Stress Analysis, Strain Analysis, and Shearing of Soils

Transcription

1 C H A P T E R 4 Stress Analysis, Strain Analysis, 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 take regard to the fact that there is no linearity condition between stresses and strains for soils. J. Brinch Hansen Foundation design requires that we analyze how structural loads are transferred to the ground, and whether the soil will be able to support these loads safely and without excessive deformation. In order to do that, we treat soil deposits as continuous masses and develop the analyses needed for determining the stresses and strains that appear in the soil mass as a result of application of loads on its boundaries. We then examine whether the stresses are such that shearing (large shear strains) occurs anywhere in the soil. Other soil design problems involving slopes and retaining structures are also analyzed by computing the stresses or strains resulting within the soil from actions done on the boundaries of the soil mass. This chapter covers the basic concepts of the mechanics of soils needed for understanding these analyses. At the soil element level, coverage includes stress analysis, strain analysis, shearing and the formation of slip surfaces, and laboratory tests used to study the stress-strain response and shearing of soil elements. At the level of boundary-value problems, coverage includes a relatively simple problem in soil plasticity (Rankine states) and stresses generated inside a semi-infinite soil mass by boundary loads. 113

2 114 The Engineering of Foundations 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 level and (2) how this behavior, appropriately described by suitable equations and ascribed to every element (that is, every point) of a soil mass subjected to certain boundary conditions, in combination with analyses from elasticity or plasticity theory, allows us to determine how boundary loads or imposed displacements result in stresses and strains throughout the soil mass (and, in extreme cases, in strains so localized and large that collapse of a part of the soil mass and all that it supports happens). At the element level, we must define precisely what the element is and how large it must be in order to be representative of soil behavior. We must also have equations that describe the relationship between stresses and strains for the element and that describe the combination of stresses that would lead to extremely large strains 1. At the level of the boundary-value problem, we must define the maximum size an element may have with respect to characteristic lengths of the problem (for example, the width of a foundation) and still be treated as a point. This is important because we use concepts and analyses from continuum mechanics to solve soil mechanics problems, and so our elements must be points that are part of a continuous mass. We must then have analyses that take into account how elements interact with each other and with specified stresses or displacements at the boundaries of the soil mass in order to produce values of stress and strain everywhere in the soil mass. Further, we must also be able to analyze cases where the stresses are such that large strains develop in localized zones of the soil mass. Concepts from plasticity theory are used for that. It is clear from the preceding discussion that the mechanics of the continuum is a very integral part of soil mechanics and geotechnical and foundation engineering. We will introduce the concepts that are necessary gradually and naturally and with 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 the concept 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 of stress in soils. Consider a small planar area A passing through point P located within a soil mass (Fig. 4-1). A normal force F N and a tangential force F T are applied on A. If soil were a continuum, the normal stress s acting normal to A at 1 Extremely large strains, particularly extremely large shear strains, are closely tied to concepts of rupture, 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 introduce appropriate terms throughout the chapter and the remainder of the text.

3 CHAPTER 4 Stress Analysis, Strain Analysis, and Shearing of Soils 115 F N P Figure 4-1 Definition of stress in soils: As the area A is allowed to shrink down to a very small value, the ratios F N /A and F T /A approach values s and t, the normal and shear stress at P, respectively. F T A point P would be defined as the limit of F N /A as A tends to the point P (that is, tends to zero, centered around P). The shear stress would be defined similarly. Mathematically, s lim AS0 F N A (4.1) t lim F T AS0 (4.2) A Because soil is not a true continuum, we must modify this definition. A point within a soil mass is defined as a volume V 0 that is still very small compared with the dimensions of the foundations, slopes, or retaining structures we analyze, but it is sufficiently large to contain a large number of particles and thus be representative of the soil. 2 To this representative volume V 0, we associate a representative area A 0 (also very small, of a size related to that of V 0 ). So we modify Eqs. (4.1) and (4.2) by changing the limit approached by the area A from zero to A 0 : s lim ASA0 F N A (4.3) F T t lim ASA0 (4.4) A The preceding discussion brings out one difference between soil mechanics and the mechanics of metals, for example. In metals, the representative elementary volume (REV) is very small. The REV for a given metal is indeed so small that, in ordinary 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, so that they too have nonzero REVs, although much smaller than those we must use in soil mechanics. In soil mechanics, our REVs must include enough particles and the voids between them that, statistically, this group of particles will behave in a way that is representative of the way larger volumes of the soil would behave. 2 Mechanicians like to use the term representative elementary volume to describe the smallest volume of a given material that captures its mechanical properties.

4 116 The Engineering of Foundations Figure 4-2 (a) Elemental representation of two-dimensional stress state at a point; (b) illustration of how the element would distort when acted upon by a positive s 13 for the simple case with zero normal stresses s 11 and s 33 ; (c) sectioned triangular prismatic element, where s u and t u depend on the angle u. Plane p 11 s 33 s 13 s 11 s s u u t u s u x 3 s 13 Plane p 33 s 13 s 33 s 33 x 1 (a) s 13 (c) Vertical leg of L parallel to p 11 s 13 (b) Horizontal leg of L parallel to p 33 Stress Analysis Stress analysis allows us to obtain the normal and shear stresses in any plane passing through a point, 3 given the normal and shear stresses acting on mutually perpendicular planes passing through the point. 4 We will see in Section 4.6 some examples of how these stresses can be calculated at a point inside a soil mass from a variety of boundary loadings common in geotechnical engineering. Later in the text, we discuss some applications requiring corrections for three-dimensional (3D) effects, but for now we will focus on two-dimensional (2D) stress analysis, which turns out to be applicable to most problems of interest in practice. Figure 4-2(a) shows a small, prismatic element of soil representing a point within the soil. The faces of the element are aligned with the directions of the reference axes x 1 and x 3. The soil element is acted upon by the normal stresses s 11 acting in the x 1 direction and s 33 acting in the x 3 direction and the shear stresses s 13 (which, due to the requirement of moment equilibrium, is identical to s 31 ). 5 The first subscript of a stress component represents the direction normal to the plane on which it acts; the second, the direction of the stress component itself. A 3 A point in the soil is indistinguishable, for our purposes, from a representative soil element, which has a very small volume (and so is a point for practical purposes) but is sufficiently large to be representative of the soil in its mechanical behavior. 4 A more proper definition of stress analysis for advanced readers would be that stress analysis aims to allow calculations of the traction (which has normal and shear components) on a plane given the stress tensor at the point. 5 Note that many engineering texts use the notation t ij instead of s ij when i j to represent shear stress. We have retained here the traditional mechanics terminology, in which s is used for both normal and shear stresses.

5 CHAPTER 4 Stress Analysis, Strain Analysis, and Shearing of Soils 117 stress component with different subscripts is a shear stress; one with identical subscripts, a normal stress. For example, s 11 is the stress acting on the plane normal to x 1 in the x 1 direction (that is, it is a normal stress), while s 13 is the stress that acts on the plane normal to x 1 in the x 3 direction (and is thus a shear stress). It is simpler (although not required) to solve problems if we adopt the practice of choosing s 11 and s 33 such that s 11 s 33. So if the normal stresses are 1,000 and 300 kpa, then s 11 1,000 kpa and s kpa. Likewise, if the normal stresses are 100 and 500 kpa, then s kpa and s kpa. The plane where s 11 acts is denoted by p 11 ; likewise, p 33 is the plane where s 33 acts. The stresses s 11, s 33, and s 13 are all represented in their positive directions in Fig. 4-2(a). This means normal stresses are positive in compression, and the angle u is positive counterclockwise from p 11. With respect to the shear stress s 13, note that the prism shown in the figure has four sides, two representing plane p 11 and two representing plane p 33. Looking at the prism from the left side, we may visualize the plane p 11 as the vertical leg of an uppercase letter L and p 33 as the horizontal leg of the uppercase letter L. We then see that a positive shear stress s 13 acts in such a way as to open up (increase) the right angle of the L formed by planes p 11 and p 33. Figure 4-2(b) shows the deformed shape that would result for the element assuming zero normal stresses and positive s 13. If we section the element of Fig. 4-2(a) along a plane making an angle u with the plane p 11, as shown in Fig. 4-2(c), a normal stress s u and a shear stress t u must be applied to this plane to account for the effects of the part of the element that is removed. While the sign of the normal stress s u is unambiguous (positive in compression), the shear stress on the sloping plane has two possible directions: up or down the plane. So we must decide which of these two directions is associated with a positive shear stress. The positive direction of the shear stress actually follows from the sign convention already discussed (that s 13 0 when its effect would be to increase the right angle of the uppercase L made up by p 11 as its vertical and p 33 as its horizontal leg). It turns out the shear stress t u is positive as drawn in the figure, when it is rotating around the prismatic element in the counterclockwise direction. We will show why this is so later, when we introduce the Mohr circle. Our problem now is to determine the normal stress s u and the shear stress t u acting on the plane making an angle u with p 11. This can be done by considering the equilibrium in the vertical and horizontal directions and solving for s u and t u. The following equations result: s u 1 2 1s 11 s s 11 s 33 2cos 2u s 13 sin 2u t u 1 2 1s 11 s 33 2sin 2u s 13 cos 2u (4.5) (4.6) where the signs of s 11, s 33, and s 13 are as discussed earlier (positive in compression for the normal stresses and determined by the L rule in the case of s 13 ). Principal Stresses and Principal Planes Equations (4.5) and (4.6) tell us that s u and t u vary with u. That means that the normal and shear stresses on each plane through a given point are a unique pair. There

6 118 The Engineering of Foundations will be two planes out of the infinite number of planes through the point for which the normal stress will be a minimum and a maximum. These are called principal stresses. They are obtained by maximizing and minimizing s u by differentiating Eq. (4.5) with respect to u and making the resulting expression equal to zero. The largest principal stress is known as the major principal stress; it is denoted as s 1. The smallest principal stress is the minor principal stress, denoted as s 3. The planes where they act are referred to as the major and minor principal planes, denoted by p 1 and p 3, respectively. When we differentiate Eq. (4.5) with respect to u and make the resulting expression equal to zero, we obtain the same expression we obtain when we make t u, given by Eq. (4.6), equal to zero. This means that the shear stresses acting on principal planes are zero. An easy way to find the angles u p1 and u p3 that the principal planes p 1 and p 3 make with p 11 (measured counterclockwise from p 11 ) is then to make t u 0 in Eq. (4.6), which leads to tan 12u p 2 2s 13 s 11 s 33 (4.7) When u p is substituted for u back into Eq. (4.5), we obtain the principal stresses s 1 and s 3, which are the two normal stresses acting on planes where t u 0 and are also the maximum and minimum normal stress for the point under consideration, given by s s 11 s s 11 s s 13 s s 11 s s 11 s s 13 (4.8) (4.9) Note that, given the definition of the tangent of an angle, there is an infinite number of values of u p that satisfy Eq. (4.7). Starting with any value of u p satisfying Eq. (4.7), we obtain additional values that are also solutions of (4.7) by repeatedly either adding or subtracting 90. Using a calculator or a computer program, the value of u p calculated from Eq. (4.7) is a number between 90 and 90. If s 11 s 33, we expect the major principal stress s 1 to be closer in direction to s 11 (the larger stress) than to s 33 (the smaller stress); so if the absolute value of the calculated value of the angle u p is less than 45, u p u p1 ; otherwise, u p u p3. Once the direction of one of the principle planes is known, the direction of the other plane can be calculated easily by either adding or subtracting 90 to obtain an angle with absolute value less than 90. For example, if u p1 is calculated as 25, then u p3 is equal to 65. Alternatively, if u p1 is found to be 25, then u p3 is calculated as Mohr Circle 1 2 Moving (s 11 s 33 ) to the left-hand side of Eq. (4.5), taking the square of both sides of the resulting equation, and adding it to Eq. (4.6) (with both sides also squared), we obtain 3s u 1 2 1s 11 s t 2 u 1 4 1s 11 s s 13 (4.10)

7 CHAPTER 4 Stress Analysis, Strain Analysis, and Shearing of Soils 119 Recalling the equation of a circle in Cartesian coordinates, (x a) 2 (y b) 2 R 2, where (a, b) are the coordinates of the center of the circle and R is its radius, Eq. (4.10) is clearly the equation of a circle with center C[(s 11 s 33 )/2, 0] and radius R 1 1 in s-t space. Each 4 1s 11 s s 13 point of this circle, which is referred to as the Mohr circle, is defined by two coordinates: the first is a normal stress (s) and the second is a shear stress (t). Figure 4-3 shows the Mohr circle for the stress state (s 11, s 33, s 13 ) of Fig. 4-2(a). According to Eqs. (4.5) and (4.6), the stresses s u and t u on the plane making an angle u measured counterclockwise from plane p 11 are the coordinates of a point on the Mohr circle rotated 2u counterclockwise from point S 1 (point S 1 represents the stresses on p 11 ). Note that the central angle of the Mohr circle separating S 1 from S 3, which is 180, is indeed twice the 90 angle separating the corresponding planes, p 11 and p 33. In Fig. 4-3, the stresses (s 11, s 33, s 13 ) plot as two points S 1 (s 11, s 13 ) and S 3 (s 33, s 13 ), diametrically opposed. The figure illustrates the case of s 11 s 33. To understand why the shear stress s 13 plots as a positive number with s 33 and as a negative number with s 11, refer to Fig. 4-2(b). With the positive values of s 11, s 33, and s 13 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 major principal stress s 1. It is clear that the major principal plane, which is normal to the direction of maximum compression, is obtained by a rotation of some angle u p1 90 counterclockwise with respect to p 11. This means that, in the Mohr circle, we must have a counterclockwise rotation 2u p1 from the point S 1 (s 11, s 13 ) associated with p 11 in order to reach the point (s 1, 0) of the circle. This implies that S 1 must indeed be located as shown in Fig. 4-3, for if we had S 1 (s 11, s 13 ) instead of S 1 (s 11, s 13 ), a counterclockwise rotation less than 90 would not take us to (s 1, 0). So this means s 13 is plotted as negative if spinning clockwise around the element, as it does for plane p 11, and as positive if spinning counterclockwise, as it does for plane p 33. This is the basis for the convention we will use for plotting points in the Mohr circle: Shear stresses rotating around the element in a counterclockwise direction are positive; they are negative otherwise (Fig. 4-4). Note that this sign convention is not independent from but actually follows directly from the shear stress sign convention we adopted for the stress s 13 appearing in Eqs. (4.5) and (4.6). Pole Method Mohr circles have interesting geometric properties. One very useful property is that the central angle of a circle corresponding to a certain arc is twice as large as an t S 3 (s 33, s 13 ) (s 3, 0) Line parallel to p 11 2u u S 1 (s 11, s 13 ) P Line parallel to p 33 S(s u, t u ) (s 1, 0) Figure 4-3 Mohr circle corresponding to state of stress shown in Fig s s Positive normal stress t Positive shear stress (for plotting in Mohr diagram) Figure 4-4 Stress sign conventions.

8 120 The Engineering of Foundations Figure 4-5 The geometric property of circles that a central angle 2u produces the same arc as an inscribed angle u. t u 2u s u inscribed angle corresponding to the same arc (Fig. 4-5). Applying this property to the Mohr circle shown in Fig. 4-3, the angle made by two straight lines drawn from any point of the circle to point S(s u, t u ) representing the stresses on the plane of interest and to point S 1 (s 11, s 13 ) is equal to u. In particular, there is one and only one point P on the circle with the property that the line joining P to point S(s u, t u ) is parallel to the plane on which s u and t u act. The point P with this property is known as the pole of the Mohr circle. Based on the preceding discussion, the pole can be defined as the point such that, if we draw a line through the pole parallel to the plane where stresses s u and t u act, this line intersects the Mohr circle at a point whose coordinates are s u, t u. In order to determine the pole, we need to know the orientation of at least one plane where the stresses are known. We can then use the known stresses (s, t) to find 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 the pole P, we can determine the stresses on any plane by drawing a straight line through the pole P parallel to the plane where the stresses are desired. This line intersects 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 the element of Fig. 4-2(a). By plotting the Mohr circle for this state of stress, we obtain the expected diametrically opposed points S 1 and S 3 shown in Fig If we look at point S 1 on the circle and consider the corresponding stresses shown in Fig. 4-2(a), we can easily determine the location of the pole. If we construct a line through S 1 that is parallel to the plane p 11 where (s 11, s 13 ) acts, we will determine 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 S 1 (also shown in Fig. 4-3). Likewise, if we look at point S 3 and construct a line through S 3 parallel to plane p 33 on which (s 33, s 13 ) acts, we can also determine the pole as the point of intersection of this horizontal line with the Mohr circle. As expected, the pole is found to be directly to the right of S 3 and to coincide with the point determined previously by examining point S 1. This shows clearly that the pole is unique for a given stress state. Figure 4-6 illustrates for the same case how the principal directions and principal stresses would be determined once the pole is known.

9 CHAPTER 4 Stress Analysis, Strain Analysis, and Shearing of Soils 121 t Line parallel to p 11 Figure 4-6 Illustration of the relationship of the principal directions and their representation in a Mohr diagram. Line parallel to p 3 (s 3, 0) u p3 P (s 1, 0) s u p1 Line parallel to p 1 EXAMPLE 4-1 A state of stress is represented by the block in Fig. 4-7(a). Determine the location of the pole and give its coordinates in the s-t system. Solution First, plot the Mohr circle [Fig. 4-7(b)]. Next, determine the location of the pole. Using point (200, 100), draw a line parallel to the plane on which s 11 (200 kpa) acts. In this case, it is a vertical line since p 11 is vertical. Likewise, if the other point is chosen (100, 100), we draw a line parallel to the plane on which s 33 (100 kpa) acts. In this case, it is a horizontal line since p 33 is horizontal. Either method produces the location of the pole: (200, 100). 200 kpa 100 kpa 100 kpa 200 kpa t (200, 100) 100 kpa s 100 kpa Pole (a) (b) Figure 4-7 (a) Stress state for Example 4-1 and (b) corresponding Mohr circle.

10 122 The Engineering of Foundations 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 s 11. For example, if the normal stresses are 300 and 100 kpa, then s kpa and s kpa. Likewise, if the normal stresses are 100 and 300 kpa, then s kpa and s kpa. The plane where s 11 acts is denoted by p 11, while the one where s 33 acts is denoted by p If using Eq. (4.5) or (4.6), assign a sign to s 13 based on whether it acts to increase or decrease the angle between p 11 and p 33 (with p 11 being the vertical leg of the L ; see Fig. 4-2). The stress s 13 is positive if it acts in a way that would tend to increase the angle. 3. Recognize that the reference plane for angle measurements is p 11 and that angles are positive counterclockwise. 4. Reason physically to help check your answers. For example, since s 11 s 33, the direction of s 1 will be closer to that of s 11 than to that of s 33. Whether s 1 points up or down with respect to s 11 now depends on the sign of s 13, for it tells us about the directions in which the element is compressed and extended. Naturally, s 1 acts in the direction in which the element is compressed the most. EXAMPLE 4-2 The state of stress at a point is represented in Fig Find (a) the principal planes, (b) the principal stresses, and (c) the stresses on planes making angles 15 with the horizontal. Solve both analytically and graphically. Solution Analytical Solution Take s kpa and s kpa. So p 11 makes an angle equal to 30 with the horizontal, and p 33 makes an angle equal to 60 with the horizontal. In order to assign a sign to s 13, we need to consider the right angle made by p 11 and p 33 ; we must look at this angle 200 kpa 50 kpa 50 kpa 50 kpa 50 kpa 200 kpa Figure 4-8 State of stress at a point (Example 4-2). 30

11 CHAPTER 4 Stress Analysis, Strain Analysis, and Shearing of Soils 123 as if it were an uppercase letter L, such that p 11 is the vertical and p 33 is the horizontal leg of the letter L. Physically rotating the page until we see the L may be helpful in this visualization. The effect of the shear stress on the right angle made by p 11 and p 33 looked at in this manner is to reduce it; accordingly, s kpa. We are now prepared to solve the problem. Principal Planes Substituting the values of s 11, s 33, and s 13 into Eq. (4.7): from which tan 12u p The absolute value of 16.8 is less than 45, so Graphically, u p1 would show as an angle of 16.8 clockwise from p 11 because the calculated angle is negative. If the answer is desired in terms of the angles that p 11 and p 3 make with the horizontal, then we need to add 30 (the angle that p 11 makes with the horizontal) to these two results: p 11 is at an angle 13.2, and p 3 is at (or 76.8 ) to the horizontal. Principal Stresses The principal stresses can be calculated using either Eq. (4.5) with u 16.8 and 73.2 or Eqs. (4.8) and (4.9). Using Eq. (4.5): Now using Eqs. (4.8) and (4.9): u p 1 2 arctan a 2 3 b 16.8 u p u p s cos sin kpa s cos sin kpa s kpa s kpa Stresses on Planes Making Angles 15 with Horizontal These planes make angles 15 and 45 with p 11, respectively. Plugging these values ( 15 and 45 ) into Eqs. (4.5) and (4.6): s u cos sin kpa t u 75 sin cos kpa Not surprisingly, the s u and t u values calculated here are very close to the values for the major principal plane (215.1 and 0). You should verify that s u 175 kpa and t u 75 kpa for the other plane (which makes an angle of 15 with the horizontal).

12 124 The Engineering of Foundations Figure 4-9 Mohr circle and graphical solution of Example 4-2. t 100 p (200, 50) (214, 6) p 1 15 P (50, 50) 200 (175, 75) s 15 Graphical Solution Using the Pole Method The solution can be seen in Fig The normal and shear stresses on plane p 11 are 200 and 50 kpa, respectively; in plane p 33 they are 50 and 50 kpa. 6 These two points are diametrically opposite each other on the Mohr circle. If we plot these two points in s-t space and join them by a straight line, this line crosses the s-axis at the center of the circle. We can then easily draw the Mohr circle using a compass. The principal stresses are now easily read as the abscissas of the two points with t 0. If we now draw a line parallel to p 11 through the point (200, 50), we obtain the pole P as the intersection of this line with the circle. Note that if we draw a line parallel to p 33 through (50, 50), we obtain the same result. The directions of p 1 and p 3 are obtained by drawing lines through P to the points (s 1, 0) and (s 3, 0) of the Mohr circle, respectively. The stresses at 15 with the horizontal are found by drawing lines through the pole P making 15 with the horizontal. These lines intersect the circle at two points with coordinates (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 stress at a point in the soil mass. If there is a nonzero pore pressure u at this point, it is the 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 a pore pressure affects only normal stresses in the soil. It is useful to examine what would 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 we substitute it into Eq. (4.10), we can see that the only effect is to have the Mohr cir- 6 Note that, for Mohr circle construction, counterclockwise shear stresses are plotted as positive, while clockwise shear stresses are plotted as negative.

13 CHAPTER 4 Stress Analysis, Strain Analysis, and Shearing of Soils 125 t Pore pressure u (s u, t) (s, t) s, s Effective-stress Mohr circle Total-stress Mohr circle Figure 4-10 Principle of effective stresses: illustration of the difference between the total and the effective stress state as represented using the concept of the Mohr circle. cle of total stresses displaced along the s-axis with respect to the Mohr circle of effective 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 of stresses. These stresses induce deformations, which are represented by strains. There are two types of strains: normal and shear strains. At a given point, a normal strain in a given direction quantifies the change in length (contraction or elongation) of an infinitesimal linear element (a very small straight line) aligned with that direction. The so-called engineering shear strain g is a measure, at a given point, of the distortion (change in shape). With respect to the two reference axes x 1 and x 3, the engineering shear strain g 13 expresses the increase of the initial 90 angle formed 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 engineering shear strain. 7 Both normal and shear strains e ij can be expressed through e ij 1 2 a 0u i 0x j 0u j 0x i b (4.11) where i,j 1, 2, 3 reference directions, x i coordinate in the i direction, u i displacement in the x i direction. The subscripts indicate the directions of linear 7 We have avoided duplication of symbols as much as possible, but there is no good alternative to using the 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.

14 126 The Engineering of Foundations differential elements and the directions in which displacements of the end points of the differential elements are considered. When i j, the strain is a normal strain; it is a shear strain otherwise. The relationship between a shear strain e ij and the corresponding engineering shear strain g ij is g ij 2e ij (4.12) EXAMPLE 4-3 A x 3 B B* dx 1 du 1 x 1 Figure 4-11 Normal strain e 11 : infinitesimal element dx 1 shown after correction for rigid body motion both before deformation (AB) and after deformation (AB*). Derive, in a simple way, the expression for the normal strain at a point in the direction of reference axis x 1. Solution Let s consider the case of Fig For an underformed soil mass, we have a differential element dx 1 aligned with the x 1 direction (Fig. 4-11). We have labeled the initial point of the segment A and the end point B. In drafting this figure, we have corrected for rigid body translation in the x 1 direction. In other words, we are plotting the deformed element as if the displacement 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 the soil mass is deformed, point B moves more in the positive x 1 direction than point A (that is, if the displacement u 1 of B is greater than that of A), as shown in Fig. 4-11, then the element has clearly elongated (this elongation is seen in the figure as B* B). Taking some liberty with mathematical notation, the unit elongation of dx 1 is the difference in displacement between points A and B (u 1B u 1A B* B du 1 u 1 ) divided by the initial length of the element ( x 1 ), or u 1 / x 1. It remains to determine whether elongation is a positive or negative normal strain. To be consistent with the sign convention for stresses, according to which tensile stresses are negative, our normal strain e 11 in the x 1 direction must be e 11 0u 1 0x 1 Note that this, indeed, is the expression that results directly from Eq. (4.11) when we make i 1 and j 1. EXAMPLE 4-4 Derive, in a simple way, the expression of the shear strain at a point in the x 1 -x 3 plane. Solution Consider two differential linear elements, dx 1 and dx 3, aligned with the x 1 - and x 3 -axis, respectively, at a point within an undeformed soil mass (Fig. 4-12). Now consider that the soil mass is deformed, and, as a result, points B and C (the end points of elements dx 1 and dx 3, respectively) move as shown (to new positions labeled B* and C*) with respect to point A (note that, as for Example 4-3, we are not representing rigid-body translation in the x 1 and

15 CHAPTER 4 Stress Analysis, Strain Analysis, and Shearing of Soils 127 x 3 directions in the figure). We can see that both point B and point C have displacements that have components in both the x 1 and x 3 directions. Here we are interested in just the distortion of the square element made up of dx 1 and dx 3, not in the elongation or shortening of dx 1 and dx 3 individually. The distortion of the element clearly results from difference in the displacement u 3 in the x 3 direction between A and B and in the displacement u 1 in the x 1 direction between A and C. Taking again some liberties with mathematical notation, we can state that the differences in the displacements of points B and A (in the x 3 direction) and C and A (in the x 1 direction) can be denoted as u 3 and u 1, respectively. Since the deformations we are dealing with are small, the angle by which the differential element dx 1 rotates counterclockwise is approximately equal to u 3 divided by the length of the element itself, or u 3 / x 1 ; similarly, the angle by which dx 3 rotates clockwise is u 1 / x 3. These two rotations create a reduction in the angle between the elements dx 1 and dx 3, which was originally 90, characterizing a measure of distortion of the element. If we add them together, we obtain the absolute value of what has become known as the engineering shear strain g 13 ; one-half the sum gives us the absolute value of e 13. It remains to determine whether this is a positive or negative distortion. Our sign convention for shear strains must be consistent with our shear stress sign convention. Recall from our earlier discussion in the chapter that a positive shear stress was one that caused the 90 angle of our square to open up (to increase), not to decrease. Therefore, we will need a negative sign in front of our sum in order to obtain a negative shear strain for the reduction of the 90 angle we found to take place for the element in Fig. 4-12: e a 0u 1 0x 3 0u 3 0x 1 b C dx 3 A C * x 3 u 1 x 3 u 3 x 1 dx 1 B x 1 B * Figure 4-12 Shear strain e 13 : infinitesimal square element shown after correction for rigid body translation both before deformation (defined by dx 1 AB and dx 3 AC) and after deformation (defined by dx 1 * AB* and dx 3 * AC*). 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 of defining strain that are more appropriate when elongations, contractions, or distortions become very large. However, Eq. (4.11) may still be used for increments of strains, even if the cumulative strains measured from an initial configuration are very 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 strain increment. As was true for stresses, there are also principal strains e 1 and e 3 (and principal strain increments de 1 and de 3 ). These are strains (or strain increments) in the directions 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 is important to understand that a point in the soil experiences distortion as long as de 1 de 3. To illustrate this, Fig. 4-13(b) and (c) show two alternative elements representing 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 immediately suggest the notion of distortion. However, there is no doubt that distortion has taken place when we observe the change of shape of the smaller, inner element as we go from the initial to the final configuration [Fig. 4-13(b) and (c)].

16 128 The Engineering of Foundations The volumetric strain increment de v, defined as minus the change in volume divided by the original volume (the negative sign being required to make contraction positive), can be easily determined in terms of a cubic element with sides with length initially equal to 1 and aligned with the principal directions (which means x 1, x 2, and x 3 are principal directions). The element is then allowed to expand as a result of elongations equal to du 1, du 2, and du 3 in the three reference directions. As the cube is aligned with the principal directions, there will be no distortion in the planes x 1 x 2, x 1 x 3, or x 2 x 3. It is apparent from Fig that dv 11 du du du Referring back to our definition of normal strain and considering that the initial length of the sides of the cube are of unit length and the initial volume of the cube is also equal to 1, we can write the following for the volumetric strain increment: de v dv de 1211 de de 3 2 which, given that the strain increments are very small (and that second- and thirdorder terms would be extremely small and thus negligible), reduces to de v de 1 de 2 de 3 (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 x 1, x 2, and x 3 were not the principal directions: de v de 11 de 22 de 33 (4.14) where de 11, de 22,and de 33 normal strains in the arbitrary directions x 1, x 2, and x 3. Initial outer element de 3 1 P P P de 1 1 Final outer element (a) Initial configuration (b) Final configuration (c) Figure 4-13 Alternative representations for the state of strain at a point: (a) an element aligned with the principal strain directions (vertical contraction and horizontal elongation); (b) the same element before deformation with an element inside it with sides oriented at 45 to the principal strain directions; (c) the same element after deformation, showing the distortion of the element with sides not aligned with the principal strain directions. x 1 x 3 x 2 1 de 2 Figure 4-14 Calculation of volumetric strain. Deformed cube Original cube

17 CHAPTER 4 Stress Analysis, Strain Analysis, and Shearing of Soils 129 Mohr Circle of Strains The mathematics of stresses and strains is the same: Normal strains play the same role as normal stresses, and shear strains, the same as shear stresses. So, just as it was possible to express the stresses at a point under 2D conditions in terms of a Mohr circle, the same is possible for strains. Many problems in geotechnical engineering can be idealized as plane-strain problems, for which the strain in one direction is zero. For example, slopes and retaining structures are usually modeled as relatively long in one direction, with the same cross section throughout and with no loads applied in the direction normal to the cross sections. Except for cross sections near the ends of these structures, it is reasonable, based on symmetry considerations, to assume zero normal strain perpendicular to the cross section. The same is valid for strip footings, which are used to support lines of columns or load-bearing walls. The result is that we can take de 2 0 and do our strain analysis in two dimensions. 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 of strains plotted in normal strain increment de versus shear strain increment 1 2 dg space. As was true for stresses, each point of the Mohr circle represents one plane through the point in the soil mass for which the Mohr circle represents the strain state. The points of greatest interest in the Mohr circle of strains are The leftmost and rightmost points, (de 3, 0) and (de 1, 0), corresponding to the minor and major principal incremental strain directions The highest and lowest points, ( 1 2 de v, 1 2 dg max ), corresponding to the directions of largest shear strain (Note that 1 2 de v 1 2 (de 1 de 3 ) is the de coordinate of the center and 1 2 dg max 1 2 (de 1 de 3 ) is the radius of the Mohr circle.) The points where the circle intersects the shear strain axis, (0, 1 2 dgz ) 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 neither extension nor contraction. It is possible to define a separate reference system for 1 2 dg A 1 (0, 1 dg z ) 2 Pole P O( 1 2 de v, 0) c c (de 3, 0) Z (de 1, 0) de A 2 (0, 1 dg z ) 2 Figure 4-15 Mohr circle of strains.

18 0 130 The Engineering of Foundations each of these two directions such that x 1 in each system is aligned with the direction of zero normal strain. To clearly indicate that x 1 is a direction of zero normal strain, we can use a superscript z, as in x 1 z. This will be useful in our discussion of the dilatancy angle, which follows. Dilatancy Angle The angle c shown in Fig. 4-15, known as the dilatancy angle, is quite useful in understanding and quantifying soil behavior. There are two ways of expressing the dilatancy angle based on the geometry of the Mohr circle of strains: sin c OZ 0OA de 1 de de 1 de 3 2 de 1 de 3 de 1 de 3 tan c OZ 0ZA 1 0 2de v 1 2dg z 0 (4.15) (4.16) where g z shear strain in the x 1 z -x 3 zp plane (Fig. 4-16), x 1 z is a direction of zero normal strain, and x 3 zp is the direction normal to x 1 z. The dilatancy angle is clearly related to the volumetric strain resulting from a unit increase in shear strain. 8 By definition, the dilatancy angle c is positive when there 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, when de v 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 volumetric strain increment resulting from a unit increment in shear strain, regardless of the orientation of the shear strain. In other words, the dilatancy angle would still be the same positive value if the element shown in Fig were sheared to the left and not to the right as shown. The direction of zero normal strain in Fig is 1 de v 0dg z 0 de v 0dg max 0 c x 3 zp g z Figure 4-16 State of strain visualized for an element with one side aligned with the direction of zero normal strain. x 1 z Direction of zero normal strain 8 Technically, both the shear and volumetric strain increments in the definition of the dilatancy angle are plastic strain increments, a distinction that for our purposes is not necessary to make.

19 CHAPTER 4 Stress Analysis, Strain Analysis, and Shearing of Soils 131 represented by x 1 z, and the direction normal to it, by x 3 zp, where the superscript zp means that the direction x 3 zp is perpendicular to the direction of zero normal strain. There are in fact two distinct directions of zero normal strain, as will be shown later. The Mohr circle of Fig corresponds to a state of dilation (expansion), as de v 0. Examining the state of deformation in the x 1 z -x 3 zp reference system, expansion implies that the normal strain in the x 3 zp direction, normal to x 1 z, is negative (that is, that elongation takes place in the x 3 zp direction, as is clearly shown in Fig. 4-16). If we locate the pole in Fig and then draw two lines (one through A 1 and one through A 2 ), the two lines perpendicular to these two lines are the directions of zero normal strain, as we will discuss in detail later. A soil element is subjected to the following incremental strains: de %, de %. Knowing that plane-strain conditions are in force (that is, the strain in the x 2 direction is zero), calculate the dilatancy angle. Solution Because we know the principal strain increments, we can immediately calculate the dilatancy angle as EXAMPLE 4-5 sin c de v 0dg max 0 de 1 de de 1 de from which c 14.5 In Problem 4-18, you are asked to continue this by plotting the Mohr circle, finding the pole for the case when the major principal strain increment is vertical and determining the directions of the potential slip planes through this element (which is the subject of a subsequent section). For a triaxial strain state, in which e 2 e 3, the dilatancy angle needs to be redefined because a maximum engineering shear strain increment dg max is no longer possible to define with clarity. In place of it, we work with de s, defined as de s de 1 2de 3 (4.17) The volumetric strain in the triaxial case follows from Eq. (4.13): de v de 1 2de 3 Thus, the dilatancy angle for triaxial conditions is written as sin c de 1 2de 3 de 1 2de 3 (4.18)

20 132 The Engineering of Foundations A single expression for it, which applies to both plane-strain and triaxial conditions, is de v where sin c de 1 kde 3 de v de 1 de 1 kde 3 de 1 kde 3 2 de v de 1 1 for plane-strain conditions k e 2 for triaxial conditions (4.19) (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 apply repeatedly the same increment of shear strain to an element of soil (this is referred to as strain-controlled loading), the increment of stress that the soil element is able to sustain decreases continuously (a process that is sometimes referred to as modulus degradation), until a state is reached (represented by point F in the figure) at which the stress increment will be zero. At this point, if we continue to increase the strain, the stress will stay the same. The other stress-strain plot shown in the figure illustrates another possible response, whereby the stress peaks at point F. This second response (referred to as strain softening) is common in soils. The limiting or peak stress associated with point F in each case is usually what is meant by 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 applying stress 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 straincontrolled loading is reached, the soil element will not be able to take any additional stress, and what will happen instead is uncontrolled deformation. It is not F Figure 4-17 Nonlinearity of stress-strain relationship for soils and failure (the onset of very large deformations at some value of stress, represented by point F). Note that the first strain increment a generates the stress increment b and that the second strain increment generates a stress increment c b b. Shear stress c b a 2a F Shear strain

21 CHAPTER 4 Stress Analysis, Strain Analysis, and Shearing of Soils 133 possible in the stress-control case to plot the stress-strain relationship after point F is reached. The Mohr-Coulomb failure (or strength) criterion has been traditionally used in soil mechanics to represent the shear strength of soil. It expresses the notion that shear strength of soil increases with increasing normal effective stress applied on the potential shearing plane. Although in stress analysis we do not have to concern ourselves with whether we are dealing with effective or total stresses because the analysis applies to both, we must now make a clear distinction. Soils feel only effective stresses (that is, any deformation of the soil skeleton happens only in response to effective stress changes); thus, the response of soil to loading and the shear strength of soil depends only on the effective stresses. Based on this consideration, we represent the Mohr-Coulomb criterion in s -t space as two straight lines making angles f with the horizontal and intercepting the t-axis at distances c 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 usually referred to as the cohesive intercept. Mathematically, the Mohr-Coulomb criterion may be represented in a simple way by s c s tan f (4.21) where s is the shear strength of the soil and f is its friction angle. For a given normal effective stress s on a plane, if the shear stress on the plane is t s as given by Eq. (4.21), shearing or what is commonly referred to in engineering practice as failure of the material occurs. Failure here means the occurrence of very large strains in the direction of that plane. This means that the soil cannot sustain shear stresses above the value given by Eq. (4.21). Equation (4.21) is a straight line in s -t space (as shown in Fig. 4-18) that is referred to as the Mohr-Coulomb strength envelope. There can be no combination of s and t that would lie above the Mohr-Coulomb strength envelope. Figure 4-19 shows the Mohr circle for a soil element (or point) within a soil mass, at failure, where s and t act on the two planes corresponding to the two tangency points between the circle and the envelope (again, only the half of the diagram lying above the s -axis is shown, as the part below the s -axis is symmetric). 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 t f c s Figure 4-18 The Mohr-Coulomb failure envelope.