TWO-DIMENSIONAL AND THREE-DIMENSIONAL LINEAR ELASTICITY SOLUTIONS BY EIGENFUNCTION EXPANSIONS AND THE BOUNDARY INTEGRAL EQUATION METHOD by Jing Chang A DISSERTATION Submitted to Michigan State University in partiai fulfiiiment of the requirements for the degree of DOCTOR OF PHILOSOPHY Department of Meta11urQY. Mechanics and Materials Science 1978 ABSTRACT TWO-DIMENSIONAL AND THREE-DIMENSIONAL LINEAR ELASTICITY SOLUTIONS BY EIGENFUNCTION EXPANSIONS AND THE BOUNDARY INTEGRAL EQUATION METHOD by Jing Chang Two numerical solutions were developed for problems in two- dimensional and three-dimensional elastostatics. The eigenfunction expansion method was based on the general solution of the Navier equations in polar coordinates and spherical coordinates and used functions which are non-orthogonal on the boundary. Numerical techniques of point matching or least-squares were used to satisfy the boundary conditions. The boundary integral equation method utilizes singular integral equations which can be solved numerically for the unknown surface tractions and displacements for the fully mixed boundary value problems. A fictitious traction applied on the boundary is used in the boundary integral equation methods. Both methods are inde- pendent of the boundary shape and the boundary conditions. Some sample problems were solved to verify the formulation. Direct comparisons were made between the eigenfunction expansion and the boundary integral equation methods in a two-dimensional plane problem and two three-dimensional problems. Discussion of the results include convergence, total cost, computation time used and storage requirements. TO MY PARENTS AND MY WIFE, SUI-HUANG C. CHANG 11° ACKNOWLEDGEMENTS The author wishes to thank his advisor, Dr. Robert N. Little, for his suggestion of the problem, counsel and encouragement which made this dissertation possible. The author would also like to thank Dr. Nicholas Altiero for his guidance and advice throughout this research. Also the author would like to recognize the other members of the Guidance Committee: Dr. David H.Y. Yen and Dr. William A. Bradley for their contribution to this dissertation and for the more specific benefits acquired from their classes. Thanks are also extended to Dr. Robert Summitt, Chairman, Department of Metallurgy, Mechanics and Materials Science, and Dr. Joseph E. Adney, Chairman, Department of Mathematics for financial assistance. Finally, to his wife, Sui-Huang, the author expresses his deep gratitude for her patience, sacrifice and encouragement. Thanks are due to Ms. Mary Grace for her excellent typing. iii TABLE OF CONTENTS List of Tables List of Figures List of Symbols Chapter I II III IV VI INTRODUCTION ON THO-DIMENSIONAL LINEAR ELASTICITY 2.l Introduction 2.2 Two-Dimensional Formulation 2.3 Exact Solution by Complex Variables 2.4 Eigenfunction Expansion Method 2.5 Boundary Integral Equation Method 2.6 Comparison of Methods Applied to Two-Dimensional Linear Elasticity THE EIGENFUNCTION EXPANSION METHOD APPLIED T0 THREE- DIMENSIONAL LINEAR ELASTICITY 3.1 Introduction 3.2 Formulation of Eigenfunction Equation 3.3 Numerical Solution THE BOUNDARY INTEGRAL EQUATION METHOD APPLIED T0 THREE- DIMENSIONAL LINEAR ELASTICITY 4.l Introduction 4.2 Formulation of Boundary Integral Equation 4.3 Numerical Solution EXAMPLES AND COMPARISONS 5.l Example: Cube Subjected to Uniform Compression 5.2 Example: Cube Subjected to Uniform Displacement 5.3 Comparison of Methods Applied to Three— 1 Dimensional Linear Elasticity SUMMARY AND CONCLUSIONS iv Page vi vii 10 13 18 23 38 38 39 6O 61 61 61 66 73 73 80 80 104 TABLE OF CONTENTS APPENDICES A - Associate Legendre Functions of the First Kind B - The Integral Values of AUij and ATij C — Fortran Computer Program of the Eigenfunction Expansion Method in Two-Dimensional Elastostatics D - Fortran Computer Program of the Boundary Integral Equation Method in Two-Dimensional Elastostatics E - Fortran Computer Program of the Eigenfunction Expansion Method in Three-Dimensional Elastostatics F - Fortran Computer Program of the Boundary Integral Equation Method in Three-Dimensional Elastostatics BIBLIOGRAPHY Page 108 110 114 120 124 134 141 LIST OF TABLES Table l Comparison of Total Cost, CPU, CMU and Matrix Size of EFE and BIE for Two-Dimensional Problem 2 Comparison of Stress Solution Away From Boundary for Cube Subjected to Uniform Compression 3 Comparison of Stress Solution Away From Boundary for Cube Subjected to Uniform Displacement 4 Comparison of Stress Solution Near Boundary for Cube Subjected to Uniform Compression 5 Comparison of Stress Solution Near Boundary for Cube Subjected to Uniform Displacement 6 Comparison of Total Cost, CPU, CMU, and Matrix Size of EFE and BIE for Cube Subjected to Uniform Compression 7 Comparison of Total Cost, CPU, CMU and Matrix Size of EFE and BIE for Cube Subjected to Uniform Displacement . . . 63 8 Comparison of Stress Solution U51ng SAPIV, EFE34. BIE48(d) and BIE48(e) for Cube Subjected to Uniform Displacement vi Page 37 81 82 83 84 100 101 102 LIST OF FIGURES Figure 1 1O 11 12 13 14 15 16 17 An Infinite Plane with an Elliptic Hole Subjected to Uniform Normal Pressure and Uniform Tangential Pressure over Part of the Hole A Conformal Mapping Which Maps the Unit Disc into the Infinite Plane with an Elliptic Hole Stress Components in Polar Coordinates The Normal and Tangential Stress Components on the Elliptic Hole A Two-Dimensional Linear Elastic Problem A Two-Dimensional Linear Elastic Region Embedded in an Infinite Plane of the Same Material Comparison of Radial Stress Solutions for the Problem of Figure l, e=0° Comparison of Radial Stress Solutions for the Problem of Figure l, 6=18° Comparison of Radial Stress Solutions for the Problem of Figure l, e=27° Comparison of Radial Stress Solutions for the Problem of Figure l, e=36° Comparison of Radial Stress Solutions for the Problem of Figure 1, e=45° Comparison of Radial Stress Solutions for the Problem of Figure l, e=54° Comparison of Radial Stress Solutions for the Problem of Figure l, e=63° Comparison of Radial Stress Solutions for the Problem of Figure l, e=720 Comparison of Radial Stress Solutions for the Problem of Figure l, 6=81° Comparison of Radial Stress Solutions for the Problem of Figure l, e=90° Comparison of Radial Stress Solutions for the Problem of Figure l, at Distance b from Boundary of Elliptic Hole vii Page 11 14 16 20 21 25 26 27 28 29 30 31 32 33 34 LIST OF FIGURES Figure Page l8 Comparison of Radial Stress Solutions for the Problem 36 of Figure l, at Distance lOb From Boundary of Elliptic Hole 19 Spherical Coordinates (R,¢,e) 42 20 A Three-Dimensional Linear Elastic Problem 63 2l A Three-Dimensional Linear Elastic Body Embedded in an 65 Infinite Medium of the Same Material 22 The Function, H, on the Singular Boundary Element 70 23 A Unit Cube with Origin at the Center Subjected to 74 Compression on Two Opposite Sides with All Other Surfaced Traction Free 24 Surface Nodal Arrangements for EFE12 and EFE24 75 25 Surface Element Arrangements for BIE 77 26 A Unit Cube with Origin at the Center Subjected to 78 Mixed Boundary Conditions 27 Surface Nodal Arrangements for EFE19 and EFE34 79 28 Percent Error Contours at Different Planes (x2=Constant) 85 for 033 of Uniform Compression Problem Using BIE48(d) 29 Percent Error Contours at Different Planes (x2=Constant) 86 for 033 of Uniform Compression Problem Using BlE48(e) 30 The Boundary Error for 0]] on x1=O.5 for Cube Subjected 88 to Uniform Displacement Using EFEgz ‘ 3l The Boundary Error for 012 on x2=0.5 for Cube Subjected 89 to Uniform Displacement Using EFEgZ 32 The Boundary Error for u3 on x3=O.5 for Cube Subjected 9O 63 to Uniform Displacement Using EFE34 33 The Contours of the Stress Solution, 033, on x3=0.5 for Cube SH Subjected to Uniform Displacement Using EFESZ 34 The Stress Solution, 033, on x2=0.5 for Cube Subjected to 92 Uniform DISPIGCement Using EFEgZ viii LIST OF FIGURES Figure Page 35 36 37 38 39 40 41 The Contours of the Stress Solution, 033, on x3=0.25 for 93 63 Cube Subjected to Uniform Displacement Using EFE34 The Contours of the Stress Solution, 033, on x3=0.000l for 94 Cube Subjected to Uniform Displacement Using EFEgi The Stress Solution, 033, on x2=O.25 for Cube Subjected 95 to Uniform Displacement Using EFEgZ The Stress Solution, 033, on x2=O.OOOl for Cube Subjected 96 to Uniform Displacement Using EFEgZ Percent Difference Contours at Different Planes (x2=Constant)97 for 033 of Uniform Displacement Problem USing BIE48(d) and 63 EFE34 Percent Difference Contours at Different Planes (x2=Constant)98 for 033 of Uniform Displacement Problem Using BIE48(e) and 63 EFE34 A Local Coordinate System with the Origin at the Center of 111 the Triangular such that Side l-2 is Parallel to the gl-axis ix LIST OF SYMBOLS Displacement in the ith direction Strain tensor Stress tensor Lame constants Poisson's ratio Young's modulus Shear modulus First invarient of strain tensor Laplace operator Airy stress function Biharmonic operator Outward normal vector Polar coordinates Stress components in polar coordinates Normal and tangential stresses Fictitious traction Analytic stress functions Conformal mapping function A matrix Vectors Boundary with specified displacements, tractions Displacement influence function Displacement influence function Particular prescribed traction or displacement component Resultant boundary data Element of the coefficient matrix X LIST OF SYMBOLS Bvij Element of the boundary value vector Re Real part of a complex number Gij Kronecker delta ¢n Homogeneous harmonic polynomial of degree n i Harmonic vector ifi Homogeneous harmonic polynomial vector of degree n (R,¢,e) Spherical coordinates an(c) The associate Legendre function of the first kind of degree n order m c = COS(¢) Ln = {2[3n+l - 2v(2n+1)]}'1 K1 = [l6nG(l-v)]'1 K2 = -(i-2v)[8n(i-v)]“ AUij Special integral of BIE in displacement ATij Special integral of BIE in traction eijk The third order absolute tensor 3 Contour integral e1j,e2j The direction Cosine of the local coordinates £1,52) in the global coordinates EFE; Eigenfunction expansion matching boundary condi- tion at m pOints and uSing n terms BIEm Boundary integral equation method in which the boundary is subdivided into m elements dPfi(Cos a) I P: (cos ¢) = d Cos¢ xi CHAPTER I INTRODUCTION In this dissertation two effective numerical methods for de- termining displacements and stresses in two-dimensional and three- dimensional linear elasticity are presented: the eigenfunction expansion method and the boundary integral equation method. It is assumed that the material is isotropic and homogeneous and that the displacements are small. The general methods of solution of two-dimensional elastostatic problems have been studied by many researchers. Solutions utilizing the Airy stress function can be expressed as a polynomial series [1], a Fourier series [2], a multiple Fourier series [3],[4], etc.. The Fourier series and multiple Fourier series solutions give orthogonal functions on rectilinear boundaries. A second approach introduces the complex potential function ¢(z) and x(z) of Muskelishvili and applies a conformal mapping function to satisfy the boundary conditions [5],[6]. Numerical techniques, such as finite difference methods [7] and the finite element methods [8], have also been used in recent years. Variational methods [9] and energy methods [lO], have been developed on the basis of calculus of variations to minimize the poten- tial energy of the system by the Rayleigh-Ritz technique. Rizzo [ll] presented an integral equation method, making use of Betti's reciprocal theory and depending on the knowledge of the singular solution to the Navier equations in two-dimensions which corresponds to a concentrated load. The Betti's reciprocal theory was developed by Betti [12], Somigliana [13], Lauricella [l4] and other geometers of the Italian School. Massonnet [15], and Altiero and Sikarskie [16] have developed a boundary integral equation method 1 2 for plane problems which also depends on the singular solution but the solution was obtained by introducing a fictitious traction along the boundary. The approaches employed in this dissertation for solution of two-dimensional elastostatic problems are the Michel solution [17] to the biharmonic equation, the Altiero and Sikarskie solution [l6] and the Muskhelishvili solution [6]. The study of three-dimensional elasticity is an area that has only recently been given attention outside of those problems which can be solved in closed form, or those problems which can be reduced to a less complicated level. A.E.H. Love in "A treatise on the mathematical theory of elasticity" [l8], stated, "The problem of determining the state of stress and strain within a solid body which is subjected to given forces acting through its volume and to given tractions across its surface, or is held by surface tractions so that its surface is deformed into a pre- scribed figure, is reducible to the analytical problem of finding functions to represent the components of displacement". The components of displacement are generally expressible in terms of a series expansion of harmonic functions. Lame [19] applied such a solution to the problem of a body bounded by a spherical surface and deformed by given surface tractions. The problem of the sphere has been examined also by Lord Kelvin [20], who sought to utilize it for the purpose of investigating the rigidity of the earth. Luré [2l] presented a simpler development for the problems of the sphere.and spherical cavity with the tractions or displacements prescribed on the boundary. The series solutions employed are expressed in terms of spherical 3 harmonics. Little [22] indicates that the importance of the Kelvin- Lure solutions lie not only in solution of non-axisymmetric spherical bodies, but also as a general solution for problems of other geometries. This is the approach used in this dissertation. Massonnet [l5], Altiero and Sikarskie [l6] introduced a new boundary integral equation method forplane problems using the influence function, the Green's function, without using the reciprocal theorem. This method can be extended to three-dimensional cases as presented in this dissertation. The other general methods of solution of three-dimensional linear elastic problems include the finite element method [23], [24], variational methods [25], [26], finite difference methods [27], [28], and the integral equation method [29] using Betti's reciprocal theorem. In chapter II, the eigenfunction expansion and boundary integral equation methods will be applied to a two-dimensional plane strain problem. The example presented is an infinite plane with an elliptic hole under normal and tangential tractions on part of the boundary. Thus both normal and tangential tractions have a jump discontinuity at some points on the boundary. The complex variable solution is used as the exact solution to which comparison is made. The different methods are compared numerically by examining the comparison of total cost, computation time used, central memory used and matrix size. The eigenfunction expansion of the three-dimensional linear elastic problems is derived in Chapter III, and the stress and dis- placement components are expressed in terms of spherical harmonic functions. The boundary integral equation method is derived in Chapter IV for the three-dimensional case and applied in numerical form. 4 Two examples, given in Chapter V, are uniform axial normal traction applied on two opposite sides of a cube, and uniform constant dis- placement on two opposite sides of a cube. Comparison of the two methods in the three-dimensional cases is also given. Chapter VI summarizes the result of both methods. All computer program lists are included and may be found in the Appendices. CHAPTER II 0N TWO-DIMENSIONAL LINEAR ELASTICITY 2.1 Introduction The application of conformal mapping and complex integration to the plane problem was developed in detail by Muskhelishvili [6]. The solutions for stretching of a plate with an elliptic hole due to forces at infinity or tractions on part of the edge of the elliptic hole was included in his book [6]. The problem of a plate with an elliptic hole subjected to uniform pressure and uniform tangential stress, can be solved by conformal mapping and complex integration, and is given in section 3. In section 4, this problem is solved by an eigenfunction expansion method of two-dimensional linear elasticity based on the Michell solution in polar coordinates. Since the eigenfunctions are non-orthogonal on non-circular boundaries, the least-squares technique [30] which involves the minimization of the square of the error on the boundary will be used to determine the coefficients of ‘the eigenfunction expansion. A boundary integral equation method has been developed [15], [16] for the plane problems of linear elasticity, and is briefly outlined in section 5. The comparison of the eigenfunction expansion method and the boundary integral equation method in two-dimensional problems is discussed in section 6. 2.2 Two-Dimensional Formulation 2.2.1. Consider the plane strain problem of a homogeneous, isotropic, linear elastic material in which body forces are neglected. The displacement in the x3 direction, u3=0, and the other displacements, u1 6 and ”2’ are independent of x3. 5.. is the ijth component of strain, 13 and is defined as: au. au. = __;L ._41 613° 1/2(ax.+ ex.) (1) j i so that differentiation yields: €13=€Z3=€33=0 (2) The stress component, Oij’ is related to strain, Eij’ as: Oij = Aedij + 2peij (3) where 6ij is the Kronecker delta, and A and u are the Lame constants. They are related to the elasticity constants E, Young's modulus, and v, Poisson's ratio by, E = u§3A+2u2 A+u __ A v - 2[A+p1 The first invariant of strain, e, is defined as: e=—+— (5) for plane problems. 7 The equilibrium equations without the body forces are o...=0 ‘ (6) To obtain the reduced Navier equation for the displacement problem, one substitutes equations (1) and (3) into equation (6), yielding: v21: + A}: 367-13) = o (7) where v2 is the Laplace operator and 3(3-6) is the gradient of the divergence of U. The equation of compatibility can be expressed in terms of stresses as: we“ + o = o (a) 22) The usual method of solving equations (6) and (8) is by introducing a new function, ¢, the Airy stress function, which satisfies, 22 axlax1 (9) 12 axlax2 8 Substituting equation (9) into the compatibility equation (8), the stress function must satisfy the equation, §:$.+ 23q¢ + 3H¢ = 0 k 2 2 4 3X] 3X13X2 3X2 v2v2¢ = O, i.e., a is governed by the biharmonic equation. 2.2.2. An Example Problem An infinite plane with an elliptic hole subjected to uniform pressure and uniform tangential traction over part of the hole is shown in Figure l. The equation of the hole is _1_ 2= + x2 1 (11) ii 2’. 5 _7_11 Onn = O o t = 0 when 05954,4 ses4fl,4 5652n " 1 11:5 7 (12) -l 0.5 when 4to+0 B AB AB where { } represents the integrand of equation (34). The limit of the second integral above for i = 1,2, respectively, can be shown to be f1 f5 lim f { }ds = ——3-—- (37) aB+0 AB 2 2 =>to+0 Thus equation (34) in final form is f? 4+ x H (p'.p1>f;(P1)n1(P')ds(P1) = t§?(P').P' on B 2 B ij;q (38) t The integral in equation (38) is to be interpreted in the Cauchy principal value sense. The solution to any boundary value problem is contained in equations (30), (31), (35) and (38). To obtain a numerical solution to equations (35) and (38), the boundary is first divided into N intervals of arbitrary length ASj,j=l,2,...,N. Resultant boundary data are then defined at the mid- point of each interval as follows: F11 = fAsjf1ds (39) = p Tij fAsjt1ds (401 = P U11 fAs.u1ds (41) J 23 Using the trapezoidal rule and equations (39), (40) and (4l), equations (35) and (38) can be written as: N E=lRMiijij = Bv11, 1=l,2, J=l,2,...,N (42) where Rwhjk are only dependent on the boundary and material constants. BV11 represents either T11 or U11, whichever is applied in direction i at boundary point j. Equation (42) represents 2N equations with 2N unknowns. Once the fictitious tractions, F11, are determined, the stresses and displacements are found from the numerical approximation of equations (30) and (31). The computer program for two-dimensional boundary integral equation method for the example is given in Appendix D. The boundary was sub- divided into 40 intervals for implementation of this program. 2.6 Comparison of Methods Applied to Two-Dimensional Elasticity Figures 7 through l6 demonstrate that the solutions of both methods converge inside the region, and that the solution of the eigenfunction expansion method also converges on the boundary although it's error is maximum there. Figures 17 and l8 show that the maximum error of both methods is near the boundary at e=90°, because the tangential stress, has a Ont! jump discontinuity there. Table 1 indicates the total cost, CPU (Computation time used), CMU (central memory used) and matrix size of both methods. 24 The boundary integral equation program, Appendix D, proved to be easy to use and possessed the nice feature that the exterior problem could be changed to the associated interior problem by simply changing the direction of the contour integral. The program for the eigenfunction expansion method, Appendix C, has been written to incorporate geometric and boundary symmetries with respect to both x1 and x2 axes. Both programs can be extended to solve any two-dimensional linear elastic problems. The CDC 6500 digit computer at Michigan State University was used for all computations. The total cost indicated in the tables is based on normal priority rate. 25 T “’9 ./6- , 0 - Exact Solution " A - BIE Solution 51 - EFE Solution .08 u \J .0/6*‘ ' ’ } :fi“*I-h l l l l i i i ~i i---1-1l r .3 5 7 9 // FIGURE 7 Comparison of Radial Stress Solutions for the Problem of Figure l, e=0° 26 0’: r l 0/0d -( ...( 0. . , .. 3 ° ; u——n-—-r.'.ié; l I I I l I I 1* 1—.P Y 2. 7 6 5. 76 /0. 76 FIGURE 8 Comparison of Radial Stress Solutions for the Problem of Figure l, e=18° 27 0’" 0.07 A 0.00 - U P V . ” 32.9....- . o " 0.07 " o J o o 237 ' 5:37 W l 8:37 I I YV‘Y FIGURE 9 Comparison of Radial Stress Solutions Figure l, e=27° for the Problem of 28 Oir l.1\ ' ~7' W IlllLlJll ILlLJI ll}! -0./8 '0./2 ’0.06 l l I l “—‘I 1 ‘ 2.40 T 5.40 8,40 ' FIGURE l0 Comparison of Radial Stress Solutions for the Problem of Figure l, e=36° ——i))’ C1" -0.90 - 0.60« - 0.30 29 FIGURE ll aée ' 4‘26 ' 6'26 Comparison of Radial Stress Solutions for the Problem of Figure l, e=45° 30 0 TT 2J6 4.16 6./6 FIGURE 12 Comparison of Radial Stress Solutions for the Problem of Figure l, e=54° 31 r ' ' a» r 5.09 4.09 5.09 FIGURE 13 Comparison of Radial Stress Solutions for the Problem of Figure 1, e=63° 32 ~ 0.75‘ - 0.50‘ . .\, V I 2.04 4.04 604 FIGURE 14 Comparison of Radial Stress Solutions for the Problem of Figure l, e=72° 33 q . . . V '7 _.Q[2. ‘ ' I I I - L, T 2.0/ 4.0! 6.0/ FIGURE 15 Comparison of Radial Stress Solutions for the Problem of Figure l, e=81° 34 - ClcSl? " CZ‘FC7 "“020 L I I I I v # 2 4 6 ’ FIGURE 16 Comparison of Radial Stress Solutions for the Problem of Figure l, e=90e 35 0% L " 0.05 - 0.25 - 0.45- - 0.65‘ I l I I I I I I I : 6 0° 45° 90 FIGURE 17 Comparison of Radial Stress Solutions for the Problem of Figure l, at Distance b from Boundary of Elliptic Hole 36 FIGURE 18 Comparison of Radial Stress Solutions for the Problem of Figure l, at Distance 10b from Boundary of Elliptic Hole 37 Table 1. Comparison of Total Cost, CPU, CMU and Matrix Size of EFE and BIE for Two-Dimensional Problem Total Cost CPU_ CMU_ Matrix Size EFE $2.48 34.77 Sec 3.22 W-H 31 BIE $2.05 28.23 Sec 2.76 N-H 80 CHAPTER III THE EIGENFUNCTION EXPANSION METHOD APPLIED TO THREE-DIMENSIONAL LINEAR ELASTICITY 3.1 Introduction Papkovich [31] in 1932 and Neuber [33] in 1934 independently developed solutions of the Navier equation of elastostatics in terms ofaaharmonic vector and a harmonic function. Lure [21] formalized the notation and presented solutions of both the solid sphere and a spherical cavity in the inifinte space under both axisymmetric and nonsymmetric loadings. The displacement components can be expressed in terms of four harmonic functions which can be related to the spherical harmonics. One of those harmonic functions can be related to the others to simplify the solution when displacements or surface tractions are specified on the surface of the sphere. Little [22] indicates that the importance of the Lure solution lies not only in solution of non-axisymmetric spherical bodies, but as general solutions for problems of other geometries. The eigen- functions are non-orthogonal on non-spherical boundaries, therefore some numerical techniques have to be used to determine the coefficients of the eigenfunction expansion. ~ A Point matching is the simplest method to satisfy boundary conditions at N points taken along the boundary which are not on the edges or at the corners, where the three displacement or traction components are prescribed using 3N terms. Truncation of the eigenfunction expansion is justified and usually use of only a few terms in the expansion gives a good approxi- mation. To improve accuracy, more points were chosen on the boundary 38 39 than terms. An overdetermined system was obtained and a least- squares method which minimizes the square of the error was used to determine the coefficients of the eigenfunction expansion. The displacements and stresses inside the body can be determined using the eigenfunction expansion after the coefficients are specified. 3.2 Formulation of Eigenfunction Expansion The material was assumed to be homogeneous, isotropic, linear elastic and the displacmenets and the strains were considered infini- tesimal. Consider a simply connected three-dimensional geometry bounded with a two-dimensional surface with no reentry points such that every point on the surface is differentiable. The solid body is subjected to the action of balanced external loads, body forces are neglected and the surface tractions are non-singular. Lord Kelvin [20] obtained a solution in terms of harmonics directly from the Navier equation which is —> +++ 1 2 _____ . = v u + 1_2v v(v u) 0 (l) -> -> -> by noting that Kim is a harmonic function and v2u is a harmonic vector. ++ In the Navier equation, v-u can be expressed in terms of the homo- geneous harmonic polynomials, ¢n’ of degree n, as follows: (2) 3P Noting that 379'15 a polynomial of degree n-l, ¢n-l’ one obtains .+ v2(Rmv¢n) = m(m+2n-1)Rm'ZV¢n (3) 40 Substituting equation (3) when m=2 into the Navier equation (1), yields: -> .1 .1 + vzu + 1:53-g 2 2n+l v2(R2v¢n) = 0 (4) Integrating equation (4) gives: + u: -1 1 + -> 1-2v fi ETEETT1RZV¢n + x (5) .+ where x is a harmonic vector. Noting that, -> -> ++ . 2 = . 2 2 = v (R v¢n) 2R v¢n + R v ¢n 2n¢n (6) the divergence of equation (5) yields: ++ ++ 1 - n , V‘“ ' ‘1-2v fi enii ¢n + V X (7) ++ Using equation (2) and equation (5), ¢n may be related to v-x as follows: ++ . _ 3n+l - 2v(2n+l) V X ' 3 (1-201(2n+17* ¢n (8) Noting that the components of x are harmonic in cartesian coordi- + nates, x may be written 41 is a homogeneous harmonic polynomial vector of degree n. + where xn Therefore, _ 0szan ‘2" 4’n ' 3n+l-2v(2n+l) V xn for "32 (10) The displacement vector is expressed as: +--> + 00 + R2 +++ “ ‘ x0 + xi + i=2 [Xn ' 2[3n+1-2v(2n+1)]v(v°xn)] (‘1) Lure [21] presented a similar' development and expressed the displace- ments as: +-ao -> 12200 .1 +++ " ‘ i=0 Xn + 2(Ro'R ) i=2 (3-4v)n-2+Zv V(V'Xn) (‘2) for the solid sphere case, and -> no -> +++ 2 Ro—R2 u = i=0 23mm ‘ ZUB-WTn’rTPZ-Zfl V(V'X-(n+1)) “3) for a spherical cavity in the infinite space, where Ro is the radius of the sphere. The eigenfunction expansion may be expressed in spherical coordinates (R,¢,e), see Figure 19, by noting that the spherical coordinates and the Cartesian Coordinates are related by 42 (3:49.91 FIGURE 19 Spherical Coordinates (R,¢,e) 43 R = / 2 2 2 R>0 X1+X2+X3 .. x3 ¢ = Arc cos ( ) n>¢>0 /T2 2 2 ’ ' x1 + x2 1 x3 x2 6 = Arc tan (;—) 2n>0>0 1 _ The divergence in spherical coordinates is 3R2F -> + . T T 1 . 1 v-(F 1R + F 1¢ + F 16) = (S1n¢ 1 2 3 RZSin¢ 3R aSin¢F 3F 2 3 ”T” 5‘9“): and the gradient is T _ T 36 T l 36 T 1 30 VG - 1R 5§~+ 1¢ fi-5$-+ 1e RSin¢ 35 + Let xn be a homogeneous harmonic polynomial vector of degree n. Expressing the vector in cartesian coordinates, + -> -> + xn = xlnil + X2n12 + X3ni3 Each component is a harmonic function, i.e., In spherical coordinates, the Laplace equation becomes (14) (16) (18) 44 2 %§'(R2 3%) + ‘ §—-(s1n¢ 3%) + ‘ a I = o. (19) RZSin¢ 30 RZSin2¢ 002 V211, = 1— R2 Using the separation of variables method, the homogeneous solutions of the Laplace equation are 11(R.¢.e) = ca"vn(¢.e> + DR'("*‘)vn(¢.e) (20) where n _ m . Yn(¢,e) - ;=O Pn(;)[AmnCos me + anS1n me] , (21) c = Cos 0, and PER) (22) are the associate Legendre functions of the first kind degree n and order m satisfying d2 d 2 _ (1-;2> EZE'P:(C) - 2; aE-Pfi(c) + [n(n+1) - $:;;1 Pfi - o. (23) The properties of the associate Legendre functions are given in Appendix A. For interior problems, 0 in equation (20) is taken to be zero to satisfy the regularity condition at the origin. 1 45 The components of Xn can be written as: n = n m ' x1“ R z 0 Pn(;) [A1mn Cos me + B1mn Sln me] (24) Let the constant Ln be Ln = [2(3n+1-2o(2o+1))]‘1 . (25) Substituting equation (24) into equation (11) and applying equations (15) and (16), the displacements may be expressed in terms of A1 n and Bimn as follows: I'll uR = A100 Cos 6 Sin 0 + A200 Sin 6 Sin ¢ + A300 Cos 1 1 ( (26) A101 R P?(C°5 4)) Cos 6 Sin 2 + ( A201 R P?(C°5 PI) Sin 6 Sin P + ( A301 R P?(C°5 4)) C05 6 + (A 1 . . _ 111 R C05 6 P1(Cos o) + B111 R Sin 6 P}(Cos ¢))Cos 6 S1n o + (A211 R Cos e P}(Cos o) + 3211 R Sin 9 P1(Cos 6))Sin a Sin 1 + 1 . (A311 R Cos 6 P1(Cos o) + B311 R Sin 6 P}(Cos ¢))Cos o w n + z z A1mn {Rn(Cos m6 Cos o Sin 6(Pm(Cos ¢)(l-n(n-1)L ) + n=2 m=0 n n l (n-1)Ln Pu (Cos o) Cos o) - m(n-1)Ln Sin m6 Sin 6 Csc o Pfl(Cos 11)} n . . + Blmn {R (Sin m6 Cos 6 S1n ¢(P:(Cos ¢)(l-n(n-1)Ln) + 1 (n-l)Ln P2 (Cos o) Cos o) 46 + m(n-1)Ln Cos m6 Sin 6 Csc 6 P:(COS ¢))} + Azmn {R"(Cos m6 Sin 9 Sin ¢(P:(Cos ¢)(1-n(n-1)Ln) + 1 (n-1)Ln P: (Cos o) Cos o) + m(n-1)Ln Sin m6 Cos o Csc ¢ Pg(Cos 41)} .1. n . , . m B2mn {R [51" m9 5‘" 9 51“ ¢(Pn(COS ¢)(l-n(n-1)Ln) + 1 (n-1)Ln P3 (Cos o) Cos o) m(n-1)Ln Cos m6 Cos 6 Csc o Pfi(Cos 01]} n m + A3mn R Cos m6(Pn(Cos o) Cos o(1-n(n-1)Ln) - 1 (n-1)Ln P2 (Cos o) $1n2¢) n . m + B mn R S1n m6(Pn(Cos o) Cos ¢(l-n(n-1)Ln) - 3 1 , (n-1)Ln P: (C05 6) Sinzo) u¢ = A100 Cos 6 Cos o + A200 Sin 6 Cos o - A300 Sin 6 + ( (27) A R P?(Cos 4)) C05 6 Cos o + ( 101 R P°(Cos 411 Sin 6 Cos o - ( A 1 201 A R P?(Cos 6)) Sin 6 + 301 (A R Cos 6 P1(Cos o) + B R Sin 6 P1(Cos ¢))Cos 6 Cos o + 111 111 (A211 R Cos e P}(Cos o) + B211 R Sin 9 P1(Cos 6))Sin e Cos o - (A311 R Cos e P1(Cos ¢) + B R Sin 9 P1(Cos 6))Sin ¢ 311 m n n m + 2 z A1mn R {Cos m6 Cos 6(P (Cos 4) Cos o ( n=2 m=0 n 1 1 + (oz-m2 CSC2¢)Ln) — P: (Cos ¢)(-n 51n2¢ + l)Ln) 1 + an Sin m6 Sin 6(P:(Cos o) Csc ¢ Cot ¢ + P: (Cos o))} 47 + B1 n Rn{Cos 6 Sin m6(P:(Cos o) Cos o ( m I 1 + (oz-m2 Csc2¢)Ln) - P: (Cos ¢)(-n Sin2¢ + 11L") 1 an Sin 9 Cos m6(P:(Cos o) Csc ¢ Cot o + P: (Cos 4))} + A n Rn{Cos m6 Sin 6(P:(Cos o) Cos ¢(1 + (n2-m2 Csc2¢)Ln) 2m 1 P: (Cos ¢)(-n Sin2¢ + l)Ln) 1 an Sin m6 Cos 0(P:(COS o) Csc o Cot ¢ + P: (Cos o))} n . . I11 2_2 2 + BZmn R Sin m6 Sin 6(Pn(Cos o) Cos ¢(l + (n m Csc ¢)Ln) m1 . -Pn (Cos ¢)(-n Sin2¢ + 1)Ln) 1 + an Cos m6 Cos 6(P:(Cos o) Csc o Cot o + P? (Cos P11} + A n Rn{Cos m6 P2(Cos 6) Sin ¢(-l + (-n2+m2 Csc2¢)Ln) 3m m1 . + nLn Pn (Cos o) Cos 6 Sin 6} + B m" Rn Sin m6{P:(Cos 4) Sin ¢(-1 + (-n2+m2 Csc2¢)Ln) 3 m1 . + nLn Pn (C05 6) Cos 6 S1n 6} - _ - _ o . - A100 S1n 6 + A200 Cos 6 A10] R P1(Cos 6) Sin 6 (28) + A201 R P$(Cos o) Cos 6 1 . 1 . - (A111 R Cos 6 P1(Cos o) + B111 R Sin 6 P1(Cos ¢)) Sin 6 + 1 . 1 (A21] R Cos 6 P1(Cos o) + 8211 R S1n 6 P1(Cos 6)) Cos 6 + m n z z A n Rn{Cos m6 Sin 0(Pm(COS ¢)(-l + (n-m2 CSCZPILn) n=2 m=0 n 1m -Pm1(Cos o) Cos o L ) n n 1 + Sin m6 Cos 0(P:(COS ¢)(nm-m CSCZo)-mP: (C05 6) Cos ¢)Ln} 48 + B R"{Sin m6 Sin 6(P:(Cos ¢)(-1 + (n-m2 Csc2¢)Ln) 1mn m1 Pn (Cos o) Cos 6 Ln) 1 Cos m6 Cos 6(P:(Cos ¢)(nm-m Csczo) - m Cos o P: (Cos 6)Ln} + A Rn{-Cos m6 Cos 6(Pfi(Cos o)(-1 + (n-m2 Csc2¢)Ln) 2mn L “I n Cos o Pn (Cos 6)) + 1 Sin m6 Sin e(P$(Cos ¢)(nm-m Csc2¢) - m Cos 1 P: (Cos ¢»Ln} n_. m _ _2 2 + BZmn R { S1n m6 Cose(Pn(Cos ¢)( 1 + (n m Csc ¢)Ln) m1 Ln Cos o Pn (Cos 6)) 1 Cos m6 Sin 6(P:(Cos ¢)(nm-m Csczo) - m Cos o P$(Cos obLn} + 1 A an Rn Sin m6(n P:(Cos o) Cot ¢ + P: (Cos o) Sin o) 3mn 1 B Rn Cos m6(n Pg(Cos o) Cot ¢ + P? (Cos 6) Sin¢)an 3mn The strain tensor in spherical coordinates becomes 3U BUR 811 U 3U 3U U = _ R 1 __J£- _JE 1 R .__9 -._9 s(R.¢.e) - TR l/ZIR— 31R + 3R R 3 1/2[R Sin o as 1 6R R ] so u au au u .l._22 .1: ¢_ l.__2.-._2 R 31 + R 1(ZER Sin 6 66 + R 34 R ¢] (29) EU U U 1 e _2_ .ii )1 R $10 1 as + R c°t 4 + R The strain-stress relationship is as follows: 3 = AeI + 2p: (30) 49 Substituting the displacement equations (26), (27) and (28) into equation (29) and applying equation (30), the spherical stress tensor components become °RR ‘ A(A101 1 A201 1 A301 ' A111 ' A211 ' A311 ' B111 ' B211 ' B311) (3‘) + zo{(A101 P1(Cos 6) Cos 6 Sin 6 + A201 P1(Cos 6) Sin 6 Sin 6 o 1 + A301 P1(Cos 6) Cos 6 + (A111 Cos 6 P1(Cos 6) + B Sin 6 P1(Cos 4)) Cos a Sin 6 + 111 1 . 1 . . (A211 Cos 6 P1(Cos 6) + 8211 Sin 6 P1(Cos 6)) Sin 6 S1n 6 + . 1 (A311 Cos e P1(Cos o) + B311 Sln e P1(Cos 6)) Cos 61 n Rn-l + Z A Z 1mn {C05 m9 C05 9 51“ ¢ [Pm(Cos 6)[n(1 +26 n=2 m=0 n + 2(nu + A)(l-n)Ln)] + Pfil (Cos 6) Cos 6 (-1 - (l-n)(2np + 21)Ln)] + Sin m6 Sin 6 P: (Cos 6) m Csc 6 (A + 2(1-n)(A + nu)Ln)} + B1mn Rn-]{Sin m6 C05 6 Sin 6[P:(Cos 6)[n(x + 26 + 2(nu +A)(l-n)Ln)] + P31 (Cos 6) Cos 6 (-1 - (l-n)(2nu + 21)Ln)] - Cos m6 Sin 6 P2 (Cos 6) m Csc 6 (A + 2(l-n)(1 + nu)Ln)} Rn-l + A {Sin 9 Cosine Sin 6 [Pg (Cos 6)[n(x + 2n 2mn + 2(nu + A)(l-n)Ln)] + Pgl (Cos 6) Cos 6 (-A - (l-n)(2nu + 21)Ln)] - Sin m6 Cos e P? (Cos 6) m Csc 6 (1 + 2(l-n)(A + nu)Ln)} BZmn Rn']{Sin m6 Sin 6 Sin 6 (P: (Cos 6)[n(A + 2p + 2(np +A) (l-n)Ln] + P31 (Cos 6) Cos s (-1 - (l-n)(2nu + 21110)] + Cos m6 Cos 6 P: (Cos 6) m Csc 6 (R + 2(l-n)(A + nu1Ln)} °¢4 50 -1 + A3mn Rn {Cos m6 (nP: (Cos 6) Cos 6 (1 + 26 + 2(1 + nu)(1-n)Ln) + Pm1 (C ’ 2 n 05 6) S1n 6 (A + 2(l-n)(1 + nu)Ln))} + B Rn']{Sin m 6(nP: (Cos 6) Cos 6 (A + 2n + 2(1 + nu)(l-n)L ) 3mn n + p31 (Cos 6) Sin26 (1 + 2(1-o)(1 + nu1Ln))} = 1(A A + A - A - B - B - B 32) 101 + 201 111 111 211 311) ( + Zp(-A101 Cos 6 Sin 6 Cos 6 - A201 Sin 6 Sin 6 Cos 6 + A Sin26 301 - A111 C0520 C0526 - A211 Sin 6 Cos 6 C0526 + A311 Cos 6 Sin 6 C05 6 - B111 Sin 6 Cos 6 C0526 - 8211 Sin26 C0526 + B 11 Sin 6 Sin 6 Cos 6) 3 a n n-l m + z z A1mn R {Cos m6 Cos 6 [Pn (Cos 6) Sin 6 n=2 m=0 (A(n + n(2-2n)Ln) + 211(n3 - n2 + n + Csc26 (-n2 - n + m3 (-(n+1) + 3 Csc26)))Ln) + Pfll (Cos 6) Cos 6 Sin 6 (-1(1 + (2-2n)Ln) + 26(-1 +( -n2 + n - 1 + (m2+2) Csc26)Ln))] + m Sin m6 Sin 6 [PE (Cos 6)(1(1 + (2-2n)Ln)Csc 6 + 26(n2 + 2 -(m2+2) Csc26) Csc 6 Ln)-P:1(Cos 6) 66Ln Cot 6]} + B Rn']{Sin m6 Cos 0[P: (Cos 6) Sin 6(x(n + n(2-2n)Ln) lmn + 211(n3 - n2 + n + Csc26 (-n2 - n + m2(-(n+l) + 3 Csc26)))Ln) + P31 (Cos 6) Cos 6 Sin 6 (-1(+1 + (2’2")Ln) + 26(-l + ( - n2 + n - 1 + (m2+2) Csc26)Ln)))] -m Cos m6 Sin6 [PE (Cos 6)(A(1 + (2-2n)Ln)Csc 6 1 + 2u(n2 + 2 - (m2+2) Csc26) Csc 6 Ln) -Pfi (C05 4) 5111-n Cot ¢]} 66 51 Rn-l A {Cos m6 Sin 0[P: (Cos 6) Sin 6 (A(n + n(2-2n)L ) 2mn n 211(n3 - n2 + n + Csc26(-n2 - n + m2(-(n+1) + 3 Csc26)))Ln) I P: (Cos 6) Cos 6 Sin 6(-A(+1 + (2-2n)Ln + 26(-1 + (-n2 + n - l + (m2+2) Csc261Ln))] m Sln m6 Cos o[Pfi (Cos 6)(1(1 + (2-2n)Ln) Csc 6 I 26(n2 + 2 - (m2+2) Csc26) Csc 6L") - P: (Cos 6) 66Ln Cot 6]} B R”"{Sin m6 Sin 6[P: (Cos 6) Sin 6 (1(n + n(2-2n)Ln) 2mn 211(n3 - n2 + n + Csc26(-n2 - n + m2(-(n+l) + 3 Csc26)))Ln) PEI (Cos 6) Cos 6 Sin 6(-A(l + (2-2n)Ln) + 26(-l + (-n2 + n - 1 (nu-2+2) CSC2¢>1L11111 m Cos m6 Cos 6[P: (Cos 6)(1(1 + (2-2n)Ln) Csc 6 211(n2 + 2 - (m2+2) Csc26) Csc6Ln) - P31 (Cos 6) 6uLn Cot 6]} n-l m A3mn R Cos m6{Pn (Cos 6) Cos 6(Rn(l + (2-2n)Ln) 1 26(n3 - n2 + n + m2(-1-n) Csc26)Ln) + P: (Cos 6) 51n26( A(l + (2-2n)Ln) + 2u(1 + (n2 - n + 1 -(m2+n) Csc 6)Ln))} B R'"'1 . m 3mn 51" m6{Pn (C05 ¢1 COS 4(Kn(1 + (2-2n)Ln) 1 2u(n3 - n2 + n + m2(-l-n) Csc26)Ln) + P: (C05 6) Sin26( 1(1 + (2-2n)Ln) + 26(1 + (n2 - n + 1 - (m2+n) Csc26)Ln))} 1(A A + A - A - B - B (33) 101 + 201 111 111 211 ' 3311) 26(-A111 Sin26 + A211 Sin 6 Cos 6 + 8111 Sin 6 Cos 6 - 8211 C0526) m n Y. )3 A n=2 m=0 n-l m 1mn R {Cos m6 Cos 6[Pn (Cos 6)(Zu Csc 6( 52 (-2n2+n) Sin26 + n2 + n + m2(n + 1 - 3 Csc26))Ln + An Sin 6 (l + (2-2n)Ln)) P31 (Cos 6)(26 Cos 6((2n-1) Sin 6 - (2+m2) Csc6)Ln + (-1 Cos 6 Sin 6)(1 + (2-2n)Ln))] m Sin m0 Sine [P2 (COS 6)(26 CSC 6(1 + (-3n + (2+m2) CSC26 )Ln+-R Csc 6(1 + (2-2n)Ln)) + P21 (Cos 6) 66Cot 6 Ln]} B Rn']{Sin m6 Cos 0[Pg (Cos 6)(Zu Csc 6 ( 1mn (-2n2+n) Sin26 + n2 + n + m2(n + l - 3 Csc26))Ln An Sin 6(1 + (2-2n)Ln)) P21 (Cos 6)(26 Cos 6((26-1) Sin 6 - (2+m2) Csc 611n + (-A Cos 6 Sin 6)(1 + (2-2n)Ln))] m Cos m6 Sin 6[P: (Cos 6)(Zp Csc 6(1 + (-3n + (2+m2) Csc26 )Ln + A Csc 6(1 + (2-2n)Ln)) + PEI (C05 6) 6uCOt¢ Ln]} Rn-l A {Cos m6 Sin 6[P: (Cos 6)(Zu Csc6((-2n2+n) Sin26 + n2 + n 2mn m2(n + 1 - 3 Csc26))Ln + An Sin6(l + (2-2n)Ln)) + P21 (Cos 6) (2n Cos 6((2n-l) Sin 6 - (2+m2) Csc6)Ln + (-A Cos 6 Sin 6) (l + (2-2n)Ln))] - m Sin m6 Cos B[P: (Cos 6)(Zu Csc 6 (l + (-3n + (2+m2) Csc26)Ln A Csc 6(1 + (2-2n)Ln)) + PEI (Cos 6) 6uCot 6 Ln]} B Rn']{Sin m6 Sin 0[Pfi (Cos 6)(Zu Csc 6((-2n2+n) Sin26 + n2 + n 2mn 1 m2(n + l - 3 Csc26))Ln + An Sin 6(1 + (2-2n)Ln)) + P: (Cos 6) (2p Cos 6((2n—l) Sin 6 - (2+m2) Csc 6)Ln + (-A Cos 6 Sin 6) 53 (l + (2-201Ln))] m + m Cos m6 Cos 6[Pn (Cos 6)(Zp Csc 6(1 + (-3n + (2+m2) Csc26)Ln 1 + 1 Csc 6(1 + (2-2n)Ln)) + P? (Cos 6) 6uCot 6 Ln]} +A 4. 4. +8 .1. 4.. 0R6 Rn" Cos m6[P: (Cos 6) Cos 6(26 Ln(-2n2 + n + (m2+nm2) CSC26) I An (1 + (2-2611111) + P: (Cos 6)(2u Ln((-2n+l) 51626 + n + 62) 1(1 + (2-2n)Ln) Sin26)] R'“1 Sin m6[P: (Cos 6) Cos 6(26 Ln(-2n2 + n + (m2+nm2) Csc26) 1 An (1 + (2-2n)Ln)) + Pg (Cos 6)(Zu Ln((-2n+l) Sin26 + n + m2) 1(1 + (2'2")Ln151n 6)] n(A101 Cos 6(C0526 - Sin26) + A201 Sin 6(COSZ6 - Sin26) (34) 2A301 Cose Sin 6 2A111 C0526 Sin 6 Cos 6 — 2A211 Cos 6 Sin 6 Sin 6 C05 6 28 1 Sin 6 Cos 6 11 Sin 6 Cos 6 - 28211 Sin26 Sin 6 Cos 6 + A311 Cos 6(Sin26 - C0526) - - 2 _ 2 B311 S1n 6(S1n 6 Cos 6)) m n n-l m 2 2 A1 R u{Cos m6 Cos 6[P (Cos 6) Cos 6(n + 2(n-l)( n=2 m=0 mn n 1 n2 - m2 Csc26)Ln) + P: (Cos 6)(51n26(-1 + 2(n2-n)Ln) + 2(l-n)Ln)] 1 Sin m6 Sin 6 x 2m x (n-l) x Ln(PE (Cos 6) + Pg (C05 6) Csc 6 Cot 6)} B R"'1 u{Sin m6 Cos 6[P: (Cos 6) Cos 6(n + 2(n-1)( 1mn 1 n2 - m2 Csc26)Ln) + P: (Cos 6)(Sin26(-l + 2(n2-n)Ln) + 2(l-n)Ln)] 54 1 Cos m6 Sin 6 x 2m x (n-l) x Ln (Pm (Cos 6) + P: (C05 6) n Csc 6 Cot 6)} A R"" u{Cos m6 Sin 6[Pfi (Cos 6) Cos 6 (n + 2(n-1)( 2mn 1 n2 - m2 Csc26)Ln) + P: (Cos 6)(Sin26(-1 + 2(n2-n)Ln) 2(l-n)Ln)] 1 Sin m6 Cos e x 2m x (n-l) x Ln (P: (Cos 6) + P: (Cos 6) Csc 6 Cot 6)} Rn-l B2mn u{Sin m6 Sin 6[P: (Cos 6) Cos 6(n + 2(n-l)( 1 n2 - m2 Csc26)Ln) + P: (Cos 6)(Sin26(-l + 2(n2-n)Ln) 2(l-n)Ln)] '1 Cos m6 Cos e x 2m x (n-l) x Ln (Pg (Cos 6) + P: (Cos 6) Csc 6 Cot 6)} A Rn"1 u{Cos m6[P: (Cos 6) Sin 6(-n + 2(n-1)(m2 Csc26 - n2)Ln) 3mn 1 P: (Cos 6) Cos 6 Sin 6(-l + 2n(n-1)Ln)]} B 3mn Rn-l u{Sin m6[Pfi (Cos 6) Sin 6(-n + 2(n-l)(m2 Csc26 _ "2)Ln) 1 Pg (Cos 6) Cos 6 Sin 6(-1 + 2n(n-1)Ln)]} p(-A101 Sin 6 Cos 6 + A201 Cos 6 Cos 6 (35) I I - o 2 - o 2 2A111 S1n 6 Cos 6 Sin 6 B111 S1n 6(Cos 6 S1n 6) A21] Sin 6(C0520 - Sin26) - 28211 Sin 6 Cos 6 Sin 6 A311 Cos 6 Sin 6 - B311 Cos 6 Cos 6) on 11 z z A1mn R"" U{COS m6 Sin 6(P: (Cos 6)(-n +2(n-1) n=2 m=0 1 (n - m2 Csc26)Ln) — 2(n-1)Ln Cos 6 P2 (Cos 6)) 66 + + 55 Sin m6 Cos 6 x m(P: (Cos 6)(-1 + 2(n-l)(n - Csc26)Ln) m1 2(n-1) Cos 6 Ln Pn (Cos 6))} B R"" {Sin m6 Sin6[Pm (Cos 6)(-n + 2(n—1) lmn “ n 1 (n - 62 CSC26)Ln) — 2(n-1)Ln Cos 6 P: (Cos 6)] Cos m6 Cos 6 x m(P: (Cos 6)(-l + 2(n-l)(n - Csc26)Ln) 1 2(n-1) Cos 6 Ln P2 (Cos 6))} A2mn Rn"1 u{-Cos m6 Cos 6(P: (Cos 6)(-n + 2(n-1)(n - m2 Csc26)Ln) 1 2(n-1) Cos 6 Ln P: (Cos 6)) Sin m6 Sin 6 m(P: (Cos 6)(-1 + 2(n-l)(n-Csc26)Ln) ml 2(n-l) Cos 6 Ln Pn (Cos 6))} BZmn Rn.1 u{-Sin m6 Cos 0(P: (Cos 6)(-n + 2(n-1)(n - m2 Csc26)Ln) 1 2(n-1) Cos 6 Ln Pfi (Cos 6)) Cos m6 Sin 6 m(P: (Cos 6)(-1 + 2(n-l)(n - Csc26)Ln) m1 2(n-l) Cos 6 Ln Pn (Cos 6))} A3mn Rn']u m Sin m6(P: (C05 6) Cot 6(2n(n-1)Ln - l) + 1 P: (Cos 6) Sin 6(2)(n-1)Ln) 63mn R""o m Cos m6(P: (Cos 6) Cot 6(2n(n-1)Ln - 1) + 1 P: (Cos 6) Sin 6(2)(n-1)Ln) n(A101 Sin 6 Sin 6 - A201 Cos 6 Sin 6 (36) - - 2 - - 2 A111 2 S1n 6 Cos 6 Cos 6 B111(Cos 6 S1n 6) Cos 6 A211 (C0526 - Sin26)(-Cos6) + 8211(-2) SineaCos 6 Cos 6 56 A311 Sin 6 Sin 6 - B311 Cos 6 Sin 6) on n z 2 A n=2 m=0 n-l , 1mn R “{COS m9 51" 6[P: (Cos 6) Cot 6 Ln(-2n2 -2n 1 6m2 Csc26) + P: (Cos 6) Csc 6(Sin26 + (4 + 2m2 - Sin26(2n+2))Ln)] Sin m6 Cos 6[P: (Cos 6) Cot 6(-m +2m(-n2 - n + Csc26(m2+2))Ln) PEI (Cos 6) Csc 6(m(6 - Sin26(2+2n))Ln)]} B Rn-l o . m 2 1mn P{S1n m6 S1n 6[Pn (Cos 6) Cot 6 Ln(-2n - 2n 1 6m2 Csc26) + P: (C05 6) Csc 6(Sin26 + (4 + 2m2 - Sin26(2n+2))Ln)] Cos m6 Cos 0[P: (Cos 6) Cot 6(-m + 2m(-n2 - n + Csc26(m2+2))Ln) Pfil (Cos 6) Csc 6(m(6 - Sin26(2+2n))Ln)]} A Rn']u[-Cos m6 Cos 0[P: (Cos 6) Cot 6 Ln(-2n2 - 2n + 6m2 Csc26) 2mn 1 P: (Cos 6) Csc 6(Sin26 + (4 + 2m2 - Sin26(2n+2))Ln)] Sin m6 Sin 6[P: (Cos 6) Cot 6(-m + 2m(-n2 + n + Csc26(m2+2))Ln) Pfil (Cos 6) Csc 6(m(6 - Sin26(2+2n))Ln)]} B Rn']u{-Sin m6 Cos 6[P: (Cos 6) Cot 6 Ln(-2n2 - 2n + 6m2 Csc26) 2mn 1 P: (Cos 6) Csc 6(Sin26 +(4 + 2m2 - $102¢(20+2))Ln)] 111 Cos m6 Sin 6[Pn (Cos 6) Cot 6(-m + 2m(--n2 + n + Csc26(m2+2))L ) n Pm1 (C05 6) Csc 6(m(6 - Sin26(2+2n))Ln)]} n A Rn-l - m 2 - 2 2 3mn u m S1n m6[Pn (Cos 6)(1 + (2n + 2n 2 Csc 6(m +n))Ln) Pfil (Cos 6)(2 Cos 6)(n+l)Ln] B Rn']u m Cos m6[P: (Cos 6)(1 + (2n2 + 2n - 2 Csc26(m2+n))Ln) 3mn P21 (Cos 6)(2 Cos 6)(n+1)Ln] 57 The displacements and stresses in Cartesinal coordinates can be expressed in terms of the spherical displacements and stresses, as follows: u1 (Cos 6 Sin 6 Cos 6 Cos 6 -Sin 6 uR u2 = Sin 6 Sin 6 Sin 6 Cos 6 Cos 6 u¢ (37) u3 1Cos 6 -S1n 6 0 u¢ = 2 033 Cos 6 0‘12 013 Case -Sin 6 Sin26 Sin 6 Sin26 0 Sin 6 C0526 -Sin 6 Cos 6 Cos 6 Cos 6 {’6111 {60520 Sin26 - 2 - 2 022 S1n 6 Sin 6 %Sin26 Sin26 Sin26 Sin26 -2Cose Sin26 \023/ 1155in6 Sin26 -‘,Sin6 Sin26 -51n26 Cos 41 Cos 6 Sin26 0 C0520 C0526 Sin26 C0526 6Cosz6 Sin26 C05 6 C0526 Sin 6 Sin 6 -Cos 6 Sin 6/ Sin26 C0526 Sin26 C0526 Sin26 Sin26 0 -Sin26 -gSin26 gSin26 Sin26 0 C056 00526 0 Sin6 C0526 58 Geometric and loading symmetries may be used to simplify these equations. A solid geometry is symmetric with respect to the x1 x3 plane if each point (x1, x2, x3) of the solid the point (x1, -x2, x3) also belongs to the solid. The loading or the displacement on the boundary of a solid which is symmetric to the x1 x3 plane, is symmetric with respect to the x1 x3 plane, if (tgfi), téfi), téfi)) or (u1, u2, U3) on the boundary point (x1, x2, x3) of the solid is such that (t1fi), -téfi), téfi)) or (u1, -u2, u3) is specified on the boundary at point (x1, -x2, x3). Consider a solid problem using the eigenfunction expansion with load and geometric symmetries with respect to the x1 x3 plane, then uR and u are even functions in 6, and u is an odd function in 6, 1.6. ¢ 0 UR(R9¢96) uR(Ra¢9‘9) u¢(R.¢.e)' u1(R.¢.-e) (39) “6(R9¢ae) ='ue(Ra¢9’e) By simple calculation, the coefficients satisfy A = B = B = 0 (40) Similarly, if loading and geometric symmetries exist with respect to the x2 x3 plane, then uR(RI¢se) = uR(RI¢9“'e) u¢(R9¢96) = u¢(RI¢I“'e) (4]) “6(R9¢ae) =‘U6(R,¢,fl-9) 59 This implies the coefficients satisfy A1mn = BZmn = 0 if m is even A3mn = 0 if m is odd. Similarly, if the problem is symmetric about the x1 x2 plane, then UR(R:¢96) = UR(R9“'¢96) U¢(R,¢,9) ='u¢(R9"'¢Ie) (43) ue(R3¢96) “9(R9fl-¢96) which implies the coefficients satisfy = B = 0 if n + m is odd lmn 2mn (44) A = if n + m i . 3mn 0 5 even The constants A100, A200 and A300 correspond to rigid body displacements at the origin and may be chosen as zeroes. At the point, 6=0 or n, the displacements and stresses are functions of R only. If the loading and geometry is symmetric with respect to x1 x2 plane, x1 x3 plane and x2 x3 plane, and rigid body displacements at the origin are set to zero, the displacement uR can be expressed at 6=0 as = - 3 - s uR(R,0,0) (AH3 + 8213)( 12L 3R ) + (A115 + 8215)( 60L 5R ) (45) + - 7 3 - (A117 + 8217)( 168 L7R ) + ... + A301R + A303R (l 6 L3) + R5(1-20 L5) + A R7(1-42 L7) + ... A305 307 60 3.3 Numerical Solution The boundary conditions will be satisfied by a point matching technique. N points on the boundary which are not on the edges or at the corners, will be selected and 3N equations obtained with 3N coefficients, A1mn and B1mn forming simultaneous linear algebraic equations. The displacements and stresses inside the body and on the boun- dary "my be calculated after A1m and B1mn are determ1ned. 11 Usually, the eigenfunction expansion converges very fast with a few terms giving good results. To improve accuracy, more points were chosen on the boundary than terms, and an overdetermined system of linear equations was obtained, i.e. (46) 11 00+ Ax + where A is an m x n coefficient matrix with mgn, B is a vector of order m and R is the unknown coefficient vector of order n. The least-squares solution of equation (46) was used to determine R such that the Euclidean norm of A; - B was minimized. That is, to minimize + + n HAx - 8|) = [4 1 6 (4.. x - B1121‘/2 (47) J A subroutine, LLSQAR [32], was used to numerically solve this system (46). CHAPTER IV THE BOUNDARY INTEGRAL EQUATION METHOD APPLIED TO THREE-DIMENSIONAL LINEAR ELASTICITY 4.1 Introduction The success of the boundary integral equation method in two- dimensional linear elasticity is motivation for the extension of this method to three-dimensional linear elasticity. In section 2 of this chapter, the appropriate three-dimensional influence functions will be formulated by Kelvin's Solution [20]. In section 3, numerical solution of the boundary integral equations will be discussed and then applied to some specific examples in Chapter V. 4.2 Formulation of Boundary Integral Equations To formulate the influence functions, H11 q(Q,P1) and I1;q(Q,P1), for three-dimensional linear elasticity, Kelvin' 5 solution to the problem of a concentrated force in an infinite medium, is used. + Consider a concentrated force F applied at the origin in an + infinite elastic medium. Let R1 be a position vector with unit + vector iR and length R1. The resultant displacement vector and 1 stress components become K ++++ 0 =§l-[(3- 4v) F + (1R 'F) 1R I (1) 1 R1 1 “2+“??? 1*“: ++:*:* onm - —3-[-(F-1R)(1n-1m) + (1n-F)(1R -i m) + (i m-F)(iR o1") (2) R1 1 R1 R1 3 ++ ++ ++ +1-2o (F'1R1H1n’1R1mm'iR1H 61 62 where the constants, K1 and K2 are 7< ll [1666(1-o)]" 7< ll —1 2 —(1-2o)[86(1-o)1 G is the shear modulus and v is Poisson's ratio. Consider a unit line load in the qth direction at a source point P1 (x1p1, x2p1, x3p1) and+let 0 (x10, x20, x30) be some f1e1d p01nt. Then the position vector R2 from P1 to Q is E_ _ _ * * * 4 2 ' Q ' P1 ’ (X10 ' x1p1’ x20 ' R2p1’ x3Q ' X3p1) ' (x1: x2’ x3) ( ) The distance between P1 and Q is R2. Using equations (1) and (2), the influence functions H1j,q(Q,P1) and I1,q(Q,P1) are found to be K x* x? x? x? x? x* ”1'- (Q’P1) = ‘%’[51° Rg’+ 51 RR" 5' 'Rl'+ 1-2 1 J R] K x? x* (5) = .1. _ _1__£1 Ii;q(Q’Pl) R2 [(3 4V) qu + 2 ] 2 where 611 is the Kronecker delta. Consider a three-dimensional linear elastic solid V with boundary S as shown in Figure 20. The boundary conditions are 63 FIGURE 20 A Three-Dimensional Linear Elastic Problem 64 The solid is embedded in an infinite medium of the same material as V for which the influence functions, H13 Sq(Q, P1 ) and I1N(Q,P1) are given by equation (5). Consider an unknown fictitious traction + = (f1,f 2, f* 3), acting along the surface S, see Figure 21. The stress and displacement components for each point 0 interior to 5 due to f: are 611(6) = g; Hij;qP1 which is integrable. Equation (8) has a singularity as P1+P1 which leads to the extracted value 1/2 fa(P]) 61q by integration over a region very close to P1 which shrinks toward a zero area. Thus, the final form of equation (8) is 1/2 f*(P ) + ff H1 (P ,P1) f*(P1) n. (P ) ds (P1) = t1 P(P ), S j;q 1 (10) P on St 65 FIGURE 2l A Three-Dimensional Linear Elastic Body Embedded in an Infinite Medium of the Same Material 66 The integral of equation (10) is to be integrated in the Cauchy Principal Value sense. 1 The boundary integral equations (9), (l0) and (7) can be used to solve any boundary value problem in three-dimensional isotropic linear elasticity. For tractions specified on S, equations (l0) are solved for f*, then equations (7) are used to find the stress and displacement components. When displacements are specified on 5, equations (9) are solved for f* and again equations (7) are used to find 011 and u1. For mixed boundary problems, equations (9) and (l0) must be used to find f* and, using (7), 0.. 11 and u1 can be determined. 4.3 Numerical Solution To obtain a numerical solution to equations (9) and (l0), the boundary S is divided into N triangular surfaces, AS1(i=l,2, ..., N), on which the surface tractions or displacements are specified. Thus N S = U AS. (ll) 1 The resultant boundary data is then defined at the center, 01, of each traingular A51, as follows: 1.. = If t§(P1) ds (P1) 1.] ASJ. U.. = ff u?(P1) ds (P1) 11 AS. 1 J 1 1 (12) F.. = If ft(P ) ds (P ) 1‘1 AS. 1 J where i=l,2,3 and j=l,2, ..., N. 67 Then equations (9) and (l0) are integrated over Ask with respect to ds(P]), using equations (ll) and (12). This leads to N 1 1 u. = 2 F If 1.. (P .Q ) ds (P ) _ N 1 1 1 (‘3) Tik - 1/2 Fik + 2_ FqL ff Hij;q(P .QL) nj (P ) ds (P ) L—l ASk where i=l,2,3 and k=l,2, ..., N. When L=k, the integrals of equations (l3) are defined as, = 1 1 AUiq ££k11;q(P ,ok) ds (P ) T H P1 (P1 d P1 (14) A iq ‘ igk ij;q( ’Qk) "j ) S ( ) where i,q=l,2,3. Both AU and ATi are dependent only on the three vertices of the iq q triangular Ask, see Appendix B. + _ - . 1 Let R3 - (x13, x23, x33) be a p051tion vector from QL to P on AS with length R3. Equations (13) can be written as: k Ulk ‘ AUllFlk + AU12sz + AU13F3k (15) N K1 xis 1 4' Z [F]L{ff R— [(3-4v) + T] (15 (P ) L=1 ASk 3 R3 Lfk K X X + F {ff §l{—l§—3§J ds (P1)} 2L ASk 3 kg K x x + F3 {ff §l{-l§—§§J ds (P1)}J L AS 3 R2 k 3 68 U2k 1 AUZlFlk 1 AU22F2k 1 A” F 23 3k N K1x 13x 23 + z [F]L{ff fi—{Z ] ds (P1 )1 L=l Kask 3 R3 Lfk + 2111 ——£(3- 4.) + 33] ds (P1 )1 Mask R3 K + F3L{xr-——[ ”:3 33] ds (P P1)1] ASkR 3 R3 U3k 1 AU3111K 1 AU32F2k+ AU33F3k N + z [F1L1ffR—_{ H13 33] ds (P1 )1 L=l ASkR 3 R3 Li‘k K + F2L1IIR ——{ H23 33] ds (P P1)} Ask: 3 R3 *2 1 + F3L{ffk-—-[(3- -4v)+ x33] ds (P )1 R3 1 = (1/2 + AT] ) Fk + AT F + 31 F lk 21 2k 31 3k 2 N x x x +21fiUn-4(? +?4 1%n1u11+(%3+%2 ”2% L=l AskRg Ra V R3 3 1 R3 Lfk 2 1 x33 3 x13x33 1 1 n, (P 1 + (R: + 1-2. R3 1 n, (P )1 as (P )1 3 X X X X 112L111 ‘311%%3 2v 23x13111111 1P11 1 (P11’1 3-2v :3 231 ASkR3 R3 R3 3 R3 1 3x13123x33 1 1 "2 (P 1 + ((l-Zv) R; 1 "3 (P )] d5 (P )1 2 3 x13x33 1 3x13x23x33 1 H1£§ E311? R3 +1-2v R3 1"1 1P 1 1 1(1-21) R3 1 k 3 2 3 3 1 x13 3x13333 1 1 n2 (P ) + (R + (1_2v) 3) n3 (P )] ds (P )l] 3 R 3 12k =H112 lk 1 11/2 1 A1221 F2k 1 AT3213K 69 N k x x x2 x x x + Z [F1L11f _g{(§g§1+ $-2v 23 13) “1 (P1) + (-§l§1+ 1-2v 13 23) L=l AskR3 3 R3 3 R3 Lfk 1 3X13x23x33 1 1 n2 (P ) + ((]_2§) R3 ) n3 (P )] ds (P )1 3 2 3 k x x x x x +FLm—2—1(—1—3-+3_ 1323M (P11+(—?3+—3_ fin (P11 2 2 R 12V 3 1 R 12v 3 2 AS R 3 R 3 R k 32 3 3 X X X 33 3 23 33 1 1 + ( + ) n (P )] ds (P )} R3 I]-2v)R§ 3 k x x x x x x2 2 3 13 23 33 1 __33_ 3 33 23 1 F3L1££ EE{(1-2v)R3 n1 1P 1 1 1 R 1 (1:237331 k 3 3 2 3 1 x23 3 x23x33 1 1 n2 (P ) + (R3 + 1-21 R3 1 n3 (P 1] ds (P )1] 3 13k 1 ATl3Flk 1 A123sz 1 11/2 1 21331 F3k N k x x x x x x + 2 [F]L{ff —-2-{(R33 + ?_2 13 33) n (P1) + (3113 23 33) = 2 V 3 1 ( V) 3 L 1 ASkR3 3 R3 R3 Lfk 2 1 x13 3 x13x33 1 1 n2 (P1+1'§;+l-2v R2 )n3 (P )] ds (P)} 3 k x x x x x x2 2 3 13 23 33 1 _33_ 3 33 23 + avg 331*12, R3 "1 1P1+1R3 1112—27151 k 3 3 2 3 1 x23 3x23x33 1 1 n2 (P)+(-§-+-(T_—27)—R-§) n3 (P )] ds (P)} k x x x2 x x x2 2 13 3 13 33 1 23 3 23 33 1 + F3L1£g EE{(R3 + (1-2v)R3) n] (P ) + (fig—1+ 11—ZV1R3) n2 (P 1 k 3 3 3 + (x33 + 3X33 ) n (P1)] ds (P1)}] R3 (1-2v)R3 3 3 The trapezoidal rule can be applied to the surface integral of equations (15) by assuming H is a function defined over the area AABC, see Figure 22. The integral of H over AABC is approximated using function values at the vertices A,B,C and the center of the triangular O; i.e., 7O HCA) [4659 fiag) 1 Hm) A 0 ’9 6 FIGURE 22 The Function, H, on the Singular Boundary Element 71 (H(A) + 11(3) + H(C))+ 311(0)] Area (AABC) (l6) souv If H ds = [ AABC Equations (15) can be written more compactly as, RM. F. = BV 3 1kL 1L ... (17) 1 ik k 7M2 —l where RMikL is the coefficient matrix and BVik are the prescribed boundary values (either Tik or Uik1' Equations (17) involve 3N equations with 3N unknowns (FiL’ i=l, 2,3, L=l W2 .,N) forming simultaneous linear algebraic equations. Now let Q(x Q,x2 Q,x3 Q) be a point inside the body and let Rj= (x*j, x2j’x3j1 be the position vector from Q. to Q, and the length of Rj be Rj. Once FiL’ the fictitious traction, is known, the displacement and stress components at Q can be computed numerically as: x2* x* x* * 'k N k x .x . u](o)=>: 4.11:] (3-4.+—,1,,J'—)+F2 j:(3. 21+F . (711—311)] (18) J=lR. j R.2 33 R.2 .1 311.12 3 N k 'k * *2 * * X .X . X . X .X . u (o) = z —1;[F1.(—1;J—2l)+ F . (3-4.4.121) + F 421—3111 2 = J 2 23 2 33 2 J lR. R. R. R. 3 3 .1 3 N k 'k 'k *2 . .X .3X X . 3(0). - 2 Jdfi- ($311+ F2 (341—31) + F3 (3-4v + $111 3=1R. J R.2 112.2 33 R.2 .1 .1 R3 3 ‘k x* * N k x 3 2x - 2 * 3 1 31 "'2 0111111 ' 2.431113 ("1j11-2v211123 1 "2 1 j(—1‘)'1'*‘1 2v 1 J- .3 R.2 R2 J *2x* J J * 3le 3' ' 1 F33“ 1 '"33'1 (_‘)’17—1-2v [2.211 _ N k2 * 3"1 "2 * 3"23 O22(01 ‘ §=12331113 ('"1j 1 1- 23 R3 *21 1 123 ("21 1 (1 237;?“1 J x*2x * J J * 3" 2 "3 1 F3j ('"3j11 2v R.2 21] j * 'k x* x2* N k2 [ * 3 1 "31 * 3" 2 ”3 033(Q) = i=lR*3 F13 ( X13 l-2v R.2) +F (-x2j+ (l- WZvSR J *3 J J * 3"3' + F33 x3j+ 11..2V1R;2 )] 3 * * N k2 * 3x12x2 °12(Q1 1 §=,E?§{:1j ("13 1 IT:%371F21112J ("11131-W23f1 3 x11: x* x* RJ "2 "3 + F3J (3 ,_ 2.. R )1 j N k * * * 3X1 X X _ 2 3 013(01'§.{2153”11(*33 112—11171 +F2 “112—11531 3 a *2 * 3" 1 "3 + F3j 1 1-2\) R 2)] j * * * o(o)=§_:21r. (I—J—yJ—J31231w2j1x’in—Lfi—1 j* X* x2* RJ CHAPTER V EXAMPLES AND COMPARISONS 5.l Example: Cube Subjected to Uniform Compression A unit cube with origin at the center is shown in Figure 23. The load on the surface x3=f0.5 is o33=-l x l06psi, all other surfaces being traction free. The material properties of the cube are taken as: E = 2.667 x l07 psi v = 1/3 (1) A = 2.0 x l07 psi u = 1.0 x 107 psi 5.l.l Eigenfunction Expansion Method The symmetric properties of the geometry and loading allow the displacements and the stresses to be expanded in terms of A11], 82]], A (n odd, m odd, min) and A (n odd, m even, m 33 X/‘EW ’9‘FF— ' .X =3 ./ X2'-'0.000/ 2 0 25 x X .7 7' '38 /V 5 A :9 x,< ' x/(f ‘ ' ' FIGURE 28 Percent Error Contours at Different Planes (x2=constant) for 033 of Uniform Compression Problem using BIE48(d) 86 x «3 70 /:;;;;;;EEEEEiE%.N9 2 I (I x, e ~ 5 "3 4.99 / 20 /0 [V 5 "/ X230250 FIGURE 29 Percent Error Contours at Different Planes (x2=constant) for 033 of Uniform Compression Problem Using BIE48(e) 87 using EFEgZ in Figure 31 is 2.94 x 103 psi at (0.375, 0.5, 0.5). The maximum boundary error for 03 using EFEgZ in Figure 3 is 3.33 x 10.5 in. at (0.5, 0.5, 0.5). It shows that the maximum boundary error is on the edges or at the corners. M.L. Williams [34] proved that the plate with mixed boundary conditions, namely,clamped-free, can have a stress singularity if the vertex angle is greater than 63 degrees. Whether the extension of the above result can be applied to the three-dimensional linear elasticity or not is not known. However, in the problem of Figure 26 the edges -2.- have displacement conditions on one face and traction conditions on (x3=to.5, x1=t0.5, -0.5 X2 9,0 /Q0 4 xI ;33 = 0.000/ 79‘170 = 7 X2 0./25 W 3% 5 x3 ><2=0.25 7 # 70 30 20 l0 . -- h l ‘ 5 X3 A 70 X5=0375 .:::::::::::::::: (7 x,v - FIGURE 40 Percent Difference Contours at Different Planes (x2=constant) for and EFE63 0 of Uniform Dis lacement Problem Usinq BIE 33 P ~ 48(e) 34 99 One can note that only eight percent of the volume is below five percent difference in Figure 39, and thirty-three percent of the volume is within ten percent difference. 5.3.3 Comparison of Total Cost, CPU, CMU and Matrix Size Tables 6 and 7 indicate comparison of total cost, CPU, CMU and Matrix Size for the eigenfunction expansion and boundary integral equation methods applied to the problems of Figures 23 and 26. 5.3.4 Comparison of Stress Solutions Using a Finite Element Program Table 8 indicates the comparison of stress solution, 633, for the problem of Figure 26 using a finite element computer program, SAPIV [35], 53 and BIE . the EFE34 48 100 Table 6. Comparison of Total Cost, CPU, CMU and Matrix Size of EFE and BIE for Cube Subjected to Uniform Compression Total Cost CPU_ QMU_ Matrix Size EFE?2 $ 3.13 35.97 Sec. 5.37 w-H 30 EFE?S $ 3.31 40.42 Sec. 5.31 w-H 36 Ethg $ 3.36 39.21 Sec. 5.69 w-H 30 EFEgi $ 4.54 54.88 Sec. 7.48 w-H 45 EFEgi $ 6.22 76.99 Sec. 10.06 w-H 63 EFEgi $ 7.64 93.58 Sec. 12.57 w-H 72 81612 $ .94 20.89 Sec. 3.29 w-H 36 811:24 $ 2.63 33.60 Sec. 3.99 w-H 72 BIE28 $ 4.48 58.14 Sec. 6.80 w-H 84 BIE48(d) $11.35 134.40 Sec. 19.73 w-H 144 BIE $11.36 134.73 Sec. 19.72 w-H 144 48(e) Table 7. EFE 101 Comparison of Total Cost, CPU, CMU and Matrix Size of EFE and BIE for Cube Subjected to Uniform Displacement Total Cost $6. $7. $6. $7. $10. $1 $4. $11 57 29 40 09 .92 .45 30 .24 .21 77. 92. 77. 91 105. 20. 36. 55. 133 132. 00 59 18 .11 70 44 11 75 .01 47 _cP_U Sec. Sec. Sec. Sec. Sec. Sec. Sec. Sec. Sec. Sec. 10. 12 10. 12. 19. 19. 19. 26 .49 28 11 70 .21 .51 .52 56 54 9m_u W-H W-H N-H N-H W-H W-H W-H N-H N-H N-H Matrix Size 63 72 63 72 63 36 72 84 144 144 102 . , , 63 Table 8. Comparison of Stress Solut1on Us1ng SAPIV, EFE34, BIE48(d) and BIE48(e) for Cube Subjected to Uniform Displacement Method 633 x 10'“ psi 63 Node ffff!_ EEE34_ BIE48(d) BIE48(e) (.125,.125,.375) 5.55 6.11 3.91 11.60 (.125,.375,.375) 5.89 6.17 12.21 3.13 ( 375,.125,.375) 5.89 6.17 12 21 3.13 (.375,.375,.375) 6.26 6.44 2.53 13.57 ( 125, 125, 125) 6.53 6.23 6.54 7.00 ( 125, 375,.125) 5.87 5.59 6.19 5.45 (.375,.125,.125) 5.87 5.59 6.19 5.45 (.375,.375,.125) 5.31 5.19 4.56 5.85 PLEASE NOTE: This page not included in material received from the Graduate School. Filmed as received. ' UNIVERSITY MICROFILMS CHAPTER VI SUMMARY AND CONCLUSIONS The stress and displacement components in the two examples of Chapter V appear to converge, using either the eignfunction expansion method or the boundary integral equation method, as the boundary points or subdivisions are increased. The convergence of the solution using the eigenfunction expansion method at points which are not located at an edge or corner is more rapid than at points on the edges and the corners. The coefficients, Aimn and Bimn (i=l,2,3,n=0,l, ... ,Osmsn), of the eigenfunction expansion method in both examples of Chapter V decrease in value as both m and n increase. The solution using the boundary integral equation method con- verges inside the body away from the boundary at which the fictitious traction is applied. The stress solution using BIE48(e) in the last example of Chapter 5 converges at some point on the boundary which is traction-free and away from the edges. It is still unknown how to explain this case. Protter and Weinberger [36] have indicated that the maximum error of the eigenfunction expansion is on the boundary. If the boundary conditions are either pure tractions or pure displacements then the maximum error of the expansion can be found without knowing the exact solution. This does not hold true, however, for the mixed boundary cases. In the first example of Chapter V, the boundary conditions are pure tractions. It indicates the maximum boundary error in the stress solution using the eigenfunction expansion is on the boundary. In the last example of Chapter V, the boundary stresses 104 105 on the fixed end changed rapidly when the number of points or terms in the eigenfunction expansion were changed, but the interior stresses were not greatly altered. The displacement boundary errors decreased, but the stress boundary values were sensitive to the number of points matched. The suitable choice of number of terms used in the expansion and the number of intervals used in the integration of the expansion for two-dimensional eigenfunction expansion to obtain maximum accuracy St111 15 not known. In the three-dimensional eigenfunction expansion, the number of points or terms needed for minimum error has not been determined. Both the point matching and the boundary integral equation methods become ill-conditioned when more points are matched or more subdivisions are taken on the boundary. Only the least-squares method, used in the two-dimensional eigenfunction expansion, avoids this ill-conditioning. The main cost of each computer program is in the solution of the system of equations, and varies directly as the cube of the order of the matrix. It is, therefore, possible to obtain improvement in the point matching solution by use of iterative techniques, see Ojalvo and Linzer [37]. Also, Massonnet [15] improved the efficiency of the traction boundary integral equation by the acceleration of the iterative methods. Let the system to be solved be Ax’=? (1) 106 where A is an non-singular matrix. Then the coefficient matrix can be split into the form A = N - P (2) where No and P0 are matrices of the same order as A. Then the one- parameter family of splittings is N(o) (l + o) No (3) P(a) (1 + a) No - A = P0 + 0N0 where A = N(o) - P(o) and o f -1. These iterates can be written as (v) _ :(v- 1) (v-1) x 1Tax m(M x + g) , v-=1,2, ... (4) where x is some arbitrary vector, Mo= NO Po and g= -No f. The con- vergence parameter a of the acceleration of the iterative methods lies between 0 - (n+m) P:_,(z)1 (3) (22-1) 3::(2) = (114-m)(n-m+l)(22--1);"2 P2'1(z) - mzP:(z) (4) varying degree, (n-m+1) P2,,(z) = <2n+1> zpfi12) - P§_,(z> (5) 122-1) 223333-= nzpfi - (n+m)P:_,=€ 36.2.1»... 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HHNNNNII(N CCEEE€_EE€.ENNN : 8 8 3 8 a = 8 =0123355xXXYYZIanu-IZXYZHNHHNNI9999NNI9IIC OGGGGGGXYZOOOXYZXXYYZZ.:EF.123B5.b7091111111N NNNNNIIIzZZNNNGGGGGGXYZOORIPT DSSSSSSUUUCODQRRRRQRRRDDDBBBBEBBBBSBBBBBESSSSSSBBBBBBUUUSSSSSSUUUCCH(NS 1 7 09 9.91 1 BIBLIOGRAPHY Neou, C.Y., "Direct method for determining Airy po1ynomia1 Stress functions", J. App1. Mech., 24(3), pp. 387-390, Sep., 1957. Mathieu, E., "Theorie de 1'e1asticite des corps so1ides", Gautheir-Vi11ars, Paris, 1890. Pickett, G., "App1ication of the Fourier method to the so1ution of certain boundary prob1ems in the theory of e1asticity", J. App1. Mech., 11, PP. 176-182, 1944. Litt1e, R.w., "Smin-infinite strip prob1em with bui1t-in edges", J. App1. Mech., 36. PP. 320-324, 1969. Timoshenko, S. and Goodier, J.N., Theory of E1asticity, 3rd edition, McGraw-Hi11, New York, 1970. Muskhe1ishvi1i, N.I., Some Basic Prob1ems of the Mathematica1 Theory of E1asticity, Noordhoff, Ho11and, 1963. 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Wi1ey, New York, 1965. A1tiero, N.J. and Sikarskie, D.L., Boundary-integra1 Equation Method: Computationa1 App1ications in App1ied Mechanics, ASME, AMD-Vo1. 11, pp. 119-141, June 1975. 141 BIBLIOGRAPHY 17. Miche11, J.H., Proc. Longon Math. Soc., 31, 100, 1899. 18. Love, A.E.H., A Treatise on the Mathematica1 Theory of E1asticity, 4th ed., Dove Pub. Inc., New York, 1944. 19. Lame, G., "Legons sur 1a theorie mathematique de e'e1asticite des corps so1ides", 1852. 20. Thomson, w. (Lord Ke1vin), "Note on the integration of the equations of equi1ibrium of an e1astic so1id", Mathematica1 and Physica1 Paper, V01. 1, Cambridge, 1882. 21. Lure, A.I., Three-dimensiona1 Prob1ems of the Theory of E1asticity, Trans1ated from the origina1 Russian by 0.3. McVean, Interscience Pub1ishers, New York, 1964. 22. Litt1e, R.w., E1asticity, Prentice-Ha11, New Jersey, 1973. 23. Zienkiewicz, 0.C., The Finite E1ement Method, McGraw-Hi11, New York, 1967. 24. Pian, T.H.H. and Tong, P., "Basis of finite e1ement methods for so1id continua", Int. J. Num. Mech. Eng., pp. 3-28, I, 1969. 25. Rashid, Y.R., "Three-dimensiona1 ana1ysis of e1astic so1ids - I, 11", Int. J. So1id Structure, pp. 1311-1331, V01. 5, 1969. 26. Me1osh, R.J., "Structura1 ana1ysis of so1ids", J. Stru. Div. Am. Soc. Civ. Engrs., 89. Pp. 205-223, 1963. 27. Hrennikoff, A., "So1ution of prob1ems in e1asticity by the framwork", J. App1. Mech., V01. 8, No. 4, Dec. 1941. 28. McHenry, D., "A 1itt1e ana1ogy for the so1ution of stress prob1ems", J. of the Int. of Civi1. Engrs., No. 2, Dec., 1943. 29. Cruse, T.A., "Numerica1 so1utions in three-dimensiona1 e1astostatics". Int. J. So1ids Structure. PP. 1259-1274, V01. 5, 1969. 30. Lanczos, C., Linear Differentia1 Operators, 1st. ed., Van Nostrand Reinho1d Comp., New York, 1961. 31. Papkovich, P.F., "Compt. Rend.", Vo1. 195, 513, p. 754, 1932. 32. LLSQAR, The IMSL Library, Internationa1 Mathematica1 and Statis- tica1 Libraries, Inc., Texas, 1974. 33. Neuber, H., "Math. Mech.", V01. 14, p. 203, 1934. 34. Ni11iams, M.L., "Stress singu1arities resu1ting from various boundary conditions in angu1ar corners of p1ates in extension", J. App1. Mech., 19, 526, 1952. 142 35. 36. 37. 38. 39. BIBLIOGRAPHY Bathe, K.J., Ni1son, E.L. and Peterson, F.E., SAPIV, A Structura1 Ana1ysis Program for Statics and Dynamics Response of Linear Systems, Co11ege of Engineering, University of Ca1ifornia, Berke1ey, June 1973. Protter, M.H. and Weinberger, H.F., Maximum PrincipJes in Differentia1 Equations, Prentice-Ha11, New Jersey, 1967. 02a1vo and Linzer, "Improved point-matching techniques", Q.J. Mech. and App1. Math., 18, 1, 1965. Abramowitz, M. and Stegun, I.A., Handbook of Mathematica1 Functions with Formu1as, Graphs and Mathematica1 Tab1es, Dove Pub1. Inc., New York, 1964. Jahnke, E. and Emde, R., Tab1es of Functions with Formu1ae and Curves, 4th ed., Dove Pub. Inc., New York, 1945. 143