# fundamentals of fracture mechanics 断裂力学基础

Fundamentals of Fracture MechanicsCRC Press is an imprint of the Taylor they will not have to spend too much time picking and choosing appropriate topics for the course from the vast knowledge presented in most fracture mechanics books available today. A professor who has never taught fracture mechanics can easily adopt this book as the official textbook for his or her course and simply follow the book chapters and sections in the same order in which they are presented. A num- ber of exercise problems that can be assigned as homework problems or test problems are also provided. At the end of the semester, if time permits, the instructor can cover some advanced topics presented in the last chapter or topics of his or her interest related to fracture mechanics. From over 20 years of my teaching experience I can state with confidence that if the course is taught in this manner, the students will love it. My teach- ing evaluation score in fracture mechanics has always been very high and often it was perfect when I taught the course in this manner. Since many students of different backgrounds over the last two decades have loved the organization of the fracture mechanics course presented in this book, I am confident that any professor who follows this book closely will be liked by his or her students. The book is titled Fundamentals of Fracture Mechanics because only the essential topics of fracture mechanics are covered here. Because I was moti- vated by my students, my main objective in writing this book has been to 8432_C000.indd 1311/30/07 4:25:42 PMkeep the materials and explanations very clear and simple for the benefit of students and first-time instructors. Almost all books on fracture mechanics available in the market today cover the majority of the topics presented in this book and often much more. These books are great as reference books but not necessarily as textbooks because the materials covered are not nec- essarily presented in the same order as most instructors present them in their lectures. Over half of the materials presented in any currently available fracture mechanics book is not covered in an introductory fracture mechan- ics course. For this reason, the course instructors always need to go through several fracture mechanics books’ contents carefully and select appropriate topics to cover in their classes. It makes these books expensive and diffi- cult for self-study. Often, instructors find that some important topics may be missing or explained in a complex manner in the fracture mechanics books currently available. For this reason, they are forced to follow several books in their course or provide supplementary class notes for clearer explanations of difficult topics. Fundamentals of Fracture Mechanics overcomes this short- coming. Since it only covers the essential topics for an introductory fracture mechanics course, it is the right book for first-time learners, students, and instructors.Tribikram Kundu8432_C000.indd 1411/30/07 4:25:42 PMThe AuthorTribikram Kundu, Ph.D., is a professor in the Department of Civil Engi- neering and Engineering Mechanics and the Aerospace and Mechanical Engineering Department at the University of Arizona, Tucson. He is the winner of the Humboldt Research Prize (senior scientist award) and Hum- boldt Fellowship from Germany. He has been an invited professor in France, Sweden, Denmark, Russia, and Switzerland. Dr. Kundu is the editor of 14 books and 3 research monographs, as well as author or coauthor of 2 text- books and over 200 scientific papers, 100 of which have been published in refereed scientific journals; 3 have received “best paper” awards. Among his noteworthy recognitions are receipt of the President of India gold medal for ranking first in his graduating class in IIT Kharagpur and the regents’ fel- lowship and outstanding MS graduate award from UCLA. He is a fellow of ASME, ASCE, and SPIE.8432_C000.indd 1511/30/07 4:25:43 PMThis page intentionally left blank 1Fundamentals of the Theory of Elasticity1.1 IntroductionIt is necessary to have a good knowledge of the fundamentals of continuum mechanics and the theory of elasticity to understand fracture mechanics. This chapter is written with this in mind. The first part of the chapter (section 1.2) is devoted to the derivation of the basic equations of elasticity; in the second part (section 1.3), these basic equations are used to solve some classical boundary value problems of the theory of elasticity. It is very important to comprehend the first chapter fully before trying to understand the rest of the book.1.2 Fundamentals of Continuum Mechanics and the Theory of ElasticityRelations among the displacement, strain, and stress in an elastic body are derived in this section.1.2.1 Deformation and Strain TensorFigure 1.1 shows the reference state R and the current deformed state D of a body in the Cartesian x1x2x3 coordinate system. Deformation of the body and displacement of individual particles in the body are defined with respect to this reference state. As different points of the body move, due to applied force or change in temperature, the configuration of the body changes from the reference state to the current deformed state. After reaching equilibrium in one deformed state, if the applied force or temperature changes again, the deformed state also changes. The current deformed state of the body is the equilibrium position under current state of loads. Typically, the stress- free configuration of the body is considered as the reference state, but it is not necessary for the reference state to always be stress free. Any possible configuration of the body can be considered as the reference state. For sim- plicity, if it is not stated otherwise, the initial stress-free configuration of the body, before applying any external disturbance (force, temperature, etc.), will be considered as its reference state.8432_C001.indd 111/17/07 2:49:13 PM Fundamentals of Fracture MechanicsConsider two points P and Q in the reference state of the body. They move to P* and Q* positions after deformation. Displacement of points P and Q is denoted by vectors u and u + du, respectively. (Note: Here and in subsequent derivations, vector quantities will be denoted by boldface letters.) Position vectors of P, Q, P*, and Q* are r, r + dr, r*, and r* + dr*, respectively. Clearly, displacement and position vectors are related in the following manner:rrurdrrdrududrdrdu****=++=+++∴=+(1.1)In terms of the three Cartesian components, the preceding equation can be written as:()()***dxdxdxdxdxdx123123eeeeee123123++=+++( ()dududu123eee123++(1.2)where e1, e2, and e3 are unit vectors in x1, x2, and x3 directions, respectively. In index or tensorial notation, equation (1.2) can be written asdxdxduiii*=+(1.3)where the free index i can take values 1, 2, or 3. Applying the chain rule, equation (1.3) can be written asdxdxu xdxu xdxu xdxdxiiiiii**=+∂ ∂+∂ ∂+∂ ∂∴11 22 33= =+∂ ∂=+=∑dxu xdxdxu dxiijjjii jj13,(1.4)Figure 1.1 Deformation of a body: R is the reference state and D is the deformed state.PP*Q*Qrr + drr*uu + dudrdr*x1x2x3RD8432_C001.indd 211/17/07 2:49:19 PMFundamentals of the Theory of Elasticity In the preceding equation, the comma (,) means “derivative” and the sum- mation convention (repeated dummy index means summation over 1, 2, and 3) has been adopted. Equation (1.4) can also be written in matrix notation in the following form:dx dx dxdx dx dx123123***=+∂ ∂ ∂∂ ∂∂ ∂ ∂ ∂∂ ∂∂ ∂ ∂ ∂u xu xu x u xu xu x u x11121321222331∂ ∂ ∂∂ ∂ u xu xdx dx dx3233123 (1.5)In short form, equation (1.5) can be written as{}{}[] {}drdrudr*T=+ ∇(1.6)If one definesεiji jj iuu=+1 2(),,(1.7a)andωiji jj iuu=-1 2(),,(1.7b)then equation (1.6) takes the following form:{}{}[ ]{}[ ]{}drdrdrdr*=++εω(1.7c)1.2.1.1 Interpretation of eij and wij for Small Displacement GradientConsider the special case when dr = dx1e1. Then, after deformation, three components of dr* can be computed from equation (1.5):dxdxu xdxdxdxu xdx1111111122111**()=+∂ ∂=+=∂ ∂ε= =+=∂ ∂=+()()*εωεω21211331131311dxdxu xdxdx(1.8)8432_C001.indd 311/17/07 2:49:24 PM Fundamentals of Fracture MechanicsIn this case, the initial length of the element PQ is dS = dx1, and the final length of the element P*Q* after deformation isdSdxdxdxdx****()()()()=++[]=+12223211121 21ε+ ++++[]≈+[]=()()εωεωε21212313121111 21 212dxdx x1111()+ε(1.9)In equation (1.9) we have assumed that the displacement gradients ui,j are small. Hence, eij and wij are small. Therefore, the second-order terms involv- ing eij and wij can be ignored. From its definition, engineering normal strain (E11) in x1 direction can be written asΕ1111111111=-=+-=dSdS dSdxdx dx*()εε(1.10)Similarly one can show that e22 and e33 are engineering normal strains in x2 and x3 directions, respectively. To interpret e12 and w12, consider two mutually perpendicular elements PQ and PR in the reference state. In the deformed state these elements are moved to P*Q* and P*R* positions, respectively, as shown in Figure 1.2. Let the vectors PQ and PR be (dr)PQ = dx1e1 and (dr)PR = dx2e2, respectively. Then, after deformation, three components of (dr*)PQ and (dr*)PR can be writ- ten in the forms of equations (1.11) and (1.12), respectively:()()()**dxdxu xdxdxdxPQPQ1111111121=+∂ ∂=+=∂εu u xdxdxdxu xdxPQ211212113311∂=+=∂ ∂=()()(*εωε3 31311+ω)dx(1.11)PP*R*Rx1x2x3QQ*90°SS*α1αα2Figure 1.2 Two elements, PQ and PR, that are mutually perpendicular before deformation are no longer perpendicular after deformation.8432_C001.indd 411/17/07 2:49:29 PMFundamentals of the Theory of Elasticity ()()()**dxu xdxdxdxuPRPR11221212122=∂ ∂=+=∂εω∂ ∂=+=∂ ∂=+xdxdxdxu xdxPR222223322321()()(*εεω3 321)dx(1.12)Let a1 be the angle between P*Q* and the horizontal axis, and a2 the angle between P*R* and the vertical axis as shown in Figure 1.2. Note that a + a1 + a2 = 90°. From equations (1.11) and (1.12), one can show thattan() ()αεω εεωεω12121111121211221=+ +≈+=+dx dx1 121212222212211tan() ()αεω εεω=+ +≈-dx dx(1.13)In the preceding equation, we have assumed a small displacement gradi- ent and therefore 1 + eij ≈ 1. For a small displacement gradient, tan ai ≈ ai and one can write:αεωαεωεααωα11221212211212211 21 2=+=-∴=+=() similarly, the other three terms in the two expressions can be shown as equal. Thus, the two sides of the equation are proved to be identical.Example 1.2 Check if the following strain state is possible for an elasticity problem:εεεε11122222223212123=+()=+()=k xxk xxkx x x,,,1 1323330===εεSolution From the compatibility condition, eij,kℓ + ekℓ,ij = eik,jℓ + ejℓ,ik, given in example 1.1, one can write e11,22 + e22,11 = 2e12,12 by substituting i = 1, j = 1, k = 2, ℓ = 2.εεε11 2222 1112 12320222,,,+=+==kkkxSince the two sides of the compatibility equation are not equal, the given strain state is not a possible strain state.1.2.2 Traction and Stress TensorForce per unit area on a surface is called traction. To define traction at a point P (see Figure 1.3), one needs to state on which surface, going through that point, the traction is defined. The traction value at point P changes if the ori- entation of the surface on which the traction is defined is changed. Figure 1.3 shows a body in equilibrium under the action of some external forces. If it is cut into two halves by a plane going through point P, in general, to keep each half of the body in equilibrium, some force will exist at the cut plane. Force per unit area in the neighborhood of point P is defined as the traction at point P. If the cut plane is changed, then the traction at the same point will change. Therefore, to define traction at a point, its three components must be given and the plane on which it is defined must be identified. Thus, the traction can be denoted as T(n), where the superscript n denotes the unit PF1F2F3F4Figure 1.3 A body in equilibrium can be cut into two halves by an infinite number of planes going through a specific point P. Two such planes are shown in the figure.8432_C001.indd 611/17/07 2:49:37 PMFundamentals of the Theory of Elasticity vector normal to the plane on which the traction is defined and where T(n) has three components that correspond to the force per unit area in x1, x2, and x3 directions, respectively. Stress is similar to traction; both are defined as force per unit area. The only difference is that the stress components are always defined normal or parallel to a surface, while traction components are not necessarily normal or parallel to the surface. A traction T(n) on an inclined plane is shown in Figure 1.4. Note that neither T(n) nor its three components Tni are necessarily normal or parallel to the inclined surface. However, its two components snn and sns are perpendicular and parallel to the inclined surface and are called normal and shear stress components, respectively. Stress components are described by two subscripts. The first subscript indicates the plane (or normal to the plane) on which the stress component is defined and the second subscript indicates the direction of the force per unit area or stress value. Following this convention, different stress components in the x1x2x3 coordinate system are defined in Figure 1.5.T(n)Tn1x1x2x3Tn2Tn3σnnσnsFigure 1.4 Traction T(n) on an inclined plane can be decomposed into its three components, Tni, or into two components: normal and shear stress components (snn and sns).x1x2x3σ22σ21 σ23σ13σ13σ11σ11σ12σ33σ32σ12σ31Figure 1.5 Different stress components in the x1x2x3 coordinate system.8432_C001.indd 711/17/07 2:49:39 PM Fundamentals of Fracture MechanicsNote that on each of the six planes (i.e., the positive and negative x1, x2, and x3 planes), three stress components (one normal and two shear stress com- ponents) are defined. If the outward normal to the plane is in the positive direction, then we call the plane a positive plane; otherwise, it is a negative plane. If the force direction is positive on a positive plane or negative on a negative plane, then the stress is positive. All stress components shown on positive x1, x2, and x3 planes and negative x1 plane in Figure 1.5 are positive stress components. Stress components on the other two negative planes are not shown to keep the Figure simple. Dashed arrows show three of the stress components on the negative x1 plane while solid arrows show the stress com- ponents on positive planes. If the force direction and the plane direction have different signs, one positive and one negative, then the corresponding stress component is negative. Therefore, in Figure 1.5, if we change the direction of the arrow of any stress component, then that stress component becomes negative.1.2.3 Traction–Stress relationLet us take a tetrahedron OABC from a continuu