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

634

5

Fig. 1. Mitered bend, scheme of loading.

3. Results and discussion In case of pressure loading in straight pipe we have the following forces

b N pR   

2 b x N pR   and

, and from

eq. (14) we get the boundary conditions:

2 Q pR  

2 L pR   

sin cos

sin sin

 

 

;

(15)

Taking into account the symmetry of the problem, the boundary condition must also be fulfilled ( 0) 0 x x    . Consider the short solution for 0 n  , in this case boundary conditions must be satisfied for , x x Q  , because in short solution they are the main variables. Thus, from (A.2) and (A.4) we define two variables 1 2 , C C , and in such way find all the forces and displacements. For parameter L we have eq. (A.7), which must be equal to L  , thus for 0 n  we have a self-contained solution, that fully meets the relevant boundary condition. In case of bending loading for a straight pipe section, we have the following boundary conditions: 1 sin sin 2 2 m L h      ,   sin 1 cos 2 2 m h Q       . (16) Again, we will start with short solution, for 0 n  from eq. (14) we get:

sin   h

0 x x    and

(17)

Q

0  

m

x x

2

0

Two constants 1 2 , C C can be defined from (A.2) and (A.4), and all the other variables. Now we will go to solution for 2 n  . Using the boundary conditions (16) we can find a short solution, which is not self-contained, because we have the additional force L , which differs from the boundary condition (16). Unbalanced part can be found as:

3 sin sin2 . 2 m h        short

long L L L

(18)

This equation, together with symmetry condition 2 n  . Defined from these boundary conditions radial and circumferential displacement have a projection on axial displacements u in oblique section: 0 long u  is boundary condition for

(19)

sin sin

cos sin

cos

obl u w 

long

long

str

v

u

 

 

2

2

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