PSI - Issue 41

Christos F. Markides et al. / Procedia Structural Integrity 41 (2022) 351–360 Christos F. Markides et al. / Structural Integrity Procedia 00 (2019) 000 – 000

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2.3 The solution of the 1 st fundamental problem for the FBD The 1 st problem of the FBD (Fig.6) is here solved by conformally mapping the actual disc of radius R on the unit disc in the mathematical ζ -plane by the function z = ω ( ζ )= R ζ (Fig.7). In that case, Muskhelishvili‟s complex potentials solving the problem in terms of the variable ζ = π e i θ and fixed points ζ j corresponding to Z j , are sought in the form:

    

    

   

   

j      

j      

1

     

n



 

 

(3)

log

(

) log

(  

) log

iP

F iP 

F iP

 

 







m

m

j

j

j

j

o

2 (1 )   

1

j



m

j

j

    

    

   

   

j      

     

     



n



 

j

log

(

) log

(  

) log

iP

F iP 

F iP





 

m

m

j

j

j

j

2 (1 )   

1

j



m

j

j

(4)

    

    

   

   

2

2

 

 

  

  

1





n



 

j

j

(

)

(  

)

iP

F iP 

F iP

 









m

m

m

j

j

j

j

o

2   2 

2   

2

2 (1 )   

     

1

j



m

m

j

j

where, φ ο ( ζ ) and ψ ο ( ζ ) are holomorphic functions in the neighborhood of ζ j , subjected to determination. Introducing φ ( z ) and ψ ( z ) (Eqs.(3, 4)) into the boundary condition (Eq.(5)) on the disc‟s perimeter ( s is the point ζ on γ ): ( ) ( ) ( ) 0 s s s s        (5) and using known properties of Cauchy integrals (Muskhelishvili 1963), φ ο ( ζ ) and ψ ο ( ζ ) are determined in closed form. Then, one should substitute φ ο ( ζ ) and ψ ο ( ζ ) (found as previously described) into Eqs.(3) and (4), and using the inverse transformation ζ = z/R , ζ j = Z j / R , φ ( z ) and ψ ( z ) are obtained as:

    

    

   

   

z Z z Z  

z Z z Z  

z Z z Z  

1

n



  z

j

j

log

(

) log

(  

) log

iP

F iP 

F iP







m

m

j

j

j

j

2 (1 )   

1

j



m

j

j

   

   



   

2 Z z R Z  2

2 Z z R Z  2

 

  

z Z z Z  

1

n

2 (  

j

j

j

j

)

(  

)

z iP

F iP 

F iP







m

m

m  

j

j

j

j

2 R Z z R Z z   2

4 R Z z 

2 2

4 R Z z 

2 2

2 (1 )   

 

1

j



m

m

j

j

(6)

      

    

2 R Z z R Z z   2 j

2 R Z z R Z z   2 j

   

   

2 R Z z R Z z   2 m

2 (1 ) 1 4 (1 )        

n



log

(

) log

(  

) log

iP

F iP 

F iP





m

j

j

j

j

1

j



m

j

j

  

z

n





2 (   

 

) F iP Z F iP Z    ( )

iP Z Z



 

m m m

j

j

j

j

j

j

2

R

 

1

j



y

j P

j P

m P

η

L 

L

j F

j F

m Z

j Z

j Z 

γ

j   m 

j 

z= ω( z)=R ζ

i e z r  

ζ = π e i θ π

r

1

R





x

Ο

ξ

Ο

j   m  j 

j Z 

m Z

j Z

j F

j F

P

j P

j P

m

Fig. 7. The conformal mapping.