## 3 Results and Discussion
### Four-Point Bending Fatigue Test
#### 3.3.2 Fatigue Equation of Two-Parameter Weibull Distribution  
Due to the discreteness of the CNFPUC material during preparation and fatigue tests, to get a more reliable result with a smaller sample, the test results could only be analyzed by means of scientific and reasonable statistical methods. In this paper, the Weibull distribution of two parameters was used to analyze the fatigue life distribution of CNFPUC [14; 29; 30; 31; 32]. The probability density function of a two-parameter Weibull distribution is:  
\[f(N)=\frac{b}{N_{a}}\Big{(}\frac{N}{N_{a}}\Big{)}^{b-1}exp\Bigg{[}-\Big{(}\frac {N}{N_{a}}\Big{)}^{b}\Bigg{]}(0\;\leq\;\mathrm{N}\;<\;\infty) \tag{1}\]  
\(b\)--Weibull shape parameter;  
\(N_{a}\)--Characteristic lifetime of Weibull distribution. \begin{table}
\begin{tabular}{c c c c c c c} \hline \hline \multirow{2}{*}{**Stress Level**} & \multirow{2}{*}{**Number**} & \multicolumn{6}{c}{**Test Temperature**} \\ \cline{3-7}  & & \(\mathbf{-40\;^{\circ}C}\) & \(\mathbf{-20\;^{\circ}C}\) & \(\mathbf{0\;^{\circ}C}\) & \(\mathbf{20\;^{\circ}C}\) & \(\mathbf{40\;^{\circ}C}\) \\ \hline \multirow{6}{*}{0.6} & 1 & 433,654 & 312,683 & 214,860 & 104,655 & 55,855 \\  & 2 & 467,852 & 355,462 & 244,379 & 116,756 & 78,624 \\  & 3 & 501,687 & 378,562 & 264,677 & 146,890 & 83,562 \\  & 4 & 526,574 & 398,565 & 304,692 & 173,216 & 106,875 \\  & 5 & 556,230 & 432,566 & 325,688 & 182,368 & 116,535 \\  & 6 & 593,782 & 457,852 & 326,985 & 219,640 & 149,856 \\ \hline \multirow{6}{*}{0.7} & 1 & 157,655 & 76,543 & 50,796 & 10,984 & 5640 \\  & 2 & 184,633 & 96,425 & 55,249 & 13,653 & 6756 \\  & 3 & 195,679 & 113,609 & 64,538 & 20,075 & 8564 \\  & 4 & 216,430 & 137,688 & 89,436 & 23,694 & 11,080 \\  & 5 & 246,582 & 143,266 & 107,855 & 27,682 & 12,568 \\  & 6 & 250,643 & 156,429 & 113,563 & 31,691 & 20,354 \\ \hline \multirow{6}{*}{0.8} & 1 & 36,215 & 10,567 & 6456 & 3286 & 676 \\  & 2 & 43,542 & 15,984 & 8237 & 4036 & 985 \\ \cline{1-1}  & 3 & 55,756 & 18,972 & 10,988 & 5033 & 1598 \\ \cline{1-1}  & 4 & 61,235 & 21,943 & 11,462 & 7862 & 2288 \\ \cline{1-1}  & 5 & 62,895 & 24,477 & 16,467 & 9236 & 3126 \\ \cline{1-1}  & 6 & 71,405 & 29,945 & 17,315 & 11,030 & 3838 \\ \hline \hline \end{tabular}
\end{table} 
Table 9: Fatigue life test results (times). According to Equation(1), the distribution function of the Weibull variable \(F\) (\(Np\)) is obtained:  
\[F\Big{(}N_{p}\Big{)}=P\Big{(}N<N_{p}\Big{)}=1-exp[-(\frac{N_{p}}{N_{a}})^{b}]\]  
The reliability \(p\) can be obtained from the failure probability distribution function \(P(N<N_{p})\):  
\[F(p)=P\Big{(}N<N_{p}\Big{)}=1-P\Big{(}N>N_{p}\Big{)}=1-exp\Bigg{[}-\Bigg{(} \frac{N_{p}}{N_{a}}\Bigg{)}\Bigg{]}^{b} \tag{2}\]  
Expression (2) can be changed: \(\frac{1}{p}=exp(\frac{N_{p}}{N_{a}})^{b}\), and the two sides can be obtained by taking the second natural logarithm: \(\ln\left[\ln\left(\frac{1}{p}\right)\right]=b\ln N_{p}-b\ln N_{a}\),  
Therefore:  
\[Y=bX-\beta \tag{3}\]  
In this formula: \(Y=\ln\left[\ln\left(\frac{1}{p}\right)\right]\); \(X=\ln N_{p}\); \(\beta=b\ln N_{a}\). Equation (3) is a linear equation; \(b\) and \(\beta\) can be obtained directly from the straight line, thus two parameters of the Weibull distribution can be obtained. Equation (3) can be used to verify the hypothesis of whether a set of test data obeys the two-parameter Weibull distribution. If the regression analysis of the test data shows that there is a good statistical linear relationship between \(Y\) and \(X\), the hypothesis will be true and vice versa. The \(k\) numbers fatigue test data obtained under a given stress level were arranged in order from small to large. The sequence number is \(i\), and the reliability factor \(p\) corresponding to fatigue life \(N_{p}\) is calculated by Equation (4):  
\[P=1-\frac{i}{K+1} \tag{4}\]  
The fatigue test results in Table 9 were tested for the Weibull probability distribution at various stress levels.