topic 1. basic characteristic of soils

7/23/2019 Topic 1. Basic Characteristic of Soils

http://slidepdf.com/reader/full/topic-1-basic-characteristic-of-soils 1/40

12/01/20

GEOTECHNICAL ENGINEERING

(CVE4201 / CVE3225)

Basic Characteristic of Soils

Lecture Week No 1 – 4

Dr Eric LOH

SCHOOL OF CIVIL ENGINEERING

LEARNING OUTCOMES

On completion of this topic successful

students will be able to:

CO1: Describe and determine geotechnical

properties of soil



12/01/20

Why study soils?

What is the difference between a soil and a rock ?

Where does soil come from ?

ACTIVITY 1

What is the difference between a soil and a rock ?

A B



12/01/20

Soil is a particulate material Rock is an intact

Where does soil come from ?

Soil is produced by weathering of rock

Weathering is the breakdown of rocks at the

Earth’s surface, by the action of rainwater, extremes

of temperature, and biological activity. Weathering

involves no moving agent of transport.

Erosion is the process by which soil and rock

particles are worn away and moved elsewhere bywind, water or ice.



12/01/20

Rock Cycle and the Origin of Soil

a) Freeze-thaw occurs when

water continually seeps

into cracks, freezes and

expands, eventually

breaking the rock apart.

Mechanical weathering is caused by the effects of

changing temperature on rocks, causing the rock to break apart.The process is sometimes assisted by water.

b) Exfoliation occurs as

cracks develop parallel to

the land surface a

consequence of the

reduction in pressure

during uplift and erosion.



12/01/20

Chemical weathering is caused by rain water reacting with

the mineral grains in rocks to form new minerals (clays) and

soluble salts. These reactions occur particularly when the water is

slightly acidic.

a) Solution - removal of rock in solution by acidic

rainwater. In particular, limestone is weathered

by rainwater containing dissolved CO2, (this

process is sometimes called carbonation).

b) Hydrolysis - the breakdown of rock by acidic

water to produce clay and soluble salts.

c) Oxidation - the breakdown of rock by

oxygen and water, often giving iron-rich

rocks a rusty-coloured weathered surface.

Biological weathering refers to the break down or

degradation of rock by living organisms.

Many animals, such as these Piddock shells, bore

into rocks for protection either by scraping away the

grains or secreting acid to dissolve the rock.

Trees put down roots through joints or cracks in the

rock in order to find moisture. As the tree grows, the

roots gradually prize the rock apart or plant acids

help dissolve rock.

Even the tiniest bacteria,

algae and lichens produce

chemicals that help break

down the rock on which they

live, so they can get the

nutrients they need.



12/01/20

Once the rock has been weakened and broken up by

weathering it is ready for erosion. Erosion happenswhen rocks and sediments are picked up and moved to

another place by ice, water, wind or gravity.

Residual Soils

vs

Transported Soils

Residual Soils hold its position of their formation, without

transporting, just above their parent rock. Residual soils show considerable

variation of engineering properties form top layer to bottom layer. The transition

is observed gradual. Relatively finer materials are found near ground surface

and they become coarser with depth to reach larger fragment of stone.

Typical weathering profiles



12/01/20

Residual Soils hold its position of their formation, without

transporting, just above their parent rock. Residual soils show considerable

variation of engineering properties form top layer to bottom layer. The transition

is observed gradual. Relatively finer materials are found near ground surfaceand they become coarser with depth to reach larger fragment of stone.

Typical weathering profiles

Transported Soils get deposited by different transporting agent at

a point that is away from its formation. These soils are found to have entirely

different engineering properties from that of the rock on/at which they are

deposited. These deposits are usually found uniform and a considerable

thickness. In foundation engineering context, they have great importance as we

deals with these deposit more frequently in founding civil engineering

structures.

a) Glacial Soils: formed by transportation & deposition of

glaciers.

b) Alluvial Soils: transported by running water & deposited

along streams.

c) Lacustrine Soils: formed by deposition in quiet lakesd) Marine Soils: formed by deposition in seas/oceans

e) Aeolian Soils: transported and deposited by the wind

f) Colluvial Soils: formed by movement of soil from its

original place by gravity (e.g. landslides).



12/01/20

ROCK-SOIL RELATIONSHIP

Igneous Rockso Granite → silty sands

o Basalts → clayey soils

Sedimentary Rockso Shales → clays and silts

o Sandstone → sandy soil

o Limestone → coarse/fine grained soils

Metamorphic Rockso

Gneiss → silty sando Slate → clayey soils

o Marble → fine grained soils

o Quartzite → coarse grained soils

What is the difference between Gravel, Sand,

Silt and Clay ?

20 µm 5 µm



12/01/20

Particle size classes of different systems

U.S.D.A – U.S. Department of AgricultureUNIFIED – Unified Soil Classification System

AASHO – American Association od State Highway and Transportation Officials

British Standard range of particle sizes



12/01/20

Particle Size Analysis(Dry Sieve Analysis)

Sieve Size (mm) Mass Retained (g)

20 0

14.4 1.7

10 2.3

6.3 8.4

5 5.7

3.35 12.9

2 3.5

1.18 1.1

0.6 30.5

0.425 45.3

0.212 25.4

0.063 7.4

Example 1

The result of a dry-sieving test are given below; plot the particle-size distribution curve

and give a classification for the soil. The original weighed quantity was 147.2g



12/01/20

Sieve Size (mm) Mass Retained (g) % Retained % Passing

20 0 0.00 100

14.4 1.7 1.15 98.85

10 2.3 1.56 97.28

6.3 8.4 5.71 91.58

5 5.7 3.87 87.70

3.35 12.9 8.76 78.94

2 3.5 2.38 76.56

1.18 1.1 0.75 75.82

0.6 30.5 20.72 55.10

0.425 45.3 30.77 24.32

0.212 25.4 17.26 7.07

0.063 7.4 5.03 2.04

Original Weight: 147.2g

Total 144.2



12/01/20

602020.60.20.06

PARTICLE SIZE DISTRIBUTION

0

10

20

30

40

50

60

70

80

90

100

0.01 0.1 1 10 100

P E R C E N T A G E P A S S

What is the approximate proportions of the soil?

TYPICAL GRADING CURVES

A flat portion of the curve indicates that little of

that particle size is present

A steep portion of the curve indicates that a lot of

that particle size is present

A well graded soil gives a smooth concavecurve



12/01/20

Effective Size

Uniformity Coefficient, Cu

Coefficient of Gradation, Cg

10d =

10

60

d

d =

( )

1060

2

30

d d

d

×

=

Uniformly graded soils will tend to have low Cu values (< 3.0 )

Well graded soils having Cu value of > 5.0.

A single sized soil would have a Cu value of 1.0.

Cg values of about 2 are ideal with values between 0.5 and 2.0

indicating a well graded soil.

Maximum size of the smallest 10% of the sample

PARTICLE SIZE DISTRIBUTION

0

10

20

30

40

50

60

70

80

90

100

0.01 0.1 1 10 100

P E R C E N T A G E P A S S

D10 D30 D60

602020.60.20.06



12/01/20

Effective Size,

Uniformity Coefficient, Cu

Coefficient of Gradation, Cg

26.010 =d

6.226.0

68.0

10

60===

d

d

( )3.1

26.068.0

48.0 2

1060

2

30=

×

=

×

=

d d

d

Uniformly graded soils will tend to have low Cu values (< 3.0 )

Well graded soils having Cu value of > 4.0 for Gravel & > 6.0 for Sand

A single sized soil would have a Cu value of 1.0.

Cg values of about 2.0 are ideal with values between 1.0 and 3.0

indicating a well graded soil.

Curve A :- poorly graded medium SAND - narrow range of sizes therefore

poorly graded

Curve B :- well graded material GRAVEL SAND - wide range of sizes

Curve C :- very silty SAND - significant silt fraction

Curve D :- sandy SILT

Curve E :- silty CLAY - typical London Clay



12/01/20

Sieve size (mm) Mass retained (g)

50 0

37.5 15.5

20 17

14 10

10 11

6.3 33

5 33.5

3.35 81

2 18

1.18 31

0.6 32.5

0.212 9

0.15 8

0.063 5.5

Problem 1

The result of a dry-sieving test are given below; plot the particle-size distribution curve

and give a classification for the soil. The original weighed quantity was 306g

DETERMINATION OF

CONSISTENCY LIMITS



12/01/20

Semi

plastic

solid

Brittle solid

As moisture is removed from these fine grained soils,

they pass through four states, solid, semi-plastic solid,

plastic, liquid, all of these states are water content

dependent.

Liquid

T o t a l V o l u m e

Water content

Plastic

This change from one state to another is a gradual

process, however for convenience we chose to definethree water content “limits” at which the changes occur.

These limits are commonly referred to as the

Consistency Limits of the soil.

The three limits in question are :-

1) Liquid Limit (WL) - this is the water content at which

the soil changes from a liquid to a plastic state. It is the

minimum water content at which the soil will flow under

it’s own weight.


Water content

WS WP WL



12/01/20


2) Plastic Limit (WP) - this is the water content at which

the soil ceases to be entirely plastic and becomes a semi-

plastic solid.

T o t a l V o

l u m e

Water content

WS WP WL


3) Shrinkage Limit (WS) this is the water content below

which further loss of moisture does not result in a

decrease in the soil volume.


Water content

WS WP WL



12/01/20

Of these three limits the most important as far as we are

concerned are the liquid limit and the plastic limit.

The range of water contents over which the soil is in a

plastic condition is referred to as the Plasticity Index (IP)

T o t a l V o

l u m e

Water content

WS WP WL

I W WP L P= −

The Liquidity Index (IL) expresses the natural water

content of the soil in terms of the consistency limits.

IL < 0 soil is in a semi-plastic or solid state

0 < IL < 1 soil is plastic

IL > 1 soil is in a liquid state and will thus flow (i.e. a

quick clay)


Water content

WS WP WL

P

P

PL

PL

I

W-w

WW

W-wI =

−

=

Natural Water Content, w



12/01/20

Dial Gauge

Cone:

Mass 80g

Length 35mm

Angle 30˚

Base

Soil in metal cup

Diameter 55mm

Depth 40mm

Manual cone

release and

locking device

(LIQUID LIMIT - penetrometer method)

DETERMINATION OFCONSISTENCY LIMITS(LIQUID LIMIT - penetrometer method)

Firstly the soil is dried and then broken up

using a pestle and mortar. The sample is

then sieved and the material passing the

425 µm sieve mixed with distilled water to

a paste of stiff consistency. This is then

left for 24 hours in an air tight container to

allow for the water to fully penetrate the

soil. After this time a portion of the soil is

placed in the penetrometer cup and the

soil struck off level with the top of the cup(care must be taken not to entrap any air

in the cup when placing the soil). The cup

is then placed on the penetrometer stand

and the point of the cone lowered such

that it just touches and marks the top

surface of the soil sample in the cup.



12/01/20

2

DETERMINATION OF

CONSISTENCY LIMITS(LIQUID LIMIT - penetrometer method)

The dial gauge reading is then taken and

noted and then the clamp released. The

cone is allowed to penetrate the soil

sample for 5 sec. (timed with a stop

watch) after which the clamp is re-

tightened and a second dial gauge

reading taken and again noted. The

difference between the second and first

dial gauge readings gives the

penetration. The same procedure is

repeated several times on the same soil

sample and an average penetrationcomputed. A small sample of the soil

sample is then taken for water content

determination.

DETERMINATION OFCONSISTENCY LIMITS(LIQUID LIMIT - penetrometer method)

The whole procedure is then repeated

five or six times with the successive

addition of amounts of distilled water (i.e

the soil sample will have an increasing

water content)



12/01/20

2

DETERMINATION OF

CONSISTENCY LIMITS(LIQUID LIMIT - penetrometer method)

A graph of cone penetration against water content is then plotted with a

“best fit” straight line drawn between the points. The liquid limit is then

the water content which corresponds to a cone penetration of 20 mm.

xx

x

x

0

0

5

10

15

20

25

C o n e p e n e t r a t i o n ( m m )

Water content (%)Liquid Limit

(WL)

Data points

Line of best fit

In a liquid limit test on a fine-grained soil, using a cone

penetrometer, the following results were recorded.

Determine the liquid limit of the soil.

Penetration (mm) 15.6 18.2 21.4 23.6

Water Content (%) 34.6 40.8 48.2 53.4

Example 1



12/01/20

2

DETERMINATION OF

CONSISTENCY LIMITS(PLASTIC LIMIT)

Take approximately 20 g of soil paste (prepared in the same way as for the

liquid limit test) and roll it into a ball in the hands until slight cracks appear in

it’s surface. Divide the ball into two halves and then one of these halves into

four equal portions. Take one of these portions, roll it into a ball and then into a

thread on a glass plate. When the diameter of the thread becomes 3 mm

knead it again into a ball, this process of handling the soil sample effectively

drying out the soil sample (i.e. decreasing the water content). Again roll the soil

ball out into a thread. Repeat the process of rolling into a ball and then into a

thread until the thread just starts to crumble at the 3 mm dia. Once this has

occurred place the thread pieces into an air tight container. The whole process

should be carried out on the remaining three portions of the first half of the 20g

sample with all thread pieces put into the same container.

The test is then repeated on the other 10g of soil sample.

The water content of the two 10g’s is then determined and the average of the

two reported as the plastic limit of the sample ( % ).

Using the relationship between the liquid limit and the

plastic limit it is possible to establish sub - groups for thefine soils. The most commonly used classification in the

UK is the British Soil Classification System and this is

based on the standard Plasticity Chart.



12/01/20

2

The liquid limit is plotted against the plasticity index of the soil and

depending where this point lies a sub - group for the soil can be

determined. The ‘A’ line on the plasticity chart gives an arbitrary

division between silts and clays with the vertical lines defining five (5)levels of plasticity:- low(L), intermediate (I), high (H), very high (V)

and extremely high (E).

SUB-GROUP SYMBOLS FOR THE BRITISHSOIL CLASSIFICATION SYSTEM

Primary letter Secondary letter

Coarse grained soils G = GRAVEL W = well graded

S = SAND P = poorly graded

Pu = uniformly graded

Pg = gap graded

Fine grained soils F = FINES L = low plasticity

M = SILT I = intermediate plasticity

C = CLAY H = high plasticity

V = very high plasticity

E = extremely high plasticity

Organic soils Pt = PEAT O = organic



12/01/20

2

SUB-GROUP SYMBOLS FOR THE BRITISH

SOIL CLASSIFICATION SYSTEM

GPu – uniformly graded GRAVEL

CV – very high plasticity CLAY

ML – low plasticity SILT

Determine the liquid limit, the plasticity index and classify the soil

Penetration (mm) 15.6 18.2 21.4 23.6

Water Content (%) 34.6 40.8 48.2 53.4

Wp = 33 %

Liquid Limit, WL = 45%

Plasticity Index, IP = WL – WP = 45 – 33 = 12%



12/01/20

2

X

Classification: MI (SILT of intermediate plasticity)

Determine the liquid limit, the plasticity index and classifythe soil

Penetration (mm) 15.6 18.2 21.4 23.6

Water Content (%) 48.6 54.8 62.2 67.4

Wp = 22 %

Problem 1



12/01/20

2

Phase Relationships

SOIL MODEL AND BASIC PROPERTIES

Soil

Solid

Air

Water

Phase Diagram



12/01/20

2

THREE-PHASE SOIL MODEL

Solid

Air

Water

Ma ≈ 0

Mw = wMs = wGsρw

Ms = Gsρw

Masses

Va= e(1-S)

Vw= Se = wGs

Vs = 1

Volumes

e

S p e c

i f i c v o l u m e ,

V = 1 + e

DEFINITIONSpecific Gravity (G s) is the ratio of the mass of a

given volume of a material to the mass of the

same volume of water

Solid

Air

Water

Ma = 0

Mw = wGsρw

Ms = Gsρw

Va= e(1-S)

Vw= Se = wGs

Vs = 1

w

s sG

ρ

ρ =



12/01/20

2

DEFINITION

Degree of saturation (S ) is the fraction of the voidvolume filled by water

Solid

Air

Water

Ma = 0

Mw = wGsρw

Ms = Gsρw

Va= e(1-S)

Vw= Se = wGs

Vs = 1

v

w

V

V S =

S = 0 for perfectly dry soil

S = 1 for fully saturated soil

DEFINITIONWater content (w ) is a measure of the amount of

water present in the soil

Solid

Air

Water

Ma = 0

Mw = wGsρw

Ms = Gsρw

Va= e(1-S)

Vw= Se = wGs

Vs = 1

sS

W

G

Se

M

M w ==



12/01/20

2

DEFINITION

Void ratio (e) is a measure of the void volume; itmay be occupied by either water and/or air

Solid

Air

Water

Ma = 0

Mw = wGsρw

Ms = Gsρw

Va= e(1-S)

Vw= Se = wGs

Vs = 1

s

wa

V

V V e

+

=

e

DEFINITIONPorosity (n) is also a measure of the void volume

to the total volume

Solid

Air

Water

Ma = 0

Mw = wGsρw

Ms = Gsρw

Va= e(1-S)

Vw= Se = wGs

Vs = 1

e

e

V

V n v

+

==

1

e



12/01/20

3

DEFINITION

Solid

Air

Water

ewGsρw

Gsρw

e(1-S)

Se = wGs

1

W S

Be

SeG ρ ρ

+

+=

1

W S

sat e

eG ρ ρ

+

+=

1

W S

d e

G ρ ρ

+

=

1

VolumeTotal Solid of Mass

_ _ _ = ρ 31081.9 −

××== ρ ρ γ g

Example 1

In a sample of moist clay soil, the void ratio is 0.788 and the

degree of saturation is 0.93. Assuming Gs = 2.7, determine the

dry density , the bulk density and the water content.



12/01/20

3

Problem 1

A specimen of clay was tested in the laboratory and the

following data were collected:

Mass of wet specimen, M1 = 148.8g

Mass of dry specimen, M2 = 106.2g

Volume of wet specimen, V = 86.2 cm3

Specific gravity Gs = 2.70

Determine:

a) the water content

b) the bulk and dry densities

c) the void ratio and porosityd) the degree of saturation

Problem 2

A cylindrical specimen of moist clay has a diameter of 38 mm,

height of 76 mm and mass of 174.2 grams. After drying in the

oven at 105oC for about 24 hours, the mass is reduced to

148.4 grams. Find the dry density, bulk density and water

content of the clay. Assuming the specific gravity of the soil

grains as 2.71, find the degree of saturation.



12/01/20

3

Problem 3

It is known that the natural soil at a construction site has a void

ratio of 0.92. At the end of compaction, the in-place void ratio

was found to be 0.65. If the moisture content remains

unchanged, determine:

a) Percent decrease in the total volume of the soil due to

compaction

b) Percent increase in the field unit weight

c) Percent change in the degree o saturation

a) 14.0% decrease

b) 16.3% increase

c) 41.5% increase

Soil Compactions



12/01/20

3

WHAT IS COMPACTION?

Compaction is the application of mechanical energy, to a

soil in order to rearrange the soil particles and thus get

them to pack closer together - reducing the void ratio.

The smallest possible void ratio is generally aimed for

when undertaking construction works on or in a soil or

when placing fill material. Why???

Increasing the shear strength of the soil which

• leads to improvements in the stability of embankments

• increases the bearing capacity of foundations, road pavements, etc.

ii. Decreasing the compressibility of the soil

• large voids can lead to the soil compacting under the imposed loads

which results in settlement

iii. Decreasing the void ratio

• reduces the permeability of the soil – usually desirable in most

construction operations

iv. Decreasing the size of any air voids, if these fill with water they may

• reduce the shear strength of the soil

• increase the potential for swelling of the soil

• increase the potential for shrinkage of the soil

• increase the potential damage from frost heave

The main objective of compaction is to improve the engineering performance of

the soil and compaction achieves this by :-



12/01/20

3

Factors Affecting Compaction

The state of compaction of a soil is measured in terms of

the DRY DENSITY of the soil. Dry density is used as this

is the mass of solids per unit volume, the higher the dry

density achieved the greater the amount of solids in the

unit volume. The degree to which any soil can be

compacted is affected by three factors:

a) the water content of the soil

b) the amount of compactive effort that is

applied to the soil

c) the type of soil and its grading

FORMULAE

Note – remember w and Av must be entered in decimal form.

w

b

d +

=

1

ρ ρ

( )( )

s

vw s

d Gw

AG

+

−

=

1

1 ρ ρ



12/01/20

3

Bulk

Density

(kgm-3)

1952 2006 2069 2099 2091 2081

Water

Content

( % )

12.5 13.4 14.8 16.2 17.4 18.4

Example 1

In a BS compaction test the following data were collected:

a) Draw the graph of dry density against water content and

from it determine the maximum dry density and optimum

water contentb) On the same axes, draw the ρd/w curve for zero (0%) and

5% air voids, and hence determine the air-voids content at

the maximum dry density

Effect of increased compactive effortThe compactive effort will be greater when using a heavier

roller on site or a heavier rammer in the laboratory. With

greater compactive effort:

a) maximum dry density

increases

b) optimum water content

decreases

c) air-voids content remains

almost the same.



12/01/20

3

Effect of soil type

a) Well-graded granular soils can be

compacted to higher densities

than uniform or silty soils.

b) Clays of high plasticity may have

water contents over 30% and

achieve similar densities (and

therefore strengths) to those of

lower plasticity with water contents

below 20%.

c) As the % of fines and the plasticity

of a soil increases, the compaction

curve becomes flatter andtherefore less sensitive to moisture

content. Equally, the maximum dry

density will be relatively low.



2



3

0.001

Equivalent Particle Size (mm)

0

10

20

3040

50

60

70

8090

100

P e r c e t F i n e

200 140 100 70 50 40 30 20 16 12 8 6 4

ASTM SIEVE SIZES

B.S. SIEVE SIZES

300 200 150 100 72 52 36 25 18 14 10 7

0.002 0.006 0.01 0.1 10.02 0.06 0.2 0.6 2

MediumSilt

Fine Coarse Medium

SandFine Coarse Fine

Clay

1 8/ " 3

16/ "

topic 1. basic characteristic of soils

Documents