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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