ME340B Elasticity of Microscopic Structures Wei Cai Stanford University Winter Midterm Exam. Chris Weinberger and Wei Cai

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1 ME34B Elasticity of Microscopic Structures Wei Cai Stanford University Winter 24 Midterm Exam Chris Weinberger and Wei Cai c All rights reserved Issued: Feb. 9, 25 Due: Feb. 6, 25 (in class Problem M. (2 Plane strain. Consider an anisotropic elastic medium with elastic stiffness tensor C ijkl under plane strain deformation. This means that the z-component of the displacement field is zero everywhere. The displacement fields in x and y directions are also independent of z. Mathematically, this can be written as, u 3 = ( u,3 = (2 u 2,3 = (3 (a What are the non-zero components of the strain field? What are the non-zero components of the stress field? (b In 2-dimension, the Hooke s law can be expressed as, σ ij = c ijkl e kl (4 where the indices now only goes from to 2. What is the expression of c ijkl in terms of C ijkl? (c Suppose the medium is subjected to body force b j (j =, 2, which is independent of z. What is the equilibrium condition in terms of the displacement fields u j? Solution: (a Since u 3 = for all z, then u 3, = u 3,2 = u 3,3 =. In combination with u,3 = and u 2,3 = we have e 33 = e 3 = e 23 = and the non-zero components of the strain are e, e 2, and e 22. For a generally anisotropic material, all of the stresses can be related to the strains, thus none of the stresses are necessarily zero. For isotropic material, σ 3 = σ 23 =. The non-zero stress components are σ, σ 2, σ 22, σ 33.

2 (b Since whenever k = 3 or l = 3, e kl =, σ ij = C ijkl e kl for i, j, k, l =, 2 (5 i.e., c ijkl = C ijkl for i, j, k, l =, 2 (6 (c The equations of equilibrium can be derived just as in the notes, no change necessary c ijkl u k,il + b j = (7 Problem M.2 ( Green s function in 2D. (a What is the equilibrium equation for the Green s function G ij (x x in terms of c ijkl, where x, x are 2-dimensional vectors? Notice that a point force in 2D corresponds to a line force in 3D. (b Solve the Green s function in Fourier space, i.e. g ij (k. Again k is a 2-dimensional vector. (c Solve the Green s function in real space G ij (x. Express the result in terms of x and θ, where x = x cos θ, x 2 = x sin θ. The final result can be expressed in terms of an integral over a unit circle. Hint: e ikx dk = 2 ln x (up to a constant (8 k Solution: (a We can derive the equilibrium for the Green s function just as in the notes, however noting that the vector x is a 2-dimensional vector and the delta function is a 2-dimensional delta function: c imns G ij,sm (x x + δ nj δ(x x = (b Note that here all of the vectors are 2-dimensional, and thus we only need to take two dimensional fourier transforms. Define 2-dimensional Fourier transform and its inverse as, g ij (k = G ij (x = = exp( ik xg ij (x dx δ(x = exp( ik xg ij (k dk exp(ik xδ(x dx exp( ik x dk 2

3 Plugging in these definitions into the above equilibrium equations gives: = (c 2 imns g ij (k + δ nj exp(ik x dk x s x m = ( cimns z m z s k 2 g ij (k + δ nj exp(ik x dk Thus where δ nj = c imns z m z s k 2 g ij (k = (zz ni k 2 g ij (k (zz ni c imns z m z s The solution for the Green s function in 2-dimension is g ij (k = (zz ij k 2 (c Taking the inverse Fourier transform, where G ij (x = = = 8π 2 = 8π = 4π exp( ik x (zz ij k 2 dk exp( ikx cos θ (zz ij k dk dθ k 2 exp( ikx cos θ (zz ij k (zz ij ( 2 ln x cos θ dθ (zz ij ln x cos θ dθ dk dθ = (zz ij (ln x + ln cos θ dθ 4π = C ln x + Φ(θ (9 C = 4π Φ(θ = 4π x = x (zz ij dθ (zz ij ln cos θ dθ x = (x cos θ, x sin θ ( 3

4 Problem M.3 (3 Inclusion in 2D. Consider an elliptic inclusion in the 2D medium that can occupies the area, ( x a 2 ( x2 2 + ( b Let its eigenstrain be e ij (i, j =, 2. Define Eshelby s tensor S ijkl and auxiliary tensor D ijkl similarly as in the lecture, but with i, j, k, l =, 2. (a Show that S ijkl and D ijkl are constants inside the inclusion (use anisotropic elasticity. (b What is c ijkl in terms of µ and ν in isotropic elasticity? (c Derive the expressions for S ijkl and D ijkl for a circular inclusion in an isotropic medium (plane strain. Solution: (a where Define D ijkl (x = G ij,kl (x x dv (x V [ = V 2 ] exp [ ik (x x ] (zz ij dk dv (x x k x l ( 2 k 2 = exp [ ik (x x ] (zz ( 2 ij z kz l dk dv (x Q(k = ( 2 V exp( ik x(zz ij z kz l Q(k dk (2 V exp(ik x dv (x (3 λ (λ, λ 2 = (k a, k 2 b, λ = λ R (R, R 2 = (x /a, x 2 /b, R = R γ = (k x/k = (λ R/k β = λ/k (4 4

5 Then Q(k exp(ik x dv (x V = ab exp(iλ R dr = ab = ab R = ab J (λ λ Therefore, D ijkl (x = ab = ab = ab where R exp(iλr cos θ dθ dr R J (λr dr exp( ik x(zz ij z J (λ kz l λ dk exp( ikγ(zz ij z J (kβ kz l k dk dθ kβ κ(γ = exp( ikγj (kβ dk β ] = iγ [ β 2 β2 γ 2 (5 (zz ij z kz l κ(γ dθ (6 (7 The derivations in Eq. (6 and (7 are carried out by Mathematica, whose outputs are given at the end of this solution for reference. Notice that D ijkl (x is real. Since (zz ij z kz l is also real, the imaginary part of κ(γ can be neglected. Therefore, as long as β > γ, we can write κ(γ = β 2 (8 which is independent of γ. Therefore D ijkl (x is independent of x. β > γ is satisfied if x is within the inclusion. This can be shown by the following. If x is inside the ellipse, then ( x a 2 ( x2 2 + = R 2 + R2 2 < (9 b which means R <. Therefore, γ = λ R /k λ R /k = λr/k < λ/k = β (2 5

6 (b The isotropic stiffness tensor for plane strain is c ijkl = λδ ij δ kl + µ(δ ik δ jl + δ il δ jk = 2µν 2ν δ ijδ kl + µ(δ ik δ jl + δ il δ jk ( 2ν = µ 2ν δ ijδ kl + δ ik δ jl + δ il δ jk (c We have shown that inside an elliptic inclusion of an isotropic medium D ijkl (x = ab (zz ij z kz l β 2 dθ For a circular inclusion, a = b, then β = a and D becomes Notice that Therefore, D ijkl (x = (zz ij z kz l dθ c ijkl = λδ ij δ kl + µ(δ ik δ jl + δ il δ jk (zz ij = µδ ij + (λ + µz i z j (zz ij = ( δ ij λ + µ µ λ + 2µ z iz j D ijkl (x = = µ ( δ ij 2( ν z iz j ( δ ij µ 2( ν z iz j z k z l dθ Notice that z = cos θ and z 2 = sin θ, D ijkl can be evaluated explicitly. Let us define and H kl J ijkl z k z l dθ z i z j z k z l dθ (2 The only non-zero elements of H kl are H and H 22, i.e., H kl = δ kl cos 2 θ dθ = πδ kl Similarly J ijkl is non-zero only when all four indices are the same or they come in pairs. J = J 2222 = cos 4 θ dθ = 3π 4 6

7 J 22 = J 22 = J 22 = J 22 = cos 2 θ sin 2 θ = π 4 therefore Thus Now, J ijkl = π 4 (δ ijδ kl + δ ik δ jl + δ il δ jk D ijkl = ( µ = µ δ ij H kl ( δ ij δ kl π 2( ν J ijkl π 2( ν 4 (δ ijδ kl δ ik δ jl + δ il δ jk = 6µ( ν ((8 8νδ ijδ kl δ ij δ kl δ ik δ jl δ il δ jk = 6µ( ν ((7 8νδ ijδ kl δ ik δ jl δ il δ jk S ijmn = 2 c lkmn(d iklj + D jkli D ikkj can be evaluated by = λd ikkj δ mn µ(d inmj + D jnmi D ikkj = 6µ( ν ((7 8νδ ikδ kj δ ik δ kj δ ij δ kk Note, that now in two dimensions, δ kk = 2 Thus D ikkj = 6µ( ν ((7 8νδ ijδ kj δ ij 2δ ij = (4 8ν 6µ( ν δ ij λ = 2µν 2ν ν λd ikkj = 2( ν δ ij (22 S ijmn = ν 2( ν δ ijδ mn + 6( ν ((6 8ν(δ inδ jm + δ jn δ im 2δ ij δ mn = 4ν 8( ν δ ijδ mn + 3 4ν 8( ν (δ inδ jm + δ jn δ im 7

8 Mathematica: In[]:= Assuming[{L>,R>}, Integrate[Exp[I*L*R*Cos[Theta]],{Theta,,2*Pi}]] Out[]= 2 Pi BesselJ[, L R] In[2]:= Assuming[ {L>}, Integrate[ R*BesselJ[,L*R], {R,,}] ] BesselJ[, L] Out[2]= L In[3]:= Assuming[{b>,g>}, Integrate[Exp[-I*k*g]*BesselJ[,k*b],{k,,Infinity}]] g Sqrt[-b + g ] Out[3]= b Problem M.4 (2 Void in 2D. Consider a 2D isotropic medium (plain strain containing a circular void with radius a, with a uniform loading σ A 22. (a What is the eigenstrain e ij of the equivalent inclusion? (b Determine the location in the matrix where the maximum stress σ22 max is reached. The stress concentration factor SCF is defined as σ22 max /σ22. A Determine SCF for a circular void. Solution: (a To solve this problem, we require the stress inside the inclusion to be zero, ie σ A ij + σ I ij = σ A ij + σ c ij σ ij = For a uniform applied stress σ A 22 the equations reduce to σ A 22 + σ c 22 σ 22 = σ c σ = σ c 2 σ 2 = Because the material is isotropic, the shear terms are completely decoupled, thus we can choose e 2 = which means that σ 2 = σ c 2 =. The problem now becomes a simple problem of two equations and two unknowns (e and e 22. 8

9 Let us first express the stress σ c ij σ ij in terms of e mn. σij c σij = (c ijkl S klmn c ijmn e mn ( = c ijkl S klmn 2 δ kmδ ln 2 δ knδ lm e mn [ ] 4ν = [λδ ij δ kl + µ(δ ik δ jl + δ il δ jk ] 8( ν δ klδ mn 8( ν (δ kmδ ln + δ kn δ lm e mn µ = 4( ν (δ ijδ mn + δ im δ jn + δ in δ jm e mn This allows us to write the two equations of this problem as: σ A 22 = 3µ 4( ν e 22 = 3µ 4( ν e Solving the second equation gives µ 4( ν e µ 4( ν e 22 e = 3 e 22 Plugging this result into the first equation gives e 22 = 3 ν 2 µ σa 22 e = ν 2 µ σa 22 Now, σ and σ 22 can be computed as σ = (λ + 2µe + λe 22 σ 22 = (λ + 2µe 22 + λe which results in σ = σ 22 = (4ν ( ν σ22 A 2ν (4ν 3( ν σ22 A 2ν (b The solution can be obtained by computing the jump of the stress over the interface. Since the stress inside the void is zero, the total jump in the stress should be equal to the stress at the inside surface of the matrix. The jump in the stress is [[σ ij ]] = σ ij c ijkl (nn km σ mnn n n l 9

10 First, lets write the jump in the constrained field [[σ c ij]] as [[σij]] c = c ijkl (nn km σ mnn n n l [ ] 2µν = 2ν δ ijδ kl + µ(δ ik δ jl + δ il δ jk µ = The total jump in stress is [ δ km [ ν ν δ ijn m n n + δ im n n n j + δ jm n n n i ν n in j n m n n ] 2( ν n kn m σmnn n n l ] [[σ ij ]] = σij + [[σij]] c [ = δ im δ jn ν ν δ ijn m n n δ im n n n j δ jm n n n i + ] ν n in j n m n n For this problem, we need to compute the SCF which is only dependent on [[σ 22 ]]. ( [[σ 22 ]] = δ 2m δ 2n ν ν δ 22n m n n δ 2m n n n 2 δ 2m n n n 2 + ν n 2n 2 n m n n = σ 22 ν ν σ mnn m n n 2σ 2nn n n 2 + ν σ mnn m n n n 2 n 2 Because only σ and σ 22 are non-zero, [[σ 22 ]] = σ 22 ν ν σ mnn m n n 2σ 22n 2 n 2 + ν σ mnn m n n n 2 n 2 = σ 22 ν ν (σ n n + σ 22n 2 n 2 2σ 22n 2 n 2 + ν (σ n n + σ 22n 2 n 2 n 2 n 2 Let n = cos θ and n 2 = sin θ, the above expression can be greatly simplified, [[σ 22 ]] = ( 4 cos 4 θ cos 2 θ σ22 A = f(θσ22 A f(θ 4 cos 4 θ cos 2 θ (23 Notice that [[σ 22 ]] is independent of ν. As shown in the figure, the maximum of f(θ occurs at θ =, with f( = 3. Therefore, the stress concentration factor is SCF = σ22 max /σ22 A = 3. σ mn σ mn σ mn

11 f(θ θ / π Figure : Stress jump [[σ 22 ]] as a function of θ. f(θ = [[σ 22 ]]/σ A 22 = 4 cos 4 θ cos 2 θ.

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