Example: Weibull Degradation Crack Propagation (Point Estimation): Difference between revisions

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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:
 
<center><math>\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></center>
 
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>y=b\cdot {{e}^{a\cdot x}}</math>  for  <math>a</math>  and  <math>b</math>  for each of the test units. Using regression analysis, these values for each of the test units are:
 
<center><math>\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></center>
 
These results are shown graphically in the next figure.
 
<math></math>
[[Image:Degradation Example 1 Plot.png|thumb|center|400px| ]]
 
These values can now be substituted into the underlying exponential model, solved for  <math>x</math>  or:
 
<center><math>x=\frac{\text{ln}(y)-\text{ln}(b)}{a}</math></center>
 
Using the values of  <math>a</math>  and  <math>b</math> , with  <math>y=30</math> , the resulting time at which the crack length reaches 30mm is then found for each sample:
 
<center><math>\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></center>
 
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>\beta =8.055</math>  and  <math>\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></math>
[[Image:Degradation Example 1 Data and Result.png|thumb|center|400px| ]]

Latest revision as of 00:05, 30 June 2015

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