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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This stress distribution is perfectly consistent with that of along the crack line in isotropic materials. Besides, the electric displacement has the equivalent distribution with same singularity. As a result, SIF and electric displacement intensity factor becomes as follows. K I = σ ∞ y √ π a , K II = τ ∞ xy √ π a , K D = D ∞ y √ π a (41) 2.4.2. Case for crack is parallel to y-axis (parallel to poling axis) In order to analyze a crack problem parallel to the y -axis of an infinite piezoelectric plate subjected to uniform far-field stress and electrical displacement using the complex potential in Eq.(32), it is important to properly set up the distribution of force and charge doublets ( T xx , T xy and Q x ). Consider the crack lies on the y -axis, which is parallel to poling axis, in the range of | y | < a , where a is a half crack length. The unknown distribution of T xx , T xy and Q x are assumed as a product of some weighting function and a fundamental density function that exhibits a crack tip singularity. Then the distribution of T xx , T xy and Q x per infinitesimal length along the crack length d η are: dT xx = t xx ( η ) ×  a 2 − η 2 d η, dT xy = t xy ( η ) ×  a 2 − η 2 d η, dQ x = q x ( η ) ×  a 2 − η 2 d η (42) Then the complex potential in Eq.(32) becomes as follows. Φ j ( z j ) = − 1 2 π i ∆  a − a   b 1 j − s 12 s 22 s 12 s 22 d 22  µ j  b 3 j  t xx ( η ) + ( µ j  b 1 j +  b 2 j ) t xy ( η ) µ j  b 2 j +  d 12 −

+  b 3 j q x ( η )  

a 2 − η 2 z j − µ j η

d η

(43)

The weight functions t xx ( η ), t xy ( η ) and q x ( η ) are usually set up as a polynomial of η . As the most simple situation, assume those weight values are all constants as before, then integral in Eq.(43) can be performed analytically as follows.

s 12 s 22 d 22  µ j  b 3 j  t xx ( η ) + ( µ j  b 1 j +  b 2 j ) t xy ( η )

1 2 i ∆ µ j   b 1 j − +  b 3 j q x ( η )    

µ j  b 2 j +  d 12 −

s 12 s 22

Φ j ( z j ) =

  z j /µ j  2

j /µ j   

− a 2 − z

(44)

and

   b 1 j µ 2

µ j 

1 2 i ∆   

    b 1 j µ j

j  

d 22   b 3 j

+  d 12 −

s 12 s 22  b 2 j µ j

+  b 2 j µ 2

s 12 s 22

Φ ′ j ( z j ) =

 t xx ( η ) +

 t xy ( η )

j −

− 1   

q x ( η )    

+  b 3 j µ 2 j

z j /µ j

(45)

 ( z j /µ j )

2 − a 2

Using this potential to find σ x , τ xy and D x on the y -axis, we will obtain the following solutions for ( | y | > a ) as   σ x τ xy D x   = 1 2 i ∆    | y |  y 2 − a 2 − 1       B 11 B 12 B 13 B 21 B 22 B 23 B 31 B 32 B 33      t xx t xy q x  

(46)

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

  

1 2 i ∆   

  t xx t xy q x

 

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

(47)

Here, B i j are all real constants. The concrete forms of these constants are shown in appendix. The unknowns, t xx , t xy , and q x , are determined from the conditions under which the stresses σ x and τ xy and electric displacement D x are