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Stresses in a circular disc containing a radial crack Wu, Qiong 1990

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STRESSES IN A C I R C U L A R DISC C O N T A I N I N G A R A D I A L C R A C K By Wu Qiong B.A.Sc. Beijing University, Beijing, P.R.China,1986  A THESIS SUBMITTED IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF  M A S T E R OF A P P L I E D SCIENCE  in THE FACULTY OF G R A D U A T E STUDIES MECHANICAL ENGINEERING  We accept this thesis as conforming to the required standard  THE UNIVERSITY OF BRITISH COLUMBIA  April 1990  ! K  Ii j  I  © Wu Qiong, 1990  In presenting degree  at the  this  thesis  in  partial  fulfilment  University of British Columbia,  freely available for reference and study. copying  of this thesis for scholarly  department publication  or  by  his  or  her  of the  for  an  advanced  I agree that the Library shall make it  I further agree that permission for extensive  purposes  may be granted  representatives.  It  is  by the  understood  that  head  of my  copying  or  of this thesis for financial gain shall not be allowed without my written  permission.  (Signature)  l he university ot British Columbia Vancouver; Canada  DE-6 (2/88)  requirements  Abstract  An analytic solution to a particular boundary value problem in linear elasticity is presented. The problem considered is that of a circular disc, in plane strain or generalized plane stress, which is loaded quite generally at the perimeter by a set of self equilibrating forces. The disc has a radial crack extending completely from the centre to the perimeter. Three particular load cases are considered: the disc is loaded by uniform radial tension; the disc is pulled by two diametrical concentrated loads with line of action perpendicular to the crack; the disc is pulled by two concentrated forces acting at the mouth of the crack. In each case the stress intensity factor is evaluated and contours of the isochromatics are presented in dimensionless form. The results agree with those derived by other workers but the method presented here is more general, more efficient, and extends the previous results for finite bodies containing cracks. The three load cases have also been applied to the disc when the tip of the crack has been removed by the presence of a concentric hole. New results are presented for the stress concentration factors for a full range of geometric parameters. Finally it is shown how the second and third of the load cases may be used for controlled laboratory tests in the study of fracture and fracture toughness of brittle materials.  u  Table of Contents  Abstract  11  List of Tables  v  List of Figures  V1  Acknowledgement 1  2  3  v m  Introduction  1  1.1  Preliminary Remarks  1  1.2  Relative Background  H  1.3  Objective of the thesis  14  1.4  Outline of the thesis  14  Mathematical Formulation  17  2.1  General plane strain solution for plate with radial crack  17  2.2  Stresses at the crack tip  21  Numerical Results  23  3.1  A disc under uniform tensile stress  23  3.2  Disc with diametrical force T  26  3.3  Disc with tangential shear forces  43  3.4  Comparison with related results  3.5  An analysis of the error  .„  47 47  iii  4  5  Application to a More General Case  54  4.1  Motivations  54  4.2  The curved beam solution in classical elasticity  56  4.3  Solution using the proposed method  61  4.4  Numerical results  64  4.5  Discussion  68  Concluding Remarks  72  Bibliography  74  appendices  77  A First mode linking between shear and normal stress  77  B Normal force requirement for crack loaded by tangential shears  80  C Normal force resultant for load case 2  83  D Fourier coefficients for normal forces; load cases 2 and 3  85  E Fourier coefficients for shear force, load case 3  88  iv  List of Tables  3.1  First ten coefficients under the uniform tension ( x l O  3)  25  3.2  The stresses on the boundary  25  3.3  Stresses in the disc under uniform tension  31  3.4  The first ten coeficients with diametrical force T  34  3.5  The stresses on the boundary  35  3.6  Stresses in the disc with diametrical force T  42  3.7  The fist ten coeficients with tangential shear force S  46  3.8  The stresses on the boundary  46  3.9  Stresses in the disc with tangential shear force S  48  3.10 Comparison with earlier results  53  4.11 The important coefficients under the uniform tension (scaled by 10" 3 ) . .  64  4.12 Stresses on the boundaries  65  4.13 The stress <re at 6 = 0 and r = a { obtained using different methods . . . .  65  4.14 The important coefficients with diametrical force (scaled by 10" 3 )  68  4.15 Stresses on the boundaries  . . . .  69  v  List of Figures  1.1  T h e three modes of fracture  3  1.2  Bueckner's Principle  5  1.3  T h e Mathematical Model and Corresponding Test Piece  16  2.4  Description of the Problem  18  3.5  Disc under the Uniform Tension  24  3.6  Isochomatics under the Uniform Tension  27  3.7  Stress a r in the Disc under the Uniform Tension  28  3.8  Stress cre under the Uniform Tension  29  3.9  Stress TT6 under the Uniform Tension  30  3.10 Disc with Diametrical Force T  32  3.11 Isochromatics with Diametric Force T ( a = 8. 5°)  36  3.12 Isochromatics with Diametric Force T (a = 6.2°)  37  3.13 Isochromatics with Diametric Force T (a = 4.9°)  38  3.14 Stress <jt with Diametric Force ( a = 4. 9°)  39  3.15 Stress a e with Diametric Force ( a = 4.9°)  40  3.16 Stress ttB with Diametric Force ( a = 4.9°)  41  3.17 Disc with the Tangential Shear Force S  43  3.18 Isochromatics with Tangential Shear Force  49  3.19 Stress a r with Tangential Shear Force  50  3.20 Stress crg with Tangential Shear Force  51  3.21 Stress r T e with Tangential Shear Force  52  vi  4.22 Disc with a hole at the crack tip . 4.23 Equivalent problem 4.24 Superposition of the three problems 4.25 Stress ae at 0 = 0, r = a; 4.26 Stress cre at 6 = 0 . . . . 4.27 Stress cre at 0 = 0, r = a { A.28 Disc under equilibrating force system B.29 Disc with tangential shear forces .  Vll  Acknowledgement  I would like to thank Dr. H. Vaughan for his strict supervision and generous help during the course of the research work. His help and guidance are appreciated very much during the preparation of this thesis. I would also like to thank Mr. gave great help with the computer work.  vm  A. Steeves, who  Chapter 1 1  Introduction  In this thesis a method is presented which may be used to determine the stresses in a circular disc with a radial crack with or without a hole at the tip of the crack.  The  crack extends from the centre of the disc to the edge and is activated by a general set of in-plane self equilibrating surface tractions acting at the perimeter of the disc.  The  method presented contributes to the small number of analytical solutions available for elastic plates containing cracks. The solution for the second problem, i.e., a disc with a hole at the crack tip, is of much interest in engineering practice.  1.1  Preliminary Remarks Fracture mechanics is a relative new branch of material science which seeks to quan-  tify the critical combination of stress and crack size for crack extension.  It is being  increasingly applied to assess the safety of engineering structures containing crack-like defects. Generally, fracture mechanics is applied to estimate the geometry independent resistance to crack extension. Several circumstances may be cited in order to justify the application of fracture mechanics theory to engineering structures. The most important circumstance is probably that all engineering structures contain cracks or crack-like flaws at some scale of examination. In recent years, fracture mechanics has received world-wide attention. T h e concepts and principles of this relatively new discipline, when properly applied, provide a  Chapter 1.  Introduction  2  sound basis for making decisions in engineering design. It offers a systematic approach to failure prevention in aircrafts, ships, rocket casings, pressure vessels, nuclear structures, bridges, pipelines, etc. In view of the rising cost and shortage of raw materials, it becomes increasingly more pertinent to optimize the use of the material with performance.  To  this end, fracture mechanics can greatly assist in the fabrication of acceptable materials, establishment of proof tests and design codes and even in making managerial decisions. It is becoming a technology that is helpful to those concerned with the developement of heavy industries. The strength of a structure which has a crack or crack-like flaw can be dangerously over-estimated. The stresses at a crack tip are very high (singularity), and the nature of the singular stress is the critical subject for fracture of brittle materials. T h e singularity is described by a parameter known as the stress intensity factor K i . T h e fracture toughness of a material is specified by a critical value of K j , denoted by K j c The stress intensity factor K j provides a one parameter characterisation of the stress field ahead of a sharp crack and the energy available to propagate the crack. Failure occurs when it attains the critical value Kjc  called the fracture toughness.  During the past three decades, the stress intensity factor has been used as a parameter to characterize the loading on cracks in practical fracture tests. Some methods of determining the stress intensity factor are presented here:  1. Effects of geometry and loading on the stress intensity factor The stress intensity factor KN  (TV = 1,2,3 corresponding to the opening, sliding  and tearing modes respectively Fig.(1.1)) can be written  Kn = Ycryfa  (1.1)  Chapter 1.  3  Introduction  Figure 1.1: Three modes of the fracture where <r is a stress, a is a crack length and Y is a geometrical factor which accounts for such things as proximity effects of boundaries or other cracks, orientations of the crack, shape of the crack and the restraints on the structure containing the crack. It is with the determination of this geometrical factor that crack analysis methods are concerned. For example, for an isolated crack in a sheet subjected to a remote uniform stress a perpendicular to the crack line, the stress intensity factor K i is given by e q n ( l . l ) with Y  2. Bueckner's principle and superposition T h e principle presented by Bueckner is very important in the determination of  stress intensity factor. It allows the problem of determining the stress intensity factor of a stress-free crack in a loaded body to be changed to that of the same crack subjected to pressures on its surfaces in the same body; the-boundary conditions for the second problem are that all applied traction and displacements are zero.  The crack surface  Chapter 1.  Introduction  4  pressures are simply determined as the stresses over the crack site in the uncracked body subjected to the originally applied boundary traction and displacements. This principle of using the distribution of stress over the crack site in the absence of the crack forms the basis of many methods, for example, of the Green's function and the weight function methods and also those based on superposition of polynomial distributions of stress. Bueckner's principle is illustrated schematically in Fig.1.2.  The linearly elastic  region inside the boundary 0 contains a stress-free crack along the line A B . T h e boundary Oi subjected to general tractions T and the part (or parts) of the boundary 02 subjected to general displacements V ; the region inside 0 is subjected to a body force distribution F . Under this system of loads and displacements the stress intensity factor is given by Kb.  In Fig.1.2(c) the body is uncracked with the same tractions T , body forces F and  displacements V applied on the boundary; <xtJ- is the resultant stress distribution along the site of the crack A B . In Fig.1.2(d) the body contains, along A B , a crack with pressure distributions s; applied to its surfaces; the boundary conditions on the body are T = V = 0 with F = 0 .  Under this pressure distribution the stress intensity factor is given by Kd.  Bueckner's principle states that if s; = a^rij then the stress intensity factor for the crack in Fig.1.2(b) and Fig.1.2(d) are identical, i.e.  Kd = Kb  at both  tips  3. Weight functions and Green's functions The determination of stress intensity factors using either Green's functions or weight functions usually involves the use of Bueckner's principle and always involves evaluating an integral of the form  Chapter 1.  Introduction  Elastic solid subjected to arbitrary tractions and displacements. \ ( a ) Cracked region before loading, (b) Cracked region subjected boundary tractions and displacements, (c) Cracked region subjected to Tand Kwith crack removed, (d) Cracked region with T =V = 0 and crack surfaces subjected to pressure distribution ptJ = atJ. { K  K  Figure 1.2: Bueckner's Principle  Chapter 1.  6  Introduction  where x' is the crack length and x represents a point on the crack. T h e stress Cij(x) is the pre-cracked stress distribution (or the pressure distribution on the crack surfaces) and H(x',x)  is a function which depends on the geometry.  This function H(x',x)  is  independent of the applied tractions, but it will depend on any imposed displacement conditions. Consideration will be limited to the opening mode Kj as this is in practice the more important. The function H(x',x)  may be determined as the stress intensity factor resulting  from a unit force on the crack surface. function g{x',x)  In this case it is proportional to the Green's  for a crack of length x' with a point force located at a coordinate x on  the crack surface. The integral in eqn.(1.2) then takes the form, with TV = 1 and <Tij(x) replaced by  c(x)  (1.3) Alternatively the function H(x', x) may be determined as a weight function G(x', x) defined as  where k = 3 — 47 for plane strain, k = ^  for plane stress and p, is the shear modulus  and 7 is Poisson's ratio. K j is a known stress intensity factor solution for the particular geometric configuration being considered and v"(x',x)  is the corresponding displacement  of the crack surfaces. Any loading condition may be used to determine Kj and Equation (1.3) thus takes the following form  v*(x',x).  Chapter 1. Introduction  5714  Kj=  rx' I Jo  cr(x)G(x',x)dx  4. Simple approximate methods for cracks at stress concentration It has been shown that, by using certain approximations to the Green's function, expressions for stress intensity factors of cracks at stress concentration can be determined. Consider, for example, a crack of length a 0 at the tip of a notch of root radius p. The notch is in a body which is subject to a uniform tensile stress cr acting in a direction perpendicular to the crack line.  The precracked stress distribution is <r(x), where the  distance x is measured along the line of the crack from the notch tip. We have, for the maximum stress method  Ki = 1.12<r(0)V™o for the tip stress method  Ki = 1.12(T(ao)v'7rao and for the mean stress method  K i = 1.12(7meon\/7ran where  1 fan CTmean = ~ / <r{x)d.T a0 Jo  Chapter 1.  8  Introduction  The stress <xmeanis the mean stress over the crack site in the absence of the crack. The maximum stress method has the advantage that only the stress concentration factor of the notch is required. Whereas, the tip stress and the mean stress methods, although more accurate, require that the entire stress distribution over the crack site be known.  5. Superposition of polynomial distributions  If the weight function or the Green's function technique are used to determine a stress intensity factor, it is necessary to evaluate an integral of the type given in eqn (1.3). Several studies have been done to evaluate such integrals, for particular configurations, with cr(x) expressed as a polynomial in x. The stress distribution over the crack site in the uncracked body is expressed in the form of  <r(x) = <rj2M-) n  (1-4)  n=0 where cr is a suitable reference stress and An are known constant coefficients. Both a centre cracked strip of width 2w and an edge cracked strip of width w have been studied for such stress distributions by substituting eqn.(1.4) into eqn.(1.3) and replacing H(x',x)  by the appropriate Green's function. In both cases the stress intensity factor  can be written in the following form  Kr = K[A0 + £  1  An(-)n[B0,n W  + 5a,n(-)  + W  B2,n{-)] W  n=1 where K is the stress intensity factor for either the centre crack or edge crack configuration subjected to the uniform stress cr. The coefficients B 0i n, i?i >n and B2, n are given for  Chapter 1.  9  Introduction  the centre cracked strip and edge cracked strip in both configurations, where a is either the half-length of the centre crack or the length of the edge crack.  6. Upper and lower bounds on the stress intensity factors  Under certain conditions, the stress intensity factor K i for an edge crack of length a in a half-plane will lie between bounds given by  [0.29cr(a) + 0.83(Tmeon]Va7r < Kj < 1.12trmean v "" where  1 fa O-mean = ~ / a Jo The conditions are that a(x),  *(x)dx  the stress over the crack site, in the uncracked  body, is a monotonically decreasing function of x, and furthermore, <r(o) > 0. The result expressed above is limited to edge cracks in a half-plane.  7. Compounding technique  Stress intensity factors for many simple configurations are available but they may not be directly applicable to the more complex configurations of typical engineering components. The compounding method is a versatile and quick way of extending these solutions to other more complex configurations for which the stress intensity factors are not known. Ofteu a cracked structural component has several boundaries, such as holes, other cracks or sheet edges all of which will influence Hie stress intensity factor at the tip of the crack. The compounding technique provides a systematic means by which the contribution of each boundary acting separately can be added to obtain a solution for all the boundaries acting simultaneously.  Chapter 1.  10  Introduction  The choice of a method for determining a stress intensity factor must be assessed in terms of the time available, the required accuracy, the cost, the use (once or many times) and how simple the real structure can be modelled. The stress intensity factor will be affected by other cracks and boundaries, regions of stress concentration, orientation of the crack relative to the applied stress, the degree of constraint on the structure and the position of the applied loading relative to the crack. The method used must take account of these effects and if an approximate method is applied care must be exercised in isolating the dominant effects to be included: it will seldom be possible to consider all when using an approximate method [9, 10]. As we know, the singular stresses, which are the main cause for brittle failure of a structure, have not been determined explicitly before.  It is one reason that the  stress intensity factor is referred to provide a one parameter characterisation of the stress field just ahead of the crack.  Methods used to determine the stress intensity factor  as mentioned above, are empirical, inaccurate and with many restrictions. In practice where a particular stress intensity factor is required, it is necessary to obtain an accurate solution. Some simple and more accurate methods are needed when the structure can not be modelled simply. With the above mentioned methods, it is impossible to consider all the effects of the boundary conditions and other restrictions which are responsible for the stress ahead of the crack.  Furthermore, more and more recent research and  experiments have shown that those parameters, such as KIC,J-,COD.,  etc, do not have  the capability to associate the local small crack behavior with the global applied tension or cyclic load. Evidence of this can be clearly seen in past efforts on retardation and crack closure. Current studies have not yet recognized that dynamic fracture is inherently load dependent and can not be characterized by a simple parameter such as K i c , e t c . One of the major shortcomings of the conventional approaches is that they failed to separate the fracture energy from other forms of energy dissipation. This is why the parameters, such  Chapter 1.  as COD.,  11  Introduction  J., Kic,  etc. are all sensitive to changes in loading and specimen geometry and  size. These arguments are not presented in this thesis. In engineering practice, it is not uncommon to drill a hole at the tip of a crack in order to "release" the stresses and to prevent the crack from propagating. This problem has received considerable attention from mathematicians.  Of particular interest is the  calculated values of the stress at 0 = 0 and r = a; (a* is the radius of the hole) to provide some theoretical justification to the empirical experience. This becomes more important for smaller holes which, to some extent, can be treated like a notch (blunt crack).  1.2  Relative Background The use of linear elasticity in the investigation of the spreading of cracks in materials  under stress is now a very well established branch of fracture mechanics.  The stress  analysis of plane cracks located in elastic material is of interest in the examination of fracture and failure initiation in brittle solids. Considerable efforts have therefore been made in the investigation. The reader is referred to section 1.1, where some techniques are described which are currently available for the determination of stress intensity factors. In order to determine the plane strain fracture toughness of a brittle material it is necessary to carry out a practical fracture test on a specimen of the material which contains a crack. In order to interpret the results of such a test it is necessary to have a theoretical formula for the stress intensity factor at the crack tip for the particular shape of the test specimen and the particular choice of loading used. It is also useful to have a corresponding formula for the opening of the crack at some chosen stations (for example, at the load-line or at the crack mouth).  Tada, Paris and Irwin [19] have given some stress intensity factors  appropriate to several test specimen configurations. However, since it is often expensive and wasteful of material to fabricate existing test specimens, the American Society for  Chapter 1.  Introduction  12  Testing and Materials Committee on Fracture Testing lias introduced the circular disc with a radial edge crack as a new standard specimen. This specimen is easily produced from a cylindrical rod. One intended application of this thesis is to offer such theoretical formula for the stresses and stress intensity factor at the crack tip for the particular shape of the test specimen and the particular choice of loading used. In the general field of linear elastic fracture mechanics, there are very few analytical solutions for bodies with finite dimensions. Most solutions involve the infinite plate under uniform tension. However even in the two-dimensional case, very few exact, closed-form solutions are known. This is particularly so in the case of edge cracks, where the only closed-form solutions known appear to be for a straight normal edge crack in a half-plane under various loadings. No non-trivial closed-form solution appears to be known for any case of an edge crack in a plate of finite dimensions. Such a solution would, however, have an importance beyond its intrinsic analytical interest, since it could act as a "model problem", against which the various approximate techniques which are available could be tested for accuracy. There are some particular solutions to the circular plate loaded in a prescribed way. Such problems haved been studied by several writers like Tweed (1971), Rooke (1971) [1-7] and Gregory (1977) [8,9]. The general solution is, howeverr, very complicated and consequently only a few numerical results were evaluated for some special cases. Tweed and Rooke [1-7] have considered a radial crack in a disc which is loaded by a uniform tension. The solution is obtained using integral transforms of the Mellin type. The stress intensity factor is given but detailed stresses are not.  This result is  valuable in that it gives a closed form solution to an interesting problem but it is of limited practical importance since it is not easy to create a biaxial uniform stress field in tension. They also have considered a number of other related disc problems [2],[3] and  Chapter 1.  13  Introduction  [4] using the Mellin transform. These include the effect of body forces due to rotation, a class of problems which obviously arise in practice for circular plates. Stress intensity factors for a circular disc are given in [14]. In the above papers the integral transform methods reduce to solving, for most cases, singular integral equations. There are three disadvantages of using such methods which make them inconvenient for the standard test. Firstly, the inverse integral transforms have to be used to get the stresses in the disc. However, these integral transforms have not been inverted. Therefore, the stresses in the disc can not be obtained explicitly. Secondly, the singular integral equations are very difficult to invert numerically. Thus, for complex boundary conditions, it would be difficult, or even impossible, to get the numerical solution.  Only special solutions for  concentrated loading and uniform tension were derived in the papers. Thirdly, it is not easy to apply uniform tension or a concentrated load to a disc in a practical laboratory test. Other studies [8,9] considered also a circular disc containing a radial edge crack, 'opened' by a constant internal pressure on the crack faces and opposite point forces normal to the crack acting at the crack mouth. The problems are expressed in terms of special stress potential developed by Jeffery using bi-polar coordinates. In this coordinate system, the special case of a disc with a fully radial crack is solved formally using the Wiener-Hopf technique.  The corresponding solution for the crack of general length is  then obtained from this special case by a transformation. The problem of determining the resulting stress field throughout the disc is solved analytically in a closed-form. However, the method is very complex and it is difficult to get the numerical stresses in the disc. There are two weaknesses in these papers. Firstly, as was mentioned above, it is difficult to set up experiments involving discs having either uniform tension or concentrated loads. Secondly, the methods fail to characterize the stress field internally. It is clearly desirable to find a theoretical method which may be used to find the  Chapter 1.  Introduction  14  stress intensity factor and the internal stresses corresponding to a simple laboratory test. A crack originating from a circular hole in a loaded body has been investigated using complex mathematical methods [13] by Morkovin (1944) [11], Buechner (1960), Bowie (1956) [10] and Wigglesworth (1958) [12]. T h e solutions, however, are very complicated and prohibitive in case of general boundary conditions and subsequently they are very limited in practical applications. T h e methodology derived during the course of this study is very efficient to solve this kind of problem.  1.3  Objective of the thesis  Referring to the above discussion, the objective of the thesis is to derive an analytical solution to some particular boundary value problems. T h e solutions can then be used to interpret the results from practical tests on specimens of similar geometry. Another intended application of this thesis is to offer theoretical formulas for the stresses and stress intensity factor at the crack tip for the particular shape of the test specimen and the particular choice of loading used.  1.4  Outline of the thesis The mathematical model and the corresponding test pieces are shown in the Fig.(1.3).  In the mathematical model (A), the forces act on the outer edge of the disc with an edge crack, a is the angle subtended at the centre by the surface loads. The corresponding test piece is a plexiglas ( C R 3 9 ) disc with a radial crack having two lugs for loading purposes. The isochromatics may be examined by loading the disc and viewing them in polarised light. Fig.1.3(B) models another very interesting experimental test piece. The forces act  >  Chapter 1.  Introduction  15  on the mouth of the crack. These two tests are very simple and can be done in a very controlled way. In this thesis, model A and B are called "diametrical concentrated force" and "tangential shear forces", respectively. In chapter 2, an exact closed-form solution for the disc under general boundary conditions is derived.  A special stress function is used to get the stresses in the disc  with an edge crack. Using Fourier series, the stresses are made to match the boundary conditions. They are obtained directly in terms of the solution of a set of simultaneously linear algebraic equations. Chapter 3 shows the numerical results for some particular boundary conditions, including a disc under uniform tension; a disc under diametrical concentrated force and a disc with tangential shear forces acting on the mouth of the crack. The closed-form of the stress intensity factors are found for each case. A comparison is made between the results of this work and other related work. Excellent agreement is found. However, the proposed method is more general and efficient in terms of solving complex boundary value problems and getting numerical values for the stresses. This chapter is concluded with an analysis of the numerical error. In chapter 4, a disc with an edge crack and a hole at the crack tip is solved using both curved beam theory and the proposed method in the thesis. The stress at r = a { (radius of the hole) and 9 = 0 is determined and the results are compared with relative algorithems along with some discussions. Finally this thesis is concluded with a discussion on the contribution of the thesis along with suggestions for futher work.  Chapter 1.  Introduction  A  B  Figure 1.3: The Mathematical Model and Corresponding Test Piece  Chapter 2  Mathematical Formulation  2.1  General plane strain solution for plate with radial crack  Figure (2.4) shows a circular disc of radius a and thickness t with a radial crack extending outwards from the centre along the full radius 9 = tt. The disc is loaded at its outer edge by surface tractions T{9)  and S{9) where T indicates a normal force and 5  is a shear force, both having dimension of "force/unit length of circumference". Both T and 5 must produce a self equilibrating force system. There are no forces acting on the faces of the crack. A stress function of rather special form is used as follow  OO  $  =  a m r r a [ m c o s ( m — 2)6 — ( m — 2) cosmO]  ^ 3  5  2 ' 2 "" OO  +  bnrn[cos(n  (2.5)  - 2)9 - cosnO]  n=2,3....  —tt < 9 < ir  0 <r  < a  The stresses are then given in their conventional form as follow:  1 d2$ ar  "  rdr  17  +  r  2  39 2  Chapter 2.  Mathematical  Formulation 5725  Figure 2.4: Description of the Problem  <92$ "  ~  dr*  Tr9  -  ~dr  3  1W r 86  The stress function is rearranged assuming a _ i = ai = bx = 0:  $  =  J cosm0[-(rn-2)r m a m + (m + 2 ) r m + 2 a m + 2 ] m=— ^ oo  + ^ cosn6[-bnrn n=l  + fcn+2rn+2] + b2r2  Using the above stress function, the stresses can ^ determined as follows:  CO  <rr  =  £  coa(m0)[m(m - 2)(m - l)a m r m " 2 - (w + 2){m + l)(m - 2)a m +  Chapter 2.  Mathematical  + f ; cos(n6)[bnrn~2n(n  n= 1  =  f ; T 7 1 - -  i  19  Formulation  - 1) - bn+2rn{n + l)(n - 2)] + 2b2  cos(m0)[—m{m — l)(Tn — 2)amrm"2  l)am+2rm]  + ( m + 2) 2 (m +  J  + f ) CO5 (n0)[-u(n - l ) 6 n ^ 2 + (n + 2)(n + l)6 n+2 r n ] + 2b2 n=l Tre  =  (2.8)  m 5 r n ( m 0 ) [ - ( m - l ) ( m - 2 ) a m r m - 2 + ( m + l ) ( m + 2)am+2rm]  £  i  m = -  5  oo  n=l By inspection it is seen  a0 = rTg = 0  at 6 = ±tt  0< r< a  Hence equation (2.5) is a stress function corresponding to a disc with a stress free crack at 6 = 7T, for all values of am and bn. Considering the general expression for the shear stress, tt8, in equation (2.8), it is written in terns of sinnd by using Fourier series for each sin{m6) as follows  n= 1  ^  Substituting (2.9) into (2.8) it is seen that the term in nsinnO in the general expression for rTg is  £  [_(m  _  1  m ——  2  _(n  _ l)6nr"-2  l)(m -  2)amrm-2  + (n + l)bn+2rn  +  (m + 1)(tt7 + 2)am+2r _  m,2rn  }—  {-l)"^  sin(miT) _ m2  (2-10)  For the normal force, <rr, the fractional trigonometric terms are also eliminated using the following equation  Chapter 2.  Mathematical  20  Formulation  . „, sin(mir) cos(m9) = — ^ TO7T  2  . .  ,  ^  mcosnd  /0 i-n  + -Wtott) ^ ( - l ) ^ 1 - — — 7T  J  (2.11)  tl — m  Substituting (2.11) into (2.8), the terms in cos(n9)  in the general expression for  <rT becomes  oo  2  £  ( —1 )n+1m  [(m. - 2 ) ( t o - l ) r n a m r m - 2 - (to - 2 ) ( m + l ) ( m + 2 ) a m + 2 r m ] - , s m ( T O 7 r ) v ^  ^  m=-| + n ( n - l)fe n r"" 2 - (n + l)(n. - 2 ) 6 n + 2 r "  (2.12)  together with a constant (9 independent) term 00  [(to - 2)(to - 1 )mamrm~2  - (to - 2)(TO + l ) ( m + 2 ) a r o + 2 r r o ]  SXTii7717T ) ; + 26z(2.13) ^  m=-| It was mentioned previously that the surface tractions at the boundaries are self equilibrating and denoted by T(9)  and S(9).  They may be expanded as Fourier series.  Also, if we take T{8) is taken as an even function and S(9) as an odd function we may write  t  00  T(9)  =  t<rT |r=B =  + X)  tncosnd  S(6)  =  trTg\r=a = J2 snsinn9  oo  (2.14)  n= 1  where t0, tn and sn are assumed to be known.  An obvious extension removes the  requirement that T(9) is even and S(9) is odd. The values of aT and r r 6 at r = a can also be obtained by substituting r = a into (2.10), (2.12) and (2.13). These values must agree with the ones in (2.14). Hence  Chapter 2.  Mathematical  Formulation  sin(nnv)  ^  =  £  2  [(m- 2)(m - l)rna - am - (m - 2)(m + l)(m + Z j a ™ ^ ] - ^ — +  __ i m  tn  ,  m 2  =  2  (_1)"+1  f ;  2m [(m - 2 ) ( m - l ) m a m - J a r a - ( m - 2 ) ( m + l ) ( m + 2 ) a r a M 2 ] 7  m=— 2 +n(n-l)M n - a -(»+!)(»- 2 ) W (-l)n+1  £ m = -  1  .  sin(mir) -~m2  n2  ^  2m sin(miv)  [ - ( " » - l ) ( m - 2)am~2am + (m + l ) ( m + 2)amam+2] —  n3  _  m2  |  -(n - 1)6„ + (n + l)fcn+2 Where n = 1,2,3...., m = - U2.12' I . .2 .' . a _ i = ai = 61 - 0 Equations (2.15) are a set of simultaneous linear algebraic equations from which the coefficients a m and bn can be found in terms of the given coefficients t0,tn and  In  the following section these equations are solved for some particular loadings.  2.2  Stresses at the crack tip The asymptotic form of the stresses near the crack tip are obtained from equation  (2.8) by retaining the singular term in a § . The following are then obtained  3  0  /c  3  39 P  , .0  . 36  P  2  ae  ft  2 3 * , 30 P ft + COS -- j -- \ cos--) oaa(3 cos- - + 2 2 atVr  =  where P has dimension of force. For 6 = 0 the asymptotic stresses are reduced In  3 ar  =  ae  =  2alat^  P  ft r  ft  ( 2 1 6 )  v ^  Chapter 2.  Mathematical  22  Formulation  Trg = 0  (2.17)  The stress intensity factor is then calculated as followed  K -  3  a32 Px/2an  2  at  (2.18)  As shown above, this method reduces to solving a set of linear algebraic equations. There are three advantages to this method. Firstly, the stresses are accurate not only near the crack tip but also over the whole disc. Secondly, for general boundary tractions acting in the same plane as the disc and for plane strain cases, the method works very well. The third advantage is that this method is not only valuable in theory because it provides a closed-form solution, but also because of its practical application.  It is  relative easy to get the numerical stresses in the disc and these may be compared with experiments. In the following chapter, it is shown that this method is more convenient as a "model problem" for a test.  Chapter 23  Numerical Results  In this chapter, the method is exemplified with different boundary conditions. Some numerical results and error analysis are presented as well.  3.1  A disc under uniform tensile stress Let T0 be the pressure of the normal force shown in Fig.(3.5). For r = a,  a T = To  Trg = 0  (3-19)  So the coefficients in the Fourier series are as follow:  tn = 0  t0 = 2T0  = 0  (3.20)  For convenience, replace am and bn by am and bn given by  „  a-m.  =  _m-2  a-mA  s u  n —  T0TT  l n—2 rri  i07T  (3.21) V '  Substituting (3.20) and (3.21) into (2.15), and for r = a, the following simultaneous equations are obtained.  23  Chapter 3.  Numerical  24  Results  Figure 3.5: Disc under tlie Uniform Tension  oo m=  ~2  °°  [(m - 2)(m - 1 ) m a m - (m - 2)(m + l ) ( m + 2)a m + z ]  SXTiTTiTX a 1 + 2b2 = -  (-l)n+1m  2  [(m - 2)(m - l ) m a m - ( m - 2 ) ( m -f l ) ( m + 2 ) d m + 2 ] - « n m i r — * - m 2  m=-\  + n ( n - l)6 n - (n + l ) ( n - 2)fc n+2 = 0 »  V  ,  (3.22) , 2 m ( —l)n+13xnm.7r  [_(m _ i)(m _ 2)am + (m + 1 m + 2 a m+2 — i Tr  r  nl — ml  -  -(n-l)6n + (n+l)6n+2 = 0 The leading coefficients found from a 45 x 45 matrix of (3.22) are shown in Table  (3-1). From Table (3.1) it is seen the values of a m and  decrease rapidly with increasing m  and n. Thus, only the first few a m and bn are important.,Table (3.2) shows the stresses on the boundary. The results from the proposed method are in good agreement with the theoretical values.  Chapter 3.  Numerical  25  Results  a?  a5  a?  a?  476.0 ^  -155.9 A h 28.3  -2.16  0.510  an 7 0.171  k -3.24  k 0.0817  b6 -0.27  b2 150.9  a 13  a is •>  an  a 19  a2i•>  0.070  -0.032  0.016  -0.009  0.005  0.101  6s -0.04  b9 0.014  bio -0.003  in -0.0001  Table 3.1: First ten coefficients under the uniform tension ( x l O  e  ( jfjcal  0.000  1.00017  (  )ideal  20.000  0.99996  40.000  1.00013  60.000 80.000  0.99986 1.00013  100.000  0.99984  120.000  1.00015  140.000 160.000  0.99977 1.00043  1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000  180.000  0.99731  1.000  (^f)ca/  0.00000 -0.00002 0.00005 -0.00007 0.00010 -0.00015 0.00021 -0.00032 0.00058  0.00000  ( ) i d e a l  0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000  Table 3.2: The stresses on Hi" boundary  3)  Chapter 3.  Numerical  26  Results  The asymptotic forms of the stresses at 8 = 0 are obtained using the calculated value of a.3 in equation (2.17). We obtain 2  <Tt = cre = 2.234T 0 y^-  (3.23)  which is in good agreement with the results given in [1]. The stress intensity factor for uniform tension is  Ki — 3.17T0V7ro  (3.24)  K i = 5.6187T 0 V®  (3-25)  or  where T0 is the pressure at the boundary. Contours of the isochromatics (constant maximum shear stress) are shown in Fig.(3.6). The dimensionless stresses  f j and s * in the disc at 6 = 0°, 8 = 45°, 8 = 90°  and 8 = 135° are shown in Fig.(3.7), Fig.(3.8) and Fig.(3.9). Selected values of crT, erg and rTg throughout the plate are shown in Table (3.3). Detailed values at 5° interval and increments dr = 0.02 are given in [19].  3.2  Disc with diametrical force T Referring to Fig.(3.10) the disc is loaded by two  self  equilibrating forces of magnitude  F with line of action through the centre of the disc and perpendicular to the crack. Each force T is the resultant of a distributed surface traction T{8)  distributed over a small  sector of angle a (for more detail see Appendix C). The actual distribution of T{8) is not  Chapter 3. Numerical  Results  Figure 3.6: Isochomatics under tlie Uniform Tension  27  Chapter 3. Numerical  Results  in  Figure 3.7: Stress crr in the Disc under the Uniform Tension  Chapter 3. Numerical  Results  Figure 3.7: StresscrrintheDiscunder the Uniform Tension  Chapter 3. Numerical  Results  Figure 3.7: StresscrrintheDiscunder the Uniform Tension  1 7.894 9 = 0° 5.928 Jr9 _ 0.000 7.833 9 = 30®" <?T 5.580 <?e 1.054 Tre 7.609 9 = 60° <rr 4.598 <?e 1.849 Tre 7.111 9 = 90° 3.181 <?e 2.164 Tr« 0 6.145 9 = 120 <*r 1.656 <?e 1.873 Tre <y r 4.454 0 = 15(f 0.456 <re 1.033 taure 1.824 0 = 180 <*> 0.000 ve 0.000 Tre Y <>"r  .2 5.397 3.363 0.000 5.328 3.191 0.596 5.113 2.697 1.074 4.732 1.953 1.322 4.130 1.093 1.239 3.198 0.335 0.773 1.749 0.000 0.000  .3 4.166 2.066 0.000 4.105 1.970 0.399 3.922 1.691 0.730 3.618 1.259 5.924 3.184 0.737 0.907 2.574" 0.243 0.610 1.670 0.000 0.000  .4 3.361 1.197 0.000 3.310 1.147 0.283 3.160 0.998 0.523 2.918 0.762 0.675 2.590 0.463 0.683 2.166 0.162 0.483 1.589 0.000 0.000  .5 2.765 0.540 0.000 2.725 0.521 0.204 2.604 0.464 0.380 2.455 0.369 0.496 2.166 0.236 0.514 1.864 0.089 0.377 1.503 0.000 0.000  .6 2.293 0.006 0.000 2.261 0.011 0.145 2.169 0.025 0.272 2.024 0.040 0.359 1.840 0.041 0.378 1.627 0.021 0,285 1.413 0.000 0.000  .7 1.899 -0.447 0.000 1.876 -0.422 0.099 1.810 -0.350 0.185 1.706 -0.245 0.247 1.576 -0.131 0.265 1.432 -0.040 0.204 1.318 0.000 0.000  " .8 1.562 -0.843 0.000 ' 1.547 -0.800 0.060 1.504 -0.680 0.114 1.437 -0.497 0.153 1.355 -0.285 0.166 1.268 -0.094 0.131 1.217 0.000 0.000  Table 3.3: Stresses in the disc under uniform tension  .9 " 1.265 -1.196 0.000 1.258 -1.139 0.028 1.237 -0.976 0.053 1.205 -0.725 0.072 1.165 -0.425 0.079 1.125 -0.140 0.064 1.109 0.000 0.000  1.0 1.000 -1.516 0.000 1.000 -1.447 0.000 1.000 -1.245 0.000 1.000 -0.934 0.000 1.000 -0.553 0.000 1.000 -0.178 0.000 0.997 0.000 0.000  Chapter 3.  Numerical  32  Results  important, but in this analysis it is chosen to be in the form of a half sine wave over an angle a where a 0 =  and K = 4m + 1, m is integer. The surface value of <rr is given  as:  The Fourier series of <rT is  a t  n=l  where f 0 and tn are derived in Appendix D and are  t0  =  2 ~  33 Chapter 3.  Numerical  Results  K 1 K+n . K+n —[ sin—-—nsin [ 7r K + n 2 2K 1 . K - n ,_K - n + — s i n — - — t t J K^n tK  =  0  For convenience, replace a m and bn by am and bn defined as  amamt Ta bnant  K  Ta r = a tlie  Replacing a m and bn by a m and bn in e q u a t i o n (2.15), and Betting following set of equations are derived.  sinmit  f; [(m - 2)(m - 1 )mam - (m - 2)(m + l)(m + 2)<im+2] mn "2 n . 2  i  1  1- 2O2 = K  {( - 1V»+1, l)"T1m 2 f) [(m - 2)(m - l)mam - (m - 2)(m + 1)(M + ^ a ^ - a m m * - ^m i + 1 )(M+2)am+2[  -  (3.26)  +n(n - l)Sn - (n -h l)(n - 2)fcn+2 = *»  2m ( —l)n+ sinm7r [_(m _ l)(m - 2)am + (m + l)(m + 2)am+2] 7T n2 — m 2  £ _ 15 m--  -(n - l)bn + (n +  =0  where  tn  =  tn  for  tif = tK = 0  v  K  Chapter 3.  rri  29  Numerical  a. 2.  a5  OI  a. 9  an  a 1•>3  axs  a it  a 1•>9  a 21  0.672  -0.069  -0.094  -0.063  0.049  0.04  -0.034  -0.029  0.025  0.023  £>2 -0.176  b3 0.051  fe4  fee -0.314  br 0.358  fe9  0.308  h 0.002  0.309  0.63  610 -0.304  as  a5  ai  a.9  an  a 13  a is  an  a 1?9  fen 0.265 a2i•>  0.672  -0.069  -0.094  -0.063  0.05  0.041  -0.034  -0.029  -0.026  0.023  b2  fe7  bx 0  0.36  bs -0.314  fe9  0.309  0.646  ai  bs 0.002 ag  be -0.317  as  h 0.051 as  an  a 13  0.673  -0.069  -0.094  -0.063  0.049  0.041  a 15 1 -0.034  an 1 -0.03  W 0.31  bs 0.002  fee -0.318  b7 0.345  bs 0.315  fe9  -0.176  37  34  Results  b2 -0.176  h 0.051  •>  "J!  fes  0.539  fen 0.389 -0.311" tt21 a 1*9 •> 0.023 0.026  x  feaO -0.314  fen -0.462  Table 3.4: The first ten coeficients with diametrical force T The ten leading coefficients 250 X 250 matrix.  and K are listed in Table (3.4) by inverting a  All the fe7,fe9 and fen must be multiplied by IO" 3 , 10" 4 and 10" B .  For a = 4.9°, 6.2° and 8.6° the value of K are 37, 29 and 21, respectively. As expected the coefficients are insensitive to a for small a. Table (3.5) shows the stresses on the boundary when K = 37 ( a = 4.9°). It is seen that the boundary conditions are satisfied well. The stresses near the crack tip for K = 37 ( a = 4.9°) are obtained by substituting the value of a j into equation (2.17).  = ,. = 1 . 0 1 ^ at V r  at  0 = 0  (3-27)  The stress intensity factor for a = 4.9° is  Kr = 2.5317  T at  r  >  (3.28)  Chapter 3.  Numerical  6  35  Results  ( jf)ca!  (h. ^  JForiour  {y^)  {j?)cal  ideal  60.00  -0.00903  0.00689 -0.00745  70.00  0.00730  0.00907  80.00  -0.01915  -0.01798  85.00  -0.10422  -0.10512  0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000  90.00  18.25948  18.26173  18.000  0.00  0.00735  0.00605  10.00  0.00543 -0.00854  0.00640 -0.00644  0.00540 -0.00854  0.00651 -0.00664  0.00551  20.00 30.00 40.00 50.00  (?nf  0.00000 -0.00013 -0.00027 -0.00041 -0.00056 -0.00071 -0.00088 -0.00107 -0.00129 0.00140 -0.00153  )ideal  0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000  Table 3.5: The stresses on the boundary or  T K i = 1.42841— y/ira at  (3-29)  where T is the total load within the loading area. Selected stresses are given in Table (3.6) and the isochromatics are shown in Fig.(3.13) with a = 4.9°. They are similar to those obtained for a = 6.5°, Fig.(3.12) and a = 8.2°, Fig.(3.11) except near the regions where the loads are applied. Since the stresses far from the loading area are not sensitive to a, for the same total loads, stresses in the disc under a concentrated tension can be determined assuming the disc is loaded with a tension applied over a very small a. Thus, the model can be used exactly to verify the experiments done on test pieces even with big lugs. The dimensionless stresses and ^  f  in the disc at d = 0°, 6 = 45°, 0 = 90° and P = 135° are shown in Fig.(3.14),  at  Fig.(3.15) and Fig.(3.16). Selected values of "f  ?  and  Tf"  throughout the disc are given  in Table (3.6). Detailed stresses for the above three values of a are also given in [19].  Chapter 3. Numerical  Results  Figure 3.11: Isochromatics with Diametric Force T (a = 8.5°)  Chapter 3. Numerical  Results  Figure 3.11: Isochromatics with Diametric Force T (a =  8.5°)  Chapter 3. Numerical  Results  Figure 3.13: Isochromatics with Diametric Force T (a — 4.9°)  38  Chapter 3. Numerical  Results  Figure 3.14: Stress c T with Diametric Force (a = 4.9°)  39  Chapter 3. Numerical  Results  Figure 3.15: Stress cre with Diametric Force (a = 4.9°)  Chapter 3. Numerical  Results  Figure 3.15: Stress cre with Diametric Force (a = 4.9°)  -r - -T. -- ... .1 T '  .3  .4  .5  .6  .7  .8  9  1.0  0.092  <Tr 2.345 <T0 2.971 TrS 0.000 3 0 * " <?r 2.630 (TH 2.494  1.371  0.944  0.696  0.288  1.234  0.770  0.528 0.384  0.398  1.853  0.051  -0.239  0.188 -0.493  -0.715  -0.009 -0.917  0.000  0.000  0.000  0.000  0.000  0.000  0.000  0.000  0.000  1.592  1.097  0.7831 0.564  0.280  0.179  0.088  0.005  1.534  1.045  0.694  0.391  0.403 0.106  -0.167  -0.426  0.798  0.623  0.471  0.334  0.216  0.124  0.060  -0.663 0.021  -0.866  1.071  0.077  -0.009 -0.774  0 = 0°  f=  .2  TrS 0 = 60°  -0.093 0.687  1.995 -0.285 0.497  2.544 -0.159  1.683 -0.206  1.268 -0.158  0.319  0.058 0.531  -0.100 0.308  -0.051 -0.134  0.024 -0.091  0.080 -0.041  -0.450  -0.388 0.002  2.136 0.687 1.132  <r« rT»  3.340 0.286 1.066  <rr <re Tr«  0 = 150° <?e Tr0 0 = 175  1.262 0.254 0.883  3.198 1.356 1.489  0 = 90°  0 = 120°  1.622 0.398 0.979  <rr <?e Tre  2.371  1.015 -0.107 -0.107  <Tr -0.429 06 -0.004 Tre -0.053  -0.001 -0.037  -0.013  0.000  0.953 0.191 0.792  0.660 0.158 0.665  0.395 0.088 0.481  0.195 -0.094 0.260  -0.405 0.079  1.845 -0.419 0.370  1.843 -0.526  1.997 -0.617  2.390 -0.695  3.310 -0.752  6.080 -0.657  0.276  0.201  0.139  0.086  0.040  0.978 -0.068 -0.217  0.726 0.047  0.486  0.268  0.158 -0.295  0.208 -0.222  0.113 0.130  -0.092  -0.008 -0.380  -0.098  -0.003  -0.003  0.050  0.029  0.011  0.002  0.008  0.085 0.046  -0.081 0.038  -0.125 -0.006  -0.297 -0.204 0.005 ' 0.005 0.009 0.025  -0.124 0.004 0.032  0.039 0.065 -0.064  0.019 -0.021  -0.007 -0.005 0.008  0.005 -0.031 0.027  0.175  -0.288 0.095 0.102  0.001 0.030  0.061 -0.026 -0.002 0.021  Table 3.6: Stresses in the disc with diametrical force T  0.037  -0.001 18.259 17.665 -0.002  Chapter 3.  Numerical  43  Results  Figure 3.17: Disc with the Tangential Shear Force S 3.3  Disc with tangential shear forces The case shown in Fig.(3.17) is now considered in which the crack is opened by two  self equilibrating forces, F, applied at the mouth of the crack. 5 is the resultant of a tangential shear distribution of S{6) applied over a small sector a. The distribution of S(8) is in the form of a half sine wave, as discussed in section 3.2, which ensures that Tre = Ter = o at 6 = tt, r = a. The surface value of tt6 is  Trg = t8  -So_K at 2  i Ke  rTe  „-(i-!)<*<* K  =0  e <  tt(1 - - )  The Fourier form of r rfl as well as the coefficient are given below (for detail see Appendix E)  Tre  S 00 = -J J2 Snsinnd a t  n=l  Chapter 3.  Numerical  K, ^  -  44  Results  1  .  .K-  1  1  .  SK  tTjr  - 1.  -\  (3.30)  To maintain the equilibrium, there must be some normal forces acting on the boundary. Let crT act on the same section as rTg and write crr as  g crT — A — sinKO at From the equilibrium equations, the coefficient A is given by Appendix C.  2  1 + cos?-  Hence  KS o  sinjf  2 at 1 +  -sinKO cosf  The Fourier form is  °V =  tn  to tK  =  a t  1  Y,tncosn0)  + n=l  A  1 , . (K + n)(K— [C0S{K + n)TT — ens 7T K + n K 1 r <w \ (K-n){I<— \C0S[K — 77. J TT — COS K — n K  44  =  j t  =  0  K 7T  1) 1)  7T TT J j (3-31'  Chapter 3.  Numerical  45  Results  The details are giveen in Appendix D. am and bn are also replaced by am and bn as follow:  a  _  A  n  am  ba k  =  ba ^  Substituting (3.30), (3.31) and (3.32) into (2.15) for r = a, the final set of equations are  sinmir  „; 2A 102 =  f) {(m - 2)(m - 1 )mam - (m - 2)(m + l)(m + 2 ) ^ ] - ^ + 262 - -Rm=— 2 oo  2m  £  . n + 1 sinmir  [(m - 2)(m - l)mam - (m - 2)(m + l)(m + 2)am+2] —(-1) 1 m—— 2 +n(n - l)Sn + (n + l)(n - 2)6n+2 = in 2m , „ Nn+1 jinnrK \{m - l)(m - 2)&m + (m + l)(m + 2)am+2] -(-!)' n2 — m 2 m = -  n2  _ m2 (3-33)  j  -(n - 1)6„ + (« + l)^n+2 = «n t n and s n are as follow:  tn = tn  sn = sn  tK = 0  aK =  when  n + K  1  ~i:  The leading coefficients are shown in Table 3.< Table (3.8) shows the stresses on the boundary winch are in good agreement with the theoretical values. The stresses at the crack tip are given by eqn.(2.17) and in this case are  Chapter 3.  Numerical  Results  0.013  a it 7 -0.01  a 19 0. 0.008  -0.006  b8 0.048  h -0.042  ho 0.038  bn -0.034  a3  (J 5  ar  a9  a 11  aii  a.15  2.047  -1.111  0.094  -0.046  0.027  -0.018  64  65  be 0.067  br -0.056  b2 1.057  63  -0.052  0.100  -0.082  a 21  Table 3.7: The fist ten coeficients with tangential shear force S  6  0.00 20.00  (  j^)cal  V T„  0.00009  -0.00001  JForiour  0.00000  -0.00008  40.00  0.00002  0.00008  60.00  -0.00002  80.00  0.00004  -0.00008 0.00008  100.00  -0.00006  -0.00008  120.00  0.00012  0.00008  140.00  -0.00028 0.00108 0.42741  -0.00010 0.00018 0.42257  moo" 176700 178.00  0.78068  0.75935  (fideal  ("jfif )cal  0.0000 0.00000 0.0000 -0.00032 0.0000 0.00050 0.0000 -0.00041 0.0000 0.00002 0.0000 0.00058 0.0000 -0.00125 0.0000 0.00166 0.0000 -0.00083 0.4165 0.7554  -9.92305 -17.6972  \ Jn  JForiour  0.00000  -0.00055 0.00077 -0.00047 0.00002 0.00100 -0.00143 0.00116 -0.00034 -9.93722 -17.7269  Table 3.8: The stresses on t l v boundary  (~T^)ideal  0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000  -9.8035 -17.783  Chapter 3.  Numerical  47  Results  (3.34)  <rT = cre — 3.07  and the stress intensity factor is  5  (3.35)  K i = 7.6953—- \/a at Where S is the total load at the crack mouth.  Selected stresses are given in Table (3.9) and the isochromatics are shown in Fieo (3.18). The dimensionless stresses \ /  at  ^at and ^at are also given in Fig.(3.19), Fig.(3.20)  and Fig.(3.21). Detailed stresses are given in [19].  3.4  Comparison with related results The comparison with related results such as the ones shown by Tweed [1-7] and  Gregory [8,9] are given in table (3.10) Cases I, II and III are for uniform tension, diametrical concentrated tension and tangential shear force, respectively. K0 = P 0 V ^ , where K0 is the stress intensity factor of a Griffith crack of lenght 2a in an infinite elastic sheet which is opened by a constant pressure P 0 , F and S are the total load for Case II and Case III respectively. The comparison shows good agreement with other related works. However, the method derived in the thesis is more general and powerful.  3.5  An analysis of the error The errors in the numerical results are introduced three ways.  Fourier series are used to approximate sin{m6) and  cos(m0)  Firstly, when the  as series in sin{nd) and  ri E =  0°  OR  Off rre E  =  30°  60°  90°  E  =  120°  150°  175°  .9  1.0  7.449 3.265  0.000  1.257  -0.106  -1.143  -1.985  -3.319  0.461 -3.871  0.000 0.000 0.000 0.000  0.000  0.000  0.000  0.000  0.000  4.015 -0.143  2.216 -1.985  1.543 -2.682  0.964  0.454  -3.287  -3.825  -4.308  0.210  3.018 -1.160 0.152  0.000 0.000  0.110  0.077  0.048  0.023  2.114  1.478 -2.641 0.117  0.926  0.437  0.000 0.000  -3.208 0.074  -3.707 0.036  1.380 -2.594 0.065  0.875 -3.132  0.417  11.320 7.123  -4.369  4.424  4.764 2.524  0.918 0.656  3.383 -0.243 0.354  2.595 -1.181 0.192  1.945 -1.953  0.039  -3.590 0.020  oe  8.218 2.991  5.191 1.871  3.719 0.945  2.800 0.037  2.159 -0.868  1.233 -2.507  0.821 -3.156  0.410 -3.669  -4.059  TRE  2.739  1.563 4.817  0.848  0.348 2.711 0.504 0.793  0.020  1.663 -1.733 -1.733  -2.507  -3.156  -3.669  -4.059  1.917 0.122 0.316  1.269 -0.565 -0.233  0.836 -1.676 -0.725  0.612  0.412  0.000  -3.135 -0.912  -4.559 -0.622  4.396 0.052 0.442  4.398 0.074  4.407 0.119  0.095  0.511  0.628  4.291 0.213 0.804  -5.483 -0.003 -0.002  -1.230 -0.894  -33.12 -0.124  OR  OR  <TR  TRE =  .8 0.977  6.028 2.429 1.233  oe E  .7 1.566 -2.698  2.865 -1.197 0.238  TRE =  .6 2.250  3.788 -0.227 0.344  oe E  .5 3.069  5.000 1.036 0.527  rTE =  .4 4.088  6.806 2.840 0.895  OR  oe 6  .3 5.441  5.333  rre =  .2  7.290 3.148 0.485  OR  oe E  .1  OR  oe Tre  11.088 6.848 0.989 10.413 6.051 1.865 9.392  6.833 1.016 1.981 4.851 0.033 0.380  0.863 1.542 4.557 0.036 0.384  1.191 0.303  3.638 0.718 1.856 4.439 0.041 0.403  -1.983 0.168  0.108  3.494 0.337 0.881  Table 3.9: Stresses in the disc with tangential shear force S  -4.151  0.000 0.000 -3.985  -0.001 0.000  Chapter 3. Numerical  Results  Figure 3.18: Isochromatics with Tangential Shear Force  Chapter 3.  Numerical  Results  Figure 3.18:Isochromaticswith Tangential Shear Force  Chapter 3. Numerical  Results  Figure 3.20: Stress erg witli Tangential Shear Force  Chapter 3.  Numerical  Results  — 0° -  45°  -  90°  Figure 3.18:Isochromaticswith Tangential Shear Force  Chapter 3.  Numerical  53  Results  Kl(Tweed)  Kl[Gregory)  _Kj(Vaughan)_  3.14  3.17  3.17  Case II  -  -  2.5317  Case III  -  7.6917  7.6953  Case I  Ki Kn  Table 3.10: Comparison with earlier results cos(n0).  Secondly, when the boundary forces are represented by Fourier series, and  finally when the simultaneous algebraic equations are solved numerically, round off error occurs.  However, it is obvious that these errors can be deduced by increasing n, the  number of terms in the Fourier series. Among these three causes the second one is found to be dominant.  Chapter 4  Application to a More General Case  4.1  Motivations In this chapter, the method is applied to a more practical and more important prob-  lem. Once a crack initiates in a structure, a hole may be drilled at the crack tip to release the high stresses and to prevent crack propagation. This problem is now explored numerically.  One motivation is to provide some theoretical justification to empirical  experience. Fig.(4.22) shows a circular disc of radius a 0 and thickness t with a radial crack extending outwards from the centre along the full radius 6 = ir. There is a hole of radius 0i  at the crack tip. The disc is loaded at its outer edge. There are no forces acting on the  crack faces and the hole. The problem of determining the stress distribution in this disc is one of great importance. For instance, the solution provides some theoretical justification of the engineering practice of "stress relief', in which circular holes are drilled at the tip of the crack with the objective of reducing the stress singularity at the tip of the crack; the problem considered here would certainly be relevant when the diameter of the hole is small compared with the length of the crack. Another motivation for considering a disc  with  a hole is that the method and  results shown in Chapter 2 and Chapter 3 are for a„ mathematically ideal crack. When r tends to zero, the stresses <rr and a g are infinite, i.e, t V stresses are singular at the crack  54  Chapter 4.  Application  to a More General  Case  55  e = ir  [ 7 crack(  ^  \  Figure 4.22: Disc with a hole at the crack tip tip. The conclusion is only valuable for an ideal crack which has zero radius at the crack tip. However, in laboratory test specimens, cracks are often fabricated or introduced by cutting with a fine saw blade. The radius of the crack tip is then not zero. To this end, the solution is of much interest in engineering practice. When the hole is not too small, the problem can also be treated as a curved beam. For uniform tension, the solution is available from classical elasticity theory. Approximate solutions of some special cases, such as the stress distribution in an infinite plate with straight radial cracks originating from the boundary of a circular hole and subjected to a prescribed load, have been derived by Morkovin (1944) [11] and Bueckner (1960), but the most complete discussion of the general case would appear to be that due to Bowie (1956) [10]; The analysis of Wigglesworth (1958) [12] is also relevant. The methods used involved mapping the problems 1' - '-.me other plane and then applying complex variable methods.  The results give the stress distribution in an infinite plate  which contains straight radial cracks originating from the boundary of a circular hole  Chapter 4.  Application  to a More General  Case  56  Tx  i - f t Mi  To  Figure 4.23: Equivalent problem and subjected to uniform tension. The references quoted above are important, but the methods become prohibitive in cases of more complicated boundary conditions.  Also,  they are very difficult to apply to a finite plate problem. This is a major weakness since, as mentioned before, the circular disc with a radial edge crack has been chosen as a new standard specimen for fracture tests. In this chapter a more elegant and general method is applied to this practical problem. Two methods are used in this regard for uniform tension. First the disc is treated as a curved beam using elastical theory; secondly the method developed in Chapter 2 is used.  The results are compared and the stress cre  at r = a.i and cr = 0 are obtained for various ratios of  ao  The case of a disc with  diametrical concentrated force is also studied, and the results are presented. some conclusions and recommendations are given.  4.2  T h e c u r v e d b e a m s o l u t i o n in classical e l a s t i c i t y  Finally,  Chapter 4.  Application  to a More General  57  Case  When the hole is not too small, the problem is similar to the one shown in Fig.(4.23). The stress, <rg, at 6 = 0 and r = a.; can be derived by superposition of three different cases shown in Fig.(4.24). Values of Ti,F\ and Mi are calculated from the equilibrium conditions as follow:  Ti  —  T  0  a  0  t  F  t  =  T  0  a  0  t  M Mi  =  ^ T  0  a  0  x  (4.36) t ( a  0  +  a;)  where T 0 is the pressure on the perimeter, Fx is the total force and Mi  is the total  moment on each end about the middle point of the wall. Theoretical solutions for each case exist. The first case is the same as the cylinder under the uniform tension at the outer boundary and free on the inner boundary as shown in Fig.(4.24)a. The stress function is thus  $  =  A l n r  +  B r  2  Inr  +  C r  2  +  D  (4.37)  the stresses are  1  =  A  + B( 1 + 2 In r) + 2c  (4.38) 0 With the following boundary conditions  Chapter 4. Application  58  to a More General Case  +  c 0)  JD  o  0)  (3  O t o O o. o Cu P  in  •if1 II  0 ) 2 b0  Chapter 4.  Application  to a More General  (rr|r=Oi = 0  59  Case  <rr|r=ao=T0  (4.39)  Substituting eqn (4.38) into (4.39), the stresses are solved as below  p 0 Wr„ i a02  T  — a^  aoVrp  -  a02  1  ~ I  r.  — af  T0a02  o  a02 - ai1 t1 Tre =  W  |  r2  0  a0l - a»  0  At 6 = 0 , r = a^ the value of a e is  ^  =  ao -  (4.40)  ai  The second case is the pure bending of the curved bar. The stress function is the same as the first case. The boundary conditions are  <rr = 0  for  r = ai and r = a0 ra o yaao o / credr — 0 / aerdr = — -M J a{ J a{ at  rrg = 0  the  boundaries  Solving the equations the stresses are  AM a2a02  aT = - — ( N  AM,  o-e  TV =  0  (  a0  In  r2 a2a02  r2  ai a0  r  2  U 0 In -  ,  a{  In—)  a0  2  r  — ln h a-o In a, a-o  aQ'  2  a;  °< In r  2a  2  o —  Chapter 4. Application to a More General Case  60  where  N A  = =  B  =  C  =  ; 2 ) 2 — 4a; 2 a 0 2 (ln — )2  (a024M N 2M N M,  CLi  <  a;  ,  2  1  „ 2\  N l-'  The stress erg at 6 = 0,r = a; is  <re =  iV  (—2a02In a ° + a 0 2 - a, 2 ) <Zi  (4.41)  The third case is bending of a curved bar by a force at the end. The stress function is defined as  $ = (Ar3 + — + Cr + Dr ln r)sin6 r  (4.42)  and the stresses are as follows  (2Ar  -  25 — )sin0 "rT + r 6 25 i  —r  +  25 r3  +  — )sin6 r  ^ j^( .... ox nr P r  (4.43)  with the following boundary conditions <rr = rTg = 0  for  r = di  and  r — d0  61  Chapter 4. Application to a More General Case  /•a 0  / rr8dr = -Tx J a{  for  9 = 0 and  9 = tt  Solving these equations for A, B, C and D  a=£< Pai2a o2  2N  D =  -^(ai2-a02)  N = a2 — a02 -j- (a2 + a 0 2 ) In —  Oi  The stress <re at 9 = 0, r = a; is thus  =  N  ^  The numerical value of ag at 8 = 0 and r =  (4.44) for different ai are shown in Table  4.13.  4.3  Solution using the proposed method The same disc as shown in Fig.(4.22) is considered where a0 is outer radius, a; is  the radius of the hole. The disc is loaded at its o n W -»dge by self-equilibrating surface tractions T(6) and S(8) where T is the normal force and 5 is a shear force. The stress function is assumed to be the same as section 2.1 except, that m and n include negative numbers.  62  Chapter 4. Application to a More General Case  $  =  J2 amrm[mcos{jn  - 2)9 - {rn - 2)cosm9]  m—TJX\ oo  + J2 Krn[cos(n  - 2)9 - cosn9}  - ir < 9 < tt  < r < a0 (4.45)  n=ni where mt = —^f1, k positive integer, ni is a negative integer,fcj.= 0. For a complete mathematical solution, k and  are both infinite. However, we terminate the series by  choosing finite values for k and n a . The actual values depend on the boundary conditions prevailing, and are obtained by examining the convergence of the numerical results. Take a m i _ 2 = a TOl -i = 0 and 6 n ,_i = 6 n i _ 2 = 0, $ may be rewritten as follows  $  =  £ m—m  cos{m9)[—amrrn{m i —2  +  £  — 2) -f {rn + 2)aTO+2rm+2]  oo  cos{n6)[—bnrn  n=ni —2  + 5n+2Tn+2]  (4.46)  The stresses are given as follow  M  cT  —  Y, cos{m9)[amrm~2 {m — 2)m(m — 1) — a m+2 r m (rn + 2){m + l)(m — 2)] m=m i —2 M  +  =  cos(n0)[fcnrn-2n(n — 1) — 6 n + 2 r n (n + l)(n — 2)]  ^  n—rij —2  cos{m9)[—m{m — l)(m — 2)amr^m_2^ + {m + 2)2(m — l)am+2rm]  m=mi —2  +  £  cos{n9)[—n{n — l)bnrn~2 + (n 4 2){n  n—nx — 2 Trg  =  M ]T  l)6n+2r,n]  m « n ( m f f ) [ - ( m - l)(m - 2)r m ~ 2 a n , 4 (m + l)(m + 2 ) r m a m + 2 ]  m=mi —2 N  +  £  nam(n0)[-(n - l)6nrn_2 + (n + ] )/\,,+2 r n ]  n=ni —2  (4.47;  Chapter 4. Application to a More General Case  63  Tlie same Fourier series are used to write cos(m9) and sin(m9) in terms of cos(n9) and sin{nO) as in Chapter 2. Also, the boundary tractions are written in Fourier series. Then after the same derivitions the final set of equation are derived as follow:  M m =  m  i  [a m r m_2 (rrz - 2)(m - 1 )m - am+2rm(rn -  •/ N. - 2)(m + l)(m + 2 ) ] ^ ^ + 262 = t0  2  SI ^ [amrm-2(rn m=mi —2 bnrn-2n{n  2  - 2)(m - 1 )m - a m + 2 r m (rn - 2)(m + 1 )(m + 2)]-5tn(m7r)(-l)" + 1 , m , T Tll-Vl  - 1) - 6 n + 2 r n (n + l)(n - 2) + n(n + l)6_ n r~ n - 2 + (n + 2 ) ( - n + l ) 6 _ n + 2 r - n = tn  £ [ - ( m - l)(m - 2)r m ~ 2 a m + (m + l)(m + m=m,-2  71  ~  771  + (n + l ) 6 _ n r - n - 2 - (n - l ) 6 _ n + 2 r - n - (n - l)5„r n " 2 + (n + l)fc n+2 r n = 0 Where a m ,_ 2 = a m ,_i = 6 n i _ 2 = b n i -i = 6 n i _ 2 = 0 and t0 — to  to - tl0 ig, t°n and  tn —  sn =  tn = tln  sn = sln  to/ien  when  r = a0  r = a{  are the coefficients in the Fourier series for the outer boundary, tl0, t%n and  are the coefficients of the Fourier series for the inner boundary. The special case of uniform tension on the outer boundary is now considered. Thus t°0 = 2 T 0  e0  =  o  t°n = . 0si = 0  = < =n  The special case of a disc with diametrical concentrated force is also considered and the coefficients are shown in Appendix D.  (4'48)  64  Chapter 4. Application to a More General Case  0.000  0.000  -0.0013  0.150  18.011  -409.105  2232.3  -807.32  dr? -36.30  b-4 -0.0003  b-3 0.0055  b-2 1.7992  b-1 36.885  bo 0.0000  k 672.003  b2 191.91  b3 -2.8605  h 0.8460  a_ £  a_ 3  a i  a3  a5  Table 4.11: The important coefficients under the uniform tension (scaled by 10 4.4  ao 3.1306 h -0.1114  3)  Numerical results  Disc under uniform tension a m and bn are replaced by <zm and bn in the same way as before:  am  _  amTo  ~  ^ H  , K T0 On = n 2 a0 ~ t  A 66 x 66 matrix was used to solve the equations. Some important coefficients are shown in Table (4.11). Table (4.12) shows how well the boundary conditions are satisfied. Fig.(4.25) shows §* at r = a{,0 = 0, for the range of fj- from 10"5 to 0.95. Fig.(4.26) is the stress a e for 0 = 0. Table (4.13) compares the stress at r =  and  0-0  obtained from curved beam theory with those obtained using the method derived in the thesis. Good agreement can be seen. Disc with diametrical concentrated force am and bn are replaced by am and bn in the following form:  Chapter 4. Application to a More General Case 5772  r  e 0.00  30.00 60.00 90.00 120.00 150.00 170.00  1.0  0.00  0.1  30.00 60.00 90.00 120.00 150.00 170.00  (Tn)caI  ( jT )ideal  (  Tn  )cai  ( Tn ) i d e a l  0.99974 1.00016 1.00009 0.99963 1.00054 0.99975 1.00155  1.000  0.00000  0.000  1.000  0.000  1.000  -0.00019 0.00028 -0.00017 -0.00024 0.00118 0.00243  0.00155 -0.00103 0.00006 0.00030 0.00002 0.00066 -0.00415  0.000  0.00000  0.000  0.000  -0.00033 0.00070  0.000  0.000 0.000  -0.00111  0.000  0.000  0.00143 -0.00009 0.00812  0.000  1.000 1.000 1.000 1.000  0.000 0.000  0.000 0.000 0.000 0.000 0.000  0.000  0.000 0.000  Table 4.12: Stresses on the boundaries  aj  oe(c.b) ae(v.w) 9-Q ae{c.b) cre(v.w)  0.05 33.4830 32.8785 0.55 62.6871 62.1084  0.10 27.2005 26.6839 0.60 78.1968 77.6275  0.30 0.20 29.9934 26.2219 29.4010 25.6424 0.80 0.70 136.1550 r302.5295 135.6084 302.H283  0.40 37.7409 37.1403 0.90 1202.32 1202.14  0.5 51.6981 51.1109 0.95 4802.36 4802.36  Table 4.13: The stress cre at d = 0 and r = arobtained using different methods  Chapter 4. Application to a More General Case  66  Chapter 4. Application to a More General Case 5774  «0  Figure 4.26: Stress <7g at 6 = 0  68  Chapter 4. Application to a More General Case  a _x  d_5  a_ 3  a_L  0.000  0.000  0.0145  1.344  b-A  b-3  6-2  0.0025  0.0000  b-!  0.1356 • 1.747  73.112  4 -806.642  -123.40  bo  k  b2  0.0000  -149.178  a5 i  ar  ao  -102.05  -62.733  49.012  k  K  h  312.722  1.850  -310.31  <23  72.059  Table 4.14: The important coefficients with diametrical force (scaled by 10  3)  A 600 X 600 matrix was used to solve the equations. Some important coefficients are shown in Table (4.14). Table (4.15) shows how well the boundary conditions are satisfied. Fig. (4.27) shows ^ at 6 = 0 and r = a ; , for the range of ^ from 0.025 to 0.95. at  4.5  Discussion From Fig.(4.25), Fig.(4.26) and Fig.(4.27), it is seen that for small holes,  < 0.16,  stress cre at 6 = 0,r = a{ increases as the ^ decreases. It means the hole should not be too small when it is drilled to release the stress singularity. Also, when  > 0.16, erg at  6 = 0 and r = a{ increases as ^ rises. It indicates that the property of the stresses are governed primarily by the local stress field at the hole. Hence, to prevent the propogation of the crack, the ratio of f j should be between 0.13-0.26. The minimum stress occurs when  ?<  ao  is 0.16 for both cases of uniform tension and diametrical force. When ^ is  greater than 0.26, the stress concentration will also b" liigh. The same conclusion can be made from Fig.(4.27), the case of diametrical concentrated force. Table 4.13 shows good agreement between the data using the curved beam theory and the proposed method for uniform tension. Consequently this problem can be treated  69  Chapter 4. Application to a More General Case  r  6  1.0  0.00 30.00 60.00 80.00 90.00 100.00 120.00 150.00 170.00  0.00000 0.00082 0.00039 -0.01906 10.58377 -0.01906 0.00039 0.00082 -0.00070  (^f ) cai 0.00000 -0.00262 0.00414 -0.00043 0.00360 0.00032 0.00009 -0.00840 -0.01771  (~fC ) ideal  0.00001 0.00332 0.00013 -0.02128 10.57990 -0.01387 -0.00776 -0.01329 0.02961  0.1  0.00 30.00 60.00 90.00 120.00 150.00 170.00  0.00001 0.00000 0.00000 0.00001 0.00002 0.00002 -0.00014  0.000 0.000 0.000 0.000 0.000 0.000 0.000  0.00000 0.00000 0.00001 -0.00001 -0.00001 -0.00004 0.00001  0.000 0.000 0.000 0.000 0.000 0.000 0.000  (n)cal  (7f,) ideal  Table 4.15: Stresses on the boundaries  0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000  Chapter 4. Application to a More General Case  70  Chapter 4. Application to a More General Case  71  as a curved beam when the hole is not too small. However, the method derived in the thesis is more general since it can deal with any kind of boundary conditions with good accuracy. Also the cases of greatest practical use are when the loads are concentrated, as shown in Figure 3.10 and 3.17. Curved beam theory cannot handle these cases.  Chapter 5  Concluding Remarks  A fairly simple method has been derived which gives the stresses in a disc with a radial crack for any distribution of surface forces. Except for the transform method of Tweed [1-7] and Gregory [8,9], this work is original. The two cases presented for diametric concentrated forces, particularly the tangential shear forces acting at the mouth of the crack, provides us with a new experimental test for studying fracture mechanics in a very controlled way. Many problems which could not be solved well before, can now be solved easily with the method shown in the thesis. Closed-form solutions for elastic bodies containing a crack are sparse, particularly for bodies of finite dimension. The solutions obtained here for circular discs add to the few available solutions. It is valuable in not only providing additional theory but also because it is of practical importance, being a representation of a "model problem". The two solutions involving concentrated loads are particularly useful for examining fracture strength since specimens may easily be machined and the load can be accurately and simply controlled. studies.  In this regard the solutions provide a useful test for comparative  The results are also particularly adapted to photoelastic methods in which  isochromatics similar to those given graphically may be obtained. It has been noted that the method will certainly give accurate values for the stress intensity factor and the stresses in the disc. It was also shown that the effect of relatively small differences in the loading area is negligible so that these formula are useful in the 72  Chapter 5. Concluding Remarks  73  calibration of a fracture test specimen. The results for the case in which a hole has been drilled at the crack tip are new. The graphs of stress cocentration factors versus « are of practical importance and significance. It is shown that there is an optimum ratio of % for stress relief in a disc but that as <n -»• 0, the stress concentration factor becomes unbounded.  Bibliography  [1] Tweed, J. and Das, S.C. "The Stress Intensity Factor of a Radial Crack in a Finite Elastic Disc", International Journal of Engineering Science, Vol. 10, 1971, pp. 323335. [2] Tweed, J. and Rooke, D.P., "The Stress Intensity Factors of a Radial Crack in a Finite Rotating Elastic Disc", International Journal of Engineering Science, Vol. 10, 1972, pp. 709-714. [3] Rooke, D.P. and Tweed, J. "The STress Intensity Factor at a Radial Crack in a Point Loaded Disc", International Journal of Engineering Science, Vol. 11, 1973, pp. 285-290. [4] Tweed, J. and Rooke, D.P., "The Stress Intensity Factor of an Edge Crack in a Finite Elastic Disc", International Journal of Engineering Science, Vol. 11, 1973, pp. 65-73. [5] Rooke, J.P. and Tweed, J. "The stress intensity Factor of an Edge Crack in a Finite Rotating Elastic Disc", International Journal of Engineering Science, Vol. 11, 1973, pp. 279-283.  [6] Rooke, D.P. and Tweed, J., "Stress Intensity Factors for a Crack at the Edge of a Pressurized Hole", International Journal of Engineering Science, Vol. 18, 1980, pp. 109-121. [7] Rooke, D.P. and Tweed, J., "Stress Intensity Factors for Periodic Radial cracks in  74  75  a Rotating Disc", International Journal of Engineering Science, Vol. 26, 1988, pp. 1059-1069. [8] Gregory R.D., "A Circular Disc Containing a Radial Edge Crack Opened by a Constant Internal Pressure", Mathematics Proceeding of Cambridge Society, Vol. 81, 1977, pp. 497-521. [9] Gregory R. D., "The Edge-cracked Disc under Symmetric Pin-loading", Math. Proc. Camb. Soc., Vol. 85, 1979, pp. 523-538. [10] Bowie, O. L., "Analysis of an infinite plate containing radial cracks originating at the boundary of an internal circular hole",J. Math. Phys., Vol. 36, 1956, pp. 60-70. [11] Morkvin, V. "Effect of small hole on stresses in uniformly loaded plate", Quart. Plli. Math. Vol. 2, 1944, pp. 350-365. [12] Wigglesworth, L.A., "Stress distribution in a notched plate" Mathamatica, 1957, Vol.5, 1957, pp. 67-80. [13] Sneddon, I. N. and Lowengrub, M., <  Crack Problem in the Classical Theory  of Elasticity > , ASME, John Wiley and Sons, INC. New York. London. Sydney. Toronto, 1971. [14] Sih, G.C., < Handbook of Stress Intensity Factors > , Lehigh University, 1973 [15] Sih, G.C. <  Outlook on Fracture Mechanics > , The Mechanism of Frature > ,  Edited by Goal, V.S. [16] Chell, G.G < Developement in Fracture Mechanics > , Galliard, Great Yarmouth, 1979.  Bibliography  76  [18] Vaughan, H. and Wu, Q.,"Stresses in a Circular Disc containing a Radial Crack Department of Mechanical Engineering, University of British Columbia, Vancouver, V6t lw5, September, 1989 [19] Tada, H., Paris, P. and Irwin, G. "The Stress Analysis of Cracks Handbook" (Del Research Corportion, 1973).  Appendix A  First m o d e linking between shear and normal stress  I n processing the algebraic equations, it was observed that when n = 1, the boundary condition on TT6 and that on crr yield identical equations. The reason for this unexpected result is now examined and explained.  Tr9 = M8),  <Tr = fi{0),  T= a  - TT < 6 < TT  Since T{6) and S(0) keep self equilibrium  £ F  y  = 0  £ M  o  = 0  (A.49)  where Fx and Fy are the resultant forces in x and y directions and M0 is the moment about O as shown in Fig.(A.28>)-  Thus  >  Jn (aTcos0 - TTesin6)ad6  = 0  j\arsin6  = 0  J* {rTea)ad9  + TrBcos6)ade = 0  ( £ My-  Expanding the integral  77  {Y,F' ^ ")  = °)  Fy = 0)  (A.50)  78 Appendix  A.  L y  Figure A.28: Disc under equilibrating force system J * CrsinBiQ + J ^ T T e c o s 6 i e = 0 J"  TrSd6 = 0  <rT and r r 6 are written in Fourier series as follow, =  a0 + Y{ancosne n=l  +  = ao + f K * n C o s n 6 + n= 1  crcos6d8  IT  /  Tresin6d8  and from ( A . 5 1 ) the following is concluded Qi =  lnsinn6)  insinn9)  (A.51)  Appendix A  Normal force requirement for crack loaded by tangential shears  In Chapter 3, it is mentioned that to maintain equilibrium, there must be some normal forces acting on the boundary in addition to the active shear forces. These normal forces are now evaluated. A disc in which the crack is opened by two self equilibrating forces, Frt applied at the mouth of the crack as shown in Fig.(B.29) is considered. The surface forces are expressed in terms of the surface stresses by  There is to be no resultant force in the x direction, hence  atTresinddd = /  atcrTcos6d8  defined in Chapter 3 as follow: 2 at  KPo  2 at  I  sinKO  smKOtuiede  79  (B.52)  80  Appendix B.  Figure B.29: Disc with tangential shear forces  =  cos(K + 1)0 - cos{K - 1)6de  HQ. [* 4 at  KPo 1 4 at K +  1 • , sin{K + 1)51^1, - Y-ZJsin(K  KPo  1  . fr,  r  4 atK + l[  Ki  K Po 1 - _ — — (—-—sin [ ~ 4at K + l Assuming crT =  Z  2  at  •  K - 1 1  WK-ji  T  K • n \ sin—) K}  A^sinKB  Facos6 = U  _ ~  * 1K^K-l  1  ~  L  Cit v  — —-—irsin(K K + l  srn{K + 1)0 + ,in(K  T " 7T  - 1 )6d6  + 1)011-., - T-^—cos{K ' k * K- 1  -  1)011-  81  Appendix B.  = -  \  a  [  i  -  cos  (K -  1 -  +  1  - ^ t  r,  1  [1 + C 0 S K ]  7T .  -  1  n ,  "  +  if  - 1)(K + 1) From eqution (B.52), A is given as A=  1 sin-j ' K -K 2 1 + cos% K  K Pn sin^ ar = — —sinKd^ 2 at l + cos% K  c o  < K - v* -  r  1  ^ r ^  Appendix A  It is discribed in Chapter 3 that T is the resultant of the distribution of T(9) in the case of diametric concentrated force. T{0) = crTt. <jr is as follow •sinK Kf) u <Tr =K T 2 at  In Chapter 3 T is assuming as follow T=  I 2 T2K crratd9 In r J  2 -  2 K  However, the real resultant force, F, is supposed to be  2 ~  2 f(  crrsin9atd6  Since a is small, the errer between T and F is very low as T — T — f2  2K  J  2  2 K  - T - —T  2  _  K 2  1 [K + 1  — —sinK9sin8atd8 2 at  +  ^2  2K  sinK9sin6d6  , 1\0I? + 21? 2  82  i ^ -  1  2 + 2K j - n/?lf 7T _ rr I 2 2K  Appendix  83  B.  K„r  1  (K + 1)TT . {K + 1)TT  r  K .  1  1  7T  1  Suppose E as the errer E = ~x  E < 0.017 %  £ < 0.053//  100%  WhenK  = 37  WhenK = 21  (K-l)ir  7T  (if - 1) 7T,  Appendix A  In this part, the coefficients in Fourier series of the normal stress defined in Chapter 3 are derived. The normal force crT is as crT —  sinKd  at 2  The Fourier series is as follow: = —XYltnCosne at  +  n=l  tn = - ^ 7T 1  Z  f  <rTcosn6d6  JO  2 at  r  TV 1  JO  Jn  2 f7+2K K t„0 = — —sinKddd t r J2; -2K ^ 2  -  [2  + 2K  sinKOdK9  \-cosK6)  ||!|  2 2K  1t ,K (K --[cos(y7r +, -^) - cos(—ir -  84  7r^  -)  Appendix D.  2 . K  2  = — sin—7T = —  7T  2  7T  For n ^ K  tn = — f*  Tr  —  +  sin(K  +  2K  ^  + n)0 + sin{K  7r J^r-^y 2 2K" 2  * [-J— n + ^ 2ir K + n Jf-Jfc  + n)0 + K  — [—— (-cos(K y 2tt K + n  K  1  1  tt^ +  +  n)d\l+A + ' '2 IK  1  "  -[cos{K  V  -  v  2  K  2/f  n)6]dd  r + fK MK j ~ j k  ,  - cos{K  2K  -  tf-n  -  SK  . K - n stn—-—irsm  2  Also, as shown in section 3.3, the a T can be written as sin:K  at 2 1 +  K  -sinK6  n)9d(K  n)6\ph »  - 1  ,  n)^-—-7r]) 2K  '  = 0  So K  -  —{-cos(k K - n  if + n . tf + n , 1 sin—-—7rsm tt + —  to  n J  K -f 1  X^n1  K — [  —sinK6cosnddd  2  ,  K - n  2K  ttJ  Appendix  86  D.  The coefficients of Fourier series are thus 2 r tr  =  Ksin%_ 2 1+ f  K « a ± r 7T 1 + £ fe.  7T1 + - F  _  sinKede  ^  s m ^ ( c o s K i t - cos(if - l)7r) 7T 1 + f  2 smj^  t  n  n  - - f — 3171 g sinK6cosn9d6 7T Jo 2 1 + f  7T 1 + f J ^ i r 2  =  2tt 1 + f V  27rl + f L  — c o ^ + n^l'. - - J—co^/r-n^l^ij V ; X + n iC-n  l—[cos(K+n)n-cos(K+n)I^^--^—{cos(K-n)n-cos{K V if + n 1 # # - ™  tK = 0  ~ ^  K  ~  3  Appendix A  In this part, the coefficients in Fourier series of the tangential shear force acting on the mouth of the crack, as shown in section 3.3, are derived. Stress rTg is defined as Tt8 = -^R — ainKO I at or in term of Fourier series S 00 r.ff = — Y^ snsinn9 n—J.  for n ^ K 1 r K , sn = — sinK9sinn9d9 7T J — 2 7T  r --[cos(K 7r Ji^-n 2  --[—l—sin(K 2 n[K + n  Kr  27r  1  K  y  + n)0\rK-, '  . (K + n)(K-  n  3 l n  ±  + n)9 - cos(K - n)9}d9  1)  £  K  s  -  —sin(K v K-n  ,  7T +  1 k  =  87  ~~  —  1  K-n  - n)9\\., 1 ' —K  . ( K - n ) ( K - 1)  SlTl-  ^  K  7T  

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