PSI - Issue 33

K. Mysov et al. / Procedia Structural Integrity 33 (2021) 365–370 Author name / Structural Integrity Procedia 00 (2019) 000 – 000

367

3



3 0 2 sin ( , ) r a    2

2 a d M J

0

,

(1.2)

   

  + + =

factor i e t  is omitted hereinafter. The upper spherical face of the cone

, , r b       = −   −   is fixed 0 r b w = =

(1.3)

The cone’s lateral surface

, , a r b        = −   is free from stress 0     = =

(1.4)

The spherical crack’s branches 1

1 , , a r b        =  −   are free from stresses ( ) 1 0 0 cot 0 Gr w w        − • =  =  = − = ,

(1.5)

( ) 

, w r



where

•

w

=

  The displacement’s jump over crack’s edges 1

, a r b        =  −   is non-zero

1 ,

( ) , 

( w r

) 0 − −

( w r

) 0 + =

( ) 1 r 

,

,

(1.6)

w r

=





here ( ) 1 r  is unknown jump. To find the solution it is required to determine the displacement satisfying boundary conditions (1.1), (1.3), (1.4), conditions at the crack (1.5), (1.6) and the torsion equation 2 2 2 2 2 2 (sin ) ( ) sin sin w w w r w r q t    • •    + − = −  , (1.7) where q c  = is the wave number ( / c G  = , is the shear wave speed). The problem is axisymmetric, thus variable  is omitted in all previous and next formulas.

2. Problems solution The problem is solved as the superposition of continuous and discontinuous solutions ( ) ( ) ( ) 1 2 , , , w r w r w r    = +

(2.1)

( ) 1 , , w r   was found in Mysov (2019) and for arbitrary function ( ) F  has the form ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 1 2 1 2 0 1 (cos ) 1 , sin (cos ) k k k k k k k k k k k k P F J qr Y qb J qb Y qr a w r l r J qa Y qb J qb Y qa P                 = − =     −    , (2.2)

here the continuous solution



( ) F d   

( ) k J x  and

( ) k Y x  are Bessel’s functions,

= 

1 0 sin (cos ) k P  

k F

1 ( ) k P x  is

where  is found from (1.2),

,

1/ 2

associated Legendre’s function of the first kind,

, k  are the roots of the transcendental equation

k k   = +

1  k

(cos ) 

P



(2.3)

1 ctg (cos ) 0 P   =

−







k

  =

To find discontinuous solution first, one must apply Legendre integral transform by generalized scheme Popov (1992): ( ) ( ) ( ) 1 sin , cos w r w r P d      =

0 

k

k

(2.4)

) 1 !   −  +    1 2 k

(

k

 

( ) , 

( ) w r P  ,1 k k k

(

)   k

1

cos ,

w r

=

=

(

) 1 !

,1

k

+

0

k

=