Issue 66
Ch. F. Markides et alii, Frattura ed Integrità Strutturale, 66 (2023) 233-260; DOI: 10.3221/IGF-ESIS.66.15
σ
z
2i β
2i β
1 k (1 k)e
Φ (z)
(1 k)e
4
2
2
z
α
(41)
(1 κ ) σ
z
δ 1 k (1 k)cos2 β i τ (1 k)sin2 β
1
2 2 z α
8 κ
σ
z
2i β
2i β
1 k (1 k)e
Ω (z)
(1 k)e
4
2
2
z
α
(42)
(1 κ ) σ
z
δ 1 k (1 k)cos2 β i τ (1 k)sin2 β
1
2 2 z α
8
In particular, the displacements of the crack lips are obtained (by adding Eqns.(9) and (16)) as: 2 2 1 2 2 1 1 2 2 (1 κ ) σ u (x) u u (1 k)cos2 β x (1 τ )(1 k)sin2 β α x 8 μ (1 κ ) σ v (x) v v (1 k)sin2 β x (1 δ ) 1 k (1 k)cos2 β α x 8 μ
(43)
and the respective contact stresses on the crack lips in the ‘general problem’ are given by Eqns.(33) and (34). Regarding the SIFs in the ‘general problem’, they are obtained by superposing Eqns.(12) to Eqns.(37) and (38), and they are written as:
2
(1 κ )
σ πα
δ 1 k (1 k)cos2 β
K
1
I
2
4 κ
(44)
2
(1 κ )
σ πα
τ (1 k)sin2 β
K
1
II
2
4 κ
Eqns.(44) should hold for any τ and δ values, including the limiting case τ = δ =1 of Eqn.(15). But in this case, since any relative displacement between the crack lips is prohibited, it holds that K I =K II =0, which (from Eqns.(44)) provides:
2
4 κ
(1 κ )
1
0
(45)
Eqn.(45) is only satisfied if the plate is under plane strain ( κ =3–4 ν ) and, in addition, the Poisson’s ratio is equal to ν =0.5 (i.e., if the material has exceeded the yield point). Thus, under the linear elastic assumption adopted here, Eqn.(45) cannot be satisfied. In this context, and in order to fulfill the condition K I =K II =0 for τ = δ =1, the SIFs in the ‘inverse problem’ (Eqns.(37) and (38)) must be redefined as follows: 2 I,(2) I,(2) σ πα 4 κ K K δ 1 k (1 k)cos2 β
(1 κ )
2 σ πα
(46)
4 κ
τ (1 k)sin2 β
K
K
II,(2)
II,(2)
2
(1 κ )
2
which upon superposition to Eqns.(12) yield the final naturally acceptable expressions:
σ πα
(1 δ ) 1 k (1 k)cos2 β
K
(47)
I
2
σ πα
(1 τ )(1 k)sin2 β
K
(48)
II
2
244
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