LIBRARY M'.Ch'98n State University PLACE IN RETURN BOX to remove thie checkout from your record. TO AVOID FINES return on or before due duo. ‘ DATE DUE DATE DUE DATE DUE IL “—71 MSU le An Affirmative ActioNEquei Opportunity Inetltution chimeras-p1 __...,,/ __ _ _. : .. .:rE:...:..::......<.z... .3: E... + ... . (.31 . Ar . c , ..,....;..:. ,. .. . . ...... , t . I: . . , ... .....i......1....A..,..r.p.«kufi.a-..>.u......A. «4.... .—_—..p o—sfi.-_—£. mt A NUMERICAL METHOD FOR THE SOLUTION OF PLANE STEADY-STATE THERMOELASTIC FRACTURE MECHANICS PROBLEMS By N engquan Liu A DISSERTATION Submitted to Michigan State University « in partial fulfillment of the requirements for the degree of DOCTOR OF PHILOSOPHY Department of Metallurgy, Mechanics and Materials Science 1991 ABSTRACT A NUMERICAL METHOD FOR THE SOLUTION OF PLANE STEADY-STATE THERMOELASTIC FRACTURE MECHANICS PROBLEMS By Nengquan Liu An efficient numerical technique is developed for plane, homogeneous, isotropic, steady-state thermoelasticy problems involving arbitrary internal smooth and/or kinked cracks. The thermal stress intensity factors and relative crack surface dis- placements due to steady-state temperature distributions are determined and com- pared to available solutions obtained by other methods. In these analyses, the ther- mal boundary conditions across the crack surface are assumed to be insulated. The present approach involves coupling the direct boundary-integral equations to newly- developed crack integral equations. This new method has distinct advantages over the Finite Element Methods (FEM) and Boundary Element Methods (BEM), since the FEM treatment requires fine dis- cretization in the vicinity of crack tips or the use of special crack-tip elements, and the BEM treatment requires division of the body into two regions and solution the cou- pled problems. Both treatments require continual remeshing. for crack propagation studies. To my father and mother. iii Acknowledgments I would like to express my appreciation to my advisor Professor Nicholas J. Altiero, Associate Dean of Engineering, for his excellent support and valuable assistance throughout this work. I also would like to thank, the members of my guidance commit- tee, Professor Thomas J. Pence, Professor Yvonne Jasiuk and Professor David Y. S. Yen, for their support and assistance. A final note of appreciation must go to my fa- ther and mother for their great support and encouragement throughout all these years. Funding of this research by the State of Michigan Research Excellence Fund is most gratefully acknowledged. iv List of Tables List of Figures Table of Contents Chapter 1 Introduction and Background ................................................. 1 Chapter 2 Mode I and Mode 11 Crack Problems 2.1 2.2 Cracks ......................................................................................... 7 2.1.1 Theoretical Development .............................................. 8 2. 1.2 Numerical Treatment .................................................... 14 2.1.2.1 A Single Smooth or Kinked Crack .................. 14 2.1.2.2 A Branch Crack or Multiple Cracks .............. 19 2. 1.3 Results ........................................................................... 33 Edge Cracks ............................................................................... 68 2.2.1 Theoretical Development ........................ A ...................... 6 8 2.2.2 Numerical Treatment ..................................................... 72 2.2.2.1 A Single Edge Crack ....................................... 72 2.2.2.2 An Edge Branch Crack .................................... 76 2.2.3 Results ............................................................................ 83 Chapter 3 Model III Crack Problems: Internal Cracks 3.1 3.2 3.3 Theoretical Development ......................................................... 92 Numerical Treatment ................................................................ 94 3.2.1 A Single Smooth or Kinked Crack ............................... 94 3.2.2 A Branch Crack or Multiple Cracks ............................. 98 Results ...................................................................................... 107 Chapter 4 Chapter 5 References Appendix A Appendix B Appendix C Appendix D Appendix E Appendix F Thermoelastic Problems Involving Cracks 4.1 Theoretical Development ........................................................ 123 4.2 Numerical Treatment .............................................................. 128 4.3 Results .................................................................................... 137 Conclusions ......................................... 147 .............................................................................................................. 148 In-plane Elasticity Influence Functions ........................ 159 Complementary Elasticity Influence Functions ........... 161 Anti-plane Elasticity Influence Functions .................... 164 Reduction of Domain Integrals to Boundary Integrals ...................................................................... 165 Heat Transfer Influence Functions ................................. 168 Computer Program for Matrices ................................... 170 vi Table 2.1 Table 2.2 Table 2.3 Table 2.4 Table 2.5 Table 2.6 Table 2.7 Table 2.8 Table 2.9 Table 2.10 Table 2.11 Table 2.12 List of Tables Stress intensity factors for a straight central crack in a finite plate. Stress intensity factors for a straight central slant crack in a finite plate 7:2259. Stress intensity factors for a straight central slant crack in a finite plate 7 = 45.0 0. Stress intensity factors for a straight central slant crack in a finite plate 7 = 67.5 0. Stress intensity factors for the Z—shaped crack of Figure 17 in a finite plate. Stress intensity factors for the Z-shaped crack of Figure 18 in a finite plate. Stress intensity factors for two equal-length collinear cracks in a finite plate. A . Stress intensity factors for two equal-length offset cracks in a finite plate, tip A. Stress intensity factors for two equal-length offset cracks in a finite plate, tip B. Stress intensity factors for two equal-length offset cracks in a finite plate, tip C. Stress intensity factors for two equal-length offset cracks in a finite plate, tip D. Stress intensity factors for a straight crack and a kinked crack in a finite plate. vii Table 2.13 Table 2.14 Table 2.15 Table 2.16 Table 2.17 Table 2.18 Table 2.19 Table 3.1 , Table 3.2 Table 3.3 Table 3.4 Table 3.5 Table 3.6 Table 3.7 Stress intensity factors for a branch crack in a finite plate, h/w=1.0, 2a/w=0.5. Stress intensity factors for a branch crack in a finite plate, 0:450, h/w=l .0. Stress intensity factors for a branch crack in a finite plate, (1:450, 28/W=0.5. Stress intensity factors for a single edge cracked plate under uniform ten- sion. Stress intensity factors for a single edge cracked plate under uniform ten— sion, L/w=1.0. Stress intensity factors for a single slanted edge cracked plate. Stress intensity factors for a single edge cracked plate as shown in Fig- ure 2.34. Stress intensity factors for a straight central crack under a uniformly dis- tributed shear stress as shown in Figure 3.1. Stress intensity factors for a straight central crack with fixed edges paral- lel to the crack as shown in Figure 3.2. Stress intensity factors for a straight central crack with fixed edges per- pendicular to the crack as shown in Figure 3.3. Stress intensity factors for a straight central crack with four fixed edges as shown in Figure 3.4. Stress intensity factors for a symmetric V-shaped crack under a uniform- ly distributed shear stress as shown in Figure 3.5. Stress intensity factors for an asymmetric kinked crack under a uniformly distributed shear stress as shown in Figure 3.7. Stress intensity factors for an anti-symmetric kinked crack under a uni- formly distributed shear stress as shown in Figure 3.9. viii )1/2 Table 4.1 Thermal stress intensity factors KII / at ET (w for an insulated straight central crack. Figtue Figure Figure Figure Figure Figure Figure Figure Figure Figure Figure Figure Figure Figure Figure Figure 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.1 l 2.12 2.13 2.14 2.15 2.16 List of Figures Source point,Y, and field point, x, in an infinite plane. Plane elastic region containing a crack. A Plane elastic region containing a branch crack. Plane elastic region containing multiple cracks. Discretized plane region containing a discretized crack. Discretized plane region containing a discretized branch crack. Discretized plane region containing discretized multiple cracks. Discretization for a straight crack. Geometry and loading for a straight central crack in a finite plate. Geometry and loading for a straight central slant crack in a finite plate. Geometry and loading for a symmetric V—shaped crack in a finite plate. Relative crack surface displacement Aul for a body containing a sym- metric V-shaped crack and subjected to pure shear. Relative crack surface displacement Auz for a body containing a sym- metric V-shaped crack and subjected to pure shear. Geometry and loading for a non-symmetric V-shaped crack in a finite plate. Relative crack surface displacement Aun for a body containing a non- symmetric V-shaped crack and subjected to uniform uniaxial tensile stress. Relative crack surface displacement Aut for a body containing a non- symmetric V-shaped crack and subjected to uniform uniaxial tensile Figure 2.17 Figure 2.18 Figure 2.19 Figure 2.20 Figure 2.21 Figure 2.22 Figure 2.23 Figure 2.24 Figure 2.25 Figure 2.26 Figure 2.27 Figure 2.28 Figure 2.29 Figure 2.30 Figure 2.31 Figure 2.32 Figure 2.33 stress. Geometry and loading for a Z—shaped crack in a finite plate, case 1. Geometry and loading for a Z-shaped crack in a finite plate, case 2. Discretization for a Z-shaped crack. Relative crack surface displacement Aul V for a body containing a Z- shaped crack as shown in Figure 2.17. Relative crack surface displacement Au2 for a body containing a Z- shaped crack as shown in Figure 2.17. Geometry and loading for two equal-length collinear cracks in a finite plate. Geometry and loading for two equal-length offset cracks in a finite plate. Geometry and loading for a straight crack and a kinked crack in a finite plate. Geometry and loading for a branch crack in a finite plate. Relative crack surface displacement Aul for a body containing a branch crack and subjected to uniform uniaxial stress. Relative crack surface displacement Auz for a body containing a branch crack and subjected to uniform uniaxial stress. Half-plane elastic region containing a single edge crack. Half-plane elastic region containing an edge branch crack. Discretized half-plane region containing a discretized single edge crack. Discretized half-plane region containing a discretized edge branch crack. Geometry and loading of a rectangular plate with a single edge crack. Geometry and loading of a rectangular plate with a single slanted edge xi Figure 2.34 Figure 3.1 Figure 3.2 Figure 3.3 Figure 3.4 Figure 3.5 Figure 3.6 Figure 3.7 Figure 3.8 Figure 3.9 Figure 3.10 Figure 4.1 Figure 4.2 Figure 4.3 crack. Geometry and loading of a rectangular plate with a single edge crack. Geometry and loading for a straight central crack in a finite sheet under a uniformly distributed shear stress. Geometry and loading for a straight central crack in a finite sheet with fixed edges parallel to the crack. Geometry and loading for a straight central crack in a finite sheet with fixed edges perpendicular to the crack. Geometry and loading for a straight central crack in a finite sheet with four fixed edges. Geometry and loading for a symmetric V-shaped crack in a finite sheet under a uniformly distributed transverse shear stress. Relative crack surface displacement Au3 for a symmetric V-shaped crack. Geometry and loading for an asymmetric kinked crack in a finite sheet under a uniformly distributed transverse shear stress. Relative crack surface displacement Au3 for an asymmetric kinked crack. Geometry and loading for an anti-symmetric kinked crack in a finite sheet under a uniformly distributed transverse shear stress. Relative crack surface displacement Au3 for an anti-symmetric kinked crack. Heat transfer problem involving a crack. Thermoelasticity problem involving a crack. Geometry and thermal loading for an insulated straight central crack in a finite plate. xii Figure 4.4 Figure 4.5 Figure 4.6 Figure 4.7 Figure 4.8 Figure 4.9 Relative crack surface thermal displacement Aut for a body containing an insulated straight central crack, alw=0. 1. Relative crack surface thermal displacement Aut for a body containing an insulated straight central crack, a/w=0.2. ~ Relative crack surface thermal displacement Aut for a body containing an insulated straight central crack, a/w=0.3. Relative crack surface thermal displacement Aut for a body containing an insulated straight central crack, a/w=0.4. Relative crack surface thermal displacement Aut for a body containing an insulated straight central crack, alw=0.5. Relative crack surface thermal displacement Aut for a body containing an insulated straight central crack, a/w=0.6. xiii Chapter 1 Introduction and Background During the last three decades powerful methods have been developed for numerical analysis. The most popular and efficient ones are Finite Element Methods (FEM), Finite Difference Methods (FDM), Boundary Element Methods (BEM) and coupled techniques. In recent years, the BEM has received considerable attention for analysis of many practical problems in science and engineering. The effort has been particularly intense in the area of stress analysis and thermal analysis of cracks where the FEM treatment requires fine discretization in the vicinity of crack tips or the use of special crack-tip elements. Such an approach is cumbersome if one seeks to model crack propagation since the special element must move with the crack tip and continual re- meshing is required. However, direct application of the BEM is not without problems. The most popular approach has been to divide the body into two regions (with the crack located along the interface of these regions) and then to solve the coupled prob— lems. This technique suffers the same shortcomings as FEM, i.e. crack propagation studies would require continual redefinition of the coupled regions and associated re- meshing. Many investigators have reported that the temperature field is disturbed and thermal 2 stresses are induced if a body having cracks or rigid ribbonlike inclusions is subjected to heat flow. The thermoelastic problem of an infinite plate with a crack [1,2] or a rigid ribbonlike inclusion [2,3] has been considered. That of an infinite solid with a penny- shaped crack [4,5] was also analyzed. In these analyses, the thermal boundary condi- tions assumed at the flaw are insulation [1,3,4], prescribed surface temperature [1,5], or prescribed heat flux across the surface [2,5]. Solutions for the corresponding exter- nal crack problems have also been developed [6- 8]. Solutions for the case in which heat is generated at the crack, giving a temperature field symmetrical with respect to the crack plane have been given [9]. Solutions were presented for the corresponding temperature field [10] and the stress field [11] in a large solid containing an imper- fectly conducting, penny-shaped crack, where the heat flux between the crack surfac- es is assumed to be proportional to the local temperature difference. Thermoelastic crack solutions have been developed rapidly because of various techni- cal applications. Solutions were presented for the thermal fracture problem of a uni- form heat flow disturbed by an insulated circumferential edge crack in an infinite circu- lar cylinder of finite radius [12], in an infinite cylindrical cavity [13], and in an infinite spherical cavity [14-16]. In these analyses, the surface of the crack and the cylinder or the cavity are assumed to be insulated and the heat flow is perpendicular to the crack surface. The heat conduction problem is first reduced to that of solving a singu- lar integral equation of the first kind. Next, by the use of the potential of thermoelastic displacement, the thermoelastic problems reduced to the isothermal elastic problem of solving a similar singular integral equation, in which the solution of the integral equa- tion for the first heat conduction problem appears as a known function. In recent years there has been great interest in the calculation of thermal stress in the neighborhood of a crack in the interior of an elastic solid, mainly because of its impor- 3 tance in the theory of brittle fracture. Most of the work has been concerned with steady thermal stress problems, but a few transient thermal stresses in an elastic solid having a crack have been considered. Solutions have been obtained for the tran- sient thermal stress field in a semi-infinite body containing an edge crack [17] and containing an internal crack [18-20]. Solutions have been given for the transient ther- mal stress problem involving a circumferentially cracked hollow cylinder [21], in an edge-cracked plate [22] and in a thin plate with a Griffith crack [23-24]. However, in [17.21.22], it was assumed that the thermal disturbance in the vicinity of the crack may be neglected. In [23], it was assumed that the surface temperature at a crack is prescribed and there was a heat exchange by convection from the flat surfaces. In [24], the elastic medium was assumed to be cooled by time- and position-dependent temperature on the external crack surfaces. Many thermal stress problems for isotropic, transversely isotmpic, orthotropic and anisotropic bodies containing many kinds of cracks have been treated. Any linear ther- moelastic problem of an infinite isotr'Opic medium can be resolved into symmetrical and antisymmetrical problems by the method of dual integral equations [45]. The steady thermal problems of a transversely isotropic materials with a penny-shaped crack [25-28] and with an annular crack [29] have been investigated. The transient thermoelastic crack problem in a transverse isotropic infinite solid with an annular crack has been developed [30], where the crack was subjected to time- and position- dependent heat absorption and heat exchange on the crack surface. The antisymmet- trical thermoelastic stress problems of an orthotropic plate containing a pair of central cracks [31] and a single central crack [32] were investigated. The symmetrical ther- mal fracture problem of an orthotropic plate containing a pair of coplanar central cracks was presented [33]. For anisotropic materials, the thermal and elastic properties were described for metallic substances [34] and for composite materials [35]. 4 The early BEM work to solve steady-state thermoelasticity went back to [36-37]. The approach consisted of two steps. In the first step, the boundary element formula- tion of steady-state heat conduction was employed to determine the surface tempera- ture and heat flux distribution. In the second step, the resulting temperatures were applied as body forces in an elastostatic BEM to obtain defamation and stress. Gen- eral properties of the steady-state temperature field were exploited to transfer the thermal body force domain integral to surface integrals involving the known boundary temperatures and heat flux. Thus, the entire two step process required only surface discretization. A collection of fundamental solutions were presented, in both Laplace transform and time domains, under the classifications of coupled, uncoupled, transient and quasistatic thermoelasticity in [3 8-42] , but there was no numerical implementa- tion. Tanaka and Tanaka [43] presented a reciprocal theorem and the corresponding boundary element formulation for the time-domain coupled problem, but kernal func- tions were not discussed and no numerical results were included. Later, the formula- tion used by Tanaka et al . [44-46] required the evaluation of a domain integral. Not only is the calculation of domain integrals undesirable, but also such schemes require that special care be taken when evaluating the integral close to the singularity. Chau- douet [47] again resorted to the volume-based approach, while Masinda [48] and Sharp and Crouch [49] have made efforts for transferring the thermal body force do- main integral to a surface integral. Masida presented some formulations for three-di- mensional problems, but stopped short of attempting an implementation. On the other hand, Sharp and Crouch developed an approach for two-dimensional quasistatic ther- moelasticity using time-dependent Green’s functions, but domain integrals were used in the time marching algorithm. Banerjee er al. presented the boundary element formu- lation in 2D for transient ground water flow [50], steady-state heat conduction [51] and time-dependent thermoelasticity [52]. Those of transient thermoelasticity were developed in [53-54]. Among these papers, domain discretization was completely eliminated in [52-54]. Several attempts have been made to determine thermal stress intensity factors (TSIFs) [1,43,55-61], Among these developments, Sladek et al. [43] determined TSIFs for various line cracks by employing BEM. Sumi [57] obtained the TSIFs for Griffith cracks with steady temperature distribution in finite rectangular plates by us- ing the modified mapping collocation method. Thereafter, Emmel et al. [5 8] applied the FEM to the same model as Sumi’s and compared his numerical solutions with Sumi’s. Sladek et al. [59] transformed area integral for body force term in BEM to line integral and calculated the TSIFs for edge cracks. Lee et al. [60] determined the TSIFs for cusp cracks in infinite bodies by using a complex variable approach. Lee et al. [61] computed the TSIFs for the same model as Sumi's in finite bodies as well as cusp cracks in infinite bodies by using BEM with the linearized body force term. The application of the BEM for solving boundary value problems with body forces, time dependent effects or certain classes of non-linearities generally lead to integral equations which contains domain integrals [62]. Although these integrals do not in- troduce any new unknowns they detract from the elegance of the formulation and af- fect the efficiency of the method as integrations over the whole volume are required. Hence, a substantial amount of research has been carried out to find a general and ef- ficient method of transforming domain integrals into equivalent boundary ones. The approaches which have so far been proposed can be divided into four groups. The first approach, the Dual Reciprocity Method, was developed in [63,64] and later extended to a variety of problems [65,66]. It has been shown that the Dual Reciprocity Method permits one to solve a wide range of problems and is very accurate [67-69]. The sec- ond approach was based on the expansion of the source term into a Fourier series to deal with potential and elasticity problems [70]. It has also been successfully applied 6 for solving neutron diffusion problems [71]. The third approach was the use of particu- lar solutions which can convert domain integrals into boundary integrals. This tech- nique offers some specific advantages over the previous approaches in some specific applications [72]. The Multiple Reciprocity Method can be thought of as an extension of the idea of Dual Reciprocity Method. However, instead of approximating the source term by a set of coordinate functions, a sequence of functions related to the fundamen- tal solution is introduced [73,74]. The Multiple Reciprocity Method is essentially dif- ferent from the Dual Reciprocity Method. While the latter applies the same fundamen- tal solution throughout the process of transformation of domain integrals into bound- ary integrals, the Multiple Reciprocity Method uses increasingly higher order fundamental solutions. The fourth approach, the method of particular integrals, was presented in [75]. The method was the use of particular integrals which is based on the well-known concept of developing the solution of an inhomogeneous differential equation by means of a complementary solution and a particular integral. This com- pletely general approach does not require any volume or additional surface integration to solve the general body force problem. In Chapter 2, the crack integral-equation representation developed in [76] is coupled to the direct boundary-element method and applied to finite plane bodies containing internal cracks and edge cracks. The method is developed for in-plane (modes I and II) loadings only. In Chapter 3, the technique is adapted to mode III problems involv- ing internal cracks. In Chapter 4, the method is extended for application to steady- state thermal fields disturbed by internal insulated cracks in finite plane regions. The numerical treatment is quite straightforward, yet the results are shown to be extreme- ly accurate. Chapter 2 Mode I and Mode II Crack Problems 2.1 Internal Cracks Recent attention has been focused on the development of integral-equation representa- tions of cracks which can be coupled to the well-known boundary-integral equations for the treatment of cracks in finite bodies. In [80,81], a crack integral-equation representation is written in terms of the crack surface tractions. The unknowns in this representation are the dislocation densities along the crack line. While this formulation is shown to be quite effective for curved cracks, the equations are shown to be invalid when the crack contains a kink. In [82-84], a crack integral-equation representation is written in terms of the resultant forces along the crack line. It is shown that, unlike the previous formulation, this one can handle kinked cracks. However, the unknowns in this representation are still the dislocation densities. Since these densities are singular at the crack tips and weakly-singular at kinks, a rather cumbersome numerical treat- ment is required. The crack integral-equation representation presented in [76] contains, as unknowns, the displacement discontinuities along the crack line. Since these are zero at crack tips and continuous at kinks, the numerical treatment of these equations need be no more com- plicated than the treatment of the boundary-integral equations themselves. In this section, the 'crack integral-equation representation developed in [76] is coupled 8 the direct boundary-integral equation method with a novel "displacement-discontinuity" representation of the internal and edge crack. It is shown that the analysis of such complex crack geometries is straightforward with this technique and several examples are reported to demonstrate method accuracy. 2.1.1 Theoretical Development Consider an infinite isotropic elastic plane in which there is a point, X, at which some "source" of stress is located and a point, x, at which the stresses are to be computed. At each of these points, we will be referring to internal "surfaces", as shown in Figure 2.1, described by unit normals It and n, respectively, and we will employ the follow- ing influence functions: (uR)iJ-(x, it) = the displacement in the i direction at x due to a unit force applied in the j direction at if in the infinite plane, (uc)ij(x, it) = the displacement in the i direction at it due to a unit displacement discontinuity applied in the j direction at i in the infinite plane, (1tR)iJ-(x, Y) = the value of rt, at x due to a unit force applied in the j direction at if _ in the infinite plane, (1tc)ij(x, it) = the value of xi at x due to a unit displacement discontinuity applied in the j direction at X in the infinite plane, where it; are stress functions defined such that the stress components are a“! E 012 = .. 351 = 3?. , (2.1.1) o =-— c =— , 11 dxz’ 22 8x1 3x1 3x2 .............. .............. ’ 10 The influence functions are given in the Appendix A. A plane elastic region Q, with external boundary I‘b, containing an internal piece-wise smooth crack I}, as shown in Figure 2.2, or an internal piece-wise smooth branch crack or multiple cracks composed of F61 and l"c , as shown in Figures 2.3 and 2.4, respectively, is loaded by prescribed tractions tj on some part of the external boundary and prescribed displacements uj on the remainder of the external boundary. Then the direct boundary-integral equations and the integral equations developed in [7 6] can be coupled as follow. For a single smooth or kinked crack: Cij(X) u,- (x) = MIR)”- (x. i) tics) dsCrE) - fugue», (x, i) ujcz) (ism + Lewis. 2) Aujm use) x on r, (2.1.2) “1(3) = firgnRhJ-(x. X) 5-6) ds(x) - Hwy-(x, x) ujco user) + [r‘(nc),j(x, a) Aujcr) ds(r‘t') x on 1“c For a branch crack or multiple cracks: cijcx) u, (x) = Mumb- (x. r) at!) dsm - fugue,- (x. X) 11,02) asap 2 + Simeon-(x. r) Anion dsm x on r, k=l MK) = Mung-(x. in am dsm - Mange. a ujm dstio ui 11 / \/, Figure 2.2 Plane elastic region containing a crack. 12 Figure 2.3 Plane elastic region containing a branch crack. l3 Figure 2.4 Plane elastic region containing multiple cracks. rug; . lit: lb 2.1. (‘4 . A 14 + £11.“ (nc)ij(xr i) Aujm (1865) x on rel (2.1.3) b1 1H(X)=§r51tR)i,-(x.i) 5-0?) asap - firSflChJ-(X, i-o ujco dso-o + iii-*(“Chfixi 7‘) Alt-(7:) (18(3) x on I‘02 k=1 In eqs.(2.l.2) and (2.1.3), i=1, 2, j=1, 2, summation on repeated indices is implied, and Auj = uj‘ - uj+ are the relative crack surface displacements. 2.1.2 Numerical Treatment 2.1.2.1 A single smooth or kinked crack A simple numerical treatment of eqs.(2.l.2) is presented here in. which the external boundary is approximated by Mb straight elements and the crack by Mc straight ele- ments, as shown in Figure 2.5. Eqs.(2.1.2) can then be written as M. ciao ujtx) = .331! ml (us),- 6:. a 5-0?) - (no),- (x. r) u,- or) 1 asap Mb+Me + 2 j (11¢)ij (X. 3) A010“) (1853) x on I‘b m=M.,+1 m (2. 1.4) Mb nix) = z ] mt (1:11),,- (x. x) 5-6) - (new. to u,- (r) 1 dsm m=1 15 Mb+Mc+l +1 k l merit in m /crac ee m x m M_ boundary element m ml 2 1 Mb xi 1 Figure 2.5 Discretized plane region containing a discretized crack. 16 Mb+Mc + 2 I m(7‘5¢)ij (x, 33) ADJ-(30 (18(3) x on I‘c . “M544 The displacements, tractions and displacement discontinuities on each element can be linearly approximated by uja) = Nl(§)uj+ N2(§)uj ii on element m of 1‘1, 5'0?) = N1(§)tj(2m—l) + N2(§)tj(2m) i on element m of 1], (2.1.5) AujCY) = N1(§)Auj(m) + N2(§)Auj(m+1) i on element m of 1"c where mo“), tic“), tic-m“) are values at the external boundary nodal point m (m=1, Mb), Aujm‘) are values at the crack nodal point m (m=Mb+2, Mb+Mc), and N1(§) = (1-§)/2 N2(§) = (1+§)/2 -1 S E S 1 . (2.1.6) Furthermore N1(§)X(m-l) + N2(§)x(m) i on element 111 of I], i = 1 (2.1.7) L N1(§)x(m) + N2(§)x(m+l) if on element m of Pd or Fez [(sé‘“) _ Sém'1))/2] d§ = [Asgnl2] (if, i on element m of 1“, (13(3) =1 ‘ (2.1.8) [(sctm+1)_ sém))/2] ‘15:: [ASE/2] d5 X“ on elemeent m of Pd or I‘d 5 17 where Ast‘,n is the length of the external boundary element in and Asg“ is the length of the crack element m. Inserting eqs.(2.l.5) through (2.1.8) into eqs.(2.l.4), we obtain Mi, ciSnmJ-(n) = gimp/4 [j m(1‘§)(uR)iJ-(X‘“’. §)d§ ] 5‘2”” M], + EIASE‘M [I m(1+§)(uR)ij(x(n)’ §)d§ ] tj(2m) Mb 1 - Elms/4 I I m(1-§)(uc)ij(x(“), gm 1 ,1..- ) M), _ 2 Asf,“/4 [j m(1+§)(“°)ij(X‘“). §)d§ ] ufm) m=1 M.+M, T 2 Asc’W4 [j m(l-§)(uc)ij(x(“), §)d§ ] Aujlm) HFM5+2 Mb‘i'Mc- l + 2 As.‘.“/4 [I m(1+§)(uc),j(x, 13).): ] Auj(m+l) DFMb'i' I (2.1.9) Mb “1(a) = ElASlgn/4 [I m(l—§)(uR)ij(x(n)’ §)d§ ] tJ(2m_1) M. + 21451?” [I m(1+§)(uR)ij(x(n)i §)d§ ] 15am) M. - ZlASénM. [I m(1‘§)(flc)ij(x(“), §)d§ ] uj(m-1) M. ' 2 AS?“ [I m(1+§)(“°)ij(X‘“). §)d§ ] ujlml m=l Mb+Me + E A8374 [jm(l-§)(1tc)ij(x(“), gag ] Aqum) m=M5+2 1 h H {IF rr1 18 Mgr-M -1 + 2 Asa/4 [I m ed: 1 Au “FMlfi'l where 111‘“) = u-(x‘n’), x‘") being the location of the external boundary nodal point 11, (n=1,. ..Mb), and rtf"): rt (x(“)), it“) being the location Of the crack element mid-point n, (n= Mb+1’ ... Mb+Mc)' Eqs.(2.l.9) can be written in the following matrix form: mini-dds) ...) ids-12111 } where [UC] is 2M, x 2Mb , [Q] is 214,, x 2(Mc-1) , [UR] is 2Mb 'x 4Mb , [PC] is 2Mc x 2Mb , [X] is 2Mc x 2(Mc—1) and [PR] is 2Mc x 4Mb. (2.1.10) Thus -[Q] [DJ [PC] {Film [0] [leiPR] [I] u u where, as in [76], we have defined a nodal force matrix on the crack by 19 Hap} where '-II 00000“ o-rroooo .61) bird—'11. I represents a 2x2 identity matrix, and [F2] is 2(Mc—l) x 2Mc. 2.1.2.2 A branch crack or multiple cracks A simple numerical treatment of eqs.(2.l.3) is presented here in which the external boundary is approximated by M, straight elements and the branch crack or multiple cracks by Mel, Mcz straight elements, as shown in Figures 2.6 and 2.7, respectively. Eqs.(2.l.3) can then be written as Ms c.,- ..,-(x) = 51] ml (um,- (x. i) so?) - (ac), (x. s) u,- (x) 1 asap M.+M.. + 2 [mung-(x, a) Aujm dsGr’) “PM 5+ 1 M5+Md+Ma+l + 2 j (“Chj (X. 75) Aujfi') dsm x on I}, m=M,+M,,+2 m M. 1ri, gag ] tj(an-u m=l M. + 2 4813“” [j m(l+§)(rtR)iJ-(x(“), mg 1 Sam) m=l M. " 2‘. AS8V4 [I m(1—.§)(rce),j(x<“’. 0dr: 1 uJKm-l) m=1 Ma .. “EIASgnM' [Im(1+§)(1§c)ij(x(n)’ §)d§ ] uj(m) Mb+Mcl + Z. Asia/4 [j m(1—§)(rtc)ij(x(“), 5X15 ] Ami“) m=Mg+2 M5+M¢1-l + 2 Asg“/4 [[ (1+§)(se),j(x, §)d§ ] Adj-(ml) . (2.1.18) m=Mg+1 m Ms+Ma+Ma+1 + 2 Ass/4 [I m<1—§>(nc).,~. ed: 1 Auj m=M5+M¢1+2 Ma+Md+Ma + 2 ASE/4 [I m(1+§)(m)ij(x(n)s §)d§ ] Aufm“) m=Mb+Md+2 Mr. Kim) = 2 Asian/4 [I m(l-§)(rr:R)iJ-(x("), §)d§ ] tj(2m-l) m=l M. + 2 483/4 [I m(1+§)(1tR)ij(x(“), gag ] ,jtzm) m=l Ma - uE1A8113n/4 [I m(l-§)(m)ij(x(n). §)d§ ] uj(m_l) 25 M. - “EIAsg‘M [] m(1+§)(rtc)ij(x(n)’ €115] slim) + in Asm/4 [j m(1—§)(rtc)ij(x(“), §)d§ 1 Au (m) m=Mg+2 M.+M,1-l + 2 Asa/4 1] m11+§> ed: 1 Au m=Mg+I Mg+Md+Md+l + 2 ASE/4 [j (1—§)(nc)ij(x<">, §)d§ 1 Allj(m) m=Mg+Mg1+2 m Mb+Mc1+M92 + )3 Asm/4 [j m(1+§)(ice),j(x, §)d§ 1 Adm“) m=Mb+Md+2 where uf") = uj(x(“)), x‘“) being the location of the external boundary nodal point n, (n=1, Mg), and 1:5”) = rti(x(“)), rd“) being the location of the crack element mid- point n. For multiple and branch cracks, n= Mb+1, Mb+Mcl, Mb+Mc1+2, Mb+Mc1+M¢2+L In addition, for a branch crack, we add a nodal point n= Mb+M¢1+2 . Eqs.(2.l.18) can be written in the following matrix form: »»1 111 »1»1 111 11111111111» 111 1 1111-1111111 11111 1 26 For multiple cracks, the matrices in eqs.(2.l.19) have the following dimensions: [UC] is 2Mb x 21111,, , [Q] is 2Mb x 2(Mc1-1) . [Q] is 2Mb x 2(Mc2-1) , [UR] is 2Mb x 4MID , [PC] is 2Mc1 x 2Mb , [x1 is 2Mcl x 2(Mc1-1) , [x12] is 2Mcl x 2(Mc2-1) , [PR] is 2Mcl x 4Mb , [PC1] is 2Mc2 x 2Mb , [x21] is 2Mc2 x 2(Mcl—l) , [x2] is 2Mc2 x 2(Mc2—1) , [PR1] is 2Mc2 x 4Mb , and r a r (Mb+l) (Maia) 1 11'. Au = [I‘zllPR] [I] [01 1F‘ * JIF21][PC1] -[I‘21][X21] -[I‘21][X7) Au2 _ [1‘21][PR1] [0] [I] _ F2 ‘LJ (2. 1.20) where, as in [7 6], we have defined nodal force mauices on the crack by 11111 28 (2.1.21) where "-1 I o o o o 0' o -I I o o o 0 [rz]=[r2J]= 9 9 f1 1 f) 9 9 (2.1.22) .11 ('1 (1 116-1 i1 and I represents a 2x2 identity matrix. For multiple cracks, [F7] is 2M1—1)x2M¢1 and [F21] is 2(Mc2_1)XZM02' For a branch crack, [F2] is 2Mc1X2(Mc1+l) and [F21] is 2Mc2x2(Mc2+1). For multiple cracks, the matrices in eqs.(2.l.21) have the following dimensions: 1 ' F‘Mb+2) W P FNb+Ms1+3) F‘Ms-i-3) {11-1 . 1 {.211 . .. F(Mb+Mcl) F(MU+MCI+MCZ+1) L J 29 For a branch crack, the matrices in eqs.(2.l.21) have the following dimensions: F(M.+2) F M1111» : 1 F(Mb+Mel+Me2+1) J L FMb+Mcl) 5 J For a branch crack, we need to employ continuity conditions and equilibrium condi- tions, i.e. Auj(M"+M""+1) = AuJ‘M~*Md+2’ (2.1.23) ij‘i-Mu-l'l) = Fj(NRHVIJJJJH) + 1..Jj(MJJ+McJ+2) , J (2124) It is well-known that one can obtain the diagonal 2x2 blocks of [UC] in eqs.(2.l.ll) or (2.1.20) from rigid-body considerations. If we apply a rigid-body displacement, i.e. uf1)= ufz) = uf3) = ....... = ulm") = 111 (2.1.25) 11»: up: up) = ....... = 119‘»): u, then the body is stress-free. Thus and eqs.(2.l.11) or (2.1.20) reduce to 111 1 n2 n1 [UCF 11.2 *={o} u1 u2 so that M UCai-rxzi-r) = _iUdZi-IXZj—l) F! jun M. . . UCm-rxzi) - _ZUCm—lxzi) F! In M UCaixzi-u ___ _iuctzixzi—I) i=1 #J UCtzixzi) = _gucwfii) J F! per (2. 1.26) (2.1.27) 31 Once eqs.(2.l.11) or (2.1.20) have been constructed, we must impose four conditions at each nodal point m on I}, involving the boundary values ufm)’ t1(2m)’ ham-+1), uém)’ 62m), t52m+1 ), and rearrange eqs.(2.l.11) or (2.1.20) accordingly to obtain 1.1.1.1.} . where {Z} contains the unknown boundary displacements and tractions on I], and the unknown matrix {Au}. Once we have obtained Au; at each crack nodal point, the displaCement discontinuities normal and tangential to the crack surfaces at that point are Aun = Aulnl + Auznz (2.1.29) Au, = Auznl — Aulnz and the nondimensional stress intensity factors can be determined from KI I =0 = V 8LEG(1+V)Alln(8)/O'J1—ta 32 Kn'w = V%G(1+V)Aut(e)/U\lfi (2.1.30) K11=1 = -8%G(1+V)Aun(l—e)/o\/E rial,=1 = —G(1+v)Au,(1-e)/ofifa where e —> 0, G, v are the shear modulus and Possion’s ratio, respectively, 1 is the length of the crack, 0 is a reference load, and a is a reference length. It should be noted that, for a problem involving a traction free crack, {F]={0] and eqs.(2.l.11) or (2.1.20) reduce to 131”“ -[Q] [mm -11‘21m [rim] ]t{ } (2.1.31) 01' [UC] -[Q] -[Qfl . u ' [UR] l [1"2][PC] —[I‘7J[X] -[1‘2][x17]1Au1>= [13mm] {I} (2.1.32) [irzliipcli 15111le1 -1r2111x21. [Auz 31311111211 _ respectively. 33 2.1.3 Results Here we employ the coupled model to find numerical solutions for some finite domain problems. 1. Straight crack. In all straight crack problems, the crack is modeled by 12 elements and the exter- nal boundary is modeled by 40 elements. The crack discretization is shown in Figure 2.8. The stress intensity factors, normalized with respect to (Nu—a, as shown in eqs.(2.l.30), are calculated and are compared to those obtained in [84]. a) Straight central crack in a rectangular plate subjected to uniform uniaxial tensile stress. A rectangular plate of height 2h and width 2b contains a central straight crack of length 2a. A uniform uniaxial stress, a, perpendicular to the crack direction, acts over the ends of the plate as shown in Figure 2.9. In Table 2.1, the stress intensity factors are given for various ratios of Nb and M. b) Straight central slant crack in a rectangular plate subjected to uniform uniax- ial tensile stress. A rectangular plate of height 2.5b and width 2b, containing a crack of length 2a, is subjected to a uniform uniaxial stress 6 at the ends. The crack is located centrally at an angle 7 to the direction of o as shown in Figure 2.10. In Tables 2.2 through 2.4, the stress intensity factors are given for various 34 Figure 2.8 Discretization for a straight crack. 35 J “ H H h 1 “ 1 h J - 11 l 1 1 1 1 i'o l‘ 2b _l 7| Figure 2.9 Geometry and loading for a straight central crack in a finite plate. 36 Table 2.1 Stress intensity factors for a straight central crack in a finite plate. KI (h/b=1.0) K1 (h/b=0.4) a/b Present Ref. [84] Present Ref. [84] .2 1.07 1.07 1.25 1.25 .3 1.12 1.12 1.51 1.52 .4 1.21 1.21 1.83 1.84 .5 1.31 1.32 2.24 2.24 .6 1.47 1.47 2.80 2.80 .7 1.67 1.67 3.66 3.66 .7sz ‘11 11““ Y f XI 2.5 b t 2 a "11111.. 1 2: 1 ‘ Figure 2.10 Geometry and loading for a straight central slant crack in a finite plate. Table 2.2 Stress intensity factors for a straight central slant crack in a finite plate, y=22.5°. Table 2.3 Stress intensity factors for a straight central slant crack in a finite plate, 1:45.08 38 KI KII a/b Present Ref. [84] Present Ref. [84] ==J .l .148 .148 .356 .358 .2 .154 .160 .367 .366 .3 .164 .164 .386 .367 .4 .180 .180 .413 .390 5 i1 .187 .188 .425 .404 .6 II .200 .200 .439 .416 KI K11 a/b Present Ref. [84] Present Ref. [84] .1 .500 .500 F .500 .500 .2 .513 .513 .506 .502 .3 .534 .538 .518 .510 .4 .550 .550 .522 .522 .5 .610 .618 .535 .538 .6 .616 .606 j .551 .551 39 Table 2.4 Stress intensity factors for a straight central slant crack in a finite plate, 1:67.52 K1 K11 a/b Present Ref. [84] Present Ref. [84] .1 0.861 0.868 .355 .351 .2 0.868 0.870 .354 .351 .3 0.900 0.900 .359 .356 .4 0.959 0.958 .374 .374 .5 1.002 1.003 .377 .369 .6 1.085 1.118 .388 .380 4.0 ratios a/b and various angles 7. Kinked crack. In the kinked crack problems, both stress intensity factors and relative crack sur- face displacements are reported The material properties selected G, v, are given in the following descriptions. a) b) C) Symmetric V-shaped crack in a square plate. This example involves a symmetric V-shaped crack in a square plate sub- jected to pure shear stress as shown in Figure 2.11. Each straight crack seg- ment is modeled by 22 elements of equal length and the external boundary is modeled by 72 elements. In Figures 2.12 and 2.13, the relative crack sur- face displacements are plotted for Glo=7 8.85, v=0.3. Non-symmetric V-shaped crack in a square plate. This example involves a non-symmetric V-shaped crack in a square plate subjected to uniform uniaxial tensile stress as shown in Figure 2.14. Each straight crack segment is modeled by 12 elements of equal length and the external boundary is modeled by 72 elements. In Figures 2.15 and 2.16, the relative crack surface displacements are plotted for G/0'=76.92, v=0.3. Z-shaped crack in a rectangular plate. The first example involves a Z-shaped crack in rectangular plate subjected to 41 —>—>—>—> —>—>—>—>O’ 60° 36 <—<—<—<— 4—4—4—4— —>—>—>-—-> -—>—>—>—> <—<—<—<—<—<—<— 36 1 Figure 2.11 Geometry and loading for a symmetric V-shaped crack in a finite plate. 4K ...oocm 0.5a 3 “030033 ecu ono noaocml> oEoEExm o mc_c._.3coo xvon o ..8 :5 EoEoooamfi oootam x080 3523. Nfiw 230E NF w v o w... m... NT. _ p b — r W 1r b F F P F p _ p w p h p 51 1. b r 00.0 1 T 1 1 T r a ... . r 12. o .. i l r l “ ‘ r. 1 a. a. .. 1 ... ... flowd ¥ ! T a. a. a t i R 1 fi ! fl ! .. i! I! 1 1 n... ...... 10nd ! l . 1 a; a... a R i 5 fl 1 inuux w . p L — p 1 b 1? . p . P1 p .1 . hillw m . _ 1p 1. L 43 ...oozm 23a 3 “030033 9.6 x080 voaocml> 05058? 0 9:53:00 been 0 ..8 95 EoEoooEmfi oootam x080 052$ 2N 230E NP m t.» o .vl ml NT. LaLLLLFl—bhthfh—Lbklfick 1000.01... 11111 41111111 111 1‘1 T 4 .. ... .. * § u. a Tonodl T r l 60° 60° .\ 36 / "1 Figure 2.14 Geometry and loading for a non-symmetric V-shaped crack in a finite plate. 45 .mmobm 0:ch _omona Eeotc: 3 ©3833 ecu x080 noaocml> oEoEExmlcoc o m££Bcoo xvon o ..8 c3 “coEoooEmE oootam x080 o>$o_om mfim 830E NF m o n o n1. 0.1 L r L 1p 1r _ L P _ b h b b h h b P 000.0 T 1 j r r I 1 TTT I § TTj o co 0. o T T i 1 .Il-Lr|\J.\.. I) .mmmbm 0:33 _onE: ELSE: 3 vBomFam vco x080 voaocml> oEoEExmlcoc 0 9:53:00 xvon 0 L8 :5 3088235 mootam x080 o>$2om Sum 959... N_. m m n o n P L— L h F F F P h IF P L L? ”I. L h L IjITIIT11j11III"IIIYIfjle #- r. 1.- 47 a uniform uniaxial tensile stress at one end and a sliding support on the opposite end as shown in Figure 2.17 and the second example involves uniaxial tensile stress as shown in Figure 2.18. The middle straight crack segment is modeled by 20 elements of equal length and each of the two other segments are modeled by 5 elements, as shown in Figure 2.19. The external boundary is modeled by 90 elements. In Figures 2.20 and 2.21, the relative crack surface displacements for the first example are plotted. In Tables 2.5 and 2.6, the stress intensity factors for the two examples are reported and compared to those obtained in [81,85] for Glo=78.85, v=0.3. Multiple cracks. In the multiple crack problems, both stress intensity factors and relative crack surface displacements are reported. The material properties, G, v, are given in the following descriptions. a) Two equal length collinear cracks in a square plate subjected to uniform uniaxial tensile stress. A square plate contains two equal length collinear cracks of length 2a. A uniform uniaxial stress, 0, acts at the ends as shown in Figure 2.22. Each straight crack segment is modeled by 9 equal elements and the external boundary is modeled by 40 elements. In Table 2.7, the stress intensity fac- tors are given for various ratios 2% and compared to those obtained in [86-88]. (GIG = 4., v = 0.25) 48 n I I I I “0' n 2a=.5 b 25 > a .4 l6 2a=.5 v 0 O O O O .\\ 20 Figure 2.17 Geometry and loading for a Z-shaped crack in a finite plate, case 1. 49 IIX2 “ I I I I ‘° —II_ 2a=.5 45° 25 ":1 10 2a=5 __‘i_ I I I I I I L 20 II Figure 2.18 Geometry and loading for a Z-shaped crack in a finite plate, case 2. 50 .02a each \< .97 a each .97a each \ \< .02a each >\\ 8=.02a Figure 2.19 Discretization for a Z-shaped crack. id 2%: c_ caocm mo x020 voaochN 0 9:53:00 xvon o ..8 :5 EoEoooEmB mootam ono 0520”. CNN 959m 51 m m 0 ml ml I c _ r . . _ I I 38.0.. fl 0.. w ... a ... x a ... ... a .. ... .. 188.0... I i T .I § * “I n ... 10000.0 r I L. T * § ‘1 I x a 10000.0 I § fl T .. .. .. 188.0 a j . _ c _1 F k Fl r 1w 0000.0 52 5.... 23?. 5 Esonm mo x098 noaocmlN o oEEBcoo xnon o ..8 N3. EoEoooamfi oootam x080 o>:o_om FNN 0.591.. o n 0 ml 0| III? P IIF. III P IF P b P IF L P 000.0 4 4 f T I x 10N0.0 . ... x 0. T i ‘ fi r ! § n .. .. $3.0 ... f W 0. ... ... 1 .. a ... j I a ... r0006 4. I l l 1 . ... x ... ... ... a .. a. a. a a. a. a I w b p . b a » _II . .I 53 Table 2.5 Stress intensity factors for the Z-shaped crack of Figure 17 in a finite plate. Present 4.516 .321 4.410 .363 Ref. [81] 4.502 .325 4.397 .405 Ref. [85] 4.555 .335 Table 2.6 Stress intensity factors for the Z-shaped crack of Figure 18 in a finite plate. =; Present 4.592 .382 Ref. [81] 4.600 .410 54 2b — I-—2a—-I - I«—2a—~I I I I I I I I I I 2b I Figure 2.22 Geometry and loading for two equal-length collinear cracks in a finite plate. Table 2.7 Stress intensity factors for two equal-length collinear cracks in a finite plate. Present Ref. [86] Ref. [87] Ref. [88] M . KIA Km KIA Km KIA Km KIA Km .1 1.0049 1.0049 1.0054 1.0053 1.0051 1.0044 1.006 1.005 .2 1.0219 1.0200 1.0223 1.0215 1.0223 1.0188 1.023 1.021 .3 1.0540 1.0480 1.0524 1.0497 1.0548 1.0458 1.056 1.048 .4 1.0999 1.0900 1.0989 1.0919 1.1066 1.0885 1.106 1.088 .5 1.1780 1.1500 1.1679 1.1521 1.1825 1.1508 1.181 1.142 .6 1.2800 1.2339 1.2699 1.2368 1.2871 1.2361 1.290 1.220 .7 1.4270 1.3500 1.4269 1.3603 1.4274 1.3488 1.450 1.340 .8 1.6779 1.5600 1.6926 1.5663 1.6244 1.5052 1.680 1.560 .9 2.2694 2.1 102 2.2777 2.1 195 b) 56 Two equal length offset cracks in a rectangular plate subjected to uniform uniaxial tensile stress. A rectangular plate of height 15. inches and width 4.5 inches, containing two equal offset cracks of lenght 0.8 inches, is subjected to a uniform uniax- ial stress a at the ends as shown in Figure 2.23. Each straight crack segment is modeled by 9 equal elements and the external boundary is modeled by 24 elements. In tables 2.8 through 2.11, the stress intensity factors are given for various angles a and compared to those obtained in [89]. (G/o = 1.257, v = 0.38) A straight crack and a kinked crack in a rectangular plate subjected to uni- form uniaxial tensile stress. A rectangular plate of height 15. inches and width 4.5 inches, containing two cracks, is subjected to a uniform uniaxial stress a at the ends as shown in Figure 2.24. Each straight crack segment is modeled by 5 equal elements and the external boundary is modeled by 24 elements. In Table 2.12, the stress intensity factors are given for various distances h. (G/o = 1.257, v = 0.38) Branch crack. In all branch crack problems, each straight crack segment is modeled by 6 ele- ments and the external boundary is modeled by 40 elements. The stress intensity factors, normalized with respect to mitt—a, as shown in eq.(2.1.30), are calculated and are compared to those obtained in [90]. (G/o = 4., v = 0.25) Fig, _l_ I I I I I I I I I. I I Figure 2.23 Geometry and loading for two equal-length offset cracks in a finite plate. 58 Table 2.8 Stress intensity factors for two equal length offset cracks in a finite plate, tip A. Present ll Ref. [89] LLKI Kn kaxmmm K fixperimenlal Klnumerical K ItItttntericai 0 1.30 0.03 1.22 0.04 1.35 0.02 15 1.20 0.21 1.18 0.31 1.31 0.15 30 0.97 0.50 0.94 0.49 0.95 0.52 45 0.70 0.60 0.64 0.59 0.79 0.65 60 0.42 0.60 0.36 0.56 0.43 0.68 75 0.13 0.51 0.14 0.45 0.11 0.54 90 0.09 0.25 0.09 0.28 Table 2.9 Stress intensity factors for two equal length offset cracks in a finite plate, tip B. Present Ref. [89] a0 KI Kn Klexperimental K fixpefimental Klnumerical Klrlrumerical I=='I 0 1.40 0.20] 1.47 0.16 1.39 0.22 15 1.18 0.25 ll 1.19 0.17 1.18 0.30 30 1.02 0.50 H 1.10 0.44 0.95 0.52 45 0.65 0.60 0.70 0.58 0.64 0.61 60 0.36 0.60 0.38 0.68 0.34 0.56 75 0.10 0.40 0.10 0.58 0.1 1 0.39 90 0.01 0.20 0.06 0.33 0.00 0.13 59 Table 2.10 Stress intensity factors for two equal length offset cracks in a finite plate, tip C. I Present Ref. [89] a0 KI Kn Klexperimental Kfixpen'mental KInurnerical KIrlmrnerical 0 1.40 0.20 n 1.42 0.21 1.39 0.22 15 1.27 0.21 1.30 0.20 1.27 0.22 30 1.25 0.19 1.27 0.19 1.22 0.19 45 1.16 0.14 1.14 0.14 1.18 0.14 60 1.14 0.10 1.14 0.08 1.14 0.10 75 1.10 0.05 1.08 0.05 1.11 0.05 90 1.06 0.03 H 1.05 0.08 1.08 0.02 Table 2.11 Stress intensity factors for two equal length offset cracks in a finite plate, tip D. Present ll Ref. [89] a0 KI K11 l] Klexperimental K fixper'imental Klnumerical Klrlrumerical TI 1.30 0.03 1.28 0.00 1.35 0.02 15 1.27 0.01 1.28 0.00 1.25 0.00 30 1.20 0.01 1.18 0.01 1.21 0.01 “$.16 0.01 1.16 0.04 1.18 0.01 60 n 1.14 0.01 1.15 0.06 1.14 0.00 75 1.10 0.00 1.06 0.02 1.12 0.01 90 1.00 0.00 1.10 0.02 1.09 0.01 15. x _- V I I I I I I I I I Ir I Figure 2.24 Geometry and loading for a straight crack and a kinked crack in a finite plate. Ea 61 Table 2.12 Stress intensity factors for a straight crack and a kinked crack in a finite plate. 11 KIA KIIA Km Km; Krc, Knc KID KIID 0.02 1.1099 0.9021 2.0836 0.01 14 0.8068 2.2328 0.0536 0.1832 0.04 1.1 176 0.9140 2.081 1 0.0148 0.7997 2.0923 0.0614 0.1764 0.06 1.1250 0.9241 2.07 84 0.0169 0.7948 1.973 1 0.0681 0. 1704 0.08 1.1321 0.9330 2.0756 0.0183 0.7901 1.8756 0.0736 0.1652 0.10 1.1386 0.9407 2.0727 0.0191 0.7852 1.7973 0.0779 0.1607 0.12 1.1447 0.9475 2.0696 0.0196 0.7800 1.7351 0.081 1 0.1568 0.14 1.1504 0.9534 2.0664 0.0198 0.7745 1.6862 0.0834 0.1535 0.16 1.1556 0.9586 2.0630 0.0197 0.7689 1.6486 0.0849 0.1508 a) b) 62 A central symmetrically-branched crack in a square plate subjected to uni— form uniaxial tensile stress. A square plate, i.e. h=w, contains a branch crack. A uniform uniaxial stress, 6, acts at the ends as shown in Figure 2.25. In table 2.13, the stress intensity factors are given for various angles a. In Figures 2.26 and 2.27, the relatve crack surface displacements are plotted. In Table 2.14, the stress intensity factors are given for various ratios Za/w. A central symmetrically-branched crack in a rectangular plate subjected to uniform uniaxial tensile stress. A rectangular plate contains a branch crack and is subjected to uniform uniaxial tensile tensile stress 6 at the ends as shown in Figure 2.25. In Table 2.15, the stress intensity factors are given for various ratios h/w. 63 2h xl I I I I I I I I I I 2w I Figure 2.25 Geometry and loading for a branch crack in a finite plate. I: 64 Table 2.13 Stress intensity factors for a branch crack in a finite plate, h/w=1.0, 2a/w=0.5. Present Ref. [90] 9° KIA KIB KIIB KIA ‘ KIB KIIB 10 0.87 0.61 0.12 0.87 0.62 0.12 20 0.92 0.58 0.24 0.92 0.58 0.24 30 0.90 0.49 0.35 0.91 0.50 0.35 45 0.87 0.33 0.50 0.87 0.33 0.50 60 0.82 0.12 0.52 0.81 0.12 0.52 70 0.77 -.08 0.50 0.77 -.08 0.50 65 .mmobm _0_x0_:: ELSE: 9. 0800.330 0:0 00.6 50:95 0 9:53:00 >000 0 .60 54 EoEoooiflv 008030 00.6 0>$0_0m 0N.N 0.50m I oooflxv 1 I om¢fl6 Elm Downs ... . oomfld E “.0.—"6 i 66 .mmobm _0_x0_:: E..0::: 3 0300.330 0:0 x030 5:05 0 0559.80 .000 0 :00 $6 208000320 mootam x080 0>:0_0m hmd 0.30m b—‘00Hfl6 oomfld oomfld omfifld OOQHG 67 Table 2.14 Stress intensity factors for a branch crack in a finite plate, 0t=45°, h/w=l.0. Table 2.15 Stress intensity factors for a branch crack in a finite plate, a=45°, 2a/w=0.5 . Present Ref. [90] 2a/w KIA Km K1113 KIA Km K1113 .10 0.69 0.31 0.32 0.69 0.31 0.32 .25 0.73 0.32 0.34 0.74 0.31 0.34 .40 0.81 0.32 0.41 0.81 0.32 0.41 .50 0.87 0.33 0.50 0.87 0.33 0.50 .60 0.93 0.35 0.60 0.93 0.35 0.60 Present Ref. [90] NW KIA Km Km; KIA Km Km, 0.8 0.99 0.38 0.55 1.00 0.38 0.55 1.0 0.87 0.33 0.50 0.87 0.33 0.50 1.2 0.80 0.31 0.43 0.80 0.31 0.43 1.6 0.74 0.30 0.40 0.74 0.30 0.40 2.0 0.71 0. 30 0.38 0.71 0.30 0.38 2.4 0.70 0.30 0.37 0.70 0.30 0.37 2.2 Edge Cracks 2.2.1 Theoretical Development A plane elastic region Q, with external boundary 1‘ = I‘ m + I‘m, containing a piece- wise smooth edge crack I‘c , as shown in Figure 2.28 or a piece-wise smooth edge branch crack, Pd and I} , as shown in Figure 2.29, is loaded by prescribed tractions ti on some part of the external boundary and prescribed displacements uj on the remainder of the external boundary. Then the direct boundary-integral equations and the integral equations developed in [76] can be coupled as follows. For a single edge crack: cijtx) ujtx) = Lat 0112).? (x. 0 9(2) — (uc)i§‘(X. an aim] dsao + Ir‘(uc)i§‘(x, r) Aujcr) (ism x on T1,] ‘ (2.2.1) nitx) = [rut («mg-«x. an gm — (mg-1x. an aim] dsm + It (might, r) Aujm dstx) x on 1“. For an edge branch crack: cijtx> ujtx) = In} (umg (x. 0 gm - «0:; (x. i) ujtr) Jason 2 + 21% (mg (x. x) Aujm dsts) x on rbl k=1 69 Figure 2.28 Half-hlane elastic region containing a single edge crack. 70 ll X2 X1 F Fbi‘ ti Figure 2.29 Half-plane elastic region containing an edge branch crack. 71 MK) = In} (nmgtx. 3) 5m — 00:30. i) am 1m) 2 + XI, (Eds-‘0‘. 3) Aujm (18(7) X on 1‘ cl (23-2) k=l “ MK) = M maggot, r) gm — (1:030, r) uJ-(x) ]ds(x) + éIr “Chg“, i) 1311,55) (18(3) x on Fez k=l ‘* In eqs.(2.2.1) and (2.2.2), i=1, 2, j=l, 2, summation on repeated indices is implied, Auj = uj‘ - uj” are the relative crack surface displacements, and ni are stress functions defined such that the stress components on=—, 022=-—, 012=-—=— . (2.2.3) It was shown in [91,92] that the half-plane fundamental solutions ()h can by represented by adding a complementary part ( )c to the well-known two-dimensional Kelvin solutions ( ) as follows: (umgtxx) = (uR).,-. ed: 1 5‘2”“) 73 1 Mb X1 Mb+l : crack e ement m\ m Mb-l-Mc-l- 1 Figure 2.30 Discretized half-plane region containing a discretized single edge crack. 74 M. + 22’3“?” U m(l+§)(uR)i§-‘(x(">, Dd: ] tiara) Mb " 2 Asg‘m [I m(1-§)(uc)i’j‘(x(“), my; ] “jm—l) m=2 Mb , ‘ “E35194 [I m(l-I-§)(uc)i§‘(x(“), §)d§ ] uj(m) M5+M¢ + 2 Ass/4 [I (1—§)(uc>,?(x.t)dt1Auj HFMVPI m Mb'i'Mc-l + 2‘. Ass/4 [I m<1+§xuc)§;(x, mg ] Auj(m+1) HFMfl'l (2.2.6) M. "im = 2 A51?" / 4 [I m(1-§)(1tR)i§-‘(x(“), gag ] tram-l) m=2 M. + Z Aslim/4 [j m(1+§)(uR),§-‘(x, {gag ] tjam) m=2 Mb - “>321:er [I m(1-§)(1tc)i'j‘(x(“), mg 1 uj M. _. z Asp/4 [j m(l+§)(m)i1jl(x(n), §)d§ ] uj(m) m=2 Mum. + )3 Asa/4 [I m(l-§)(1cc)i§-‘(x‘“’. §>d§ 1 Auj(m) m=Mb+l M.+M.-l + 2 AS?” [I m(l+§)(1tc)i§‘(x(“), §)d§ ] Auj(m+l) ID=M~+1 where u}"", 552'“), 5‘2”” are values at the external boundary nodal point m (m=1, Mb), Aujo“) are values at the crack nodal point m (m=Mb-I-1, Mgr-MC); 111.01) = uj(x = £sz (uni? (x. r) tic!) - (nos? (x. x) ujtso] also-o M5+Md + 2 I m(uc)i,-*‘(x, it) Aujcr) dsCit‘) m=Mb+l M.+M¢1+Md+1 + 2‘. I m(uc),§‘(x, r) Aujor) dsoz) m=Mb+M¢1+2 Mb nitx) = 2 1 ml (1:11)}; (x. 0 5m -(1tC)i,-“(X.X) u, an] dsm m=2 Mb‘l’Md + E Jm(1w)i‘j‘(x. 92) Aujm dsm (2.2.11) me'i’ 1 M5+MQ+M4+1 + z j m(sc),*j|(x, x) Aujm ds(x) x on I‘c1 m=Mb+Md+2 AX2 77 Mb+Mc1+Mc2+2 Mb+Mcl+l c ack efement m m und e emegiym m- Figure 2.31 Discretized half-plane region containing a discretized edge branch crack. 73 M1, n.(x) = 2, I m[ («mg-‘0‘. x) 5m —(sc),§1(x,sz) uj (2)1 ds(i') m=2 Mb+Me1 + 2 I m(nc),§-‘(x. s) Aujor) (ism ”My? 1 M.+M,,+Ma+l + 2 I mtnc).'}(x.x1 Aujtx) dsm x on 11,. m=Mb+Md+2 Following the same procedures as in previous section, eqs.(2.2.11) become M. Ciin)“i(") = 2213914 [Im(1-§)(un)g(x, §)d§ ] tj(an-1) M. + 22mg‘I4 [Im(l+§)(uR)i§1(x(n)’ §)d§ ] tjam) M. " 22138104 [I m(1—§)(uc),'}(x, gag ] uj(rs-1) Mb 111 h (I!) (m) - "EzAsb I4 [I m(1+§)(lIC)ij(x . §)d§ ] uj Mb+Mel + 2: Ass/4 [ I m(1-§)(uc)1§-‘(x‘"’. eat 1 AuJ-(m) 313M544 Mb+Md-1 + 2 Ass/4 [Im(1+§)(uc)i‘j‘(x‘“’. 0.11:1 Auj(m+1) ”M544 Mb+Mel+Md+l + 2‘. Ass/4 [ I m(l-§)(uc)i’,-‘(X‘“’. eat 1 Aujo“) m=Mb+Md+2 M5+Md+Md + Z Asa/4 [ I m(1+§)(uC)i‘J-‘(x(“). eat 1 Auj m=Mb+Ma+2 M. xi“) = 2248104 [I “Ia-00111530“), mg 1 5‘2”” 79 M. + 2 Ase/4 [ I m(1+§)(1tR)1’J-‘(x(“). §)d§ 1 .1211» m=2 M. - 2 As1‘.“/4 [I m(l'§)(“°)5("(n)' gm: 1 slim-1) m=2 M. _ "Elfin/4 [I m<1+§)(“°)i§’(x("). §)d§ ] ufm) Mb'i‘Mcl + 2 As.”/4 [Im(1-§)(1t0)i§-‘(X‘“’. F.1d: 1 Auj (2.2.12) HFMfi'l M.+M,1—l + 2: Asa/4 [I mt1+§). an: 1 Asj m=M 5+ 1 Ms+Ma+Ma+1 + )3 Ass/4 [I (1-§)(1w),’,-‘(x‘“). out; 1 Auj m=Mb+M¢1+2 m Mb+Mci+Mc2 + 2 Mir/4 [I m(l+§)(1;c)i:.\(x(n)’ a”: ] Auj(m+l) m=Mb+Md+2 M. Kim) = “E2A56n/4 [Im(1‘§)(1tR)ig-‘(x(“), §)d§ ] thm—l) M. + 22481?“ [I m(l+§)(nR)§}(x “3M3? l Mfi'Md-l + 2: Ass/4 [ I mtl+§>,';(x. ed: 1 Asj, gm: 1 Auj m=M5+Md+2 where ujm), 59‘“), 552““) are values at the external boundary nodal point m (m=l, Mb), Aujm‘) are values at the crack nodal point m (m=Mb+1, Mb+Mc1 , Mb+Mcl+2, Mb+Mc1+Mc2+l) ; uj(“) = uj(x(“)), x0” being the location of the external boundary nodal point n, (n=1, Mb), and 1:5“) = ni(x(“)), x‘“) being the location of the crack ele- ment mid-point n, (n= Mb+1, Mtg-Mel, Mb+Mcl+2, Mb+Mc1+Mc2+1), in addi- tion, we add two nodal points n= Mb+l, Mb+Mc1+2. For an edge branch crack, we need to employ continuity conditions and equilibrium conditions, i.e. AuijW’ = Auj‘Mhmdiz’ (2.2.13) ijs+Mu+l) = Fj(M"+M“‘+1) + Fj(M"+M“+2) . (2.2.14) Then, eqs.(2.2.12) can be written in the following matrix form: min}-[iAull-Iollwlwmlt} 0.1-2.1}-.11.}...{11q ...{u}-{.11.}-..{s}-..,1.}-1.} 81 where [UC] is 2Mb x 2Mb , [Q] is 2Mb x 2Mcl , [Q] is 2Mb x 2M,2 , [UR] is 2Mb x 4(Mb-1), [PC] is 2(Mc1+1) x 2M1, , [x1 is 2(M,1+1) x 2Mcl , [x121 is 2(M,1+1) x 2Mc2 , [PR] is 2(M,1+1) x 4(Mb—1) , [PC1] is 2(Mc2+1) x 2Mb , [x211 is 2(Mc2+1) x 2Mc1 , [)9] is 2(Mc2+1) x 2Mc2 and [PR1] is 2(Mcz+1) x 4(Mb—1). Thus [UC] -[Ql -le - u ' EUR} [01 [01' t [mm -[I‘2][X] -[I‘21[x121 121.1»: [1‘2][PR] [I1 [01 11:1 * _[r21][PC1] -[I‘21][X211 £2.11an Au2 _[1‘21l[PR1] [01 [11‘ F2 (2.2.16) where, as in [76], we have defined nodal force matrices on the crack by (2.2.17) -1 I 0 0 0 0 0 0 -I I 0 0 0 0 [1‘2] = [r21] = f) 9 f1 I f) 9 9 (2.2.18) . ('1 ('1 (1 ('1 ('1 -'r i. 82 1 represents a 2x2 identity matrix, [P2] is 2Mclx2(Mc1+1), and [F21] is ZMczx2(Mc2+l). It is well-known that one can obtain the diagonal 2x2 blocks of [UC] in eqs.(2.2.8) or (2.2.16) from rigid-body considerations, as in the previous section. Once eqs.(2.2.8) or (2.2.16) have been constructed, we must impose four conditions at each nodal point m on I], involving the boundary values “10.0, “(2.11), t1(2m+1), u ém)’ témn)’ t £21m), and rearrange eqs.(2.2.8) or (2.2.16) accordingly to obtain .1414 ‘ where {2} contains the unknown boundary displacements and tractions on I}, and the unknown matrix {Au}. Once we have obtained Aui at each crack nodal point, the displacement discontinuities normal and tangential to the crack surfaces at that point are 83 Ann = Aulnl + Anznz (2.2.20) Au. =3 Allan —' Aulnz and the stress intensity factors can be determined from KIIFO = \/ gleoawmunm) Kn!“0 = \j -81-ce-G(1+V)Aut(e) (2.2.21) rills=l = %G(1+V)Aun(l-e) K111,=1 = 81t—80(1+V)Au.(1-e) where e -> 0, G, v are the shear modulus and Possion’s ratio, respectively, and l is the length of the crack. 2.2.3 Results Here we employ the coupled model to find numerical solutions for some finite domain problems. b) C) 84 Single edge crack in a rectangular plate subjected to uniform uniaxial tesile SUCSS. A rectangular plate of height h and width w with a single edge crack of length a is loaded by a uniform tension acting over an interval of length L. The total force acting on the interval is denoted by constant 0, as shown in Figure 2.32. The crack line is modeled by 20 equal elements and the external boundary part PM is modeled by 45 equal elements. In Tables 2.16 and 2.17, the stress intensity fac- tors, normalized with respect to 6‘5, are given for various ratios of a/w and L/w and compared to those obtained in [93-96]. Single slanted edge crack in a rectangular plate subjected to uniform uniaxial ten- sile stress. A rectangular plate of height 2.5w and width w, containing a single slanted edge crack of length a, is subjected to a uniform uniaxial tesile stress 6 at the ends. The crack is located eccentrically a distance w from one end and inclined at an angle [3 towards the other end, as shown in Figure 2.33. The crack line is modeled by 25 equal elements and the external boundary part PM is modeled by 45 equal elements. In Table 2.18, the stress intensity factors, normalized with respect to %, are reported for various ratios a/w and various angles [3 and compared to those obtained in [84]. Single edge crack in a rectangular plate subjected to mixed mode loading. A rectangular plate of height h and width w with a single edge crack of length a is subjected to a unit uniform tensile stress 0' at one end and a sliding support on 85 9 x2 x1 ‘ 11 A +— —> <— a —> L _‘l_ <—— ———> W 11 6 ‘— —— _—> O 1 ‘ Figure 2.32 Geometry and loading of a rectangular plate with a single edge crack. ...LIIrwv [IIIIL1I— ..Ic _II I ILIIIIII. ..I..I.H 111.111.! 01.-.”..1 x .-.”... . g h- Infill. ...bILI LL.\ 86 Table 2.16 Stress intensity factors for a single edge cracked plate under uniform tension. a/w 0.1 0.2 0.3 0.4 present [93] present [93] present [93] present [93] 1.0 1.22 1.23 1.48 1.49 1.83 1.85 2.30 2.32 0.9 1.41 1.43 1.70 1.72 2.10‘ 2.13 2.64 2.67 0.8 1.65 1.67 1.98 1.99 2.42 2.45 3.01 3.05 0.7 1.93 1.95 2.28 2.31 2.80 2.82 3.45 3.48 0.6 2.29 2.31 2.66 2.70 3.21 3.25 3.90 3.95 0.5 2.75 2.78 3.16 3.19 3.73 3.76 4.43 4.48 0.4 3.34 3.38 3.75 3.76 4.27 4.32 5.00 5.03 0.3 4.05 4.09 4.40 4.43 4.87 4.92 5.52 5.57 0.2 4.84 4.88 5.13 5.16 5.50 5.54 6.10 6.13 0.1 5.60 5.64 5.82 5.88 6.12 6.16 6.61 6.68 Table 2.16 Stress intensity factors for a single edge cracked plate under uniform tension. L w 0.5 0.6 0.7 0.8 /W present [93] present [93] present [93] present [93] l 0 2.99 3.01 4.11 4.15 6.34 6.40 11.2 12.0 .9 3.41 3.45 4.68 4.73 7.18 8.22 12.8 13.4 .8 3.88 3.91 5.28 5.31 8.00 8.05 14.4 14.8 .7 4.36 4.40 5.90 5.92 8.85 8.88 15.9 16.2 .6 4.90 4.93 6.50 6.54 9.66 9.71 17.1 17.6 .5 5.45 5.48 7.15 7.17 10.45 10.50 18.6 19.0 .4 6.00 6.03 7.73 7.78 1 1.33 l 1.40 20.0 20.4 .3 6.52 6.57 8.35 8.39 12.14 12.20 21.2 21.8 .2 7.10 7.12 9.00 9.01 12.96 13.00 22.8 23.2 .1 7.61 7.67 9.60 9.65 13.84 13.90 24.0 24.7 Table 2.17 Stress intensity factors for a single edge cracked plate under uniform tension, I./w=l.0. U W=L0 W present [93] [94] [95] [96] .125 1.28 --- 1.26 1.27 1.299 0.150 1.35 --- 1.30 1.34 1.362 0.200 1.48 1.49 1.40 1.48 1.505 0.300 1.83 1.85 1.67 1.82 1.867 87 I XZ w l 1.5w : : X1 1 “ >-—> ...— .__.. ‘— ——. . 1——> ‘— —> 0' <— "_> C «— —_> ‘— ___. <—-— I—p _ v : Figure 2.33 Geometry and loading of a rectangular plate with a single slanted edge crack. -ILIL IL L-IIHIL 88 Table 2.18 Stress intensity factors for a single slanted edge cracked plate. L... [3 = 45° [3 = 67.50 present [84] present [84] KI/ov‘nT tin/own? K1051? XII/00E ' KI/oficsT Kn/ci/nTKI/offi Kn/oficT 0.30 0.35 0.40 0.45 0.50 11.55 0.60 035 0.443 0.15 0.443 1.43 0.342 1.43 0.342 0.95 0.472 0.95 0.473 1.58 0.370 1.58 0.370 1.01 0.502 1.02 0.504 1.76 0.402 1.77 0.404 1.08 0.534 1.10 0.536 1.99 0.440 2.00 0.441 1.20 0.570 1.20 0.571 2.27 0.491 2.28 0.494 1.30 0.608 1.32 0.612 2.60 0.562 2.62 0.565 1.40 0.658 1.43 0.662 3.03 0.658 3.06 0.662 89 the other opposite end as shown in Figure 2.34. The crack line is modeled by 30 equal elements and the external boundary part PM is modeled by 46 equal ele- ments. In Table 2.19, the stress intensity factors are calculated and compared to those obtained in [97-99]. h/2 J h/2 II—_ _‘ II II _>< I I I W= 10 I I I Figure 2.34 Geometry and loading of a rectangular plate with a single edge crack. 91 Table 2.19 Stress intensity factors for a single edge cracked plate as shown in Figure 2.34. present [97] [98] [99] KI 32.6 34.0 33.20 33.1 Kn 4.38 4.55 4.50 4.36 Chapter 3 Mode III Crack Problems : Internal Cracks 3.1 Theoretical Development Consider an infinite isotropic elastic plane in which there is a point, X, at which some "source" of stress is located and a point, x, at which the stresses are to be computed. We will employ the following influence functions associted with antiplane strains: (uR)33(x, x) = the displacement in the x3 direction at it due to a unit force applied in the x3 direction at Y in the infinite plane, (uc)33(x, Tr) = the displacement in the x3 direction at x due to a unit displacement discontinuity applied in the x3 direction at i in the infinite plane, (ER)33(x, i) = the scalar 1:3 in the x3 direction at it due to a unit force applied in the x3 direction at “i in the infinite plane, (nc)33(x, it) = the scalar 1:3 in the x3 direction at it due to a unit displacement discontinuity applied in the x3 direction in the infinite plane, where 113 is a stress function defined such that the stress components are 811: 0.. = 5;}. 023 = — {2%. (3.1) 92 93 The influence functions are given in the Appendix C. In Figure 2.2, suppose that I‘c represents a crack, the surfaces of which are subjected to equal and opposite traction t3(s) given by t3 = — ( 013111 + 023112 ) = _ _ (3.2) A plane elastic region Q, with external boundary I‘b, containing an internal piece-wise smooth crack I}, as shown in Figure 2.2, or an internal piece-wise smooth branch crack or multiple cracks composed of Pd and Peg, as shown in Figures 2.3 and 2.4, is loaded by prescribed tractions t3 on some part of the external boundary and prescribed displacements u3 on the remainder of the external boundary. Then the direct boundary-integral equations and the crack integral equations developed as in [76] can be coupled as follows. For a single smooth or kinked crack: cm 113 (x) = IrguRbs (x. r) 600 dsCi) - 1.5m)” (x. s) .1300 ass) + Ir,(uc)33 (X1 3) A9307) dsfi) x 0“ Pb (3'3) 160:) = 1.0181330. 10 1.00 dstio 4.9.6133“, 8) 113m dsm + Irc(“°)33(xv i) A1135) dsm x on I‘c . 94 For a branch crack or multiple cracks: 030!) 1130!) =I1rsuR)33(X. 8) I300 (1860 - £91033 (X. 8') 113(8) €155) 2 + 21rd(UC)33 (X, Y) AU3(Y) (1803) X _ on I}, k=l m) = 1.010330. 10 1300 dson — Irgncwx. r) .1300 ass) 2 + z I 1:. (01.30:. s) Augm ass) x on I}. (3.4) k=1 193(x)=I>r§nR)33(x.Y) 13(8) dsGl) — Irgucxatx. a) 113m dsm 2 + Z I r. (masts m 411.00 dst) x on 112 #1 In eqs.(3.3) and (3.4), Au3 = u; — u; is the relative crack surface displacement in the x3 direction and c3 is a coefficient which depends on the smoothness at x on 1],. 3.2 Numerical Treatment 3.2.1 A single smooth or kinked crack A simple numerical treatment of eqs.(3.3) is presented here in which the external boundary is approximated by M, straight elements and the crack by Mc straight ele- ments, as shown in Figure 2.5. For this model, eqs.(3.3) can then be written as M. 030‘) 1130‘) = z I Int (1111).. (x. i) t. (s) - (uc)33 (x. in 113 (i1 1 dstio m=l \U U] M.,+Mt + 2‘. I (6.1,, (x. r) M301 dsm x on rb m=Mwl m (3.5) M. 103(11): 2 I m[ (1:103. (x. in .3 (r1 - (n.1,, (x. 81 u. 00 1 dstro m=l Ms+M. + z I more)” (x. :1 411.0) dsm x on rc . th+ l The displacements, tractions and displacement discontinuities in the x3 direction on each element can be linearly approximated by 113(2) = N1(§)u§m-1) + N2(§)u§m> r on element m of 1“b 135:) = N1(§)t§2m-1)+ N2(§)t§2m> x on element m of I], ‘ (3.6) A630.) = N,(§)Au]m> + N2(§)Au§m+1> ii on element in of r, where uI’“), 62‘“), $261+» are values at the external boundary nodal point m (m=1, Mb), Aué‘“) are values at the crack nodal point 111 (m=Mb-I-2, Mb+Mc), and ME) = (1—0/2 N2(§) = (1+§)/2 -1 s s s 1 . (3.7) Furthermore N1(§)x(’“’l) + N2(§)x(m) x on element in of I], N1(§)X(m) + N2(§)X(m+l) it" on element m of I‘c 96 “signal _. sém'1))/2] (1:: [As t511,2] (I: i on element m of I], 95(3) = (3.9) [(Sémfi) ' Scmb/Z] d9 = [A531 / 2] d5.» x on element m of R where Ast‘,n is the length of the external boundary element m and As;n is the length of the crack element m. Inserting eqs.(3.6) through (3.9) into eqs.(3.5), we obtain M1, cénluy‘) = )3 Asf,“/4 [I m(1—§)(uR)33(x(“), gm: 1 19"“) m=l Mb 2m + 2 8.11/4 II m<1+§>ss. ed: 1 ti ’ m=1 M. - z Ass/4 [Im(1'§)(uc)33(x(n). 01:1 .11“) m=l M1, - 2 Ass/4 Ilm<1+§>. ed: 1 aim) m=l Mb'i'M‘ + 2; Ascm/4 [I m(l-§)(uc)33(x(“), gm: 1 Aug?” me‘I'z M5+M¢- l + )3 Asg‘m [I m(1+§)(uc)33(x(“), 00:16.11“) HFMfi' 1 (3.10) M. 11;“): zlAsgnm [ I m(1‘§)("R)33("(n)1 §)d§ ] t§2m-l) M. + 2 A81%“74 [I m(14.5,)(1110.30.00, gag ] ti“) m=1 M. - 2148104 [I mU-txmsstx‘“). 0d: 1 aim-1’ 97 M. — z Asp/4 [I m(1+§)(1cc)33(x‘“’. §)d§ ] 113‘“) m=l M.,-t-Mc + z ASén/4 [I m(1—§)(KC)33(X(n), §)d§ ] Auém) “FMF+2 M5+M¢-l , + 2 Asgn/4 [ I m(1+§)(nc)33(x(n>, 13d: ] AuIm+1> [FMV‘I’ l where u?” = u3(x(“)), x‘“) being the location of the external boundary nodal point 11, (n=1, M), and it?) = 1:301“). x‘“) being the location of the crack element mid- point n, (n= Mb+1, Mb-I-Mc). Eqs.(3.10) can be written in the following matrix form: [Uc31Iu3I- [6.1I .., I = 16.,1I.I misl-mimstIeI-(I where [UC3] is Who [Qil is M6X(Mc"1). M31 is MbXZMb. [PC3] is McXMb9 IX3] is M,x(Mc-1) and [PR3] is Mcx2Mb. (3.11) Thus r [U03] '[Q3] 113 I [1‘3] [PCs] -[F3][X3] (3.12) WR3] [0] I t3 Au3 [F3] [PR3] [I] L F3 98 where, as in [76], we have defined a nodal force matrix on the crack by {F3 } = {@4103} (3.13) where I-1100000‘ 0-110000 ...= 991%??? «34> .0 0 000—111 and [F3] is (MC-1) x M,. 3.2.2 A branch crack or multiple cracks A simple numerical treatment of eqs.(3.4) is presented here in which the external boundary is approximated by Mb straight elements and the branch crack or multiple cracks by Mel, M02 straight elements, as shown in Figure 2.6 and Figure 2.7, respec- tively. Eqs.(3.4) can then be written as M, cstx) u3. ad: 1 (gm-11 m=l M. + 2 AS?” [I m331x33(x<"’.§>d§1 Any“) m=M,+M.,+2 Myl'Mcl +M¢2 + 2 Asa/4 [j m(1+§)(uc)33(x(n). ode 1 Auémm m=Mb+M¢1+2 M. “gm = 2,111ng [I m(1—§)(nR)33(x‘"’. §)d§ 1 tim'” M. + 2 Asgn/4 [j m(1+§)(1:R)33(x“”’. §)d§ 1 tal“) m=l M. — glassy/4 [j m(1—§)(1cc)33,(x“”). §)d§ l vim-1’ M. - Elissa/4 [I m<1+§xnc1331x. oat 1 uém’ Mb'i'Ma + 2 Asa/4 11m(1-§>3a(x. am: 1 Au?” m=Mb+2 Mb'i'Md-l + z 21.31/41] (1+§)(1tc)33(x(“), gm: 1 Au§m+1> (3.19) m=Mb+1 m M5+Md+Ma+l + )3 mom/4 [j m(1—§)(nc)33(x(“), gm: 1 Auém) m=Mb+Mfl+2 M5+M¢1+Ma + 2 Assn/4 1] m331x‘n>. ed: 1 Auémm IIFM5+M¢1+2 M. my!) = )3 As§‘/4 [j m(1—§)(nR)33(x‘“’. §)d§] 62"“) m=l . Mb + )3 MIT/4 [j m(1+§)(nR)33(x‘“’. W: 1 ti“) m=l M. ' - Elma/4 limo-€100.30“). ed: 1 uém'” 103 M. — zAsgm [j m(1+§)(1ec)33(x"‘°. §)d§ ] vim) m=l Mb+M¢l + 2 Asa/4 [I m<1—§)3333. ed: 1 Auém” m=Mb+1 M.+M,1+M,2+1 + 2: Asa/4 [I m<1—§>33. ed: 1 Au?” m=Mb+M¢1+2 Mb+Mcl+Mc2 + )3 Ass/4 11m<1+txvcc133. oat 1 Au?” m=Mb+Md+2 where u?” = u3(x(“)), x‘“) being the location of the external boundary nodal point n, (n=1, Mb), and né") = 1:3(x(“)), it“) being the location of the crack element mid- point n. For multiple and branch cracks, n= Mb+l, Mb+Mc1, Mb+Mc1+2, Mb+Mc1+Mc2+L In addition, for a branch crack, we add a nodal point n= Mb+Mc1+2. Eqs.(3.19) can be written in the following matrix form: [UC3l{ u3 } - [Q3 A“; } — [Q311{ Au32 } = [UR31{ ‘3} [PG,){ .13 } - [x3]{ A1131 } - [x3,){ A1132 } = [PR3]-{ t3 } -{ 1:31 } (3.20) [Posll{ u3 } - [X23}{ Allal } - [X33}{ Au32 } = [PR31}{ t3 } ‘ { 7‘2 } 103 For multiple cracks, the matrices in eqs.(3.20) have the following dimensions: [UC3] is Mb x Mb , [03] is Mb x (Mel-1) , [Q31] is Mb x (Ma-1) , [UR3] is Mb x 2Mb , [PC3] is Mcl x Mb , [x3] is Mc1 x (Mel-1) , [x311 is Mc1 x (Ma-1) , [PR3] is Mei 3 2Mb » [PC31] is Mc2 x Mb . [X23] is Mc2 X (Merl) . [X33] is Mc2 X (Merl) and [PR31] is M62 x 2Mb. For a branch crack, the matrices in eqs.(3.20) have the following dimensions: [UC3] is Mb x Mb , [(2,] is Mb x (Mel-1) , [Q31] is M, x Mc2 , [UR3] is Mb x ml, , [PC3] is M...1 x Mb , [x3] is Mc1 x (Mel-1) , [x321 is Mc1 x Mc2 , [PR3] is M,l x 2Mb, [PC3311 is (Mc2+1) X Mb . [X23] is (Me2+1) X (Mei-1) . [X33] is (Mez‘i'l) X Mc2 311d [PR31] is M2+1) x 2114,. Thus F [UC3] —[Q3] ’[Q31] - “3i - [UR3] [0] [0] q t3 [F3][PC3] “[r3llx3] "lr3llx37J iAu31> = [F3][PR3] [I] [0] i F31 _[I‘31][PC31] -[F 311331] -[F31][X33]‘ Au2 _W31][PR31] [O] [I] j P} Y where, as in [76], we have defined nodal force matrices on the crack by 104 (3.22) {F232}=[r311{7‘32} i-1 1 0 0 0 0 0' 0 -1 1 0 0 0 0 [32232 9 9 399 9 9 333 For multiple cracks, [F3] is (Mel-1) x Mel and [F31] is (Md-1) x M62. For a branch craCk, [F3] is Mel X (Mcl+1) and [F31] is Mcz X (Mez‘i'l). For a branch crack, we need to employ continuity conditions and equilibrium condi- tions, i.e. Au3‘M9th-km = Au3(M"+M“+2) (3.24) F 3(M5+M&+l) = F 1:Mmeil) + F 3(M5+Mel+2) (3.25) It is well-known that one can obtain the diagonal entries of [UC3] in eqs.(3.12) or (3.21) from rigid-body considerations. If we apply a rigid-body displacement, i.e. uél) = u?) = u?) = ....... = u3(M”) = u3 (3.26) 105 then the body is stress-free. Thus (1+1 131131 and eqs.(3.12) or (3.21) reduce to r ‘13 1 I13 [UC31i ' 9 ={ 0} (3.27) L “3 J so that . . M. . . UC§‘X‘) = -2UC§00> (3.28) j=1 133i Once eqs.(3.12) or (3.21) have been constructed, we must impose two conditions at each nodal point m on I], involving the boundary values 11:9"), té’lm)’ téZnH-l) and rearrange eqs.(3.12) or (3.21) accordingly to obtain 1013 [A{ Z} ={ B } (3.29) where {Z} contains the unknown boundary displacements and tractions in the x3 direc- tion on I}, and the unknown matrix {Au}. Once we have obtained Au3 at each crack nodal point, the stress intensity factors can be determined from Kmlg0 = Via/1.13m) (3.30) '1:— Kmls.1 = Emma—e) where t: —> 0, G is the shear modulus, and l is the length of the crack. It should be noted that, for a problem involving a traction free crack, {F3} = [0} and eqs.(3.12) or (3.21) reduce to 1 = [F 3] [PCs] -[F3] [X3] A03 [F3] [PR3] L [UC3] "lel 113 [UR3] {. } . (3.31) Of 107 I [UC31 —1031 -[Q31] ' .3 ' [UR31 " [1‘3][PC3] '[rsllxsl {1‘31 [X32] [F3IIPR3] { t3 } (3-32) L[I‘31][PC31] "ll-311331] -[I‘31][X33]_ A1132 _U31][PR31] J A1131 V II _A respectively. 3.3 Results Here we employ the coupled model to find numerical solutions for some finite domain problems. 1. Straight crack. In all straight crack problems, the crack is modeled by 20 equal elements and the external boundary is modeled by 40 elements. The stress intensity factors, nor- malized with respect to o‘la, are calculated. a) Straight cenu'al crack in a rectangular sheet subjected to anti-plane shear SUCSS. A rectangular sheet of height 2h and width 2b contains a straight central crack of length 2a. A uniform shear stress, 0’, acts over the ends of the plate as shown in Figure 3.1. In Table 3.1, the stress intensity factors are given for various ratios of a/b and a/h and are compared to those obtained in [100]. 108 X2 __@@@ @636)“ ll h X1 __1-1_ a l 23 | h _!_ 1: 9 4 Figure 3.1 Geometry and loading for a straight central crack in a finite sheet under a uniformly distributed shear stress. Table 3.1 Stress intensity factors for a straight cenu'al crack under a uniformly distributed shear stress as shown in Figure3.l. 109 a: o 1:0.25 1:0.5 1: 1 1:2 1:4 1:00 1.897 1.723 1.689 1.686 1.686 1.687 1:1.2 . | H 1.900 1.780 1.725 1.691 1.689 1 .689 1.689 1.460 1.369 1.359 1.358 1.358 1:1.4 .1. 1.782 1.463 1.370 1.361 1.360 1.360 1:1.6 iu‘ ‘ I 0.. 1.771 1.773 1.399 1.401 1.254 1.256 1.233 1.235 1.233 1.235 1.234 1.235 1:2.0 1.770 1.377 1.176 1.127 1.126 1.127 1.1 1.772 1.379 1.178 1.130 1.128 1.128 ~ ... dl.| 1.770 1.375 1.147 1.046 1.012 1.000 H 1.772 1.377 1.149 1.047 1.013 1.000 110 b) Straight central crack in a rectangular sheet subjected to anti-plane shear stress on the crack. A rectangular sheet of height 2h and width 2b contains a straight central crack of length 2a. The crack is subjected to anti-plane shear stress 0. Three examples are considered for the problem of a rectangular sheet with fixed edges parallel to its crack, with fixed edges perpendicular to its crack and with four fixed edges as shown in Figures 3.2 throught 3.4, respectively. In Tables 3.2 throught 3.4, the stress intensity factors are given for various ratios of a/b and a/h and are compared to those obtained in [101]. Kinked crack. In all kinked crack problems, both stress intensity factors, normalized with respect to (Fla, and relative crack surface displacements are reported. G/o = 200. a) Symmetric V-shaped crack in a rectangular sheet subjected to anti-plane shear stress. This example involves a symmeuic V-shaped crack in a rectangular sheet subjected to a uniform anti-plane shear stress, as shown in Figure 3.5. The crack is modeled by 40 equal elements and the external boundary is modeled by 44 equal elements. For various angles or, the relative crack sur- face displacements are plotted in Figure 3.6 and the suess intensity factors are reported in Table 3.5. 111 X2 7 __ / 2 h w 8%_ ' ' 6 iii 11 ‘ 6 2a 7 h __"__ 4 Figure 3.2 Geometry and loading for a straight central crack in a finite sheet with fixed edges parallel to the crack. 112 11 7 h ——11 V = .6 / 2a h _4 A 11 1 Figure 3.3 Geometry and loading for a straight central crack in a finite sheet with fixed edges perpendicular to the crack. 113 Figure 3.4 Geometry and loading for a straight central crack in' a finite sheet with four fixed edges. 114 Table 3.2 Stress intensity factors for a straight central crack with fixed edges parallel to the crack as shown in Figure 3.2. g azb 1:1.2 1:1.4 1:1.6 1:2.0 1:00 3 L111 11 111 Present 111 Present 101 '- nt 101 ' - nt 111 1:0.251.398 0.400 0.397 0.399 0.397 0.399 0.397 0.399 0.397 0.399 1:0.51.573 0.575 0.560 0.564 0.560 0.563 0.561 0.563 0.561 0.563 1:11.830 0.833 0.778 0.780 0.767 0.769 0.762 0.764 0.762 0.764 1:2 1.111 1.114 0.989 0.991 0.948 0.950 0.920 0.923 0.912 0.914 1:4 1.339 1.342 1.145 1.147 1.072 1.074 1.014 1.016 0.974 0.975 1:00 1.687 1.689 1.358 1.360 1.234 1.235 1.127 1.128 1.000 1.000 Table 3.3 Stress intensity factors for a straight central crack with fixed edges perpendicular to ithe crack as shown in Figure 3.3. . 1 '_r‘ .' 1:0.25 1:0.5 1: 1 1 :2 1 :4 1:00 1 .650 1. 177 0.844 0.772 0.767 0.768 1:1.2 0 1.653 1. 180 0.847 0.775 0.770 0.770 111‘ ' 1.759 1.295 0.981 0.868 0.856 0.857 1:1.4 0 1.760 1 .298 0.984 0.87 1 0.859 0.859 u. ‘_ 1.769 1.351 1.055 0.920 0.899 0.900 1:1.6 1 1.77 1 1 .354 1 .058 0.922 0.901 0.901 1:2.0 'u' ‘ 1.770 1.373 1. 120 0.977 0.941 0.942 O 1.772 1.375 1.122 0.979 0.943 ’ 0.942 1:00 311‘ 1.770 1.375 1.147 1.046 1.012 1.000 x 1 0 1.772 1.377 1.149 1.047 1.013 1.000 Table 3.4 Stress intensity factors for a straight central crack with four fixed edges as shown in Figure 3.4. 1:00 Present [101] 1:1.6 Present [101] 1:2.0 Present [101] 11.2 1:1.4 [101] Present [101] 115 11 Figure 3.5 Geometry and loading for a symmetric V-shaped crack in a finite sheet under a uniformly distributed transverse shear stress. .xoouo voaocml> 05088.3 0 ..8 034 EmEmooEflv mootam x020 m>$20m 0.0 950E ig‘i Dmpud Gowns omens oomnud On.“ "5 OGCEEBK 00.0 15.0 1N0.0 100.0 10.0 00.0 b) C) 117 Asymmetric kinked crack in a rectangular sheet subjected to anti-plane shear SUCSS. This example involves an asymmetric kinked crack in a rectangular sheet subjected to a uniform anti-plane shear stress, as shown in Figure 3.7. The crack is modeled by 30 equal elements and the external boundary is modeled by 44 equal elements. For various ratios L/Za, the relative crack surface displacements are plotted in Figure 3.8 and the stress intensity fac- tors are reported in Table 3.6. Anti-symmetric kinked crack in a rectangular sheet subjected to anti—plane shear stress. This example involves an anti-symmetric kinked crack in a rectangular sheet subjected to a uniform anti-plane shear stress, as shown in Figure 3.9. The crack is modeled by 35 equal elements and the external boundary is modeled by 44 equal elements. For various angles a, the relative crack sur- face displacements are plotted in Figure 3.10 and the stress intensity factors are reported in Table 3.7. 118 X2 ___ooooooo 11 x1 11 _L_ 11 10 4 Figure 3.7 Geometry and loading for an asymmetric kinked crack in a finite sheet under a uniformly distributed transverse shear stresss. 119 Table 3.5 Stress intensity factors for a symmeuic V-shaped crack under a unifome distrib- uted shear stress as shown in Figure 3.5. (X 15 30 45 60 75 Km 1 min? 2.0700 1.8743 1.5755 1.1965 0.7496 Table 3.6 Stress intensity factors for an asymmetric kinked crack under a uniformly distribut- ed shear suess as shown in Figure 3.7. 11221 0.2 0. 16 0. 12 0.08 0.04 Km / 6317:? 3.4359 3.9220 4.5877 5.6497 7.9809 Table 3.7 Stress intensity factors for an anti-symmetric kinked crack under a uniformly dis- tributed shear stress as shown in Figure 3.9. (X 15 30 45 60 75 Km/ min—a— 4.5605 4.3281 3.9757 3.5427 3.0664 120 .1111! I 03< EoEoooEmfi mootam xoouo o>02om 0.0 0.59... .ono 09E: EtoEExmo :0 cos. 1* B i § 0 m—fi T 0.x* ‘03 0 ‘0 0 4 0 0 1 0 0 4 01 B 0 4 0 fl! 5 0 0 B x 01‘ 0 E f 0¢l|8 X 0‘00!!! OWI0EI¥ 0WI051¥ OH‘0HI¥ N0"... 1 .31.. 0.0.1.4 j w.O..1u1_ 0. T1.._ O.,-(x.8>b,(8)48(8) = 1’11[(00)13(X.70T®-(11R)i3(xsm(§)1ds®-§rb(uR)ij(X,i)'YnJ-(i‘)T(‘i‘)dsCi) — 1.533311308408448) x on 1“,. s 1. a (4.3) “10" + 1., [(8c):j(x.s)uj(8)—(8R).j(x.8)t,-(8)st(8)- j rc(nc)ij(x,i)AuJ-(‘i)ds(‘x’) = 10(8R).,-(x,8)bj(x)dsm) = 1r,[(801.3(8.8)T18)-(8R)s(x.8)q.(8)143(8) 'frb(fiR)ij(X.WJ-(X)T(X)ds(i) - jrc(nc)i3(x,x)AT(i)ds(i') x on 1“,, 11 in a where i=1, 2, j=1, 2, summation on repeated indices is implied, and Auj = uj’ — uj” are the relative crack surface thermal displacements. In eqs. (4.3), we have inserted b]: — ya” and applied the divergence theorem as shown in Appendix D. It should be noted that the right sides of eqs.(4.3) are known functions. 4.2 Numerical Treatment A simple numerical treatment of eqs.(4.3) is presented here in which the external boundary is approximated by M, straight elements and the crack by Mc straight ele- ments, as shown in Figure 2.5. Eqs.(4.3) can then be written as Mb 33,03) 13,031 + “a; mi (1331.03. 8) u,- (8) - (13R).- (:3. 8) 35(8) 1 43(8) 129 Mb-t-M, - 2 I munch-(x. 8) Au,-(8) 43(8) HFMb'i'l M. = 2: I m[(uc),3(x. 8) T(8)—(uR).3(x. 8) 31.001 43(8) m=1 M1. - 2‘. I m(48).,(8. 811111)me 43(8) m=1 Mb'i'M, _ Z Im(uc)i3 (x! i) ATm dsci) X on Pb m=Mb+1 (4.4) M. 8.03) + 2 I m1 (333).,- (x. 8) u,- (8) - (88).,- (x. 8) 5(8) 1 43(8) m=1 Mb+Mc - )3 I m<8c>.,-(x. 8) 415(8) dsCr'r‘) m=Mb-1-1 . M. = 2 I m[(1tC)13(x. 8)T(8) — (11R)13(X.i)qn(i)] 43(8) m=1 Mb - 21m<8R>s<8.8)1n,-(8)T(8)43(8) m=1 M..+Mc - )3 I m(10,301, 8) AT® (ism x on 1“. . lfl=Mh+1 The displacements, tractions, displacement discontinuities, temperatures, heat fluxes and temperature jumps on each element can be linearly approximated by 15(8) = N,(§)uj + N2(§)uj 8 on element m of 1‘1. 5(8) = N1(§)tj(2m‘1) + N2(§)tj(2‘“) Y on element m of I], AujO'l) = N1(§)Auj(m) + N2(§)Auj(m+1) if on element m of 1'} 130 1(8) = N1(§)r + N2(§)r 8 on element m of I‘b (4.5) anl') = N1(§)q§2‘“‘1) + N2(§)q§2‘“) i on element m of I], AT(X) = N1(t_-,)Ar + N2(§)AT"“+1) x on element 111 of P. where uji‘“), 51210-1), 59’“), 1(8)), qISZm-l), gym) are values at the external boundary nodal point m (m=1.... Mb), Aujim) and A11”) are values at the crack nodal point (m) (m-- M,+2,... Mb+M,), and N1(§) = (1—§)/2 N2(§) = (1+§)/2 -1 5 § 5 1 . (4.6) Furthermore if = N1(§)x(m“l) + N2(§)x(m) i on element m of I], and I‘c (4.7) dsGt") = [(309) - s)/21 d§ = [11399121 dE, 8 on element m of 1‘1. and I; (4.8) where As“1 is the length of the element m. Inserting eqs.(4.5) through (4.8) into eqs.(4.4), we obtain Mb 983891“) + £14394 1 I m(1—§)(uc).,-(8°°. :14: 1 uIm'” M. + 23 A3974 [ I nu(1+§)(11e),j(x<9), gm: ] slim) m=1 131 M1, - 351”” [I m(1—§>(nR),,-(8<9>. £143 1 4239-1) Mr. - m2:31Asm/4 [I m(1+§)(“R)ij(X(n)3 §)d§ ] tj(2111) M5+M¢ . - 2 (ism/4 1 I m(1—3.)(u<:).,(x<9>. §)d§ 1 2.1m) m=M5+2 M,+M, " 2 [ism/4 [I m(1+§)(u0),j(8‘“’. §)d§ ] Auj(m+1) IIFMfl-l M. = 21mm/4 [j n"(1—§)(uc)i3(x(n), §)d§ ] T(rn-1) M. + 2148mm [I m(1+§)(uc),3(1s<9>, gm; 1 311m) M1. _ zlAsm/4 [I m(1-§)(uR)ij(x(n)’ §)'Ynj(§)d§ ] T(m-1) M. - 21439914 I I m(1+§>(uR).,-(x<9>. gmjemg 1 143» M.,+Mg - 2‘. 439/4 1 I m(1—§)(uc),.(x<9>. 04:1 8183) m=Mb+2 M5+M¢ - 2 11.3314 1 I (10031.09). 0481 8183+» m=Mb-1-1 111 Mb - 2143914 1 I m(181081.309). 5141; 1 (11233-11 M. - 2: 439/4 1 I "108108181891 08: 1 .1133) m=1 (4.9) M. 11,00 + 23 A3974 [I m(1-§)(1ce),j(x<9>, §)d§ 1 111-(m4) m=1 132 M. + 2 Asm/4 [I m(1+§)(m)11(X‘"). E.)d§ ] uj(...) m=1 M. - Z A~°>""/4 [I m(1—§)(rtR)ij(x("), mg 1 tJ(233-1) m=1 Mr. _ "Erma“ [I mil+§)("R)iJ-(X‘“). 9d: 1 11‘2“” Nib-1M: _ E4: 2ASm/4 [I m(1—§)(KC)ij(x(n)’ §)d§ ] Auj(m) tn: ‘,1. M1,+M, ._ 2 Asm/4 [I m(1+§)("c)ij(xi“). §)d§ ] AuJ-(m’rl) m=Mb+l Mr. = 2 Asm’4 [1m(1-§)(8c),3(x, 0d: 1 1033-1) m=1 Mb + 2 AW [1 m<1+§>s(x<">. :14: 1 1831 m=1 Me .. z Asm/4 [I m(l-§)(1cR)ij(x(n)’ gmjmmg ] T(m_1) m=1 M. - 2 11.3314 [Im(1+§)(33R).,-(x<9>.3111.104“ 1831 m=1 Mb+M¢ - {4; 243974 [I m(1-§>(3c)..(x<9>. £14: 1 mm) M5+M, - )3 11.3314 [I m. :14: 1 A'I'im“) m=M5+1 M1. - 2 Asm/4 [I m(1—§)(1:R)i3(x(n)’ §)d§ 1 q12113-1) m=1 M. ‘ 2 Asm/4 [I m(l+§)(8R),3(x, 9d,; 1 q12331) m=1 133 where 111‘“) = uj(x(“)), xi“) being the location of the external boundary nodal point 11, (n=1, M), and 8,00 = «1649), xi") being the location of the crack element mid- p01nt n, (n= Mb'i'l, ... Mb+Mc)' Eqs.(4.9) can be written in the following matrix form: 2311- 3131- .1}. 211- 31131 411 .11. .41.}...11 . 1.1. .21.}- ...,1..1- ....1.1 where [UC] is 2M1, x 2M1, , [Q] is 2M1, x 2(Mc-1) , [UR] is 2M1, x 4M1, , [PC] is 2Mc x 2M1, , [X] is 2Mc x 2(Mc-1) , [PR] is 2Mc x 4M1, , [UC13] is 2Mb x Mb , [Q13] is 2Mb x (Mc -1), [UR13] is 2M, x 2Mb , [PC13] is 2Mc x Mb, [X13] is 2Mc x (MC—1) and [PR13] is 2M0 x 2Mb. ' (4. 10) Thus -[Q1 [F21 [PC] -[I‘2][X] [0] 1:1-“J“ m {L} ‘ [UC131 '1Q13] {2‘} [UR13] I I = — q. (4.11) [FZHPCIB] -[l‘2][X13] T [I‘;][PR13] 134 where, as in [76], we have defined a nodal force matrix on the crack by 1.1.1.} where ”-1100000' 0—110000 [1.2]: 0 041000 (4.13) .0 0 000-11. I represents a 2x2 identity matrix, and [1‘2] is 2(Mc-1) x 2M? It is well-known that one can obtain the diagonal 2x2 blocks of [UC] in eqs.(4.11) from rigid-body considerations. If we apply a rigid-body displacement, i.e. ujm = “112) = uj(3) = = um‘) = u (4.14) then the body is suess-free. Thus {.11.} 1.1.1.} and eqs.(4.1l) reduce to n1 n2 111 112 ' [UCF . *={0} (4.15) u2 so that M UC(21-1)(2i—l) = _iudzi-lxzj—l) i=9 1*] M UC(21-l)(2i) = _iUCQi-IXZj) 1:9 1:31 (4. 16) M ucaixzi-l) = _iUCaixzi-I) F] M UCaixzi) = -2‘:UC‘2‘X25) . [=9 per Once eqs.(4.11) have been constructed, we must impose four conditions at each nodal point m on I], involving the boundary values u 1(m)’ t1(2m)’ t1(2m+1)’ 11 ém)’ 62m), té2m+l)’ and rearrange eqs.(4.11) accordingly to obtain [11.1-1.1 where {Z} contains the unknown boundary thermal displacements and tractions on I], and the unknown matrix {Au}. Once we have obtained Aui at each crack nodal point, the thermal displacement discontinuities normal and tangential to the crack surfaces at that point are Alln = Aulnl + Auznz (4.18) Aut = Auznl - Aulnz and the thermal stress intensity factors can be determined from K11s=0 = \I%G(1+V)Aun(e) K111,=0 = \I %G(I+V)Aut(e) 137 (4.19) K11“=1 = -8%G(l+v)Aun(l—e) K1113:1 = %G(1+V)Aut(1—e) where e —) 0, G, v are the shear modulus and Possion’s ratio, respectively, and 1 is the length of the crack. It should be noted that, for a problem involving a traction free crack, {F}={0} and eqs.(4.11) reduce to [UC] -[Q1 {1:} [UR] 1‘} [13] [PC] -[I‘2][X] [F21 [PR] [UC13] -[Q13] {AT} [URla] I I ' = — qn (4.20) [F2111’C13] -[F2J[X13] T [FflIPRn] 4.3 Results Here we employ the coupled model to find the numerical solution for a finite domain problem. 138 A rectangular plate of height 2h and width 2w contains a su'aight central crack of length 2a. The surface of the crack are assumed to be free of tractions and ther- mally insulated. Mechanical and thermal boundary conditions are shown in Frgure 4.3. The crack is modeled by 20 equal elements and the external boundary is modeled by 40 elements. In Table 4.1, the thermal stress intensity factors are given for various ratios a/w and compared to those obtained in [57] and [61]. The relative crack surface thermal displacements are plotted in Figures 4.4 through 4.9 for G = 8.411104, v = 0.3 and ot = 1.67x10-5. 139 11 x2 T==100 _11_' h q=0 I: 2 a :1 q=0 -X , I I $20m 4.4 830E 2.0 85 mod No.01 8.01 2.01 . b I — . L P b F . p P I p - 1— b p I “Clues—.1 fl 1 t I fi . x a 1 § § u 1 .. ..eo1moe1 r i ... X 1 1 0.. 1 x 1001m0.Nl r ... 9 .. a 1 .. . 1eo1mo.m fl .. H r 1 ¥ 1 100lm0.0 w I I 1 1 fl fl 2 eons} x a a ... 4 p p p h r P p L . p p P b p h — . L p mOIUOoP .N.0H;\o .ono .9550 290.3% 03.239: :0 9:53:00 300 o .68. U5Q EoEoooEmfi _octofi oootzm x080 020230 0.4 830E 0N0 N 70 LV0.0 ¢0.0.1 9.0.1 0N0... . _ . . _ . . — L p — b . . mouwoo¢l - n p u - 142 1 N.O"3\O * fi fi 1 b h _ . . p _ p . p L b - b _ . b - mOIMOov 143 .0.ou;\o .ono .9200 EBB? 03239: :0 mEEEcoo >000 o ..8 054 “CoEoooamfi _oEuofi oootzm x096 023.23”. 0.4 959... omd w To 00.0 00.01 m v.01 00.01 . . h b1 . . L r F P p h b _ p _ h b . mOIIMOoml m x x - x 1 a. a. 1 .. 181M481 1 fi fi 1 x 1001mm. Fl 1 ! fl I x Imolmmé 1 l 1 1 . 481.4% a. a. r I I Lfln mug—111m. I . . . _ _ . . . a m a. . molmod .¢.on3\o .ono 65:00 298% 83.35 :0 9:53:00 xvon o ..2 ¢N.0 h P b h 3.0 9.0... 954 EmEoocamfi _octof 80.56 x080 3523* NJ. 0.39.”. 00.0! . L F .4 4 1 L a a. a. x ~11: La: {- *olmw. _.l fl 0. éfimolmodl ¢olmoé .0.0H;\0 $0000 .0508 £90.50 080305 :0 9:53:00 >000 0 ..00 0:4 0.005.000.0000 _octofi mootam x0000 0>$0_0m 04. 0.59.“. 140 00.0 00.0 0 P0 0 70.... 00.0! 00.0.... b P p — 0 p b P F p I? b b . . m b F P .Voulmonomll .. a x w I X fl T ... * I i .. .. 13:08. T. . i I i 1 T fl T 0.. 100100001 1 ! u 1 n 10010000 n .. m I * fi .. .. 15:08; T l 1 a x x a .. 0.0“;\O * i I .. p . p m n b . P F . p h p 1“ P _ m P I. .VOlImomom 146 .o.ou;\0 £0000 6.550 0090.50 030305 :0 90:00:00 >000 0 ..00 054 0c0E000_0m_0 65.05 008.50 0.000 0200.01 mg“. 0.590 0.0 0.0 Nd 0.0 Nd! 0.0! 0.0! r *5 f p h L F L b {p [P b Nnooooo.l. .. fl . 1 X .I .l I I w .. _ .. 1208.0... . . I 0 ... 0 7000000! 1 i I I x T000000 a 0. .. x 1 1 1 1 w. .. 1280.0 ' ‘ a ! 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Brebbia, Boundary element solution for half-plane prob- lems, International Journal of Solid Structure 17, 1149-1158 (1981). O. L. Bowie, Solutions of plane crack problems by mapping techniques, Methods of Analysis and Solutions of Crack Problems (George C. Sih ed.), Noordhoff, 1- 55 (1973). B. S. Annigeri and M. P. Cleary, Quasi-static fracture propagation using the sur- face integral fmite element hybird method, Computational Fracture Mechanics - Nonliear and 3-D problems (P. B. Hiltio and L. N. Gifford, eds.), PVP -Vol. 85, AMD - Vol. 81, ASME (1984). T. A. Cruse and R. B. Wilsion, Advanced applications of boundary-integral equation methods, Nuclear Engineering and Design 46, 223-234 (1987). N. Fares and V. C. Li, An indirect boundary element method for 2-D fi- nite/infinite regions with multiple displacement discontinuities, Engineering F rac- ture Mechanics 26, 127-141 (1987). W. K. Wilson, Combined mode fracture mechanics, Ph.D dissertation, Universi- ty of Pittsburgh (1969). M. Stern, E. B. Becker and R. S. Dunham, A contour integral computation of mixed-mode stress intensity factors, International Journal of Fracture 12, 359- 368 (1976). G. Ogen and B. Schiff, Stress intensity factors for two-dimensional crack prob- lems using constrained finite elements, International Journal of Fracture 28, 55- 68 (1985). [100]S. W. Ma, A central crack in a rectangular sheet where its boundary is subject- ed to an arbitrary anti-plane load, Engineering Fracture Mechanics 30, 435-443 (1988). 158 [101]S. W. Ma, A central crack of mode III in a rectangular sheet with fixed edges, In- ternational Journal of Fracture 39, 323-329 (1989). [102] W. Zang, Crack kinks and crack closure modeled by integral equation methods, Ph.D Dissertation, Royal Institute of Technology, Sweden, (1990). 159 Appendix A In-plane Elasticity Influence Functions The Kelvin influence functions for plane stress can be written as in [77] (uR)iJ- = [—(3—v) Sij logp + (1+v) qi qj ]/81tG . (a.l) (non = 12mm. qi-fiz 93110—091 QI+(3+V)fiZQZ WP (110,, = [2(1+v)(-1‘1'2q?-n'1q§)+(l+3v)fizq1+(3+v) a, q2 ]/41tp (a.2) (UC)21 = [2(1+V)(-fizq13-fil qgm3+V)fi2 ql+(1+3V) El qz ]/41tp (“022 = [‘2(1+V)(fit (If-52 C123 )+( 14052 92+(3+V)fir ‘11 V4715!) (11101,: {-26 - (1+v) q1q2]/41t (1:1012 = [(l-v) logp + (1+v) (1121/41: (93) (111021: [—(l-v) logp - (1+v) q22 1/41: (1:1022 = {-26 + (1+v) q1q2]/41t (M911 = (3;?) [252 (I? + 231923 " i'1'2 c11 " 3 fi1‘12] (“012 = (75021 = Gé—lgl [2519? ‘ 23293 ' i71 (11 + iT2‘12] (3.4) (1:022 = 9‘31)- r—2r2q13— 25. qt + 3'9qu + a. <12] Zrtp 160 where Pr=x1-§1 Pz=X2-3?2 p = [(x1 - r.)2 + (x2 - r921“? _ x1- x1 _ x2 " i2 ql p r c12 p .. fi - ii 9 = arctan[ 92—1 c11.2 ] c11111+ (1292 (8.5) (96) (a7) (9.8) and G, v are the shear modulus and Poisson’s ratio, respectively. For plane strain, v must be replaced by v/( l—v). 161 Appendix B Complementary Elasticity Influence Functions The complementary expressions for plane stress can be written as in [102] (uR)f1= —[[(3-v v)-—119gp N+[(3—v)-2(1+V) xi * .2 i2f'+1q+2<1+v)—2.,{"—q221 p p c 1 (111012“ - fi[4%i-Y-H(1+V)x :32 BQ1¢12+(3-V)CI1‘I7J (b.1) xx2 2 ur- xt- 9192+(3-v)<11<12] (uR)f1- - —[+4-1'i—v-6+4(1+V) p‘2 R ° -—1—- 3— -—8—1 221+ “2 ‘2+ 3— 21+ L259 *7— (1)22- 81:0 [1( v) 1+..ng < 0:391 [( v)+( v) p.210] (uc)f1=_L;-[3(1-v)+2(1+v)q1"2-r-4(1-!-v)x222{(1-4q2)-I-4(1-v)—?,-q2]qn 41w p p —2—p22 qfn,+2(1-v)—n,J P P 4np' — - 2 1 x2X2 . . x2 .1 . x2 _ . p' [[2(l+V)(1-3 p.2 )91‘12‘4(1-V)-p7¢11lqn+[(1-V)—4(1+V)F]nzqu (we)c = 12 41: Xziz _ . x2. _ —v)+4(1+v) p‘z ]n1q2-2(l-v)?-nl] (b.2) (uc)§1 = p 22 )q1q24-4[(1+v)-p——(1-v)—p-]q1]qn-2(l-v)—n1] x22 . x259 9- 162 —v)+2(l+v)q2"2+4(1+v)22 p2 —22f(1-4q 2H(1-V)%CI2]C1112 (UC)° = 22 42 x292 0.. 2‘2 q, qn1-2( l-V) 7112] 247129 P Xziz t 0 X2 0 (711011: 4—2n'[29+(1+V)(1'4';.—2')Q1 Q2-2(1-V)?h=()+()° where ( ) are given in Appendix A 164 Appendix C Anti-plane Elasticity Influence Functions The influence functions associted with antiplane strains can be written as in [79] (111033 = -§IG- logp 2 (0.1) (UC)33 = iii-[El ql+h-ZQZ] (C.2) 1 .. (“mm = ‘311—2 (c.3) G _ _ (nc)33 = 31;;an q2+n2q1] (CA) where p = [(111 — r02 + (112 -- @211” (11.5) x - i x - i c11=1p1. q2= 2p 2 (c.6) .. fi - fi 6=arctan[chl ql2] (c.7) (1151 + 9252 and G is the shear modulus. 165 Appendix D Reduction of Domain Integrals to Boundary Integrals Here we remove the domain integrals involving the temperature gradient. According to the Divergence Theorem and Green’s Second Identity, i.e. jag dS =41]; 11j ds W) 3 8f I 9(ka - We as = (51.15% — 115;) as (1L2) the domain integral terms in eqs.(4.3) can be transformed to boundary integral terms as follows, I dumb-(xxx — mom dsm) = jQ[(uR)1,-J(x.swr(s) — [(uR)1j(x.swr(2)1J-] «18(2) = I arrmvzokmxs) — (magmas/"ran 113(1) - 531—. 1j1 dsm — {11. (uR)1,-(x,:)y11jrc11) 111(1) (11.3) where 166 (amines) = v2(1111),,(11,11) Wm) = o (11.4) (1-1) = in?) (11). (xx) = i(11R)- (xx) q" an ‘3 8n ‘3 ° All the problems discussed in this Dissertation involve a finite region with a crack. The boundary I‘ consists of two parts, namely the external boundary I], and the crack . surface I}. The upper face F: and the lower face I"; of the crack surface differ only in their normal, so that 11"“ = - n’. We make use of the apparent relations of symmetry: Ta) = Tux) 11:0?) = — (11.26:) (115) (UC)13'(x.x = - (1101326.?) (uR)13'(x50 = (uR)13*(x3) . The first integral in eqs.(d.3) can be expressed as follows, (51,.11(uc)131,-.1(x.s)= V211111),,(11,11) (111),,(11,11 = %(11R),3(11,1-1) . (as) 168 Appendix E Heat Transfer Influence Functions The influence functions for plane stress can be written as in [59] (1+v)ott (UR)i3 '—" 8 [1+210gp]pi (6.1) It 1+v)0tl (uc)i3 = 8n [(l+210gp)fi,+2qiqkfik] (e.2) (111013 = —§7;[4p1é+(4vlogp - (l—v) >132] (6.3) (111% = +—8‘%t(4vlogp — (l-v) >p1+4pzél (“0)13 = "8%K4V92‘lfi49‘49192)51+(4V1°gP-(1-V) +4V92Q2’f4912)52] (e.4) (“023 = +§Y;[(4VIOSP’(1-V) +4VPIQ1‘4922)51+(4VP1Q2Mé+49192)fi7j where Pl = x1 ‘ i1 P2 = x2 ' i2 (65) p = [(x1 — x1)? + (x2 — 5(2)211/2 (6.6) x - i x - if ‘11: 1 1’ Q2: 2 2 (6.7) p p 169 .1 fi - 'fi 6 = arctan[ Q2: (11 2 ] (e8) (1101 + C12112 = 2031 (e 9) 2, l-v ' and G, v , a, are the shear modulus, Poisson’s ratio and thermal expansion coefficient, respectively. For plane strain, v must be replaced by v/(l-v) and or1 must be replaced by (1+v)a,. (ER)i3 = —G I [(UR)i3'n + (uR)k3.i3nk + TZEIJOJRMGBIH] d5 0 (e.10) (1:013 = aa—nOtRm . 170 Appendix F Computer Program for Matrices 171 Ammmz MUGMU mo BZHOmIQHEV mBZHOm DamHh. 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