PSI - Issue 65
Anvar Chanyshev et al. / Procedia Structural Integrity 65 (2024) 56–65 Anvar Chanyshev / Structural Integrity Procedia 00 (2024) 000–000
63
8
2 A b iHz
2 2 b b iHz )
4 3(
) 4 (
.
(23)
( ) z
2 z b
2 2
2 z b
2 3
1
(
)
(
)
If we introduce the notations
2 2 2 x y b ,
2 2 2 6 3 x y yH b ,
2 2 2 x y yH b , 2
2 xy ,
then, by the Kolosov-Muskhelishvili formulas, we will obtain expressions for stresses and displacements within the half-plane:
2 b Hy
2 2
2
A
Hx
8
(
)(
) 2
2
,
x
y
2 2
2 2
1
(
)
2 2 ( A
2 ) 4 ( b
2
2 3 )
8
,
(24)
y
x
2
2 2
2
2 3
1 (
)
(
)
2 2 ( b
2
2 3 )
A
8
,
xy
2
2 2
2
2 3
1
(
)
(
)
Ax
2 6 yH
y
2 [ ( b H y H x y 2 2 ) (
2
2
2
)]
u
y
8
,
x
2 2
2
2 2
1
(
)
(25)
[( A y H x
2 H y x y b 2 ) ] 3 (3
2
2 ) 2[ (
2 b H y H x y 2 ) (
2
2 2
)](
)
u
.
y
2 2
2
2 2
1
(
)
From (24), taking into account , , , , it follows that the stresses at infinity tend to zero. Next, on the basis of (24), let us try to determine what causes the displacements (21) at the boundary
0 y . We
0 y loaded by a constant internal pressure p . To calculate
initially assume that there is a hole in the half-plane the pressure p , we have the following equation:
p
x
xy
,
0
p
xy
y
after expansion of the determinant, it has the form
2 xy p p 2 ( ) x y x y
0.
(26)
Solving (26), we find
2
2
x
y
x
y
.
2
(27)
p
xy
1,2
2
From the two roots (27) we must choose the one that is constant and which value corresponds to the constant pressure within the cavity.
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