Kroner

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Similarly: M 33 = Thus, for: ∫ π ∫ 2π M 22 = ∫ 2π sin 2 ΘdΘ ∫ π 0 0 cos 2 ΦsinΦdΘdΦ = (−1)2π 0 0 0 sin 3 ΦdΦ = 4 3 π (250) ∫ π cos 2 Φd(cosΦ) = 4 3 π (251) ijkl = 1111, 1122, 1133, 2211, 2222, 2233, 3311, 3322, 3333 (252) M ijkl = 4 3 π (253) These results clearly show that, given all the non-zero terms, M ijkl does have major symmetry as well as minor symmetry. Now that we have shown that both N ijkl and M ijkl have both major and minor symmetry, we can construct the following model for D ijkl which assumes both major and minor symmetry: D ijkl = γδ ij δ kl + ω(δ ik δ jl + δ il δ jk ) (254) where γ is given by any member of the list in Eq. 252, say 1122 for which: = ∫ π ∫ 2π 0 0 N 1122 = ∫ π ∫ 2π 0 0 ( z 2 1 z 2 2sinΦ ) dΘdΦ (255) ( sin 2 Φcos 2 Θsin 2 Φsin 2 ΘsinΦ ) dΘdΦ (256) = ∫ 2π 0 cos 2 Θsin 2 ΘdΘ ∫ π 0 ∫ π ( ∫ 1 2π ) ( ) 4 = sin 2 (2Θ)d(2Θ) sin 3 ΦdΦ 8 0 5 0 ( [ ] ) 2π ( 1 1 = 8 2 (2Θ) ( 4 [ 0 5 ) − 1 ] π ) 3 (cosΦ)(2) = 4π 0 15 sin 5 ΦdΦ (257) (258) (259) so that: D 1122 = − 1 (M 1122 + 2κ 1 N 1122 ) 4πµ 1 (260) = − 1 ( 4 4πµ 1 3 π + 2κ 4π 1 15 ) (261) = − 1 3µ 1 (1 + 2 5 κ 1) (262) γ = D 1122 = − 1 3µ 1 (1 + 2 5 κ 1) (263)
Finally, ω represents any D ijkl term for which M ijkl = 0 but N ijkl ≠ 0, that is, for the following list of (ijkl) indices for which (i ≠ j and k ≠ l) but for which (i = k and j = l) or (i = l and j = k): 1212, 2121, 1313, 3131, 2323, 3232, 1221, 2112, 1331, 3113, 2332, 3223 (264) For D 1313 , M 1313 = 0 and: = ∫ 2π 0 = = ∫ 2π 0 ∫ π ∫ 2π 0 N 1313 = 0 ∫ 2π = 0 ∫ π ∫ 2π 0 0 ( z 2 1 z 2 3sinΦ ) dΘdΦ (265) ( sin 2 Φcos 2 Θcos 2 ΦsinΦ ) dΘdΦ (266) cos 2 ΘdΘ ∫ π 0 sin 3 Φcos 2 ΦdΦ (267) (∫ π cos 2 ΘdΘ sin 3 Φ ( 1 − sin 2 Φ ) ) dΦ (∫ π cos 2 ΘdΘ sin 3 ΦdΦ − 0 0 ∫ π 0 (268) ) sin 5 ΦdΦ = 4 15 π (269) so that: ω = D 1313 = − 2κ 1 N 1313 = − 2κ 1 4π 4πµ 1 4πµ 1 15 = − 2 15 κ 1 µ 1 (270) giving our model as: D ijkl = − 1 (1 + 2 3µ 1 5 κ 1)δ ij δ kl − 2 κ 1 (δ ik δ jl + δ il δ jk ) (271) 15 µ 1
Page 1 and 2: Derivation and role of Kroner’s c
Page 3 and 4: 1 Abstract The pioneering work of H
Page 5 and 6: Such a tensor can be decomposed as
Page 7 and 8: For a constant point force F acting
Page 9 and 10: or, defining: we can write: ∫ u i
Page 11 and 12: 7 Symmetry properties and model of
Page 13 and 14: where: [k] = k 2 − k 1 (105) [µ]
Page 15 and 16: Markov now describes the critical c
Page 17 and 18: 11 Rotational averages Before proce
Page 19 and 20: Note that from Eqs. 10, 11, and 174
Page 21 and 22: Since I ′ and I ′′ are orthog
Page 23 and 24: value at the arithmetic average of
Page 25: 14 Major symmetries of tensor D Wri
Page 29 and 30: otherwise there is no solution (Mor
Page 31: [15] P. Sisodia and M.P. Verma. She

Kroner

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