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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