lecture 3 physical properties of soil

Al-Rafidain University Collage Civil Engineering Department

Soil Mechanics Lectures for Third Year Students By Dr. Ahmed Al-Adly

Ph. D Civil / Geotechnical Engineering

Lecture 3: Physical Properties of Soil 21

Lecture 3

Physical Properties of Soil

3.1 Introduction

Soil is complex physical system, soil composed of various solid particles with number of

voids in between them, the voids between its solid particles filled with water and air.

Because of these particulate natures of soil, the development of models to describe and

predict the soil behaviour of soil is a difficult.

3.2 Physical State of Soil Sample and Phase Diagram

In natural, soil mass is generally a three phase-system. It consists of solid particles, liquid,

and gas as the following:

❶ Solid phase: represents the soil particles or grains

❷ Liquid phase: represents the water filling part of the voids between the soil particles

❸ Gaseous phase: represents the air filling that other part of the voids

Figure (3-1) below shows as soil sample has the volume (V) and weight (W) as they

would exist in a nature state.

Figure (3-1) Soil Mass in Natural State





To study the physical properties of soil, we need to make idealization to the soil element

by means of the so—called soil phase diagram or block diagram. We can idealize the three

phases of soil as shown in Figure (3-2).

Figure (3-2) Idealization of Soil Mass by Three-Phase Diagram

❶ When the soil voids are completely filled with water, the gaseous phase being absent,

it is said to be "fully saturated", and the soil will be two phase (Figure 3-3 a).

❷ When there is no water at all in the voids, the voids will be full of air, the liquid phase

being absent; the soil is said to be "completely dry", and the soil will be two phase also

(Figure 3-3 b).

(a): Fully Saturated Soil (b): Completely Dry Soil

Figure (3-3) Idealization of Soil Mass by Two-Phase Diagram

Solid Air

Idealized

Water

Three and Two Phase Diagram of Soil





3.3 Weight-Volume Relationships of Soil Figure (3-4) below show the three-phase diagram of soil sample with volume (V) and

weight (W) consists of three components of soil (solids, water, and air). The weights or

masses of the three components are presented on the right side of the diagram, while the

volumes are presented on the left side as shown in the Figure.

Figure (3-4): Three-Phase Diagram of Soil (volumes and weights of phases)

The total volume of soil sample can be expressed as:

퐕퐭 = 퐕퐚 + 퐕퐰 + 퐕퐬 = 퐕퐯 + 퐕퐬 … … … … … … … (ퟑ − ퟏ) Where:

퐕퐯 = 퐕퐚 + 퐕퐰 … … … … … … … (ퟑ − ퟐ)

Assuming that the weight of the air is negligible (W = 0), we can give the total weight of

the sample as:

퐖퐭 = 퐖퐬 + 퐖퐰 … … … … … … … (ퟑ − ퟑ)

Note: the right side of phase diagram may be in terms of masses instead of weights, e.g.

(M , Ms, M instead of W , W , W ).

Where:

V = Volume of air

V = Volume of water

Vs = Volume of solids

V = Volume of voids

V = Total volume of soil sample

W = Weight of air (W ≈ 0)

W = Weight of water

W = Weight of solids

Vt Wt





The following relationships and parameters can be defined based on three-phase

diagram of soil showing above:

❶ Water or Moisture Content: is defined as the ratio of the weight of water (W ) to

the dry weight of soil (W ). It is denoted as (W ) and expressed as a percentage:

퐖퐜 = 퐖퐰

퐖퐝 × ퟏퟎퟎ… … … … … … … (ퟑ − ퟒ)

The water content can be expressed as:

퐖퐜 = 퐖퐭 −퐖퐝

퐖퐝 × ퟏퟎퟎ… … … … (ퟑ − ퟓ)

Where:

W : represents the total weight or mass of soil in natural state (before drying).

W : represents the dry weight or mass of soil (after drying).

❷ Porosity: is defined as the ratio of the volume of voids (V ) to the total volume (V ). It

is denoted as (n) and expressed as a percentage:

퐧 =퐕퐯퐕퐭

× ퟏퟎퟎ… … … … … … … (ퟑ − ퟔ)

❸ Void Ratio: is defined as the ratio of the volume of voids (V ) to the volume of solids

(V ). It is denoted as (e) and expressed as a number without units:

퐞 =퐕퐯퐕퐬

… … … … … … … … … … … … (ퟑ − ퟕ)

The water content of soil (W ) ranged as (W ≥ 0), (W = 0) for dry soil and

(W 100 %) for softy clay below the water table level

The porosity of soil generally ranged as (0 < 푛 < 100 ) %, note that, the porosity of

soil cannot exceed (100 %)

Typical values of void ratio for natural sand may be ranged as (0.5 – 0.8) and for silt

and clay ranged as (0.7 – 1.1). The void ratio may be more than (1) for clayey soil





❹ Degree of Saturation: is defined as the ratio of the volume of water (V ) to the

volume of voids (V ). It is denoted as (S) and expressed as a percentage:

퐒 =퐕퐰퐕퐯

× ퟏퟎퟎ… … … … … … … (ퟑ − ퟖ)

❺ Air Content: is defined as the ratio of the volume of air (V ) to the total volume (V ).

It is denoted as (A ) and expressed as a percentage:

퐀퐜 =퐕퐚퐕퐭

× ퟏퟎퟎ… … … … … … … (ퟑ − ퟗ)

❻ Unit Weight and Density: the unit is defined as the weight of soil (W) per unit

volume (V) and denoted as (γ) Gamma:

후 =퐖퐕

… … … … … … … … … … … … (ퟑ − ퟏퟎ)

Density is defined as the mass of soil (M) per unit volume (V) and denoted as (ρ) Rho:

훒 =퐌퐕

… … … … … … … … … … … . . . (ퟑ − ퟏퟏ)

In SI (System International) Units, the unit weight (γ) is expressed by (kN/m3) and density

(ρ) is expressed by (kg/m3) or (gm/cm3). The unit weight can be related with density as the

following, where (γ) in (kN/m3) and (ρ) in (gm/cm3):

후 = 훒 × 퐠… … … … … … … … … … (ퟑ − ퟏퟐ) Where:

g: is gravity acceleration = 9.81 m/s2 10 m/s2

For example: Soil has a density of (ρ = 1.9 gm/cm3). What is the unit weight (γ) in (kN/m3)?

ρ = 1.9 gm/cm3 γ = ρ × g = 10 × 1.9 γ = 19 kN/m3

The degree of saturation (S) of soil ranged as (0 S 100) % for partially saturated

soil, (S = 0) for dry soil and (S = 100 % or 1) for saturated soil. The degree of saturation

of soil cannot exceed (100 %)





Cases of Unit Weight and Density of Soil

6-1: Total or Bulk or Moist Unit Weight of soil: is defined as the ratio of total weight

(W ) of soil per unit volume (V ) and denoted as (γ ):

후퐭 =퐖퐭

퐕퐭… … … … … … … … … … (ퟑ − ퟏퟑ)

In addition, the total or bulk density (ρ ) is:

훒퐭 =퐌퐭

퐕퐭… … … … … … … … … … (ퟑ − ퟏퟒ)

6-2: Dry Unit Weight of soil: is defined as the ratio of dry weight of soil or weight of

solids (W ) per unit volume (V ) and denoted as (γ ):

후퐝 =퐖퐝

퐕퐭… … … … … … … … . … (ퟑ − ퟏퟓ)

In addition, the dry density (ρ ) is:

훒퐝 =퐌퐝

퐕퐭… … … … … … … … . … (ퟑ − ퟏퟔ)

6-3: Saturated Unit Weight of soil: is defined as the ratio of fully saturated weight of soil

(W ) per unit volume (V ) and denoted as (γ ):

후퐬퐚퐭 =퐖퐬퐚퐭

퐕퐭… … … … … . … … … (ퟑ − ퟏퟕ)

6-4: Submerged or Buoyant Unit Weight of soil: is defined as the ratio of submerged

weight (W ) of soil per unit volume (V ) and denoted as ( γ ):

후 =퐖퐬퐮퐛

퐕퐭… … … … … … … … … (ퟑ − ퟏퟖ)

(W ) is equal to the weight of soil solids in air minus the weight of water displaced by

the solids. This leads to the following: 후 = 후퐬퐚퐭 − 후퐰 … … … … … … … (ퟑ − ퟏퟗ)

Where (γ ) is denoted as the unit weight of water





6-5: Unit Weight and Density of Water: unit weight of water is defined as the ratio of

weight (W ) to a given volume of water (V ) and denoted as (γ ):

후퐰 =퐖퐰

퐕퐰… … … … … … … … … (ퟑ − ퟐퟎ)

In addition, the density of water (ρ ) is:

훒퐰 =퐌퐰

퐕퐰… … … … … … … … … (ퟑ − ퟐퟏ)

In general, the unit weight of water (γ ) and density (ρ ) has constant values and

dependent in all soil mechanics calculations, these are:

후퐰 ≈ 10 kN/m3

훒퐰 = 1000 kg/m3 or 훒퐰 = 1 gm/cm3

❼ Specific Gravity of Soil Particles: is defined as the ratio of unit weight of soil solids

(γs) to the unit weight of water at a stated temperature (4 c°) and denoted as (Gs):

퐆퐬 =후퐬후퐰

… … … … … … … … … (ퟑ − ퟐퟐ)

And

퐆퐬 =퐖퐬

퐕퐬 × 후퐰… … … … … … … (ퟑ − ퟐퟑ)

Soil Type Specific Gravity (퐆퐬) Gravel 2.65-2.68 Sand 2.65-2.67 Silt 2.67-2.7

Clay 2.7-2.8

Nearly all soils have (Gs) ranged as (2.6-2.8), which is a very narrow range. A value of

(2.67) is commonly used for sandy and gravely soils and a value of (2.70) for clayey

soil. Table below list the typical values of (Gs) for common soil types





Soil Property Symbol Relationship Unit

Void Ratio 퐞 퐞 =퐕퐯퐕퐬

Without Unit

Porosity 퐧 퐧 =퐕퐯퐕퐭

× ퟏퟎퟎ %

Degree of Saturation 퐒 퐒 =퐕퐰퐕퐯

× ퟏퟎퟎ %

Air Content 퐀퐜 퐀퐜 =퐕퐚퐕퐭

× ퟏퟎퟎ %

Water Content 퐖퐜 퐖퐜 = 퐖퐰

퐖퐝× ퟏퟎퟎ %

Total (Bulk) Unit Weight

후퐭 후퐭 =퐖퐭

퐕퐭 kN/m3

Dry Unit Weight 후퐝 후퐝 =퐖퐝

퐕퐭 kN/m3

Total (Bulk) Density 훒퐭 훒퐭 =퐌퐭

퐕퐭 kg/m3 or gm/cm3

Dry Density 훒퐝 훒퐝 =퐌퐝


Saturated Density 훒퐬퐚퐭 훒퐬퐚퐭 =퐌퐬퐚퐭


Specific Gravity 퐆퐬 퐆퐬 =후퐬후퐰

=퐖퐬

퐕퐬 × 후퐰 Without Unit

Unit Weight of Water 후퐰 후퐰 =퐖퐰

퐕퐰 ≈ 10 kN/m3

Density of Water 훒퐰 훒퐰 =퐌퐰

퐕퐰

= 1000 gm/cm3 = 1 kg/m3

Summary of Basic Physical Properties of Soil





3.4 Interrelationships between Different Soil Properties

By combining the different physical properties which given by phase diagram, we can

derive A number of new useful relationships of soil as the following:

Useful Correlations between Different Soil Properties

퐒 × 퐞 = 퐖퐜 × 퐆퐬

퐧 =퐞

ퟏ + 퐞 퐚퐧퐝 퐞 =

퐧ퟏ − 퐧

후퐭 = 후퐝 (ퟏ + 퐖퐜) 퐚퐧퐝 훒퐭 = 훒퐝 (ퟏ + 퐖퐜)

후퐭 =퐆퐬 + 퐒 × 퐞ퟏ + 퐞

× 후퐰 퐚퐧퐝 훒퐭 =퐆퐬 + 퐒 × 퐞ퟏ + 퐞

× 훒퐰

후퐭 =퐆퐬 (ퟏ + 퐖퐜)

ퟏ + 퐞× 후퐰 퐚퐧퐝 훒퐭 =

퐆퐬 (ퟏ + 퐖퐜)ퟏ + 퐞

× 훒퐰

후퐝 =퐆퐬ퟏ + 퐞

× 후퐰 퐚퐧퐝 훒퐝 =퐆퐬ퟏ + 퐞

× 훒퐰

후퐝 =후퐭

ퟏ + 퐖퐜 퐚퐧퐝 훒퐝 =

훒퐭 ퟏ + 퐖퐜

후퐬퐚퐭 =퐆퐬 + 퐞ퟏ + 퐞

× 후퐰 퐚퐧퐝 훒퐬퐚퐭 =퐆퐬 + 퐞ퟏ + 퐞

× 훒퐰

퐞 =퐆퐬 × 후퐰후퐝

− ퟏ 퐚퐧퐝 퐞 =퐆퐬 × 훒퐰후퐝

− ퟏ

퐀퐜 = 퐧 (ퟏ − 퐒)

후 = 후퐬퐚퐭 − 후퐰





3.5 Relative Density or Density Index

Relative density is property of granular soil (cohesionless soil); it is commonly used to

describe the density of granular soil and measures the in situ denseness or looseness of

sand and gravel. Relative density is denoted as (Dr) and defined as:

퐃퐫 =퐞퐦퐚퐱 − 퐞

퐞퐦퐚퐱 − 퐞퐦퐢퐧× ퟏퟎퟎ… … … … … … … … (ퟑ − ퟐퟒ)

Where:

e : Maximum void ratio of soil in loosest condition

e : Minimum void ratio of soil in densest condition

e: Natural or in situ void ratio of soil

In addition, the relative density can be defined in terms of unit weight or density of soil as:

퐃퐫 =후퐝 − 후퐝(퐦퐢퐧)

후퐝(퐦퐚퐱) − 후퐝(퐦퐢퐧)×후퐝(퐦퐚퐱)

후퐝… … … … (ퟑ − ퟐퟓ)

Where: γ ( ): Maximum dry unit weight or density of soil in densest state 퐞퐦퐢퐧

γ ( ): Minimum unit weight or density of soil in loosest state 퐞퐦퐚퐱

γ : Natural or in situ dry unit weight or density of soil 퐞

Relative Density (Dr %) Description of Soil Deposit 0-15 Very loose

15-35 Loose 35-65 Medium 65-85 Dense 85-100 Very dense

The values of (Dr) of soil may vary from a minimum of (0 %) for very loose soil to a

maximum of (100 %) for very dense soil. Soil engineers describe the granular soil

deposits according to their relative densities, as shown in the table below:





Example (1): A soil sample has a volume of (105) cm3 and weight of (201) gm. After oven

drying, the weight reduced to (168) gm. If the specific gravity of soil solids is (2.7),

determine the following: water content, total density and total unit weight, dry density and

dry unit weight, void ratio, porosity, and degree of saturation.

Solution:

1- Water content (퐖퐜):

W =WW

× 100 W =W − W

W× 100 =

201 − 168168

× 100 퐖퐜 = ퟏퟗ.ퟔퟒ %

2- Total density (훒퐭) and total unit weight (후퐭):

ρ =WV

=201105

훒퐭 = ퟏ.ퟗퟏ 퐠퐦/퐜퐦ퟑ γ = 1.91 × 10 후퐭 = ퟏퟗ.ퟏ 퐤퐍/퐦ퟑ

3- Dry density (훒퐝) and dry unit weight (후퐝):

ρ =WV

=168105

훒퐝 = ퟏ.ퟔ 퐠퐦/퐜퐦ퟑ γ = 1.6 × 10 후퐝 = ퟏퟔ 퐤퐍/퐦ퟑ

4- Void ratio (e): ρ =

Gs 1 + e

× ρ e =Gs × ρρ − 1 =

2.7 × 1 1.6

− 1 퐞 = ퟎ.ퟔퟖ

5- Porosity (n):

n =e

1 + e=

0.681 + 0.68 퐧 = ퟎ.ퟒퟎퟒ = ퟒퟎ.ퟒ %

6- Degree of saturation (S):

S × e = W × Gs S =W × Gs

e=

19.64 × 2.70.68 퐒 = ퟕퟕ.ퟗퟖ %

Solved Examples

Given:

Total volume of soil sample, 퐕퐭 = 105 cm3

Total weight of soil sample, 퐖퐭 = 201 gm

Dry weight of soil sample, 퐖퐝 = 168 gm

Specific gravity of soil solids, 퐆퐬 = 2.7





Example (2): A soil sample of sand has a volume of (1150 cm3) and wet weight (23 N).

After oven drying, the weight is reduced to (20 N). If the specific gravity of soil solids is

(2.65), compute the following: moisture content, total unit weight, dry unit weight, void

ratio, porosity, degree of saturation, and air content.

Solution:


W =WW

× 100 W =W − W

W× 100 =

23 − 2020

× 100 퐖퐜 = ퟏퟓ %

2- Total unit weight (후퐭):

γ =23 × 10

1150 × 10 후퐭 = ퟐퟎ 퐤퐍/퐦ퟑ

3- Dry unit weight (후퐝): γ =

20 × 10 1150 × 10

후퐝 = ퟏퟕ.ퟑퟗ 퐤퐍/퐦ퟑ

4- Void ratio (e):

γ =Gs

1 + e× γ e =

Gs × γ γ − 1 =

2.65 × 10 17.39

− 1 퐞 = ퟎ.ퟓퟐ

5- Porosity (n):

n =e

1 + e=

0.521 + 0.52 퐧 = ퟎ.ퟑퟒퟐ = ퟑퟒ.ퟐ %


S × e = W × Gs S =W × Gs

e=

15 × 2.650.52 퐒 = ퟕퟔ.ퟒퟒ %

7- Air content (Ac):

Ac = n × (1 − S) Ac = 0.342 × (1 − 0.7644) 퐀퐜 = ퟎ.ퟎퟖퟎퟓ = ퟖ.ퟎퟓ %

Given:

Total volume of soil sample, 퐕퐭 = 1150 cm3 퐕퐭 = 1150×10-6 m3

Total weight of soil sample, 퐖퐭 = 23 N 퐖퐭 = 23×10-3 kN

Weight of drying soil sample, 퐖퐝 = 20 N 퐖퐝 = 20×10-3 kN

Specific gravity of soil solids, 퐆퐬 = 2.65





Example (3): A sample of soil was compacted into a cylindrical mould of (10 cm)

diameter and (15 cm) high. The weight of the moist sample was (2360 gm), after oven

drying, the weight of sample was (2200 gm). If the (Gs = 2.7), calculate the following: water

content, dry unit weight, moist unit weight, void ratio, and degree of saturation.

Solution:

Firstly, calculate the total volume (V ) of specimen, the volume of specimen can be calculated as:

V =π4

× D × h V =π4

× (10) × 15 퐕퐭 = ퟏퟏퟕퟖ.ퟏ 퐜퐦ퟑ


W =WW

× 100 W =W − W

W× 100 =

2360− 22002200

× 100 퐖퐜 = ퟕ.ퟐퟕ %

2- Dry unit weight (후퐝):

ρ =WV

=2200

1178.1 훒퐝 = ퟏ. ퟖퟔ 퐠퐦/퐜퐦ퟑ γ = 1.86 × 10 = ퟏퟖ.ퟔ 퐤퐍/퐦ퟑ

3- Moist unit weight (후퐦퐨퐢퐬퐭):

ρ =WV

=2360

1178.1 훒퐦퐨퐢퐬퐭 = ퟐ.ퟎퟎ 퐠퐦/퐜퐦ퟑ γ = 2.00 × 10 = ퟐퟎ 퐤퐍/퐦ퟑ

4- Void ratio (e): γ =

Gs1 + e

× γ e =Gs × γ

γ − 1 =2.7 × 10

18.6− 1 퐞 = ퟎ.ퟒퟗ


S × e = W × Gs S =W × Gs

e=

7.27 × 2.70.49

퐒 = ퟒퟎ.ퟎퟓ %

Given:

Sample diameter, D = 10 cm

Sample length, L = 15 cm

Moist (total) weight, 퐖퐭 = 2360 gm

Dry weight (after drying), 퐖퐝 = 2200 gm

Specific gravity, 퐆퐬 = 2.7





Example (4): A saturated soil sample of clay has a water content of (29 %) and bulk unit

weight of (19.3 kN/m3). Determine the dry unit weight, void ratio, and specific gravity of soil

particles.

Solution:

1- Dry unit weight (후퐝):

γ =γ

1 + W=

19.31 + 0.29

후퐝 = ퟏퟒ.ퟗퟔ 퐤퐍/퐦ퟑ

2- Void ratio (e) and specific gravity (퐆퐬):

γ =Gs

1 + e× γ e =

Gs × γγ − 1 =

10 × Gs 14.96

− 1

퐞 = ퟎ.ퟔퟔퟖ × 퐆퐬 − ퟏ… … … … … … (ퟏ)

We known that,

S × e = W × Gs

for saturated soil 퐒 = ퟏ, then e = W × Gs

퐞 = ퟎ.ퟐퟗ × 퐆퐬… … … … … … … … (ퟐ)

Solving eq. (1) & (2) as:

0.29 × Gs = 0.668 × Gs − 1 0.668 × Gs − 0.29 × Gs = 1 0.378 × Gs = 1

∴ 퐆퐬 = ퟐ.ퟔퟒ

Sub. in Eq. (2) to ind:

e = 0.29 × Gs = 0.29 × 2.64 ∴ 퐞 = ퟎ.ퟕퟔ

Given:

Saturated soil sample , S = 1

Water content, 퐖퐜 = 29 %

Bulk unit weight,후퐭 = 19.3 kN/m3





Example (5): A sample of sandy soil has a total volume of (2830 cm3), the following data

are obtained from laboratory tests: 퐖퐝 = 4450 gm, 퐆퐬 = 2.66, emax = 0.8, emin = 0.3. Calculate

the relative density of sample. State the sand loose or dense.

Solution:

The relative density (Dr) of soil can be calculated as:

Dr =e − e

e − e× 100

We need to calculate the natural void ratio (e) of soil sample; firstly, we must calculate the

dry density of soil sample as the following:

ρ =MV

=44502830

훒퐝 = ퟏ.ퟓퟕ 퐠퐦/퐜퐦ퟑ

Now, can be calculating the void ratio of soil sample as the following:

ρ =Gs

1 + e× ρ e =

Gs × ρρ − 1 =

2.66 × 1 1.57

− 1 퐞 = ퟎ.ퟔퟗ

Then, the relative density of soil sample will be:

Dr =e − e

e − e× 100 =

0.8 − 0.690.8 − 0.3

× 100 퐃퐫 = ퟐퟐ %

Not that, the relative density (Dr = 22 %), by back to the table of relative density, the

relative density of soil between (15-35) and in loose state, so Loose Sand

Given:

Total volume, 퐕퐭 = 2830 cm3

Dry weight, 퐖퐝 = 4450 gm

Maximum void ratio, 퐞퐦퐚퐱 = 0.8

Minimum void ratio, 퐞퐦퐢퐧 = 0.3

Specific gravity, 퐆퐬 = 2.66





Proof the Interrelationships Example (1): Prove that:

퐒 × 퐞 = 퐖퐜 × 퐆퐬 Proving: Left Side:

S × e =VV

×VV

=VV

Right Side:

W × Gs =WW

×W

V × γ =W

V × γ =W

V × WV

=VV

∴ Left side = Right side Okay

Example (2): Prove that:

퐧 =퐞

ퟏ + 퐞 Proving: Left Side:

n =VV

Right Side:

e1 + e

e =VV

∴

VV

1 + VV

=

VV

V + VV

=V

V + V=

VV

= n 퐎퐤퐚퐲


후퐭 = 후퐝 (ퟏ + 퐖퐜) Proving: Left Side:

γ =WV

Right Side:

γ (1 + W ), γ =WV

, W =WW

sub. in right side:

WV

× 1 +WW

=WV

×W + W

W=

W + WV

=WV

= γ 퐎퐤퐚퐲





Example (4): Prove that: 후퐝 = 퐆퐬

ퟏ + 퐞 × 후퐰

Proving: Left Side:

γ =WV

Right Side: Gs

1 + e× γ , Gs =

WV × γ

, e =VV

, γ =WV

, sub. in right side:

WV × γ

1 + VV

× γ =WV

V + VV

= WV + V

=WV

= γ 퐎퐤퐚퐲


후퐭 = 퐆퐬 + 퐒 퐞ퟏ + 퐞 × 후퐰

Proving: Left Side:

γ =WV

Right Side:

Gs + S e1 + e

× γ , Gs =W

V × γ, S = V

V, e =

VV

, γ =WV

, sub. in right side:

WV × γ + V

V × VV

1 + VV

× γ =W

V × γ + VV

V + VV

× γ =WV + V

V × γ

V + VV

=

W + V × γV

V + VV

=W + V × γ

V + V=

W + V × WV

V=

W + WV

=WV

= γ 퐎퐤퐚퐲





Q (1): A soil specimen of clay has a total volume of (60 cm3). In natural state, the weight of

sample was recorded as (105 gm), after oven drying, the sample had a weight of (88.2 gm). Take

(Gs = 2.7). Calculate the following: water content, total density and total unit weight, dry density

and dry unit weight, void ratio, porosity and degree of saturation.

[Ans. 19.04 %, 1.75 gm/cm3, 17.5 kN/m3, 1.47 gm/cm3, 14.7 kN/m3, 0.83, 45.35 %, 61.93 %]

Q (2): A cylindrical soil sample has a diameter of (10 cm) and height (15 cm). The moist weight

of sample is (2179.3 gm), after completely drying, the weight reduces to (1884.8 gm). The

specific gravity of soil is (2.66). Compute the following: moisture content, dry unit weight, moist

unit, weight, void ratio, porosity, degree of saturation, and air content.

[Ans. 15.625 %, 15.99 kN/m3, 18.49 kN/m3, 0.66, 39.76 %, 62.97 %, 14.72 %]

Q (3): The following information was obtained from laboratory test conducted on a soil sample

of clay: total volume = 510.2 cm3, wet weight = 960.3 gm, dry weight = 810.3 gm, specific

gravity = 2.7. Determine the dry unit weight, void ratio, and degree of saturation.

[Ans. 15.88 kN/m3, 0.7, 71.4 %]

Q (4): In the natural state, a moist soil has a volume of (0.3 m3) and weight (5500 N). The oven

dry weight of the soil is (4910 N), if (Gs = 2.74), calculate: moisture content, moist unit weight,

dry unit weight, void ratio, porosity, and degree of saturation.

[Ans. 12.01 %, 18.33 kN/m3, 16.36 kN/m3, 0.67, 40.11%, 49.11 %]

Q (5): A (152.46 gm) of wet clayey soil has a volume of (80 cm3). The soil is put into the oven

and dried to a constant weight of (132 gm). Assume the specific gravity (Gs = 2.7). Determine:

water content, dry unit weight, total unit weight, void ratio, porosity, and degree of saturation.

[Ans. 15.5 %, 16.5 kN/m3, 19.05 kN/m3, 0.63, 38.65 %, 66.42 %]

Q (6): A saturated soil sample has a water content of (25 %) and bulk unit weight of (20 kN/m3).

determine the dry unit weight, void ratio, and specific gravity. What would be the bulk unit

weight of the soil at the same void ratio but at a degree of saturation of (90 %)?

[Ans. 16 kN/m3, 0.665, 2.66, 19.6 kN/m3 at S = 90 %]

Problems and Questions





Q (7): A specimen of sand has a natural moisture content of (22 %) and moist unit weight of

(18 kN/m3). Calculate the void ratio and degree of saturation. Take (Gs = 2.65).

[Ans. 0.79, 73.79 %]

Q (8): a cylindrical sample of soil has a (38 cm2) cross-sectional area, (8 cm) height. The wet

weight is (650 gm), dry weight is (500 gm), and specific gravity of the soil solids is (2.65).

Calculate the water content, dry unit weight, and void ratio. If the sample was decreased in

volume by (10 cm3), what is the change in void ratio of soil?

[Ans. 30 %, 16.45 kN/m3, 0.61, 0.05]

Q (9): The natural dry unit weight of sand is (17.5 kN/m2). Find the relative density of the sand if

the maximum and minimum dry unit weight is (18.5 kN/m2) and (16 kN/m2) respectively. State

the sand loose or dense. [Ans. 63.4 %]

Q (10): A sandy soil has a natural water content of (24.5 %) and bulk unit weight of (19 kN/m3).

The minimum and maximum void ratios are (0.51) and (0.97) respectively. Assume (Gs = 2.67),

determine the dry unit weight, relative density, and degree of saturation.

[Ans. 15.26 kN/m3, 50 %, 88.39 %]

Q (11): Prove that: γ =

Gs × (1 + Wc)1 + e

× γ

Q (12): Prove that:

Dr =γ − γ ( )

γ ( ) − γ ( )×γ ( )

γ

Q (13): Prove that: γ = γ + n × γ At saturation





Q (14): Multiple-Choice Questions (MCQ) — Select the Correct Answer: 1- The ratio of volume of voids to the total volume of soil is:

(a): Void ratio (b): Degree of saturation (c): Air content (d): Porosity

2- A soil sample has a specific gravity of (2.6) and void ratio of (0.78). The water content to

fully saturate the soil at that void ratio will be:

(a): 20 % (b): 30 % (c): 40 % (d): 60 %

3- If a dry sand deposit has a porosity of (35 %) and (Gs = 2.66), then the percentage air voids

of the sand deposit is:

(a): 30 % (b): 40 % (c): 35 % (d): 45 %

4- A soil has a bulk unit weight of (22 kN/m3) and water content (10 %). The dry unit weight

of soil is:

(a): 18.6 kN/m3 (b): 20.5 kN/m3 (c): 22 kN/m3 (d): 23.2 kN/m3

5- If the voids of a soil mass are full of water only, the soil is termed as:

(a): Partially saturated soil (b): Dry soil (c): Saturated soil (d): None of these

6- If the volume of voids is equal to the volume of solids in a soil mass, then the porosity of

soil is:

(a): 1.0 (b): 0.0 (c): 0.5 (d): None of these

7- If the degree of saturation of a partially saturated soil is (60 %), then air content of the soil

is:

(a): 40 % (b): 60 % (c): 80 % (d): 100 %

8- A soil sample has a specific gravity of (2.6) and void ratio of (0.78). the water content

required to fully saturate the soil at that void ratio will be:

(a): 20 % (b): 40 % (c): 30 % (d): 60 %





Q (15): Answer with True (T) or False (F) and correct the false ones: 1- The void ratio of any soil cannot exceed (1.0).

2- If the water content of a fully saturated soil mass is (100 %), then the void ratio of the soil

is equal to its specific gravity.

3- A fully saturated soil is said to be three-phase system.

4- Water content of soil can be greater than (100 %).

5- The porosity of any soil cannot exceed (100 %).

6- The sandy soil at densest state has maximum dry density.

7- The sand deposit with relative density of (75 %) consider dense.

8- The soil below the water table level called partially saturated soil.

9- The specific gravity of particles of coarse-grained soil is always greater than (2.7).

lecture 3 physical properties of soil

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