ABSTRACT

The mapping function of a crack into outside of a unit circle can found by substituting c a=

2 and m=1 into 6.5.14, that is,

z a= +  

 2

1 ζ

ζ (7.1.3)

Exercise 7.1.1: for a crack find ζ versus z

Solution: by squaring both sides of Eq. (7.1.3) we have, z a2 2 2

24 1 2= + +

 

 ζ ζ

z a2 2

2 24 1 2= + +

 

 ζ ζ , and then by deducting a2 from both sides we have: z a a a2 2

4 1 2

4 1

− = + − 

 

  = −

 

 ζ ζ

ζ ζ

z a a a2 2 2

4 1 2

4 1

− = + − 

 

  = −

 

 ζ ζ

ζ ζ

and taking square root of this provides:

z a a2 2 2

1 − = −

 

 ζ ζ

Then from Eqs. (7.1.3) and (7.1.4) ζ can be found versus z, that is,

ζ = − +z a z a

ζ =

− −z z a a

By substituting Eqs. 7.1.5) into (7.1.1) we can change φ ζ( ) into φ z( ), that is,

φ σz a z a z

a z z a

a ( ) = − + − − −

 

 8

which can be simplified to:

φ σz z a z( ) = − −( )4 2 2 2 (7.1.6)

Exercise 7.1.2: Convert ψ ζI ( ) into ψ I z( ) Solution: Similarly we substitute Eq. (7.1.5) into Eq. (7.1.2) and we have:

ψ σ

I z a

a z a z a z a z

a z z a z

a z a z ( ) =

− +( ) − − +( ) − − − +( )

− +4

1 4

2 2 2( ) −

       

       

2 1

and can be simplified to,

ψ σI z a a z a z a z a z a z z a

z a z a z ( ) =

− +( ) − − +( ) − − −( ) − − +(4

1 4

  

  

Further simplification ψ σI z z a

z a z a a z z a z a z

( ) = −

− +( ) − − − − − +

  

  8

ψ σ

I z z a

z a z a a z z a z a z

( ) = −

− +( ) − − − − − +

  

  8

2 2 this gives, ψ σI z

z a z a z a z z a( ) =

− − +( ) − − − −( )

  8

ψ σ

I z z a

z a z a z z a( ) = −

− +( ) − − − −( ) 

 8

the expression inside square bracket can be simplified as

well, ψ σI z z a

z z a a( ) = −

− −  8

4 4 2 2

ψ σI z z a

z a ( ) = −

 

 

Now that we have Eqs. (7.1.6) and (7.1.7), we can rewrite Eq. (6.4.27) in terms ψ I z( ):

2µ κφ φ ψu i v z z z zI+( )= ( ) − ( ) − ( )' (7.1.8)

Differentiation of Eq. (7.1.6) versus z provides φ ' z( ), that is,

φ σ σ' z z z a

z z a

( ) = −

− 

 

  =

− −

 

 4

2 1 2

φ σ' z z z a

( ) = −

− 

 

 2

1 22 2

From Eq. (7.1.7) we need to find:

ψ σI z z a

z a ( ) = −

 

 

The objective is finding the displacement v of the face of the crack. It is obvious that we need to change z into x and the crack region can be designated by − ≤ ≤a x a.