Issue 49

V. Matveenko et alii, Frattura ed Integrità Strutturale, 49 (2019) 225-242; DOI: 10.3221/IGF-ESIS.49.23

displacements

=

=

=

u x

v x

w x

0 ,

0 ,

0 ,

( )

( )

( )

(13)

k

k

k

stresses

  

k du x

     

  

 1 x x 



   v x



  



1

0

k

dx

   k dv x



   

  S x x  1

 

1  





 

S

S u x

2

k

dx

 





 k S  1





S

1

1 2

  

  

 

 

 





x

k v x

k w x

0

(14)

 x x  1



 x x 



2 1



   

   

 





1 2 

k dw x

x

k

 1 x x 



 

 









k w x

k v x

0

 x x 



 x x 



dx

2 1

2 1

 

k du x

 1 x x 



 

 







k v x

k w x

0 ;

0 ;

0.

(15)

dx

C ONSTRUCTION OF PARTIAL SOLUTION FOR ZEROTH HARMONIC OF THE FOURIER SERIES

A

solution for zeroth harmonic of the Fourier series is considered separately, because it does not follow explicitly from the algorithm for construction of partial solutions to the system of differential Eqns. (10)-(12) for any 0 k  . At 0 k  there are two problems: axisymmetric rotation and axisymmetric deformation. In the first problem, the component of the displacement vector 0 w is defined by Eqn. (12). In the problem of axisymmetric deformation the components of the displacement vector 0 0 , u v are defined by Eqns. (10), (11). The solutions for the function 0 w are constructed in the form of the generalized power series

    





 

 





m

(16)

0 w x

m R x



m

0

where m R are the coefficients of the series,  is the characteristic index. The possibility of constructing a solution in the form of (16) was proved in [21]. For Eqn. (12) the point 0 x  is a regular singular point. In this case, one of the partial solutions is written as a power series (16), for which the range of convergence lies in the interval 0 1 x   , because the value 1 x  is the zero of the high-order derivative. function closest to the point 0 x  . The coefficients of the series m R and the characteristic index  are found by substituting (16) into (12). Setting to zero the expressions with like powers of x , yields a recurrence relation for m R :            1 2 2 2 1 2 2 1 4 1 2 2 1 1 4 2 1 0,( =0, if 0 ) m m m m m m R m m R m m R R m                                      (17)

From the condition of existence of non-zero solution for 0

R we obtain the characteristic equation:

(2 1)(2 1) 0     

(18)

1/ 2   and 2

  

are the roots.

for which 1

1/ 2

229