PSI - Issue 80
Akihide Saimoto et al. / Procedia Structural Integrity 80 (2026) 352–367 A. Saimoto et al. / Structural Integrity Procedia 00 (2023) 000–000
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366
b 1 j
b 1 j
µ j γ j
j ϵ 22 )
3 j = 1 3 j = 1 3 j = 1 3 j = 1 3 j = 1 3 j = 1
3 j = 1
ϵ 22 s 12 − d 12 d 22 ϵ 22 s 11 − d 2
ϵ 22 s 12 − d 12 d 22 ϵ 22 s 11 − d 2
µ j ( ϵ 11 + µ 2
12
12
µ j b 2 j −
µ j b 2 j −
= − 2
(B.4)
A 21 = − 2
3 j = 1
µ j ( ϵ 11 + µ 2
µ j γ j ( µ j b 1 j + b 2 j + d 61 b 3 j ) = − 2
j ϵ 22 )( µ j b 1 j + b 2 j + d 61 b 3 j )
A 22 = − 2
(B.5)
b 1 j b 1 j
b 1 j
µ j γ j
j ϵ 22 )
3 j = 1
d 12 ϵ 22 d 11 − d 2
d 12 ϵ 22 d 11 − d 2
µ j ( ϵ 11 + µ 2
12
12
µ j b 3 j −
µ j b 3 j −
A 23 = − 2
= − 2
(B.6)
r j
ϵ 22 s 12 − d 12 d 22 ϵ 22 s 11 − d 2
µ j b 2 j −
12
A 31 = − 2
b 1 j
j ( ϵ 11 d 12 + ϵ 22 d 61 ) }
3 j = 1
ϵ 22 s 12 − d 12 d 22 ϵ 22 s 11 − d 2
{ ϵ 11 d 22 + µ 2
µ j b 2 j −
12
(B.7)
= 2
3 j = 1
{ ϵ 11 d 22 + µ 2
r j ( µ j b 1 j + b 2 j + d 61 b 3 j ) = 2
j ( ϵ 11 d 12 + ϵ 22 d 61 ) } ( µ j b 1 j + b 2 j + d 61 b 3 j ) j ( ϵ 11 d 12 + ϵ 22 d 61 ) } d 12 ϵ 22 d 11 − d 2
A 32 = − 2
(B.8)
r j
b 1 j
b 1 j
3 j = 1
d 12 ϵ 22 d 11 − d 2
{ ϵ 11 d 22 + µ 2
µ j b 3 j −
12
µ j b 3 j −
12
A 33 = − 2
= 2
(B.9)
Referring to Eq.(A.5) and Eq.(A.8), as ℑ 3 j = 1 µ 2 k = 0, for k = 1 ∼ 3, A 11 and A 13 are found to be real number not complex immediately. On the other hand, regarding to Eq.(B.2), we need simple verification. From Eqs.(A.4), (A.6) and (A.7), we found ℑ 3 j = 1 ( ϵ 11 + µ 2 j ϵ 22 ) µ j b 1 j = i ∆ 2 , ℑ 3 j = 1 ( ϵ 11 + µ 2 j ϵ 22 ) b 2 j = − i ∆ 2 , ℑ 3 j = 1 ( ϵ 11 + µ 2 j ϵ 22 ) b 3 j = 0 therefore, finally we may find that A 12 is also a real number not a complex number. As for the remaining terms, A 22 , A 31 and A 33 are easy to find to be real numbers, but for A 21 , A 23 and A 32 requires some calculation to confirm that they are real number not complex. Constants B i j defined in Eq.(47) are as follows. j b 1 j = 0 and ℑ 3 j = 1 µ 2 k − 1 j b 2 j = ℑ 3 j = 1 µ 2 k − 1 j b 3 j
µ j b 2 j + d 12 −
γ j b 1 j −
d 22 µ j b 3 j
3 j = 1 3 j = 1 3 j = 1
s 12 s 22
s 12 s 22
B 11 = 2
(B.10)
γ j µ j b 1 j + b 2 j
B 12 = 2
(B.11)
γ j b 3 j
B 13 = 2
(B.12)
γ j b 1 j γ j
d 22 b 3 j
b 2 j + d 12 −
3 j = 1 3 j = 1
s 12 s 22 µ j
s 12 s 22
B 21 = − 2
(B.13)
µ j −
b 1 j + b 2 j
B 22 = − 2
(B.14)
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