PSI - Issue 19

Jean-Gabriel SEZGIN et al. / Procedia Structural Integrity 19 (2019) 249–258 Jean-Gabriel Sezgin et al./ Structural Integrity Procedia 00 (2019) 000 – 000

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presented some analogies with JIS-SUJ2, whose failure mechanism is time-dependent, resulting in both an unbounded FCG acceleration ratio and an IG fracture. These points imply that the H-assisted FCG acceleration may be out of scope of the HISCG mechanism. For these reasons, the assumption of a failure mechanism issued from the combination of two elementary mechanisms was investigated. A combination rule for the HISCG model (cycle dependent and bounded at 30) and a time-dependent model (unbounded) was established in regard with experimental measurements. Figure 8 illustrates the combination of the mechanisms using the fatigue life (a), the translated FCG rate (b), and the FCG acceleration ratio (c). The data related to the cycle-dependent mechanism were extrapolated from the literature of JIS-SCM435 (Matsuoka et al. 2017) as represented in Figure 5. The value at a frequency of 10 Hz being unavailable, the value of FCG acceleration ratio for H1150 at a stress amplitude of 200 MPa were considered since the IG failure was not significantly observed for this condition. In the case of a purely time-dependent mechanism, the FCG rate and the FCG acceleration ratio can be expressed as (with the nomenclature introduced in Figure 8):

Figure 6 – Proportion of IG surface highlighted in red given for two frequencies at a stress amplitude of 300 MPa: a) f=10 Hz and b) f=10 -3 Hz

Figure 7 – Percentage of IG failure measured on the basis of fractographic analysis for five frequencies (fitted by the expression in the frame) ( ) = ( ) + 1 ( ) (1) ( ) = ( ) / ( ) (2) In case of cycle-dependent mechanism, these expressions of the FCG rate and FCG acceleration ratio are expressed by (3) and (4). ( ) = ( ) + ( ) (3) ( ) = ( ) / ( ) (4) From the expression (1) and (2), the FCG acceleration ratio related to time-dependent mechanism was expressed in (5) as a function of the testing frequency. ln (( ) − 1) = − ln + ln ( ) − ln ( ) = − ln + ln 0 (5)