PSI - Issue 3

5

Nataly Vaysfeld et al. / Procedia Structural Integrity 3 (2017) 526–544 Author name / Structural Integrity Procedia 00 (2017) 000–000

530

0   0  

dx

  ,     t

    1 1 2 2 2  J x J t x

,

(13)

k

x



  k

dx

  ,     t

    1 1 2 2 2  J x J t x

(14)

,

k

x

  k



where   1 J x is Bessel function. For the problem №1 with regard to the correspondences

         xI x I x xI x xK x K x xK x         ,  

1

1

2

1

1

2



      k

1   j j N k j cos    

K

Hp



  

2

2   k

k

from the boundary conditions (9) one finds

.

A

I

  t

t t dt





1

  k

k

k

j

k G I

I

 

2

2

k j

For the problem №1 from the boundary conditions (10) one finds              2 2 2 2 1 2 cos                      j j N k k k k k k j k k k j Hp K K A I K t K G k k 1  

  1      k k j I t 

 

t t dt













1   j j N k j j cos    

k Hp I

I

  k

  k

  

    K t K I 



 

2

2 2

B

I

    t

t t dt

 





  k

 



2

1

2

1

k

k

k

k

k

j

 

  k

 k G k 



   K





   

   

where K One should substitute the found values of the integration constants in the corresponding equalities (11) and (12) and use the inversion formula (7). As a result, the expressions of the displacement will be constructed for Problem №1 2 2 2 2 .      k  k k k k I I

 j

0  



     k k

N

 

I

2

1 2

H

x

  





 





       t t dt J x J t x ch

1

,  

sin

1

u

p





    

  k

 j

1

1

k

j

G

  I



2

k

1

1

k

j





 j



      k k

1     j j N j k 1      

K

  x ch

x

  





2

2

sgn

1

sech

2

cos      k j k sin k

xdx

  

 





 







j

j

I





2



  1      k k j I t 

 

(15)

I

t t dt





1

 j

0  

  

N

 

F

2

1 2

H

x

  





 

for Problem №2  u



       t t dt J x J t x ch

k

(16)

,  

sin

1

p





    

  k

 j

1

1

  k

k

j

G

 k





1

1

k

j





 j



    ,  k

1    j j N j k 1        

   

N t

x

x

  









 

k

2

sgn

1

sech

2

cos      k j k sin k

.

ch

xdx

t t dt



  

 





 



 j

j









  1     k k I 



  1 K



  

Here

F

K

I





  k

 

2

2

k

k

          K I I   2  2 1 k k k k

  1 K t K  



  1    k k I



  , 

N t

t

 





2

k