## 3 Proposed Method
### Weibull Stress Family Estimation and Its Random Behavior Analysis  
Before determining the random behavior of the addressed \(\sigma_{1}\) and \(\sigma_{2}\) values, it is necessary to mention that they should be validated by using them with the material yield strength \(S_{y}\) value, and the desired safety factor (_SF_) in the yielding maximum shear stress theory (_MSS_) and the distortion energy theory (_DE_) criteria [25] as follows:  
\[MSS\ theory=(\sigma_{1})<\frac{S_{y}}{SF} \tag{11}\]  
and,  
\[DE\ theory=\Big{(}\sigma_{1}^{2}-\sigma_{1}*\sigma_{2}+\sigma_{2}^{2}\Big{)}^{ 0.5}<\frac{S_{y}}{SF} \tag{12}\]  
If Equations (11) and (12) do not hold, then a material with higher \(S_{y}\) for which Equations (11) and (12) holds, must be selected.  
**Note 1.** Note that from Equations (11) and (12), we only know whether the design is safe or not, but at this point it is not possible to determine either the designed reliability or the corresponding expected failure times.  
Based on the above, the expected behavior of \(\sigma_{1}\) and \(\sigma_{2}\) is determined as follows:(i) From [24], determine the sample size \(n\) value to be used in the analysis.  
\[n=\frac{-1}{ln(R(t))} \tag{13}\]  
where \(R(t)\) is the desired _reliability of the analysis_, here \(R(t)\) could be seen as the equivalent of a confidential interval level used in the quality field. Note that the reason why \(R(t)=0.9535\) is used because for this value, \(n\) is an integer (\(n=21\)), but any desired \(R(t)\) index can be used.  
(ii) Using the \(n\) value from Equation (13) in the median rank approach function [26] given in Equation (14), determine the corresponding cumulated failure percentile \(F(t_{i})\) elements as:  
\[F(t_{i})=\frac{i-0.3}{n+0.4} \tag{14}\]  
(iii) by using the \(F(t_{i})\) elements in the linear form of the reliability function defined in Equation (7), determine the corresponding \(Y_{i}\) elements as shown in Equation (15). Then, from the \(Y_{i}\) elements, determine its corresponding mean value, obtained by using Equation (16).  
\[Y_{i}=LN(-LN(1-(i-0.3)/(n+0.4)))) \tag{15}\]  
and,  
\[\mu_{y}=\sum_{i=1}^{n}\frac{Y_{i}}{n} \tag{16}\]  
(iii) by inserting the \(\mu_{y}\) value and the principal stress \(\sigma_{1}\) and \(\sigma_{2}\) values into Equations (8) and (9), determine the corresponding _Weibull stress parameters_ (\(\beta_{s}\), \(\eta_{s}\)). These estimated (\(\beta_{s}\), \(\eta_{s}\)) parameters represent the _Weibull stress_ family used to model the random behavior of the estimated principal stresses \(\sigma_{1}\) and \(\sigma_{2}\) values that represent the minimum required strength the used material must present in order for it to have the designed reliability. Thus, \(\sigma_{1}\) and \(\sigma_{2}\) are the applied stresses from which the \(\sigma_{\alpha\beta}\) value was determined; \(\sigma_{\alpha\beta}\) is the equivalent constant applied stress value, that by its cyclical application demands the material presents at least the minimum strength of \(\sigma_{2i}\) value. Based  
Consequently, by using the \(t_{0 max}\) and the \(\beta_{s}\) values, the corresponding \(Y_{1}\) value is established.  
\[Y_{1}=ln(t_{01})*\beta_{s} \tag{21}\]  
Finally, the reliability index corresponding to the \(Y_{1}\) value is obtained.  
\[R(t)=exp\{-exp\{Y_{1}\}\} \tag{22}\]  
**Note 2**. It is important to mention that the \(R(t)\) index obtained in Equation (22) corresponds to an element which has a strength equivalent to the \(\sigma_{1}\) value. Thus, if \(S_{y}\
eq\sigma_{1}\), the reliability of the designed element is found by replacing the \(S_{y}\) value with \(\sigma_{1}\) in Equation (20).  
However, note that while the reliability of the designed element can be obtained from Equation (22), the corresponding expected cycles to failure remain unknown. In this paper the cycle to failure are determined as follows: