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 ρMV

ρ =

Weight WW Mg=

Unit weight γWV

γ =

So that

W Mg gV V

γ ρ= = =

Unit weight: γ = ρgUnit 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

γ ρ ργ ρ ρ

= = =

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=VwγwWw – 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) γ :

t s w

t t

W W WV V

γ += =

Dry unit weight γd :

,sd t s d s

t

W V VV

γ γ γ= > ∴ <∵

Saturated unit weight (when saturation S=1) γsat :

tsat

t

WV

γ =

Moisture unit weight γ:

t s w

t t

W W WV V

γ += =

Solid unit weight γs

dry unit weight γd

ss

s

WV

γ =

,sd t s d s

t

W V VV

γ γ γ= > ∴ <∵

Since

1

s t w t w w sd d

t t t t s t

d d d

W W W W W W W wV V V V W V

so that w andw

γ γ γ γ

γγ γ γ γ

−= = = − = − = −

+ = =+

Thus, the dry unit weight γd can be approximated as:

(1 )d wγ γ= −

From the original form of the dry unit weight

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

1d wγγ =+

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

wγγ γ γ= = − + − + − + ≈ −+

Because the moisture content w is a number always smaller than one, i.e., w<1.

Relationships among S, e, w, and Gs

, , 1ww v s

v

VS then V SV Se given VV

= = = =

A simple way to get Das, Equation 3.18

, 1, 1w w w ws v

s s s s w s

s

W V eS eSw V VW V G G

thus Se wG

γ γγ γ

= = = = = ∴ =

=

∵

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

se wG=

Relationships among γ, 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

γ γγ

= = − = −

= = −

So that the dry unit weight γd is

And the moist unit weight γ is

(1 ) (1 )1

s s wd s w

t

W G n G nV n n

γγ γ−= = = −

− +

(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

γ γγ

γ γ

+ − + −= = =

= + − = − +

Relationships among γ, 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

γ γγ γ+ − += = = − +

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

γγ

= = = =− −

= = =∵

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, γd

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

γs =2.65

109.0g

341.1cm3

109.0cm3243.9cm3

134.9cm3

γW =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, γd

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 ===γ

%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

ρρρρρ

)1(10

−=−=−=−=−

Thus, the submerged density is

eG

eG

VMM

VM wsws

t

sw

t

tsubmerg +

−=

+−+

=+

=′

=′=1

)1(1

)1(0 ρρρρ

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

ρρρρρ

=====

Thus, the saturated density is

eGe

eGe

VMM swwsw

t

swsat +

+=

++

=+

=1

)(1

ρρρρ

and the dry density is

eG

VM ws

t

sd +

==1ρρ

Recall that the saturated density is

eGe

VMM sw

t

swsat +

+=

+=

1)(ρρ

If we do the following

wsat

wsws

wwsw

wswsat

eie

Ge

eGee

eGee

Ge

ρρρ

ρρρ

ρρρρρρ

−=′

′=+−

=+−−+

=

++−+

=−+

+=−

.,.1

)1(1

)1(1

)1()(1

)(

If you have got the submerged density ρ’ and sure you know water density ρw

eGe

VMM sw

t

swsat +

+=

+=

1)(ρρ

You can calculate the saturated density ρsat.If you know ρw then you can calculate the void ratio e. if you think you can know e from the dry density ρd

eG ws

d +=

1ρρ

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

eSeG ws

+−+−

=′1

)1()1( ρρ

Reading Assignment:

Das, Ch. 3

Homework:

3.4, 3.5