stresses, strains, and elastic deformation of soils

CHAPTER 4
STRESSES, STRAINS, AND
ELASTIC DEFORMATIONS OF
SOILS
1

Copyright © 2009 Huawei Technologies Co., Ltd. All rights reserved.
1. Introduction
➢ Definitions
➢ Fundamentals of Stress-Strain
➢ Material Responses to Normal Loading
➢ Stresses in Soil Mass
3

Introduction Definitions
Stress: or intensity of loading, is the load per unit area.
Strain: or intensity of deformation, is the ratio of the change in a
dimension to the original dimension or the ratio of change in length to
the original length.
Stiffness : resistance to deformation
Strength : ability to withstand an applied load without failure or
plastic deformation
Poisson‘s ratio : ratio of transverse contraction strain to
longitudinal extension strain in the direction of stretching force
Modulus of Elasticity: describes tendency of an object to deform along
an axis when opposing forces are applied along that axis
Nominal stress = actual load/original cross-sectional area of specimen,
i.e. no allowance is made for reduction in area, due to necking, as the
load is increased.
4

Introduction Definitions
Stress (strain) state at a point is a set of stress (strain) vectors
corresponding to all planes passing through that point.
Mohr’s circle is used to graphically represent stress (strain) state for
two-dimensional bodies.
Isotropic means the material properties are the same in all directions,
and also the loadings are the same in all directions.
Anisotropic means the material properties are different in different
directions, and also the loadings are different in different directions.
Elastic materials are materials that return to their original configuration
on unloading and obey Hooke’s law.
Plastic materials do not return to their original configuration on
unloading.
5

Introduction cntd
Soils are not homogeneous, elastic, rigid bodies, so the
determination of stresses and strains in soils is a
particularly difficult task.
One may ask: “If soils are not elastic materials, then why
do I have to study elastic methods of analysis?”
❑ An elastic analysis of an isotropic material involves only two
constants—Young’s modulus and Poisson’s ratio—and thus if
we assume that soils are isotropic elastic materials, then we
have a powerful, but simple, analytical tool to predict a soil’s
response under loading. We will have to determine only the
two elastic constants from our laboratory or field tests.
6

Introduction cntd
Stresses and Strains – A Recap of the Fundamentals
Normal Stresses and Strains
Consider a cube of dimensions 𝑥 = 𝑦 = 𝑧 that is subjected to forces
𝑃𝑥 = 𝑃𝑦 = 𝑃𝑧 , normal to three adjacent sides.
7

Introduction cntd
The normal stresses:
𝜎𝑧 =
𝑃𝑧
𝑥𝑦
, 𝜎𝑥 =
𝑃𝑥
𝑦𝑧
, 𝜎𝑦 =
𝑃𝑦
𝑥𝑧
Consider a cube of dimensions 𝑥 = 𝑦 = 𝑧 that is subjected to forces
𝑃𝑥 = 𝑃𝑦 = 𝑃𝑧 , normal to three adjacent sides.
Let us assume that under these forces the cube compressed by
∆𝑥, ∆𝑦, ∆𝑧 in the X, Y, and Z directions.
The strains in these directions, assuming they are small (infinitesimal):
𝜀𝑧 =
∆𝑧
𝑧
, 𝜀𝑥 =
∆𝑥
𝑥
, 𝜀𝑦 =
∆𝑦
𝑦
Volumeteric strain:
𝜀𝑝 = 𝜀𝑥 + 𝜀𝑦 + 𝜀𝑧
8

Introduction cntd
Shear Stresses and Shear Strains
Let us consider, for simplicity, the XZ plane and apply a force F that
causes the square to distort into a parallelogram, as shown below.
9

Introduction cntd
The force F is a shearing force, and the shear stress is
𝜏 =
𝐹
𝑥𝑦
Simple shear strain is a measure of the angular distortion of a body by
shearing forces. If the horizontal displacement is ∆𝒙, the shear strain or
simple shear strain, 𝜸𝒛𝒙, is
𝜸𝒛𝒙 = tan−𝟏
∆𝒙
𝒛
For small strains, tan 𝜸𝒛𝒙 = 𝜸𝒛𝒙, and therefore
𝜸𝒛𝒙 =
∆𝒙
𝒛
10

Introduction cntd
Material Responses to Normal Loading / Unloading
If we apply an incremental vertical load, ∆𝑃, to a deformable cylinder of
cross-sectional area A, the cylinder will compress by, say, ∆𝑧 and the
radius will increase by ∆𝑟.
The loading condition we apply here is called uniaxial loading.
The change in vertical stress is
∆𝜎𝑧 =
∆𝑃
𝐴
The vertical and radial strains are, respectively,
∆𝜀𝑧 =
∆𝑧
𝐻𝑜
∆𝜀𝑟 =
∆𝑟
𝑟𝑜
11

Introduction cntd
Material Responses to Normal Loading / Unloading
The ratio of the radial (or lateral) strain to the vertical strain is called
Poisson’s ratio, 𝜐, defined as
𝜐 =
−∆𝜀𝑟
∆𝜀𝑧
12

Introduction cntd
Linear elastic materials
▪ For equal increments of ∆𝑃, we get the same value of ∆𝑧.
▪ If at some stress point, say, at A , we unload the cylinder and it
returns to its original configuration.
Non-linear elastic materials
▪ For equal increments of ∆𝑃, we get the different value of ∆𝑧.
▪ If at some stress point, say, at A , we unload the cylinder and it
returns to its original configuration.
Elasto-plastic materials
▪ do not return to their original configurations after unloading.
▪ The strains that occur during loading consist an elastic or
recoverable part and a plastic or unrecoverable part.
13

Introduction cntd
14

Introduction cntd
Moduli
❑ The elastic modulus or initial tangent elastic modulus
(E) is the slope of the stress–strain line for linear
isotropic material .
❑ The tangent elastic modulus (Et) is the slope of the
tangent to the stress–strain point under
consideration.
❑ The secant elastic modulus (Es) is the slope of the
line joining the origin (0, 0) to some desired stress–
strain point.
15

Introduction cntd
Stresses in Soil Mass
If we consider an elemental cube of soil at the point considered,
then a solution by elastic theory is possible. Each plane of the
cube is subjected to a stress, σ, acting normal to the plane,
together with a shear stress, τ, acting parallel to the plane.
16

Introduction cntd
Stresses in Soil Mass
➢ There are therefore a total of six stress components acting on the
cube.
➢ Once the values of these components are determined then they can
be compounded to give the magnitudes and directions of the
principal stresses acting at the point considered.
➢ Many geotechnical structures operate in a state of plane strain, i.e.
one dimension of the structure is large enough for end effects to be
ignored and the problem can be regarded as one of two dimensions.
17

Introduction cntd
Stresses in Soil Masses
➢ Geostatic stresses: from self weight of soil
➢ Additional stress: from imposed load
❑ Vertical stress
❑ Horizontal stress
❑ Shear stress
18

Introduction cntd
Vertical pressure at a point below ground surface is normally
induced in four ways:
❑ Geostatic Pressure/Overburden: from self weight of soil;
increases linearly with depth
❑ Hydrostatic Pressure: from ground water; increases linearly
with depth
❑ Additional stress: from an imposed structural load; decreases
with depth in a non-linear manner
❑ Surcharge load: a load of infinite extent on the surface; does
not vary with depth.
19

TOTALAND EFFECTIVE STRESSES
 Effective Stress Principles
 Effect of water table fluctuations on effective stress
 Effective stress in a soil mass under hydrostatic conditions
 Effective stresses in soils saturated by capillary action
 Effective stress and surcharge
 Effective stress and seepage pressure
 Effective stress in partially saturated soils
 Stress distribution in soils due to surface loads
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Effective Stress and Pore Water Pressure
Introduction
Stress is defined as the load divided by the area over which it acts.
As in other materials, stresses may act in soils as a result of an external load and the
volumetric weight of the material itself.
Soils, however, have a number of properties that distinguish it from other materials.
1. Soils can only transfer compressive normal stresses, and no tensile stresses.
2. Shear stresses can only be transmitted if they are relatively small, compared to the
normal stresses.
3. Furthermore, it is characteristic of soils that part of the stresses is transferred by the
water in the pores.
In geotechnical engineering, a compressive stress is considered positive and tensile
stress is negative.
Stress and pressure are often used interchangeably in geotechnical engineering.
In the International System of Units (SI), the units for stress are kPa.
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Effective Stress and Pore Water Pressure
Total Stresses(σ) –stresses in the particles generated partly by the concentrated forces
acting in the contact points between the particles, and partly by the pressure in the
water, that almost completely surrounds the particles.
For the condition of a uniform soil and a level ground surface (geostatic condition), the
total vertical stress (σ) at a depth (z) below the ground surface is:
σ = γz…………………..4.1
where γ the total unit weight of the soil.
For soil deposits having layers with different total unit weights, the total vertical stress
is the sum of the vertical stress for each individual soil layer.
Pore pressure(u)- the pressure transmitted by the water surrounding the soil particles
or pressure due to pore water filling the voids of the soil.
For the condition of a hydrostatic groundwater table (i.e., no groundwater flow or
excess pore water pressures), the static pore water pressure (u) is:
u = γw zw………………….4.2
Where zw = depth below the groundwater table, γw =unit weight of water
Effective stress(σ’)-is a measure for the concentrated forces acting in the contact points
of a granular material.
σ’= σ- u………………..4.3
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3.1. Effective Stress Principles
Definition of Effective stresses
Fig. 3.1 shows a soil mass which is fully saturated. Let us consider a prism of soil with
a cross-sectional area A. The weight P of the soil in the prism is given by
P= γsathA…………………..4.4
Figure 3.1 Saturated soil mass.
Where γsat is the saturated weight of the soil, and h is the height of the prism
Total stress (σ) on the base of the prism is equal to the force per unit area. Thus
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Effective Stress Principles
Pore water pressure (u) is the pressure due to pore water filling the voids of the soil.
Thus
u = γwh……………………4.6
Pore water pressure is also known as neutral pressure or neutral stress, because it
cannot resist shear stresses.
Pore water pressure is taken as zero when it is equal to atmospheric pressure, because
in soil engineering the pressures used are generally gauge pressure and not absolute
pressures.
The effective stress (σ’) at a point in the soil mass is equal to the total stress minus the
pore water pressure. Thus
σ’ = σ-u……………………4.7
For saturated soils, it is obtained as
σ’=γsath-γwh ……………………4.8
or
σ’=(γsat-γw)h
or
σ’=γ’h
Where γ’ is the submerged unit weight.
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Effective Stress Principles
Importance of Effective Stress
The effective stress controls the engineering properties of soils. Compression and shear
strength of a soil are dependent on the effective stress. Thus
compression =f(σ’)
and shear strength = φ(σ’)
Where f and φ represent some functions.
As the effective stress in a soil increases, the compression of the soil occurs.
This causes settlement of structure built on soils.
The shear strength of a soil depends on its effective stress.
As the effective stress changed, the shear strength changes.
The slope stability of slopes, the earth pressures against retaining structure and the
bearing capacity of soils depend upon the shear strength of the soil and hence, the
effective stress.
The importance of shear strength in soil engineering problem cannot be over-
emphasized. It is one of the most important properties of soil.
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3.2. Effect of WaterTable Fluctuations on Effective Stress
Let us consider a soil mass shown in figure 4.2. The depth of the water table (W.T.) is
H1 below the ground surface.
The soil above the water table is assumed to be wet, with a bulk unit weight of γ. The
soil below the water table is saturated, with a saturated unit weight of γsat .
Figure 4.2.
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Effect of WaterTable Fluctuations on Effective Stress
Total down ward force (P) at section X-X is equal to the weight (W) of the soil.
Thus,
P = W = γH1A+ γsatH2A
Where A is the area of cross-section of the soil mass .
Dividing by A throughout,
Where and
The pore water pressure (u) is given by Eqn. 4.6. as
u = γwH2
From equation 4.7
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1 2
P
H H
sat
A
 
 
P
A
  1 2
H H
sat
  
 

Effect of WaterTable Fluctuations on Effective Stress
Equation 4.8 gives the effective stress at section X-X . Figure 4.2 also shows the
direction of σ’ and u at X-X.
a. If the water table rises to the ground surface, the whole of the soil is saturated, and
As γ’<γ, the effective stress is reduced due to rise of water table.
b. If the water table is depressed below section X-X,
σ’ = γH……………………4.11
In this case, the effective stress is increased.
Thus, it is observed that the fluctuation in water table level cause changes in the pore
water pressure and the corresponding changes in the effective stress.
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3.3. Effective Stress in a Soil Mass under Hydrostatic
Conditions
Figure 4.3 shows a soil mass under hydrostatic conditions, wherein the water level
remains constant.
As the interstices in the soil mass are interconnected, water rises to the same elevation
in different piezometers fixed to the soil mass.
Figure 4.3. water table above soil
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Effective Stress in a Soil Mass underHydrostatic Conditions
The effective stress at various sections can be determined using Equation 3.6.
1) Water Table above the soil surface A-A:
Section A-A σ=γwH, u= γwH
Therefore, σ’ = σ - u= 0……………….(a)
Section B-B σ=γwH+ (γsat)1 H1, and u = γw (H+H1)
Therefore, σ’=[(γsat)1 - γw ]H1 = γ1’H1……………….(b)
Where γ1’ is the submerged unit of soil-1.
Section C-C
σ=γwH + (γsat)1 H1 + (γsat)2 H2
u= γw(H+H1+H2)
σ’= γ1’H1 + γ2’H2……………….(c)
Where γ2’ is the submerged unit weight of soil -2
2) Water table at the soil surface A-A
Figure 4.3(b) shows the condition when the depth H of water above the section A-A is
reduced to zero.
In this case, the effective stresses at various sections are determined as follows
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Effective Stress in a Soil Mass under Hydrostatic
Conditions
Section A-A σ = u= σ’ = 0 ……………….(d)
Section B-B σ = (γsat)1H1 , and u =γwH1
and σ’ = γ1’H1 ……………….(e)
Section C-C σ = (γsat)1H1 + (γsat)2H2
u =γw(H1+H2)
Therefore, σ’ = γ1’H1 + γ2’H2 ……………….(f)
From above, the depth H of water above the soil surface does not contribute to the
effective stress at all. In other words, the effective stress in a soil mass is independent
of the depth of water above the soil surface.
3) Water Table in soil-1
Figure 4.4(a) shows the condition when the water is at D-D in the soil-1 at depth H1’.
The effective stresses at various sections are determined as follows.
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Figure 4.4. Water table in (a) soil-1 and (b) soil-2
Section A-A σ = u= σ’ = 0
Section D-D σ = γ1H’1 ,
Where γ1 is the unit weight of soil above D-D. u = 0
Therefore, σ’ = γ1H1’
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Section B-B σ = γ1H1’ + (γsat)1H1” (Note. H1’+H1”=H1)
u = γwH1”
Therefore, σ’ = γ1H1’ + [(γsat)1 - γw]H1” = γ1H1’ + γ1’H1”
Section C-C σ = γ1H1’ + (γsat)1H1”+ (γsat)2H2
u = γw(H2+H1’’)
Therefore, σ’ = γ1H1’ + γ1’H1” + γ2’H2
When H1” = 0, σ’ = γ1H1’ + γ2’H2
4) Water table in soil-2
Figure 3.4(b) shows the condition when the water table is at EE in Soil-2 at depth H2’.
The effective stresses at various section are as under:
Section A-A σ = u= σ’ = 0
Section B-B σ = γ1H1 , u = 0, σ’ = γ1H1
Section E-E σ = γ1H1 + γ2H2’ , u = 0 (Note. H2’ +H2’’ = H2)
σ = γ1H1 + γ2H2’
Section C-C σ = γ1H1 + γ2H2’+ (γsat)2H2” , u = γwH2’’
σ’ = γ1H1 + γ2H2’+ γ2’H2”
where γ2’is submerged unit weight of soil-2
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Effective Stress in a Soil Mass under Hydrostatic
Conditions
5) Water table below C-C
Fig. 4.5 shows the condition when the water table is below C—C.
As the pore water pressure is zero everywhere, the effective stresses are also equal to
the total stresses.
Figure 4.5. Water Table below soil.
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Section B-B σ = σ’ = γ1H1
Section C-C σ = σ’ = γ1H1 + γ2H2
The following points are worth noting in the five cases studied above.
1. The effective stress at any section goes on increasing as the water table goes down.
2. The effective stress depends upon the bulk unit weight above the water table and
the submerged unit weight below the water level.
3. The effective stresses in a soil mass can be determined from the basic definitions,
without memorizing any formula.
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3.4. Effective Stresses in Soils Saturated By CapillaryAction
If the soil above the water table is saturated by capillary action, the effective stresses
can be determined using equation 4.6
However, in this case the pore water pressure above the water table is negative. Figure
4.6 (a)
Figure 4.6 Effect of capillary rise
The water table is at level B-B. let us consider two cases:
1) Soil saturated up to surface level A-A. Figure 4.6(a)
The pore pressure diagram is drawn on the right side.
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Effective Stresses in Soils Saturated By CapillaryAction
The stresses at various sections are determined as under.
Section A-A σ = 0, u = -γwH1
Therefore, σ’ = 0-(-γwH1) = γwH1
If the soil was not saturated with capillary action, the effective stress at section A—A
would have been zero. Thus, the capillary action has increased the effective stress by
γwH1. In other words, the negative pressure acts like a surcharge (q)
Section D-D σ = (γsat)1H1’ (Note. H1’ + H1’’ = H1)
u = -γwH1’’ = -γw(H1-H1’)
Therefore, σ’ = (γsat)1H1’ – γwH1’+γwH1
or
σ’ = γ1’H1’+γwH1
If the soil had been saturated due to rise in water table to A-A, the effective stress at
section D-D would have been γ1’H1’. Thus, the effective stress is increased by γwH1
due to capillary action.
Section B-B
Therefore,
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If the soil above B—B had been saturated due to rise in water table to A—A, the
effective stress would have been γ1’H1 . Thus, the effective stress is increased by γwH1
by capillary action.
Section C-C
Therefore,
At this section also, the effective stress has also increased by γwH1
It may be noted that the effective stress at all levels below the plane of saturation A-A,
due to capillary water, is increased by γwH1. The capillary water pressure acts as if a
surcharge.
2) Soil saturated up to level D-D figure 4.6 (b)
Let us now consider the case when the soil above the water table B-B is saturated only
up to level D-D up to a height H1’’. The soil above level D-D is wet and has a unit
weight of γ.
The capillary rise in this case is H1’’
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The stresses at various sections can be determined as under.
The effective stress is increased by γwH1" due to capillary action
The following points may be noted from the study of both cases
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The following points may be noted from the study of both cases
1. The capillary water above the water table causes a negative pressure γwH where H
is the capillary rise. This negative pressure causes an increase in the effective
stresses at all levels below the saturation level. The increase is equal to γwH. The
capillary action is equivalent to a surcharge q = γwH.
2. If the soil is saturated due to rise in water table, the effective stress depends upon
the submerged unit weight; whereas for the soil saturated with capillary water, the
effective stress depends upon the saturated unit weight. In the latter case, the water
docs not contribute to hydrostatic pressure.
3. If the water table rises to the top soil surface, the meniscus is destroyed and the
capillary water changes to the free water, and the effective stress is reduced
throughout.
4. Equation 4.3 is applicable in all cases. However, it should be remembered that the
pore water pressure in the capillary zone is negative.
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3.5. Effective Stress and Surcharge
Let us consider the case when the soil surface is subjected to a surcharge load of
intensity q per unit area.
Let us, assume that the water table is at level B-B (Fig.4.7). The stresses at various
sections are determined as under.
Figure 4.7. Effect of Surcharge
Section A-A
i.e., all the points on the soil surface are subjected to an effective stress equal to q.
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Effective Stress and Surcharge
Section B-B
Therefore,
Section C-C
Therefore,
From the above illustrations, it is clear that the effective stress throughout the depth is
greater than the case with no surcharge discussed in the preceding section. The
difference is equal to the intensity q. In other words, the effective stress is increased by
q throughout.
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3.6. Effective Stress and Seepage Pressure
As the water flows through a soil, it exerts a force on the soil. The force acts in the
direction of flow in the case of isotropic soils.
The force is known as the drag force or seepage force. The pressure in the soil is
termed as seepage pressure.
• The seepage force affects the interparticle forces and hence the effective stresses.
• The effective stress is increased when the flow is downward, as the seepage force
increases the interparticle forces.
• When the flow is upward, the effective stress is decreased as the seepage force
decreases the interparticle forces.
The two cases are discussed separately below.
a) Downward Flow.
Let us consider the case when the flow is downward (Fig.4.8). The head causing flow
in h. The pore water pressure at sections A-A and B-B are indicated by the
piezometers.
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Effective Stress and Seepage Pressure
The effective stresses at various sections determined using Equation 4.3.
Figure 4.8. Downward Flow
Section A-A
Therefore,
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25

Section B-B
Therefore,
For hydrostatic conditions, the effective stress is γ1’H1.
The second term indicates the effect due to flow.
As (Hw+H1)>Hw1, the effective stress is increased due to downward flow.
Section C-C
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A comparison with the effective stresses corresponding to hydrostatic conditions shows
that the effective stress is increased by γwh.
The conclusion that the effective stress is increased due to downward flow can also be
from intitutive feeling that as the water flows downward, it exerts a drag force in the
downward and causes an increase in the interparticle forces.
b)Upward Flow
Fig 4.9. shows the case when the flow is upward. The piezometers at various elevations
indicate the pore water pressure.
Figure 4.9. Upward Flow
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Section A-A
Therefore,
Section B-B
Therefore,
or
As Hw1 > (H1 + Hw), the term γw (H1 + Hw – Hw1) is negative, and the effective stress is
less than that from the corresponding hydrostatic conditions.
Which shows decrease in effective stress.
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Section C-C
Thus, the effective stress is reduced by γwh from the corresponding hydrostatic
conditions.
In general, the upward flow decreases the effective stresses in a soil mass.
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3.7. Effective Stress in Partially Saturated Soils
In partially saturated soils, air is also present along with water.
It is assumed that air is in sufficient quantity such that there is continuity in both the air
phase and the water phase.
Because of meniscus formation, the air pressure is greater than the water pressure.
It is assumed that the air pressure and water pressure are constant throughout the void
spaces.
Thus, there are three measurable stresses in partially saturated soils, namely, total stress,
σ , pore water pressure uw and air pressure ua.
Let us consider the forces acting on the wavy plane X-X, shown in Figure 3.10(a).
Figure 4.10.Partially
saturated soil
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Effective Stress in Partially Saturated Soils
The wavy plane passes through the points of contact of solid particles. The wavy plane
can be taken as a plane as already mentioned.
Figure 4.10(b) shows the forces acting on the plane.
From equilibrium in the vertical direction.
Where σ = total stress, A total area of the plane, Aw = area of the plane passing through
water,
Aa = area of the plane passing through air, and
ΣN = summation of normal forces acting at the particle to particle contact points.
Dividing throughout by A
Where σ’ = effective stress =
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...........................( )
A N u A u A a
w w a a
   

 
' 1 ...........................( )
w a
w
w
N A A
u u
w a
A A A
or
u a u a b
w a

 

  
   
𝑁
𝐴

aw= Aw/A, and aa = Aa/A = 1 – aw
The area at the points of contact is neglected as it is very small as compared to the area
through the water and that through the air.
Equation (b) can be written as
Equation 3.11 cannot be verified experimentally as it is difficult to measure the area aw
Bishop et al (1990) conducted a large number of tests and give the following equations
for the effective stress in partially saturated soils.
Where χ represents the fraction of the area of soil occupied by water.
It depends mainly on the degree of saturation S. Its value is zero for dry soil and unity
for fully saturated soil.
The value of χ also depends upon the soil structure, the cycle of wetting and drying,
and stress changes.
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Figure 4.11 Variation of χ with S
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33

Worked Example
Example 4.1
A sand deposit is 10m thick and overlies a bed of soft clay. The ground water table is
3m below the ground surface. If the sand above the ground water table has a degree of
saturation of 45%, plot the diagram showing the variation of the total stress, pore water
pressure, and the effective stress. The void ratio of the sand is 0.7. Take Gs = 2.65
Fig . E.4.1(a)
Solution
Bulk density
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1
G Se
s
w e
 

 
  

 
3
3 2
3 2
2.65 0.45 0.70
1000 1744.12 /
1 0.70
1744.12 / 9.81 / 10 17.11 /
x
x kg m
or
kg m x m s x kN m

 

 
 
 

 
 

Worked Example
For saturated soils, S = 0, and
Fig E 3.1 shows the soil profile. The stresses at section B-B and C-C are as under:
Section B-B
σ = 17.11 x 3 = 51.33 kN/m2, u = 0
σ’ = 51.33 kN/m2
Section C-C
σ = 17.11 x 3 +19.33x7 = 186.64 kN/m2
u= 7mx9.81 kN/m3 = 68.67 kN/m3, σ’ = 186.64-68.67 = 117.97 kN/m2
Fig E 4.1 (b) shows the variation of stresses
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35
3
3 2
3 2
2.65 0.70
1000 1970.59 /
1 0.70
1970.59 / 9.81 / 10 19.33 /
x kg m
or
kg m x m s x kN m

 

 
 
 

 
 

Worked Example
i) Total stress , ii) Pore water pressure, iii) Effective stress
Fig E 4.1 (b) shows the variation of stresses
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36
51.33 kN/m2
186.64kN/m2
68.67kN/m2
51.33 kN/m2
117.97 kN/m2

Worked Example
Example 4.2
Determine the increase in the effective stress at section C-C in Example 4.1. when the
water table is lowered by 5 m. Assume that the soil above the water table has the degree
of saturation of 45% throughout.
Solution
σ = 8 m x 17.11 kN/m3 + 2 m x 19.33 kN/m3 = 175.54kN/m2
u = 2 m x 9.81 kN/m3 = 19.62 kN/m2, σ’ = 175.54 – 19.62 = 155.92 kN/m2
the increase in effective stress due to water table lowering is
= 155.92 – 117.97 =37.95 kN/m2
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37

STRESS DISTRIBUTION IN SOILS DUE TO SURFACE
LOADS
Introduction
• The determination of the stress distributions created by various
applied loads has occupied researchers for many years.
• When a load is applied to the soil surface, it increases the
vertical stresses within the soil mass.
• The increased stresses are greatest directly under the loaded
area, but extend indefinitely in all directions.
• Many formulas based on the theory of elasticity have been
used to compute stresses in soils.
• They are all similar and differ only in the assumptions made to
represent the elastic conditions of the soil mass.
• The formulas that are most widely used are the Boussinesq and
Westergaard formulas.
• These formulas were first developed for point loads acting at
the surface.
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38

• These formulas have been integrated to give stresses below
uniform strip loads and rectangular loads.
The extent of the elastic layer below the surface loadings may
be any one of the following:
1. Infinite in the vertical and horizontal directions.
2. Limited thickness in the vertical direction underlain with a
rough rigid base such as a rocky bed.
• The loads at the surface may act on flexible or rigid footings.
• The stress conditions in the elastic layer below vary according
to the rigidity of the footings and the thickness of the elastic
layer.
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STRESSDISTRIBUTIONINSOILSDUETOSURFACELOADS

BOUSSINESQ’S FORMULA FOR POINT LOADS
Figure below shows a load Q acting at a point O on the surface of a
semi-infinite solid.
A semi-infinite solid is the one bounded on one side by a horizontal
surface, here the surface of the earth, and infinite in all the other
directions.
4.12. Vertical pressure within an earth mass
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40
Semi-infinite solid

The problem of determining stresses at any point P at a depth z as a
result of a surface point load was solved by Boussinesq (1885) on the
following assumptions.
1. The soil mass is elastic, isotropic, homogeneous and semi-
infinite.
2. The soil is weightless.
3. The load is a point load acting on the surface
The expression obtained by Boussinesq for computing vertical stress
σz, at point P on figure 3.11 due to a point load Q is
Where, r = the horizontal distance b/n an arbitrary point P below the
surface and the vertical axis through the point load Q.
z = the vertical depth of the point P from the surface.
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41

The values of the Boussinesq coefficient IB can be determined for a number
of values of r/z.
The variation of /„ with r/z in a graphical form is given in Fig. 3.12. It can
be seen from this figure that IB has a maximum value of 0.48 at r/z = 0, i.e.,
indicating thereby that the stress is a maximum below the point load.
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42
Figure 4.13.Values of IB or Iw for use in
the Boussinesq or Westergaard formula

• WESTERGAARD'S FORMULA FOR POINT LOADS
Boussinesq assumed that the soil is elastic, isotropic and
homogeneous for the development of a point load formula.
However, the soil is neither isotropic nor homogeneous.
• The most common type of soils that are met in nature are the water
deposited sedimentary soils.
• When the soil particles are deposited in water, typical clay strata
usually have their lenses of coarser materials within them.
• The soils of this type can be assumed as laterally reinforced by
numerous, closely spaced, horizontal sheets of negligible thickness
but of infinite rigidity, which prevent the mass as a whole from
undergoing lateral movement of soil grains.
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43

Westergaard, a British Scientist, proposed (1938) a formula for the
computation of vertical stress σz by a point load, Q, at the surface
in which µ, is Poisson's ratio. If µ, is taken as zero for all practical purposes,
the above equation simplifies to
where is the Westergaard stress coefficient.
The variation of Iw with the ratios of (r/z) is shown graphically in Fig. 6.12
along with the Boussinesq's coefficient IB.
The value of Iw at r/z = 0 is 0.32 which is less than that of IB by 33 per cent.
Geotechnical engineers prefer to use Boussinesq's solution as this gives
conservative results.
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44

Example 3.3
A concentrated load of 1000 kN is applied at the ground surface.
Compute the vertical pressure (i) at a depth of 4 m below the load, (ii)
at a distance of 3 m at the same depth. Use Boussinesq's equation.
Given
Q = 1000 KN, r = 0 , r = 3m, z = 4m
Required σz =? @ r = 0, r = 3m at the same depth
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45

Example 3.4
A rectangular raft of size 30 x 12 m founded at a depth of 2.5 m below the
ground surface is subjected to a uniform pressure of 150 kPa. Assume the
center of the area is the origin of coordinates (0, 0). and the corners have
coordinates (6, 15). Calculate stresses at a depth of 20 m below the
foundation level by the methods of (a) Boussinesq, and (b) Westergaard at
coordinates of (0, 0), (0, 15), (6, 0) (6, 15) and (10, 25). Also determine the
ratios of the stresses as obtained by the two methods. Neglect the effect of
foundation depth on the stresses (Fig. Ex. 3.4).
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46

LINE LOADS
The basic equation used for computing a, at any point
P in an elastic semi-infinite mass is Eq. (6.11) of
Boussinesq.
By applying the principle of his theory, the stresses at
any point in the mass due to a line load of infinite
extent acting at the surface may be obtained.
The state of stress encountered in this case is that of a
plane strain condition.
The strain at any point P in the Y-direction parallel to
the line load is assumed equal to zero.
The stress σy normal to the XZ-plane (Fig. 6.13) is the
same at all sections and the shear stresses on these
sections are zero.
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48

By applying the theory of elasticity, stresses at any point P (Fig. 6.3)
may be obtained either in polar coordinates or in rectangular
coordinates.
The vertical stress a at point P may be written in rectangular
coordinates as
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49
where, Iz is the influence factor equal to 0.637 at x/z = 0.
Figure 4.14. Stresses due to
vertical line load in rectangular
coordinates

STRIP LOADS
The state of stress encountered in this case also is that of a plane
strain condition.
Such conditions are found for structures extended very much in one
direction, such as
Strip and wall foundations,
Foundations of retaining walls,
Embankments, dams and the like.
For such structures the distribution of stresses in any section (except
for the end portions of 2 to 3 times the widths of the structures from
its end) will be the same as in the neighboring sections, provided that
the load does not change in directions perpendicular to the plane
considered.
Fig. 4.14(a) shows a load q per unit area acting on a strip of infinite
length and of constant width B.
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50

Figure 4.14. Strip Load
The vertical stress at any arbitrary point P due to a line load of qdx
acting at x = 𝑥 can be written as
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51

Applying the principle of superposition, the total stress σz at point P
due to a strip load distributed over a width B(= 2b) may be written
as
The non-dimensional values of σz /q are given graphically in Fig.
6.15. can be expressed in a more convenient form as
The principal stresses σ1 and σ3 at any point P may be obtained
from the equations
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52

Figure 4.15. Non-dimensional values of σz /q for strip load
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53

Example 3.5
Three parallel strip footings 3 m wide each and 5 m apart center to
center transmit contact pressures of 200, 150 and 100 kN/m2
respectively. Calculate the vertical stress due to the combined loads
beneath the centers of each footing at a depth of 3 m below the base.
Assume the footings are placed at a depth of 2 m below the ground
surface. Use Boussinesq's method for line loads.
Fig. Ex 4.5. Three parallel footings
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54

Example 3.5
Solution
According to Boussinesq's method the equation to determine vertical
stress beneath the center of footing
The stress at A is
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55

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