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    Reliability maths problems and solutions

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    Hello Sir,
    In order to save time and effort. You don't need to provided a detail solution like you did on the last two questions. Please use the direct formula and provide a brief solution and answer.
    Thank you for all your assistance.

    1. A component experiences failures at a constant rate (CFR) with an MTTF of 1100 hours. Find the reliability for a 200-hr mission. Please round your answer to 2 decimals.

    2. Wearout (failure) of an oil-drilling bit is normally distributed with mean 110 of drilling hours and a standard deviation of 15 drilling hours. Drilling occurs for 24 hours each day. How many hours can drilling continue before the operation is stopped in order to replace the drill bit? A 95 percent reliability is desired. Please round your answer to 2 decimal places.

    3. Fatigue wearout of a driveshaft has a lognormal distribution with tmed = 4500 hr. and
    s =0.3. What is the MTTF? Please round your answer to the nearest integer.

    4. Wearout (failure) of a Formula 1 car engine is normally distributed with a mean of 20 racing hours and a standard deviation of 2 racing hours. How many racing hours can an engine operate before it has to be replaced? A 97.5 percent reliability is desired. Please round your answer to 2 decimal places.

    5. A component experiences failures at a constant rate (CFR) with an MTTF of 1100 hours. Find the design life (in hours) for a 0.9 reliability. Please round you answer to 2 decimals.

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    A component experiences failures at a constant rate (CFR) with an MTTF of 1100 hours. Find the reliability for a 200-hr mission. Please round your answer to 2 decimals.
    We use the fact that for a constant failure rate the MTTF is given as the inverse of the failure rate constant or
    MTTF=1/λ
    So
    λ=1/1100

    For such a distribution the cumulative failure probability is F(t)=1-e^(-λt) and the reliability
    Probability function being 1 minus the cumulative failure probability is
    R(t)=e^(-λt)
    So when t=200 hr

    R(200)=e^(-200/1100)=0.83

    Wearout (failure) of an oil-drilling bit is normally distributed with mean 110 of drilling hours and a standard deviation of 15 drilling hours. Drilling occurs for 24 hours each day. How many hours can drilling continue before the operation is stopped in order to replace the drill bit? A 95 percent reliability is desired. ...

    Solution Summary

    A number of reliability maths/engineering questions based on constant failure ad normal and lognormal failure distributions

    $2.19