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| '''Competing Failures with Two Failure Modes Example'''
| | #REDIRECT [[Complex_Failure_Modes_Example]] |
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| From Meeker & Escobar [[Appendix: Weibull References|[27]]], the following table gives failure times for an electric component that has two failure modes.
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| 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).
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| Considering that these are competing failure modes, determine the overall reliability for the component at 100,000 cycles.
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| <center><math>\begin{matrix}
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| Number & Failure & Failure & Number & Failure & Failure \\
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| in State & Time* & Mode & in State & Time* & Mode \\
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| \text{1} & \text{2} & \text{V} & \text{1} & \text{147} & \text{W} \\
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| \text{1} & \text{10} & \text{V} & \text{1} & \text{173} & \text{V} \\
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| \text{1} & \text{13} & \text{V} & \text{1} & \text{181} & \text{W} \\
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| \text{2} & \text{23} & \text{V} & \text{1} & \text{212} & \text{W} \\
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| \text{1} & \text{28} & \text{V} & \text{1} & \text{245} & \text{W} \\
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| \text{1} & \text{30} & \text{V} & \text{1} & \text{247} & \text{V} \\
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| \text{1} & \text{65} & \text{V} & \text{1} & \text{261} & \text{V} \\
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| \text{1} & \text{80} & \text{V} & \text{1} & \text{266} & \text{W} \\
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| \text{1} & \text{88} & \text{V} & \text{1} & \text{275} & \text{W} \\
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| \text{1} & \text{106} & \text{V} & \text{1} & \text{293} & \text{W} \\
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| \text{1} & \text{143} & \text{V} & \text{8} & \text{300} & \text{suspended} \\
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| \end{matrix}</math></center>
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| *Failure times given are in thousands of cycles.
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| '''Solution'''
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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:
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| [[Image:SnV.png|thumb|center|400px| ]]
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| 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> .
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| 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.
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| [[Image:VoW.png|thumb|center|400px| ]]
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| 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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| We can now use Eqn. (sysrel) to determine the overall system reliability at 100,000 cycles:
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| ::<math>\begin{align}
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| & {{R}_{sys}}(100)= {{R}_{V}}(100)\cdot {{R}_{W}}(100) \\
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| & = 0.694\cdot 0.995 \\
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| & = 0.69053
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| \end{align}</math>
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| Or the reliability of the unit (or system) under both modes is <math>{{R}_{sys}}(100)=69.053%.</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.
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| [[Image:300s.png|thumb|center|400px]]
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