2phase Diagram Of Soil.pdf

PHASE DIAGRAM OF SOIL

Soil: A 33-Phase Material Air

Water

Solid grain

Geotechnical Engg Engg--I/Civil Engg./SKCET/ Coimbatore

Vv

Va

Air

Vw

Water

Ww

Vw=Vv

Vs

Soil solids

Ws

Vs

Water

Soil solids

Va=Vv

Vs

Air

Soil solids

Ws

Three phase diagram

Two phase diagram

Two phase diagram

Partially saturated soil

Fully saturated soil

Fully dry soil

Geotechnical Engg Engg--I/Civil Engg./SKCET/ Coimbatore

Phase

Volume

Mass

Weight

Air

Va

0

0

Water

Vw

Mw

Ww

Solid

Vs

Ms

Ws

Geotechnical Engg Engg--I/Civil Engg./SKCET/ Coimbatore

PROPERTIES OF SOIL Volumetric relationships •Void ratio (e) •Porosity (n) •Degree of saturation (S) or (Sr) •Air content (ac) •Percentage air voids (na) Geotechnical Engg Engg--I/Civil Engg./SKCET/ Coimbatore

Void ratio (e) The ratio between volume of voids to that of total volume of soil solids Vv

It is expressed in decimal

Va

Air

Vw

Water

Ww

Vs

Soil solids

Ws

For some soils its value can be more than 1 It’s a measure to indicate the denseness of soil

Vv e  Vs Geotechnical Engg Engg--I/Civil Engg./SKCET/ Coimbatore

Porosity (n) or percentage voids The ratio between volume of voids to that of total volume of soil

Vv

Va

Air

Vw

Water

Vs

Soil solids

It is expressed in percentage (%) It can not exceed 100% It is a measure to indicate the denseness of soil

Vv n  100 V

Geotechnical Engg Engg--I/Civil Engg./SKCET/ Coimbatore

Porosity in terms of void ratio

V v n  V 1 V  n V v 1 n



V

1  1   e e n  1  e

v

 V V v

s

e  1 or e

Geotechnical Engg Engg--I/Civil Engg./SKCET/ Coimbatore

Hence porosity “n“ is given by,

e n 1 e Geotechnical Engg Engg--I/Civil Engg./SKCET/ Coimbatore

Void ratio in terms of porosity

1 1 wkt ,  1  or n e 1 1 1 n  1  or e n n n e  1 n Geotechnical Engg Engg--I/Civil Engg./SKCET/ Coimbatore

Degree of saturation (S) or (Sr) The ratio between volume of water Vv

to the volume of voids

Va

Air

Vw

Water

Ww

Vs

Soil solids

Ws

It is expressed in percentage Its value ranges between 00- 1

V

0 for dry soil and 1 for saturated soil Can it exceed 100% ?

Vw S  100 Vv Geotechnical Engg Engg--I/Civil Engg./SKCET/ Coimbatore

Air content (ac)

Va ac  Vv

Vv

Percentage air voids (na)

na

Va

Air

Vw

Water

Ww

Vs

Soil solids

Ws

V

Va  V

The values of ac and na are zero for a saturated soil Geotechnical Engg Engg--I/Civil Engg./SKCET/ Coimbatore

Water content (w) or moisture content

Mw w  100 Ms For some soils water content may be more than 100% Geotechnical Engg Engg--I/Civil Engg./SKCET/ Coimbatore

Volume mass relationship Bulk mass density

 

Dry mass density



Saturated mass density Submerged mass density

d

M V M V



 sat   '

s

M

s at

V

M

s ub

V

Geotechnical Engg Engg--I/Civil Engg./SKCET/ Coimbatore

Where V = Unit total volume and the unit for mass density is kg/m3 (or) g/cm3

Volume weight relationship Bulk unit weight Dry unit weight Saturated unit weight Submerged or buoyant unit weight Unit weight of soil solids

W This is also known as total or wet   3 V unit weight. Its unit is kN/m W d  s V NOTE : V = Unit

 sat   '

W s at V

W s ub

s 

V

Ws Vs

Geotechnical Engg Engg--I/Civil Engg./SKCET/ Coimbatore

total volume

SYMBOLS AND UNITS Description Mass density (g/cm3)

Symbol Weight density or unit weight ( kN kN/m /m3)

Total or bulk

t

Saturated

sat

sat

Submerged

sub or ( (’)

sub or (’)

Dry

d

d

or

b

t

Geotechnical Engg Engg--I/Civil Engg./SKCET/ Coimbatore

NOTE  = g Where g = Acceleration due to gravity and its value is 9.81 m/s2 Geotechnical Engg Engg--I/Civil Engg./SKCET/ Coimbatore

Example 134.9cm3 243.9cm3 109.0cm3

Air

Wa~0

W =1.00

Water

109.0g 1013.0g

585.0cm3 s =2.65 341.1cm3

Solid

Volumes

904.0g

Weights Geotechnical Engg Engg--I/Civil Engg./SKCET/ Coimbatore

Example 134.9cm3 243.9cm3 109.0cm3

Air w=1.00

Water

585.0cm3 G=2.65 341.1cm3

Solid e

Volumes

Vv 243.9   0.72 Vs 341.1

Vv 243.9  100%   100  41.7% V 585.0 V 109.0 S (%)  w  100%   100  44.7% Vv 243.9 Geotechnical EnggEngg-I/Civil Engg./SKCET/ n(%) 

Coimbatore

Three phase diagram in terms of void ratio (e)

ea

e

ew= Se

Ma=0

Air Water

Mw=S e w

ea

e ew=Se

Air

Wa=0

Water

Ww=Sew

Soil solids

Ws=Gw

1+e 1

Vs 

Ms Gs  w



Soil solids

Ms=Gw

Ws Gs  w Geotechnical Engg Engg--I/Civil Engg./SKCET/ Coimbatore

1

The massmass-volume relationship can be written directly from the fig. Porosity (n)

Vv e n  V 1 e

e n 1 e

Degree of saturation (S (Sr) S

ea

e

ew= Se

Air

Ma=0

Water

Mw=S e w

Soil solids

Ms=Gw

1+e

Vw ew  Vv e

1

Dry density “d “

M s G w d   V 1 e

Geotechnical Engg Engg--I/Civil Engg./SKCET/ Coimbatore

Bulk density or total density “”

M s  M w G w  Se w (G  Se)  w    V 1 e 1 e Saturated density “sat”

 sat

(G  e)  w  1 e

Submerged density “sub” fully saturated soil sub =

e

ew= Se

or ’ for

Air Water

Ma=0 Mw=S e w

sat - w 1+e

(G  e)  w '  w 1 e

Ms=Gw 1

Submerged density “sub” or ’ for partially saturated soil sub = sat-

(G  Se)  w '  w 1 e

ea

w

Geotechnical Engg Engg--I/Civil Engg./SKCET/ Coimbatore

Soil solids

The weight volume relationship can be written directly from the fig. Porosity

Vv e n  V 1 e

ea

e

ew= Se

Air

Wa=0

Water

Ww=S e

w

Degree of saturation 1+e

V e S w  w Vv e

1

Bulk unit weight or total unit weight “ ”

Ws  Ww G w  Se w    V 1 e (G  Se) w 1 e Geotechnical Engg Engg--I/Civil Engg./SKCET/ Coimbatore

Soil solids

Ws=Gw

Saturated unit weight or “ sat”

Ws  Ww G w  Se w  sat    V 1 e (G  Se) w (G  e) w    sat 1 e 1 e

ea

e

ew= Se

Air

Wa=0

Water

Ww=S e

w

1+e Dry unit weight “ dry” 1

d 

Ws G w  V 1 e

Geotechnical Engg Engg--I/Civil Engg./SKCET/ Coimbatore

Soil solids

Ws=Gw

Submerged unit weight for saturated soil

sub = sat - w

Ws  Ww G w  Se w    V 1 e (G  Se) w (G  e) w    w   sub 1 e 1 e

ea

e

ew= Se

Air

Wa=0

Water

Ww=S e

w

1+e Submerged unit weight for partially saturated soil sub = sat - w

Ws  Ww G w  Se w    V 1 e (G  Se) w   w   sub 1 e

Geotechnical Engg Engg--I/Civil Engg./SKCET/ Coimbatore

1

Soil solids

Ws=Gw

Three phase diagram in terms of porosity

n

na nw=S n

1

Air Water

Ma=0 Mw=S n w

na

Air

Wa=0

nw

Water

Ww=S n w

Soil solids 1-n

Soil solids

Ms=Gw(1-n) 1-n

Geotechnical Engg Engg--I/Civil Engg./SKCET/ Coimbatore

Ws=Gw(1-n)

Three phase diagram in terms of porosity Void ratio “e” na

n

Air

nw=Sn

Water

Ma=0

e 

V V

v

n 1  n



s

Mw=S n w Dry density “d”

1



Soil solids Ms=Gw(1-n)

1-n

 

d



M V

s



Ms Gs  w



w

1 

n



1

Bulk or total density “” Ms  M M  V V

w

 G  w 1  n   Sn  w

  G 1  n   Sn  w

Vs 

G

Ws Gs  w Geotechnical Engg Engg--I/Civil Engg./SKCET/ Coimbatore

Three phase diagram in terms of porosity

n

na nw=Sn

Air Water

Ma=0 Mw=S n w

Saturated density “sat”  sat 

Ms M M  V V

w

 sat  G 1  n   n  w

1

Soil solids 1-n

Ms=Gw(1-n)

Geotechnical Engg Engg--I/Civil Engg./SKCET/ Coimbatore

 G  w 1  n   Sn  w

Three phase diagram in terms of porosity na

n nw

1

Air Water

Soil solids 1-n

Ma=0 Mw=S n w

Submerged density for saturated soil “sub”  sat 

Ms M M  V V

w

 G  w 1  n   Sn  w

 sat  G 1  n   n  w   w

Ms=Gw(1-n) Submerged density for partially saturated soil “sub”

Ms Mw M   V V  G  w 1  n   Sn  w   w

 sat 

Geotechnical Engg Engg--I/Civil Engg./SKCET/ Coimbatore

These relationships can be obtained in terms of unit weight also

Geotechnical Engg Engg--I/Civil Engg./SKCET/ Coimbatore

Three phase diagram in terms of porosity Void ratio “e” na

n nw

Air Water

Ma=0

e 

V V

v

n 1  n



s

Mw=S n w Dry unit weight “d”

1



Soil solids 1-n

Ms=Gw(1-n)

d



M V

s



G

w

1 

n



1

Bulk or total unit weight “”  

Ms M M  V V

w

  G 1  n   Sn  w

Geotechnical Engg Engg--I/Civil Engg./SKCET/ Coimbatore

 G

w

1  n  

Sn 

w

Three phase diagram in terms of void ratio & water content Va

Vv Vw

Air Water

Ma=0

e

Mw=Vww

ea

Air

Ma=0

ew

Water

Mw=Sew

1

Soil solids

Ms=Gw

V Vs

Soil solids Ms=VsGw

Water content, w w 

Geotechnical Engg Engg--I/Civil Engg./SKCET/ Coimbatore

M M

w s



Se  W G W



Se  wG

Se=wG Product of degree of saturation and void ratio is equal to product of water content and specific gravity

Geotechnical Engg Engg--I/Civil Engg./SKCET/ Coimbatore

Three phase diagram in terms of air voids

Va

Vv Vw

Air Water

Ma=0

Vv

Mw=w Ms

Va

Air

Vw

Water

Vs

Soil solids

Wa=0 Ww=w Ws

V Vs

Soil solids

Ms

Geotechnical Engg Engg--I/Civil Engg./SKCET/ Coimbatore

Ws

V

 V

1 

V V

1  n

a

 V

s

s





 V

w

V w  n V M s G  V

a



V V

a



1  n M

w

1  n 1  n 1  n



d



a

a

a

  

1

d

G 



s



V

w



w 

d



G 

w

 

1    w    G  

d w

 n a G  1  wG



w

s



a



w

V



d

W



wM



V V



w

w

 V V

V w V s





V a V

V w V







 

d w

  w  



Geotechnical Engg Engg--I/Civil Engg./SKCET/ Coimbatore

1    G 



List of formulae Mass density (kg/m3)

(G  Se)  w  1 e

 sat

(G  e)  w  1 e

(G  1)  w '  1 e e

n 1 n

se  wG

Unit weight (kN/m3)

(G  Se) w  1 e

 sat

(G  e) w  1 e

(G  1) w  ' 1 e n e 1 n

se  wG

Mass density (kg/m3)

G w d  1 e

d  d 

G w d  1 e

 1 w

(1  na )G w 1  wG

na  nac n

Unit weight (kN/m3)

e 1 e

Geotechnical Engg Engg--I/Civil Engg./SKCET/ Coimbatore

d  d 

 1 w

(1  na )G w 1  wG

na  nac n

n

e 1 ee

1 e

Determination of water content of soil  By

oven drying method

 Pycnometer

method

Geotechnical Engg Engg--I/Civil Engg./SKCET/ Coimbatore

Determination of water content of soil By oven drying method   

 

Clean the empty container, dry it and weigh it with lid (M1). Take the required quantity of the wet soil specimen in the container and close it with lid. Take the mass (M2) Place the container, with its lid removed in the oven at a temperature of 110oC  5oC for 24 hours till the mass becomes constant. When the soil has dried, remove the container from the oven, using tongs and cool it in a desiccator with the lid. Find the mass (M3) of the container with lid and dry soil sample. Geotechnical Engg Engg--I/Civil Engg./SKCET/ Coimbatore

S. No.

Observation and calculation

Trial Numbers 1

2

3

Observations 1. 2. 3. 4.

5. 6. 7.

8.

Container number Mass of empty container, with lid, M1 Mass of container with wet soil, M2 Mass of container with dry soil, M3 Calculations Mass of water, Mw = M2 – M3 Mass of solids, Ms = M3 – M1 Water M   (M  M )  content, = w   M   100 = w   M  M   x100 w

2

3

s

3

1

Average water content (w) Geotechnical Engg Engg--I/Civil Engg./SKCET/ Coimbatore

%

Determination of water content of soil using pycnometer Mass of empty pycnometer with lid = M1

Mass of pycnometer & soil = M2

Mass of pycnometer, soil and water = M3

Mass of pycnometer and water = M4

Water Water Soil solids

M1

M2

Soil solids

M3 Geotechnical Engg Engg--I/Civil Engg./SKCET/ Coimbatore

M4

Determination of water content of soil by Pycnometer method Mass of empty pycnometer with lid = M1 Mass of pycnometer and wet soil =M2 Mass of pycnometer soil and water=M3 Mass of pycnometer and water=M4 Mass M4 is equal to mass M3 minus mass of solids , Ms plus mass of equal volume of water M4  M3  Ms 

Ms M w  M 3  M s  s G w G

1  M 4  M 3  M s 1    G  G  M s  M 3  M 4   G  1   Geotechnical EnggEngg-I/Civil Engg./SKCET/ Coimbatore

Mass of wet soil = (M2-M1) Mass of water = mass of wet soil minus mass of solids (Ms) M

w

 (M

M w  M

w s

2

 M 1 )  M

  (M x 100     

2

3

 M

G   G 1

 4 

M

3

 M  (M 2  M 1 ) w     G   M  M 3  M 4  G  1    (M w    M

2 3

G   3  G  1   x 100  G    M 4    G 1 

 M 1 )  M

 M

3

 M

 4 

3

 M

4

  

  4 

G   G  1   x 100 G   G 1

  M 1)  G  1 100    1  xEnggGeotechnical Engg-I/Civil Engg./SKCET/  M 4  G   Coimbatore

Hence, water content

 (M w    M

2 3

  M 1)  G  1    1  x 100  M 4  G  

Geotechnical Engg Engg--I/Civil Engg./SKCET/ Coimbatore

Determination of specific gravity of soil using pycnometer Mass of empty pycnometer with lid = M1

Mass of pycnometer & soil = M2

Mass of pycnometer, soil and water = M3

Mass of pycnometer and water = M4

Water Water Soil solids

M1

M2

Soil solids

M3 Geotechnical Engg Engg--I/Civil Engg./SKCET/ Coimbatore

M4

Mass of empty pycnometer with lid = M1 Mass of pycnometer and soil = M2

M

4

 M

3

 M

Mass of pycnometer pycnometer,, soil and water = M3

M

4

 M

3

 M

M

4

 M

3

 M

M

3

 M

4

 M

M

3

 M

4

 M

Mass of pycnometer and water = M4

Mass M4 is equal to mass M3 Minus mass of solids ( Ms ) plus mass of equal volume of water

Geotechnical Engg Engg--I/Civil Engg./SKCET/ Coimbatore

s

M s   Gw

M s s  G 1    s 1  G   1    s 1  G   G 1  s   G 

w

Mass of solids (Ms)= M2- M1 M M

M

M

3

 M

4

3

 M

4

3

 M

3

1  G

G 

 G  1   eqnI s  G    G  1  M 2  M 1   G  

4

 M

4



M

2

M

2

 M

1

G



M

2

 M

1



G

M

2

 M

1   M

 M 1  (M 3  M M 2  M 1 

M 2

Substitute in equation I

 M



M

,

 M

 M 1  (M 3  M

2

4

 M

1  1

)

2 1

4

)

Geotechnical Engg Engg--I/Civil Engg./SKCET/ Coimbatore

G

Hence, specific gravity G =

G 

M

M

2

 M 1  M 1  (M 3  M 2

Geotechnical Engg Engg--I/Civil Engg./SKCET/ Coimbatore

4

)

Numerical:

1. The mass of a chunk soil is 20 kg and its volume is 0.011 m3. After drying in an oven the mass reduces to 16 16..5 kg kg.. Determine the water content (w), dry density, density of moist soil, void ratio, porosity and degree of saturation saturation.. Take G = 2.70 70.. 2. A soil specimen has a water content of 10 % and a wet unit weight of 20 kN kN/m /m3. If the specific gravity of solids is 2.70 70,, determine the dry unit weight, void ratio, porosity and degree of saturation saturation.. Take w = 10 kN kN/m /m3.

Geotechnical Engg Engg--I/Civil Engg./SKCET/ Coimbatore

Numerical: 3. A wet soil sample weighs 3.52 N. After drying in an oven its weight reduces to 2.9 N. the specific gravity of solids and mass specific gravity of soil are 2.65 and 1.85 respectively. Determine water content, void ratio, porosity and degree of saturation. Take w = 10 kN kN/m /m3.

4. A soil has a porosity of 40 % , the specific gravity of solids is 2.65 65,, and a water content of 12 %. Determine the mass of water required to be added to 100 m3 of this soil for complete saturation saturation.. Geotechnical Engg Engg--I/Civil Engg./SKCET/ Coimbatore

Numerical:

5. There are two borrow areas A and B which have soil with void ratios of 0.8 and 0.7 respectively respectively.. The in place water content is 20 20% % and 15 15% % respectively respectively.. The fill at the end of construction will have a total volume of 10000 m3, bulk density of 2 g/m3 and a placement water content of 22 %. Determine the volume of soil required to be excavated from both areas areas.. Take G = 2.65 65.. If the cost of excavation of soil and transportation is Rs Rs.. 200 per 100 m3 for area A and Rs Rs.. 220 per 100 m3 for area B. Which of the borrow area is more economical?

Geotechnical Engg Engg--I/Civil Engg./SKCET/ Coimbatore

Numerical: 6. An airport runway fill needs 600000 m3 of soil compacted to a void ratio of 0.75. There are two borrow pits A and B from where the required soil can be taken and transported to the site. Which of the borrow pit would be more economical?

Borrow In situ Transp pit void ortatio ratio n Cost A

0.80

Rs. 10 m3

B

1.7

Rs. 5 m3

Geotechnical Engg Engg--I/Civil Engg./SKCET/ Coimbatore

Numerical:

7. An embankment having total volume of 2000 m3 is to be constructed having a bulk density of 1.98 g/cm3 and placement water content of 18 %. The soil is to be obtained either from borrow area A or borrow area B which have void ratio of 0.78 and 0.69 and water content 16 16% % and 12 12% % respectively.. Take specific gravity for both the soil as 2.66 respectively 66.. If the cost of excavation is Rs Rs.. 35 35// m3 in each area, but the cost of transportation is Rs Rs.. 32 and Rs Rs.. 36 per m3 from areas A and B respectively, which of the area is more economical?

Geotechnical Engg Engg--I/Civil Engg./SKCET/ Coimbatore