PSI - Issue 18

Yaroslav Dubyk et al. / Procedia Structural Integrity 18 (2019) 630–638 Yaroslav Dubyk et al. / Structural Integrity Procedia 00 (2019) 000–000

632

3

To solve system (1) two types of solutions for cylindrical shell are presented below: short which quickly decay along the axis, and long one that behave opposite. 2.1. Short solution Suppose that any geometrical or force function in circumferential direction changes much less than in the axial one:   2 2 2 2 d dx d Rd     (4)

Then accounting for (2)

x Q Q   , we can neglect Q  in eq. (1). Excluding L from eq. (1), we can get that

2        , according to our function expansion (4) we can write that 2 2 2 / / N x N

x N N   , so it can be concluded

x

that:

(5)

(1 2 ) N H   

/ u x      

Eh

 

 





x









Substituting eq. (4) to second eq. (1), we get:

  

  









2

2

2

2 G v 

2

u

v

v

w

v



 

 









(6)





0

E

G

G

E G G   















2

2 x R

2

2 x R

2

2 

R

R x

R

E G  

x





 















w     , and

, so:

According to eq. (4), we can infer from eq. (6) that / v

/ w R

  

(7)

/ N Ehw R   

Now using only the last equation in (1) and Fourier expansion (3), we get the following equation:



 2

1

 

4

4 d x 

2 d x 

( )

( )

2

  

2

0    where B

A



s B n R   

n

n

12

and

(8)

2 s A n R  

n

4

2

2 2

dx

dx

R h

 

The solution of 4 th order ODE is well-known, so neglecting the increasing terms, we obtain:

  n d x С e  2

  n d x

n c x

n c x





( ) x  

cos

sin

(9)

1 С e

n





Here cos . For a radial displacement we can simply , x x x M Q  , and their explicit form is write ( , ) w x n     , the other main variables for short solution , cos( ) n presented in Appendix A1 as well as for variables , , , x u v N L , which are initiated by a short solution. Note, that for 0 n  , short solution is a simple axisymmetric end effect, which is considered in most Strength of Materials textbooks. 2.2. Long solution Assume that   2 2 2 2 d dx d Rd     , then from (1) and (2) it follows that / Q N       . Then, we exclude the tangential force from (1), respectively, we obtain that 2 2 2 2 / / x N N x        . According to our assumption x N N   . Now we can say that x      and:   2 (1 ) / x N H Eh u x         (10) n c    , sin n d    , 2   s B , 2 arccos / s   s A B