Issue 57

E. Sgambitterra et alii, Frattura ed Integrità Strutturale, 57 (2021) 300-320; DOI: 10.3221/IGF-ESIS.57.22

Therefore, if a further term, n =2, is added to the Eqn. (8), the new equations can be written as follows:

    11 21

 

  

       ψ u U

T

12

 

I K T

(10)

22

where

r

 

  

cos

 2 1 

12

 vr

v

    

(11)

  

sin

 2 1 

22

 

v

The theoretical displacement fields, calculated from Eqn. (10), are illustrated in Fig. 1 in the form of contour maps. Case study 2: displacement fields of a brazilian disk The displacements field of a disk, under plane stress condition, subjected to a compression load P along the y direction (see Fig. 2), can be expressed as follows [56]:

    11 21

  

  

       ψ u U

  / T P E P v E  /

12

(12)

22

where

  x

            

2

  

  

1

1 2

1 2

   1

   2

  1 2

 

 

sin 2

sin 2

11

 t

D

  x

2

  

  

1

1 2

1 2

   1

   2

   12

  1 2

 

sin 2

sin 2

 t

D

(13)

  

  

y

2

      2 1 r

1

1 2

1 2

  

   1

     

ln r

2

cos 2

cos 2

21

2

 t

D

  

  

y

2

1 1

1 2

   1

     

 

cos 2

cos 2

22

2

 t

D

2

In Eqn. (12) coordinates   , x y are referred to the center of the disk ( x 0 , y 0 ), t is the disk thickness, D is the disk diameter, ( r 1 ,  1 ) and ( r 2 ,  2 ) are the polar coordinates of the generic point with respect to the contact points A and B reported in Fig. 2. These latter can be expressed as follows:

  1

/ 2 x D y

 

 1 tan

(14)

   

  x D y / 2

  1

 2

  tan

(15)

   

2

    

  x D y    2 / 2

r

(16)

1

304

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