PSI - Issue 28

Alla V. Balueva et al. / Procedia Structural Integrity 28 (2020) 873–885 Author name / Structural Integrity P o edi 00 (2019) 000–000

879 7

for each crack contour, we are solving stationary equation (1.9), namely 2 0 c   . After subtraction of the equilibrium state for the medium without the crack, the solution of this boundary value problem can be found e.g. in Sneddon [1972], and the expression for the flux density, q =  D  c /  z ( z = 0, 0  r < a ), can be written as follows:

2

c D

( , 0, )

,

( )

q r t

r a t 

(1.11)



0



2

2

( )

a t

r



Using the expression (1.11) now to find the full flux Q(t) , we get the following expression:

1

a



(1.12)

( ) 4 

4 ( ) rdr c Da t 

Q t

c D

0

0

2 a r 

2

0

So we have the following expression of molar mass m in terms of a as follows:

t

t





( ) m Q t dt   





4

( ) c D a t dt

(1.13)



0

0

0

From here, we substitute our expression for molar mass (1.13) into the formula (1.6), and we come up with a kinetic equation for the crack radius a ( t ):

5 2

a

t



( ) a t dt 



(1.14)





a





0



3

2 2 c B K 

0 c DE RT K  2

and

where

(1.15)





 

c

We reduce the integral equation (1.14) into a differential equation by differentiating both sides with respect to t :

1 2

5

a

a

(1.16)

[

]

da dt  



2

2(

) 2(

)

a

a









which leads to the close-form solution as follows:

1





2 2 ln

3  

2 3 a 

(1.17)

a   

a

t C

 

  

(

)

a





and taking into account that a (0) = 0, thus we obtain the analytical solution for the radius of the internal crack a ( t ) as it depends on t , as follows:

  

  

  

  

1

1

a

(1.18)

2 2 ln 1

3 

2   

2 3 a 

a   

t

 





(

)

a



