SOILMECHANICS I

Asst. Prof. Dr. Qassun S.

Mohammed Shafiqu
MECHANICS I SOIL

Introduction
Definition of Soil

In civil engineering, soil is any uncemented or weakly cemented accumulation of

mineral particles formed by the weathering of rocks, the void space between the
particles containing water and/or
air.

Based on its method of formation, a soil

is:

1- ​Residual : when the products of weathering remain at their original

location.

2- ​Sedimentary : when the individual particles were created at one location,

transported, and fully deposited at another
location.

3- ​Fill : when it is a man-made soil deposit.

What is Soil Mechanics ?

The term “ Soil Mechanics” was coined by Late Dr. Karl Terzaghi, who is

recognized as the father of soil mechanics. Soil mechanics deals with the

engineering properties and the behavior of soils under

stress.

Why Studying Soil Mechanics

?

As the civil engineering projects are either seated on soil (e.g. buildings
and

roads) or retaining soil (e.g. walls) or constructed of soil (e.g.

embankments)

or passing through soil (e.g. tunnels). Soil mechanics becomes an

important
 subject in civil engineering studies.

CHAPTER ONE

BASIC CHARACTERISTICS OF SOILS

1.1 The Nature of Soils

The nature of soils depends upon the type of destructive process in the formation
of

soil from rock. This process may be physical or

chemical.

In the physical process the resultant soil particles retain the same composition as
that

of the parent rock. The structural arrangement of particles is described as “Single

Grain Structure”, each particle being in direct contact with adjoining particles,

without there being any bond or cohesion between them as shown

below.
Chemical weathering results in the formation of groups of crystalline particles of

colloidal size (<0.002mm) known as the clay minerals. Most clay mineral
particles

are of plate-like form having a high specific surface (i.e., a high surface area to
mass

ratio).

The basic structural units of most clay minerals consist of a silica tetrahedron
and an

alumina octahedron. The basic units combine to form sheet structures (a silica
sheet

or an alumina sheet). The various clay minerals (i.e., Kaolinite, Illite and

Montmorillonite) are formed by the stacking of combinations of the sheet

structures

with different forms of bonding between the combined

sheets.
Figure (1.1) Single grain structure
2
The surface of clay mineral particles usually carry residual negative charges
which

result in cations (positive ions) present in the water in the void space being
attracted

to the particles. Also, layers of water molecules are held around a clay mineral

particle by hydrogen bonding and by attraction to the negative charged

surfaces.
There are two types of forces act between adjacent clay
particles:

1- Repulsive forces due to like charges on the surface of particles. Net

repulsive

gives a dispersed structural form of

particles.

2- Attractive forces (Van Der Waals forces) due to electrical moments existing

within the units. The forces of attraction are responsible for the cohesive
nature
Figure (1.2) Clay minerals basic units

Figure (1.3) Clay minerals: (a) kaolinite, (b) illite and (c)
montmorillonite.
 2
of soils containing a significant proportion of clay mineral particles. Net

attraction gives a flocculent structural form of

particles.

Figure (1.4) Particle size ranges ​Particles sizes in soils can vary from over
100mm to less than 0.001mm. In British

Standards the size ranges as detailed below. According to this classification the
terms

“clay”, “silt”, etc. are used only to describe the sizes of particles between
specified

limits regardless of their mineralogical

condition.

All clay size particles are not necessarily clay mineral particles. If clay mineral

particles are present, they usually exert a considerable influence on the

properties of

soil, an influence out of all proportion to their percentage by weight in the

soil.

1.2 Particles Size

Analysis

The particle size analysis of a soil sample involves determining the percentage
by

weight of particles within the different size

ranges.

Soils consisting of particles mostly in the gravel size are sand size ranges are
referred

to as coarse-grained. The particle size distribution of such soils can be

determined by

the method of sieving. The soil is passed through a series of standard test sieves

having successively smaller mesh

sizes.

Soils consisting of particles mostly in the silt size and clay size ranges are
referred to
 as fine-grained. The particle size distribution of such soils can be determined by
the

2
method of sedimentation using the hydrometer analysis. The method is based on
stockes’ law which governs the velocity at which spherical particles settle in a
suspension; the larger the particles the greater are the settling velocity and vice
versa.
The size of a particle is given as the diameter of a sphere which would settle at
the
same velocity as the particle.The particle size analysis is usually presented in the
form
of distribution curve, this curve is obtained by plotting particle diameter against
percent finer.
Figure (1.5) Particle size distribution curves (Example 1.1)
The size such that 10% of the particles are smaller than that size is denoted by

D​10​. Other
​ sizes such as ​D​30 ​and D​60 ​can be defined in a similar way. The general

slope and
​ shape of the distribution curve can be described by means of the
coefficient of

uniformity, ​C​u ​and the coefficient of curvature, ​C​z ​defined as:

60 ​
CU ​ ​D​10​
​ ​= D ,
2​
C ​Z ​= DD​D ​106030 ​ 1.3 Plasticity of Fine-Grained Soils
The term plasticity describes the ability of soil to undergo unrecoverable
deformation
at constant volume without cracking or crumbling. Plasticity is due to the
presence of
clay minerals or organic material.
Depending on its water content a soil may exist in the liquid, plastic, semi-solid or
solid state. The water contents at which the transitions between states occur are
called
Atterberg limits.
2
Note: All plastics soils at liquid limits passes a constant value of shearing
resistance.
It has been found by means of direct shear tests of different types of clays that
the liquid limit corresponds to a shearing strength of about 27gm/cm​2 ​(2.65kPa). It
is
noticed that the constant shear strength occurs at 25 blows, that way the 25
blows is
chosen.
The upper and lower limits of the range of water contents over which a soil exhibit

plastic behavior are defined as the liquid limit, w​L​, and the plastic limit, w​P​,

respectively. The water content range itself is defined as the plasticity index, I​P​,
wwI ​ = ​ - ​
P​ L ​ P ​The natural water content, w, of a soil relative to the liquid and
plastic limits can be

represented by means of the liquidity index, I​L​,

I ​L ​=
ww - ​I
P P ​The degree of plasticity of the clay size fraction of soil is expressed by the
ratio of the
plasticity index to the percentage of clay size particles in the soil; this ratio is
called
the activity.
The transition between the semi-solid and solid states occurs at the shrinkage
limit,
defined as the water content at which the volume of the soil reaches its lowest
value
as it dries out.
1.4 Soil Classification
The object of soil classification is to divide soils into groups such that all the soils
in a
particular groups have similar characteristics, by which they may be identified,
and
exhibit similar behavior in giving engineering situations. Soil classification,
however,
should be regarded as the first step only in the evaluation of a soil.
Most soil classification systems divide soils into a number of groups each
denoted by
a group symbol. A soil is identified and allocated to the appropriate group on the
basis of particle size distribution and plasticity, these characteristics being
determined
either by standard tests in the laboratory or by simple, and less accurate, visual
and
manual procedures in the field. In addition, a general description of the soil and
its in-
2
situ condition should be given depending on the experience of the site engineer.
For

example;

‘Reddish-brown, dense, homogeneous, well-graded, clean sand consisting

of

subangular particles’.

‘Dark-grey, firm, silty clay of low plasticity with small fissures and silt

inclusions’.

In the unified soil classification system (USCS) the group symbols consist of a

primary and secondary descriptive

letter;

Primary Letter Secondary Letter

G Gravel W Well graded

 S Sand P Poorly graded

M Silt M Non-plastic fines

C Clay C Plastic fines

O Organic L Low plasticity

Pt Peat H High plasticity

The unified soil classification system refers to the plasticity

chart.

A procedure is recommended for field classification which includes the manual

tests;

dilatancy (reaction to shaking), dry strength, and

toughness.

Figure (1.6) Plasticity chart

2
Figure (1.7) Flowchart of Unified Soil Classification
System.
 2
Example 1.1

The results of particle size analyses of four soils A, B, C and D are shown in
Table

below. The results of limit tests on soil D

are:

Liquid limit Cone penetration (mm) 15.5 18.0 19.4 22.2 24.9 Water content (%)
39.3 40.8 42.1 44.6 45.6 Plastic limit Water content (%) 23.9 24.3 The fine
fraction of soil C has a liquid limit of 26 and a plasticity index of 9. (a)

Determine the coefficients of uniformity and curvature of soils A, B and C. (b)

Allot

group symbols, with main and qualifying to each

soil.
solution:

The particle size distribution curves are plotted in Figure below. For soils A, B
and C

the sizes D​10​, D​30 ​and D​60 ​are read from the curves and the values of C​u ​and C​z

are calculated:
​

2
For soil D the liquid limit is obtained from Figure below: The percentage water

content, to the nearest integer, corresponding to a penetration of 20mm is the LL

and

is 42. The PL is the average of the two percentage eater contents, and is 24.
PI=LL-

PL= 18.

Soil A: Coarse fraction=99%, coarser than 2mm=76%, 76/99=77%

˃50%
Gravel GW Soil B: Coarse fraction=98%, coarser than 2mm=2%,
2/98=2% <50%
Sand SP Soil C: Coarse fraction=66%, coarser than 2mm=41%,
41/66=62% ˃50%
Gravel GM Soil D: Using the plasticity
chart CL

1.5 Phase
Relationships
 Soils can be either of two or three-phase composition. In dry soil there are two

phases, namely the solid particles and pore air. A fully saturated soil is also two

phase, being composed of solid particles and pore water. A partially saturated
soil is

three phase, being composed of solid soil particles, pore water and pore air.The

components of a soil can be represented by a phase diagram as shown in Figure

(1.8)

below.

2
Figure (1.8) Phase diagrams
The water content, w or moisture content, m: ratio of the mass of water to the
mass of
solids in the soil
M​
w=​ M​w​s​The water content is determined by weighing a sample of the soil
and then drying the sample in an oven at a temperature of 105​o​-110​o​C and
reweighing. A drying period of
24hr is normally adequate for most soils.

Degree of saturation, S​r​: ratio of the volume of water to the total volume of void
space.
w​
S r​ ​= V ​
​ V​ v​Void ratio, e: ratio of the volume of voids to the volume of solids.
V​
e = ​ V​v​s​Porosity, n: ratio of the volume of voids to the total volume of the soil.
V​
n = ​ V​v ​n =
​
e1

​ ​e, =
+e ​ n1 ​- n
​ ​Air content, A: ratio of the volume of air to the total volume of
the soil.
V​
A = ​ V​a ​Bulk density, ρ: ratio of the total mass to the total volume.
2
M​
V​ρ ​= ​ Specific gravity of the solid soil particles, G​s​.
=
G ​s ​

V​M ​ws
=​
ρ ​s ​ ρ s​ ​ρ​w​ρ ​w =
​ ​ρ ​of water
Assume the volume of solids is 1 unit, then the volume of voids would be e units.
M,GM ​ = ​ ρ ​ = wG ​ ρ ​V, = wG ​ = wG ​ ​
s​ ws ​ w​ wws ​ s ​S r​ ​ e​s For fully saturated

soil, S​r​=1, then e=wG​s

A = wGe
​ - ​e1+​s ​= S1n (​ ​- r​ )​ ​ρ ​=
​ 1
w1G s​ ​( e

( ​+​+ ​) ) ​ρ w​ ​ρ ​= eSG
​ e1 rs ​
​ + ​+​ρ w
​ ​For a saturated soil: ​ρ s​ at ​= eG
​ ​e1 s​

+​+ ​ρ ​w For
​ a dry soil:

ρ d​ ​=
s​
G ​
e1​ + ​ ρ w
​
2
 W​ Mg
V​γ ​= ​ V= ​

the above equations for ρ are valid for γ.

e.g. ​γ

= w1G
​ ​s ​e1

(​ )​
+​+ ​ γ ​w ​= eSG
​

​ ​+​+ ​γ ​ww ​, ​γ ​= m/kN8.9 ​3 ​For a fully saturated soil, the buoyant uit
e1 rs
weight, ​γ​′ i​ s,

γ ​′
wws
​ ​
=G

γ ​- ​γ ​+= s​ ​- e1 ​+​γ w ​ ​Relative density, RD, is used to express the

​ ​= γ​ ​sat ​- ​γ w
actual density of a soil relative to the
maximum and minimum possible densities;
max ​
e ​max​e ​min ​1G e1

​ ​-​- e ​Procedures are recommended for determination of the maximum

RD = e
and minimum void
ratio of a soil sample.
Example 1.2 ​In its natural condition a soil sample has a mass of 2290gm and a

volume of 1.15×10​- ​3​m​3​. After being completely dried in an oven the mass of the

sample is 2035gm. The

value of G​s ​for the soil is 2.68. Determine the bulk density, unit weight, water
content, void ratio, porosity, degree of saturation and air content.
 ρ=M​ =​
solution: ​Bulk density, ​ V​ 1015.1 290.2
​
× ​-
=​
3​ )m/Mg99.1(m/kg1990 ​3 3 ​Unit weight, ​γ = Mg
​ ​V= 1990
​ ​× 8.9
​ ​=

m/kN5.19m/N19500 3​ ​= 3​ ​Water content, ​w

=
M​ -​
M​w ​s​= 2290 ​2035​ 2035 = %5.12or125.0 ​Void ratio, w1Ge
​ =

s​ ( ​+ )​ ​ρ ​ρ​w ​- 125.168.21 = ⎛​ ​│⎝× × 1000

​ ​1990 ⎞​ -│​⎠​52.01 = ​Degree
of saturation, Air content, ​%1.12or121.0355.034.0S1nA =
S ​r ​( -​ ​=

r​ ​ ​= ​es​ ​= 68.2125.0
) wG
×​=​
52.0 ​ × = ​ %5.64or645.0
2
1.6 Soil Compaction

Compaction is the process of increasing the density of a soil by packing the

particles

closer together with a reduction in the volume of air; there is no significant

change in

the volume of water in the soil.

Field ​compaction means

are:

1. Rollers:

a. Smooth-wheeled

b. Sheepsfoot Depending on the type of soil.

 c. Pneumatic-tyred

d. Drum

2. Vibrators coarse grained soils.

3. Rammers small areas and where access is difficult.

Figure (1.10) a drum-type roller

Figure (1.9) a sheepsfoot roller ​

The degree of compaction of a soil is measured in terms of dry density, i.e. the
mass

d=
​ ​ ρ​
of solids only per unit volume of soil. ​ρ ​ w1 + ​ The dry density after

comaction depends on the type of soil, the water content and the

effort supplied by the compaction

equipment.

Laboratory ​compaction tests

are:

1. Proctor test.

2. Modified AASHTO test.

 2
3. Vibrating hammer test.

Dry density is plotted against water content as shown

below.

The curve shows that for a particular method of compaction (i.e., a particular

compactive effort) there is a particular value of water content, known as the

optimum

water content, w​opt​, at which a maximum value of dry density is obtained. If

​ all
the air in a soil could be expelled by compaction, the soil would be in a state of

full saturation. The maximum possible value of dry density is referred to as the
‘zero

air voids’ dry density or the ‘saturation’ dry density so

that,
In general:
Figure (1.11) Dry density-water content
relationship

The calculated relationships between ρ​d ​and w for different air contents can be
obtained.

2
Figure (1.12) Dry density-water content curves for different compactive
efforts

Figure (1.13) Dry density-water content curves for a range of soil

types.
 Laboratory compaction tests are not directly applicable to field compaction; they

provide only a rough guide to the water content at which the maximum dry
density

will be obtained in the field. The main value of laboratory tests is in the
classification

and selection of soils for use in fills and

embankments.

2
Relative compaction: ​percentage of the field dry density after compaction
relative to
the maximum dry density in a particular laboratory test.
A number of methods for measuring bulk density in the field are recommended
elsewhere. The required standard of field compaction may be specified in terms
of
relative compaction; for example a specification may state that the dry density
should
not be less than 95% of the maximum dry density in the BS 2.5kg rammer test.
Example 1.3 ​Natural water content of a soil in the borrow areas is 8% and its
bulk density is 1.6 gm/cm​3​. This soil is to be used in construction of an
embankment. The specification
for embankment compaction require its water content to be 10% and dry density
of 1.65 gm/cm​3​. Compute the quantity of soil to be excavated per 100m​3​of the
s
embankment. The value of G​s ​is 2.7. solution:
​ ​ρ = w1G
​ ​

e1 ​( +
​ +​ ​) ρ​ ​w ​6.1
=
08.017.2 ​e1 (​ ​+​+ ​b
)
e
=
V​ V
V ​V​v ​s​V, ​s =
​ ​ e​ v ​ but V​ is constant ​( ​
s ​

) f​ or the borrow and embankment 6363.0

​ 8225.0​ρ ​w ​⇒ 8225.0e ​b ​= ρ ​d

= ​e1​G
s​
+ ​ ρ ​w ​65.1, =
​ ​e1 7.2
​ ​+ ρ
​ ​w ​⇒ 6363.0e ​e ​= ​e )V(
​ ​ev ​= v​ b ​, )V(
​

(V
​
bv v

) e​ ​=
8225.0 ​6363.0 = 2932.1
​ ​n
=

e1

​ ​n, e​
+e
= ​6363.1
V​
6363.0 = 3888.0 = ​( ​ V​v ​) ​e ​=

8225.1
V​
= 4513.0 = ​(
V​
​ V​b )​​ for 100m​3 ​of embankment ​( ​ v​) ​e =​ m88.38 ​3 ∴
​
 ​ × = m28.50 ​3 ​
( 2932.188.38V
​ v ​) ​b ​ 4513.0 = ​
28.50 V​
b​∴ V ​b =
​
m41.111 ​3 ​e

n
b
8225.0 ​v b
2

CHAPTER TWO
SEEPAGE
2.1 Phase Relationships
The pressure of the pore water is measured relative to atmospheric pressure, and
the
surface within the ground at which the water is at atmospheric pressure is called
the
water table or more generally the phreatic surface. Artesian conditions may exist
sometimes.
Above the water table, water can be held at negative pressure by capillary
tension and
the zone affected by it is called the capillary zone. The soil between the ground
surface and the capillary zone forms the zone forms the zone of aeration. In this
zone
the soil is also able to retain small droplets of water, surrounded on all sides by
air.
This water known as contact moisture. The negative pressure of water in the
capillary
zone results in attractive forces between particles; this attraction is referred to as
soil
suction.
Bernoulli’s theorem applies to the pore water seeping through soil:
h
V​
= ​ g2​2 ​+ ​γ
p

w​
​ ​The velocity head can be neglected, thus
+Z
h
u​ =​ +​
Z ​w ​ γ ​ h=total head, u=pore water pressure, ​γ ​w​=9.8 kN/m​3​, Z=elevation
head above a
chosen datum.
2
2.2 Permeability
Darcy’s empirical law assuming a fully saturated soil and in one dimension.

​ ​A​ki q
vq
=​
Aki == ​q=volume of water per unit time, A=cross sectional area of soil ,
k=coefficient of
permeability, i=hydraulic gradient, v= discharge velocity.
k depends primarily on the average size of pores and for a given soil k is a
function of
void ratio.
k varies with temperature upon which the velocity of the water depends.
γ​
k = ​ η​w​K ​η​=viscosity of water, K= absolute coefficient depending only on the

soil skeleton.
Table (2.1) Coefficient of permeability (m/s) (BS 8004; 1986).
k values for different soils are within expected ranges for sands approximately.

mminD),s/m(D10k ​= 2-
​ ​10 2​​ 10 The
​ average velocity at which the water flows
the soil pores is obtained by dividing
the volume of water flowing per unit time by the average area of voids. Av, on a
cross
section normal to the macroscopic direction of flow, this velocity is the seepage
velocity, ​V′​:
v
q​ v
nA​
ki ​
​ ​A​v
nn​v q
V​ A​
An, ​ v ​V′ ==≈​ v ​′ = = = ​Determination of the coefficient of permeability:
2
1. Laboratory methods:

a) Constant head permeability test for coarse-grained

soils:

• ​Disturbed specimen at the appropriate density is contained in a Perspex

cylinder of cross-sectional area A; the specimen rests on a coarse filter

or a

wire mesh.

• ​A steady vertical flow of water, under a

constant total head, is maintained through
the

soil and the volume of water flowing per unit

time (q) is measured.

• ​Tappings from the side of the cylinder enable

the hydraulic gradient (h/l) to be
measured.

• From Darcy’s law:

kiAq =

ql ​
k​ ​ h ​lkq, =
Ah​i = ​ h ​l​A =
b) Falling head permeability test for fine-grained
soils:
 • ​Undisturbed specimens are normally tested and the containing cylinder in
the
test may be the sampling tube
itself.

• ​The length of the specimen is (l) and the cross-sectional area

A.

• ​Acoarse filter is placed at each end of the specimen and a standpipe of

internal
area (a) is connected to the top of the
cylinder.

• ​The water drains into a reservoir of constant level.

• ​The standpipe is filled with water and a measurement is made of the time (t​1​)

for the water level (relative to the water level in the reservoir) to fall from h​o ​to

h​1​.

• ​At any intermediate time t the water level in the standpipe is given by h and
its
rate of change by (-dh/dt).

• ​At time t the difference in total head between the top and bottom of the
specimen is h. Then applying Darcy’s
law:

2
l-
h​
dh ​
a ​ dt​ = Ak ​ -
a
h​∫ ​1​∫ ​1 ​dt ​h
o

∴ = = ​• ​Precautions must be taken to ensure that the degree of saturation

remains close to 100%.
• ​A series of test should be run using different

values of h​o ​and h​1 ​and/or standpipe of different

​ diameters.
c) Consolidation test:
2. Insitu methods:
a) Well pumping test
• ​This method is most suitable for use in homogeneous coarse soil strata.
• ​The procedure involves continuous pumping at a constant rate from a well,
normally
at least 300mm in diameter, which penetrates to the bottom of the stratum under
test.
• ​A screen or filter is placed in the bottom of the well to prevent ingress of soil
particles.
• ​Perforated casing is normally required to support the sides of the well.
• ​Steady seepage is established, radially towards the well, resulting in the water
table
being drawn down to form a cone of depression.
• ​Water levels are observed in a number of boreholes spaced on radial lines at
various
distances from the well.
• ​The test enables the average coefficient of permeability of the soil mass below
the
the cone of depression to be determined.
• ​Analysis is based on the assumption that the hydraulic gradient at any distance r
from the center of the well is constant with depth and is equal to the slope of the
water table.
dh ​
h= Ak

l​t​0

k
At​al ​1
o
h ​
ln ​ h​
h ​ o​
1​3.2 At​ al ​
1 l
​ og ​ h​ 1
2

rh2A, ​i
r

= dh ​dr​r ​= π rhk2q
= π dh ​
dr​q r​​ ∫ ​2​r 1​

dr ​r​=hdhk2 π ​h​2​∫ ​h
1

lnq
=

r ​r​12​

- ​( ​( ​2​2 1 ​2​- ​ For a confined stratum of thickness H, the

π=​
​ π ​2​2 hhk
( hh ​
)​
area through which seepage takes place is ​rH2​π ,​ where r is variable and H is
constant. Then
()
( ) ( ​12 )​​ 1 2​​ ) ​k
dh ​
r/rlogq3.2 ​12 ​) r​ Hk2q = π ​ dr∴
q
r​∫ ​2​r 1​

dr ​r​= dhHk2 π ​h​2​∫ ​h

1

lnq
r ​r​12​
∴
= hhHk2 π ​12 ​- ​

k
=​
r/rlogq3.2 hhH2 π ​12 ​- ​b) Bore hole tests
• ​The general principle is that water is either introduced into or pumped out of a
borehole which terminates within the stratum, the procedures being referred to as
inflow and out flow tests respectively.
• ​A hydraulic gradient is thus established causing seepage either into or out of the
soil
mass surrounding the borehole and the rate of flow is measured.
• ​In a constant head test (Fig, a), the water level is maintained throughout at a
given
level.
• ​In a variable head test (Fig, b) the water level is allowed to fall or rise from its
initial
position and the time taken for the level to change between two values is
recorded.
2
• ​The tests indicate that the permeability of the soil within a radius of only 1-2m
from
the center of the borehole.
• ​A problem in such tests is that clogging of the soil face at the bottom of the
borehole
tends to occur due to the deposition of sediment from the water. And to alleviate
the
problem, the borehole may be extended below the bottom of the casing (Fig, c),
increasing the area through which seepage takes place. The extension may be
uncased or supported by perforated casing depending on the type of soil. Another
solution is to install within the casing a central tube perforated at its lower end
and
set within a pocket of coarser material.
• ​Expressions for k depend on whether the stratum is unconfined or confined, the
position of the bottom of the casing within the stratum and details of the drainage
face in the soil.
• ​If the soil is anisotropic with respect to permeability and if the borehole extends
below the bottom of the casing (Fig, c), then the horizontal k tends to be
measured.
• ​If the casing penetrates below soil level in the bottom of the borehole (Fig, d)
then
vertical k tends to be measured.
• ​For constant-head test
Fhq ​
k=​ c ​F=intake factor with values published by Hvorslev

• ​For a variable head test, ​

k= ​
​ ttF
h​
( A3.2
​ ​12 ​- ​) log
​ ​ 1

h​
2​.
• ​k for a coarse soil can also be obtained from insitu measurements of seepage
velocity. The method involves excavation uncased boreholes or trial pits at two
points A and B (Fig, e). Seepage taking place from A towards B. (i) is given by
the
difference in the steady-state water level in the boreholes divided by the distance
AB. Dye or any other suitable traced is inserted in to A and the time taken for the
Dye to appear in B is measured. The seepage velocity is then the distance AB
divide
 nv ′ ​
by the time. ​k =​ i​, n=porosity from density tests.
2
2.3 Seepage theory
Assume soil is homogeneous and isotropic and the seepage is in two dimension
(X-Z
h​ h​
plane). ​x​ k kiv ​x x ​∂∂​ -​ = = ​z​ k kiv ​z z ​∂∂​ ​- = =

Total head h decreasing in the direction of v​x and

​ v​z​. Assume
​ a fully saturated
element dx dy dz.

q(entering the element)=v​x​dydz+v​z​dxdy q(leaving

​ the element)= ​dxdy dz ​z​v ​v

dydz dx x​v ​v z​ ​z x​ ​x ​│⎠​⎞ │⎝​

​ ⎛
∂​
∂​ + │​⎠​⎞ ​│⎝​⎛
∂​
∂​ +
Assume no volume change in the element and incompressible water then:
q(entering)-q(leaving)=0
v​
0 ​z​v x​ zx​
= ∂​∂ ​+ ∂​∂ ​(equation of continuity) ...(1)
2

dV ​Assume a volume change in the element per unit time= ​dt :​ ⎛​ ​│⎝​dV dt
v​ ∂​
∂​ x ​∂​x+ ​ ∂v ​z​z ​⎞ ​│⎠dxdydz = ​Consider a function ​φ ​(x,z) called the

potential function such that,

∂​∂​φ ​x
∂h​
= kv x​ ​= - ​ ∂x∂ ​φ​∂z
∂h​
= kv ​z​= - ​ ∂z​....(2) ​∂​∂ ​zx 2​
φ ​2​+ ∂​ ​∂​2 ​φ ​2​= 0 ​Upon integration of equation (2): ​φ ​)z,x(kh)z,x( = -
+constant
If the ​)z,x(​φ i​ s given a constant value it will represent a curve on the x-z plane
along which the value of total head is constant. If ​)z,x(​φ i​ s given a series of

constant values ​( ​φ ​321 ​etc...,,, φ

​ φ ​) ​a family of curves is specified along
each of
which the total head is a constant value (h1,h2, h3,...etc). These curves are
called
equipotentials. Consider a second function ​ψ ​( )​z,x ​called the flow function
such that:

-
ψ​
∂ ​∂​ x
∂h​
= kv z​ ​= - ​ ∂z​-
∂h​
∂ ​∂​ψ​z =
​ kv ​x​= - ​ ∂xd ​ψ​=

∂ ​∂​ψ ​x dx
​ ​+ ∂​ ​∂​ψ ​z dzvdxvdz
​ = - ​z ​+ ​x ​If ​ψ ​( )​z,x ​is given a constant
dz ​ v
​ 0 ​∴
value ​ψ ​1​, then d​ψ = dx​ = v​
z​
x​ ...(3)
2

The tangent at any point on the curve represented by ​ψ ​( )​z,x ​= ​ψ ​1​specifies

the direction of the resultant discharge velocity at that point: the curve represents
the flow path. If the function ​ψ ​
​( )​z,x ​is given a series of constant values ​( ψ
321 etc...,,,
​ ​ψ ψ ​) ,​ a second family of curves is specified, each representing a
flow path. These curves are called flow lines.
∆
q
= ​ψ​∫​2​ψ ​1
( ​- dzvdxv ​z
+ ​x ​)

=
ψ​
∫ ​2​ψ ​1

dx ​ ∂ ​ dz ​
⎛ ​│⎝​∂ ​∂​ψ ​x ​ + ​ ∂​ψ ​z ​ =

ψ
∂​ =
​ψ =
12 ​- ​ constant​ dzvdxv ​
⎞ │​ d ​ ∂ ​ φ ​
⎠​ φ ​= ​ ∂​ x ​ dx + ​ ∂​ φ ​ dz
z​ ​
I f ​

+​
)z,x(​φ ​x ​ is ​z ​constant then ​φ​d =​ 0
dz ​ v
dx= - ​
v​x​z​...(4) from equations (3) and (4) the flow
lines and the equipotentials interest
each other at right angles.
∂​ x​ ∂​
sinvv,cosvv ​x = ​ α ​ =
​ s​ z​ s​​ α ​ ∂ ​ φ
∂​ ​
x s= ​ ∂​ φ ​
∂ ​ ∂​ zs+ ​ ∂​φ ​( )()
s
z​ ∂
∂s​= ​
=
cosv
x​α ​+ sinv ​z ​α ​=
cosv
s
 =
2 ​α ​ sinv ​s 2​ ​α ​
+
v

∂​ψ
s ​∂​

∂​ x​ ∂​ ∂​ z​
xn= ​ ∂​ψ ​∂ ​∂​ zn+ ​ ∂​ψ ​∂​ n​ =
- sinv
z
- ​α ​+ cosv ​x ​α ​=
- sinv ​s

α ​- sin ​α ​= cosv ​s 2​ ​α ​v
2
∂​
sn ​∴ ∂​​ ∂​ψ ​= ​ ∂​φ ​or approximately ∆
​ ​∆​ψ ​sn = ∆
​ ​∆​φ ​2.4 Flow Nets
A flow net is a family of flow lines and a family of equipotentials. The flow net is
used to solve seepage problems.
In construction of a flow net:
1. Every intersection between a flow line and an equipotential must be at right
angles.
2. It is convenient that ​ψ​∆ =
​ constant, ​φ​∆ =​ constant
3. It is also convenient that:
∆ ns = ∆ ​(curvilinear square formed by flow lines and equipotentials).
h​
Since ​∆ ​ ∆​ ∆∴ ​ψ ​= ∆ ​φ ​ ​ s∆ ​ψ ​= ∆ ∆ ​φ ​= ∆ ∆∴ = ∆
​∆​ψ ​sn = ∆ ​ φ ​ i
h=difference in total head between the first and last equipotentials.

N​d​=number of equipotentials drops, each representing the same total head loss

h∆ ,​
Nh ​
N​f​=number of flow channels, each carrying the same flow ​q∆ ,​ ∆
​ h=​ d​,q
 ∆​
hkqhk = ​ ∆​and
h ​f ​
kNhkNqNq = f​ ​∆ = f​ ​∆ = ​ N​d ​∴
N ​f​
khq = ​ N​d
2
Example (1.2)
The figure shows a line of sheet piling driven 6m in to a stratum of soil 8.6m thick,
underlain by an impermeable stratum. On one side of the piling the depth of water
is
4.5m; on the other side the depth of water (reduced by pumping) is 0.5m. Draw
the
flow net and calculate the seepage
rate (q).
solution​:
∆
h
h​
= N​ d

h=​ ​12​4 = m33.0 ​khq

=

N f​ ​N​d​= 36.04k × × ​3
γ​
( ( ) ) ​=
γ​ppw

​
( zh
+​
​
) impervious boundary

upstream downstream
=
k44.1
m​
s​stream

h
p
=
d​
N​ ​ d​h =
n ​ 10 ​12​× 4 = m33.3 u
pwp
=
h - - z ​p ​1st​ ​trial
after several trials
not squares but the length to breadth ratio should be constant within the flow channel
2
Example 2.2
The section through a sheet pile wall along a tidal estuary is given in figure below.
At
low tide the depth of water in front of the wall is 4m; the water table behind the
wall
lags 2.5m behind tidal level. Plot the net distribution of water pressure on the
piling.
solution​:
At level 4:

​ ​12​m21.05.2 ​h​b ​m83.15.2 = ​12​8 × = ​f​= × = ​The net pressure on the

h1
back of the piling is

​ ​-
9.155.521.08.95.583.18.9uu fb
= (​ ​+ )​ ​- (​ ​+ )​ ​= kN
​ ​m​2
2
Example 3.2
A river bed consists of a layer of sand 8.25m thick overlaying impermeable rock;
the
depth of water is 2.5m. Along cofferdam 5.5m wide is formed by driving two lines
of
sheet piling to a depth of 6m below the level of the river bed and excavation to a
depth of 2m below bed level is carried out within the cofferdam. The water level
within the cofferdam is kept at excavation level by pumping the flow of water into
the cofferdam is 0.25m​3 ​/hr per unit length. Find k of sand and I below excavated
surface.
solution​:

​ ​df
​ N/Nh ​( q
25.0 ​k =
) ​= 6010/65.4
s/m106.2 ​
×× ​ ​=
2 ​ × -​ 5

s∆ b​ etween the last two equipotentials=0.9m

i ​=
h​
∆ ​ ∆​s= ​9.010

5.4 ​× =
​ 5.0
2
Example 4.2
The section through a dam is shown in figure below. Determine the quantity of
seepage under the dam and plot the distribution of uplift pressure on the base of
the
dam. The coefficient of permeability of the foundation soil is ​s/m105.2 ​ ​.
× 5-
solution​:
)mper(s/m101.3 ​khq
=

N f​​ N​d​= 105.2 × ​-

7.4 ​
5 ​× 0.4 × ​ 15=
× ​-
35
2
2.5 Anisotropic soil conditions

k​max​=maximum k value in the direction (x) of stratification k​​ min​=minimum k value in

the direction (z) normal to that of stratification k​​ x​=k​max ​and k​z​=k​min
h​
ikv ​x ​= ​xx ​= - k ​x ∂​ ​∂​h ​x​kikv, z​ ​= ​zz ​= - ∂​ ​ z ​∂z​
​ in a direction s inclined at
angle ​α t​ o the x direction
h​ h​ ∂​ h
kv ​s ​= - ∂​ ​ s ​∂​s∂ ​ ∂​ s= ​ ∂​
x∂
x​ ∂​ h
∂​ s+ ​ ∂​
∂
z​
z ​ v ​s ​
∂s​ k​s​=

k​v
x​ z​
x​cos α
​ ​
+ ​ k​ ​ z​sin ​α ​sinvv,cosvv ​x =
v ​ ​z =
​ ​s α ​ ​∴
​ ​s α
1​
k​ s ​= cos
2α
​ ​
+ ​x2​
α ​k ​ s ​2
z​∴ k​
x​ 2
s​= k​ x​

z
+ k​
2

v​
z​the directional variation of permeability represent an ellipse ∂
​ ​ x​
∂​x​+ ∂​ ​∂​v

​ ​equation of continuity
z​z ​= 0
k ​k​x ​z​sin k
2
2

x ​2​z 2​⎛ │​│⎝​h 0 ⎞ │​│⎠​∂​∴ k ∂​ ​∂x​h ​+ k ∂​ ​∂z+ ∂​ ​∂​= h = 0 ∂ ​2

h
2

2 2x​assume:

z​
xx t​ ​= k​ ​ k​x​∂ ​2
h​ 2
∂​x​2​t​+ ∂​ ​∂z​
h ​2​= 0 ​equation of continuity for an isotropic soil in the x​t​, z
2
This equation provides a scale factor in the x-direction which transforms a given
anisotropic flow region into a fictitious isotropic flow region.
A flow net may be drawn for the transformed section and then obtained for the
natural
section by applying the inverse of the scaling factor if the latter is necessarily
required.
Consider an elemental flow net field through the x-direction drawn to the
transformed
and natural scales.

kv
x
∂h​ h​
= - ′ ∂​∂ ​x​h ​t =
​ - k ​x ​ ∂x∂ ​ ∂​x​t

= ∂​ h
k z​ ​k​x​∂

​ ​k
∂=

​ ​z ​k​x=
xk ​ kk ​zx ​k/k

xz ​kk ′ = x​ k​ ​z ​k​x∂​ x ​kk

′
x
= ∂​ x/h ∂ ​
∂
z​
/h k​ ​ ∂ x x/h ​= kk
k ​
x ​ ​zx ​k′​is the equivalent isotropic coefficient which is the
value of the permeability
coefficient applying to the transformed section.
2.6 Non-homogeneous soil conditions
Assume that the two layers to be a single homogeneous anisotropic layer of
thickness

(H​1​+H​2​) in which the coefficients in the directions parallel and normal to that of

stratification are ​kandk ​x z respectively.

​
=

k
x
∂ ∂ x/h
2
1) Horizontal seepage

the equipotentials in each layer is vertical on the boundary H​1​=H​2,​ then the

vertical line
​ through the two layers should be a common potential, thus i​x ​is also

common. ikHkHikHHq
​ ​x =
​ ​( 1​ +
​ ​xx2 ) ​ ​(
​ = ​11 +​ ​x22 )​ ​k x​ ​= kHkH
​ ​11 ​HH

​ 2 ​+​2 ​2) Vertical seepage

1 ​+ 2
The discharge velocity through each layer and the equivalent single layer should
be
the same to satisfy continuity:
ikikikv ​ = ​ = = ​i ​ is the average I over (H​
z​ 2211zz ​ z​ 1​+H​ 2​)

i 1​ ​= k​ ​z ​k​1i​ ​z ​and ​i 2​ ​= k​k ​2​z ​i ​z loss

​ in head over (H​1​+H​2​) is equal to sum of

HiHi)HH(i ​ + ​ = ​ +​
losses in H​1 ​and H​2 ​ 1z ​ 2 ​ 2211 ​ =

ik
⎛ │​ H​ ⎞ │​
zz ​ │⎝​ k​11​ ​+ H ​k​2​2 ​ │⎠​∴
/HHk
H ​ 1​
1 ​ 2 ​) ​⎛ │​
​ 2​
z​ (
=​ +​ ​ ​k​2​ ⎞ │​│⎠​Similar procedure may be applied for
│⎝​ k​1​ + H
any number of layers. Note that ​k ​x​must

always be greater than ​k ​z​k ​x =

​
 ​ kHkHk ​2211 +​ + 33
H​1 ​( H ​ ​+ ...... )​ .​ ...HHHH = ​1 +
​ ​2 +
​ ​3 +
​ ​k

z ​=

( ​k/Hk/Hk/H ​11 ​) ​+ ​( ​H ​22 ​) ​+ ​( ​33 ​) ​+ ...... ​Tranfer condition:

If seepage takes place diagonally across the boundary between two isotropic
soils 1
and 2 having coefficient of permeability k1 and k2 respectively.
2
φ ​ = - ,hk ​ φ ​= - hk ​
22 ​ 1​ 211 ​ at B , h1=h2: ​φ ​kk
1​
1​=

φ 2​
1​
2​differentiating along the boundary, s ​ sk​1 ​∂ ​∂​φ ​1
1​
= sk​ 2

∂​ ∂​ v​ v
φ ​2 ​ ∴ ​ k​s1 ​1​= ​
1v
k​s2 ​2​for continuity of flow across the boundary ; vv
​ ​n1 =
​ ​n2 ​
s1 ​
k​1 ​v​n1​=
1
k​
vv
2​
s2
n2​

∴ tan
​

tan ​α​α
1​ k
2​= k​​
1

2​∆ ​ ​= ∆
ψ ​ ​∆​n
n ​ hk ​
s​ ∆ φ
​ ​
, ​ ∆∴ ​ = ​ ∆​ s​ ∆ ​If on both sides of the boundary each of ​q∆ ​and
q ​ ∆ ​

h∆ ​are equal.
∆
⎛ ​│⎝​1 ​1 ​
∆​
n ∆​s​2 2​ ​⎞ │​⎠k = ​ ⎛ ​ │⎝​ n ∆​s​⎞ ​│⎠​k ​curvilinear squares ​⎛
∆​
│​⎝​ ∆n
⎞ │​ = ​
s​ ​ ⎠​ are possible in only one soil ​∴ ⎛
1 1
​

∆n​ kk
│⎝​ ∆s⎞ │​ ⎠​ ​
2 =
​
1

2​2.7 Seepage through earth dams

In previous problems the boundary conditions were completely defined by the
physical boundaries of the problem. Sometimes the seepage is unconfined and
one
boundary of the flow region being a pheriatic surface (free surface flow) on which
the
pressure is atmospheric.
For example, the homogeneous isotropic earth dam on an impermeable
foundation.
2
On CD (top flow line) the total head is equal to the elevation head and there must
​ etween points of intersection with equipotentials.The boundary
be equal ​z∆ b
condition
of the flow region.
The function of the filter is to keep the seepage entirely within the dam; water
seeping
out onto the downstream slope would result in the gradual erosion of the slope.
2.8 Flow net of earth dams
Since the uppermost flow line (line of seepage) is phreatic, it cannot be
determined by
the physical boundaries. Therefore it should be located before the flow net can be
constructed.
Assume the flow lines to be confocal parabolas with the same focus at A, and the
equipotentials lines are conjugate confocal parabolas. The two families of
parabolas
satisfy the requirements of a flow net.
To locate the phreatic surface CD, make use of the property that the distance AP
from
the focus to any point P is equal to the distance PM from the point P to the
directrix.
xx2zx
22

+ = ​o ​- xxx4x4zx
22

+ = ​o 2​​ - ​o +
​ ​2 ​zx4xx4

o
= ​o 22​
​ -∴
xx
= ​o
z ​2 ​
- x4​ o
2
The known point of this parabola is at C because of the boundary condition. At C:

​ ​2​
x1 ( )​lhl lx2lh,lxandhz
= = - 22​
​ + = ​o ​+ ∴

o= ​ 2 ​+
​ 2 - ​with ​x ​o​known, the various points on the basic parabola may be
calculated and the
flow net then drawn.
kx2q = ​
Using the complex variable theory: ​ o ​It is recommended to take the initial

point at G where GC=0.3HC. The coordinates of

point G is then substituted in the equation to find ​x ​o​and then plot the basic
parabola.
A correction CJ is made using personal judgment. The flow net can then be
completed.
If the discharge surface AD is not horizontal then the filter is neither a flow line
nor
an equipotential. A correction to the basic parabola must be made as in the figure
and
table below.
Table (2.2) Downstream correction to basic parabola. Reproduced from A.
Casagrande (1940) ‘Seepage through dams’, in Contributions to Soil Mechanics
1925-1940. by permission of the Boston Society of Civil Engineers
2
2.9 Seepage control in earth dams
High hydraulic gradients caused by seepage through the earth dam may erode
channels within the dam. The process of internal erosion is referred to as piping.
Thereby, a central core of low permeability may be used to reduce the volume of
seepage.
Erosion may take place at the downstream boundary if the core due to the high
exit
gradient. A chimney drain (filter) may provide a barrier to soil particles from the
core
also serves as an interceptor keeping the downstream slope in an unsaturated
state.
If the foundation soil is more permeable than the dam, use an impermeable
cut-off
such as grout curtain or an impermeable upstream blanket.
Most earth dams are non-homogeneous due to zoning therefore the transfer
condition
(

tan tan ​α​α 1​

k
2​= ​
1​
k​2​) must be satisfied at all zone boundaries.
2
2.10 Filter requirements
1) Small size for the pores to prevent particles being carried in from the adjacent
soil.
2) High k to allow rapid drainage.
Criteria
D
f=soil of filter , s=soil adjacent to filter ​( ​
) ( ​D​85
​
) ​s 5to4
​ ​ ​
15 f ​〈 ​⇒​to prevent piping (
D

) ( ​D​15
​ D
) ​s 15 f ​〉 5to4
​ (​
) ( ​D​50
​
) ​s 25
​ to ensure a sufficiently high k ​50 f ​〈
for drainage purposes
Filters comprising two or more layers with different grading can also be used:
such
arrangement is called a graded filter.
2.10 Frost heave
• Increase in volume of water due to freezing is 9% ​⇒ ​V ​v​increase 9%, the
overall increase in volume of soil is 2.5%-5% depending on e.
• A greater increase in V may occur due to the formation of ice lenses.
• Water freezes initially in large pores, higher soil suction develops and water
migrates towards the ice in the larger voids where it freezes and adds to the
volume of ice causing the formation of ice lenses. The process continuous only
if the bottom of zone of freezing is within the zone of capillary rise.
• After thawing the soil contains an excess of water making it soft with a
reduced strength.
• The worst conditions for water migration occur in soils having a high
percentage of silt-size particles.
2
2.11 Grouting
• Grouting is the process used for reducing the permeability of coarse-grained
soils.
• The process consists of injecting suitable fluids, known as grouts, into the pore
space of the soil; the grout subsequently solidifies, preventing or reducing the
seepage of water.
• Grouting also results in an increase in the strength of the soil.
• Fluids used for grouting include mixes of cement and water, clay suspensions,
chemical solutions, such as sodium silicate or synthetic resins, and bitumen
emulsion.
• Injection is usually effected through a pipe which is either driven into the soil
or placed in a borehole and held with a packer.
• The particle size distribution of the soil governs the type of grout that can be
used.
Example 5.2
A homogeneous anisotropic embankment dam section is detailed in figure below,
k in
the x and z directions being

105.4 × ​- 8 ​and ​s/m106.1 ​ ​, respectively. Construct the flow net and

× 8-
determine the quantity
of seepage through the dam. What is the pore water pressure at point P.
solution:
2
k​z ​
6.0 ​ k​x​= 6.1 ​5.4​= ​The equivalent isotropic permeability is
k ′ = ​( ​kk ​zx ​) ​= ​( ​106.15.4 × ​) ​× -​ 8 ​= s/m107.2 × -​ 8
m1.80.273.0HC3.0GC = = × = ​the coordinates of G are ​0.18z,8.40x

= - = + -∴ ​x8.40

= ​o ​- 0.18 ​x4 o​ 2​ ​Hence: ​m9.1x​o =

​
using equation:

xx =
2​
o​
- x4​z ​ o​x 1.9 0 -5.0 -10.0 -20.0 -30.0

z 0 3.8 7.24 9.51 12.9 15.57

hkq =
​
′ N f​ ​N​d​= ​107.2 × -​
​ ​ 8 ​× ​18 ×
-​
​ 8.3 ​18​= s/m100.1 × ​ 37
s/m100.19.1107.22xk2qor = ′ ​o ​= × × -​ 8 ​× = × -​ 37 ​Level AD is selected
as datum. An equipotentials RS is drawn through P. By
inspection the total head at P is 15.6m. At P the elevation head is 5.5m, so the

pressure head is 10.1m and the pore water pressure , u​P ​is

m/kN991.108.9u ​P =
​
× = 2​
2

CHAPTER THREE

EFFECTIVE STRESS
3.1 Introduction

solid particles incompressible

Soil water incompressible

 air compressible

• Volume reduction is through escape or compression of air in the voids

provided there is scope for particle

rearrangement.

• In fully saturated soil, reduction is only possible if water is allowed to escape

from the voids.

• Shear stresses are resisted only by solid particles through forces developed
at

the inter-particle contacts.

• Normal stresses are resisted by solid particles and also by water in fully

saturated soil.

3.2 The principle of effective

stress

The principle which applies only to fully saturated soils relates the following three

stresses: 1) the total normal stress (​σ )​ on a plane within the soil mass, being
the force per unit

area transmitted in a normal direction

across

the plane. 2) the pore water pressure (​u )​ ,

being the

pressure of the water filling the void space

between the solid particles. 3) the effective

normal stress (​σ​′​) on the

plane representing the stress transmitted

 P​
through the soil skeleton only. ​A​ , ​A​N ​=

′ ​= ′ ∑​ ​σ σ ​If point contact is assumed

between

particles, the pore pressure will act on the

2
plane over the entire area A (the total contact area is normally 1-3% of A)
∴ ,uANP
P​ ′​
= ​∑​+′ ​ A= ∑​ ​A​N ​ + u ​It ​σ s​ hould =
​

​ ​be u
σ +′ ​

understood that ​σ​′​does not represent the true contact stress between two
​ here a
particles, which would be the random but very much higher stress ​a/N′ w
is
the actual contact area between the particles.
Effective vertical stress due to self-weight of soil: ​σ ​v =
​ ​γ ​sat z
​ ​u will be
hydrostatic since the void space between the solid particles is continuous, ​zu
=

γ
w
σ​
GS WT ​∴​ v

σ​
′=​ v ​- u
=
( ​γ​sat
- ​γ ​w ​) ​zz
= ​γ ​′ ​z
where ​γ​′​is the buoyant unit weight of soil
3.3 Response of effective stress to a change in total stress
lateral confinement
increment of an increase in total vertical stress ​
immediate equal increment of increase in pore water pressure (excess u)
(no arrangement of particles due to lateral confinement) (undrained
condition)
transient flow (drainage) of pore water towarissipation of u)ds a
free-draining boundary until reaching a steady-state pore water pressure
rearrangement of particles with a resulting increase in the inter- particle
forces (increasing ​σ​′​)
​ ill be carried entirely by the soil skeleton (equal
increament of ​σ w
incremental increase in ​σ​′​) (drained condition)
2
• Particle rearrangement is largely irreversible upon a reduction in ​σ (​ small

elastic strain is ignored), however limited expansion may occur specially if

appreciable proportion of clay minerals exist, this cause a negative u, the

process called swelling.

3.4 Consolidation
analogy

• The mechanics of the one-dimensional consolidation process can be

represented by means of a simple spring

analogy.
• The spring representing the compressible soil skeleton, the water in the

cylinder the pore water, the bore diameter of valve the permeability of the
soil.

• The cylinder itself simulates the condition of no lateral strain in the soil.

Assuming water to be
incompressible.

• If load applied on the piston with the valve closed, the load will be carried by

the water. The situation with the valve closed corresponds to the
undrained

condition in the soil.

• If the valve is now opened, water will be forced out through the valve at a
rate

depends on bore diameter. The piston will move and the spring will

compressed as load gradually transferred to

it.

• Eventually all the load will be carried by the spring and the piston will come
to

rest, this corresponding to the drained condition in the

soil.

• At any time, the load carried by the spring is the effective stress in the soil,
the

pressure of water in the cylinder the pore water pressure and the load on
the

piston the total stress.

• The movement of the piston represents the change in volume of the soil and
is

governed by the compressibility of the

spring.

2
3.5 Partially saturated soils

• In the case of partially saturated soils part of the void space is occupied by

water and part by air.

• The pore water pressure (u​w​) must always be less than the pore air pressure

(u​a​) due to surface tension.

• Unless the degree of saturation is close to unity the pore air will form

continuous channels through the soil and the pore water will be
concentrated in

the regions around the interparticles

contacts.

• The effective stress equation for partially saturated soils: ​( ) ​wa a uu

​ u--

+′ = ​χ σ σ ​where ​χ ​=a parameter, to be determined experimentally, related

to S​r ​of the

soil. ​( ) ​wa uu
​ - ​=a measure of the suction in
the soil.
• For a fully saturated soil​( )​1S​r =​ ​, ​χ ​=1; and for a completely dry soil, ​χ ​=0, ​(
)​0S​r =​ ​. ​Example 1.3
A layer of saturated clay 4m thick is overlain by sand 5m deep, the water table
being

3m below the surface. ​( ) ​sat γ​ ​of the clay and sand are 19 and 20 3​​ m/kN ​,
respectively; above the water table ​( ) ​γ ​of the sand is 17 3​​ m/kN ​. Find the
values of

total vertical stress and effective vertical stress against depth. If sand to a height
of

1m above the water table is saturated with capillary water, how are the above
stresses

affected.

solution:

The alternative calculation of ​v​σ​′ ​at depths of 5 and 9m is as

follows:

3​
γ γ ​′ = ​buoyant of
​ sand = ​ m/kN2.108.920 = -
2
3​
γ γ ​′ = ​buoyant of
​ clay = ​ m/kN2.98.919 = -

= × + × = ′​σ ​
At 5m depth: ​( ) ( ) 2​ ​v m/kN4.712.102173
​ At
9m depth: ​( ) ( ) ( ) 2​ v​ m/kN2.1082.942.102173
​
=×+×+

× = ′​σ

Effect of capillary rise: The only effect of 1m capillary rise, therefore, is to

increase

3​
​ f sand between 2 and 3m depth from 17 to 20 ​
the ​t​γ o m/kN ​, an increase of 3

3​m/kN .​ ​vσ ​ nd ​v​σ​′ ​below 3m depth are therefore increased by

​ a

2​
m/kN0.313 = × ,​ ​u b​ eing unchanged.

3.6 Influence of seepage on effective

stress

u​ u​
h z ​ z ​ ,h hh ​A w​A ​B w​B ​A B ∆-
​ + = + ∆-
 = ​γ γ

│​ ⎞ ​ │​ ⎛ ​ u​
│​ ⎠​ │​ ⎝​ ∆- - + = hzz ​ u BA
​

w​A ​wB ​γ γ​ ​( )​hzz uu ​BAwA B ∆-

​ -

​ - ​( )​h
+ = ​γ ​θ ​sinbzz ​BA =
sinb uu ​wA B ∆-
​ + = ∴ ​θ γ

( ​uu ​C =​ ​wA +​ ​γ ​sinb ​θ ​+ cosb ​θ ​∆- ​)​h uu ​D =​ ​wA +​ γ​ ​cosb ​θ ​∴ uuuu

AB -​ = ​wDC -​ = ​γ ​( s​ inb ​θ ​∆- ​)h​ uuuu ​AD -​ = ​wBC -​ = ​γ ​cosb ​θ ​boundary
water force on ​BC = ​γ
=
w​ ( ​
s inb θ
​ ​
∆ - )
​ ​
b h ​
= γ
​ ​ w​ sinb ​
2 θ
​ ​
∆ - bh γ
​ ​ w​ boundary water force on C
​ D ​
γ​
​ ​
γ ​w ​cosb 2​ ​θ ​if there is no seepage, the forces would be ​γ ​w ​sinb​2 ​θ and w

​ ​
cosb​2 θ
the resultant ​
γ​
w​
242​sinb
( θ​ ​+ cos ​2 θ​ ​) =​ ​γ ​w b​ ​2 ​∴​seepage
​

force, ​bhJ = ∆ ​γ ​w​(acting in the direction of flow)

seepage force: the frictional drag on the solid particles, acting in the direction of
flow,
caused by the water seeping through pores.

i=∆​ ​b​h ​J, = ∆ ​ ​2 ​VibiJ = ​γ ​w 2

​ ​b​h ​γ ​wb ​ ​= ​γ ​w ,​ ​V​is the volume of soil
element seepage pressure ​j i​ s the
ij = ​γ ​ , ​j d​ epends on ​i
seepage force per unit volume ​ w​

The resultant body force:

1) Total (saturated) weight + resultant boundary water force, or
2) Effective (buoyant) weight + seepage force
The quick condition:
Critical hydraulic gradient, ​i ​c​: the value of hydraulic gradient corresponding to
zero

​ γ
resultant body force. ​VVi wc ​ ​= γ​ ​′ ​∴ ​i c​ ​= ​γ​γ
′​
w​ = 1G e1
-
s ​+​
2

if ​i ​c​is reached the soil would be in the quick condition and the sand surface will
appear to be boiling, the effective normal stress on any plane will be zero.
Conditions adjacent to sheet piling:

i
m

=
hh ​DC ​- d ​AB ​= 0h
​
=h​ i​ i
DC ​d - ​i ​m ​ d​m ​F = i​ c​
m​ exit ​ii ​ ​∆​h ​s​)AEFG( ​F = ​
= ​e​i​e ​= ∆
i​c​e​Practically no significant difference between
the two factors of safety (average h along the

boundary CD is h​m​) calculation

​ of the resultant body force
(combination of gravity and seeping forces):
1) total weight (ABCD) ​=
1​ d ​2 ​
2​γ ​sat​ CDonu =​γ ​mw (​ )​dh + ​Boundary water force on ​CD = d
​ ​γ ​2​w
)dh( ​m ​+ ∴​
​ Resultant body force of
2

( ​ABCD =

​ ​2​(
1 ​2​γ ​sat ​d 2​ ​- d ) ​dh ​m ​+ (​ ​) ​γ ​w )​
1 ​2d 1
​ ​2​dh=
1 ​2​γ

+′ ​γ ​w ​d 2​ ​- 1 ​2​ddh ​m ​+ 2​ ​γ ​w ​=
γ
′ 2​ ​- ​wm ​γ ​or

2) effective weight ​)ABCD( =

2​
1 ​2​γ​′ d
​ ​ average ​i ​through ​ABCD = h
​ ​d​m ​Seeping force on ​ABCD =
h​ d​
d​m ​γ ​w ​ 2​2
=1

2​dh​wm γ ​ ​Resulting body force of ​ABCD = 1 ​ ​2​γ ​′ d 2​ ​- 1

​ ​2​dh​wm ​γ ,​ heaving
1​ d ​2 ​ γ​
will take place
​ when ​1 ​2​dh​wm γ
​
​= ​ 2​γ ​′ ​ ∴ F = d5.0 ​ ​ ′ ​2 ​dh5.0
wm

γ​= γ​
′dh
i
wm ​
γ​= i​​
e
m​
Example 2.3 ​The flow net for seepage under a sheet pile wall is shown in figure
below, the
 3​
saturated unit weight of the soil being 20 ​m/kN ​ . Determine the values of ​σ​′ ​v​at
A
and B. Determine also F against failure by heaving adjacent to the downstream
face
γ​ = m/kN20 ​3 ​of the soil.
of the piling. ​ sat ​
2
solution:

1- ​considering the combination of total weight and resultant boundary water

force. ​m5.58 ​ A ​= × = ​m7 z​A -​ = ​, ​(

​12​2.8 h​ ) ( ) 2​ ​AAwA
m/kN12275.58.9zh u = + = - =​γ ​
boundary water force on

bottom surface= ​kN122

net vertical boundary water force= ​kN8339 122 =

-

total weight of the column= ​kN220

vertical component of resultant body force= ​kN137 83 220

=-

2 ​A m/kN137
​ = ′​σ

or

2 ​w sat A m/kN259
​ 39 220 4 11 = + = + = ​γ
γσ

2 ​A m/kN122
​ u = ​2 ​AA A m/kN137
​ 122 259 u =
- = - = ′ ​σ σ ​2 ​wat sat B m/kN1308.9120
​ 16=
 - = + = ​γ γ σ ​m7 h​B -​ = ​( ) ( ) 2​ ​BBwB
m/kN8476.18.9zh u = + = - =​γ

2 ​BB B m/kN4684
​ 130 u = - = - = ′ ​σ σ
2- ​considering the combination of effective weight and seepage
force.

number of equipotential drops between D and

A=3.8

loss in total head between D and A ​m5.28 ​12​8.3

=×=
Average value of vertical component of seepage pressure between D and A,
acting in

3​
the same direction as gravity ​ m/kN3.28.9 ​11​5.2 ​=

×=

γ​′​of soil, ​′​γ ​= m/kN2.108.920 - = 3​​ for column AD, of unit area, resultant body
force ​= ​ ​) =​ ​′​σ ​A =​ m/kN137 ​2 ​for point B:
kN1373.22.1011 ​( +

loss in total head between B and C ​= 4.2

​ ​12× m6.18
​ ​= Average
​ value of
vertical component of seepage pressure between B and C, acting in
3​ ′​σ ​ =
the opposite direction to gravity ​= 6.1
​ ​ 6× ​ m/kN6.28.9 ​ = ​ hence, ​ B​

m/kN466.22.106 ​ - ​ = ​2 ​
(​ )​ Now the stability of the soil mass EFGH is
analyzed
c​
​ 3.2 ​6​= 39.0 ​i ​ = ​γ​γ
​ 5.3 ​12​× m3.28 = ​i​m =
h​m =
′=​ =​
w​ 2.10 ​8.9 ​ 04.1 ​F
=
i​
i​ m​c ​= 04.1 ​39.0
= 7.2
2