CE 240 Soil Mechanics & Foundations Lecture 2lanbo/CE240LectW022weightvolume.pdf · Soil Mechanics...
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CE 240Soil Mechanics & Foundations
Lecture 2.2
WeightVolume 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.Weightvolume 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
= = = =
= = =

WeightVolume 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 twophase 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 twophase system, i.e., if S=100%, and we let Vs=1, we then have:
eVVeV vwt ==+= since,1
Consider the notsubmerged case, i.e., the twophase 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 DiagramWeightVolume Relationships (Table 3.1)Example:ExampleResults