Adiabatic equation for cracks. Department of Physics and Center for Nonlinear Dynamics, The University of Texas at Austin, Austin, Texas 78712, USA

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1 PHILOSOPHICAL MAGAZINE B, 1998, VOL. 78, NO., 03± 14 Adiabatic equation for cracks By Michael Marder Department of Physics and Center for Nonlinear Dynamics, The University of Texas at Austin, Austin, Texas 7871, USA Abstract This article describes dynamical equations for cracks in strips. After rst nding the stress and strain elds around a crack moving at constant velocity, the equations are then extended to cracks which accelerate very slowly. Assuming that the motion of cracks is restricted to a line, these calculations show them to be stable, unless the fracture energy that they require to advance is a decreasing function of their velocity. 1. Introduction The purpose of this article is to record some calculations concerning motion of cracks in strips, as shown in gure 1. The calculations are in terms of continuum linear elasticity, and were performed as part of an analytical search for instabilities that had been observed experimentally in the motion of fast cracks (Fineberg et al. 199). A good approximate expression summarizing the results of the calculations is C (v) < ( c l ). (1) G 1 b Çv 1 [1 (v/c R ) ] Here C (v) is the fracture energy that a crack devours per length, Çv is the acceleration, G is the energy stored per unit length in the strip far ahead of the crack, b is the half the height of the strip, c l is the longitudinal wave speed and c R is the speed of the Rayleigh waves. The predictions of equation (1) depend upon how C (v) behaves with velocity. (1) If C (v) increases with increasing v, then the crack will speed up until the energy G available to it per unit length stored in the strip equals C (v). () If C is independent of v, then, when G= C, the crack can travel in steady state at any speed while when G>C, the crack will accelerate perpetually towards the Rayleigh wave speed, storing more and more energy in singular fields around its tip. (3) If C is a decreasing function of v, the crack will rapidly accelerate until it arrives at velocities where C begins to increase again. There are some quasibrittle polymers where fracture energy decreases with increasing velocity over a limited range. However, the more usual case is that fracture energy is a monotonically increasing function of crack velocity; this was clearly the case in the experiments where crack tip instabilities were observed (Irwin et al. 1979). The continuum calculations insisted that a crack travelling along a line should be stable in this case. One possibility was that the crack could not be treated as a branch cutting moving along a line, and this argument has been pursued at length by 0141± 8637/98 $1.00 Ñ 1998 Taylor & Francis Ltd.

2 04 M. Marder Figure 1. A semiin nite crack travel at velocity v in a strip of halfwidth b. Ching et al. (1996a,b,c). I followed a more drastic path and abandoned continuum analyses altogether. In the course of trying to advance beyond the unexpectedly stable continuum equations, the results were never properly reported. The following pages ll in most of the details omitted in a previous publication (Marder 1991). Equation (1) has not yet been tested experimentally, since a sample needs to be gripped with great precision to separate e ects of adiabatic acceleration from variations in boundary conditions.. Steadystate crack in a strip Consider a crack moving at steady velocity v in a strip whose top and bottom edges are rigidly held at height 1, and with a constant stress s applied to the crack faces, causing them to open by an amount d = s /( + ¹), where and ¹ are Lame  constants for the elastic strip. The boundary conditions for crack motion are u x (x,1) = 0, () u y (x,1) = 0, (3) s xy(x,0) = 0, (4) u y (x,0) = 0, for x >0, (5) s yy(x,0) = s, for x <0. (6) The dynamical equation for the strain eld u of a steady state in the moving frame is ( + ¹)Ñ (Ñ u) + ¹ Ñ u = q v u x. (7) Divide u into transverse and longitudinal parts so that u = u t + u l, (8) with u l = Ñ u l, u t = It follows immediately that ( + ¹)Ñ q v ( x ) u l + ¹Ñ u t (, u t y x ). (9) q v ( x ) u t = 0. (10)

3 a a a Taking a dot product with the operators ( / x, / y) and ( / y, / x) in turn gives where x + ( y ) Ñ b x + ( y ) Ñ = 1 b = 1 q v Therefore the general form of the potentials is u l = 0, (11) u t = 0, (1) q v + ¹, (13) ¹. (14) u l = u 0 l (z) + u 0 l (z) + u 1 l (x + ia y) + u 1 l (x + ia y), (15) u t = u 0 t (z) + u 0 t (z) + u 1 t (x + ib y) + u 1 t (x + ib y), (16) where all functions on the righthand sides of equations (15) and (16) are harmonic, and z = x + iy. In fact, the purely harmonic pieces u 0 l and u 0 t disappear entirely from the expressions for u. They result from a freedom one has to add a harmonic function to u l and u t simultaneously, and can be neglected. Thus, to nd the potentials u that give displacements u through equation (9), one has to solve b u l x + u l y = 0, (17) u t x + u t y = 0. (18) Rather than working in a strip of height b, more compact expressions can be obtained by setting b to unity. There is no di culty in scaling distances by b at the end. In a strip of unit halfheight, it is convenient to use the solutions of equations (17) and (18): u l = A sl sinh(a ky) + A cl cosh(a ky), (19) u t = A st sinh(b ky) + A ct cosh(b ky). (0) In terms of the kdependent coe cients A, one has Adiabatic equation for cracks 05 u x = A ct b ksinh(b ky) + A st b kcosh(b ky) ik [ A sl sinh(a ky) + A cl cosh(a ky)], (1) u y = ik [ A st sinh(b ky) + A ct cosh(b ky)] + a A cl k sinh(a ky) + a A sl kcosh(a ky), () s xy = k ¹ {Ast (b + 1) sinh(b ky) + A ct (b + 1) cosh(b ky) ia [A cl sinh(a ky) + A sl cosh(a ky)]}, (3)

4 s 06 M. Marder s xx = ia ct b k ¹ sinh(b ky) ia st b k ¹ cosh(b ky) + [ A sl b (a + [ A cl b (a + 1)A sl ]k ¹ sinh(a ky) s yy = ia ct b k ¹ sinh(b ky) + ia st b k ¹ cosh(b ky) + 1)A cl ]k ¹ cosh(a ky), (4) + (A sl b + A sl )k ¹ sinh(a ky) + (A cl b + A cl )k ¹ cosh (a ky). (5) Three of the coe cients A cl,...,a st can be found from the boundary conditions which apply to all x, and hence all k: u x (k,1) = u y (k,1) = 0, (6) s xy(k,0) = 0. (7) In order to specify all four of these constants in terms of a quantity with more physical content, de ne Then with the abbreviated notation with u y (k,0) º u 0 y(k). (8) s a = sinh(a k), sb = sinh(b k), c a = cosh(a k), cb = cosh(b k), (9) A st = i(a b s a sb c a cb + b + 1)u 0 y(k) B(k), (30) A cl = [(b + 1)s a sb (a b 3 + a b )c a cb + a b ]u 0 y(k) a B(k), (31) A ct = iu0 y(k) (b 1)k, (3) A sl = (b + 1)u 0 y(k) (a b a )k, (33) B(k) = (b 1)k(c a sb a b s a cb ). (34) One now uses the Wiener± Hopf trick. De ne 0 + yy = s yy(z) exp(ikz) ò dz, s yy = s yy(z) exp(ikz) 0 ò dz, (35) with u + y and u ȳ de ned similarly. Note that s yy has no poles in the lower halfplane, + and s yy no poles in the upper halfplane. Similarly u + y has no poles in the upper halfplane, and u ȳ has no poles in the lower halfplane. Write s yy(k) = s + yy + s yy = s + yy s, (36) ik u y (k) = u ȳ. (37)

5 gives Then de ning F(k) = s yy u y (38) F(k)u ȳ = s + yy s ik. (39) The point of de ning F(k) lies in the fact that, since it is a ratio of two quantities expressible in terms of the A cl,...,a st, the unknown function u 0 y(k) which appears in all these cancels out. The function F(k) is F(k) = k¹ s a sb [ (b + 1) + 4(a b ) ] c a cb a b [ (b + 1) + 4 ] + 4a b (b + 1) a (b 1)(a b s a cb c a sb ) Suppose that F(k) can be written as Adiabatic equation for cracks 07 (40) F(k) = F (k) F + (k), (41) where F has no poles in the lower halfplane, and F + has no poles in the upper halfplane. Then kf + (k)u ȳ = ks yyf + (k) + is F + (k). (4) One has set equal an expression with no poles in the upper half plane to one with no poles in the lower halfplane. Therefore, both must equal a constant. The constant can be xed by examining the behaviour of equation (4) for k 0. Note that lim u d y. k 0 ik ; (43) this statement follows from the fact that u y vanishes for large positive x and goes to d for large negative x. So u 0 y(k) = u (k,0) ȳ = u y (k,0) = d F (0) ikf (44) (k). The problem is now solved, apart from the di culties of decomposing F into F + and F. In fact most quantities of physical interest can be obtained from integrals over F itself, without the need ever to compute the decompositions. 3. Adiabatically acceler ating planestress cracks Consider next a crack in a strip which accelerates slowly. The location of the crack tip is given by l(t), and the dimensionless parameter indicating when acceleration is su ciently slow is bèl c! 1, (45) t where b is the half width of the strip. The velocity of the crack is required to change slowly over the time scale in which sound communicates with the boundaries. Both powers of acceleration higher than the rst and higherorder time derivatives will be discarded in the calculation that follows. Then to leading order the potentials for the crack are given by u a(x,y,t) = u s a (x l(t),y,v), (46)

6 08 M. Marder where u s a is the steadystate potential appropriate for a crack moving at constant velocity v = Çl, and a = l,t can give either the longitudinal or the transverse potential. However, to be consistent to order Çv = Èl, for an accelerating crack, one must write u a(x,y,t) = u s a (x l(t),y,v) + ÇvD u a(x l(t),y,v). (47) In the accelerating frame of reference x  = x l(t) one has t x = t xâ v(t) xâ so the second time derivative of the potentials is given by ( ) t xâ v(t) xâ = u a t v u a t Çv xâ u a xâ ; (48) u a (49) + v u a xâ. (50) In the accelerating frame, the potentials depend upon time only through their dependence upon v(t). Therefore t Çv v. (51) Inserting equation (46) into equation (50) and working only to lowest order in Çv gives u a t x = v [ xâ u s a (x Â,y,v) + D u a(x,y,v)] ( ) u s a (x Â,y,v) + O b Çv c t Çv v v xâ + xâ ( ). (5) We now are ready to write the wave equation to order Çv. Fourier transforming so that xâ k, one nds that 1 c v k s (u a + D u a) + ikv Çv a v u s s a + ik ( Çvu a = ) y k ( ) (u s a + D u a). (53) Since u s a (k,y,v) is de ned by the fact that it obeys the wave equation for each velocity v, one can simplify this last expression to read y k [ ( 1 ) ] v c D u a a = ik Çv c a v ( v ) + 1 u s a. (54)

7 Adiabatic equation for cracks 09 One may construct a solution of this inhomogeneous equation as follows: consider y k [ ( 1 ) ] v c u s a = 0, (55) a y k [ ( 1 ) ] v c u s a = 0, (56) a y k [ ( 1 ) ] v c a v u s a = k c a ( v v ) + 1 u s a. (57) v Comparison of this last equation with equation (54) shows that D u i a = 1 ik v u s a (58) is a solution of the inhomogeneous equation (54). The complete solution of the problem D u a = D u i a + D u h a (59) is obtained by adding a function which satis es the homogeneous wave equation on the lefthand side of equation (54) so as to bring the result into accord with the boundary conditions. The boundary conditions must be written fairly carefully. Let the sample extend from L to L in the laboratory frame, and choose a small convergence factor ² such that ²L! 1. The boundary conditions in the accelerating frame are u x (x Â,1) = 0, (60) u y (x Â,1) = 0, (61) s xy(x Â,0) = 0, (6) u y (xâ,0) = 0, for xâ >0, (63) s yy(x Â,0) = s exp(²x  ), f or x  <0. (64) The inclusion of ² makes all the Fourier transforms well de ned but causes the elds to decay negligibly within the physical boundaries of the sample. All elds u and s can be obtained from the potentials u a, by action with linear operators. Therefore, the elds u and s may be written in the form u a = u s a + ÇvD u a, (65) s a b = s s a b + ÇvD s a b, (66) where the rst term is the steadystate result appropriate for velocity v, and the second is derived from the D u a. Since the steadystate elds already obey the

8 10 M. Marder boundary conditions (60) through equation (64), the perturbations must obey D u x (x Â,1) = 0, (67) D u y (x Â,1) = 0, (68) d s xy(x Â,0) = 0, (69) D u y (xâ,0) = 0, for xâ >0, (70) d s yy(x Â,0) = 0, for x  <0. (71) Since the inhomogeneous solution (58) is formed from the steadystate solution by the action of a linear operator, one can nd all the elds D u i i a, D s a b which result from the inhomogeneous solution merely by acting on the steady state solutions with the same operator: D u i a = 1 ik v us a, (7) D s i a b = 1 ik v s s a b. (73) These elds are close to being the needed solution. The steadystate solutions satisfy all the boundary conditions and it is easy to verify that the solutions (73) satisfy equations (67)± (69). However, they do not obey equation (70) or (71); so one needs to add some solution of the homogeneous problem. Before doing so, recall the solution of the steadystate problem, but taking care with ². After transforming the boundary conditions into Fourier space one nds that for elds, as a function of k and evaluated at y = 0, F(k)u s y = s ik + + ² s s+ yy, (74) F (k)u s y + F+ (i²)s ik + ² = F+ (k) F + (i²) ik + ² ( s ) + F + (k)s s yy. (75) The lefthand side is free of poles in the lower halfplane; the righthand side is free of poles in the upper half plane; so the two must equal a constant. Checking the asymptotic behaviour of either side as k 0 one nds the constant to be zero. So u s y = u s y = F+ (i²)s (ik + ²)F (k), (76) = d F (i²) ik + ² F (k), (77) to order ². With this background we now return to the problem of nding D u y. As before, we evaluate all functions at y = 0. We have that D u y = D u i y + D u h y. (78) Since elds with superscript h obey the homogeneous equations we must have that F(k) D u h y = F(k) D u h h+ h y = D s yy = D s yy. (79) Thus we nd immediately that F(k)k(D u ȳ D u i y ) = k(d s + yy D s i+ yy). (80)

9 d Multiplication by k on both sides is necessary to get rid of the pole at k = 0 which appears in equation (73). As before, we nd that kf (k)d u ȳ kf (k)d u i y = C, (81) D u ȳ = D u i y + C kf (k). (8) Using equations (77) and (73), one has that D u y = 1 d F (i²) ik v ik + ² F (k) + C kf (k). (83) To nd the constant C, one can look at the behaviour of D u y as k 0. There can be no pole there; so lim D u 1 d F (i²) y = lim k 0 k 0 ( ik v (²) F (0) + C kf (0)). (84) C = d F (i²) i v ² Since F (0) is independent of v, one can write to order ² that with ( ) = F (0) F (i²) v ² H(v) = i k lnf (k) ( ). (85) v H(v), (86) k=0. (87) The general rule concerning ² is that it must be kept in any term such that for some value of k the term can become in nite. Finally we have that where D u y (k,0) = 1 ik M = 1 ik v us y(k,0) F (0) F (k) ik v H(v), (88) = M us y(k,0), (89) = 1 ik F (0) v F ik + ² (i²) ik ( v ) H(v) (90) ik + ² v ik ( v ) H(v) (91) to relevant order in ². In general, one can nd any eld by application of the operator M to the appropriate steadystate eld. For example, s yy(k,y) = s gives the full stress eld to rst order in acceleration. Adiabatic equation for cracks 11 s yy(k,y) + ÇvM s s yy (k,y) (9)

10 1 M. Marder 4. Energy flux Laboratory experiments for cracks in strips do not apply stresses to the crack faces while keeping upper and lower boundaries rigid. Instead, they displace upper and lower boundaries by an amount d and leave the crack faces stress free. There is no need to solve anything new to obtain solutions for this alternative geometry. The stress and strain elds are obtained from all the existing solutions simply by adding the stress and strain elds of an unbroken and uniformly strained plate, whose displacement eld is d y = u y, and which is at stress s yy = s everywhere. With the understanding that elds can be obtained in such a fashion, the entire remaining discussion will be in terms of this experimental geometry. The total energy J owing to the tip of a moving crack may be determined by considering the time rate of change in energy contained inside any region in the strip bounded by a curve S. It is J = d dt (K + P) = d dx dy dt ò q Çu a Çu a + 1 u a s xb a b ( ). (93) The spatial integral must be taken over a region which is static in the laboratory frame. So d dt (K + P) + ò dx dy q Èu a Çu a + Çu a s a b ( xb ), (94) where the symmetry of the stress tensor under interchange of indices is used for the last term. Using the equation of motion gives q Èu a (x) = xb J = ò dx dy [ = ò dx dy xb s a b (x) (95) s a b xb ]Çu a + Çu a s xb a b ( ), (96) (s a b Çu a ), (97) = ò S Çu a s a b nb, (98) where the integral is now over the boundary S, and ^n is an outward unit normal. Choose now a boundary S which runs along the x axis at a height a very nearly equal to zero from L <x <L, where L is very large, runs vertically from (L,a ) to (L,1), back to ( L,1) and nally closes as ( L,a ). Only the lower boundary of S has the chance to contribute to J, and equation (98) becomes J = v ò dx u y x s yy. (99) Since u y vanishes for x >0, and s yy vanishes for x <0, only very near the crack tip at x = 0 can there be any contributions to J. That is, all the contributions come from small x in real space, or large k in Fourier space. The integrand of equation (99) is a delta function, which means that its Fourier transform becomes constant for

11 ò s large k. By working out the asymptotic forms of u y and s yy in Fourier space one nds that J = lim k [p vk u y (k)s yy( k)], (100) It is nally possible to determine the equation of motion for a crack. Equation (88) shows that, for large k, ( ), (101) u y. u s y 1 Çv v H(v) s yy. s yy 1 Çv ( v ) H(v). (10) Therefore the energy ux from the tip of an accelerating crack must be 1 Çv ( v ) H(v) (103) times greater than the ux from the nonaccelerating crack at the same velocity. When a crack moves in steady state, the ow of energy out of the strip and into the crack is d s v; (104) so the corresponding energy ux in the presence of acceleration is ( ) (105) d s v 1 Çv v H(v) = d s v + d s Çv v v v 1 which implies that the total energy of the plate is l(t) d s + d s v v 1 ( ) H(v), (106) ( ) H(v), (107) where l(t) is the total length of the crack. The desired equation of motion is equation (106). To obtain useful results, it is necessary to evaluate H(v). The following computation is helpful dk p Adiabatic equation for cracks 13 F (0) ( v (ik + ²)F (k)) = p i F (0) p ( v ²F (i²)) F + (0) ( ik + ²)F + ( k) F + (0) ( i)f + (i²) = 1 (108) H(v). (109) v

12 ò 14 Adiabatic equation for cracks On the other hand, one can also write this integral as dk F (0) p ( v (ik + ²)F (k) ) dk 1 1 = ò p k + ² F(k) v dk 1 1 = ò p k F(k) 1 v F(k) = 1 dk 1 p k ln 1 F(k) so that v ò F(k)F + (0) ( ik + ²)F ( k) (110) F (0)F + (0) ( F (k)f ( k) ), (111) ( ), (11) ( ), (113) 1 k ( ln F(k,0) F(k,v) ). (114) dk H(v) = ò p Carrying out the integrals in equation (114) and tting the results to a polynomial (Liu and Marder 1991) gives to about 10% accuracy, for a strip of height b and Rayleigh wave speed c R, ( v v ) 1 H(v) = b c t 0.5v /c t c l 1 (v/c R ). (115) One nds therefore, dividing the energy ux rate by the crack velocity v, that the energy C consumed per unit extension is C (v) < d s 1 b Çv 1 ( c l ). (116) [1 (v/c R ) ] Identifying the energy G stored ahead of the crack with d s gives equation (1). ACKNOWLEDGEMENTS I am grateful for nancial support from the Texas Advanced Research Program, by the National Science Foundation under grant DMR , and by the Exxon Education Foundation. References Ching, E. S. C., Langer, J. S., and Nakanishi, H., 1996a, Phys. Rev. E. 53, 864; 1996b, Phys. Rev. Lett., 76, 1087; 1996c, Physica A, 1, 134. Fineberg, J., Gross, S., Marder, M., and Swinney, H., 199, Phys. Rev. B, 45, Irwin, G. R., Dally, J. W., Kobayashi, T., Fourney, W. L., Etheridge, M. J., and Rossmanith, H. P., 1979, Exp. Mech., 19, 11. Liu, X., and Marder, M., 1991, J. Mech. Phys. Solids, 39, 947. Marder, M., 1991, Phys. Rev. Lett., 66, 484.