Soil Composition

Terms, Definitions and Relationship

Engineering vs. Geology

Geologists class all items of Earth’s crust as rock, whether hard or soft deposits.

Engineers consider rock and soil separately

Engineering Definition• Soil is considered to be

any loose sedimentary deposit, such as gravel, sand, silt, clay or a mixture of these materials. They are readily separated into relatively small pieces in which it may contain air, water or organic material (from vegetation).

• Rock is the naturally occurring material composed of mineral particles so firmly bonded together that great effort is required to separate the particle.

Main Soil Group

Peat, organic clays, organic silt.

Organic soils

Silt and ClayFine-grained soils(Cohesive)

Sand and GravelGranular Soils/Coarse-grained soils(Frictional)

SOLID AND VOID

VOIDSOLID

The soil is:

• Dry if the voids are full of air• Saturated if the voids are full of water• Partially saturated if the void contains

both.

Two-phase soilThree-phase soil

air water

Soil Analytical Representation• Soil is made up of various

sized particle packed together. Spaces between particles are known as voids.

• The voids usually contain a mixture of water and another part air or other gas. Or it can be completely water or air.

Air

Water

Solid

Void

Weight/Mass-Volume Relationship

Volume of air = Va

Volume of water = Vw

Volume of solid = Va

Volume of voids= Vv

VT

Air

Water

Solids

Wt. of water = Ww

Mass of water= Mw

Weight of air = 0

Wt. of solids = WS

Mass of solids= MS

Weight/Mass-Volume Relationship

WT = WS+WW

MT = MS+MW

VT=VS+VW+Va=VS+VV

Ws = VsGsγw

Ms = VsGsρw

Ww = VwGwγw=Vwγw

Mw = VwGwρw = Vwρw

Gw = 1Gs range from 2.6-2.75

Unit Weight and Density

• Unit Weight(γ)(kN/m3)

Bulk unit weight:

Dry unit wt.

• Density(ρ) (kg/m3)

Bulk density:

Dry density:

T

T

VM

=ρ

T

T

VW

=γ

T

Sdry V

W=γ

T

Sdry V

M=ρ

Water

9.81 kN/m3 (SI)1 Mg/m3 (SI)Or1000 kg/m3

Unit Wt.Density

Water content, w% = %100×S

W

S

W

MM

orWW

Water content of a soil sample

A moist sample of soil in a cup had a mass of 25.24g. The empty cup had a mass of 14.2

After drying in an oven for 24 hrs, the cup and soil had a mass of 21.62. Find water content.

Let: m1 = mass of cup

m2=mass of cup + wet soil

m3= mass of cup + dry soil

w = [(m2-m3)/(m3-m1)] x 100

= [(24.24-21.62)/(21.62-14.20)] x 100

= 49%

Void Ratio and Porosity

• The ratio of volume of voids to volume of solids is known as the void ratio.

• The ratio of volume of voids to total volume of sample is called porosity.

S

V

VV

e =T

V

VV

n =

een+

=1

Degree of saturation, S

• The term for the portion of the void spaces in soil that is filled with water is –degree of saturation. Or simply the ratio of volume of water to volume of voids.

V

W

VV

S =

Soil Below Water Table

• Void is completely filled with water• Degree of Saturation = 1• Percentage of saturation = 100%

• In fine grain soil such as silt, water will rise because of capillary action, soil in some distance above water table may be saturated.

Example 1

• Determine wet density, dry unit wt., void ratio, water content, and degree of saturation for a soil sample with the following data:

Total mass = 18.18 kgTotal volume = 0.009 m324hr oven dried = 16.13kgSpecific gravity = 2.7Gravity = 9.81 m/s2

Example 1: solution• Wet density:

• Dry unit wt.:

32020009.018.18 mkg

VM

T

Twet ===ρ

( )( )

3

3

2

/58.17009.0

/81.913.16

mkNm

smkgV

gMVW

T

S

T

Sdry

=

=×

==γ

Example 1: solution cont’d

• Water content

• Void ratio

%7.12%10013.16

)13.1618.18(%100% =×−

=×=kg

MM

wS

W

STV

WS

SS

S

V

VVVG

MV

VV

e

−=

=

=

ρ

Example 1: cont’d

• Void ratio……. • Degree of saturation:

( )[ ]

53.00059.00031.0

0031.00059.0009.0

0059.010007.213.16

3

3

===

=−=

−=

=

=

=

S

V

STV

WS

SS

VV

e

m

VVV

m

GM

Vρ

%7.6453.0

)7.2%)(7.12(

%%

=

=

=eGw

S S

Application: Sand Replacement Method

• ASTM D1556 or BS 1377

100%1 w

WW field

dry

+=

Compacted fill

Test hole filled with std. sand

VWdry

dry =γ

In situ density of a soil sample• Mass extracted

from hole = 4.0 kg

• Mass of dry sand to fill the hole = 3.1 kg

• Mass of dry sand to fill container of 4.2 liters = 5.8 kg.

3/38.12.48.5 mMg

sandofvolumesandofmasssandofDensity ===

sandofdensityholefilltosandofmassleVolumeofho =

333

1025.238.1101.3 m−

−

×=×

=

holeofvolumeexcavatedsoilofmasssoilofdensitybulkorwet =

33

3

/78.11025.2100.4 mMg=××

= −

−

Some special soil categories

• Collapsible soil• Liquefaction

• Expansive clays• Laterites

In construction industry, special soil implies a soil type with property or behavior that is considered unusual and capable

of causing problems and therefore requires special treatment.

Collapsible soils

• Soil deposit that experience significant decrease in volume when exposed to water.

Liquefaction• Liquefaction can occur

when saturated cohesionless sand deposit exist in relatively loose condition.

• When subjected to vibration or shock wave, the soil grain will quickly move into denser arrangement

• However water prevented the particle to particle contact causing shear strength lost.

Expansive clay• Clays containing

montmorillonite expand in volume when in contact with water.

• The expansive force created by clay undergoing an increase in water content is capable of lifting heavy structure and imposing excessive lateral pressure.

• Swell forces can go beyond 500kPa

Laterites• Residual soil formed from

weathering of igneous rock under conditions of high temperature and high rainfall.

• Usually reddish in color but not always.

• Agriculturally – not suitable because of low nutrients.

• Acceptable construction material for road bases and embankment or foundation.