Example: Weibull Degradation Crack Propagation - Point Estimation

Five turbine blades were tested for crack propagation. The test units are cyclically stressed and inspected every 100,000 cycles for crack length. Failure is defined as a crack of length 30mm or greater.

Following is a table of the test results:

[math]\displaystyle{ \begin{matrix} Cycles (x1000) & Unit A (mm)& Unit B (mm) & Unit C (mm) & Unit D (mm)& Unit E (mm) \\ 100 & 15 & 10 & 17 & 12 & 10 \\ 200 & 20& 15 & 25 & 16 & 15 \\ 300 & 22 & 20 &26 & 17 & 20 \\ 400 & 26 &25 & 27 & 20 & 26 \\ 500 & 29 & 30 & 33 &26 & 33 \\ \end{matrix} }[/math]

Using degradation analysis with an exponential model for the extrapolation, determine the B10 life for the blades.

Solution

The first step is to solve the equation [math]\displaystyle{ y=b\cdot {{e}^{a\cdot x}} }[/math] for [math]\displaystyle{ a }[/math] and [math]\displaystyle{ b }[/math] for each of the test units. Using regression analysis, these values for each of the test units are:

[math]\displaystyle{ \begin{matrix} {} & a & b \\ Unit A & 0.00158 & 13.596 \\ Unit B & 0.00271 & 8.272 \\ Unit C & 0.00140 & 16.435 \\ Unit D & 0.00177 & 10.361 \\ Unit E & 0.00294 & 7.931 \\ \end{matrix} }[/math]

These results are shown graphically in the next figure.

[math]\displaystyle{ }[/math]

These values can now be substituted into the underlying exponential model, solved for [math]\displaystyle{ x }[/math] or:

[math]\displaystyle{ x=\frac{\text{ln}(y)-\text{ln}(b)}{a} }[/math]

Using the values of [math]\displaystyle{ a }[/math] and [math]\displaystyle{ b }[/math] , with [math]\displaystyle{ y=30 }[/math] , the resulting time at which the crack length reaches 30mm is then found for each sample:

[math]\displaystyle{ \begin{matrix} {} & Cycles-to-Failure \\ Unit A & \text{500,622} \\ Unit B & \text{475,739} \\ Unit C & \text{428,739} \\ Unit D & \text{600,810} \\ Unit E & \text{452,832} \\ \end{matrix} }[/math]

These times-to-failure can now be analyzed in the conventional manner. Assuming a two-parameter Weibull distribution and using the MLE estimation method, the distribution parameters are calculated as [math]\displaystyle{ \beta =8.055 }[/math] and [math]\displaystyle{ \eta =519,555. }[/math] Using these values, the B10 life is calculated to be 392,918 cycles. The degradation analysis tool in Weibull++ performs this type of analysis for you. The following figure shows the data as entered in Weibull++ for this analysis.

[math]\displaystyle{ }[/math]