Lecture Note 3. Eshelby s Inclusion II

Transcription

1 ME340B Elastcty of Mcroscopc Structures Stanford Unversty Wnter 004 Lecture Note 3. Eshelby s Incluson II Chrs Wenberger and We Ca c All rghts reserved January 6, 004 Contents 1 Incluson energy n an nfnte sold Incluson energy by work method 5 3 Incluson energy n a fnte sold 7 4 Colonett s theorem 9 5 Fnte sold wth appled tractons 1 1

2 1 Incluson energy n an nfnte sold So far we have obtaned the expressons for the stress, stran and dsplacement feld both nsde and outsde the ncluson. An mportant queston s: what s the total elastc energy E of the sold contanng an ncluson? In ths and subsequent sectons, we derve the expressons for E, whch we refer to as the ncluson energy for brevty. However, we emphasze that E s the total elastc energy of the sold contanng an ncluson. E ncludes the elastc energy stored both nsde and outsde the ncluson. For example, f we obtan E as a functon of the ncluson sze, then the dervatve of E provdes the drvng force for the expanson (or shrnkage) of the ncluson. Notce that ths s the case only f E s the total elastc energy, not just the energy stored nsde the ncluson. There are two ways to obtan the expresson for the total energy E. Frst, we can ntegrate the elastc energy densty both nsde and outsde the ncluson, usng the feld expressons we have already obtaned. Second, we can obtan the elastc energy E by measurng the work done n a vrtual experment that transforms a sold system wth zero elastc energy to the sold contanng an ncluson. In ths secton, we take the frst approach. The work method s dscussed n the next secton, whch leads to dentcal results but may provde more physcal nsght. For clarty, let us ntroduce some symbols to descrbe the elastc felds nsde and outsde the ncluson. Let the elastc (stress, stran, dsplacement) felds nsde the ncluson be denoted by a superscrpt I, and the elastc felds outsde the ncluson (n the matrx) be denoted by a superscrpt M. Notce that whenever the superscrpt I or M s used, the felds only nclude the elastc component. For a homogeneous nfnte sold, the elastc felds n the matrx and the ncluson are, matrx ncluson e M = e c e I = e c e M = c I = c u M = u c u I = u c e x j Therefore, the total elastc energy s, E = 1 e I I d + 1 M e M d (1) Rewrtng E n terms of dsplacements, we have E = 1 4 (u I I,j + u I j,) d + 1 M (u M,j + u M 4 j,) d () and notng the symmetry of the stress tensor E = 1 u I I j, d + 1 M u M j, d (3) Now, the dervatve can be factored out usng the followng rule u,j = ( u j ),, u j (4)

3 E = 1 ( I u I j), I,u I j d + 1 ( M u M j ), M,u M j d (5) The body s assumed not to have any body forces actng on t, thus the dvergence of the stress tensor,,, s zero. Thus E = 1 ( u I I j), d + 1 ( M u M j ), d (6) We wsh to now use Gauss s theorem on ths equaton. We need to careful about the sgn of the unt normal vector that ponts outsde the ntegraton volume. Let the normal vector pontng out of the ncluson volume be n out. Let the unt normal vector pontng out of the outer surface of the matrx (at nfnty) be n. Applyng Gauss s theorem, E = 1 u I I jn out ds 1 M u M j n out ds + 1 M u M j n ds (7) S We expect that the surface ntegral over S should vansh as t approaches nfnty. To show ths, let S be a sphercal surface whose radus R approaches nfnty. Notce that D kl (x) = G,kl (x x )d (x ) (8) Because G kl (x x ) R 3 where R = x, for large R, then D kl (x) R 3. Therefore, ( ) 1 e M = O (9) R ( 3 ) 1 M = O (10) R 3 ds = O ( R ) (11) Thus S M u M j n out d 0 as R (1) Combnng the two ntegrals over, E = 1 ( ) I u I j M u M j n out ds (13) Although the stress across the ncluson nterface does not have to be contnuous, the tracton force across the nterface must be contnuous,.e., I n out whch leads to E = 1 = M n out I ( u I j u M j (14) ) n out ds (15) 3

4 From the defnton of (elastc dsplacement felds) u I j and u M j, we have Thus u I j u M j = (u c j e jkx k ) u c j = e jkx k (16) E = 1 I n out e jkx k ds (17) Therefore, we have expressed the total elastc energy E n terms of a surface ntegral over, the ncluson nterface. We can further smplfy ths expresson by transformng the ntegral back nto a volume ntegral (over the ncluson volume ). E = 1 ( ) I e jkx k d, = 1 ( ) e jk I, x k + x I k, d = 1 e jkkj I d = 1 e I d = 1 ( ) e c d (18) For an ellpsodal ncluson, the stress nsde s a constant, thus E = 1 ( ) c e = 1 I e (19) If the volume s not an ellpsod, we can stll wrte the energy n terms of the average stress n the ncluson E = 1 I e (0) where I 1 c (x) d (x) (1) Suppose that we wsh to account for how much of the energy s stored nsde the ncluson and how much s stored n the matrx. The energy store nsde the ncluson s E I = 1 e I I d For ellpsodal ncluson, the stress and stran are constant nsde, hence E I = 1 I e I = 1 I ( ) e c e 0 4

5 Snce the total elastc energy s E = 1 I e the elastc energy stored nsde the matrx must be, E M = E E I = 1 I e c Incluson energy by work method In ths secton, we re-derve the expressons n the prevous secton concernng the ncluson energy usng a dfferent approach. Rather than ntegratng the stran energy densty over the entre volume, we make use of the fact that the stored elastc (potental) energy n the sold must equal to the work done to t. By consderng a vrtual experment that transforms a stress-free sold nto a sold contanng an ncluson and accountng for the work done along the way, we can derve the total elastc energy (or the elastc energy stored wthn the ncluson or the matrx) usng consderably less math than before. To better llustrate ths method, let us consder a smple example. Consder a mass M attached to a lnear sprng wth stffness k. Let E 0 be the equlbrum state of the system under no appled force. Obvously E 0 = 0. Defne the orgn as the poston of the mass at ths state. Suppose we gradually apply a force to the mass untl the force reaches F 1. At ths pont the mass must have moved by a dstance x 1 = F 1 /k. Let the energy of ths state be E 1. The work done n movng the mass from 0 to x 1 equals to the average force F appled to the mass tmes the dstance travelled (x 1 ). Because the ntal force s 0 and the fnal force s F 1, the average force s F = F 1 /. Therefore, the work done n movng the mass from 0 to x 1 s, Hence W 01 = F x 1 = 1 F 1x 1 = 1 kx 1 () E 1 = E 0 + W 01 = 1 kx 1 (3) Suppose we further ncrease the force to F and the system reaches a new state at x = F /k wth energy E. Snce the ntal force durng ths transformaton s F 1 and the fnal force s F, the average force s F = (F 1 + F )/. The mass moves by a dstance of x x 1 under ths force. Therefore the work done s 5

6 Hence W 1 = 1 (F 1 + F )(x x 1 ) = 1 k(x x 1) (4) E = E 1 + W 1 = 1 kx (5) Now, lets apply ths method to Eshelby s ncluson problem. Let us consder the four steps n Eshelby s constructon of a sold contanng an ncluson. Recall that after step 1, the ncluson s outsde the matrx. The ncluson has undergone a deformaton due to ts egenstran. No forces are appled to ether the ncluson or the matrx. Obvously, the total elastc energy at ths state s E 1 = 0. In step, we apply a set of tracton forces on the ncluson surface. At the end of step, the tracton forces are T j = n j and the dsplacements on the surface are u j = e kj x k. Therefore, the work done n step s W 1 = 1 T j (x)u j (x) ds(x) = 1 n e kjx k ds(x) = 1 e kj x k n ds(x) (6) and usng Gauss s theorem W 1 = 1 e kj x k, d (x) = 1 e kj δ k d (x) = 1 e In step 3, the ncluson s put nsde the matrx wth the tracton force unchanged. No work s done n ths step,.e. W 3 = 0. In step 4, the tracton force T j s gradually reduced to zero. Both the ncluson and the matrx dsplace over a dstance of u c j. Snce the tracton force s T j at the begnnng of step 4 and 0 at the end of step 4, the average tracton force (7) 6

7 s, agan, T j /. The work done to the entre system (ncluson + matrx) s W 34 = 1 T j (x)u c j(x) ds(x) = 1 n u c j(x) ds(x) = 1 u c j,(x) ds(x) = 1 e c ds(x) = 1 e c (8) The total elastc energy at the end of step 4 s, E = E 1 + W 1 + W 3 + W 34 = e c e = 1 (c )e = 1 I e (9) whch s exactly the same as Eq. (19). The same approach can also be appled to obtan the elastc energy stored nsde the ncluson (E I ) or nsde the matrx (E M ). In step 4, the matrx also exert force on the ncluson, whch does work as the nterface moves. Ths leads to a transfer of elastc energy from the ncluson to the matrx. 3 Incluson energy n a fnte sold Let us now consder a problem wth an ncluson n a fnte sold. Agan, the stress-stran felds n ths case can be solved by superpostons. Suppose the fnte sold assumes the stress-stran felds of an nfnte sold contanng an ncluson, as solved prevously. Then we must apply a set of tracton forces T j to the outer surface S ext of the sold to mantan equlbrum. To obtan the soluton of a fnte sold wth zero tracton on ts outer surface, we need to remove T j on S ext. Ths s equvalent to apply a cancellng tracton force F j = T j on the outer surface S ext of the fnte sold. Let e m, m and u m be the stran, stress and dsplacement felds n response to the surface tracton F j. They are called mage felds. In ths case, the elastc felds nsde the matrx and the ncluson are, matrx e M M u M = e c + e m = c + m = u c + u m ncluson e I = e c e + e m I = c + m u I = u c e x j + u m 7

8 Obvously, the mage felds satsfy the condton, e m (x) = 1 (um,j(x) + u m j,(x)) (30) m (x) = C kl e m kl (x) (31) Notce that the mage felds are generally not unform wthn the ncluson. The free tracton boundary condton on the outer surface S ext can be expressed as, M n ext = 0 (on S ext ) (3) Smlar to Eq. (7), the total elastc energy n the sold can be expressed n terms of surface ntegrals, E = 1 ( u I I j M u M j )n out ds + M u M j n ext ds (33) S ext Because of Eq. (3), the second ntegral does not contrbute. Usng the tracton contnuty argument (n I out = M n out ) as before, we get E = 1 (u I I j u M j )n out ds (34) Agan, usng u I j u M j = e k x k, we get E = 1 e I jkx k n out ds (35) Ths s the same as Eq. (17) except that the stress feld nsde the ncluson now contans the mage component. Defne I, c (36) as the stress feld nsde the ncluson n an nfnte medum. Then I (x) = I, Smlarly, defne E 1 + m (x) (37) I, e jkx k n out ds = 1 I, e (38) as the ncluson energy n an nfnte sold. Then the ncluson energy n a fnte sold s, E = E 1 m e jkx k n out ds (39) 8

9 Convertng the second ntegral nto volume ntegral, we have E = E 1 ( m e jkx k ), d = E 1 m e d = E 1 e d where m m = E 1 m e = E + E m 1 m (x) d (x) (40) s the averaged mage stress nsde the ncluson. E m 1 m e s the mage contrbuton to the total ncluson energy. ncluson s, I I, + m (41) The average stress nsde the (4) Thus the total ncluson energy s stll related to the averaged stress nsde the ncluson as E = 1 I e (43) The results of the total ncluson energy for ellpsodal ncluson under varous boundary condtons are summarzed below. total elastc energy nfnte sold E = 1 I e fnte sold wth zero tracton E = 1 I e 4 Colonett s theorem I = I, + m We now wsh to study the energy of a sold contanng an ncluson and also subjectng to appled forces at ts outer surface. Before we do that, let us frst prove Colonett s theorem, whch s very helpful to study such problems. Colonett s theorem [1] states that There s no cross term n the total elastc energy of a sold, between the nternal stress feld and the appled stress feld. 9

10 However, there s an nteracton energy term between the nternal and appled felds when the energy of the appled loads s ncluded. Colonett s theorem can greatly smplfy the energy expressons when we apply stress to a fnte sold contanng an ncluson. To apprecate Colonett s theorem, we need to be specfc about the meanng of nternal and appled stress felds. Let us start wth a stress-free homogenous sold wth outer surface S ext. Defne nternal stress felds as the response to a heterogeneous feld of egenstran nsde the sold wth zero tracton on S ext. Defne appled stress felds as the response to a set of tractons on S ext when there s no egenstran nsde the sold. Let us consder two states of stress. State 1 s purely nternal, and state s appled. The total elastc energy nsde the sold for these two states are, E (1) = 1 (1) e(1) d E () = 1 () e() d Now consder a state 1+ whch s the superposton of state 1 and. Its total elastc energy should be, E (1+) = 1 ( (1) + () )(e(1) + e () ) d = 1 ( (1) e(1) + (1) e() + () e(1) + () e() ) d where E (1 ) 1 = E (1) + E () + E (1 ) ( (1) e() + () e(1) ) d (44) s the nteracton term between state 1 and state. Colonett s theorem states that E (1 ) must be zero, whch we wll prove below. Frst, we note that so that () e(1) = C kl e () kl e(1) (1) e() = C kl e (1) kl e() () e(1) = (1) e() E (1 ) = (1) e() d = (45) () e(1) d (46) Snce there s no body force, (1), = 0. Therefore, E (1 ) = ( (1) u() j ), d (47) 10

11 Now, we wsh to apply Gauss s theorem to convert the volume ntegral nto a surface ntegral. However, to use Gauss s theorem, the ntegrand must be contnuous nsde the entre volume. However, ths s not necessarly the case f the egenstran feld e (x) s not suffcently smooth. For example, n Eshelby s transformed ncluson problem, e (x) s not contnuous at the ncluson surface. As a result, the nternal stress feld (1) (x) s not contnuous at the ncluson surface ether. However, for clarty, let us assume for the moment that the egenstran feld e (x) and the nternal stress feld (1) (x) are suffcently smooth for the Gauss s theorem to apply. Ths corresponds to the case of thermal stran nduced by a smooth varaton of temperature nsde the sold. In ths case, E (1 ) = ( (1) ), d = u() j S ext n ext (1) u() j ds (48) Snce (1) s a purely nternal stress state, Hence (1) next = 0 (on S ext ) (49) E (1 ) = 0 (50) whch s Colonett s theorem. Let us now consder the case where the egenstran feld e (x) and the nternal stress feld (1) (x) are pecewse smooth nsde varous ncluson volumes K as well as n the matrx K K. Let n out,k be the outward normal vector of ncluson volume K. We can apply Gauss s theorem n each ncluson and the matrx separately, whch gves, E (1 ) = (1) u() j n ext ds + ( (1),K (1) )u() j n out,k ds (51) S ext K S K where (1),K s the stress nsde the volume k and (1) s the stress n matrx. The tracton force across the ncluson nterface must be contnuous,.e., ( K (1) )nout,k = 0 for any K, (5) where the summaton s not mpled over K. Therefore, agan we have, E (1 ) = n ext ds = 0 (53) (1) u() j S ext whch s Colonett s theorem. Colonett s theorem only deals wth the elastc stran energy that s stored nsde the sold (nternal energy). When the system s under appled load, ts evoluton proceeds 11

12 towards mnmzng ts enthalpy, whch s the nternal energy subtractng the work done by the external force. For example, the enthalpy of a system under external pressure p s H = E + p. The enthalpy for the sold under study s, H = E (1+) W W s the work done by the loadng mechansm, W = j + u () j S ext () (u(1) whch can be wrtten as where W (1 ) W = W (1 ) W (1 ) = W () = (54) )n ext ds (55) + W () (56) () u(1) j n ext S () u() j n ext S ds ds can be regarded as the cross term between the two stress states n the total enthalpy. 5 Fnte sold wth appled tractons We now apply Colonett s theorem to our problem of an ncluson n a fnte sold under a set of appled tractons. We wll use superscrpt A to denote the felds n response to the appled tractons when the egenstran vanshes (no ncluson). Let superscrpt F denote the felds of an ncluson n a fnte sold under zero external tractons (as n secton 3). From Colonett s theorem, where E = E A + E F E A = 1 E F = 1 (I, The enthalpy of the system s H = E W (57) A e A d (58) + m )e (59) where the A and F felds do have nteracton terms n the work term W,.e., W = W A + W A F (61) W A = u A A j n ext ds = e A A d = E A (6) S ext W A F = u A F j n ext ds (63) S ext 1 (60)

13 We would lke to express W A F result s, where n terms of an ntegral over the ncluson volume. The W A F = e A (64) A 1 A d (65) To show that ths s the case, frst note that n F ext S ext. Thus W A F = (u A F,M j S ext F,M u A j ) n ext ds = 0 on the surface where the superscrpt M denotes the felds n the matrx. Consder a volume ntegral of the same ntegrand over the matrx volume M =, 0 = (e A F,M F,M e A ) d M = (u A F,M j F,M u A j )n ext ds (u A F,M j F,M u A j )n out ds S ext whch means (u A F,M j S ext F,M u A j )n ext ds = (u A F,M j F,M u A j )n out ds (66) Hence W A F = (u A F,M j F,M u A j )n out ds (67) Notce that the ntegral s on the matrx sde of the ncluson nterface. We can smlarly wrte out the volume ntegral nsde the ncluson 0 = (e A F,I F,I e A ) d 0 = (u A F,I j F,I u A j )n out ds whch means that u A F,I j n out ds = F,I u A j n out ds (68) 13

14 Substtutng ths nto Eq. (67) and notng the tracton contnuty condton F,I n out = F,M n out, we have, W A F = (u A F,M j u F,I j )n out ds S 0 = e A jkx k n out ds (69) Usng Gauss s theorem, we get, W A F = e A d = e A d = e A (70) The major results of ths lecture note are summarzed below. total elastc energy nfnte sold E = 1 I e fnte sold wth zero tracton E = 1 I e E = E A fnte sold + E F E A = 1 wth tracton E F = 1 I e References A e A d I = I, I = I, + m + m total enthalpy H = E W = E A + E F W A A F W = E F E A Ae [1] G. Colonett, Per una teora generale delle coazon elastche, Att R. Acad. Sc. Torno: Cl. Sc. fs. mat. natur., 56, , (191). 14