PSI - Issue 3

Christos F. Markides et al. / Procedia Structural Integrity 3 (2017) 334–345 Christos F. Markides, Stavros K. Kourkoulis / Structural Integrity Procedia 00 (2017) 000–000

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6

Then, introducing Eq.(23) in Eqs.(5b), substituting in the left-hand sides of Eqs.(20) and performing the inte grations, the left-hand sides of Eqs.(20) are written in series form in terms of e ikθ . Properly passing from e ikθ to e imθ and comparing terms of e imθ on both sides of Eqs.(20), a m are obtained as (all other terms are found to be zero):

a b a b

1 P R 2 sin 2       

  

sin 2 cos 2

i2  

e



 

 

  

o

1

o   i2  

(24)

2

sin 2 e

sin 2

c

o

o

o

o



 

  



i 2          

o

o

o

o

2

2

2sin

2sin

2



 

1

o

o

1 P R

  

 

sin 2 cos 2

sin 4

1



 



 

 

3

i2  

sin 2

sin 2

e

c

o

o

o



 

   



i 6         

o

o

o

o

2 2sin sin 4 2cos 2 sin 4 sin 2 cos 4           o 2

4

3

 

i2  

e

 

o

i 2  

e

o

o

o

o

o





o



2

2

3

2sin

 

 

o









    

a b

sin m 1

sin m 1

1 P R

 

 

 

1

          

m

o

o

c



i 2 m m 1  

2 m 1 m 1 cos 2 sin m 1 2sin 2 cos m 1    o 2sin 

m

(25)















       



  

o i m 1   

o

o

o

o

1 cos m e



   





2

4 m 1  

 









c 1 P R           i 2 m m 1   sin m 1

sin m 1

 

 

 

1





o

o

m 5, 7, ... 





2 2sin m 1 cos 2 sin m 1 2sin 2 cos m 1      o m 1















       



   i m 1  

o 

o

o

o

o

1 cos m e



   





2

4 m 1  

 

Eventually, combining Eqs.(25) and (22), two systems are induced providing the real  and imaginary  parts of the non-zero coefficients A m and B m (m=3, 5, 7,…) of Φ 1 and Φ 2 , as:         m m 1 m 2 m m m m 1 1 m 2 2 m m 1 t A 1 t B a 1 t A 1 t B b                 and         m m 1 m 2 m m m m 1 1 m 2 2 m m 1 t A 1 t B a 1 t A 1 t B b                 (26)

2.2. The displacement field on the orthotropic disc Combining Eqs.(14), (16) and (26) the displacement components are written as:             1 0 1 1 1 2 0 2 1 2 1 3 13 1 m 1m 1 2 3 23 2 m 2m 2 m 5,7,9,... m 5,7,9,... u x, y 2 p A p A z p B p B z 2 p A P z A P z p B P z B P z                                    

(27)

  1 0    v x, y 2 q A q A z q B q B z 1 1 1 2 0   2 1 2 





(28)

   

    

 

  

  

  





  1 3 13 1 2 q A P z   

  A P z q B P z    m 1m 1 2 3 23 2

 

m 2m 2 B P z





m 5,7,9,... 

m 5,7,9,... 

and it is seen that are not completely determined due to the indeterminacy in A 0 , B 0 , A 1 and B 1 . In this direction, it is mentioned that in the present problem the disc’s center should remain fixed, i.e. it should hold that u(0,0)=v(0,0)=0. But Eqs.(17), for z 1 =z 2 =0, yields P 1m (0)=P 2m (0)=0 (m=3,5,7,9,…); hence, by Eqs.(27), (28) it follows that the real parts of the expressions p 1 A 0 +p 2 B 0 and q 1 A 0 +q 2 B 0 should also be zero, which could be the case if:

(29)

0 0 A B 0  