SOIL MECHANICS - CHAPTER 4 ppt

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Chapter 4
STRESSES IN SOILS
4.1 Stresses
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. Firstly, a special property is that soils can only transfer compressive normal stresses,
and no tensile stresses. Secondly, shear stresses can only be transmitted if they are relatively small, compared to the normal stresses. Furthermore

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x
y

σ
xx
σ
xy
σ
y y
σ
y x
σ
xx
σ
xy
σ
y y
σ
y x
Figure 4.1: Stresses.
it is characteristic of soils that part of the stresses is transferred by the water in the pores. This will
be considered in detail in this chapter.
Because the normal stresses in s oils usually are compressive stresses only, it is standard practice to
use a sign convention for the stresses that is just opposite to the sign convention of classical continuum
mechanics, namely such that compressive stresses are considered positive, and tensile stresses are
negative. The stress tensor will be denoted by σ . The sign convention for the stress components is
illustrated in Figure 4.1. Its deﬁnition is that a stress component when it acts in a positive coordinate
direction on a plane with its outward normal in a negative coordinate direction, or when it acts in
negative direction on a plane in positive direction. This means that the sign of all stress components
is just opposite to the sign that they would have in most books on continuum mechanics or in applied
mechanics.
It is assumed that in indicating a stress component σ
ij

the ﬁrst index denotes the plane on which the
stress is acting, and the second index denotes the direction of the stress itself. This means, for instance,
that the stress component σ
xy
indicates that the force y-direction, acting upon a plane having its normal in the x-direction is F
y
= −σ
xy
A
x
,
where A
x
denotes the area of the plane surface. The minus sign is needed because of the sign convention of soil mechanics, assuming that the
sign convention for forces is the same as in mechanics in general.
4.2 Pore pressures
Soil is a porous material, consisting of particles that together constitute the grain skeleton. In the pores of the grain skeleton a ﬂuid may be
present: usually water. The pore structure of all normal soils is such that the pores are mutually connected. The water ﬁlls a space of very
complex form, but it constitutes a single continuous body. In this water body a pressure may be transmitted, and the water may also ﬂow
25
Arnold Verruijt, Soil Mechanics : 4. STRESSES IN SOILS 26
through the pores. The pressure in the pore water is denoted as the pore pressure.
In a ﬂuid at rest no shear stresses can be transmitted. This means that the pressure is the same in all directions. This can be proved
by considering the equilibrium conditions of a small triangular element, see Figure 4.2, bounded by a vertical plane, a horizontal plane and a

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. . . . . . . . . .
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pA
pA
Figure 4.2: Pascal.
sloping plane at an angle of 45
◦
. If the pressure on the vertical plane at the right is p, the force on that plane
is pA, where A is the area of that plane. Because there is no shear stress on the lower horizontal plane, the
horizontal force pA must be equilibrated by a force component on the sloping plane. That component must
therefore also be pA. Because on this plane also the shear stress is zero, it follows that there must also be
a vertical force pA, so that the resulting force on the plane is perpendicular to it. This vertical force must
be in equilibrium with the vertical force on the lower horizontal plane of the element. Because the area of
that element is also A, the pressure on that plane is p, equal to the pressure on the vertical plane. Using a
little geometry it can be shown that this pressure p acts on e very plane through the same point. This is often
denoted as Pascal’s principle.
If the water is at rest (i.e. when there is no ﬂow of the water), the pressure in the water is determined by the location of the point considered
with respect to the water surface. As shown by Stevin the magnitude of the water pressure on the bottom of a container ﬁlled with water,

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Figure 4.3: Hydrostatic water pressure depends upon depth only.
depends only upon the height of the column of water and the volumetric weight of the water, and not upon the shape of the container, see
Figure 4.3. The pressure at the bottom in each case is
p = γ
w
d, (4.1)
where γ
w
is the volumetric weight of the water, and d is the depth below the water surface. The total vertical force on the bottom is γ
w
dA.
Only in case of a container with vertical sides this is equal to the total weight of the water in the container. Stevin showed that for the other
types of containers illustrated in Figure 4.3 the total force on the bottom is also γ
w
dA is. That can be demonstrated by considering equilibrium
of the water body, taking into account that the pressure in every point on the walls must always be perpendicular to the wall. The container at
Arnold Verruijt, Soil Mechanics : 4. STRESSES IN SOILS 27
the extreme right in Figure 4.3 resembles a soil body, with its pore space. It can be concluded that the water in a soil satisﬁes the principles of
hydrostatics, provided that the water in the pore space forms a continuous b ody.
4.3 Eﬀective stress
On an element of soil normal stresses as well as shear stresses may act. The simplest case, however, is the case of an isotropic normal stress,
see Figure 4.4. It is assumed that the magnitude of this stress, acting in all directions, is σ. In the interior of the soil, for instance at a
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Figure 4.4: Isotropic stress.
cross section in the center, this stress is transmitted by a pore pressure p in the water, and by stresses in
the particles. The stresses in the particles are 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. It can be expected that the deformations of the particle skeleton are almost
completely determined by the concentrated forces in the contact points, because the structure can
deform only by sliding and rolling in these contact points. The press ure in the water results in an equal
pressure in all the grains. It follows that this pressure acts on the entire surface of a cross section, and
that by subtracting p from the total stress σ a measure for the contact forces is obtained. It can also be
argued that when there are no contact forces between the particles, and a pressure p acts in the pore
water, this same pressure p will also act in all the particles, because they are completely surrounded
by the pore ﬂuid. The deformations in this case are the compression of the particles and the water caused by this pressure p. Quartz and water
are very stiﬀ materials, having an elastic modulus about 1/10 of the elastic modulus os steel, so that the deformations in this case are very small
(say 10
−6
), and can be disregarded with respect to the large deformations that are usually observed in a soil (10
−3
to 10
−2
).
These considerations indicate that it seems meaningful to introduce the diﬀerence of the total stress σ and the pore pressure p,
σ

= σ − p. (4.2)
The quantity σ

is denoted as the eﬀective stress. The eﬀective stress is a measure for the concentrated forces acting in the contact points of

a granular material. If p = σ it follows that σ

= 0, which means that then there are no concentrated forces in the contact points. This does
not mean that the stresses in the grains are zero in that case, because there will always be a stress in the particles equal to the pressure in
the surrounding water. The basic idea is, as stated above, that the deformations of a granular material are almost completely determined by
changes of the concentrated forces in the contact points of the grains, which cause rolling and sliding in the contact points. These are described
(on the average) by the eﬀective stress, a concept introduced by Terzaghi. Eq. (4.2) can, of c ourse, also be written as
σ = σ

+ p. (4.3)
Terzaghi’s eﬀective stress principle is often quoted as “total stress equals eﬀective stress plus pore pressure”, but it should be noted that this
applies only to the normal stresses. Shear stresses can be transmitted by the grain skeleton only.
Arnold Verruijt, Soil Mechanics : 4. STRESSES IN SOILS 28
It may be noted that the concept is based upon the assumption that the particles are very stiﬀ compared to the soil as a whole, and also
upon the assumption that the contact areas of the particles are very small. These are reasonable assumptions for a normal soil, but for porous
rock they may not be valid. For rock the compressibility of the rock must be taken into account, which leads to a small correction in the formula.
To generalize the subdivision of total stress into eﬀective stress and pore pressure it may be noted that the water in the pores can not
contribute to the transmission of shear stresses, as the pore pressure is mainly isotropic. Even though in a ﬂowing ﬂuid viscous shear stresses
may be developed, these are several orders of magnitude smaller than the pore pressure, and than the shear stresses than may occur in a soil.
This suggests that the generalization of (4.3) is
σ
xx
= σ

xx
+ p, σ
y z
= σ

y z

,
σ
y y
= σ

y y
+ p, σ
zx
= σ

zx
, (4.4)
σ
zz
= σ

zz
+ p, σ
xy
= σ

xy
.
This is usually called the principle of eﬀective stress. It is one of the basic principles of soil mechanics. The notation, with the eﬀective stresses
being denoted by an accent, σ

, is standard practice. The total stresses are denoted by σ, without accent.
Even though the equations (4.4) are very simple, and may seem almost trivial, diﬀerent expressions may be found in some publications
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p
σ

σ

Figure 4.5: Eﬀective stress.
especially relations of the form σ = σ

+ np, in which n is the porosity. The idea behind
this is that the pore water pressure acts in the pores only, and that therefore a quantity np
must be subtracted from the total stress σ to obtain a measure for the s tress es in the particle
skeleton. That seems to make sense, and it may even give a correct value for the average stress
in the particles, but it ignores that soil deformations are not in the ﬁrst place determined by
deformations of the individual particles, but mainly by changes in the geometry of the grain
skeleton. This average granular stress might be useful if one wishes to study the eﬀect of stresses
on the properties of the grains themselves (for instance a photo-elastic or a piezo-electric eﬀect),
but in order to study the deformation of soils it is not useful. Terzaghi’s notion that the soil
deformations are mainly determined by the contact forces only leads directly to the concept of
eﬀective stress, because only if one writes σ

= σ −p do the eﬀective stresses vanish when there
are no contact forces. The pore pressure must be considered to act over the entire surface to obtain a good measure for the contact forces, see
Figure 4.5.
The equations (4.4) can be written in matrix notation as
σ
ij
= σ

ij
+ p δ
ij
, (4.5)
in which δ
ij
is the Kronecker delta, or the unit matrix. Its deﬁnition is

δ
ij
=

1 als i = j,
0 als i = j.
(4.6)
Arnold Verruijt, Soil Mechanics : 4. STRESSES IN SOILS 29
Calculating the eﬀective stresses in soils is one of the main problems of soil mechanics. The eﬀective stresses are important because they
determine the deformations. In the next chapter the procedure for the determination of the eﬀective stress will be illustrated for the simplest
case, of one-dimensional deformation. In later chapters more general cases will be considered, including the eﬀect of ﬂowing groundwater.
4.4 Archimedes and Terzaghi
The concept of eﬀective stress is so important for soil mechanics that it deserves careful consideration. It may be illuminating, for instance, to
note that the concept of eﬀective stress is in agreement with the principle of Archimedes for the upward force on a submerged body.
Consider a volume of soil of magnitude V , having a porosity n. The total weight of the particles in that volume is (1 −n)γ
p
V , in which γ
p
is
the volumetric weight of the particle material, which is about 26.5 kN/m
3
. Following Archimedes, the upward force under water is equal to the
weight of the water that is being displaced by the particles, that is (1 −n)γ
w
V , in which γ
w
the volumetric weight of water, ab out 10 kN/m
3
.
The remaining force is

F = (1 −n)γ
k
V − (1 − n)γ
w
V,
which must be transmitted to the bottom on which the particles rest. If the area of the volume is denoted by A, and the height by h, then the
average stress is, with σ

= F/A,
σ

= (1 −n)γ
p
h −(1 − n)γ
w
h = (1 − n)(γ
p
− γ
w
)h. (4.7)
The quantity (γ
p
− γ
w
) is sometimes denoted as the submerged volumetric weight.
Following Terzaghi the eﬀective stresses must be determined as the diﬀerence of the total stress and the pore pressure. The total stress is
generated by the weight of the soil, whatever its constitution, i.e. σ = γ
s
h, in which γ
s

is the volumetric weight of the soil. If the ground water
is at rest the pore pressure is determined by the depth below the water table, i.e. p = γ
w
h. This means that the eﬀective stress is
σ

= γ
s
h −γ
w
h. (4.8)
Because for a saturated soil the volumetric weight is
γ
s
= nγ
w
+ (1 − n)γ
p
,
this can also be written as
σ

= (1 −n)γ
p
h −(1 − n)γ
w
h = (1 − n)(γ
p
− γ
w

)h. (4.9)
This is identical to the expression (4.7). Terzaghi’s principle of eﬀective stress appears to be in agreement with the principle of Archimedes,
which is a fundamental principle of physics. It may be noted that in the two methods it has been assumed that the determining factor is the
force transmitted between the particles and an eventual rigid surface, or the force transmittance between the grains. This is another basic aspect
of the concept of eﬀective stress, and it cannot be concluded that Archimedes’ principle automatically leads to the principle of eﬀective stress.
Arnold Verruijt, Soil Mechanics : 4. STRESSES IN SOILS 30

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Figure 4.6: Archimedes.
Terzaghi’s approach, leading to the expression (4.8), is somewhat more direct, and especially more easy to
generalize. In this method the porosity n is not needed, and hence it is not necessary to determine the
porosity to calculate the eﬀective stress. On the other hand, the porosity is hidden in the volumetric weight
γ
s
. The generalization of Terzaghi’s approach to more complicated cases, such as non-saturated soils, or
ﬂowing groundwater, is relatively simple. For a non-saturated soil the total stresses will be smaller, because
the soil is lighter. The pore pressure remains hydrostatic, and hence the eﬀective stresses will be smaller,
even though there are just as many particles as in the saturated case. The eﬀective principle can also be
applied in cases involving diﬀerent ﬂuids (oil and water, or fresh water and salt water). In the case of ﬂowing
groundwater the pore pressures must be calculated separately, using the basic laws of groundwater ﬂow. Once
these pore pressures are known they can be subtracted from the total stresses to obtain the eﬀective stresse s.
The procedure for the determination of the eﬀective stresses usually is that ﬁrst the total stresses are determined, on the basis of the total

weight of the soil and all possible loads. Then the pore pressures are determined, from the conditions on the groundwater. Then ﬁnally the
eﬀective stresses are determined by subtracting the pore pressures from the total stresses.
Problems
4.1 A rubber balloon is ﬁlled with dry sand. The pressure in the pores is reduced by 5 kPa with the aid of a vacuum pump. Then what is the change of
the total stress, and the change of the eﬀective stress?
4.2 An astronaut carries a packet of vacuum packed coﬀee into space. What is the rigidity of the pack (as determined by the eﬀective stresses) in a
spaceship? And after landing on the mo on, where gravity is about one sixth of gravity on earth?
4.3 A packet of vacuum coﬀee is dropped in water, and it sinks to a depth of 10 meter. Is it harder now?
4.4 A treasure hunter wants to remove a collection of antique Chinese plates from a s unken ship. Under water the divers must lift the plates very carefully,
of course, to avoid damage. Is it important for this damage to know the depth below water of the ship?
4.5 The bottom of a lake consists of sand. The water level in the lake rises, so that the water pressure at the bottom is increased. Will the bottom of the
lake subside by deformation of the sand?

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