Deformation kinetics
- 格式：pdf
- 大小：1.15 MB
- 文档页数：44
University of Cambridge Department of Materials Science and Metallurgy NATURAL SCIENCES TRIPOS Part III MATERIALS SCIENCE & METALLURGY M5 – Deformation kinetics W.J. Clegg Michaelmas Term 2009Aims and synopsis of the courseWhatever the primary function, it is generally the onset of flow and failure processes that limit the lifetime and application of a material. Such processes are usually treated in terms of constants, such as a flow or failure stresses, where time is unimportant. The aim of this course is to show that flow and cracking are rate-dependent processes and that by considering the kinetics, a much fuller picture of a material’s resistance to applied or internal forces can be obtained.Does the lattice give a resistance to dislocation motion?Concept of the lattice resistance. Estimating its magnitude. Atom misalignments around a dislocation. Relation of misalignment to interatom potential. The work required to move a dislocation. Peierls stress. Effect of crystal structure and bonding. How do the predictions compare with observations?c.c.p. metals, e.g. Cu, Ni, estimate of slip plane spacing and Burgers vector. Existence of other obstacles to glide: forest dislocations. Importance in materials with a low lattice resistance.Slip in other materials. Diamond (also Si, GaAs), glide and shuffle planes; TiC, TiN;b.c.c. metals, e.g. Fe. Overall comparison with observations.How is dislocation motion possible below the Peierls stress?Observation that flow can take place below Peierls stress. Deformation as a thermally activated process. The rate of flow and the dislocation velocity. Magnitude of the activation energy. Effect of τ on the activation energy. Importance of the magnitude of the energy barrier.How important is temperature when other obstacles control dislocation motion? Other obstacles to dislocation motion: forest dislocations. Comparison of magnitude of effect of forest dislocations and lattice resistance. Estimate of energy required to overcome forest dislocation obstacles. Comparison with observed behaviour in Fe and Ni. Effect of recovery on yield stress.How fast can a material deform?Regimes of defect movement. Flow at high rates, e.g. Cu. Phonon, electron drag of dislocations. Rate dependence of flow.Why does discontinuous flow occur?Yield drops and Lüders bands. Criteria for yield drop behaviour in crystals. Change in dislocation density during yielding, e.g. Cu. Effect of strain-rate exponent, e.g. Fe. Solutes. Lüders bands and forming, e.g. nylon.How can deformation occur without work hardening?Recovery, its importance in processes at high temperatures. Rate of recovery. Friedel equation, e.g. LiF.Glide and recovery together: steady-state dislocation creep. Predictions and observations of creep rates. Role of diffusion. Comparison with experimental observations.Why limit glide if diffusion is rate controlling?Dislocation glide in particle hardened systems. Effects of increasing temperature on glide. Creep-resistant alloys.Stability against failureInhomogeneous flow in solid materials. Superplasticity. The onset of tensile failure and necking. Observations of flow in superplastic alloys. Interacting flow processes. Mechanisms of superplastic flow. Superplastic materials and forming, e.g. Ti, SiC. Dynamic crackingThe importance of dynamic cracking. The Mott extension to Griffith. Fracture surfaces: Mirror, mist and hackle. The need to dissipate energy. The importance of kinetic energy of the sample. The limits.Reading listYou might like to read chapter 2 of Deformation Mechanism Maps by Frost and Ashby gives a good overview of different deformation processes, including the lattice resistance. You should also read chapter 17, which is concerned with how materials fall into groups with similar properties. Another book to skim through is J.J. Gilman, Electronic Basis of the Strength of Materials. This covers elastic more than plastic behaviour, but you might like to read from Chapter 16 on, although chapter 15 is also interesting (and short). By far the best book on fracture is B.R. Lawn, Fracture in Brittle Solids.Why are some materials hard and others very weak?Estimating the changes in misfit energy of a moving dislocation Basic approachTo estimate the misfit energy, we must find(i) the atom positions to give the spacings between atom pairs (andthe difference from the equilibrium values),(ii) the potentials of each atom pair.The potentials of each atom pair are then summed to give the overall misfit energy. We only want the changes in misfit energy. This allows a great simplification, namely that we need only consider the atom planes on either side of the slip-plane.Atom positions around a dislocation in an unstressed crystalFigure 1 The atom positions in (a) two perfect crystals before bonding and (b) after the two crystals have been joined and the atoms relaxed formingthe dislocation. The grey lines in (b) give the original atom positions.To make an edge dislocation, take two undistorted crystals of the same material. The crystals have a repeat distance b in the horizontal direction and d in the vertical direction and the upper crystal has one more crystal plane than the lower one.The initial atom positions, figure 1a, in plane A , x A , arenb x =A (1a) In plane B , x B 2B b nb x m = (1b) where n is an integer denoting the plane, and the plane numbered 0 is the plane thatbecomes the edge dislocation when the crystals are joined, figure 1b. (Note in eqn 1bthat b /2 is subtracted for n > 0 and added for n < 0.)Looking at figure 1b, it can be seen that the displacements of the atoms on the A -plane are symmetrical on either side of the dislocation line and that the atom at thedislocation core is not move when the crystals are joined, so0)( 0==x uFigure 2 The variation of the displacement u with distance from the dislocationline.Far from the dislocation, the atoms will be aligned, so that ()()−∞=+=+∞=−=x b u x b u 44 So the atom displacements (from their positions in figure 1a) decrease in magnitudefrom ±b /4 at the ends of the sample to zero at the dislocation, see figure 2, suggestingthat u might vary with tan -1x , givingwx k x u A 1A tan )(−−= (2a) where k and w are numerical constants. When x A,B = ±w , |u | = b /8, that is one-half ofits maximum value, so w is called the half-width of the dislocation .A similar argument applies to the atoms on theB -plane except that the direction of thedisplacements is different, giving wx k x u B 1B tan )(−+= (2b) The final position of the atoms, x ′A , x ′B , is then given by the sum of the initialposition, x A , x B , and the displacements, u A , u B , so for the atoms on the A -planeA A Au x x +=′ (3) From the final positions the in-plane spacings and misalignments between any pair ofatoms can be estimated. For instance the in-plane spacing of the atoms labelled (A,1)and (A,2) in the A-plane, figure 1b, ∆x ′(A,1;A,2), is)1A,()2A,()2A,;1A,(x x x ′−′=′∆For the lattice shown in figure 1 the atoms across the slip-plane and in the samevertical plane are aligned when their x -co-ordinates are equal, so that themisalignment of, say, the atoms in plane 3, ϕ(A,3;B,3), is)3B,()3A,()3B,;3A,(x x ′−′=ϕWe only need k and w to obtain the final atom positions. Using the values of u at x = 0and ∞, gives k = b /2π. Only the value of w is now required to completely describe theatom positions around the dislocation.How wide is a dislocation?The dislocation half-width, w , can be determined by finding the value of w thatminimises the misfit energy.To do this, we need functions that describe the contributions to the misfit energy from(i) the in-plane separations between atoms in the same plane (parallel to the slip-plane) and (ii) the misalignment of atoms across the slip-plane.Misfit energy due to the in-plane strainsFor simplicity assume that the bonds are linear elastic. If a line of pairs of atoms iscompressed or stretched from its equilibrium value in a direction parallel to the slip-plane, the change in potential for a single line of atom pairs of unit length, U i,n , isgiven by ) ()1(212i 2i,d b E U n εν−= (4) where the in-plane strain associated with a given atom pair is εi = δ/b and δ is thedisplacement of the bond-length from its equilibrium value (in either tension orcompression). These can be obtained from the final atom positions.Taking the material to be isotropic and setting E = 2G (1+ν), gives U i,n as 22i, )1()1(δ⎟⎠⎞⎜⎝⎛ν−ν+=b d G U n (5)Misfit energy due to the misalignments between atoms across the slip-planeAny function must vary with a period of b and give Hooke’s law for smalldisplacements. One function is that used by Frenkel in his estimate of the theoreticalshear strength of a solid (see Pt IA M&MS). He assumed that the potential per unitarea, U A , when two planes of atoms each of unit area are displaced with respect toone another by an amount ϕ is given by()⎥⎦⎤⎢⎣⎡πϕ−π=ϕb d Gb U 2cos 1422A (6) In our case the atoms are not displaced uniformly over a whole plane. We have linesof atoms being displaced, with each line being displaced by an amount which dependson its distance from the dislocation. The energy due to the misalignment of a givenpair of atom lines each of unit length , U n , is therefore ⎦⎤⎢⎣⎡πϕ−⎟⎠⎞⎜⎝⎛π=b d b Gb U n 2cos 1 822m, (7) where φ is the misalignment between the lines of atoms (calculated from the positionsof adjacent atoms across the slip plane). Here we count the misfit associated with themisalignments between an atom and both its nearest neighbour across the slip-planeand its next-nearest neighbour. This is why the numerical constant in eqns 6 and 7differs by a factor of 2.Total misfit energyThe misfit energy, ∆U T , of the dislocation is the sum of the potentials due to the in-plane strains and to the misalignments for a chosen number of atoms pairs, that is∑∑+=n n U U U m,i,T00.20.40.60.81012345U T , total misfit energyU m , misalignment energyU i , in-plane strain energyU /G b 2w/b U mU iFigure 3 The variation of the misfit energy with dislocation half-width, w , in acrystal lattice where d/b = 1 and ν = 0.2. For a very wide dislocation, e.g. w /b = ∞, the only contribution to the misfit energy isfrom the misalignment of atoms across the slip plane. If w /b is allowed to decrease,the energy due to the misalignment of atoms across the slip-plane, U m , decreases , butthe in-plane strain energy, U i , increases. This gives a minimum misfit energy at agiven value of w /b .Changes in misfit energy as the dislocation movesNow we need to estimate the energy changes as the dislocation moves from the lowenergy position.If the dislocation moves a distance αb (with respect to the lattice) to the right, then theinitial positions of the atoms, x A and x B , are now given byb nb x α−=A (8a) and in plane B b b nb x α−=2B m (8b) The final positions of the atoms on planes A and B can now be calculated, as before,from the displacements given by eqns 2 using the appropriate values of x A and x B forany given value of α. We can then calculate the in-plane separations andmisalignments and the resulting potentials in the same way as before, setting α tovalues between 0 and 1.For simplicity we shall assume that w remains constant as the dislocation moves. Wecan use the method here to check this, by finding the value of w that gives a minimummisfit energy at each different value of α. The variation is extremely small and weshall ignore it here. What do we mean by the dislocation?Normally we take the edge dislocation to be the extra half-plane of atoms, but we arenearly always thinking of the case in figure 1b, that is where α = 0. However whenthe dislocation has moved through b , that is α = 1, what was initially the extra half-plane of atoms will be the plane adjacent to the new half-plane. The dislocation willhave moved about twice as far as the extra half-plane.The dislocation is the origin from which all the displacements are calculated usingeqns 2. In the analysis above, the dislocation remains fixed (at the origin) and thelattice moves around it.The overall energy changesWe can now calculate the misfit energy at each value of α, and so determine thechange in misfit energy, ∆U T (α), as the dislocation moves.-2-112345-0.5-0.2500.250.5∆U /G b 2 (x 105)αFigure 4 Energy changes as the dislocation moves through a crystal with d /b = 1and ν = 0.2. All the calculations here have been done for 1,000 atompairs on either side of the dislocation line. Increasing the number ofatom pairs gives only a negligible change to the results. The total misfit energy varies sinusoidally and has a minimum at α = 0, likewise themisfit energy associated with the in-plane strains, ∆U i . However the misfit energyassociated with the misalignments, ∆U m , is greatest at the positions of lowest overallenergy. The period is b /2, not b as we had expected. This gives ∆U T (α) as(πα+∆=α∆4sin 1 21)(p T U U ) (9) where ∆U p is the maximum change in the total misfit energy, ∆U T (α). ∆U p iscommonly known as the Peierls energy . It is the energy required to move unit lengthof dislocation line over the resistance of the lattice.What determines the magnitude of ∆U p ?We know that the total misfit energy, ∆U T , scales with Gb 2. As we might expect, thechanges in misfit energy do too. The total misfit energy also varies with dislocationwidth so we expect the changes in misfit energy to depend on the dislocation widthtoo.10-1010-810-610-410-200.51 1.52ν = 0ν = 0.2ν = 0.3∆U p /G b 2w/bFigure 5 Effect of the dislocation width, expressed as w /b , on the Peierls energy,expressed as ∆U p /Gb 2. Note that the Peierls energy variesexponentially with w /b . Repeating our estimates for different values of w /b , shows that the Peierls energy,∆U p , is very strongly dependent on w /b , see figure 5, which shows the predictions formaterials with Poisson ratios of 0, 0.2 and 0.3. Figure 5 gives ∆U p as⎟⎠⎞⎜⎝⎛−=∆b w k k Gb U 212p exp (10) k 1 varies slightly with Poisson ratio and can be written approximately as)1( 811ν−π≈k whereas k 2 is given approximately byπ≈ 45.32kStress to move a dislocation through a crystal, Peierls stressThe force required to move the dislocation is()()b U F α∂α∆∂=T giving the stress, τ, required to move the dislocation by αb as ()α∂α∆∂=τT 21U b(11) Substituting for ∆U T (α) from eqn 9 gives the maximum stress required to move thedislocation against the lattice friction, τp , from one low energy position to the next, ⎟⎟⎠⎞⎜⎜⎝⎛∆π=2p p 2τGb U G (12) and substituting for ∆U p from eqn 10, gives ()⎟⎠⎞⎜⎝⎛π−ν−≈τb w G 45.3 exp 1 41p(13) τp is called the Peierls stress after Rudolph Peierls, who made the first estimate of thelattice resistance in 1940. Dislocations had not yet been observed and Peierls assumedthat the dislocation was sufficiently wide that continuum elasticity could be used todescribe the in-plane strains. This means they do not change as the dislocation movesand therefore do not contribute to the Peierls energy or stress. The only change istherefore due to the misalignment energies, U i , which are a maximum at α = 0.In the estimate here we have replaced the continuum treatment with a (rather simple)atomistic one. Despite this the general form of the expression is the same as thatderived by Peierls and others, but has a slightly lower magnitude.What determines w /b ?w /b was fixed by the energies associated with (i) the in-plane separations and (ii) themisalignments of atoms across the slip-plane. Both of these depend not only on G andb but also on the ratio d /b .Take two crystals with the same elastic constants, G and ν, and Burgers vectors, b .One has d /b = 1, the other has d /b = 0.5. We have already seen that decreasing d /bwill decrease the in-plane strain energy, U i , see eqn 5, but increase the misalignmentenergy, U m , see eqn 7. (Note these expressions are for single atom pairs.)00.20.40.60.81012345U T , d/b = 0.5U mU iU T , d/b = 1.0U mU i U /G b 2w/b d/b = 0.5d/b = 1.0Figure 6 The misfit energies for edge dislocations in two crystals with thevalues of G , ν and b , but where one has d /b = 1 the other 0.5. As d /b isdecreased U m increases for a given w /b , whilst U i decreases, causingthe width at which the minimum misfit energy occurs to diminish. ν istaken as 0.2.The effect is geometrical. If d /b decreases (b remaining constant), there will be morebonds per unit area in a plane transverse to the slip-plane. The stiffness of each bondparallel to the slip-plane must decrease with d /b if the crystals are to have the sameelastic constants. It is this reduction in E that causes a reduction in the in-plane strainenergy, U i , as d /b is decreased.The increase in U m with decreasing d /b occurs because the shear strain associatedwith the misalignment increases, approximately as the square of d /b , whilst thevolume decreases only linearly. The overall effect is to increase U m approximatelylinearly for a decrease in d /b . Figure 6 shows that for a given w /b this is the case.00.20.40.60.811.21.400.51 1.5ν = 0ν = 0.2ν = 0.3w /b d/bFigure 7 The variation of the dislocation half-width expressed as w /b with d /b . Decreasing d /b reduces w /b . Our estimates show that the dislocation half-width isrelated to the ratio d /b (the ratio of the atom spacing normal to the slip-plane with thatparallel to it) as shown in figure 7. The variation is approximately linear. However asw /b is in the exponential term in eqns 10 and 13, changing d /b can vary the latticeresistance, expresses as a stress or an energy, by several orders of magnitude.Copper (and other c.c.p. metals)Diamond (or silicon or germanium)Titanium carbideb.c.c. crystals, e.g iron, tungsten, potassium and solid heliumLattice resistance–comparison with observationsThe expression for the Peierls stress can be compared with experiments by plotting log (τp /G ) against (d /b ). Using measurements of the Peierls stress as well as the relevant shear moduli and crystal structure data, the graph below is obtained.10-610-410-2100.00.51.0 1.5τP/Gd/bα-quartzα-alumina Diamond, SiOlivine CdTeCuBrGaAsMgO, (hard)TiC NiAl K Nb α-Fe,Ta α-TiAlNiCu NaClMgO (soft)LiF KI, KBrKClWInP,InSbHgSe Mo Hescrewdislocationν = 0.2ν = 0.3Effect of crystal structure and bondingThe variables influencing the Peierls stress:(i) Crystal structure:•Ratio of the atom spacing across the slip plane to theBurgers vector (d/b)•Burgers vector, b(ii) Elastic properties:•Shear modulus, G•Poisson ratio, νTemperature dependence of the lattice resistanceMeasurements of the flow stress in materials where the flow stress is dominated by the lattice resistance show that above a certain temperature the flow stress decreasesrapidly with temperature.10-510-410-310-210-1050010001500F l o w s t r e s s /GTemperature (K)The rate of plastic flowWe have assumed that a material’s resistance to dislocation motion can be completely characterised by a flow stress, τY , without considering the deformation rate.Consider a cubic crystal of side-length h with slip occurring on a plane parallel to oneset of the cube faces.When a single dislocation with a Burgers vector b moves across the whole slip-plane, it produces a displacement of b and a strain, γSD , equal tohb SD =γ If N dislocations move across the whole slip plane, the strain, γ, ishNb =γ If the dislocations are moving at a velocity v so that they travel across the full-widthof the crystal in a time t , then the strain-rate is given by γ& v v //3b h Nh h h Nb t×⎥⎦⎤⎢⎣⎡==γ=γ& where N h /h 3is the length of dislocation line that is moving per unit volume, knownas the mobile dislocation density, ρm . Substituting for ρm givesv m b ρ=γ&Dislocation glide and thermal activationWhen τ = τp all of the energy required to overcome the lattice resistance is supplied by the applied stress. If τ is less than this, flow can occur if the extra energy required is supplied by thermal activation. Treating dislocation motion as we would any other thermally activated reaction, then below τp the velocity, v, is()kTG b τ∆−×ν×=exp v AHow does ∆G vary with τ?When τ = 0Now apply a shear stressAs dislocation segment moves from 1 to 2, work, W, is done by the applied stressNeglecting movement of the dislocation in the backward direction, giveskTVF b τ−∆−νρ=γexp A 2m & which giveskT V F b kTVF b τ−∆−=⎥⎥⎦⎤⎢⎢⎣⎡νργτ−∆−=νργA 2m A 2m ln exp )(&&⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧⎥⎥⎦⎤⎢⎢⎣⎡νργ+∆=τA 2m ln 1b kT F V&00.0010.0020.0030.0040.0050.00601002003004005τ/GT (K)00TaFeChange in flow stress with temperature for Ta and Fe, from T. Suzuki et al, “Plastic homology of BCC metals”, Philos. Mag. A , 79[7] (1999) 1629-42.How fast can dislocations move?Effect of strain-rate0.0010.1101000200300400500S t r a i n -r a t e , s-1Stress, MPaTaken from figure 2, the effect of strain-rate on the flow stress of OFHC copper prestrained by varying amounts. Note that m is high at lower strain-rates and then decreases above some critical value.0500100015002000200300400500S t r a i n -r a t e , s-1Stress, MPaTaken from figure 3, the linear variation of stress with strain-rate above the critical stress, τc .Effect of temperature5001000150020002500200300400500S t r a i n -r a t e , s-1Stress, MPaTaken from figure 7, the linear variation of the flow stress on the strain-rate at temperatures of 300 K, 420 K and 590 K. Note that the slope of the line decreases as the temperature increases, indicating that the drag coefficient, B , decreases, assuming a constant mobile dislocation density.See: A. Kumar and R.G. Kumble, “Viscous drag on dislocations at high strain rates in copper”, Journal of Applied Physics , 40[9] (1969) 3475-3480.Why do we get yield drops?Effect of ρU and m on the magnitude of the yield drop123410610710810910101011τU/τLρU, m-2m = 1m = 5m = 15m = 35m = 100(Assumes ρm remains constant.)Yield drop in steel1020020*******% c h a n g e i n f l o w s t r e s sTemperature, KSee P.J. Worthington, Acta metallurgica , 15 (1967) 1795.Yield drop in Mo20406080010*******% c h a n g e i n f l o w s t r e s sTemperature, KI.O. Hall, Yield Point Phenomena in Metals and Alloys , Plenum Press, New York, 1970.Effect of solutes on temperature dependence of the flow stress10203040500200400600F l o w s t r e s s , M P aTemperature, KMitchell, Progress in Applied Materials Research , 6 (1964).How can a material keep deforming?Dislocation density on annealing LiF crystal after a 1% prestrain0.00.51.01.52.0010203041/ρ x 10-10 , m2time, hoursTaken from J.C.M. Li, in Recrystallization, grain growth and texture , ASM, Metals Park, Ohio and Chapman and Hall, London, p 45.Variation of creep stress exponent, n , with the Dorn constant, A024681012141612345678910l o g Acreep stress exponent, nFrom the compilation of data by H.J. Frost and M.F. Ashby, Deformation Mechanism Maps , Pergamon Press, Oxford, 1982.Superplastic formingFigure 1 A cross-section through a Hollow Wide Chord Fan Blade. The blade is made up of 3 layers of a superplastic Ti alloy. The inner layer is screenprinted with stripes of Y2O3 preventing the inner Ti sheet bonding tothe outer layers on diffusion bonding. The blade is then hot twistedinto the outer shape before blowing to give the inner sandwichstructure. (Courtesy Rolls-Royce Ltd.)Figure 2 The bonded joint and the ligaments that are superplastically formed.(Courtesy Rolls-Royce Ltd.)Estimating the failure strain of a superplastic material0.0010.010.111101001,00010,000s t r a i n -r a t e s e n s i t i v i t y , mElongation, %The variation in elongation of materials with different strain-rate sensitivities. We have shown in lectures that the large failure strains observed in superplastic materials are associated with the stability of any necks that form in the material, perhaps at regions in the material that are initially thinner. We have also shown that the rate of reduction of cross-sectional area, d A /d t , is given by()m m mAK P t A −⋅⎟⎠⎞⎜⎝⎛−=1/11d d (1) where P is the aplied load, K is a constant in a given test, A is the cross-sectional area at any given time and m is the strain-rate sensitivity. The time taken for the cross-sectional area of the sample to decrease from A o to A can be obtained by integrating this, so thatA At K P mm AA mtt d d -11oo⋅=⋅⎟⎠⎞⎜⎝⎛−∫∫ (2)If failure occurs by the necking of some region that is initially thinner than A , say aA o , where a < 1, then when the sample is stretched so that the overall cross-sectional area is reduced from A o to A , the cross-sectional area of this thinner region will have decreased from aA o to bA o in the same time, so thatA At K Pmm bAaA mtt d d -11oo⋅=⋅⎟⎠⎞⎜⎝⎛−∫∫ (3)Integrating gives,bAaAm A A m AA m m mA mA A m m oo o][][11111)11(==⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡⋅⎟⎠⎞⎜⎝⎛−+−+ (4) Som m mm aA bA A A 1o 11o 1)()(−=− (5)From eqn 5 an expression for A o /A can be obtained in terms of a , b and m . Assuming that the volume of the sample remains constant during deformation, that is Al = A o l o , then1)1()1(1/11o o o −⎥⎥⎦⎤⎢⎢⎣⎡−−=⎟⎠⎞⎜⎝⎛−=−mm ma b A A l l l (6)A criterion for failure is that b = 0, that is the cross-sectional area of the locally thinned region is 0, giving the failure strain, expressed as the change in length divided by the original length, εf , as1)1(1f −−=ε−m m a (7)Kinetic crack growthMott extension of GriffithConsider a sheet of infinite size under an applied stress σa and containing an internal crack of length 2c .Then,)(2K 2a2c U Rc Ec U ++σπ−= (1)To find U K , we need expressions for U and R . At equilibrium, d U /d c = 0 and c = c o , soR Ec 2202ao +σπ−= (2)M5 Deformation kinetics WJC To obtain the constant U , use the condition the kinetic energy in the sample is zero,i.e. U K (c ) = 0, so o 2a 2o 2Rc Ec U +σπ−= (3) Substituting for R from eqn 2, gives U as Ec U 2a 2o σπ= (4) As the crack grows from c o to c , energy is dissipated driving new crack face, the crackdriving force required to keep the crack growing diminishes. The excess workbecomes kinetic energy of the sample. Rearranging eqn 1 and substituting for R and Ugives U K (c ) as 2o 2a 21)(⎥⎦⎤⎢⎣⎡−σπ=c c E c c U K。
