Competing Failure Modes Example - Revision history

Lisa Hacker: Redirected page to Complex Failure Modes Example

2012-08-08T07:37:15Z

Redirected page to Complex Failure Modes Example

← Older revision		Revision as of 07:37, 8 August 2012
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	[[~~Category: For Deletion~~]]		#REDIRECT [[Complex_Failure_Modes_Example]]

Kate Racaza: Replaced content with 'Category: For Deletion'

2012-07-24T04:00:32Z

Replaced content with 'Category: For Deletion'

← Older revision		Revision as of 04:00, 24 July 2012
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	~~'''Competing Failures with Two Failure Modes Example'''~~		[[Category: For Deletion]]

	~~From Meeker & Escobar~~ [[~~Appendix~~: ~~Weibull References\|[27]]], the following table gives failure times for an electric component that has two failure modes.~~

	One failure mode is due to random voltage spikes which cause failure by overloading the system (denoted as a <math>V</math> in the table). The other failure mode is due to wear-out failures which usually happen only after the system has run for many cycles (this failure mode is denoted as a <math>W</math> in the table).

	~~Considering that these are competing failure modes, determine the overall reliability for the component at 100,000 cycles.~~


	~~<center><math>\begin{matrix}~~
	~~Number & Failure & Failure & Number & Failure & Failure \\~~
	~~in State & Time* & Mode & in State & Time* & Mode \\~~
	~~\text{1} & \text{2} & \text{V} & \text{1} & \text{147} & \text{W} \\~~
	~~\text{1} & \text{10} & \text{V} & \text{1} & \text{173} & \text{V} \\~~
	~~\text{1} & \text{13} & \text{V} & \text{1} & \text{181} & \text{W} \\~~
	~~\text{2} & \text{23} & \text{V} & \text{1} & \text{212} & \text{W} \\~~
	~~\text{1} & \text{28} & \text{V} & \text{1} & \text{245} & \text{W} \\~~
	~~\text{1} & \text{30} & \text{V} & \text{1} & \text{247} & \text{V} \\~~
	~~\text{1} & \text{65} & \text{V} & \text{1} & \text{261} & \text{V} \\~~
	~~\text{1} & \text{80} & \text{V} & \text{1} & \text{266} & \text{W} \\~~
	~~\text{1} & \text{88} & \text{V} & \text{1} & \text{275} & \text{W} \\~~
	~~\text{1} & \text{106} & \text{V} & \text{1} & \text{293} & \text{W} \\~~
	~~\text{1} & \text{143} & \text{V} & \text{8} & \text{300} & \text{suspended} \\~~
	~~\end{matrix}</math></center>~~


	*Failure times given are in thousands of cycles.

	~~'''Solution'''~~

	We will begin by performing a Weibull analysis of the voltage spike ( <math>V</math> ) failure mode. In order to do this, we must consider all of the failures for the wear-out mode to be suspensions. The input data for the analysis are shown next:

	~~[[Image:Competing Failiure Mode V Mode Data.png\|thumb\|center\|250px\| ]]~~

	Analyzing this data set using the maximum likelihood method (recommended due to the number of suspensions in the data) and assuming a Weibull distribution, we obtain the parameters <math>{{\beta }_{V}}=0.6711</math> and <math>{{\eta }_{V}}=449.4</math> . The reliability for this failure mode at <math>t=100</math> is <math>{{R}_{V}}(100)=0.694</math> .

	We follow an identical procedure for the wear-out failure mode, counting only the <math>W</math> entries as failures and assuming the <math>V</math> entries are suspensions. This is shown next.

	~~[[Image:Competing Failiure Mode W Mode Data.png\|thumb\|center\|250px\| ]]~~

	Once again, analyzing with a Weibull distribution with maximum likelihood estimators, we obtain the parameters <math>{{\beta }_{W}}=4.337</math> and <math>{{\eta }_{W}}=340.4</math> . The reliability for this failure mode at <math>t=100</math> is <math>{{R}_{W}}(100)=0.995</math> .

	~~We can now use the system Reliability Equation to determine the overall system reliability at 100,000 cycles:~~

	~~::<math>\begin{align}~~
	~~& {{R}_{sys}}(100)= {{R}_{V}}(100)\cdot {{R}_{W}}(100) \\~~
	~~& = 0.694\cdot 0.995 \\~~
	~~& = 0.69053~~
	~~\end{align}</math>~~

	~~Or the reliability of the unit (or system) under both modes is <math>{{R}_{sys}}(100)=69.053%.</math>~~

	Note that Weibull++ can perform this analysis for you automatically. To accomplish this, the data would be entered in a single data sheet and competing failure modes chosen as the analysis method. This is shown in the next graphic.

	~~[[Image:Competing Failiure Mode Example 1 Data.png\|thumb\|center\|250px~~]]

Kate Racaza: moved Template:Example: Competing Failures with Two Failure Modes Example to Competing Failure Modes Example

2012-07-23T10:20:09Z

moved Template:Example: Competing Failures with Two Failure Modes Example to Competing Failure Modes Example

← Older revision	Revision as of 10:20, 23 July 2012
(No difference)

Nicolette Young at 22:01, 25 April 2012

2012-04-25T22:01:55Z

← Older revision		Revision as of 22:01, 25 April 2012
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	We will begin by performing a Weibull analysis of the voltage spike ( <math>V</math> ) failure mode. In order to do this, we must consider all of the failures for the wear-out mode to be suspensions. The input data for the analysis are shown next:		We will begin by performing a Weibull analysis of the voltage spike ( <math>V</math> ) failure mode. In order to do this, we must consider all of the failures for the wear-out mode to be suspensions. The input data for the analysis are shown next:

	[[Image:Competing Failiure Mode V Mode Data.png\|thumb\|center\|~~400px~~\| ]]		[[Image:Competing Failiure Mode V Mode Data.png\|thumb\|center\|250px\| ]]

	Analyzing this data set using the maximum likelihood method (recommended due to the number of suspensions in the data) and assuming a Weibull distribution, we obtain the parameters <math>{{\beta }_{V}}=0.6711</math> and <math>{{\eta }_{V}}=449.4</math> . The reliability for this failure mode at <math>t=100</math> is <math>{{R}_{V}}(100)=0.694</math> .		Analyzing this data set using the maximum likelihood method (recommended due to the number of suspensions in the data) and assuming a Weibull distribution, we obtain the parameters <math>{{\beta }_{V}}=0.6711</math> and <math>{{\eta }_{V}}=449.4</math> . The reliability for this failure mode at <math>t=100</math> is <math>{{R}_{V}}(100)=0.694</math> .
Line 37:		Line 37:
	We follow an identical procedure for the wear-out failure mode, counting only the <math>W</math> entries as failures and assuming the <math>V</math> entries are suspensions. This is shown next.		We follow an identical procedure for the wear-out failure mode, counting only the <math>W</math> entries as failures and assuming the <math>V</math> entries are suspensions. This is shown next.

	[[Image:Competing Failiure Mode W Mode Data.png\|thumb\|center\|~~400px~~\| ]]		[[Image:Competing Failiure Mode W Mode Data.png\|thumb\|center\|250px\| ]]

	Once again, analyzing with a Weibull distribution with maximum likelihood estimators, we obtain the parameters <math>{{\beta }_{W}}=4.337</math> and <math>{{\eta }_{W}}=340.4</math> . The reliability for this failure mode at <math>t=100</math> is <math>{{R}_{W}}(100)=0.995</math> .		Once again, analyzing with a Weibull distribution with maximum likelihood estimators, we obtain the parameters <math>{{\beta }_{W}}=4.337</math> and <math>{{\eta }_{W}}=340.4</math> . The reliability for this failure mode at <math>t=100</math> is <math>{{R}_{W}}(100)=0.995</math> .
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	Note that Weibull++ can perform this analysis for you automatically. To accomplish this, the data would be entered in a single data sheet and competing failure modes chosen as the analysis method. This is shown in the next graphic.		Note that Weibull++ can perform this analysis for you automatically. To accomplish this, the data would be entered in a single data sheet and competing failure modes chosen as the analysis method. This is shown in the next graphic.

	[[Image:Competing Failiure Mode Example 1 Data.png\|thumb\|center\|~~400px~~]]		[[Image:Competing Failiure Mode Example 1 Data.png\|thumb\|center\|250px]]

Harry Guo at 18:26, 2 March 2012

2012-03-02T18:26:50Z

← Older revision		Revision as of 18:26, 2 March 2012
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	We will begin by performing a Weibull analysis of the voltage spike ( <math>V</math> ) failure mode. In order to do this, we must consider all of the failures for the wear-out mode to be suspensions. The input data for the analysis are shown next:		We will begin by performing a Weibull analysis of the voltage spike ( <math>V</math> ) failure mode. In order to do this, we must consider all of the failures for the wear-out mode to be suspensions. The input data for the analysis are shown next:

	[[Image:Competing Failiure Mode W Mode Data.png\|thumb\|center\|400px\| ]]		[[Image:Competing Failiure Mode V Mode Data.png\|thumb\|center\|400px\| ]]

	Analyzing this data set using the maximum likelihood method (recommended due to the number of suspensions in the data) and assuming a Weibull distribution, we obtain the parameters <math>{{\beta }_{V}}=0.6711</math> and <math>{{\eta }_{V}}=449.4</math> . The reliability for this failure mode at <math>t=100</math> is <math>{{R}_{V}}(100)=0.694</math> .		Analyzing this data set using the maximum likelihood method (recommended due to the number of suspensions in the data) and assuming a Weibull distribution, we obtain the parameters <math>{{\beta }_{V}}=0.6711</math> and <math>{{\eta }_{V}}=449.4</math> . The reliability for this failure mode at <math>t=100</math> is <math>{{R}_{V}}(100)=0.694</math> .

Harry Guo at 18:25, 2 March 2012

2012-03-02T18:25:36Z

← Older revision		Revision as of 18:25, 2 March 2012
Line 31:		Line 31:
	We will begin by performing a Weibull analysis of the voltage spike ( <math>V</math> ) failure mode. In order to do this, we must consider all of the failures for the wear-out mode to be suspensions. The input data for the analysis are shown next:		We will begin by performing a Weibull analysis of the voltage spike ( <math>V</math> ) failure mode. In order to do this, we must consider all of the failures for the wear-out mode to be suspensions. The input data for the analysis are shown next:

	[[Image:~~SnV~~.png\|thumb\|center\|400px\| ]]		[[Image:Competing Failiure Mode W Mode Data.png\|thumb\|center\|400px\| ]]

	Analyzing this data set using the maximum likelihood method (recommended due to the number of suspensions in the data) and assuming a Weibull distribution, we obtain the parameters <math>{{\beta }_{V}}=0.6711</math> and <math>{{\eta }_{V}}=449.4</math> . The reliability for this failure mode at <math>t=100</math> is <math>{{R}_{V}}(100)=0.694</math> .		Analyzing this data set using the maximum likelihood method (recommended due to the number of suspensions in the data) and assuming a Weibull distribution, we obtain the parameters <math>{{\beta }_{V}}=0.6711</math> and <math>{{\eta }_{V}}=449.4</math> . The reliability for this failure mode at <math>t=100</math> is <math>{{R}_{V}}(100)=0.694</math> .
Line 37:		Line 37:
	We follow an identical procedure for the wear-out failure mode, counting only the <math>W</math> entries as failures and assuming the <math>V</math> entries are suspensions. This is shown next.		We follow an identical procedure for the wear-out failure mode, counting only the <math>W</math> entries as failures and assuming the <math>V</math> entries are suspensions. This is shown next.

	[[Image:~~VoW~~.png\|thumb\|center\|400px\| ]]		[[Image:Competing Failiure Mode W Mode Data.png\|thumb\|center\|400px\| ]]

	Once again, analyzing with a Weibull distribution with maximum likelihood estimators, we obtain the parameters <math>{{\beta }_{W}}=4.337</math> and <math>{{\eta }_{W}}=340.4</math> . The reliability for this failure mode at <math>t=100</math> is <math>{{R}_{W}}(100)=0.995</math> .		Once again, analyzing with a Weibull distribution with maximum likelihood estimators, we obtain the parameters <math>{{\beta }_{W}}=4.337</math> and <math>{{\eta }_{W}}=340.4</math> . The reliability for this failure mode at <math>t=100</math> is <math>{{R}_{W}}(100)=0.995</math> .

	We can now use ~~Eqn. (sysrel)~~ to determine the overall system reliability at 100,000 cycles:		We can now use the system Reliability Equation to determine the overall system reliability at 100,000 cycles:

	::<math>\begin{align}		::<math>\begin{align}
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	Note that Weibull++ can perform this analysis for you automatically. To accomplish this, the data would be entered in a single data sheet and competing failure modes chosen as the analysis method. This is shown in the next graphic.		Note that Weibull++ can perform this analysis for you automatically. To accomplish this, the data would be entered in a single data sheet and competing failure modes chosen as the analysis method. This is shown in the next graphic.

	[[Image:~~300s~~.png\|thumb\|center\|400px]]		[[Image:Competing Failiure Mode Example 1 Data.png\|thumb\|center\|400px]]

Harry Guo at 17:20, 21 February 2012

2012-02-21T17:20:26Z

← Older revision		Revision as of 17:20, 21 February 2012
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	::<math>\begin{align}		::<math>\begin{align}
	& {{R}_{sys}}(100)= & {{R}_{V}}(100)\cdot {{R}_{W}}(100) \\		& {{R}_{sys}}(100)= {{R}_{V}}(100)\cdot {{R}_{W}}(100) \\
	& = & 0.694\cdot 0.995 \\		& = 0.694\cdot 0.995 \\
	& = & 0.69053		& = 0.69053
	\end{align}</math>		\end{align}</math>

Harry Guo at 17:19, 21 February 2012

2012-02-21T17:19:40Z

← Older revision		Revision as of 17:19, 21 February 2012
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	'''Competing Failures with Two Failure Modes Example'''		'''Competing Failures with Two Failure Modes Example'''

	From Meeker & Escobar [27], the following table gives failure times for an electric component that has two failure modes.		From Meeker & Escobar [[Appendix: Weibull References\|[27]]], the following table gives failure times for an electric component that has two failure modes.

	One failure mode is due to random voltage spikes which cause failure by overloading the system (denoted as a <math>V</math> in the table). The other failure mode is due to wear-out failures which usually happen only after the system has run for many cycles (this failure mode is denoted as a <math>W</math> in the table).		One failure mode is due to random voltage spikes which cause failure by overloading the system (denoted as a <math>V</math> in the table). The other failure mode is due to wear-out failures which usually happen only after the system has run for many cycles (this failure mode is denoted as a <math>W</math> in the table).

Harry Guo at 17:18, 21 February 2012

2012-02-21T17:18:55Z

← Older revision		Revision as of 17:18, 21 February 2012
Line 1:		Line 1:
	'''Competing Failures with Two Failure Modes Example'''		'''Competing Failures with Two Failure Modes Example'''

	From Meeker & Escobar [27], the following table gives failure times for an electric component that has two failure modes.		From Meeker & Escobar [27], the following table gives failure times for an electric component that has two failure modes.

Line 27:		Line 28:

	'''Solution'''		'''Solution'''

	We will begin by performing a Weibull analysis of the voltage spike ( <math>V</math> ) failure mode. In order to do this, we must consider all of the failures for the wear-out mode to be suspensions. The input data for the analysis are shown next:		We will begin by performing a Weibull analysis of the voltage spike ( <math>V</math> ) failure mode. In order to do this, we must consider all of the failures for the wear-out mode to be suspensions. The input data for the analysis are shown next:

Harry Guo: Created page with ''''Competing Failures with Two Failure Modes Example''' From Meeker & Escobar [27], the following table gives failure times for an electric component that has two failure modes. …'

2012-02-21T16:54:05Z

Created page with ''''Competing Failures with Two Failure Modes Example''' From Meeker & Escobar [27], the following table gives failure times for an electric component that has two failure modes. …'

New page

'''Competing Failures with Two Failure Modes Example'''
From Meeker & Escobar [27], the following table gives failure times for an electric component that has two failure modes.

One failure mode is due to random voltage spikes which cause failure by overloading the system (denoted as a <math>V</math> in the table). The other failure mode is due to wear-out failures which usually happen only after the system has run for many cycles (this failure mode is denoted as a <math>W</math> in the table).

Considering that these are competing failure modes, determine the overall reliability for the component at 100,000 cycles.

<center><math>\begin{matrix}
Number & Failure & Failure & Number & Failure & Failure \\
in State & Time* & Mode & in State & Time* & Mode \\
\text{1} & \text{2} & \text{V} & \text{1} & \text{147} & \text{W} \\
\text{1} & \text{10} & \text{V} & \text{1} & \text{173} & \text{V} \\
\text{1} & \text{13} & \text{V} & \text{1} & \text{181} & \text{W} \\
\text{2} & \text{23} & \text{V} & \text{1} & \text{212} & \text{W} \\
\text{1} & \text{28} & \text{V} & \text{1} & \text{245} & \text{W} \\
\text{1} & \text{30} & \text{V} & \text{1} & \text{247} & \text{V} \\
\text{1} & \text{65} & \text{V} & \text{1} & \text{261} & \text{V} \\
\text{1} & \text{80} & \text{V} & \text{1} & \text{266} & \text{W} \\
\text{1} & \text{88} & \text{V} & \text{1} & \text{275} & \text{W} \\
\text{1} & \text{106} & \text{V} & \text{1} & \text{293} & \text{W} \\
\text{1} & \text{143} & \text{V} & \text{8} & \text{300} & \text{suspended} \\
\end{matrix}</math></center>

*Failure times given are in thousands of cycles.

'''Solution'''
We will begin by performing a Weibull analysis of the voltage spike ( <math>V</math> ) failure mode. In order to do this, we must consider all of the failures for the wear-out mode to be suspensions. The input data for the analysis are shown next:

[[Image:SnV.png|thumb|center|400px| ]]

Analyzing this data set using the maximum likelihood method (recommended due to the number of suspensions in the data) and assuming a Weibull distribution, we obtain the parameters <math>{{\beta }_{V}}=0.6711</math> and <math>{{\eta }_{V}}=449.4</math> . The reliability for this failure mode at <math>t=100</math> is <math>{{R}_{V}}(100)=0.694</math> .

We follow an identical procedure for the wear-out failure mode, counting only the <math>W</math> entries as failures and assuming the <math>V</math> entries are suspensions. This is shown next.

[[Image:VoW.png|thumb|center|400px| ]]

Once again, analyzing with a Weibull distribution with maximum likelihood estimators, we obtain the parameters <math>{{\beta }_{W}}=4.337</math> and <math>{{\eta }_{W}}=340.4</math> . The reliability for this failure mode at <math>t=100</math> is <math>{{R}_{W}}(100)=0.995</math> .

We can now use Eqn. (sysrel) to determine the overall system reliability at 100,000 cycles:

::<math>\begin{align}
& {{R}_{sys}}(100)= & {{R}_{V}}(100)\cdot {{R}_{W}}(100) \\
& = & 0.694\cdot 0.995 \\
& = & 0.69053
\end{align}</math>

Or the reliability of the unit (or system) under both modes is <math>{{R}_{sys}}(100)=69.053%.</math>

Note that Weibull++ can perform this analysis for you automatically. To accomplish this, the data would be entered in a single data sheet and competing failure modes chosen as the analysis method. This is shown in the next graphic.

[[Image:300s.png|thumb|center|400px]]