# Vibration Fatigue Life Reliability Cable Trough Assessment by Using Weibull Distribution
## 4 Numerical Application
### Cycle Random Analysis  
In this section, the cycle to failure (\(N\)) value that corresponds to the stress amplitude \(\sigma_{a}\) is determined. The steps are as follows. Step 1. By using Equation (27) the cycle to failure (\(N\)) value that corresponds to the stress amplitude \(\sigma_{a}\) is determined. Equation (27) according to [11] represents an efficient option to determine the fatigue life. \[N=A\times\sigma_{a}^{-m} \tag{27}\]  
where the stress amplitude \(\sigma_{a}\) is given by Equation (28) [26]. \(A\) and \(m\) are given by Equations (29) and (30), respectively. \[\sigma_{a}=\frac{\sigma_{1}-\sigma_{2}}{2} \tag{28}\]  
\[A=0.5\times a^{m} \tag{29}\]  
\[m=-\frac{1}{b} \tag{30}\]  
In Equation (29) \(a\) is the fatigue strength coefficient and in Equation (30) \(b\) represents the fatigue strength exponent. They are given by Equations (31) and (32), respectively. \[a=\frac{(fS_{ut})^{2}}{S_{end}} \tag{31}\]  
\[b=-\frac{1}{3}log\left(\frac{fS_{ut}}{S_{end}}\right) \tag{32}\]  
The fatigue strength factor \(f\) is taken from [26], \(S_{ut}\) is the material ultimate stress, and \(S_{end}\) is the material endurance limit. Now, let us use the data stated in the Section 4 numerical application, starting by substituting the data into Equations (31) and (32) [27], respectively. \[a=\frac{(0.9\times 7250)^{2}}{3625}=11,745\]\[b=-\frac{1}{3}log\left(\frac{0.9\times 7250}{3625}\right)=-0.085\]  
The \(a\) and \(b\) values are included in the Equations (29) and (30), respectively. \[A=0.5\times 11,745^{11.764}=3.8\times 10^{47}\]  
\[m=-\frac{1}{(-0.085)}=11.764\]  
The amplitude stress \(\sigma_{a}\) is given by the application of the component submitted to random vibration, where \(\sigma_{1}=1188.00\) and \(\sigma_{2}=330.00\). \[\sigma_{a}=\frac{\sigma_{1}-\sigma_{2}}{2} \tag{28}\]0  
Then, the \(\sigma_{a}\), \(A\) and \(m\) values are included in the Equation (27). \[\sigma_{a}=\frac{\sigma_{1}-\sigma_{2}}{2} \tag{28}\]1  
This \(N\) value corresponds to the cycle to failure \(N\) at the stress amplitude \(\sigma_{a}\) stated. Step 2. By using the \(N\) value from Equation (27) and the Weibull element \(t_{0a}\) value that corresponds to the stress amplitude \(\sigma_{a}\) value in Equation (21) and Table 4, the corresponding Weibull scale cycle to failure \(\eta_{t}\) parameter is obtained by Equation (32) [27]. \[\sigma_{a}=\frac{\sigma_{1}-\sigma_{2}}{2} \tag{28}\]2  
\[\sigma_{a}=\frac{\sigma_{1}-\sigma_{2}}{2} \tag{28}\]3  
Step 3. The Weibull cycle \(\eta_{t}\) parameter is employed to determine the corresponding expected cycle to failure \(N_{i}\) values that belongs to each \(Y_{i}\) elements in Equation (16). It is performed by using Equation (34) [27]. \[\sigma_{a}=\frac{\sigma_{1}-\sigma_{2}}{2} \tag{28}\]4  
About the Weibull shape parameter, the \(\beta_{s}\) value is used in the Weibull cycle to failure family since the failure mode (random vibration) remains constant. Hence, the Weibull cycle to failure family is W (1.7087, 7.700 \(\times\) 10\({}^{16}\)). The corresponding expected cycle to failure values for each one of the \(Y_{i}\) elements are determined (see the ninth column in Table 5).