Problem Set 4. Eshelby s Inclusion

Transcription

1 ME34B Elasticity of Microscopic Structures Wei Cai Stanford University Winter 24 Problem Set 4. Eshelby s Inclusion Chris Weinberger and Wei Cai c All rights reserved Problem 4. (5 Spherical inclusion. (a Derive the expressions for the auxiliary tensor D ijkl for a spherical inclusion in an isotropic medium with shear modulus µ and Poisson s ratio. Hint: many components of D ijkl are zero, unless there are repeated indices. (b Derive the corresponding expressions for Eshelby s tensor S ijkl. Solution (a The auxiliary tensor inside the inclusion can be expressed in terms of the following surface integral (see lecture notes, where D ijkl abc β λ/k (zz ij z k z l sin Φ β 3 dθ dφ ( (a 2 cos 2 Θ + b 2 sin 2 Θ sin 2 Φ + c 2 cos Φ (2 z k/k (sin Φ cos Θ, sin Φ sin Θ, cos Θ (3 For a sphere a b c and β a. Therefore, D ijkl For isotropic material, (zz ij D ijkl µ (zz ij z k z l sin Φ dθ dφ (4 has an explicit form, so that, ( δ ij λ + µ µ λ + 2µ z i z j ( δ ij z k z l λ + µ λ + 2µ z i z j z k z l z k z l sin Φ dθ dφ We can write the three components of z in terms of Φ and Θ sin Φ dθ dφ z sin Φ cos Θ z 2 sin Φ sin Θ z 3 cos Φ (5

2 Let us consider the first integral H kl z k z l sin Φ dθ dφ This integral will be zero if k l. By symmetry, H H 22 H 33. H 33 is the simplest to evaluate in the present form, i.e., H 33 2π 3 Therefore, cos 2 Φ sin ΦdΘdΦ s 2 ds (6 H kl 3 δ kl (7 Now let us consider the second integral, J ijkl z i z j z k z l sin Φ dθ dφ The J ijkl is non-zero only if the indices are all the same, or come in pairs. There are two types of terms, J J 2222 J 3333 and J 22 J 33 J 2233 J J 33 Thus we can write J ijkl as cos 4 Φ sin 3 Φ dθ dφ 5 cos 4 Φ sin Φ dθ dφ 5 J ijkl 5 (δ ijδ kl + δ ik δ jl + δ il δ jk (9 The auxiliary tensor D ijkl is D ijkl ( δ ij H kl λ + µ µ λ + 2µ J ijkl ( Substituting the solutions for H kl and J ijkl, we have D ijkl ( 5δ ij δ kl λ + µ 5µ λ + 2µ (δ ijδ kl + δ ik δ jl + δ il δ jk ( δ ij δ kl 3µ (δ ijδ kl + δ ik δ jl + δ il δ jk 3µ( (9 δ ijδ kl (δ ik δ jl + δ il δ jk ( 2 (8

3 using the fact that λ+µ λ+2µ 2(. (b The Eshelby s tensor S ijkl is related to the auxiliary tensor D ijkl through, S ijmn 2 C lkmn(d iklj + D jkli For isotropic medium, thus, C lkmn λδ lk δ mn + µ(δ lm δ kn + δ ln δ km (2 S ijmn λ 2 (D ikkj + D jkki δ mn µ 2 (D inmj + D jnmi + D imnj + D jmni When D ijkl is given in Eq. (, it satisfies both major and minor symmetries, i.e. D ijkl D klij D jikl D ijlk. Thus, S ijmn λd ikkj δ mn µ(d inmj + D imnj (3 Noticing that Thus D ikkj 5 3µ( δ ij (4 λ 2µ (5 2 λd ikkj 3( δ ij (6 S ijmn 3( δ ijδ mn + 3( (9 δ ijδ kl (δ ik δ jl + δ il δ jk 3( δ ijδ mn + 3( (9 (δ inδ jm + δ im δ jn (δ im δ jn + δ ij δ mn + δ in δ jm + δ ij δ mn 3( ( 2δ ijδ mn + (9 (δ in δ jm + δ im δ jn 5 5( δ ijδ mn ( (δ inδ jm + δ im δ jn (7 Problem 4.2 (5 Dilation field. The constrained dilation of a transformed inclusion (not necessarily ellipsoidal is, u c i,i σkjn k (x G ij,i (x x ds(x S σkjg ij,ik (x x dv (x (8 V 3

4 (a Show that if e ij εδ ij (pure dilational eigenstrain, then in isotropic elasticity the constrained dilation is constant inside the inclusion and independent of inclusion shape. (b What is u c i,i inside the inclusion in terms of ε? Hint: The Green s function G ij (x can be expressed in terms of second derivatives of R x. G ij (x δ ij 2 R 2( i j R (9 Notice that 2 R 2 R (2 2 R δ(x (2 Solution In the case of e ij εδ ij (22 the eigenstress is σkj C kjmn e mn ε C kjmm (23 where C kjmm λδ kj δ mm + 2µδ km δjm (3λ + 2µδ kj (24 Hence, σ kj ε(3λ + 2µδ kj Substituting this into the equation for constrained dilatation u c i,i σkjg ij,ik (x x dv (x V ε(3λ + 2µG ij,ij (x x dv (x V Notice that G ij (x δ ij R,kk 2( R,ij (25 (26 4

5 Therefore, G ij,ij (x In other words, δ ij R,kkij 2( R,ijij R,kkii 2( R,ijij 2 6πµ( R,iijj 2 2 6πµ( 2 R 2 6πµ( 8πδ(x 2 δ(x (27 2µ( G ij,ij (x x 2 2µ( δ(x x (28 Hence when x is inside V, u c i,i 2 ε(3λ + 2µ V 2µ( δ(x x dv (x 2 ε(3λ + 2µ 2µ( ε + When x is outside V, the constrained dilation field must be zero. (29 5