-
Stress Analysis, StrainAnalysis, and Shearing of Soils
Ut tensio sic vis (strains and stresses are related
linearly).Robert Hooke
So I think we really have to, first, make some new kind of
theories in which we takeregard to the fact that there is no
linearity condition between stresses and strainsfor soils.
J. Brinch Hansen
Foundation design requires that we analyze how structural loads
are transferred tothe ground, and whether the soil will be able to
support these loads safely andwithout excessive deformation. In
order to do that, we treat soil deposits as contin-uous masses and
develop the analyses needed for determining the stresses andstrains
that appear in the soil mass as a result of application of loads on
its bound-aries. We then examine whether the stresses are such that
shearing (large shearstrains) occurs anywhere in the soil. Other
soil design problems involving slopesand retaining structures are
also analyzed by computing the stresses or strainsresulting within
the soil from actions done on the boundaries of the soil mass.
Thischapter covers the basic concepts of the mechanics of soils
needed for understand-ing these analyses. At the soil element
level, coverage includes stress analysis,strain analysis, shearing
and the formation of slip surfaces, and laboratory testsused to
study the stress-strain response and shearing of soil elements. At
the levelof boundary-value problems, coverage includes a relatively
simple problem in soilplasticity (Rankine states) and stresses
generated inside a semi-infinite soil massby boundary loads.
113
C H A P T E R 4
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4.1 Stress Analysis
Elements (Points) in a Soil Mass and Boundary-Value Problems
In this chapter, we are concerned with (1) how soil behaves at
the element leveland (2) how this behavior, appropriately described
by suitable equations andascribed to every element (that is, every
point) of a soil mass subjected to certainboundary conditions, in
combination with analyses from elasticity or plasticity the-ory,
allows us to determine how boundary loads or imposed displacements
result instresses and strains throughout the soil mass (and, in
extreme cases, in strains solocalized and large that collapse of a
part of the soil mass and all that it supportshappens).
At the element level, we must define precisely what the element
is and howlarge it must be in order to be representative of soil
behavior. We must also haveequations that describe the relationship
between stresses and strains for the ele-ment and that describe the
combination of stresses that would lead to extremelylarge strains1.
At the level of the boundary-value problem, we must define the
max-imum size an element may have with respect to characteristic
lengths of the prob-lem (for example, the width of a foundation)
and still be treated as a point. This isimportant because we use
concepts and analyses from continuum mechanics tosolve soil
mechanics problems, and so our elements must be points that are
part ofa continuous mass. We must then have analyses that take into
account how ele-ments interact with each other and with specified
stresses or displacements at theboundaries of the soil mass in
order to produce values of stress and strain every-where in the
soil mass. Further, we must also be able to analyze cases where
thestresses are such that large strains develop in localized zones
of the soil mass. Con-cepts from plasticity theory are used for
that.
It is clear from the preceding discussion that the mechanics of
the continuumis a very integral part of soil mechanics and
geotechnical and foundation engineer-ing. We will introduce the
concepts that are necessary gradually and naturally andwith a
mathematical treatment that is kept as simple as possible.
Stress
A significant amount of the work we do as foundation engineers
is based on theconcept of stress. Stress is a concept from the
mechanics of continuous bodies.Because soil is not a continuous
medium, it is useful to discuss the meaning ofstress in soils.
Consider a small planar area A passing through point P
locatedwithin a soil mass (Fig. 4-1). A normal force FN and a
tangential force FT areapplied on A. If soil were a continuum, the
normal stress s acting normal to A at
114 The Engineering of Foundations
1 Extremely large strains, particularly extremely large shear
strains, are closely tied to concepts ofrupture, yield, and failure
(which term is used depends on the publication and the subject it
deals with).None of these three terms perfectly describes the range
of problems we deal with, so we will introduceappropriate terms
throughout the chapter and the remainder of the text.
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point P would be defined as the limit of FN/A as A tends to the
point P (that is,tends to zero, centered around P). The shear
stress would be defined similarly.Mathematically,
(4.1)
(4.2)
Because soil is not a true continuum, we must modify this
definition. A pointwithin a soil mass is defined as a volume V0
that is still very small compared withthe dimensions of the
foundations, slopes, or retaining structures we analyze, but itis
sufficiently large to contain a large number of particles and thus
be representa-tive of the soil.2 To this representative volume V0,
we associate a representativearea A0 (also very small, of a size
related to that of V0). So we modify Eqs. (4.1)and (4.2) by
changing the limit approached by the area A from zero to A0:
(4.3)
(4.4)
The preceding discussion brings out one difference between soil
mechanicsand the mechanics of metals, for example. In metals, the
representative elementaryvolume (REV) is very small. The REV for a
given metal is indeed so small that, inordinary practice or
introductory courses, we tend to think of it as being a
point,forgetting that metals are also made up of atoms arranged in
particular ways, sothat they too have nonzero REVs, although much
smaller than those we must usein soil mechanics. In soil mechanics,
our REVs must include enough particles andthe voids between them
that, statistically, this group of particles will behave in away
that is representative of the way larger volumes of the soil would
behave.
t limASA0 FTA
s limASA0 FNA
t limAS0 FTA
s limAS0 FNA
C H A P T E R 4 Stress Analysis, Strain Analysis, and Shearing
of Soils 115
FN
P
FTA
Figure 4-1Definition of stress in soils: As the area A is
allowed toshrink down to a very small value, the ratios FN/A
andFT/A approach values s and t, the normal and shearstress at P,
respectively.
2 Mechanicians like to use the term representative elementary
volume to describe the smallest volumeof a given material that
captures its mechanical properties.
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Stress Analysis
Stress analysis allows us to obtain the normal and shear
stresses in any plane pass-ing through a point,3 given the normal
and shear stresses acting on mutually per-pendicular planes passing
through the point.4 We will see in Section 4.6 someexamples of how
these stresses can be calculated at a point inside a soil mass
froma variety of boundary loadings common in geotechnical
engineering. Later in thetext, we discuss some applications
requiring corrections for three-dimensional (3D)effects, but for
now we will focus on two-dimensional (2D) stress analysis,
whichturns out to be applicable to most problems of interest in
practice.
Figure 4-2(a) shows a small, prismatic element of soil
representing a pointwithin the soil. The faces of the element are
aligned with the directions of the ref-erence axes x1 and x3. The
soil element is acted upon by the normal stresses s11acting in the
x1 direction and s33 acting in the x3 direction and the shear
stressess13 (which, due to the requirement of moment equilibrium,
is identical to s31).5The first subscript of a stress component
represents the direction normal to theplane on which it acts; the
second, the direction of the stress component itself. A
116 The Engineering of Foundations
(a) (c)
(b)
s11 s11
sutu
us11
s13
s13
s13
s13
s13
s33 s33
s33
x3
x1
u
Vertical legof L parallel to p11
Plan
e p
11
Plane p33
Horizontal legof L parallel to p33
Figure 4-2(a) Elemental representation of two-dimensionalstress
state at a point; (b) illustration of how theelement would distort
when acted upon by apositive s13 for the simple case with zero
normalstresses s11 and s33; (c) sectioned triangularprismatic
element, where su and tu depend onthe angle u.
3 A point in the soil is indistinguishable, for our purposes,
from a representative soil element, which hasa very small volume
(and so is a point for practical purposes) but is sufficiently
large to be representativeof the soil in its mechanical behavior.4
A more proper definition of stress analysis for advanced readers
would be that stress analysis aims toallow calculations of the
traction (which has normal and shear components) on a plane given
the stresstensor at the point.5 Note that many engineering texts
use the notation tij instead of sij when i j to represent shear
stress.We have retained here the traditional mechanics terminology,
in which s is used for both normal andshear stresses.
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stress component with different subscripts is a shear stress;
one with identical sub-scripts, a normal stress. For example, s11
is the stress acting on the plane normalto x1 in the x1 direction
(that is, it is a normal stress), while s13 is the stress thatacts
on the plane normal to x1 in the x3 direction (and is thus a shear
stress). It issimpler (although not required) to solve problems if
we adopt the practice ofchoosing s11 and s33 such that s11 s33. So
if the normal stresses are 1,000 and300 kPa, then s11 1,000 kPa and
s33 300 kPa. Likewise, if the normalstresses are 100 and 500 kPa,
then s11 100 kPa and s33 500 kPa.
The plane where s11 acts is denoted by p11; likewise, p33 is the
plane wheres33 acts. The stresses s11, s33, and s13 are all
represented in their positive direc-tions in Fig. 4-2(a). This
means normal stresses are positive in compression, andthe angle u
is positive counterclockwise from p11. With respect to the shear
stresss13, note that the prism shown in the figure has four sides,
two representing planep11 and two representing plane p33. Looking
at the prism from the left side, wemay visualize the plane p11 as
the vertical leg of an uppercase letter L and p33as the horizontal
leg of the uppercase letter L. We then see that a positive
shearstress s13 acts in such a way as to open up (increase) the
right angle of the Lformed by planes p11 and p33. Figure 4-2(b)
shows the deformed shape that wouldresult for the element assuming
zero normal stresses and positive s13.
If we section the element of Fig. 4-2(a) along a plane making an
angle u withthe plane p11, as shown in Fig. 4-2(c), a normal stress
su and a shear stress tu mustbe applied to this plane to account
for the effects of the part of the element that isremoved. While
the sign of the normal stress su is unambiguous (positive in
com-pression), the shear stress on the sloping plane has two
possible directions: up ordown the plane. So we must decide which
of these two directions is associated witha positive shear stress.
The positive direction of the shear stress actually followsfrom the
sign convention already discussed (that s13 0 when its effect would
beto increase the right angle of the uppercase L made up by p11 as
its vertical andp33 as its horizontal leg). It turns out the shear
stress tu is positive as drawn in thefigure, when it is rotating
around the prismatic element in the counterclockwisedirection. We
will show why this is so later, when we introduce the Mohr
circle.
Our problem now is to determine the normal stress su and the
shear stress tuacting on the plane making an angle u with p11. This
can be done by consideringthe equilibrium in the vertical and
horizontal directions and solving for su and tu.The following
equations result:
(4.5)
(4.6)where the signs of s11, s33, and s13 are as discussed
earlier (positive in compres-sion for the normal stresses and
determined by the L rule in the case of s13).
Principal Stresses and Principal Planes
Equations (4.5) and (4.6) tell us that su and tu vary with u.
That means that the nor-mal and shear stresses on each plane
through a given point are a unique pair. There
tu 12 1s11 s33 2sin 2u s13 cos 2u su 12 1s11 s33 2 12 1s11 s33
2cos 2u s13 sin 2u
C H A P T E R 4 Stress Analysis, Strain Analysis, and Shearing
of Soils 117
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will be two planes out of the infinite number of planes through
the point for whichthe normal stress will be a minimum and a
maximum. These are called principalstresses. They are obtained by
maximizing and minimizing su by differentiating Eq. (4.5) with
respect to u and making the resulting expression equal to zero.
Thelargest principal stress is known as the major principal stress;
it is denoted as s1.The smallest principal stress is the minor
principal stress, denoted as s3. The planeswhere they act are
referred to as the major and minor principal planes, denoted byp1
and p3, respectively. When we differentiate Eq. (4.5) with respect
to u and makethe resulting expression equal to zero, we obtain the
same expression we obtainwhen we make tu, given by Eq. (4.6), equal
to zero. This means that the shearstresses acting on principal
planes are zero.
An easy way to find the angles up1 and up3 that the principal
planes p1 and p3make with p11 (measured counterclockwise from p11)
is then to make tu 0 inEq. (4.6), which leads to
(4.7)
When up is substituted for u back into Eq. (4.5), we obtain the
principal stresses s1and s3, which are the two normal stresses
acting on planes where tu 0 and are alsothe maximum and minimum
normal stress for the point under consideration, given by
(4.8)
(4.9)Note that, given the definition of the tangent of an angle,
there is an infinite
number of values of up that satisfy Eq. (4.7). Starting with any
value of up satisfy-ing Eq. (4.7), we obtain additional values that
are also solutions of (4.7) by repeat-edly either adding or
subtracting 90. Using a calculator or a computer program,the value
of up calculated from Eq. (4.7) is a number between 90 and 90. If
s11 s33, we expect the major principal stress s1 to be closer in
direction to s11 (thelarger stress) than to s33 (the smaller
stress); so if the absolute value of the calcu-lated value of the
angle up is less than 45, up up1; otherwise, up up3. Once
thedirection of one of the principle planes is known, the direction
of the other planecan be calculated easily by either adding or
subtracting 90 to obtain an angle withabsolute value less than 90.
For example, if up1 is calculated as 25, then up3 isequal to 65.
Alternatively, if up1 is found to be 25, then up3 is calculated
as25 90 65.
Mohr Circle
Moving (s11 s33) to the left-hand side of Eq. (4.5), taking the
square of bothsides of the resulting equation, and adding it to Eq.
(4.6) (with both sides alsosquared), we obtain
(4.10)3su 12 1s11 s33 2 4 2 tu2 14 1s11 s33 2 2 s13212
s3 12 1s11 s33 2 214 1s11 s33 2 2 s132
s1 12 1s11 s33 2 214 1s11 s33 2 2 s132
tan 12up 2 2s13s11 s33
118 The Engineering of Foundations
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Recalling the equation of a circle in Cartesiancoordinates, (x
a)2 (y b)2 R2, where (a, b)are the coordinates of the center of the
circle and Ris its radius, Eq. (4.10) is clearly the equation of
acircle with center C[(s11 s33)/2, 0] and radius
in s-t space. Eachpoint of this circle, which is referred to as
the Mohrcircle, is defined by two coordinates: the first is anormal
stress (s) and the second is a shear stress (t).Figure 4-3 shows
the Mohr circle for the stress state(s11, s33, s13) of Fig. 4-2(a).
According to Eqs. (4.5)and (4.6), the stresses su and tu on the
plane makingan angle u measured counterclockwise from planep11 are
the coordinates of a point on the Mohr circlerotated 2u
counterclockwise from point S1 (point S1represents the stresses on
p11). Note that the centralangle of the Mohr circle separating S1
from S3,which is 180, is indeed twice the 90 angle separat-ing the
corresponding planes, p11 and p33.
In Fig. 4-3, the stresses (s11, s33, s13) plot as two points
S1(s11, s13) andS3(s33, s13), diametrically opposed. The figure
illustrates the case of s11 s33. Tounderstand why the shear stress
s13 plots as a positive number with s33 and as anegative number
with s11, refer to Fig. 4-2(b). With the positive values of s11,
s33,and s13 shown in Fig. 4-2(a), the prismatic element deforms as
shown in Fig. 4-2(b), with the direction of maximum compression
associated with the majorprincipal stress s1. It is clear that the
major principal plane, which is normal to thedirection of maximum
compression, is obtained by a rotation of some angle up1 90
counterclockwise with respect to p11. This means that, in the Mohr
circle, wemust have a counterclockwise rotation 2up1 from the point
S1(s11, s13) associ-ated with p11 in order to reach the point (s1,
0) of the circle. This implies that S1must indeed be located as
shown in Fig. 4-3, for if we had S1(s11, s13) instead ofS1(s11,
s13), a counterclockwise rotation less than 90 would not take us to
(s1,0). So this means s13 is plotted as negative if spinning
clockwise around the ele-ment, as it does for plane p11, and as
positive if spinning counterclockwise, as itdoes for plane p33.
This is the basis for the convention we will use for plottingpoints
in the Mohr circle: Shear stresses rotating around the element in a
counter-clockwise direction are positive; they are negative
otherwise (Fig. 4-4). Note thatthis sign convention is not
independent from but actually follows directly from theshear stress
sign convention we adopted for the stress s13 appearing in Eqs.
(4.5)and (4.6).
Pole Method
Mohr circles have interesting geometric properties. One very
useful property is thatthe central angle of a circle corresponding
to a certain arc is twice as large as an
R 114 1s11 s33 2 2 s132
C H A P T E R 4 Stress Analysis, Strain Analysis, and Shearing
of Soils 119
t
s
u
Line parallel to p33
S3(s33, s13)
2u
S1(s11, s13)
S(su, tu)
(s3, 0) (s1, 0)
P
Line
par
alle
l to p
11
Figure 4-3Mohr circle corresponding to state of stress shown in
Fig. 4-2.
Positive normalstress
Positive shear stress(for plotting in Mohr
diagram)
s
t
Figure 4-4Stress signconventions.
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inscribed angle corresponding to the same arc (Fig. 4-5).
Applying this property tothe Mohr circle shown in Fig. 4-3, the
angle made by two straight lines drawnfrom any point of the circle
to point S(su, tu) representing the stresses on the planeof
interest and to point S1(s11, s13) is equal to u. In particular,
there is one andonly one point P on the circle with the property
that the line joining P to pointS(su, tu) is parallel to the plane
on which su and tu act. The point P with thisproperty is known as
the pole of the Mohr circle. Based on the preceding discus-sion,
the pole can be defined as the point such that, if we draw a line
through thepole parallel to the plane where stresses su and tu act,
this line intersects the Mohrcircle at a point whose coordinates
are su, tu.
In order to determine the pole, we need to know the orientation
of at least oneplane where the stresses are known. We can then use
the known stresses (s, t) tofind the pole by drawing a line
parallel to the plane acted upon by these stresses.This line
intersects the circle at a point; this point is the pole. Once we
know thepole P, we can determine the stresses on any plane by
drawing a straight linethrough the pole P parallel to the plane
where the stresses are desired. This lineintersects the circle at a
point whose coordinates are the desired stresses.
We can use Figs. 4-2 and 4-3 to illustrate the concept of the
pole. Consider theelement of Fig. 4-2(a). By plotting the Mohr
circle for this state of stress, weobtain the expected
diametrically opposed points S1 and S3 shown in Fig. 4-3. Ifwe look
at point S1 on the circle and consider the corresponding stresses
shown inFig. 4-2(a), we can easily determine the location of the
pole. If we construct a linethrough S1 that is parallel to the
plane p11 where (s11, s13) acts, we will deter-mine the pole as the
point where this line intersects the Mohr circle. In this
case,since we are dealing with u 0, the line is vertical, and the
pole (point P in Fig.4-3) lies directly above S1 (also shown in
Fig. 4-3). Likewise, if we look at point S3and construct a line
through S3 parallel to plane p33 on which (s33, s13) acts, wecan
also determine the pole as the point of intersection of this
horizontal line withthe Mohr circle. As expected, the pole is found
to be directly to the right of S3 andto coincide with the point
determined previously by examining point S1. Thisshows clearly that
the pole is unique for a given stress state. Figure 4-6
illustratesfor the same case how the principal directions and
principal stresses would bedetermined once the pole is known.
120 The Engineering of Foundations
t
s
2uu
u
Figure 4-5The geometric property of circles that acentral angle
2u produces the same arc asan inscribed angle u.
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C H A P T E R 4 Stress Analysis, Strain Analysis, and Shearing
of Soils 121
P(s3, 0)(s1, 0)
up1
up3
Line parallel to p11
Line parallel to p3
Line parallel to p1
t
s
Figure 4-6Illustration of the relationship of the principal
directionsand their representation in a Mohr diagram.
EXAMPLE 4-1
A state of stress is represented by the block in Fig. 4-7(a).
Determine the location of thepole and give its coordinates in the
s-t system.
SolutionFirst, plot the Mohr circle [Fig. 4-7(b)]. Next,
determine the location of the pole. Usingpoint (200, 100), draw a
line parallel to the plane on which s11 (200 kPa) acts. In this
case,it is a vertical line since p11 is vertical. Likewise, if the
other point is chosen (100, 100),we draw a line parallel to the
plane on which s33 (100 kPa) acts. In this case, it is a
horizon-tal line since p33 is horizontal. Either method produces
the location of the pole: (200,100).
(a)
100 kPa
100 kPa
100 kPa
100 kPa
200 kPa200 kPa
Figure 4-7(a) Stress state for Example 4-1 and (b) corresponding
Mohr circle.
(b)
100 200
Pole
(200, 100)
t
s
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Solving Stress Analysis Problems
The steps in solving a 2D stress analysis problem can be
outlined as follows:
1. Choose the largest normal stress as s11. For example, if the
normal stresses are 300 and 100 kPa, then s11 300 kPa and s33 100
kPa. Likewise, if the normal stresses are 100 and 300 kPa, then s11
100 kPa and s33 300 kPa. The plane where s11 acts is denoted by
p11, while the one wheres33 acts is denoted by p33.
2. If using Eq. (4.5) or (4.6), assign a sign to s13 based on
whether it acts toincrease or decrease the angle between p11 and
p33 (with p11 being the verticalleg of the L; see Fig. 4-2). The
stress s13 is positive if it acts in a way thatwould tend to
increase the angle.
3. Recognize that the reference plane for angle measurements is
p11 and thatangles are positive counterclockwise.
4. Reason physically to help check your answers. For example,
since s11 s33,the direction of s1 will be closer to that of s11
than to that of s33. Whether s1points up or down with respect to
s11 now depends on the sign of s13, for ittells us about the
directions in which the element is compressed and
extended.Naturally, s1 acts in the direction in which the element
is compressed the most.
122 The Engineering of Foundations
30
200 kPa
200 kPa
50 kPa
50 kPa
50 kPa
50 kPa
Figure 4-8State of stress at a point(Example 4-2).
EXAMPLE 4-2
The state of stress at a point is represented in Fig. 4-8. Find
(a) the principal planes, (b) theprincipal stresses, and (c) the
stresses on planes making angles 15 with the horizontal.Solve both
analytically and graphically.
SolutionAnalytical SolutionTake s11 200 kPa and s33 50 kPa. So
p11 makes an angle equal to 30 with the hor-izontal, and p33 makes
an angle equal to 60 with the horizontal. In order to assign a
signto s13, we need to consider the right angle made by p11 and
p33; we must look at this angle
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as if it were an uppercase letter L, such that p11 is the
vertical and p33 is the horizon-tal leg of the letter L. Physically
rotating the page until we see the L may be helpful inthis
visualization. The effect of the shear stress on the right angle
made by p11 and p33looked at in this manner is to reduce it;
accordingly, s13 50 kPa. We are now preparedto solve the
problem.
Principal PlanesSubstituting the values of s11, s33, and s13
into Eq. (4.7):
from which
The absolute value of 16.8 is less than 45, so
Graphically, up1 would show as an angle of 16.8 clockwise from
p11 because the calcu-lated angle is negative. If the answer is
desired in terms of the angles that p11 and p3 makewith the
horizontal, then we need to add 30 (the angle that p11 makes with
the horizontal) tothese two results: p11 is at an angle 13.2, and
p3 is at 103.2 (or 76.8) to the horizontal.Principal StressesThe
principal stresses can be calculated using either Eq. (4.5) with u
16.8 and 73.2 orEqs. (4.8) and (4.9). Using Eq. (4.5):
Now using Eqs. (4.8) and (4.9):
Stresses on Planes Making Angles 15 with HorizontalThese planes
make angles 15 and 45 with p11, respectively. Plugging these
values(15 and 45) into Eqs. (4.5) and (4.6):
Not surprisingly, the su and tu values calculated here are very
close to the valuesfor the major principal plane (215.1 and 0). You
should verify that su 175 kPa and tu 75 kPa for the other plane
(which makes an angle of 15 with the horizontal).
tu 75 sin 32 115 2 4 50 cos 32 115 2 4 5.8 kPa su 125 75 cos 32
115 2 4 50 sin 32 115 2 4 214.9 kPa
s3 12 1200 50 2 214 1200 50 2 2 150 2 2 34.9 kPas1 12 1200 50 2
214 1200 50 2 2 150 2 2 215.1 kPa
s3 12 1200 50 2 12 1200 50 2cos 32 173.2 2 4 150 2sin 32 173.2 2
4 34.9 kPa s1 12 1200 50 2 12 1200 50 2cos 32 116.8 2 4 150 2sin 32
116.8 2 4 215.1 kPa
up3 16.8 90 73.2
up1 16.8
up 12
arctan a23b 16.8
tan 12up 2 2 150 2200 50 23
C H A P T E R 4 Stress Analysis, Strain Analysis, and Shearing
of Soils 123
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Graphical Solution Using the Pole MethodThe solution can be seen
in Fig. 4-9. The normal and shear stresses on plane p11 are 200and
50 kPa, respectively; in plane p33 they are 50 and 50 kPa.6 These
two points are dia-metrically opposite each other on the Mohr
circle. If we plot these two points in s-t spaceand join them by a
straight line, this line crosses the s-axis at the center of the
circle. Wecan then easily draw the Mohr circle using a compass. The
principal stresses are now easilyread as the abscissas of the two
points with t 0.
If we now draw a line parallel to p11 through the point (200,
50), we obtain the pole Pas the intersection of this line with the
circle. Note that if we draw a line parallel to p33through (50,
50), we obtain the same result. The directions of p1 and p3 are
obtained bydrawing lines through P to the points (s1, 0) and (s3,
0) of the Mohr circle, respectively.The stresses at 15 with the
horizontal are found by drawing lines through the pole Pmaking 15
with the horizontal. These lines intersect the circle at two points
with coordi-nates (214, 6) and (175, 75), respectively.
Total and Effective Stresses
When we plot Mohr circles (as in Fig. 4-10), we are representing
the state of stressat a point in the soil mass. If there is a
nonzero pore pressure u at this point, it isthe same in every
direction, and thus it does not affect the equilibrium of the
point.We should remember that water cannot sustain shear stresses,
so the presence of apore pressure affects only normal stresses in
the soil. It is useful to examine whatwould happen if we plotted
both a total stress and an effective stress Mohr circle.Let us
assume Eq. (4.10) applies to total stresses. Referring back to Eq.
(3.3), if wesubstitute it into Eq. (4.10), we can see that the only
effect is to have the Mohr cir-
124 The Engineering of Foundations
15
15
P
p3
p1
50
100
200
(200, 50)(214, 6)
(175, 75)(50, 50)
100
t
s
Figure 4-9Mohr circle and graphical solution ofExample 4-2.
6 Note that, for Mohr circle construction, counterclockwise
shear stresses are plotted as positive, whileclockwise shear
stresses are plotted as negative.
saL00581_ch04.qxd 9/14/06 4:54 PM Page 124
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cle of total stresses displaced along the s-axis with respect to
the Mohr circle ofeffective stresses by an amount equal to the pore
pressure u.
4.2 Strains*
Definitions of Normal and Shear Strains
Analysis of geotechnical problems cannot always be done only in
terms ofstresses. These stresses induce deformations, which are
represented by strains.There are two types of strains: normal and
shear strains. At a given point, a normalstrain in a given
direction quantifies the change in length (contraction or
elonga-tion) of an infinitesimal linear element (a very small
straight line) aligned with thatdirection. The so-called
engineering shear strain g is a measure, at a given point, ofthe
distortion (change in shape). With respect to the two reference
axes x1 and x3,the engineering shear strain g13 expresses the
increase of the initial 90 angleformed by two perpendicular
infinitesimal linear elements aligned with these axes.A shear
strain, as it is defined in mechanics, is one half the value of the
engineer-ing shear strain.7
Both normal and shear strains eij can be expressed through
(4.11)
where i,j 1, 2, 3 reference directions, xi coordinate in the i
direction, ui displacement in the xi direction. The subscripts
indicate the directions of linear
eij 12a 0ui
0xj
0uj0xib
C H A P T E R 4 Stress Analysis, Strain Analysis, and Shearing
of Soils 125
Effective-stressMohr circle
(s u, t) (s, t)
Total-stressMohr circle
Pore pressure u
t
s, s
Figure 4-10Principle of effective stresses: illustration of the
difference between the total and the effectivestress state as
represented using the concept of the Mohr circle.
7 We have avoided duplication of symbols as much as possible,
but there is no good alternative to usingthe traditional notation
for engineering shear strain, which of course is the same as used
for unit weight.The reader should observe the context in which
symbols are used to avoid any confusion.
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differential elements and the directions in which displacements
of the end points ofthe differential elements are considered. When
i j, the strain is a normal strain; itis a shear strain
otherwise.
The relationship between a shear strain eij and the
corresponding engineeringshear strain gij is
(4.12)gij 2eij
126 The Engineering of Foundations
EXAMPLE 4-3
Derive, in a simple way, the expression for the normal strain at
a point in the direction ofreference axis x1.
SolutionLets consider the case of Fig. 4-11. For an underformed
soil mass, we have a differentialelement dx1 aligned with the x1
direction (Fig. 4-11). We have labeled the initial point of
thesegment A and the end point B. In drafting this figure, we have
corrected for rigid bodytranslation in the x1 direction. In other
words, we are plotting the deformed element as if thedisplacement
of A were zero for easier comparison with the original, undeformed
element.This way, every displacement in the figure is relative to
the displacement of A. If, after thesoil mass is deformed, point B
moves more in the positive x1 direction than point A (that is,if
the displacement u1 of B is greater than that of A), as shown in
Fig. 4-11, then the elementhas clearly elongated (this elongation
is seen in the figure as B* B).
Taking some liberty with mathematical notation, the unit
elongation of dx1 is the differ-ence in displacement between points
A and B (u1B u1A B* B du1 u1) dividedby the initial length of the
element (x1), or u1/x1. It remains to determine whether elonga-tion
is a positive or negative normal strain. To be consistent with the
sign convention forstresses, according to which tensile stresses
are negative, our normal strain e11 in the x1direction must be
Note that this, indeed, is the expression that results directly
from Eq. (4.11) when we makei 1 and j 1.
e11 0u10x1
dx1 du1
x3
AB B*
x1
Figure 4-11Normal strain e11:infinitesimal elementdx1 shown
aftercorrection for rigidbody motion bothbefore deformation(AB) and
afterdeformation (AB*).
EXAMPLE 4-4
Derive, in a simple way, the expression of the shear strain at a
point in the x1-x3 plane.
SolutionConsider two differential linear elements, dx1 and dx3,
aligned with the x1- and x3-axis,respectively, at a point within an
undeformed soil mass (Fig. 4-12). Now consider that thesoil mass is
deformed, and, as a result, points B and C (the end points of
elements dx1 anddx3, respectively) move as shown (to new positions
labeled B* and C*) with respect to pointA (note that, as for
Example 4-3, we are not representing rigid-body translation in the
x1 and
saL00581_ch04.qxd 9/14/06 4:54 PM Page 126
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x3 directions in the figure). We can see that both point B and
point C have displace-ments that have components in both the x1 and
x3 directions. Here we are interested injust the distortion of the
square element made up of dx1 and dx3, not in the elongationor
shortening of dx1 and dx3 individually. The distortion of the
element clearly resultsfrom difference in the displacement u3 in
the x3 direction between A and B and in thedisplacement u1 in the
x1 direction between A and C.
Taking again some liberties with mathematical notation, we can
state that the dif-ferences in the displacements of points B and A
(in the x3 direction) and C and A (inthe x1 direction) can be
denoted as u3 and u1, respectively. Since the deformations weare
dealing with are small, the angle by which the differential element
dx1 rotates coun-terclockwise is approximately equal to u3 divided
by the length of the element itself,or u3/x1; similarly, the angle
by which dx3 rotates clockwise is u1/x3. These tworotations create
a reduction in the angle between the elements dx1 and dx3, which
wasoriginally 90, characterizing a measure of distortion of the
element. If we add themtogether, we obtain the absolute value of
what has become known as the engineeringshear strain g13; one-half
the sum gives us the absolute value of e13. It remains to
deter-mine whether this is a positive or negative distortion. Our
sign convention for shearstrains must be consistent with our shear
stress sign convention. Recall from our earlierdiscussion in the
chapter that a positive shear stress was one that caused the 90
angleof our square to open up (to increase), not to decrease.
Therefore, we will need a nega-tive sign in front of our sum in
order to obtain a negative shear strain for the reductionof the 90
angle we found to take place for the element in Fig. 4-12:
Note that this equation results directly from Eq. (4.11) when we
make i 1 and j 3(or vice versa).
Strains as expressed by Eq. (4.11) are small numbers; there are
other ways ofdefining strain that are more appropriate when
elongations, contractions, or distor-tions become very large.
However, Eq. (4.11) may still be used for increments ofstrains,
even if the cumulative strains measured from an initial
configuration arevery large. It is appropriate in that case to use
a d (the symbol for differential)before the strain symbol (as in de
and dg) to indicate that we refer to a strainincrement.
As was true for stresses, there are also principal strains e1
and e3 (and principalstrain increments de1 and de3). These are
strains (or strain increments) in the direc-tions in which there is
no distortion, and the shear strains (or shear strain
increments)are equal to zero. Distortion happens any time the shape
of an element changes. It isimportant to understand that a point in
the soil experiences distortion as long as de1 de3. To illustrate
this, Fig. 4-13(b) and (c) show two alternative elements
repre-senting a point P. The larger, outer element is aligned with
the principal directions,and the deformed shape of the element
[shown in Fig. 4-13(a)] does not immediatelysuggest the notion of
distortion. However, there is no doubt that distortion has
takenplace when we observe the change of shape of the smaller,
inner element as we gofrom the initial to the final configuration
[Fig. 4-13(b) and (c)].
e13 12a 0u1
0x3
0u30x1b
C H A P T E R 4 Stress Analysis, Strain Analysis, and Shearing
of Soils 127
dx1
dx3
x3
A BB*
C*
C
x1
u1x3u3x1
Figure 4-12Shear strain e13:infinitesimal squareelement shown
aftercorrection for rigid bodytranslation both beforedeformation
(defined bydx1 AB and dx3 AC)and after deformation(defined by dx1*
AB*and dx3* AC*).
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The volumetric strain increment dev, defined as minus the change
in volumedivided by the original volume (the negative sign being
required to make contrac-tion positive), can be easily determined
in terms of a cubic element with sides withlength initially equal
to 1 and aligned with the principal directions (which meansx1, x2,
and x3 are principal directions). The element is then allowed to
expand as aresult of elongations equal to du1, du2, and du3 in the
three reference directions. Asthe cube is aligned with the
principal directions, there will be no distortion in theplanes
x1x2, x1x3, or x2x3. It is apparent from Fig. 4-14 that
Referring back to our definition of normal strain and
considering that the initiallength of the sides of the cube are of
unit length and the initial volume of the cubeis also equal to 1,
we can write the following for the volumetric strain increment:
which, given that the strain increments are very small (and that
second- and third-order terms would be extremely small and thus
negligible), reduces to
(4.13)The volume change at a point is clearly independent of the
reference system
and of any distortion, so the following equation would also
apply even if x1, x2,and x3 were not the principal directions:
(4.14)where de11, de22, and de33 normal strains in the arbitrary
directions x1, x2, and x3.
dev de11 de22 de33
dev de1 de2 de3
dev dV1
1 11 de1 2 11 de2 2 11 de3 2dV 11 du1 2 11 du2 2 11 du3 2 1
128 The Engineering of Foundations
Initial configuration
P
Final configuration
P
(a)
P
Initial outer element
Final outer element(b) (c)
Figure 4-13Alternative representations for the state of strain
at a point: (a) anelement aligned with the principal strain
directions (verticalcontraction and horizontal elongation); (b) the
same element beforedeformation with an element inside it with sides
oriented at 45 tothe principal strain directions; (c) the same
element afterdeformation, showing the distortion of the element
with sides notaligned with the principal strain directions.
x3
x2x1
de3
de1
de2
Deformed cubeOriginal cube
1
1
1
Figure 4-14Calculation of volumetric strain.
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Mohr Circle of Strains
The mathematics of stresses and strains is the same: Normal
strains play the samerole as normal stresses, and shear strains,
the same as shear stresses. So, just as itwas possible to express
the stresses at a point under 2D conditions in terms of aMohr
circle, the same is possible for strains. Many problems in
geotechnical engi-neering can be idealized as plane-strain
problems, for which the strain in one direc-tion is zero. For
example, slopes and retaining structures are usually modeled as
rel-atively long in one direction, with the same cross section
throughout and with noloads applied in the direction normal to the
cross sections. Except for cross sectionsnear the ends of these
structures, it is reasonable, based on symmetry considera-tions, to
assume zero normal strain perpendicular to the cross section. The
same isvalid for strip footings, which are used to support lines of
columns or load-bearingwalls. The result is that we can take de2 0
and do our strain analysis in twodimensions.
In the case of the Mohr circle of strains, incremental strains,
not total strains,are plotted in the horizontal and vertical axes.
Figure 4-15 shows a Mohr circle ofstrains plotted in normal strain
increment de versus shear strain increment 12dgspace. As was true
for stresses, each point of the Mohr circle represents one
planethrough the point in the soil mass for which the Mohr circle
represents the strainstate. The points of greatest interest in the
Mohr circle of strains are
The leftmost and rightmost points, (de3, 0) and (de1, 0),
corresponding to theminor and major principal incremental strain
directions
The highest and lowest points, (12dev, 12dgmax), corresponding
to the directionsof largest shear strain (Note that 12dev 12(de1
de3) is the de coordinate of thecenter and 12dgmax 12(de1 de3) is
the radius of the Mohr circle.)
The points where the circle intersects the shear strain axis,
(0, 12dgz)The two points with zero normal strain increment
correspond to the two
directions along which de 0, that is, the directions along which
there is neitherextension nor contraction. It is possible to define
a separate reference system for
C H A P T E R 4 Stress Analysis, Strain Analysis, and Shearing
of Soils 129
A1
A2
dg
cc
Pole P
(de3, 0) (de1, 0) de
(0, dgz)
Z
O( dev, 0)12
12
(0, dgz)12
12
Figure 4-15Mohr circle of strains.
saL00581_ch04.qxd 9/14/06 4:54 PM Page 129
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each of these two directions such that x1 in each system is
aligned with the direc-tion of zero normal strain. To clearly
indicate that x1 is a direction of zero normalstrain, we can use a
superscript z, as in x1z. This will be useful in our discussion
ofthe dilatancy angle, which follows.
Dilatancy Angle
The angle c shown in Fig. 4-15, known as the dilatancy angle, is
quite useful inunderstanding and quantifying soil behavior. There
are two ways of expressing thedilatancy angle based on the geometry
of the Mohr circle of strains:
(4.15)
(4.16)
where gz shear strain in the x1z-x3zp plane (Fig. 4-16), x1z is
a direction of zero nor-mal strain, and x3zp is the direction
normal to x1z.
The dilatancy angle is clearly related to the volumetric strain
resulting from aunit increase in shear strain.8 By definition, the
dilatancy angle c is positive whenthere is dilation (volume
expansion). This is apparent from Eqs. (4.15) and (4.16),for the
dilatancy angle clearly results positive when volume expands, that
is, whendev 0. Note that the denominators of Eqs. (4.15) and (4.16)
are always positive(hence the absolute values taken), for the
dilatancy angle is related to the volumet-ric strain increment
resulting from a unit increment in shear strain, regardless ofthe
orientation of the shear strain. In other words, the dilatancy
angle would still bethe same positive value if the element shown in
Fig. 4-16 were sheared to the leftand not to the right as shown.
The direction of zero normal strain in Fig. 4-16 is
tan c OZ0ZA1 0 12 de v0 12 dgz 0 dev0dgz 0
sin c OZ0OA1 0 12 1de1 de3 212 1de1 de3 2 de1 de3de1 de3
dev0dgmax 0
130 The Engineering of Foundations
Direction of zero normal strain
x3zp
x1z
c
gz
Figure 4-16State of strain visualized for anelement with one
side aligned withthe direction of zero normal strain.
8 Technically, both the shear and volumetric strain increments
in the definition of the dilatancy angle areplastic strain
increments, a distinction that for our purposes is not necessary to
make.
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represented by x1z, and the direction normal to it, by x3zp,
where the superscript zpmeans that the direction x3zp is
perpendicular to the direction of zero normal strain.There are in
fact two distinct directions of zero normal strain, as will be
shownlater.
The Mohr circle of Fig. 4-15 corresponds to a state of dilation
(expansion), asdev 0. Examining the state of deformation in the
x1z-x3zp reference system, expan-sion implies that the normal
strain in the x3zp direction, normal to x1z, is negative(that is,
that elongation takes place in the x3zp direction, as is clearly
shown in Fig.4-16). If we locate the pole in Fig. 4-15 and then
draw two lines (one through A1and one through A2), the two lines
perpendicular to these two lines are the direc-tions of zero normal
strain, as we will discuss in detail later.
C H A P T E R 4 Stress Analysis, Strain Analysis, and Shearing
of Soils 131
EXAMPLE 4-5
A soil element is subjected to the following incremental
strains: de1 0.03%, de3 0.05%. Knowing that plane-strain conditions
are in force (that is, the strain in the x2 direc-tion is zero),
calculate the dilatancy angle. SolutionBecause we know the
principal strain increments, we can immediately calculate the
dila-tancy angle as
from which
In Problem 4-18, you are asked to continue this by plotting the
Mohr circle, finding thepole for the case when the major principal
strain increment is vertical and determining thedirections of the
potential slip planes through this element (which is the subject of
a subse-quent section).
For a triaxial strain state, in which e2 e3, the dilatancy angle
needs to beredefined because a maximum engineering shear strain
increment dgmax is nolonger possible to define with clarity. In
place of it, we work with des, defined as
(4.17)The volumetric strain in the triaxial case follows from
Eq. (4.13):
Thus, the dilatancy angle for triaxial conditions is written
as
(4.18)sin c de1 2de3de1 2de3
dev de1 2de3
des de1 2de3
c 14.5
sin c dev0dgmax 0 de1 de3de1 de3 0.03 0.050.03 10.05 2 0.25
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A single expression for it, which applies to both plane-strain
and triaxialconditions, is
(4.19)
where
(4.20)
4.3 Failure Criteria, Deformations, and Slip Surfaces
Mohr-Coulomb Failure Criterion
Soils are not elastic. Referring to the top stress-strain plot
of Fig. 4-17, if we applyrepeatedly the same increment of shear
strain to an element of soil (this is referredto as
strain-controlled loading), the increment of stress that the soil
element is ableto sustain decreases continuously (a process that is
sometimes referred to as modu-lus degradation), until a state is
reached (represented by point F in the figure) atwhich the stress
increment will be zero. At this point, if we continue to
increasethe strain, the stress will stay the same. The other
stress-strain plot shown in thefigure illustrates another possible
response, whereby the stress peaks at point F.This second response
(referred to as strain softening) is common in soils. The lim-iting
or peak stress associated with point F in each case is usually what
is meantby the shear strength of the soil, as it is the maximum
stress the soil can take.
Alternatively, we could have started loading the soil element by
applyingstress increments to the element (which is referred to as
stress-controlled loading).In this case, when the point F at which
the peak stress was observed during strain-controlled loading is
reached, the soil element will not be able to take any addi-tional
stress, and what will happen instead is uncontrolled deformation.
It is not
k e1 for plane-strain conditions2 for triaxial conditions
sin c de1 kde3de1 kde3
dev
de1 kde3
devde1
2 devde1
132 The Engineering of Foundations
2a Shear strain
Shea
r stre
ss
a
F
F
b
c
Figure 4-17Nonlinearity of stress-strain relationship for soils
and failure (theonset of very large deformations at some value of
stress, representedby point F). Note that the first strain
increment a generates the stressincrement b and that the second
strain increment generates a stressincrement c b b.
saL00581_ch04.qxd 9/14/06 4:54 PM Page 132
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possible in the stress-control case to plot the stress-strain
relationship after point Fis reached.
The Mohr-Coulomb failure (or strength) criterion has been
traditionally usedin soil mechanics to represent the shear strength
of soil. It expresses the notion thatshear strength of soil
increases with increasing normal effective stress applied onthe
potential shearing plane. Although in stress analysis we do not
have to concernourselves with whether we are dealing with effective
or total stresses because theanalysis applies to both, we must now
make a clear distinction. Soils feel onlyeffective stresses (that
is, any deformation of the soil skeleton happens only inresponse to
effective stress changes); thus, the response of soil to loading
and theshear strength of soil depends only on the effective
stresses. Based on this consid-eration, we represent the
Mohr-Coulomb criterion in s-t space as two straightlines making
angles f with the horizontal and intercepting the t-axis at
distancesc and c from the s-axis. Figure 4-18 shows only the line
lying above the s-axis, since the diagram is symmetric about the
s-axis. The distance c is usuallyreferred to as the cohesive
intercept. Mathematically, the Mohr-Coulomb criterionmay be
represented in a simple way by
(4.21)where s is the shear strength of the soil and f is its
friction angle. For a given nor-mal effective stress s on a plane,
if the shear stress on the plane is t s as givenby Eq. (4.21),
shearing or what is commonly referred to in engineering practice
asfailure of the material occurs. Failure here means the occurrence
of very largestrains in the direction of that plane. This means
that the soil cannot sustain shearstresses above the value given by
Eq. (4.21). Equation (4.21) is a straight line ins-t space (as
shown in Fig. 4-18) that is referred to as the Mohr-Coulombstrength
envelope. There can be no combination of s and t that would lie
abovethe Mohr-Coulomb strength envelope.
Figure 4-19 shows the Mohr circle for a soil element (or point)
within a soilmass, at failure, where s and t act on the two planes
corresponding to the twotangency points between the circle and the
envelope (again, only the half of thediagram lying above the s-axis
is shown, as the part below the s-axis is symmet-ric). All other
points, representing all the planes where (s, t) do not satisfy
Eq.(4.21), lie below the Mohr-Coulomb envelope. If we know the
directions of the
s c s tan f
C H A P T E R 4 Stress Analysis, Strain Analysis, and Shearing
of Soils 133
t
s
c
f
Figure 4-18The Mohr-Coulomb failure envelope.
saL00581_ch04.qxd 9/14/06 4:54 PM Page 133
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principal stresses, we can determine the pole for this circle.
Assuming a verticalmajor principal stress and a horizontal minor
principal stress, the pole P lies atpoint (s3, 0). By simple
geometry, the central angle corresponding to the arcextending from
(s1, 0) to the point of tangency is 90 f. This means the
planescorresponding to the points of tangency lie at (45 f/2) with
the horizontal.The direction of a real shear (slip) surface passing
through the soil element can beapproximated by one of these two
possible directions. Note that this refers to thedirection of the
slip surface at the point under consideration, and that at
otherpoints of the soil mass the slip surface direction may be
different because the prin-cipal directions at those points may be
different. This means that a slip surfacethrough a soil mass is a
surface tangent at every point to the direction estimated
asdescribed earlier, a direction making an angle of (45 f/2) with
the directionof the major principal plane at the point. In a
subsequent subsection, we discuss inmore detail the issue of slip
surfaces, their nature, and their geometry.
It is possible to find the relationship between the principal
stresses s1 and s3at failure from the geometry of Fig. 4-19. It is
easier to proceed if we define new,transformed normal stresses s*
through
(4.22)Taking Eq. (4.22) into Eq. (4.21) leads to
(4.23)This transformation (sometimes referred to as Caquots
principle after the per-
son who first made use of it) is represented graphically in Fig.
4-20. A Mohr circle atfailure is also represented. We can write sin
f in terms of the ratio of the radius ofthe Mohr circle to the
distance from the center of the Mohr circle to the origin of
thetransformed system of stress coordinates (that is, where the s*-
and t*-axes cross):
sin f s1* s3*
s1* s3*
s s* tan f
s* s c cot f
134 The Engineering of Foundations
s
t
f
f
t s
45 f/2
90 f
(s1 s3)
(s1, 0)(s3, 0)
c cot f
c P
12 (s1 s3)12
Figure 4-19Mohr circle at failure.
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The principal stress ratio s1*/s3* easily follows:
(4.24)
where N is known as the flow number, given by
(4.25)
If the relationship between the principal stresses in its
original form is needed,we just need to use Eq. (4.22) to rewrite
s1* and s3* in Eq. (4.24):
which can be rewritten as(4.26)
This expression can be further rewritten as
(4.27)by recognizing that
Note that, for c 0, Eq. (4.26) reduces to(4.28)
Slip Surfaces*
Figure 4-21(a) shows a slip surface and an element of it, which
is expanded in Fig.4-21(b). A slip surface (also referred to in the
literature as a failure surface or the
s1 Ns3
cot f B1 sin2 f
sin2 f
s1 Ns3 2c2N
s1 Ns3 1N 1 2c cot fs1 c cot fs3 c cot f
1 sin f1 sin f
N
N 1 sin f1 sin f
s1*
s3* N
C H A P T E R 4 Stress Analysis, Strain Analysis, and Shearing
of Soils 135
t* t
f
c cot f
(s1 s3)
s, s*
* *12
(s1 s3)* *12
Figure 4-20Caquots principle: transformednormal stresses.
saL00581_ch04.qxd 9/14/06 4:54 PM Page 135
-
more technical shear band) develops when the shear stresses on
every point ofit reach the corresponding shear strengths. The slip
surface is a surface alongwhich a soil mass slides with respect to
another, usually stationary, soil mass.These soil masses behave
very much like rigid blocks, not undergoing any defor-mation.
Accordingly, an acceptable model for the slip surface element shown
inFig. 4-21(b) has the element bounded by two rigid blocks. In
truth, slip surfacesare not truly surfaces, but very thin zones or
bands of highly concentrated shearstrains, hence the representation
of a slip surface element in Fig. 4-21 as havingnonzero
thickness.
The slip surface is bounded by rigid blocks that are on the
verge of sliding pastone another. On the onset of failure, these
rigid blocks are connected to the slip sur-face and, being rigid,
prevent any contraction or elongation in the direction of theslip
surface. This means the normal strain in the direction x1z of the
slip surface isequal to zero. The slip surface x1z is, accordingly,
a zero-extension line. The state ofstrain at a point of the slip
surface, referred to the axis x1z parallel to the slip surfaceand
another x3zp normal to it, is represented by Fig. 4-16. Taking the
thickness of theslip surface as being equal to 1, the state of
strain for the slip surface element ofFig. 4-21 can be expressed as
follows:
dev de11z de33zp dx3
0dg z 0 dx11
dx1
de33zp dx31
dx3
de11z 0
136 The Engineering of Foundations
(a)
(b)
x3zp
x1z
Q
Expanded in (b)
Slip surface Zero-extension line
Rigid blocks
1
dx1
dx3
Figure 4-21Details of slip surfaceelement.
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Recalling the definition of the dilatancy angle c given in Eq.
(4.16), we canwrite
(4.29)
Equation (4.29) shows that the dilatancy angle represents the
angle of the motionat the top of the element with respect to the
horizontal. Equation (4.29) is instru-mental in understanding the
geometry of slip surfaces. Take, for example, the soilslope of Fig.
4-22. It is common to see such slopes failing along curved slip
sur-faces. The shape of a potential slip surface can be determined
if we take a center ofrotation such that every point of the surface
is at a variable distance r from the cen-ter of rotation. Once a
relationship between r and the angle of rotation u aroundthe center
of rotation is defined, the shape of the slip surface becomes
known.
An infinitesimal rotation du with respect to the center of
rotation correspondsto a tangent displacement rdu along the slip
surface and to a possible smallincrease of the radius by an amount
dr. So dr is analogous to the dx3, and rdu isanalogous to the dx1
of Eq. (4.29). Taking these into Eq. (4.29) gives
(4.30)
With knowledge of one point of the slip surface, defined by r0
and u0, integra-tion of Eq. (4.30) leads to
(4.31)where r0 is the radius corresponding to a reference angle
u0.
Equation (4.31) is a geometrical shape referred to as a
logarithmic spiral (orlog-spiral, for short). The implication is
that a homogeneous soil mass, with thesame dilatancy angle
throughout, is expected to fail along a curved slip surface withthe
shape of a log spiral. We often approximate these curved surfaces
by planes orcylinders (straight lines or circles in cross section).
Both are strictly applicable onlyif c 0, for then Eq. (4.31)
reduces to that of a circle, and a straight line is nothingmore
than a circle with infinite radius; but the approximation is
sometimes justifiedwhen the maximum value (umax u0) of u u0 in Eq.
(4.31) is small. We analyze
r r0e1uu02 tan c
tan c drrdu
tan c dev0dg z 0 dx3dx1
C H A P T E R 4 Stress Analysis, Strain Analysis, and Shearing
of Soils 137
du
rdu
dr
Center of rotation
Figure 4-22Slip surface geometry.
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soils that shear under undrained conditions using totalstresses
and c 0. We often do that for clays. The use ofa slip surface with
circular cross section then followsdirectly from Eq. (4.31).
Slip Surface Direction*
The Mohr circle can be used to determine the two direc-tions of
the two potential slip surfaces at a point within asoil mass. The
direction of the slip surface at a point is thedirection of its
tangent at the point. The pole method canbe used to do this
graphically.
Assume that the major principal strain increment isvertical,
therefore normal to the horizontal plane. In suchcase, the pole P
is located at the leftmost point of the cir-cle, as shown in Fig.
4-23. We start by drawing lines PA1and PA2 from P through the
points A1 and A2 correspon-ding to zero normal strain. These lines
are parallel to theplanes that are perpendicular to the directions
of zero nor-
mal strain (that is, the slip surface directions). So lines PB1
and PB2, normal to PA1and PA2, respectively, are the directions of
the two potential slip surfaces throughthe point under
consideration.
From the geometry of the Mohr circle of Fig. 4-23, it can be
shown that theslip surfaces make an angle 45 c/2 with the direction
of the minor principalstrain (which is horizontal in this case). If
c f (a common assumption, even iftacitly made, in many geotechnical
analyses), then the slip surface directions corre-spond to the
directions of the planes where s and t satisfy the Mohr-Coulomb
fail-ure criterion, which are at 45 f/2 to the horizontal. Even
when c differs fromf, so long as the difference is not large, the
direction of the slip surfaces can stillbe approximated as being 45
f/2 with respect to the horizontal. But, in dilativesands (sands
that exist in a dense state and/or in a state of low effective
confiningstresses), the deviations from reality from assuming c f
in analyses may besubstantial and should not be ignored. This point
will be illustrated in Chapter 10in the context of bearing capacity
analyses of footings.
The Hoek-Brown Failure Criterion for Rocks
Although the Mohr-Coulomb failure criterion is also used for
rocks, the Hoek andBrown (1980, 1988) criterion is more practical,
particularly for fractured rock, asits parameters have been related
to quantities usually measured in site investiga-tions in rock. The
criterion is written as
(4.32)
where qu uniaxial, unconfined compressive strength of the rock;
m and s model parameters that depend on the degree of fracturing of
the rock. The parame-
s1 s3 quBm s3qu
s
138 The Engineering of Foundations
B2
B1
A1
A2
2 dlg
dede3 de1
PB1 PA1PB2 PA2
45 c/2 90 c
45 c/2
cc
c
P
Figure 4-23Determination of direction of slip planes for asoil
element.
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ters m and s can be expressed in terms of the type of rock,
degree ofweathering, and frequency of discontinuities. This is
covered inChapter 7.
4.4 At-Rest and Active and PassiveRankine States
At-Rest State
Let us consider a semi-infinite mass of homogeneous, isotropic
soil[Fig. 4-24(a)]. The word homogeneous means the soil has the
samephysical properties (for example, unit weight, hydraulic
conductivity,shear modulus, and shear strength) at every point, and
the wordisotropic means the directional properties at any point of
the soil massare the same in every direction. This soil deposit has
a free surfacethat is level (horizontal); the soil deposit extends
to infinity downwardfrom the free surface and in the directions
parallel to the free surface.Vertical and lateral stresses within
this deposit are principal stresses,as the shear stresses in
horizontal and vertical planes are clearly zero.
This soil deposit idealization of Fig. 4-24(a) allows us to
obtainvery important results. If we assume that the soil deposit
has notbeen disturbed after formation, the soil is said to be in a
state of rest.The ratio K0 sh/sv of the lateral to the vertical
effective stress atany point in a soil mass in a state of rest is
referred to as the coeffi-cient of lateral earth pressure at rest.
Note that K0 is a ratio of effec-tive, not total stresses. It is a
very important quantity that appearsoften in geotechnical design.
For a purely frictional soil, withstrength parameters c 0 and
nonzero f, Jaky (1944) established anempirical relationship between
K0 and f, given as9
(4.33)An alternative expression for K0 can also be obtained from
elas-
ticity,10 in which case K0 is related to the Poissons ratio of
the soilthrough
(4.34)
Equation (4.33) applies only to normally consolidated soils
(thatis, soils that have never experienced a larger vertical
effective stressthan the stress they currently experience). In a
normally consolidated
K0 n
1 n
K0 1 sin f
C H A P T E R 4 Stress Analysis, Strain Analysis, and Shearing
of Soils 139
45 f/2
45 f/2
(a)z
g, c, f
(b)
(c)
(d)
(e)
Infinite plane
Active
Extension
Compression
Passive
Figure 4-24(a) A semi-infinite soil mass; (b) asemi-infinite
vertical plate separatingtwo halves of the semi-infinite soilmass;
(c) vertical plate with soilcompletely removed from the left
side;(d) soil in the active Rankine state, asillustrated by drawing
the two familiesof possible slip surfaces at angles of
(45 f/2) with the horizontal; (e)soil in the passive Rankine
state, asillustrated by drawing the two familiesof possible slip
surfaces at angles of
(45 f/2) with the horizontal.
9 For clays, some prefer to use 0.95 instead of 1 in Eq.
(4.33).10 To derive this equation, we can write the equation for
the lateral strain in terms of itscoaxial normal stress (that is,
the lateral effective stress) and the two transverse normalstresses
(the vertical and again the lateral effective stress, but now in
the other lateraldirection) and make the lateral strain equal to
zero.
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soil, if the vertical effective stress is increased by an amount
dsv, the lateral effec-tive stress increases by an amount dsh K0
dsv in order to keep the ratio of sh tosv constant and equal to K0.
To consider the effects of stress history, assume thatthe vertical
effective stress is then reduced by dsv down to the original value;
if thatis done, the lateral effective stress does not go back to
its previous value, retaining aconsiderable fraction of the
increase dsh it experienced when sv was increased. Itfollows that
soils that have experienced a larger vertical effective stress
previously,referred to as overconsolidated soils, have higher K0
than normally consolidatedsoils. Physically, the reason lateral
stresses get locked in is the change in the fabricand density of
the soil required to accommodate the increase in vertical
effectivestress, which is to some extent inelastic in nature and
thus irrecoverable.
Brooker and Ireland (1965) investigated the effects of stress
history on K0,arriving at the following equation:
(4.35)where OCR overconsolidation ratio, defined as
(4.36)
where svp preconsolidation pressure, which is the maximum
vertical effectivestress ever experienced by the soil element and
sv is simply the current verticaleffective stress.
Rankine States
Level Ground In order to investigate the behavior of soil
deposits when sub-jected to relatively large strains, it will help
us now to go through an imaginaryexercise. Let us consider that we
could insert a smooth, infinite, infinitesimallythin plane
vertically into the soil deposit without disturbing the soil [Fig.
4-24(b)],thus keeping vertical shear stresses equal to zero. Let us
consider further that wecould remove all the soil from one side of
the plane [Fig. 4-24(c)] so that we couldnow either push or pull on
the plane in the horizontal direction, thereby eithercausing the
soil on the other side of the plane to compress or extend in the
hori-zontal direction.
When we pull the vertical plane horizontally, it allows the soil
to expand inthe horizontal direction, which leads to a drop in the
horizontal effective stress sh.If we continue to pull on the plane,
sh continues to drop, while sv remainsunchanged. This process is
illustrated by the Mohr circles in Fig. 4-25, all with thesame
major principal stress s1 sv but a decreasing minor principal
stress s3 sh. This process, by which sh decreases while sv remains
unchanged, cannot goon indefinitely; it in fact comes to an end
when sv/sh becomes equal to the flownumber N s1/s3 (1 sin f)/(1 sin
f), at which point the Mohr circletouches the Mohr-Coulomb failure
envelope (assumed here with c 0), the failurecriterion is
satisfied, and slip surfaces can potentially form anywhere in the
soilmass. This state in which the whole soil mass is in a state of
incipient collapse isknown as an active Rankine state: active
because the self-weight of the soil con-
OCR svpsv
K0 K0, NC2OCR 11 sin f 22OCR
140 The Engineering of Foundations
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tributes or is active in bringing it about; Rankine because it
was first identified byLord Rankine in the 19th century (Rankine
1857, Cook 1950).
The coefficient KA of active earth pressure is the ratio of the
lateral to the ver-tical effective stress in a soil mass in an
active state, given by
(4.37)
The vertical effective stress is the major principal effective
stress in the activecase, which means the leftmost point of the
circle is the pole PA (Fig. 4-25). Thedirection of potential slip
surfaces is the direction of a line through the pole PApassing
through the point of tangency of the Mohr circle with the failure
envelope.As shown in Fig. 4-25, this direction makes an angle of
(45 f/2) with the hor-izontal. Figure 4-24(d) shows the two
families of potential slip surfaces associatedwith the active
Rankine state.
While the active state provides a lower bound to the value of
the lateral earthpressure coefficient K, the passive Rankine state
is on the other extreme, cappingthe possible values of K. The
passive state is obtained by pushing the vertical planeof Fig.
4-24(c) toward the soil mass, which increases sh until sh/sv
becomes equalto N. The progression from the at-rest to the passive
Rankine state can be visual-ized through Mohr circles as shown in
Fig. 4-25. The coefficient KP of passiveearth pressure is the ratio
of the lateral to the vertical effective stress in a soil massin a
passive state, given by
(4.38)
Recognizing that the lateral effective stress is the major
principal stress, wecan find the pole PP (see Fig. 4-25).
Connecting the pole to the points of tangencyof the Mohr circle for
the passive state and the strength envelope, we find that
thedirections of the potential slip surfaces in the passive Rankine
state make angles of
KP shPsv
s1s3
N 1 sin f1 sin f
KA shAsv
s3s1
1N
1 sin f1 sin f
C H A P T E R 4 Stress Analysis, Strain Analysis, and Shearing
of Soils 141
PPPA
shP s1
s
f
f
Mohr circleat rest
Mohr circlefor active
Rankine state
Mohr circlefor passive
Rankinestate
45 f/2
45 f/2
45 f/2
45 f/2
t
shA s3
sv sh
Figure 4-25Sequence of Mohr circles illustrating the
horizontalunloading of a soil mass until the active Rankine stateis
reached and the horizontal compression of a soilmass until the
passive Rankine state is reached.
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(45 f/2) with the horizontal (Fig. 4-25). Figure 4-24(e) shows
the directionsof the potential slip surfaces in the soil mass.
Figure 4-26 shows the Mohr circlesfor the three states we have
examined and how the corresponding lateral effectivestresses relate
to the vertical effective stress through the corresponding
coefficientof lateral earth pressure. Note that K0, KP, and KA are
ratios of effective, not total,stresses.
Active and passive earth pressure analysis can be easily
extended to c-f mate-rials (a good reference for that is Terzaghi
1943), but its usefulness in practice israther limited. We will
therefore not discuss this extension in this text.
142 The Engineering of Foundations
t f
f
s
shP KPsv
sv gz
sh A KAsv
sh 0 K0sv
Passive
At rest
Active
Figure 4-26Mohr circles representing theat-rest state and the
active andpassive Rankine states.
EXAMPLE 4-6
For a normally consolidated cohesionless soil with f 30,
calculate the at-rest, active, andpassive values of the earth
pressure coefficient K.
SolutionThe coefficient K0 of lateral earth pressure at rest is
calculated using Eq. (4.33):
The active and passive lateral earth pressure coefficients are
calculated using Eqs.(4.37) and (4.38):
Note how the ratio of passive to active pressures is 9 for a 30
friction angle.
KP 1 sin 301 sin 30
3
KA 1 sin 301 sin 30
13
K0 1 sin f 1 sin 30 0.5
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Sloping Ground* Consider a soil mass sloping at an angle ag with
respect to thehorizontal [Fig. 4-27(a)]. For simplicity, consider
the soil mass to be free of water.Focusing on the prism of unit
width shown in Fig. 4-27(a), we see that the weightof the prism per
unit length into the plane of the figure is equal to gz. The
weightof the prism is the only reason its base is subjected to a
normal and a shear stress.By projecting the weight normally and
tangentially to the base of the prism anddividing the component
forces by the area of the base (which is 1/cos ag per unitlength of
prism normal to the plane of the figure), we obtain the following
expres-sions for the normal and shear stresses at the base of the
prism:
(4.39)(4.40)
Note that the effective traction T (the resultant of the normal
and shear effec-tive stresses) on the base of the prism of Fig.
4-27(a) is vertical and given by
(4.41)It is important to understand that T is vertical but is
not sv (that is, it is not
the vertical effective stress on the horizontal plane) and does
not act on the hori-zontal plane.
T gz cos ag
t gz sin ag cos ag
s gz cos2 ag
C H A P T E R 4 Stress Analysis, Strain Analysis, and Shearing
of Soils 143
s gz cos2 ag
s
ag
f
f
ag
ag
1
(a)
(b)
0
gz cos2 ag
T gz cos ag
t gz sin ag cos ag
gz cos ag sin ag
z
F
PE
A
CD
BG
ag
t
Figure 4-27Active Rankine state in sloping ground: (a) a prism
withunit width with base at depth z in a soil deposit withsloping
surface; (b) Mohr circle corresponding to activestate at base of
prism.
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We can plot the state of stress defined by Eqs, (4.39) and
(4.40) in a Mohr dia-gram as point A, as illustrated in Fig.
4-27(b). Note that the distance OA from theorigin of the s-t space
to point A is equal to gz cos ag (that is, T). When OA isprojected
onto the s- and t-axes, we get Eqs. (4.39) and (4.40). The Mohr
circlecorresponding to the active Rankine state is drawn in the
figure going through A.Because point A defines the stresses on a
plane making an angle ag with the hori-zontal, the pole P lies on
the intersection of the Mohr circle with the OA line.
Now let us say we wish to determine the resultant stress on a
vertical plane.Drawing a vertical line through the pole, we obtain
point D as the intersection ofthis line with the Mohr circle. The
resultant stress (traction) TA (which, note, is notthe horizontal
effective stress and is not even horizontal) is thus given by the
lineOD; that is,
Let us now find the magnitude of the ratio of OD to OA and then
its normaland tangential components acting on the vertical plane.
We do that by analyzingthe geometry of the Mohr diagram of Fig.
4-27(b). We first note that, because CEis perpendicular to OA,
We can now write
(4.42)
But OE is simply
(4.43)
To find EP, we note that AEC is a right triangle and that CA CF.
We cannow write
So we have
which leads to
and that, in turn, leads to
(4.44)EA OC2sin2 f sin2 ag OC2cos2 ag cos2 f EP
EA2 CA2 CE 2 OC 2 sin2 f OC 2 sin2 ag
CA2 CE 2 EA2
CA CF OC sin f
EC OC sin ag
OE OC cos ag
T AT
ODOA
OPOA
OE EPOE EP
EP EA
TA OD
T OA
144 The Engineering of Foundations
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Taking Eqs. (4.43) and (4.44) into Eq. (4.42), we obtain an
expression for theactive Rankine earth-pressure coefficient for
sloping ground:
(4.45)
The horizontal stress is obtained by projecting the resultant
stress TA acting onthe vertical plane onto the horizontal
direction:
(4.46)while the shear stress is obtained by projecting TA on the
vertical direction:
(4.47)The directions of slip planes can be determined in much
the same way as we
found directions for level ground, with lines drawn from P
through F and G yield-ing these directions.
4.5 Main Types of Soils Laboratory Tests for Strength and
Stiffness Determination
Role of Stiffness and Shear Strength Determination
In the preceding sections, we examined stresses, strains, and
the relationshipbetween strains and stresses in soils, particularly
shearing (or failure). We did sobecause they are an integral part
of the calculations we do in the analysis of foun-dations, slope,
retaining structures, and other geotechnical systems. In this
section,we examine the important issue of how to measure
stress-strain properties properlyand, in particular, the shear
strength of soils. This issue is fundamental both forwork done in
practice and for research on soil load response.
Stress Paths in s-t Space
It is not practical to trace the loading of a soil element or
soil sample in a labora-tory test by using Mohr circles because an
infinite number of Mohr circles wouldbe required and no discernible
representation of the loading process would result.A diagram that
achieves the same purpose is based on plotting only the
highestpoints of the Mohr circles, producing a continuous line. The
highest point of theMohr circle has coordinates s and t (as shown
in Fig. 4-28) given by
(4.48)(4.49)
where s corresponds to the center of the Mohr circle, and t is
equal to the radius ofthe Mohr circle and is thus a measure of
shear stress.
t 12 1s1 s3 2s 12 1s1 s3 2
tvA KAT sin ag KA gz sin ag cos ag
shA KATcos ag KA gz cos2 ag
KA T AT
cos ag 2cos2 ag cos2 fcos ag 2cos2 ag cos2 f
C H A P T E R 4 Stress Analysis, Strain Analysis, and Shearing
of Soils 145
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A stress path is a plot in s-t space of the progression of (s,
t) points represent-ing the loading process for a soil element or
laboratory soil sample. Figure 4-29shows the four possible general
directions a stress path can take from a point (s0,t0) depending on
whether only s1 or only s3 changes, and whether the change isan
increase or a decrease.
Referring to Eqs. (4.48) and (4.49), we can write the
differentials of s and t as(4.50)(4.51)
When only s1 changes, ds3 0 and
dt 12ds1
ds 12ds1
dt 12 1ds1 ds3 2ds 12 1ds1 ds3 2
146 The Engineering of Foundations
(s3, 0) (s1, 0)
t
s
(s, t)
12
12
(s1 s3)
(s1 s3)
Figure 4-28Definition of stress variables sand t.
t
s
s3
s1 unchanged
s1
s3 unchangeds3
s1 unchanged
s1
s3 unchanged
(s0, t0)
Figure 4-29Possible directions for s-t paths from a starting
point (s0, t0).
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When only s3 changes, ds1 0 and
It follows that all possible stress paths when only one of s1 or
s3 changes arestraight lines at 45 with the horizontal, with dt/ds
1 when s1 alone changes anddt/ds 1 when s3 alone changes.
Naturally, other directions are possible if s1and s3 are allowed to
change simultaneously.
Stress plots can also be done in terms of effective
stresses:
(4.52)
(4.53)where u pore pressure (see Chapter 3).
Effective stress paths clearly coincide with total stress paths
if loading isapplied under drained conditions. Under undrained
conditions (u 0), effectivestress paths do not coincide with total
stress paths and are not linear.
Stress Paths in p-q Space
The stress variables s and t are two-dimensional variables that
do not capture theeffects of s2. The mechanical response of soil is
more rigorously expressed interms of p and q, which are defined as
follows in terms of the principal stresses:
(4.54)
(4.55)
Those readers with a background in mechanics will recognize p
and q as beingrelated to the octahedral normal and shear stresses
sm soct and toct through
(4.56)
(4.57)
where the octahedral normal stress is also known as the mean
stress sm. Thus, pand q are normal and shear stresses that are
representative of the three-dimensionalstress state at a point.
Effective stress versions p and q of p and q can also be
defined:
(4.58)(4.59)q q
p p u
q 223
toct
p sm soct
q 1
222 1s1 s3 2 2 1s1 s2 2 2 1s2 s3 2 2p 13 1s1 s2 s3 2
t 12 1s1 s3 2 ts 12 1s1 s3 2 s u
dt 12ds3
ds 12ds3
C H A P T E R 4 Stress Analysis, Strain Analysis, and Shearing
of Soils 147
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Main Laboratory Tests
The laboratory soil tests of greatest importance for what we
cover in this text arethe one-dimensional compression test (more
routinely called the consolidationtest11), the direct shear test,
the triaxial compression test, and the unconfined com-pression
test. (See Figs. 4-30 to 4-32.) The consolidation test is a
one-stage test,
148 The Engineering of Foundations
Figure 4-30Consolidometer:(a) photo,(b) schematic diagram.
Q
Porousstones
WT
Sample
Consolidationring
(a) (b)
Figure 4-31Direct shear machine:(a) photo,(b) schematic
diagram.
N
N
Slip surface
Before applicationof shear force T
After applicationof shear force T
T 0T 0
T 0T 0
Soil sample
Soil sample
(a) (b)
11 Note that soil consolidation can be accomplished in many
different test setups; however, we will retainthe more traditional
name for this test in the remainder of the book.
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while both the direct shear test and the triaxial compression
test have a consolida-tion stage (in which the sample is taken to a
desired stress state) and a shearingstage (in which the shear
stress level is increased within the sample, usually untilvery
large shear strains, of the order of 20%, develop). The unconfined
compres-sion test is a test that can be done only on soil samples
that dont fall apart whenunconfined and dont have major defects
(fissures, seams of different materials cut-ting across the soil,
and the like). Many clays can be tested in this manner, as wewill
see in Chapter 6. The sample is not subjected to a confining
stress, that is, thetotal stress s3 0. It has only a shearing
stage, in which an axial load is appliedon the sample until
failure.
The consolidation (one-dimensional compression) test is
performed by apply-ing a vertical load on a cylindrical sample of
soil restrained laterally by a steel ring(Fig. 4-30) and measuring
the resulting vertical deflection. If we recall our discus-sion of
the at-rest state and of the coefficient of lateral earth pressure
at rest as aratio of sh to sv for a normally consolidated, level
soil deposit, it is easy to under-stand why the stress path imposed
by this test is as shown in Fig. 4-33(b). Thesample is kept under
water to ensure full saturation. Although there is more thanone
size of consolidometers, the most typical sample size is 70 mm
(diameter) by25 mm (thickness). The consolidation test is
particularly useful in the study ofclays, and we will therefore
rediscuss it in Chapter 6.
The direct shear test is performed on a soil sample that is also
restrained hori-zontally by a split box constituted of a top and a
bottom part that can move hori-zontally with respect to each other
(Fig. 4-31). The sample is first loaded vertically
C H A P T E R 4 Stress Analysis, Strain Analysis, and Shearing
of Soils 149
Figure 4-32Triaxial test system: (a) photo,(b) schematic
diagram.
Axial force
Top platen
Soilsample
Bottom platen
Cell pressure s3
Drainage
Drainage
Rubbermembrane
(a) (b)
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without the possibility of lateral expansion (much as in a
one-dimensional com-pression test); then, a horizontal force is
applied on the upper part of the box, caus-ing the sample to shear
in the horizontal direction. The stress path for this test isshown
in Fig. 4-33(c). The consolidation portion (AB) of the stress path
is just likethe stress path of a one-dimensional compression test.
The shearing is a verticalline BC, indicating that the shear stress
added to the sample horizontally does notchange s, just t. In
truth, s may increase slightly because of one of the shortcom-ings
of the direct shear test, which is that some passive pressures
develop as someof the push imposed in the upper part of the sample
ends up in part as a pushagainst the lower part of the box on the
other side of the sample. This also meansthat the shear strength of
the soil is not the only resistance to failure; this
passiveresistance also contributes to the resistance to shearing of
the sample. Anothershortcoming of this test is that it forces
shearing to develop horizontally instead ofallowing it to develop
naturally along the optimal shearing plane. Nonetheless, thedirect
shear test is used in practice often and does provide a reasonable
estimate ofshear strength.
In a triaxial (TX) test, a cylindrical sample, typically either
70 or 35 mm indiameter and at least twice that in height, is
wrapped in a close-fitting, cylindrical,impervious membrane and
placed between two platens. The bottom platen isattached to the
base of the triaxial chamber and is stationary. The top platen
isattached to a piston to which an axial force may be applied.
During the test, thesample, platens, and piston are located within
a plexiglass or aluminum cylinderthat is sealed at the top and
bottom. So, in the course of a test, the sample is firstsubjected
to an all-around stress by increasing the air pressure around the
sampleand then to a change in the axial stress. The first stage is
called consolidation(although the soil may or may not undergo
volume change); the second, shearing.The impervious membrane allows
transmission of the applied pressure to the sam-ple as a total
stress. Drainage may be either allowed or prevented during either
or
150 The Engineering of Foundations
(a)s
t
Figure 4-33Stress paths for (a) isotropic compression, (b)
one-dimensional consolidation, (c) direct shear test, and (d)
triaxialcompression and extension tests.
(b)
K0 line
s
t
(c)
AB 1D consolidationBC Shearing
K0 lineC
AB
s
t
AB Isotropic consolidationBC Shearing in compressionBD Shearing
in extension
(d)
11
B
D
A
C
s
t
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both the consolidation and shearing stages. The effective stress
state at the end ofthe consolidation stage will be different from
the initial state only if drainage isallowed. Tests in which
drainage is allowed during the consolidation stage arereferred to
as consolidated or C tests; when drainage is prevented, they are
knownas unconsolidated or U tests. Tests in which drainage is
allowed during the shear-ing stage are known as drained (D);
otherwise, tests are referred to as undrained(U). Table 4-1
summarizes the drainage conditions for the three main types of
tri-axial tests.
Table 4-2 shows the evolution of the effective stress state
acting on a triaxialtest sample tested under drained conditions.
Samples are usually saturated withwater. One of the advantages of
doing so is that sample volume change can be eas-ily measured by
measuring the volume of water either expelled from within thesample
or sucked in by the sample. When water is sucked in, the void ratio
of thesoil increases, weakening the so