Right-handed screw dislocation in an isotropic solid

Transcription

1 Dislocation Mechanics Elastic Popeties of Isolated Dislocations Ou study of dislocations to this point has focused on thei geomety and thei ole in accommodating plastic defomation though thei motion. We now tun to anothe impotant aspect of dislocations: thei elastic fields. The emakable thing about dislocations is that while they leave a cystal intenally pefect afte they have passed though the cystal, they poduce elastic distotions in the cystal as long as they ae pesent. Thei elastic fields detemine, to a lage extent, how they inteact with each othe and with othe stuctual defects in the cystal. Ou study of the elastic popeties of dislocations will assume that the mateial in which they ae found is elastically isotopic. While most cystals ae not elastically isotopic, the famewok that emeges by assuming elastic isotopy is still quite useful and even easonably accuate fo most cystals. Scew Dislocation in an Infinite Cylinde We wish to descibe the elastic field of a ight-handed scew dislocation in an elastically isotopic solid. The diagam shows an infinitely long cylinde containing a scew dislocation unning along its axis. The outside adius of the cylinde is R while the adius of the small hole at the coe is o. The coe mateial is assumed to be emoved to avoid singula stesses and stains, as discussed befoe. Right-handed scew dislocation in an isotopic solid The elastic field of a scew dislocation in an isotopic elastic solid, such as shown in the diagam, is a poblem of anti-plane stain, a two dimensional elastic field which is chaacteized by the conditions that 154

2 u z = f (x, y) g(z), u x = u y = 0 that is, thee ae only z displacements, u z, and they depend only on the in-plane coodinate positions, (x, y), and not on z. The govening equation fo anti-plane stain elasticity poblems is Laplace s equation, u z = u z x + u z y = 0. We need a displacement field u z (x, y) that satisfies this equation and also satisfies the Buges condition: u ( / s)ds = b. By inspection one can guess that the ight solution is u z = b = b y tan1 x. It is easy to demonstate that this solution satisfies Laplaces s equation: u z x = b 1 1+ y x u z y = b 1 1+ y x u z x u z y y x = b y x + y 1 x = b x x + y = b (x + y )(0) y(x) (x + y ) = b (x + y )(0) x(y) (x + y ) = b xy (x + y ) = b xy (x + y ), so that Laplace s equation is satisfied: u z = u z x + u z y = 0. Since we know the displacement field, it is a simple matte to compute the elastic stains and stesses aound the dislocation. By diffeentiating the displacement field we can calculate the stains: 155

3 xz = u x z + u z x = b y x + y yz = u y z + u z y = b x x + y. These ae the only non-zeo stain components. Now we use Hooke s law to compute the coesponding stesses xz = μ xz = μb y x + y yz = μ yz = μb. x x + y In cylindical coodinates we could calculate the stains and stesses as follows: z = 1 u z + u z = z = μ z = μb b. A simple way to deive this basic esult is to conside a thin shell of adius and thickness d suounding a unit length of dislocation. If this shell is unwapped and flattened out we have a ectangula sheet of height and unit width with a shea stain of z = b /. The coesponding shea stess is then z = μ z = μb /, the same esult obtained above. We note that the stesses nea a dislocation line depend linealy on the shea modulus and Buges vecto and invesely on the distance fom the dislocation line. Thus the elastic field is singula, with a singulaity of the fom ij 1. It is the singula natue of this field that equies us to emove the coe fom consideation. Obviously, the solution beaks down in the coe of the dislocation whee the stesses cannot go to infinity and must be bounded. Linea elasticity beaks down in the coe as atoms slide past each othe. It is sometimes useful to convet the stesses fom one coodinate system to anothe. Using the coodinates shown below, the shea stess in the cylindical coodinate system, z, can be witten as z = a zk a l kl = a zx a z xz + a zz a x zx + a zy a z yz + a zz a y zy, whee a zx is the cosine of the angle between the diections z and x and whee a z is the cosine of the angle between and z, etc., ie. the diection cosines. It is easy to see that the diection cosines fo this tansfomation ae 156

4 so that Cylindical-Catesian coodinate tansfomation a zx = 0,a zy = 0,a zz = 1 a x = sin,a y = cos,a z = 0, z = sin zx + cos zy, which, using cos = x / x + y and sin = y / x + y and xz = μb y x + y yz = μb, x x + y leads back to z = μb 1 x + y = μb. Scew Dislocation in a Finite Cylinde The above solution fo the stesses about a scew dislocation is valid if the cylinde is infinitely long and constained at its ends so that it does not twist. If, instead, the cylinde is of finite length and toque fee, then the cylinde will twist when the dislocation is ceated and the stess field will be modified. The following diagam shows a dislocated cylinde in a constained state and the twist associated with the toque-fee state. The twist that occus when the cylinde is made toque-fee is called the Eshelby twist, afte J.D. Eshelby who fist studied these popeties in the 1950s. As shown in the diagam, when the cylinde is constained, the shea stesses, z, that act on any section of the cylinde add up to a toque, T z, that acts in 157

5 the cylinde. The supescipt is meant to denote an infinitely long cylinde. If that toque is eleased (the equivalent of applying an equal but oppositely signed toque) then the cylinde will twist as shown in the diagam. The toque acting in the constained cylinde can be expessed appoximately as T z = R o 0 z ddi whee the fist tems in the integal epesent the shea stess and aea element, constituting a shea foce, and the last tem epesents the moment am length fom the oigin. Using the basic solution fom above, z = μb we have, fo R >> o, T z = R o 0 ( μb )ddi R = μb d = μb R o o μbr. If the end of the cylinde is fee (o if the cylinde is of finite length) a toque cannot exist in the cylinde. To satisfy this end condition the above solution must be coected by supeimposing an image toque, T z = T z = μbr /, which is equal and opposite to the above toque. Supeimposing this image toque causes the cylinde to twist, as shown in the figue. Scew dislocation in an infinitely long, constained cylinde and in a cylinde that is unconstained and toque fee. 158

6 Tosion of a cylindical ba. We can calculate the additional o image displacements, stains and stesses associated with the image toque by finding a solution fo pue tosion. By inspection we can guess that the displacement field fo pue tosion is u = Az whee A is a constant to be detemined and the displacements ae linea in both and z. The coesponding elastic stains ae z = u z = A = u u + 1 u = 0 with all othe components zeo. The coesponding shea stesses ae then z = μ z = μa. The coesponding image toque is then T z which is T z R = z ddi = (μa)ddi = μa 3 d o 0 = μa R4 4 4 o 4 R whee again we use R >> o. o 0 μa R4 R o 159

7 By setting this toque equal and opposite to the oiginal toque we can find the constant A, T z = T z = μbr = μa R4, so that, A = b R. We now have the coection o image solution fo a scew dislocation in a cylinde of finite length. u = b R z z = b R z = μb R. The full stess field fo a scew dislocation in a cylinde of finite length is now z = z + z = μb μb R = μb 1 R. We see that nea the coe of the dislocation, whee << R, the standad solution is ecoveed. But the image solution becomes moe and moe impotant fa away fom the coe. Note that the shea stess changes sign nea the oute bounday of the cylinde. One inteesting aspect of this solution is that a finite cylinde containing a scew dislocation along its coe is expected to twist. Using the image displacement field, we can calculate the citical length fo which one complete evolution would be expected by setting u ( = R) = R : u ( = R) = b R z c = R z c = R. b This kind of twisting has been obseved in PbSe nanowies ecently by J. Zhu, Y. Cui et al. Natue Nanotechnlogy (008). 160

8 Eshelby twist in a PbSe nanowie. A scew dislocation is found along the axis of the main nanowie. The nanowie banches ae fee of dislocations and ae not twisted. (J. Zhu, Y. Cui et al. Natue Nanotechnology (008)) Edge Dislocation in a Cylinde We now descibe the elastic field of a positive edge dislocation in an elastically isotopic solid. The diagam below shows an edge dislocation in a cylinde of adius R with a small hole at the coe of adius o. Edge dislocation The elastic field of an edge dislocation is a poblem of plane defomation, specifically, plane stain. Note that when the dislocation is ceated, the slab of mateial lying in the xy plane emains plana when the dislocation is fomed and the thickness of that slab neithe inceases of deceases when the distotion is ceated. So it is a poblem of plane-stain defomation (the stains ae all inplane). Fo two dimensional plane defomation poblems, including plane stain, 161

9 the govening equation that detemines the elastic field is the bi-hamonic equation 4 = x + y x + y = 0, whee is a potential, o stess function, the deivatives of which poduce the stesses xx = y yy = x xy = xy. It can be shown that stess functions that satisfy the bi-hamonic equation lead to displacements that ae compatible and stesses that satisfy the conditions of equilibium. So any stess function that satisfies the bi-hamonic equation and also the bounday conditions fo the poblem povides the coect elastic field fo that poblem. In tems of pola o cylindical coodinates, = = = Solution fo an Edge Dislocation 1. The geneal solution to the bi-hamonic equation fo an edge dislocation takes the fom = f( )sin f( )= B 3 A ln, whee B and A ae constants to be detemined fom the bounday conditions. Because the field is peiodic in the dependence of the stess function on sin can be anticipated. It is easy to show that this function satisfies the bi-hamonic equation. Since fo a dislocation in an infinite solid the stesses ae expected to go to zeo at, it follows that B = 0 fo a cylinde of infinite adius R. Also, it can be shown that the Buges condition, u ( / s)ds = b equies that the constant A be selected as A = μb ( 1 ), whee is Poisson s atio. To show this we would need to compute the stesses fom the stess function, calculate the stains using Hooke s law, integate the 16

10 stains the get the displacements and then, finally, equie that they satisfy the Buges condition. So the stess function fo a positive edge dislocation in an infinite solid is = μb ln sin = A ln sin, ( 1 ) o in the Catesian coodinates = Ayln x + y. Stesses aound a Positive Edge Dislocation Using the stess function, we can now calculate the stesses aound the dislocation: = = 1 ( Asin ) [ 1+ ln ]+ 1 (A ln )(sin) = Asin = = Asin [ 1+ ln ] = Asin = 1 = 1 ( A ln )cos = Acos 163

11 The moe familia expessions fo the stesses about a positive edge dislocation ae the ones expessed in the Catesian coodinate system. Using the equations above in the Catesian coodinates we can easily find xx = y 3x + y y = A (x + y ) yy = x ( ) ( ) =A y x y (x + y ) ( ) xy = xy = A x x y (x + y ). Fo this plane-stain poblem we can also compute the out-of-plane stesses zz = ( xx + yy )= Ay x + y, which, of couse, comes fom the condition that zz = 0. We may also want to know the hydostatic pessue at any point in the solid aound the dislocation p = 1 ( 3 xx + yy + zz )= A (1 + )y 3 (x + y ). Visualization of the Stess Field It is helpful to develop an intuition about the stesses aound a dislocation, so that one can anticipate how dislocations inteact with each othe and with othe defects. The diagam below shows the in-plane stess components about a positive edge dislocation, as descibed by the equations above. We note that the signs of some of the components can be guessed fom the natue of the distotion about the exta half plane. Notice that the shea stess, xy, is natually positive on the ight hand side of the dislocation while that stess component is negative on the left hand side, consistent with the solution. Those signs ae expected fom the way the exta half plane displaces mateial at its edge, as shown in the figue. Since the exta half plane is expected to ceate hydostatic compession whee it is pesent, we can anticipate that both xx and yy (and zz fo that matte) ae all negative above (and positive below) the dislocation. The hydostatic compession and tensions expected ae shown in the diagam. Because an exta half plane is inseted, one can futhe guess that xx would be negative (compession) eveywhee above the slip plane while it would be positive (tension) eveywhee below the slip plane. Othe components of the stess field cannot be easily guessed and the fact that some of the stess components change sign along the 45 degee lines cannot be pedicted intuitively. Fo example, yy on eithe side of the dislocation has a sign that is opposite that of xx in those 164

12 same egions. One might emembe this by noting that a shea stess and not a hydostatic stess is found on eithe side of the dislocation. Visualization of the stess field about a positive edge dislocation. Edge Dislocation in a Cylinde of Finite Radius The solution we have descibed is valid fo a positive edge dislocation in an infinite elastic solid. If we want to have a solution of an edge dislocation in a cylinde of finite adius, then coections need to be made to account fo the taction-fee cylindical sufaces, ( = R) = ( = R) = 0. We do this by choosing a non-zeo value fo the constant B in the above stess function, so that we have a coection stess function descibing the coection field: = B 3 sin. Fist calculate the coection stesses 165

13 = = Bsin = B sin 3 + B3 ( sin ), and = 1 = 1 B3 cos. = B cos We equie that ( = R)= ( = R) + ( = R) = 0 ( = R)= ( = R) + ( = R) = 0. Consideing the component, we have ( = R) + ( = R) = 0 Asin R + brsin = 0, which gives B = A R. Consideing the component, we have ( = R) + ( = R) = 0 Acos, BRcos = 0 R which, again, gives B = A R. 166

14 Thus, this value of B leads to a coection stess field that makes the tactions on the cylindical suface zeo. With this coection the shea stess field fo a positive edge dislocation in a cylinde of finite adius is = + = Acos Acos R = Acos 1 R. The coection is simila in fom to the coection fo the finite length of the cylinde fo a scew dislocation, except in this case the stess goes continuously to zeo and does not change sign with inceasing adial distance fom the coe. Stess Field fo a Staight Mixed Dislocation Using the pinciple of supeposition of solutions fo linea elastic poblems, we can easily descibe the stess field fo a mixed dislocation by supeimposing the fields of the scew and edge pats of that dislocation. Buges vectos fo a mixed dislocation In the z coodinate system the stess field of the mixed dislocation in an infinite solid can be expessed as whee 0 ij = z, 0 z zz = = Asin = = Acos 167

15 and A = μb e (1 ) = μbsin (1 ) z = z = μb s = μb cos. The supeposition of elastic solutions is justified fo linea elasticity. The displacements, stains and stesses fo diffeent solutions can be added o supeimposed without any inteactions. The enegies of diffeent elastic fields depend quadatically on the stesses o stains and cannot be added in this way. 168