PSI- Issue 9

Ch.F. Markides et al. / Procedia Structural Integrity 9 (2018) 108–115 Author name / Structural Integrity Procedia 00 (2018) 000–000

111

4

2 2 1 2

) 1 r 

1 cos 

R R

(    

( )   

(7c)







 

 

r

2 ( 2 ) r R R R R r       2 2 3

2 2

1

1

2

2 2 1 2

) 1 3r 

1 cos 

R R

(    

( )   

(7d)







 

 



2 ( 2 ) r R R R R r       2 2 3

2 2

1

1

2

2 2 1 2

) 1 r 

1 sin 

R R

(    

r ( )    

(7e)







 

 

2 ( 2 ) r R R R R r       2 2 3

2 2

1

1

2

It is easily seen that the set of Eqs.(6) represent the case of a curved beam (θ 1 ≤θ≤θ 2 ) under bending by couples at its straight ends, whereas Eqs.(7) represent the case of bending of the same beam by transverse forces applied at its straight ends. According to Muskhelishvili (1963), the above solutions were firstly reached by Kh. Golovin (1882). Regarding now the problem discussed here, i.e., that of the CSR, it is readily seen that its solution may be obtained by the superposition of the two preceding ones. Namely, substituting in the configuration of Fig.1c the shaded parts of the specimen |(π/2+θ ο )|≤|θ|<|π/2| (Fig.1), by the respective forces, ±P, and couples,  Pc (where c denotes the eccentricity), applied at the straight segments θ=  π/2 respectively, one obtains, at least at points far from these imaginary edges (Saint-Venant’s Principle), the case of the neat CSR, subjected to bending by the couples  Pc and the transverse forces ±P at its straight edges (θ=  π/2). In this context, the moment M=Pc is obtained as:

R

2

( )     

(8)

Pc 2h

rdr

R

1

where σ θ (ε) is the value of transverse stress on the straight edges θ=  π/2 expressed by Eq.(6d). Substituting now in Eq.(8) from Eq.(6d), and solving for ε, yields:

 R R R 4R R og R       2 2 R R 2 1 2 2 2 2 2 2 1 2 1

2Pc ( 2 )    

(9)

0

 



2

h (

)

   

  



2



1

On the other hand, the transverse force P is obtained as:

R

2

r      ( )

(10)

P 2h

dr

R

1

where τ rθ (β) is the value of shear stress on the straight edges θ=  π/2 given by Eq.(7e). Substituting in Eq.(10) from Eq.(7e), e.g. for θ=π/2, and solving for β, yields:

 ) R R og R R R      2 R R 2 2 1 2 2 2 2 2 1 R

P ( 2 )    

(11)

0

  



2h (

2     

2

1

1

where for θ=π/2, a negative sign should be assigned to P. Thus, for a given value of the overall load P, exerted from the loading frame to the specimen (Fig.1b) and a known eccentricity c, ε and β are knownthrough Eqs.(9) and (11). In order, now, for the components of the displacement field of the CSR to be determined one should first integrate Φ ε,β (z) and Ψ ε,β (z) obtaining, respectively: