Stress Strain Analysis of Soils

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


(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


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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


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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


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.


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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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.


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


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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-


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.

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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


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


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

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


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


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


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.

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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


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.


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


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.

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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


t

s

c

f

Figure 4-18The Mohr-Coulomb failure envelope.

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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


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


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


(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


du

rdu

dr

Center of rotation

Figure 4-22Slip surface geometry.

saL00581_ch04.qxd 9/14/06 4:54 PM Page 137

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


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.

saL00581_ch04.qxd 9/14/06 4:54 PM Page 138

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


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


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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


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.


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


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


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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


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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


(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


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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,


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


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


(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

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Stress Strain Analysis of Soils