TM_chap15_strain - CHAPTER To further our understanding of...

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Unformatted text preview: CHAPTER To further our understanding of the origin and signif- icance of the folds, foliations, and lineations discussed in the lasr four chapters, we need 1‘0 become more fa- miliar with the nature of strain, as manifested in rocks. We introduced some concepts of strain in Chapters 7, 9, 12, and 14, but we need a more thorough and sys- tematic understanding in order to evaluate theoretically the models proposed for formation of ductile structures, as well as to test these models against observations of natural deformation. Our approach is largely geometric and qualitative, because our intent is to provide intuition into the phys- ical characteristics of deformation, and strain lends itself easily to geometric description. The quantirative anal— ysis of the ideas discussed in this chapter requires a rigorous mathematical treatment of strain, which we introduce in Box 15.1, and which is developed in depth in more advanced boo s on continuum mechanics and its geologic plicatio 8 (see the list of readings at the end this. chap er). Readers interested in this mation. size e srrain is homogeneous if the changes in shape are proportionately identical for each small part of the body and for the body as a whole (Figure 15.1A, B). A consequence of these c0nditions is that for any homogeneous strain, planar surfaces remain planar, straight lines remain straight and parallel planes and lines remain parallel. The strain is inhomogeneous (Figure 15.1A, C) if the changes in size and shape of small parts of the body are proportionately different from place to place and different from that of the body as a whole. Straight lines become curved, planes become curved surfaces, and parallel planes and lines generally do nor remain parallel after deformation. The strain must be inhomogeneous during folding, because in such a deformation, planes and lines do not generally remain planar, straight, or parallel. Within very small volume-elements, however, the strain is sta- ristically homogeneous, and we describe an inhomo- geneous strain as a variation of homogeneous strain from place to place in the Structure. We discuss how big such a “small” volume-element musr be in Section 15.7. The progressive deformation of a body refers to the motion that carries the body from its initial unde- formed state to its final deformed state. The strain states through which the body passes during a progressive deformation define the strain path. The state of strain of a body is the net result of all the deformations the body has undergone. Although all states of strain are the result of progressive deformation, the final state of strain provides no information about the particular strain path that the body experienced. A. Undeformed state B. Homogeneously deformed state C. inhomogeneously deformed state Figure 15.1 Homogeneous and inhomogeneous plane defor- mation of a material square. A and B. Homogeneous Strain. The small black square is strained in exactly the same way as the whole square and as all the other squares. Lb is the angle of shear. A and C. Inhomogeneous strain. The small black square is sufficiently small that its srrain is essentially ho- mogeneous, but it is not identical to the strain of the whole square or to that of any of the other small squares. Strain in general must be described in three di- mensions, because the size and shape of a body are three-dimensional characteristics. In much of our dis- cussion, however, we consider only a two-dimensional deformation called plane strain, in which the strain is completely described by changes in size and shape in a single orientation of plane through the body, and no deformation occurs normal to that plane. Although plane strain is commonly used to analyze deformation, its application to many situations in natural rock de— formation is, strictly speaking, unjustified. Nevertheless, the geometry of two—dimensional deformation is intu- itively easier to understand, and the generalization to three dimensions adds considerable complexity but lirtle insight into the geometric characteristics of deforma— tion. For these reasons we concentrate on the properties of two-dimensional strain. In discussing the geometry of strain, we refer to geometric objects such as lines, planes, circles, and elm lipses. Such geometric objects are called material objects if they are always defined by the same set of material particles. A bedding plane, for example, is a material plane because no matter how it moves and deforms, it is always defined by the same set of material particles. A coordinate plane defined by two reference axes, on the other hand, is a nonmatetial plane because as a body deforms, its material particles can move through the coordinate plane and, consequently, different sets of material particles occupy the coordinate plane at dif— ferent times. This distinction is important in the sub- sequent discussion. Lz' ear Strain ' e of a body is measured by its volume, which in turn is proportional to the product of three characteristic lengths of the body. For example, the volume V of a rectangular block that has edges of lengths [1, f2, and {’3 is V = {16263, and the volume of an ellipsoid that has semiaxes of lengths r1, r2, and Q is V = [4/3] rt r1r2r3. In Cartesian coordinates, the description of the change in size requires specification of the change in length of line segments in the three coordinate direcrions. The change in absolute length is an inadequate measure of the deformational state of a line segment, because for a given change in length, the intensity of the change is much greater for a short line segment than for a long one. Thus the lengthening is expressed as a propOrtion of the original line length. Two measures in common use are the stretch 5,, and the extension e,,, which was int the beginning of Chapter 9. f I; to the rs when r-ferring (152) value for extension measures a lengthening, whereas a positive value for stress measures a compression. We thus end up with a positive stress Causing a negative extension. i'his incompatibility does no: arise with the engineering sign convention for stress, which is why it is generally used in analytic applications of continuum mechanics. Geometry of Homogeneous Strain 293 294- m A More Quantitative View of Strain A homogeneous transformation of any material point from the undeformed state to the deformed state is represented mathematically by a linear relationship between the coordinates of any point in the unde- formed state (X1, X3) and its coordinates in the de- formed state (11, x3), where we use upper-case letters to describe the undeformed state and lower-case let- ters to de5cribe the deformed state. If we restrict our analysis to plane deformation, the general form of such a transformation is x1=AX1+BX3+C I3=DX1+EX3+F (l5.l.l) where A, B, C, D, E, and F are constants. The parts of the transformation defined by C and F are the same for all particles, and therefore these constants describe a rigid-body translation. If any or all of these constants vary with time, then these equations describe the mo- tion of the material particles. The equations say that given the original lo cation of any material particle in the undeformed state (X 1, X3), we can calculate its final location in the deformed state (x1, x3). The equations may be solved for X1 and X3 so that given the deformed location of a material particle (11, 13), we can also calculate its original location (X1, X3). These equations define the inverse transformation. X1=axl+bx3+c X3=dx1+ex3+f (15.1.2) where E —B BF—CE as b5 :5— AE—BD AE—BD AE—BD d= —D e— A f_DC—AF _AE—BD _AE—BD _AE—BD (15.1.3) and where, again, (2 and f describe a rigid body trans- lation. As examples of such a transformation and its inverse, the following equations describe a pure shear, which transforms a square with sides parallel to the principal coordinates into a rectangle (Figure 1598) x1=AX1 X1=(1/A)11 x3 =(1/AlX3 X3 = AX3 (15.1.4) A simple shear, which transforms a square into a parallelogram (Figure 15.11B), and its inverse are de- scribed by xl=XJ+BX3 I3=X3 (15.1.5) X1=x1—Bx3 X3 =13 When the constants A in Equation (15.1.4) and B in Equation (15.1.5) are linear functions of time, the motions are steady and these equations describe DUCTILE DEFORMATION progressive pure shear and progressive simple shear, respectively (see Section 15.4). With Equations (15.1.2), it is easy to show that a homogeneous deformation transforms a circle into an ellipse. A circle of unit radius in the undeformed state is represented by the equation (X02 + (X3)2 =1 (15.1.6) If we substitute for X1 and X3 from Equations (15. 1.2), we find the locus in the deformed state of all material particles that lie on the circle in the undeformed state. Because a rigid-body translation does not contribute to the strain, we assume c = f: 0. Then, making the substitution, we find (a2 + d2)(x1)2 + 2(ab + de)x1x3 + (b2 + el)(x3)2 =1 (15.1.7) Equation (15.1.7) is the equation of an ellipse with its principal axes tilted with respect to the coordinate axes, and it is, in fact, the strain ellipse. The components of the strain tensor are related to the displacement vectors for the material particles. A displacement vector connects the position of a par- ticle in the undeformed state to its position in the deformed state. The vector and its components (U1, U3) parallel to the X1 and X3 coordinate axes are (Figure 15.1.1A) st—x (15.1.3) U1 =X1 —X1 U3=X3—X3 (15.19) When a material deforms, the displacement vec- tors for two neighboring material points are different. If they were the same, the ”deformation” would be a rigid body motion. The difference in these displace- ment vectors therefore describes the deformation. Thus we consider two neighboring points A and Bthat are displaced by the deformation to a and b, respec- tively. The displacement vectors for the two points are UV” and [1(8), and the difference between them is dU (Figure 15.1.1B). The material line segment dX connecting A to B is deformed into dx connecting a to b. The change in that line segment due to the deformation AdX is also described by the vector dU (Figure 15.1.18). Thus, dU a U03) — UM) = Adx 2 dx i ax (15.1.10) The relationship between the first and last terms in this equation is just the differential of Equation (15.1.8). ‘ . We can consider the components Xm and dX3 of the line segment dX to be two material line seg— ments that are initially perpendicular to each other and parallel to the coordinate axes X1 and X3 respec- tively. If we restrict our analysis to infinitesimal strain, Deformed Displacement position vector U Undetermed position characterized by the conditions dUl << 1 and d U3 << 1, the displacement associated with each of these line segments due to the deformation can be expressed using Equation (15.1.10) and the chain rule of dif- ferentiation for dU 6U 6U Adx=AdX AdX=dU=——dx —dX 1+ 3 6X1 1+aX3 3 (15.1.11) Thus the changes :3Xm and AdX3 in each of the line segments due to the deformation is given in terms of the components of the displacement vect0r U by (Fig- ure 15.1.2). 6 3U Adxl = _Us dX1= _“ de + .6231 mg ax, 6X1 ax, (15.1.12) 5U 5U, 6U3 AdX =_dX =_ X -—d 3 6X3 3 axsd 3 +ax3 X3 For each of the material line segments 01X 1 and dX3, the extensional strains are labeled e11 and egg, respectively, and each one is the change in length divided by the initial length, as defined in Equation (15.2). For in, for example, the change in length is (am/6X1) fig, and the initial length is Xm (Figure 15.1.2). Similar relations hold for dX3. Thus 1 1 av, _ 5U, 6“ ‘ dX, [6X1 dX‘] ‘ ax, _ l 5U}, EU; 833 = (EL—X3 dXs] = 5Y3 The shear strain of Xm relative to dX3 and vice versa are labeled (213 and e31, respectively, and are defined in Equation (15.7) to be half the tangent of the shear angle (1113 = $31: 11'; = or + 13. For very small strains, or << 1 and ,6 << 1, and the standard trigonometric iden- tity for the tangent of the sum of two angles gives (15.1.13) tana+tanfi mtetm+m=mm astanct+tanfi (15.1.14) Figure 15.1.1 The displacement vector. A. The displacement vector connects the p0v sition of a material particle in the unde- formed state to its position in the deformed state. B. If a material is deformed, the dis- placement vectors for two neighboring points are different. Point A is deformed re the position a; B is deformed to the position b. The difference in the displacement vectors dU describes the deformation of the mate rial. because the product tan at tan ,3 is negligibly small. The tangent of an angle is the length of the side 0p- posite the angle divided by the length of the adjacent side. For infinitesimal strains, the side opposite the angle at is approximately (6 U1/3X3)dX3, and the ad— jacent side is dX3 (Figure 15.1.2). Similar relationships hold for the angle ,8. Thus we have tmam;[fllldX3]=flll ng, 5X3 6X3 (15,15) 1 5U3 0U3 :— — X =__. a” dxllax, d 1] ax, (Continued) : ax. maxi ax, ‘ tau, e___ 1) 3X1 Figure 15.1.2 The geometrical interpretation of the com- ponents of infinitesimal strain for two-dimensional strain. For clarity, the strain is greatly exaggerated in the diagram. The vectors dX, dx, and dU are the same as the vectors having the same labels that appear in Figure 15.1.18. The strain components thns are defined by the change in the displacement vector dU for two neighboring points. Geometry of Homogeneous Strain m 295 BOX 15.]. (Continued) Then using the definition of the shear strain (Equation 15.7) with Equations (15.1.14) and (15.1.15) gives 6U 6U 213 2 231: 0.5 tan (tr z 0.5(—fil+——:5 6X3 5X1) (15.1.16) These relations for the extensions and shear strains associated with the material line segments Xm and dX3 are the components of the infinitesimal strain tensor.'In shorthand component notation, we sum- marize Equations (15.1.13) and (15.1.16) by 0U), art) a 0.5 — 8’“ (ax, + 5X), ' k,Z= 1,2, 3 (15.1.17) This expression for ck). remains exactly the same if k and f are interchanged, which shows that ek, = 8m and that the strain tensor is a symmetric tensor (com- pare Equation 15.12). Thus 51:; is the symmetric part of the displacement gradient tensor 6Uk/8X). The antisymmetric part of the displacement ga- dient tensor can be shown to be the infinitesimal ro- tation tensor, defined by 5 a rhEO.5(fl—i), k,[=1,2,3 (15.1.18) The antisymmetric character of rk, is evident from this equation, because interchanging the subscripts k and 3’ gives the relation rkr="’zk (15.1.19) Components on the principal diagonal of the matrix rk, must therefore be zero. In two-dimensional strain, there is only one independent off-diagonal component n3 = — r31. Thus from Equations (15.1.15) and (15.1.18) we can see that rl3:0.5(tanot—tan ,8) (15.1.20) For very small angles, the tangent of the angle is ap proximately equal to the angle measured in radians, so we can write ruzO-SW—B) (15.1.21) Thus r13 is half the difference in the components of the shear angle, and no. is thus a measure of the net rotation of the material line segment dX. The displacement components (U3, U3) can be expressed solely in terms of the coordinates of the material point in the undeformed state by substituting Equations (15.1.1) into (15.1.9), assuming the rigid translations are zero (C = F : 0) U1=(A —1)X1+ 8X3 U3 = ox) + (15— nx3 (15.1.22) Using Equations (15.1.22) in (15.1.17), we find the values of the strain components in terms of the con- stants that define the motion of the material particles: [811 943] =I: (A— i) 0.5(B + 0)] (15.1.23) E31 (333 0.5(D + B) (E - 1) As indicated above, the relationships given here are correct only for very small strains. The analysis of large strains is considerably more complex, al- though this geometric interpretation of the strain components remains intuitively useful. For a line segment of arbitrary orientation in the undeformed state, given by the angle 6 with respect to the principal coordinate axis 21, it can be shown that the extension and the shear strain for infinitesimal plane strain are given in terms of the principal exten- sions by en=elc0328+é3 5111119 (151 15) as = (1‘21 — €33) sin Bees 3 These equations are identical in form to Equations (8.36), which we found for the stress components, and the mathematical characteristics of the stress and the infinitesimal strain tensors are identical, including the possibility of deriving a Mohr circle for infinites— imal strain. The relationships for large deformations are somewhat more complex, but a Mohr circle that is useful in solving strain problems can nevertheless be defined for large strains. We refer the reader to books containing more quantitative analyses (see the works by Means, Ramsay and Huber, and Eringen in the list of additional readings at the end of this chapter). Comparing Equations (15.1) and (15.2} shows that these two measures of extensional strain are related: 5 —1 (1531 Values of 5,, > 1 and of e” > 0 represent increases in the length ofrnaterial lines, and values where 0 < 5,, < 1 and en < 0 represent decreases in length (Table 15.1). 296 DUCTILE DEFORMATION Other measures are also used, including the qua- dratic elongation and the natural strain. The quadratic elongation is simply the square of the stretch, and it is often given the symbol 2., although some authors use this symbol to designate the stretch. The natural strain E”, also called the logarithmic strain, is the integral of all the infinitesimal increments of extension required to make up the deformation, where the reference length Table 15.1 Extensional Strain of a Material Line .._s_.gr.flaf..cfi-cz,s.:.~fimega Mum, - ,.__.,..,."‘._... Length Change Stretch Extension AL 5,15 {IL 3,, a (f — L)/L Undeformed L AL = 0 5,1 = 1 an = 0 Shortened l—Q li— AL AL=£—L<0 0<sn<1 en<0 ,- Lengthened l—£_i:ml AL = f — L > 0 s,t > 1 3,, > 0 for each increment in length (if is taken to be the in- stantaneous deformed length 5. 5r _ d€_ fr _ 5,, {7—111 (I) 1115,, (15.4) where L is the initial length, ff is the final length, and in indicates the natural logarithm. Notice that the nat- ural strain is the natural logarithm of the stretch. The natural strain is sometimes convenient for discussion of strain history (see Figure 15.7.0). It also provides a sym— metric measure of shortening and lengthening.2 The time derivative of the natural strain is also often used as a measure strain rate (see Box 18.1). Volu et‘ric Srr in We can n eonsider measures of the volumetric strain, whic e refer to as the volumetric streteh (5”) and the vo‘ umetric extension3 (6,). If the undeforrned volume is V and the deformed volume is U, v — V AV v V V =7=s,—1 (15.5) A rectangular block thar undergoes only volumetric. strain has undeformed sides (L1, L2, and L3) and de— formed sides (f1, {1, and {3). The volumetric stretch is [1’25] 51* = £1LG 5,, = 515253 =(e1+ 1)(e2 + 1)(53 +1) (155) We consider further aspects of volumetric strain in the next sectio / Shear Sr A dy can also change shape without changing volume. or example, a cube can deform into a rhombohedron, or a sphere into an ellipsoid. Changes in shape are 2 For example, for a line segment stretched to twice its initial length and one shortened to half its initial length, 5,, = 2 and 0.5, and en = 1 and 0.5, but 3,, : 0.693 and —0.693, respectively. 3The volumetric extensiOn is commonly given the symbol A and called the dilation, or even the dilatation. We reserve A to indicate the change in a variable. described by the changes in the angle between pairs of lines that are intialiy perpendicular (Figure 15.2). The change in angle is called the shear angle (1'1, and the shear strain e_. is defined by as E 0.5 tan If/ (15.7) As defined here, a5 is the tensor shear strain. It differs from another common measure of the shear strain, the engineering shear strain 3;, by a faCtor of 2 (y E tan (1'1 = 255). For two material line segments originally ori- ented along the positive co...
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