41

2. PHASE RELATIONSHIPS OF SOILS Because soil is particulate in nature as opposed to more or less solid as is the rock from which it comes, any sample of soil obtained in the field has more than just soil particles in it. Every sample of soil will have voids between the particles because soil cannot be compacted into a solid mass and still remain soil! The voids between the soil particles can contain only two types of substances beside soil particles: liquids (usually water) and gas (usually air) as illustrated in Figure 2.01 below.

FIGURE 2.01 – Phases of Elements in a Soil Sample If the different elements in the soil sample on the left are rearranged and grouped together, the three different phases of matter in the sample appear as on the right side. This is the basic structure of a Phase Diagram which is used to express the relationships that exist between the volumes and masses of each element of the soil sample. The volumes of each component of the diagram are shown on the top of the diagram and the masses are shown on the bottom as illustrated below. Volume (cm3 or m3) w = 1 g/cm3 or = 1000 kg/m3 Mass (g or kg) where V = total volume of sample M = total mass of sample Vs = volume of solids Ms = dry mass or mass of solids Vw = volume of water Mw = mass of water Va = Volume of air Vv = Volume of voids

FIGURE 2.02 – Phase Diagram of Soil Sample Elements The mass of air compared to the other elements is negligible and so is never included. From this diagram it is easy to see the following relations:

SOLIDS WATER AIR

SOLIDS WATER AIR

V

Vs Vv

Vw Va

M

Ms Mw

42

In addition, the density of water, which can be considered a constant for most calculations, can also be expressed as

It is absolutely essential that the units used in the diagram be expressed explicitly as shown. Masses should be in grams (g) or kilograms (kg). If grams are used for masses then cubic centimetres (cm3) MUST be used for volumes. If kilograms are used for masses then volumes MUST be expressed in cubic metres (m3). These conditions are non-negotiable! There are also a number of phase-related properties that can be determined once all volumes and masses on the phase diagram are known: Density (a.k.a. Wet Density, a.k.a. Bulk Density, g/cm3 or kg/m3):

Dry Density (g/cm3 or kg/m3):

Void Ratio (decimal, no units):

vs VVV

awv VVV

ws MMM

w

ww V

Mρ

V

Mρ s

d

V

Mρ

s

v

V

Ve

43

Porosity (usually expressed as a percentage):

Saturation Rate (usually percent):

Air Content (usually percent):

Water Content (usually percent):

Specific Gravity of Soil Solids (dimensionless):

The specific gravity of any substance is simply the density of that substance divided by the density of water. This is also sometimes known as the relative density. Example Problems Example 2.1 A 125.6 g sample of soil was obtained from a split barrel sampler and its volume was found to be 65.34 cm3. The sample was then oven dried to a constant mass of 102.5 g. If a specific gravity test determined that the soil has a specific gravity of solids of 2.672, find the water content, density and dry density (kg/m3), void ratio, porosity, saturation rate and percent air content. Solution:

1. List all given quantities with related equations GIVEN: M = 125.6 g (= Mw + Ms) V = 65.34 cm3 (= Vs + Vv) Ms = 102.5 g (= M – Mw)

Gs = 2.672 (ws

s

ρV

M )

2. Draw Phase Diagram showing all given values.

V

Vn

v

v

w

V

VS

V

VA

a

s

w

M

Mw

ws

ss

ρV

MG

44

Volume (cm3) w = 1 g/cm3 Mass (g) 3. Mw = M – Ms = 125.6 g – 102.5 g = 23.1 g

4. Vw = 3

w

w

g/cm 1

g 23.1

ρ

M 23.1 cm3

5. Vs =

3

ws

s

g/cm 12.672

g 102.5

ρG

M 38.36 cm3

6. Vv = V – Vs = 65.34 cm3 – 38.36 cm3 = 26.98 cm3

7. Va = Vv – Vw = 26.98 cm3 – 23.1 cm3 = 3.88 cm3

8. w = g 102.5

g 23.1

M

M

s

w 0.225 = 22.5%

9. = 3cm 65.34

g 125.6

V

M1.922 g/cm3 =1922 kg/m3

d = 3

s

cm65.34

g 102.5

V

M1.569 g/cm3 = 1569 kg/m3

11. e = 3

3

s

v

cm 38.36

cm 26.98

V

V 0.703

12. n = 3

3v

cm65.34

cm 26.98

V

V 0.413 = 41.3%

13. S = 3

3

v

w

cm 26.98

cm 23.1

V

V0.856 = 85.6%

SOLIDS WATER AIR

65.34

38.36

3.88

125.6

102.5 23.1

23.1

26.98

45

14. A = 3

3a

cm65.34

cm 3.88

V

V 0.059 = 5.9%

Example 2.2 A saturated soil has a moisture content of 25.7% and a void ratio of 0.668. Determine the density and specific gravity of solids. Solution: Whenever volumes and masses are not known, the calculations are

performed based on an assumed volume or mass. Which volume or mass to assume depends on the information given. If the void ratio is given then assuming a volume of solids of 1 m3 or 1 cm3 is most advantageous. If water content is know, assuming a dry mass of 1 kg or 1g is most advantageous. Two solutions therefore are presented here (A and B) based on these two possibilities. A: Assume volume of solids, Vs = 1 cm3

1. List all given quantities with related equations

GIVEN: w = 25.7% = 0.257 (s

w

M

M )

e = 0.668 (s

v

V

V )

S = 100% = 1.000 (v

w

V

V )

2. Draw Phase Diagram showing all given/assumed values. Volume (cm3) w = 1 g/cm3 Mass (g)

3. Vv = eVs = 0.668(1 cm3) = 0.668 cm3

4. V = Vv + Vs = 0.668 cm3 + 1 cm3 = 1.668 cm3

SOLIDS WATER AIR

1.668

1

0

3.267

2.599 0.668

0.668

0.668

46

5. Vw = SVv = 1.000(0.668 cm3) = 0.668 cm3

6. Va = Vv – Vw = 0.668 cm3 – 0.668 cm3 = 0 cm3

7. Mw = wVw = (1.000 g/cm3)(0.668 cm3) = 0.668 g

8. Ms 0.257

g 0.688

w

Mw = 2.599g

9. M = Ms + Mw = 2.599 g + 0.688 g = 3.267 g

10. 3cm1.668

g 3.267

V

M = 1.959 g/cm3 = 1959 kg/m3

11. Gs )g/cm 1.000 x cm (1

g 2.599

ρV

M33

ws

s = 2.599

B: Assume mass of solids, Ms = 1 g

1. List all given quantities with related equations

GIVEN: w = 25.7% = 0.257 (s

w

M

M )

e = 0.668 (s

v

V

V )

S = 100% = 1.000 (v

w

V

V )

2. Draw Phase Diagram showing all given/assumed values. Volume (cm3) w = 1 g/cm3 Mass (g)

SOLIDS WATER AIR

0.6417

1

0

0.3847

1.257

0.257

0.257

0.257

47

3. Mw = wMs = (0.257)(1 g) = 0.257g

4. M = Mw + Ms = 0.257 g + 1 g = 1.257 cm3

5. Vw 3

w

w

g/cm 1.000

g 0.257

ρ

M = 0.257 cm3

6. Vv = 1.000

cm 0.257

S

V 3w 0.257 cm3

7. Va = Vv – Vw = 0.257 cm3 – 0.257 cm3 = 0 cm3

8. Vs = 0.668

cm 0.257

e

V 3v 0.3847 cm3

9. V = Vs + Vv = 0.3847 cm3 + 0.257 cm3 = 0.6417 cm3

10. = 3cm0.6417

g 1.257

V

M = 1.959 g/cm3 = 1959 kg/m3

11. Gs )g/cm 1.000 x cm (0.3847

g 1

ρV

M33

ws

s = 2.599

Special Cases In some problems some manipulation of formulae is necessary before the solution can begin. Three of these special circumstances are considered here. I. Given Total Mass, M and Water Content, w M = Ms + Mw (1)

w s

w

M

M or Mw = wMs (2)

Substitute (2) into (1): M = Ms + wMs and factoring Ms: M = Ms(1+w) Rearranging:

w)(1

MMs

48

II. Given Density, and Water Content, w If you divide both sides of the equation above by the total volume,

V, the following equation results:

III. Given Total Volume, V and Void Ratio, e V = Vs + Vv (1)

e s

v

V

V or Vv = eVs (2)

Substitute (2) into (1): V = Vs + eVs and factoring Vs: V = Vs(1+e) Rearranging:

IV. Algebraic Manipulation

Some combinations of given values will require formula manipulation in order to begin a solution. This can happen when specific gravity and water content are not known. Example 2.3 below illustrates such a case.

Example 2.3

A sample of soil has a total volume of 0.0282 m3, a saturation rate of 56% and a water content of 18.5%. If the specific gravity of the soil is 2.529, determine the values of the wet and dry densities and void ratio of the soil. Solution:

List all given quantities with related equations GIVEN: V = 0.0282 m3 (= Vs + Vv)

S = 0.56 (v

w

V

V )

w = 0.185 (s

w

M

M )

e)(1

VVs

w)(1

ρρd

49

Gs = 2.529 (ws

s

ρV

M )

w = 1000 kg/m3 (w

w

V

M )

Instead of manipulating the given information into one formula to solve for, say Vs or Ms, this solution will use the phase diagram to express the unknown quantities in terms of only one unknown value.

Volumes (m3)

0.0282

Vs 0.0282 - Vs 0.015364 0.012836

0.015792 – 0.56Vs 0.012408 – 0.44Vs 0.007188 0.005648

SOLIDS WATER AIR

Masses (kg)

2529Vs 467.865Vs 38.855 7.188

2996.865Vs 46.043

Starting with the unknown Vs, Vv = V – Vs = 0.0282 – Vs then, Vw = SVv = (0.56)(0.0282 – Vs) = 0.015792 – 0.56Vs and, Va = Vv – Vw = 0.0282 – Vs –(0.015792 – 0.56Vs) = 0.012408 – 0.44Vs Using Gs, Ms = wVsGs = (1000)(Vs)(2.529) = 2529Vs then, Mw = wMs = (0.185)(2529Vs) = 467.865Vs and, M = Ms + Mw = 2529Vs + 467.865Vs = 2996.865Vs Now, Mw can also be found using Mw = wVw So, Mw = (1000)(0.015792 – 0.56Vs) = 15.792 – 560Vs Since Mw can have only one value, 467.865Vs = 15.792 – 560Vs Rearranging, Vs = 0.015364 m3 Substituting this into the formulae for the remaining unknowns completes the diagram. Therefore, the required parameters are calculated:

50

3m 0.0282

kg 46.043

V

M = 1633 kg/m3

� 3

s

m 0.0282

kg 38.855

V

M = 1378 kg/m3

e = 3

3

s

v

m 0.015364

m 0.012836

V

V 0.835

51

Relative Density When preparing granular soil for the loads of structures it is important that the soil remains in a stable condition. If no effort was expended to prepare the soil then it would be reasonable to expect the soil under the structure to compress and densify under the weight of the structure. If this happens uniformly across the support soil then the entire structure will undergo settlement which, of itself, is undesirable. However, it is more likely that this densification will not be uniform and that the settlement will be differential from one part of the foundation to another. When this happens structural damage is caused. It is in everyone’s best interest therefore to compact granular soils (subgrade) prior to placing structures. Compaction reduces the magnitudes of differential settlement and increases the shear strength of the soil. To understand how well a soil has been compacted, relative density is used to compare the volume reduction that has been achieved by compaction to the maximum volume reduction possible. Consider a portion of soil in three states (Figure 2.03): its loosest, current and densest states. Compaction does not change the volume of solids but rather reduces the volume of voids by forcing the soil particles into more efficient arrangements. Therefore the total volume reduction is the same as the reduction in the volume of voids. The volume reduction achieved by compaction is the volume of voids after compaction subtracted from the volume of voids the soil would have in its loosest state. This is then compared to the maximum possible volume reduction which would be the volume of voids in its densest possible state subtracted from the volume of voids the soil would have in its loosest state.

FIGURE 2.03 – Volume Reduction from Compaction of Granular Soil

Loosest State emax dmin

Hv m

ax

Densest State emin dmax

Intermediate State e d

Hvmin

Hv

Hs

Soil Solids

Soil Solids

Soil Solids

Voids Voids

Voids Hi

H

Ms Ms Ms

52

The relative density, Dr is the parameter described above that compares the volume reduction achieved by compaction to the maximum possible volume reduction. This can be expressed in terms of void ratios or dry densities. Note that the volume of voids is the height of voids, Hv multiplied by cross-sectional area, A and the volume of solids, Vs is the height of solids, Hs multiplied by the same area, A. The relative density formulated based on % volume reduction would therefore be:

100%

VV

VVD

minmax

max

vv

vvr

If numerator and denominator are divided by Vs, then each term becomes a void ratio which results in the final void ratio expression of this relationship:

100%e(e

e)(eD

minmax

max

r

)

If total volume was used to calculate % volume reduction, the same result would be produced because Vs would cancel in the subtractions (for example, (Vmax – V) = (Vs +Vvmax) – (Vs + Vv) = Vvmax – Vv). Hence:

100%V(V

V)(VD

minmax

max

r

)

If numerator and denominator are divided by Ms, then each term becomes the inverse of a dry density:

100%

ρ

1

ρ

1

ρ

1

ρ

1

D

dmaxmind

dmindr

Rearranging, the equation becomes:

100%ρ

ρ

)ρ(ρ

)ρ(ρD

d

dmax

dmindmax

dmindr

It is also useful to have a relationship between the void ratios and settlement (H). If the initial depth of the soil element is Hi with a void ratio of ei and the soil is then compacted to a void ratio of ef , a settlement of H results. If we first visualize the problem in terms of volume reduction:

53

i

fi

i

fi

i

f

ivs

fvs

i

fi

ii e1

ee

e1

)e1e(1

)e(1

)e(11

)V(V

)V(V1

V

VV

V

V

AH

AH

The final form of the settlement equation is:

i

fii e1

eeHH

Relative Density Example Problems

Example 2.4

A soil with a maximum void ratio of 0.800 and a minimum void ratio of 0.400 was compacted to a void ratio of 0.500. Determine the relative density after compaction. Solution:

1. List all given quantities emax = 0.800 emin = 0.400 e = 0.500

2. Equation: 100%e(e

e)(eD

minmax

max

r

)

3. Substitution: 75%100%0(0.800

)0(0.800Dr

)400.

500.

4. Answer: Dr = 75%

Example 2.5

A sand has maximum and minimum void ratios of 0.700 and 0.300 respectively. A 3 m thick deposit of this sand has a relative density of 40% and is compacted to a relative density of 75%. By how many millimetres would the sand surface settle?

Solution:

2. List all given quantities emax = 0.700 emin = 0.300 Hi = 3 m Dri = 40% Drf = 75%

2. Equation: 100%e(e

)e(eDr

minmax

imax

)i

54

3. Substitution: ii e0.7000.16 ,

0.300)(0.700

)e(0.7000.4

ei = 0.540

4. Equation: 100%e(e

)e(eDr

minmax

fmax

f

)

5. Substitution: ff e0.7000.30 ,

0.300)(0.700

)e(0.7000.75

ef = 0.400

6. Equation: i

fii e1

eeHH

7. Substitution: mm 27372272.0.5401

0.4000.5403000ΔH

8. Answer: Settlement = 273 mm

55

Phase Relationship Problems 1. A chunk of moist soil has a volume of 0.02135 m3 and has a mass of 41.54 kg. The specific gravity of the solids is 2.722 and the moisture content is 23.2%. Determine the void ratio and the degree of saturation. 2. A chunk of moist soil has a volume of 0.3020 m3 and a mass of 634.2 kg. A 250.6 g sample from it was dried in the oven and its mass found to be 225.8 g. The specific gravity of the solids was 2.663. Determine the moisture content, dry density, wet density, void ratio and degree of saturation. 3. A soil chunk has a volume of 1.354 L and a mass of 2.565 kg. A 185.2 g sample was dried in the oven to a constant dry mass of 171.3 g. The specific gravity of the solids was 2.668. What was the void ratio and degree of saturation. 4. A cylindrical sample of clay 50.8 mm in diameter by 102 mm long has a mass of 400.3 g in its natural state. It is dried in an oven and found to weigh 304.0 g. Since the soil came from below the water table it is assumed to be saturated. a) Determine the void ratio and the specific gravity of the soil. b) If the soil was only 95% saturated what are the void ratio and specific gravity? 5. A saturated soil has a wet density of 2050 kg/m3 and a dry density of 1650 kg/m3. Find the specific gravity of the solids and the water content of the soil. 6. A 87.3% saturated soil sample has a void ratio of 0.823 and a moisture content of 36.2%. Find the density and specific gravity of the soil. 7. A 76.6% saturated soil has a moisture content of 38.4% and a specific gravity of 2.683. Find the void ratio and density in kg/m3. 8. A 63.7% saturated sample of soil has a dry density of 1680 kg/m3 and a porosity of 37%. Find the void ratio and the specific gravity of the solids. 9. A cylindrical sample of clay 25.1 mm in diameter by 51.86 mm long has a void ratio

of 0.722, a water content of 20.9% and a saturation rate of 82.5%. Determine the sample’s wet and dry densities in kg/m3 and the specific gravity of solids. 10. A chunk of soil has a mass of 42.53 kg, a specific gravity of solids of 2.501, a water

content of 2.4% and a saturation rate of 17.6%. Determine the soil’s wet and dry densities in kg/m3 and the void ratio.

11. A saturated chunk of soil has a mass of 53.27 kg, a volume of 0.0300 m3 and a

water content of 40.4%. Find the wet and dry densities in kg/m3, the void ratio and specific gravity of solids for this soil.

56

12. A soil sample has a mass of 40.90 g, a volume of 23.11 cm3, a void ratio of 0.849 and a specific gravity of solids of 2.672. Find the wet and dry densities in g/cm3, the water content and the saturation rate.

13. A 29.9% saturated sample of soil with a dry mass of 31.41 g has a void ratio of

0.796 and a specific gravity of solids of 2.962. Determine the water content and the wet and dry densities in g/cm3.

14. A 24.8% saturated sample of soil with a volume of 29.67 cm3 is oven dried to a

constant mass of 37.09 g. If it has a void ratio of 1.048, find the wet and dry densities of the sample in g/cm3 along with the water content and specific gravity of solids.

15. A chunk of soil has a mass of 40.73 kg and a volume of 0.0242 m3. If it is 41.4%

saturated and the oven dried mass is 35.66 kg, determine the wet and dry densities in kg/m3 along with the water content, void ratio and specific gravity of solids.

16. A saturated soil sample has a water content of 22.9% and a void ration of 0.601.

Determine the wet and dry densities in g/cm3 and the specific gravity of solids. 17. A nuclear densometer determined that the dry density of a soil mass is 1125 kg/m3

with a water content of 23.3%. Tests on a sample of the soil determined that the specific gravity of solids for the soil is 2.534. Find the wet density, void ratio and saturation rate for the soil mass.

18. A sample of soil with a dry density of 1.330 g/cm3 has a void ratio of 0.855 and a

water content of 26.4%. Find the wet density, saturation rate and specific gravity of solids of this sample.

19. Field tests show that a subgrade soil has a wet density of 1813 kg/m3 and a water

content of 12.0%. If the void ratio is estimated to be 0.585, determine the dry density, saturation rate and specific gravity of solids.

20. The dried mass of a soil sample is 5.489 kg and the water content is 12.4%. If the

wet density of the sample is 2068 kg/m3 and the void ratio is 0.443, find the dry density, saturation rate and specific gravity of solids.

21. A 34.32 kg chunk of soil has a volume of 0.0224 m3, is 23.7% saturated and has a

specific gravity of solids of 2.551. Find the wet and dry densities in kg/m3, the water content and void ratio.

22. A 12.69 kg chunk of soil has a porosity of 34.3%, a specific gravity of solids of

2.754 and is 82.5% saturated. Find the wet and dry densities in kg/m3, the water content and void ratio.

57

Relative Density Problems 23. The in-situ void ratio of a soil is 0.621. In its densest state, the void ratio is 0.416 and in its loosest state the void ratio is 0.815. Determine the in-situ relative density of the soil. 24. If a construction contract specification requires that the soil at the site be compacted to a relative density of 85%, determine the maximum acceptable void ratio given that the minimum and maximum void ratios found in tests on samples of the soil are 0.482 and 0.827 respectively. 25. A sand deposit has an in-situ dry density of 1894 kg/m3. Laboratory tests on samples of the sand indicated that the sand has maximum and minimum dry densities of 2115 kg/m3 and 1829 kg/m3 respectively. Determine the in-situ relative density of the sand. 26. The average depth of loose sand at a site is 10 m and the relative density is 25%. The sand is compacted by “Vibro-Flotation” to a relative density of 75%. The maximum and minimum void ratios of the sand are 0.800 and 0.400 respectively. How much additional fill in m3/m2 will have to be brought in to maintain the original grade at the site? 27. A compressor was located on a loose sand that settled 38 mm over several months in service before being shut down. On further investigation it was determined that the sand deposit was originally 2.60 m thick, the average void ratio when the compressor was shut down was 0.621 and the maximum and minimum void ratios for the sand are 0.806 and 0.411 respectively. What were the values of relative density at the beginning and end of the compressor’s time in service? If the compressor had not been taken out of service, how much would it ultimately have settled? 28. Studies have shown that when Granular A is spread (dumped) in a 200 mm lift, it has a relative density before compaction of 20% with a void ratio of 0.515. If a Standard Proctor test performed on Granular A produced a maximum dry density of 2075 kg/m³ and a specific gravity of 2.720 and the 200 mm lift is compacted to 100% compaction, find the minimum and maximum void ratios, the initial and minimum dry densities (kg/m³) and the expected settlement (mm).

58

Phase Relationship Problem Answers Question e Gs S w d

1 0.723 87.3%

2 0.408 71.9% 11.0% 2100 kg/m3

1892 kg/m3

3 0.522 41.3% 4a 0.872 2.754 4b 0.963 2.887 5 2.750 24.2%

6 1.985

1.483 g/cm3 or

1483 kg/m3

7 1.345 1584 kg/m3

8 0.587 2.667

9 2.850 2001 kg/m3

1655 kg/m3

10 0.342 1908 kg/m3

1863 kg/m3

11 1.045 2.586 1776 kg/m3

1265 kg/m3

12 70.7% 22.5% 1.770 g/cm3

1.445 g/cm3

13 8.0% 1.782 g/cm3

1.650 g/cm3

14 2.560 10.2% 1.377 g/cm3

1.250 g/cm3

15 1.024 2.983 14.2% 1683 kg/m3

1474 kg/m3

16 2.624 2.014 g/cm3

1.639 g/cm3

17 1.253 47.1% 1387 kg/m3

18 2.467 76.2% 1.681 g/cm3

19 2.566 52.6% 1619 kg/m3

20 2.655 74.3% 1840

kg/m3

21 0.755 7.1% 1532 kg/m3

1430 kg/m3

22 0.522 15.6% 2092 kg/m3

1809 kg/m3

59

Relative Density Problem Answers Question ANSWERS

23 Dr = 48.6% 24 emaxacceptable= 0.534 25 Dr = 25.4% 26 Volume of extra fill = 1.176 m3/m2 27 Dri = 40.8%, Drf = 46.8%, Hult = 370 mm 28 emin = 0.311, emax = 0.566, ρdi=1795 kg/m3, ρdmin=1737 kg/m3, ∆H = 27 mm