PSI - Issue 18

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

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

Differentiate equation (1c) with respect to the angular coordinate  once and subtract from equation (1b). Differentiating the obtained equation once with respect to the coordinates  and replacing it with the relation (1a), which was differentiated with respect to the axial coordinate x , we obtain the following equation:

3 Q Q   1 

2

3 Q x R x R       2 2 1 1 x x N





0





(11)

2

2

R











Substituting the values of the transverse forces (2) and expressing

x N from (10), the equation (11) will be written

through the functions   , u x  ,   , v x    , w x  . In this case, according to assumptions / v w      and / / u R v x       , the problem is expressed only through tangential displacements   , , v x t  :

2

2

    

   

    

    

4 4 2 v R   

2

2

3

2

4 2 4 2     

   

   

   

   

   

12(1 )  

1

1

v



 





(12)

2

0.



v

v

v















4 4 

2

2 2

3

2 4 





2

x   

h

R x 

R













We represent the displacement   , v x  as   , v x 

  n 

( ) sin x

  

, thus the equation (12) will be rewritten:





2

1

n







 2

6 n n

4  

4 2

1

n n



4 d x 

2 d x A B x  ( )

( )

1

1

2 4 12(1 )  

2

( ) 0    where



and

(13)

l

B

l



A



4

2

dx

dx

4 R n R 4

2

2 12(1 )  

2 R n R 4

2





2

2

h

h

Again, we have to solve 4 th order ODE, and once more neglecting the increasing terms, we obtain:

a x n

a x n





  b x C e 

  b x n

( )  

cos

sin

x C e

n

3

4





2  

Here cos

n a    , sin

n b    ,

,

. For a radial displacement we can simply , x x u N L  , and their explicit form is presented

l B

2 arccos / l  

l A B

write ( , ) v x n     , the other main variables for short solution , cos( ) n in Appendix A2 as well as for variables , ,

1, 2 n 

, x x x w M Q  , which are initiated by a long solution. Note, that for

, long solution doesn’t exist. 2.3. Boundary conditions To form the boundary conditions for the miter we need to project forces from the straight section on the oblique section, see Fig. 1 and Orynyak et al. (2016). For small values of the miter angle  the general boundary conditions have the form by Orynyak (2016):

b N N L N Q N         x x b

b

cos

sin cos ;  

L





(14)

b

b

cos sin

cos sin

;

N

L





 

 

x

b

sin sin .  

x

x

0 x   , now we have four

Additionally, from the symmetry condition we can write that in oblique section

boundary conditions, which completely define our problem.