PSI - Issue 2_A

M. Nourazar et al. / Procedia Structural Integrity 2 (2016) 3423–3431 Author name / Structural Integrity Procedia 00 (2016) 000–000

3425

3

where k is the spring constant quantifying the bonding imperfection at the interface

0 y = . Solutions of Eqs. (4) and (5) with

the help of Fourier transforms are taken in the form: * 1 2 * 1 2 ( , ) ( ) ( ) 1 ( , ) [ ( ) cosh( ) w s y C s C s s y D s sy D s ϕ +∞               = +

∫

i x sy e dx ω −

y h

( ) sinh( )]

0

< <

  

  

  

  

1

2

π

  

  

1 E s

2 E s

( )

( )

*

−∞

( , ) s y

ψ

+∞

1 A s B s 1

2 A s B s 2

y < <

( ) ( )

( ) ( )

0

ξ

  

  

  

  

1

∫

i x sgy e dx ω −

*

w s y

cosh( ) sgy

( , )

(

sinh( ))

=

+

(7)

h y − < <

ξ

2

π

2

−∞

, , , i i i i A B C D and

, ( 1, 2) i E i = which may be determined from the

Thus, there are all together ten unknown functions following boundary, continuity and limiting conditions: 1 1 1 ( , ) 0, ( , ) 0, ( , ) 0, yz y y x h D x h B x h σ = = =

+

+

D x

B x

( , x h σ yz

( , 0 ) 0, =

( , 0 ) 0, =

) 0, − = ,

y

y

2

−

+

w x

w x

( b H x wz

( , ) ξ

( , ) ξ

),

−

=

−

η

,

( , ) x x σ ξ σ ξ + − = (8) where wz b is the displacement jump across the slip plane, (.) H is the Heaviside step-function. Substituting coefficients into (7) with the aid of constitutive equations after some routine manipulations, we find that ( , ) zy zy

0 ∫ ∞

G b

sinh( ) sgy

cosh( ) sinh( ( sgy χ

+

x wz

sg h ξ +

s x

ds

)) cos( (

))

=

−

σ

η

zx

2

sgh

sgh

cosh(

)

sinh(

)

−

π

χ

2

2

0 ∫ ∞

gG b

cosh( ) sgy

sinh( ) sinh( ( sgy χ

+ −

y wz

sg h ξ +

s x

ds

y < <

)) sin( (

))

0

=

−

σ

η

ξ

zy

2

sgh

sgh

cosh(

)

sinh(

)

π

χ

2

2

0 ∫ ∞

G b

cosh( ) sg

sinh( ) cosh( ( sg ξ

ξ χ +

x wz

sg y h +

s x

ds

)) cos( (

))

=

σ

η

−

zx

2

sgh

sgh

cosh(

)

sinh(

)

π

χ

−

2

2

0 ∫ ∞

gG b

cosh( ) sg

sinh( ) sinh( ( sg ξ

ξ χ +

x wz

(9)

sg y h +

s x

ds

h y − < <

)) sin( (

))

=

−

σ

η

ξ

zy

2

2

sgh

sgh

cosh(

)

sinh(

)

−

π

χ

2

2

where

G G

1 44 2 15 3 15 cot ( ) k gh sh e h α α + +

y x

.

(10)

s

[

]

χ

=

−

k

c

The difficulty in the solution of the problem comes from the evaluation of integrals (9), which are represented by infinite integrals that contain singularities, on the path of integration. In order to determine the possible singular behavior of (9), the behavior of the integrands (9) at x η = and y ξ = needs to be examined. For this, it is sufficient to determine and separate those leading terms in the asymptotic expansion of kernels for s → ∞ that would lead to unbounded integrals. After performing this separation it can be shown that ( ) 2 2 2 2 2 2 0 ( ) sinh( ) cosh( ) [ sinh( ( )) ]cos( ( )) , 0 cosh( ) sinh( ) 2 2 [ ( ) ( ) ] sg y x wz x wz zx G b g y G b sgy sgy e sg h s x ds y sgh sgh g y x ξ ξ χ σ ξ η ξ π χ π ξ η ∞ − − + = − + + + − < < − − + − ∫ ( ) 2 2 2 2 2 2 0 ( ) cosh( ) sinh( ) [ sinh( ( )) ]sin( ( )) , 0 cosh( ) sinh( ) 2 2 [ ( ) ( ) ] sg y y wz y wz zy gG b x gG b sgy sgy e sg h s x ds y sgh sgh g y x ξ η χ σ ξ η ξ π χ π ξ η ∞ − − + = + + − − < < − − + − ∫