Issue 59

M. Shariyat, Frattura ed Integrità Strutturale, 59 (2022) 423-443; DOI: 10.3221/IGF-ESIS.59.28









*





, N R

, N R

Σ

/ Σ

 

 

a

a

1

2

1

1

2

2

(26)









*





, N R

, N R

Σ

/ Τ

 

 

a

a

1

1

1

Therefore, according to Eqns. (22) and (26), one may deduce that:

2

2

 







   



    , , m u m u 1 1 2 2

  , m u











    

        

 1

 1

a

a

1

1

1 1





R

R

1

1

2

  t

    2 t

    t

  



 1

 





1

1









 

  , m u





a

a

2





R

R

1

1

  t

2

12





(27)





2

eq

 







    , , m u m u 1 1 2 2

 1

a

1



R

1



  t

( ) ( ) t



1



1

2







a

2



R

1

2



2

The time history of    eq t stress has to be plotted and the contents of the cycle histograms may be determined based on the Rainflow procedure; so that Miner’s damage accumulation rule can be applied to the stochastic time history sample to evaluate the resulting fatigue damage:

( ) n

   1 i

i eq

D

(28)

N



eq

where

m

   

    



   m

 

 

1

* 1 Σ Γ

 u

R

,

, eq



1

eq

 

N

(29)

  



eq

a eq

 eq N for the i th Rainflow cycle counting stage of the histogram of the equivalent

( ) i eq n is the number of cycles associated

stress. To check the matrix failure, one may start from Eqn. (23), expressing all fatigue strengths in terms of Y instead of X . Using a procedure similar to that led to Eqn. (27), the following equivalent stress expression may be proposed for checking the matrix cracking:

2

2

 







   



    , , m u m u 1 1 2 2



  , m u

    









        





a

a

2

2

2 2





R

R

1

1

2

  t

    1 t

2

2



  





( ) t

 





2

2







2





 

  , m u

 1



a

a

1





R

R

1

1

  t

12





(30)





1

eq

 







    , , m u m u 1 1 2 2



a

2



R

1

2



  t

( ) ( ) t



2



1

2





 1

a

1



R

1



1

Similarly, starting from Eqn. (24), shear failure may be checked by using the following equivalent stress expression:

2

2









    

          

    

  , m u

  , m u

 

 





a

a

    1 t

    2 t





R

R

1

1

   2 ( ) t

12

12









  , m u

  , m u













a

a

1

1 1

2

2 2





R

R

1

1

1

2

  t





1

2





(31)





eq

2





  , m u

 



a



R

1

       1 2 t t

12







   

  , m u

  , m u









a

a

1

2

1 1

1 1





R

R

1

1

1

1





1

2

So that, Eqn. (28) may be replaced by:

430