Weight and Volume Relationships

8/10/2019 Weight and Volume Relationships

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Weight-Volume Relationships


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Outline of this Lecture

1.Density, Unit weight, andSpecific gravity (Gs)

2.Phases in soil (a porousmedium)

3.Three phase diagram

4.Weight-volume relationships


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For a general discussion we have

density

V=

Weight W

W Mg=

Unit weightW

V

=

So that

W Mg

gV V = = =


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Unit weight: = gUnit weight is the product of density and

gravity acceleration. It is the gravitational

force caused by the mass of materialwithin a unit volume (density) in the unit of

Newtons per cubic meter in SI system.


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Specific Gravity (Gs)

Specific gravity is defined as the ratio of

unit weight (or density) of a given materialto the unit weight (or density) of water since

s

w w w

gG

g

= = =


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Three Phases of Soils

Naturally occurred soils always consist of

solid particles, water, and air, so that soilhas three phases: solid, liquid and gas.


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Soil model

Volume

Solid Particles

Voids (air or water)


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Fully Saturated Soils (Two phase)

Water

Solid

Mineral Skeleton Fully Saturated


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Dry Soils (Two phase) [Oven Dried]

Air

Solid

Dry SoilMineral Skeleton


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Three Phase Diagram

Solid

Air

WaterWT

Ws

Ww

Wa~0

Vs

Va

Vw

Vv

VT

Weight Volume


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Phase relationship: the phase diagram

Wt: total weight

Ws: weight of solid

Ww: weight of water

Wa: weight of air = 0

Vt: total volume

Vs: volume of solid

Vw: volume of water

Vv: volume of the void

Solid

Air

WaterWT

Ws

Ww

Wa~0

Vs

Va

Vw

Vv

VT


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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; and

Wt = Ws + Ww, since W=Mg


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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 ne

V V V V V n

V 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.


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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 solid

Water content is measured by the ratio of weight.

So that w can be greater than 100%.


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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 solid

Water content is measured by the ratio of weight.


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Definition of 3 types of unit weight

Total unit weight (moisture unit weight, wet unitweight) :

t s w

t t

W W W

V V

+

= =

Dry unit weight d :

,sd t s d s

t

W V VV

= >


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Moisture unit weight :

t s w

t t

W W W

V V

+= =

Solid unit weight s

dry unit weight d

ss

s

WV

=

,s

d t s d s

t

W

V VV = >


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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 thefirst 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


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Relationships amongS, e, w, and G

s

, , 1ww v s

v

VS then V SV Se given V

V= = = =

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=


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Relationships among, n, w, and G

s

(1 ), 1

(1 )

s s s s w s

w s s w

W V G n given V n

W 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 n

V V

w G n G n w

+ + = = =

= + = +


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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

= = = =

= = =


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Weight-Volume Relationships (Table 3.1)

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


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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


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Solid

Air

Water

Wa~0

1013.0g585.0cm3

904.0g

s =2.65

109.0g

341.1cm3

109.0cm3

243.9cm3

134.9cm3

W =1.00

Example

Volumes Weights


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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

715.01.341

9.243

3

3

===cm

cm

V

Ve

s

v

%7.41100)(0.585

)(9.2433

3

===cm

cm

V

Vn

T

v

%1.12100)(904

)(109===

g

g

W

Ww

s

w

355.1

585

904

cm

g

V

W

T

sd ===

%7.441009.243

109 ===v

w

VVS


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Measurement of the submerged density (or unit weight)


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Now consider the submerged case, i.e., the two-

phase system has been put into the water:

wswwss

www

GGbuoyancyM

andeebuoyancyM

)1(1

0

==

==

Thus, the submerged density is

e

G

e

G

V

MM

V

M wsws

t

sw

t

t

submerg +

=+

+

=

+

=

== 1

)1(

1

)1(0

In a two phase system i e if S=100% and we let


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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

GVGVM

andeVM

===

==

Thus, the saturated density is

e

Ge

e

Ge

V

MM swwsw

t

sw

sat +

+=

+

+=

+=

1

)(

1

and the dry density is

e

G

V

M ws

t

sd

+== 1


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Recall that the saturated density is

e

Ge

V

MM sw

t

swsat

+

+=

+=

1

)(

If we do the following

wsat

wsws

wwswwswsat

eie

G

e

eGe

e

eGe

e

Ge

=

=

+

=

+

+=

+

++=

+

+=

.,.1

)1(

1

)1(

1

)1()(

1

)(

If you have got the submerged density and


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If you have got the submerged density and

sure you know water density w

e

Ge

V

MM sw

t

swsat

+

+=

+=

1

)(

You can calculate the saturated density sat.If youknow 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.

e

SeG ws

+

+=

1

)1()1(

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Weight and Volume Relationships