PSI - Issue 39

Jesús Toribio et al. / Procedia Structural Integrity 39 (2022) 484–487 Author name / Procedia Structural Integrity 00 (2021) 000–000

485

2

Fatigue crack growth based on a Paris law type approach has been numerically studied in a CCRB with several types of misalignments, by using three-dimensional (3D) solutions for the stress intensity factor (SIF). Generally, as the misalignment increases, the asymmetric crack growth accelerates with a consequent decrease in the time taken to reach the critical situation (Yngvesson, 2000; Zhao et al ., 2011; Kim et al ., 2013). In this paper, a numerical approach is developed to study (on the basis of the Paris law) fatigue propagation paths associated with eccentric external cracks in a CCRB subjected to fixed axial displacement. 2. Numerical modelling Fig. 1a shows the sample used in the numerical analysis with a length 2 L of twenty times its diameter D . The eccentric CCRB geometry (Fig. 1b) was determined by the following parameters: round bar diameter D , maximum crack depth a max , minimum crack depth a min , ligament diameter d and ligament eccentricity ε :

max 2 a a ε − =

min

(1)

The point A on the crack front is associated with the maximum crack depth a max and the point B with the minimum crack depth a min (Fig. 1b). The basic hypothesis of this modeling is to consider that the crack propagates by fatigue maintaining its circular geometry and according to the Paris law:

d

a C K N = ∆

m

(2)

d

m and C being the Paris parameters of the material. In the calculation of the fatigue crack path, carried out in successive iterations, only the points A and B at the circular crack front are taken into account.

D

u/2

a min

a max

d

A

B

2L

D

u/2

(a)

(b)

Fig. 1. Eccentric CCRB subjected to fixed axial displacement (a) and detail of the cracked surface (b).

For the first point the advance Δ a A is the same throughout the calculation, while for the second point the obtaining of Δ a B is performed with the Paris-Erdogan law through the following expression:

* K a a K   ∆ = ∆     IB B A IA

* m

(3)