PSI - Issue 5

Ahmed Bahloul et al. / Procedia Structural Integrity 5 (2017) 430–437 Ahmed Bahloul et al. / Structural Integrity Procedia 00 (2017) 000 – 000

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(ii) Fig.3 shows the crack growth path simulation of the cracked attachment lug. A good agreement is found between experimental and numerical result In order to highlight the effect of residual stress distribution near the crack tip on the stress intensity factor (SIF) range. Tab.1 shows the evolution of ∆ ⁄ ∆ versus crack length for the attachment lug. Two different stress ratios are considered (R=0.1 and R=0.5). It is observed that the ratio ∆ ⁄ ∆ decreases as the crack length increases. The difference between ∆ ∆ can reach 20% for the attachment lug as illustrated in Tab.1. This result can be explained by the fact that the plastic zone surrounding the crack tip increases as the crack size increases. Accordingly, a compressive residual stress occurs and reduces the effect of the tensile stress during crack propagation.

Fig. 3.Crack growth path of the attachment lug

Table 1. Effect of residual stress distribution on the SIF.

∆ ⁄ ∆

a (mm)

R=0.5 0.957 0.939 0.933 0.918 0.914 0.897 0.903

R=0.1 0.923 0.891 0.879 0.853 0.845 0.816 0.826

3 5

7.5 10

12.5

15

17.5

(iii) In order to validate the proposed approach’s performance for predicting fatigue life, cracked attachment lug subjected to different load ratios are considered. The effect of residual stress distribution surrounding the crack tip is evaluated using the RC-SIF as previously mentioned. The material dispersions are assumed to be normally distributed with coefficient of variance equals to 1.75%, having for R=0.1: C=2.1E-08 and m=3.86 and for R=0.5: C=11.3E-08 and m=3.7. due to the random aspect in fatigue life data for cracked lugs, the suggested method consist in determining the iso-probabilistic a-N curves corresponding respectively to 5%, 50% and 95% fatigue reliability. Fig.4 shows a comparison between the proposed model and the relevant experimental data for various loading conditions (R=0.1 and