A COMPARISON OF DISCRETE AND CONTINUUM MODELS OF AN EDGE DESLOCATEON IN A RECTANGULAR SIMPLE CUBE CRYSTAL Thesis far the Degree of Ph. D. MICH!GAN STATE UNiVERSlTY Edware‘i M. Sahel! 1965 TH ESIS 111111111111111111111 L 12_93 01006 1314 Michigan State University This is to certify that the thesis entitled A COMPARISON OF DISCRETE AND CONTINUUM MODELS OF AN EDGE DISLOCATION IN A REC- TANGULAR SIMPLE CUBIC CRYSTAL presented by EDWARD M. SCHALL has been accepted towards fulfillment of the requirements for IL degree in MOS l . (Mmfh Date March 30, 1965 0-169 100 ABSTRACT A COMPARISON OF DISCRETE AND CONTINUUM MODELS OF AN EDGE DISLOCATION IN A RECTANGULAR SIMPLE CUBIC CRYSTAL by Edward M. Schall A single edge dislocation in a rectangularly bounded simple cubic crystal is simulated by two distinct model 5 . The first of these is a discrete lattice in which each nodal point is connected to its four nearest neighbors and its four next nearest neighbors in the plane by linear springs. An edge dislocation is intro- duced into an otherwise undisturbed lattice by placing a partial plane of lattice points between two existing full planes and then considering the springs which connect the extra plane lattice points to their neighbors to be held in a compressed configuration. This constraint is then removed and the strain energy initially concentrated in the springs of the extra plane of lattice points is dis- tributed through the lattice displacing each point from its original position. The second model is an elastic continuum into which a Volterra edge dislocation is introduced. The calculation of stresses and displacements requires the Edward M. Schall superposition of an analytic solution which satisfies the displacement boundary condition and a numerical solution which satisfies the stress free condition on the rectangular boundaries of the continuum. In order to achieve elastic similarity so that corresponding results from the two models may be compared, it is necessary to use anisotropic elasticity theory for the continuum analysis. Displacements for the two models are compared and the stresses of the continuum analysis are compared with analogous discrete quantities. These comparisons show that the results of the two models are remarkably similar. Corresponding values of the problem variables begin to differ appreciably only within four lattice spacings of the dislocation. Since the discrete model is a more physically correct means of describing the dislocation than the continuum model, this indicates that the con- tinuum description may be accurately used at points which are much closer to the dislocation than the generally accepted value of ten lattice spacings. ,A COMPARISON OF DISCRETE AND CONTINUUM MODELS OF AN EDGE DISLOCATION IN A RECTANGULAR SIMPLE CUBIC CRYSTAL BY A .\ EdwardM. Schall A THESIS Submitted to Michigan State University in partial fulfillment of the requirements for the degree of DOCTOR OF PHILOSOPHY Department of Metallurgy, Mechanics and Materials Science 1965 ACKNOWLEDGMENTS I wish to thank Professor T. Triffet for his lualp in selection and solution of this problem. Special iflmanks are also due to Professor L. E. Malvern for sxeveral key suggestions which contributed substantially le the problem solution. Thanks also to the other members of my graduate cxxmnittee, Professors w. A. Bradley, G. B. Mase, and C. P. Wells. ii TABLE OF CONTENTS ACKNOWLEDGMENTS O O O 0 O O 0 0 O 0 O O O O O O 0 LI ST OF TABLES I O O O O O O O O O O O O O O O O O ILIST OF ILLUSTRATIONS . . . . . . . . . . . . . . LIST OF APPENDICES. O O O O O O O O 0 O O O O O 0 Chapter I. INTRODUCTION. . . . . . . . . . . . . . . II. THE DISCRETE MATERIAL MODEL . . . . . . . Description of the Undislocated Lattice . . . . . . . . . . . . . . Introduction of an Edge Dislocation into the Undisturbed Lattice. . . . Bond Forces on the Extra Atom Plane . . . . . . . . . . . . . . . Equilibrium Equations for Lattice Points in the Crystal . . . . . . . Calculation of Interplanar Forces . III. CONTINUUM MODEL OF AN EDGE DISLOCATION IN A RECTANGULAR CRYSTAL WITH CUBIC SYWETRY. O 0 0 O O O O O 0 o O O 0 O O O 301. 3.2. 3.3. Problem Formulation . . . . . . . . Analytic Solution . . . . . . . . . Numerical Solution. . . . . . . . . IV. DISCUSSION OF RESULTS . . . . . . . . . . 4.1. 4.2. 4.3. Results of the Discrete Analysis. . Results of the Continuum Analysis . Comparison of Discrete and Continuum Analyses. . . . . . . . . V. POSSIBLE EXTENSIONS OF RESEARCH . . . . . APPENDICES. O O O O O O O O O 0 0 0 O O O C O O BIBLIOGRAPHY. o o o o o o o o o o o o o o o o o 0 iii Page ii iv vi ix 15 16 23 39 46 46 49 62 97 97 107 117 127 131 205 Table 12. 13. 14. 15. 16. 17. 18. 19. 20. LIST OF TABLES u Displacement for the Discrete Model. v Displacement for the Discrete Model. Lattice Forces Acting across a Vertical Section. . . . . . . . . . . . . . . . u Displacements for Analytic Solution. v Displacements for Analytic Solution. 0x for Analytic Solution . . . . . . . oy for Analytic Solution . . . . . . . Txy for Analytic Solution. . . . . . . Stress Function for Numerical Solution oX for Numerical Solution. . . . . . . o for Numerical Solution. . . . . . . Y Txy for Numerical Solution . . . . . . u Displacements for Numerical Solution v Displacements for Numerical Solution u Displacement for Combined Continuum Solution . . . . . . . . . . . . . . . v Displacements for Combined Continuum SOlution O O O O O 0 O O O O O O O O O 0x for Combined Solution . . . . . . . oy for Combined Solution . . . . . . . T for Combined Solution. . . . . . . XY Discrete and Continuum Values of u Displacement at Corresponding Points in Units of b . . . . . . . . . . . . . . iv Page 40 41 44 63 64 65 66 67 79 81 82 83 86 89 9O 91 92 93 94 118 Table 21. 22. List of Tables--Continued Page Continuum and Discrete Stresses at Corresponding Points . . . . . . . . . . . 122 Non-zero Components of the Constant Vector for Finite Difference Equations. . . . . . 173 Figure 1. 10. 11. 12. 13. 14. 15. 16. LIST OF ILLUSTRATIONS Physical Appearance of a Positive Edge Dislocation. . . . . . . . . . . . . . . . Positive Elastic Edge Dislocations . . . . Simple Cubic Lattice in the Undisturbed Configuration with Detail of Lattice Point i,j and its Neighbors. . . . . . . . . . . Neighboring Lattice Points in a Displaced POSition O O O O O O O O O 0 O O O O I O O Constrained Lattice after the Addition of the Extra Atom Plane . . . . . . . . . . . Neighboring Lattice Points on the Extra Atom Plane 0 O O O O 0 0 O O O I O O O O O F/B Versus Relative u for Constant REIa-tive V O O O O O 0 O O 0 O O O O O 0 0 Force Bonds for Lattice Points on and to the Left of the Extra Atom Plane . . . . . Lattice Forces Acting on the Right Hand Portion of the Lattice Due to Interaction with the Left Hand Portion . . . . . . . . Formation of an Elastic Dislocation. . . . Element Lying on the Closed Boundary, C, of an Elastic Continuum. . . . . . . . . . Boundary Stresses for Finite Difference Solution . . . . . . . . . . . . . . . . . Finite Difference Mesh for the Approxi- mation of the Compatibility Equation . . . Mesh for Numerical Solution. . . . . . . . u Displacement for Discrete Case Versus j. u Displacement for Discrete Case Versus 1. vi 10 12 17 18 22 27 43 47 47 71 75 77 98 100 Figure 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. List of Illustrations--Continued v Displacement for Discrete Case Versus j. v Displacement for Discrete Case Versus i. F versus j. 0 O 0 O O O O O O O O 0 O O O x F Versus j . . . . . . . . . . . . . . . XY Diagrammatic Representation of the Distorted Discrete Model . . . . . . . . . u Displacement for Combined Continuum Versus yl. . . . . . . . . . . . . . . . . u Displacement for Combined Continuum versus X 0 O O O O O O 0 O O O O O O O O O l v Displacement for Combined Continuum versus yl O D O O O O O O O O O O 0 O O O 0 v Displacement for Combined Continuum Versus x1. . . . . . . . . . . . . . . . . Ox for Combined Continuum Versus yl. . . . TXY for Combined Continuum Versus y1 . . . o and I for Combined Continuum Y XY Versus x1. . . . . . . . . . . . . . . . . Regions of Similarity for the Continuum and Discrete Models. . . . . . . . . . . . Complex Representation of p2 . . . . . . . Mapping of the xy Plane onto the w(2) Plane by the Function w(2) = Log 2(2). . . Mapping of xy Plane on the w(l) Plane by the FunCtion w(l) = 1.109 2(1) 0 o o o o o 0 Relation of (x,y) Coordinate to (r,s) Coordinates. C O C C O C O 0 O O O O O O O u Displacement for Analytic Solution Versus yl O O O O O O O O O 0 O 0 O O O O 0 u Displacement for Analytic Solution Versus x1. . . . . . . . . . . . . . . . . vii Page 103 104 105 106 108 109 110 111 112 113 114 115 124 142 149 151 155 196 197 List of Illustrationsn-Continued Figure Page 36. v Displacement for Analytic Solution Versus yl. . . . . . . . . . . . . . . . . 198 37. v Displacement for Analytic Solution versus x1. 0 0 O O O O O O 0 O O O O O O O 199 38. Txy for Analytic Solution Versus yl. . . . 200 39. Ox for Analytic Solution Versus yl . . . . 201 40. u and v Displacements for Numerical Solution Versus y1 . . . . . . . . . . . . 202 41. o and T for Numerical Solution Y XY versus x1. 0 0 O O O O O O O O O O 0 O O O 203 42. Ox and Txy for Numerical Solution versus yl° O O O O O 0 O 0 O O 0 O O O O O 204 viii Number II. III. IV. V. VI. VII. VIII. IX. LIST OF APPENDICES EQUILIBRIUM EQUATIONS FOR THE DISCRETE MODEL IN A PARTICULAR EXAMPLE. . . . . . . DETERMINATION OF THE FORM OF DISPLACEMENTS FOR THE CASE OF PLANE STRAIN AND CUBIC SYMMETRY AND WITH A DISCONTINUOUS DISPLACEMENT . . . . o . . . . . . . . . . EXPANSION OF LOG 2(1) AND LOG 2(2) INTO REAL AND IMAGINARY PARTS . . . . . . . . . SIMILARITY OF DISCRETE AND CONTINUUM MODELS O O O O O O O O 0 9 O O O O O O O 0 FORM OF THE STRESS FUNCTION EQUATION . . . CALCULATION OF THE BOUNDARY VALUES OF qb FOR THE NUMERICAL SOLUTION . . . . . . . THE SYSTEM OF FINITE DIFFERENCE EQUATIONS. FORTRAN PROGRAMS . . . . o . . . . . . . . GRAPHICAL REPRESENTATION OF THE ANALYTIC AND NUMERICAL CONTINUUM SOLUTIONS. . . . 0 ix Page 131 139 148 154 160 162 168 174 195 I. INTRODUCTION In the more than thirty years since its origin, the theory of crystal dislocations has been successfully applied to many problems in the plastic deformation of crystalline solids. On the basis of dislocation theory reasonable physical explanations have been supplied for such experimentally observed phenomena as low mechanical strength, work hardening, annealing, and many others. MUCh of this theory has been of a qualitative nature, but that part which is quantitative has been, for the most part, derived from a material model which although it yields readily to mathematical analysis gives a rather poor description of the physical situation. This model is the elastic continuum and its mathematical key the theory of elasticity either anisotropic or more commonly isotropic. The elastic dislocation as conceived by Volterra [1] and applied to the case of crystal dislo- cations by Cottrell [2] and others has the inherent dis— advantage of describing a crystal defect which is atomic in nature by a model which is continuous. Even with such a serious shortcoming the continuum model of the crystal dislocation has been extensively used to determine stress 1 ”- I'.I n-V . .— .....- n ,p-. - 0"" u.... ~ vuuu- Viv-v .- O! - A "I‘Iov a... ‘ V‘ "vv. 0“. (II 'f’ u “~.._ . C“ . . M" ‘ '1. F ‘p: , u” I § - ‘\ \‘\ if -2 " t .~ ‘- Q g b - . '.“§ \ i.‘ . . \_I I. . 5 ._‘ I v S Q I \.u ‘I‘ § 2 and displacement fields for such problems as isolated edge and screw dislocations, pairs of dislocations in proximity, arrays of dislocations with a common slip plane, dislocations in proximity with impurity atoms, dis- locations near coated boundaries, and many other specific examples [2], [3]» All of these analyses possess the same essential weak points. These weak points may be emphasized by a consideration of the problem of a straight edge dislo- cation in terms of its description in the elastic dislo- cation theory. Referring to Figure 1 the physical description of this defect in a simple cubic lattice is that there is one more vertical plane of atoms above plane AA' than there is below it. What displacements then are necessary to mate the half plane above AA' to the half plane below it assuming that far away from the center of displacement the atomic planes such as 88' and CC' are completely lined up? The elastic dislocation which corres- ponds to this physical description is formed by cutting an elastic continuum along the plane OS in Figure 1 or Figure 2a and then displacing the adjacent faces of the Cut by one atomic spacing, b, relative to one another, in the horizontal or x direction. To correspond to the symmetrical case shown in Figure 1 each face would be dis- placed by amount b/2 in opposite directions. The elasti- City solution for the stress and strain fields in this problem is found for the isotropic case in Read [4] and for the anisotropic case in a paper by Eshelby [5]. Both C S B D' D A. ,0 l A \ ‘1/‘6 E' E C. B! Fig. 1.--Physical appearance of a positive edge dislocation A! (a) T :- b/Z‘ b/2 A' .1. _ i (b) Figo 2.--Positive elastic edge dislocations 5 contain a singularity at the root of the cut, the point 0 in Figure l or Figure 2a, i.e., the stresses and strains are infinite here. For practical purposes a cylinder of material surrounding the point 0 in which stresses are too large to be legitimately described by an elastic analysis is also omitted from the region of the solution. Alternately the elastic edge dislocation may be con- structed by cutting the continuum along the plane 0A in Figure l or Figure 2b and displacing the opposite faces of the cut by the relative amount b as shown in Figure 2b. The justification for the application of the elastic dislocation theory to describe the crystal dislo- cation is that for small relative atomic displacements the crystal lattice displacements and interatomic forces nay be considered to be proportional. However, in the region close to the point 0 in Figure 1 where the rela- tive atomic displacements are large the continuum model is completely unrealistic. The extent of the region in Mmich the elastic continuum analysis is inapplicable is (fifficult to determine since it must depend on the Characteristics of the atomic bond forces in this region, but a conservative estimate excludes a cylinder of about 10 atomic spacings [2]. Some attempts have been made to develop material nmdels which may describe the region of a large deformation nmre realistically. Peierls [6], Nabarro [7], and Foreman [8] have described the edge dislocation in 6 Figure l in a way which takes some account of the atomic structure on the slip plane while assuming an elastic continuum elsewhere. They treat the problem as two elastic continua, one above the plane DD' and the other below the plane EE', in which the two planes exert a mutual periodic shear stress tending to compress the material above DD' and expand the material below 83'. The shear stress is a function of the relative displace— ment, E , between corresponding points on DD' and EB' and is assumed to be a maximum when E = b/2 and zero when 67 = 0 so that the period of the shear stress is b. This model avoids the singularity encountered in the elastic dislocation analysis and uses a non-linear force displacement representation on the slip plane and especially in the region of high distortion. However, it assumes no displacement normal to the slip plane and also assumes that the only large deflections occur on the slip plane. This analysis indicates that the length of the slip plane over which the quantity E? is more than .25b is only 1.5b, i.e., the region of large deformation is 1.5b, for the case when the periodic shear force is a sine function. This appears to be a low estimate of the region of large deformation. By equating the potential energy change due to an infinitesimal displacement of the dislocation to the work done by the shear stress on the slip plane, Nabarro and Foreman have determined the shear stress which is necessary to cause the dislocation to propogate leaving slipped material in its wake. This stress is large if the region of large disturbance is confined to a small length of the slip plane and becomes smaller if the disturbance is spread out over a large length of the slip plane. The actual atomic displacements for any real crystalline substance containing dislocations, at least in the region of high atomic misalignment, can only be found by a quantum mechanical analysis of the problem. This approach is prohibitively complex and must immediately be abandoned. The next simplest possibility would be to consider the atoms as particles which exert atomic forces on one another. But again the details of interatomic forces are to a large degree unknown and even simple models of these forces are mathematically complex for large atomic displacements. It is true, however, that material models using the interatomic force concept do exist which are still simple enough to be mathematically practical. Hopefully these models are closer to reality than those assuming the elastic continuum. Such a simple atomic force model was used by Maradudin [9] to determine the displacements in an infi- nite crystal lattiCe due to the presence of a screw dislocation. His model consists of a simple cubic lattice in which each lattice point exerts a shearing force on its four nearest neighbors. This force is linearly propor— tional to the relative shearing motion of that lattice point with respect to each of its neighbors. For this particular case he is able to find an analytic function 8 which satisfies the force equilibrium equations of all lattice points in the crystal. The displacements pre— dicted from this analysis seem to be more realistic than those resulting from the elastic dislocation solution. A somewhat similar material model of a crystal lattice, in this case bounded rather than infinite, is used in this thesis. An edge dislocation is introduced into the otherwise undisturbed and unloaded lattice and the resulting displacements are determined by the simul- taneous solution of the force equilibrium equations for each lattice point. As a comparison with the results of this analysis, a similar calculation is made for the case of an elastic dislocation. The material parameters and the geometry of both models are as similar as possible, to the extent of using cubic anisotropic elasticity for the continuum solution. The similarity of the two models is discussed in detail in Appendix IV. II. THE DISCRETE MATERIAL MODEL 2.1. Description of the Undislocated Lattice The discrete model of the crystal lattice will consist of a simple cubic arrangement of atoms for which the crystal boundaries will be parallel to the cubic axes. The atomic spacing is b and the x and y dimensions of the lattice are mb and nb respectively. Since the eventual problem involves only plane deformations, the z dimension is redundant. Suppose that this lattice is in its minimum potential energy position as shown in Figure 3 and that in this position all atomic forces are zero. Now suppose that, in response to some arbitrary internal or external disturbance, the lattice undergoes a small plane defor- mation from its minimum potential configuration. What relationships exist between the displacement of the lattice point (1,3) and the displacements of all other lattice points in.the crystal? The answer to this question depends on the form Of the atomic bond forces between atom (i,j) and all other atoms in the system. These bond forces will be approximated in the following way: (1) the displace- ment of each lattice point is affected directly by its nearest and next nearest neighbors, (2) the relative 9 10 n o o o o o o o o o o o o o o o o o o O o 6 o o o o O o o o o o :[b 5 o o o O O o O 4 o o o o o o o Y i j 9 3 o o o o o o o 2 o o o o o o o o o o X l o o o o o o o O o c j i l 2 3 4 S 6 8 99 m (a) i-l,j+l i,j+l i+l,j+l O O O i-l,j o o o i+l,j i-l,j~l O O Oi+1,j-l 1,341, (b) Fig. 3.--Simple cubic lattice in the undisturbed configuration with detail of lattice point i,j and its neighbors ll displacement of two adjacent lattice points results in a restoring force wnich is linearly proportional to the displacement, and (3) the restoring force is directed along the line joining the two lattice points. With these bond force characteristics the crystal may be visualized as an array of points each of which is connected to its nearest and next nearest neighbors by a linear spring. In the absence of a disturbance all the springs are unstretched. If the lattice is subjected to a plane disturbance, each lattice point is displaced to a new position. The mutual force between that atom at (i,j) and its nearest neighbor at point (i+l,j) is pro— portional to the length d = r-b in Figure 4a where d is given by the equation d = r-b = [(b + u. - u. .) + ( V .)2]l/2 - b. 1+1,j 1,3 vi+l.j ” i.) (2.1) if the quantity (v - vi j) is small enough so , i+laj that its square may be neglected, the length d in linearized form is given by d = ( ). (2.2) u. '- 11 .. o i+l,j 1,3 Assuming a force constant a, the force acting on the atom (i,j) is Fi,j = OL(ui+l j - ui,j)' (2.3) By virtue of the assumption that (v. ) is small 1+1,j ‘ Vi,j compared to[b + (u the rotation of the bond i+laj D uiaj)], force may be neglected so that the x and y components of 12 i.j id “1’3 i+1,j i.) L‘ b =4 ui+l,j (a) Nearest Neighbors // ’i£l,jtl i+1,j+1 1,1 L ' b J (b) Next Nearest Neighbors Fig. 4.--Neighboring lattice points in a displaced position 13 force acting on atom (i,j) due to its interaction with atom (i+1,j) are Fx = O“hugs ' i,j (2.4) F = O. Y In a similar manner the force between atom (i,j) and its next nearest neighbor at (i+l,j+l) is proportional to the distance d1 = r1 - «2 b. Referring to Figure 4b d = r — J? b = [(b + u - u )2 + 1 l i+1,j+l i,j (2.5) 2 1/2 . (b + Vi+l,j+1 ~ Vi,j) J - f2 b. . . . . -_ , 2 Neglecting the terms anOlVlng (ui+l,j+l - hi,j) and (v - v )2 gives i+l,j+1 i,j ' _ 2 , 1/2 . d1 _ [2b + 2b(ui+1’j+l _ ui,j + Vi+l,j+l — vi’j)] - J? b. (2.6) Expanding equation (2.6) in a binomial series gives d ='l—[(u - u ) + (v . — v, )1. (2 7) 1 J2 i+1,j+l i.j ” i+1,j+1 i,j ' If the force constant is B, the mutual force on these two atoms is u .) + (v. (2.8) -2. .. _ F - J§[(ui+1,j+l i,J i+1,j+l Vi,j)]° Neglecting the bond rotation, the force components on atom (i,j) due to its interaction with atom (i+l,j+l) are - ui,j) + (v. v. .)]. (2.9) -2 .. X y - 2[(ui+1,j+1 i+l,j+1 1,] 14 In the case of static equilibrium all such force interactions between atom (i,j) and its neighbors as shown in Figure 3b must produce a zero resultant. The equili- brium equations for atom (i,j) are ZF=0 x 0L(“14,1 ‘ 2”i,j + ui-l,j) +'§(ui+l,j+l + Vi+l,j+l + + + + ui+l,j-l ‘ Vi+l,j—l ui-l,j+l ‘ V1-1,j+1 ui-l,j-l vi-l,j-l — 4ui,j) = 0 (2.10) ZFy=O 2v. + v. . ) + E( a(vi,j+1 - i,j 1,3—1 2 ui+1,j+l + vi+l,j-l ui+1,j—1 + vi+l,j-l ‘ ui-1,j+1 + vi-l,j+1 + ui-l,j-l * vi-l,j-l "' 4vi,j) = 00 (2.11) These equations are similar to those presented for a more general case by Kittel [10]. The equilibrium equation for lattice points on the boundary of the crystal are similar to equations (2.10) and (2.11) except that the effects of the missing neighbors are omitted. A lattice point on the lower boundary at point (i,l) has for example the equilibrium equations + v. + .E _ 2 ui+1,2 1+1,2 u1-1,2 — 2u. ) = 0 (2.12) 15 - u + v i+l,2 i-l,2 1-1,2 ‘ 2vi,1) = O. (2013) These equations assume no external force acts on the lattice point. 2.2. Introduction of an Edge Dislo- cation into the Undisturbed Lattice An edge dislocation will now be introduced into the initially undisturbed lattice shown in Figure 3a. At the lower boundary of the crystal between lattice point (r,l) and the succeeding point the atomic bonds, i.e., the linear springs, are severed up to a height j = s. A new plane of atoms, 5 atomic spacings high, is inserted into the gap resulting from this process. While the lattice is constrained in its undisturbed position, atomic bonds are established between atoms on the extra plane and their neighbors. These bonds possess a high potential energy, i.e., the linear springs connecting the atoms on the extra plane with their neighbors are compressed, while the potential energy elsewhere remains zero because of the constraint of the lattice. The lattice constraint is now released and the potential energy concentration around the extra atom place is distributed through the lattice, resulting in the displacement of each lattice point from its original position. The force bonds between atoms on . .u 16 the extra atom plane and its neighbors are shown in Figure 5. According to the sign convention given by Cottrell [2], this extra atom plane produces a negative edge dislocation in the crystal lattice. 2.3. Bond Forces on the Extra Atom Plane The most important factor in determining the lattice deformation is the form of the interatomic forces between atoms on the extra plane and their neighbors. These forces will now be discussed in detail. Consider a lattice point (r+l,j) on the extra atom plane and its nearest neighbor at point (r,j). When the lattice is in the constrained position, these points are separated by half an atomic spacing, .Sb. Assume that the mutual bond force is zero when the separation is the normal atomic spaCIng b and that the mutual atomic force is linearly related to the difference between the zero force separation and the actual separation. In the constrained position then, there is mutual repulsive force of .Sba on atoms (r+l,j) and (r,j) where the force con- stant is assumed to be the same as that for normal nearest neighbor interactions. Upon removal of the constraints the points (r+l,j) and (r,j) will move to their displaced positions as shown in Figure 6a. The difference between the displaced separation of these two points and the assumed zero force separation is _ 2 2 1/2 d2_b- [(05b+ur+l,j -ur,j) + (vr+1,j -vr,j)] 0 (2.14) n o o o o o O o o o o o O o o o o o o o o c o o o o o o o o o o o o o o o O o s o o o o o o o o o o o o o o 0 P0 17 O O O O O O O O O O O O O O O O 0 O O O O O O O O O O O O 0 O O O O O o o O o q o o O o o r+2 l 2 r r+l (extra atom plane) Fig. 5.--Constrained lattice after the addition of the extra atom plane m+1 b JEb/z Vr+1 ,j 1,—— r+l,j ur+1,j b/2 (b) ¢§b/2 b r,j-l L—b/2 _.J (c) Fig. 6.--Neighboring lattice points on extra atom plane 19 . . 2 Neglecting the quantity (vr+1,j _ Vr,j) as small compared with other terms in the bracket gives d2 b - [.5b + (u u .)] (2.15) r+l,j - r,J d = .5b + (u (2.16) r,j ‘ ur+1,j)° Assuming a linear relation between d and the bond force 2 results in a mutual repulsive force acting on atoms (r+1,j) and (r,j) of magnitude .5bd + d(u . - (2.17) r,3 ur+l,j)° If the rotation of the bond is small, and this seems to be a reasonable assumption for the atom pairs of which this is a typical example, the components of force on atom (r+1,j) are F I In (I Q + Q C: I "’ . u . X r33 r+laJ (2.18) F :00 Y The interaction between the atom (r+1,j) and its next nearest neighbor at (r,j+l) may be similarly determined. In the constrained position of the lattice these atoms J? are separated by the distance §#b. If it is assumed that the zero force separation for all next nearest neighbors is JE-b and that a linear relationship exists between atomic force and the difference between actual separation and zero force separation, then a mutual repulsive force J" of B( J2'b —-72b) acts on atoms (r+1,j) and (r,j+l) when the lattice is in the constrained position. After 2O relaxing the constraints the difference between actual and zero force separation is seen from Figure 6b to become _ _ _ _ _ 2 d3 — J2'b p2 _ J2 b [(.5b + ur+1,j ur,j+1) + (2.19) 2 1/2 (b + vr,j+l — vr+1,j) 1 ° Neglecting orders of (ur+l,j — ur,j+l) and (vr,j+1 - vr+l,j) of two and higher leaves the linearized form .5 1 2 d3 = J5 b ’ [Tb + T§(ur+l,j " ur,j+l) + T§(Vr,j+l ' vr+l,j)]° (2.20) Assuming a linear relationship between the atomic force and d3 results in a mutual repulsive force on atoms (r+1,j) and (r,j+1) of \/E F = ( J2 --§—)b8 - ) 2 (v - L4 _ .. Jg'ur+1,j ur,j+l ¢§ r,j+1 Vr+l,j)' (2.21) It will be initially assumed that the bond rotation between atom pairs of this type may be neglected. This assumption, however, for this particular type of atom pair may not be justified. Whether some different bond orientation should be used will be determined after the displacement solution of a particular numerical example has been found and examined. In any case, according to the initial assumption, the x and y force components on atom (r+1,j) are 21 F 2 1 Fx = 75 = (£— 2>bB M 5(ur+l,j _ ur,j+l) -2§(v - v ) r,j+l r+l,j (2.22) 2F 2 1 2E Fy = - (3': -29/5--'2)b8 + 5(ur+l,j - ur,j+1) + £E(v — v ) r,j+l r+1,j ° The force components on atom (r+1,j) due to its interaction with atom (r,j-l) which are shown in Figure 6c are ’11 ll 2 l 2 x 2/5.-'2)b6 +"5(ur,j-l - ur+l,j) + 5(Vr,j-l — Vr+l,j) (2.23) 2 l 2 4 y ZS/g - §)bB + _E(ur,j-l - ur+l,j) + _5(vr,j-1 - vr+l,j)° The assumptions involved in defining the bond forces in F this way are formidable. The linearity between force and displacement up to a relative displacement of half an atomic spacing is the most unrealistic of all assumptions. The simplification in calculations which results from such an idealization is, of course, the justifying factor. The neglect of second powers of the relative displacements is a good approximation up to values of relative dis- placement of about .3b which may be seen from Figure 7. The neglect of bond rotation is certainly reasonable up to angles of five or six degrees and the values of dis- placement involved in any foreseeable application of this theory should be such that most bond rotations are within this limit. A correction can be applied to any bond rotations which exceed the limiting value. 22 > m>wumamu pampmcou mom 3 O>Hpmamu mamum> mbmllé. 833853 Tut/m + u3 mm... .. lhl >+ Ninfim N H H am»; H M m S + A SN + SVQ + Q + QQL p a). 23 Despite the formidable assumptions which must be made to facilitate a solution, this model does possess, in a simple way, the discrete character of a real crystal lattice and is superior to the ordinary continuum model in this regard. 2.4. Equilibrium Equations for Lattice Points in the Crystal The equilibrium equations for those atoms which are surrounded by a normal complement of nearest and next nearest neighbors and do not lie on or next to the extra atom plane are given by the equations (2.10) and (2.11). Referring to Figure 5a all lattice points for which i and j satisfy the relationships 2 5 i 5 m 2< - j S n-l where i fi r, r+l,r+2 for j 5 5+1 have equations (2.10) and (2.11) as equilibrium equations. All lattice points on the lower boundary except at the corners and on or next to the extra atom plane, that is, points for which i and j satisfy the relations 2 5 i 5 m i # r, r+l,r+2 J=‘1. have equations (2.12) and (2.13) as equilibrium equations. The equilibrium for all other lattice points will now be listed along with the appropriate values of the indices 1 and j. a) b) c) 24 At the lower left-hand corner of the crystal 1 = 1 j = 1 23FX = O a(u2,1 - ul,l) + %(u2,2 + V232 — u1,l _ V1,1) = O (2.24) EZFY = O “(V1,2 - V1,l) + g4u2,2 + v2,2 — u1,l _ V1,1) = 0° (2.25) For the left-hand boundary i = l 2 5 j 5 n-l EZFX = O “(u2,j - u1,j) + 2&u2,j+1 + v2,j+1 + u2,j~l - V2,j-l - 2ul,j) = 0 (2.26) EZFY = 0 “(V1,j+l - 2V1,j + Vl,j-1) + §4u2,j+1 + V2,j+1 — u2,j-1 + v2,j-l — 2u1,j) = O. (2.27) At upper left-hand boundary 1 = l j = n ZFX = 0 “(u2,n " ul,n) + 2(u2,n-1 V2,n--1 u1,n + Vl,n) = O (2.28) ZFy = 0 O“(Vim-.1 Vl,n) + %(-u2,n-l * v2,n-l + ul,n - vl,n) = 0. (2.29) 25 d) For the upper boundary noting that there is no lattice plane for i = r+l so that i skips directly from i = r to i = r+2 25i§m i;ér,r+l,r+2 j=n 22);: O “(ui+1,n ”lli,n + ui-l,n) + 2(ui+l,n-l vi+1,n_1 + ui~l,n~l + Vi-l,n-l 2ui,n) = O (2.30) 23Fy = O amU mQmU hqmr BLMH mom .Hv HMQOZ mmeUmHQ mi? mom BZMZMU m mqm the relative displacements of (k,j) and its three rueighbors (k-l,j+l), (k-1,j), and (k-l,j~1). The compo- ruent of this resultant force which is normal to the 'Vtertical plane is analogous to OX and will be called I?)c(j). The tangential component is analogous to Txy and Vii.ll be called ny(j). These lattice forces are shown in E”j_gure 9 and referring to this figure the forces on the ILEittice points on i = k are r~ = 1 - E. _ ><(l) O“(ukh uk_1,1) + 2(uk,l uk-1,2 F‘ _ fl. _ _ >cy‘1) ' 2(uk,l uk-1,2 Vk,1 + Vk-1,2 FxU) = “(hm ' uk—Lj) “ 2(uksj "’ uk-l.3+l ‘ de Vk—l,j+l) + g‘uk,j ' uk~1,j—l + Vk,j ' Vk-l,j-l vk,j + Vk—l,j+1) ‘ g‘uk,j ‘ FXy(j) = 2(uk,j - uk-l,j+1 uk-l,j-l + vk,j ‘ vk-1,j-l) _ _ P. _ Fx(n) ’ a) - v(-§E) 0. (3.7) It may be seen from the mapping of the xy plane onto the w(2) = log z(2) plane in Figure 310f Appendix III that a positive (counter-clockwise) angle change of 2m in the xy plane causes a change in the value of w(2) of amount -2ni. Similarly, a positive change of angle 2m in the xy plane causes w(l) = log 2(1) to change by amount +2ni as may be seen from the mapping in Figure 32. With this in mind, a substitution of equations (3.5) into the boundary con- ditions (3.6) and (3.7) leads to the equations D (2) — mm = b A (3.8) A D 0. (3.9) (2) (2) ‘ A(1)”(1) 2 The imaginary parts of these two equations have no significance for reasons already discussed. Therefore, only the two real equations associated with equations (3.8) and (3.9) are available for the solution of the two complex constants D( and D(2) where l) D(1) = Ru) * 5"1(1) (3.10) R (2) = (2) + 311(2)" 51 Hence, two additional equations are necessary to determine Ru)’ 1(1)’ R(2)’ and I<2)° b) Stress Boundary Conditions Consider the equilibrium of that portion of a large body in static equilibrium which is included inside the contour C in the Figure 11a. If the element shown in Figure llb is isolated from the larger body, equilibrium requires that the forces de and dFy which are transmitted across the elemental length dc must satisfy the equations dF x -U dy + T dx x xy (3.11) dF Y Y XY If these expressions are integrated around the contour C, the result is the net x and y components of force trans- mitted by the material inside C to the material outside C. However, if the entire region is in equilibrium, then the material inside C is also in equilibrium so that the net force on C is zero. .de jF-G dy + T dx = O x x xy dF Ud-Td=00 fy jgyx xyy 5N1e vanishing of these integrals regardless of the path C (3.12) “as long as it is closed) implies the existence of two Potential functions be and qu such that {d dx=¢ {gait dx + gg—é dy =f—o‘xdy + Txydx (3.13a) 52 ad; qu d = —a—1 d —a——Y-d= de-Td. 3.13b {#3}, f x X+ y y y ny ( ) Equating coefficients of dx and dy in equations (3.13) gives __da_.5¢> G = (a) I”: = - 7—91 (b) (3.14) 0y = .31; (C) The functions qu and 95y may be found by integration of the perfect differentials of be and Qby along the path C1C2 in Figure 11a. To find qu integrate the expression d¢x= ade dx + (993x d along ClCZ a Y (fix = six—BEE— dx + 35:31 dy. (3.15) b yV = b I i‘i x I The first integral in (3.15) is a function of x only since y = b along path C1, while the second is a function of both x angqg which may be found by an indefinite inte- x gration of —E;§— with respect to y. The function be then may be written cpx = @(x) + F(x,y) (3.16) Where a F(x,y) = l—Eééi dy. (3.17) 53 Differentiating equation (3.16) with respect to x gives c) 73335 = 9'”) +3; or d Q'(x) = 5:" J3; (3.1.8) Equation (3.18) may now be integrated indefinitely with respect to x to find e(x) dqu (SF. 6(X) = ('3;— —E)-§)dx. (3019) This procedure is outlined in Advanced Calculus by Taylor [12]. For the present situation 695x _ T _ c (21.1 + 61) x - xy — 44 dy (TX (3.20) é¢x - nO' - _. (C (2.2 + C él) 3y ‘ x — lldx 120y where u and v are given by equations (3.5). Substituting the appropriate derivatives into equations (3.20) gives 1 ei (“bx _ c D(1) e 7? + A(1) 1D(2) 77-). A(2) (3 21) _ ,— . -— —"E'T"“' -—)-(-—:-:I-— o . 44 211:1. x + eifly 2711 72y ei ’ 1 d¢x = _ D(1) c11 + CA12 (1)e 7? D(2) C:11 ‘ 512A(2)e 7? 3y 21ti x + €1.77Y 2m. x _ eiTIy (3.22) Substituting expression (3.22) into equation (3.17) gives 54 D _ (1) i]? dy “X’Y’ - ‘ 2751‘ ‘-C11 + c1.2%,)e J; + e177,, " D (2) i d W (C11 - c12A(2)e TI)/V——XW1T (3.23) x - e y i F(X ) - - D(ll [C11 + C12A(l)e n] 10 (X + e1 ) + ’Y ' 2ni in g Y e i D(2) [C11 ‘ c12A(2)e NJ log (x - elny). (3.24) 2ni eifl Differentiating equation (3.24) with respect to x 1 dF(x ) = _ D(1) [511 + C12A(1)e ”1 1 + x 2ni efn x + eifiy i D(2) [C11 ‘ C12"(2)e7?J 1 211:1— e177 x _ elny Substituting the expressions (3.21) and (3.25) into (3.25) equation (3.19) 1 -1U em) D(1) [c4463 77 + <:44A(1) * c:11e + " 3?)? x . .isz C12A(1)J i -i [c en-c A +c11e 7.I—chzAmfl dx (3.26) D . . Goo =§%%—)- [c44e177+ c A + c 517?. D in (2) 'U log (X + e y) -'§fi$‘ [C448 - C44A(2) 11 i c12A(2)] log (x - e fly). Now substituting 9(x) and F(x,y) into the equation (3.16) gives 55 D .. (1) i ' . 1; $3; = 2711'“ C44[e T, + 1“(1)]l°g(3‘c + e 7g) '" 7%)— c44[ei7( .- A(2)] log (x - e11). (3.27) In a similar manner the potential function <¢&_is found to be D l » i <251! = 271%; Lclle UA(1)12] log (X + 8 my) + .giél [~c1 leinA(2) + c123 log (x - einy). (3-28) Equilibrium requires that the integral of these functions (fix and Sb}, about any closed path be zero. If, for example, the path starts at 9!: —.% and ends at the same point at W: fad; gbx ($35) - CIDX (- 12‘") {6% = Sty (€35) — (by (- 35‘). Substituting equations (3.27) and (3.28) into the boundary -%2, the conditions for static equilibrium are (3.29) conditions (3.29) and noting that log (x + einy)changes value by the amount +2ni and log (x - einy)changes value by the amount -2ni as 4! changes by +2n, the following equations result: 1 i 77.1 ,[e77-1(2,1 =0 (3.30) ”(1Ne (1)3 + D(2 - 00 (3031) 1 17(1).“ " C12J ‘ D(2)["C11e mm) " c:12] ‘ D(1)[511e It is shown in Appendix II that 56 A(2) = ‘A(1)' Making this substitution into equations (3.30) and (3.31) gives the equations 1 [D(l) + D(2)]Le n + A(l)] = O (3.32) - 1 (0(1) — D(2)]LcllA(l)e U + 612] _ o (3.33) which, together with equations (3.8) and (3.9) may be solved for the real and imaginary parts of D(1) and 0(2). The simultaneous solution of the four boundary condition equations using only the real parts of these equations gives 0(1) = -D(2) = -.5b(l - ikl) (3.34) where the constant kl is given by c 2 + c c - c 2 - c c cos 2 12 12 44 11 ll 44 In k = U o (3.35) 1 cllc44 Sin5271 The value of D and D(2) given by equation (3.34) with (1) k1 given by equation (3.35) are different from those found by Eshelby [S]. The displacements for an edge dislocation in an unbounded continuum with cubic anisotropy are given by the real parts of the following equations: u z'Zfi (k1 + i)[log z(1) - log 2(2)] + H l (3.36) b . v z'IE (k1 + i) A(1) [log z(1) + log z(2)] + H2. (3.37) The constants H1 and H position of the unbounded medium. For comparison with 2 are determined by the rigid body the displacements of the discrete model, the u displace- ment will be symmetrical about the y axis and the v 57 displacement will be made zero at some point on the negative y axis. The boundary conditions for displace- ment are u(- 12‘.) -u(-§-E) = b/2 (3.38) v(O,L) I O (3.39) Referring to the transformation diagrams, Figures 31 and 32 in Appendix III, a point (O,y) on the negative y axis maps on the point log y + i(72+ n) in the log 2(1) plane and on the point log y + in in the log 2(2) plane so that the condition (3.39) leads to the equation -% =-§fi (k1 + i) [log y + i(7? + n) - log y - in] + H1. Expanding this expression and ignoring the imaginary portion gives H1 = 3b/4. (3.40) The condition (3.39) leads to the equation b . O =-Zfi (k1 + i)A(1)[log L + i(‘n + E) + log L + 1 J + H2 where -i 1 C11‘3 7? + c:44"2 7? C11 + C44 . A(l) = - c + c = I c + c C0572 + l 12 44 12 44 c - c C1]. + (:44 sinTI 12 44 or making the substitutions c 11 44 ll 44 k — cosn , K = Slnn 2 c12 + C44 3 C1 + C44 Substituting this value of A( into (3.41) gives 1) H2 = —-§fi (kl + i)(-k2 + ik3)[2 log L + 1(272+ n)l. Expanding this expression and discarding the imaginary portion gives b 2 H2 = 4% [(klk2 + k3) log L + (klk - k2)(2U-+)n]. (3.42) 3 Then for the case of symmetrical u displacement and a zero v displacement at the point (0,-L) the displacement field of the edge dislocation in an infinite continuum is; given by the equations b - u = Re Zfi‘k1 + i)[1og 2(1) - log 2(2)] + 3b/4 (3.43) v - Re b [(-k k - k ) + i(k k - k )][lo 2 + ‘ I? 1 2 3 3 1 2 9 (1) b 2 log z(2)] + zfi[(klk2 + k3) log L + (klk3 - k2) (27] + n)l . (3.44) If the logarithmic functions of the complex variables are expressed in complex form, that is 109 Z(1) ‘ r(1) * is(1) 109 2(2) ‘ r(2) * 15(2)’ and if the equations (3.43) and (3.44) are expanded, u and v may be written in the form The the 59 b . 4R (k1[r(1) ‘ r(2)J ‘ LS(1) ‘ 5(2)] + 3”) (3°45) EL. [—k k - k ][r + r - 2 10 L3 - 4): 1 2 3 (1) (2) ‘3 [k3k1 - k2][s(l) + 5(2) - 2n - WJ) . (3.46) real portions of log 2(1) and log 2(2) are given by equations =-% log (x2 + y2 + 2xy cosn ) (3.47) = % log (x2 + y2 - 2xy cosn ) (3.48) while the imaginary portions may be determined with the help of the transformation diagrams, Figures 31 and 32, in Appendix III. The stresses associated with the dis- placements given in equations (3.43) and (3.44) may be determined by the application of the definitions of stress in terms of displacement. The _ 611 dV’ Ox ‘ C311 ‘33? + C112 ’6? (3’49) _ 511 <5v Cy — C12 OX 4" C11 W (3°50) _ OV‘ (Bu Txy - C44( dx + ‘5‘?)0 (3051) resulting stresses are y(Clx2 + C2y2) x + Dx y + y4 C3y(y2 - x2) ( x4 + szy2 + Y4 6O x(C4x2 + C5y2) Txy = b 4 2 2 4 (3.54) x + Dx y + y where 2 2 c = .1. _ (C12 + c:12‘344 " <:11 " C11544 COS 27?) l 2 cllc44 Sin 27f (511544 ‘ H12 44 C05 4277):: CC:12 44C 451,127? c057? c + c4 12 c2+cc -c2-cc cos27? ( 12 12 44 11 11 44 )(;12: 44 M) C11C44 s1n 27'! c c - c c cos 27? sin 27? — 11 44 c 13 :4 sinTI (3.55) 12 44 2 2 c = _1__ (C12 + cE2544 ' C11 " C:12‘:44 COS 27?) 2 2n cllc44 Sln 27? c c - c c cos 271 c c ( 11 44 12 44 12 44 sin 27.1 (2057? c12 + C44 C12 * c44 c2+cc -c2-c c052 ( 12 12 44 11 12c 44 7(HC; 12: 44 C11C44 Sin 27'"? c c - c c cos 2 sin 27?) _ 11 44 C 12 :4 7? sin” 12 44 2 2 2 C = 1 (c12 + clzc44 — C11 - c12c44 cos 27?) + 3 fi - (c12 + C44)(C11C44 $111 27?) c c cll :14c sin 27? c0572 12 44 2 2 C = C44 _ (C12 * c212544 " c:11 2:12:44 “5 27?) 4 2n (C11C44 sin 27? T (C11 - C12) cos - C11 + €12 sin (3 57) (c + c T T( .c + c 7? ' 12 44 12 44 61 c (c 2 + c c - c 2 - c c cos 2n ) 44 _ 12 12 44 11 12 44 C = ——— r _ 5 2n (C11C44 Slnfi2n ) (C11 C12) C11 + C12 2 (c + c ) COS” + c + c Sifin [l — 2 cos 7?] — 12 44 12 44 (C2+cc —c 2--acc c0527?) 2 12 12 44 11 12 44 (cllc44 sin 2n ) (C + C ) C — c (C11 (21% sinTI _ 11 c12 c057? sinn cosn (3.58) 12 + 44 C12 + 44 i D = 2(1 - 2 cosZTI). (3.59) The similarity between the equations for stress in the anisotropic case and the isotropic case may be seen by comparing equations (3.52) through (3.54) with the equations given on page 116 of Read [12]. The isotropic case requires the special condition c057? = 0 or 7?: 13g. In Appendix IV it is shown that the relationship between the constants a and B of the discrete model and the constants Cll’ c12, C44 of the continuum model are 2 C11 = b (a + B) 2 C44 = b B c + c - 2b2B 12 44 - ' For the special case a = B which is used for the discrete model solution these conditions reduce to 2 c 2b a ll 12 44 :— 54» \ 62 or .11, (3.60) c12 = C44 = 2 ° When the elastic moduli are related by the relationship (3.60), the various constants in the solution have the following values -025 cos 2n sin 27? = .96824584 7]: .91173829 rad .79056942 (.0 Ho :3 43 H .61237244 0 o (I) .3 I! = .07549382 c 5 11 -o 750 (.9375)”2 -k k = 0.316227 1 2 ' k 3 k3k1 - k2 = -1.224744. The displacements and stresses for this particular case are shown in Tables 4 through 8. Curves showing the variation of displacements and stresses are shown in Appendix IX. 3.3. Numerical Solution Suppose that the crystal shown in Figure 10 is square and that the dislocation is symmetrically placed. The geometrical parameters of the problem may then be simplified to 63 .H mmm. oom. mum. 0mm. mNH. o a» mmmad. mahmfi. fivmmd. mumbv. mmwmd. ommmv. oooom. .HI momovo mmmvfi. fiammv. fimfibv. mommfi. vmmmv. oooom. mum. I mvmmm. mommfi. .mommv. wfimmfi. hHHm¢. fimamw. oooom. Omb. I mammm. mmmav. vmmmv. ommmfi. vabfi. hvmmv. oooom° mum. I mobfim. Hommm. mmmavo Obfiqfi. fivmwvo mmmmvo oooom. com. I bmvmmo vomwm. mwmmm. mmmHv. mommqo baamfi. oooom. mum. I mooomo mammm. mwbdm. mmmhm. mmmflvo vdwmv. oooom. OmN. I mambmo Naomm. MOOOMo mmmHm. mmemo mmmav. oooom° mmHo I ooommo 000mm0 000mm. 000mm. ooommo 000mm. I o hmfimmo mmmom. bmmmao mommao bmmma. boamo. o mNH. bmmmao mebH. bmmmao NH¢NH° hOHmO. mmamo. 0 0mm. mwmbag omvma. mmHHH. boamoo mmmvo. mmmao. o mum. bmmmao mm¢oa. boawo. ommmoo mmHmo. bvmao. 0 com. mmomao boamo. mfiomo. omovo. mmmmo. mmOHo. o mmm. mmHHH. mammoo Nmmdo. omamoo mmmHo. mmmoo. o omb. mmfimo. bmflmo. mmbmo. wbmmoo mmmao. omeooo o mum. boamo. bmmdo. mmamo. mhamo. bfiMHo. .Hfimoo. o .H |lll ‘|I\|.l ‘ly! ll‘ “.II‘ d mAm .nc oneoqom UHewq m mqm<fi (O ..I. 3° 0 . . .. o . H a mug am: mmm oom men 0mm mNH o VV\mM Iwmlo.- ammmo I mslnan mmmmo.I Hasmoon mmmmoII aamsooI mosmoII mamno.u .HI iuonoi mmbwc I OamTZCI Hmcoaol NNHOHI! nbthII mdwmool 0mmmo.! mwmmcal mom. I .pwcaII mommo,I mosofl I mmmaa,I mHmHfiII ommsfi I srmofl I vfimoaos mocoHII on». I wfibwch mbflmoefl mmCQHII mNONHoI mmwma.! vomvH.I momma”! mmmmfiofl thmHoI mmw. I mm mm32 HmOBOCI HmHoQ.I mmmafl.l GOOmHII vNVhaoi mwvhaol mammaofl mmommol m. I sowvoII HmmmoII saves I mmmoA,I mfioqaos mmqomII mmmmm I vamHmII mmHONII mam. I mommrou mm~n9.e mwomaoi wmmbooi mNHHHII ammmaal mmficm.8 momvmoa mmHOm.I 0mm. I wmv OI! vmmaoui ommmc.I mvwmo.! wmmmO.I flmHQHII Hmmmmol mmmooo! mmmowol mNH. I O o O o O O O O o | HI .1 VA 0 ‘H . VA A0 I > OH Bummmmm mEHZ UumBmZZNmHBZ< WH b HH v ZOHBDAOW UHBNA46- 71 Y fi\ bc .075494 L11 ” 3hr):— | l 1 \r ‘T‘ ”L L 1 L L rpmn bc \\ 11 “~ .075494 L I I I {JV Fig. 12.--Boundary stresses for finite difference solution 72 ()6) 32¢ ay( 0x = - x y (3.73) I = - = _ xy 17;;( '_?f§) O = d( ): &2¢0 y W73? 7;?" It is shown in Appendix V that the function QB, for the case of cubic anisotropy must satisfy the compatibility equation 2 2 C 64$ + c:11 " 2C12‘244 ' c:12 24¢ + C d4d> = o (3.74) which for the special case c12 = C44 = cll/2 becomes 2392+): (94¢ +—-d—f$= O. (3.75) dx dxzdy2 dy Note that if the relationship C44 = %(Cll ’ C12) which holds for isotropic materials is substituted into equation (3.75) the biharmonic compatibility equation of isotropic elasticity is the result. The boundary values of the function 6b may be found by integration of the boundary stresses given by equations (3.66) through (3.71). This integration is carried out in Appendix VI. The boundary values of the stress function Qb and its normal derivatives which result from this integration are given by the following equations. On the boundary y1 = l 73 . 2 O¢ = bL c (1 + .59) tan—d 2x1 + D _ 51,1 2 (4 - 132)”2 (4 — 132)”? 4 2 .25C2 log (Xl + DX1 + l) (3.76) 25C x 2 - 2x cost + 1 ° 2 1 l = bL 5 x1 log 2 - CO x1 + 2x1 c057? + l .25C2 log (x1 + Dxl + l) + C2 2 1/2 tan (4 - D ) 2x12 + D . (3.77) (4 - 02)172 On the boundary x1 = l d .25C y — 2y cos + l l 7? y1 + 2yl cos” + 1 ([5 cl - .5DC2 23712 + D = bL - y tan - .25C y log (4 _ D2)1/2 1 (4 _ D2,1/2 2 1 (DC - C - C ) y + 2y cos + (yl + Dyl2 + l) + 28 c0672 2 log 1 1 In y1 - 2y1 cos” + (C2 + C2D - Cl) 1 2yl Slnn C1 + C2 -1 4 sin tan 2 + C2+ 241/2 tan ‘n l — y.l (4 - D ) 2 + D + C2 + DC2 ‘ C1 10 1 - cos + (4 _ D2)l/2 yl 8 c0572 9 1 + cos (C2 + C2D - Cl)“ - C (3.79) 8 sinyz 2 On the boundary y1 = -l 74 2 395 =bL c (1 + .50) tam—1 2x1 + D _ dyl 2 (4_D2)l72~ (4 _D2)172 25c lo (x 4 + Dx 2 + 1) (3 80) ' 2 g 1 1 ' 2 qb bL -.25C2 xl - 2x1 cosTZ+ 1 = -—————— x log + cos” 1 x 2 + 2x c057? + l l 1 .25C2 log (x1 + Dxl + l) - C2 2 I72 tan (4 - D ) 2 2x1 + D + C2 + DC2 - Cl log 1 _ coin (4 _ D2;I72 4 c0571 I + cosn (C + C D - C )n 4 sinTI 2 Noting the impossibility of finding an analytic solution of equation (3.75) which will satisfy the conditions (3.76) through (3.81), a finite difference approximation will be made. b) Finite Difference Approximation The difference approximation of the equation (3.75), referring to Figure 13a, is 149130 " “((31 +432 +953 +954) + ($5 ““957 +759 +9511) + .5(§/>6 “be ”1310 +9512) = o. (3.82) 31f the point 0 is one or two mesh spacings from a boundary, Cuie or more of the points 0 through 12 fall on the boundary Cir one mesh spacing outside the boundary. The values of<fi a1; points which fall on the boundary are evaluated directly fi?om the equation (3.77), (3.79), or (3.81). The value of 75 _l_ 8 2 6 K— 6 l9 3 o 1 [5 10 4 12 .___1_1_ (a) X1 8 2 I9 3 o | I ' ' ' l y1 1o 4 ' 1 ) Y1 ' __11 yl (b) Fig. 13.--Finite difference mesh for the approximation of the compatibility equation 76 at points which fall outside the boundary are, by the central difference approximation, assumed to equal the unknown value of (P at the adjacent interior mesh point plus the product of the normal derivative at the boundary as given by equation (3.76), (3.78), or (3.80) and twice the mesh spacing 6. On the lower boundary the sign of 6 is negative since yl is positive in the upward direction. Referring to Figure 13b, where point 0 is one mesh spacing from the side boundary and two from the top boundary, the finite difference equation (3.82) at 0 is " d '0 555.- 5555. >+52+535545 5.525 923: ’ ¢(X1)+§b9 +due to the last iteration was 1.82 x 10_12. This was believed to be sufficiently small to define convergence. The running time on the CDC 3600 computer for the program which determined.(b and.also stresses and displacements in the region was 4 minutes and 55 seconds. This Fortran program may be found in Appendix VIII. Values of the stress function at interior, boundary, and exterior points are shown in Table 9. It should be noted that a direct solution of this system of equations, which involves 120 members, is possible on the CDC 3600 if the quasi-diagonal property of the coefficient matrix is taken into account. (If the entire coefficient matrix including all zero values is input, the available computer memory is exceeded.) Because of the questionable accuracy of the direct solution, due to loss of significant figures, and because of the programing difficulties involved, the iteration method was used instead. A direct solution was determined for a mesh of 28 interior points. When single precision arithmetic (12 significant figures) was used, large errors appeared in the solution. However, a double precision arithmetic (24 significant figures) solution agreed exactly with an iterated solution, both having errors on the order of 10-12. Values of stresses at the interior and boundary FNDints were determined by the finite difference approxi- mations of the second derivatives of ¢). Referring to F'ig’ure 13a the equations from which the stresses at mesh point 0 were determined are 79 l I mwa.a OOO.H mow. Onh. mmw. oom. mom. 0mm. mma. o x H» vammoma.l mmmmmvH.I «Ommbvfl.l mmOhHmH.I Hb¢mmma.l mamwmma.l Ommmowd.l mmmmmwa.l omommwa.l mmH.HI vmbwmma.l vmmavma.l mmwmeH.I mwmmvva.l mmmmm¢H.I ommfivma.l vowammH.I mwbmowH.I mmabmmfl.l 0H0mm®H.I .HI mmmhvma.l mnOmOma.I mmmwmma.l hammova.l wbmmmvH.I bomnoma.l hwmmva.I bvmwbma.l wmoammH.I mmmbmmH.I ohm. I ObomeH.I Homomma.l mmmHOMH.I HmHNmMH.I Hmmmmma.l mammvwa.l mmvmbva.l mommomH.I mmOmNMH.I wmmwNmH.I omb. I Hmbwmaa.l ovMbmHH.I mvvamH.I whmmbma.l me0mma.l Hmwmmma.l mbqommH.I ovwvHvH.I mmbmmvH.I mamvmvd.l mum. I Havaboa.l bmmmHHH.I ebovnaa.l ommamHH.I mommeH.I mmammma.l ommvmmH.I OhNVOmH.I QO>©HMH.I mvaNMH.I com. I momhmmo.I vamamoa.l mbmmooa.l mmammoH.l mmHONHH.I bmav¢HH.I mmavwaa.l vmmmbHH.I mammmHH.I homflmHHII mbm..l mbmmHmo.I vamvmo.I mhmmwmo.l vmvmmmo.l Hmmmooa.l QHOONOH.I mmvmmoa.l momvaH.I HammVOH.I mmommoH.I Cam. I bmommmo.I mmmOmwo.I ¢w¢H0m0.I wmrabmo.l mmoammo.l amammwo.l mommmmo.l Hmmoomo.l N¢Hv0m0.I hHNmOmO.I mmH. I mmmvmho.l mmmvmho.l mmmvmbo.l mmmvmho.l mmavmbo.l mmmfimbo.l mmmvmbo.l mmmebo.I wmmfimho.l mmmvmbo.l o ommabwo.l vmwmmwo.l mavmvwo.l wmammmo.l mmwmmwo.l meOmwo.I mummawo.l mmmwomo.l fimnnowo.l owwvowo.l mmH. vomommo.l wmmwwmo.l Homvvmo.l mmmvmmo.l mmO©OmO.I ommmmvo.l vambvo.l mwmmmvo.I wmmmmvo.l bamhmvo.l 0mm. HhmHHmo.I mmmbbqo.I moawvvo.l cabwavo.l vwommmo.l mmbmwmo.l mvbmvmo.l mm00mmo.l HmHHmm0.I mmmbamo.l mum. wmmmmvo.l mbmmmmo.l mmbmmmo.l bmmmeo.I whammmo.l ombammo.l bmvmmmo.l oowmomo.l mwammao.l mmmmmHO.I oom. wmaabmo.l mmmmmmo.l «membmo.l oom0mm0.I maommao.l meHmHO.I wmmmHHO.I wmmmmoo.l MHHomOO.I vwmvboo.l mmw. bomHHmo.l wmvmmmo.l membowo.l mmbbmao.l memOHHO.I Hombwoo.l mmvamoo.l mnmmooo.l mommaoo. bmomaoo. Omb. maOmwmo.I vowbomo.l mammnao.l Ommooao.l oommvoo.l Ohmvooo.l ommmmoo. OBOMwoo. mamawoo. bbmrwoo. mew. NQOmmmo.I mmmbwao.l mammHHo.I vmmmmoo.l mammooo.l mww¢moo. mmmaboo. mammmoo. mmwbaao. fimammao. .H mommmao.l bmmwmoo.l momemoo.l mmmbOOQ. mommvoo. bmwmboo. vrvmmoo. mmdmaao. omeHHo. mmH.H AA\HHUQV ZOHBDAOm A0¢H0. Ohmao. mbmao. mmmmo. maamo. mmwmo. bmmmo. HHHVO. mum. I mmaao. 00HHO. thHO. 00¢H0. mvbao. vHHNO. mvvmo. mmmmo. ombmo. 0mm. I 0m000. 00000. mmwoo. emuoo. H0000. bmoao. wmmao. mmmao. mbmao. mNH. I 0 0 0 0 0 0 0 0 0 % A0 u H» OH Bummmmm mBHS A3’j - 2962,j + (phi) . Simplifying and noting that u 1 j = 0 because of symmetry 3 gives u3’j = €255 L¢22j+1 + ¢2.j-1 " '5¢3,j " ”5191.1" ¢2,j]° (3.94) For the point i,j the finite difference equation.is _ 8 ui+1.j‘ Sc'lTl L E¢i,j+1 ” (bid-1 ' °5¢i+l,j " '5¢1-1,j " qbiqj] + ui_1,j° (3.95) Using the equation (3.95) at successive points on the horizontal mesh line allows the calculation of all u dis- placements on that line. Use of the equations (3.93) through (3.95) on every horizontal mesh line will yield u displacements at all mesh points in the region. Values of u for the mesh shown in Figure 14 appear in Table 13. It is known that the value of the v displacement at the point {0,-1) in the x plane is zero. 1,Y1 V(O,-l) = 0 If the central difference approximation of equation (3.92) is written for the mesh point i = l, j = 1 the result is the following: r0 8 .H mam. oms. mmm. oom. mam. omm. mma. 0 ax » mmeom. ssama. mmosa. mesma. mamma. mmmoa. msmso. masmo. 0 .HI mmasa. mmmma. enmea. enema. maeoa. mmomo. oqmmo. mammo. o mam. I ended. momma. mmvaa. mmmmo. whomo. mwamo. mmaeo. maamo. 0 one. I mveaa. ommoH. mmomo. 66660. mommo. Homeo. meamo. msmao. o mmm. I mammo. soomo. mmmmo. mmmmo. mmmeo. mmemo. mammo. mmaao. 0 com. I 6mmmo. ommmo. emomo. cameo. wmmmo. mnemo. emmao. Hamoo. o msm. I memeo. mmmmo. mommo. mmemo. mmamo. mmmHo. mvoao. samoo. 0 0mm. I mHHmo. mmmao. emmflo. ommao. Hmoao. owsoo. cameo. mmmoo. o mNH. I o o o o o o o o o o lo I a» 09 summmmm maHs q1,2 +¢1,1) . 62 Since v(0,-l) = 0 this simplifies to 8 v1,2 " ”EELS [2732,2 ‘ '5¢1,3 ' °5¢1,1 '¢1,2]° (3'96) The difference equation at point i = l,j where j is an odd integer is 8 V1,j+1 ‘ ”NHL [29b2,j " °5¢l,j+l ‘ °5¢1,j-1 " ¢l,j] + V1,j—1° (3.97) Successive use of this equation for all even values of j will give the v displacements at all mesh points on the yl axis for which j is even. The finite difference form of equation (3.88) for the pOint i,j for the case C44 = Cll/2 is ) C V - V . u. _ u . 11 i+1,j i-l,3 1,j+1 1,3-1 2 (7’ 26’ + 25 _ i(¢i+l,j+l +¢i-l,j-l _¢i+l,j-l ‘¢1—1,j+1) L 462 SOIVing for vi+l j gives the equation 3 1 ax dx 0 O 89 TABLE 14 v DISPLACEMENTS FOR NUMERICAL SOLUTION (b, V I5 SYMMETRICAL WITH RESPECT TO y1 = 0) fl. 0 -.00465 .00716 .04167 .09655 .16885 - .25 —.00557 .00676 .04235 .09779 .16917 - .50 -.00739 .00603 .04485 .10239 .17147 - .75 *p.00843 .00717 .05037 .11148 .17910 —1. 0 .01605 .06082 .12379 .19190 Y x o .25 .50 .75 1. 90 .H mew. ems. mmm. com. mam. 0mm. mma. o Hx H» 6mmmo. mmmmo. ommmo. H66H6. mmmom. Hmmmm. Hmmmm. mnemm. oooom. .HI Hsmam. massm. mmmsm. camsm. mmomm. mammm. mammm. owomm. oooom. mam. I mammm. mmamm. memmm. 668mm. emOmm. Nmomm. oammm. mmmam. oooom. ems. I momma. ommme. mmome. smmme. mmaom. Hemom. smsom. mmmom. oooom. mam. I Hmomv. amovq. aemee. 66mme. Nmmmv. mmmae. ooflmv. momme. oooom. com. I 666mm. Hammm. smmmm. «H605. mmame. momqe. mammv. mamme. oooom. mum. I Hmmam. Nemam. memvm. Hmmmm. mmmmm. omamm. memmq. Hmmsa. oooom. 0mm. I mmmmm. bosom. Hmmmm. Nmmom. emoam. mflwmm. msmmm. mvflma. oooom. mma. I 000mm. 000mm. 000mm. 000mm. 000mm. 000mm. 000mm. 000mm. 0 o wsmom. mmmom. maoom. mmmma. mmmmfl. mmmefl. amsefl. mmmso. o mNH. masmfl. mmema. smOmH. memea. whoma. ommHH. Nmoso. mmmmo. 0 0mm. mwoaa. mmmoa. msHoH. mammo. samba. Hmmmo. mmmmo. whoao. o mum. mfimmo. msmmo. mmvmo. ammeo. mavmo. mvomo. oamoo. mmaoo. 0 com. emmao. ommao. memoo. meeoo. mmaoo.I Hemoo.I smsoo.I mmmoo.I o mmo. maomo.I mmflmo.I mvmmo.I eqemo.I emmmo.I mmOmo.I oammo.I mmmao.I o oms. Hemso.I masso.I mmmso.I onso.I mmmmo.I mammo.I Hemmo.I omomo.I o mam. mmmmH.I NmmmH.I mNHmH.I H64HH.I mmmoa.I Hmmmo.I Hmmmo.I Abomo.I o .H ”no ZOHBqum zDDZHezoo omZHmzoo mom ezmzmo) are of d). 2 y 2 0 y $1! The v displacements on the y axis may be found by an integration of equation (3.90) from point (0,-L), at which v = 0 to (O,y). Y v( ) - C11 «fliggd - C12 y — c 2 - c 2‘ (fix y c 2 - c 2 ll 12 -L 11 12 a ( ) _ acI5(-L) (3 101) y 3y ' After having found the values of v on x = 0 equation (3.88) may be rearranged in the form _é_x___1_82¢ -212. 0x" (M4 5x5y dy ° Integrating this equation from 0 to x gives X . 1 020:) 605(0) bu I -| = — —_ —’ - . Q 3. O The calculation of several values of u using equation (3.100) and a Simpson's two strip integration gave values about 2% higher than the u's determined by the method of Section 3.3c. Since this difference is rela- tively small, no attempt was made to recalculate dis— placements using the more accurate method. IV. DISCUSSION OF RESULTS 4.1. Results of the Discrete Analysis The variation of u on various vertical atom planess defined by i = const is shown in Figure 15. Note that the values of u in this figure and for all succeeding figures of this section are for the right half of the lattice. This is for purposes of comparison with the continuum analysis which applies to the right half crystal. For this reason u values in these figures are opposite in sign to the values calculated by the system of equations in Appendix I and displayed in Table 1. The index i = 18 still corresponds to the extra atom plane and i = 1 still corresponds to the side boundary, however, in this case the right rather than the left side boundary. Considering the atom plane immediately adjacent to the extra atom plane for which index i = 17, the u displacements of lattice points below the slip plane, j = 17, deviate only slightly from a value of .58b Inhile those above the slip plane differ very little from zero. Most of the transition from .58b to 0 occurs Tbetween the three lattice points j 16, 17, and 18. The 16 and 17 and between relative u displacements between j 97 Fig. 29— 27— 23— 21— 15— 13— 98 1"" ll l7 l3 u(b) 15.-—u displacement for discrete case versus j 99 j = 17 and 18 are .1774b and .2690b respectively. These large relative displacements are quickly distributed. On the vertical atom plane with index i = 13 the largest relative u displacement is approximately .05b. As the side boundary is approached the relative u displacement becomes more and more uniform until it becomes a virtually constant value of .025b at the boundary. The largest u displacement, .7lb, occurs at the lower corner while the smallest algebraic value of u, —.122b, occurs at the upper corner. An examination of Figure 16 which shows the variation of u on various horizontal atom planes defined by j = const also indicates that the region of large relative displacements is definitely limited to the region surrounding the lattice point i = 18, j = 17. Relative u displacements with values of .05b or greater occur only for certain pairs of lattice points lying in the region W 17 3 i 14, 21 E j Z 14. The antisymmetrical property of the u displacement is apparent after examining Figures 15 and 16. The value of u on the slip plane j = 17 is a nearly constant value of .27b and any two lattice points symmetrically located relative to the line j = 17 have approximately equal but opposite displacements relative to the displacement on the row j = 17. The positive slope of a curve in Figure 16 indicates that the distance between two adjacent lattice points has increased and hence the mutual bond force is tensile. Similarly, a negative slope indicates a decrease 100 Fig. 16.--u displacement for discrete case versus i 101 in spacing and a compressive bond force. In the region adjacent to the lower boundary the lattice is in hori- zontal tension. Moving upward in the lattice a region of compression is encountered, followed by a region of tension and then a region of compression adjacent to the upper boundary. It may also be noted that for any constant j the slope is either always positive or always negative so that every row is either in tension or com- pression and never changes from one to another. Figure 18 shows the variation of v along several horizontal atom planes. These curves indicate that from the vertical plane 1 = 13 to the side boundary v increases at an approximately constant slope for all values of j. In this region a horizontal plane of lattice points remains approximately straight but has been tilted upward, with i = 18 as the approximate pivot point. The general deformation of any particular horizontal row of lattice points involves an upward rotation of the row plus a varying tension or compression of the row. This tension or compression is maximum at i = 17 (next to the extra atom plane) and decreases in magnitude to zero at the side boundary. Most of the decrease takes place in the first four lattice spacings. A crude picture of the distorted lattice appears in Figure 21. The general description of the deformed crystal lattice neglects some less important local variations. The most pronounced of these is the general increase in v as the horizontal lattice plane j = 17 is approached from 102 j less than 17 and the decrease as j becomes greater than 17. This characteristic is indicated by the bulge which appears in Figure 17 around the horizontal plane j = 17, and is most significant at the point (17,17) where the v displacements relative to lattice points (17,16) and (17,18) are .059b and —.062b respectively. This abrupt departure from the otherwise gradual variation of v may be explained by an examination of the bond forces acting on point (17,17) in Figure 8b. The vertical component of the force between point (17,17) and point (18,16) is not approximately cancelled by a symmetrically located point at (18,18) (since [18,18] does not exist). This vertical component, which is directed upward since the force is compressive, causes the increase in v. The magnitude of the relative v displacements associated with the bulge in Figure 17 decrease quickly and on the vertical lattice plane i : 14 the greatest relative vertical displacement is down to a value of .02b. Distributions of lattice force components, FX and ny, which act across vertical sections in the lattice are shown in Figures 19 and 20 for several different sections. Since there are no external forces acting on the lattice these distributions of force must be selfwequilibrating. Summation of forces in the horizontal and vertical directions proved that the force resultant of the various distributions was zero since the resultants had a magni- tude less than 10-7ab. No calculations were made to determine resultant couples of the FX distributions, but 103 33F 31"— 29'— 27’— 25*- 15" 13" 11" v(b) Fig. l7.--v displacement for discrete case versus j 104 .5,— 04"— j = 17 .30 v(b) 02'— j = l \—j = 33 .1- 0 l I I l I I I I I 17 15 13 11 9 7 5 3 l i Fig. 18.——v displacement for discrete case versus 1 33 31 29 27 25 23 21 19 17 15 13 11 105 -.03 ‘002 "oOl O 901 F (db) X Fig. 19.—55FX versus j .02 106 33 — 31,. 29 — ///__.i = 5 9 _ 7 I_ 5 _ 3 _ 1 l I l -003 ‘002 “-001 0 .01 ny (ab) Fig. 20.-—Fxy versus j 107 a qualitative examination indicates that these couples are approximately zero. 4.2. Results of the Continuum Analysis The continuum analysis for the edge dislocation appears to give reasonable results for all system variables. The analytic portion of the solution, which solves the problem of the edge dislocation in an infinite medium, is very similar to the wellmknown isotropic solution given in Cottrell L2]. The superimposed numerical solution required to cancel the boundary stresses of the analytic solution comes very close to accomplishing its purpose. An exami- nation of Tables 17, 18, and 19 shows that the boundary stresses remaining after the superposition of the two solutions have magnitudes less than 7 x 10'4. The dis— tributions of 0X, 0y, and fxy (shown in Figures 26, 27, and 28) are such that the net stress resultants on various sections appear to be zero. This is necessary since no external loads are applied. Like the isotropic case discussed in Cottrell the solution contains a singularity at the origin. The u displacements on the line x = 0 are .5b for yl <:0 and 1 zero for yl >-0. At the origin itself u has the value (90 - 9‘9). The v displacement approaches negative infinity at points approaching the origin. However, large negative values of v occur only at points very near the origin. The v displacement at the point (0,.0625L) in the xy plane has the relatively small value of -.l395b. 108 Fig. 21.-~Diagrammatic representation of the distorted discrete model .75 P 109 l l l I I . J I Figs) 22.-"'11 O .2 .4 .6 u(b) displacement for combined continuum versus y1 110 Y], = -.25 O .25 .50 .75 1 Fig. 23.-~u displacement for combined continuum versus xl 111 .75” .50— .25~ Y1 - .25—- - .75.. \ -1. l \l J J 0 .1 .2 .3 v(b) Fig. 24.--v displacement for combined continuum versus yl 112 .3— .2'- Y1 = 0 v(b) , yl = i l .1‘” 0 + Y1 =1 ’05 -,1I— I I I 0 .25 .50 .75 1 x1 Fig. 25.n-v displacement for combined continuum versus x1 .75 .50 .25 113 .75 ‘01 O ox (bcll/L) for combined continuum versus yl .75 P .50 — 114 - .75 Y1 - .50 — ~ .75- _1. I I I -03 -02 "'01 0 01 TXY (bcll/L) for combined continuum versus yl 115 H x mamum> ESDCHucou CmcaQEou Mom ax 05. 00. _ _ _ xx p 0:0 oII.mN .00..“ 116 The values of stresses OX, Oy’ and Txy all approach either plus or minus infinity as the origin is approached. However, these stresses also have relatively small values at points near the origin. At the point (.0625L,0) Txy has the value -l.20790bc11/L and at the points (0, : .0625L) Ox and 0y have values of $1.20790bc11/L. For purposes of comparison of the continuum and discrete models the length .0625L in the xy continuum corresponds to the one lattice spacing, b, in the dis- crete material model. Therefore, the above mentioned points are located just one atomic spacing from the origin. Making the substitution b = .0625L into the values of stress given above results in values of stress in terms at c11 of 1.20790(.0625L)c11/L = .07549c11. This value is large compared with macroscopic values of the elastic limit which are in the neighborhOod "4 J of /10 ; however, it may not be large with respect to C11 atomic scale elastic limits. The values of stresses drop very quickly and a glance at Tables 17 through 19 shows that at points on the rectangle x = .25L, y = :.25L the stresses are all below .2756bcll/L = .01723c11. It should be recognized that the equations of the theory of elasticity are derived on the assumption that strains are small. This assumption is necessary for geometric linearization of the equations. It is possible that the size of the displacement discontinuity relative to the dimensions of the continuous body (The discontinuity 117 is one—thirtysecond of the length of the square body.) is such that geometric small strain is exceeded. 4.3. Comparison of Discrete and Continuum Analyses In general a remarkable similarity exists between the results of the continuum and discrete dislocation models. Comparing the graphs of the discrete values of ud in Figures 15 and 16 with the corresponding represen- tation of uC in Figures 22 and 23, it is apparent that the respective variations of u are practically identical. Table 20 shows corresponding values of 11C and ud in adja- cent columns. An examination of this table shows that while the respective variations of u are similar the mag- nitude of u is in general 10% to 20% greater than u . d c This is due, at least in part, to the fact that the repulsive bond forces between next nearest neighbors on the extra atom plane, 1 = 18, and the neighboring plane, 1 = 17, are greater than they should be because of the linearization approximation described in Section 2.4. These larger forces tend to increase the u displacements whereas the v displacements, because of an approximate cancellation of vertical forces, should be relatively unaffected by this error. However, the similarity between corresponding values of v for the two solutions is by no means as striking as that shown by the u displacements. A comparison of Figure 17 with Figure 24 shows that a great disparity in results exists, especially in region near the extra atom plane, 1 = 18. However, in 118 amas. mmmm. mmmm. mmom. 0mmm. mamm. mmmm. momm. a .aI mmmm. mmmm. .mmmm. momm. Noam. ammm. Namm. mwam. m me. I omaa. mmma. amom. mmma. emam. mama. mamm. amme. 0 mm. I ammm. mmam. mmmm. Nmmm. mama. mama. omam. mmba. ma mm. I ooam. comm. mmsm. comm. aomm. comm. mmmm. comm. .a o mesa. msma. mmma. mama. mmmo. moso. maao. ammo. am mm. mma . ammo. Noam. memo. asao. ammo. mmoo. maoo. mm om. ammo. momO.I mmmo.I mmmo.I ammo.I ammo.I mmao.I mmao.I mm ms. 8mma. mmma.I meoa.I mmoa.I Eamo.I mmmO.I memo.I mom0.I mm .a aIa aIax mIa m.Ia maIa mm.Ia maIa mma.Iax a a s 0: on Us 0: Us Us 05 Us Q mo mBHZD 2H mBZHOm wZHQZOmmmmmOU B4 Bzmsz EDDZHHZOU 324 mBmmumHQ 119 the region of the discrete model which lies between Ihe vertical plane i a 13 and the side boundary, i 2 1, the vertical variations of v are very similar. The variation of v along horizontal planes, j = constant, in this region are also nearly identical. Figures 18 and 25 show these variations graphically and indicate an approximately constant change in v as the side boundary is approached. This constant change is nearly the same for both models. The major difference then between the values of v for the two models in the region between i = 13 (x1 2 .25) and i = 1 (x1 = l) is that vd is greater than vC by an amount which varies from .16b to .2lb. Since the discre- pancy between respective values of v originates in the region adjacent to the extra atom plane, it is probable that these differences are primarily due to the fact that the dislocations were introduced into the two material models by slightly different methods. In the method used to form the continuum dislocation, the displacement discontinuity, b, across the cut on the negative y axis is caused by normal stresses acting across this cut. The faces of the out are assumed to be free from shear stress. In the discrete dislocation tangential forces are transmitted between the extra atom plane, i = 18, and its adjacent plane. This difference may well be the cause of the discrepancy in corresponding values of v for the two models. It is possible to construct a continuum edge dislocation by a method which corresponds more closely to 120 that described in Section 2.2 for the formation of the discrete dislocation. This procedure first requires the cutting of a notch of width b and height h in the continuum. Then a strip of material of width 2b and height h is elastically compressed to a width b and inserted into the notch and allowed to expand. It is necessary to make some reasonable assumption regarding the Poisson effect on the height, h, of the strip during the compression of its width, and also to choose a reasonable condition of tangential continuity across the notch-strip interface. It is obvious that the solution of such a model is much more difficult in a mathematical sense than the model analyzed in Chapter III. Figures 26 and 27 show the distributions of Ox and TXY on various vertical sections of the continuum model. Their discrete analogs, FX and ny, are shown in Figures 19 and 20. Qualitatively, the distributions of corresponding values from the two solutions are almost exactly identical. In order to compare the magnitudes of discrete and continuum quantities at corresponding points, it is necessary to convert the respective values to similar units. Suppose that at a particular point Ox has magnitude A and its discrete counterpart FX has a value B. All stresses are written in terms of the factor bcll/L while discrete quantities include the factor ab. These factors include the dimensions of the respective quantities bc 11 F. = Babe (402) 121 If FX is assumed to be uniformly distributed over an area one lattice spacing square, a quantity which is dimen- sionally similar to stress is the result F X _ .95. 7 — B b. (4.3) b Geometric similarity of the two models requires that and in Appendix IV it is shown that elastic similarity requires that 11 (4.5) bc11 F bc x 2 11 E7 = B L = 8B L . (4.6) 16 TherefOre in order to compare the magnitudes of Fx and nyImith oX and Txy’ the latter should be multiplied by the fiactor eight. Table 21 gives values of discrete and contiJTuum stress at corresponding geometric positions in adjacent columns. An examination of this table shows that the magnitudes of corresponding quantities are quite similar. In summary it may be said that the u displacements for theItwo solutions are very nearly the same. If the linearization error of the discrete solution could be Cornazted, the discrete values of u would be reduced and Similadrity would most probably be improved. It is notable that tile similarity between uc and ud extends almost to the exfl:ra atom plane. A comparison of uc values on x1 = 122 TABLE 21 CONTINUUM AND DISCRETE STRESSES AT CORRESPONDING POINTS (bcll/L) F F X X OX 1:? Txy -71b yl j x=.25 i=13 x=.25 i=13 l. 33 -.1588 -.1231 .0004 .0275 .75 29 —.O450 -.0176 .0406 .0445 .50 25 .0811 .1023 .0712 .0649 .25 21 .2575 .2345 .0261 -.0411 0 l7 0 -.0106 -.2756 -.2310 - .25 13 -.2575 -.2446 .0261 -.0308 — .50 9 -.0811 —.1119 .0712 .0858 - .75 5 .0450 .0558 .0406 .0835 —l. l .1588 .1271 .0004 .0395 F F X X OX I)? Txy —§Xb yl j x1=.5 1:9 x1=.5 i=9 1. 33 -.0927 -.0679 .0007 .0182 .75 29 -.0126 -.0384 .0493 .0452 .50 25 .0647 .0644 .0409 .0253 .25 21 .0682 .0545 -.O489 -.0518 0 l7 0 .0044 -.1034 -.0891 - .25 13 -.0682 -.0610 -.0489 -.O436 - 050 9 -0064? “00593 .0409 00387 - .75 5 .0126 .0127 .0493 .0529 -1. 1 .0927 .0612 .0007 .0172 123 .125 with u values on i = 15 (see Table 20) shows that d differences between 11C and u on a vertical plane only d two lattice spacings from the extra atom plane are no larger than for points on the side boundary. Respective values of v for the two solutions show a much greater disparity than the u values. This dis- parity is believed to result from the fact that no shear stress acts on the sides of the cut in the continuum dislocation while in the discrete dislocation shearing forces are transmitted between the extra atom plane and its neighboring plane. In the region .25 5 x1 5 l of the continuum model which corresponds approximately with the discrete region where 13 2 i 2 1, the relative v displace- ments for the two solutions show great similarity. This situation is certainly a manifestation of the principle of Saint Venant which states that the effects of different but statically equivalent boundary stresses are virtually identical except at points near the boundary. Figure 29 pictorially illustrates three regions of similarity common to the two dislocation models. Figure 29a indicates with cross-hatching the region in which the u displacements, both relative and absolute, for the two models show a particular similarity. This region excludes only the strip of material lying within a distance 2b = .125L of the vertical axis of symmetry. Figure 29b shows the region in which the relative values of vC and v closely resemble one another. Figure 29c d 124 Y1 .125 = 2b _~L:[°25 = 4b (a) Region in which u (b) Region in which relative Ylil— and u are similar values of v and v are d d similar Y1 .1875 = 3b 7/1 N N '*— ‘ .0625 = (c) Region in which relative displacements of both the continuum and discrete solution are less than .05 .1875 x .1875 r--1 Fig. 29.--Regions of similarity for the continuum and discrete models (cross-hatched) 125 shows the region in which both the relative displacements for the discrete solution and the strains for the con- tinuum solution are all less than .05 or 5%. The appro— ximate values of the continuum strains were determined by scaling slopes from Figures 22 through 25. The signi— ficance of the value 5% is that for a relative displace- ment of 5% or less from the equilibrium atomic spacing a linear force displacement relationship is approximately correct. Johnston and Gilman, in tests of single crystals of lithium flouride reported linear elastic strains of up to 5% [13]. Neither of the two models can be expected to accurately represent the configuration of a real edge dislocation in the non-crosshatched region of Figure 29c, simply because the linear force (stress) relative dis- placement (strain) approximation is not true for relative displacements (strains) of the size encountered in this region. It is certainly apparent, however, that the dis- crete solution, even though it predicts linear relative displacements of nearly 60% for certain lattice pairs, is much closer to reality than a continuum solution which predicts infinite displacements, strains, and stresses at a particular point. Stresses and lattice forces on several corresponding vertical sections are practically identical both with regard to distribution and magnitude. The comparisons between stress and lattice force were all made on vertical sections which lie in the cross-hatched portion of Figure 29b. In this region relative values of u and v for l I: 6 both solutions .xe Very similar and they are also swull (under 5%). Since stlesses and lattice forces deptrd directly on relative displacements, and since the acturacy of these quantities requires that the relative displace~ ments be small, good similarity between lattice forces and stresses is assured in this region. The most surprising result of the comparison of the discrete and continuum solutions is the great simiw larity of all system variables at points which are very close to the center of the dislocation at x1 = O, y1 : O. Cottrell has estimated that the region to which the elastic dislocation solution applies lies outside a cylinder of radius lOb centered on the point (0,0) [23. Comparison of the discrete and a continuum solutions here indicates that they show great similarity at points which lie only 4b from the dislocation center. If the discrete solution is accepted as giving a reasonable description of an edge dislocation, this indicates that the elastic dislocation solution may be accurately applied at points far closer to the dislocation center than the previously accepted limiting distance of lOb. V. POSSIBLE EXTENSIONS OF RESEARCH In Section 4.3 two causes for differences in the results of the two solutions were mentioned. The most obvious extension of the work of this dissertation is to eliminate these two causes. A technique to construct a continuum dislocation in a way which approximates the method used in Chapter II to produce the discrete dislocation was described in Section 4.3. An actual theory of elasticity analysis of this problem is difficult, perhaps too difficult to jus- tify an attempted solution since it is questionable whether the tangential effects between the strip and the notch may be made similar to the analogous discrete effect. Similarity in the construction of the two dis- locations may also be achieved by creating a discrete dislocation by the method described in Section 3.1 for the creation of a continuum dislocation. This involves the insertion of a rigid strip of material of width 2b and height sb (see Figure 5) into the gap between i = r and i = r + 1. It is then assumed that no tangential forces are transmitted between the rigid block and the lattice points on i z r and i = r + l. 127 ~ 128 Equilibrium equations may then be written for each lattice point with ui,’ and vi,j as unknowns at all points except those in contact with the rigid block. Here the u displacement is a known value, .5b for the symmetri- cal case, and the normal force between the block and the lattice point is unknown. The vertical displacement is the other unknown associated with such points. If the discrete dislocation is formed in this way, both of the conditions mentioned in Section 4.3 as contributing to the differences in the solutions would be corrected. If the method described in Section 2.2 for the formation of the discrete edge dislocation is retained (The author believes that this is physically a more realistic methodd, then an attempt may be made to improve the poor approximation involved in linearizing the forces between next nearest neighbors on the extra atom plane and the adjacent plane. As a first improvement these forces may be left in the non-linearized form. Forces in the rest of the lattice may still be expressed in linearized form since relative displacements there are small enough so that the linearizing approximation is not seriously inaccurate. Solution of this mixed system of linear and non-linear equations may then be attempted by an iteration technique. If convergence occurs, some improvement in results is expected. If the iteration does not converge, the attempt would have to be abandoned. Another possible improvement might be to assume a simple Lennard—Jones force potential between lattice 129 pairs on the extra atom plane and the adjacent plane while retaining a linear force (linearized in terms of u and v) elsewhere. Here again, a solution must be attempted before there is any assurance that it can be found. In Kittel it is pointed out that the central force condition c12=c 44 is not true for most of the common metals such as copper, aluminum, or iron [10]. A discrete model more nearly approximating a metal must take into account forces which are transverse to the line joining the individual lattice points. More complex edge dislocation problems might be attempted either with the model used in this dissertation or with an improved model. These problems may include a non—symmetrical case of the problem solved here or a lattice containing more than one dislocation. The limiting factor in attempting a solution of more involved dislocation problems is that a reasonably large crystal lattice is necessary in order that the solution have any physical significance. The large lattice, of course, involves a large number of unknowns, and the solution of such large systems, even if possible, requires a prohi— bitively large amount of computer time. The convergence of the iteration for the problem considered here, which involved 1139 unknowns, took 30 minutes of computer time. It may be possible to use a composite material model to describe the more complex dislocation problems. 130 Since the results of the continuum and discrete analyses are very close except in the region near the dislocation, a model consisting of a continuous region with discrete lattices in the vicinity of dislocations may be a possibility. Departing completely from the consideration of dislocations in crystal lattices, the similarity of results for the discrete and continuum models indicates that it may be advantageous to directly simulate the elastic continuum by a ball and spring system. This technique has been used for analyses, usually dynamic, of such complex composite structures as building frames and airplane wings but has found little use in continuous bodies. If this simulation could be done more or less by inspection, an advantage would be gained over the accepted practice of formulating the problem with the partial differential equations of elasticity and then approxi- mating them by difference equations. This technique has been used in the solution of problems involving torsion of prismatic bars. A membrane analogy is used initially and then the stretched membrane is directly simulated by a net of nodal points connected by stretched strings with pressure loads concentrated at the nodal points. This direct simulation may be especially advantageous for materials with complex stress-strain relationships. It may also be possible to simulate visco-elastic materials by nodal points connected by means of spring—dashpot combinations. APPENDIX I EQUILIBRIUM EQUATIONS FOR THE DISCRETE MODEL IN A PARTICULAR EXAMPLE A. For the case a = B m = 34 n = 33 r = 17 s = 17 (Parameters m, n, r, s are defined in Figure 5 .) The equilibrium equations (2.10) through (2.13) and (2.24) through (2.49) have the form 2 5 i 5'16 2 S j S 32. z}%(::ti -4ui,j + ui+l,j + ui—l,j + °5(ui+l,j+l Vi+l,j+l + ui+l,j~l ‘ Vi+l,j-l * ui~1,j+l ‘ Vi-l,j+l + ui~l,j~l + Vi~l,j-l) ‘ 0° ZF&,= O --»4vi’j + Vi,j+l + v10.“1 °5(ui+l,j+l Vi+l,j+l " “1+1,j~1 + vi+1,j~l ‘ ui—l,j+l + vi—l,j+l + ui-l,j~l + Vi~l,j~1) 2 0° 131 Ein II ll 0 4| 1. I1°5u1,l + “2,1 + .5(u2,2 + --1.5v1,1 + V1,2 * '5(u2,2 + 1 j s 32 -Qul, + u2,j + '5(u2,j+l + V2,ju1 O. -3Vl,j + v1,j+l + Vl,j-l + - u2,j-1 + V2.jwl) 0' l 33 -1.5u1,33 + u2’33 + 5(u2,32 V1,33) 0° ~--l.5v1,33 + v1332 + .b(v2,32 “1,33) 2 0° 1 5 16 33 -3ui,33 + ui+1,33 + ui~1,33 V1.1,32 + ui-l,32 + Vi-1,32) —2vl,33 + V1.32 + 05(Vi+1,32 vi-1,32 + ui-1,32) 2 0° VZ’Z __ 'Vl,1) :: Oo V2,2 .- 111,1) = O. V2,j+l + u2aj’1 - 5(u2,j+1 + V2,j+l + 05(ui+1,32 .- = O ‘ ui+1,32 + O 133 -U (-1 5‘ J17,32 5(v + l7,j+1 + l6,j—l ) + v17,j-1 ' u18,j+1 7-.- O. '5(ul6,l9 - ,17 * v16,17 .13246b 19 + V17,17) +V ,17 16,17 1 = 17 j = 33 ‘4u17,33 + u16,33 + ' u16,32 + Vl6,32) = O ‘2V17,33 + v17,32 + ° V16,32 + ”18,32) = 0' i = 17 19 s j s 32 -5ul7,j +016,j + .5( u16,j+1 ‘ v16,j+1 + u‘ u17,j-1 ‘ Vl7,j~l) ‘ 0 -4Vl7,j + 1.5(v17,j+1 °5(-ul7,j+l + Vl6,j+l u16,j-1 + ”17,j_1) 1 = 17 j = 18 ‘5°2u17,18 + u16,18 + u17,19 * v17,19 + u16 °4 H ’U N H, OJ X N ‘< 0.5 L< ll 3, [.4 "O H] 52 H ——-7 = A f 0)< 2 4 82v " W = A210f ° Substitution of these derivatives into the equations (1) and cancellation of the common factor f" shows that the equations (2) are solutions of (1) if A1, A2, and p satisfy the equations Al(c11 + c44p2) + A2(c12 + c44)p = 0 (3) 2 Al(cl2 + c44)p + A2(Cllp + C44) = 0. If Al and A2 are non-zero the determinant of their coefficients must be zero. 2 C11 + c44p (cl2 + c44)p| = c c p 2 ll 44 (c12 + c44)p (Cllp + C44. 2 2 + (c + C44 11 2 2 2 c12 ‘ 2C12C44 ’ C44 )9 + c11C44 = 0 (2c c + c - c ) p _ 12 44 12 11 p + 1 = 0. (4) C11C44 Solving the characteristic equation (4) for p2 gives 2c c + c 2 - c 2 2c c + c 2 — c 2 2 _ 12 44 12 11 i [( 12 44 12 ll )2 _ p I 2c c 26 c 11 44 ll 44 111/2. (5) Let 2c c + c 2 ~ c 2 12 44 12 11 = C 2C11C44 Then equation (5) becomes p2=4-(€2 «1):L (6) For the case of real cubic crystals Eshelby [5] points out that 42 < l (for the example in Chapter III 42 = 1/16), in which case p2 is a complex number p2 4+ 1(1 -§2)1/2 (7) p2 §-1<1 - g 2)1/2. (8) Referring to Figure 30a the equation (7) may be written in polar form p2 = cos(27? + 2nn) + i sin(27? + 2nn). (9) The angle 7? is defined by the following equations: 2C C + 2 - 2 cos 27? = g -1 12 44 C12 11 7?=-l co s-léz =-§ cos . 2 2C11C44 142 2 | imag Ip 2 l imag 2 21/2 l (1.; ) 272 : 5 [real 7 I real (a) 1(192)> o (b) I(p2) < 0 Fig. 30.--Complex representation of p2 Applying DeMoivre's theorem to equation (9) to find p gives p = cos(T? + nn) + i sin(T? +nn). The two independent values of p correspond to n = 0 and n = 1 respectively p(l) 2 COST? + i sinTz (lO) p(2) = cos(‘n + n) + i cos (7? +n) = -cos7? - i sin72. (11) In exponential form equations (10) and (11) become 1 p(1) = e T? (12) i I (D H. 43 p(2) . (13) 143 A similar analysis for the case when p2 is given by equation (8), with reference to Figure 30b, gives p2 = cos (—27? + 2nn) + i sin (—27? + 2nfi) p=COS (-7?+ mt) + i sin (~77 + nTI). For n = 0 p(3) ; cos (-7?) + i sin (-7?) = c057? - i sinTz. (14) For n = 1 p(4) = cos (-7? + n) + i sin (—‘n + n) = -cos7z + i sinTZ . (15) Expressing equations (14) and (15) in exponential form gives —i 13(3): e 7? (16) —1 p(4) = -e n . (17) If the constant Al is arbitrarily chosen to equal unity the constant A which will henceforth be called A, may 2 3 be found by solution of either of equations (3) for any one of the four values of p. c c 2 _ 11 44p(i) "F LC12 + C4433(1) + A 1 = 1, 2, 3, 4 (18) (i) The displacements in equations (2) may now be written as a linear combination of the solutions corresponding to each individual value of p u = f[x + p(1)y] + f[x + p(2)yj + f[x + p(3)y] + f[x + p(4)y] (l9) 144 V = A(l)f[x + p(l)y] + A(2)fLX + p(2)y] + A{3)fLX + It is noted, however, that p(3) = e-in and p(4) = —e-in are complex conjugates of the values p(l) = ein and p(2) = -ein respectively so that the f[x + p(3)y] and f[x + p(4)y] are complex conjugates of fo + p(l)y] and f[x + p(2)y] respectively. Similarly, A(3)fo + p(3)y] and A(4)fo + p(4)y] are complex conjugates of A(l)f[x + p(l)y] and A(2)fo + p(2)y] respectively. The imaginary parts of equations (19) will therefore be equal and of opposite sign and will cancel while the real parts are of like sign and will add. (It is physically necessary that u and v be real functions.) In consequence of this only the two roots p(l) = e177 and p(2) = -ei7? will be consi— dered to be independent, and only the functions corres- ponding to them will be considered in the solution. The imaginary parts of u and v and of any variable derived from u and v, such as stress or strain, will be ignored since they do in fact cancel. The general form of the displacements for any boundary value problem which satisfies the conditions of plane strain and cubic anisotropy is C II Re f[x + ein y] + f[x — e17? y]} < n Re or 145 Re u : _ (20) V = Re (INDfLZu)J + A(2)1c'[z(2)J where from equation (18) c e-in + c e177 11 44 Am = “(2) = ‘ c + c (21) 12 44 and 2c c + c 2 - c 2 1 -l 12 44 12 11 7?: 7 cos 2c c . (22) ll 44 The form of the function f is determined by the boundary conditions for the specific problem. Consider the case in which the boundary condition involves a dis— placement discontinuity. The evaluation of the stresses from the displacements in the equations (20) involves expressions of the type . 81f x 11 dx 12 dy' The derivatives in these expressions contain the terms df[z df[z 1 J (1) and (2) . (23) dz"(1) ”2(2) Because of physical considerations the stresses must be single valued and continuous. This condition will cer- tainly be satisfied if the derivatives of (20) are analytic in those portions of the 2(1) and z ) planes (2 respectively to which the region of interest of the xy plane maps. If these derivatives are analytic in some portion of their respective complex planes, they may be 146 expanded in a power series about some point in the region of analyticity. Suppose that the origin of xy coordinates is the point at which the displacement discontinuity ends. For strictly physical reasons the stresses at this point must be infinite. This point is therefore a singular point in the region. If the point (0,0) is the point of expansion in the 2(1) and z(2) planes, the expansions of the derivatives (23) must be of the Laurent type since (0,0) in both these planes is an isolated singularity. Expanding df[z(l)]/dz( in a Laurent series about the l) origin gives dez ] (l) 1 1 dz = . . . + G_2 -———§ + G-l z + GO + 612(1) + (1) z (1) (l) G22(2) + . . . (24) Integrating this expression gives f[z ] - — G 1 + G 10 z. + G 2 + (1) ’ ° ° ’ -2 2(1) -1 '9 (1) 0(1) 2 G 2 i-é'fi—El- o o o (25) All terms in equation (25) are continuous and single valued in the 2(1) plane except the log 2(1) term which is infinitely many valued. The many-valued function may be considered a single-valued function by introducing a branch line in the z ) plane so that no circuit of the (1 origin may cross the branch line. In such a case the value of log 2(1) on one side of the branch differs from its value on the other side by the amount 2Wi. This 147 discontinuity across the branch line is exactly that property which is necessary to satisfy the discontinuous displacement boundary condition; therefore D(1) ”-1 109 Z(1) = ‘EI‘ 109 Z<1) f[z (1)J 13(2) (26) f[2(2)] ’ ”—1 109 Z<2) = 2E1‘ 109 Z(2) is chosen for the particular solution. APPENDIX III EXPANSION OF LOG 2 AND LOG Z INTO (1) REAL AND IMAGINARY PARTS (2) Expanding log 2‘ 2) into real and imaginary parts gives w(2) = r(2) + is(2) = log 2(2) 2 log (x - ei7zy) = log (x - y cosT? - iy sin7?) = log (x. + iy ) = I! £5109 (x"2 + y"2) + i[tan‘1(Y—,7) + 2th1 =%10g (x2 + x y2 - 2xy c057?) + :I_[t<':1n-l(my sin7z ) + 2kfi1 (l) x - y cos ' The multivalued log function is being used to represent a single valued but discontinuous displacement. Which branch, then, of the equation (1) will be used to repre- sent the single-valued function? The equation (1) may be Q ' ' thought of as a mapping of the xy plane onto the x' y plane onto the r(2)s(2) plane. The Figure 31 shows the mapping of the negative y axis from the xy plane to the x y plane and then to the r(2)s ) plane in the heavy (2 black line with the arrow pointing in the direction of increasing absolute value of y. The positive y axis is shown in the dotted black line. Note that increasingAP in the xy plane corresponds to decreasing w” in the . 0 x 'y' plane. The negative y axis maps onto the infinite 148 149 (a) xy plane (b) x"y" plane 5(2) k = 1 77+27T ._ ______ L. __________ —11- .. -— 71+7T k = O N\\\\\\\\\ ‘ \ NN\ WNN - 1" (c) r(2)s(2)[w(2)] plane Fig. 31.—-Mapping of the xy plane onto the w(2) plane by the function w(2) = log 2(2) 150 number of lines in the r(2)s(2) plane for which _ " _ -1 sin 1 _ , 5(2) *' \y - tan 358%- + Ck”: -- 72 + 2k“. The entire xy plane maps on the strip between any two successive values of k. The mapping may be made single valued by arbitrarily choosing one strip to which the xy plane maps. If the line 9!: -n/2 in the xy plane is arbitrarily chosen as corresponding to the line 5(2) :7? (k = O) in the r(2)s(2) plane, then since \y" decreases as 9! increases the line Y: 37r/2 in the xy plane corresponds to the line 5 7?- 2n (k = -l). The (2) : branch of the r(2)s(2) plane to which the xy plane maps is crosshatched in Figure 31. In a similar manner expand log 2(1) w”) = r“) + 15(1) = log 2(1) = log (x + e177 y) = log (x + y c057? + iy sin7z) =«%1og (x.2 + y'2) + ' . i[tan_l(-”-/T) + 2km] = %:1C>g(x2 + y2 + 2xy COST!) + X 1( Y 51”” ) + 2th]. <2) 1_[tan X + y C057? The mapping from the xy plane to the x'y' plane to the r(l)s(l) plane is shown in Figure 32. Here an increasing Nicorresponds to increasing HJ'. The negative y axis is mapped onto the infinite number of horizontal lines -sin' ' -l 5(l) = ”:10 = tan -cos +2k1t= 77+TE+2k7t and the entire xy plane maps onto the horizontal strip 151 y \ __3_7_r " \1/ ‘11-. H ‘i’ X 71 X. \\\»~—«//<:6=‘“%; Y \ifl'=7?+3fl' (a) xy plane (b) x'y' plane S(1) / / k = 1 // ////" 1 a , / 73+” 177 . r __ ' (1) 77 W k = -1 (c) r(l)s(l)[w(1)] plane Fig. 32.-~Mapping of xy plane on the w(l) plane by the function w(1) = log 2(1) 152 between any two successive values of k. To make the mapping single valued the line 4!: -n/2 in the xy plane will be arbitrarily chosen to correspond to 5(1) = 72+ n (k = O) in the r(1)s(l) plane in which case the line 4!: 3n/2 corresponds to the line 5(1) = In + 3n (k = 1) since increasing 1y corresponds to increasing 9!'. The branch between k = O and k = l is crosshatched in Figure 32. As an example of the evaluation of the functions log 2(1) and log 2(2), find the points in the 2(1) and 2(2) planes to which the point x = .5 and y = —.25 maps, for the case sin n _ .790569 cosTZ ...612372 T] = .911738. These values correspond to the case cl2 = C44 = cll/2. 109 Z(1) = r(1) + is(1) 1, , 2 2 1 . — r(l) = Elog(x + y + 2xy 6657]) = 516gL.25 + .0625 + 2(.5)(-.25)(.61237)J _ 1. , .c g 1 _ - , 1 - _ 17(1) 45109..15841) - '12“( 1654256) — .92128 tar ' _ y sin _ ~.25(.79057) _ -.79057 ‘ I x + y cos 7 .5 w .25(.61237I I 1.38763 tan Y' = -.56973 O U 0 This angle 4! is in the fourth quadrant of the x y plane )1], = —.51787 rad 5(1) = 2n - .51787 = 6.28319 - .51787 = 5.76532 rad 153 log z( = -.92128 + 5.76532i l) 109 Z(2) = r(2) * is(2) 1 2 2 l r(2) = Elog(x + y — 2xy c0572) = 710g[.25 + .0625 - 2(.5)(-.25)(.61237)] = %log(.46659) = %(-.76232) = -.38116 " _ y sin _ -(-.25)(.79OS7) _ .79057 tan\y _ x - y cos _ .5 — (—.253(.61237) - 2.61537 tan EV' ' = .30263 ' ' ' lies in the first quadrant of the x y The angle \y plane Y" S(2) .29387 rad .29387 log 2(2) 2 ~.38116 + .29387i. APPENDIX IV SIMILARITY OF DISCRETE AND CONTINUUM MODELS The response of the discrete material model to mechanical disturbance is characterized by the constants a and B, while the response of the continuum model is 11’ c12, and C44. In order to compare the response of these two models to the same determined by the constants c type of stimulus these models must be made in some sense similar. Depending on how this similarity is defined a relationship will exist between the constants of the two models. The similarity between the two models will now be defined and the relationships between the two sets of constants will be determined. Following the example of Kittel [10] consider the x component of the force equilibrium equation for an atom which is surrounded by a normal complement of neighbors. a4 u 02C» m u o 157 Substitution of these derivatives into equation (4) gives fidzu = cos2@————-§a u + 2 sinGcosGT—Tazu + sin2G——§d u (5) or ox X y dy and for G = 45° this equation becomes é2u __I 6211+ dzu + 1 <32u (6) drE - 2 Ox: dxdy 7 dy o In a similar manner since 6x x=-s sinG as=-sin@ y = 5 cost) ‘33; = cos@ 2 2 2 2 d u . 2 d u . a u 2 d u —-—2- = Sin 6 . - 2 Sinecosg + cos 9 ds 6x2 OX5), O—y-g and for 6: 45° 62u : l, @2u _ OZu + l <52u (7) as2 2 ax? dXOY Ox: Equation (2) may now be rewritten in terms of x and y derivatives as ézu 2 2 1 (321.1 (52?; l 6211. 2 l 52v b“32*b5‘?‘52+m+3 .yz’ +bB(3‘é‘;.7+ (32v +1 (3%) + 102%; 52.. _ 2u .l 52..) _ 2 2 2 2 l d v 6v 1 a v b (- -- +- f.) = 0- 62 (3X2 3x y 2 (3Y2 2+bBOZ+2bBW=Oo (8) 158 The x component of the static equilibrium equation for plane strain in an elastic continuum with cubic anisotropy is 32.. azu 32., C11 W + C44 $17-2- + (C12 + C44) m = O. (9) Dividing equation (8) by b3 to achieve dimensional similarity and then comparing with equation (9), it may be seen that the two equations are exactly the same if their respective coefficients satisfy the relationships C = M2— 11 b _ E C344 ‘ b (10) 2 C312 + C:44 " "E or C = 3‘9. 11 b The condition c12 = c44 results from the central force assumption for the discrete model [see Kittel]. For the special case a = B Cll/2 = c12 = C44. - (11) It should be noted that the relationships (10) and (11) do not conform to those which describe an isotropic solid. In Love [1] it is shown that the relationship between the elastic constants for an isotropic material is 1 C44 —-§(c11 - C12)’ (12) But by the relationship (10) c must equal C44; therefore 12 159 Substituting this relationship into equations (10) gives 3c12 = (a + B)/b C12 = B/b BB/b = (a + B)/b 28 = a. (13) Therefore for a central force model elastic isotropy corresponds to the specific relationship (13) between the discrete force constants a and B. APPENDIX V FORM OF THE STRESS FUNCTION EQUATION In Chapter III it is shown that for the case of plane strain 0' X 0 Y xy _ a function QD exists such that (32¢ dyz (32¢ d z (32915 X ' dxdy° Expressing the stresses in terms of displacements leads to the equations .612 2.1___§ 11 6X 12 ()y ay (92 for (l) 2 av... 32: 49 11 dy 12 x "‘"2'dx dV' c5u C52 C44(‘53‘<+_a‘i7) 2" dxdy the case of and (2) for cubic anisotropy. Solving equations du/ dx and dv/dy gives anb 2 c5x2 _ c:11 c3295_ C12 - c 2 - c 2 <5 2 c 2 - c 11 12 y Y 11 __ c311 d2¢_ C12 _ c 2 - c 2 ()x2 c 2 - c 11 12 11 160 (l) (2) (3) (4) (5) 161 Differentiating equation (3) with respect to both x and y gives —————-a2-a3u + ()3v_ = - l a4

. 4 2 log(xl + DXl + l) + K7 (7) 2 l m x 2 l. a 9D _ G _ _t_> c 1 L2 (5 2 ‘ y — L 2 4 2 x X + DX. + l 1 1 1 d2 1 - x12 ‘5‘“? = ch2 4 2 x x. + DX + 1 1 l l (34) -.2SC2 x12 - 2xl cos]? + l —5-)-(—- " bL COS log 2 4' K8 (8) l H x1 + 2xl cosn + l 165 2 4) b ~025C2 x1 - 2x1 c057? + 1 = L X log . + cos 1 X 2 + 2X COST? + l l l 4 2 (1 + 05D) -1 log(x + Dx + l) - C, 'e-tan 1 l 2 (4 - D2)l/2 K8X1 + K9 . If (D is to be symmetrical with respect to X the constant K2 must be taken as zeroo The constants K O 1 and K3 are arbitrarily set equal to zero. All other inte- gration constants may now be found from the continuity requirement at the two cornerso d Equate (jg) from the top and side boundaries at 1 the intersection of these boundaries '25C2 l - cosn “25C2 1 - cosn cos]? log 1 + cos)? = 30577“ log 1 + cos)? + K4 K4 = 00 w , Equate ——_— at the top corner by; 1 2 . -1 2 + D _ . tan - .25C log(2 + D) + K (4 _ D2)1/2 (4 _ D2)1/2 2 C + .SDC 2 2 -1 2 + D , tan . 4» - 025C loglz + D) (4 - D§)l]2 (4 - D2)l/2 2 K = C1 + C2 tan-1 2 + D 5 (4 - D2)l/? (4 - 132)“2 5 (10) (ll) 166 Equate (D at the top corner “C + OSDC 1 2 -l 2 + D 7 tan w 025C. log (2 + D) + (DC2 m Cl - C2) 10 1 + COSEL; - (C2 + C2D - C1) + Ecosn g'l - cosn , 4 sinTI '2 C1 + C2 -1 2 + D ‘25C2 C + . tan + K = ————— 2 (4 _ D2)l/2 (4 _ D2)1/2 6 cosn lo 1 - cosn _ 225C lo -- cosn C:2 + '5C2D g l + cosn ’ g l + cosn _D2)172 —l 2 + D tan (4 - D 2)”2 C + DC - K _ .2 2 Cl lo 1 cos + 6 — 8cosT? g l + cos C2 + C20 - Cl 4 sin]? m c mi: 20 (12) Equate ———— at the lower corner dxl -.25C2 ___, it) _, COSD_+ - 42532” 1 + C0872 cosn“ 9'1 + cosn _ cosn l.~’cosn K8 = 00 (13) d1,3 +¢2fl = 81,1 i=2 j=l , 16<132,1 " 5(¢3,1 +452,2 +§b1,1) +4,1 +4923 " '5(¢3,2 “131,5: B2,1 3Sise,j=1 ”951,1 " 5<¢i+l,l +¢i-1,1 +¢i,2) +¢i+2,1 +8,2 +¢6,2) = 137,1 i=8,j=1 16¢8,1 " S<¢8,2 +¢7,1) +¢8,3 +¢6,1 + °5¢7,2 = B23,1 168 '6- u.) bu + e W L l [1‘ F. 7v 15Cp2,;2 "' 5(¢3,2 $13 " (132,3 + (152,1) + Cp4,2 + C132,4 + °S‘i¢3,1 * (133,3 “ Cblfi + C131,3) z B2,2 3 si s 6, j = 2 l4¢i,2 " Smbr’ulfl + (13.-14,2 * (131,3 + (191,1) + q>i+2,2 * C131-4,2 * (131,4 * °S(¢i+1,3 " ¢i.+1,1 + (Pi—1,3 + <131-“1.,1) = 81,2 + i=7 j=2 14Cb7,2 ‘ 5(438,2 “ (peg * <137,3 * $2.1) + ¢S,Z ‘“ (p7’4 + °5(Cb8,3 + (p8,1+ (136,3 + (136,1) 2 B7.)2 i-8,j=2 15¢8,2 ' 5(437,2 " Cbafi + (Mal) + (13692 + $834 + '5‘¢7,3 + (1)7,1) 2 B8,2 121,35j513 14¢1,j '” 1091324 ~ 5<¢1,j+1 + 431,34) + 24%,], + 21,342+ <¢>1,j..2+ $25.“ + <22,j~1=0 149514 i+1,j’ (1)123 + $143 ¢i+1,j 1 + (13.-p1 i=7,35j513 1491375: " “(bad + (137, _ ”(131,14 '” $2,153 +Cb...+gb...g+ 1~1,3 1,3+l 3.) (be 1: + ¢7fi+1 ”$2,141 “ + . . + 7,15 + (157,13) " C135,14 + C1>7,12 + °5(¢8,1S + 438,13 * C136,15 + (136,13) = B7,14 i = 8, -< = 14 15958,.14 ‘" 5((1)7314 * $8,153 + (138,13) + C156,14 + $8,212 * °5"'¢7,15 + (1)-7,11} "“ 88,14 :1 = 1, j z :15 ”$1,115" j3115,15 ‘“ 5951,14 * 2193,15 * 431,13 ” (132,14 z B1,15 = 2, j =2 15 1615,15 ”" 5“133,123 + C131,15 + (152,14) * 954,115 + <152,13 + ”($3,141 * ¢1,14) : B2,15 3 :5 i s 6, j = .15 ”$1,133 “ “431111, 13 * (13,141,155 *‘ (131,14) + $144,155 * (bi-2,15 * (Pam + “5(¢1+1,14 + (bi-1,14) = B1,15 15 115@-7,15 " S<¢8 1.5 + (be 15+d>7,14) + $5 ,15 + (1)7113 + H N \J u L» H 15 1.1 H 03 LJ H + 16(b8 15 “” 5‘91)? 15 + $8,14 +<138,15 @8313 9 Table 22 shows the value of all nonmzero components of the constant vector, NON-ZERO COMPONENTS OF THE CONSTANT VECTOR FOR FINITE DIFFERENCE EQUATIONS TABLE 22 1 j B 1 j E i j B 1 1 _.6507132 7 5 01031994 8 10 -.2218186 2 1 ~,6482036 7 6 ,0943348 8 11 —.1842433 3 1 _,6408067 7 7 ,0850253 8 12 -.1496600 4 1 -,6289199 7 8 ,0754938 8 13 -.1189605 5 1 -.6131988 7 9 ,0659624 8 14 —.0815076 6 1 -,5945097 7 10 ,0566528 8 15 -.1319692 7 1 -,4436136 7 11 ,0477882 1 14 -.0123134 8 1 41,151426 7 12 ,0395879 2 14 -.0117259 1 2 ,1633010 7 13 ,0322536 3 14 -.0099912 2 2 ,1627135 7 14 ,0319480 4 14 «.0071928 3 2 ,1609788 7 15 ~,0093493 5 14 -,0034683 4 2 ,1581804 8 3 _,4849900 6 14 .0009988 5 2 ,1544560 8 4 -,4542905 1 15 ,0467626 6 2 ,1499888 8 5 4,4197072 2 15 ,0442531 7 2 ,2700273 8 6 ~03821320 3 15 ,0368562 8 2 -,3714553 8 7 -,3425526 4 15 .0249694 7 3 ,1187348 8 8 _,3619753 5 15 ,0092483 7 4 ,1113997 8 9 «02613979 6 15 “00094409 APPENDIX VIII FORTRAN PROGRAMS The three FORTRAN programs used for the solution of the discrete problem, the analytic continuum problem, and the numerical continuum problem are shown below. The letter C in the first column of the line indicates that the line is a comment statement and that it is not executable. In the PROGRAM CRY, which solves the discrete problem, the lattice points are located by the two indices I,J. These indices are identical with the i,j indices in Figure 5. The PROGRAM IDIS determines the solution of the analytic continuum problem. This program determines u, v,G x’ 0y, and TXY at points in the region 0 5 x s'L, -L 5 y 5 L. The points at which these quantities are determined are the nodes of a rectangular grid with spacing .0625L. The nodes are identified by two indices I and J. I and J are both one at x = O, y = -L and I increases to the right while J increases upward. PROGRAM NUT solves the numerical continuum problem. This program determines values of u, v, OX, 0 and IX Y’ y at nodal points of the rectangular grid shown in Figure 14. Each nodal point is located by a single index I. 174 175 I = 1 corresponds to the point i = l, j = l in Figure 140 Each nodal point is then numbered in ascending order of I from left to right with the point I = 8 corresponding to i = 8, j = l in Figure 140 The point I = 9 in the FORTRAN program corresponds to i = l, j = 2 in Figure 14 and then increases to I = 16 at the nodal point located 2 in Figure 140 The nodal point indicated by i 8. j by i = 8, j = 15 in Figure 14 is denoted by I = 120 for purposes of PROGRAM NUT. IO 20 176 PROGRAM CRY PROGRAM CRY DETERMINES DISCRETE VALUES OF DISPLACEMENT DIMENSION U(18933)0V(18033)oDIFFV‘lBoSB)ODIFFU‘18933) READ IOOOOU READ lOOOoV PRINT 200 PRINT 205.((IOJQ(U(IQJ))9(V(IQJ))9J31933)9I31918) PRINT 2 FORMAT(21H NUM BIG) NUM=O DO 3 I=1918 DO 3 J=Ic33 DIFFU(IQJ)=OoO DIFFV(IOJ)=OoO UI=(U(291)+.5*(U(292)+V(292)-V(191)))/105 DIFFU(101)=ABSF(UI‘U(191)I U(1!1)=U1 VI=(V(102)+o5*(U(292)+V(292)*U(l9l)))/Io5 DIFFV(101)=ABSF(V1*V(IQI)) V(lol)=Vl DO 10 J=2932 U1=(U(20J)+05*(U(29J+I)+V(29J+1)+U(29J‘l)‘VCZoJ-l)))/20 DIFFU(IOJI=ABSF(U1“U(IcJ)I U(10J)=Ul Vl=(V(qu+l)+V(loJ‘1)+.5*(V(29J+1)+U(29J+I)+V(29J-1) 1-U(29J-1)))/3o DIFFV‘IOJ)=ABSF(VI‘V(IOJ)I V‘IQJ)=VI UI=(U(2933)+05*(U(2932)’V(2932)+V(1933))I/Ios DIFFU‘I933)=ABSF(UI’U(IO33)) U(1933)=Ul Vl=(V(Ic32)+05*(V(2932)‘U(2932)+U(1033)))/105 DIFFV(IO33)=ABSF(VI‘V(I033)) V(Io33)=V1 DO 20 1:2016 UI=(U(I+I933)+U(1-1933)+05*(U(I+lo32)‘V(I+1932)+U(I-1932) 1+V(I*1932)))/30 DIFFU(I¢33)=ABSF(U1~U(1933)) u<1.33)=u1 v1:+U+.5*(V(17.32)+U(17.321+V(16.32)+U(16.32))1/2. DIFFVt17.33)=ABSF+u<16.J+1)-V(16.J+1) 1+U(l6oJ-l)+V(169J-l)‘U(17.J-1)“V(170J*I)))/5o DIFFU(179J)=ABSF(UI~U(I7oJ)) U(179J)=Ul 177 VI=(105*(V(I79J+I)+V(I7oJ-I))+.5*("U(l70J+l)+V(160J+I) I“U(160J+I)+V(160J“I)+U(16vJ‘-I)+U(I7OJ‘IIII/4o DIEFVII7QJ)=ABSF(VI“V(I70JII 30 V(17¢J)=VI UI=(U(16918)+05*(U(16919)“V(16919)‘U(I7919)+V(I7919) I+U(16917I+V(16017)"U(I7917)"V(17917)) 2+.4*V(I79I8)-04*V(18917)“013246)/502 DIFFUC17018):ABSF(UI*U(17818)I UII7QIBI=UI VI=(I05*V(I7019)+V(I70I7)+.5*(‘U(16019)+V(16$I9)‘U(I7919) I+U(16017)+V(16017I+U(I7917)“V(I79I7)I 2+o4*U(I7918)+08*V(18917I+026491I/408 DIFFVII7QIBI=ABSF(VI“’V(17918)I VII7QIBI=VI UI=(U(169l7)+05*("U(I7918)+V(I7018)+U(16916)+V(16916) l +U(16018)‘V(16918))+005872*V(I7917)‘.55872 2 *V(18916)‘068507)/3o77936 DIFFU(I79I7)=ABSF(UI“U(I7QI7)I U(I7917)=U1 Vl=(I05*V(I7918)+V(l7916)+05*(-U(I7918)+U(16016)+V(16016) I -U(16018)+V(16018))‘015080*U(I7QI7I+06984I*V(18015) 2 +023127)/401984I DIFFV(I7QI7)=ABSF(VI*V(17917)I V(I7117)=VI DO 40 K=I793I J=33-K UI=(U(169J)+05*(U(160J+I)‘V(169J+I)+U(169J-I)) I ‘063246*(V(189J-II“V(189J+I)I‘091886)/3063246 DIFFUII79JI=ABSF(UI’U(I79JII U(I70J)=UI V1=(V(17qJ+I)+V(I79J-l)+.5*(U(169J*I)+V(169J-1I‘U(I69J+I) I +V(169J+I))+063246*(V(189J-I)+V(189J+I)))/4026492 DIFFV(I79J)=ABSF(VI“V(I70JII 4O VII79JI=VI U1=(U(1691)+05*(U(1692)‘V(16‘2)+V(I7OI))+063246 I *(V(1892)‘V(1791))‘070943I/2081623 DIFFU(I7QI)=ABSF(UI"U(I791I) UCI7QII=UI VI=(V(I792)+05*(-U(1602)+V(1602))+018377*U(I7QI) I +.63246*V(1802)‘020943)/20I3246 DIFFV(I791I=ABSF(VI‘V(I791)) V(1791)=V1 DO 50 J=2915 VI=(V(189J+I)+V(189J“I)‘063246*(U(I70J+I)-U(I7oJ-I)) I +1.26492*(V(I79J+I)+V(l7oJ-I)))/4052984 DIFFV(J)=ABSF(VI~V(180J)I 5O V(189J)=Vl VI=(V(IBQI7)+V(IBOISI+063246*U(I7915)+1.26492*V(I7915) I “.6984I*U(I7QI7I+1039682*V(I70I7)‘.O¢I7O)/4066174 DIFFV(18016)=ABSF(VI‘V(18916)) V(18916I=VI VI=(V(18016)+063246*U(17916I+1026492*V(17916)‘08*U(I7018) I +106*VII7OIBI+019764)/3086492 DIFFV‘IB!I7)=ABSF(VI‘V(18017)I 60 70 500 80 90 110 120 130 140 144 145 160 150 195 2200 2205 300 1 1 l 178 V(18917I=VI DO 60 K=lo15 I317-K UI=(U(I+IOI)+U(I‘Iol)+.5*(U(I+102)+V(I+102)+U(I“IOE)‘V(I‘192)))/3o DIFFU‘IOI)zABSF(UI‘U(IQI)) U(IOI)=U1 VI=(V(I.2)+.5*(U(1+192)+V(I+102)‘U(I‘192)+V(I‘192)))/20 DIFFV(I9I)=ABSF(VI*V(I¢1)) V(IQI)=V1 DO 70 I32916 DO 70 J=2932 UI=(U(I+19J)+U(I‘I0J)+05*(U(I+19J+I)+V(I+loJ+l)+U(I-IQJ+I) -V(I‘IQJ+I)+U(I’IQJ“1)+V(I‘19J~I)+U(I+IQJ'I)‘V(I+IOJ'I))I/4o DIFFU(IQJ)=ABSF(Ul-U(IQJI) U(IOJ)=UI VI=(V(IOJ+I)+V(I9J“I)+.5*(V(I+IQJ+I)+U(I+19J+II+V(I‘IQJ+I’ -U(I“IQJ+I)+V(I’IQJ‘I)+U(I‘IvJ-l)+V(I+loJ-l)’U(I+IQJ“I))I/4o DIFFV‘IQJ)=ABSF(VI“V(IQJ)I V(IoJ)=Vl NUM=NUM+I IFINUM-SBOO)16091609SOO BIGU=OOO DO 90 I=1018 DO 90 J=Io33 IF(BIGU“DIFFU(I9J))80090990 BIGU=DIFFUII9JI CONTINUE BIGV=0.0 DO 120 I=1918 DO 120 J=Io33 IF(BIGV-DIFFV(I9J))11091209120 BIGV=DIFFV(IQJ) CONTINUE IF(BIGV‘BIGU§I3OQI3OQI4O BIG=BIGU GO TO 144 BIG=BIGV PRINT 1459NUMOBIG FORMAT(169E1506) IFCNUM-6000ISOISOQISO PRINT 195 FORMAT(22H DISPLACEMENT SOLUTION) PRINT 200 FORMAT(4OH . I J U V) PRINT 2059‘(IOJ§(U(IOJ))Q(V(IQJ))‘J:1033)OI=1018) FORMATIZISQEEISoé) FORMAT(3EIS¢6) PRINT 2050((I9J9(DIFFU(I9J))0(DIFFV(I9J))9J=Io33) 0I=1918) PUNCH 1000. U PUNCH 1000. V 1(300 FORMAT<3E20¢6I K=5 1 500 CALL DXST(K9UoV) iII-‘I'ul1‘lIII-II I'll“! II.[IIIII I [ll I It." I II .II .III 1 1 1| I", 7 2000 2001 20 25 4O 50 60 70 179 IF(K‘I3)2000QZOOIQZOOI K=K+4 GO TO 1500 CONTINUE END SUBROUTINE DXST‘K-UQV) SUBROUTINE DXST DETERMINES THE LATTICE FORCES ACROSS A VERTICAL PLANE DIMENSION U(18c33)9V(18933)0FX(33)0FY(33) KK=K-I FX(1)=105*U(K¢I)-U(KK01)“05*(U(KK02)+V(K91)‘V(KK92)) FY(1)305*(U(K01)“U(KK02)'V(K01)+V(KK92)) DO 20 J=2032 FX(J)=2.*U(K9J)"U(KK9J)".5*(U(KKOJ+I) -V(KK9J+I)+U(KK9J‘I)+V(KKQJ-I)) FYIJ)=‘V(K9J)‘05*(U(KK6J+I)-V(KK9J+I)-U(KK9J“I)*V(KK9J-I)) FX(33)=105*U(Ko33)‘U(KK933)“05*(U(KK933) -V(K933)+V(KK932)) FY(33)3-05*(U(K933)‘UIKK932)+V(KQ33)’V(KK!32)’ SUMX=OQO SUMY=OOO DO 25 J=Io33 SUMX=SUMX+FX(J) SUMY=SUMY+FY(J) PRINT 300K FORMAT(5H K = I2) PRINT 4O FORMATI44H J FX FY) PRINT SOQIJoCFX(J))9(FY(J))9J=1933) FORMAT(I492E2006) PRINT 609SUMX FORMAT(8H SUMX = E2006) PRINT 7OOSUMY FORMAT(8H SUMY = E2006) END 180 PROGRAM 1015 ' PROGRAM 1015 DETERMINES ANALYTIC VALUES OF STRESS AND DISPLACEMENT DIMENSION Y(33)9X(I7)9XP(I7917)951(330I7)052(33QI7)9 1 X2P(17917)9U(33ol7)9V(33017)9R1(33017)0R2(33017) vSX(33917)QSY(33OI7)QSXY(33QI7) C THIS PROGRAM GIVES DISPLACEMENTS FOR THE ANALYTIC SOLUTION IO 15 2O 25 31 32 33 30 C 'THE 40 45 51 52 53 50 1W4E TWfiE I Z=o3208 EI=ATANF(Z) E2=ATANF(*Z) T3000 Y‘1)=-100 XII)=OOO DO IO J=2933 Y(J)=Y(J~I)+00625 DO 15 I=20l7 X(I)=X(I-I)+00625 DO 20 J=18933 DO 20 I=IQI7 SI(J91)=ATANF((0790569*Y(J))/(X(I)+0612372*Y(J))) +60283185 DO 25 I=2917 51(179I)=60283185 DO 30 J=1916 DO 30 1:1017 XP(J9I)=X(I)+o612372*Y(J) IF‘XP(JOI)‘T)3I932933 SI(J9I)=ATANF((0790569*Y(J)I/(XP(J9I)))+3.I41593 GO TO 30 SICJvI)=IoS*30141593 GO TO 30 SI(JoI)=ATANF((0790569*Y(J))/(X(I)+0612372*Y(J)I) +60283I85 CONTINUE DO LOOPS 20925¢AND 3O GIVE ALL IMAGINARY VALUES OF LN 2(1) DO 40 leclé DO 40 1:1017 SEIJvII=ATANF((“.790569*Y(J))/(X(I)‘0612372*Y(J))) DO 45 I=2917 52(17OI)=000 DO 50 J=18933 DO 50 I=1917 X2p(JvI)=X(I)*0612372*Y(J) IF(X2P(J0I)-T)51952953 SZIJQI)=ATANF((‘0790569*Y(J))/(X2P(J0I)I)‘3QI4I593 GO TO SO 52(Jol)=~3.141593/20 GO TO 50 52(J01)=ATANF((‘o790569*Y(J))/(X2P(JQI))) CONTINUE DO LOOPS 409459AND 50 GIVE THE IMAGINARY PARTS OF LN 2(2) VALUES OF SIII7¢I)AND SZ(1791)ARE INFINITY. THEREFORE ASSIGN IUQBITRARILY LARGE VALUES TO THEM 51(1791)=100**6 181 52(17.I)=100**6 C THE DO LOOPS 60 AND 65 GIVE THE REAL PARTS OF LN 2(1) AND LN 2(2) DO 60 J=1933 DO 60 I=2o17 RI(JcI)=.5*LOGF(X(I)**2+Y(J)**2+2.*o612372*X(I)*Y(J)) 60 R2(JoI)=.5*LOGF(X(II**2+Y(J)**2-2o*o612372*X(I)*Y(J)) DO 65 J=1916 RI(J91)=o5*LOGF(Y(J)**2) 65 R2(J91)=o5*LOGF(Y(J)**2) DO 66 J=18933 RI(J91)=o5*LOGF(Y(J)**2) 66 R2(Jol)=05*LOGF(Y(J)**2) C THE REAL PARTS OF LN 2(1) AND LN 2(I) ARE UNDEFINED AT THE POINT C (17.1)THEREFORE GIVE THEM ARBITRARILY LARGE VALUES THERE RI(I7¢I)=10o**12 R2(I79I)=IOO**I2 C THE DO LOOP 70 GIVES THE DISPLACEMENTS FOR THE ANALYTIC SOLUTION DO 70 J=Io33 DO 70 I=IQI7 U(JQI)=(((*075)/o968246)*(RI(J9I)‘R2(JQI))‘(SI(J9I) I “52(JOI))+3o*3oI41593)/(4o*3ol4l593) 7O V‘JOII=(0316227*(RI(JOI)+R2(JOI))+10224744*(51(J91) I +SZIJ9I)‘20*o9II738-3ol41593))/(4o*3ol41593) PRINT 75 OEIOEa 75 FORMAT(2E1506) PRINT 95 95 FORMAT(//63H J I X Y SI I 82) PRINT IOOQ((JQIQ(X(I))9(Y(J))0(SIIJQI))o($2(J9I))9I=IQI7) I 9J=Io33) PRINT 96 96 FORMAT(//65H J I X Y RI I R2) PRINT IOOoI(J919(X(II)9(Y(J))9(RI(J0I))0(R2(J9I))0I=IOI7) I 9J=1933) PRINT 85 85 FORMAT(//36H DISPLACEMENTS FOR ANALYTIC SOLUTION) PRINT 80 80 FORMAT(//62H J I X Y U I V) PRINT 1009((J9I0(X(I))9(Y(J))o(U(JoI))o(V(J9I))9 I I=IQI7)0J=1033) 100 FORMATIZI402E12049251506) SX(1791)=109**12 SY(17¢I)=100**12 SXY(1791)=100**I2 L131 L2=16 113 DO 110 J=LIoL2 DO 110 I=Iol7 SX(J9I)=Y(J)*(.II324073*X(I)**2+.07549382*Y(J)**2) 1 /(X(I)**4+.5*X(I)**2*Y(J)**2+Y(J)**4) SYIJQI)=.07549382*Y(J)*(Y(J)**2-X(I)**2)/(X(I)**4 I +05*X(I)**2*Y(J)**2+Y(J)**4) 182 110 SXYIJQI)=OO7549382*X(I)*(Y(J)**3“XII)**2)/(X(I)**4 I +05*X(I)**2*Y(J)**2+Y(J)**¢) IF(L2‘33)IIIOII2OIIZ III LI=I7 L2=33 GO TO 113 112 D0 120 I=2917 SX(I71I)=OoO SY(17OI)=OOO 120 SXY(I7OI)=“0O7549382/X(I) PRINT I30 I30 FORMAT(68H J I SX 1 SXY) PRINT 1409((Jolc(SX(JoI))9(SY(J9I))o(SXY(JoI))OI=IQI7) I 9J=1933I 140 FORMAT¢21593E2O¢6I END SY (1 ,5 L I 400 L- 11 I3 900 2 PROGRAM NUT PROGRAM NUT DETERMINES THE STRESS FUNCTION. STRESSESQ AND DISPLACEMENTS FOR THE NUMERICAL SOLUTION DIMENSION BFEE(I7)1SFEE(3I)9UFEEII7)QBBFEE(I7)C SSFEE(3I)vUUFEEII7)oB(496)9X(496)¢TBXY(I7)9TSXY(3I)9TUXY(I7) oDBFEE(I7)9DSFEE(3I)0DUFEE(17) READ IQNQNZODEL FORMATI2I48F604) NI:(N-I)*N2 CALL FCAL(DEL9N9N208FEEQDBFEEQSFEEO DSFEEOUFEEQDUFEEODBSODUSI CALL BCAL(NoNEQBFEEQSFEEQUFEEQDBFEE9 DSFEEODUFEEOBODEL) READ 4009X FORMAT(4EI7-8) CALL SFIT(NONBOBQXOSUSQSBSQDUSODBSO BFEEOBBFEEODBFEEOSFEEOSSFEE‘DSFEE’ UFEEQUUFEEQDUFEEvDEL) CALL STRESS( NQN29DELQXQBFEEOBBFEEOSFEEOSSFEEQ UFEEOUUFEEQTBXYQTSXYQTUXYQSUSQSBS) CALL DISP(DELQNQNzQXOBFEEQBBFEEQSFEEQSSFEEO UFEEOUUFEE) PUNCH 4000(X(I)9I=10NI) END SUBROUTINE FCAL(DEL9N9N2oBFEEoDBFEEOSFEEoDSFEEQ SUBROUTINE FCAL DETERMINES BOUNDARY VALUES OF THE STRESS FUNCTION AND ITS DERIVATIVES UFEE.DUFEE.DBS.DUS) DIMENSION 21(17)o22(31)QBFEE(I7)ODBFEE(I7)OUFEE(I7)9 DUFEE(I7)QSFEE(31)cDSFEEI3I)GSIGX(3I)9RSIGX(3I)0 SIGY(I7)9PSIGY(I7)QB(50) ZI(I)=OoO DO 11 J=29N ZICJ)=ZI(J-I)+DEL Z2(I)=-I.O+DEL DO 6 J=29N2 22(J)=22(J*1)+DEL H=SQRTF(.9375) G=.5*ACOSF(-025) P=6o2831853 SG=SINF(G) CG=COSF(G) D=2o-4o*(CG**2) Cl=-(I(‘0750/H)*05625+.25*H)*CG+((“0750*025)‘05625)*SG)/P C2=(((-0750/H)*05625+025*H)*CG*((“0750*025)‘05625)*SG)/p DO 13 J=ION SIGY!J)=C2*(ZI(J)**2-1.)/(Zl(J)**4+D*21(J)**2+1.) DO 900 J=IQN2 SIGXIJI={-CI*22(J)-C2*22(J)**3)/(Z2(JI**4+D*Z2IJ)**2+Io) DD=SQRTF(4o-D**2) DO 5 J=I9N UFEE(J)=(.25*C2/CG)*ZI(J)*LOGF((21(J)**2-2.*CG*ZI(J)+Io)/ (ZI(J)**2+2.*CG*ZI(J)+IO))‘025*C2*LOGF(ZI(J)**4+D*ZI(J)**2+lo) 2 +(C2*(lo+o5*D)/DD)*ATANF((2o*ZI(J)**2+D)/DD) I l 5 GUIACJNH 7 1 I2 14 21 I 23 I I IO 20 1 184 BFEE=~UFEE+<(C2+D*C2—C1)/(4.*GG))*LOGF((1.~CG)/(1.+CG)) +<<62+C2*D-CI)/(4.*SG))*(P/2.)-2.*cz DUFEE(J)=((C2*(I.+.5*D)/DD)*ATANF((2.*ZI(J)**2+DD/DD) ~.25*c2*LOGF(ZI(J)**4+D*21{J>**2+1.)> DBFEE(J)=DUFEE(J) DO 7 J=IoN2 SFEE +((C2+D*C2-CI)/(8.*CG))*LOGF((1.-CG)/(1.+CG)) +(C2+C2*D~CI)*P/(16.*SG)-C2 DSFEE(J)=((.25*C2/CG)*LOGF((ZZ(J)**2-2.*CG*22(J)+1.) /(ZZ(J)**2+2o*CG*22(J)+lo))) OBSrC.25*C2/CG)*LOGF((1.+CG)/(1.-CG)) DUS=(.25*C2/CG)*LOGF((lo+CG)/(l.-CG)) PSIGYII)=2.*(UFEE(2)-UFEE(I))/DEL**2 M42N—1 DO 8 J=2.M4 pSIGY(J)=(UFEE(J-1)~2.*UFEE(J)+UFEE(J+1))/DEL**2 PSIGX+1 LL2=N1¢3*M4+3 M6=2*M4 NUM=O PRINT 70 FORMAT(/13H NUM BIG) DO 71 I=IoNI DIFF(I)=0.0 L=2*M4+3 LL=L+M4~5 DO 90 I=L.LL x1(I)=(S.* NL4=M4*(N2-3) NL5=M4*(N2-3)-l NL6=M4*(N2-2)-2 NL7=M4*(N2-I)-l NL8=M4*N2-2 NL9=M4*N2-3 NJ=M4*(N2-1)*3 X1(N1)=(B(Nl)+5.*(X(L11)+X(M3))-X(NL8)-X(LL10) I ~.5*X(NL7))/16o DIFF(N1)=ABSF(X1(N1)-X(N1)) X(N1)=X1(N1) X1(M3)=(B(M3)+5.*(X(NI)+X(NL7)+X(LLIO))~X(NN2)-X(NL4) 1 -.5*(X(LII)+X(LL8)))/15. DIFF(M3)=ABSF(X1(M3)-X(M3)) X(M3)=X1(M3) X1(Lll)=(B(L11)+5.*(X(NL8)+X(NL7)+X(N1))‘XINL9)—X(LL8)— 1 .5*(X(NN2)+X(M3)))/15. DIFF(L11)=ABSF(X1(Lll)-X(Lll)) X(L11)=X1(Lll) X1(NL7)=(B(NL7)+5.*(X(LII)+X(NN2)+X(M3)+X(LL8))-X(NJ) I-X(NL5)-.5*(X(LL10)+X(N1)+X(NL8)+X(NL6)))/14. DIFF(NL7)=ABSF(X1(NL7)-X(NL7)) XINL7)=XI(NL7) NJI=M4*(N2-1)+2 NJ2=NJ1+2 NJ3=M4*(N2-2)+I NJ4=NJ3+I NJ5=NJ3+3 NJ6=M4*(N2-3)+I NJ7=NJ6+I NJ8=NJ6+2 NJ9=M4*(N2—4)+1 NK=NJ9+1 XI(L5)=(B(L5)+IO.*X(NJI)+5o*X(NJ3)-2.*X(NN3) l -X(NJ6)-X(NJ4))/15o DIFF(L5)=ABSF(XI(L5)*X(L5)) X(L5)=XI(L5) X1(NJI)=(B(NJ1)+5o*(X(NN3)+X(NJ4)+X(L5))‘X(NJ2)-X(NJ7) I -.5*(X(NNI)+X(NJ3)))/160 III IIIII 0(3 189 DIFF.I=1.M4) FORMAT114.2E20.6) PRINT21.(I.(SSXY(I)).I=1.N2) FORMAT114.E20.8) PRINT 22 FORMAT(39H ACTUAL BOUNDARY VALUES OF SHEAR STRESS) PRINT 20.11.1TBXYII)).(TUXY(II).I=1.M4> PRINT 21.(I. I H H H I h II mo. 0 H H% 10H... ImH. 200 .75“— .SOL— _l. J l I l I 1 § 1 J -06 -04 -02 0 .2 T (bell/L) Fig. 38.--'l:Xy for analytic solution versus y1 IllilIIll‘Il.llllll{"-iill .75 050 .25 201 - .50 ._ / X1 = 05 / x1 = .125 x1 = l x1=.25 1 J 1 1 l L l I I "'06 -04 -02 O 02 04 .6 OX (bcll/L) Fig. 39.--Ox for analytic solution versus y1 202 0 I I I I : I . . P I I ?\——x1 = 75 I I I‘\—X1 = 51 \x1 = 1 - 25 — | I I I I ' l I I I _ I I I I . I \ . I I I I " 5° ‘— I I I I : I 1 x1 = 1 II Y1 _ I ‘ II ‘I ‘ I ‘ I ‘ I .. 75 __ I ‘ I ‘ \ \I I I I I \ _ I x = O I \ x/_ l \ xl=.25 \ x1 = 5 \ - I -1 C: I \ l1 \1 l I I2 u or v Ib) Fig. 40.--u and v displacements for numerical solution versus y1 (u is antisymmetric and v is symmetric with respect to y = 0)' 1 , 203 xx h ax mamum> OORPDHOm HOUHROEDO Row P Dam OLI.Hv .mHm Hun H mb. om .IMN. fl . a . _ \\ \\\\\ “MW IIIIIII.I! \\\\ I H II \\\\ HH I h .IIII. \\ \\ , \ \ III I IIII.|I|:I II.II 11111 I u a» ”V, 11111, mbm + \\ \\ \ \\ \ \ \ \\ m I u HM \\ \ \\ \ \\ \ \\ \\ \\ H H \\ O" % HI." N \\ xx P111 % Rx 110,0 204 0' O ‘ * I —_— xy I\' ' I I i /\\\——-x1 = 5 __ " I 25 ' /f\\¥—— x1 = 1 I / I / .... 50L... II/ yl // 4] x1 = 0 // // - 75- // I I l/y x1 = 1 x1 = 5 / / / -1 / I I I I 4 0 .05 .10 .15 .20 .25 Ox or Txy (bcll/L) Fig. 42.--—OX and Txy for numerical solution versus yl (Ox is antisymmetrical and Txy is symmetrical with respect to y1 = O) 11. 12. 13. Love, A. E. H. of Elasticity. BIBLIOGRAPHY A Treatise on the Mathematical Theory New York: Dover, 1944. Cottrell, A. Crystals. Conners, G. H. 1962. Read, W. T. H. Dislocations in Crystals. Dislocations and Plastic Flow in Oxford: Clarendon Press, 1953. Ph.D. Thesis, Michigan State University, New York: McGraw-Hill, Eshelby, J. D. 1953. , Read, W. T., and Shockley, W. Acta Metallurgica, 1, 253 (1953). Peierls, R. E. (1940). Nabarro, F. R. 256 (1947) A. J. Soc. Foreman, Phys. Maradudin, A. A. Kittel, c. Introduction to Solid State Physics. Proc. Phys. Soc. of London, 22, 23 N. Proc. Phys. Soc. of London, 59, , Jaswon, M. A., and Wood, J. K. Proc. of London, £4, 156 (1951). J. Phys. Chem. Solids, 2, 1 (1958). 2nd ed. Faddeeva, V. Algebra. Taylor, A. E. Johnston, W. New York: N. New York: G., and Gilman, J. J. Wiley, 1961. Computational Methods of Linear Dover, 1959. Advanced Calculus, Ginn, 1955. J. Appl. Phys., 29, 129 (1959). 205 "‘IIIIIIIIIIIIIIIIES