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CE 240 Soil Mechanics & Foundations Lecture 2.2 Weight-Volume Relationships (Das, Ch. 3)

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  • CE 240Soil Mechanics & Foundations

    Lecture 2.2

    Weight-Volume Relationships(Das, Ch. 3)

  • Outline of this Lecture1.Density, Unit weight, and

    Specific gravity (Gs)2.Phases in soil (a porous

    medium)3.Three phase diagram4.Weight-volume relationships

  • For a general discussion we have

    density MV

    =

    Weight WW Mg=

    Unit weight WV

    =

    So that

    W Mg gV V

    = = =

  • Unit weight: = gUnit weight is the product of density and gravity acceleration. It is the gravitational force caused by the mass of material within a unit volume (density) in the unit of Newtons per cubic meter in SI system.

  • Specific Gravity (Gs)

    Specific gravity is defined as the ratio of unit weight (or density) of a given material to the unit weight (or density) of water since

    sw w w

    gGg

    = = =

  • Specific Gravity

    Expected Value for Gs

    Type of Soil Gs

    Sand 2.65 - 2.67

    Silty sand 2.67 2.70

    Inorganic clay 2.70 2.80

    Soils with mica or iron 2.75 3.00

    Organic soils < 2.00

  • Three Phases of Soils

    Naturally occurred soils always consist of solid particles, water, and air, so that soil has three phases: solid, liquid and gas.

  • Soil model

    Volume

    Solid Particles

    Voids (air or water)

  • Three Phase Diagram

    Solid

    Air

    Water

    Idealization:Three Phase Diagram

    Mineral Skeleton

  • Fully Saturated Soils (Two phase)

    Water

    Solid

    Mineral Skeleton Fully Saturated

  • Dry Soils (Two phase) [Oven Dried]

    Air

    Solid

    Dry SoilMineral Skeleton

  • Three Phase Diagram

    Solid

    Air

    WaterWT

    Ws

    Ww

    Wa~0

    Vs

    Va

    Vw

    Vv

    VT

    Weight Volume

  • Phase relationship: the phase diagram

    Wt: total weightWs: weight of solidWw: weight of waterWa: weight of air = 0

    Vt: total volumeVs: volume of solidVw: volume of waterVv: volume of the void

    Solid

    Air

    WaterWT

    Ws

    Ww

    Wa~0

    Vs

    Va

    Vw

    Vv

    VT

  • Solid

    Air

    WaterWT

    Ws

    Ww

    Wa~0

    Vs

    Va

    Vw

    Vv

    VT

    Vt = Vs + Vv = Vs + Vw + Va;It is convenient to assume the volume of the solid phase is unity (1) without lose generality.

    Mt = Ms + Mw; andWt = Ws + Ww, since W=Mg

  • Void ratio: e = Vv/Vs; Porosity n = Vv/Vt

    /1 / 1

    /1 / 1

    v v v t

    s t v v t

    v v v s

    t s v v s

    V V V V neV V V V V nV V V V enV V V V V e

    = = = =

    = = = =+ + +

    Apparently, for the same material we always have e > n. For example, when the porosity is 0.5 (50%), the void ratio is 1.0 already.

  • Solid

    Air

    WaterWT

    Ws

    Ww

    Wa~0

    Vs

    Va

    Vw

    Vv

    VT

    Degree of saturation: S =Vw/Vv x 100%Saturation is measured by the ratio of volume.

    Moisture content (Water content): w = Ww/Ws, Ww weight of water, Ws weight of solidWater content is measured by the ratio of weight.So that w can be greater than 100%.

  • Solid

    Air

    WaterWT

    Ws

    Ww

    Wa~0

    Vs

    Va

    Vw

    Vv

    VT

    Degree of saturation: S =Vw/Vv x 100%Saturation is measured by the ratio of volume.Moisture content: w = Ww/Ws, Ww=VwwWw weight of water, Ws weight of solidWater content is measured by the ratio of weight.

  • Definition of 3 types of unit weightTotal unit weight (moisture unit weight, wet unit weight) :

    t s w

    t t

    W W WV V

    += =

    Dry unit weight d :

    ,sd t s d st

    W V VV

    = >

  • Moisture unit weight :

    t s w

    t t

    W W WV V

    += =

    Solid unit weight s

    dry unit weight d

    ss

    s

    WV

    =

    ,sd t s d st

    W V VV

    = >

  • Thus, the dry unit weight d can be approximated as:

    (1 )d w =

    From the original form of the dry unit weight

    By taking the Taylor expansion and truncated at the first order term:

    1d w =+

    2 3 4 5(1 ...) (1 )1d

    w w w w w ww

    = = + + + +

    Because the moisture content w is a number always smaller than one, i.e., w

  • Relationships among S, e, w, and Gs

    , , 1w w v sv

    VS then V SV Se given VV

    = = = =

    A simple way to get Das, Equation 3.18

    , 1, 1w w w w s vs s s s w s

    s

    W V eS eSw V VW V G G

    thus Se wG

    = = = = = =

    =

    When the soil is 100% saturated (S=1) we have, Equation 3.20

    se wG=

  • Relationships among , n, w, and Gs(1 ), 1

    (1 )s s s s w s

    w s s w

    W V G n given V nW wW wG n

    = = =

    = =

    So that the dry unit weight d is

    And the moist unit weight is

    (1 ) (1 )1

    s s wd s w

    t

    W G n G nV n n

    = = = +

    (1 ) (1 )1

    (1 ) (1 ) (1 )(1 )

    t s w s w s w

    t t

    s w s w

    W W W G n wG nV V

    w G n G n w

    + + = = =

    = + = +

  • Relationships among , n, w, and Gs (cont.)

    When S=1 (fully saturated soil)

    (1 ) [ (1 ) ]1

    s w s w wsat s w

    t

    W W G n n G n nV

    + += = = +

    the moisture content w when S=1 can be expressed as

    (1 ) (1 ), 1

    w w

    s s w s s

    s s

    W n n ewW G n G n G

    recall Se wG thus e wG S

    = = = =

    = = =

  • Weight-Volume Relationships (Table 3.1)

    The 3rd column is a special case of the 1st column when S = 1.

  • Example: Determine moisture content, void ratio,

    porosity and degree of saturation of a soil core sample. Also determine the dry unit weight, d

    Data: Weight of soil sample = 1013g Vol. of soil sample = 585.0cm3

    Specific Gravity, Gs = 2.65 Dry weight of soil = 904.0g

  • Solid

    Air

    Water

    Wa~0

    1013.0g585.0cm3

    904.0g

    s =2.65

    109.0g

    341.1cm3

    109.0cm3243.9cm3

    134.9cm3

    W =1.00

    Example

    Volumes Weights

  • Results

    From the three phase diagram we can find:

    Moisture content, w

    Void ratio, e

    Porosity, n

    Degree of saturation, S

    Dry unit weight, d

    715.01.3419.243

    3

    3

    ===cmcm

    VVe

    s

    v

    %7.41100)(0.585)(9.243

    3

    3

    ===cmcm

    VVn

    T

    v

    %1.12100)(904)(109

    ===gg

    WWw

    s

    w

    355.1585904

    cmg

    VW

    T

    sd ===

    %7.441009.243

    109===

    v

    w

    VVS

  • Measurement of the submerged density (or unit weight)

  • Now consider the submerged case, i.e., the two-phase system has been put into the water:

    wswwss

    www

    GGbuoyancyMandeebuoyancyM

    )1(10

    ====

    Thus, the submerged density is

    eG

    eG

    VMM

    VM wsws

    t

    sw

    t

    tsubmerg +

    =

    ++

    =+

    =

    ==1

    )1(1

    )1(0

  • In a two-phase system, i.e., if S=100%, and we let Vs=1, we then have:

    eVVeV vwt ==+= since,1

    Consider the not-submerged case, i.e., the two-phase system has been just put in the air:

    wsswssss

    wwww

    GVGVMandeVM

    =====

    Thus, the saturated density is

    eGe

    eGe

    VMM swwsw

    t

    swsat +

    +=

    ++

    =+

    =1

    )(1

    and the dry density is

    eG

    VM ws

    t

    sd +

    ==1

  • Recall that the saturated density is

    eGe

    VMM sw

    t

    swsat +

    +=

    +=

    1)(

    If we do the following

    wsat

    wsws

    wwsw

    wswsat

    eie

    Ge

    eGee

    eGee

    Ge

    =

    =+

    =++

    =

    +++

    =+

    +=

    .,.1

    )1(1

    )1(1

    )1()(1

    )(

  • If you have got the submerged density and sure you know water density w

    eGe

    VMM sw

    t

    swsat +

    +=

    +=

    1)(

    You can calculate the saturated density sat.If you know w then you can calculate the void ratio e. if you think you can know e from the dry density d

    eG ws

    d +=

    1

    You can also calculate the submerged density when the sample is not 100% saturated.

    eSeG ws

    ++

    =1

    )1()1(

  • Reading Assignment:

    Das, Ch. 3

    Homework:

    3.4, 3.5

    CE 240Soil Mechanics & FoundationsLecture 2.2 Specific GravitySoil modelFully Saturated Soils (Two phase)Dry Soils (Two phase) [Oven Dried]Three Phase DiagramWeight-Volume Relationships (Table 3.1)Example:ExampleResults