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CV45: Lecture Notes for Module 1 by Prof. S.K. Prasad 1

CV45 Geotechnical Engineering – 1

Module 1

Introduction, Phase Diagrams, Index Properties & Soil

Classification

Dr. S.K. Prasad Professor of Civil Engineering

Sri Jayachamarajendra College of Engineering Mysuru 570 006

Email: [email protected]

February 2018


CV45 Geotechnical Engineering – 1

Module 1

Syllabus

Introduction

Origin & Formation of Soil, Phase Diagram, Phase relationships, Definitions and their

relationships.

Determination of Index Properties

Specific Gravity, Water Content, Insitu Density, Particle size analysis (Sieve Analysis and

Sedimentation Analysis), Atterberg’s Limits, Consistency Indicies, Relative Density Activity

of Clay

Soil Classification

Plasticity Chart, Unified and BIS Soil Classification

ACKNOWLEDGEMENT

A number of pictures and some materials have been downloaded from internet over years. I

wish to acknowledge all those who have indirectly contributed. The lecture material is only

intended to help the students. Hence, I do not claim anything as original.

INTRODUCTION

1.1 Geotechnical Engineering – Why?

1. We are unable to build castles in air (yet) !

2. Almost every structure is either built on or built in or built using soil or rock.

3. Mechanics of Soils and Rocks is the basis of Geotechnical Engineering

4. Geotechnical engineering problems involve:

– Strength

– Stability

– Deformation

– Water flow

Soil is an un-cemented aggregate of mineral grains and decayed organic matter (solid

particles) with liquid and gas in the empty space between the solid particles formed by

weathering of rocks in the top surface of earth crust. Fig. 1 represents a portion of soil mass

comprising of solid particles and void space. Void space is made up of liquid (water) and/or

gas (air).


Fig. 1 : Soil mass, a conglomeration of solid particles and void space

Soil is an important construction material that is

1. Oldest.

2. Cheap or available free of cost many a times.

3. Most complex, yet having interesting properties.

4. Modified to suit the requirements, many a times.

Soil is used

1. to manufacture Bricks, Tiles or Earthenware.

2. as foundation material.

3. to construct dams and embankments.

4. to fill hollow zones behind retaining walls, low lying areas etc.

5. to construct landfills and other environmental protection systems

1.2 Why soil is complex?

The following properties of soil make it perhaps the most complex construction material. 1. Porous

2. Polyphasic

3. Permeable

4. Particulate

5. Heterogeneous

6. Anisotropic

7. Non-Linear

8. Pressure Level Dependent

9. Strain Level Dependent

10. Strain Rate Dependent

11. Temperature Dependent

12. Undergoes volume change in shear

Yet, soil possesses some Interesting properties that relate with human beings, namely,

1. Colorful

2. Sensitive


3. Possesses Memory

4. Changes its properties with time

1.3 What is Geotechnical Engineering?

Geotechnical Engineering is an integration of Physics, Earth Science, Solid Mechanics,

Geology and Hydrogeology. The basic science of Geotechnical Engineering is called Soil

Mechanics. Geotechnical Engineering provides the theoretical basis for describing the

mechanical behaviour of earth materials. It involves application of theory of soil mechanics

to a variety of field problems. For most other engineering disciplines, the material

properties are well-defined or can be controlled. But, in Geotechnical Engineering, material

properties are highly variable and difficult to measure with a reasonable degree of accuracy. Geotechnical Engineering is among the younger branches of Civil Engineering. Yet, it has

evolved over centuries.

1. Geotechnical Engineering is probably one of the most challenging engineering disciplines.

2. For a geotechnical engineer, no two days at work are going to be similar. 3. Geotechnical engineering expertise is required in a vast variety of disciplines that

includes the oil and offshore industry. 4. Being a relatively new discipline, there is ample scope for innovation. 5. For a geotechnical engineer, achieving job satisfaction is never a problem. 6. Material properties must be measured for each new construction site. 7. Remember that geotechnical engineers deal with natural materials and there can be

no quality control. 8. Ground consists of innumerable variety of particle sizes and minerals. 9. To make matter worse, engineering properties of earth materials are strongly

influenced by their past geological history that is normally unknown. Climatic conditions also influence these properties.

1.4 Sub-branches in Geotechnical Engineering

The following are the sub-branches of Geotechnical Engineering.

1. Foundation Engineering

2. Deep Excavation

3. Tunneling

4. Earth Pressure and Retaining Structures

5. Earth embankments

6. Stability of Slopes

7. Environmental Geotechniques

8. Earthquake Geotechnical Engineering

9. Ground Improvement technique

10. Rock Mechanics

11. Engineering geology


Fig. 2 represents the family of Geotechnical engineering.

Fig. 2 : Family of Geotechnical Engineering

1.5 Distinctions between Fine and Coarse Grained Soils

Soil can be broadly classified in to two types, namely Fine grained soil and coarse

grained soil based on the size, shape and behaviour. Table 1 provides the important

distinctions.

Table 1 : Distinctions between Fine grained soil and coarse grained soil

Fine Grained Soil Coarse Grained Soil

Size of particle is less than 75 microns Size of particle is mores than 75 microns

Silt & Clay belong to this group Sand, Gravel, Cobble, Boulder etc. belong to this group

Properties are influenced by surface area Properties are influenced by gravity

Attraction and bonding between particles enable strength

Dense packing, particle to particle contact enable strength

Mostly plate-like Mostly round, sub-round, angular

Void ratio & water content can be very high Void ratio & water content can not be very high

Possess consistency (liquid, plastic & shrinkage) limits

Consistency (liquid, plastic & shrinkage) limits are absent

Possess Cohesion Possess Angle of Internal friction

1.6 Soil formation, a geologic Cycle

Soil is formed from rock due to erosion and weathering action. Igneous rock is the basic rock

formed from the crystallization of molten magma. This rock is formed either inside the earth

or on the surface. These rocks undergo metamorphism under high temperature and

pressure to form Metamorphic rocks. Both Igneous and metamorphic rocks are converted in

Geotechnical

Engg.

Foundation

Engg

Deep

Excavations

Tunneling

Retaining

Structures

Slopes

Embankments

Environmental

Geotechnics

Earthquake

Geotechnical

Engg.


to sedimentary rocks due to transportation to different locations by the agencies such as

wind, water etc. Finally, near the surface millions of years of erosion and weathering

converts rocks in to soil.

Fig. 3 : Geological Cycle of Soil


PHASE SYSTEM

1.7 Soil Mass, a three phase system

If you take a lump of soil in hand, it looks like the one in Fig. 4. When seen through

microscope, you can observe that the soil mass is made up of innumerable number of soil

particles in contact with each other. Each particle can have different shapes and sizes. In

between particles, there will be voids which can be filled with air and / or water. Hence, soil

mass comprises of solid particles of soil, water and air. The three of them form a three phase

system. However, it is very difficult to consider their interaction precisely. Hence, to

simplify the understanding, the three phases are divided in to three volumes by combining all

materials belonging to one phase. They are further explained on volume and weight basis. We

call this as three phase system.

A typical soil mass Idealisation as three phase system

Fig. 4: Soil mass idealized as three phase system


W = Weight

V = Volume

s = Soil grains

w = Water

a = Air

v = Voids

Basic terminologies in a three phase system of soil mass

Conversion from three to two phase system

Fig. 5 : Soil mass as three phase system

1.8 Basic Definitions

The following are the basic definitions of soil.

1. Water Content (ω)

2. Void Ratio (e)

3. Porosity (n)

4. Degree of Saturation (S)

5. Air content (Ac)

6. Percentage air voids (na)

7. Bulk Density (γb)


8. Dry density (γd)

9. Density of soil solids (γs)

10. Saturated density (γsat)

11. Density of water (γω)

12. Submerged density (γsub)

13. Specific Gravity of Soil Solids (G)

14. Specific Gravity of Soil Mass (Gm)

15. Relative density (Dr)

Each of the above definition is defined with soil represented as three phase diagram.

1.8.1 Water Content (ω)

1. It is defined as ratio of weight of water to weight of solids.

2. It is also called Moisture Content.

3. It has no unit. It is expressed in percentage or decimals (for calculation purpose).

4. It indicates the amount of water present in the voids in comparison with weight of

solids.

5. In dry soil, water content ω = 0.

6. Clayey soil may possess very large water content leading to unfavourable situation.

7. Water content of soil mass changes with season, being close to zero in summer and

maximum during rainy season.

8. It represents the amount of water present in soil mass. In dry soil, water content ω = 0

9. Higher the water content, greater will be the vulnerability, especially in clayey soil.

1.8.2 Void Ratio (e)

1. It is defined as the ratio of volume of voids to volume of solids

2. It has no unit. It is normally expressed in decimals.

3. It indicates the amount of voids present in a soil mass in comparison with the

amount of solids.

4. Normally, void ratio of clayey soil will be large.

5. The more the void ratio, more loose will be the soil mass and hence, less strong and

less stiff.

6. It is not possible to determine void ratio in the laboratory. Hence, it is computed

from other properties.


1.8.3 Porosity (n)

1. It is defined as the ratio of volume of voids to total volume of soil mass.

2. It has no unit. It is expressed in decimals or percentage.

3. Its value ranges from 0 to 100 % (0 < n < 1).

4. Similar to void ratio, it indicates the amount of voids in comparison with the total

volume of soil mass.

5. In some countries, it is more familiar than void ratio. But either can be used

interchangeably in calculation.

6. Like void ratio, porosity is computed and can not be directly determined in the

laboratory.

1.8.4 Degree of Saturation (S)

1. It is defined as the ratio of volume of water to volume of voids.

2. It has no unit. It is usually expressed in percentage.

3. Its value ranges from 0 to 100 % (0 < S < 100 %)

4. It represents the amount of water present in the void space of soil mass.

5. In dry soil, S = 0 and in fully saturated soil S = 100 %. Hence, during summer S is close

to zero, while during rainy season, S is close to 100 %. In partially saturated soil, S lies

between zero to 100 %.

6. It is computed and cannot be directly determined in the laboratory.

1.8.5 Air content (Ac)

1. It is defined as the ratio of volume of air to volume of voids.

2. It has no unit. It is usually expressed in percentage.

3. Its value ranges from 0 to 100 % (0 < Ac < 100 %).

4. It represents the amount of air present in the void space of soil mass.

5. In dry soil, Ac is 100 % and in fully saturated soil Ac is 0 %. In partially saturated soil Ac

lies between 0 and 100 %.

6. S + Ac = 1

7. It is computed and cannot be directly determined in the laboratory.


1.8.6 Percentage air voids (na)

1. It is defined as the ratio of volume of air to total volume of soil mass.

2. It has no unit. It is expressed in percentage.

3. Its value ranges from zero to 100 % (0 < na < 100 %).

4. It represents the amount of air present in the total volume of soil mass.

5. Always na < Ac.

6. It is computed and can not be directly determined in the laboratory.

1.8.7 Bulk Density (γb)

1. It is defined as the ratio of total weight to total volume of soil mass.

2. In SI units, it is expressed as kN/m3.

3. Its value normally ranges from 12 to 24 kN/m3.

4. It includes the weights of air, water and solids as a function of total volume of soil

mass. It changes with season, being maximum during rainy season and minimum in

summer.

5. Bulk density of soil mass can be determined experimentally. It is therefore used to

compute other properties such as dry density and void ratio.

1.8.8 Dry density (γd)

1. It is defined as the ratio of weight of soil solids to total volume of soil mass.


3. Dry density will always be less than or equal to bulk density of soil mass.

4. Dry density is independent of season. Hence, it is used in many design calculations.

5. Knowing water content and bulk density, dry density can be computed.

1.8.9 Density of soil solids (γs)

1. It is defined as the ratio of weight of soil solids to volume of soil solids.


3. It is always greater than dry density of soil.


4. It can not be determined experimentally. Hence, it is computed from other

parameters. It is used to calculate other properties such as specific gravity of soil

solids.

1.8.10 Saturated density (γsat)

1. It is defined as the ratio of total weight to total volume of soil mass when the soil is

fully saturated. Hence, it is the bulk density of soil mass when S = 1.


1.8.11 Density of water (γω)

1. It is defined as the ratio of weight of water to volume of water.

2. In SI units, it is expressed in kN/m3 and can be taken as 9.8 kN/m3.

3. It is used in computation of other quantities.

1.8.12 Submerged density (γsub)

1. It is defined as the net weight of weight per volume of soil mass in water.


3. It is equal to saturated density minus density of water.

4. satsub

5. It is also called buoyant density.

6. In saturated soil, water exerts upward pressure on soil. Net weight of soil particles

acting downward will be actual weight of soil minus weight of water.

1.8.13 Specific Gravity of Soil Solids (G)

1. It is defined as the weight of soil solids to weight of equal volume of water.

2. Hence, it is the ratio of density of soil solids to density of water.

3. It has no units and is expressed in decimals.

4. Normally, G of most soils varies from 2.6 to 2.75. Organic soils may have G up to 2.

5. G is determined in the laboratory and is used to compute other parameters such as

void ratio.

6. Many a times, specific gravity means G


1.8.14 Specific Gravity of Soil Mass (Gm)

1. It is defined as the weight of soil mass to weight of equal volume of water.

2. It is also called Apparent Specific Gravity.

3. It has no units and is expressed in decimals.

4. Its magnitude is always smaller than that of G.

5. It is less commonly used in calculations.

1.8.15 Relative density (Dr)

1. It is also called Density Index.

2. It has no unit. It is expressed in percentage.

3. Dr ranges from 0 to 100 %.

4. It is applicable for coarse grained soil such as sand and gravel.

5. It indicates whether the insitu density of soil is close to loosest or densest state.

6. minmax

max

ee

eeDr

7. In terms of dry density, relative density is given as follows.

minmax

minmax

dd

dd

d

d

rD

8. When Dr = 1, soil in its densest state and when Dr = 0, soil is in its loosest state.

Table 2 : Influence of Relative density on Soil State

Relative Density (%) State of Soil

0 to 20 Very Loose

20 to 40 Loose

40 to 60 Medium dense

60 to 80 Dense

80 to 100 Very Dense


1.9 Problems and Solutions related to Phase Diagram

Problem 1

A natural soil mass has a bulk density of 18 kN/m3 and water content of 8 %. Calculate the

amount of water required per cubic meter of soil to raise the water content to 18 %. What

will be the degree of saturation at this water content? Assume void ratio to be constant and

take G = 2.7. (July 2006) 8 Marks

Data

γb = 18 kN/m3

ω = 8 %

G = 2.7

γω = 9.8 kN/m3 (assumed)

3/67.161

mkNb

d

587.01d

Ge

V

Ws

d

Hence, for 1 m3 of soil mass, Ws = 16.67 kN

If ω = 8 %, weight of water = 0.08Ws = 1.33 kN

If ω = 18 %, weight of water = 0.18Ws = 3 kN

Hence, amount of water required per cu.m of soil = 3 – 1.33 = 1.67 kN = 167 lt.

%8.82828.0

S

SeG

Problem 2

How many cu.m of soil can be formed with void ratio of 0.5 from 100 m3 of soil having void

ratio of 0.7? (Jan 2006) 5 Marks

Data

e1 = 0.7

Vv1 = 0.7Vs

V1 = 100 m3 = Vv1 + Vs = 1.7 Vs

Vs = 58.82 m3

e2 = 0.5

Vv2 = 0.5Vs = 29.41 m3

V2 = Vv2 + Vs = 88.23 m3


Problem 3

A dry soil has a void ratio of 0.65 and specific gravity is 2.8. Find its unit weight. Water is

added to the sample so that its degree of saturation is 55 % without any change in void

ratio. Determine the water content and unit weight. The sample is then submerged in

water. Determine the unit weight when the degree of saturation is 90 % and 100 %. (Jan

2008) 10 Marks

Data

e = 0.65

G = 2.8

γω= 9.8 kN/m3 (assumed)

3/63.161

mkNe

Gd

When S = 55 %, ω = 12.77 % 3/75.18)1( mkNdb

When S = 90 %, ω = 20.89 %

When S = 100 %, ω = 23.21 %

Problem 4

In an earth dam under construction, the bulk unit weight is 16.5 kN/m3 at water content 11

%. If the water content has to be increased to 15 %, compute the quantity of water to be

added per cu.m of soil. Assume no change in void ratio. Determine the degree of saturation

at this water content. Take G = 2.7. (Jan 2009)

10 Marks.

Data

γb =16.5 kN/m3

ω = 11 %

G = 2.7


3/86.141

mkNb

d

Wd = 14.86 kN per unit volume

Ww = 1.63 kN per unit volume

78.01d

Ge

%92.51e

GS


Problem 5

An undisturbed specimen of clay was tested in a laboratory and the following results were

obtained. Weight = 2.1 N, Oven dry weight = 1.75 N. Specific Gravity of soil solids = 2.7.

What was the total volume of original undisturbed specimen assuming that the specimen

was 50 % saturated? (Model QP) 8 Marks

Data

W = 2.1 N

Ws = 1.75 N

Ww = 0.35 N

G = 2.7

S = 0.5


%20sW

W

08.1S

Ge

3/72.121

mkNe

Gd

72.121075.1 3

V

X

V

Ws

d

V = 0.1376 X10-3 m3

Problem 6

The maximum and minimum dry unit weights of sand determined in the laboratory are 21

kN/m3 and 16 kN/m3 respectively. If the relative density of sand is 60 %, determine the in-

situ porosity of the sand deposit. Take G = 2.65. (June 2008) 6 Marks

Data

Dr = 60 % =0.6

γdmax = 21 kN/m3

γdmin = 16 kN/m3

γd = 18.67 kN/m3

minmax

minmax

dd

dd

d

d

rD

78.01d

Ge

e

en

1 0.44


Problem 7

For a soil in its natural state, void ratio, water content and specific gravity are respectively

0.8, 24 % and 2.68. Determine bulk density, dry density and degree of saturation. If the soil

is completely saturated by adding water, what would be its water content and saturated

density.

Data

e=0.8

ω = 0.24

G = 2.68


3/59.141

mkNe

Gd

3/09.18)1( mkNdb

%4.80e

GS

If S =1, ω =29.85 % 3/95.18)1( mkNdsat

Problem 8

In its natural condition, a soil sample has a mass of 22.9 N and a volume of 1.15 X 10-3 m3.

After being completely dried in the oven sample weighs 20.35 N. Find bulk density, water

content, void ratio, porosity, degree of saturation, air content, dry density and percentage

air voids.

Data

W = 22.9 N

V = 1.15 X 10-3 m3

Ws = 20.35 N

G = 2.7


3

3/91.19

1015.1

9.22mkN

XV

Wb

ω = 12.53 %

γd = 17.7 kN/m3

495.01d

Ge

e

en

10.331

%35.68

e

GS


Ac = 1 – S = 31.65 %

G

Gna

d

1

)1(

na = 0.105

Problem 9

A moist sample of soil has a weight of 6.33 N and a volume of 300000 mm3 at a water

content of 11 %. Taking G = 2.68, determine e, S and na. Also determine water content at

which soil gets fully saturated. What will be the unit weight at saturation?

Data

W = 6.33 N

V = 3X105 mm3

G = 2.68

ω = 11%

γw = 9.8 kN/m3

3/1.21 mkNV

Wb

3/191

mkNbd

38.01d

Ge

%6.77e

GS

If S = 1, then %18.14G

e

3/69.21)1( mkNdsat

Problem

A sample of soil has a volume of 1X106 mm

3 and weighs 19 N. The water content of the soil

is 15% and G = 2.7. Find dry density, void ratio, degree of saturation and water content at

saturation. Take γw = 9.81 kN/m3

W = 19 N = 0.019 kN

V = 1X 106 mm

3 = 1X10

-3 m

3

γb = W/V = 19 kN/m3

If S = 1, then

3/52.161

mkNbd

60.01d

Ge

%5.67e

GS

%2.22G

e


Problem

A sample of soil weighing 19 kN/m3 has a water content of 30%. The specific gravity of soil

particles is 2.68. Find void ratio, porosity and degree of saturation.

γb = 19 kN/m3 w = 30% G = 2.68

Problem

A partially saturated sample of soil from a borrow pit has a natural water content of 14% and

bulk unit weight of 19 kN/m3. The specific gravity of soil solids is 2.7. Determine void ratio

and degree of saturation. What will be the unit weight of sample on saturation

γb = 19 kN/m3 w = 14% G = 2.7

= 20.3 kN/m3 when S = 1

Problem

At a site, the dry density of soil is 16 kN/m3, weight of soil filled in container of 1 lt in

loosest and densest states are respectively 15 N and 17 N. Determine the maximum and

minimum void ratios of the soil. What is the relative density at the site. Take G = 2.7

%64e

GS

59.01d

Ge

3/62.141

mkNbd

80.01d

Ge

0.4441

e

en

%100e

GS

3/67.161

mkNbd


1.10 Distinction between Mass Density and Weight Density

In geotechnical engineering, it is common to use weight density or unit weight than mass

density. It should be noted that weight density depends on acceleration due to gravity (g)

and hence it is place dependent. However, g does not change considerably from place to

place within the engineering limits. Commonly, symbol ρ is used to represent mass density

and symbol γ is used to represent weight density. Further, γ = ρ * g.

Table 3 : Common mass densities and Unit weights in Geotechnical Engineering

Mass Density (kg/m3)

Unit weight (kN/m3)

Bulk ρb γb

Dry ρd γd

Soil Solids ρs γs

Water ρω γω

Saturated ρsat γsat

Submerged ρsub γsub

ρω = 1000 kg/m3

γω = 9800 N/m3 = 9.8 kN/m3


INDEX PROPERTIES OF SOIL

1.11 GENERAL

Index properties are the properties of soil that help in identification and classification of soil.

Water content, Specific gravity, Particle size distribution, In situ density (Bulk Unit weight of

soil), Consistency Limits and relative density are the index properties of soil. These

properties are generally determined in the laboratory. In situ density and relative density

require undisturbed sample extraction while other quantities can be determined from

disturbed soil sampling.

The following are the Index Properties of soil.

1. Water content

2. Specific Gravity

3. In-situ density

4. Particle size distribution

5. Consistency limits

6. Relative Density

1.12 DETERMINATION OF INDEX PROPERTIES OF SOIL

This section explains the important methods approved by Indian Standards for the

determination of Index properties of soil both in laboratory and in field.

1.12.1 Water Content

Oven drying method and Calcium carbide method are the two popular methods of

determination of water content. Refer to IS 2720 – Part 2- 1973 for more detail.

a) Oven Drying Method

It is an accurate method of determining water content of soil in the laboratory. The

procedure is as follows.

1. Collect a representative sample of soil in a steel cup carrying a lid.

2. Find the weight of cup and lid along with soil (W1)

3. Keep the cup with lid open in a thermostatically controlled oven for 24 hours at

around 105oC. Free water in the soil evaporates.

4. After cooling the cup, find the weight of cup and lid along with dry soil (W2)


5. Find the empty weight of cup and lid (W3)

Why Steel cup with a lid ?

Lid should be covered after taking the sample to avoid movement of water, dust air etc.

Steel (Stainless steel) cup is used to avoid interaction between material of cup, water

and minerals & salts (if any) present in soil.

A sample calculation for the determination of water content in the laboratory by oven

drying method is presented in the table below.

Table 1: Determination of water content by Oven drying method

Sl No

Description Determination No.

I II III

1 Weight of cup + Wet Soil (W1) g 44.12 44.11 46.10

2 Weight of cup + Dry Soil (W2) g 41.18 41.16 43.01

3 Weight of cup (W3) g 20.12 20.08 20.00

CALCULATIONS

1 Weight of water = W1 – W3 2.94 2.95 3.09

2 Weight of solids = W2 – W3 21.06 21.08 23.01

3 Water content 13.96 13.99 13.43

Average Value 13.79 %

b) Calcium Carbide method or Rapid Moisture Meter Method

It is a rapid method of determination of water content of soil. It consists of an air tight

container with a diaphragm and a calibrated meter. About 6 g of soil is mixed with fresh

Calcium carbide. The mixture is rigorously shaken. Water in the soil reacts with Calcium

Carbide to release acetylene gas. The amount of gas produced depends on available water.


This gas creates a pressure on sensitive diaphragm and water content is directly recorded on

the calibrated meter. The method is not very accurate, but is extremely rapid.

Fig. 6 : Typical picture of Rapid Moisture meter and its assembly

1.2.2 Specific Gravity of Soil Solids

Specific gravity of soil solids is commonly determined by Pycnometer method. Refer to IS

2720 – Part 3- Sections 1 & 2 - 1981 for more detail.

a) Pycnometer method 1. Use pycnometer for the determination of specific gravity of coarse grained fraction

and density bottle for that of fine grained fraction. 2. Find the weight of clean, dry and empty pycnometer (W1). 3. Put dry soil (about one third the height) in the pycnometer and find the weight (W2). 4. Add water till the top such that the air bubbles are completely removed and find the

weight (W3). 5. Empty the soil and fill water up to the top in the pycnometer and find the weight

(W4).

1234

12

2314

12

)()( WWWW

WW

WWWW

WW

waterofvolumeequalofWeight

SoilofWeightG

If the specific gravity is determined using any other liquid (say kerosene), other than water,

then, Kosenein GGG *ker where GK is the specific gravity of the liquid.

Further, the temperature influences specific gravity. At higher temperature viscosity of

water will be less and vice versa. Viscosity of the medium affects G. The effect of

temperature should be considered as follows.

CatWaterofGravitySpecific

CtatWaterofGravitySpecificGG

o

o

t27

27


From Indian Standard point of view 27oC is standardized.

W1 W2 W3 W4 Fig. 7 : Stages in determination of Specific Gravity of soil using Pycnometer or Density Bottle

b) Why pycnometer for coarse grained soil and density bottle for fine grained

soil?

Coarse grained soil requires larger size container owing to the bigger particle size. Large

quantity of fine grained soil is difficult to handle especially expelling out air. Hence smaller

density bottle of 50 ml capacity is used.

c) Why pycnometer has a conical top with a opening ?

The purpose of conical top is to reduce the cross section gradually to a minimum such that

any difference in level of water in different stages should not cause serious error. Further,

the opening will allow any air present in the voids of soil to be expelled out.


1.2.3 Bulk Unit weight of soil

The density of soil (especially in coarse grained portion) is the direct indication of strength

and stiffness. The following are the popular methods of finding bulk density in field.

1. Core Cutter Method

2. Sand Replacement method


Fig. 8 : Apparatus for field density determination of soil using Core Cutter

a) Determination of In-situ unit weight of soil by Core Cutter Method

The following are the steps involved in determination of bulk density and hence dry density

of soil at a site.

1. Refer IS 2720 – Part 29 – 1975

2. Main apparatus include Core cutter with a sharp edge, dolly or Collar and a rammer.

3. This method is applicable for soil that sticks to the surface of cutter (Clayey soil) and

that is not very stiff (where cutter can be penetrated in to ground by ramming).

4. Sampling is done vertically by ramming downwards.

5. The inner surfaces of core cutter and dolly are greased.

6. The ground surface is leveled after removing the top soil.

7. Core cutter with collar on top and sharp edge at bottom is placed on the ground.

8. It is then driven in to the ground using the rammer till the soil collects up to the

collar.

9. It is carefully taken out by loosening from outside such that the soil inside remains in

tact.

10. Dolly is carefully removed. The soil surface in the core cutter is trimmed from both

the ends.

Let, V = Volume of the core cutter

W1 = Empty weight of core cutter

W2 = Weight of core cutter with soil

Then, Bulk unit weight of soil at the site


A representative sample of soil from the middle of core cutter is placed for the

determination of water content.

Determination of In-situ unit weight of soil by Sand Replacement Method

Fig. 9 : Apparatus for field density determination of soil from Sand Replacement approach


1. This method is particularly suitable for granular soil that does not stick to the surface

of sampler.

2. Refer IS 2720 – Part 28 – 1974.

3. The equipment consists of sand pouring cylinder with cone, calibrating container and

a tray with a central hole.

4. The work involves two stages, namely, calibration of container and determination of

volume of hole.

5. For the purpose of calibration of container, clean, dry, uniformly graded medium

sized sand (called standard sand) is filled in to Sand Pouring Cylinder (SPC)up to the

top, shutter is closed and weight of SPC is noted (W1).

6. The SPC is placed on a smooth glass plate and the sand is allowed to run out of cone.

Weight of SPC after running sand in to cone is noted (W2).

7. SPC is refilled and placed on calibrating container and sand is allowed to run out. The

weight of SPC is noted after running sand in to calibrating container (W3).

8. Weight of sand in cone Wcone = W1 – W2

9. Weight of sand in calibrating container Wcc = W1 – W3 – Wcone

10. Volume of calibrating container is determined (Vcc).

11. SPC with sand up to W1 is taken to the field where bulk unit weight is required. The

surface is cleaned and leveled.

12. The tray with central hole is placed in position. A hole of approximately same size as

calibrating container is dug.

13. The removed soil is collected on the sides of tray and its weight is noted (Wsoil).

14. The SPC is placed on the hole and sand is allowed to run in to hole. Weight of SPC is

noted (W4).

15. Weight of sand in the hole W5 = W1 – W4 – Wcone


1.2.3 Particle Size Distribution

1. Soil in nature exists in different sizes, shapes and appearance. Depending on these

attributes, the soil at a site can be packed either densely or loosely. Hence, it is

important to determine the percentage of various sized soil particles in a soil mass.

This process is called particle size distribution analysis. For this purpose, a particle

size distribution curve is plotted. Packing of soil, amount of voids present influence

the strength and stability of soil mass.

The distribution of grain size influences packing. Good distribution of all sizes reduces voids

if compacted well.

Soil Type Description Average grain size

Gravel Rounded and/or angular bulky hard rock Coarse: 80 mm to 20 mm

Fine: 20 mm to 4.75 mm

Sand Rounded and/or angular bulky hard rock Coarse: 4.75 mm to 2 mm

Medium: 2 mm to 0.425 mm

Fine: 0.425 mm to 0.075 mm

Silt Particles smaller than 0.075 mm, exhibit

little or no strength when dried

0.075 mm to 0.002 mm

Clay Particles smaller than 0.002 mm, exhibit

significant strength when dried; water

reduces strength

<0.002 mm

Importance of Particle Size Distribution

1. Used for the soil classification.

2. Used to design drainage filter.

3. Used to select fill materials of embankments, earth dams, road sub-base materials.

4. Used to estimate performance of grouting, chemical injection and dynamic

compaction.

5. Effective Size, D10, can be correlated with the hydraulic conductivity.

6. More important to coarse-grained soils

a) Particle size distribution curve

It is a graph with percentage finer (N) plotted along the vertical axis (ordinate) in normal

scale and particle diameter (D) along the horizontal axis (abscissa) in logarithmic scale. It

gives an idea about the size and gradation of soil. A soil sample is well graded if it has a good

distribution of all sized particles, otherwise, it is called poorly graded. Uniformly graded

sample is a special category of poorly graded samples in which all soil particles have the


same size. Gap graded samples possess different proportions of same sized particles in

increasing sizes. Intermediate size particles are missing. For achieving good density, well

graded soils are most suitable.

D10, D30, D60 are the important particle sizes. D10 represents a size in mm such that 10 % of

particles are finer than this size. It is some times called effective diameter. D30 and D60

represent sizes in mm such that 30 % and 60 % of particles respectively are finer than this

size obtained on the particle size distribution curve. The following terminologies are

defined.

Fig. 10 : Particle size distribution curve

Uniformity coefficient is a measure of range of particle sizes is given by,

10

60

D

DCu

Coefficient of Curvature represents the shape of particle size distribution curve and is given

by,

6010

2

30

XDD

DC c

For well graded soil, Cc should be between 1 and 3 and Cu should be greater than 4 (if gravel)

or 6 (if sand). Even if either of the two conditions is not satisfied, the sample is poorly

graded.

Particle size distribution is obtained by sieve analysis for coarse grained fractions and

sedimentation analysis for fine grained fractions. Hydrometer analysis is a popular

sedimentation analysis.


b) Sieve Analysis

Fig. 11 : Sieve analysis for particle gradation

Refer IS 2720 – Part 4 – 1975 for more details.

A set of IS sieves are arranged in order with one having largest aperture at the top and that

with smallest aperture at the bottom. A lid at the top and a receiver at the bottom complete

the assembly.

A known weight of representative sample of soil (say 1000 g) is placed in the top sieve.

The assembly is vibrated on a sieve shaker for at least 10 minutes.

Depending on the particle size, soil is collected in different sieves. Weight of soil in each

sieve is measured.

Table 2 gives the details of calculation. Particle size distribution curve is plotted. The soil

sample is classified as well graded or poorly graded.

Table 2 : Particle size distribution Analysis

IS sieve number

Sieve size (mm)

Weight of Soil on each sieve (g)

Cumulative weight retained

% Cumulative weight retained

% Finer

4.75 4.75

2.36 2.36

1.18 1.18

600 0.600

425 0.425

300 0.300

212 0.212

150 0.150

75 0.075

Receiver

Total



c) Sedimentation Analysis

It is also called wet analysis and is applicable for fine grained soils

The analysis is based on Stoke’s law which states that the velocity at which soil particles

settle in a suspension depend on shape, size and weight of particles. Assuming spherical

particles and same density, 2Dv

)1(

18

1

18

1 22

GDDv s

Here, v = Terminal velocity of sinking soil particle

D = Diameter of soil particle

γs = Unit weight of soil solids

γw = Unit weight of water

G = Specific Gravity of Soil Solids

η = Viscosity of water

The particles are assumed to settle independent of each other and without the interference

of wall of container.


Based on the velocity of settlement, the diameter of soil particles may be computed. For the

purpose of evaluating velocity of settlement of soil particles, hydrometer analysis is

performed.

d) Hydrometer analysis

A hydrometer is a device made of glass, consisting of a bulb with calibrated weight and stem

with calibrated readings such that when placed in pure water it floats at a level giving

reading 1.000. If it floats in a denser fluid, the readings are greater than 1 (up to 1.030).

Fig. 13 : Determination of grain size distribution from Hydrometer test

Effective depth He of hydrometer depends on volume of hydrometer (Vh), height of bulb (h),

cross sectional area of jar (A) and hydrometer reading (Rh). Height above the ease of the

stem (H) is a function of Rh. He is given by, )(

2

1

A

VhHH h

e . Hence, calibration curve for

hydrometer is prepared prior to the test as shown.




e) Correction for Hydrometer reading

The recorded hydrometer reading may require the following corrections depending in the

situation.

Cm : Correction for meniscus

The solution containing soil will be opaque. Hence, it is difficult to obtain reading

corresponding to lower meniscus. Rh is taken to upper meniscus and correction is applied.

The correction is always positive.

Ct : Correction for temperature


Hydrometer is calibrated at 27oC. If the laboratory temperature is higher, the liquid will be

less viscous and hydrometer sinks more. Rh will be less than actual needing positive

correction. Similarly, low laboratory temperature requires negative correction.

Cd : Correction for dispersing agent

A dispersing agent is added to water to disperse the soil particles and allow them to settle

individually. Dispersing agent increases the density of water and hydrometer floats higher.

Therefore the correction is always negative.

Hence, corrected hydrometer reading dtmh CCCRR

Here, Rh is the observed hydrometer reading. The three corrections are known as composite

correction (C).

dtmhh CCCRCRR

Here, dtm CCCC

From the hydrometer reading R, diameter D of soil particle and corresponding percentage

finer are computed.

Limitations of Hydrometer Analysis

Assumption Reality

Sphere particle Platy particle (clay particle) as D 0.005mm

Single particle (No interference between particles & wall)

Many particles in the suspension

Known specific gravity of particles

Terminal velocity Average results of all the minerals in the particles, including the adsorbed water films.

1.2.4 Consistency Limits

Engineering Characterization of Soils


Consistency is the relative ease with which soil can be deformed. It is applicable to fine

grained soils whose consistency depends on water content. Relative consistency can be

expressed as very stiff, stiff, medium stiff, soft, very soft with increasing water content.

Atterberg, a Swedish agriculturist in 1911 observed four states of consistency, namely,

Liquid state, Plastic state, Semi solid state and Solid state in clayey soil with changing water

content. He set arbitrary limits for these states called consistency or Atterberg’s limits.

Fig. 16 : Consistency Limits in soil

Liquid Limit (wL), Plastic Limit (wP) and Shrinkage Limit (wS) are the atterberg limits. These

limits are most useful for engineering purpose in order ro classify the soils.

a) Liquid Limit (wL) : It is the water content corresponding to an arbitrary limit between liquid and plastic states of consistency of a soil. It is a minimum water content at which soil is still in liquid state, but possessing a small shear strength and exhibiting some resistance to flow.

b) Plastic Limit (wP) : It is the water content corresponding to an arbitrary limit between

plastic and semi-solid states of consistency of a soil. It is a minimum water content at which soil will just begin to crumble when rolled in to a thread of approximately 3 mm diameter.


c) Shrinkage Limit (wS) : It is the water content corresponding to an arbitrary limit between semi-solid and solid states of consistency of a soil. It is the lowest water content at which soil is fully saturated. It is also the maximum water content at which any reduction in water content will not reduce the volume of the soil mass.

Figure explains the three limits. Water content is plotted along the horizontal axis and volume of soil mass is plotted on the vertical axis. It can be seen that the soil mass in dry state possesses some volume. Any addition of water will not immediately increase the volume as water occupies the void space till wS is reached. Further increase in water content will continuously increase the volume of soil mass.

Determination of Liquid Limit from Casagrande approach

Fig. 17 : Determination of Liquid Limit

The water content at which a groove cut in a soil paste will close upon 25 repeated drops of

a brass cup with a rubber base.

A required amount of soil passing through a 425µm sieve is mixed with water until a

uniform consistency is achieved.

The liquid limit device is calibrated such that the height fall is 10 mm. The soil is then taken

into the cup such that the surface is parallel to the horizontal. Then using a grooving tool,

the soil is taken out such that the grooving tool is always perpendicular to the cup at the

time of contact.

Blows are given such that handle revolves at 2 revolutions per second. The number of blows

required for the soil to fail is noted and a sample along the failure plane is taken for


moisture content determination. Best results are obtained for blows ranging between 15 to

35.

Determination of Liquid Limit by FALL CONE Method

The fall cone test is more popular in Europe. This method attempts to eliminate the

shortcomings of Casagrande test

Fall cone test for liquid limit determination consists of a Standard cone that is brought into

contact with soil and clamped. At time zero cone is released and penetration at end of 5

seconds is recorded. Water content of soil is determined, cup emptied, and test repeated as

necessary at other water contents to produce a semi-log curve of w versus penetration.

Test Procedure for Fall Cone Test

1. Obtain the soil passing through 425 micron sieve, just as for the liquid limit test. The

difference is that it is a larger quantity and is placed in the cup.

2. The cone tip is brought into contact with the soil surface and time-initialized.

3. The cone is released for a free-fall into the soil and the penetration at the end of 5

seconds is recorded.

4. Since the water content for a penetration depth of 20 mm would be a chance

occurrence, several trials at different water contents on both sides of 20-mm

penetration are used to construct a semi-log plot of penetration versus water

content.

5. In one point test, the liquid limit is obtained as follows

where x = Penetration in mm

ω = Water content corresponding to penetration x

Table 3: Distinctions between Casagrande and Fall Cone methods

Casagrande Method Fall Cone Method

Professor Casagrande standardized the test and developed the liquid limit device.

Developed by the Transport and Road Research Laboratory, UK.

Multipoint test. Both Multipoint & One-point test.

Resistance to flow. Resistance to penetration.

At low ωL slightly lower values At low ωL slightly higher values


Determination of Plastic Limit

The moisture content at which a thread of soil just begins to crack and crumble when

rolled to a diameter of 3 mm. The plastic limit (ωP) is the water content (w%) where soil

starts to exhibit plastic behavior. It is the water content at which a soil changes from a

plastic consistency to a semi-solid consistency. It was developed by Atterberg in 1911. The water content at which a 3 mm thread of soil can be rolled out but it begins to crack

and cannot then be re-rolled was arbitrarily defined as Plastic limit.

Fig. 18 : Determination of Plastic Limit

Determination of Shrinkage Limit

It is the water content where further loss of moisture will not result in any more volume

reduction. It is much less commonly used than the liquid limit and the plastic limit. It is the

water content at which the soil volume ceases to change.

Fig. 19 : Determination of Shrinkage Limit


Soil volume: V, Soil weight: W Soil volume: Vd, Soil weight: Wd

Before drying After drying

Here. W = Initial weight of soil

Wd = Dry weight of soil

V = Initial volume of soil

Vd = Dry volume of soil

γω = Unit weight of water

The following are a few important definitions.

d) Plasticity : It is the property of soil that allows it to deform rapidly without rupture,

elastic rebound and volume change. The thin film of water adhering to clay particles is

responsible for its plastic behavior.

e) Plasticity Index (IP) : It is the difference between liquid and plastic limits. A soil with IP = 0

is called non-plastic (NP). Higher the IP, more plastic will be the soil.

PLPI

f) Consistency Index (IC) : It is the ratio of liquid limit minus natural water content to

plasticity index of soil.

P

LC

II


Here, w is the natural water content. When, IC < 0, soil is in liquid state. When IC = 0, soil is at its plastic limit and when IC > 1, soil is in semi-solid state and beyond and it is stiff. Hence, more the IC, the stiffer will be the soil. g) Liquidity Index (IL) : It is the ratio of natural water content minus plastic limit to plasticity

index of soil.

P

PL

II

Here, w is the natural water content. When, IL = 1, soil is at its liquid limit. When IL = 0, soil is

at its plastic limit and when IL < 0, soil is in semi-solid state and beyond. Hence, more the IL,

the softer will be the soil.

Flow Index (IF)

It is the slope of flow curve.

Toughness Index (IT)

1. It is the ratio of plasticity index to flow index.

2. It indicates the stiffness of soil.

3. More the toughness index, greater will be the stiffness and resistance to deformation of

soil

Activity of Clay

It is the ratio of Plasticity Index to percentage clay fraction.


Type of Clay Activity

Inactive Clay < 0.75

Normal Clay 0.75 – 1.25

Active Clay > 1.25

1. It was proposed by Skempton (1953).

2. Plasticity Index is influenced by both the amount and type of clay. Activity separates

this effect.

3. Higher the activity of clay, greater will be the volume increase when wetted, and

greater will be shrinkage when dried.

Determination of Density Index

Please refer to IS 2720 – Part 14 – 1983.

To find emax

1. Fill 1000 ml graduated glass cylinder up to 500 ml with distilled water.

2. Place a large funnel on top such that its bottom tip is in water.

3. Slowly place dry soil in to cylinder through the funnel such that particles settle uniformly

& independently.

4. Measure the weight and volume of soil to find dry density and hence emax

To find emin

1. Place the soil in compaction mould with collar in wet state.

2. Using needle vibrator compact it and add more soil to fill beyond collar.

3. Remove the collar and level the soil. Find the volume and dry weight and hence dry

density. emin can be determined later.

or

mmfractionclay

weightfractionclay

PIA

002.0:

)(%


1.13 Numerical problems on Index Properties Determination

Problem 10

A sample of soil was tested for water content determination and the following were the

results. Find water content.

Weight of empty cup = 55 g

Weight of cup + wet soil = 88 g

Weight of cup + soil after oven drying = 82 g

Solution

Weight of water = 6 g

Weight of solids = 27 g

Water content = 22.2 %

Problem 11

A sample of clayey soil was collected for the determination of water content. The following

are the details of measurement. Find water content. If the soil is fully saturated find

porosity. Take G = 2.7.

Weight of empty cup = 60 g

Weight of wet soil + cup = 120 g

Weight of dry soil + cup = 100 g (after drying in oven)

Solution

Weight of water = 120 – 100 = 20 g

Weight of dry soil = 100 – 60 = 40 g

Water content = 50%

If the soil is fully saturated, e = ωG = 1.35

Porosity n = 0.574

Problem 12

A Specific Gravity test is performed on river sand and the following are the results.

Weight of pycnometer = 900 g

Weight of pycnometer + dry soil = 1300 g

Weight of pycnometer + soil + water = 3050 g

Weight of pycnometer + water = 2800 g

The test was performed at 5oC during winter. Find G

Specific Gravity of water at 5oC = 1.000

Specific Gravity of water at 27oC = 0.997

Solution


Problem 13

The following are the results of specific gravity test on locally available sand.

Weight of pycnometer = 800 g

Weight of pycnometer + soil + water = 2800 g

Weight of pycnometer + water = 2600 g

G = 2.7

Find the weight of dry soil taken in the pycnometer in the beginning.

Problem 14

A density bottle was used to find specific gravity of clay and kerosene was the liquid

medium instead of water. Find G given the following details.

Weight of density bottle = 80 g

Weight of density bottle + dry soil = 110 g

Weight of density bottle + soil + kerosene = 141 g

Weight of density bottle + kerosene = 120 g

Specific Gravity of kerosene = 0.8


Problem 15

A core cutter of 100 mm internal diameter and 128 mm height was used to determine

the unit weight of soil at a site. The total weight of core cutter and soil was 3015 g and

empty weight of core cutter was 1374 g. A representative sample of soil was kept in a

steel cup of weight 60 g for the determination of water content. Weights of cup and soil

before and after placing in oven were respectively 106.4 g and 92 g. Find dry unit weight

of soil and degree of saturation if G = 2.7.

Weight of core cutter + soil = 3015 g

Weight of core cutter = 1374 g

Weight of soil = 1641 g

Volume of core cutter = 0.785D2H = 1.048 X 10-3 m3

Bulk Unit weight of soil = 15.66 kN/m3

Weight of water = 14.4 g

Weight of solids = 32 g

Water content = 45%

Dry unit weight = 10.8 kN/m3

Void ratio e = 1.45

Degree of Saturation S = 83.8%

Problem 16

The following are the results of test for in situ bulk unit weight at a site by sand replacement

method. Find γb and ρb.

Initial weight of sand and SPC W1 = 10550 g

Weight of sand in SPC after pouring in to cone W2 = 10105 g

Weight of sand in SPC after pouring in to calibrating container W3 = 8655 g

Weight of sand in SPC after pouring in to hole W4 = 8512 g

Diameter of Calibrating container = 100 mm

Height of Calibrating container = 127.3 mm

Weight of soil Wsoil = 1820 g


Problem 17

A sample of 1000 g of soil from a site was performed sieve analysis. The weights of soil

collected on each sieve are presented in the tabular entry. Find effective diameter, D30, D60

and coefficients of uniformity and curvature.

IS sieve

number

Sieve size

(mm)

Weight of Soil on

each sieve (g)

Cumulative

weight retained

% Cumulative

weight retained

% Finer

4.75 4.750 60 60 6 94

2.36 2.360 110 170 17 83

1.18 1.180 150 320 32 68

600 0.600 170 490 49 51

425 0.425 110 600 60 40

300 0.300 120 720 72 28

212 0.212 90 810 81 19

150 0.150 60 870 87 13

75 0.075 80 950 95 5

Receiver 50 1000 100 0

Total 1000



Problem 18

A soil sample of particle size ranging from 0.075 mm to 0.003 mm is put on top of the water

surface in a tank. Depth of water in the tank is 3 m. Estimate the time required for the first

particle and also the entire sample to reach the bottom of tank. Assume μ = 0.01 poise and

G = 2.7

Note: μ = Absolute viscosity (Poise)

1 Poise = 1 N-s/m2/g = 1 N/m-s

Viscosity η = μ/g 1 N/m-s/9.8 m/s2 = 0.102 N-s/m2

Data

μ = 0.01 Poise

η = 0.00102 N-s/m2 = 1.02X10-6 kN-s/m2

G = 2.7

γω = 9.8 kN/m3

First particle to settle has diameter 0.075 mm

Velocity v = 5.104 X 10-3 m/s

In a tank of water depth 3 m, time taken for settling = 587.77 s

Last particle to settle has diameter 0.003 mm

Velocity v = 8.17 X 10-6 m/s

In a tank of water depth 3 m, time taken for settling = 367197 s

Problem 19

In a shrinkage limit test the following data was obtained.

Initial weight of soil = 1.928 N

Initial volume of soil = 106000 mm3

Weight of soil after drying = 1.46 N

Volume of soil after drying = 77400 mm3.

Find shrinkage limit and initial and final void ratios. Take G = 2.7.

Data

W = 1.928 N

V = 106000 mm3

Wd = 1.46 N

Vd = 77400 mm3

G = 2.7

γω = 9.8 kN/m3 = 9.8X10-6 N/mm3

ωS = 12.86 %

ω = 32.05%

)1(

18

1 2

GDv


Since soil is saturated beyond shrinkage limit, S = 1

ωSG = Sefinal efinal = 0.35

ωG = Seinitial einitial = 0.87

Problem 20

A clayey soil was tested for liquid and plastic limits and the following were the results. Find

plasticity index, flow index, toughness index, consistency index and liquidity index. Plastic

Limit was found to be 28% and natural water content was 35%.

Number of Blows 34 23 18 12

Water Content 44.6 49.4 51.4 55.6

ω = 35%

ωP = 28%

ωL = 48.5%

IP = 20.5%

ω1 = 55.6%

ω2 = 44.6%

n1 = 12

n2 = 34

IF = 24.3%

IT = 0.84

IL = 0.34

IC = 0.66


SOIL CLASSIFICATION

1.14 Introduction

Soil classification is the arrangement of soils into different groups such that the soils

in a particular group have similar behavior. As there are a wide variety of soils

covering earth, it is desirable to classify the soils into broad groups of similar

behavior. Soils, in general, may be classified as cohesionless and cohesive or as

coarse-grained and fine-grained. These terms, however, are too general and include

a wide range of engineering properties. Hence, additional means of classification is

necessary to make the terms more meaningful in engineering practice.

1.15 The need for Soil Classification

Natural soil deposits are never homogeneous in character; wide variations in

properties and behavior are commonly observed. Deposits that exhibit similar

average properties, in general, may be grouped together, as a class. Through

classification of soils one can obtain an appropriate, but fairly accurate, idea of the

average properties of the soil group or a soil type, which is of great convenience in

any routine type of soil engineering project. From engineering point of view,

classification may be made based on the suitability of a soil for use as a foundation

material or as a construction material. For complete knowledge of behavior of soils,

all the engineering properties are determined after conducting a large number of

tests. However, an approximate assessment of the engineering properties can be

obtained from the index properties after conducting classification tests. A soil is

classified according to index properties, such as particle size and plasticity

characteristics. A classification system thus provides a common language between

engineers dealing with soils. It is useful in exchanging of information and experience

between the geotechnical engineers. It is important to note that soil classification is

no substitute for exact analysis based on engineering properties. For final design of

large structures, the engineering properties should be determined by conducting

elaborate tests on undisturbed samples.


1.16 Requirements for a Soil Classification System:

For a soil classification system to be useful to the geotechnical engineers, it should

have the following basic requirements.

1. It should have a limited number of groups

2. It should be based on the engineering properties, which are most relevant for

the purpose for which the classification has been made.

3. It should be simple and should use the terms, which are easily understood.

Any soil classification must provide with information about the probable engineering

behavior of a soil. Most of the classification systems developed satisfy the above

requirements.

1.17 Soil Classification Systems

Several classification systems were evolved by different organizations having a

specific purpose as the object. A. Casagrande (1948) describes the systems

developed and used in highway engineering, airfield construction etc. The two

classification systems, which are adopted by the US engineering agencies and the

State Departments, are the Unified Soil Classification (UCCS) and the American

Association of State Highway and Transportation Officials (AASHTO) system. Other

countries, including India, have mostly the USCS with minor modifications.

For general engineering purposes, soils may be classified by the following systems

1. Particle size classification

2. Textural classification

3. Highway Research Board (HRB) classification

4. Unified Soil Classification

5. Indian Soil Classification

1.18 Indian Standard Classification

As per I.S. Classification (IS: 1498-1970) the soil is divided into six groups:

(a) Boulders, particle size greater than 300 mm

(b) Cobble, particle size between 80 mm to 300 mm

(c) Gravel, particle size between 4.75 mm to 80 mm

(d) Sand, particle size between 0.075 mm to 4.75 mm


(e) Silt (size), particle size between 0.002 mm to 0.075 mm

(f) Clay (size), particle size smaller than 0.002 mm (2 m)

1.19 Indian Standard Soil Classification System:

Indian Standard Soil Classification system is in many respects similar to the Unified

system. However, there is one basic difference in the classification of fine-grained

soils. The fine-grained soils in this system are sub-dived into three categories of low,

medium and high compressibility instead of two categories of low and high

compressibility in Unified soil classification system.

Soils are divided into three broad divisions:

1) Coarse-grained soils, when 50 % or more of the total material by weight retained

on 75 m IS sieve.

2) Fine-grained soils, when more than 50 % of the total material passes 75 m IS

sieve.

3) If the soil is highly organic and contains a large percentage of organic matter and

particles of decomposed vegetation, it is kept in a separate category marked as peat

(Pt)

In all, there are 18 groups of soils

Coarse-grained soils – 8 groups

Fine-grained soils – 9 groups

Peat - 1 group


Coarse-Grained Soils

The classification of coarse-grained soils is done on the basis of their grain and

gradation characteristics as illustrated in Table, when the fines (75 m) present in

them are less than 5 % by weight. Coarse-grained soils are sub-dived into gravel

and sand. The soil is termed gravel (G) when more than 50 % of coarse fraction (>75

m) is retained on 4.75 mm IS sieve, and termed sand (S) if more than 50 % of

coarse fraction is smaller than 4.75 mm IS sieve. These are further sub-divided as

given in Table into 8 groups.

Coarse-grained soils which contain more than 12 % fines (< 75 m) are classified as

GM or SM if fines are silty in character (meaning, the limits plot below the A-line on

the plasticity chart). On the other hand they are classified as GC or SC if the fines

are clayey in character (meaning the limits plot above the A-line on the plasticity

chart).

Fine-grained Soils

The fine-grained soils are classified on the basis of their plasticity characteristics

using the I.S Plasticity chart shown in Fig. The fine-grained soils are further divided

into three sub-divisions depending upon the values of the liquid limit:

a) Silts and clays of low compressibility – These soils have a liquid limit less than 35

% (represented by a symbol L)

b) Silts and clays of medium compressibility – These soils have a liquid limit > 35 %

but < 50 % ( represented by a symbol I)

c) Silts and clays of high compressibility – These soils have a liquid limit > 35 %

(represented by a symbol H)

Fine-grained soils are further sub-divided, as given in table in 9 groups.

Boundary Classifications:

Sometimes, it is not possible to classify to classify a soil into any one of 18 groups

discussed above. A soil may possess characteristics of two groups, either in particle

distribution or in plasticity. For such cases, boundary classifications occur and dual

symbols are used.


a) Boundary classification for coarse-grained soils:

Coarse-grained soils having 5 % to 12 % fines are borderline cases and given a dual

symbol. While giving dual symbols, first assume a coarse soil and then a finer soil

i.e., the first part of the symbol is indicative of the gradation of coarse fraction, while

the second part indicates the nature of fines.

For example, a soil with the dual symbol SW-SC is a well-graded sand with ‘clayey’

fines that plot above A-line.

(i) Boundary classifications within gravel or sand groups can occur. The following

classification are common

GW-GP, GM-GC, GW-GM, GW-GC, GP-GM,

SW-SP, SM-SC, SW-SM, SW-SC, SP-SM.

(ii) Boundary classifications can occur between the gravel and sand groups such as

GW-SW, GP-SP, GM-SM, and GC-SC

Note: The rule for correct classification is to favour the non-plastic classification.

For eg. A gravel with 10 % fines, Cu = 20 and Cc = 2 and IP= 6 will be classified as

GW – GM, and not GW – GC.

b) Boundary classification for fine-grained soils:

Fine-grained soils also can have dual symbols.

(i) If the limits plot in the hatched zone on the plasticity chart, i.e., IP between 4 and

7, the soil has a group symbol CL – ML.

(ii) If the position of the soil on the plasticity chart falls close to the A-line, dual

symbol is used, such as CI – MI , CH – MH

(iii) If the liquid limit is very close to 35 % or 50 %, dual symbols are used, such as

ML – MI, MI – MH, CL – CI, CI – CH, OL – OI, OI – OH.

c) Boundary classification between coarse-grained and fine-grained soils:

Dual symbols can also be used when the soils have about equal percentage of

coarse – grained and fine – grained fractions. The possible dual symbols in this case

are GM – ML, GM – MI, GM – MH, GC – CL, GC – CI, GC – CH, SM – ML, SM – MI,

SM – MH, SC – CL, SC – CI, SC – CH.


Step-by-step procedure:

A step-by-step procedure for classifying the soils as per IS: 1498 –1970 is illustrated

below:

1. Determine whether the given soil is of organic origin or coarse-grained or fine

grained.

An organic soil is identified by its colour (brownish black or dark) and characteristic

odour. If 50 % or more of the soil by weight is retained on the 75 m sieve, it is

coarse-grained, if not, it is fine-grained.

2. If the soil is coarse-grained:

(a) Obtain the GSD curve from a sieve-analysis. If 50 % or more of the coarse

fraction (>75 m) is retained on the 4.75 mm sieve, classify the soil as gravel (G); if

not, classify it as sand (S)

(b) If the soil fraction passing through the 75 m sieve is less than 5 %, determine

the gradation of the soil by calculating Cu and Cc from the GSD curve. If well graded

(according to the criteria laid down), classify the soil as GW or SW; if poorly graded,

classify as GP or SP.

(c) If more than 12 % passes through the 75 m sieve, perform the liquid limit and

plastic limit tests on the soil fraction passing though the 0.425-mm sieve. Use the I.S.

plasticity chart to determine the classification (GM, SM, GC, SC, GM–GC or SM–SC)

(d) If between 5 % and 12 % passes through the 75 m sieve, the soil is assigned a

dual symbol appropriate to its gradation and plasticity characteristics. (GW–GM,

GW–GC, GP–GC, GP–GM, SW-SM, SW-SC, SP–SC, SP-SM)

3. If the soil is fine-grained (inorganic):

(a) Determine wL and wP on the minus 0.425 mm sieve fraction and determine the

plasticity index.

(b) If the limits plot below the A – line, classify as silt (M). Further, if wL is less than

35, classify as ML; if wL is between 35 – 50 %, classify as MI; if wL is greater than 50, classify as

MH.

(c) If the limits plot above the A – line, classify as clay ©. Assign the group symbol

CL or CI or CH, depending on the value of liquid limit, as in (b).

(d) If the limits plot in the hatched zone, classify as CL – ML. If the limits plot close to

the A – line or close to wL= 35 % or wL= 50 % lines, assign dual symbols as outlined

earlier.


4. If the soil is of organic origin, the plasticity chart is used after determining wL and

wP and the soil classified as OL, OI or OH.

5. If the soil has about 50 % each of fines and coarse – grained fractions,

(a) Determine whether the coarse – grained fraction is gravel (G) or sand (S)

(b) Determine wL and wP on the minus 0.425 mm fraction

(c) Depending on whether the limits plot above the A – line or below the A – line,

classify as C or M

(d) Based on wL, classify as L, I or H

(e) Assign the dual symbol from the information obtained in steps (a), (b), (c), and (d)

as for example, GM – ML, GM – MI, etc.,

Fig. Plasticity Chart as per IS 1498 - 1970


Major Divisions Group

Symbol Typical Names Field Identification

procedures Information

required for

describing soils

CO

AR

SE

-GR

AIN

ED

SO

ILS

More

than

hal

f of

mat

eria

l is

lar

ger

75 μ

m s

ieve

size

.

The

75 μ

m s

ieve

size

is

about

the

size

par

ticl

e vis

ible

to t

he

nak

ed e

ye

GR

AV

EL

S

More

than

hal

f of

coar

se f

ract

ion i

s sm

alle

r th

an

4.7

5 m

m s

ieve

size

Cle

an g

ravel

s (l

ittl

e or

no f

ines

)

GW Well graded gravel, gravel-sand mixtures, little or no fines

Wide range in grain size and substantial amounts of all intermediate particle sizes

For undisturbed soils and information on stratification, degree of compactness, cementation, moisture conditions and drainage characteristics. Give typical name;

indicate approximate percentages of sand, gravel, maximum size, angularity, surface condition and hardness of the coarse grains, local or geologic names and

other pertinent descriptive information and symbol in parenthesis. Example: Silty and gravelly, about 20 % hard angular gravels 10

mm maximum size, rounded and sub-circular sand grains about 15 %. Non-plastic fines with low dry strength, well compacted and moist in place alluvial sand,

(SM)

GP Poorly graded gravels or gravel-sand

mixtures, little or no fines

Predominantly one size or a range of sizes with

some intermediate sizes missing

Gra

vel

wit

h f

ines

(Appre

ciab

le a

mount

of

fines

)

GM Silty gravels, poorly graded gravel-sand-silt mixtures

Non-plastic fines or fines with low plasticity (for identification procedures see ML and

MI below)

GC Clayey gravels, poorly graded gravel-sand-silt mixtures

Plastic fines (for identification procedures see CL and CI below)

SA

ND

S

More

than

hal

f of

coar

se f

ract

ion i

s sm

alle

r th

an

4.7

5 m

m s

ieve

size

Cle

an s

ands

(Lit

tle

or

no

fines

)

SW Well graded sands, gravelly sands, little or no fines

Wide range in grain size and substantial amounts of all intermediate

particle sizes

SP Poorly graded sands or

gravelly sands, little or no fines

Predominantly one size

of a range of sizes with some intermediate sizes missing

San

d w

ith f

ines

(Appre

ciab

le a

mount

of

fines

)

SM Silty sands, poorly graded sand silt

mixtures

Non-plastic fines or fines with low plasticity

(for identification procedures see ML and MI below)

SC Clayey sands, poorly graded sand-clay mixtures

Plastic fines (for identification procedures see CL and CI below)

Symbols Laboratory Classification Criteria

GW Cu Greater than 4 Cc Between 1 and 3

Determine percentage of gravel and sand from grain size curve depending on percentage of fines (fraction smaller than 75 μm sieve size); coarse-grained soils are classified as follows: Less than 5 %: GW, GP, SW,SP More than 12 %: GM, GC, SM,SC 5 % to 12 % : Border line cases requiring use of dual symbols Uniformity coefficient, Coefficient of Curvature,

GP Not meeting all gradation requirements for GW

GM Atterberg limits below “A” line or Ip less than 4

Limits plotting above “A” line with Ip between 4 and 7 are border line cases requiring use of dual symbols

GC Atterberg limits above “A” line with Ip greater than 7

SW Cu Greater than 6 Cc Between 1 and 3

SP Not meeting all gradation requirements for SW

SM Atterberg limits below “A” line or Ip less than 4

Limits plotting above “A” line with Ip between 4 and 7 are border line cases requiring use of dual symbols

SC Atterberg limits above “A” line with Ip greater than 7


Major Divisions Group

Symbol

Typical Names Identification procedures (on

fraction smaller than 425 μm sieve

size)

Information

required for

describing soils

Dry

strength

Dilatancy Toughness For undisturbed soils, add information on

structure, stratification, consistency in undisturbed and remoulded states, moisture and drainage conditions Give typical names,

indicate degree and character of plasticity, amount and maximum size of coarse grains, colour in wet condition, odour, if any, local or geologic name and

other pertinent descriptive information, and symbol in parenthesis. Example: Clayey silt, brown, slightly plastic, small percentage of fine-

sand, numerous vertical root holes, firm and fry in place; losses, (ML)

FIN

E-G

RA

INE

D S

OIL

S

More

than

hal

f of

mat

eria

l is

sm

alle

r th

an 7

5 μ

m s

ieve

size

.

The

75 μ

m s

ieve

size

is

about

the

size

par

ticl

e vis

ible

to t

he

nak

ed e

ye

SIL

TS

& C

LA

YS

of

low

com

pre

ssib

ilit

y;

liquid

lim

it

< 3

5 %

ML Inorganic silts and very fine sands, rock flour, silty or

clayey fine sand or clayey silts with none to low plasticity

None to low

Quick None

CL Inorganic clays, gravelly clays, sandy clays, lean clays of low plasticity

Medium None to very slow

Medium

OL Organic silts of low plasticity

Low Slow Low

SIL

TS

& C

LA

YS

of

med

ium

com

pre

ssib

ilit

y:

Liq

uid

lim

it

35 a

nd 5

0 %

MI Inorganic silts, silty or clayey fine-sand or clayey

silts of medium plasticity

Low Quick to slow

None

CI Inorganic clays, gravelly clays, sandy clays, silty

clays, lean clays of medium plasticity

Medium to high

None Medium

OI Organic silts and organic silty clays of medium plasticity

Low to medium

Slow Low

SIL

TS

&C

LA

YS

of

hig

h

com

pre

ssib

ilit

y

Liq

uid

lim

it

> 5

0 %

MH Inorganic silts of high compressibility, micaceous or diatomaceous fine sandy or silty soils, elastic silts

Low to medium

Slow to none

Low to medium

CH Inorganic clays of high plasticity, fat clays

High to very high

None High

OH Organic clays of medium to high plasticity

Medium to high

Slow to very slow

Low to medium

High organic soil Pt Peat and other highly organic soils with very high compressibility

Readily identified by colour, odour, spongy feel and frequently by fibrous texture

1.20 Unified Soil Classification System

Originally developed by Casagrande (1940), the Unified Soil Classification System was

used for construction during World War II. It was modified in 1952 by Bureau of

Reclamation & Corps of Engineers (USA) for the construction of geotechnical structures.

It was later adopted by ASTM & BIS (1970).

• Two broad divisions – Coarse grained and Fine grained. • Coarse grained – 8 groups

(GW, GP, GM, GC, SW, SP, SM & SC) • Fine grained – 7 groups

(ML, MH, CL, CH, OL, OH & Pt) • A Line – same eqn : Ip = 0.73 (wL-20%) • However medium plasticity range is not available here.


1.21 Numerical problems on Soil Classification

Problem 21

Classify the following three soils as per IS S N % G % S % F Cu Cc wL wP

1 1 98 2 6.5 1.8 - -

2 1 9 90 - - 60 20

3 0 86 14 - - - -

Cu > 6, 1 < Cc < 3 1. SW 2. IP = wL – wP = 40

IP,th = 0.73(wL-20) = 29.2 < 40, Point above A Line, Clay wL > 60, Soil of High Plasticity CH 3. S Problem 22 The following results were obtained from soil test in lab. % of soil passing 0.075 mm IS sieve = 25% % of soil passing 4.75 mm IS sieve = 75% Liquid Limit = 60% Plastic Limit = 25% Uniformity Coefficient = 8 Coefficient of Curvature = 1.5 Classify the soil according to IS %G = 25%, %S = 50%, %F = 25% - Coarse Grained & Sand IP,th = 0.73(wL-20%) = 29.2% < IP, Above A Line SW-SC Problem 23 Classify the following soils as per IS

S N %G %F Cu

Cc

wL

wP

A 20 60 - - 50 35

B 60 30 - - 52 38

C 40 2 5 1.2 - -

A: MH B: GM C: S Problem 24 The following data refers to a sample of soil. % passing 4.75 mm IS sieve = 64% % passing 0.075 mm IS sieve = 6% Uniformity Coefficient = 7.5, Coefficient of curvature = 2.7 Plasticity Index = 2.5 Ans: SW-SM

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