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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Fig. 2. Three basic loads at infinity for line crack of free of traction and electric displacement

The weight functions t yy ( ξ ), t xy ( ξ ) and q y ( ξ ) are usually set up as a polynomial of ξ . As the most simple situation, let us consider those weight values are all constants, then integral in Eq.(34) can be performed analytically as follows.

b 1 j    t yy

1 2 i ∆   

   µ j  b 2 j −

ϵ 22 s 12 − d 12 d 22 ϵ 22 s 11 − d 2

12 

Φ j ( z j ) =

b 1 j    q y

    z 2

+ ( µ j  b 1 j +  b 2 j + d 61  b 3 j ) t xy +  

j 

d 12 ϵ 22 s 11 − d 2

 µ j  b 3 j −

12 

2 − z

(35)

j − a

and

b 1 j    t yy

1 2 i ∆   

   µ j  b 2 j −

ϵ 22 s 12 − d 12 d 22 ϵ 22 s 11 − d 2

Φ ′ j ( z j ) =

12 

   z j  z 2

1   

b 1 j    q y

  

+ ( µ j  b 1 j +  b 2 j + d 61  b 3 j ) t xy +  

d 12 ϵ 22 s 11 − d 2

 µ j  b 3 j −

12 

(36)

2 −

j − a

Using this potential to find σ y , τ xy and D y on the x -axis, we will obtain the following expression for ( | x | > a ) that   σ y τ xy D y   = 1 2 i ∆  | x | √ x 2 − a 2 − 1     A 11 A 12 A 13 A 21 A 22 A 23 A 31 A 32 A 33      t yy t xy q y  

(37)

and for ( | x | < a ) that   σ y τ xy D y   = −

  

1 2 i ∆   

  t yy t xy q y

 

A 11 A 12 A 13 A 21 A 22 A 23 A 31 A 32 A 33

(38)

Here, A i j are all real constants. The concrete forms of these constants are shown in appendix. The unknowns, t yy , t xy , and q y , are determined from the conditions under which the stresses σ y and τ xy and electric displacement D y are zero on the crack face. According to Fig.2, the boundary condition becomes   σ ∞ y τ ∞ xy D ∞ y   − 1 2 i ∆    A 11 A 12 A 13 A 21 A 22 A 23 A 31 A 32 A 33      t yy t xy q y   =   0 0 0   (39) After solving unknowns ( t yy , t xy , q y ) from the above equation then substitute to Eq.(37), stresses and electric dis placement on the x -axis (outside of the crack) becomes

x |

x |

x |

|

|

|

σ ∞ y , τ xy =

τ ∞ xy , D y =

D ∞ y

(40)

√ x 2

√ x 2

√ x 2

σ y =

− a 2

− a 2

− a 2