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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(