©Noor Azalina Khalil, Nadhirah Nor Zelan, Mohd Syafaril Jamaludin 2021

First Published 2021

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Perpustakaan Negara Malaysia Cataloguing-in-Publication Data

Noor Azalina Khalil, Nadhirah Nor Zelan , Mohd Syafaril Jamaluddin

Basic Characteristic of Soil

Published by POLITEKNIK TUANKU SULTANAH BAHIYAH, Kulim Hi-Tech Park, 09000 Kulim,

Kedah.

ALHAMDULILAH…First and foremost, we would like to praise Allah the Almighty,

the Most Gracious, and the Most Merciful for His blessing given to we during our

writing and in completing this eBook.

Our completion of this eBook could not have been accomplished without the support

of our family and friends. We express our deep gratefulness to all of you.

Soil consist of grain with water and air in the void between grains. The water and air

contents are readily changed by changes in conditions and location. Although the size

and shape of the solid content rarely changes at a given point, they can vary

considerably from point to point.

All soil contains mineral particles, organic matter, water and air. The combination of

these determine the soil properties. Some important physical and chemical properties

of soil are mineral content, texture, action exchange capacity, bulk density, structure,

porosity and organic matter content.

Soil may be described in different ways by different people for their purposes.

Engineers are primarily in soil’s mechanical properties such as strength, stiffness

and permeability. These depend primarily on the nature of the soil grains, the

current stress, the water content and unit weight.

The book has illustrated liberally to help students understand the material.

A number of exercises are provided at the end of each sub topic for homework assignment.

Noor Azalina Khalil

Nadhirah Nor Zelan

Mohd Syafaril Jamaluddin

1 BASIC CHARACTERISTICS OF SOIL

11 SOIL CLASSIFICATION

16 ATTERBERG LIMITS

27 ENGINEERING CLASSIFICATION

31 COMPACTION

38 SAND REPLACEMENT

TOPIC BASIC CHARACTERISTICS OF SOIL

1.0 BASIC CHARACTERISTICS OF SOILS

Soil properties

All soils contain mineral particles, organic matter, water and air. The combinations

of these determine the soil’s properties – its texture, structure, porosity, chemistry

and colour. The spaces between the solids (soil particles) are called voids. If all the

voids are filled by water, the soil is saturated. Otherwise, the soil is unsaturated.

If all the voids are filled with air, the soil is said to be dry.

Figure 1: Soils contain water, solid and air.

The Three Phases of Soil

Referring to figure 2, soil is generally three phases material which contains solids,

water and air.

Figure 2: Three phases of soil

1

The Two Phases of Soil

Figure 3: Two Phase (Fully saturated)

Figure 4: Two Phase (Dry soil)

Physical properties of soil

*The Phases Relationship in Mass Soil and Volume Soil

To develop the weight/mass volume relationship, soil must separate the three phases

(solid, water and air). The total of given soil sample can be expressed as:

VT = VS + VV = VS + VW + Va

Assuming that the weight of the air is negligible, the total weight of the sample as:

W = WS + WW

Where; WS = weight of soil solids

WW = weight of water

2

Relationship Volume Mass Weight

Air VA 0 0

Water VW

MW

WW

Solid VS M

S W

S

Water content (w) is the ratio (expressed as a percentage) of weight of water to weight

of solids or the ratio of mass of water to mass of solids.

w = WW X 100% = MW X 100% WS MS

Unit weight () is total weight (weight of solid plus weight of water) divided by total volume

(volume of solid plus volume of water plus volume of air)

= WT (kN/m3) VT

Soils engineers sometimes refer to the unit weight as the moist unit weight.

To solve earthwork problems, one must know the weight per unit volume of soil,

excluding water. This weight is referred to as the dry unit weight. Dry unit weight (d)

is weight of solids divided by total volume.

d = WS (kN/m3) VT

Unit mass/density () is total mass divided by total volume.

= MT (kg/m3) VT

The relationship of unit weight, dry unit weight, and moisture content can be given as

d = / (1 + w)

In SI, unit weight is Newton per cubic meter (N/m3). The SI unit of density kilograms per

cubic meter (kg/m3). Dry unit mass/dry density (d) is mass of solids divided by total volume.

d = MS (kg/m3) VT

Density of solids,s is the ratio of the mass of the solids to the volume of the solids.

s = Ms

Vs

3

The unit weight in kN/m3 can be obtained from densities in kg/m3

= .g

and d = d . g

Where;

g = acceleration due to gravity = 9.81 m/sec2.

In terms of volume, the following new parameters are important – void ratio, porosity,

and degree of saturation.

Void ratio (e) is the ratio (expressed as a decimal fraction) of volume of voids to volume

of solids.

e = VV VS

Porosity (n) is the ratio (expressed as a percentage) of volume of voids to total volume.

n = VV X 100% VT

The relationship between void ratio and porosity,

e = n / (1-n)

n = e / (1 +e)

Degree of Saturation (Sr) is the ratio (expressed as a percentage) of volume of water

to volume of voids.

Sr = Vw X 100% VV

4

Relationship Among Unit Weight, Void Ratio, Moisture Content, Specific Gravity

and Air Content.

To obtain a relationship among unit weight/density, void ratio, and moisture content,

consider a volume of soil which the volume of the soil solids is one, VS.. If the volume

of the soil solids is one, then the volume of voids is numerically equal to the void ratio, e.

e = VV

VS

e = VV / (VT - VV)

The weights of soil solids and water can be given as

WS = GSW

WW = wWS = wGSW

Where GS = specific gravity of soil solids

w = moisture content

w = unit weight of water = 9.81 kN/m3

Specific gravity solid (GS) is the ratio of unit weight of solids (weight of solids divided by

volume of solid) to unit weight of water or of unit mass of solids (mass of solids divided

by volume of solids) to mass of water.

Note: The specific gravity of soil solids is often needed for various calculations in soil

mechanic. It can be determined accurately in the laboratory. Most of the values fall

within a range of 2.6 to 2.9. The specific gravity of solids of light colored sand, may

be estimated to about 2.65, for clayey and silty soils, it may vary from 2.6 to 2.9.

Gs = WS/VS = WS

W VS W Gs = MS/VS = MS

W VS W

5

Where W and w are the unit weight and unit mass of water. Now, using the definition of

unit weight and dry unit weight,

= W = WS + WW = GSW + wGSW = (1 + w) GSW V V 1 + e 1 + e

d = WS = GSW

V 1 + e

e = GSW - 1

d

From the definition of degree of saturation,

Sr = VW = wGS

VV e Sr = wGS e

If the soil sample is saturated, that is the void spaces are completely filled with water,

Sr = 1, so;

e = wGS

Air ratio (Ar)

Ar = n (1 – Sr)

Relationship between density, unit weight, water content and void ratio can be given as

Density, = (1 + w )GSW

1 + e

Dry density,d = GSW 1 + e

Saturated density,sat = (GS + e)w

1 + e Note: Unit weight of water varies slightly with temperature, but at normal temperatures,

it has a value of around 62.4 pounds per cubic foot (lb/ft3) or 9.81 kN/m3. The unit mass

(density) of water is 1000 kg/m3 or 1 g/cm3.

6

EXAMPLE 1

For a moist soil, given that volume = 6500 cm3; mass = 11.70 kg; moisture

content = 11.50%; Gs = 2.65, calculate the following:

i. Moist density (kg/m3)

ii. Dry density (kg/m3)

iii. Void ratio

iv. Porosity

iv. Degree of saturation (%)

SOLUTION:

VT = 6500 cm3 = 6.50 x 10-3 m3

MT = 11.70 kg

w = 11.5%

Gs = 2.65

(i) T = MT/VT

= 11.50/6.50 x 10-3

= 1769.23 kg/m3

(ii) ρdry = ρ / (1+w)

= 1769.23 / (1 + 0.115)

= 1586.75 kg/m3

(iii) Void ratio,e

Use formula d = Gsw / 1 + e

e = Gsw - 1

d

e = (2.65)(1000) - 1 1586.75

= 0.67

7

(iv) Porosity, n = e / 1 + e

= 0.67 / 1 + 0.67 = 0.4

(v) Sr = wGs/e

= 0.115(2.65) / 0.67 = 0.46

EXAMPLE 2

Given: -

1.The moist mass of a soil specimen is 25.2kg.

2.The specimen’s volume measured before drying is 0.026m3

3.The specimen’s dried mass is 18.3kg

4.The specific gravity of solids is 2.67

Determine

a. Void ratio, e

b. Degree of saturation, Sr

c. Wet unit mass (Wet density), ρb

d. Dry unit mass (Dry density), ρd

e. Wet unit weight, b

f. Dry unit weight, d

SOLUTION:

1. Gs = Ms

Vsρw

Vs = Ms / Gsρw = 18.3 kg / (2.67)(1000kg/m3) = 6.85 x 10-3 m3

Vv = VT – Vs = 0.026 - 6.85 x 10-3 = 0.019 m3

e = Vv /Vs = 0.019 / 6.85 x 10-3 = 2.796

2. Sr = Vw /Vv

Vw =25.2 – 18.3 / 1000 = 6.9 x 10-3 m3

Sr = Vw /Vv = 6.9 x 10-3 m3/0.019 m3 = 0.36

8

3. ρb = MT / VT = 25.2 / 0.026 = 969.23 kg/m3

4. ρd = 18.3 / 0.026 = 703.85 kg/m3

5. b = (969.23 x 9.81) / 1000 = 9.508 kN/m3

6. d = (703.85 x 9.81) / 1000 = 6.9 kN/m3

EXERCISE

1. Sketch the three phases of soil elements.

2. A specimen of clay was tested in the laboratory and the following data were collected:

Mass of wet specimen, M1 = 163.2g

Mass of dry specimen, M2 = 122.2g

Volume of wet specimen, V = 85.2cm3

Determine:

a. bulk density, ρb (Ans. ρb =1.916 g/cm3)

b. dry density, ρd (Ans. ρd =1.434 g/cm3)

3. A specimen of clay was tested in the laboratory and the following data were collected:

Mass of wet specimen, M1 = 156.8g

Mass of dry specimen, M2 = 132.4g

Volume of wet specimen, VT = 105.3cm3

Specific gravity of particles, GS = 2.69

Determine:

i. Water content (Ans.: w = 0.16)

ii. Bulk and dry density (Ans.: 1.50 g/cm3, 1.30 g/cm3)

iii. Void ratio and porosity (Ans.: 1.07, 0.52)

9

4. The mass of a moist soil sample having a volume of 0.0057m3 is 10.5kg. The moisture

content, m and the specific gravity of soil solids, Gs were determined to be 13% and

2.68, respectively. Determine:

a. Moist density, ρb (Ans.1842kg/m3)

b. Dry density, ρd (Ans.1630kg/m3)

c. Void ratio, e (Ans.0.64)

d. Porosity, n (Ans.0.39)

e. Degree of saturation, Sr (Ans.54.4%)

5. For a moist soil, given that

Volume of moist soil = 0.25 ft3

Weight of moist soil = 30 lb

Weight of dry soil = 26.1 lb

Specific Gravity, Gs = 2.63

Calculate

a. Moisture content, m (Ans.14.9 %)

b. Moist unit weight, b (Ans.120 lb/ft3)

c. Dry unit weight, d (Ans.104.41 lb/ft3)

d. Void ratio, e (Ans.0.57)

e. Porosity, n (Ans.0.36)

f. Degree of saturation, Sr (Ans.0.69)

6. For a moist soil, given that volume = 5660cm3, mass = 10.4 kg, moisture

content = 10%, Gs = 2.7, calculate the following:

a. Moist density, b (kg/m3) (Ans.1838 kg/m3)

b. Dry density, d (kg/m3) (Ans.1670 kg/m3)

c. Void ratio, e (Ans.0.62)

d. Porosity, n (Ans.0.38)

e. Degree of saturation, Sr (Ans.0.44)

10

2.0 SOIL CLASSIFICATION

Particles – Size Distribution Curve

The particles-size distribution curves can be used for the comparison of different

soils. Also, there are three basic soil parameters that can be determined from

these curves, and can be used to classify granular soils. These parameters are:

a. effective size

b. uniformity coefficient

c. coefficient of gradation.

Types of particles-size distribution curves

a. well graded soil

b. non-uniform graded soil

c. poorly graded soil

Uniformity coefficient is given by the relation;

Cu = D60

D10

The coefficient of gradation may be expressed as;

Cc = D230

D60 x D10

11

A well-graded soil will have a uniformity coefficient greater than about 4 for gravel

and 6 for sands, and a coefficient of gradation between 1 and 3 (for gravels and sands).

Figure 5: Types of particles-size distribution curves- well graded soil

Figure 6: Types of particles-size distribution curves- non-uniform graded soil

12

Figure 6 shown curve for non-uniform graded, a soil might have a combination of two or more

uniformly graded fraction.

Figure 7: Types of particles-size curves – poorly graded soil

EXAMPLE

Total mass of soil = 450 g

Sieve No

#

Diameter

(mm)

Mass of soil retained

on each sieves (g)

Mass of

passing (g)

Percent passing

(%)

10 2.000 0 450 100

16 1.180 9.90 440.10 98

30 0.600 24.66 415.44 92

40 0.425 17.60 397.84 88

60 0.250 23.90 373.94 83

100 0.150 35.10 338.84 75

200 0.075 59.85 278.99 62

Pan -- 278.99 0 0

TOTAL 450

Column 4 = Total – column 3

Column 5 = (Column 4 /Total mass of soil) x 100

13

Figure 8: Particles-size distribution curve

From the Particles-size distribution curve

% Sand = % passing # 2 mm - % passing # 0.06 mm

= (100 – 63)% = 37 %

% Silt = % passing # 0.06 mm - % passing # 0.002 mm

= (63 – 13)% = 50 %

% Clay = % passing # 0.002 mm - %passing # 0 mm

= (13 – 0)% = 13 %

So, sample is the silt soil.

Cu = D60

D10

Cu = 0.06 / 0.0015 = 40 > 4

Cc = D230

D60 x D10

Cc = (0.007)2 / (0.0015 X 0.06) = 0.54

So sample from poorly graded soil.

14

EXERCISE

1. For a soil, given:

D10 = 0.12mm, D30 = 0.43mm, D60 = 0.65mm

Calculate the uniformity coefficient, Cu and the coefficient of gradation

of the soil, Cc. (Ans.: Cu = 5.42, Cc = 2.37)

2. The result of s sieve analysis of a soil was as follows:

Sieve size (mm) Mass Retained (g)

20 0

12 1.7

10 2.3

6.3 8.4

4.75

5.7

2.8 12.9

2 3.5

1.4 1.1

0.5 30.5

0.355 45.3

0.180 25.4

0.075 7.4

The total mass of the sample was 147.2g. Plot the particle-size distribution curve

and describe the soil. Comment on the flat part of the curve.

15

3.0 ATTERBERG LIMITS

In the early 1900s,a Swedish scientist named Atterberg developed a method to

describe the consistency of fine grained soils with varying moisture contents. At very

low moisture content, soil behaves more like a solid. When the moisture content is

very high, the soil and water may flow like a liquid. The behavior of soil can be divided

into four basic states – solid, semisolid, plastic, and liquid.

The moisture content, in percent, at which the transition from solid to semisolid state

takes place is defined as the shrinkage limit. The moisture content at the point

of transition from semisolid to plastic state is the plastic limit, and from the plastic

to liquid state is the liquid limit. These parameters are also known as Atterberg limits.

There are three Atterberg Limit:

a. Liquid limit

b. Plastic limit

c. Shrinkage limit

Liquid limit

a. minimum water content of the soil can be drained by its own weight

b. is the moisture content above which the soil water mixture passes to a liquid state.

c. at this state the mixture behaves like a viscous fluid and flows under its own weight.

Plastic limit

a. is defined as the moisture content, in percent, at which the soil, when rolled into threads

of 3mm in diameter, crumbles. Plastic limit is the lower limit of the plastic stage of soil.

b. is the moisture content above which the soil water mixture passes to a plastic state.

c. below this moisture content, the mixture is in semi solid state.

d. any change in moisture content at either side of the PL produces a change in volume

of the soil.

16

Figure 9: Illustrating Plastic Limit Test

Source: Geotechnical Test Method, New York State Department of Transportation

Shrinkage limit

a. is defined as the moisture content, in percent, at which the volume of the soil mass

ceases to change.

b. is the moisture content above which the mixture of soil and water passes to semi

solid state.

c. below this moisture content, the mixture is in a solid state.

d. any increase in moisture content is associated with volume change but a decrease

in moisture content does not cause volume change. This is the minimum moisture

content that causes full saturation of the soil water mixture. The volume remains

constant as the mixture goes through the dry state to the SL moving from zero

saturation to 100% saturation.

17

Figure 10: Shrinkage Limit Test

Consistency of Soils Atterberg limits are the limits of water content used to define

soil behavior. The consistency of soils according to Atterberg limits gives the following diagram.

Figure 11: Consistency States

18

3.1 Casagrande Method

Figure 12: Casagrande cup with soil pat and grooving tool.

The liquid limit (LL) is the water content where a soil changes from plastic to

liquid behavior. The original liquid limit test of Atterberg's involved mixing a pat

of clay in a little round-bottomed porcelain bowl of 10-12cm diameter. A groove was cut t

hrough the pat of clay with a spatula, and the bowl was then struck many times against

the palm of one hand.

Casagrande subsequently standardized the apparatus and the procedures to make the

measurement more repeatable. Soil is placed into the metal cup portion of the device

and a groove is made down its center with a standardized tool. The cup is repeatedly

dropped 10mm onto a hard rubber base during which the groove closes up gradually

as a result of the impact. The number of blows for the groove to close for 13 mm

is recorded. The moisture content at which it takes 25 drops of the cup to cause the

groove to close is defined as the liquid limit.

19

Another method for measuring the liquid limit is the Cone Penetrometer test.

It is based on the measurement of penetration into the soil of a standardized cone

of specific mass. Despite the universal prevalence of the Casagrande method, the cone

penetrometer is often considered to be a more consistent alternative because it

minimizes the possibility of human variations when carrying out the test

Figure 13: Graph Moisture Content Versus Number of Blows

3.2 Cone Penetration

In this method the liquid limit is the moisture content of a soil water mixture placed

in a standard cylinder when a standard cone penetrations 20mm into the soil paste in

five seconds after it is released freely from the surface of the sample. The 60° cone is

manufactured from metal and has a height of 32mm whilst the cylinder has a diameter of

53mm and a height of 40mm.The test has to be repeated at least four times using the

original cured sample. Results are plotted in log linear scale, with moisture content on

linear vertical axis and cone penetrations on logarithmic horizontal axis (or graph

penetration vs moisture content). A line of best fit is drawn through the data and

moisture content corresponding to a cone penetration of 20mm is determined.

20

Figure 12: Cone Penetration Apparatus

Figure 13: Graph Cone Penetration Test

Plasticity Index (PI)

Plasticity index is the difference between the liquid limit and plastic limit of a soil.

PI = LL – PL

Liquidity Index (LI)

Liquidity index is the relationship between consistency (stiffness) of a remolded clayey soil.

LI = w – PL / PI

21

When the LI equals 1, the water content of the soil is at the liquid limit, and the soil is

very soft.

When the LI equals 0, the water content of the soil is at the plastic limit and the soil is

stiff or hard. This applies only to saturated soil in the remolded state.

Activity Index (AI)

The plasticity of soil is caused by the adsorbed water the surrounds the clay particles,

so that the type of clay minerals and their proportional amounts in a soil will affect

the liquid and plastic limits. Skempton (1953) observed that the plasticity index of a soil

increase linearly with the percentage of clay size fraction (% finer than 2 µm by weight).

The various clay mineral types above have different volume changes from the liquid limit

to the plastic limit for equal amount of clay sizes in the fraction finer than 0.42mm sieve

– as the activity of the clay.

When the clay content, the percentage of the fraction finer than 0.42mm that is finer 2 µm

(0.002mm) obtained from the hydrometer test, the AI is defined as:

AI = PI / % < 0.002mm

When the clay content is less than 40% the AI become:

AI = PI / % < 0.002mm – 5mm

The AI for various clays are:

a. kaolinite, 0.3 – 0.5

b. illite, 0.5 – 1.3

c. Ca-montmorillonite, 1.5

d. Non- montmorillonite, 3-7

22

A high AI is associated with those clay minerals that can absorb large amount of water.

Clays with an AI < 0.75 are defined as inactive clays.

AI = 0.75 – 1.25 are defined are normal activity clay

AI > 1.25 are defined as active clays.

EXAMPLE 1

Following are the results from the liquid and plastic limit test for a soil: Liquid

Limit test:

Number of blows, N Moisture content (%)

17 42.1

22 38.2

27 36.2

32 34.1

Plastic limit = 22%

i. Draw the flow curve and obtain the liquid limit

ii. What is the plasticity index of the soil?

23

SOLUTION:

(i)

(ii) PI = LL – PL = 37 – 22 = 15 %

EXAMPLE 2

A saturated, remolded clay soil has a water content 32%, a liquid limit of 35%,

plastic limit of 20%, and an amount of clay sizes (fine than 0.002mm) of 17%.

a. What are the liquidity index?

PI = 35-20 = 15

LI = 32-20 / 35 – 20 = 0.8

b. Is the sample likely to be soft or stiff?

Because the LI is near one, the sample will be soft

c. What is the activity for this soil and which clay mineral does it possess?

A = 15 / (17-5) = 1.25. This indicates that the dominant clay mineral is an illite or that the

clay is a mixture of several clay minerals including kaolinite, Illite and montmorillonite.

24

Mo

istu

re c

on

ten

t (%

)

Number of blows

Moisture content vs Number of blows

EXERCISE

1. In a liquid limit test on a fine-grained soil, using a cone penetrometer, the following

results were recorded:

Cone penetration (mm) 15.9 17.7 19.1 20.3 21.5

Water content (%) 32.6 42.9 51.6 59.8 66.2

In a plastic limit test on the same soil the plastic limit was found to be 25%. Determine

the liquid limit and plasticity index of the soil. (Ans.: LL = 55%, PI = 25%)

2. Following are the results from the liquid and plastic limit tests for a soil:

Liquid limit test:

Number of blows, N Moisture content (%)

15 42.0

20 40.8

28 39.1

Plastic limit = 17.2%

a. Draw the flow curve and obtain the liquid limit. (Ans. LL = 39%)

b. What is the plasticity index of the soil? (Ans. PI = 21.8%)

c. Determine the liquidity index of the soil when the in situ moisture content

is 30%? (Ans. LI = 0.58)

3. The following results were obtained from a liquid limit test using the cone

penetrometer apparatus.

No. of test 1 2 3 4 5

Mass of container 11.40 11.20 10.20 10.40 10.42

Mass of container +

wet soil

21.30 21.20 19.45 19.76 21.32

Mass of container +

dry soil

19.05 18.70 16.76 17.23 16.90

Cone Penetration

(mm)

12.60 15.50 16.80 18.50 23.40

25

From the data above:

a.Determine the liquid limit of soil sample. (Ans. LL = 19%)

b.Calculate the plasticity index if the plastic limit is 21 %.(Ans. PL = 2%)

4. In a liquid limit test using a cone penetrometer method, the following readings

were recorded below.

Cone

penetration

(mm)

14.44 16.4 18.2 21.1 22.3

Moisture

content (%)

30.8 42.2 51.2 68.1 77.4

In a plastic limit test on the same soil, the plastic limit was found to be 23 %.

Determine the liquid limit and plasticity index of the soil, and suggest classification

according to with the British Classification System (BSCS). (Ans.: LL= 62%, PI = 38%, CH)

5. After a series of laboratory tests, the following data were established for fine soil:

Liquid limit = 48%

Plastic limit = 32%

Clay content = 24.2 %

a. Calculate the activity of the soil. (Ans.: 0.66)

b. Determine the liquidity index of the soil when its natural moisture content

is 39%. (Ans.: 0.44)

6. If the saturated, remolded clay of example 1 is dried to water content of 22%,

a. What is it LI? (Ans. LI = 0.13)

b. Is the sample likely to be soft or stiff? (Ans.: Because the LI is near zero, the sample

will be stiff to hard.)

26

4.0 ENGINEERING CLASSIFICATION OF SOIL

Soils are classified according to criteria that are based on specific identification

tests and observations. A soil classification system, as used in geotechnical engineering

practice, is grouping or rating that provides an indication of expected behavior in

a specific engineering application. Three soil classification system are found in the

geotechnical engineering, there are:

a) The Unified Soil Classification System (USCS)

b) The American Associated of State Highway and Transportation Official (AASHTO)

c) U.S Department of Agriculture (USDA)

4.1 Unified Soil Classification System

Table 1: Symbols USCS

27

First Symbols Second Symbols

G – Gravel W – Well graded

S – Sand P – Poorly graded

M – Silt M – Silty

C – Clay C – Clayey

O – 0rganic L –Low plasticity, LL<50%

Pt – Peat soil H – High plasticity, LL>50%

Table 2: Group Symbols for Gravely and Sandy Soils USCS

28

Table 3: Group Symbols for Silty and Clayey USCS

29

EXERCISE

1. Classify each types of soil using British Soil Classification System

2. Classify the following soils by the Unified Classification System.

Percent Passing

Size A B C D E

Sieve No.4 94 98 100 100 100

Sieve No.200 3 10 77 86 45

Liquid Limit - - 63 55 36

Plasticity Index NP NP 25 28 22

30

5.0 COMPACTION

Compaction is the densification of soil by removal of air. There are five principle reasons

to compact soil:

a. Increases the shear strength and load-bearing capacity of the soil.

b. Prevents soil settlement and frost damage

c. Provides stability

d. Reduces water seepage, swelling and contraction

e. Reduces settling of soil

5.1 Factors Affecting Compaction

a. Effect of soil type

The soil type that is grain size distribution, shape of the soil grains, specific gravity

of soil solids, and amount and type of clay minerals present has a great influence on

the maximum dry density and optimum moisture content. Generally, coarse grained soils

can be compacted to higher dry density than fine grained soils.

b. Effect of compaction effort

As the compaction effort is increased, the maximum dry density of compaction is also

increased. As the compaction effort is increased, the optimum moisture content in

decreased to some extent.

c. Effect of moisture content

At low water content, the soil is stiff and offers more resistance to compaction. As the

water content is increased, the soil particles get lubricated. The soil mass becomes

more workable and the particles have closer packing.

The dry density of the soil increase with an increase in the content till the optimum

water content in reached. At the stage, the air voids attain approximately a constant

volume. With further increase in water content, the air voids do not decrease, but the

total voids increase and dry density decrease.

After the optimum water content is reached, it becomes more difficult to force air out

and to further reduce the air voids.

31

Saturation Line

Saturation line is also known as zero air voids line. It represents the dry density obtained

if all the air in the soil can be removed, so that after compaction of the soil samples

to be fully saturated. However, this may not be achieved through compaction either in the

laboratory or on-site. On land usually have 5% or 10% air voids after compaction.

Dry density (kg/m3)

10% air 5% air 0% air

Water content (%)

Figure 15: Compaction Curve and Saturation Line

Curve saturation line can be drawn in the graph compaction by using the following formula:

d = )1(

)1(

wGs

ArGs w

Where,

Gs = Specific Gravity

w = Water content

d = Dry Density

w = Density of water

Ar = Air ratio

32

5.2 Standard Proctor Compaction Test.

The soil is compacted in a mold that has a volume of 943.3cm3. The diameter of the mold

is 101.6mm. During the laboratory test, the mold is attaché to a base pla te at the bottom

and to an extension at the top. The soil is mixed with varying amounts of water and then

compacted in 3 layers by hammer that delivers 25 blows to each layer. The hammer

weighs 2.5kg and has a drop of 304.8mm.

For each test, the moist density of compaction,, can be calculated as

b = M / V

Where;

M = mass of the compacted soil in the mold

V = volume of the mold

For each test, the moisture content of the compacted soil is determined in the laboratory.

With the known moisture content, the dry density can be calculated as

d = b / 1 + (w%/100)

Where;

w% = percentage of moisture content

The values of d can be plotted against the corresponding moisture contents to obtain the

maximum dry density and the optimum moisture content for the soil.

Figure 16: Standard Proctor Compaction Test

33

5.3 Modified Proctor Compaction Test.

With the development of heavy rollers and their use in field compaction, the standard

Proctor test was modified to better represent field conditions. This revised version

sometimes is referred to as the Modified Proctor test.

For conducting the modified Proctor test, the same mold is used with a volume of

943.3cm3, as in the case of the standard proctor test. However, the soil is compacted

in 5 layers by hammer that has a mass of 4.54kg. The drop of the hammer is 457mm.

The number of hammer blows for each layer kept at 25 as in the case of standard Proctor

test.

Because it increases the compactive effort, the Modified proctor test results in an

increase of the maximum dry density of the soil. The increase of the maximum

dry density is accompanied by decrease of the optimum moisture content.

Table 4: The difference between Standard Proctor Test and Modified Proctor Test

Item Standard Proctor Test Modified Proctor Test

Weight of hammer 2.5kg 4.5kg

Height of hammer drop 304.88mm 457.2mm

Number of layers of

compaction

3 5

34

EXAMPLE

Experiment of compaction test have done in laboratory test, the result was shown on

the table below

Plot a dry density vs moisture content curve, and determine optimum moisture content

and dry density optimum.

SOLUTION:

35

Test no. 1 2 3 4 5

Moisture content,

%

12.9 14.3 15.7 16.9 17.9

Bulk Density, kg/m3 2092 2164 2152 2127 2043

Moisture

content

(%)

Bulk

Density

(kg/m3)

Dry Density

(kg/m3)

12.9 2092 1853

14.3 2164 1893

15.7 2152 1860

16.9 2127 1820

17.9 2043 1733

Graph Dry Density vs Water Content

EXERCISE

1. Give the description of compaction in terms of civil engineering discipline.

2. List THREE (3) types of compaction test.

3. State FIVE (5) objective of compaction.

4. Describes THREE (3) factors affecting compaction.

5. List THREE (3) types of compaction machines used on construction sites and specify the

types of soil suitable for the machines.

6. Compaction test result obtained from the laboratory cannot be directly used for

compaction processes on site. Explain the above statement.

7. If the bulk density of a soil sample is 2.15 Mg/m3and water content is 15.5 %, what is the

value of dry density after standard proctor compaction test? (Ans.: 1.86 Mg/m3)

36

Max dry density = 1895 kg/m3

Dry Density

(kg/m3)

Water Content (%)

Optimum water content = 15 %

8. Experiment of compaction test have done in laboratory test; the result was shown on

the below.

Test no. 1 2 3 4 5

Moisture content, % 12.9 14.3 15.7 16.9 17.9

Bulk Unit Weight, kN/m3 20.92 21.64 21.52 21.27 20.43

i. Plot a dry unit weight vs moisture content curve, and determine optimum

moisture content value. (Ans.: γdry =21.00 kN/m3, mopt. = 16%)

ii. Determine a void percent, Ar in maximum of dry unit weight optimum of water

content (use Gs=2.67) (Ans.: Ar = 0.14%)

9. The following data were recorded during a BS compaction test:

Volume of mould =0.945 x 10-3 m3, Gs = 2.70

Mass of wet soil in

mould (kg)

1.791 1.937 2.038 2.050 2.022 1.985

Water content (%) 8.4 10.6 12.9 14.4 16.6 18.6

Plot the curve of dry density against water content and from it obtain the maximum dry

density and optimum water content for the compacted soil.

On the same axes, plot the dry density vs water content curve for 0% and 5% air voids

and so obtain the air voids content of the soil at the optimum water content.

(Ans.: ρdmax = 1900kg/m3, mopt. = 14%)

37

6.0 SAND DISPLACEMENT METHOD

The method is performed by removing a soil sample from the earth and measuring its

weight. The volume of the excavated soil is then determined by measuring the volume

of fine sand required to fill sand required to fill the hole. This method is suitable for

granular soils and involves the use of a sand pouring cylinder as shown in Figure 17.

Firstly, a small hole is dug about 100mm in diameter and not more than 150mm in depth

and the soil removed carefully weighed. The volume of the hole is then determined by

pouring sand into it from the pouring cylinder. The sand pouring cylinder is weighed

before and after this operation, and the mass of sand filling the hole determined.

Since the density of the sand is known the volume of the hole can be determined, and

hence the bulk density of the in situ soil. Two sizes of sand pouring cylinder are

recommended: a small version suitable for fine and medium soils and a large

version for fine, medium and coarse soils.

Figure 17: Sand Displacement Test

38

EXAMPLE

The following data were recorded during a sand replacement test:

Mass of soil removed from hole = 1.914kg

Mass of soil after oven drying = 1.664kg

Initial total mass of sand pouring cylinder = 3.426kg

Mass of sand pouring cylinder after running sand into hole = 1.594kg

Density of pouring sand = 1.62Mg/m3

Mass of sand in cone in sand pouring cylinder = 0.248kg

Determine the bulk and dry unit weights of the soil in situ.

SOLUTION:

Mass of sand run out of cylinder = 3.426 – 1.594 = 1.832kg

Mass of sand in cone of cylinder = 0.248kg

Mass of sand required to fill the hole = 1.832 – 0.248 = 1.584kg

Volume of the hole = 1.584 / Density of pouring sand = 9.778 x 10-4m3

Bulk density of in situ soil = 1.914 / 9.778 x 10-4m3= 1957.456 kg/m3

Bulk unit weight = (1957.456 x 9.81) / 1000 = 19.20 kN/m3

w = (1.914 – 1.664) / 1.664 = 0.15

Dry unit weight = 19.20/(1 + w) = 19.2 / 1 + 0.15 = 16.7 kN/m3

39

REFERENCE

Braja (2010). Principle of Geotechnical Engineering (7th ed.). Cengage.Learning.

United States of America.

Donald P. Coduto (1999). Geotechnical Engineering (Principles and Practices).

Prentice-Hall. Inc. Upper Saddle River. New Jersey. US

Marcus M. Truitt (1983). Soil Mechanics Technology. United States of America.

Prentice – Hall. Inc. Englewood Cliffs, New Jersey. US

Muni Budhu (2011). Soil Mechanics & Foundation (3rd ed). John Wiley & Sons Inc.