liquid and gaseous matter.
The solid phase generally enclosed with open spaces called voids. The liquid
phase generally fills partially or wholly wit the voids. The gaseous phase (Usually
air) occupies the voids not filled by water.

• Phase rule normally classified in to 3- phase system (Partially saturated soil), 2 –

Phase system (Saturated, Dry soil).
 3 – Phase System
The volume and the Weight of Soil mass is
designated as V & W.
The subscriptions referred to a, w, s, and v stands
for Air, Water, Solids and Voids respectively.
The weight of air for all practical purposes will be
assumed as zero.
The combined Volume of air (Va ) and Volume of
Water (Vw) is called as Volume of Voids (Vv).

Total Volume
V = Va + Vw + VS
V = Vv + Vs
Total Weight
W= Wa +Ww +Ws
W = Ww +Ws
 Two phase Diagram
(SATURATED SOIL)

The total Volume

V = V s + Vw

The total Weight

W = W s + Ww
Two phase Diagram
(DRY SOIL)

The total Volume

V = Vs + V a

The total Weight

W = Ws
 DEFINITIONS
WATER CONTENT (w)
Water Content (w) or the Moisture content is defined as the Weight of water
to the Weight of Solids. Expressed in terms of Percentage.
w=

VOID RATIO (e)

Void ratio e, of a sample is defined as the ratio of Volume of Voids to the
Volume of Soil.
e=
 DEFINITONS (CONT)
POROSITY (n)
Porosity n, is defined as the ratio of Volume of voids to the Total
volume of the solids.
n=

Degree of Saturation (S)

Degree of saturation is defined as the ratio of volume of water to the
Volume of Voids.
S=
PRECENTAGE AIR VOIDS
Percentage air voids (na) is the ratio of volume of air voids expressed as a
total volume of soil mass.
na =

UNIT WEIGHT
Unit weight of the soil is its Weight per Unit Volume.

Bulk Unit Weight

Bulk Unit weight t ) is the total weight of soil mass per unit volume
t=
DRY UNIT WEIGHT )
It is the ratio of Weight of Solids per Unit soil mass
=

SATURATED UNIT WEIGHT )

It is ratio of Total weight of fully saturated soil mass to the total Volume of
the soil mass.
=
SUBMERGED UNIT WEIGHT (
When a soil mass is submerged below the ground water table, a buoyant force acts on
the soil mass, that will reduce the weight of soil mass equivalent to that of water.

= – ( = 9.81 KN / m3 or 1 g / cm 3 )

SPECIFIC GRAVITY (G or Gs)

Specific gravity of solids, is defined as the ratio of Weight of given volume of the solids to
the weight of an equivalent volume of water at 4. Celsius.
Gs = =
 Relationships
• Relationships between Ws, W and w

W= Ws+Ww = Ws( 1 + )
= Ws (1 + w)

WS =
• Relation ship between e & n

•n=

• e=
• Relationship between e, w, Gs and S

e=

Relationship between Bulk Unit Weight t , In terms of G, e, w and

=[ ] OR =
Relationship between Saturated Unit Weight , In terms of G, e, w and

=[ ]

Relationship between Dry Unit Weight , In terms of G, e, and

= , This equation can also be re written as

e= -1
• Relationship between Submerged Unit Weight , In terms of
G, e, and
• =[ ]
 Tutorial 1

• I cumec of wet soil weights 20KN, Its dry weight is 18KN , Sp gr

is 2.67. Determine Water content, Porosity, Void ratio, Degree of
saturation. Draw a phase diagram.
Solution:
Given data V=1 m3 W = 20KN Ws = 18KN G = 2.67
Ww = W – Ws = 20 -18 = 2 KN

G = = = = 0.6879 m3
• = Ww/Vw
• Vw = = 2/9.81 = 0.2041 m3

Vv = V - Vs = 1 – 0.6879 = 0.3121 m3

Water Content w = = = 0.1111 = 11.1 %

Porosity, n = = 0.3121 / 1 = 0.3121 = 31.2 %
Void Ratio, e = = = 0.45
Degree of Saturation = S = = = 0.6539 = 65.4 %
Tutorial 2
Tutorial 3
 Tutorial

A clayey soil has the saturated moisture content of 15.8 %, the specifc
gravity is 2.72. its saturation % is 70.8 %. The soil is allowed to absorb
moisture, After some time the soil saturation increased to 90.8 %. Find
the water content of later case.
• Given w = 15.8 % G = 2.72 S = 70.8 %
Initially, e = = = 0.607

Even though the change in the moisture content, will not have any
effect in the void ratio.
e = , w = = = 0.2026 = 20.26 %
 Tutorial

• The void ratio and the specific gravity of the soil sample are 0.73 and
2.7. if the voids are 92 % saturated, find the bulk density, dry density,
and the water content. What will be the water content, if the soil is
completely saturated and the void ratio remain the same.
 Solution
• Given e = 0.73 , G = 2.7, S = 92 %
w = = = 0.2487 = 24.87 %

γd = = = 15.31 KN / m3

=[ ] = 9.81 = 19.11 KN / m3
or
γd = = = (1+w) γd = (1+0.2487)15.31 = 19.11 KN / m3

e = , w = = = 0.2703 = 27.3 %
 Tutorial

An undisturbed soil sample has a volume of 100 cm3 an mass of 190g.

On oven drying mass has reduced to 160g. Sp gr of soil is 2.67.
Determine void ratio, water content and degree of saturation.
Solution

Given total weight of the soil mass = 190 g, Volume = 100 cm3
Dry weight = 160g. Sp gr = 2.67

Density = = = 1.9 g / cm3 . = 18.64 KN / m3

Dry density = = 1.6 g / cm3 . = 15.7 KN / m3

γd = = w = - 1 = - 1 = 18.8 %

γd = = e = - 1 = - 1 = 0.67

e = = S = = = 0.75
 Grain Size Distribution

• Grain size distribution or the percentage of various sizes of soils grains

present in a given dry soil sample.

• Grain size analysis or coarse grained soils is carried out by sieve analysis
and whereas the fine graded soils are analysed by the Hydrometer
analysis or Sedimentation analysis.
 SIEVE ANALYSIS (Procedure)
• Sieves are wire screens having square openings
• As per IS code (IS: 460 - 1962) sieve sizes are expressed from larhe sizes
(mm) to small sizes (microns).
• The sieve starts with a mesh opening of about 4.75 mm and end up with
75µ size.
• The set of IS sieves consist of sizes of 4.75mm, 2mm, 1mm, 600µ, 425µ,
212µ, 150µ and 75µ.
• In a dry sieve analysis, a suitable quantity of soil (Known weight 500g or
1000g) can be taken.
• Sieved through the set of sieves arranged in sizes, with the largest opening
at the top and the smallest opening at the bottom.
• The shaking of the sieve can be done manually or by mechanical means.
• The amount of soil retained on each sieve is weighed.
• On the basis of total weight of sample taken and the weight of soil retained
on each sieve, The percentage weight of total soil passing through each
sieve can be calculated.
 Calculations
% retained on the particular sieve =
× 100

Cumulative % retained = Sum of % retained in all sieves of larger sizes and

the % retained on that particular sieve.

Percentage finer than Sieve under reference = 100 % - Cumulative %

retained.
Grain size distribution curves
 Grain size distribution curves
• The grain size distribution curve is a curve plotted against the % finer than
weight vs Particle size diameter.

• The position and the shape of the slope of the curve indicates the type and the
gradation of the soil.

• The range of the particle sizes of gravel, sand, silt clay sizes are fitted on the
absicca with sand size further divided to Coarse (C), Medium (M) and Fine (F).

• The Diameter D10 corresponds to 10% of the sample finer in weight of the GSD

curve. The diameter D10 is called as the effective size.

 Grain size distribution curves
The coefficient for Uniformity Cu is a shape parameter and is defined as the
Cu =
Where D60 = grain diameter (mm) corresponding to 60% finer than.

Another shape parameter is Coefficient of curvature Cc and is defined as

Cc =

Where D30= grain diameter (mm) corresponding to 30 % finer than, for a soil to

be well graded, Cc must be lies between 1 and 3, in addition, Cu must be greater

than 4 for gravels and greater than 6 for sands.

 Tutorial 4 - Sieve Analysis
500g of soil was subjected to sieve analysis. The weight of soil retained on
each sieve are as follows. Plot the grain size distribution curve and also
Determine the following. A) Percentage of gravel, Coarse sand, medium
sand, fine sand, silt and clay As per IS:1498-1970. b)Effective size. C)
Uniform coefficient D) Coefficient of curvature E) The gradation of the soil.
IS sieve Weight of Soil (g) IS sieve Weight of soil (g)

4.75mm 10 425 µ 85

2.00 mm 165 212 µ 40

1.0 mm 100 150 µ 30

75 µ 50
IS sieve Dia of the grain Weight of Soil Cumulative % Retained % finer than
(mm) (g) Weight retained,
g

4.75mm 4.75 10 10 2.0 98.0

2.00 mm 2.00 165 175 35.0 65.0
1.0 mm 1.0 100 275 55.0 45.0
425 µ 0.425 85 360 72.0 28.0
212 µ 0.212 60 420 84.0 16.0
150 µ 0.15 20 440 88.0 12.0
75 µ 0.075 40 480 96.0 4.0
Pan 20 500 100 0
 CALCULATION (SIEVE ANALYSIS)
Is grain size classification

Percentage of Gravel ( > 4.75mm) = 100 – 98 = 2 %

Percentage of coarse sand ( 4.75mm – 2.0 mm) = 98 – 65 = 33 %

Percentage of medium sand ( 2.0 mm – 0.425 mm) = 65 – 28 = 37 %

Percentage of Fine sand ( 0.425 mm – 0.075mm) = 28 – 4 = 24 %

Percentage of silt clay (> 0.075mm) = 4 %

From the Figure

D10 = 0.13

D30 = 0.5

D60 = 1.8
 CALCULATION (SIEVE ANALYSIS)
• Effective size D10 = 0.13mm

The coefficient for Uniformity Cu = = = 13.8

Coefficient of curvature Cc = = = 1.1

Since the Cu value is greater than 6 and Cc value lies between 1 and 3
The soil is classified as well graded soil.
 Sedimentation Analysis
• The sedimentation analysis Is the most convenient method of
grain size distribution of the soil fraction is finer than 75 µ size.
• The analysis is based on the STOKE’S law
• According to which the velocities of free fall spherical, fine
particles through a liquid medium of different sizes.
• The particles with high mass will have more velocity under the
action of gravity.
• Thus the particle size can be identified based on their Settling
velocities
• This phenomenon is explained by using stoke’s terminal
velocity theory
 Sedimentation Analysis
• Stoke’s terminal velocity ‘v’ is given by

V = D2
Where

Unit weight of soil grains (g / cm 3)

= Unit weight of the liquid (g / cm 3)
= Viscosity of the liquid ( g-s / cm 2 ) = µ / g
µ = Viscosity dyne s / cm2 or poise
G = Acceleration due to gravity (cm / s 2)
D = Diameter of the grain (cm)

Usually water will be the medium of suspension, Thus can be

consider as 1 g / cc. hence the equation can be modified as

2
 ATTERBERG LIMITS
Consistency of a Soil: The consistency of a fine-grained soil refers to its firmness,
and it varies with the water content of the soil.

A gradual increase in water content causes the soil to change from solid to semi-solid
to plastic to liquid states. The water contents at which the consistency changes from
one state to the other are called consistency limits (or Atterberg limits).

The three limits listed out based on the consistency of the soil are

 Plastic Limit (Wp)

 Liquid Limit (WL)

 Shrinkage Limit (WS)

Plastic Limit (Wp)

It is the minimum water

content at which a soil will
just begin to crumble
when it is rolled into a
thread of approximately 3
mm in diameter
Laboratory Determination
of Plastic Limit (Wp)
 Prepare paste of soil finer than 425
micron sieve.
 Hold the Soil in the palm and roll in
to the form of ball
 Keep the ball in the glass plate and
roll in to the thread of uniform
diameter of 3mm is reached.
 Continue this process until the 3mm
dia thread starts crumbling
 Some of the crumbled portion is
kept in oven for moisture content
determination.
 The test is repeated for three
different samples.
 The average of three values of
water content is taken as the plastic
limit.
 Liquid Limit (WL) Determination
• In the Laboratory Liquid limit test is carried out with the help of
Liquid limit Apparatus.
• The apparatus consist of a vulcanised rubber compound base
with a brass cup suitably mounted.
• The brass cup may be lifted and allowed to fall of the rubber base
through a arrangement operated by handle. Before starting the
test falling height of the brass cup must be adjusted to10 mm.
• A grooving tool, is used to cut the pat in the soil placed in the cup
• About 120g of soil air dried soil, is sieved through 425µ sieve and
mixed with water, so that the soil will attain a patty consistency.
• A portion of sand is placed in the brass cup and is levelled so as
to have the maximum depth of 10mm.
 Liquid Limit (WL) Determination
• A groove is cut in the soil placed in the cup, using the grooving tool.
• In cutting the groove, the grooving tool is drawn through the axis along the
symmetrical axis of the cup, holding the tool perpendicular to the cup.
• The handle is rotated at the rate of 2 revolutions per second and the number
of blows necessary to close the groove for a distance of 13 mm is noted.
• By altering the water the water content of the soil, repeating the operations,
mentioned above four to five readings of water content in the range of 10 to
40 blows are obtained. The graph is plotted between Log no of blows (N)
and the Water content.
• A semi Logarithmic straight line called a flow curve is obtained.
• The liquid limit is determined by reading the water content corresponding to
25 blows on the flow curve.
 Shrinkage Limit (Ws)
• Shrinkage limit is defined as the moisture content, in percent, at which
the volume of the soil mass ceases to change is defined as the
shrinkage limit.

• Shrinkage limit tests [ASTM (2007)—Test Designation

• The shrinkage limit is defined as the water content at which the soil
changes from a semi-solid to a solid state.
 Shrinkage limit determination
• About 30 g of soil passing through IS:425 micron sieve, is taken in an
evaporating dish. The soil is mixed with the sufficient quantity of water, to bring a
soil consistency it may flow.

• The soil sample is placed in the three shrinkage dish, so as to fill the dish. The
dish is weighed with the soil.

• The pat is allowed to dry till the color of the pat changes from dark to light.

• The dish is kept in the oven and dried and the dry weight of the sample is noted.

• The volume of the dry pat of the soil is measured by mercury displacement
method.
 Shrinkage limit determination

Shrinkage Limit (Ws) = × 100

Where W1 = Initial wet mass of soil

W2 = Final dry mass of soil
V1 = Initial volume of the soil
V2 = Final volume of the soil
 Tutorial
• A saturated clay has a volume of 18.9 cm3 and a mass of 18.9 cm3 and
a mass of 30.2 gm. On Oven drying the mass reduces to 18 gm. The
volume of dry specimen as determined by mercury displacement
method is 9.9 cm3 . Determine shrinkage limit and the shrinkage ratio.

W1 = 30.2 gm
W2 = 18 gm
V1 = 18.9 cm3
V2 = 9.9 cm3
• Shrinkage Limit (Ws) =
= { × 1 } × 100 = 17.8 %

Shrinkage Ratio =
Dry Density = = = 1.81 g / cc

Shrinkage ratio = = 1.81

 Index properties

• Plasticity Index (Ip) : It is the numerical difference between the Liquid Limit and the

Plastic limit. (WL difference Wp )

• Liquidity Index (IL) : It is the ratio of difference between Natural water content of the

soil and its plastic limit to its plasticity Index.

IL =

• Flow Index(IF) : It is the slope of the flow curve obtained between the number of

blows and the water content in determination of Liquid Limit.

Tutorial 5
Solution
 Solution
• Wp for soil A is 25 and WL for soil A is 38 %
• Wp for soil B is 30 and WL for soil B is 60 %

Plasticity index for Soil A I P = 38 - 25 = 13

Plasticity index for Soil B I p = 60 - 30 = 30
Soil B has the higher level of plasticity.

Liquidity Index for Soil A I L = (40 – 25)/ 13 = 1.15

Liquidity Index for Soil B I L = (50 -30) / 30 = 0.67
Liquidity index < 1 will serve as a better foundation material, hence soil
B will serve as a better foundation material.
Flow index for Soil A If = 42.0 - 32.0 = 10 (considering water contents at 10
and 100 blows).
If for Soil B = 62.6 – 55.5 = 7.1.

Toughness Index It = Ip/If = 13 / 10 = 1.3

It for Soil B = 30 / 7.1 = 4.22.

Soil B has the higher toughness index, hence produces a higher shear
strength.
 Indian standard classification system (ISCS)
IS:1498 - 1970
FOUR MAJOR DIVISIONS : Coarse Grained, Fine grained, Organic
Soil, peat.

Classification of Groups
The soil is classified into 18 groups (iscs).

Each group is designated a symbol consisting of two capital letters.

The first letter is based on main soil type.

The second letter is based on gradation and plasticity

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