Issue 63

P. Livieri et alii, Frattura ed Integrità Strutturale, 63 (2023) 71-79; DOI: 10.3221/IGF-ESIS.63.07

and   0 0 0.5 y L

 

     2 3 Q L y ,

For example, we explain Eqn. (6) on an equivalent triangle with side L, where

as shown

 0

in Fig. (3):

   

   

   

   

   

   

0.4

0.4

0.4

y L

y

y

1 2

3 2

1 2

  

  

  

  

  

  

   Q’

0

0

0









·

(7)

1.046tanh 4.14

tanh 4.14

·tanh 4.14

L

L

Clearly, in the r.h.s. of the Eqn. (7), the only significant contribution is given by

   

   

0.4

y L

1 2

  

  

   Q’

0





(8)

1.046tanh 4.14

Figure 3: Equilateral triangular crack.

Figure 4: Square crack.

and   0 0 0.5 y L

     1 , 2 Q L y

 

As a second example, Fig. 4 shows the case of a square crack of side L, where

. Then

 0

   

   

   

   

0.4

0.4

y

y

1 2

3 2

  

  

  

  

0

0





1.033·tanh 4.34

· tanh 4.34

·

L

L

   Q’



(9)

   

   

   

   

0.4

0.4

y

y

3 2

1 2

  

  

  

  

0

0





tanh 4.34

·tanh 4.34

L

L

In the r.h.s. of the Eqn. (8), the only significant contribution is given by

   

   

0.4

y L

1 2

  

  

   Q’

0





(10)

1.033tanh 4.34

75