PSI - Issue 28

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

773

4

where

φ n , 0 ( ρ, ν ) = N µ ( a ν ) J µ ( νρ ) − J µ ( a ν ) N µ ( νρ ) X n ( r , z ) = X n ( a 1 ρ, a 1 ζ ) = X n ( ρ, ζ )

(11)

Eigenvalues ν 0

k are positive roots of transcendental equation

N µ ( a ν ) J µ ( ν ) − J µ ( a ν ) N µ ( ν ) = 0

(12)

We get one-dimensional equations

∗ a

2 c − ν 2 W ∗ ν 2 Z Z ∗ ,, nk + a

,, nk ( ζ ) + µ − 1

, nk ( ζ ) = a 1 α µ µ − 1 ∗ T , , nk ( ζ ) = − α µ ν 2 T

2 1 ω

nk ( ζ ) + a 1 µ − 1

nk ( ζ )

W

∗ µ 0 Z

,, nk ( ζ ) + a 2

2 c − µ

nk ( ζ ) − a − 1

2 W

n ( ζ )

(13)

Z

1 ω

1 µ 0 ν 2 c − ν 2 Z ∗

2 1 ω

nk = 0

2. Formulating of vector boundary problem

The solution will be built so that the boundary conditions (3) are satisfied. To write boundary conditions (3) in notation (6), we consider similar combinations for shear stresses τ zr and τ z φ (considering (5)). τ ( r , φ, z ) τ ∗ ( r , φ, z ) = 1 r   r τ zr r τ z φ ± τ z φ τ zr ˙   (14) So, we can write down the boundary conditions

nk h 1 + µ Z nk h 1 = − (1 − 2 µ ) p nk − α µ T nk h 1 a − 1 1 ν 2 W nk ( h 1 ) − Z , nk ( h 1 ) = 0 W nk ( h 0 ) = Z nk ( h 0 ) = 0

1 (1 − µ ) W ,

a − 1

(15)

And homogenious equation for function Z ∗ n

Z ∗

n ( ρ, ζ ) = 0

(16)

if we introduce functions as unknowns of the system (13)

y 0 ( ζ ) = W nk ( ζ ) , y 1 ( ζ ) = W ,

nk ( ζ ) , y 2 ( ζ ) = Z nk ( ζ ) , y 3 ( ζ ) = Z ,

nk ( ζ )

(17)