PSI - Issue 28

H. Lutsenko et al. / Procedia Structural Integrity 28 (2020) 770–775 H. Lutsenko / Structural Integrity Procedia 00 (2020) 000–000

774

5

which represents components of vector y ( ζ ), then we can write down system in form

y ( ζ ) − P k y ( ζ ) = f ( ζ ) , ¯ h 0 < ζ < ¯ h 1

y ( ζ ) = y 0 ( ζ ) y 1 ( ζ ) y 2 ( ζ ) y 3 ( ζ )

, P k = µ − 1

, f ( ζ ) = a 1 µ −

0

1

0 0 0

0

0

∗ ν

2 c 0 0

, nk ( ζ )

2 − a 2

1 ∗ µ 0

1 ∗ α µ T

1 µ −

− a

(18)

1 ω

0

0 0

1 0

2 T

2 µ

2 − a 2

2 c

a − 1

nk ( ζ )

a 1 α µ ν

1 ω

1 µ 0 ν

∗ ν

Boundary functional and boundary conditions have form

U y ( ζ ) = Ay h 0 + By h 1 = − (1 − 2 µ ) γ 0 0 0

A =

, B = 0 a − 1 1 ν 0 0

, γ =

0 0 0 p nk − α µ T nk h 1

1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0

(19)

2

0 0 − 1 0 0 0

a − 1 1 (1 − µ ) µ 0

2.1. Solution of vector-boundary problem

The solution of vector boundary problem was obtained by Green’s matrix method Popov G. Ya. (1999)

α µ µ ∗

h 1

y ( ζ ) =

G ( ζ, s ) f ( s ) ds − (1 − 2 µ ) Ψ ( ζ ) γ

(20)

h 0

where

G ( ζ, s ) = Φ ( ζ − s ) − Ψ ( ζ ) U Y ( Φ ( ζ − s )) Φ ( y ) = 3 j = 0 ∆ 3 − j d dy j ψ ( y ) , y = ζ − s Y ( ζ ) = 3 j = 0 ∆ 3 − j d dy j ψ ∗ ( y ) ψ ( y ) = λ 2 e −| y | λ 1 − λ 1 e −| y | λ 2 2 λ 1 λ 2 ( λ 2 1 − λ 2 2 ) , ψ ∗ ( y ) = λ 2 sinh λ 1 ζ − λ 1 sinh λ 2 ζ λ 1 λ 2 ( λ 2 1 − λ 2 2 )

(21)

k − λ 2 − a 2

k − λ

2 2 I , ∆ 3 = P 3 ∗ , λ 2 = ν

2 2 P k = λ 2

∆ 0 = I , ∆ 1 = P k , ∆ 2 = P 2

2 1 + λ

2 1 + λ

2 2 P − k

1

1 λ

λ 1 = ν

2 c µ − 1

2 − a 2

2 c

1 ω

1 ω

2.2. Deriving of exact solution of the original problem

Thus, displacement transformation W nk ( ζ )

W ( a 1 ρ, φ, a 1 ζ ) 2 G

W ( ρ, φ, ζ ) 2 G

u z ( r , φ, z ) =

(22)

=