## 3 Proposed Stress-Strength Method
### Numerical Application of the Proposed Method
#### 3.3.1 Stress Distribution Based on the Time-Dependent Stress
_Step 7.2._ By using the \(S_{a-max}\) values and by taking the minimum stress \(\sigma_{2}=S_{a-min}\), the estimated values of the maximum and minimum strain (\(\epsilon_{max},\epsilon_{min}\)) that define the hysteresis loops given in Figure 3 are:  
\[\epsilon_{max}=\frac{S_{a-max}}{E}+\left(\frac{S_{a}}{H}\right)^{1/n} \tag{30}\]  
\[\epsilon_{min}=\epsilon_{max}-2\left[\frac{S_{a-max}-S_{a-min}}{2E}+\left( \frac{S_{a-max}-S_{a-min}}{H}\right)^{1/n}\right] \tag{31}\]  
As a result, by using data obtained in step 7.2, the determined values of the strain amplitude \(\epsilon_{a}\) are shown in Table 4. Here it is important to highlight that the value of the determined \(\epsilon_{max}\) for \(N_{f}=10^{6}\) is related to the determined time-dependent strain \(\epsilon_{i}\) value that defines the equivalent stress \(\sigma_{ \begin{table}
\begin{tabular}{c c c c c c c} \hline \multirow{2}{*}{\(N_{f}\)} & \(S_{a}\)=\(S_{max}\) & \(\epsilon_{max}\) & \(\epsilon_{min}\) & \(\epsilon_{a}\)=\(\epsilon_{max}\) & \multirow{2}{*}{\(\epsilon_{aa}\)} & \multirow{2}{*}{\(\epsilon_{aa}\)} & \multirow{2}{*}{\(\epsilon_{aa}\)} \\  & **Equation (28)/MPa** & **Equation (30)** & **Equation (31)** & & \(-\epsilon_{min}\) & \\ \hline
1 & 665.2646 & & & & & \\
1000 & 349.8785 & 0.001993 & 0.000632 & 0.001361 & 0.000810 & 0.000551 \\
10,000 & 282.4174 & 0.001351 & 0.000307 & 0.001044 & 0.000654 & 0.000390 \\
100,000 & 227.9637 & 0.001072 & 0.000284 & 0.000789 & 0.000528 & 0.000261 \\
1,000,000 & 184.0094 & **0.000864** & 0.000282 & 0.000582 & 0.000426 & 0.000156 \\
10,000,000 & 148.5300 & 0.000700 & 0.000282 & 0.000416 & 0.000344 & 0.000072 \\
100,000,000 & 119.8915 & 0.000560 & 0.000282 & 0.000281 & 0.000278 & 0.000004 \\ \hline \end{tabular}
\end{table} 
Table 4: Strain-life curve data. Figure 3: Hysteresis loop of the \(N_{f-10^{9}}\) stress amplitude. Note: Notice that hysteresis loops analysis of Table 4 for each row is needed. By using both the plastic strain amplitude \(\varepsilon_{pa}\) data and the defined cycles to failure \(N_{f}\) values, the log-linear analysis constant \(c\) value is \(c=-0.3643\). The generated strain-life curve is shown in Figure 4.  
_Step 8_. : As the objective is to define the cycles to failure that correspond with the strain of \(\varepsilon_{max-106}=0.000864\), then the determined value of the strain given in Table 2 of step \(5\)\(\varepsilon_{(t=10.5)}=0.000864\) was interpolated. To this interpolated value, the corresponding time-dependent stress, as shown in Table 2 is \(\sigma_{i}=74.28\) MPa. Nevertheless, because the time-dependent stress \(\sigma_{i}\) represents the generated maximum principal stress \(\sigma_{1}\) at \(t=i\), then the equivalent supposed load \(W_{i}\) that generates the determined deflection \(f(\delta_{i})\) at this time can be estimated as  
\[W_{i}=\frac{f(\delta_{i})8 EI}{L^{4}} \tag{32}\]  
By using this \(W_{i}\) value, the corresponding equivalent stress is \(\sigma_{eq-i}=184.03\) MPa. The main goal of this section is to estimate the stress distribution by using the determined equivalent stress \(\sigma_{eq-i}\), therefore, the next normal distribution analysis is also necessary. _Step 9_. : By using the equivalent stress \(\sigma_{eq-i}=184.03\) MPa value as a mean of the normal distribution \(\mu_{s}\) and its estimated standard deviation \(\sigma_{dev}\)\(s=18.40\) value, the normal stress distribution parameters are \(N_{s}\)(184.03, 18.40). Now, the strength distribution analysis is given.