Issue 58

K. Benyahi et alii, Frattura ed Integrità Strutturale, 58 (2021) 319-343; DOI: 10.3221/IGF-ESIS.58.24

The mean strain in Eqn. 6 can be transformed as follows:

1

 = ij

− u n u n dS )



(21)

(

i

j

j i

j

V

( ) S

j

In order to evaluate the effective properties ij C , it is first necessary to evaluate the mean of the stress and the strain of Eqns. 5 and 6 respectively, by the application of different outline conditions. Then to introduce them into the constitutive relation as follows:

 

ij

=

C

(22)

ij

ij

with: ij C : The stiffness modulus corresponding to the application of the strain mode. From the relation between the flexibility matrix S, and the elastic property, we can calculate the effective elastic properties from the stiffness matrix of the RVE.





1

                 

                 

31

21

-

0

-

0 0

E

 E E 1 - 11 12

33

22



32

-

0 0 0

11 E E E 1 22

33

0 0 0

 

13

23

-

-

E

33

 

E E

S =

(23)

11

22

1

0

0

0

0 0

G

12

1

0

0 0

0 0 G

13

0 0

0 0

1

0

G

23

There are only five independent material constants (

)   11 33 12 31 32 , , , , E E G for a transversely orthotropic / isotropic

material. Where  13 depends on (

)   11 33 12 31 , , , E E and can be calculated by the following relation:

  31 13 = E E

(24)

33

11

A material is isotropic when its property is the same in all three directions. In this case, (

) = = = 11 22 33 E E E E ,

( )     = = = 12 31 32 and (

) = = = 12 31 32 G G G G . There are only two independent material constants (

)  , E for an

isotropic material. However, anisotropic materials have non-zero values in the upper right and lower left parts of their flexibility and stiffness matrices. There exists a relation between the components of the flexibility matrix, with the components of the rigidity matrix, which is as continuation: ( ) ( ) 2 11 33 13 11 2 2 2 2 11 33 11 11 13 12 33 12 13 C .C -C 1 S = = E C .C -2.C .C -C .C +2.C .C (25)

325