PSI - Issue 3

5

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

338





m 2  

m 2  

  z A A z     1 1 0 1 1

  A P z , m 1m 1

  z B B z     2 2 0 1 2

 

(16)

m 2m 2 B P z

where

 R 1 i  m 1

m

m

  

  









 

  

     

  

(17)

2 z R 1          2 2 2 z R 1 2 z

2

P z

z





1m,2m 1,2

1,2

1,2

1,2

1,2

1,2

1,2

m

 

1,2

The coefficients of Φ 1 and Φ 2 of Eqs.(16) will be determined by fulfilling the boundary conditions of the problem. In this direction, combining Eqs.(7), (12) and (13), yields:             2 2 x 1 1 1 2 2 2 y 1 1 2 2 xy 1 1 1 2 2 2 2 z z , 2 z z , 2 z z                                          (18) Introducing Eqs.(18) in Eqs.(5a) and integrating along L with respect to S one obtains the Cartesian components of the resultant force per unit thickness over an arc of length S of L as:         S S 1 1 2 2 n 1 1 1 2 2 2 n 0 0 2 z z Y dS, 2 z z X dS                        (19) Substituting in the left-hand sides of Eqs.(19) from Eqs.(16) (taking into account that when z is on L it holds that z 1,2 =R(1–iμ 1,2 )s/2+R(1+iμ 1,2 )s/2, so that from Eqs.(17) it is P 1m,2m = – s m – t 1,2 m s – m , where t 1,2 =(1+iμ 1,2 )/ (1–iμ 1,2 )), considering small s=e iθ , and writing the right hand sides of Eqs.(19) in Fourier series form:     S S m m m m n 0 m m n 0 m m m 1 m 1 0 0 Y dS a a s a s , X dS b b s b s                  (20) And then comparing the coefficients of s ±m of the same order on both sides of the resulting expressions, the following systems are obtained for the coefficients of Φ 1 and Φ 2 (Lekhnitskii 1968):               1 1 1 1 1 1 1 1 1 2 1 1 1 2 1 1 1 1 2 2 2 2 1 1 1 2 1 1 1 2 1 1 A B A B a a R A B A B a a Ri b b R , m 1 A B A B b b Ri                                (21)



m

m

m m m 1 A B A t  

B t

a



 

m 2

m

m                        m m 1 A B A t m 2 m 1 1 m 2 2 B t m m m m m m 1 A B A t m 2 m m m m 1 A B A t m 2 m 1 1 m 2 2 m b B t      a B t b 





,

m 2,3,... 

(22)

A 0 and B 0 (for m=0) remain arbitrary, while as it can be seen from Eqs.(21) 1 1 1 1 A , B , A and B are determined apart from an arbitrary real constant. Thus, the problem of obtaining the coefficients of Φ 1 , Φ 2 is reduced to the determination of a m and b m . In this direction the left-hand sides of Eqs.(20) must be expressed in series form also. Bearing in mind Eqs.(5b), σ r of Eq.(1) is first expanded in Fourier series form, as:

  

  

  

P

2 sin 2

sin 2 cos 2

i 2  

e

  

 

 

  

o

o    i2   sin 2 e

i 2

c      2

sin 2

e



o

o

o

o

   

r

o

o

o

o

2

2 2sin sin 2 cos 2

2

2sin





 

o

o

  

   

sin k   

i2



e

k 3   

 

o

o    i2 sin 2 e 

i 2  

sin 2

e

o

o

o

   

 

 

 



o

o

o

2 2sin sin k k c  o

2

k

 



(23)

     

os 2 sin k 2sin 2 cos k

sin k

1



     

    

  





 

ik   

ik  

 

ik

e

e

e



o

o

o

o

o

o









o

    

o

k 4 k sin k k cos 2 sin k 2sin 2 cos k        2

2

k

2sin



o

    

1







 

ik

 

o e e   ik

ik  

e

o

o

o

o

o





o

 

2

2

k

2sin

4 k 



o