- Email: [email protected]
Appendix C
Appendix C An alternate method for evaluating the eigenfunctions of an elastic wave propagating in the [111] direction is as follows. The equations of motion for the displacement of elastic waves in the [111] direction, from Eqs. (5.116), (5.118), (5.119), can be written in matrix form as 0 1 0 1 1 1 + C 3L C + C + C C C u1 44 12 44 12 44 B 11 C 2 2 B CB C B CB C 1 B C12 + 1 C44 C B u2 C ¼ 0 (C.1) C11 + C44 3L C12 + C44 B CB C 2 2 B C@ A @ A 1 1 u3 C12 + C44 C12 + C44 C11 + C44 3L 2 2 For the eigenvalue L2, given by Eq. (5.128), Eq. (C.1) becomes 0 1 0 1 1 1 + C 3L C + C + C C C 44 2 12 44 12 44 B 11 C u1 2 2 B CB C B CB C 1 B C12 + 1 C44 C B u2 C ¼ 0 C11 + C44 3L2 C12 + C44 B CB C 2 2 B C@ A @ A 1 1 u3 C12 + C44 C12 + C44 C11 + C44 3L2 2 2 Substituting the value of L2 from Eq. (5.128) into Eq. (C.2), we 0 1 1 C12 + C44 B 2 C12 + 2 C44 B 2 B 1 B C12 + 1 C44 2 C12 + C44 B 2 2 B @ 1 1 C12 + C44 C12 + C44 2 2
(C.2)
obtain 1
0 1 C u1 CB C CB C C B u2 C ¼ 0 CB C @ A C A 1 u3 2 C12 + C44 2
1 C12 + C44 2 1 C12 + C44 2
(C.3)
The above matrix equation gives the following three equations 2u1 + u2 + u3 ¼ 0 u 1 2 u2 + u3 ¼ 0
(C.4)
u1 + u2 2u3 ¼ 0 Solving Eq. (C.4), one immediately gets u1 ¼ u2 ¼ u3
(C.5)
According to Eq. (C.5) the eigenfunction corresponding to L2 is in the [111] direction. Hence the displacement is in the direction of propagation (K) of the wave, which means that it corresponds to a longitudinal wave.
597
598
Appendix C
Now for the first solution given by L1, Eq. (C.1) becomes 0 1 0 1 1 1 C C + C 3L C + C + C u1 11 44 1 12 44 12 44 B C 2 2 B CB C B CB C 1 B C12 + 1 C44 C B u2 C ¼ 0 C11 + C44 3L1 C12 + C44 B CB C 2 2 B C@ A @ A 1 1 u3 C12 + C44 C12 + C44 C11 + C44 3L1 2 2 Substituting the value of L1 from Eq. (5.134) into Eq. (C.6), 0 1 1 B C12 + 2 C44 C12 + 2 C44 B B B C12 + 1 C44 C12 + 1 C44 B 2 2 B @ 1 1 C12 + C44 C12 + C44 2 2
one gets 1 C12 + C44 2 1 C12 + C44 2 1 C12 + C44 2
(C.6)
1
0 1 C u1 CB C CB C C B u2 C ¼ 0 CB C C@ A A u3
(C.7)
The three equations given by Eq. (C.7) yield u1 + u 2 + u 3 ¼ 0 (C.8) Let one of the eigenfunctions corresponding to L1 be given by 110 , which satisfies Eq. (C.8). The other eigenfunction must be perpendicular to both [111] and 110 and can be shown to be 112 .
(C.2)
obtain 1
0 1 C u1 CB C CB C C B u2 C ¼ 0 CB C @ A C A 1 u3 2 C12 + C44 2
1 C12 + C44 2 1 C12 + C44 2
(C.3)
The above matrix equation gives the following three equations 2u1 + u2 + u3 ¼ 0 u 1 2 u2 + u3 ¼ 0
(C.4)
u1 + u2 2u3 ¼ 0 Solving Eq. (C.4), one immediately gets u1 ¼ u2 ¼ u3
(C.5)
According to Eq. (C.5) the eigenfunction corresponding to L2 is in the [111] direction. Hence the displacement is in the direction of propagation (K) of the wave, which means that it corresponds to a longitudinal wave.
597
598
Appendix C
Now for the first solution given by L1, Eq. (C.1) becomes 0 1 0 1 1 1 C C + C 3L C + C + C u1 11 44 1 12 44 12 44 B C 2 2 B CB C B CB C 1 B C12 + 1 C44 C B u2 C ¼ 0 C11 + C44 3L1 C12 + C44 B CB C 2 2 B C@ A @ A 1 1 u3 C12 + C44 C12 + C44 C11 + C44 3L1 2 2 Substituting the value of L1 from Eq. (5.134) into Eq. (C.6), 0 1 1 B C12 + 2 C44 C12 + 2 C44 B B B C12 + 1 C44 C12 + 1 C44 B 2 2 B @ 1 1 C12 + C44 C12 + C44 2 2
one gets 1 C12 + C44 2 1 C12 + C44 2 1 C12 + C44 2
(C.6)
1
0 1 C u1 CB C CB C C B u2 C ¼ 0 CB C C@ A A u3
(C.7)
The three equations given by Eq. (C.7) yield u1 + u 2 + u 3 ¼ 0 (C.8) Let one of the eigenfunctions corresponding to L1 be given by 110 , which satisfies Eq. (C.8). The other eigenfunction must be perpendicular to both [111] and 110 and can be shown to be 112 .