PSI - Issue 52

T. Profant et al. / Procedia Structural Integrity 52 (2024) 455–471 T. Profant et al / Structural Integrity Procedia 00 (2023) 000 – 000

465 11

( ) ,0

l r

0 im →

2

K

rE r 

=

(32)

E

r

0  = :

with boundary conditions at

K

. E

0

,

= = t r

E

(33)

=

r

2

r 

0 r E = Ǥ

On the crack faces,   = , traction and higher-order traction are zero and For insulating crack they defined the intensity factor K D as

( )

0 lim 2 r →

,0 ,

D K

rD r  

=

(34)

0  = :

with boundary conditions at

2 D K

0,

= = t r

D

=

(35)

.



r



where 0 D E P    = + is the component of the electric displacement. Mao and Purohit (Mao & Purohit, 2015) considered mixed modes when pure mechanical Modes I and II are combined with electrical modes D and E . They showed that the boundary conditions for the individual modes must be compatible so that the respective combination of the mechanical mode and the electrical mode is admissible. For example, in a flexoelectric solid, Mode I (pure insulating crack) cannot be mixed with Mode D. Similarly, Mode II (pure conductive crack) cannot be mixed with Mode E under the current framework. On the other hand, since the Mode I solution does not exclude the possibility of non-zero K E , a crack in which Mode I pure conducting conditions are mixed with Mode E is admissible. Then, the boundary condition is different from that of the Mode I pure conducting case specified in Eqs. (21) and (22). Namely, the electrical boundary conditions for a conductive crack read

K

( E r r

)

( ) ,0

E

(36)

,

0,

E r

  =

=

r

2

r



Asymptotic form of the electric potential then is given as

1 2

 

  

1 2

1

1 2

3 2

1 2

f

  

  

( ) , r  

(

)

cos

sin

5 cos A

3 5 cos l A

cos

,

r C 

D

A

=

+

−

+ +

+











(37)

1

1

1

12

3

4

2 2

a

and the components of the asymptotic electric field follow from Eqs. (2) 3 and (37) as

  

  

1

1 2

1

1 2

3 2

1 2

f

  

  

( ) , 

(

)

cos

sin

5 cos A

3 5 cos l A

cos

,

E r

C

D

A











=

+

−

+ +

+

1

1

1

12

3

4

r

1/2

2 2

a

2

r

(38)

  

  

1

1 2

1

1 2

3 2

1 2

f

  

  

( ) , 

(

)

sin

cos

5

sin

12 3 3 5 l

sin

sin

.

E r 

C

D

A

A

A











=−

−

−

+ +

+

1

1

1

3

4

1/2

2 2

a

4

r

From the boundary condition (36) 1 one gets