geotechnical engineering

53
UNIT 1 SOIL PHASES, STRUCTURE, CBNSISTENCU AND CLASSIFICATION Structure 1.1 Introduction Objectives 1.2 Origin and Nature of Soil 1.2.1 The Cieological Cycle 1.2.2 Definition of Soil 1.2.3 Three-phase Con~position 1.3 Phasc Relationships 1.3.1 VolurneRelationships 1.3.2 Weight Relationships 1.3.3 Intcrrdationships 1.3.4 Relative Dcnsity 1.4 Gradation 1.4.1 Particle Size 1.4.2 Particle Sip=Classification 1.4.3 Particle Size Dist~ibution 1.4.4 Significance of Gradation 1.5 Soil Structure 1.5.1 Stluctures in Cohesianlcss Soils 1.5.2 Structures in Cohesive Soils 1.5.3 Composite Structures 1.6 Consistency 1.6.1 Consistency 1,imits 1.6.2 Ddernlination of Consistency Lirllits 1.6.3 Significance of Consistency Limits 1.6.4 Activity of Cli~yey Soils 1.6.5 Engineering IJse of Allcrberg Limits 1.7 Clay Mineralogy 1.7.1 Structure:of Clay Minerals 1.7.2 Sheet Structures 1.7.3 Conlnion Clay Minerals 1.7.4 Natu1.c: of Water in Clay 1.7.5 Itlentiiication of Clay Mincrids 1.8 Soil Classification 1.8.1 Geological Classification 1.8.2 Pedologicd Classification 1.8.3 Textunl Classification 1.8.4 Engineering Classification 1.8.5 Cornparisol~ of AASHO and UNIFIED Systenls 1.9 Summary 1.10 Key Words 1.1 1 Answers to S AQs 1.12 Further Reading 1.1 INTRODUCTION This unit seeks to introduce you to several basic concepts necessary lo u~idersland the behaviour of soils. To begin wih, the idea of soil as a product ot'natural geological processes is prese~iled. The representation of such a nlalerial using a three-phase rllotlel is inlroduced. This is followed by its quantitative chiuacterization tluough weight-volume relationships between the constituent phases. Next, concepts of structure of soil, consistency and struclure oC clay are explained. Finally, the various attempts a1classifying

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Page 1: Geotechnical engineering

UNIT 1 SOIL PHASES, STRUCTURE, CBNSISTENCU AND CLASSIFICATION

Structure 1.1 Introduction

Objectives

1.2 Origin and Nature of Soil 1.2.1 The Cieological Cycle 1.2.2 Definition of Soil 1.2.3 Three-phase Con~position

1.3 Phasc Relationships 1.3.1 VolurneRelationships 1.3.2 Weight Relationships 1.3.3 Intcrrdationships 1.3.4 Relative Dcnsity

1.4 Gradation 1.4.1 Particle Size 1.4.2 Particle Sip= Classification 1.4.3 Particle Size Dist~ibution 1.4.4 Significance of Gradation

1.5 Soil Structure 1.5.1 Stluctures in Cohesianlcss Soils 1.5.2 Structures in Cohesive Soils 1.5.3 Composite Structures

1.6 Consistency 1.6.1 Consistency 1,imits 1.6.2 Ddernlination of Consistency Lirllits 1.6.3 Significance of Consistency Limits 1.6.4 Activity of Cli~yey Soils 1.6.5 Engineering IJse of Allcrberg Limits

1.7 Clay Mineralogy 1.7.1 Structure: of Clay Minerals 1.7.2 Sheet Structures 1.7.3 Conlnion Clay Minerals 1.7.4 Natu1.c: of Water in Clay 1.7.5 Itlentiiication of Clay Mincrids

1.8 Soil Classification 1.8.1 Geological Classification 1.8.2 Pedologicd Classification 1.8.3 Textunl Classification 1.8.4 Engineering Classification 1.8.5 Cornparisol~ of AASHO and UNIFIED Systenls

1.9 Summary

1.10 Key Words

1.1 1 Answers to S AQs

1.12 Further Reading

1.1 INTRODUCTION

This unit seeks to introduce you to several basic concepts necessary lo u~idersland the behaviour of soils. To begin wih, the idea of soil as a product ot'natural geological processes is prese~iled. The representation of such a nlalerial using a three-phase rllotlel is inlroduced. This is followed by its quantitative chiuacterization tluough weight-volume relationships between the constituent phases. Next, concepts of structure of soil, consistency and struclure oC clay are explained. Finally, the various attempts a1 classifying

Page 2: Geotechnical engineering

Soil Water System soils are traced and engineering classification systems are explained. Some illustrative examples and problems are given to reinforce the concepts presented.

Objectives The main objective of this unit, is to present lo the reader the conceptual model and the mathematical relationships used to represent soil and the scientific approach required to characterize its behaviour.

After studying this unit, you should be able to :

o appreciate the need for a three-phase conceptual lnodel and to make computations using weight-volumne relationships,

e know the importance of consistency limits and evaluate them,

describe the basic building blocks constituting a clay mineral and their significance, and

0 classify a soil for engineering purposes.

1.2 ORIGIN AND NATURE OF SOIL

In a btoad sense, soil may be thought of as an incidental material in the vast geological cycle which has been going on continuously for millions of years of geological time. It is useful to have an idea of the cycle in order to appreciate what soil is.

1.2.1 The Geological Cycle This cycle consists of three principal phases. These u e erosion; transportation and deposition; earth movement. The cycle is schematically illustrated in Figure 1. I.

E A R T H M O V E M E N T

Figure 1.1: The Geological Cycle

The cycle starts with the erosional phase in which there is degradation of exposed rock by weathering processes. The weathering processes may be mecl~anical or chemical. In mechanical weathering alternate seasonal heating and cooling oS rock leads to cracking, disintegration and breakdown of massive rock into smaller blocks. Freezing and expansion of water in pores and joints accelerates disintegration in cold regions. Disintegration could also be caused by wind or ice. In chemical weathering water aid carbon dioxide from the atmosphere react with the salts and acids in the rock and form new nlinerals and salts. Minerals formed at high temperatures are less resistait to weathering than those formed at low temperatures. These two processes (mechanical a11d chemical) act often in combination. The resulting disintegrated material is termed as soil.

In the second phase the fragmented material is &ansported by agents such as wind, water or ice to new locations, The fragmented materials x e further eroded. Then sorting md mixing take place resulting in the deposition and forlnatioil of ncw soil.

The erosionand deposition processes eventually lead to new la~dlbrms and, consequently disturbance of equilibrium in the earth's crust. As a result, Ihe third phase, earth movement, occurs and rock formations are again exposed and subjected to renewed erosion. After several such cycles of varying intensity a helerogeneous soil mass is

8 , evolved. I

1

I

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1.2.2 Definition of Soil AS you know very well, the earliest interaclion of lnan wilh soil was for agriculture. In

applications soil is collsidered as that part of the earth's crust which supports life. On Ihe othcr hand, the interesls of geologists include rocks :uld oher maleriiils

d ~ c p in the earth's interior. engineer's prirnary inlerest is in strength md delormiibility soils and rocks. Neverhelcss, lhe basic pri~lciplcs ofscicnce used in these allied fields

ilre useful to engineering as well.

n u s , thcre is a formal definilioli for soil. For engineering purposes, soil is defined ils millera1 maltel at or near thc earth's surhce contai~li~ig water, orga~lic

matter, air and olher substances which may be present in gaseous or liquid fonns in Ihc void spaces. This broad definition of soil includes a wide rmgc of materials from thc hart1 coarse grained gravels to soft fine-grained peat soils.

1.2.3 Three-phase composition From the above definition you will llotice that soil ciui consist or materiiils in solid, li~luid ;md gascous states. These iue known as Phases. The solids consist of mineral matter. The liquid phase is usuillly inilde up ol'water allliough dissolved salt solutions ;u.c also possible. Thc gaseous phase is generally made up of air Lhough, occas~onally, pases l~kc methane may idso be present.

The bellaviour of soil is ilfluencecl by tllc rcliltive proportions zu~d chenlical compositions ot' the three phases. Hence, in Soil Mech:ulics, the concept of il Ihrce-ph;lse milteriill is used for representing soil. Actually, Lhesc phases coexist in rcal soils in a mixed maluler as depicted in Figurc 1.2 (a). Bul, li)r purposes ol'cngincering studies these phases are co~lsidered to be independcat and separate. The conceptual lllrce-phiwc model used in Soil Mechanics is shown in Figure 1.2 (b). Allliough conccptu;il, the model is extremely uscful in explaining certain observed pheno~~lena in soils such iLS ('low, compressibility, stress transfer-and shear strenglli.

A l R

S O L I D S

W A T E R

W A T E R

S O L I D S

Figure 1.2 (a) : Soil Three-ptiirsc systcln Figure 1.2 (b) : Coliceptuol Thrcc-pl~nsc Modcl

1.3 PHASE RELATIONSHIPS

It is useful lo express Ulc influence of thc phascs quantitatively. For Lhis purpose, relationships betwccn weights and volumes are used. Let us designale the weights and the volunles ofthe thrce phases as shown in Figurc 1.3.

Figurc 1.3 : Wcigllt-Volulnc Relolio~~ships

St~il Phnscs, Structure, Corisistc~icy rind

Cklssificntion '

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Snil Water Systcm The sy~nhols in the Figure 1.3 havc Ule followirlg meaning :

V = Total volulne of soil sample

V, = Volume of solids

V,, = Volume of voids

V,, = Volume of water in the voids

V, = Volunle of air in the voids

W = Total weight of given sol1 sample

W, = Weight of solids

W,, = Weight of water in the voids

W, = Weight of air in the voids

Obviously, the total volulne and weight must be equal to the sum of the volumes and weighls, respectively, of the consliluents. Therefore, two relationships which must bold are

and, neglecting the weight of air,

1.3.1 Volume Relationships The volume relationships co~nmo~lly used for the hree phases in a soil element are void ralio, porosity arid degree of saturation.

Void ratio e is defined as the ralio of the volulne o l voids to the volume of solids, or

9 I Generally volume of voids will be less thim volume of solids. Tllercfore e is usually expressed as il decinlal fraction. But it can attain v;llues greater than unity in soil suspensioils and in macro-porous soils like loess, where V,, could be mucl~ greater than V,. ' 1 I

Porosity n is defined as the ratio of the volume of voids to the total volulne, or 1

This is usually expressed as a percentage. I Degree of saturation S is define8 as thc ratio of the voluinc of water to thc volume of voids, or I

This is also usually expressed as a percentage and varies from 0 for a dly soil to 100 for a saturated soil.

The relationships between void ralio and porosity call be derived from Equalions (1.3), (I.]), and (1.4) as follows :

Also, from Equation (1.6),

You should note that porosity is detii~ecl wit11 respect to V while' void ratio is defined with respect to V,. The total volume V is a variable quantity. Bul, sincE solids are incompressible, V, remains invariant in the total volumc V of Ihe soil. Thus, in engineerii~g studies, void ratio serves as a useful reference puanleter for rcprcsenling cllallpe ill

10 volume of soil undcr compression.

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1.3.2 Weight Relatior~ships The common weight relationships used are moisture conlent, unit weight and specific gravity. Moisture ~01ltcnt f i t is also referred to often as water content and denoted as W. ~t is defined as the ratio of the weight of water to the weight of solids in a given volume of soil, or

Unit weighty is the weight of soil per unit volume ~md three differerll values are normally . used depending on the state of Lhe soil. The bulk or moisl unit weight yl is defined as the weight of the soil including b a t of waler per unit volume, or

W Yr = 7 ...( 1.9)

Dry unit weight Yd is the weight of solids alone per unit volume of soil, or

ws Yd = 7 ...( 1.10)

Using Equations (1.9), (1.21, (1.8) and (1.10) we can relate y, with yd as follows :

In addition we frequently use the submerged or buoyant unit weight ts,b for soils submerged under water. This is obtained by subtracting the upward thrust of water on unit volumc of soil, which is nothing but the unit weight of water.

where, y,,, = unit weight of water. :*,

Specific gravity of a substance is the ratio of its weight to the weight of an equal volume of water. The specific gravity of the solids, G,r, (excluding air and water) is expressed by

Specific gtavily is a dimensionless quantity. An average value of Gs usually adopted for soils is 2.65.

1.3.3 Interrelationships There are several uscful interrelationships between the parameters defined above.

The water contei~t and void ratio of a soil are interrelated as shown below :

Rom Equations (1.5) and (1.3), we get,

Similarly, from Equations (1.8) ruld (1 .l3),

where,

Therefore,

Soil Phuscs, Strocture, Col~sistcncy and

Clsssificatio~~

When soil is saturated, S = 1 and we get,

e = mGs ...( 1 .16a) *

It is much easier to measure weights accurately than volumes. For example, the water Lolltent of a soil sample is easily determined by weighing a little soil in its nat~ral moist

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Soil Water System state and after drying in the oven at llO°C and computing the difference. The difference is the weight of water, except in organic soil or soils containing additives such as asphalt. Hence, Equation (1.16) is very useful for solving problems involving three-phase relationship.

It is now possible to express the unit weights in terms of Gs, e, m and S as follows

Similarly,

For a saturated soil S =I. Therefore, the saturated unit weight of soil, y,,,, will be

1.3.4 Relative Density It will be shown in sub-section 1.5.1 that in soils such as sands Ule packing or arrangement of grains can vary over a wide range. Correspondingly, the void ratio also varies, Therefore, it is convenient tdiexpress density of packing at any stage in terms of void ratio using Relative Density D, wliich is defined as

where, D, = relative density, usually a percentage, - ,, ei, = vdid ratio of soil in loosest condition,

enaL = in situ void ratip of soil, and

e = void ratio of soil in densest condition.

This is now known as Density Index and denoted as ID.

It may also be expressed in terms of dry unit wejghtsas a percentage as given below :

The relative density D, or Density Index ID may vary from a value close to zero for a very loose soil to a maximum of one for a very dense soil. Some typical values of relative densities are given in Table 1.1. Some typical values of minimum and maximum void ratios of granular soils are given in Table 1.13. I

Table 1.1 : Relationship between Relative Density.and State of Granular Soil

The relationships given above are fundamental and important and the student should understand them well and learn their application. 'Ihese are used in calculations of volume change, compaction, settlement and stability. A few examples are given below to illustrate the application.

Relative Density (%)

0 t11 15

15 to 38

35 to 65

65 to 85

8 5 b 1 0 0

I Example 1.1

Description of Soil Deposit

very loose

loose

medium

dense

very dense

The weight of a soil in its dry state is 150.5 N. When some water is added lo it, its weight increases to 180.3 N and it has a volume of 0.01 m3. If Gs = 2.65, calculate

12 the moisture content, void ratio, dry unit weight and moist unit weight.

Page 7: Geotechnical engineering

Solution

1) From Equations (1.8) and (1.2),

2) From Equations (1.13), (1.1) and (1.3), taking y,,, = 9.81 x I O ~ N / I ~ ~

3) From Equation (1. lo),

4) From Equation (1.9),

yt =-=-- 180a3 - 18030 N / I - ~ ~ = 18.03 kN/m3 v 0.01

Example 1.2

At a certain construction site the natural moisture content is 25% and tlie void ratio is 0.7. If the specific gravity is 2.66, calculate Ule porosity, moist unit weight, dry unit weight and degree of saturation.

Solution

1.) From ~ ~ u a t i o n (1.7),

2) From Equation (1.16) and (1.17),

(1 + 0.25) 2.66 x 9.81 - 19.19 irN,md 'Yt = -

1 + 0.7

3) From Equation (1.18),

4) From Equation (1.1 6),

Example 1 3

A soil is saturated at 52 per cent moisture content and has a unit weight of 16.5 k~lrr?. Calculate its void ratio, specific gravity, dry unit weight and submerged unit weight.

Solution

1) Assume volume of soil to be V = , 1 m3

2) From given data W=l6.5 kN; rn = 0.52;.S = 1

3) From Equations (1.2) and (1.Q we ctm write,

Suil I'l~uses. Structure, Consistency and

Clnssificntio~r

Page 8: Geotechnical engineering

Soil Wnter System

4) Solving these equations,

5 ) SinceThe soil is saturated, using gati ions (1.5f and (1.15a),

6) From Equation (1. I),

V, = 1.00 - 0.57 = 0.43 m3

7) From Equation (1.3),

8) From Equation (l.l3),

9) Froin Equation (1.18),

Gs YW 2.57 x 9.81 = kN/mR 'yd=-- - l t e 2.33

10) From Equation (1.12),

Example 1.4

The unit weight of a slightly moist soil is 1.75 g/cmband its moisture content as determined in the laboratory is 8%. On adding some water to 10 rn3 of the soil its water content rises to 17%. Let the specific gravity be 2.65. Determine the quantity of water added. (This example iIlustrates tlie use of CGS units.)

Since the soil is not remulded or disturbed in aily way its void ratio remains unchanged.

1) Let the volume of soil be V = 1 cm3

2) Then we have from Equations (1.2) and (1.8),

W = W, + Ww = 1.75g.

3) Solving above equations we have,

(I + 0.08) W, = 1.75 .

W, = 1.620g ; W, = 0.129g.

4) After adding water, m = 0.17.

Since remains unchanged we have,

W, = 0.17 (1.62) = 0.275g. 1

5 ) Weight of water to be added to 1 c m h f soil is :

14 6) +Volulne of water to be added Lo 1 cm%f soil is : i

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Vwadd = 0.146 cm3 = 0.000146 m3

(Since unit weight of water is 1 g/cm3)

7) Volume of water required to be added to 10 m3 of soil is :

7 Vw,r, = 1.46 m- . Example 1.5

The porosity of a sand sample is 0.4. Assu~ning a specific gravity of 2.68, compute void ratio, dry unit wcight and moist unit weight at 60% saturation. Unit weight of water may be taken as 9.81 kN/n~.?.

Solution

1 j Using Equation (l,6),

e = n41- n) = 0.4/(1 - 0.4) = 0.667

2) Frorn Equation (1 . I 8),

3) Using Equation (1.17) now,

Soil Pha.:cs, Structure, C msistency and

ClassifEcation

(2.68+0.6X0.667)(9.81) = 18.13m/m3 At, S = 0.6, yt = (1 + 0.667)

SAQ 1

i) A moist soil weighs 10.1 N and has a volume of 600 cm3. 11s dry weight is 9,18 N and Gs is 2.7. Determine void ratio, porosity, moisture content, degree of saturation, bulk unit weigh1 and saturated unit weight. Take unit weight of water as 10 k ~ / r n ~ .

ii) The initial moisture content of a soil is 10% at a porosity of 0.4. The specific gravity of solids is 2.65. Determine the quantity of water to be added to 1 m3

of the soil in order to saturate it completely.

iii) A soil in a borrow pit has a moist unit weight of 1.8 g/cm3 and water content of 15%. The soil is broughl to a construction site, placed in the form of an embankment and compacted. Tile dry unit weight of the compacted soil is - 1.6 g/cm3 and water content is 18%. Specific gravity of solids is 2.65. (Note that unit weight of water is 1 g/cm3.) CaIculate :

a) void ratios of the soil in the borrow pit and in the embankment,

b) volume of the soil to be excavated to make 100 m3 of compacted soil, and

c) volume of water to be added during compaction.

(This problem uses CGS units. The student is advised to solve it using the same units and get familiar with this system of units.)

iv) Values of certain parameters ai-e given below for three soils. Compute tlle parameters against tile (1) mark using the appropriate expressions derived above. Take y, = 10 k ~ / r n ~ .

Table 1.2 : Data for SAQ 1 (iv)

m Gd

- 2.70

45 2.70

20 2.70

Unit wcights are in k ~ l m 3 ; S and m nre in %.

S

7

100

7

e

0.84

7

-

Soil No.

1

2

3

'Yd

7

- -

71

18

?

19

3

-

?

-

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Soil Water S y s t c ~ ~ ~ v) The moist unit weight of a soil at a moisture content of 11% is 1.82 glcm-'. The soil is now dried and remoulded and its unit weight increases to 1.87 g/crn3 and the void ratio decreases 0.460. What is the water content of the dried soil? Take G, = 2.67.

(Students may note that this exercise also uses CGS units.)

1.4 GRADATION

1.4.1 Particle Size Soil is made up of particles of different sizes. This is referred to as gradation. Soil, as defined in sub-section 1.2.2, includes particles varying in size from mm, to even a metre or more.

Certain names such as gravel, sand and clay have been in use in practice for long to describe, somewhat arbitrarily, certain size ranges. Although widely used, the words did not have a fixed connotation. These names and their corresponding size ranges have now been somewhat standardized. A number of systems have been evolved for this purpose. The size ranges, as used in Soil Mechanics, are given in Table 1.3. There are minor differences between the various systems used. The values outside the parentheses in the table are as per Bureau of Indian Standards while those wilhin are according to the internationally used UNIFIED system (see sub-section 1.8.4).

Sand

Silt

Qay

Table 1.3 : Particle Sizes

0.06 (or 0.076 rnm) to 2.0 mm (or 4.76 mm)

0.002 to 0.06 mm (or 0.076 mrn)

< 0.002 mm

Type of Soil

Boulder

Pab ble

Gravel

1.4.2 Particle Size Classification

Size

> 30 cm

15 to 30 cm.

2.0 mm (or 4.76 mm) to 15 cm.

There have been many altempts to define particle, size ranges corresponding to the terminology gravel, stuld, silt and clay. Several Particle Size Classification Systems have been developedto describe soils by their particle size. Figure 1.4 shows the particle-size limits suggested by some of the more important systems, namely, those devcloped by Massachusetts Institute of Technology (MIT), United S tates Department of Agriculture (USDA) and Bureau of Indian Standards (BIS).

1.4.3 Particle Size Distribution The distribution of particles of various sizes in a given soil is h o w n as particle-size distribution. It is also known as grain-size dislribution. It is determined by a process known as Mechanical Analysis. This involves the use of two methods :

i) Sieve Analysis, and

ii) ' Hydrometer Analysis.

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Soil Phases, Structure, 1 0 0 - 0 Consistency and

Classification

Figure 1.4 : Particle Size Classification Systems

Sieve analysis is used for analysing particle sizes greater than 0.075 mm. Since it is difficult to make sieves of openings lesser than 0.075 mm, the hydrometer method is used for identifying smaller particles.

1) Sieve Analysis

Sieve analysis consists of passing Ule given soil successively lhrough a set of sieves that have progressively smaller openings. In sieve analysis, the 'size' of a particle is the side dimension of the square opening in the sieve through which it just passes. Indian Standard Code No. 2720-Part IV (1975) suggests a set of sieve numbers with lhe sizes of openings as given in Table 1.4. The equivalent American and British Standards are also given alongside.

Table 1.4 : I.S. and Equivalent Sieve Sizes

I.S. Sieve No. Sieve Size Equivalent B.S.S. (mnl) A.S.T.M

570 5.66 3 - 480 4.76 4 3/16 " 400 4.00 5 - 340 3.353 6 5

3 20 .3.180 - 118 " 280 2.8 18 7 6

240 2.399 8 7 200 2.032 10 8

170 1.676 12 10

1 60 1.600 - 1/16"

140 1.405 14 12

120 1.201 16 14

100 1,000 18 16

85 0.842 20 18

80 0.790 - 1/32 " 70 0.708 25 22

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Soil Wutcr Systc111 I.S. Sieve No. Sievc Size Equivalent, B.S.S. (mm) A.S.T.M

A typical sieve analysis is carried out as explained below. About hall' a kilogranl ol' soil is dried in the oven. If it is a cohesive soil it may contain lumps and clods. These may give an erroneous idea about Lhe true particle size. Thereforc, the clods are broken with a wooden mallet. The soil is then passed through Ihc sieves successively and weight of soil retained on each sieve is recorded. If the sieve opeiling is small, the soil is washed through the sieve and weight of soil retained is obtained after drying in the oven.

The weight of soil relained on each sieve is expressed as a percentage of Ule total dry weight of the soil taken for the analysis. I1 represeilts the percentage finer than the size oS the larger sieve just preceding the onc uildcr consideration. Ylese percentages are plotted against the sieve sizes to obtain Ihe gradation or grain-size distribution curve, shown in Figure 1.6.

li) Hydrometer Analysis

This analysis makcs use of an instrument known as tllc Hydrometer (Figurc 1.5). This ii~strument has a glass stem which expands into a bulb at its lower end. The lower part of Ule bulb is made heavy so tllat the instrument can tloat vertically when introduced in a liyuid. Tile analysis is based on the principle of sedimentalion, It ~llakes use of the phcnonlenon that when a small quantity of soil is put into water it disperses and the particles settle. If wc assume Uiat Uie cluanlity of soil is small enough so Ihat the particles do not collide with each other, they inay be taken to bc settling with uniform velocities.

Tlle velocity actually gradually increases and reaches llle uniform value known as lhe terminal vclocity. The settling velocities of tlle parliclcs depend on their shape, size and weight. If we again assume that the particles arc spherical, Ihen thc terminal velocity v of Ule soil particles can be expresscd by Stoke's law as :

wllc:e, y. = unit weight of soil particles, kN/m3,

yw = unit weight of water, kN/m3,

n = viscosity of water, k~-s/rn2, and

D = diameter of soil particles, m.

The equatioil above shows Ulat velocity is proportional to square of the diameter of the particle. Therefore, larger particles in ll1e soil suspensioii will settle faster than smaller piuticles. For exainple, it can be shown h a t the time lcaken by particles of

Page 13: Geotechnical engineering

different sizes to drop through 10 cm (assuming G,T=2.65 and n=0.01 poise) arc given in following Table 1.5.

Table 1.5

Diarnetre (mm) I Timc (nlin)

It1 this test, about 50 g of oven-dry soil is put illto 1000 cmhof distilled water in a graduated measuring cylinder. A dispersing agent such as Hydrogen peroxide or Sodium hexalnetaphosphale is added to disperse the soil better. The cylinder is shaken thoroughly so that Ihe soil is nearly uiliformly distributed ovcr llle volume and the density is also uniform. Howevcr, as parlicles of differen1 sizes bcgi~l to settle with different velocities, the density of thc suspeilsion starts varying with time and depth.

When the hydrometer is genlly introduced into such a suspension il comes to equilibrium at a depth where the density is such as to produce a buoyult force which will balance the weight ol'the hydrometer. Thus, the hydrometer will come lo equilibrium at different depths as time progresses. This depth h of the levcl of thc centre of its bulb is measured by introducing the hydroineler at tliffercnl values of time t. Now, a three-way relatiorlship exists:

i) The velocity of the particles which would have seltled through a dcpth h in time t is given by

'

v = h/t ... (1.23)

ii) Corresponding to this velocity, there is a dia~lielcr D, as per Equalion (1.27 ), such that all particles of size equal to or greater thul Illis diameter would have settled through the depth h, while particles filler thal1.I will be in suspellsioll within the depth h. From Equations (1.22) and (1.23) Chis clinmclcr is given by

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Soil Water System iii) The specific gravity of the suspeiision at ally doptll h is proportional to the percentage of particles filler than D, which are still in suspension within that depth.

Hydrometers are, therefore, calibrated such that they give the specific gravity of the suspension within the depth of immersion h. From Ulis. the percentage of particles finer than the corresponding diameter D can be calculated. By taking readings of h at different values of time t we can find the percentages finer Ulan different values of D.

This diameter is known as the equivalent diameter. This is the diameter of an equivalent spherical particle which would settle with the same velocity as the real soil particle.

1t must be mentioned here that the hydrometer reading has to be corrected for tllree effects : the opacity of the soil suspension, expailsioil of the hydrometer bulb due to temperature and the effect of the dispersion agent. Let us see thein one by one.

In clear water the lower meniscus is visible and i t is taken as the water level. But, in the suspension, its opacity prevents reading of the lower melliscus and, therefore, only the upper meniscus is read. 31e correction, C,,, known as the meniscus correction is equal to the difference between the upper and Iower meniscus readings in clear water. This is positive and is to be added to the hydrometer reading since the hydrometer readings increase downwards.

If the test is conducted at a temperature higher than the calibration temperature, the bulb expands. This raises the bulb and the reading is lower. The correction, C,, known as the temperature correction, is positive and is to be added to the measured reading. This is obtained by taking readings in distilled water at the same temperature as the suspension.

The dispersing agent increases the density of the liquid and hence a negative correction has to be applied to the measured readings. This correction, C, is obtained by taking the readings in distilled water co~~taining the dispersing agent.

The use of the method is illustrated in the form of a solved example.

Particle-size Distribution Curve

Particle-size distributior?. of a soil as obtained from sieve ,and hydrometer 'analysis is usually.represented in the form of a graph known as the particle-size distribution curve. The particle diameters are plotted against the correspol~ding percentage filler on a semi-log paper. The particle diameters are plotted to log scale along the abscissa and the percent finer to natural scale. The semi-log plot helps to accommodate within a small space the wide range of particle sizes found in soils, Also soils of same uniformity coefficient (see below for definition) have similar curves displaced horizontally with respect to each other. Typical particle-size distribution curves for four soils are shown in Figure 1.6.

l?k particle-size distribution curve shows the range of particle sizes present in a . soil and also their relative proportions. The diameter in the particle-size distribution curve corresponding to 10% finer is defined as the Effective Size arid denoted as, Dlo, Similar definitions are given to DRO and D60 also. The variations in particle-size distribution curves can be expressed in terms of some constants such as effective size, uniformity coefficient, and coefficient of gradation.

The slope of the straight line portion of the curve is expressed by the Uniformity Coefficient C, given by

The Coefficient of Gradatiou C, is expressed as

In Figure 1.6, the vertical segment of the curve for Soil I shows that particles of diameter approximately 0.4 m m . ~ s t i t u t e about (80-20) or 60% of the soil. Owing

Page 15: Geotechnical engineering

10 I' 0.1 0.01 0.001 o-door

P A R T I C L E D I A M E T E R , r n m

C - Coarse M - M e d i u m F - F i n o

Figure 1.6: Typical Particle Size Distribution Curves

to the limited variation in grain size, this soil is known as a poorly-graded soil. On the other hand, curve of Soil I1 shows t l~e presence of a wider range of particle sizes and this soil is called a well-graded soil. Curve for Soil 111 shows two horizontal segments. The first one signifies the absence of particles in the range of 0.1 nun to 0.01 rnm. The second one implies the nor]-existence of 0.002 mm Lo 0.0006 nlm sizes. This soil is termed as gap-gradcd soil. Of course, it is not essential for a soil to liave two horizonlal segments to be called gap-graded, o ~ ~ e is enough. Soil IV is also a well-graded soil. Curve for Soil IV is similar to tllal of Soil 11, because its uniformity coefficient is same. But i t is displaced 11orizont;llly because the range of particle sizes present in it are different (check it yourself).

1.4.4 Significance of Gradation I1 is known from experience that the behaviour of a cohesioliless soil can often be related to its density or degree of packing of its particles and hence to the distribution of particle sizes in it. On the other hand, the hehaviour of a col~esive soil usually depcnds much illore on its stress history and structure tll'm on particlc size. Tllerefore, particle size curves, particularly those of gravels, sands and silts, have practical significance from poiill of view of thpir properties, which are dependent directly on gradation. Examples of such properties are capillarity, permeability, compactioil and shear strength (these are described in the Units, which follow). Experiments show that permeability and capillarity are related Lo the effective size.

One of Ule important applications of this relationship is in Uw design of inverted tiltcrs for earth damns. Based on the relationship of particle size to permeability, the particle size of filter material is so chosen as to prevent h e migration or outflow of particles of the soil from the body of the earth dam when water Plows through it. It is also used to choose a soil for an embankment in such a way that good density is obtained on conlpaction. A soil wilh a greater range of particle size will achieve greater density since finer particles will rill the voids in the coarser material, *

Example 1.6

From the gradation curve for Soil I1 in Figure 1.6, compulc perceqt gravel, sand, silt and clay according to MIT, USDA and BIS particle-size classification systcms. Also compute the parameters Dlo (Effective size), aild lleilce C,, (Uniformity Coefficient) and Cc (Coefficient of Gradation).

Solution

The computations involved are shown in Table 1.6. Two typical calculations are explained. Others are similar and the student is advised lo work thein out himself.

Soil Phases, Structure, 1

Consistency and Clad~?fiention

I I

Page 16: Geotechnical engineering

I

Soil Water System Let us compute the percentage of gravel and sand in the given soil according to the MIT system. According to MIT system (Figure 1.4), gravel corresponds to a particle size 2 mm. From the gradation curve we find that percent finer than 2 mm is 91, which means that percent greater than 2 mm is (100 - 91) or 9%. Similarly, sand corresponds to the size range 2 to 0.06 rnrn. The values for percent finer for these sizes are 91 and 59, respectively. The percentage of sand in the soil is (91 - 59) or 32%.

Table 1.6 : Percentage of Various Size Fractions

Let us now compute the other parameters required.

Dlo = 0.0003 mrn

D60 = 0.07 mm

C,, = D6fl10 = 0.07/0.0003 = 233.3

C, = = (0.006)~/(0.07*0.0003) = 1.71

Example 1.7

BIS

(> 4.76mm) 100-91~9

(4.76-0.075 mm) 91- 61=30

(0.075-0.002 mm) 6 1-20=41

(<0.002 mm) 20

Results of sieve analysis for a soil are given in Table 1.7 in columns 1 to 3. Weight of oven-dry sample used for the analysis is 482.0 g. It is necessary to compute percent finer and plot the gradation curve.

Solution

USDA

(> 2 mm) 100-91=9

(2-0.05 mm) 9 1 - 5 7 ~ 3 4

(0.05-0.002 mm) 57 - 20=37

(e0.002 rnm) 20

Type of Soil

Gravel (%)

Sand (%)

Silt (%)

Clay (%)

The computations involved are also shown in the same Table 1.7. Incidentally. these are typical of the calculations required in a sieve analysis.

MIT

(> 2 mm) 100 -91= 9

(2- 0.06 mm) 91-59=32

(0.06-0.002 mm) 59-20=39

(<0.002 mm) 20

Table 1.7 : Sieve Analysis Calculations

The gradation curve based on the percent finer calculated above is the upper part of curve for Soil I1 in Figure 1.6. (The lower par1 pertains to particles less than 0.075 mm in size and is obtained by Hydrometer analysis.)

1.S Sieve No.

(1)

200

100

60

40

20

I0

8

Diameter (mm)

(2)

2.032

1 .OOO

0.592

0.420

0.21 1

0.104

0.075

Percent Filler

(5)

9 1.50

88.80

82.15

79.40

73.20

64.60

59.48 .

Weight of Soil Retained

B) (3)

40.97

13.01

32.05

13.26

29.89

41.45

24.68

in % (4)

8.50

2.70

6.65

2.75

6.20

8.60

5.12

Page 17: Geotechnical engineering

- - Example 1.8

Derive an expression for calculating percent finer than a given diiuncter from the readings of hydrometer analysis.

Solution

Initially, at time t = 0, when the soil is thoroughly mixed in the suspension. the unit weight will be uniform. It can be written as :

where, ywt is unit weight of water at the temperature of Lhe experiment.

Analogously at any time f and depth h we call write,

wl~hre,

N = Required percent tiller by weight than Ule diameter corresponding to the time t and depth h.

Therefore,

Since the hydrometer reads tlle specific gravity, the unit weights of thc suspension and water in the above equation can be written in term of thc hydrometer reading corresponding to time t and deptl~ as follows:

Hence, we can now write the following expression for N as a percent mtl for the corrected value K (96) :

where,

Pf = Percent finer than IS Sieve No.8, from wlliclr 50 g were taken for hydrometer analysis.

Example 1.9

The data frotn a typical hydrometer a~lalysis on a clay siumple is given below. It is necessary:

i) to compute the particle sizes and the percent finer correspo~ldi~~g to the ti~lle intervals given, and

ii) to plot the gradation curve.

Solution

For convenience, the computations involved are presented in a secluential nlalrricr with the help of Tables 1.7 and 1.8 :

The basic data is:

Weight of oven-dry soiI taken (N'J = 5 0 g

Specific Gravity of solids (G.$,) = 2.65 ' li\.

).

Srbil I'l~ascs, Structutc. Consistency on11

Clurssiflcrtio~~

Page 18: Geotechnical engineering

Soil Water System Area of cross-section of cylinder = 19.61 crn2

Volume of suspension (V) = 1000 cm3

Volume of Hydrometer bulb (Vh) = 65 cm 3

Hydrometer reading in clear water (r,) = 0.9965

Room temperature = 27OC

Viscosity of water @ room temp. (n) = 0.008545 poise

Dispersing Agent correction (Cd) = - 0.0004

Meniscus correction (C,) = + 0.0004

Volume correction (Vh/2A) = - 1.657 cm

The hydrometer readings are taken at close intervals in the beginning and at larger intervals later as shown in Table 1.7 (Can you guess why?). Meniscus correction is applied first. Knowing the distance in cm from the centre of the bulb to the bottom reading on the I stem and also the distance bf the measured reading from the bottom-most reading, the depth 2, is calculated. This is then corrected for volume of the hydrometer. From the corrected value h the velocity is computed.

i I

In the second stage D and N are calculated as shown in Table 1.9. Firstly dispersirlg agent ~ correction and then temperature correction are applied. Diameter is then calculated from Equation (1.24) written as follows with a slight change of units : I I

e l

where,

D is in m, n is in poises (1 poise = kN/m2), h is in cm and t is in min.

Percent finer is then calculated using Equation (1.31 a). The corrected percentage is calculated using Equation (1.31 b). Here the value of Pf is assumed to be 0.50 for the purpose of illustratjng the calculations.

Table 1.8 : Cornputation of Depth and Velocity

Time

( min.)

1 I2

1

2

5

15

30

60

1 20

1440

Hydro. Rdg.

(4

1.0155

1.0135

1.0120

1.0095

1.0070

1.0060

1.0050

1.0030

1.0020

Rdg. after Men. cow.

(r=rcCm)

. 1.0159

1.0139

1.0124

1.0099

1.0074

1.0064

1.0054

1.0034

1.0024

zr

(cm)

12.20

12.75

13.20

13.95

14.65

14.80

15.25

15.80

16.20

Vh h =Zr - ( - )

24 (cm)

10.543

1 1.093

1 1.543

12.293

12.993

13.143

13.593

14.143

14.543

v = h/t

(cmlmin)

21.086

1 1,093

5.772

2.459

0.866

0.438

0.227

0.1 18

0.010

Page 19: Geotechnical engineering

Table 1.9 : Computation of Diameter and Percent Finer

Rdg. aftter Disp. Corr. (R=F - Cd )

1.0155

1.0135

1.0120

1.0095

1.0070

1.0060

1.0050

1.0030

1.0020

Rdg. after Tenip. Corr.

(R-~w)

Dia. D

(111111)

Corr. 70

(N)

9AQ 2 i ) Sieve ;inalysis 01.500 g of i111 oven-clry soil k~~owu ils niurruni gilve Lhc

following results (Table 1.10). Draw tlie pnrticle size tlistrihulion curvc iuitl

cleter~ni~ie the cfl'eclivc: size, unil'oni~ity cocfficicnl iuld cocl'ficicnt of gradation for the soil.

Table 1.10 : Di~tit for SAQ 2 (i)

1.S.Sieve No.

Wciglit of Soil Kctiriocd ( g )

i i ) The particle-size cliauacterislics ol' two soils ilrc givcn hclow (Tablc 1 . 1 1). Draw Uic pnrliclc size distribution curvcs. Firid Llic ~ ~ S C C I ~ ~ ~ I ~ C S ol' gravel, said, sill and clay li,r eacli soil according to ( 1) MIT, (2) USDA illit1 (3) BIS systems.

Soil Pllases, Str~lcture. Consistency slitl

Clessificn ti011

'I'able 1.11 : Data for SAQ 2 (ii)

Page 20: Geotechnical engineering

iii) The results of sieve analysis on a sandy soil showed Ulal percent finer by weight than 4.76 mm was 78%, than 2.399 mm was 57%, than 1 mm was 37%, than 0.1 mm was 11% and 0.075 mm was 9%. Determine the percentages of gravel, s'md, silt and clay according to MIT, USDA and BIS systems, effective size, uniformity coefficient and coefficient of gradation.

iv) In a hydrometer analysis 50 g of oven-dry soil p ~ s i n g 0.075 rnrn sieve is dispersed in 1000 cm3 of water. For a reading of 0.0070 .(corrected value) taken after 120 minutes, the depth of the centre of the hydrometer bulb after applying all necessary corrections is found to be 145 mn. Specific gravity of' soil grains is 2.65 and viscosity of water is 0.001 ~ - s / m ~ . What is the parlide size whose percentage can be estimated from this reading and what is the percentage ?

v) Construction debris from a site is dumped in a nearby lake. The depth of Ule bottom of the lake is 4 m. Estimate the time required for the silt particles of diameter 0.005 mrn in the debris to reach Ule bed. Take the s ecific gravity of 4 the particles as 2.65 a11d coefficient of viscosity 0.001 N-s/m . I

vi) Plot n semi-log graph showing grain-size distributions of the two soils whose I sieve and sedimentation analysis data are given below (Table 1.12). 1

Table 1.12 : Data for SAQ 2 (vi) C

I * From hydrometer analysis

No.

480

200

85

40

25

I5

8

- - -

I -

1.5 SOIL STRUCTURE

As pointed out in sub-section 1.2.3, soil is a three-phase material and in real soils the phases coexist as shown in Figure 1.2 (a). The arrangement of the particles and the voids between them is known as structure. The structural arrangement evidently depends on the shape and size of the particles. What is, however, not so evident is the fact that the mineralogical composition of &he grains (see section 1.7) and the nature and composition of the water in the soil (see Unit 2) also affect the interaction between the particles and hence their arrangement. For understanding the structures, soils can be divided into two groups :

1) cohesioi~less, and

(lnm)

4.76

2.032

0.842

0.420

0.25 1

0.151

0.075

0.050*

0,020*

0.005*

.2) cohesive.

The structures developed in soils of each group are described in forthcoming sections.

0.001~ 1 2.15 1 4.25 1

Soil I

94.50

60.55

42.00

31.95

21.75

17.00

11.80

10.78

33.00'

4.75

Soil I1

100.00

100.00

72.75

70.35

59.00

47.95 a

3 1 .OO

23.82

13.25

8.00

Page 21: Geotechnical engineering

1.5.1 Structures in Cohesiorkless Soils Cohcsionless soils gcncrally havc bulky grains hat can be represenlcd by spheres For all praclical purposes. Cohcsionless soils exhibil two types oT struclures :

1) single-grained structure, and

2) honeycombed structure.

1) Single-grained Structure

This structure is co~nrnonly found in sands and gravels. P,ulicles of scud or gravel rue lleposited onc above Lhe other during their formation. Each ~~arlicle is in contact with three lo I'our surrounding particles as shown in Figure 1.7.

If we assume the parlicles to be spherical two types of ;lrrangenlents are possible as shown in Figure 1.8. The first arrangement is known as cubic packing. The second, wlierc particles of one layer occupy lhc spaces bctween particlcs of Lhe lower laycr, is lulown as hexagonal packing.

(a) Cubic l'acking (b) Ilrrxugo~ml Pnckin~g Figure 1.8 : Pnckings in Cohesionless Soils

However, in real soils, the particles are non-spherical ruid of unequal size. So, two contradictory effects take place. Smaller particles occupy the voids belween larger particles to give a close packing, while angular shape of particles prevenls close packing. While the former decreases the void ratio, the latler increases it. Nevertheless, the combined effect is such that the void ratios agree closely with those observed in idealized spherical parlicle arrangeinents.

Depending on the relative positions of the grins, it is possible to have a wide range of void ratios. For example, the void ratio of Ule cubic packing can be shown to be equal to 0.91, while that of hexagonal packing is 0.35. The inaximum arid ~ilini~ilurn void ratios in real soils, however, will be different since the particles are neilher spherical nor uniform in shape and size. The void ratios are called e,,,, and enin, The former is determined by pouring dry soil of known weight loosely into a measuring jar and noting ils volunie. The lalter is determined after compacting the soil to Lhe maximurn possible extent without crushing the particles. Some typical values of minimum (md maxinlum void ratios and unit weights are given in Table 1.13.

Table 1.13 : Typical Void Ratios and Unit Weights of Cohesionless Soils

Void Ratio ~ r y unit ~ t . ( ~ m ~ ) Description of Soil Min Max

Min Max

Well graded mixture of gravel, 0.20 0.75 15 22 sand and silt

Well graded sand 0.30 0.80 14 20

Uniform sand 0.50 0.85 14 17

Micaceous sand-silt mixture

Page 22: Geotechnical engineering

S I ~ I Water System The actual state of a cohesionless soil may be expressed in terms of its maximum and mini~num void ratios using the para~neter Relative Density D, as already explained in sub-section 1.3.4.

The loose structure is prone to instability especially under shock and vibration because particles have freedom to move and occupy new positions and attain a stale of increased density and stability.

2) Honeycombed Structure

Iloneycombed structures arise when fine-grained cohesionlcss soils, like fine sand or silt, settle slowly in water and get deposited at the bed level. The inter-particle attraction between such particles depends on the unbalanced electric charges on their surfaces, which, in turn, depend on the specific surface. Specific surface of a particle is defincd as surface area per unit volume. Therefore, the specific surface for a spherical particle of diameter D is given by :

s = arealvolume = 6 / 0 ... (1.32)

Thus, for a typical fine sand or silt particle of size 0.05 rnm, s = 6/(0.05/1000)= 120000 m2

per unit volume. Hence, the inter-particle forces of attraction exceed the gravitational forces in such particles. Therefore, the particles form arch-like shapes instead of moving under their own weight into spaces between neighbouring particles. This is shown in

1

I

Figure 1.9.

Agurc 1.9 : Honeycombed Slmcture I

In such soils the void ratio exceeds the maximum possible value for single-grained structures. The relative density, consequently, takes on negative values. The structure is normally stable under loads due to the arch formation, but may collapse under vibration.

11.5.2 ~tructuies in Cohesive Soils It will be seen in sub-section 1.7.1 that clay particles are negatively charged on their surfaces. When clay particles suspended in water come close to each other they expericnce the well known Van der Waals forces of molecular atlraction and forces of electrical repulsion due to tlle negative charges. Two types of structures may result from this force

i system depending on the concentration of the suspension :

I ) Dispersed structure, and I

2) Flocculent structure.

1) Dispersed Structure

In a dilute suspension the particles will be widely spaced and the forces of repulsion will 1 far exceed the forces of attraction. The forces of gravity are negligible owing to the fineness of the particles. The structure resulting from the sedimentation of such par'.icles is known as dispersed structure. The particles will be nearly parallel to each other as shown in Figure 1.10.

Dispersed structure is found in micaceous soils where particles are flaky. The void ratio is usually low if the flakes are oriented. But, if the flaky particles are not oriented but randomiy po;n:c2, the void ratio can he high. Oriented mica particles are dense and anisotropic. Soils having dispersed structure are likely to be quite watertight. The void

28 ratio may vary from 0.5 to 2.. ,

Page 23: Geotechnical engineering

2) Floccuienl Structure

On the other hand, in a concentrated suspension the particles will be closely spaced and the forces of attraction will exceed those of repulsion. Sometimes the edges of the particles possess local positive charges. This leads to development of edge to surface contacts between particles. The electrostatic attraction attains significant magnitudes and leads to the formation of flocs or aggregates of particles bound to one another. The resulting structure is known as flocculent structure (Figure 1.1 1).

Soil Phases, Structure, I

j I

Consistency and ' Classificntion I

1

figure 1.11 : Nonnnl Flocculent Structure 1 I

Sometimes flocculation is induced in a clay suspension by the presence of salt in the water or by external addition of salt. The salt neutralizes the charges on the clay particles and the forces of repulsion are reduced. The particles then aggregate together to form salt-induced flocculent structure (Figure 1.12). However, the soil can be dispersed by the addition of sodium silicate or sodium hexametaphosphate as already explained under Hydrometer analysis in sub-section 1.4.3.

Figure 1.12 : Sdt-induced Flocculent Structure

Flocculent structure is found in sediments under sea due to salt induced aggregation. Sediments under fresh water exhibit flocculation to a much less extent. Void ratios may go as high as 4 in such soils.

Due to the strong edge to face electrostatic force of attraction between particles in a flocculent structure, the structure is strong and cru~ withstand vibrations, However, these are hi@ly compressible. If the soil is remoulded it tends to mix wilh the adsorbed water inside and become soft. The softening effect is termed as Sensitivity,

1.5.3 Composite Structures Composite structures constitute an arrangement of coarse grairis cemented together by a binding agent. Common binding agents are dried stiff compressed clay and calcium carbonate. They are usually strong but may be weakened by water. Iron oxides and silica also act as binders and are resistant to softening by water.

Coarse grains cemented in a matrix of fines is known as matrix structure (Figure 1.13 (a)). When coarse grains constitute a large part, two types of structures develop. They are :

1) contact bound (Figure 1.1 3 (b)), and

2) void bound (Figure 1 .I3 (c)). In contact-bound structure, the coarse grains are held together at points of contact by the binder. In the void-bound structure, the binder occupies the voids between the coarse grains. Contact-bound composite structures are usually strong, but their strength may reduce if the binder is prone to softening when it comes into contact with water. If the structure is loose it may collapse or settle, if relatively dense. In void-bound structure, the binding is stronger and the packing is dense. The binder fills almost the entire void space. The void ratio may be as low as 0.2. Softening effect due to water is less.

Example 1.10

Compute the void ratio of cubic packing theoretically.

Page 24: Geotechnical engineering

Soil Water Systern

Figure 1.13 : Conlposile Structures

Solution

Consider a unit volume of soil, i.e., a volume of soil bounded by a cube of unit length. If the particles are assumed to be uniform spheres of diameter D, U~en we I

have :

Volume of each sphere = 3.1416 ~ ~ / 6 I i

No. of sphereslunit volulne = ( 1 ~ ) ~ I I

Volume of solids

Volume of voids

Void ratio

= Total volunlc of spheres

= 0.5236

= 1-0.5236 = (1 - 0.5236)/0.5236 = 0.91

Example 1.11

Calculate the typical specific surfaces of sand, silt and clay size particles.

Solution

Let us assume typical values of particle sizes of sand, silt and clay as 1.2 lnm, 0.02 Inn1 aid 0.002 nun, respectively. Assurnirig h e particles to be spherical, the specific surfaces will be,

2 3 forsand s = 6 / ( 1 . 2 x l ~ ~ ) = 5000 m / m 5 2 3 forsilt s = 6/ (0 .02~10-~) = 3 x 1 0 1nIm

I

6 2 3 for clay s = 6/(0.002 x 1 0 - 9 = 3 x 10 m /m I

Example 1.12 1 Estimate the approximate dry unit weight of a cohesiollless soil made up of mif form I

I

spllerical particles of diarncter D. I , I

Solution

Assume the specitic gravity of solid particles as 2.65. It was shown above that volume of solids in unit volume of soil is 0.5236. Then, we have, from the

I definition of specific gravity, ,

SAQ 3 i) What is the effect of particle size on void ratio and unit weight of

cohcsionless soils made up of uniform spherical particles in single-grained loose packing'?

ii) Compute the specific surfaces of 5-nlicron clay 'and typical colloidal particles.

iii) What is the difference between iiormaI and salt-induced ilocculation?

Page 25: Geotechnical engineering

iv) Under what external effects cnn a honeycombed or floccule~it soil struclure collapse ?

V) Calculate the typical unit weights of Uie lbllowing by assunli~~g suitable values of void ratio. Take G, = 2.67.

a) Well-graded dry simd.

b) Well-graded subangular saturated said.

c) Loose uniform rounded dry sand.

d) Honeycombed saturated silt.

1.6 CONSISTENCY

Try to do this siniple test. T'alce a little clry soil in your hand. You will ~~ot ice Uliit yo11 cannot mould it into a paste, Now adtl a little water. You will find t1i;it the soil hccomes n little sticky and plastic. You will be able to k~lcad it ant1 sliilpc it. Go 011 adding water. Tlic soil will soon flow like a liquid.

This is hiown as consistency. In the case of coarse grained soils the intcr~llediate stage or plasticity is not pronounced. But in Cine-grained soils Ule plastic state is secn tlistinctly uld the soils can be ~iloulded to any shape without crmnbling. Illis is presumably due to the presence of clay minerals and adsort~ed water (you will learn Inore ;\bout clay miiieri~ls in sub-section 1.7 ;u~d about soil water including adsorbed water in U~iit 2).

Swedish scientist A. Atterberg studied this phenon~eiion of consistency of ;I soil, iuld in 191 1, suggested a method to describe its variatiorl lkom solid to semi-solid to plastic ;uicl finally liquid state.

1.6.1 Consistency Limits Atterberg defined the boundaries or linlits Cor llle states of consistency ill terms of tlle moisture content at which the soil attains a given state. Thus tile lnoisturc conte~~t at which the soil changes over from solid to senli-solid consistency is defined ;IS Ihe Sliri~~kagc Limit: that at which the soil attains plastic state is known as Plastic Limit; Ulat at wllicli it begins to flow is called Liquid Limit, Between iu~y two limnits Ule soil remains in the state defined by the lower limit. Table 1.14 depicts the stages of consistency iuld b e Consisleiicy Limits.

Table 1.14 : Atterberg Limits

1 Stage I Description I

I'lastic

Semisolid

1

I Can he re~nouldcd without cracking

Cannot he remoulded without cl.ncking I

Lirluid

I Solid Cannot he rernoulcled: Fnils on deforrnntion I I

Flows like a viscous lirluirl

Liquid lhnit (LL)

I'lustic lhnit (PL)

It is difficult to determine the stages of consistency basctl on the above gclieral definitions, Therefore, definitions based on methods of determination of I l ~ c consislcncy limits are Inore commonly used.

Soil Phuscs, Structure, Col~sistcncy nncl

Ulsssific~tior

The tests are carried out on fine grained soils passing IS sieve no, 40, having 31 opening s i x of 0.42 mrn (See Table 1.4).

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1.6.2 Determination of Consistency Limits i ) 1,iquid 1,imit (LL)

You must have often heard the statement that a liquid has no shear strength. Thus, the liquid state of a soil is evidently one at which the soil has no shear strength. Therefore, liquid limit is the moisture content at which the sheu strength of the soil just becomes zero. However, since his is not measurable, the liquid limit is reckoned as the moisture

I ) ) content at which the shear strength is close to zero, but, just measurable (about 2.5 kNlm2),

Thc device used for determining Liquid Limit, commonly known as the Liquid Limit Dcvice, is shown in Figure 1.14 (a). It consists of a hard rubber base on which is mounted

R A S S D I S H

D R U B B E R B A S E

(a) Liquid Limit Device

a brass cup, which can be raised by 10 mrn and dropped with the help of a crank-cam arrangement. A soil pat is made inside the cup and a groove is cut in i t with a standard grooving tool. Two types of grooving tools and their approximate dimensions are shown in Figure 1.14(b). Tile cup is raised and dropped until the groove (Figure 1.14 c) closes along

(b) Grooving Tool

R A D I U S = 5 4 rnm / (c) Soil Pat Before Test (d) Soil Pat nfter Test (Closed Groove)

Figure 1.14 : Liquid Limit Test

its bottom by about 12.7 mrn as shown in Figure 1.14 (6). The moisture content at which the cIosure occurs in 25 blows is defined as the Liquid Limit.

In order to determine this moisture content, about four or more tests are conducted wilh varying moisture contents and the corresponding number of blows required for closure of Ule groove are noted. A curve known as the Flow curve (Figure 1.15) is plotted between

32 number of blows and moisture content on a semi-log paper with number of blows along

Page 27: Geotechnical engineering

~ h c logarillrnlic axis. The graph is plotted as a straight line. The liquid limit, which corresponds lo 25 blows, is read off l'rom this straighl line.

N U M B E R O F B L O W S , N

( l o g s c a l e )

figurc 1.15 : H o w Curve

Thus, Lhc cquillion of lhc [low curve c:ul be written as

t~ = -IF log N + C where, IF = slopc of Lhc curve, : u ~ l

The slope ol' fhc flow line is defined as the flow index (Id and may be wrilleml as

I ~ I ,. 1/12 = nioislurc co~itcnts of soil, in pcrccnt, corresponding Lo N I and N2 blows, rcspcclivcly.

ii) Plastic Limit (PI,)

The plastic slalc is reckoned as one in which soil c:ui be rcmouldcd wilhoul breaking. Thus. llic plastic limit is dcl'inctl ;IS the n~oisturc content nl which a 3 mm Llircatl of soil crm jusl be rollctl by hand wihoul cn~nihling.

This is determined by a simple lcsl in which il soil pat is rollcd into threads of 3 11l11l

diameter on a ground glass plate. Sli~rting with a high nloislurc conlcnt, tesls iUC done at progressively lowcr ~lloisture contents hy just acltling a little tlry soil each lime, The lowest moisture content at which Lhc threads can jusl be rollcd wilhoul cnlnlbling is the Plaslic Li~nil.

iii) Shrinkage Limit (SI,)

A1 shrinkage liniil Ulc soil is s;liil lo Ililv~ rcachctl LIIC solid slate. This is tlefined as Ulc moisture conLen1 at which Uic volume ol'soil reaches its minimum valuc ant1 remains consl:unt oe li~rlhcr drying. Furlher rc~noval of rnoislurc from the soil docs not result in rciluclioti in volume as clcpiclcd in Figure 1.16. In olhcr words, il is llic moisture corltcnt corrcsponiling to the slate whcn Iherc is just c~iough walcr LO fill all Lhc voids in h e dry ~011.

111 lhd laboratory, a porcelain dish about 45 null in diameter and 12.5 mm in height is coaled on the inside wiUi grgase and l'illed wilh we1 soil ruld Ulc lop is levelled off. The Inass of the soil is nolcd. The soil is now oven-dried, weighed ruld U I ~ I I its volunle is determined by displacement of mercury. The Shrinkage Limit is now conlputed as I'tAlows :

Soil Phnscs, Structure, Consistcl~cy a ~ i d

Classificetio~~

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Soil Watcr Systcrn

2 - 0 m k 0

W Z

3 SHRINKAGE 0 v

MOISTURE CONTENT ( ' l r )

Figure 1.16 : Vdume Change During Shrinkage

where,

wl = weight of the wet soil pat in the shrinkage limit dish

w2 = weight of the oven-dried soil pat

V1 = volume of the wet soil pat (inside volu~ne of the dish)

V2 = volume of the oven-dried soil pat

y, = Unit weigh1 of water

A number of useful soil indices are derived from the above consistency limits. l l ~ e y are Plasticity Index, Liquidity Index, Shrinkage Ratio and Degree of Shrinkage. These are cxplaincd below.

Plasticity Index (PI)

The Plasticity Index is the difference between liquid limit and plastic limit i.c. the range ol' moisture content over which the soil remains in a plastic state.

PI = LL-PL . j)

1,iquidity Index (1,I)

Thc consistency of a cohesive soil in its natural state can be expressed in relation to its limiting consislencies by a ratio known as the Liquidity Index (Li),

I = in situ moisture content of soil.

Liquidity index varies from 0 at plastic limit to 1 at liquid limit. Some 1 unconsolidated soils may have moisture contents greater than the liquid liinit and a ' liquidity indcx grcatcr that1 one, Heavily precompressed soils may have natural moisture contcnts less than Ule plastic limits and hence negative values of liquidity index, LI is used to define the state of a soil as follows :

Scmi-solid to Solid LI <O

Plastic O < L I < l

Liquid LI > 1

And wihin Ule ~las t i c state the soil is termed as follows depending on the LT :

Stiff 0 c LI c 0.25

Medium soft 0.25 c LT <0.5

Soft 0.5 < LI < 0.75

, Very soft 0.75 < LI < 1

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Shrinkage Ratio

This is defined as the ratio of volume change ~ I I 9i of dry soil volume V, to the corresponding change in moisture content from the initial value to the shrinkage limit. Using the notations of equation (1.39,

Degree of Shrinkage

This is the change in volume p r unit original volume and is expressed as, again using notations of Equation (1.35),

Fro111 SL aid SR it is possible to calculate G, using the followir~g relationship, which tl~e student is lidviscd to derive I~inlself.

Soil Phnses, Structure, Consistency and

Classification

1.6.3 Significance of Consistency Limits The Attcrbcrg lilllits arc no doubt arbitrarily and empirically dcfined. Interpreting them fundamentally, therefore, is not advisable. However, they are useful illdices for understiu~ding tl~e states in which a tile grained soil can exist. 'I'hcy are also useful for identification and classification of fine-grained soils.

As shown later in Unit 2, water in a clayey soil is in the f o m ~ of adsorbed films around the particles. Tliis water, which is bound to Uie clay particles by the force of adsorption, is not frcc to movc. When the ~noisture content increases and the adsorptior~ capacity is fully satisfied the water tends to lnove as free water under gravity, Thus, clays which have greater tendency to adsorb writer will l~ave higher moisturc contents at liquid limits. So, also for plastic limits. Thus, the consistency limits exprcss Ule clay-watcr behaviour at different nloisture contents.

1.6.4 Activity of Clayey Soils I As mentioned abovc t l~c adsorbed water in a claycy soil is bound to Ule surface of me particles, The quantity of adsorbed watcr, therefore, depends on the surhce area of UIC particles. It was shown in sub-section. 1.5.1, that i f Ule particles arc:issumned to be spl~erical, the surface area per unit volume will bc 6/D, where D is tllc dia~netcr of h e particle. Thus, as Ule size of the particle decreases Uie surfacc area increases. Also, as the percentage of clay size particles increases, ll~e available surface arca increucs. Tlds fact has been uscd by Skelnpton to express lhe water-holding capacity of clay minerals ill ternls of a parameter called Activity. Activity is defined as :

A = -- Plas t i c i ty~@~ ...( 1.41) Percenlagc of clay - size fraction, by weight

Activity is used to distinguish between clay minerals. Activity valucs for Uirce of the most commonly occurring clay nlinerals are :

Kaoliliite = < l

Illite = 1- 2

1.6.5 Engineering Use of Atterberg Limits i/ a I The Atterberg limits and related indices arc very usel'ul for soil idcilti'ficatio11 aid classificatiai.

3 5

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Snil Wnlcr Syslc111 Plasticity Chart

A. Casagraidc studicd a large number of soils and established a relationship between the Plaslicity Index arid thc Liquid Limit. Since both these parameters can be determined by simple laboratory techniques, this relationship has found favour with all soil engineers. Thc relationship is used as a criterion for classifying fine-grained soils in the Unified C!assification Sysle~n (See Table 1.21 in sub-section 1.8.4). It is presented in the form of a graph known as the Plasticity Chart. Tlie relationship is given by

PI = 0.73 (LL - 20) ... (1 .42)

TIic chart is uscful in criginrering practice for classifying fine-grained soils. Soils falling above Lhc A line are inorganic clays and Uiose below are inorganic silts. The plasticity of thc inorgallic clays and Lhc compressibility of the inorganic silts vary from low (LL < 30) througll ~nediurn (30 < LL < 50) to high (LL > 50). Organic silts fall below A line in the ~nedium compressibility range while organic clays fall in the high compressibility range.

Example 1.13

Given Ihe following data from a liquid limit test on a soil, compute lhe liquid limit and the tlow index.

Table 1.15 : Data for Example 1.13 - -

Solution

No. of Blows

32

27

22

18

16

First, plot lhe given data on a semi-log paper with No. of blows along Uie X axis and Moisture content along the Y axis. The curve so obtained, as already stated, is known as the tlow curve. This curve is represented in Figure 1.15 as Curve No. 1. From the plot it can be seen hat moisture content corresporlding to 25 blows, is 56%. Therefore,

Moisture Content(%)

51.2

55.0

58.9 61.4

66.6

Now, from Equation (1.34), the flow index can be calculated using any two points on the graph as

Example 1.14

The plastic limit test on a soil sample gave the following moisture contents. Deternline the plastic limit.

Table 1.16 : Data for Example 1.14

1 Trial No. Moisture Content(%) 1

Solution

From the description of the plastic limit test in sub-section 1.6.2 we see that the given readings represent the moisture contents at which a soil thread of 3 mm dia is formed without crumbling. Therefore, the plastic limit is given by h e average of the four readings as

,.?

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Exarn~le 1.15 Soil Pl~ases, Structure, Consistency and

Tl~e natural moisture content of a soil is 32.5'70. Col~slslc~lcy li~nil lests on Lhe soil gi~vc: CIiassitication

tl~e liquid linlit as 53.3% arid plnstic limit ns 23.7LX1. Whal 1s Ulc pl;aticily inclex oL' the soil ? Would the soil be stiff or soft in its ~~nlural stale '! How would thc soil behave if Ihere is a light rainfall ?

Solution

From Equation (1 36 ) and Equalion (1 -37) wc gel

PI = 53.3 - 23.7 = 29.6; LI = (32.5 - 23.7)/29.(1= 0.297

Thus, Lhe soil is medium soft in its natural slalc. IL' there is ;i lighl rainfall the moisture content or UIC soil will increase i111cl 111c soil would lend to become soft Lo very soft. 111 non-engineering lenns, Uie soil woulcl tend to he sluslly ;uld difficult to walk on.

Example 1.16

The readings from liquid lilnil and plaslic linlillcsls on a soil ilrc given bclow. What is tlie plasticity index ? Classify thc soil using Plaslicily Chi~rt.

Table 1.17 : Data for 1l:~iltnple 1.16 i-. I I I

Tcst No. Wt. of Dish (g) Wt. of IJisll + Wt. of ])is11 + Wet Soil (g) IJry Soil (g) 1 (21 1 (3) 1 (4)

- Solution

Liquid limit test

1

2

3

Plastic limit test

No. ot'I5lows I

The moisture contcnt corresponding to e ;~c l~ liqui tl limit test is c;~lculnlcd as :

20.52

20.33

24.23

The con~puled moisture contents and lhe N o . ol'blows arc Ii~l~ul;~tctl hclow.

Table 1.18

39.03

40.09

42.69

The liquid limit test dala is plotted as Curve No.2 in Figure 1.15. Fro111 Uie curve, we get

LL = 4n.5%

32.45

33.66

37.18

Using the same formula the moisture contents lix the plilstic l i ~ n i t test data are 28.23% and 27.64%. Therefore, taking Ule average,

PL = 27.9%

Thus, PI = 46.5 - 27.9 = 18.0. Fro111 Ulc Plasticity Chiut of U~ill'iccl Classil'icalion Syslem CTable 1.20), the given soil lies almosl on Ulc A line, Thcrclhrc, h e given soil is classified as CL - ML.

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Soil Wntcr Systcm Example 1 .I 7

A shrinkage linlit test on a clay gave a value of 19.5%. Tie initial iliolsture content as determined in the field was 41 3 8 . Take thc sacific gravity rs 2.65. On evaporatio~l the moisture content comcs tlown to 17.Xf&. Calculate Ulc corresponding decrease in voluriae per unit volurne and the shrinkace ratio.

.>

Solution

I t can he shown Uiat Lhc following relationships bohl.(TI\e studc~lt shouhl try to prove these from fundamentals.)

where, nt,,, = natural moisture content and all other notations are sarne as in Equation (1.38) to Equation (1.40).

Here, S L = 1 9 . 5 ; ~ ~ , , , = 4 1 . 3 ; G . ~ = 2 . 6 5 ; V ~ = 1 c m 3

Substituting Lhese above and solving for V2, we get, V2 = 0.48. Thus, the expected change in volume = 1 - 0.48 = 0.52 crn3,

SAQ 4

i) The following data were obtained from the liquid and plastic limit tests for a soil.

Liquid limit test : The number of blows N ant1 the moisture content 6. respectively, are 37,42.5; 33,462; 21, 52.3 and 13,61.5.

Plastic limit test : Moisture content = 18.7%.

Draw the flow curve from the liquid limit test data and find the liquid limit and flow index. What is the plasticity index of the soil ?

ii) A. soil whose liquid limit is 598 , Plastic limit is 26%, is excavated from a borrow pit for placement in an embankment. The natural moisture content of

\ the excavated soil is 32%. Calculate

a) pfasticity index

b) is Ule sail stiff or soft when compacted at its existing moisture content in the embankment?

iii) If the in situ moisture content of the soil in SAQ 4 (i) is 22% what is its liquidity index ? What will be the state of the soil in Ule natural condition ?

iv) The flow curves for three soils are shown in Figure 1.15. Which soil possesses least shear strength ?

v) A saturated soil with a volume of 18.00 cm%as a mass of 34 g. After drying, the soil had a volume of 13.9 crn? Its rnms was 24 g. Deterrninc the shrinkagc limit and shrinkage ratio of the soil.

1.7 CLAY MINERALOGY

It was pointed out in sub-section 1.2.1 that soil and rock are subjected to chemical weathcring. In chemical weathcring, the minerals present in Ule rocks undergo alteration by chemically reacting with the water and carbon dioxide, which are present in the atmosphere. 73ere is r large number of minerals in nature but promineot among them are about a dozen minerals which are present in most of the rocks. Of these, feldspars, micas and ferromagnesium minerals are especially important since Uley give rise to complex alumino silicates which are known as Clay Minerals.

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hcigl~l ot'a lclmllc(~n~n is 5 ArlgsVonl units (i.e., 5 A equal to 5 x J.O-' mm). Tllerel'ora, LC

thickness of a silica sheet is 5 A.

Bigurc 1.19 : Silica Shcct

When alumina ocl;~t~ctlra co~nb~ne in a similar lnauuler by sharing the unbalimced hydroxyl units, a shcet slruclure know11 as Ule Gibbsite sheet resulls (Figure 1.20). This sheet is relativcly morc conlplex. Adjacent ;ilun~inums share llydroxyls as well as oxygcn units obtai~~ed by a~lleratioii of one to the other. The thickness of this sheet is ills(.) 5 A. In case of 1n;lgncsium octahednIn Ule corresponding shcct is ci11led Ule Brucite sheet.

0 HYDROXVL

@ A L U M I N U M

Figl~rc 1.20: Thc Gibhsilc Sl~cet

The silica tetrahedron is elcclrically active while the sil~ca shcct is elcctric;llly ncutr:rl bccausc h e unbal;u~ced oxygcn atoms arc sal~sfied ilfer combination. Likewise, a Gibbsite shcct is also usually electrically neutral, However, in certain insuulces complcLo balancing iliay no1 Lake place and the sheel rnay be non-unil'orrn. In the case ol' brucite non-un~l'ornlity is be more because of substitution of trivalenl aluminum by bivalcnl magnesium. Subst~tutions are also done by olller atoms such as iron, sodium or potassium

In real clay nlinerals tllc silica and Ulc alun1in;l sheels combinc lo l'orm a morc complex Lhree-dimensional struciure. The silica shecls are stacked over Ule gibbsilc sheets. The unbalanced oxygcns of the siIica shect arc satisfied by unbalarlced a lu rn i r~~~m of the gibbsitc sheet. This gives rise to a platy structure. In some cases atoms arc shaucd by two plales and thus a bond develops which hold the plalcs togclher unlil substiluted by mother

I atom or dislurbcd by exlcmal lncans. Hydrogen atoms may swilch back ancl forth l~clwce11 plates. This shared attraction is known as Hydrogen Bond. It holds the plates togeher in stacks. 11' ~nag~les iu~n takes the plilcc ol' alu~ninum the shect is called a brucitc shcet. A typical silica-gibbsite shcet struclurc is shown in Figure 1.21. !

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1.7.3 '~oinmon Clay FMincrals Soil Phases, Structure, Consistency and

~t follows from above that clay particles carry a net unbalanced negative charge on them. CIassitication The negative charges are distributed over the surface. The surface area of a particle per unit volume, or sometimes, unit mass is known as specific surface. Thus, as specific surface increases negative charges also increase. The negative charges are balanced by Isomorphous Substitution. Substitution of one element, for example aluminum, by 'mother, say magnesium, without changing the crystalline structure is known as isomo~hous substitution.

~ h u s , there is a large number of clay minerals which vary in their degree of electrical activity and the extent of isomorphous substitution by various cations. But the most common among them are Kaolinite, Illite and Montrnorillonite.

Kaolinite

Kaolinite consists of repeating layers of elemental silica-gibbsite sheets (Figure 1.22 (a)) that are about 7.2 A thick. Kaolinite occurs as platelets. The sheets are tightly bonded in the plates by hydrogen bonding. Lateral dimensions of the plates may be 1000 A to 20,000 A and thickness 100 8. to 1000 A. Tliey stack in the form of layen as thi;ck as 0.01 rnm. The surface area of the kaolinite particles per unit mass is about 15 m Ig. There are several members of the kaolulite family, depending on variations in the alumina sheet. In general they are relatively well balanced electrically 'and exhibit only limited isomorphous substitulion.

&timzma LzzzzD Sl LlCA SHEET A L U M I N A SHEET POTASSIUM ATOM

WATER VARl ABLE

T

(n) Knolu~ile (b) Montn~orillu~~ile (c) mile E'igrre 1.22 : Stmelureof Clny Minernls

Montmorillonite

Montmofillonite has a structure in which a gibhsite slieet is sruidwiched between two silica sheets (Figure 1.22 (b)). There is extensive isornorphous substitution of niagiiesium and iron for aluminurn in Uie tetrahedral sheets. Each substitution produces a different nlineral, for example, nontronite, sauconite and saponite. Rut, from the engineering point of view, this differentiation is not important. 'lliese units do not stack together easily, and when stacked, break apart easily. A large amount of water is attractcd into the space between the layers. Particles of montmorillonitc have lateral dimensions of 1000 A to 5000 A aid thickness of 10 8. to 50 8.. The specific surface is about 800 m2/g.

Generally, montmorillonites form in regions rich in ferro-mi~griesium rocks (the source of magnesium) sucli as volcanics, and particuli~rly in areas of high temperature arid iritetlse rainfall. In India the black cottoii soils of Karnataka aid nearby regions contain this clay mineral. Ingress of water into sucli soils in nionsoori ciluses extensive swelling leading to cracking of buildings and heaving of the fou~~dation.

Illite

Illite also consists of a gibbsitc sheet bonding wilh two silica slicets likc montmorillonite. The difference is Uiat there is isomorphous substitution by potassium. ?he layers share Uie potassium aloms and form tightly bonded stacks (Figure 1.22 (c)). 111m'e is a limited isomorphous substilution in the alumina slieet. The negativc cliargc to balance Uie potassium atonk comes from Ule substilution of silicon by alunlinuni in tlie telrdiedral sheets. Illite particles generally have lateral dinlcnsiolis ranging from 1000 to 5000 A aid thickness froin 50 A to 500 A. The specific surface of the particles is about 80 m21g. Illites are often present in shales and other deposits that llave bee11 subject lo n changing environment; Uiey appear to be the alteratioll products of otlicr clay minerals.

Page 35: Geotechnical engineering

Suil Watcr System Other Clay Minerals

Besides kaolinitc, montmurillonile ruld illite, some of the ohe r clay minerals gefierally found are halloysile, vernuculite, and attapulgite.

Halloysite is a member ol' the kitolinile family. 11 $ontains a sheet of water between adjacenl clay units. Tlle clay unit lhicla~ess is 10 A rather thrul7.2. Halloysite exhibits shrinkage and reduction in thickness when dried. Rewetting does not restore the original stale. Its bchaviour is co~~lpletely chrulged by dehydration. Therefore, there have been problems in embankment construction, wllere halloysite has been tested in the laboratory under dry conditions but usell a1 site in Ule moist condition.

Vermiculite is collsidered to be a part of the montiriorillonite family. However, like halloysite it coiltai~ls sheets of watcr betweell the vert~uculite unit sheets. Bolh minerals exhibit isomorphous substitution bul not to Ule same degree as montn~orillonites.

The altapulgite mineral is different in that it has a double layer chain of silica Letrahedra linked by magnesium and alunlinum atoms. The crystal is in the form of a long ribbon lhal can contain water molecules.

1.7.4 Nature of Water in Clay I

As mentioned above, clay particles carry anet negative charge on their surfaces which is caused mainly by isomorphous substitution. Some positive charges also occur at the edges of h e particles. In dry clay when no waler is present the negative charge is balanced by exchangeable cations (such as cat+, Mg++, Nat, and Kt) which are held by eleclrostatic i attraction. However, when water is added to clay, these cations are released. These and a small nunlbcr of anions float around the clay particles and form what is known as iL

Diffuse Double Layer (Figure 1.23 (a)). The cation coricenlration is high close to the surface of the particle and il decreases as Ule distance from thc particle increases (Figure 1.23 (b)).

0 c ~ ~ l o ~ 0 WATER DIPOLE

L O W C A T I O N C O N C E N T R A T 1 0 N

H I G H C A T I O N C O N C E N T R A T I O N k,

l. I I % - 0 5 0 100

D I S T A N C E F R O M C L A Y S U R F A C E , A

(n) Adsorbed Waict- m ~ d Catior~ Distrihutio~~ (b) Cation Concer~tration

Figure 1.23 : Diffuse Double Layer

In order to understand this one must look at the structure of a water nlolecule. Watcr molecules are polar molecules. Hydrogen atonis are arranged in an unsyrnrnetric manner around the oxygen atom, at a bond 'angle of 105' (Figure 1.24). Therefore, a water lnolecule behaves like a small rod with a positive charge at one end and a negative charge at the other end. Hence it is known as a Dipole.

There is a three-way atlraction which takes place in the presence of water & shown in Figure 1,23 (a). Firstly, the cations in thc diffuse double layer are attracted to the clay pacticle surfaces. The positive end of the water dipole is also attracted to the clay particle surfaces, T l ~ e negative end of the dipole is attracted by Ule cations. Thus, water is held to clay and the force of attraction decreases with distance from the surface of the particles.

All the water held to clay particles by force of attraction is known as double layer water. The inner layer of double layer water, which is held very strongly by clay, is known as adsorbed water. The property of this water is very different from ordinary water. It is more viscous than ordinary water, and in the first layer of adsorbed water the density c'm be as

42 high as 1400 k ~ / i n ~ (See Unit 2 : Soil Water for more details).

Page 36: Geotechnical engineering

S V M B O L

H Y D R O G E N H Y D R O G E N

111 view ol' llicsc properlies cl:ly riiincr;lls liavc very high v;~lucs of liquid and plastic limits. Sorile ~ypicill vi~lucs ;11 '~ gi VCI I in 7';tbIc 1.19.

Table 1.19 : Atterberg Limits of Cl;ry Minel-ills --

1Sxchangcablc bliucral

1.7.5 Identificatio~~ ol' Clay Millerals The cliarilctcr 01' die cIi~y niincr;~ls g c n ~ r i ~ l l y influznces Llic behi~viour of the soils which conlain Llicm. Tllcy c:ln a11ect ti~.i~iliage, slrcngh ;~xid co~nprcssibili~y. I~iiprovcnicnl ol' Uic behav~ot~r of iL soil c o ~ i l i ~ i ~ i i ~ i g c.l;~y ~ninel-als requires ;i knowledge of Ll~c type of clay 1nincr;ll a~itl its S L I ~ U C L L I ~ ~ . Sillce L'I;L)I ~ ~ i i ~ i ~ r : ~ l s arc s111:111 ill size illid arc si~nilar in size uld wciglil Llicir iilenlifici~lion ruid scpi~riltion in m y soil is tlil'ficull.

One niclliotl ol' iclcn~il'icnlioi1 ol' lilixcti cloy I I U ~ C ~ ~ L ~ S is Dil'l'crcnliol Tlicr~ual A ~ i i i l y ~ i ~ . The '

cl;~p is lica~cil slowly. At difl'crcnl I C I ~ I ~ L ! ~ ~ L L U ~ C S , clcpcnding on Ihc clay n i i ~ ~ e r ; ~ l , cithcr slructu~i~l or 1,llasc changes or rcle;\se ol' waler takes place. The Iicat rcquirctl !o release tbc water protluccs ii cliiu'i~ctcristic cl i ;~~ige ill l(1c L.;I~C oI ' I i~~;~tr~ig . Tlic clay is itlc~il~fied eliipiricillly ilnd , s u ~ n i - q i ~ a ~ ~ ~ i t a t i v ~ l y by coli~piuing tlic clil'l'crc~iliel rate ol'licnling ol'llie clay will1 ~cspccl lo iI i~ i c r l hubslancc such i ~ s calcincil illt~lllin;~.

Page 37: Geotechnical engineering

Soil Water System . Another method is known as X-ray Diffraction. The soil is subjected to an X-ray beam at varying ruigles. The shadows produced by the atoms in tlie structure can be deduced. This is probably the most reliable method for identification and can give both the clay mineral and the approximate amount present.

Yet another technique is the use of the Electro~i Microscope, which makes a shadow photograph of the mineral: The well-stacked kaolin particles can be easily identified, but the montmorillonites, whose unit plates separate easily, sometimes produce no identifiable pattern.

1.8 SOIL CLASSIFICATION

Measurement of fundamentaI soil properties like permeability, compressibility and strength can be difficult, time consuming and expensive. For these reasons, sorting soils into groups showing similar behaviour may be very helpful. In many soil engineering problenls, empirical expressions are used for analysis and rigorous determination of soil properties is usually not required in the preliminary stages of analysis. In such cases again sorting of soils is quite useful. This is called Soil Classification.

Hence, soil classification consists of assigning a soil to a group of soils all of which exhibit similar behaviour. The classification system is usually an empirical one developed through

-. considerable experience. Soil classification helps in making preliminary evaluation of inany simple soil engineering problems. It also helps in planning detailed test programs for difficult and important problems.

The earliest attempl to classify soils was by geologists, who were interested in it as a product of the geological evolution process. Side by side, agriculturists also developed their own system considering soil as that part of the earth's crust, which supports plant life. The system is known as Pedological Classification System. Engineers required more rigorous systems and evolved methods based on index properties of soils. All soil classification systems make use of index-type tests to obtain the soil characteristics needed for the classification. These index tests are necessarily simpler than the standard methods used for determination of the fundamental soil properties. The most commonly used characteristics are particle size and plasticity.

Here the geological, pedological and the engineering classification techniques are briefly described in their clu-onological order of development.

1.8.1 Geological Classification The basis for this system of classification is Genesis. As stated in the very beginning, soils .

are products of weathering. These products may he stationary or transported to other places by water, wind, ice or gravity.

Soils formed at the place of their origin are called Residual Soils. An important characteristic of residual soil is the presence of fine-grained soil at the surface: grain sizes increase with depth dowli to the bedrock. At larger depths, angular rock fragments may be found.

Transported soils may be classified into several groups, depending on their mode of transportation and deposition, such as :

1) glacial soils - formed by transportation and deposition of glaciers

2) alluvial soils - transported by running water and deposited along a stream

3) lacustrine soils - formed by deposition in quiet lakes

4) marine soils - formed by deposition in the seas

5) aeolean soils - transported and deposited by wind

6) colluvial soils -formed by movement of soil by gravity such as larldslides, from its original place

1.8.2 Pedological Classification A soil derived or deposited by the geological processes described in sub-section 1.2.1, undergoes physical and chemical changes due to climatic factors prevalent in the locale in . the geological era subsequent to its deposition. Vegetation grows in the newly deposited

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material; rainfall contributes to leaching and eluviation of the surface and development of topography, which in turn exerts influence on the process of further changes. n e s e changes continue to take place over geological time. This leads to a characteristic external appearance of the soil.

Thus, the pedological classification system is based on Genesis and Morphology. From the above discussion five Genetic Factors, which contribute to soil formation, can be identified as :

1) Parent material

2) Climate

3) Vegetation

4) Relief or topography

5) Age or Geological time

Moqhology refers to the external appearance of the soil. The factors used to express this are Texture uld Colour.

In this system soil is considered as the end product of the combined action of the five genetic factors which is manifested externally in the Soil Profile. And morphology is used to characterise the profile. This system is said lo have had its origin in the late 19th century in the former USSR due to a soil scientist by name Dokuchaev. Now let us see what a soil profile is.

The Soil Profile

The sciil profile is a natural succession of soil layers below theground surhce and represents the effects of tile wealhering processes. These layers are known as Horizons. The profile may extend to various depths, and Ule horizons may have various thicknesses. Generally, Uree distinct horizons can be seen in a natural soil profile. These are called from top downwards as A, B hid C horizons, respectively. However, this number may be more in old and mature soils.

A typical profile is shown in Figure 1.25. The A horizon is rich in humus and organic plant residues. The layer following i~lxnediately below, called the B horizon, has a contrasti~lgly different colom. The relatively unwealhered parent material lying below, from which the upper horizons were derived is designated as the C horizon. From the figure it can be seen that the A niid B horizons could have several sub-horizons designated as A], Az, ... and B1, Bz, ... etc.

Horizon Description 7' D u k colour, ~ r g n n i c matter and minerals

Light-colour, leaclled zone

Leached, but trnrlsitional to B

Accumulatior~ zone, transitional to A

Deep colour, Accumulatioll ~nile, clay formalion

Transitional to C, Deeper cblour than C, Carbonate accumulalion

Parent mnlerial, Salt accumulation, slight weathering

Unaltered rock

Rg111r 1.25 : The Soil Profile

To classify Uie soils in an area the soil profiles are observed at regular inlervals. If necessary, pits are excavated to expose the profiles. The number, sequence, thickness, colour and texture of the soil in each horizon are recorded. Soils with similar profiles ace considered to be formed by the sane processes 'md hence belonging to the same class,

On a continental level soils ace classified into :

i) Great Soil Group,

ii) Zonal Group,

iii) Intra-zonal Group, and

iv) Azonal Group.

1 '

1,

Soil Phases, Structure, , , , Consiste~~cy and I

Classiflcntion

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Soil Water System Those with simillzs profiles are members of the same great soil group. Soils within the zonal group are identified on the basis of variations in climate and vegetation irrespective of the parent material, as similar profilcs can develop from a variety of different parent materials. Those profiles in which local topography and drainage are the dominant factors are included in the intra-zonal groups that cross climatic bo$daries. Some soils exhibit little profile development becaus of insufficient time for development. These are azonal soils.

On a regional level, for agricultural purposes, smaller classes are used. Soils with similar profiles and derived from the same parent material are grouped into Series. They are generally named after the area, where they are first identified. To have a finer level of classification, the Series are subdivided into Types and Phases. Types are classified based on the texture of the upper materials. Phases are identified within Types by considering variations in erosion, stoniness or rockiness, depth to bed rock etc.

In the US, in particular, Pedologic mapping has been carried out exlensively for evaluating the agricultural potential. In many cases series, and even types, have been mapped in considerable detail. Empirical correlations have been developed even for engineering use between pedologic map units (types) and drainage, plasticity, texture and potential use as construction materials. Some general description of the major soil classes is given below.

In cool or temperate and humid climates, vegetation grows abundantly and leads to formation or organic matter. The downward moisture movement adds to the creation of a profile with the three distinct horizons. The A-horizon is characterised by an acidic cl~emical alteration environment. The upper part is dark coloured and rich in organic matter followed by sandy and light coloured sub-horizon due to downward leaching of clays. The B horizon contairls the leached materials and deep coloured. It is thicker and rich in clay minerals, iron and carbonates, often partly cemented. This is underlairk by the C horizon, which is the slightly weathered parent material. The unweathered D horizon is the last.

In hot, humid regions the top layer undergoes alternate wetting, drying and downward leaching. The climate causes decay of organic matter and the environment is basic. The soluble silica and insoluble oxidised aluminum (md iron give rise to cemented quartz, which is a stiff rock-like material called laterite. The ratio of silica to oxides of aluminum and iron is low. The colour ranges from light t.0 bright red. Laterites are strong and relatively incompressible although often light weight and porous. Some laterites may soften on wetting. They are used as aggregate for road construction and as building material.

In dry regions vegetative grow* is insignificant and hence there is very little organic matter. Upward movement of moisture due to evaporation brings soluble carbonates up and the top layer consists of partly cemented soil. Sometimes carbonate c~ncretions occur as lenses in the mass. When dry, these exhibit good strength. But, when stlturated Uiey become weak or collapse due to sudden loss of strength. During monsoon rainfall the surface becomes alkaline and fonns a white crust. They may be good construction nlaterials depending on the parent material.

In a wet climate, organic decay and organic matter accumulation are rapid. When I

associated with fluctuating water levels, muck and peat formation takes place. In cool climate the peat deposits are thick because the decay tends to be especially slow. Thick peatdeposits can also be found in tropical regions in inundated areas such as lakes.

The pedological classification system has undergone significant and dramatic changes in the twentieth century with the advent of aerial photography and, more recently, satellite remote sensing. It is now possible to prepare a soil map in the office using air-photos and satellite imagery. Field work is limited to selective confirmatory ground-truth verification. A detailed description of this technique is beyond the scope of this block.

1.8.3 Textural Classification Texture of a soil refers to its surface coarseness or fineness. Particle size is a measure of texture. We have already seen in sub-section 1.4.2 particle-size classification of soils into classes such as gravel, sand, silt and clay. However, natural soils are mixlures of several size fractions. In the textural classification system, natural soils are classified and named after their principal components. A textural classification chart, known as the triangular I

ch'art, developed by the U.S. Bureau of Public Roads, is shown in Figure 1.26. The I

percentage of sand, silt and clay are calculated according to following size limits :

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sand 2.0 mrn to 0.05 mm

silt 0.05 mm to 0.0005 mm

clay < 0.005 mm

Based on this chart, a soil C with 20% sand, 40% silt, and 40% clay is classified as "clay", as.shown in Figure 1.26. If gravel or larger size particles are present, the percentages of sand, silt and clay are to be corrected by subtracting the gravel fraction and taking the sum of the other three fractions as 100%.

Figure 1.26 : Texturnl Classification Chart

1.8.4 Engineering Classification Since textural classification system is based entirely on particle-size distril..!tion and does not take plasticity illto account, it is inadequate for most engineering purposes, Therefore, classification schemes for engineering purposes were developed to consider both gradation and plasticity characteristics. The Bureau of Public Roads, USA, developed a system for soils for highway construction. A. Casagrande developed a general purpose sclierne. The Corps of Engineers and CAA developed systems for airfield construction. In 1952, the Bureau of Reclamation arid the Corps of Engineers evalilated all available systems and adopted two systems. They are

1) the AASHO classification system, and

2) the Unified Classificatioii System.

1) AASHO Classification System

This system of soil classification began in 1929 and was called the Public Road Administration classification system. Since then it has gone through several revisions, and the present version was proposed in 1945. The AASHO classification in its pre8ent form is given in Table 1.20. According to this system, soil is classified into seven major groups, A-1 through A-7. Soils classified under groups A-1, A-2 and A-3 are graiulaf materials with 35% or less passing through a IS Sieve No, 8 (0.075 mm). Soils with more than 35% passing No. 8 sieve are classified under groups A-4, A-5, A-6 and A-7. These soils are mostly silt and clay-type materials. To classify a soil by Table 1.20, one must proceed from left to right with the required lest data available, By the ptocess of elimination, the first group from the left into which the test datil will flt gives the correct classification.

Soil Phases, Structure, Consistency and

Classification

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Sail Wn ter Sysle~n

PERCENT PASSING N 0 . 2 0 0 S I E V E

OR LESS OR MORE

7 5 7 0 6 5 60 56 5 0 L5 LO 3 5

OR MORE OR LESS

PERCENT PASSING N0.200 SIEVE

Figure 1.27 : Group Index Chart

Group A-1 A-3 A-2 A-4 A-5 A-6 A-7 Clussification

A-I-a A-I-b A-7-5' A-2-4 A-2-5 A-2-6 A-2-7 A-7-sb

Table 11.20 : AASHO Classification System

l'e~.cent I'assi~ig Nil. 10 50 max No. 40 30 max 50 max 51 min No. 200 15 max 35 maw lOmax 35 rnax 35 max 35 max 35 max 36 min 36 min 36 min 36 min

(;enera1 Classification

1 Group Index 0 0 0 4 m u 8max 12max 1 6 m u 20max (

Granular Materials (5 35% Total Sample Passing No. 200)

Chxaete~~istics of fractii~n passing No. 40 :

Liquid Limit 40max 41 mi11 40max 41 mi11 40max 41 min '40max 41 min Plasiticity Index 6 max NP' IOrnax 10max Ilrnin I lmin 10max IOrnax l l m i n I lmin

Usual sigliificallt stone, silty soils clayey soils Collstitue~lt fragments, fine silty or clayey gravel and .sand Materials gravel and sand sand '

General Subgrade rating excellent to good fair to poor

Silt-Clay Materials (z 35% Total Sample

Passing No. 200)

I

I I I , Source : Sowers, G. B, and Sowers, G . K, lntroductory Soil Mechm~icsand Foundnlion Etzgineeritzg. I

To evaluate the performance quality of a soil as a highway subgrade material under the system, a number called the Group Index is included with the groups and subgroups of the soil; it is written in parentheses after the group or subgroup designation. The group index of a soil may range from 0 to 20 and is expressed as a whole number. The approximate subgrade performance quality of a given soil is inversely proportional to its group index, and it can be expressed by the followin&empirical relation,

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GI = 0.2 a + 0.005 ac +0.01 bd

where,

GI = group index,

a = 0 for % passing #8 I 35; incremented by 1 for every additional percent upto 40 for % passing 2 75%,

b = 0 for % passing #8 5 15; incremented by 1 for every additional percent upto 40 for % passing 2 55%

c = 0 for,LL I 40; incremented by 1 for every 1% increase in LL upto 20 for LL 2 60, and

d = 0 for PI < 10; incremented by 1 for every 1 % increase in PI upto 20 for PI 2 30.

The group index may also be evaluated from Figure 1.27 by adding the vertical readings obtained from the two charts.

2) Unified Classification System

This system is a revised version of the Casagrande system proposed in 1948. The unified classification system is presented in Table 1.21. This system is also used by Bureau of Indian Standards. Under this system, the soils are first broadly classified into two categories based on percent passing IS Sieve No. 8 : Coarse-grained and Fine-grained. If percent passing is less than 50% the soil is coarse-grained, otherwise fine-grained. Let us consider the classification of the coarse-grained and the fine-grained soils separately.

Table 1.21 : Unified Soil Classification System

Soil Phases, Structure, Consistency and

Classi6cation

I Major Division I Group I Typical Names ( Laboratory Classification Criteria 1 Symbol

Well-graded gravels, - D60 GW gravcl-sand mixture, little Cu = - greater than 4 : C - (~30)'

or no fines Dlo '- DIOXDGO

'C between 1 and 3 4)

- wl

Poorly graded gravels, 2 4) 3 X

gravel-su~d ~nixlures, littlc g 2 % Not meeting all gradation requirements for GP or no fines GW

2'z B d Silty gravcls. 3 Atterburg limits

gravel-smd-silt mixtures below "A.' line or Above "A" line - f i , PI less than 4 with P I belwecn 4

U bi ta ~r and 7 arc

Atterburg limits bordcrline cases requiring use of

Clayey gravels, above "A" line dual sylllbols gravel-sand-dry m i x t e with PI greater

Q M than 7 .B 8 m @

kh -3

8 a. n Well-grad&dsands, 3 D60

gravelly sands, little or no Cu = - greater than 6 : C c = b 3 0 1 2

6 3 SW 8 " Dl0 DID x D30 .z fines

', S 'u between 1 and 3

J a Poorly graded sands, SP gravelly sands, little or no Not meeting all gradation requirements for

fines SW

- d z B%z '" Atterburg limits

SM* - Silty sands, sad-silt below "A" line or Limits plotting in mixtures PI less Ulan 4 hatched zone will1

U PI betwcen 4 and 7 are borderlit~e

Z * Atterburg limits cases requiring use Clayey sands, sand-clay a 5 above "A" line of dual symbols

' S C lnixtureb wit11 PI greater I . than 7

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Soil Water Syste~n Coruse-grained soils are sub-divided illlo gravels a11d sands on the basis of percent passing IS Sieve No. 480 (4.76 innl). If it is less than 50 % h e soil is gravel, otherwise sluld. The soils of U~ese groups are represented by sy~nbols stating with G and S, respectively. Thesc arc furll~er classified on the basis of unitbrnlity coefficient and coefficient of gradation inlo well-graded (GW or SW) or poorly graded (GP or SP). If i i~les are present, the soil is classified according to the procedure for fine-grained soils. If the soil is silly then GM or SM is used as the symbol. If clayey tllen GC or SC is used. When the percent passing No. 8 sieve is between 5 and 12, dual symbols such as GW-GM, GP-GM, GW-GC, SW-SM, SW-SC, SP-SM, and SP-SC are adopted.

Fine-grained soil is classified inlo inorg'mic oi sill or clay by plotling the liquid limit and plasticity index of soil in the Plasticity Chart give11 at Lhe bollom right-hand pc~rlion ol' Table 1.21. The liquid limit and plasticity index of that portion of soil which passes IS Sieve No. 40 is used. The d iago~~al line drawn in the plasticity chart is called the A line and is given by the equation,

PI = 0.73 (LL - 20) ... (1.47)

A soil wih LL less th'm 30 is Low plastic. between 30 and 50 is Mediuinp1;islic and greater tha.1150 is High plastic. A soil lying below A line is silt and thilt lying above A linc is clay. Differentiation betwcer~ inorganic and org'mic silts and clays is made on the basis of the orgarlic content. To designate these groups the symbols used are M for illorganic silt, C for inorglmic clay, O for organic sills ,uld clays, L for low plasticity, M for mediunl plasticily and I1 S G ~ high plaslicily. For pent, visual-mlulunl idenlificalion is ncccssary. Other details for classificatioil are provided in Table 1.21.

The purpose of classificatioil is to identify the group lo which a soil belongs imd to gain a general idea of its behaviour. Table 1.22 gives a general indication of Uie permeabilily, strength and compressibility of the various soil groups along with their relative desirability for use in earth dims, canal seclions, foulldations and runways.

Table 1.22 : Relative Desirability of Soil Groups for Engineering Uses

Group Symbol

GW

G P

Relative 1 Probable Range ~ e n n c a ~ l i t y I of L IC' cdsuc)

Pervious

Relative Piping Kwistance

1000 to 100,000

Pervious to very pervious

High

High to medium

5000 to 10,000,000

Relative Ease of Shear Moisture

Strength Dmsity Control , Very l ~ i g l ~

High

Very goocl

Very good

GC I Impervious 1 0.01 ti) I0 I Vesy l ~ i g l ~ I High 1 Very gmd I CiM 1 Semi Pervious 0.1 to 100 High to High Very good 1 medium 1 1 1

Pervious

I ' sW I sM I Semi-pervious to 0.1 to 500 Medium io High Chad to firir

impervious 1 low

500 to High to Very high Very good 50.000 1 medium 1 1

Pervious to semi-pervious

Impervious

Impervious

Impervious

hnpervious

Very Impervious

Very Impervious

Low to very low

50 to High 5(%.000

High

Medium

C~ood to fair

Medium to high

Very high

High to medium

Mcdium to low

Medium

Low

Low

Low to Medium

Fair to very poor

Good to fair

Fair to Poor

Poor to very poor

Very poor

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1.8.5 Comparison of AASWO and UNIFIED Systems 130th the systems divide the soils into coarse and fine grained categories based on percent pzssing IS Sieve No. 8 and consider both gradation and plasticity cllcuacteristics. Practice shows that with about 35% fines, a coarse-grained soil will behave as a finc-grained material, indicating that the AASHO system is Inore appropriate. So is the case in separating gravel from sand. In the unified system the gravelly and simdy soils ;Ire separated better. Symbols used in the unified system describe soil properties better. The classification of organic soils such as OL, OH and Pt is not possible with the AASHO system. A correspondence between the groups of the two systems is presented in Table 1.23.

Table 1.23 : Comparison of AASWB and UNIFIED Systems

AASHO Systetu

A-l-a

A-l-b

I UNIFIEQ Systelll

GW, GP

SW, SP,

GM, SM

SP

GM, SM

GM. SM

GC, SC

GM, GC

ML, OL

OH, MH

ML, OL

CL

OH, MH

I A-7-6 I CH, CL I I I I

(All examples and SAQs givcn below use US Sieve Nos.)

Example 1.18

The results of the particlc-size analysis of a soil are as follows :

percent passing No. 10 sieve = 100

percent passing No. 40 sieve = 80

pcrcent passing No. 200 sieve = 58

The liquid linlil imd plasticity index of ininus No. 40 fraction of the soil are 30 ;md 10, respectivcly. Classify the soil by AASHO system.

Solution

Using Table 1.20 (since 58% of Ule soil is passing through No. 200 sieve), it falls under the silt-clay type of material, Lo., A-4, A-5, A-6, A-7. Proceeding from lefl lo right, it falls under group A-4. From Equation (1.41),

GI= 0 . 2 ~ + 0 . 0 0 5 ~ ~ + O.0lbd

b = 4 0

c=(30-40)=-10<0;takec=O,d=(10-20)=O

GI= 0.2(23) + 0.005(25)(0) + 0.01(40)(0) = 4.6 = 5 Therefore, the soil will be classified as A-4(5).

Example 1.19

The portion of a soil passing through No. 200 sieve is 95% and it has a liquid limit of 60 and plasticity index of 40, Classify the soil by the AASHO system.

1

Soil lDllases, Structure. Consislc~~q ; I I I~

Classifica ti011

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ReJi\~.ri~~g lo T:rhlc 1.20 i~ ld proceeding as in Example 1.18, this soil fillls under group A-7. Siticc PI > (LL - 30), ( i.e. 40 > (68 - 30)) his soil is A-7-6.

ISxample I .20

Cli~ssil'y llic soil described in Example 1 .I8 by the Unified Classification System.

Solutioo

Siricc 58% ol'llic soil is passing through No. 200 sieve, it is a fine-grained soil. Rcferritlg lo lllc pli~sticily chart in Table 1.21 for LL = 30 and PI = 10. it can be , classil'icd as CL.

ISxirnlple 1.21 I

Considcr Soil I I of SAQ 2 (vi) and Soil I of SAQ 2 (ii). The particle-size I

I

dislrihulions of LRe lwo soils are given in Figure 1.28. The liquid and plaslic limits of nlitlus No. 30 fraclion of dlr soils are as follows : I

Soil IIofSAQZ(vi): LL=31,PL=16

Soil 1 of SAQ 2 (ii) : LL = 27, PL = 21 I

Classify Ihc soils by the unified classificaticm system.

D I A M E T E R 0 ( m m ) --to

Figure 1.W : I;ridutic~i~ Curves Solution

Soil I1 of SAQ 2 (vi) : From the particle-sizc distribution curve, about 31% or the soil is finer than 0.0'74 rnm in diameter (No. 200 sieve), Hence, this soil is coarse- grained; since 31 is greater than 12, dual symbols need not be used. Also, 100% of the soil is finer than 4.76 mm (No. 4) sieve. Therefore, it is a sandy soil. From Figure 1.28, Dlo = 0.005 rnm, D30 = 0.08 rnm, and D60 = 0.25 mrn, Thus,

C, = DGdDlo = 0.2510.005 = 50 > 6

Cc = (~ '30)~1(~10 X D60)

= (0.08)~/(0.005 x 0.25) = 5.123 Hence P.

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With LL = 3 1 :uld PI = 3 1 - 16 = 15 (7). this soil plots i~bovc r l ~ c "A" linc. So thc cliissil'ication is SP-SC.

Soil I of SAQ 2 (ii) : More than 50741, i.e., 57% is piissi~lg No. 200 sieve (0.074 rnln diameter). Therefore, it is a fine-grained soil.

The LL = 27 arid PI = 27 - 21 = 6. It falls inside the hatched area in Ule plasticity chart, so the classificalion is CL-ML.

SAQ 5 i) Classify the followil~g soils, using 1ri;ulgular charl.

Table 1.24 : Data for' SAQ 5 (i)

1 Particle-size Distribution (%) 1

ii) Sieve analysis data for two soils are givcn in Table 1.25. The liquid and plastic limits of Ule fraction passing a No. 40 sieve are also given. Clilssify thc soils according to the AASHO classil'icalion system.

Table 1.25 : Data for SAQ 5 (ii)

iii) I

Sieve Analysis, Percent Finer

Soil'No. No. 10 No. 40 No. 200 LL PI,

1 8 1 5 7 28 35 27

2 94 8 0 62 47 3 8

Classify Soil I of SAQ 2 (vi), Soil I1 of SAQ 2 (ii) (see Figure 1.28) arid Soi A in Table 1.26 according to the U N I F I ~ system,

Table 1.26 : Data for SAQ 5 (iii)

Percent Passing Sieve Size

Soil ~ S A Q 2(vl) Soil IISAQ 2 (ii) A

No. 4 97.2 100 100

No. 10 62.5 100 100

No. 20 41.5 100 98

No. 40 33.8 86 93

No. 60 22.0 43 88

No. 200 12.0 32 7 7 LL NP 37 2 0

PI NP 23 8

iv) Classify the above soils of SAQ 5 (iii) according to the AASHO system,

1.9 SUMMARY

In this Unit fundamentals of soil formation, index properties, weight volume relationships, effect of water on consistency of soil, clay minerals, soil grabtion and soil classification have been dealt with.

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Soa Water System After going through this unit the student must satisfy himself that he can answer the following questions :

1) What is soil and how is it formed?

2) What is conceptual three phase soil model?

3) What are weight vol~lme relationships and how do they help in engineering computations?

4) What is soil consistency, what are Atterberg Limits and how are they determined?

5 ) What is the significance of Atterberg Limits?

6) What are clay minerals, how are they formed and how do they influence soil behaviour?

7) How is soil gradation determined?

8) How is soil classified for engineering purposes?

1.110 KEY WORDS

Soil : Soil is defined as unconsolidated mineral matter, derived by disintegration of rocks at or near the earth's surface and containing water, organic matter, air and other substances which inay be present in gaseous or liquid forms in the void spaces.

Three-phase : For engineering studies soil is considered to be a three-phase Material material consisting of solids, liquid (usually water) and gaseous

matter (usually air). Although in real soils the three phases are mixed, a conceptual model, in which these are considered to be ,

independent and separate, is used in soil mechanics.

Void Ratio : This is the ratio of volume of voids in the soil to the volume of the solids.

Unit Weight : This is defined as the weight of soil per unit volume. Since the weight of soil varies according to the state of the soil, we have three ' values qf unit weight. The dry unit weight corresponds to the dry state of the soil, the bulk unit weight to the moist state (a special case of which is the saturated un!t weight corresponding to the saturated condition) and the buoyant or submerged or effective unit weight to the submerged state (equal to the difference between the bulk unit weight and the unit weight of water).

Grain-size : The distribution of the various particle sizes in a soil is known as Distribution particle-size distribution or grain-size distribution. It is

determined by a process known as mechanical analysis, which involves the use of (i) sieve analysis for particles larger than 0.074 rnm and (ii) hydrometer analysis for smaller particles. The distribution is represented in the form of a plot between percent finer than a given diameter and the logarithm of diameter (in rnm) on a semi-log graph.

Cohesionless Soil : A soil which does not possess cohesive strength. Usually, coarse- grained soils such as gravel, sand and coarse silt are cohesionless soils.

Cohesive Soil : A soil which possesses cohesive strength. Usually medium to fine silts and clayey soils are cohesive.

Structure of Soil : This term is used in soil mechanics to describe the geometry of the particle-void formation or the geometric arrangement of mineral particles with respect to each other. This is affected by shape, size and mineralogical composition of the soil particles and the nature and composition of soil water. The structures encountered in cohesionless soils ate single-grained or

54 honey-combed, while those encountered in cohesive soils are '

-- -

:t

Page 48: Geotechnical engineering

Consistency

dispersed or flocculent. Bulky grains held together by a matrix of binding agent such as clay give rise to composite structures.

: On addition of water a soil passes through the semi-solid, plastic and finally liquid state. When i t is dry or at a low moisture content i t behaves like a solid. On tile other hand, when the moisture content is high soil flows like a liquid. This is known as consistency.

Clay minerals : Minerals in soil such as feldspar, mica and ferno-magnesium conlpounds, which are complex alumino-silicates, when subjected to chenucal weathering, give rise to new minerals lulc.rwn as clay minerals.

Soil classification : Soil clnssilicalion is the process of sorting soilds illto groups showing sinular 1whavic.rur bi~scd 011 si~nple index properties. The 1110~1 coounonly used characteristics arc gradation a id plasticity.

1.11 ANSWERS TO SAQs

SAQ 1

i) From Equatio~l ( l.C)),

Using Equalion ( l . l8 ) ,

Froni Equation ( 1.7): n = 0/ ( I +e) = 0.765/( 1 -: 0.765) = 0.433

Wc gcl liom Equatinn (1.1 I ) ,

Fillally using Equalio~l( I. 19) we gel,

ii) From Equatio~l (1.6), e = n/(l - n) = 0.4/(1 - 0.4) = 0.67.

At saturation Ule unit weight woulti be, EquaLion ( I . 17)

Tile unit weight at 10% ~noisturc content is, Equalions (I. 16) uld (1.17) :

Mass 01' water to bc added to 1 rn3 of soil is : (ysnt - Y[)/~ = (19.5 - 17,12)/9.81 = 0.24 kg

iii) (Only calculalio~~s are show11,The student sllould idenlify the equations involved from section? 1.3).

Soil Phases, Stnacture, Consistency and

Classification

Tlr subscripts 11 m d a are used to indicate borrow pit anb enlhankment. respectively.

Page 49: Geotechnical engineering

Snil Water Systct~i

G, = 2.65, yw = 1 glcm 3

The sLeps in the calculations are :

a) i) y~~~~=1.801(1+0.15)=1.57g/cm3.

ii) y,, = 1.67(1+0.18) = 1.97 g/cm3.

iii) yd,, = 2.65/(l+eb) or eh = (2.6511.57) - 1 = 0.688

iv) yd,, = 2.651(1+ee) or e, = (2.6511.67) - 1 = 0.587

b) i ) W, in 100 rr? of compacted soil = (1.67)(100)= 167 t (Here the equivalence lglcm3 = lt/m3 is used.)

ii) Since W, cannot change, we have Wl, of soil to be excavated = (167).(1+0.15) = 192.05 t

iii) V/) of sail to be excavated=192.05/1.8 =106.69 m3

C) i) Vw to be added during compaction = (We- Wb)lyw Bul, W, in 100 m3 = ((y,)(100) =I97 L

ii) Vw to be added = (197 - 192.05)ll.O = 4.95 rn3

iv) Soil No. 1

ii) yf = (G, + S e ) I + e Yw

Soil No. 2

ii) y, = (2'7+ lo = 17-67 rnIm3 1 + 1.215

iii) y,, = y - y,,, = 7.67 k~lrn"

Soil No. 3

iii) We also know that

Solving above equations (a), (b) and (c), e = 0.704; S = 76.7%

.. . (a)

(Only calculations are given. The student should identify tlie equations usf

1) Assume the volume of soil to be V=.l cm3

Page 50: Geotechnical engineering

2) Therefore yd= 1.82/(1+0.11) = 1.639 glcm'

3) Weight of dry soil in 1 cm" Wd = 1.639 g

4) The initial void ratio is given by, e = (2.6711.639) - 1 = 0.629

5) Volume of solids is obtained as V, = 1/(1+0.629) = 0.614 cm"

6) Since volume of solids cannot change, the new volume of the soil is given by :

7) The new value of yci will be equal to : yci = 1.63910.896 = 1.829 glcrn"

8) Using tlle new values of y, and yd wc get m = (1.8711.829) - 1 = 0.022 = 2.2%

SAQ 2

i) From Table 1.4 we get Ule sieve sizes corresponding to the Sieve numbers given. We then calculate the cumulative weight retained. Knowing that the weight of soil takcn is 500 g we obtain thc weight passing nnd the percent finct as shown in Table 1.27.

1.S Sieve No.

The sieve sizes ilnd Ulc percent I'incr arc plottetl in Figurc 1.28. Frorn thc figure we gct

Table 1.27 : Cotnpr~tatiorl of Percent I'iner

D10 = 0.08 mol; D30 = 0.27 mm; D60 = 1.5 Illm

c,, = 1.510.08 = 18.75; Ci.=(0.27)21(l.S~0.08)=0.41

ii) Thc particle sizes ;u~d pcrccnl fincr arc given directly. The particle-size dislributio~i curvcs ilre plotted in Figurc 1.28. The percentages of Ihc va~.ious fractions ire calculrrlcti in Tables 1.28 and 1.29.

1)iamcter (nim)

(2)

4.76

2.399

1.201

0.502

0.420

0.296

0.151

0.075

Table 1.28 : Table for Soil I

Weight of Soil (g) Percelit Finer

Retained (3) Passing (4) (5)

133.48 366.52 73.30

29.06 337.46 67.49

49.48 287.98 57.59

36.94 25 1.04 50.21

21.22 229.82 45.96

61.55 ' 168.27 33.65

74.77 03.50 18.70

46.93 46.57 , 9.31

Soil Ty pc

Grwcl (%I)

Sand (941)

Sill ((a)

Clay (%I)

MIT

(> 2 rnm) I(X) - 100 = 0

(2-0.06 nun) 100-51=46

(0.06 - 0.002 mrn) 54-22=32

( < 0.002 mm) 22

(> 4.76 mm) loo- l 0 0 = 0

Soil Phases, Structure, Consislel~cy and

Cli~ssific~tion

Page 51: Geotechnical engineering

Suil Water gystem

I Sand (%) I (2 - 0.06 nim) (2 - 0.05 mm) (4.76 - 0.075 nun) 100-31 =69 lob-30=70 100-32=68

Table 1.29 : Table for Soil I1

Soil Type

Gravel (%)

I I Silt(%) (0.06 - 0.002 mm) (0.05 - 0.002 rnm) 31 - 13 = 18 30- 13 = 17

iii) The grain-size distribution curve for the given sand is also plotted in Figure 1.28. The percentages of gravel, sand, silt and clay are computed in Table 1.30#

MIT

0 2 mm) 100- 100=0

(0.075 -0.002 mm) 32 - 13 = 19

Clay (%)

Table 1.30 : Table for Percentages of Fractions

USDA

0 2 100- 100=0

( < 0.002 mm) 13

BIS

(> 4.76 mm) 100-100=O

Soil Type

(< 0.002 mm) 13

Gravel (%)

Silt (%) 1 (0.06 - 0.002 m) I (0.05-0.002 mm) 1 (0.075-0.002 nun) (

(< 0.002 mm) 13

I I I MIT

Sand (%)

I (Data not available for sizes 0.075 mm)

0 2 mm) 100-SO= 50 100-50=50

USDA

(> 4.76 mm) 100-78=0 I

(2-0.06 mm)* 5 0 - 8 ~ 4 2

I - -- -

(Data not available for sizes 0.075 mm) I

BIS

Clay (%)

* Extrapolated value.

(2-0.05 mm)* 50-8=42

The values of the gradation curve parameters are :

Dlo = 0.08 mm; D30 = 0.6 mm; D60 = 2.7 mrn c, = 2.7/0.08 = 33.75; C, = (0.612/(2.7x 0.08) = 1.67.

iv) To calculate diameter of the particles corresponding to given time and depth we shall use Equation (1.24 a). Keeping in mind the units to be used in the equation we note : I

(4.76-0.075 mm) 78-9=69

( < 0.002 mm)

t =120min;h=145mm=14.5cm;

n = 0.001 ~ - s / r n ~ = 0.001 x / lo4 poise

= 0.01 poise

Now substituting these values in Equation (1.24 a) we have, I

(< 0.002 mm)

The percent finer than this diameter is now computed from Equation (1.29) a s follows :

(c 0.002 mm)

v) Again using Equation (1.24 a) and taking care to adopt appropriate units,

t = 2916 min. = 48 hrs. 36 min.

vi) The gradation curves are plotted on semi-log paper in Figiirc 1.7-8.

Page 52: Geotechnical engineering

SAQ 3

i) As long as particles are uniform spheres the void ratio for a cohessionless soil in single grain loose packing remains constant at 0.91 as demonstrated in Example 1.10. Similarly the unit weight also remains unchanged.

ii) As per Equation (1.32), 6 2 3 For 5-micron clay: s = 6i0.005 x 1 o - ~ = 1.2 x 10 rn /m

For colloids, assuming a diameter of 0.0002 rnrn,

s = 6/0.0002 x 10-" 3 x lo7 m2/m"

iii) See sub-section 1.5.2.

iv) External effects must be such as to break the bonds between the particles. Such effects are

1) ingress of water which will electrically neutralise the bonds,

2) heat, which will cause release of adsorbed water, 'and

3) vibrations, which will physically dislodge the particles.

V) i e t us assume the following typical void ratios :

1) e=0 .5 2 ) r = 0 . 4 3 )e=0 .8 4 )e=1 .2

Using Equation (1.18):

a) yd = (2.67 x 9.81)/(1+0.5) = 17.46 k ~ / n ? .

Similarly, we have for the remaining cases,

b) ysa, = 2 1 .5 1 k ~ / i n ~ .

C) y,i = 14.55 k ~ / m ~ .

SAQ 4

i) From the given data.U~e flow curve is obtained as shown by Curve No. 3 in Figure 1.15. From there we get,

LL = 50.5% ; PL = 18.7% (given); PI = 31.8%

lf= (52.3 - 46.2)/10g(33/21) = 3 1.08

~ i ) a) PI = 59 - 26 = 33

b) LI = (32 - 26)/(59 - 26) = 0.18

From sub-section 1.6.2 li)r LI = 0.1 8 the co~lsisterlcy of Ule soil is STIFF.

iii) LI = (22 - 18.7)/(50.5 - 18.7) = 0.104. Under natural inoislure content the soil is STIFF.

iv) For a given reduction in moisture content, a soil wilh smaller shear strength will require a lesser number of blows. Thus, a flow curve wilt1 a steeper slope will represent a soil with lesser shear strength. Curve No. 1 has the steepest slopc in this case. Therefore, the soil represented by Ulis curve has thc least shear slrenglh.

v) From Equation (1 35) wc have,

SL = (18 - 13.9)/(34 - 24) = 0.41 or 41%

SAQ 5

i) Using Figure 1.26 we have,

Soil Phwes, Sti-ucturc, Consistency and

Classificntion

Soil A : CLAY

Soil B : SILTY LOAM

Soil C : CLAY

Soil D : SANDY CLAY LOAM

(All Sieve Nos. in Ule following SAQs are as per US system - see Table 1.4)

Page 53: Geotechnical engineering

i i ) Soil 1

Co~isidering percentages passing Sievc Nos. 10.40 & 200 we find that the soil hclongs LO group A-2 in Table 1.20. Considering that LL = 35 and PI = 35 - 27 = 8, thc soil is assigned to group A-2-4. GI works out to 0, which i ~ l ~ t ) salisfies the classification.

Soil 11

Considering pcrcerilages passing Sieve Nos. 10,40 & 200 we find that the soil hcloligs 10 one of the groups A-4 to A-7. Considering that LL = 47 and PI = 47 - 38 = 9, the soil is assigned to group A-5. GI = 6.35, which also salisl'ies Lhe classificalion.

iii) Soil I of SAQ 2 (vi)

Considering percentage passing Sieve Nos. 200 & 4, the soil can be classified as SAND. Froin Figure 1.28,

Dlo = 0.03 mrn: D3o = 0.3 mrn; D60 = 2.00 111111.

Therefore, c,, = 66.7 > 6 : C, = 1.5. Thus the soil is classified as SW.

Soil 11 of S 4 Q 2 (ii) \ Corisicleri~ig percentage passing Sieve Nos. 200 & 4, the soil can be classified

as smd. From Figure 1.28,

Therclhre, c,, = 3 10 > 6 ; C, = 1 1.6 > 3. Thus the soil does not satisfy gradation requirements fully. Since the soil has inore than 12% fines (96 passing Sieve No. 200 = 32), the Plasticity chart of Table 1.21 is used for further classification. Since LL = 37 and PI = 23 the soil is clay.

Illerefore, the soil is classified as SC.

Soil A

Corisidering percentage passing Sieve No. 200, the soil can hc classified as a fine-grained material. Therefore the Plasticity chart is uscd. Since LL =20 and PI = 8 the soil is classified as CL-ML.

iv) Soil I of SAQ 2 (vi)

Considering percentages passing Sieve Nos. 10.40 & 200 we find that lhe soil belongs to group A-1-b in Table 1.20. The soil is non-plastic and GI = 0. This agrees with the classificalion.

Soil I1 of SAQ 2 (ii)

Considering percentages passing Sieve Nos. 10.40 & 200 we find that the soil belongs to group A-2 in Table 1.20. Considering that LL = 37 and PI = 37 - 23 = 14, the soil is assigned to group A-2-6. GI works out to 2, which also satisfies Ule classification.

SoiI A

Considering percentages passing Sieve Nos. 10,40 & 200 we find that the soil belongs to one of the groups A-4 to A-7. Considering that LL = 20 and PI = 8, the soil is assigned to group A-4. GI works out to 8, which also satisfies Ule classification.