University of Wasit

College of Engineering
Civil eng. Division
Karl Terzaghi (18831963)

Soil Mechanics
Lecture notes
Prepared by
Dr. Eng. Asad Hafudh Al-Defae

Grading
Attendance, Class Participation and quizzes
10% for each term
Midterm Exam

(20%)

Laboratory attendance, participation and experiments

reports
Final Exam

(10%)
(60%)

Description of sections
Origin of Soil, soil composition and index

properties

Weight-Volume Relationships, Plasticity and

Structure of Soil

Engineering Classification of Soil

Permeability
Fluid flow in soil
In Situ Stresses (Effective Stress Concept)

Stresses in a Soil Mass

Compressibility of Soil
Shear Strength of Soil
Soil Compaction

Ref.
Textbook- Das, B.M., Principle of Geotechnical
Engineering 5th
edition.
Soil mechanics laboratory manual 6th ed. Braja M. Das
2000
Soil mechanics, R.F. Craig

 The term Soil has various meanings, depending upon the general field in which
it is being considered.
To a Pedologist ... Soil is the substance existing on the earth's surface, which
grows and develops plant life.
To a Geologist ..... Soil is the material in the relative thin surface zone within
which roots occur, and all the rest of the crust is grouped under the term ROCK
irrespective of its hardness.
To an Engineer .... Soil is the un-aggregated or un-cemented deposits of mineral
and/or organic particles or fragments covering large portion of the earth's
crust.
Soil Mechanics is one of the youngest disciplines of Civil Engineering involving the
study of soil, its behavior and application as an engineering material.
According to Terzaghi (1948): "Soil Mechanics is the application of laws of

mechanics and hydraulics to engineering problems dealing with sediments and

other unconsolidated accumulations of solid particles produced by the mechanical
and chemical disintegration of rocks regardless of whether or not they contain an
admixture of organic constituent."

 Soil Mechanics is the branch of civil

engineering that concerns the application of
the principles of mechanics, hydraulics and to
smaller extent, chemistry, to engineering
problems related to soils.

The study of the science of soil mechanics

equips a civil engineer with the basis
scientific tools needed to understand soil
behavior.

Define: Soil
Naturally occurring particulate material.
Formed, directly or indirectly, from solid rocks (i.e.
weathering)
Composition of soil particles depends on composition of

parent rock
The void spaces between the particles contain water and/or
air.

Basic points:
Every civil engineering structure, whether it be a
building, a bridge, a tower, an embankment, a road
pavement, a railway line, a tunnel or a dam, has to
be founded on the soil (assuming that a rock
stratum is not available) and thus shall transmit
the dead and live loads to the soil stratum.
Applications
Engineer uses the SOIL to build:
On it: e.g. buildings

In it: e.g. tunnels

With it; e.g. earth dams

Example: Bridge foundations

Example: Tunnels

Example: Earth dams

Example: Failure

Soil formation
A- Geoloical origin (by nature).
1- mechanical weathering process (disintegration of rocks).
Physical forces.
Impact of sand grain carried by high winds or water or glaciers.
2- Chemical weathering of rocks
Due to water, acids or alkaline
Oxidation union of oxygen with minerals in rocks forming
another minerals
Hydration water will enter the crystalline structure of
minerals forming another group of minerals
Hydrolysis the release Hydrogen from water will union with
minerals forming another minerals
Carbonation when Co2 is available with the existence of
water the minerals changed to Carbonates
B- Fill (man made)

Residual soil: the products of rock weathering remain at their

original location.
Transported soil: the products are transported and deposited in
a different location. (gravity, wind, water).
Marians soil: formed by deposition in the SO4.

Soil-Particle Size or Grain Sizes

Between boulder (D>75 mm) to Ultra fine grain (D<0.001mm)\
Coarse-grained, Granular or Cohesionless Soils
Excellent foundation material for supporting structures and roads.
The best embankment material.
The best backfill material for retaining walls.
Might settle under vibratory loads or blasts.
Dewatering can be difficult due to high permeability.
If free draining not frost susceptible
Fine-Grained or Cohesive Soils
Very often, possess low shear strength.
Plastic and compressible.
Loses part of shear strength upon wetting.
Loses part of shear strength upon disturbance.
Shrinks upon drying and expands upon wetting.
Very poor material for backfill.
Poor material for embankments.
Practically impervious.
Clay slopes are prone to landslides.

Silts
Characteristics
Relatively low shear strength
High Capillarity and frost
susceptibility
Relatively low permeability
Difficult to compact
Compared to Clays
Better load sustaining qualities
Less compressible
More permeable
Exhibit less volume change

Solutions of soil engineering problems

Soil Mechanics
Geological exploration
Experience
Economic

Engineering Judgment
solution of soil
engineering problem

Soil may be divided into three main classes:

1. Coarse grained or non-cohesive soil
2. Fine cohesive soil
3. Organic soil
Coarse grain soil

Fine grain soil

Boulders
Cobble
Gravel
Sand
Silt
Clay

> 300mm
150-300mm
2-150mm
0.06-2mm
0.002-0.06mm
<0.002mm

 Building block of clay mineral

Silicon Oxygen tetrahydra (Sio4) Silica sheet
Aluminium or Magnisium sheet (Al2(OH)3, Mg2(OH)3- Gibbsite sheet

Common clay minerals of (two layers sheet)

Kaolinite

and

Halloysite

 Common clay minerals of (Three layers sheet)

Muscovite (Mica)

and

Illite (Hydrous mica)

Consistency: When clay mineral are present in fine grained soil, and soil can be
remolded in presence of some moisture without rumbling, this cohesive nature is due
to the absorbed water surrounding the clay minerals.
Atterberg limits: the limits are
based on the concept that the fine
grain soil can exist in any four state
depends on its water content.

L.L. = Liquid limit

P.L. = Plastic Limit.
S.L. =Shrinkage limit.
P.I. = Plastic index
P.I.= L.L. P.L.

= . . =

( )

. .
= . . =
. .
. .
= . . =
. . . .

If L.I. < 0

< . .

Soil in solid or semi-solid state

If L.I. = 0

= .

Soil plastic limit

If L.I. > 0

> .

Soil in plastic state

If L.I. = 1

= .

Soil liquid limit

If L.I. < 0

> .

Soil in liquid state

Activity of Clay: because of great increase of surface area per mass (specific surface) it
may be expected that the amount of attached water will be influenced by the amount of
clay minerals that is present in the soil.

Activity of Clay =

. .
% 0.002

Activity of Clay depends on :

Specific surface
Amount of clay minerals
Type of clay mineral

The physical state of a soil sample

Total Volume = Vt = Vs + Vw + Va
Total Weight = Wt = Ws + Ww
Porosity (n): is the ratio of void volume.
n = Vv/Vt
Void Ratio (e):is the ratio of void volume to solid
volume. e = Vv/Vs
now n = Vv/Vt = Vv/ Vv+ Vs = Vv/Vs/ Vv/Vs+1
e=e+1

Note:
n < 1 and is expressed as %
e may be any value greater or smaller than unity.
Example: A soil has a total volume of 250ml and a void ratio of 0.872. What
is the volume of solids of the sample?

Example: A soil has a porosity of 0.45. What is the value of its void ratio?

Water Content, W:
is the ratio of water weight in a soil sample to the solids weight

Ww
wc
100
Ws
Degree of Saturation, S:
is the ratio of water volume to void volume.

Vw
S
100
Vv

Bulk unit weight:

Dry Unit Weight

Submerged Unit Weight

Unit weight
Unit Weight of Water

1.0g/cc =1000kg/ m3
9.8 kN/m3

Specific gravity (GS):

specific gravity of soil solids of a soil is defined as the ratio of
the density of a given volume of the solids to the density of

any equal volume of water at 4 C.

Soil type
G
Gravel
2.65-2.68
Sand
2.62-2.65
Silt
2.66-2.7
Clay
2.68-2.8
Organic soils may be less than 2.0

In term of Gs, e and w

Relationship between n and e

Vv
Vv
V v/ Vs
n

V Vs Vv 1 Vv / Vs
e
n
i.e. n
or e
1 e
1 n
Relationship between w, e, S, Gs

Vv Vv Vw Vv Ww / w
e
Vs Vw Vs Vw Ws / s
Vv Ww Gs w 1
wcGs

.wc.Gs
Vw Ws
w
S
S

W Ws Ww Ws (1 Ww / Ws )
t

V
Vs Vv
Vs (1 Vv / Vs )
Ww
V
W
w, v e, and w s Gs w ,
Ws
Vw
Vs
Gs w (1 wc )
1 e
Se
wc
Gs

G Se
t s
w
1 e

In term of Gs, e and w

W Ws Ww Gs w e w
sat

V
Vs Vv
1 e
G e
or sat s
w

1 e

d in terms of Gs, e and w

d in terms of Gs, w, S and w
e in terms of d, Gs and w
Ws
W Ws
d

V V Vs Vv
Gs w
Gs w
d

1 e 1 wc Gs
S
Gs w
if S 1, d
1 wcGs
Zero Air Void unit weight

Ws
W Ws
d

V V Vs Vv
Gs w
Gs w
d

1 e 1 wc Gs
S

In term of Gs, e and w

d

Ws
W Ws

V V Vs Vv

(1 e) d Gs w
e

Gs w

Gs e
sat w
w w
1 e
Gs 1
or
w
1 e

Relationship between dry and total Unit weight

Example 1: A moist chunk of 25kg soil had a volume of 12000 cm3.

After it is dried in an oven, the weight of the dry soil became 23kg.
The specific gravity of the soil material is 2.65.
Using a phase diagram, determine:
Water content (wc),
Unit weight of moist soil,
Void ratio (e),
Porosity (n),
Degree of Saturation (S)
Solution:
Given:
W=25kg, V=12000 cm3, Ws=23kg, Gs=2.65
w=1g/cm3

Calculate water content:

Ww=W-Ws, Ww=25-23=2kg,
so, wc=Ww/Ws=8.7%

Calculate unit weight of moist soil:

=W/V=25000/12000=2.08g/cm3
Calculate void ratio:
s=Gs w=(2.65)(1)=2.65g/cm3, Vs=Ws/s=8679.2cm3
Vw=Ww/w=2000/1=2000cm3, Vv=V-Vs=12000-8679.2=3320.8cm3
e=Vv/Vs=3320.8/8679.2=0.38

Calculate porosity:
n=Vv/V=3320.8/12000=0.28
Calculate degree of saturation:
S=Vw/Vv=2000/3320.8=0.60=60%

Example 2: A 0.01 m3 of saturated soil sample has a unit weight of

1.6g/cm3. After it is dried in the oven the weight of the soil sample
reduced to 13.5kg.
Use a phase diagram, determine:
Weight of water for the saturated soil, Ww
Water content, wc
Void ratio, e
Porosity, n
Solution:
Given:
Ws=13.5kg, V=0.01 m3, =1.6g/cm3,
w=1g/cm3

0.01 m3

13.5kg

Calculate weight of the saturated soil:

W= V=(1.6)(0.01)(1000000)=16kg

Calculate water content:

Wc=Ww/Ws=(W-Ws)/Ws=[(16-13.5)/13.5](100)=19%

0.01 m3

13.5kg

Calculate void ratio:

Vw=Ww/w=2500/1=2500cm3=0.0025 m3
Vs=V-Vw=0.01-0.0025=0.0075 m3
For saturated soil: Va=0,
e=Vw/Vs=0.0025/0.0075=0.33
Calculate porosity:
n=Vv/V=0.0025/0.01=0.25
Example 3: A soil sample has a porosity of 0.30 and specific gravity of
2.50. Use a phase diagram, determine:
Void ratio, e
Dry unit weight, d
Unit weight if the soil is 50% saturated, 50
Unit weight if the soil is completely saturated, sat

Solution:

Given: n=0.3, Gs=2.5, w=1g/cm3=1000kg/m3

Calculate void ratio: assume V=1 m3
Vv=nV=(0.3)(1)=0.3 m3
Vs=V-Vv=1-0.3=0.7 m3
e=Vv/Vs=0.3/0.7=0.43
Calculate dry unit weight:
s=w Gs=(1000)(2.5)=2500kg/m3
Ws=sVs=(2500)(0.7)=1750kg
d=Ws/V=1750/1=1750kg/m3
Calculate moist unit weight if the soil is 50% saturated:
Vw=SVv=(0.5)(0.3)=0.15 m3
Ww= w Vw=(1000)(0.15)=150 kg
W=Ww+Ws=150kg+1750kg=1900 kg
50=1900/1=1900 kg/m3
Calculate saturated unit weight:
Vw=SVv=(1)(0.3)=0.3 m3
Ww= w Vw=(1000)(0.3)=300 kg
W=Ww+Ws=300+1750=2050kg
sat=2050/1=2050 kg/m3

Example 4: A soil sample has a unit weight of 1.5g/cm3. The moisture content
of this soil is 20% when the degree of saturation is 50%.
Determine Void Ratio, Specific Gravity of soil solid, and Saturated Unit Weight.

Solution:
Given:
=1.5 g/cm3, Wc=20%, S=50%,
w=1g/cm3
Step1: Assume V=1 cm3
W=V=1.5g
Wc=Ww/Ws, so, Ww=0.2 Ws
W=0.2Ws+Ws, so, Ws=W/(1+0.2)
Ws=1.25g
Ww=W-Ws=0.25g
Vw=Ww/w=0.25cm3
Vv=Vw/S=0.5 cm3
Vs=V-Vv=0.5 cm3
e=Vv/Vs=1

W=1.5g

V= 1 cm3

Gs=s/w,
s=Ws/Vs=1.25/0.5=2.5 g/cm3
Gs=s/w=2.5
For the saturated unit weight, Vv=Vw,
so, Ww=wVv=0.5g
sat=(Ww+Ws)/V=1.75g/cm3