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    1111.... SOIL STRUCTURESOIL STRUCTURESOIL STRUCTURESOIL STRUCTURE

    The structure of the soil represents one the important proprieties of the mineral grains. Soil

    structure is responsible for the integrity of the system and for the response to externally applied

    and internally induced sets of the forces and fluids.

    Soil structure can be defined as the property of soil, which provides the geometricarrangement of the particles or mineral grains and the antiparticle forces which may act upon

    them (Figure 2.2. a, b, c, d).

    a) Grained

    structure b) Honeycomb structure c) Floccules structure

    2 SOIL TEXTURE2 SOIL TEXTURE2 SOIL TEXTURE2 SOIL TEXTURE

    Soil texture may be defined as the visual appearance of a soil based on a qualitative

    composition of soil grain sizes in a given soil mass.

    The relative sizes and shapes of particles, and their distribution, define the soil texture.

    Soil texture is used for the classification of soils based on a visual grain description and

    connection of the particles, which compose them (cohesionless or cohesive).

    Soils are divided into coarse grained and fine grained soils on the basis of their texture,and the dividing reference size is that which is visible to the naked eye (about 0.05 mm).

    Sands and gravels, in this respect, appear to be coarse textured.

    a) homogeneous b) inhomogeneous

    Figure 2.3 Texture of the cohesionless soil

    Gravels and sands are coarse materials, whereas silts and clays are termed fine-grained

    materials. These are cohesionless materials and the grain size distribution is a relevant parameter.

    Silt and clays appear as fine textured soils, because they are composed of particles that are

    invisible to the naked eye.

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    3. GRAIN SIZE DISTRIBUTION3. GRAIN SIZE DISTRIBUTION3. GRAIN SIZE DISTRIBUTION3. GRAIN SIZE DISTRIBUTION

    The solid phase of soil is an inhomogeneous material consisting of three different phases.

    To properly classify a soil one must know the grain size distribution on it.

    To obtain the grain size distribution of a soil in laboratory such methods are used:

    The sieve analysis of coarse grained soil: The hydrometer analysis for fine grained soil.This method is used to determine the relation between percentages of the grain size fraction

    distribution.

    4. Hydrometer AnalysisThis laboratory method is based on the principle of sedimentation of soil particles in water.

    The use an immersion hydrometer to measure the specific weight of the liquid is well known.

    The principle can be extended to the measurement of the varying specific weight of a soil

    suspension as the grains settle, thereby determining the grain size distribution diagram.

    The diameter of the soil particle still in suspension at time t can be determined by Stoke`s law[

    ] .

    The ternary diagram is useful to identify the soil, giving a name to each of them and to classify

    the soils.The finer percentage in each sieve determined by a sieve analysis is plotted on semilogarithmic

    graph paper as shown in Figure 2.6.

    The grain diameter, d, is plotted on the logarithmic scale, and the finer percentage is plotted on

    the arithmetic scale.

    From the grain size distribution curves of coarse grained soil two parameters can be

    determined:

    The uniformity coefficient (Un or Cu)

    %10

    %60

    d

    dU

    n= (2.1.)

    The coefficient of gradation (Cv or Cz)

    ( ) ( )%10%60

    2

    v

    %30

    dd

    dC

    = (2.2.)

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    Where %10d , %30d and %60d are the diameters corresponding to percentages finer than 10,

    30 and 60 respectively. The ternary diagram is useful to identify the soil, giving a name to each

    of them, and to classify the soils.

    2.4.1 Definition geotechnical physical indices

    In nature, soils are three phase systems consisting of solid soil particles, water and air (or

    gas). In order to develop the weight relationships for a soil, the three phases can be separated as

    shown in Figure. 2.9.

    Based on this, the volume relationships can be defined as follows:

    To determine the physical properties of soils we have to know three simplest characteristics:

    specific weight, s , of solid particles of the soil; unit weight, , of the soil of natural structure; natural moisture content or water content w, of the soil. the specific weight, s , can be defined as follows:

    S

    Ss

    V

    W= (2.3.)

    where:

    WS the weight of the solid particles;

    VS specific volume or volume of soil solids.

    unit weight, , of the soil of natural structure , can be defined as follows:V

    W= (2.4)

    where:

    W = Ws + Ww + Wa total weight of a soil specimen;

    V = Vs + Vv total volume of soil.

    the moisture content, w, can be defined as follows:100=

    s

    w

    W

    Ww (2.5)

    where:

    wW - weight of water;

    sW - weight of the soil solids.

    The weight relationships are moisture content, moist unit weight, dry unit weight and

    saturated unit weight.

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    More useful relations can now be developed by considering a representative soil specimen in

    which the volume of soil solids is equal to unity.

    Porosity, n, is ratio of the volume of voids to the volume of the soil specimen, or

    V

    Vn v= [%] (2.6)

    where:

    V = total volume of soil.

    vV = volume of voids;

    sV = volume of solids soil.

    Void Ratio, e, is the ratio of the volume of voids to the volume of solids soil in a givensoil mass and can be written as,

    v

    s

    Ve

    V= (2.7)

    where:It can also be seen that:

    e1

    e

    V

    V

    V

    V

    V

    V

    VV

    V

    V

    Vn

    s

    v

    s

    s

    s

    v

    vs

    vv

    +=

    +

    =+

    == (2.8)

    Degree of Saturation, S, is the ratio of the volume of water in the void spaces to thevolume of voids, and it is generally expressed as a percentage. So:

    v

    w

    rV

    VS = (2.9)

    where: wV = volume of water and vV - total volume

    2.4.2 General Relations for Calculus Relationships

    the natural unit weight:( )

    e1

    w1G

    VV

    WW

    V

    W ws

    vs

    wo

    +

    +=

    +

    +== (2.10)

    the dry unit weight:w1

    e1

    G

    VV

    W

    V

    W

    ws

    vS

    ss

    d +=+=+==(2.11)

    The saturated unit weight of soil can now be determined as:

    e1

    eG

    VV

    WW wws

    vs

    ws

    sat+

    +=

    +

    += (2.12)

    The unit weights:

    G g 9 .8 1 G= = where: G density of soils [g/cm3]

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    g gravitational acceleration=9,81 [m/s2]

    ( )n1 sd = (2.13)

    ( ) ( ) ( )w1n1w1 sd +=+= (2.14)

    ( ) ( )n1 ws = (2.15)

    ( ) wnn1n swdsat +=+= (2.16)

    6. CLASSIFICATION OF CLAY MINERALSCLASSIFICATION OF CLAY MINERALSCLASSIFICATION OF CLAY MINERALSCLASSIFICATION OF CLAY MINERALS

    The clay minerals are commonly found in soils belong to the larger mineral family termed

    PHYLLOSILICATES, which also contains other layer silicates. The clay minerals usually occur

    in small particle size.

    The two basic units in the clay minerals structures are the SILICA TETRAHEDRON, with a

    tetrahedral silicon ion coordinated with four oxygen atoms, and the ALUMINIUM orMAGNESIUM OCTAHEDRON, wherein aluminum or a magnesium ion is octahedral

    coordinated with six oxygen atoms or hydroxyls.

    Table 2.2 The properties of some clay minerals

    KAOLINITE is composed of alternating silica and octahedral sheets. The tips of silica

    tetrahedral and one of the planes of atoms in the octahedral sheet are common.

    HALLOSYTEis a particularly interesting member of the kaolinite subgroup.

    Two distinct forms of this mineral exist: one a no hydrated form having the same structural

    composition as kaolinite and the other a hydrated form consisting of unit kaolinite layers

    separated from each other by a single layer of water molecules.

    ILLITE perhaps the most commonly occurring clay mineral found in the soil encountered in

    engineering practice has a structural similar to that of mica and is termed illite or hydrous

    mica. The basic structural unit for illite is the three-layer silica-gibbsite-silica sandwich that

    forms pyrophyllite.

    MONTMORILLONITE has a structural consisting of an octahedral sheet sandwiched

    between two silica sheets.

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    7. THE FORMS OF WATER IN SOILTHE FORMS OF WATER IN SOILTHE FORMS OF WATER IN SOILTHE FORMS OF WATER IN SOIL The first criterion refers to the state of aggregation of water .Using this criterion we can

    distinguish the following forms of water:

    - fluid water (liquid)- solid water (ice)- gaseous water (water vapors)

    Another criterion is given by the forces which on the water molecules namely the natureof the force fields which act on these molecules.

    The structural form of water molecule is:

    - chemical bound water in two forms: hydratation water (constitution water, crystallizationwater) and zeolitic water;

    - physical bound water in two forms: hydrocolloid water and gravitational water;- free water in two forms: - capillary water(surface tension) and gravitational water.

    hygroscopic water

    water vapors

    coat water

    contact capillary water

    capillary water

    ground water

    observation pipe

    Figure 2.11 Forms of water in soil

    The physical bound water is due to electro molecular forces that develop between the soils

    mineral grains and the water molecules called also hydrocoloidal forces.

    These forces are the result of the interaction between the solid grains and the water

    molecules and give peculiar properties to the soil.

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    The forces which act on the capillary water are the gravity and the surface tension, forces

    that in a certain moment can balance each other.

    The gravitational water is influenced by the forces of gravity.

    The chemical bound water may be:- water of hydration (water of constitution and water of crystallization)- zeolitic water.Clay particles in soils are always hydrated. These are surrounded by layers of water

    molecules called adsorbed water. The layer is at least two molecules thick and is called the

    diffuse layer, the double diffuse layer or the double layer.

    This presence of adsorbed water has as result that clayey soils have several characteristics

    reflected in general clay mineral properties.

    Clay particles interact through the layers of adsorbed water through the diffuse layers of

    exchangeable cautions in some cases through direct particle contact.

    10. CHACHACHACHARACTERISTICS STATES OF COHESION SOILSRACTERISTICS STATES OF COHESION SOILSRACTERISTICS STATES OF COHESION SOILSRACTERISTICS STATES OF COHESION SOILS ---- ATTERBERGSATTERBERGSATTERBERGSATTERBERGS

    LIMITS.LIMITS.LIMITS.LIMITS.

    To characterize a clayey soil from this physical point of view, we use following property

    indices:

    Liquid limit wL, plastic limit wp and shrinkage limit ws, called also Atterbergs limits or

    characteristic humidities.

    The liquid limit (wL) is the water content about which the soil behaves as a viscous liquid (a soil,

    water mixture with no measurable shear strength).

    The plastic limit (wp) is the water content below which the soil no longer behaves as a plastic

    material and when it is worked crumples.

    Shrinkage limit (ws) is the water content defined as the humidity below which no further soil

    volume change occurs with . drying.

    The behavior of cohesion soils depends on its mineral composition the water content, the degree

    of saturation and its structure.

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    Using this method the upper plastic limit can be defined as being the humidity at which a

    slot made in the soil paste in the bowl of the device is closed on a length of 12 mm after 25 drops

    of the cup, from a height of 10 mm with a frequency of 120 drops

    per min

    11. CAPILLARITY SOIL AND CAPILLARY EFFECTSCAPILLARITY SOIL AND CAPILLARY EFFECTSCAPILLARITY SOIL AND CAPILLARY EFFECTSCAPILLARITY SOIL AND CAPILLARY EFFECTS

    2.9.1 Definition surfaces tension.

    We suppose that the capillary menisci are an elastic membrane. If on its two faces act the

    same forces, its shape its plane. If the external pressing, ep is bigger than the internal pressing,

    ip with u , the menisci tensions ST appear (figure 2.16 a, b).

    The capillary action is attributed to electro mechanical forces existing between the water

    molecules. If the adhesion forces between a liquid and any other material are larger than the

    intermolecular attraction of the liquid, the surface of the dissimilar material will be wetted by

    liquid (Figure 2.16).

    T i T i

    p e

    p e = p i

    r

    p i= p e * u

    E l a s t i c

    e m b r a n e

    T s T s

    - T s -T s

    u T a

    T vp e

    a )

    b )

    Figure 2.16 The phenomenon of the surface tension

    Any quantity of liquid will behave as the surface of a tightly stretched skin due to the

    intermolecular attractive forces in the interior. This phenomenon is termed surface tension. Since

    surface tension is a material property of liquids and depends on intermolecular attraction it will

    be temperature dependent.

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    2.9.2 Determination capillary height

    In the capillarity rise in a tube is considered (Figure 2.17) there is an atmospheric pressure. It

    is well-known fact that when capillary tube is placed in water, the water level in the tube rises.

    This is caused by the surface tension effect.When in a hallow open ended tube is inserted into a container of liquid and if the liquid wets

    the contact surface it will climb the inside walls of the tube because of surface tension. The water

    in the capillary tube is submitted to stresses.

    r

    T

    r

    T uSSu

    cos**2*2

    1

    ==

    (2.28)

    TsTv

    Th

    hc

    hc

    d

    TsTv

    Th

    Ts

    r1 r1

    r r

    u

    2r

    r

    R=0

    r

    R? 0

    Figure

    2.17 Capillary tube and capillary menisci

    The weight of the capillary water is transmitted to the skeleton so that its weight increases

    the weight of the skeleton with the weight of the water comprised in the capillary rise. Thus

    results the capillary pressure.

    12 DARCY LAWDARCY LAWDARCY LAWDARCY LAW

    If the adhesion forces between a liquid and any other material are larger than the

    intermolecular attraction of the liquid the surface of the dissimilar material will be wetted by

    liquid.

    In 1856, Darcy established this conclusion:

    The flow on the unit surface is function of the height of the two examined points.

    Darcy found that the macroscopic flow velocity was proportional to the hydraulic head or

    hydraulic gradient.

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    h1

    h2

    dh

    l

    1 - sample

    23

    Porousstone

    Figure 2.19

    Darcy law proposed the following equation for calculating the velocity of flow of water

    through a soil:

    ikx

    hk

    tA

    Qv =

    =

    = (2.35)

    where: i hydraulic gradient is measure of the resistance of the soil to flow of water and

    has the dimension of velocity;

    v - velocity (cm/sec)

    k coefficient of permeability of soil (cm/sec)

    The hydraulic gradient (i) may be defined as:

    i =

    (2.36)

    where: h - piezometric head difference between the sections at AA and BB

    L - distance between the sections AA and BB;

    Figure 2.20. Geometrical significance of the Darcys law

    These sections are perpendicular to the direction of flow. [Note: Eg. (2.36)]

    v

    iii=0 is

    v=ki

    flow

    const. turbulent

    application area of

    Darcy's law

    a) sand

    v

    iii isi00

    v=k1(i-i0)

    v=kzi

    modified applicationarea of Darcy's law

    Application area of

    Darcy's law

    a) clay

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    Figure 2.21 Graphic representation of Darcys law

    Darcys law is valid for a wide range of soil types. The coefficient of permeability it can

    be determined in the laboratory by means of constant head or variable head permeability

    tests.[46], [48].The value of this coefficient varies greatly.The value Coefficient of permeability Table 2.8

    Type of soilCoefficient of permeability, k

    (cm/sec)

    Medium to coarse level Greater than110

    Coarse to fine sand 110 to 310

    Fine sand, silt sand 310 to510

    Silt, clayey silt, silt clay 410 to 610

    Clays 710 or less

    13131313COMPRESSION AND CONSOLIDATION DEFINITION PHENOMENADEFINITION PHENOMENADEFINITION PHENOMENADEFINITION PHENOMENA

    Under load application due to buildings or other kinds of superstructures, physical

    deformations of the subsoil will occur.

    The nature and amount of deformation occurring is a function of not only the applied load but

    also of the soil properties and time.

    a) stiff structure b) elastic structureThe mechanisms will be evoked the deformations of soils are:- volume changes due to the extrusion of pore air and pour water;- shear distortions producing particle and fabric unit displacement or yield phenomena with

    or without measurable pore water extrusion and with development of slip planes.

    Compressibility describes the volumetric response behaviour of soil mass and recent and

    common usage of the term has restricted it to describing behaviour characteristics under

    compression.

    The change in volume of a soil mass with time due to the extrusion of pore water is said to

    be a process of consolidation

    The overall load-volume change performance is identified as a stress - strain -time

    phenomenon and can be called rheological behaviour.

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    0 p

    e

    e1

    e2

    p1 p2

    (1)

    (2)

    p

    e

    logp

    e

    av = - e p = - tg

    a b

    c

    logp2logp1

    ei

    e

    e

    log (logp)

    log( p+ i)

    Cc = - tg =- e (logp)

    (a)e - p

    (b)

    e - log p

    0

    14THE ESTABLISHING OF COMPRESSIBILITHE ESTABLISHING OF COMPRESSIBILITHE ESTABLISHING OF COMPRESSIBILITHE ESTABLISHING OF COMPRESSIBILITY "IN SITU" WITH PLATETY "IN SITU" WITH PLATETY "IN SITU" WITH PLATETY "IN SITU" WITH PLATEballast

    rod

    jack

    loading platebench mark rods

    Df

    D

    Figure 3.6 Loading plate

    The test results are plotted in a diagram which contains:

    - The variation of clear pressure (pn) functions of time;

    - The variation of settlement (s) functions of time;

    - The variation of settlement (s) functions of clear pressure.

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    p l0

    s l

    p

    s

    p i0

    si

    p

    s

    pi-1 pi+1

    si-1

    si+1

    = = / = = /

    Figure 3.7 - The establishing of the limit pressure of proportionality.

    With this diagram, we establish the limit pressure of proportionality (p1) until there is

    proportionality betweenp and s.

    n ef f p p D= (3.16)

    - rigidity factorl

    S

    l

    pk tg

    s= = (3.17)

    iS

    i

    pk tg

    s= = (3.18)

    1 11.5( )

    i i i is s s s+ > (3.19)

    - linear deformation module2

    2

    s

    (1 ) DE=k D(1 ) l

    l

    p

    s

    = (3.20)

    = form factor

    15-16-17. 3.1.2 The compression testsThe tests that characterise the soil deformations are:

    - Laboratory compressibility tests;- In situ tests (with plate).3.1.2.1 Laboratory testsA one-dimensional consolidation process can be simulated in laboratory by compressing a

    soil specimen in a special testing apparatus called oedometer or consolidometer.

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    Porous stone

    Soil specimen

    Brass ring

    Porous stone

    Water level

    Load

    Dial gauge

    Figure 3.2 Oedometer

    This apparatus models the behavior of a soil volume at a certain depth beyond the axis of a

    foundation.

    The load is applied step by step and after the application of a certain load one waits until the

    deformation due to this load stops.

    The result of odometer test is plotted in compression - settlement curve. Compressibility

    tests of are soils are carried out in devices with rigid walls (oedometers) in order to ensure that

    the soil is compacted in one direction only (to prevent lateral expansion).

    The relationship between moisture content and pressure can be represented in the diagram

    which is called the compression curve (Figure 3.3).

    0

    p

    p

    p

    p1 p2 logp(daN/cm )

    1

    2

    M=tg p

    2p

    Figure 3.3 Compression-settlement curve

    The mechanism will be evoked the deformations of soils are:

    - volume changes due to the extrusion of pore air pour water- shear distortions

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    2 1

    p2 p1

    pM=

    p p

    h

    h

    =

    [kPa] (3.1)

    The compression curve can be easily reconstructed into void ratio pressure coordinates

    (figure 3.4).Result the compression void ratio curves (e p and e log p).

    0 p

    e

    e1

    e2

    p1 p2

    (1)

    (2)

    p

    e

    logp

    e

    av = - e p = - tg

    a b

    c

    logp2logp1

    ei

    e

    e

    log (logp)

    log( p+ i)

    Cc = - tg =- e (logp)

    (a)e - p

    (b)

    e - log p

    0

    Figure 3.4Compression - void ratio curve18.Yield and Field Criterion3.2.1.The General Introduction

    By applying the Mohr Coulomb failure criteria one can determine the shear strength of a

    soil (s).

    The effective stress is calculated with the equation:

    ' tans c= + (3. 36)where:

    - effective normal stress on plane of shearing;

    c - cohesion(or apparent cohesion);

    - internal friction angle.For the most day to day work, the shear strength parameters of a soil are determined by

    standard laboratory tests: the direct shear test, the triaxial test,the unconfined compression test.

    The normal and shear stresses at failure can be determined as:

    =

    and = s =

    (3. 37)

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    where:

    A = area of the failure plane soilFor sand

    for s tg = (3. 38)

    clayfor s tg c = +

    dz

    ds

    90-

    dx

    dx

    dz

    dzds

    ds

    M

    o +

    convention of sign

    xz

    z

    f

    zx

    z

    Figure 3.13 Stress state in the soil mass

    The locus of all (, ) referring to all plans passing through L point is Mohrs circle.Several tests of this type can be conducted by varying the normal load. The angle of friction

    of the sand can be determined by plotting a graph of s vs. sigma. As shown in Figure..

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    s(or )

    o

    P

    o

    90-o

    s(or ')

    ''1

    P2

    '3o

    '

    90-o

    u u( 1- 3)=( '1- '3)

    '3

    3

    1

    '1

    Figure 3.14 Mohrs circle

    If we have to know the main directions knowing the efforts, we plot the Mohrs circle: Aand B points have these efforts as co ordinates. Then we plot point by drawing parallels to the

    co ordinate.

    (s)

    1

    P

    303 1

    xz

    xz

    zx

    zx

    z

    x

    x

    z

    B( )zxz

    A xzx

    3

    3

    1

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    Figure 3.15 The establishing of the principal stresses

    This theory may be applied also to the case spatial state of stresses.

    o

    B

    C

    R( )

    23 1

    1

    3

    1

    1

    3

    *

    Figure 3.16 Mohrs representation of stresses in three dimensional systems

    For sands , the angle of friction usually ranges from 260 to 450 . It increases with relative

    density of compaction .The approximate range of relative density of compaction and the

    corresponding range of the angle of friction for various sands are given in Table 2.4,chapter two.

    Mohr envelopes are often curves.

    O

    Circle of

    breach

    Figure 3.17 Mohrs representation of stresses in the state of rupture

    The analytical form of the failure criterion can be expressed in several manners (forms).- Skempton, Bishop(gives straight line);- Rendulic, Henkel(gives straight line);Coulomb expressed in its most general form the failure surface:

    - The total stress failure envelope can be defined by the equation:

    20. THE DIRECT SHEAR TESTTHE DIRECT SHEAR TESTTHE DIRECT SHEAR TESTTHE DIRECT SHEAR TEST

    This test is simple on principle. The soil specimen is encased in a shear box (Figure 3.18).

    This has been divided into two halves, in order to allow relative movements. After a normal load

    is first applied to the specimen, one half of the box is pushed horizontally, and the other half

    remains fixed. This way the vertical and the horizontal displacements can be measured.

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    Normal stress

    s= Shearstress

    s4s3s

    2

    s1

    1 2 3 4

    ( )s=c+ tanN

    RT

    (a) (b)

    Figure 3.18 Schematic diagram of the direct shear test

    The major disadvantage of this test is that it is impossible to control the drainage. The state

    of tress cannot be completely determined because the only stresses that are known are the normal

    and the shear stresses on the horizontal plane.

    21. THE UNCONFINED COMPRESSITHE UNCONFINED COMPRESSITHE UNCONFINED COMPRESSITHE UNCONFINED COMPRESSION TESTON TESTON TESTON TEST

    The unconfined compression test is a special type of unconsolidated-undrained test. In this test

    , the major principal total stress is applied vertical to cause failure (Figure 3.21 ,a ) and stresses

    = = 0.

    The corresponding Mohrs circle is shown in (Figure 3.21, c)

    Normal

    stress

    Shear

    s() stress

    3=0sr Degree ofsaturation

    Unconfined compressive

    qu strength

    1=f=qu

    (b) (c)(a)

    f

    =qu

    =1

    Figure 3.21 Unconfined compression test

    a) soil specimen b)Variation of qu with the degree of saturation

    c) Mohrs circle for the test;

    The shear strength of saturated clays can be given as:

    s = c =

    for = 0

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    where: q= the axial stress at failure

    s = shear strength

    c= cohesion force

    The unconfined compression strength can be used in the specifications the consistency of

    clays in the table below (Table 3.1)Table 3.1

    Standard

    penetration

    number N

    Consistency of clays

    Unconfined compression

    strength

    q[kN

    m]

    0 - 2 Very soft 0 25

    2 - 5 Soft 25 50

    5 - 1 Medium soft 50 100

    10 20 Stiff 100 20020 - 30 Very stiff 200 400

    >30 Hard >400

    As an indicator to relate the consistency of clays. This is shows in Table on page.

    Are sometimes conducted on unsaturated soils. With the void ratio of a soil specimen remaining

    constant, the unconfined compression strength rapidles de creases with degree of saturation

    (Fig.).

    22. THE PRINCIPLE OF TRIAXIAL TEST:THE PRINCIPLE OF TRIAXIAL TEST:THE PRINCIPLE OF TRIAXIAL TEST:THE PRINCIPLE OF TRIAXIAL TEST:

    By referring to a triaxial test, the strength parameters can be obtained by means of:

    Consolidated Drained test (C.D.) Consolidated Undrained test(C.U.) Unconsolidated Undrained test(U.U.)

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    1

    3

    3

    1

    Lucitechamber

    Chamber

    fluid

    Porous

    stone

    Piston

    Porous

    stone

    Rubber

    membrane

    Soil

    specimen

    Chamber

    fluidTo drainage and/or pore

    water pressure device

    Base

    plate

    a) b) Figure 3.19 Schematic diagram of triaxial test equipmentFor clays , three main types of test can be conducted by means of triaxial equipment:

    Consolidated drained tests:

    Normalstress

    Shears stress

    '3 3 '1 1'3

    '1

    c'

    '

    Figure 3.20 a) Consolidated drained test

    The shear strength parameters ( ) can now be determined by plotting Mohrs circle

    at failure as shown in Figure 3.20 (a).

    + = = (3. 40)

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    = - major principal effective stress

    = - minor principal effective stressConsolidated Undrained Testes:

    The total stress Mohrs circles of test this type and effective stress can be presented in Figure

    3.20, (b).

    '

    Normal stress

    Shears stress

    '3 3 '1 1

    Effective stress

    failure envelope

    Figure 3.20 (b) Consolidated undrained test

    = + = ( + ) (3. 41)

    =

    Normalstress

    Shear

    s stress

    '3 3 '1 1

    ccu

    cu

    Total stress

    failure envelope

    Figure 3.20 (c) Unconsolidated Undrained testwhere: ( ) are the shear strength parameters from the consolidated undrained

    cohesion. Kenny(1959) has given a correlation between the friction angle , and the plasticity

    index (PI) of normally consolidated clays. This correlation is shown in Figure 3.20, (c).

    The Unconsolidated Undrained Test:The total stress Mohrs circle at failure can be determined with the relation:

    + - the major principal total stress

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    - the minor principal total stress

    when :

    = = constant =0 and Coulombs line is horizontal

    The shear stress for this condition can be given as:

    = =

    (3. 42)

    Normalstress

    Shears stress

    '3 3 '1 1

    Total stress

    failure envelope

    ( =0 )0

    s=cu

    3

    1

    Figure 3.20 (d) Unconsolidated Undrained Ttest

    24 STRESSES IN SOILSSTRESSES IN SOILSSTRESSES IN SOILSSTRESSES IN SOILS

    4.1 State of stresses in half spaceBoussinesq, in 1885, developed the mathematical relationships for the determination of the

    normal and shear stresses at any points. It is considered that the medium is homogeneous elastic,

    and isotropic and due to concentrated point load located at the surface.

    In figure 4.1 are according to his analysis for a vertical punctual load (at point M).

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    zxxz

    xz

    yxzy

    xy

    x

    z

    y

    x

    y

    z

    y

    x

    z

    a) b) Figure 4.1 -

    State of stresses in half-space vertical punctual load

    4.1.1 Vertical punctual load

    The stresses in a point of the half space are:

    ( )

    ( )

    ( )

    ( )

    ( )

    ( )

    22

    25 33

    22

    25 33

    23 1 2 1

    2 3

    23 1 2 1

    2 3

    x

    y

    R z xP x y z

    R R R z RR z R

    R z yP y z z

    R R R z RR z R

    + = +

    + +

    + = +

    + +

    (4.1)

    3

    55 22 2

    3 3 1

    2 2

    1

    z

    P z P

    R zr

    z

    = =

    +

    2

    5

    2

    5

    3

    2

    3

    2

    xz zx

    yz zy

    P xz

    R

    P yz

    R

    = =

    = =

    ( )

    ( )25 3

    23 1 2

    2 3xy yx

    R z xyP xyz

    R R z R

    += =

    +

    It can be determined by coordinates z, and r:

    xr

    zR M(x,y,0)

    P

    M(x,y,z)

    y

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    2 2r x y= +and

    2 2 2 2 2 2R x y z r z= + + = +

    where: P = vertical punctual load and

    x, y, z = coordinates of the point M.(figure 4.1,c)

    Figure 4.1 (c) State of stresses in half space vertical punctual load

    We have from the first line of relation:

    2z

    Pk

    z = (4.2)

    where:5

    2 2

    3

    2 1

    rK f

    zr

    z

    = =

    +

    (4.3)

    The value of the coefficient K gives in Table 4.1.

    Table 4.1

    Figure 4.2Variation stresses for vertical load

    z

    x

    y

    P1

    P2

    P3

    0r1 r2

    r3

    z

    z

    Figure 4.3 Verticals pressures distribution function of depth

    If a number of concentrated forces Pi (fig 4.3) are applied to the surface of soil, then the

    compressive stress in any point of soil for the horizontal planes parallel to the boundary plane

    can be found by soil summation.

    31 21 2 32 2 2z

    PP PK K K

    z z z = + +

    or (4.4)

    21

    1 n

    z i i

    i

    K Pz

    =

    =

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    Where the coefficients iK are found from the tables for the corresponding ratios irfz

    .

    4.1.2 The horizontal force

    In the case that half-space is loaded in its origin by a horizontal force, (Fig. 4.4) the stress z

    , may be computed whit the expression:2

    5

    3

    2z

    xzQ

    R

    = (4.5)

    x

    y

    z

    R

    0Q

    M

    Figure 4.4 Horizontal punctual forces on half space.

    25. 4.1.3 DISTRIBUT4.1.3 DISTRIBUT4.1.3 DISTRIBUT4.1.3 DISTRIBUTION LOADION LOADION LOADION LOAD

    The elementary stress, for the infinitesimal surface (d d) will be:

    ( ) ( )

    ( ) ( )

    3 3

    552 2 2 2

    3 , 3 ,

    2 21

    z

    p d d p d dz zd

    Rx y z

    = =

    + +

    (4.6)

    zy

    x

    p

    L

    Bd

    d

    0

    Figure 4.5 Rectangular distributed force on half-space

    For the whole surface the stress will be:

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    ( )

    ( ) ( )

    3

    52 2 2 2

    3 ,

    21

    z

    p d d z

    x y z

    =

    + +

    (4.7)

    where: p(,)-the average load on the surface A = d *d;

    , - coordinates of the load;

    z - depth of the point;

    R - distance between the elementary load and the considered point;

    x,y,z - coordinates of the point considered.

    Using these expressions, it is easy to calculate the maximum compressive stress under the

    centre and the corner of surface.

    The values of0z

    andcz

    can be determined using tables with the relations:

    0 0zp = (4.8)

    and

    cz cp =

    where: p = the uniform distributed load

    And 0 0

    c c

    K

    K

    =

    == coefficients taken from the tables 4.2 and 4.3 for the ratio

    (4.9)

    - L andB are the sides of the surface rectangle;- z is the depth of the point.

    ,L z

    fB B

    =