Soil PhaseSoil Phase RelationshipRelationship

updated March 11, 2008

Haryo Dwito Armono, ST, M.Eng, PhD

Phase of SoilPhase of Soil

S lidSolid particles

Water

Air

Volumetric RatiosVolumetric RatiosVv = Volume of void

Va = Volume of Air

Vw = Volume of water

Vs = Volume of solid

V l f id VVoid ratio (e) 0,65

v

s

Volume of voids Ve = = Volume of solids V

Porosity (n) 65% vVolume of voids Vn = = Total volume V

Degree of saturation (S) 65%wVolume of water VS = =

V l f id Vg (S) 65%vVolume of void V

Void ratio – Porosity RelationshipVoid ratio – Porosity RelationshipvV

1v v

vs v

V V nVeVVV V V n

V V

= = = =− −−

V Vv

v v s

V

V V V en = = = =

Typical Values

1s vs v

s s

nV VV V V eV V

+ ++

1

ne

n=

1 ne

n

−

=1

ne

=+

(Lambe and Whitman, 1979)

Engineering Applications (e)Engineering Applications (e)

Si l bi (SC) 0 91 CSimple cubic (SC), e = 0.91, Contract Cubic-tetrahedral (CT), e = 0.65, Dilate

Volume change tendency

Strength changeStrength change

Link: the strength of rock joint

)itan(strengthShear n +φσ=i

Engineering Applications (e)Engineering Applications (e)

Hydraulic conductivityHydraulic conductivityWhich packing (SC or CT) hashigher hydraulic conductivity?

SC

e = 0.91g y y

CT

The fluid (water) can flow more easily through the soil with higher hydraulic conductivity

CT

e = 0.65

Engineering Applications (e)

Filter

Engineering Applications (e)

SC

0 91

Filter

e = 0.91

The finer particle cannot pass

Clogging

CT

e = 0.65

The finer particle cannot pass through the void

Critical state soil mechanicsCritical state soil mechanics

Engineering Applications (e)Completely dry soil S = 0 %

Engineering Applications (e)

Completely saturated soil S = 100%

Unsaturated soil (partially saturated soil) 0% < S < 100%

%100)V(voidsofvolumeTotal

)V(watercontainsvoidsofvolumeTotalS

v

w ×=

Effects of capillary forces. Capillary action is responsible for moving groundwater from wet areas of the soil to dry areas.moving groundwater from wet areas of the soil to dry areas.

Engineering applications:

Slope stability

Underground excavation

Engineering Applications (e)

Most of landslides are due to erosion

Engineering Applications (e)

Most of landslides are due to erosion and “loss in suction”

The slope stability is significantly ff t d b th f taffected by the surface water.

(Au, 2001)

Density and Unit Weight

• Mass is a measure of a body's

Density and Unit Weight

• Mass is a measure of a body s inertia, or its "quantity of matter". Mass is not changed at different places

,Mass

DensityVolumeWeight Mass g

ρ =

⋅different places.

• Weight is force, the force of gravity acting on a body. The value is different at various

,

:

Weight Mass gUnit weight

Volume Volume

for example

γ ⋅= =

value is different at various places (Newton's second law F = ma)

3

:

, 1000w

for example

kgDensity of waterm

Unit weight of Water g

ρ

γ ρ

=

= •• The unit weight is frequently

used than the density is (e.g. in calculating the overburden

3 2

3

,

1000 9.8sec

9 8

w wUnit weight of Water g

kg mm

kN

γ ρ •

= •

=gpressure).

39.8m

=

Weight RelationshipWater Content w (100%)

Weight Relationship

%100)(

)(⋅=

s

w

MsolidssoilofMass

MwaterofMassw

for some organic soils w>100%, up to 500 %

or quick clays w>100%or quick clays, w>100%

Density of water

Temperature Density (at 1 atm)

°C kg/m³

0.0 999.84Density of water (slightly varied with temperatures)

31 /g cmρ

4.0 999.98

15.0 999.10

20.0 998.20

3

1 /

1000 /

w g cm

kg m

ρ =

=25.0 997.05

37.0 993.33

50.0 988.0431 /Mg m=

50.0 988.04

100.0 958.37

Density of SoilDensity of Soil

a. Dry density sd

MMass of soil solidsρ = =d Total volume of soil sample Vρ

b. Total, Wet, or Moist density (0%<S<100%, Unsaturated) s wM MMass of soil sample

Total volume of soil sample Vρ += =

c. Saturated density(S=100%, Va =0) s w

sat

M MMass of soil solids water

T t l l f il l Vρ ++

= =a

d. Submerged density (Buoyant density)

sat Total volume of soil sample V

wsat' ρ−ρ=ρg y ( y y) wsat ρρρ

Weight RelationshipSubmerged unit weight: wsat' γ−γ=γ

Weight Relationshipg g

Consider the buoyant force ( )W V W V Vγ γ− ⋅ − − ⋅yacting on the soil solids:

( )( 100%)s s w s w w

s w w

W V W V VS

V VW V W

V

γ γ

γ

= =

− ⋅ +=

s w w

VW W V

V

γ+ − ⋅=

sat wγ γ= −

Archimedes’ principle:Archimedes principle:The buoyant force on a body immersed in a fluid is equal to the weight of the fluid displaced by that object.

Cartoon by Ken Otter, 1997

Engineering Applications• For fine-grained soils, water

Engineering Applicationsg ,

plays a critical role to theirengineering properties(discussed in the next topic)(discussed in the next topic).

• For example,

The quick clay usually has a water content w greater than100 % and a card house100 % and a card house structure. It will behave like a viscous fluid after it is fully disturbeddisturbed. Clay

particle

Water (Mitchell, 1993)

Other RelationshipOther Relationship

(1) Specific gravityProof:

sGweS ⋅=⋅

w

s

w

ssG

γγ

=ρρ

=s

w

s

v

v

w

s

V

V

V

V

V

VeS =⋅=⋅

ww γρ

ws

s

wsw

ssv

VVM

MMGw ==

ρ=(2)

s

sw

GweS

weS

⋅=⋅ρ⋅=⋅⋅ρ

s

w

wswss V

VMMM

Gw =⋅=ρ⋅=⋅

Gs typical valuesGs typical values

(Lambe and Whitman, 1979)

(Goodman, 1989)

Remember the following simple rules

1 Remember the basic definitions of w e ρ S etc

g p

1.Remember the basic definitions of w, e, ρs, S, etc.

2.Draw a phase diagram.

3 A ith V 1 V 1 if t i3.Assume either Vs=1 or V = 1, if not given.

4.Often use ρwSe=wρs, Se = wGs

(Holtz and Kovacs, 1981):

Example

1. A saturated clay has a water content of 35% and a dry unit weight

dγ dγ

y y gof 13.6 kN/m3. Find the void ratio, the specific gravity and total unit weight

You have three ratios, S = 100%, w = 35% and γd = 13.6 kN/m3.

Assume a value say V = 1 00Assume a value, say V 1.00.

Then Ws = 13.6 kN.

( )( )Ww = (0.35)(13.6 kN) = 4.76 kN

Vw = 4.76 kN/9.81 kN/m3 = 0.4852 m3

Vv = 1.0 Vw = 0.4852 m3v w

Vs = V - Vv=1 - 0.4852 = 0.5148 m3.

e = 0.4852/0.5148 = 0.945,

G = 13 6/((0 5148)(9 81)) = 2 69Gs = 13.6/((0.5148)(9.81)) = 2.69,

γ = (13.6 + 4.76)/1.0 = 18.36/1.0 = 18.36 kN/m3