## 3 Probabilistic Model to Explain the Statistical Fatigue Property in Interior-Induced Fracture  
### Effects of Size and Depth of Inclusions on the Fatigue Strength  
According to the \(\sqrt{area}\) model proposed by [PERSON] [9], the fatigue limit of a metallic material, \(\sigma_{w}\), was well provided by the defect size \(\rho\) and the Vickers' hardness \(HV\) (kgf/mm\({}^{2}\)), as follows;  
\[\sigma_{w}=\alpha(HV+120)/\left(\sqrt{area}\right)^{1/6}, \tag{1}\]  
where \(area\) indicates the area of the defect projected perpendicular to the longitudinal direction of the specimen, and the factor of \(\alpha\) was given by \(\alpha=1.43\) for surface defect and \(\alpha=1.56\) for interior defect, respectively.  
When this concept of \(\sqrt{area}\) was applied to the fatigue strength at \(N=10^{9}\) cycles as an attempt, assuming a spherical interior inclusion with a radius of \(\rho\), the fatigue strength at \(N=10^{9}\), \(\sigma_{w9}\), was given by the following expression,  
\[\sigma_{w9}=1.56(HV+120)/\left(\sqrt{\pi\rho^{2}}\right)^{1/6}. \tag{2}\]  
\begin{table}
\begin{tabular}{c c} \hline Tensile strength & 2316 (MPa) \\ \hline Vickers’ hardness & 778 (HV) \\ \hline \end{tabular}
\end{table}
Table 1: Static mechanical properties of the material.  
Figure 2: SEM observations of fracture surfaces in surface-induced fracture and interior inclusion-induced fracture.  
Since the Vickers' hardness of this bearing steel (JIS:SUJ2) was reported as \(HV\) = 778 [3], Equation (2) was reduced as follows;  
\[\sigma_{\text{\tiny{uv9}}}=1273\rho^{-1/6} \tag{3}\]  
Equation (3) indicates the relationship between the fatigue strength at \(N\) = 10\({}^{9}\) and the inclusion size (radius) for this steel.  
Such a relationship is depicted in Figure 3. When the inclusion size is small, those inclusions have no harmful effect on the fatigue strength, as reported by many researchers [24; 25]. From this point of view, if the critical small size is assumed as \(\rho\) = 3 \(\upmu\)m in this study, the upper bound of the fatigue strength given by Equation (3) becomes \(\sigma_{\text{\tiny{uv9}}}=1060\) MPa, as indicated in Figure 3. On the other hand, an inclusion larger than 40 \(\upmu\)m (\(\rho\) > 40 \(\upmu\)m) is seldom found in the present steel. In such an assumption, the lower bound of the fatigue strength given by Equation (3) becomes \(\sigma_{\text{\tiny{uv9}}}=689\) MPa, as indicated by the horizontal dotted line in Figure 3. Thus, the distribution range of the fatigue strength at \(N\) = 10\({}^{9}\) due to the distribution of the inclusion size would be 689 MPa\(\sim\)1060 MPa.  
In the case of the rotating bending, the stress distribution across the specimen section has a distinct gradient, such that the stress on the surface gives the maximum and the stress becomes zero at the center of the section. Accordingly, the inclusion depth is another important factor to govern the fatigue strength of \(\sigma_{\text{\tiny{uv9}}}\), even if the inclusion size is fixed to a certain value. When the fatigue strength at the depth \(\xi\) is denoted as \(\sigma_{\text{\tiny{uv9}}}\), the fatigue strength evaluated by nominal bending stress on the specimen surface, \(\sigma_{\text{\tiny{uv9}}}^{*}\), is given as follows [26];  
\[\sigma_{\text{\tiny{uv9}}}^{*}=\frac{r}{r-\xi}\sigma_{\text{\tiny{uv9}}}=\frac {1.5}{1.5-\xi}\sigma_{\text{\tiny{uv9}}} \tag{4}\]  
where \(r\) indicates the radius of the critical portion of the specimen. Here, substituting Equation (3) into Equation (4), we have the following expression;  
\[\sigma_{\text{\tiny{uv9}}}^{*}=\frac{1.5}{1.5-\xi}\times 1273\rho^{-1/6}=\frac {1910}{1.5-\xi}\rho^{-1/6} \tag{5}\]  
Consequently, the fatigue strength of \(\sigma_{\text{\tiny{uv9}}}^{*}\) for the specimen with an inclusion size of \(\rho\) and a depth of \(\xi\) could be calculated by Equation (5).