13th International Conference on Fracture June 16–21, 2013, Beijing, China -2- sets corresponding to three specific specimen geometries of the same material: the aluminium alloy AlMgSi1-6082. 2. Model proposal In this section a new method to evaluate fatigue data is presented which extends the applicability of the probabilistic fatigue model presented by Castillo and Fernández-Canteli [6] for a more general description of the fatigue behaviour considering specimen geometry, i.e. size effect and variable stress state. 2.1. Probabilistic model The Weibull regression model described in [6] is based on physical and statistical assumptions. The compatibility condition between the probability distributions present in the Wöhler field, i.e. the probability distribution Pf (Δσ| Ν) of Δσ for constant N and the probability distribution Pf (N|Δσ) of N for constant values of Δσ, plays an important role, manifesting that the values of the failure probability for every combination of Δσ and N must be equal for Pf (Δσ| Ν) and Pf (N|Δσ). The model describes the Wöhler field in a probabilistic way by means of percentile curves, i.e. curves representing a constant failure probability, and computes the failure probability Pf (N, Δ σ) for a combination of stress range Δσ and number of cycles N by ,Δ 1 exp ln lnΔ , (1) which corresponds to a three-parameter Weibull distribution of the variable ln lnΔ with location parameter λ, shape parameter β, and scale parameter δ [7]. V represents a normalizing variable and could be interpreted as a damage parameter. B and C are the threshold parameters for lifetime and stress range, respectively. Fig. 1 gives an example of the model depicting the SN field on the left and the normalized variable on the right. A detailed description of the model can be found in [6]. Figure 1. SN field and normalized variable V 10 4 10 5 10 6 10 7 10 8 180 200 220 240 260 280 300 320 340 N [cycles] Δ σ [MPa] Threshold lifetime Endurance limit P failure = 0% P failure = 1% P failure = 50% P failure = 99% Data 0.2 0.4 0.6 0.8 1 1.2 0 0.2 0.4 0.6 0.8 1 V = (ln N - B) ⋅ (ln Δσ - C) P f Δ σ (MPa) 220 227 230 235 240 245 250 260 270 280 290 300 330
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