Example: Weibull Degradation Crack Propagation (Point Estimation) - Revision history

Lisa Hacker: Redirected page to Weibull++ Examples

2015-06-30T00:05:08Z

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2012-07-17T11:51:45Z

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	~~\| valign="middle" align="left" bgcolor=EEEDF7\|[~~[~~Image~~: ~~Weibull-Examples-banner.png\|400px\|center]]~~
	~~\|}<br>~~

	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 cycles-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\| ]~~]

Lisa Hacker at 18:30, 29 March 2012

2012-03-29T18:30:36Z

← Older revision		Revision as of 18:30, 29 March 2012
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	\end{matrix}</math></center>		\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.		These cycles-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>		<math></math>
	[[Image:Degradation Example 1 Data and Result.png\|thumb\|center\|400px\| ]]		[[Image:Degradation Example 1 Data and Result.png\|thumb\|center\|400px\| ]]

Lisa Hacker at 18:29, 29 March 2012

2012-03-29T18:29:55Z

← Older revision		Revision as of 18:29, 29 March 2012
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	<center><math>\begin{matrix}		<center><math>\begin{matrix}
	Cycles-to-Failure \\		{} & Cycles-to-Failure \\
	Unit A & \text{500,622} \\		Unit A & \text{500,622} \\
	Unit B & \text{475,739} \\		Unit B & \text{475,739} \\

Lisa Hacker at 18:29, 29 March 2012

2012-03-29T18:29:21Z

← Older revision		Revision as of 18:29, 29 March 2012
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	<center><math>\begin{matrix}		<center><math>\begin{matrix}
	~~{} &~~ Cycles-to-Failure \\		Cycles-to-Failure \\
	Unit A & \text{500,622} \\		Unit A & \text{500,622} \\
	Unit B & \text{475,739} \\		Unit B & \text{475,739} \\

Lisa Hacker at 18:28, 29 March 2012

2012-03-29T18:28:14Z

← Older revision		Revision as of 18:28, 29 March 2012
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	These values can now be substituted into the underlying exponential model, solved for <math>x</math> or:		These values can now be substituted into the underlying exponential model, solved for <math>x</math> or:

	::<math>x=\frac{\text{ln}(y)-\text{ln}(b)}{a}</math>		<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:		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:

Chuck Smith: Created page with '{| class="FCK__ShowTableBorders" border="0" cellspacing="0" cellpadding="0" align="center"; style="width:100%;" |- | valign="middle" align="left" bgcolor=EEEDF7|[[Image: Weibull-…'

2012-03-22T22:40:36Z

Created page with '{| class="FCK__ShowTableBorders" border="0" cellspacing="0" cellpadding="0" align="center"; style="width:100%;" |- | valign="middle" align="left" bgcolor=EEEDF7|[[Image: Weibull-…'

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{| class="FCK__ShowTableBorders" border="0" cellspacing="0" cellpadding="0" align="center"; style="width:100%;"
|-
| valign="middle" align="left" bgcolor=EEEDF7|[[Image: Weibull-Examples-banner.png|400px|center]]
|}<br>

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:

::<math>x=\frac{\text{ln}(y)-\text{ln}(b)}{a}</math>

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| ]]