Issue 65

S. M. J. Tabatabee et alii, Frattura ed Integrità Strutturale, 65 (2023) 208-223; DOI: 10.3221/IGF-ESIS.65.14

where ' ij C s are the components of the compliance matrix. In orthotropic material with a Cartesian coordinate system, this matrix can be written as:      

1

yx

zx

0 0 0

                   

                   

E E E

x

y

z









1

xy

zy

0 0 0

                        x y yz zx xy  z   

        x y                  z yz zx xy   

E E E

x

y

z

 

  xz

1

yz

0 0 0 1

E E E

x

y

z



(12)

0 0 0

0 0 1

G

yz

0 0 0 0

0 1

G

zx

0 0 0 0 0

G

xy

For plane stress conditions in the x-y plane, only the 11 C , 22 C , 12 C , 21 C , and 66 C will remain. The components of this matrix are constant, and the matrix is symmetric. For - plane strain cases, four components of this matrix should be changed as below:    3 3 33 ( , 1,2) ' / ij ij i j i j C C C C C (13) To include the effect of the reinforcement factor, we repeat the previous procedure with a slight change. This time we are using Eq. (1-3) Instead of Eq. (7-9), and the result can be written as [26]:

4

4

4

 













 





(14)

C

n C

1 12 (1 ) ) n C

nC

(

0

22

6 66

1 11

4

2 2 x y

4

 



x

y

The reinforcement factor should have also followed the below configuration to be used as isotropic material.

11 x y C E C E

  22

(15)

n

1

) xy G

  

   C C C C C C 12 22 11

   12 2

22

   (2

  yx

n

(16)

xy

6

E

y

66

For a cracked orthotropic material generalized elastic moduli defined as [27]:



1/2

   

   

   

 

12 C C

2

11 22 C C C

66

22 11





(17)

E

  

I

C

C

2

2

11



1/2

   

   

   

2

 

12 C C

2

11 C C

66

22 11





(18)

E

  

II

C

C

2

2

11

218