Brown 断裂力学课程 8.4

EN175: Advanced Mechanics of SolidsDivision of Engineering Brown University8.4 Energy methods in fracture mechanicsEnergy methods provide additional insight into fracture, and also provide a foundation for a range of analytical and numerical methods in fracture mechanics. In this section, we outline some of the most important results.8.4.1 Definition of crack tip energy release rate for cracks in linear elastic solidsThe crack tip energy release rate quantifies the rate atwhich strain energy is dissipated as the crack grows.To make this precise, consider an ideally elastic solid with elastic constants subjected to some loading (applied tractions, displacements, or body forces). Suppose the solid contains a crack (the figure shows a circular crack with radius a as a representative example). Define the potential energy of the solid in the usual way asSuppose the crack increases in size, so that the crackadvances a distance with loading kept fixed, wheres measures position around the crack front. The principle of minimum potential energy (sect 4) shows that , since the displacement field associated with is a kinematically admissible field for the solid with a longer crack. The energy release ratearound the crack front is defined so thatEnergy release rate has units of (energy per unit area).For the special case of a 2D slit crack with length a , the energy release rate iswhere is now the potential energy per unit out-of-plane distance.8.4.2 Energy release rate as a fracture criterionPhenomenological fracture (or fatigue) criteria can be based on energy release ratea(s)σσδaarguments as an alternative to the K based fracture criteria discussed earlier.The argument is as follows. Regardless of the actual mechanisms involved, crack propagation involves dissipation (or conversion) of energy. A small amount of energy is required to create two new free surfaces (twice the surface energy per unit area of crack advance, to be precise). In addition, there may be a complex process zone at the crack tip, where the material is plastically deformed; voids may be nucleated; there may be chemical reactions; and generally all hell breaks loose. All these processes involve dissipation of energy. We postulate, however, that the process zone remains self-similar during crack growth. If this is the case, energy will be dissipated at a constant rate during crack growth. The crack can only grow if the rate of change of potential energy is sufficient to provide this energy.This leads to a fracture criterion of the formfor crack growth, where is a property of the material. Unfortunately is often referred to as the fracture toughness of a solid, just like defined earlier. It is usually obvious from dimensional considerations which one is being used, but its an annoying source of confusion.8.4.3 Relation between energy release rate and stress intensity factorThe energy release rate G is closely related to the stressintensity factors defined in Sect 8.3. The result isDerivationA neat argument due to Irwin provides theconnection.A crack of lengtha can be regarded as a crack withwhich is being pinched closed by an appropriate distributionof traction acting on the crack faces betweenand. We can therefore calculate the change in potential energy as the crack propagates by distance by computingthe work done as these tractions are progressively relaxed tozero.The asymptotic crack tip field (Sect 6.3.1) gives the tractions acting on the upper crack face as(equal and opposite tractions must act on the lower crack face). As the crack is allowed to open, the upper crack face displaces bywhere we have assumed plane strain deformation. The total work done as the tractions are relaxed quasi-statically to zero isa δ a δ a x 2x 2x 1(the work done by tractions acting on the upper crack face per unit length is , andthere are two crack faces). Evaluating the integrals givesThe same result can be obtained by applying crack tip energy flux integrals, to be discussed shortly.8.4.4 Relation between energy release rate and complianceEnergy release rate is related to the compliance of a structure or specimen. Consider the compact tension specimen shown in the picture. Suppose that the specimen is subjected to a load P , which causes the point of application of the load to displace by a distance in a direction parallel to the load. The compliance of the specimen isAs the crack grows, the compliance of the specimen always increases, so C is a function of crack length. The energy release rate is related to compliance C byThis result is useful for two reasons:(i) It can be used to measure energy release rate in an experiment. All you need to do is to measure the crack length as it grows, and at the same time measure the compliance of your specimen.(ii) It can be used to calculate stress intensity factors, as outlined in the next section.Derivation: The load induces a total strain energyin the specimen.Now, suppose that the crack extends by a distance . During crack growth, the loadincreases to and displaces to . In addition, the strain energy changes to, while the compliance increases to . The energy released during crack advance is the change in strain energy, less the work done by applied loads, so thatNote thatδaΦΦ+δΦx 0+δxx 0P+δPPBawhenceThe energy release rate therefore is related to compliance by8.4.5 Calculating stress intensity factors using complianceThe relation between compliance and energy release rate can be used to determine energy release rates, and possibly also stress intensity factors, for structures whose rate of change of compliance with crack length can be easily determined. One example is the cantilever beam specimen shown belowThe deflection d of the loaded point must be twice that of a cantilever beam, length a , width B and height h , clamped on its right hand end and subjected to a load P at its left hand end. From elementary strength of materials theorywhere E is the Young’s modulus of the specimen. ThusThe energy release rate therefore follows asBy symmetry, the crack must be loaded in pure mode I. We can therefore deduce the stress intensity factor using the relationThus8.4.6 Integral expressions for energy flux to a crack tipIn this section we outline a way to compute the energy release rate for a crack, which applies not only to linear elastic solids under quasi-static loading conditions, but is completely independent of the constitutive response of the solid, and also applies underhBhPaddynamic loading (it is restricted to small strains, however). The approach will be to find an expression for the flux of energy through a cylindrical surface enclosing the crack tip, which moves with the crack. We will get the energy release rate by shrinking the surface down onto the crack tip.Energy flux across a surface in a solid: Consider anarbitrary surface S, which encloses some volume V in asolid. The surface need not necessarily be a material surface– it could move with respect to the solid. Let denote the displacement, (infinitesimal) strain and stress field in the solid, and let denote the velocity of a material point with respect to a fixed origin. Assume that the solid isfree of body forces, for simplicity. The energy flux across S can be calculated as follows:1. Define the rate of change of mechanical energy density at a point inside V aswhereis the rate of work done against stresses, andis the kinetic energy density. 2. Define the work flux vector. ThenTo see this, note thatwhere we have used the linear and angular momentum balance equations. 3. Now, integrate over the volume V and apply the divergence theorem to see that4. Finally, apply the Reynolds transport theorem to the term on the right hand sidewhere denotes the velocity of the surface . The term on the right hand side clearly represents the total rate of change of mechanical energy in V . Consequently, the term on the left hand side must represent the mechanical energy flux across . This is the result we need.Energy flux to a crack tip. We can use the energy flux integral to obtain an expression for the energy flux to a crack tip. Suppose the crack tip runs with steady speed v in the direction. Let denote a cylindrical surface enclosing the crack tip, which moves with the crack tip. The energy flux through follows aswherex 1x2mVSis the net work done on the solid per unit volume by stresses, and is the kinetic energy density. The energy flux to the crack tip follows by taking the limit as shrinks down onto the crack tip.Contour integral formula for energy release rate. To obtain an expression for the energy release rate, assume that the crack tip fields remain self-similar (i.e. an observer traveling with the crack tip sees a fixed state of strain and stress). In addition, assume that the crack front is straight, and has length L in direction perpendicular to the plane of the figure. Under these conditions , and . Consequentlywhere C is a contour enclosing the crack tip. (Equivalent results can be derived for general 3D cracks, but these details are omitted here).This result is valid for any material response (including plastic materials), and applies to both static and dynamic conditions.8.4.7 Rice’s J integralThe result derived in the preceding section becomes particularly useful if we make two further assumptions:1. Quasi-static loading2. The material is elasticIn this case T=0 and is simply the strain energy density in the solid - e.g. for a linear elastic solid with no thermal stress,The expression for energy flux through a surface surrounding the crack tip reduces tox 1x 2Γmx 1x2ΓmThis is the famous J integral . It has the following properties:1. The crack tip energy integral is path independent , as long as the material enclosedby the contour is homogeneous. There is no need then to shrink the contour down onto the crack tip – we get the same answer for any contour that encloses the crack tip.2. J=G for an elastic solid - so the contour integral gives an elegant way to calculate the crack tip energy release rate.Path independence of J : To show this, we first show that if the J integral is evaluated around any closed contour that does not enclose the crack tip, it is zero. To see this, applythe divergence theoremwhere A is the area enclosed by . To see that the integral on the right hand side is zero, note thatwhere we have used the equilibrium equation .Now, evaluate the integral around the closed contour shown on the right. Note that the integrand vanishes on and so thatNow reverse the direction of integration around (note that m = -n ) to getshowing that the integral is equal for any two contours that start and end on the two crack faces.8.4.8 Calculating energy release rates using the J integralThe J integral has many applications. In some cases it can be used to compute energy release rates. For example, consider the problem shown below. A long linear elastic cracked strip is clamped between rigid boundaries. The bottom boundary is held fixed; the top is displaced vertically by a distance . Calculate the energy release rate for the crack.x 1x 2m AΓx 1x2C 1C 3C 2C 4mmnFor this case J=G and we can easily evaluate the J integral around the contour shown. The integrand vanishes everywhere except the segment . On this segment it is easy to see thatwhile can be made arbitrarily small by taking far ahead of the crack tip. Therefore, evaluating the integral, we getSymmetry conditions show that the crack must be loaded in pure mode I, so the stress intensity factor can also be computed.x 1x2Γ1Γ2Γ3Γ4Γ5h/2h/2© A. F. Bower, Division of Engineering, Brown University, Providence RI 02912. These notes are intended for students at Brown University. You are welcome to print or reproduce these notes for individual study, but please seek the author’s permission before using them for any other purpose.。