PSI - Issue 75

Sgamma M. et al. / Procedia Structural Integrity 75 (2025) 709–718

712

4

Author name / Structural Integrity Procedia 00 (2025) 000–000

stress / strain components –or more precisely, their standard deviations (i.e., the square roots of those variances) to preserve dimensional consistency– leading to:

σ m + √ v σ F ( N ) σ y

P FS = 2 √ v γ 1 + k

(4)

where v γ denotes the variance of the maximum shear strain acting on the candidate plane, v σ is the variance of the normal stress, and σ m is the global mean of that normal stress. The term F ( N ) represents a statistical factor for the maximum value in a random process, derived following Davenport’s approach [13]:

F ( N ) = 2 ln( N ) +

0 . 5772 √ 2 ln( N )

(5)

Since this formulation relies on variances rather than direct cycle counts, the number of cycles N remains unspeci fied. A practical recommendation is to adopt a fixed value such as N = 10 6 when evaluating all possible orientations, ensuring a consistent basis for comparing planes and identifying the plane that maximizes equation 4.

3.2. Defining a joint probability density function for shear strain and normal stress

Adapting the Fatemi–Socie (FSC) criterion to the frequency domain requires understanding how fatigue damage accumulates through numerous pairs of shear strain amplitudes, γ a , and corresponding maximum normal stresses, σ n , max . In a time-domain analysis, these pairs are identified via a multiaxial rainflow procedure, which isolates strain cycles and identifies the highest normal stress occurring within each cycle—an approach traditionally performed at a single, fixed orientation of the critical plane. When employing spectral methods, the distribution of shear strain amplitudes can be estimated by leveraging models designed for random processes. Among the various techniques, the Dirlik model [14] is one of the most widespread. Its probability density function (PDF) for amplitudes is commonly expressed as:

1 √ m γ 0

2 ,

Z R 2

K 1 K 4

Z 2 2 R 2 + K 3 Ze −

Z 2

Z K 4 + K 2

e −

e −

p γ ( γ a ) =

(6)

where

1 − I − K 1 + K 2 1 1 − R ,

2( x m − I 2 ) 1 + I 2 ,

γ a √ m γ 0

Z =

K 1 =

K 2 =

,

m γ 1 m γ 0

m γ 2 m γ 4

I − x m − K 1 1 − I − K 1 + K 2 1 ,

x m =

K 3 = 1 − K 1 − K 2 , R =

,