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',.|3, .73 l'ni ' " din. . .3 i‘»“‘. 3366"“ "3 Ive. - 4:3 9/“:31“ ’33, I114; ' ’ I5 N? 3 ?E‘:\\:\ 33.3. g 3'.” '51: 5-,? W31"? '. 3:338}. $33,. 3‘.‘ . , -33, 3 Etc-(1‘. .. \)3.5§‘ (\J§‘ I 3,4332%? sEWN-Lt 33 “3‘3““. 0. v.3} 6 I . igcts' 23-1: , ”q- «33% 5'3 3 ‘- x "' "$351“ 33'“; 'i‘fi'L‘ (5:23 $fi§§£¥E ‘ ‘ ‘S'é‘ul‘: ”‘3 th'q‘. .2...” v. \l3‘3 5 LIBRARY l\\\ \\\\\\\\ \\\\ \\\\\\\\\\\\\\\\\lllllllll\\\\\\\\l ' M52312? S3 w, 3 1293 10396 4999 This is to certify that the thesis entitled ON THE PROBLEM OF A PLANE, FINITE, LINEAR-ELASTIC REGION CONTAINING A HOLE OF ARBITRARY SHAPE: A BOUNDRY INTEGRAL APPROACH presented by Ali Reza Mir Mohamad Sadegh has been accepted towards fulfillment of the requirements for Ph.D. . Metallur Mechanic degree in gy ’ S and Material Science Date §A%&8 0-7 639 gJ , “ . ifs partqu In , . _. “.wntfi ._ n . . l. Easlmk~FlASTlC ' i - "1" ‘-\:._ 9.1!.3 OE.- Metallurgv, nr .AHigs JL; fitterinl miiance - . _ 19'8 Lc:?,' 3;“9N THE PROBLEM OF A PLANE, FINITE, LINEAR- -ELASTIC REGION CONTAINING A HOLE OF ARBITRARY SHAPE: A BOUNDARY INTEGRAL APPROACH ‘1‘.” BY Mir Mohamad Sadegh Ali Reza 4': , '. ‘P‘ ' .A . . 'A ‘ < A.“ 13“" V -, 'v‘f‘lv .. {7. M- '.“ I I v A DISSERTATION ' ‘ Submitted to Michigan State University in partial fulfillment of the requirements for the degree of H , DOCTOR OF PHILOSOPHY 1‘ i 1978 G 'E \ h:‘~ huib Wu .5 .\. . 1A a» aha. t 1‘ L w I ‘ n HIM VIM I‘ it“ ~' m 1 - ' A) x, O .\_‘ . ABSTRACT ON THE PROBLEM OF A PLANE, FINITE, LINEAR—ELASTIC REGION CONTAINING A HOLE OF ARBITRARY SHAPE: A BOUNDARY INTEGRAL APPROACH By Ali Reza Mir Mohamad Sadegh Previous boundary integral equation methods have been developed for problems of two-dimensional elastostatics which yield excellent results everywhere except near the boundary. This presents a major disadvantage for problems in which a hole, slot or sharp crack is present, since such an opening must be considered as boundary. Thus, results in the vicinity of the hole are not reliable. In this dissertation a new formulation of the boundary inte- gral method is presented which eliminates this inaccuracy on and near the opening. This is done by replacing the kernel of the integrand (the influence function) with one which includes the effect of the opening. This influence function is determined in terms of the complex potential functions for an infinite elastic plane containing the opening and subjected to a concentrated line load at an arbitrary point. This is accomplished using the Muskhelishvili method of plane elasticity. Potential functions are found for the cases of a circular hole, an g Ali Reza Mir Mohamad Sadegh ptical hole and a sharp crack. The determination of ii! functions for other opening shapes is also discussed. z'bOundary integral equation method is then applied to finite regions containing either a circular hole, an ‘iptical hole, or a sharp crack. The results are pre- ted and compared with exact solutions and experimental .‘ults where available. To my brother, my mother and my father .. ' . k ‘ , M '21. an if} ACKNOWLEDGEMENTS . It is my pleasure to take this opportunity to express my deepest appreciation and gratitude to my major advisor, Professor Nicholas J. Altiero, for his contributions, guidance, and encouragement during the course of this investigation, and also for his friendship and painstaking review of this manuscript. Grateful appreciation is expressed to Professor flatacommittee, is appreciated. Thanks are extended to Professor ifirilf Financial support for this research was provided by ‘ V i;;§{¢he National Aeronautics and Space Administration under i , “;grant number NSC-3101. Partial financial aid was provided Special thanks are due to my family, especially my V'her, for their encouragement and moral support, and for 5genuine interest they have always shown in my work. TABLE OF CONTENTS ,r Page ;-"ACKNOWLEDGEMENTS . . . . . . . . . . . . . . . . . . iii ... LIST OF TABLES . . . . . . . . . . . . . . . . . . . vi =:,; LIST OF FIGURES. . . . . . . . . . . . . . . . . . . viii ; LIST OF APPENDICES . . . . . . . . . . . . . . . . . xi <.¢a LIST OE SYMBOLS. . . ._. . . . . . . . . . . . . . . xii ‘- INTRODUCTION . . . . . . . . . . . . . . . . . . . . 1 “a ’CHAPTER I - BACKGROUND AND PRELIMINARIES . . . . . 5 1.1 AN INTEGRAL EQUATION METHOD . . s 1.2 THE MUSKHELISHVILI METHOD: A COMPLEx- VARIABLE METHOD IN ELASTICITY. . . . . . . . 25 1.3 CAUCHY INTEGRALS AND RELATED . THEOREMS. . . . . . . . 34 :“CHAPTER II - GENERAL SOLUTION AND A MAPPING _ .**. TECHNIQUE. . . . . . . . . . . . . . . 38 11.1 INTRODUCTORY REMARKS. . . . . . 38 11.2 A MAPPING TECHNIQUE . . . . . . 38 :3 11.3 GENERAL SOLUTION. . . . . . . . 43 ' GHAPTER III - CIRCULAR HOLE IN A FINITE TWO- ,. DIMENSIONAL REGION . . . . . . . . . . 52 111.1 INTRODUCTION. . . . . . . . . . 52 III.2 DERIVATION OF THE INFLUENCE FUNCTIONS USING KNOWN POTEN- TIAL FUNCTIONS. . . . . . . . . 53 111.3 DERIVATION OF THE INFLUENCE FUNCTIONS USING A MAPPING TECHNIQUE . . . . . . . . . . . 58 iv H O “'7'? Cmt ll. P'v .|v. . ‘4 h, L.“ .' .‘L- _ .CHAPTER IV CHAPTER V CHAPTER VI III.4 THE BOUNDARY INTEGRAL EQUATION METHOD APPLIED TO PLANE FINITE REGIONS WEAKENED BY A CIRCULAR HOLE. . . . . ELLIPTICAL HOLE OR SHARP CRACK IN A FINITE TWO- DIMENSIONAL REGION. . . . IV.1 INTRODUCTION. . . . . . . . . . IV.Z DERIVATION OF THE INFLUENCE FUNCTION USING THE MAPPING TECHNIQUE: THE ELLIPTICAL HOLE PROBLEM. . . . . . . . IV.3 DERIVATION OF THE INFLUENCE FUNCTION: THE SHARP CRACK PROBLEM . . . . . . . . . . . . ON THE PROBLEM OF AN ARBITRARILY- SHAPED HOLE IN A TWO-DIMENSIONAL REGION . . . . . . . . . . . . . . . . v.1 INTRODUCTION. . . . V.2 THE CONTOUR OF AN ARBITRARILY- SHAPED HOLE AND THE MAPPING ’ FUNCTION. . . . . . . . . . . . v.3 ON THE INFLUENCE FUNCTION OF A PARTICULAR CLASS OF OPENING, CASE 1. . . . . . . . . ... . . V.4 ON THE INFLUENCE FUNCTION OF A PARTICULAR CLASS OF OPENING, CASE 2. . . . . . . . . . . v.5 ON THE INFLUENCE FUNCTION OF A MORE GENERAL CLASS OF OPENING . . . . . . . . . . . ~ CLOSURE. . . . . . . . . . . . . . .,. Page 73 92 92 93 116 156 156 157 167 181 202 206 210 253 Table 3.1 LIST OF TABLES Stresses and displacements in a rectangular region containing a circular hole at the origin, Case 1 . . . . . . . Rectangular region containing a circular hole at the origin, Case 2 . . Rectangular region containing a circular hole at the origin, Case 3 . Stresses and displacements in a rectangular region containing a nonsymmetrically located circular hole. Stress and displacement in a circular plane containing a circular hole at the origin, Case 1 . . . . . . . . Circular plane containing a circular hole at the origin, Case . . . . . . . Circular plane containing a circular hole at the origin, Case 3 . . . . . . . Stress and displacement of a circular plane containing a nonsymmetrically located circular hole. . . . . . . . Stress and displacement of a rectangular plane containing an elliptical hole at the origin . . . . . . . . . . . . . . . . Rectangular plane containing an elliptical hole (inclined major axis) at the origin, Case 1 . . . . . . . . . . . . . . . Rectangular plane containing an elliptical hole (inclined major axis) at the origin, Case 2 . . . . . . . . . . . . . . Rectangular plane containing a nonsymmet- rically located elliptical hole (inclined major axis). . . vi Page 81 83 84 85 88 89 90 91 136 137 138 139 Page Stress and displacement of a circular plane containing an elliptical hole at the origin. Circular plane containing an elliptical hole (inclined major axis) at the origin, Case 1. 144 Circular plane containing an elliptical hole (inclined major axis) at the origin, Case 2. 145 Circular plane containing an elliptical hole (inclined major axis 30°) nonsymmetrically located. . . . . . . . . . . . . . . . . . . 146 Stress and displacement of a rectangular plane containing a sharp crack at the origin . . . . . . . . . . . . . . . . . . . 149 Rectangular plane containing an inclined sharp crack at the origin, Case 1. . . . . . 151 Rectangular plane containing an inclined sharp crack at the origin, Case 2. . . . . . 152 Rectangular plane containing an inclined nonsymmetrically located sharp crack . . . . 154 Sharp crack (notch) on the side of a rectangular plane. . . . . . . . . . . . . . 155 LIST OF FIGURES Figure Page 1.1 Finite two-dimensional region with pre- scribed traction all around the boundary . . 9 Half plane subjected to a concentrated force on the boundary. . . . . . . . . . . . 9 Region of Figure 1.1 embedded successively in half planes . . . . . . . . . . . . . . 10 Boundary value problem in plane elasticity . 10 Region R embedded in an infinite plane . . . 12 Concentrated line load . . . . . . . . . . . 39 The problem of interest expressed as the superposition of two problems. . . . . . . . 41 Mapping the auxiliary problem to the disc. . 42 Fundamental problem expressed as super- position of two problems . . . . . . . . . . 60 Mapping the auxiliary problem to a unit disc . . . . . . . . . . . . . . . . . . . . 60 A unit circular hole in a plane finite region with prescribed traction on the boundary 0 C I O I I O I I O O O I C I I O O 7 4 Region R embedded in an infinite plane con- taining a circular hole at the origin. . . . 74 Circular hole symmetrically placed in a finite rectangular plate under uniaxial tension. . . . . . . . . . . . . . . . . . 80 Circular plane, containing a circular hole, subjected to radially uniform tension over a portion of the boundary. . . . . . . . . . 87 viii § A Figure Page 4.1 Superposition of the problem of ellip- tical hole . . . . . . . . . 95 4.2 Mapping the auxiliary problem to a unit disc . . . . . . . . . . . . . . . . 96 4.3 A sharp crack in an infinite plane . . . . . 117 4.4 An elliptical hole in a plane finite region with prescribed traction on the boundary, Be . . . . . . . . . . . . . . . . 123 4.5 A horizontal slit in a plane finite region with prescribed traction on the boundary, BS . . . . . . . . . . . . . . . 123 4.6 The finite region Re and RS, with sub- divided boundary and c0ncentrated line loads. . . . . . . . . . . . . . . . . . 124 4.7 Region Re, embedded in an infinite plane containing an elliptical hole at the origin . . . . . . . . . . . . . . . 125 4.8 Region RS, embedded in an infinite plane containing a horizontal slit at the origin . 126 4.9 Rectangular plane weakened by an ellip- tical hole at the origin . . . . . . . . . . 134 4.10 Circular plane weakened by an elliptical hole at the origin . . . . . . . . . . . . 141 .11 Rectangular plane weakened by a sharp crack at the origin. . . . . . . . . . . . . 148 Different contours for 2 and c = 1 . . . 159 2 and o . The first application of the methods of poten- 49.- tial theory to classical elasticity theory was introduced ti. —~ by E. Betti [7] in 1872. Later this work was expanded by :é; Somigliana [8], Lauricella [9] and others. In particular, , -Bettifs contribution, i.e., the general method of inte- ITS; grating the equations of elasticity, was simply a develop- ;Ii:ment of the potential methods of Green and Poisson. Thus, gsome fundamental results from potential theory should first be discussed. Let a function 6 be the solution of #Laplace 5 equation throughout a region R. V2¢ = o in R (1-1) 6 = f on GR; %%= g on 8R2 (1.2) 1" 3R; + 8R2 is the boundary of R. Note that ’17Fere r is the distance between two points in R, is a,“ L A”) . L1 A . _ I» if;' found. ‘ ¢ 6 . I ‘,;a,singular solutIOn to Laplace's equation. Combining 1/r 11;”with the solution ¢ in the classical Green's theorem of integral calculus [10] results in the identity, 4(2) 1 Where 2 is 1 l 3 1 If j:R[g(z°)FTZ:TFT - f(zo) 55 (TIITIITJ ]' d5(zo) (1.3) any point in R and Z0 any point on 8R; Since f and g are both needed everywhere on 3R, only one of To " them is known at each point, then the other has to be accomplish this [ll-13], consider taking the limit of equation (1.3) as Z approaches a boundary point '-T: Z}, on 3R. Ami) - 24,—, ’nfthe lim "eted in zf. g g is The result is: [[gczo) mi—m - H20) 5392—; (mf—Jw] d5(zo)=0 BR (1.4) it [11]. Thus, this integral is to be inter- the Cauchy principal value sense. Equation Solving the integral equation (1.4) for either given, or g, if f is giVen, leads to the car lea Een' 6t .‘ .1“. n\nv for: and - i 7 In spite of this classical foundation of the boun- , dary integral equation method, the literature contains at __£5 least two seemingly distinct formulations for the treat- ment of elasticity problems. One of these, due to Rizzo et a1. [14-16] and Cruse [l7], follows directly from Somigliana's identity of elasticity [18]. The other _formu1ation due to Massonnet [19] and extended by Altiero and Sikarskie [20] attacks the problem by embedding the region in an infinite plane and distributing a layer of body force on the proposed boundary in such a way that the desired solution is produced within the region of interest. Both approaches will be discussed in this chapter and the latter will be employed in the subsequent analysis. The formulation of the boundary integral equation method due to Rizzo [14] is based on Somigliana's identity: \ . 'Uq(Z) = [B Ii;q(Z,Zo) ti(Zo) dS(Zo) - LHij;q(Z,zo) Ui(zo) nJ-(Zo) dSCZo) (1-5) + where U is the displacement vector, Ii_q(Z,Zo) is the l ' 9 [,ith component of the displacement at 2 produced by a ’gnit force applied in the q direction at 20 in an infinite ‘Qedium, and ti and Hij°q(z’z°) nj(Zo) are the components ‘1 i'q’ respectively. In three dimensions, ( . , AJEEBESI 7",. Q 6 f ! prete tract 9 '12“. 31.2“ 8 Taking the limit as Z approaches a point 2; on B from the inside leads to % Uq(zi) + -/; Hij;q(zr.zo) Ui(zo) nj(zo) d5(zo) = L11;q(21120) ti(zo) (15(20) (106) where the integral on the left-hand side is to be inter- preted in the Cauchy principal value sense. If the traction is prescribed everywhere on B, then the right- hand side of equation (1.6) is known and a system of singular integral equations can be solved for the boundary displacement U. The interior displacements can then be found-from equation (1.5). If the displacements are prescribed everywhere on B, then the left-hand side of equation (1.6) is known but the resulting set of integral equations are not singular. For the mixed boundary condi- tions, some of the equations are singular and some are not . The second formulation of the boundary integral method, i.e., that of Altiero and Sikarskie [20], is an extension of the work of Massonnet [19]. Massonnet intro- duced a method for solution of traction boundary value problems in which the real body is embedded in a series of "fictitious" half planes which are sequentially tangent to the real boundary. To demonstrate the idea, consider a finite two dimensional region with a prescribed traction all around the boundary, Figure 1.1. Choose the «LA Figure 1.1. Finite two-dimensional region with pre- in ha} 10 '~< Figure 1.3 Region of Figure 1.1 embedded successively Thalf planes. T 1-9'; l f” I .‘ :‘T‘l “ m ,3... . I . , i X '7 ' a, 1" i a" Er 1.4 Boundary value problem in plane elasticity. 11 simple radial stress distribution, i.e., a half plane subjected to a concentrated line load on the boundary, as a fundamental singular stress field, Figure 1.2. This solution is well known [21]. Then, draw the tangent to a point, Z0, of the real boundary and consider the half plane extending indefinitely below this tangent, Figure 1.3. In other words, the body has been embedded in a succession of half planes. An unknown "fictitious" line load is introduced at each point of tangency. A vector boundary integral equation for the unknown fictitious tractions results when one forces satisfaction of the traction boundary conditions of the original problem. An approach somewhat similar to Massonnet's has been developed [22] for anisotropic regions subject to traction boundary conditions. This approach, later, was extended by Altiero and Sikarskie [20] to mixed boundary value problems. Consider a two dimensional, linear elastic region R with boundary B as shown in Figure 1.4. For prescribed boundary conditions, i.e., tractions and/or displacements on B, the stress field and displacement field in the region R are to be determined. The region R will be embedded in an infinite (fictitious) plane of the same material and thickness as R, Figure 1.5. The influence function which satisfies the equations of elasticity, i.e., H (Z,Zo) and Ii,q(Z,Zo), are known ij;q [23], where H. (Z,Zo) is the ijth stress component at UN a field point 2 due to a unit line load in the q direction . -—.- w an infinite plane. acti sclu .\~ DC a \ll r y 13 at a source point 20 and Ii;q(Z,Zo) is the displacement in the i direction at Z due to the unit line load at Z0. Con— sider now a fictitious layer of body force P* (unknown) acting along the contour B, see Figure 1.5. Since the problem is linear, then the superposition of fundamental solutions leads to the determination of stresses and dis- placements at a point 2 as follows: 0.. Z = H.. Z,Z P* Z d Z 13() [B 13;q( o) q( o) $( 0) Ui(Z) = [B Ii;q(Z,Zo) Pane) d5(zo) (1'7) where Z0 is now on the bOundary B and ds(Zo) is an element of length along B at 20. Then all equations of linear elasticity are satisfied by equations (1.7) since they represent the superposition of fundamental solutions. In order to solve the boundary value problem of interest, the boundary conditions on B are yet to be satisfied. These conditions are: U. = U: on B (1.8) where nj is the j—component of the outward unit normal to a point on B and Pi, U: are the specified traction and displacement components, respectively. Note that one may also specify one traction component and one displacement compc are l ‘0 Ant 14 component at a particular boundary point, provided they are mutually orthogonal. Let the interior point Z approach a boundary point 21, on B, Figure 1.5. Then the stresses and displacements, equations (1.7), must satisfy the boun- dary conditions of equations (1.8). Thus, substitution of equations (1.7) into equations (1.8) leads to I P§(Z1) + S¢i Hij;q(zl’z°) Pa(ZO) njCZi) ds(Zo) = Pfitzi) Z1 on B1: (1.9) 9531,;qczi,zo)1>;(zo)dsczo) = Uiczl) Z1 on Bu (1.10) Note that the subscript 1 refers to a co-ordinate direc- tion at a boundary point 21. Equations (1.9) and (1.10) represent coupled integral equations in the unknown fic— titious traction P*. Note that the singularity has been extracted from equation (1.9) and the integral of this equation is to be interpreted in the Cauchy principal value sense. Equations (1.9) and (1.10) contain several types of problem. For the first fundamental problem of IJlane elasticity, i.e., traction boundary conditions only, 12he vector equation (1.9) is to be used. For the mixed boundary value problem, both equations appear but not in 15 the same direction at the same point, i.e., if a traction is specified in the i direction at Z1, then equation (1.9) holds; and if a displacement is specified, then equation (1.10) holds. Like the Rizzo formulation, equations (1.9), i.e., the traction boundary value problem, are singular. However, for displacement boundary value problems, equations (1.10) are used and these are not singular. For mixed boundary value problems, some of the equations will be singular and some not. It is felt that the formulation of Altiero and Sikarskie is preferable for the following reason. In the method of Rizzo, one must first perform integration around the boundary before the required integral equations are defined. This is clearly not necessary in the Altiero and Sikarskie formulation, where one merely needs to specify the tractions and displacements themselves and the right-hand sides of the required integral equation are immediately known. Therefore, the Altiero and Sikarskie formulation will be used here. Note the fact that the Altiero and Sikarskie formu- lation is not restricted to embedment in an infinite plane. Massonnet, as discussed earlier, used embedment in a succession of half planes. However, to obtain singular equations for the traction problem, and there- fore more numerically efficient equations, this approach requires tangency of the half plane to the embedded body successively around the boundary. 16 The Massonnet approach is therefore somewhat cumber- some and particularly inconvenient, particularly for the solution of anisotropic elasticity problems [22]. Also, it is very difficult to apply this method to multiply connected regions. Whichever formulation is used, the fundamental solutions for the fictitious region should be simple. This is best satisfied by the infinite plane. Once P* has been determined, the stresses and dis~ placements at any field point can be determined by sub- stituting P* into equations (1.7). These stresses and displacements represent the solution to the boundary value problem of interest within R. The influence functions, i.e., the stress and displacement fields in component form, due to a concentrated line load P*ds in an infinite plane are given by Love [23]. These are: * = _ 1 * 2 2 2_ 2 Hxx;q Pq 4rr“ [erx(a1rx + azry) + P;ry(a3rX azry) 1 2 2 2 2 *=- * _ * Hyy;qpq 4nru [erx(a3ry azrx) + Pyry(a1ry+a2rx)] 1 2 2 2 2 *=-— * * xy;qpq 4nr“ [ery(a1rx + azry) + Pyrx(a2rx+a1ry)] (1.11) P* = - 1 [P*(a.rzlog r + asrz) — P*a5r r x;qq 4,,2 x y y xy 1 a = - _ * * 2 2 y;qPq 4rr2 [ ansrxry + Py(aur log r + asrx)] (1.12) where rx and ry are the x,y components of the radius vector from Z to 20 and the constants a1 through as for the problem of plane strain are a1 = (3-2v)/(1-v) a2 = (l-Zv)/(1-V) a3 = (l+2v)/(l-v) (3-4v)(1+v)/(1-v) 9: t l (1+v)/(1-v) (1.13) as and, for plane stress, v is replaced by v* in all the coefficients of equation (1.13) where: \J * = V 1——+v The influence functions found in equations (1.11) and (1.12) can be obtained using the complex potential func— tions associated with the concentrated line load in an infinite plane. This will be discussed further in the next section. The solution to any boundary value problem of plane elasticity is contained in equations (1.9), (1.10), and (1.7). For tractions specified everywhere on B, equations (1.9) are to be solved for P*. These values of P* are then substituted into equations (1.7) to find the stresses and displacements at any field point. Equations (1.7) may give the displacement field to within a rigid body displacement. The rigid body displacement, however, can A rq/A—J O-r-«I 'V 18 be eliminated by suitably prescribing sufficient boundary displacement information. For displacements specified everywhere on B, equations (1.10) are solved for P* and equations (1.7) again used to find the stress and dis- placement fields. For mixed conditions at a point either the x component equation (1.9) and the y component of equations (1.10) or the converse must be satisfied. To obtain a numerical solution to equations (1.9) and (1.10), the boundary is first replaced by an N-sided polygon with sides of arbitrary length ASi. The resultant boundary data over the interval ASi.is now defined at the midpoint of each interval as follows: P*. = / P*d , P*. = f P*d X1 AS. x S Y1 AS. Y s 1 1 P . = f Psds , P . = f Psds *1 AS. X Y1 AS. Y 1 1 Uxi = f Uids , U i = / Usds (1.14) ASi V Asi Y Note that the superscript * implies the fictitious com- ponent. Multiplying equations (1.9) and (1.10) by ds(Z1) leads‘to %P€(ZI) dS(ZI) + fiHij;q(zl’Z°) Pa(Zo) (15(20) . nj(21)dS(Zi) = Pfcz.) ds(Z.) 19 9% Ii;q(zi,zo) Pa(zo) d5(zo) ' dS(Zi) = U§(Z1) ds(21) (1.15) Integrating equations (1.15) over boundary interval ASi and assuming that the influence functions are independent of 20 over a particular interval yields: 1 P*PZ +¢Hn Z,Z P*Z -.2 d2 1‘ 1( I) B 1];q( 1 a) q( a) n3( 1) S( I) g Pi(zl) YE Ii;q(Zi,Zo) Pa(Zo) dSCZI) = Ui(Zi) (1~16) 71§Eppint of an interval by the interval length. This is 2sufficient for all intervals except for the second equa- r7 ‘ w‘.‘ .. c-.. ' ,J—_‘ )-- ' . “4.1-z".- . ..‘ a- ~—e-—-———-“ .. u. where over bound repre nunbe (cap—1 ’73 l 20 Si P§(Z1) + 1 Ii.q(zl!ZO) Pafzo) AS(ZI) o= ’ 20192, = Ui(z.) (1.17) where g. = _/- 1., (21,20) ds(Z1) (1.18) 1 As. 1’q 1 over the interval which includes 2; = 20. Note that the boundary points Z0 and Z; in the discrete terms will now represent the center-point location of the intervals, numbered counterclockwise. Separating x and y components, one obtains l 2' P;(Zl) + :1[H H’XX q(Zi,Zo) P*(Zo) nx(Z1) 2,921 + ny qCZl,Zo) P*(Zo ) ny (Z )] AS(ZI) = PXCZI) -]2'-P;(Zl) + :1[H H'X)’ q(zi,zo) P*(Zo) 11x (21) 0: Z0#Z1 (21.20) Pa(Zo) ny(21)] AS(Z;) = PYCZI) (1.19) M EXP;(ZI) + ;§;1 [Ix;q(21’z°) Pa(zo)] A5(Z1) = Ux(zl) H YYBQ gyp;(zl) + ::;1 [Iy;q(zl,zo) PaCZOI] AS(ZI) ' Uy(zl)(1-20) where the influence functions can now be written as Um I. liq s” ' D'J 3.4x- ; l 3'4 21 (21,20) P;(Zo) q(zi,z o) P*(Zo) = Hij' 'x y(21,20) P;(Zo) ’ ' Ii q(21,Zo) Pa(Zo) = Ii;X(Z1,Zo) P;(Zo) y(zi,zo) P;(Zo) i,j = x,y (1.21) .th .2IASubstituting equations (1.21) into (1.19) and (1.20) and ,¥,$‘ rearranging the equations leads to .‘.,.: N . LIZ!) + :E; { [H “xx x(21,20) nx (2,) a: Zo¢Zx + MXY .x(zi,zo) n y(zi)] P;(Zo) + [Hxx;y(zl’z°) DXCZI) + H (21,20) ny(Zl)] P;(Zo) } AS(Zl) = Px(ZI) XYEY A) f _1 g [H H,,. x(21.2 o) nxtzil 2 22.2. (21,20) HXCZII x(zi,zo) ny(zr)] P;(Zo) + [ny;y Hyy;y(21.zo) ny(zi) ] P;(Zo) 2 ASCZI) = Px(ZI) (1-22) *, gx'ka (JO ‘~< I b—rJ ‘ < :0- xi vi LL) ii ‘33.. u '3‘ 22 ';;5 ‘n;”.P;(Z1) + g { Ix;x(zl’z°) P;(Zo) 20’21 k. I:(u + Ix;y(zl,zo) P;(Zo) }AS(Z1) = Ux(zl) éiz * .'r.'?v"’§‘z” + XX W‘) PM " 20‘21 + 1y;y(21,zo) P;(Zo) } 5(21) = UYCZI) (1-23) Qiyrfi-quations (1.22) and (1.23) can be written in the compact . 9}" "7: form: i=1,...N (1.24) : , my(21.20) nxczz) + ny;y(zl.zo) nycz,)]Ascz.) fl-- 4 v - “‘AfD- C.. 1] 0.. 1] *4 C) 23 Cij = [ny;x(zl,zo) nx(zl) + H yy;x(21.zo) ny(Zx)]AS(z1) Dij = [ny;y(zl,zo) nX(Z1) + Hyy;y(zl,zo) ny(zl):]AS(Z1) = = 1 Kxi Kyi 2 Bvxi = Px(Zl) , vai = Py(Zl) (1.25) for a traction condition specified and Aij = IX;X(Z1,ZO) AS(Z1) Bij = Ix;y(Zl,Zo) AS(Z1) Cij = Iy;x(Zl,Zo) AS(Z1) Dij = Iy;y(ZI,Zo) AS(Z1) Kxi = gx Kyi = gy Bin = UX(Z1) BVyi = Uy(Z1) (1.26) or a specified displacement condition. There are several methods for solving equation (1.24) for the fictitious traction. The first and simplest method is iteration. Iteration works particularly well for traction boundary value problems, equations (1.22). An initial choice of fictitious fractions equal to the actual tractions pro- duces fairly rapid convergence. For the mixed problem, equation (1.23), the iteration, in general, does not con- verge. A second method is matrix inversion or elimina- tion. Equation (1.24) can be written in the matrix form r1 XX x}. X‘.‘ 24 [[Kx] - [AH [B] {p;} €=%{Bvx} where all sub-matrixes can be found using equations (1.25) and/or (1.26). Equation (1.27) is written more compactly W =33) Once the fictitious tractions are determined, the as: stresses and displacements are found from the numerical approximation of equations (1.7): N . = * ‘ * 0xx 5%; [H xx; X(F’ i) Px xi + Hxx;y(F’1) Pyi] i5 0 = F P*i + H F ' P*. U i=1 [Hm KC ’1) yy;y( ’1) >71] 'Mz _ *i ' * XY [Hm X(F’ i) PX + ny;y(F’1) Pyi] 1=1 N = - * - * Ux 1:21 [Ix;x(F’l) Pxi + Ix;y(F’1) Pyi] N = I ' P*. + I F,' P*.] 1.29 Uy E [y;x(F’1) x1 y;),( 1) Yl ( ) where F is a field point and P*xi’ P;i are the components of the known fictitious traction at the interval 1. 25 It is clear from equations (1.29) that the stresses and displacements can be found at small expense anywhere in the field by simple summation. It is important to note that the embedment in an infinite plane can also be used for multiply connected domains, such as a region containing a hole. However, the hole would need to be treated as boundary. Discreti- zation of the boundary would therefore cause inaccuracy of the solution near the edge of the hole where the solu- tion is most important. The goal of this dissertation is to eliminate the contour of the hole from the boundary and to find a new influence function for the problem, which contains the effect of the hole, thus improving the accuracy of the solution along and near the hole. To accomplish this goal, the Muskhelishvili method will be employed. I.2 THE MUSKHELISHVILI METHOD: A COMPLEX VARIABLE METHOD IN ELASTICITY After the formulation of the linear theory of elasticity had been largely completed (by the middle of the nineteenth century), functions of a complex variable were introduced into plane elasticity problems in 1909 by Kolossoff [24] who, together with Muskhelishvili [25], developed the theory. However, nearly forty years elapsed before the theory, based on Kolossoff's idea, was brought to a successful conclusion. This was accomplished, in the main, by a group of Russian mathematicians inspired by the work of Muskhelishvili. The development has been 26 described by Muskhelishvili in two works [26,27]. The general solution of the fundamental biharmonic boundary- value problem can be made by means of two analytic func- tions of a complex variable. Consider the biharmonic boundary-value problem vzv2 U(x,y) = o in R U 6 = fe(s) on GR (1.30) ’ and let V2U = X(x,y) (1.31) Then, clearly, the function X is harmonic in R. Note that a harmonic function is a single-valued function of class C2 which satisfies Laplace's equation in R, i.e., V2X=0. For every harmonic function there is a conjugate harmonic function which satisfies V2Y=o where the function X + iY is an analytic function. Every analytic function is a C°° function because it has a series expansion. Also, an analytic function satisfies the Cauchy-Riemann equations and the Cauchy integral formulae. Thus, every analytic function is a harmonic function [28]. The complex conjugate of function X, i.e., Y(x,y), can be easily found by the Cauchy-Riemann equations to within an arbitrary constant. Thus, an analytic function of a complex variable 2 = X + iy can be constructed 27 F(Z) X + iY. Let ¢(2) =21ff1=(2)dz = X° + iY° (1.32) where X° and Y° are the integrated functions of X and Y. Then, ¢(Z) is analytic and its derivative is ax° . ay° + ¢'(Z) = M 1 fi— = %(X+iY). From the Cauchy-Riemann equations, it is clear that o = o = 1 X,X Y,y IX 0 = _ o = l x,y Y,x 4Y Let H(x,y) = U - X9x - Y9y (1.33) Then it is easy to verify that H(x,y) is a harmonic function, because v2 (U - x‘Zx-Y‘iy) = VZU - v.V(X‘3x) - v-V(Y‘2y) and the fact that X°, Y° are harmonic leads to Solri The d, 4“ 28 x - 2vx°- 2VY° l 1 1 1 X - 7X + 2'Y - 7X - '2-Y Thus, H(x,y) is a harmonic function, the complex conjugate of which can be easily found. Calling this conjugate K(x,y), function can be constructed so that: X(Z) = H(x,y) + iKCX,y) (1 34) Solving equation (1.33) for U leads to: U = XQX + YQV + H(x,y) and substituting the analytic functions of equations (1.34) and (1.32) into the above equation, the biharmonic function is obtained in terms of the two analytic functions, ¢(Z) and x(Z): U = Re[7¢(2) + x(2fl (1.35) Since ¢(Z) and x(Z) are analytic functions, it follows that U(x,y) is of class Cm in R. Denoting the complex conjugate values by bars, the equation can also be written as 2U = 7¢CZ1 + Z¢(21 + XCZ) + Xizj (1-36) The determination of stresses and displacements in terms of the two analytic functions will now be discussed. The 29 stresses can be written in terms of the biharmonic function: 0 = U XX ,YY Oyy = U,xx oxy = -U,xy (1.37) This leads to: Oxx + ioxy = -1(U, + 1U, ),y - ' = U + 'U 1.38 xx loxy ( .x 1 ,y).x ( ) Let W(Z) = x'(Z) Then, from equation (1.36), the expression U x + iU y can ’ , be written U + iU y = ¢(Z) + Z¢'121 + WZZ) (1.39) ,X Calculating the derivatives of equation (1.39) with respect to x and y and substituting into equations (1.38) leads to: - ¢'(Z) + ¢'(Zi - Z¢"IZ) - @TTZT' Q + H0 Q I ¢'(Z) + ¢'(21 + Z¢"ZZi + W‘ZZ) Q I Ho Q II Stresses in terms of two analytic functions, ¢(Z) and U(Z), can now be written as: 30 OXX + Oyy = 4Re [¢'(Z)] ny - Oxx + ZioXy = 2[7¢"(Z)+W'(Z)] (1.40) Finally, displacements in terms of the two analytic func- tions in the compact formula will be 2u(UX + in) = a¢(Z) - z¢'izi - W225 (1.41) where a = 3-4v for plane-strain problem or _ 3-v _ a - l+v for plane stress problem Stresses and displacements can be found individually, using equations (1.40) and (1.41), and are: on = Re [2¢'(Z) - Z¢"(Z) - mm] or”, = Re [2¢'(2) + Z¢"(Z) + ‘1“(Z)] ox), = Im [7W (2) + \v'm] u = Im [mm - 2W - m)”. U = Im [a¢(Z) - ZCD'iZS - (NJ/2p (1.42) 31 Now that stresses and displacements have been formulated in terms of the two analytic functions, ¢(Z) and 9(2), the structure and arbitrariness in the definition of the two functions is an issue to be discussed. If the state of stress in the region R is specified, from equations (1.40), one can prove that the single-valued analytic functions ¢(Z) and U(Z) could be determined to within a linear function Ci+y and a constant B, respectively [29]. In addition, if the displacements are prescribed, follow— ing equation (1.41), one can find that and aY - B = 0 Hence, when the stresses are given, the three constants c, y, B will be chosen in such a way that NC) = 0 Im¢'(0) = O 1(0) = 0 (1.43) and when the displacements are given, a suitable choice of y will be assured by the condition Mo) = 0 (1.44) Thus, using the conditions (1.43) and (1.44), the func- tions ¢(Z) and U(Z) will be determined uniquely [27]. 32 The structure of the two analytic functions for a finite and infinite simply connected regions has been discussed in [27]. Since the state of stress and the displacements can be expressed by means of the two complex functions ¢(Z) and f(Z), the fundamental boundary-value problems of plane elasticity lead to the determination of these func- tions from prescribed values of certain combinations of these functions on the boundary of the region. Beginning with the first boundary-value problem in which tractions are prescribed on the boundary, the biharmonic function in terms of applied tractions, f(s), can be written as U,x + iU,y = f(s) on SR The equation (1.39) leads to: ¢(Z) + Z¢'(Zj + W125 = f(s) on ER (1.45) The corresponding boundary conditions of the second boundary-value problem follow from equation (1.41): a¢(Z) - Z¢'125 - [25 = g(s) on 3R (1.46) where g(s) is a prescribed displacement function on the boundary. From either equations (1.45) or (1.46) one can obtain the two complex functions. However, mapping the region R into the inside or outside of a unit circle makes the determination of the two functions much simpler. Suppose the mapping function 33 Z = w(2;) (1.47) maps point in the region R, Z plane, into a unit circle Iclsl. The mapping function for a finite region where the origin is taken in the interior can be represented as a power series Z=chn Iclsl n=l n whereas for an infinite region, where the origin is an exterior point, the function is given by: n Z=%+Zn=o kn: lclsl The boundary conditions, equations (1.45) and (1.46), can then be written as ¢1(c)+ w“) “m, c + m. r. = Fm aqua) - W?) m; c - ‘77. c = cm (1.48) w' 2; where ¢[w(c)] = ¢1(C) and w[w(;)] = vim Equations (1.48) can now be solved for the two functions ¢1(C) and W1(;) by a power series expansion method or integrodifferential equations using Cauchy integral formulae [27]. Since the solution of the integrodifferential 34 equation reduces to the solution of the standard Fredholm integral equation, then the existence of a solution of equations (1.48) would follow, almost directly, from the Fredholm theory [27]. 1.3 CAUCHY INTEGRALS AND RELATED THEOREMS Since the integrodifferential equations method will be used to determine the two complex functions, it is important to discuss Cauchy integrals and related theorems briefly. The proof of the following theorems has been presented in [30] and in [27]. Suppose R+ is a finite open simply connected region enclosed by the contour B described in a counterclockwise sense. Denote the region exterior to (R++B) by R- and the points on the boundary B by t. Let f(Z) be a complex function analytic (holomorphic) in R+ and continuous on C. Then 2%]; 131,1". = £(2) for 2512* (1.49) and "2'31?ij fifgfi dt = fwm for 2512* 2%i./; %L%l dt = 0 for ZER- (1.50) Equation (1.50) is a necessary and sufficient condition that the continuous function f(t) defined on B can be the boundary value of a function analytic in R+. Let f(Z) be a complex fu an may Khol the 35 function analytic in R- including the point at infinity and continuous on B. Then 2%1'1; «f—E? dt = f(°°) for ZcR+ (1.51) 2%1']; i—(él dt = f(w) - f(Z) for ZeR' (1.52) The condition (1.51) that the Cauchy integral have a con- stant value in R+ is both necessary and sufficient for the continuous function f(t), defined on B, to be the boundary value of a function analytic in R-. Let ¢(t) be a complex function which satisfies the Holder condition on an arc L. Then the Cauchy integral <1>1(Z)= 7%1/14 $4? dt (1.53) may be shown to be a sectionally analytic function in the whole plane cut along the arc L. Further, the limiting values ¢+(t), o-(t) may be shown to exist on L and satisfy the relations ¢1+(to) - ¢1'(to)= we) ¢1+(to) + ¢1-(to) =‘#T./: isig dt (1-54) where to is a point on L and the integral in equation (1.54) is represented as a principal value. The assumption that ¢(t) satisfies the Holder condition is sufficient for the existence of the principal value. These results are referre [26]. Si determi followi used fr and let for the This is U5, the 36 referred to as the Plemelj formulae. They are derived in [26]. Since the unit circular region will be used in the determination of principal value of some integrals in the following chapters, the following integral form will be used frequently. Let a and b be constants where b/a°[w(c)] = ¢>°(Z) 119(5) = W°[w(c)] = “(2) and the derivatives are 44 * ¢ '(Z) ° w'(C) * ¢1'(C) 19m 1W2) - w'(c) * * * ¢1"(c) = d2 "(2) - w'ZU.) + cb '(Z) ~ w"(c) (2.2) To find the influence function, the derivatives of equa- tions (2.1) are needed, so ¢°'(2) + ¢*'(Z) ¢'(Z) = ‘1“(2) = w°'(z) + v*'(2) ¢"(2) = ¢°"(Z) + ¢*"(2) (2.3) i: where ¢ '(Z), ¢*"(Z) and W*’(Z) can be easily found from equations (2.2): Q) * *1 = ¢1'(C) <1 (2) C ¢*vc(z) = ¢I"(C) _ ¢I'(C)'w"(C) w'ZICI 45T3(C) * w*'(2) = 54% (2.4) Substituting equations (2.4) into equation (2.3) and recon- sidering equations (2.1), the requirements for the influ- ence function become: * ¢(Z) = ¢°(Z) + mm 1(2) = W2) + vim ¢*'(c) I _ 01 + 1 4) (Z) r 49 (Z) W I -— i w "(C 1(Z)-‘P°(Z)+—fiyrfijl * * ¢1v(z) = ¢0!1(Z)+ W .. (bl US$2.62) (C) (2.5) The complex potential functions for an infinite plane with a concentrated force, P, at a point 20, ¢°(Z) and W°(Z), are known (Maskhelishvili [27], Sokolnikoff [29], Green and Zerna [31]). (130(2) = ' 2% ln(Z-Zo) W0(Z) = C1 mim- ln(Z-Zo) + 7&1?- ' 2?.20 (2.6) To find the complex potential functions for the problem of Figure 2.2(A), it is necessary to find the complex potential function for the problem of Figure 2.3(D), i.e., ¢I(c) and 11(c). Since ¢:(c) and WT(§) have to be analytic in the domain, then as mentioned in Chapter I, just one boundary condition is necessary to find the complex potential functions, i.e., either of equations (1.45) or (1.46). Note that in the original problem, Figure 2.2(A), the boundary of the hole is traction-free. Recall equation (1.45) for the traction boundary condition [29]: 46 f(s) = £1 + if2 + const. = 0 Then equation (1.45) becomes: ¢(t) + t o'it) + Wit) = 0 on C (2.7) where t represents the values of Z on the contour C. Sub- stituting equations (2.1) into (2.7) leads to ¢*(t) + t ¢*‘(t) + P*(t) = - ¢°(t) + t ¢0'(t) + w°(t) on C (2.8) Clearly, the left-hand side of equation (2.8) is in the form of the traction boundary condition for the problem of Figure 2.2(C), since ¢*(t) and W*(t) are the boundary values of 6*(2) and P*(Z). Also, the right-hand side of equation (2.8) is known, since ¢°(t) and W°(t) are the values of ¢°(Z) and W°(Z) on the fictitious contour in the problem of Figure 2.2(B). Since the boundary condition for the problem of Figure 2.3(C) is known (equation 2.8), then the boundary condition for the problem of Figure 2.3(D) can be obtained by transforming (2.8) to the c plane using the transforma- tion functions Z = w(§) so that the boundary transforms by: t = w(0) where 0 represents the values of c on the circumference of 47 the disc. Hence, equation (2.8) becomes: ¢T(o) + 949-1- ¢’1"(o) + We) = (We) + ”(0) <12 0 “m o m + ngETZ) (2.9) * i: 9: where ¢1(o) and W1(o) are the boundary values of ¢1(;) and WT(;), respectively. Also, ¢?(o) and W9(o) are the boundary values of ¢9(c) and W?(c), respectively. The right-hand side of equation (2.9) is known, so let F(0) = -[p9(0) + “(0) ¢?'(0) + 19(0)] (2.10) (0'10) then, equation (2.9) becomes: ¢1'(0) + 11(0) = F(0) (2.11) ¢I(0) + “(0) :3 This is the mixed boundary condition for the problem of Figure 2.3(D) from which the two analytic functions ¢:(c) and PT(;) will be found. It is necessary to point out some characteristics of ¢t(c) and W?(;) before proceeding. As mentioned in section 1.2, ¢:(;) and Wt(c) must be analytic (holomorphic) inside y, the unit circle. Also, without loss of generality, it can be assumed that ¢I(0) = 0. Thus, ¢:(c) and WT(;) may be developed for Icl<1 in power series of the form 00 ¢I(5) = g;; akck , vI(c) = ;E% bkck (2.13) 48 where, in the first series, the constant term is absent * because of the condition ¢1(0) = 0. Furthermore, _ —k * —k ¢TZC1 = Ea; akC , V1(C) = 2E; HR: (2.14) Let g approach the boundary 7, i.e., c+o. Note that since the radius of the disc is equal to one, then: 00 = l (2.15) Equations (2.13) and (2.14) are valid for the boundary 9: i: values ¢1(0) and 11(0). Substituting equation (2.15) into equations (2.14) along with equations (2.13) for the boundary values, these become: ¢I(o) =2; akok , 111(0) = 1:0 bkok (2.16.a) ¢§105 = E:; aka-k , Wfloi = 25% Ska-k (2.16.b) Equations (2.16.a) show that ¢:(o) and Wf(o) have poles at infinity, so they are analytic functions inside the unit circle. Also, equations (2.16.b) show that 57(3) and TTTET have poles at the origin, so they are analytic outside of the unit circle. Using this analysis and employing the Cauchy integral formulas, the complex potential functions for the problem of Figure 2.3(D), ¢:(c) and W?(;), can be computed. To find ¢:(§), let both sides of equation (2.11) be multiplied by 49 l . do Zni o-c where C is a point inside 7, the unit circle. Integrating both sides of the equation counterclockwise around the unit circle leads to: Li ¢1(0)d0+7L_¢w(01¢1“105d0 2N1 n1 +2111¢W1T§Td=2:1;;552 it 3! Since ¢1(o) is analytic inside y and 91(0) is analytic (2.16) outside y, then due to Cauchy integral formulas (section 1.3), equation (2.16) can be written as: * l w(o) ojioi —;T—7-_ l dg F(0) ¢1(C) + TIT—{é m —O'C d0 + WI 0 - W Y O-C do (2.17) Note that the third integral of equation (2.16) becomes: Y 0:... where I???) is a constant. Equation (2.17) is an intergodifferential equation for ¢T(§). It contains an unknown constant WTTET, which can be determined by letting c = O and imposing the condi- * tion ¢1(0)= . Thus, if the value of W1(0) in equation 50 (2.17) is chosen arbitrarily and the corresponding solution * for ¢1(;) is found, then the actual value of WfIUI can i: be computed from the condition ¢1(0) = 0. This is due to the fact that if ¢f*(;) is any solution of (2.17) for a given wf(0), and if ¢f*(0) = auto, then ¢f*(;)-ao is a solu- tion of (2.17) with T¥TUT replaced by TTTUT+ao. Thus, WT(0) can be tentatively fixed, say VfTUT = 0. Also, as mentioned in section 1.2, in order to have a unique solu- tion for ¢T(§) and WT(;), the following conditions must be satisfied: 3! it ¢1(0) = 0 11(0) = 0 (2.18) * . To find W1(c), take the conjugate of equation (2.11) and multiply both sides of the equation by 11 do 2“ C where g is a point inside y. Integrating both sides of the equation counterclock- wise around the unit circle leads to: a 1 f dbin 5 £517 ¢ '(01 Zfll Y o- C do + 21w'(o) 3- C do 1 P*(o) _ l o mi 'L‘e-r, «5-me 55;?“ (2'19) An argument similar to that presented for reducing equation (2.16) to equation (2.17) can also be presented here to obtain: 51 it 1 m7 ¢1'(0) * - 1 f 13:" W*m mwdw‘mm'my e-.; d“ (2.20) Substituting condition (2.18) into equations (2.17) and (2.20) and rearranging leads to * = 1 f F(o) _ 1 $1.1m W o ¢1(C) "ZN—f Y 0"; (10 W Y w 0 0-; C10 (2.21) i: * _ l Fioi l 5‘. (oi o '(o) *1“) ‘ mi 0-; do ' 273 Y 31(0) é-r, do “-2“ It is easy to reduce the solution of the integrodifferen- tial equation (2.21) to the solution of the standard Fredholm integral equation. The existence of a solution of equation (2.21) would then follow, almost directly, from the Fredholm theory. The second integrals of the right-hand sides of equations (2.21) and (2.22) are left in general form since function w(C) has not yet been Specified. CHAPTER III CIRCULAR HOLE IN A FINITE TWO-DIMENSIONAL REGION 111.1. INTRODUCTION The effect of a circular hole on the stress distri- bution in an elastic region has attracted considerable attention for the past seventy years. The effect of a circular hole on an infinite plate subjected to uniaxial tension was first solved by Kirsch [32]. This work was extended to other load conditions by Bickley [33]. Howland [34] solved the problem of a long strip weakened by a circular hole subjected to uniaxial tension. Other load conditions were considered by Savin [35]. The effect of a circular hole on the stress distribution in a finite elastic region has been treated numerically and experi- mentally using several methods. In this chapter, the solution of the problem of a finite plane elastic region containing a circular hole and subjected to traction boundary conditions is presented. This is the first implementation of the mapping technique and boundary integral equation method. In section 2 of this chapter, some known complex potential functions [36] are used to find the influence function for an infinite domain weakened by a circular hole. In section 3, the 52 53 Muskhishvili method is used and the mapping technique is employed to determine the influence function directly. It is shown that the two results are identical. In the last section some example problems are considered and the results are compared to some known solutions, where available. The computer program for the computation is included in Appendix B. 111.2 DERIVATION OF THE INFLUENCE FUNCTIONS USING KNOWN POTENTIAL FUNCTIONS Consider an infinite elastic plane with a circular cavity of radius a centered at the origin. Let a force P act at a point 20 where IZoI>a. Bhargava and Kapoor [36] have constructed the potential functions, ¢(Z) and W(Z), for this problem. They assume, following Green and Zerna [31], that the complex potentials are of the form: ¢(2) = P {-1n(z-z ) - a 1n (z-iz—M A (z-fi)’1 2nio+li ° 7? Z? + a 1n 2} (3.1) (2) = F a 1n (Z-Z ) + 2 2 (2-2 )-1 + In (2-2:) “7—HT a+ O 0 p 0 7; + B (z-f‘-—:—)'1 + c (Zia-Y2 - ln 2 + 112'1 + E z‘z}(3.2) 20 o This choice clearly gives prOper singularities at the point of action of the concentrated force 20. Also, it satis- fies the condition of zero stresses at infinity. The 54 unknown constants can be found from the condition that and 0 normal and tangential stresses, orr re’ are zero at the boundary of the hole, i.e., (0 rr - 10re)2=o - The boundary condition can be written in terms of ¢(Z) and W(Z) following Muskhelishvili [27] as follows: 0,, - iOre = ¢'(Z) + 67177 - eZietf ¢"(2) + 1'(Z)1 (3.3) Substituting equations (3.1) and (3.2) into equation (3.3) leads to: 2- A=§B 120 (1286 2- B=3a20-B 120 F 82 2- c=B 12% B“ 2- n=8 12. --I:(Ol'-L)Zo 82 P- 82 E = - E a a2 P where 52 - Z°Z° Th f ' 1 d ° f p - -7;T' US, or an 150 ate p01nt- OTCC acting at the point 20, the complex potentials at the point Z are: SS 2 ¢(2) — 2“ 5+1 3 - ln(Z-Zo) - a 1n(2-%:) + a ln 2 0 F 2020-1 a2 -1 + (z-—) ZTTTOH'I) ; 322-: n P' 0.2 202-0-1 0.2 '1 1(2) = {a ln(Z-Zo) + ln(2-——) - 1n 2 - ——————(Z-——) ZTTiOH'1) 0 70 To z Z -1 o2 -2 2020-1 . 1 P — _ -1 + 02?: (2-2%) + ‘20 Z) + 71W“ 3 2°“ 2°) 2 - __ _ _ _ + a 20(Z-%E) l - a 202 1 - i%—- 2 1 - a 322 2: Without loss of generality, assume a = 1. Then 6(2) and U(Z) become: P 22 -1 MD = —(——T)- 3 - ln(Z-Zo) - a 1n( ° 1; 2” 0+ 270 + P 2020-1 . 1 Z"(“+1) 23 220-1 P 22 -l 2 2 -l W(2) ='_-T—__I 3a ln(Z-Zo) + 1n(——£——) - —£—£——- 211 CHI 22-0 Zia-l + ZoZofl + Zofo’l$+ P To + 02% (220-1)2 220 2"E“+15 z'z° 270-1 _ aZogz'tl _ 292'; (3.3) 56 Let these potential functions be written as: ¢(z) = 7FTETTj-{¢I(Z)} + 2FTSTTT {¢II(Z)} “1) = 113+ {‘1’1(Z)}* man) {111(2)} where 61, ¢II’ WI and W11 can be found by comparison to equations (3.3). Since ¢'(2), ¢"(Z) and w'(Z) will be needed, they will be listed here: 4"”) = minim} * Wis—WWI} 1v .. P 11 P .. ¢ (2) _ n a+ {¢I } + n a+ {¢II} w'(z) = fingTT {wi(z)} + 21(g117 {111(2)} (3.4) l a Z(ZZo‘1) 1 222 -1 ¢"(2)=+m_ + ° I ° 22(220-1)2 ¢' (2) = - - II 7% (220-1): 57 _ 2 ¢ii(z) = 202-0 1 . 2.2.0 7% (220-1)3 W'(Z) = _ 2b - __EZ§___ + o202o+1 + 2o I (Z'Z°) (220-1)2 7°22 7? , _ l (Zo2o-l)Zo 2(Zo2o—l)2o w (23-n-‘3‘ +_____+ - 11 ° 2(220-1) (220-1)2 (220-1)3 - 2335;}. (3.5) 2022 Hence, the influence function can be easily found as described in section 1.2. They are Hxx;qP3 = Re [2¢'(Z) - Z¢"(2) - w'(2)] *= 1 I! v Hyy;qPq Re 12¢ (Z) + 2¢ (2) + W (2)] *= 1! v ny;qpq 1m [2o (2) + W (2)] l * = _ - Ix:qpq Zfi-Re [o¢(Z) Z¢'125 W125] *=_l_ - ' - Iyquq 2m 1m [a¢(2) 2¢ 125 W125] (3.6) Substituting (3.4) into (3.6) leads to Hxx;q(Z,Zo)Pa(Zo) = 2115:17'512¢i(z) - Zei'(2) - wi(2)] p* + 12¢],(2) - 7111(2) - wi1(2) 1 P3] 58 R g - H + 1 + 12¢i1(z) + Z¢]i(Z) + wil(2) 5 Pt] Im “xy;q‘z’z°)P3(Z°) = 2113:1111 7¢i'(z) + 11(2) 1 P* Wham + 111(2) 5 W1 Ix;q(z,z )P;(zo) — 153%5117-[1a¢1(2) - 26:177‘- 1125 1 P* + 18811”) + z11157:j ’ 115Zj 15“] 1y;q(z,zo)1>a(zo) - fimy [{e¢1(2) - z—(‘Tei 2 — T71 z 113* +1e¢n(z) + 2‘1—(‘74511 z - mu 2 [15*] (3.7) where ¢I(Z), 11(2). ¢i'(2), ¢II(21. ¢i,(2), ¢ii(z)» w,(2). Wi(2), WII(Z) and WiI(Z) are defined by equations (3.4). These influence functions will be used to solve a simple problem by the boundary-integral method in section 4 of this chapter. 111.3 DERIVATION OF THE INFLUENCE FUNCTIONS USING A MAPPING TECHNIQUE The mapping technique which is presented in Chapter II is now employed to obtain the influence functions. Con- sider the problem of an infinite plane having a circular 59 hole of radius a at the origin and a concentrated point force P acting in the plane at some points 20 where IZol>a. This problem can be expressed as the superposition of two problems, Figure 3.1. In the problem of Figure 3.1(B), the concentrated point force P acting at the point 20 in an infinite plane is considered. In the problem of Figure 3.1(C), the infinite plane contains a circular hole with a prescribed traction acting on its circumference. The traction on the circular hole is equal in magnitude and opposite in direc- tion to the generated traction on a circular contour in the problem of Figure 3.1(B). By adding the solutions to the problems of Figure 3.1(B) and 3.1(C), the zero traction on the hole of the problem of Figure 3.1(A) is obtained. The solution of the problem of Figure 3.1(B) is well known (Muskhelishvili [27]) so that the required traction can be found. Also, the problem of Figure 3.1(C) may be handled by mapping into a unit circle (disc), Figure 3.2. Clearly, the mapping function for this problem is 2=w(c)=% which is conformal and one to one [37]. Without loss of generality, let a = 1. Let the complex potential function . for the problems of Figure 3.1(A), 3.1(B), and 3.1(C) be ¢(Z), W(Z); ¢°(Z), W°(Z); and ¢*(Z) and W*(Z), respectively. 60 (C) Figure 3.1 Fundamental problem expressed as super- position of two problems. Figure 3.2 Mapping the auxiliary problem to a unit disc. 61 Then the complex potential functions for the problem of Figure 3.1(A) are M2) ¢°(Z) + ¢*(Z) NZ) ‘1’°(Z) + W*(Z) where the derivatives are given by equations (2.2) through (2.5) in section 11.3. To find ¢*(Z) and W*(Z), the transformed complex potential functions which are given by equations (2.21) and (2.22) will be used. It is first necessary to calculate the integrals of equations (2.21) and (2.22). I; = 7.13% :37: ‘bgéoj do (3.8) :2 mi 273%.. (>313) .. .3... Taking the derivative of equation (2.16) leads to M'CO) = Z: kakok (3.10) k=1 and the complex conjugate is: ¢’1"(o§ = k2; kEkEk-l Since 03 = 1, then 62 W o = 2:1 szko’k"1 (3.11) The mapping function and its derivative, evaluated on the boundary, are = l ' = - __ MO) 0 , w (o) 02 so that “(0) = - 3— (3.12) w'ioi o3 Multiplying equations (3.11) by (3.12) gives: (0(0) . W; z: -kako'k'z (3.13) w'io) k=1 From equation (3.13), it is clear that the right-hand side is an analytic function outside of y, the unit circle. The value of the right-hand side at infinity is zero. Thus, due to the Cauchy integral formulas, the principal value of the integral of equation (3.8) leads to: _ 1 (0(0) W - I1" mi 00 O 0-; d0 - 0 (3.14) To calculate the integral of equation (3.9), consider the complex conjugate of equation (3.13), which is w 0 = - 03 (3.15) Multiplying equations (3.10) by (3.15) leads to: 63 oo ¢f'(o) = kzl - kakok+3 (3.16) + Clearly, the right-hand side of equation (3.16) is analytic inside the unit circle. Hence, following the Cauchy integral formulas, the principal value of the integral of (3.9) leads to: 12=m5€ 3‘30) ‘1’??? do = - 4: ewe) (3.17) Substituting integrals (3.14) and (3.17) into the general formulation for ¢f(g) and W§(;) (equations [2.21] and [2.22]), the complex potential functions for a circular disc with a specified boundary value, F(o), are obtained: M02) - 7.13% if) do (3.18) ‘P’fm = 7,13% SC: do + c3¢*'(c) (3.19) Y , For the case considered, F(o) and Fig) will now be calcu- lated. Rewriting equation (2.10) and taking the conjugate leads to: 13(0) = - («33(0) + %W1 o + W o] (3.20) F105 = - [¢‘,’(0) + % ¢‘{'(o) + 13(0)] (3.21) Substituting equation (3.7) into (2.6), the transformed complex potential functions ¢2(c) and Wg(§) are: 12 F l .773/ 64 ¢2(c) = - 71%ETT) 1n (ligii) (3.22) _ P' l-ZC P 2c 12(c) - a ZFTETTj-ln ( o ) + 2fi(a+1) 1320; (3.23) Taking the derivative of equation (3.22) and substituting c = 0 into equations (3.22) and (3.23) leads to: P (133(0) = " m 1H (1-3500) . _ P 1 ¢3 (O) ' ’ 2nCa+l) (1-200) 0 _ fi 1 ZOO P _ 700 WI(O) - a Znia+I$ 1n ( ) + n a+ 1-Zoo Taking the complex conjugates of equations (3.24) along with equations (3.12) and (3.15) will provide all the terms on the right-hand side of equations (3.20) and (3.12). Then F(o) and Fioi will be = l'ZoO _ _ + — l‘ZoC (3.25) No) Q{1n ( O ) a In (a 70)} Q{_o(o-70)} FTE) = Q‘{g§%%§%l-} + o'{1n (0-20) - a 1n.l;§19} (3.26) _ P where Q - 7ET6:T73 Substituting equation (3.25) into (3.18) leads to 73795 ”(12°C 9 1( J) l. _ n 0 0 $1 0- c do Zni Y -C do U+-§$I.¢§ (1‘Z°°) do (3.27) Y 0(0'70) (O'C) 65 (l;%12) is an Recalling the discussion in section 1.3, In analytic function outside of y, the unit circle. Because Zo>l, the function has two essential singular points at on and o = 1/20. The function is defined at infinity as: l'ZoC [1n ( O [It—J = In (’20) 0V then 1 1 200 1 9;. C o ) _ 2?: Y O_C do - 1n (~20) (3.28) Clearly, ln(o-Zo) is an analytic function inside y, the unit circle. Hence, the Cauchy integral formulas lead to: mf l——:—-—n(0; 20) d0 = ln(C' “2.0) (3.29) Also { l'ZoO 1 (1-Z°O) do = Residu 7F? 36 oco-zo)(o-c) Y 0(0'Zo)(0'5) O=C + Residu I { 1'Z°O 0:0 0(O'ZO)(O'C) or 1 (l-Zoq) = 1-20; 1 'TFI _______.+ ___. (3.30) Y 0(o-Zo)(o-c) c(c-fo) Zoc 66 Substituting the integrals of equations (3.29) and (3.30) into equation (3.27), the complex potential function can be obtained: cptcc) = Q{1n(-zo) - a ln(c-m} + M 1'Z°7° } (3.31) 70(C‘Zo) To obtain Wf(c) it is necessary to find the integral of equation (3.19), i.e., = 1 .96 Flo) I3 2N1 Y 0—; do Then from equation (3.26): _ 6(0-7 )do — 1 (0-2 ) I3 ’ 733$ (1-zoo§(o-;) 1 2%? Y no-; 0 do l-Zoo -. ln( ) - 3?%'5§ 0-: do (3.32) where —l+ C(O'Z°) do = Residu 0(O-Z°) ZNI SE (1 2003(0-C) Ulc{ -Zo(o-;;)(o-C) } + Residu C(Oiz°) 1 -Z°(O-ZF)(O-C) O: 7? After some simplification: 1 0(0-7 ) _ g 1-Z 2 my: (I-zoofio-c) ‘10 ‘ ' 2‘; ‘ _°—'°‘ (3.33) 67 Substituting equations (3.33), (3.29) and (3.28) into equation (3.32) leads to: 2%"? it?“ = Q{‘ 2;: $23} + U{1n(c°Zo) - a ln(-20)} (3.34) Taking the derivative of equation (3.31) gives: ¢T1(C)= Q{ '01 } + Q’{1_'Z_£E£ . _'1_____2.} (3.35) C'Zb Zo (C‘Zo) Substituting equations (3.35), (3.34) into equation (3.19) and following some simplification, the other complex potential function is: me) = Q { 7;“; - lit-71 - 953—}+Q{1n(c-Zo) - a ln(-z.) _ 1-z.Zo_ . ____C3 } (3.36) To (C’Zo)2 As is discussed in section 1.2, the two complex potential functions ¢f(§) and ?f(c), expressed by equations (3.31) and (3.36), are not a unique set of functions. Since the origin of the coordinates is within 7, then, following section 1.2, the uniqueness conditions for ¢f(c) and Wf(c) are: 68 ¢1(0) = 0 , WT(0) = 0 (3.37) Conditions (3.37) lead to the unique complex potential functions: We) Q{a ln(-7.) - a ln(a-70)}+ q{___1'zozo 702 1 1-2020 . 1 } (3.38) 70 6-2-0 _ _ c _ 6:3 + — _ - - wt) - Q{ 73— th} Q{1n(c 2,) ln( To) _ l;ZQZh.. __£3___.} (3.39) 2. (c-Zo)2 These can be rewritten as: ma) Q {mm} + 6{¢f1(c)} 8 {W0} + 8 {18102)} 81%;) where ¢f(c), 6f1(c), W?(;) and W¥I(c) can be obtained by comparison to equations (3.38) and (3.39) and are given in Appendix A. Then ¢§'(c) = Q {¢f‘(c)} + Q {¢§i(c)} ¢f"(c) = Q {¢f"(c)} + — {¢fi'(c)} Q wt'cc) = Q {wg'm } + Q {mm} (3.40) 69 where ¢’f'(c:) = - °‘_ C‘Zo me.) = - —_1'§°7° . _1_ Z0 (C'Zo)2 ¢*H(C) = a I (C‘70)2 cb’fi'hz) = ———1'Z°Z° - ____2 79 (C‘zh)3 2 _ _ wg'cc) = - .21. - 8.; L24 3:0) o (C'Zo) W’I‘i(z;) = __1_ - 1-ZoZo , r.’-(c-3Zo) 5'20 Z. (C'Zo)3 (3.41) Complex potential functions of equation (2.6) can be rewritten as: ¢°(Z) = -Q.ln(Z-Zo) 11°(z) = Q- 7° + (5.111(2-20) (3.42) Z-Zo and the derivatives are: ¢°'(Z)='Q°2T0 ¢°"(2) =+Q-—————1 (Z-Zo)2 ‘1’“(2) = -Q 7° +Q- 0‘ (3.43) (Z-Zb)2 2’78 70 Substituting equations (3.40), (3.42) and (3.43) into equations (2.5) leads to: 4(2) = Q {- 1n(2-zo) + ¢I(2) }+ 6{¢II(z.-)} 8(2) = Q {272 + WI(2)}+ Ma 1n(2- 2.) + 8* II(:)} - 0 ¢*'(c) (2) , _ -1 + I — ‘1’ (Z) ' Q{z-zo W2.)— (25%} ' _ -Z wf'(C) a + w¥i(C) w (Z) - Q{m+m}+q{z-zo W} 4"(2) _ Q{ 1 ¢I"(2) ¢I'(2;)w"(2)} (Z-Zo)2 (u'2(c) w'3CE) ¢*"(c) ¢*'(c)w"(c) + 6{—L———- II } (3.44) w'2(C) w'3(c) where ¢I'(c), ¢I"(c). 4* II(2). ¢II'(2), WI'CC) and VI I(c) are defined by equations (3.41). Substituting equations (3.44) into equations (3.6) along with the mapping function, w(§) = 1/C, and its derivatives, leads to the influence functions for a circular opening: Hxx;q(z’z°)Pa(Z°) = Re: Q*<;q(z’7‘°)1’q(zo)nx1 + ny;q(Z,Zo)Pq(Zo)ny1 A 1 j#i = Pxi (1:1,..N) N p. . Z{ 1 9. , + H z z P* z . A8. _7%_ + j=1 ny;q(Z’Z°)Pq(Z°)nX1 yy;q( ’ 0) q( °)ny1 1 j¢i = Pyi (1:1,..N) (3047) where the influence functions Hij,q(Z,Zo)Pa(Zo) given by equation (3.44) and the resultant fictitious traction on a given interval is represented by: * = * . * .= Pqi PXi + 1Pyi (1 1,..N) In equation (3.47) nxi and nyi are the components of the unit normal to the interval i. Also, Pxi and Pyi are the x and y component of the real resultant traction applied to the mesh i. Considering the influence functions, equation(3.44) and splitting each equation into two components of P*, i.e., P; and P;, leads to: II I: 'U a. + I: ’U a. * . Hxx;q(z’z°)Pq(Z°) xx;x X XX;y II I! ’U a. 4. IE ’U 3&- H y;q(Z.Zo)1’c"I(Zo) y yy;x ° x YY3Y ° II V 3} .4. ”U a- “ y;q(Z,Zo)Pa(Zo) x ny;x ° x ny;x 76 t = . * . * Ix;q(Z,Zo)Pq(Zo) Ix;x PX + Ix;y Py I Z,Z P* Z — I ° P* + I - P* 3.48 ysqc °) q( °) y;x x yzy y ( ) where Hij;q and Ii;q can be easily found by comparing equations (3.48) with (3.44), see Appendix A. Substituting equations (3.48) into equations (3.47) leads to: N P*. X1 * * * + :E:: 01 , - P + H , P ) - n . + (H , - P 2 j= xx,x x xx,y y x1 xy,x x j#i . it o = °= + ny;y Py) nyi :ASi PXi (1 1,..N) N 12*. Z: _§£ . * . * . . k + i=1 (nyzx PX + ny;y P y) ”xi + (Hyy;x PX j¥i . a . = .= + Hyy;y Py) nyi }ASi PXi (1 1,..N) (3.49) Rearranging equations (3.49): N P*. XI + [H - 11 . + H ° n ] ° P* +- HI ' n 7 .=1 xx;x x1 xy;x yi xi xx;y xi jfi * = + HXY.Y nyi] Pyi :AS xi ' N +” Z} 1 . , . * . + j=1 [ny;x nxi + Hyy;x nyi] Pxi + [ny;y nxi i=1 . . * = . . + Hyy;y nyi] Pyi :ASi Py1 (3 50) N 1 * . it . * = °= 7 PXi + ;§;(Aij pXi + Bij Pyi) BVXi (1 1,2..N) j¥i N % P;i + :E:(C.. - P*. + D.. ' P*.)= BV . (i=1,2..N)(3.51) .= 13 x1 13 y1 y1 where: ij xx;x xi xy;x yi B.. = H , ° n . + H , ° n . 1] XX:Y X1 xy,y Yl C.. = H , - n . + H , - n . 1] XV,X X1 yy,x Yl D.. = H - . + H - n . 1) xy;y nx1 yy;y y1 (i,j=l,...N) Euqations (3.51) represent a set of 2N equations with 2N unknowns, i.e., P* t ' = xi and Pyi for 1 1,...N. Methods for obtaining the solution have been discussed in section 1.1. 78 Writing equation (3.51) in matrix form: .. B.. P*. BV . 1) 13 x1 x1 C.. D.. P*. BV . _ 1] 1] d y1 y1 (3.52) Note that the diagonals of submatrices [Aij] and [Dij] are 1/2 and the diagonals of submatrices [Bij] and [Cij] are zero Equation (3.52) can be solved by matrix inversion, iteration, or elimination (Faddeeva [38]). Once the fictitious tractions are found, then the stress and dis- placement at the point F can be easily found following section 1.2. These are: . * , * (HXX;X(F,20) P . + Hxx;y(F.2o) PyI) xx . x1 1 N o = :E: [H ;X(F,Zo) - P§i - H (F,z,) - p*.] YY i=1 yy yy;y Yl N OxY = g;; [ny;x(F’Z°) ' P§i + HXY;Y(F’Z°) . p;i] N Ux = i=1 [IX;X(F,ZO) - p;i + Ix;y(F,Zo) - P;i] N Uy = 22; [Iy;x(F,Z ) - P;i + 1y;y(F,z,) - P;i] (3.53) 79 EXAMPLE III.l A Rectangular Plane Weakened by a Circular Hole Consider the rectangular region (10cm x 20cm) of unit thickness (h = 1cm) which is weakened by a circular hole of radius r = 1cm at the origin, see Figure 3.5. A uniformly distributed traction (w = 1.0 MPa) is applied to the t0p and the bottom of the rectangular region as shown. The boundary has been subdivided into sixty equally-spaced meshes, each of length 1.0cm, i.e., 10 meshes are defined on each of the top and bottom edges and 20 meshes on each vertical edge. The field points, the points where the stress and displacement are calculated, are chosen along the x,y axis and include points on the edge of the hole. These are also shown in Figure 3.6. The data, i.e., the coordinates of the nodal points, X(I) and Y(I), the resultant of the traction on each sub- division (calculated by the trapezoidal rule), BVX(I) and BVY(I), and the coordinates of the field points, XF(I) and YF(I), are read into the program (Appendix B). The results are presented in Table 3.1. The results are compared to the theoretical solution of a long strip weakened by a circular hole subjected to uniaxial tension (Howland [34] and Savin [35]). The program required 35 seconds of CPU time on a CDC 6500 computer. u b 6 5 7 a 5 4 3' 2 | 42 001 H3 * n4 ‘ Ins 4 FIG " -I7 .35 .. no . ‘ “K #20 [J ‘ 20 cm I 3' ‘:_\ “w“: 5 ¥ X IHHIHH q- Figure 3.5 Circular hole symmetrically placed in a finite rectangular plate under uniaxial tension. 81 Table 3.1. Stresses and displacements in a rectangular region containing a circular hole at the origin, Case 1 Geometry: rectangular plane (10 x 20cm2) (1 cm thickness) Load: w = 1.0 MPa Eccentricity: X0 = 0.0 Y0 = 0.0 ' E = 70000 MPa, u = 26315.79 MPa, v = 0.33 Field Coordinates U U Point X Y Oxx Oyy xy x y N0. cm cm (MPa) (MPa) (MPa) microns microns 1 1.0 0.0 70.0 3.13128 0.0 -o.oo14 0.0 2 1.2 0.0 0.32793 2.1486 0.0 -o.oo152 0.0 3 l 4 0.0 0.38043 1.70146 0.0 -0.00154 0.0 4 1.8 0.0 0.31226 1.33672 0.0 -0.00155 0.0 5 0.0 1 0 -1.ll98 0.0 0.0 0.0 0.00281 6 0.0 1 2 -0.44908 -0.0235 0.0 0.0 0.00291 7 0.0 l 4 -0.18639 0.10502 0.0 0.0 0.00298 8 0.0 1.8 -0.02031 0.36858 0.0 0.0 0.00314 9 -1.0 0.0 0.0 3.13128 0.0 0.0014 0.0 10 0.0 -l.0 -1.1198 0.0 0.0 0.0 -0.00275 AY Available Solution: 1 f f I I t_f ‘ Field Point No. References 1 0 =3.14 MP 34 35 S oxx= “1'11 MPa [341.[351 \J V 9 =3.14 MP 34 35 a”, a ( 1.( 1 10 0xx= -l.11 MPa [341,[35] 1 H I I I 4 \w 82 To see the effect of the size of the plane on the stress and displacement solutions, smaller rectangular planes, 8cm x 16cm and 6cm x 12cm, weakened by the circu- lar hole of radius r = 1cm at the origin were considered. The results are presented in Tables 3.2 and 3.4. The program has been written in such a way that, if different dimensions of the rectangular plane are needed, only one character, WR, is to be changed. Note that the proportionality of the long side to the small side remains constant and equal to 2.0. Also, for different locations of the hole, the new coordinates of the center of the hole XO,YO must be read into the program. Finally, the example of the problem of a rectangular plane (9cm x 19cm) weakened by an unsymmetrically located circular hole is solved and the results are presented in Table 3.4. Again, the CPU time was 35 seconds for each run on a CDC 6500 computer. EXAMPLE 111.2 A Circular Plane Weakened by a Circular Hole Let a circular plane of radius R = 6cm and unit thickness (h = 1cm), which is weakened by a circular hole of radius r = lcm at the origin, be considered, see Figure 3.6. A radially uniform distributed load (w = 1.0 MPa) is partially applied to the tOp and the bottom of the outer circumference, as shown. The boundary has been subdivided into sixty equally spaced meshes each of which covers 6 degrees of angle (0.6283cm) numbered from the top and 83 Table 3.2 Rectangular region containing a circular hole at the origin, Case 2 eometry: rectangular plane (8 x 16cm2) (1 cm thickness) oad: w = 1.0 MPa ccentricity: X0 = 0 Y0 = 0 = 70000 MPa, u = 26315.79 MPa, v = 0.33 ield Coord1nates oint X Y 0xx ofiy XY UX UY NO- cm cm (MP3) (M a) (MP3) [microns hicrons 1 1.0 0.0 ‘ 0.0 3.2212 0.0 -0.218 0.0 2 1 z 0.0 0.3339 2.2013 0.0 -0.229 0.0 3 1.4 0.0 0.382 1.738 0.0 -0.232 0.0 4 1.8 0.0 0.3040 1.3619 0.0 -0.234 0.0 5 0.0 1.0 —1.1821 0.0 0.0 0.0 0.404 6 0.0 1.2 -0.4851 -0.0294 0.0 0.0 0.418 7 0.0 1. -0.2090 0.0992 0.0 0.0 0.427 8 0.0 1.8 -0.0293 0.3662 0.0 0.0 0.450 9 -1.0 0.0 0.0 3.2212 0.0 0.218 0.0 10 0.0 -l.0 -1.1821 0.0 0.0 0.0 -0.392 0V 4111111“ ([[[[[10 K. 84 Table 3.3 Rectangular region containing a circular hole at the origin, Case 3 Geometry: rectangular plane (6 x 12cm2) (1 cm thickness) Load: w = 1.0 MPa . Eccentricity: X0 = 0.0 Y0 = 0.0 E = 70000 MPa, u = 26315.79 MPa, 0 = 0.33 Field Coordinates U U Point X Y Oxx O y Oxy X y No. cm cm (MPa) (Mga) (MP8) microns microns 1 1.0 0.0 ' 0.0 3.4341 0.0 -0.244 0.0 2 1.2 0.0 0.3456 2.3206 0.0 -0.257 0.0 3 1.4 0.0 0.3827 1.8198 0.0 -0.260 0.0 4 1.8 0.0 0.2724 1.4061 0.0 0.265 0.0 5 0.0 1.0 -1.306 0.0 .0 0.0 0.406 6 0.0 1.2 -0.555 -0.0425 0.0 0.0 0.429 7 0.0 1.4 -0.2486 0.0852 0.0 0.0 0.439 8 0.0 1.8 -0.0361 0.3580 0.0 0.0 0.460 9 -1.0 0 0 0.0 3.4341 0.0 0.244 0.0 10 0.0 -l.0 -1.306 0.0 0.0 0.0 -0.390 IY ( no ‘111111‘1 ([[[[[v 85 Table 3.4 Stresses and displacements in a rectangular region containing a nonsymmetrically located circular hole Geometry: rectangular plane (9 x l8cm2) (1 cm thickness) Load: w = 1.0 MPa Eccentricity: X0 = -0.5 cm Y0 = 1.5 cm - E = 70000 MPa, u = 26315.79 MPa, 0 = 0.33 Field Coordinates o o 0 U U Point X Y xx y xy x y No. cm cm (MPa) (MPa) (MPa) microns microns 1 1.0 0.0 0.0 3.16445 0.0 -0.3267 0.6244 2 1.2 0.0 0.3313 2.16904-0.0010>0.3376 0.6248 3 1.4 0.0 0.3841 1.71526-0.0013 -0.340 0.6251 4 1.8 0.0 0.31466 1.34361-0.0015-0.3415 0.6253 5 0.0 1.0 ~1.1624 0.0 0.0 -0.1193 1.033 6 0.0 1 2 r0.4736 -0.0269 -0.0038 -0.1180 1.044 7 0.0 1.4 ~0.2027 0.10251-0.0047’—0.1184 1.053 8 0.0 l 8 ~0.0294 0.3697 -0.0046 -0.1192 1.0763 9 -l.0 0 0 0.0 3.19990 0.0 0.0989 0.626 10 0.0 -1.0 [1.15149 0.0 0.0 -0.11812 0.2181 AY’ I Y 4 1H t I d :; / h \J 1 1 [4 [I 1) 86 counterclockwise. The field points are chosen along the x,y axis and include points on the edge of the hole. These are also shown in Figure 3.6. The data, i.e., the coordinates of the nodal points, X(I) and Y(I), the resultant of the traction on each sub— division (calculated by the trapezoidal rule), BVX(I) and BVY(I), and the coordinates of the field points, XF(I) and YF(I), are read into the program (Appendix B). The results are presented in Table 3.5. The program required 36 seconds of CPU time on a CDC 6500 computer. The effect of the radius on the stress and displace- ment solution has also been considered by solving the problem for R = 4.8cm and 3.6cm and r = 1cm. The results are presented in Tables 3.6 and 3.7. To obtain the solution for different radii of the plane, one has to change the character WR which is the ratio of the desired radius to the R = 6cm. Also, for a different location of the hole, the new coordinates of the center of the hole, XO,YO must be read into the pro- gram. To see the effect of eccentric placement of the circular hole on the stress and displacement field, the example of a circular plane (R = 5.4cm) weakened by an unsymmetrically located circular hole of radius r = lcm is solved and the results are presented in Table 3.8. Again, the CPU time was 36 seconds for each run on a CDC 6500 computer. 87 X 1‘ ‘I 1 I ” “ -‘ 3 ~\ J/f \b 5 I I ." \ 16" 4 \ _- a, \\ {3 \ .0 \\ \ ll \ ' \ ‘0 \u/”££“\y o \ I ‘I! b \ I \ I .M \\ I \ I 5‘5 \ I 0‘ 1'6 \ \\ ’fi 4" —X> . , J . / \ 6‘ I 6- / \ v / \ \ I, \ I... o I R' I l/ b I {/1 I, /l V I // I v I / / / I‘ I ._ ‘ 1’ ~. 36 \3 // / ‘1 “In...” \, Figure 3.6 Circular plane, containing a circular hole, subjected to radially uniform tension over a portion of the boundary. 88 Table 3.5 Stress and displacement in a circular plane containing a circular hole at the origin, Case 1 A ‘— Geometry: circular plane R = 6 cm (1 cm thickness) Load: w = 1.0 MPa Eccentricity: X0 = 0.0 Y0 = 0.0 E = 70000 MPa, u = 26315.79 MPa, v.= 0.33 Field Coordinates 0 U U Point X Y Oxx yy Oxy x y No. cm cm (MPa) (MPa) (MP8) microns microns 1 1.0 0.0 0.0 2.948 0.0 -0.272 0.0 2 1.2 0.0 0.2975 1.9438 0.0 -0.284 0.0 3 1.4 0.0 0.32771 1.4742 0.0 -0.287 0.0 4 1.8 0.0 0.2345 1.06002 0.0 -0.290' 0.0 5 0.0 1.0 r1.70816 0.0 0.0 0.0 0.323 6 0.0 1.2 ~0.8751 -0.0846 0.0 0.0 0.380 7 0.0 1.4 I0.5090 0.0332 0.0 0.0 0.388 8 0.0 1.8 -0.2547 0.31029 0.0 0.0 0.409 9 -1.0 0.0 0.0 2.948 0.0 0.267 0.0 10 0.0 -1.0 (1°7081 0.0 0.0 0.0 -0.318 =‘X. 89 Table 3.6 Circular plane containing a circular hole at the origin, Case 2 Geometry: circular plane R = 4.8 cm (1 cm thickness) Load: w = 1.0 MPa . Eccentricity: X0 = 0.0 Y0 = 0.0 E = 70000 MPa, u = 26315.79 MPa, v.= 0.33 Field Coord1nates o 0 U U Point X Y xx gy xy x y No. cm cm (MPa) (M a) (MP3) microns microns 1 1.0 0.0 0.0 3.0268 0.0 -0.288 0.0 2 1.2 0.0 0.3077 1.9808 0.0 -0.3011 0.0 3 1.4 0.0 0.3414 1.4806 0.0 -0.304 0.0 4 1.8 0.0 0.2502 1.0181 0.0 -0.306' 0.0 5 0.0 1. -1.8281 ‘0.0 0.0 0.0 0.347 6 0.0 1.2 -0.9085 -0.0886 0.0 0.0 0.362 7 0.0 1.4 -0.528 0.0397 0.0 0.0 0.412 8 0.0 1.8 -0.232 0.339 0.0 0.0 0.434 9 -1.0 0. 0.0 3.0268 0.0 0.282 0.0 10 0.0 -l.0 51.8281 0.0 0.0 0.0 -0.340 =‘X 90 Table 3.7 Circular plane containing a circular hole at the origin, Case 3 Geometry: circular plane R = 3.6 cm (1 cm thickness) Load: w = 1.0 MPa Eccentricity: X0 = 0.0 Y0 = 0.0 E = 70000 MPa, 0 = 26315.79 MPa, v-= 0.33 Field Coordinates o 0 U U Point X Y xx . yy xy x y No. cm cm (MPa) (MPa) (MPa) microns microns l 1.0 0.0 0.0 3.2127 0.0 -0.327 0.0 2 1.2 0.0 0.3298 2.0625 0.0 -0.341 0.0 3 0.0 0.3686 1.4878 0.0 -0.344 0.0 4 1.6 0.0 0.2748 0.9105 0.0 -0.344 0.0 S 0.0 1.0 -2.093 0.0 0.0 0.046 0.352 6 0.0 1.2 -1.0129 -0.0980 0.0 0.0 0.407 7 0.0 1.4 -0.550 0.0526 0.0 0.016 0.417 8 0.0 1.8 ~0.136 0.3982 0.0 0.0 0.485 9 -1.0 0.0 0.0 3.2127 0.0 0.319 0.0 10 0.0 -1.0 (2-0937 0.0 0.0 0.042 -0.343 IIY‘ —.—X 91 Table 3.8 Stress and displacement of a circular plane containing a nonsymmetrically located circular hole Geometry: circular plane R = 5.4 cm (1 cm thickness) Load: m = 1.0 MPa Eccentricity: X0 = -1.0 Y0 = 2.0 E = 70000 MPa, u = 26315.79 MPa, v-= 0.33 Field Coordinates o 0 U U Point X Y xx yy xy x y No. cm cm (MPa) (MPa) (MPa) microns microns 1 1.0 0.0 0.0 3.239 0.0 0.0 0.919 2 1.2 0.0 0.325 2.161 -0.019 -0.0116 0.9163 3 1.4 0.0 0.358 1.667 F0.0346 -0.0166 0.9118’ 4 1.6 0.0 0.259 1.239 -0.0629 -0.0232 0.898 5 0.0 1.0 -1.532 0.0 0.0 0.358 1.226 6 0.0 1.2 r0.6160 -0.0417 0.05507 0.344 1.239 7 0.0 1.4 80.2224 0.11601 0.0629 0.332 1.247 8 0.0 1.8 0.1352 0.4443 0.04002 0.2523 1.309 9 - 0.0 0.0 2.779 0.0 0.509 0.782 10 -l.0 1.5716 0.0 0.0 0.1903 0.5704 [1,, CHAPTER IV ELLIPTICAL HOLE OR SHARP CRACK IN A FINITE TWO-DIMENSIONAL REGION IV.1 INTRODUCTION Problems associated with stress concentration around holes in structures have motivated the effort to solve problems of plane elastic regions weakened by elliptical holes or sharp cracks. The solution for the stress near an elliptical hole in an infinite plane subjected to a uniform load was first obtained by Inglis [39] using complex potentials. Later this problem was examined experimentally by Durelli and Murray [40]. A method for the determination of stresses and displacements near the tip of a sharp crack in an infinite plane subjected to in plane load was developed in an infinite series form by Westergaard [41]. The effect of holes of more general shape on infinite planes has received considerable atten- tion, most notably by Muskhelishvili [27]. The problem of an elliptical hole or a sharp crack in a long strip subjected to uniform tension and compression has been treated experimentally and numerically using several methods and techniques. Yet, no solution for an arbitrary plane region weakened by an ellipse or crack is available. 92 93 In this chapter the solution for the problem of an arbitrary, finite, two dimensional elastic region, weakened by an arbitrarily located and oriented ellipse or sharp crack, is presented. This is the extension of the imple- mentation of the mapping technique and boundary integral equation method. In section 2 of this chapter, the Muskhelishvili method is used and the mapping technique is employed to determine the influence functions for an elliptical hole. In section 3 this influence function is extended to a sharp crack. Finally in the last section, some example problems are solved for an elliptical hole and a sharp crack at different orientations. These solutions are compared to some available experimental [42,43] and analytic [44] results. The computer programs are included in Appendices C and D. IV.2 DERIVATION OF THE INFLUENCE FUNCTION USING THE MAPPING TECHNIQUE: THE ELLIPTICAL HOLE PROBLEM In this section, the influence function for an infinite plane region containing an elliptical hole is derived. Consider an infinite plane containing an ellip- tical hole at the origin and a concentrated point force P acting in the plane at some point 20, where 20 lies on or outside of the ellipse, i.e., x o "< o (4.1) ”I N + °'| N \V ._n 94 where Z0 = Xo + iY0 and a,b are the semi-major and semi- minor axes of the elliptic hole. The problem can be expressed as the superposition of two problems, see Figure 4.1. The problem of Figure 4.1(B) is simply that of a concentrated point force P applied at 20 in an infinite plane and the problem of Figure 4.1(C) is that of prescribed traction acting on an elliptic hole in an infinite region. This applied traction on the elliptic hole is equal in magnitude and opposite in direction to the traction generated on an elliptic contour in the problem of Figure 4.1(B), by the concentrated point force P. Adding the solutions of Figures 4.1(B) and 4.1(C), the zero traction condition on the hole of the problem of Figure 4.1(A) is obtained. The solution to the problem of Figure 4.1(B) is known (Muskhelishvili [27]). Thus, the required traction can be found. To solve the problem of Figure 4.1(C), it is neces- sary to map this problem into a unit circle (disc), see Figure 4.2. It is easy to verify that the mapping function 2 = R(%+ m) for R > 0 and 0 < m s l (4.2) transforms the region exterior to the ellipse into a unit circle Iclgl (Churchill [37]), provided R and M are taken as: 95 .oHo; Hmowpawaao mo EoHn0ha on“ mo :ofiufimompoasw R: «mC H.v ohsmfla 3c 6261 -N N \J [L t x 96 .omfiv was: m ow EoHLOHQ xpmwfiflxsm ecu mcwmmwz N.v owsmflm «m: «E ocmi -N -\/ T «u 97 a+b R = T and m - 3—4'5 (4.3) where a and b are the semi—major and semi-minor axes of the ellipse, respectively, and are equal to: a = R(1+m) , b = R(1-m) The mapping function is conformal for, if w'CC) is con- sidered, i.e., w'(c) = -¥L-+-n1 o< II R(1+m) Cos 0 *< ll R(l-m) Sin 0 (4.4) If m = o, the ellipse becomes a circle and the transforma- tion function equation (4.2) becomes w(c) = R/c. How- ever, it will be seen that several expressions derived in 98 this chapter will be singular when m = o and therefore the analysis is invalid for the case of the circular hole. Since the case of the circular hole has already been treated, the following restrictions will be placed on m: o ll moz-Zoo+1 (4.26) U7 ll oZ-Zoo+m (4.27) Solving equation (4.5[a]), the mapping function, for c, yields: mcz — Z; + 1 = O (4.28) As discussed earlier, the mapping function, equation (4.5[a]), represents a conformal mapping, i.e., for every point Z exterior to the ellipse, there exists only one 104 corresponding point in the a plane interior to the circle. Since equation (4.28) is of the quadratic form and has two complex roots, then one root has to fall inside ; and the other root has to fall outside y, the unit circle. The two quardratics, equations (4.26) and (4.28), have the same coefficients. Thus, equation (4.26) has two roots, one inside and one outside y. Denote the root inside Y by ri and the root outside y by r0. Then equa- tion (4.26) can be written as A = moz-Zoo+l = m(o-ri)(o-ro) (4.29) where For examination of equation (4.27), consider the following mapping function: 2 = 4(2) = 2 +13 (4.30) This function maps points in the plane exterior to the ellipse onto points in the plane exterior to the unit circle. This mapping function is also conformal. For, if w'(§) is considered w'(;) = 1 - IL of"(c:) = Q (¢1§"(c))+ Q (¢’f'1'(c) ) 9’1"“) = Q (W§'(c))+ Q(‘¥’fi(c)) where ¢f‘(c), ¢f"(c), ¢fi(c), ¢ff(c), Wf'(c) and WfiCC) are given in Appendix C. The complex potential functions of Figure 4.1(B) and their derivatives are given by equations (3.42) and (3.43). Applying superposition and adding the two sets of poten- tial functions, expressed by equations (3.42) and (4.49), leads to the potential functions of the problem of Figure 4.1(A): 113 M2) (23- ln(Z'zo) + ”(03+ 53¢f1u): Y(Z) Q g z' + Wf(c) £+ Q';a ln(Z-To) + Wf1(c)£ (4 50) Z-o Since ¢'(Z), ¢"(Z) and W'(Z) are also needed for the influence functions, they are written here: -1 ¢*'(C) ¢¥i(§) ¢'(Z) = Q3* Z-Zo +'—TG:f—+ W3 and) - Q3 ——°——Z+w71—WC)+Q:M)°‘ +51%“) (Z-Zo)2 “" ¢*"(c) ¢*(c)w"(c) ¢*"(c) ¢"(Z) = 3 1 + I _ I £+ Q: I (Z-Zo)2 w'ZCC) w'3(C) w'2(c) (4.51) ¢*i(c)w"(c) : w'3(c) Finally, substituting equations (4.50) and (4.51) along with the mapping function and its derivatives, equations (4.5), into equation (3.6) leads to the influence functions for an infinite region weakened by an elliptic hole: Hxx;q(Z,Zo)Pa(Zo) H z,z P* yy;q( o) q(Zo) + 4. + + c“ n -———————— ¢f (c) - 114 Re Q*<:ZE%F.+ 2:2 mcz—l ¢f'(c) - 7 [———l—-— (2'20)2 2:3 (mCZ-l)3 W§'(c)) 3'” WW .1 on] To (z-Zo)2 C2 mgz-l 1 n z [EETEI§? + ¢§ (c) ¢§'(c)] - ff(c) (mcz-l)“ To 2:3 (Z'Zo)2 (mCZ-l)3 6* “2 mcz-l Z .___£:___ ¢ (mcz-l)“ . :2 mcz-l wg'm) *n C) _‘ 2§3 (mcz-l)3 II ¢ii(g)]* 115 2 a + C who) 2'70 mcz-l Im Q* Z'[ 1 + ———£:——- ¢*"(C) (. (Z’Zo)2 (ch-l)“ I * ny;q(Z,Zo)Pq(Zo) 2; *1 2-0 ‘ —-——- (b (C) - (m<:,2-1)3 I ] (Z-Zo)2 3 2 - ———3£——— ¢¥i(C3 + a + C W§'(c) (mcz-l)3 Z-To mcz-l I 2,2 P* z x;q( o) q( 0) Re %(- 0L ln(Z'Zo) + 0‘ 43?“) _ —2 z i + -§e—— ¢g'(c) - 2° Z-Zo m; -1 {-20 3! :2 ‘Y*IZC5)+ %(a ¢§I(c) - ZLEz-l d) CO] d ln(Z—Zo) - V¥ETE{) H» H" 116 I (Z Z )P*(Z ) = Im 91- - a ln(Z-Z ) + a ¢*(c) y;q 9 0 q 0 2U 0 I _ '1 :2 ! Z0 Z + _ ¢I C ' ' TIE)— Z'z—o 111C, ’1 Z + it - '52 9E a ¢¥I(C) Z [:m32_1 ¢II:C:] a ln(Z-Zo) - WTIlCi (4.52) Hence, the influence functions for an infinite plane with an elliptical cavity at the origin are found. IV.3 DERIVATION OF THE INFLUENCE FUNCTION: THE SHARP CRACK PROBLEM Considering m = l, a sharp crack along the x-axis between x = 2 and x = -2 is obtained, Figure 4.3. The transformation function _ _ 1 Z - m(C) - E'+ C (4.53) transforms the whole region exterior to the crack into a unit circle |c|<1. Substituting m = 1 into equations (4.52) leads to the influence functions for the crack problem: ’K (-2,0) (2,0) V>< Z .plane Figure 4.3 A sharp crack in an infinite plane. * HXX;q(Z’Z°)Pq(ZO) ll 5U (D O 8' I I N o + N W N '9- H} r-\ n \J 2 2 - ci-l wf‘(c))+ 5* (cf: ¢fi(c) - 7' __£:L___ ¢*n(;) - _£¥::___ ¢*'(;):] [(cZ-l)“ II (c2-1)3 II 2 - 0L - “-9—— $1102) z-zo c2—1 Re Q* 7:5— + 2C2 ¢*'(c) + 7 '20 C2 I ___1__ (Z'Zo)2 + c“ n 2:3 . (CZ-1)“ ¢f (c) - (;2_1)3 ¢f (c)] 20 + C2 (11* ( 2C2 - -——————— ' c) + 6* ¢*'(c) (2-7.)2 Cz-l I c2-1 II H Z,Z P* Z yy;q( o) q( o) -1 4. H xy;q(z,zo)Pg(zo) Ix;q(Z,Zo)Pa(Zo) 119 = Im Q" '2' ————l——— + C“ *H (Z-Zo)2 (C‘°'-1)"CbI (C) - 2:3 (3,...” 7 ' (CZ-1 3 I C - 0 + Cl ' ) (Z‘Zo)2 Cz'l W? (C) +‘Q* 2’ -—£:L——— ¢*"(c) -—3¥::—— (2:2-1)“ II - (c2-1)3 Hm + a C2 Z-ZO + 2 WT'CC)) II 50 (D - Z[ -1 + E2 ¢g Z z-Zo 52-1 I C - Z 2 - I C - 0 7:- + %(0 ¢fI(C) ' 2 ° _Ez ¢IIEC3 1 :2- a ln(Z-zo) - W§}TETJ)‘ 120 Iy;q(Z’Z°)Pa(Z°) - a ln(Z-Zo) + a ¢f(c) ll H 5 r0 _ —2 Z-Zo Cz'l 7’20 9—5.. -_Lij. Zu a ¢II(C) Z 32-1 ¢Ii C 4. a ln(-2'20) " 15511;; g (4.54) where ¢f(c). ¢f1(c), ¢f'(c). ¢fi(c), ¢f”(c). ¢ff(c). Wf(c). Wf1(c), Wf'(;) and Wfi(c) are given in Appendix D. Hence, the influence functions for an infinite plane with a horizontal sharp crack lying on the x-axis at the origin are found. One must exercise some care when using these influence functions in that a singularity will occur when Z = 70. This case will now be considered. First solve the trans- formation function, equation (4.S3), for C. c = g 1 //Z2/4 - 1T (4.54) Then write the two roots inside and outside 7, equations (4.29) and (4.34), for the case (m = l): ri,o = %} : //Z%/4 - 1 (4.55) t. = Zo/Z i J/Zfi/4 - 1 (4.56) 121 If the field point, Z, is equal to the complex conjugate of the load point, 20, then equations (4.54) and (4.56) will be identical (; = ti), since lclsl and Itil<1. Note that this makes the right-hand side of integral III, equation (4.41), infinite. Thus, integral III has to be reevaluated. Substituting ti = C into the integral leads t0: M(C-r.)(0-r ) 111 = 7%; I. 1 0 do ‘Y (O-to)(o-z;)2 Thus, m(o-r.)(o-r ) m(o-t.)(o_r ) III = Residuel 1 0 = é% oft o | o=C (O-toMO'C)2 o 0:; or III — 71—. ¢m(o'ri)(°‘ro) do = (ZmC'Zo)(C'to)'(mC2-ZOC+1) fll Y (U'to)(O-C)2 (C'to)2 This change modifies the complex potential function ¢§(C), equation (4.47), to: r0 0 r ‘C t -C ¢’;(<:)=Q{1n(° )-a1n(‘§ )} _.:(ZmC'Zo)(C‘to)'(mC2'ZoC+1) + Q (c-to)2 Zot -1 - ——9-——} (4.57) 2 t0 122 Clearly, this change affects the other complex potential function and all the derivatives. The modified functions, ¢f(c). ¢fI(c). ¢f'(c). fi(c). ¢f”(c), ¢ff(c). Wch). Wf1(c), Wf'(c) and T?i(c) for this special case are given in Appendix D. Now that the influence functions have been obtained it is important to notice that at the tips of the crack where Z = :2 the stress influence functions will become infinite as expected. These two points are the only singular points in the plane. IV.4 THE BOUNDARY INTEGRAL EQUATION METHOD APPLIED TO A PLANE FINITE REGION WEAKENED BY AN ELLIP- TICAL HOLE OR A CRACK In this section, two classes of problems will be con- sidered. These are: (1) a plane finite region subjected to traction boundary condition t and weakened by an ellip- tical hole, Figure 4.4; and (2) a plane finite region sub- jected to traction boundary condition t and weakened by a crack, Figure 4.5. Solutions will be obtained by embedding the regions Re (region with elliptical hole) and Rs (region with a sharp crack) in infinite (fictitious) planes of the same material as Re and Rs’ containing an elliptical hole, Figure 4.7, or a crack, Figure 4.8, respectively. In the treatment of either of these problems, the boundary is divided into a finite number of divisions, N, of equal or unequal length. A concentrated line load, which is the resultant of the traction on each division, is then applied at the center of the division, i.e., 123 ““L \ v \m/ Re t8, Figure 4.4 An elliptical hole in a plane finite region with prescribed traction on the boundary, B .- Y A r Rs k/ KB. Figure 4.5 A horizontal slit in a plane finite region with prescribed traction on the boundary, Bs' 124 45% v' I .J/'—d——“\ __—/’ \ ¢x \7RLX f2» («~44 =~< Figure 4.6 The finite regionsRe and Rs, with sub- divided boundary and concentrated line loads. 125 Figure 4.7 Region Re, embedded in an infinite plane containing an elliptical hole at the origin. 126 Figure 4.8 Region Rs, embedded in an infinite plane containing a horizontal slit at the origin. 127 P . = p ds x1 A51 x P . = ds y1 {5, py 1 and for the fictitious tractions * = * PXi AS pX ds i P*. = p ds yl Asi Y where A81 is the ith interval and i = 1,...N, see Figure (4.6). The trapezoidal rule is used to approximate these integrals. Following section 1.1, the fictitious traction P* around the fictitious boundary can be found from: N P*. 2 x1 . . o T .j_1(Hxx;q(z,zo)pa “xi + “xy;q(z’z°)Pa nyi)Asl jv‘i = P _ X]. N P*. ' Z( 1 * * . . + + 1 ny;q(Z,Zo)Pq nxi + Hyy;q(z’z°)Pq ny1)ASi i gnu. 1L II = Pyi for 1 = 1,...N (4.58) 128 where the resultant fictitious traction on a given interval is represented by 13* . = P*. + i P* q1 x1 yi (i = 1,...N) (4.59) and the influence functions Hij'q(z’z°) are given by equa- tions (4.54), for a crack. Note that, in equatlon (4.58), nxi and nyi are the components of the unit normal to the division i and Pxi and Pyi are the x and y component of the real resultant traction applied to the division i, i.e., P . = P . + i P . (i = 1,...N) Substituting the components of the resultant fictitious traction, equation (4.59), into the influence functions for an elliptical hole or a slit, equations (4.52) or (4.54), and rewriting them leads to: Hxx;q(Z,Zo)Pa(Zo) = Hxx;x - p; + Hxx;y - p; Hyyzq(z’z°)P3(z°) = Hyy;x p; + Hyy;y ' P; ny;q(z’z°)Pa(Z°) = ny;x P; + nysy ° P; Ix;q(Z,Zo)Pa(Zo) = Ix;x ° P; + 1*;y P; 1y;q(z,zo)Pa(zo) = 1y;x P; + 1y;y P; (4.60) where Hij-q(Z’Z°)’ which represents the ijth stress com- ponent at a point 2 due to a unit load in the q direction 129 at a source point 20, and Ii;q(Z,Zo), which represents the ith displacement component at the point 2 due to the unit load in the q direction at the source point 20, can be easily found by comparing equations (4.60) with either equations (4.52) for the elliptical hole (see Appendix C) or equations (4.54) for the crack (See Appendix D). Substituting equations (4.60) into equations (4.58) and rearranging leads to: N 2: o o . * 2 + j=1 [Hxx;x nxi + ny;x nyi] Pxi jti * = + [Hxx,y nx1 + ny;y nyi] Pyi) AS1 Pxi N Pic ;1 * + 32:; ([HXY,X x1 + “max ny1] PM j¢i . . . * = + [ny;y nxi + Hyy;y nyi] Pyi)ASi Pyi (4.61) or writing equation (4.61) in the form of equations (1.24) leads to: ‘1 * * ' * = szi * ;(Aij Pxi + Bij Pyi) 3in N 1 * t t - fpyi " Zj=1(cij Pxi ” Dij Pyi)‘ vai j#i for i=l,2,...N (4.62) where ij = Hxx;x nxi + ny;x nyi ii = Hxx;y “xi + ny;y nyi C.. = H n . + H n 13 xy;x y1 yy;x xi = H ii xysy “xi + H n . YY3Y Yl Equations (4.62) are a set of 2N linear algebraic equations with 2N unknowns, i.e., P;i and P;i for i = 1,...N. The methods for obtaining the solution have been discussed in section 1.1. Clearly, from equation (4.61),'one can conclude that 1 1 Aij-Z Dij‘z B.. 0.0 C.. 13 13 0.0 for i j. In matrix form, equations (4.62) become: 131 P o o B- o q P*. BV U 13 13 x1 x1 * This system of equations can be solved for Pxi and P*. by matrix inversion as follows: -1 * Pxi ’ Aij Bij ‘ Bin P*. _ c.. D.. . BV . y1 13 13 y1 (4.64) or by other methods for solving systems of linear equa- tions (Faddeeva [38]). Let F be a field point at which the stresses and displacements are to be found. Then, using Appendix C or Appendix D, the stresses and displacements at the field point due to a unit load at a boundary point such as 20, for the plane finite region containing either the ellipti- cal hole or the slit, i.e., (F,Z°) and Ii,q(F,Zo), H.. 13:4 can be found. The known fictitious tractions, i.e., equa- tion (4.64), will now be applied to find the real stresses and displacements at the field point. These stresses and displacements are n1 132 N <3 == 2: [H * * xx . l xx;x(F’Z°)Pxi + Hxx;y(F’z°)Pyi] * ;X(F,ZO)P;1 + H y;y(F,ZO)Pyi] O YY i=1 YY Y . * * (F,Zo) Pxi + H y”(lazo) Pyi] Oxy i=1 xy;x x . * . * [IX;X(F,ZO) Pxi + Ix;y(F,Zo) P -] N U = .23 [I * * i=1 Y:X(F’Z°) Pxi + 1y;y(F’z°) Pyi] Some example problems will now be considered. The plane stress or plane strain problem can be considered by choosing the apprOpriate value for a in the complex potential function. For generalized plane stress: 3-v a = +V pa and for generalized plane strain: a = 3-4v 133 EXAMPLE IV.1 A Rectangular Plane Weakened by an Elliptical Hole Consider the rectangular region (10cm x 20cm) of unit thickness (h = 1cm) which is weakened by an elliptical hole described by x = (1+m) Cos 0 y = -(l-m) Sin 6 om 1|- X r L 2.3 1 1. a 1 an L4 , _36 ._ _ , 1 3; 1 1 1 1 1 1 1 1 1 1 L IOCM :1 Figure 4.9 Rectangular plane weakened by an ellip- tical hole at the origin. 135 coordinates of the field points, XF(I) and YF(I), are read into the program as the data input (Appendix E). The results are presented in Table 4.1. The results are compared to the theoretical solution of an infinite plane weakened by an elliptical hole sub- jected to a uniaxial tension [39] and solution of a long strip weakened by an elliptical hole subjected to uniform tension [42]. The program required 41 seconds of CPU time on a CDC 6500 computer. Two angles of inclination of the ellipse, 6 = 30°, 60°, are also considered and these results are presented in Tables 4.2 and 4.3. Note that the rectangular boundary has been embedded in the infinite domain at inclination 0 to the "horizontal" ellipse. The program has been written in such a way that if different angles of inclination are desired, only one character, THETA, is to be changed. Also, different sizes of the ellipse, i.e., different a and b, can be obtained in each case by changing the character, M, in the program. For different locations of the center of the hole, the new coordinates of the center of the hole, Xo,Yo, must be read into the program. The problem of a rectangular plane subjected to uniform load and weakened by an elliptical hole with major axis 2a = 3.6cm and minor axis 2b = 0.4cm centered at X0 = 1.5cm, Y0 = 2.0cm and inclined at an angle of 6 = 30° is solved. The results are presented in Table 4.4. Again, the CPU time was 42 seconds for each run on a CDC 6500 computer. 136 Table 4.1 Stress and displacement of a rectangular plane containing an elliptical hole at the origin Geometry: rectangle 10 x 20 cm2 (1 cm thickness) Load: w = 1.0 MPa , Eccentricity: X0 = 0 Y0 = 0, Angle: e = 0.0 E = 70000 MPa, u = 26315.79 MPa, v = 0.33, m = 0.5 ‘ Coordinates Field Point X Y 0xx 0 y Oxy Ux Uy No. cm cm (MPa) MPa (MPa) Juicrons microns 1 1.5 0.0 0.0 7.4638 0.0 -0.359 0.0 2 1.7 0.0 1.0963 2.50561 0.0 -o.291 0.0 3 2.3 0.0 0.4496 1.4263 0.0 -0.251 0.0 2. 0.25124 1.2630 .0 -0.253 0.0 5 0. 0.5 -l.107l 0.0 .0 0.0 0.545 6 0.0 0.8 -0.5836 -0.0139 0.0 0.0 0.547 7 .0 1.7 -0.0108 0.30356 0.0 0.0 0.564 8 0.0 2.4 0.0566 0.5381 0.0 0.0 0.605 9 -l.5 0.0 0.0 7.4638 0.0 0.359 0.0 10 0.0 -0.5 -l.1071 0.0 0.0 0.0 -0.545 1V Available Solution: 4 t f f f f f H Field Point No. Reference 1 =7.4 oyy MPa [42) 1 0 =7.0 MPa [39] <::::> a; yy (infinite plane) 5 Oxx= -l.lSMPa [42] v 1 11 1 1 1 v 137 Table 4.2 Rectangular plane contaihing an elliptical hole (inclined major axis) at the origin, Case 1 Geometry: rectangular plane (10cm x 20cm x 1cm) Load: w = 1.0 MPa - Eccentricity: X0 0 Y0 = 0, Angle: 6 = 30° E = 70000 MPa, u = 26315.79 MPa, v = 0.33, m = 0.5 Field Coordinates 0 U U Point X Y xx yy xy x y No. cm cm (MPa) (MPa) (MPa) microns microns l 1.5 0.0 0.0 5.34611 0.0. -0.1923 0.2319 2 1.7 0.0 0.8876 1.8630 0.9115 -0.1345 0.2241 3 1.9 0.0 0.7232 1.3653 0.8148 -0.1120 0.2313 4 2.8 _0.0 0.4109 0.9319 0.6010 -0.075 0.295 5 0.0 0.5 -0.4309 0.0 0.0 0.2514 0.3946 6 0.0 0.8 -0.0934 0.0103 0.2562 0.214 0.3905 7 0.0 1.1 0.1047 0.0784 0.4023 0.1983 0.3878 8 0.0 2.4 0.2976 0.4357 0.5113 0.2319 0.4190 9 -1.5 0.0 0.0 5.3461 0.0 0.1923 -0.2319 10 0.0 -0.5 -0.4309 0.0 0.0 -0.2514 -0.3905 W” .{ 1 \111111“ f // *w 1111111 138 Table 4.3 Rectangular plane containing an elliptical hole (inclined major axis) at the origin, Case 2 Geometry: rectangular plane (10cm x 90cm x 1cm) Load: w = 1.0 MPa - Eccentricity: X0 = 0 Y0 = 0, Angle: 0 = 60° E = 70000 MPa, u = 26315.79 MPa, v = 0.33, m = 0.5 Field Coordinates 0 o 0 U U Point X Y xx yy xy x y No. cm cm (MPa) (MPa) (MPa) microns microns 1 1.5 0.0“ 0.0 1.0337 0.0 0.1539 0.2344 2 1.7 0.0 0.453 0.5588 0.9191 0.1903 0.2263 3 1.9 0.0 0.5660 0.4398 0.8190 0.2086 0.2334 4 2.8 '0.0 0.69189 0.30451 0.5902 0.2853 0.2951 5 0.0 0.5 0.97538 0.0 0.0 0.2564 0.0818 6 0.0 0.8 0.9229 0.0593 0.2649 0.2191 0.0653 7 0.0 1.1 0.8810 0.1063 0.4156 0.2031 0.05381 8 0.0 2.4 0.7922 0.2095 0.5315 0.2385 0.0321 9 -1.5 0 0 0.0 1.0337 .0 -0.1539 -0.2344 10 0.0 -0. 0.9753 0.0 0.0 -0.2564 -0.0818 AY’ .y ‘111111‘ .1 “O w111111 139 Table 4.4 Rectangular plane containing a nonsymmetri- cally located elliptical hole (inclined major axis) Geometry: rectangular plane (9cm x 18cm x 1cm) Load: w = 1.0 MPa . Eccentricity: X0 = 1.5 Y0 = 2.0, Angle: = 30° E = 70000 MPa, u = 26315.79 MPa, v = 0.33, m = 0.8 Field Coordinates 0 U U Point X Y xx yy xy x y No. cm cm (MPa) (MPa) (MPa) microns microns l 1.8 0.0 1 0.0 17.1905 0.0 0.2434 0.3892 2 1.9 0.0 2.0736 2.9680 1.6635 0.6380 0.5744 3 2.1 0.0 1.1269 1.7636 1.1255 0.7245 0.6193 4 3.2 0.0 0.4972 0.9748 0.6480 0.8071 0.7640 5 .8 0.0 0.2913 1.2050 0.4385 0.8116 0.8020 0.0 0.2 -0.5871 0.0 0.0 1.1892 0.7932 0.0 0.25 -0.5355 -0.0015 0.0332 1.1808 0.7930 8 0.0 0.3 -0.4849 -0.0027 0.0656 1.1728 0.7927 9 0.0 0.4 -0.3873 -0.0030 0.1277 1.1585 0.7920 10 0.0 0.7 -0.1350 0.0151 0.2847 1.1261 0.7898 AY’ §\‘1HH‘1 1"” A 1 1111\11 140 EXAMPLE IV.2 A Circular Plane Weakened by an Elliptical Hole Let a circular plane of radius R = 6cm and of unit thickness (h = 1cm) be weakened by an elliptical hole at the origin described by: X = (1+m) Cos 0 Y = -(l-m) Sin 0 o2 in equation (5.5) or n>2 in equation (5.2), it is very cumbersome to find the inverse Of the transformation func- tion even though some numerical technique could be employed, but the following two methods appear to be the most power- ful methods for determining the inverse transformation function of equation (5.2). Method 1: Power Series Expansions for the Inverse Transformation Function The transformation function Of equation (5.2) is written in the following form: N _ R(1+c) . -1 l-c 2 26 Z _. n+1 Z — __2———' C { 1 + 1+c C + 1+c n=1 (dn lhn)C } (5'6) and letting W R(1+c) Z 01ka } (5.7) 164 where (11:0 _ 28 . l-c a2 - 1+c (dl-lhl) + 1+c a = 26 (d -ih ) for k=1 3 4 N k 1+c k-l k-l ’ ’ "" ak = 0 for k>N Equation (5.7) is a special case of: where 173—2126 (kV+Q)/P)1al(< ll R(Cos 0 + e Cos 20) ..< I! R(c Sin 0 - 6 Sin 20) (5.10) where Ofiwpmpufinpm cm wcficwmucoo ocmHm muflcwmcfi mo EoHnowm one «my a.m owsmfia «S .6ch N x? 184 .omfiw pmfisopwo was: m oucfi coauowwu wowfiamm ow pouoonn3m oHo: may mo anpoum may Meagan: m.m owzmflm «m: «S 6:30. a ..U mcmBTN U 185 let 9 = (Mm/£)/§. Then equation (5.48) can be written as follows: 4:17? 9 93 or w = 9(1 + (2'2 + m'“) (5.49) where and >a II (In - r)/£2 Finally, equation (5.49) can be written in the following form: w = (2% + Z ako‘Zk} (5.50) k=1 where a; = 1, a2 = A and ak = 0 for k>3. The inverse power series of equation (5.50) can now be obtained following equation (5.8) where, in this case, P = q = l and V = -2. Thus, (5.51) :0 II 2 MIN H 4. M 8 m w 2I N w W where 186 and the afin)'s can be formulated as n aén) = ”In k “E n (k-n) Since ok = 0 for k>3 and a = 1, then (n) = n k-n Gk (k-n) A so that k e = 1 2 (2H) (“MM1 k l-ZE n=l 11 k-n (5.52) which leads to: 81 = '1 82 = '1 - A 83=‘2'4}\ s. = -s - 15x - 3A2 -14 - 56A - 28).2 85 It is clear that the Bk's are not converging very rapidly, a deficiency of this method. However, the Bk's are all functions of powers of A, which indicates that there would be some general closed form for equation (5.51): 9 W = 1 - p, - m - 92x2 - 93x3 ........ (5.53) 187 The coefficient pi's will now be found. To find Do, let A = 0 in equation (5.53), then % = 1 - p, (5.54) Also, when A = 0, the transformation function, equation (5.49), becomes 92 - WQ + 1 = 0 or Qi-Q+i=0 w2 W w2 . Q S01v1ng for W leads to: Q = l i Vl-4/W2 (5.55) W 2 The minus sign is not valid since the limit of the solu- tion, equation (5.55), must approach unity to satisfy conformality of the transformation function, i.e., infinity is transformed to infinity. Equating the right-hand sides of equations (5.55) and (5.54) leads to: l - Vl-4/W2 00 = T (5'56) Now let n = k in equation (5.52). Then 1 2k-1 k 0 (Zk-Z)! Bk ‘ T77? ( 1c ) (o) A ' ' ETTFTITT‘ 188 or 2p 1 BP+1 = ' (P) PTI Substituting into the inverse transformation function, equation (5.51) leads to: 2 2P w-zP-z Comparing (5.57) and (5.54), it is found that z 2P w-zp-z Co = (p) W— (5-58) Now let k - n = l in equation (5.52). Then 1 Zk-l k-l 1 Zk-l k Bk=—2‘k‘1_(k-1)(1)>\+—2-E1_ (k)(o)x° Substituting Bk into the inverse transformation function, equation (5.51), yields: _2 w 2P'2 E 2k-2 -2k (2 P) - A ( ) w pO= P I I k=1 k-Z Comparing this equation with equation (5.53) leads to: =1) = (I?) ' PTT'(I)) Substituting k = P + l and the binomial coefficient above into equation (5.59) leads to: 2P -2P-2 z ZP _ _ 01=-Z (p)w———+ (P)W2P2 (5.60) Taking the derivative of po, equation (5.58), and con- structing % paw 2P 1 _ E -2P-2 Substituting equations (5.61) and (5.58) into the right- hand side of equation (5.60) leads to the expression: _ 1 . 01 - ' Do ‘ 7 DOW (5-62) Taking the derivative of pa in its closed form, equation (5.56), and calculating the right-hand side of equation (5.62) leads to: 190 2 _ 0° pl — 1_Zpo (5.63.61) Similarly, one can find oz in terms of be, which will lead to better convergence of equation (5.53). Thus, the inverse transformation function becomes 2 n = w (1 - p0 - 19390 A— ...) (5.63) where _ 1 - ./1-4/w2 00" 2 This was an example of using the power series expansion to- find the inverse transformation function, even though the method did not prove efficient. The second method, con- tinued fractions, will also give the coefficient of 12, i.e., oz. The second method will now be applied. Rewrite equation (5.49) in the following form: n“ - w n3 + oz + A = 0 Dividing the equation by W“ leads to n . n 3 n 2 , 1 A = (w) ‘ (w) + (w) F+T 0 or _L g ($3 c%-1)+ (€342 53+ w = o (5.64) In order to check the accuracy and efficiency of this method with the previous method, let the following till I11]! 191 assumption be made: r = 1 - 1%, (5.65) and _ 1 - Vl-4/W2 o0 - 7 (5.66) Constructing (po - pi) using equation (5.66) leads to: Substituting these results into equation (5.64) leads to: f(T) = I‘(1"‘1)3 + (T-l)2 (Do-0%) + Mm-offlz = 0 (5-67) Equation (5.67) is a polynomial in F of the form of equa- tion (5.9), which can now be solved by the continued fractions method. Let the first estimate to the required root F of the equation f(F) = 0 be 60. Then, after some simplification, the function and the derivatives of the function at the estimated root are: f(Oo) = (pa-p312 f'(oo) = (co-1)2 (200-1) f"(po) = 2(500-3) (00'1) f'"(Do) = 6(4 po-3) f""(po) = 24 192 f(n)(po) = 0 for n>4 Thus 2 n = _ f(Do) = 400 E'Zpoj ‘ f"(oo) _ Spa-3 5 "' 77—72 ' Do ' -(1-po)(2po-1) .= f'" 00 = 400'3 T 77—13 " Do (Spa-31Tpo-1) 6 = fun(po) = 1 ail (Do) 2{Do-3 Then d1 = +n, d2 = fig and d3 = U(E'Y) = _t(T-6) d“ U(g €_.Y Substitution into the continued fractions leads to: - d1 T - 00 + + 2 01' O “M'W I-nlY-Ei For simplicity, consider the first numerator. Then expand- ing as a binomial series and just choosing the first two terms, the equation leads to: 193 I‘=oo-n-n2€+ ..... Substituting equation (5.65), n and a into the above equa- tion leads to 02 01(50 3) $2 = W(l - p - ° 1 - ° °' ~12 ..... ) (5.68) ° 1'2“ (1'Zoo)3(1'oo) where _ 1 - Vl-4/W2 pO‘ 2 The inverse transformation function has now been obtained by the two methods. Comparison of the two equations (5.63) and (5.68) shows that the first three terms of both equa- tions are exactly the same except that the second method, continued fractions, provides one more term. It is also clear that the second method was more efficient. The first method, however, can be more useful in some special cases. Returning to the solution of the problem of Figure 5.7(A) and equation (5.46), the complex potential functions will now be obtained following the general procedure pre- sented in section 11.3. Let ¢°(Z) and W°(Z) be the complex potential functions for the problem of Figure 5.7(B) and ¢*(Z) and 1*(2) be the complex potential functions for the problem of Figure 5.7(C). Hence, the potential functions for othe problem of Figure 5.7(A) are 194 ¢°(Z) + 4*(2) ¢(Z) W(Z) W°(Z) + P*(Z) (5.69) where ¢°(Z) and W°(Z) are known [27]. The transformed complex potential functions of ¢*(Z) and P*(Z) can be obtained in a manner similar to that presented in section 11.3. These are 1 5'1 1 £2) (10 ' mfi‘ 20_(£2_ (b1 0 (10 (5.70) _ 1 F ¢T(C) - 7;? Y o w'(0) O-C _ 1 FIG) 1 1 wio) ¢*'(o) WT(C) - W Y 0'; (10 mi (”'(0) 0'; (10 (5.71) where F(o) and ET?) are given by equations (3.20) and (3.21). Equation (5.70) is an integral equation in which the second integral can be evaluated. Let the transforma- tion function of equation (5.48) and its complex conjugate be evaluated at c = 0: (11(0) =§+ 26 + r03 1' (1)10) mo+%+— 0.3 Then the derivative and conjugate derivatives are w'(o) = - 1L + 1 + 3ro2 2 o m=-m62+2+§£ 0,2 g(o) = o(m+2oz+ro“) w'ioi 3r+Roz-mo“ 2.3111211. Zako-Zk m m2 02 k=2 = - % O - Z 616'21‘” (5.72) k=1 where the ak's are the coefficient of the expansion. Similarly, mo“+202+r (3ro“+£oz-m) wloi = (L) 10') = X Bk6‘2k'1 (5.73) k=0 where the Bk's are the coefficient of the expansion. Note that ¢f(o) and of'(o) are analytic inside Y and $¥TTET is analytic outside 7. Thus, following section 111.3: 196 ¢*'(0) = z kakok'1 (5.74) 1 k=1 M'io) = kzl kit—ko-l<+1 (5,75) Multiplying equation (5.72) by (5.75) leads to: (0(0) r z — -k+2 2 — O-k+1 Z on o-Zk+1 ETTTET = - a - kako - kak - k w'io) k=1 k=1 k=1 = - % 510 - %§ 52 - 2;; end.“ where en is a coefficient of a power series expressed in 00 n terms of 5k and ak. Since the summation 2 e o- is an n e=l analytic function outside y, and the value of the summation at infinity is zero, then, following section 1.3, the value of the second integral in equation (5.70) is _ 2 _ — - %alz; - 71-]?— a2 (5.76) Note that the 51‘s are the complex conjugates of the coef- ficients of the expansion series of ¢f(c) which are unknown. Since -(2r/m)az is a constant term and does not have any effect in obtaining ¢f(§), it will be omitted. Substituting equation (5.76) into (5.70) leads to: l F(o) ZFT Y o-c r — 2r — “—1.81; ' F32 (5.77) do + ¢f(c) = 197 In order to obtain a closed form for ¢f(c), 51 has to be evaluated. This can be easily determined. Since n 1 1 1 C C —=——=_+__+ + + 0.: 0(1-%) G 02 omI then _L 1% ”0) do = Z c k (5.78) where the ck's can be found by substituting the expansion of l/(o-c) into (5.78): _ 1 CR - Zni F o) Sé. géero The Ck's could also be obtained by taking the kth deriva- tive of equation (5.78) and setting C = 0. The function ¢§(;), however, is analytic inside 7 and has the series form given by equation (2.13). Hence, substitution of equations (5.78) and (2.13) into (5.77) leads to: z 1.: k=1 00 where 198 Finally, equating the real parts and the imaginary parts leads to evaluation of 51: (5.79) 51 = m{Re(c1) , . Im(c;) m-r m+r } Substituting equation (5.79) into (5.77) leads to the closed form expression for the transformed complex poten- tial function: ¢§(c) = %f :52) do + r{B;—E§f1—)— - i .Ilmn10;_1)} c (5.80) In order to find Wf(c), the second integral of equation (5.71) must first be calculated. Multiplying equation (5.73) by equation (5.75) w'ioi ¢*'(o) = mo"+SLoz+r . 2 ka Ok-l w (o) 1 (3ro“+£oz-m) k=1 k It is clear that the numerator of the equation above is analytic inside 7 and the denominator has five roots, one of which is zero and four of which can be found by solving the biquadratic 1, 3ro + £02 - m = 0 Some of these roots are inside 7 and some outside. Thus, the following integral can be evaluated by the Cauchy integral theorem presented in section 1.3: 199 *1 lb 2 ..L, ¢ 31%?) ¢0~ go) do = "“1 ”C +1" WE'VE) * 89?) Y 4(3rc“+8cz-m) where g(c) represents the sum of the residues of (m0“+202+r) ¢f'(0) 0(3r0“+£02-m)(0-c) at the roots which are inside y. Thus, the other trans- formed complex potential function, i.e., equation (5.71), becomes: E is 2 WT(C) = 2%T.9£.8 do - (m; +£§ +r) - ¢f'(c) - g(C) C(3rc“+£C2-m) (5.81) Recall that F(o) and ET?) were given by equations (5.25) and (5.26). The transformed complex potential functions 02(0) and 92(0) can be easily found by substituting the transformation function, equation (5.48), into equations (2.6). The results are _ 2 4 02(0) = _ Q 1n m ZOU+§U +r0 (5.82) - _ 2 1+ 1112(0) = Q ZOO + Q 1 a 1n "1 2004'th +1.0 (5.83) m-200+£02+r0“ O For simplicity, let 200 A(0) m — 200 + £02 + r0“ (5.84) r + £02 - 2003 + m0“ (5.85) B(o) Taking the derivative of equation (5.82), obtaining the complex conjugate of equations (5.82) and (5.83), and sub- stituting into equations (5.25) and (5.26) leads to the expressions for F(o) and FIE). The calculation is omitted and the results are F(O) = Q {1n Ago) - a 1n Biz) }+ Q{62 -A(.§)} (5.86) 16 {322141133 - . 1,, Am where A(o) and B(o) are given by equations (5.84) and (5.85). The two remaining integrals in the potential func- tions of equations (5.80) and (5.81) can now be calculated since F(o) and ET?) are known. To evaluate these two integrals, note that the con- formality of the mapping function m(c) implies that the equation A(o) = 0 has one root inside and three roots outside Y (since the polynomial of Z-w(§) = 0 and A(0) = 0 are identical). Denote the inside root by ri and the three outside roots by r01, r02, and r03. Root ri has a significant role in the calculation of the two integrals whereas ro , 1 r02, and r03 need not be calculated. Root ri can be 201 calculated by the inverse transformation function using either of the two methods presented at the beginning of this section. Now A(o) may be rewritten as: A(o) = r(0-ri)(0-r01)(0-r02)(0-r03) Similarly, equation B(o) = 0 is identical to Z-w(l/c) = 0, one root of which is outside Y and three of which are inside. Denote the outside root by to and the inside roots by til’ tiz and tia' Thus B(0) may be rewritten as: B(o) = m(o-til)(0-tiz)(0-tia)(0-to) Using these forms of A(o) and B(o), the two integrations may be computed quite simply in a manner similar to that presented in the previous section. The results are: 7%; §§%l do Q{111r(c-r01)(c-r02)(c-r03)-aIh1m(c-to))1 'Y 3 (0-t. )t? A(t. ) + c2A(c) 1k 1k 1k 8 {—BTEI— + k=1 B(tik)(tik-;) } (5'88) 1 F10) _ d. B(o) + B C 2N1 0-c d0 - Q‘130'A(0)-(0-;)| 2C ) 7 0:0 6 A(c) B(ri)(0-ri) r;A(r1)(r1-;) + }+ o{1nm(c-to) a 1nr(c-rol)(c-r02)(c-r03)} (5.89) 202 Substituting equations (5.88) and (5.89) into equations (5.80) and (5.81) leads to the complex transformation functions ¢f(;) and wf(;). The procedure for obtaining the influence functions from ¢f(§) and Wf(§) is very straightforward. This is done by substitution of these functions into equation (5.69) and subsequent substitution of this result into equation (3.6). v.5 ON THE INFLUENCE FUNCTION OF A MORE GENERAL CLASS OF OPENING In the previous two sections, two cases of opening were considered and the influence functions were found. In this section a general form of this special kind of opening is discussed. . Consider an infinite p1ane bounded by the contour given by equations >< II R( Cos 0 + e Cos N 0) ..< ll R(c Sin 0 - 8 Sin N 0) where o(0 0 \D\XPO ~90 XMNH )(Nt‘ '0 o '0 Nan“ - ammo“ O-b O r'mcm mx . Nat-09". ovaxxv a xnu- '\O- \\M \N\r“ \N\HU|\P \O O. z‘ HXOKC O W 1" C'fl'fl C °fl\1l"l .1!" VFW-tn” “1". 1"" “”0.“ I" N. Uh. 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C.....L XYYXY 89... 3.87.8 U212233CCAF92123233 992299 DHKsaSS ZZZZAA XYXUUsToTaT: F282=23=88812DO1212 88:33: 83:12:: :ssssa HHH338N¢13N8 EIHHCCC12120000TTDO 121212 111TT12 123h56 GGGXYaOoRao 2F QQOOOXXPPPPPPZZZZ HRSSwa PHKZZSS QRQQRR SSSUU8COH 8C8 2 1 czc czc END APPENDIX C THE POTENTIAL FUNCTIONS AND THE INFLUENCE FUNCTIONS FOR AN INFINITE PLANE REGION CONTAINING AN ELLIPTICAL HOLE The following complex functions were used in the influence functions for an elliptical hole, equations (4.52). * ro-C to-g ¢Ic:)-1n(ro)- aln(to) 2 ¢* ( ) m;2-z; +1 (mti‘z°ti*1)c 1 c = _ _ . _ *, = 1 _ a 2 z +1 ¢*'(c) = PD - mti- Oti II (c2-70;+m)2 (to-tim-tiv ¢f"(c) = '1 + °‘ («z-r0)2 (c-to)2 226 227 [2m(c2-foc+m)-2(mc2-Zoc+1)](CZ-Zoc+m)-2(2c-Zo)[PD] ¢*"(c) = II (CZ-Zoé+m)3 2 + 2 (mti'ZOti+1) (to-tiMc-ti)3 2 _ (r.-Z r.+m)C 2 wf(c) = 52 7°C*m + 1 ° 1 Elé—ifll-- ¢f‘(c) - m + mcz-Zoc+1 rim(ro'ri)(‘3'rij 1-mr,2 2 VEICC) ‘ 1n ( 2 ) - a 1n ( 0 ) + C(C +m) ¢*1(C) O o 1-mC2 rI-Torwm 2 V*'(C) = 'PD - 1 1 + g(c +m) . ¢*"(c) I (mCZ‘Z°C+1)2 m(ro-riMc-ri)2 1-m;2 I + (352+m)(1-mc2)+2mc2(cz+m) ¢f'(c) (l-mcz)2 ' " 1 __(1 C(CZ'HII n W* (C) - __ - _ + _______1 ¢* (C) II C to C re 1'mC2 II + L3gz+m)(1-mc2)+2mc2(c2+m) ¢*'(c) (1-mc2)2 II PD = (ZmC'Zo)(C2'ZoC+m) ‘ (ZC‘Zo)(mC2‘ZoC+1) Substituting the components of the resultant fictitious traction, equations (4.59), into the influence functions for an elliptical hole, equations (4.52), 228 xx-x = Re 2E2— + lg:- ¢§'(C) ' 7 —-1—'— + Cu ¢*"(C) ’ ° mc2-1 (Z'Zo)2 (mcz-l)“ ‘ - 253 ¢I'(C)] + £9— - 0‘ (mC2'1)3 (2'7032 Z'70 2 - (m:2-1) WI'(C) /2"(“+1) xx.y = Re i zlé— + 2C2 (¢f'(c) - ¢*'(c)) - 7' ———l——— , mC2_l II (Z'ZO)2 2:3 + (mcij1)“ (¢f”(c) - ¢fgcc))- m;2-1)3 (¢f‘(c) - ¢fi(c)) ] + EE%§:3—-+ EIEZ - EEEETIS (v? (C) - W¥i(q))‘) /2w(a+1) Hyy;x = Re 73%; + m:§f1 ¢§'(c) + 2'[ng%:33-+ (m::-1)“ ¢§"(c) _ (m::j1)3 ¢§'(c)] - 13%;??? + 2p;o + C 2 v ' m;2-1 w; (c) l/(2w(a+1)) 229 HWY = Re 1(2320 + 2‘32 (¢’f’(c) - ¢fi(c))+ 4 1 ’ mCZ-l (Z-Zo)2 + (mcff1)“ (¢f"(c) - ¢fg(c))- (m:fj1)3 (¢f‘(c) _ 11(0)] - (2302 - 2:170 + ME; (wf'm - mun) /2n(a+1) ny;x = Im 2‘ [(z-lzo)2 + (mg-1)“ ¢’I‘"(c) - (m::jl)3 ¢’f'(c)] - (2;?)2 + zjzo + "IS-1 wi'cc) /(2n(a+1)) nyw = Im 1(7[(Z-:o)2 + (mail-.1)“ ( I /2n(a+1) X§Y YSX YSY 235 _ —2 R€<{‘<11n(z‘zo) + a ¢f’(c) - Z [ 1' + _C ¢§’(c5] Z-Z'o cz-l Zo Z-Zo ’ a ln(Z‘Zo) ‘ W¥icj‘}//¢ynp(a+u Re{i (- on ln(Z-Zo) + a 013?“) ‘ ¢fi(§)) ' Z [i;‘ Z-Zo 32 (W - ¢fi(c))1 - 2° + a INT-2°) - ”1*“? cz-l WIIECJ) ) } /4Ufl(a+l) Im{- a ln(Z-Zo) + a ¢§'(c) - z [-'—1—+ 32 m :1 2-70 32-1 _ZZ—z— - a 1n(7-zo) - Wilt) 5/4"P‘4*" - 0 Im{i(a ln(Z-Zo) + a (cb’f'UE) - mm) - 2 [Eli + a 1n(7-Zo) 32 (¢*'(c) - ¢*'(c))] - 2° 32.1 I II 7-20 (wflzgi - *IIICI))}/47TUCG+1) APPENDIX E COMPUTER PROGRAM FOR PLANE, FINITE REGION CONTAINING AN ELLIPTICAL HOLE OR A SHARP CRACK A computer program was employed for the numerical computation of the stresses and the displacements at the field points of a two-dimensional region containing an elliptical hole or a sharp crack. A listing of that program for the rectangular region subjected to uniaxial tension (w = 1.0 MPa) and containing an elliptical hole or a sharp crack and for the circular p1ane subjected to a uniformly radially tension (w s 1.0 MPa) and containing an elliptical hole are presented in this appendix. A. INPUT DATA The following information must be provided as input (this is the order of appearance in the program). PR - Poisson's ratio of the material BMUD - modulus of elasticity of the material NML - total number of subdivisions on the boundary NFP - the number of field points at which the stresses and displacements are to be computed M - parameter used to describe semi-major axis (a elliptical hole 1+M) and semi—minor axis (b = l-M) of 236 THETA XO’YO " WR - X(I).Y(I) I=1,NML IPLANE BVX(I): ' BV (I+NML) I= ,NML XF(I)9 ' YF(I) I=1,NFP 237 angle of inclination of the elliptical hole or the sharp crack with respect to the x-axis location of the center of the rectangular or circular plane with respect to the center of the ellipse or crack "width ratio" for the rectangular p1ane (WR width/10 cm) and for the circular plane (WR radius/6.0 cm) coordinates of the outer boundary points, specified counterclockwise (for circular p1ane case, this input has been included in the program) this character specifies the type of the plane problem, i.e., for plane stress, IPLANE=1 and for plane strain, IPLANE=2 x and y components of boundary tractions speci- fied at each subdivision coordinates of the field points at which the stresses and displacements are to be computed B. OUTPUT DATA The following information is obtained as output (this is in the order of the appearance in the output). PR,BMUD - SHMUD - NML,NFP - Poisson's ratio and modulus of elasticity shear modulus of elasticity total number of subdivisions on the boundary and number of field points at which the stresses and displacements are computed THETA Xo,Yo WR X(I),Y(I) I=1,NML BV (I) - BVx(I+NML) I=I,NML PSX(I) - PSY(I) I=1,NML XF(I)2 ' YF(I) I=1,NFP SIGMAXX, - SIGMAYY, SIGMAXY, UX,UY 238 angle of inclination of the elliptical hole or the sharp crack with respect to the x-axis location of the center of the rectangular or circular p1ane magnification of the size of the plane for the rectangular plane (WR = width/10 cm) and for the circular plane (WR = radius/6.0 cm) coordinates of the outer boundary points, listed counterclockwise x and y components of boundary tractions speci- fied at each subdivision components of the fictitious traction on the boundary, represented by the concentrated loads Pii’ P;i at the center of each one of the sub- divisions. These are computed by solving a system of linear equations (4.61) (LEQTlF, computer library). location of the field point at which the stresses and displacements are computed components of stress and displacement at each of the field points. These are computed using equation (4.65). C. THE COMPUTER PROGRAM FOR A RECTANGULAR PLANE CONTAINING AN ELLIPTICAL HOLE OR A SHARP CRACK (follows) ---------- R1 R2 vRJ .Rh.R5.R6 . ZRO.ZTOZ¢ZR02 acn .ufiz,uoo 29: o 2 ; 1’32 ITO Bibi. ‘;ZP(60).ZZZI60).ZEC(66) é . HO. NO 0 S» S: ZT R. £5: 5 ( 1 x o z 6 g ‘0‘ 2C... 191. 239 Z I N Z O 3 Z l A 5 § E E E E E E 5 IT _ . _ . - . . - - _ _ O 0 . _ . I A D . . s I L v J v _ 8 T o. E O E 3 . W N X T. o N v o v . . I 5 I .PX A O 6. . . I 0 I 1 ) I. 5 A K S 8 F: . . c P I I .D I 1 a .l O 9 . . T. o O \I l‘ E I N .31. . . T o J o 10 .P ”I z A To .1089 . _ S I. I 0 X1 Y V. T 0 X2 N _ . A E I l. 5 o I V .1 E P A v0 . . L I ) I C v A 0. 2) IV I . I. F I 8 u 2 X FN TG I v. R. or. o. . 1| o o... 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