1_Combined Notes_Engineering Geology and Soil Mechanics_Sch_Chapter 1 & 2

7/25/2019 1_Combined Notes_Engineering Geology and Soil Mechanics_Sch_Chapter 1 & 2

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CBE2021 SOIL MECHANICS AND GEOLOGY

INTRODUCTION

Page 1 of 5

INTRODUCTION

1 Definition of Engineering Soil

Engineering soil is defined as an un-cemented aggregate of mineral grains and decayed

organic matter (solid particles) with liquid and gas in the empty spaces between the solidparticles. Engineering soil is used as a construction material in various civil engineering

projects, and it supports structural foundations.

2 Definition of Soil Mechanics, Soil Engineering and Geotechnical Engineering

Soil Mechanics: It is the branch of science that deals with the study of the physical properties

of soil and the behaviour of soil masses subjected to various types of forces.

Soil Engineering: It is the application of the principles of soil mechanics to practical problems.

Geotechnical Engineering: It is defined as the sub-discipline of civil engineering that involves

natural materials (soils and rocks) found close to the surface of the earth.

3 Geotechnical Engineering Problems

There are three basic problems:

(a) Stability Problem: it involves whether a structure will stand or collapse. A structure

may be: a building foundation

an earth slope (cut/fill slope or an embankment)

an earth retaining structure

a buried structure

There are short term and long term stability problems.

(b) Deformation Problem: it involves whether a structure performs satisfactorily even

without collapse.

building/structure settlements (magnitude and rate of settlement, differential

settlement) buried structure and road pavement deformation

(c) Water Flow Problem: it involves whether the movement of water in the soil causes

problems.

dam leakage

rate of settlement (consolidation)

dewatering excavations

landfill (contamination transport)

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CHAPTER 1

ENGINEERING GEOLOGY AND SOIL MECHANICS


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INTRODUCTION

Page 2 of 5

4 History of Soil Mechanics

Coulomb (1776) made one of the first attempts to solve the problem involving soil when

he examined the earth pressures acting on retaining walls. With some restrictions his

work is still in use.

Rankine (1862) contributed further to earth pressure theory when he attempted to examinethe stresses within a soil mass.

The Swedish Physicist, Atterberg, advanced the understanding of basic soil properties at

the beginning of the 20th Century.

Development of modern soil mechanics began from the publication of Soil Mechanics

by Dr. Karl Terzaghi in 1925. Many present-day theories stem from Terzaghis work.

5 Origins and Modes of Formation of Soils

All soils originate, in one way or the other, directly or indirectly, from solid rocks that areclassified according to their mode of formation as follows:

IGNEOUS ROCKS: rocks formed by the solidification of magma (hot molten material)

ejected (either by fissure eruption or volcanic eruption) from deep in the earths mantle.

Sometimes, the magma cools on the earths surface to form extrusive igneous rock. At

other times, the magma cools below the earths surface to form intrusive igneous rocks.

SEDIMENTARY ROCKS: rocks formed in layers from soil sediments derived from

weathering actions. The deposits of gravel, sand, silt and clay formed by weathering may

go through a lithification process and becomes sedimentary rocks.

METAMORPHIC ROCKS: rocks formed by changing the composition and texture of

existing rocks, without melting, by heat and/or pressure.

Soil is composed of loose, uncemented collections of mineral particles of various shapes and

sizes with relatively large void spaces between them or with a high value of ratio of void

space volume to solid particle.

Rock, however, is an intact, usually tough material, consisting of mineral particles and

crystals cemented or welded together. The void spaces in rock are normally small. Even

when they are large they occupy only a small part of the total volume. In engineering terms,soils may be considered as materials that can be worked without drilling and blasting.

Figure 1 shows diagrammatically how soil is derived from rock.

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CHAPTER 1



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INTRODUCTION

Page 3 of 5

Fig. 1.1 Soil formation process: Weathering, erosion, transportation and deposition.

Figure 1: Soil formation process: weathering, erosion, transportation & deposition

6 Some Factors Affecting the Formation of Solid Rocks to Soils

Nature and composition of parent rocks.

Climate conditions, particularly temperate and humidity

Topographic and general terrain conditions such as degree of shelter or exposure, density

and type of vegetation, etc.

Length of time the weathering processes prevail.

Interference with other agencies, e.g., earthquakes, cataclysmic storms, actions of man.

Modes and conditions of transportation.

Colluvium

Alluvium

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CHAPTER 1

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INTRODUCTION

Page 4 of 5

7 Some Engineering Soil Terminology

ROCK. Hard rigid coherent deposit forming part of the earths crust. Rocks require some

forms of blasting or drilling techniques to facilitate excavation.

SOIL. Engineering soils are materials that may be used in some ways in engineeringprocesses, i.e., worked on, worked in and worked with.

SUBSOIL. This is essentially an agriculture term describing an inert soil layer between the

topsoil layer and bedrock. Strictly speaking the use of this term shall be avoided in

engineering, but nowadays it has been used collectively to refer to soil layers beneath the top

soil.

ORGANIC SOIL. This is a mixture of mineral grains and decomposed organic material of

mainly vegetable origin. Most organic soils have their origins in lakes, bays, estuaries,

harbours and reservoirs. Soil with a smooth touch usually signifies presence of organicmaterials that may be characterized by a dark colour and unpleasant odour.

PEAT. Spongy, highly compressible and combustible soil made up of organic matters with

presence of inorganic materials. If percentages of organic matter are large, soils tend to

become organic.

RESIDUAL SOILS. Weathered down remains of rocks that have undergone no transport

and do not retain any structures of the parent rocks. They are usually sandy or gravely with

high concentrations of oxides resulting from leaching processes.

ALLUVIAL SOILS (ALLUVIUM). Materials such as sands and gravels deposited fromrivers and streams. They are usually well sorted and often occur in discontinuous and

irregular formations.

COLLUVIAL SOILS (COLLUVIUM). Materials such as deposits of sand and gravel

originated from the movement of weathered residual soils short distances down slope by

gravity, landslips and running water. They usually consist of angular to round blocks of rock

fragments varying in size from a few millimetres to several meters in a mixture of clayey,

silty matrix.

COHESIVE SOILS. Soils containing clay/silt particles which process some forms ofcohesion and plasticity.

COHESIONLESS SOILS. Soils such as sands and gravels consisting of non-flaky particles,

and which do not exhibit plasticity and cohesion.

DRIFTS. A geological term used to describe superficial unconsolidated deposits of recent

origin, such as alluvium, glacial deposits, wind blown sand, etc.

MECHANICAL WEATHERING. It is the disintegration of rock by purely physical means,

such as differential expansion and contraction of rock masses due to temperature changes.

This leads to a loosening of the coherent structure (block disintegration) and also cracking

(sun cracking). Cracks in rock may also be caused by the rebound effect as the upper rock

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CHAPTER 1



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INTRODUCTION

Page 5 of 5

layers are unloaded and stress relief occurs. This process is called sheeting and causes

horizontal cracks or joints to develop. In temperature and cold climates, the expansion of

water upon freezing also disintegrates the rock through a wedging action in the rocks cracks.

This is called frost weathering.

CHEMICAL WEATHERING. It is the action of water, atmospheric gases and organicchemicals from vegetation, which corrodes rock minerals by the processes of solution,

oxidation and hydrolysis to produce new minerals such as clay. These processes are

comparable to the rusting of steel.

9 Distribution of Soils in Hong Kong

Figure 2 shows grammatically typical distribution of soils in Hong Kong.

Fig. 2 Soil distribution in Hong Kong.

Bedrock

10 Rock Mass Weathering Profile in Hong Kong

In Hong Kong, with its generally wet semi-tropical climate, chemical weathering penetrates

very deep, sometimes to a depth of 60 meters. The rock profile is usually described by six

mass weathering zones as shown in Figs. 3a, 3band 3c.Page 5 of 8

CHAPTER 1



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CHAPTER 1


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Page 7 of 8

CHAPTER 1


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S.D.R.

M.D.R.

Page 8 of 8

CHAPTER 1


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CBE2021 SOIL MECHANICS AND GEOLOGY Dr. Paul Ho

PHASE RELATIONSHIPS Sept/2009

Page 1 of 12

PHASE RELATIONSHIPS

1. MASS VOLUME RELATIONSHIPS

A volume of soil consists of the volume making up the soil solids and the volume of the

voids or pores(empty spaces) between the solid particles.

2 Soil System

Soilis usually modelled as a THREE-PHASE material, consisting of:

Solidmineral particles (solid phase) -

quartz, feldspars, carbonates, mica / clay minerals, organic matter

Pore fluid (liquid phase) - normally water

Pore gas (gas phase) -

normally air

3 Phase Diagram

To quantify the properties of soil, it is useful to introduce some definitions and terminology to

describe the three-phase system (the Unit Solid Volume Model):

Volume Mass

Fig. 2.1 Unit Solid Volume Model

Go to Appendix 1 for explanation of symbols.

The soil is represented as a fixed volumeof soil solid grains associated with various

amounts of water and air. The soil solid grains are assumed to be incompressible; the volume in the soil not

occupied by soil solid grains is the voidspace.

In a perfectly drysoil there is no water and the void space is entirely air. In a saturated

CHAPTER 2

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Also refer to Table 1



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Page 12 of 12

12 Appendix 1: Symbol Used

Symbol Designation Relationship

Ms Mass of soil solid particles or grains in soil

Mw Mass of water in soil

MT Total mass of soil = Ms+ Mw

Ws Weight of soil solid particles or grains in soil

Ww Weight of water in soil

WT Total weight of soil = Ws+ Ww

Vs Volume of soil solid particles or grains in soil

Vw Volume of water in soil

Va Volume of air in soil

Vv Volume of void in soil = Vw+ Va

VT Total volume of soil = Vs+ Vv

e Void ratio = Vv/ Vs

w Moisture content = Mw/ Ms

n porosity = Vv/ VT

Av Air void ratio = Va/ VT

S Degree of saturation = Vw/ Vv

s Density of soil solid particles or grains

w Density of water = [1000 kg/m3or 1Mg/m

3]

b Bulk density of soil = MT/ VT

d Dry density of soil = Ms/ VT

sat Saturated density of soil = (Ms+ Vvw) / VT

subor Submerged density of soil = sat- w

w Unit weight of water = [9.81kN/m3]

b Bulk unit weight of soil = WT/ VT

d Dry unit weight of soil = Ws/ VT

sat Saturated unit weight of soil = (Ws+ Vvw) / VT

subor Submerged unit weight of soil = sat- w

Gs Specific gravity of soil solid particles or grains = s/ w

CHAPTER 2

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11

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o

igure

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Page 2 of 12

soil the void space is full of water. Between the two extremes, the soil is partially

saturated.

The model provides an easy means to identify what are known and the relationship

between known and desired (wanted) quantities.

The total volume VT, the mass of water MW, and the mass of soil solid grains MS are

usually measured, the rest of the values can be calculated.

It is useful to consider each phase individually as shown in Table 1.

Phase Volume Mass Weight

Air VA 0 0

Water VW MW WW

Solid VS MS WS

Table 1 Distribution by Volume, Mass, and Weight

4 Units

For most engineering applications the following units are used:

Length metres (m)

Mass kilogram (kg)

Density (mass/unit volume) kg/m3

Weight kilonewtons (kN)

Stress kilopascals (kPa) 1 kPa = 1 kN/m2

Unit Weight kN/m3

5 Volumetric Relationship

Void Ratio, e:

VV= volume of voids

VS= volume of soil solid grains

Expressed as a decimal.

May be larger than 1.

Typically:

Sands 0.4 < e < 1.0 very loose sand e 1.0

Clays 0.3 < 1.5 soft clay e > 1, organic clay e > 3

Much of the soil behaviour is related to e:

As e decreases densityincreases.As e decreases strengthincreases.

As e decreases permeabilitydecreases

S

V

V

Ve=

CHAPTER 2

Page 3 of 19

0.3 < e < 1.5

Figure 2.1

easy

NGINEERING GEOLOGY AND SOIL MECHANICS


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Porosity, n:

VV= Volume of voids

VT= Total volume

Expressed as a decimal or percentage (usually percentage).

Can notbe larger than 100% (1).

Relationship with e can be expressed as:

For example, for a very loose sand with e = 0.8,

(Note: When doing calculation involving %, you need to change the percentage values

back to decimal values. This applies to all other quantities.)

Degree of Saturation, S:

VW= volume of water

VV= volume of voids

Expressed as a percentage.

Tells the percentage of the total volume of voids that contain water.

Range is from 0% to 100%.

Sr= 0%, soil is completely dry.

Sr= 100% soil is fully saturated.

%)100x(V

Vn

T

V=

e1 en +=

n-1

ne =

44%100%)(x0.81

0.8n =+

=

100%xV

VS

V

W=

CHAPTER 2

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Page 4 of 12

Air Void Ratio, AV:


Tells the percentage of volume of air void relative to the total volume of soil.

6 Mass Relationship

Moisture Content, w:


Tells the amount of water in a soil relative to the mass of soil solid grains.

More important quantity for clay soils.

The higher the value usually the weaker the clay soil is.

Density, :

M = mass

V = volume

Expressed as kg/m3or Mg/m

3

Density of Soil Solid Grains, S:

S

Ss

V

M=

MS= mass of soil solid grains

VS= volume of soil solid grains

V

M=

(%)100xAT

A

V

VV =

100(%)xMMw

S

W=

CHAPTER 2

Page 5 of 19

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Page 5 of 12

Specific Gravity of Soil Solid Grains, GS:

W

SSG

=

Defined as the mass (weight) of the soil solid grains divided by the mass (weight)

of an equal volume of water at 20C.

Typically GS= 2.6 to 2.8 for the solid minerals in soil.

Can use it to calculate the mass of mass or volume of soil solids if either one is

known.

Density of Water, w:

MW= mass of water

VW= volume of water

Bulk Density of Soil(also known as moist, wet, or total density), b:

MT= total mass of soil

VT= total volume of soil

Dry Density of Soil, d:

S = 0% (i.e., MW= 0)

Saturated Density of Soil, sat:

S = 100%, (i.e., VA= 0 or Vw= Vvand therefore MT= MS+ VVx W)

Submerged (buoyant) Density of Soil, sub or:

sub(or ) = sat- w

3kg/m1000W

VW

M

W ==

T

T

V

Mb =

T

S

V

Md =

T

WS

T

T

V

VvM

V

M

sat

+

==

CHAPTER 2

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Page 6 of 12

7 Weight Relationships

Unit Weight, :

The relationships just defined in terms of masses (or densities) can be expressed interms of weights and are called unit weights. In most applications it is not the mass that isimportant, but the force due to the mass, and the weight, W, is related to the mass, M, by the

relation:

W = M g

where g is the acceleration due to gravity. g = 9.81 m/s2

Because the force is usually required it is often convenient in calculations to use the unit

weight, (weight per unit volume).

=W

V

=M g

V= g

Hence the unit weight of water,

w= 1000 kg/m3x 9.81 m/s

2 = 9810 (kg-m/s

2)/m

3= 9810 N/m

3= 9.81 kN/m

3

For soil, say if = 2100 kg/m3,

= 2100 kg/m3x 9.81 m/s

2= 20601 ((kg-m/s

2)/ m

3) = 20600 N/m

3= 20.6 kN/m

3

(Note: 1 kg-m/s2= 1 N (Newton))

CHAPTER 2

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8 Useful Mass-Volume Relationships

If one assumes the solid grains occupy 1m3 (or one unit volume, based on the unit

solid volume model), then the following relationships hold:

Phase Volume Mass

Air e (1 - S) 0

Water e S e S w

Solid 1 Gsw

Note that the above table assumes a solid volume Vs= 1 m3, All terms in the table should be multiplied by the

actual Vsif this is not the case.

Unit Solid Volume Model

(Make sure you can derive all the relationships by yourself by following the

procedures shown in the next page.)

Gsw

eSw

soil

air

water

1

eS

e

e(1-S)

Volume Mass

CHAPTER 2

Page 8 of 19

Mass

Volume x Density

Vs/Vs

Vv/Vs

ENGINEERING GEOLOGY AND SOIL MECHANICSReasons for using the model:1. d, b, e and Gsare known.

2. To express M and V in terms of

e, S, G and win a way to simplify

the expressions


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Page 8 of 12

VS= 1

VV= e (since: e = VV/ VS, and VS= 1)

VT= 1 + e (since VT= VS+ VV, VS=1, and VV= e)

VA= e (1 - S) (since S = VW/ VV, VW= VV- VA, and VV= e)

VW= e S (since S = VW/ VV, and VV= e)

VW= w GS (since VW= MW/ W, and MW= w MS= w GSW)

MA=0 (mass of air is negligible compared with other phases)

MS= GSW (since: GS= S/ W, S= MS/ VS, and VS= 1)

MW= wMS= wGSW (since: w = Mw/ Msand Ms= Gsw)

It then follows:

b = [(GS+ Sre) W] / (1 + e)

d = (GSW) / (1 + e)

sat = [(GS+ e) W] / (1 + e)

subor = [(GS+ e) / (1 + e) - 1] W

n = e / (1 + e)

e = n / (1 - n)

e = (wGs) / S

w = (S e) / GS

S = (w GS) / e

GS = (S e) / w

CHAPTER 2

Page 9 of 19



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Page 10 of 12

If the sample were saturated there would need to be an additional 14.9 10-6m3of water. This would have a

water mass of 0.0149 kg and thus the saturated density of the soil would be

3

6/1941

101.157

)0149.0290.0(mkg

sat =

+=

3/94110001941 mkgsub ==

Example 2 Calculation of Unit Weights

Note: This time you work with weight rather than mass.

A soil has a voids ratio of 0.7. Calculate the dry and saturated unit weight of the material. Assume that the solid

material occupies 1 m3

, then assuming Gs= 2.65 the distribution by volume and weight is as follows.

Phase Volume

(m3)

Dry Weight

(kN)

Saturated Weight

(kN)

Voids 0.7 0 0.7 9.81 = 6.87

Solids 1.0 2.65 9.81 = 26.0 26.0

Dry unit weight drykN

mkN m= =

26 0

17153

3

3.

.. /

Saturated unit weight sat kN m= +

=( . . )

.. /

26 0 687

1719 3 3

If the soil were fully saturated the moisture content would be

Moisture content %4.26264.00.26

87.6===w

CHAPTER 2

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Add Mw=Va x 1000

MwMs



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12 Further Worked Examples

CHAPTER 2

Page 12 of 19


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CHAPTER 2

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mw

ms


m1

m2

m3

m2-m1

m4


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CHAPTER 2


W1 W1 SW1+w

W1

S

w

sand

Mass of Soil

- Mass of water to fill jar with soil in it

Mass of water with vol. having the

e as the soil

M1

M2


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12 Further Worked ExamplesCHAPTER 2

(d, e, Gs)

s

A.

B.

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In case is

known

In case is

not known


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12 Further Worked ExamplesCHAPTER 2

Use the model (a)

when V is known.

Use s

Vv

Vs

Use s

Mt

Use s

Use the model (b)

when V is unknown.

Vs


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w 0.33

M, 2.68

M

w

= 0.33

x

2.68 0.884 g

S 0.884 = 0.997

r

0.887

M,+M

w

x

9.81

Vs V

13.93 kN/m

3

2 68 0 884

x

9.81 18.53 kN/m

3

I 0.887

2.68

d x

9.81

1.887

oil

4

n

e

0.85

0.85

1 0 85

V

v

0.85 x I 0.85 m

3

0.459

14.60 kN/m

3

Sr

= 0.90

: V

w

= 0.90

x

0.85 = 0.765

m

3

M

w

0.765 g

M, = 2.71 x I x I = 2.7 g

w

=

0.765

0.283

2.7

= 0 765 2 7 x 9.81 18.37 kN/m

3

1 0 85

2.7

d x 9.81

1.85

~

A

class is

a group

of

soils which may

be

related

to

one another

on the basis

o Nhe

grading of their constituent particles and the plasticity of

the fraction w h i c h ~ l t p a s s a 425

Lm

sieve.

The description

of

a soilwil ive detailed information about its grading

plasticity colour particle characte .

cs as well as its fabric and state of

bedding. Few soils

will

have identical

d e s r i p t i e n ~ e

purpose of classification

therefore

is

to place a soil in one

of

a limited

n u m o e ~ f

groups on the basis

of

the grading and plasticity

of

a disturbed sample.

S i n c e t n : e s ~ r a c t e r i s t i C S

are independent

of

the particular conditions

in

which a soil

o u r s ~ i v e s

a good guide to how the disturbed soil will behave

as

a construction m a t e ~

~ i f i c t i o n

a

soil ~ a

Explain the purpose

of

classifying a soil sample.

b)

Discuss the bases on which soil classification systems are devised.

SOLVING PROBLEMS IN SOIL MECHANICS


Page 17 of 19

Vw

Use soil model (b)

VsMw

Vs

CHAPTER 2

Vv

n=e/(1+e)

Mw+Ms

Vs+ Vv


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CHAPTER 2

Page 18 of 19

b, b, w, e, n, Sr, A

Refer to Page 9/19

VvVt

Vv-Vw

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IVE(TY) Chapter 2 Soil Phases and Index Properties Soil Mechanics & Geology

Soil Mechanics_Chapter 2_Class Practice_2014

Chapter 2 Soil Phases and Index Properties

Class Practice

Q1. A soil sample had a volume of 9.15x10-5

m3and weighed 153g. After completely

drying out in the oven, its weight was reduced to 136g. The specific gravity of

soil sample was found to be 2.65.

Determine the following physical properties of the soil sample:-

(i) Bulk density and unit weight (2 marks)

(ii) Dry density and unit weight (2 marks)

(iii) Moisture content (in %) (2 marks)

(iv) Saturated density (4 marks)

(v) Submerged density (1 mark)(vi) Air void (in %) (3 marks)

(vii) Void ratio (1 mark)

(viii) Porosity (1 mark)

(ix) Degree of saturation (in %) (1 mark)

soil particles

Gs=2.65

water

air

Mt=153gVt=9.15x10-5 m3

oven dried

soil particles

Gs=2.65

Mt=136gVt=9.15x10-5 m3

air


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IVE(TY) Chapter 2 Soil Phases and Index Properties Soil Mechanics & Geology

Soil Mechanics_Chapter 2_Class Practice_2014


Class Practice

Q2. A natural Hong Kong soil sample was collected on site and tested in laboratory.

The following data were measured:

Bulk density, b = 2.20Mg/m3

Water content, w = 9.2%

Specific gravity, Gs = 2.69

Calculate the below physical properties of the soil

(i) dry density (d), (3 marks)

(ii)

void ratio (e), (2 marks)(iii) porosity (n), (2 marks)

(iv) degree of saturation (Sr), (2 marks)

(v) air content (Av), (2 marks)

(vi) fully saturated unit weight (sat), (2 marks)

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HNC in Construction and the Built Environment (Civil Engineering)

Engineering Geology & Soil Mechanics

Assignment No.1

Page 3 of 5

3. Determine basic soil properties (P6)

(a) For laboratory testing, a sand sample collected has a total volume of 1.5 m3, a total mass of

2850 kg and a dry mass of 2478 kg. The specific gravity of the soil solids is 2.65. The

maximum and minimum void ratios are 0.8 and 0.4 respectively. You, are asked to

compute with detail calculation for the following:

(i) the mass of water

(ii) the water content

(iii) the volume of the soil solid

(iv) the volume of the water

(v) the volume of air

(vi)

the void ratio

(vii) the porosity

(viii) the air void ratio

(ix)

the degree of saturation

(x) the bulk density

(xi) the dry density

(xii)

the saturated density

(xiii) the submerged density

(xiv)

the corresponding water content if the soil is fully saturated

(b) A natural soil sample was retrieved from a Ground Investigation trial pit. It had a mass of

0.183 kg and a volume of 86.2 x 10-6m. After being completely dried in an oven, the dry

density dry of the soil sample was found to be 1830 kg/m. The specific gravity of the

soil grains was 2.72.

You are requested to determine the following physical properties for this natural soil

sample:

(i) bulk density, bulk

(ii) moisture content, m or w

(iii) void ratio, e

(iv) porosity, n

(v) degree of saturation, Sr

(vi) saturated density, sat


Class Practice

Q3.


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Student Name:______________________

CPE - Engineering Geology and Soil Mechanics


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Student Name:______________________

CPE - Engineering Geology and Soil Mechanics


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11 Home Exercises:

(1) Dry soil and water are to be used to reconstitute a soil to a total volume (VT) of

2.356 x 10-3

m3. The water content (w) and air void ratio (AV) are to be 17%

and 10% respectively. Assuming the specific gravity (GS

) of the soil solidgrains (GS) is 2.7 and the density of water (W) is 1000 kg/m3, determine, for

the reconstituted soil, the following:

(a) the masses of soil solids (MS), water (MW) and air (MA);

(b) the volumes of soil solids (VS), water (VW) and air (VA);

(c) the void ratio (e), porosity (n), and degree of saturation (Sr);

(d) the dry density (d), bulk density (b), saturated density (sat), and

submerged density (sat).

(Hint: Begin by assuming the volume of soil solid is 1 m3and then calculate therelative volume and mass distributions based on the phase diagram.

Then determine the actual volumes and masses using the given VT of

2.356 x 10-3

m3.)

(2) The bulk density bof a soil is 1850 kg/m3. Given that w = 15% and GS= 2.7,

determine:

(i) the dry density, d

(ii)

the porosity, n

(iii)

the degree of saturation, Sr

(iv) the mass of water, in kg, to be added to the soil to reach 100% saturation

(3) Assume a Gs of 2.65, determine the moisture content (w) and the porosity (n)

of a soil sample that has a total volume (VT) of 0.645 x 10-3

m3and a dry mass

(MS) of 1.050 kg, assuming the sample is 90% saturated.

(4) (a) In order to determine the specific gravity Gs of the soil solids in a soil sample,the following data were recorded in a specific gravity test:

Mass of density bottle 0.040 kg

Mass of density bottle + dry soil 0.485 kg

Mass of density bottle + dry soil + water 0.946 kg

Mass of density bottle +water 0.674 kg

(b) Calculate the moisture content (m) and the porosity (n) of the soil sample in (a)

if its total volume was 0.645 x 10-3 m3 and its dry mass (after oven drying) was 1.050kg) and mass (before oven drying) was 1.4 kg.

CHAPTER 2

Ans: 1609 kg/m3Ans: 0.404

Ans: 59.7%

Ans.: 163.1kg per 1m3 of soil sample

Ans: 1.453x10-3, 6.67x 10-4,

2.356x10-4m3

Ans: 3.923, 0.67, 0 kg

Ans:0.62, 0.38 & 0.74

Ans: 1665.2, 1949.5, 2048.2, 1048.2kg/m3

Ans: n=0.386 w=21.3%

Ans.: G=2.57

Ans.: m=33.3%, n=0.367


1_Combined Notes_Engineering Geology and Soil Mechanics_Sch_Chapter 1 & 2

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Transcript of 1_Combined Notes_Engineering Geology and Soil Mechanics_Sch_Chapter 1 & 2