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Fatigue question ?

  1. Nov 29, 2006 #1
    Im having a problem with the following question and my understanding of it...

    An aluminium alloy obeys the fatigue law S = 895N^-0.12, where S is the stress amplitude and N is the number of cycles. An airdraft componentfabricated from this alloy, undergoes the following stress amplitude history in a typical flight: 1000 cycles where S = 130; 450 cycles where S = 190; and 2 cycles where S = 290. Calculate the number of flights corresponding to the fatigue life of this component?

    My method was to use miners law and so the number of flights would be N1(S1/895)^1/0.12 + N2(S2/895)^1/0.12 + N3(S3/895)^1/0.12 = the number of flights but this answer comes out very large at something to the power 16!!

    Is this the correct method?
  2. jcsd
  3. Nov 29, 2006 #2


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    Start with a simpler problem: suppose the flight cycle just has 1000 stress cycles at S = 130 and nothing else.

    (1) What is the fatigue life at S = 130, from the fatigue law?

    (2) What fraction of that life does 1000 stress cycles use up?

    How many flight cycles does it take to use up all the fatigue life? (as a check, I got roughly 10000 flight cycles)

    Then repeat (1) and (2) for the other stress levels and use Miners law to find the amount of the fatigue life used in one flight cycle.
  4. Nov 29, 2006 #3
    Ok i took the simple case and my answer was approimately 9600 flight cycles. Do I just do the same for the next 2 and add them together?
  5. Nov 29, 2006 #4
    Please could you tell me what answer you get for the total number of flights so I can check?
  6. Nov 29, 2006 #5


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    No, you don't add the number of cycles together, Miner's law says you add the amounts of life used up in one flight cycle. Think about it - if the component sees more fatigue (from the other two stress conditions) the fatigue life must be less than the 9600 cycles from the first condition, not more.

    In the S = 130 case, the fraction of the life used up in 1 flight cycle was 1/9600 = 1.04e-4 (Hence the total life of 9600 flights).

    Work out what the corresponding fraction of the life is for the other parts and add up the three. If the sum came to 4.0e-4 (that's just an example - I haven't done the sums for the other parts!) that means (according to Miner) you used up 4.0e-4 of the life in one flight cycle, so the number of flight cycles would be 1/4.0e-4 = 2500.
  7. Nov 29, 2006 #6
    ok....im working it out now
    Last edited: Nov 29, 2006
  8. Nov 29, 2006 #7
    725 flights in total??? as my sum was 1.379x10^-3.................
    Last edited: Nov 29, 2006
  9. Nov 29, 2006 #8


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    Yep I get 725 as well.

    I saw your earlier question "what does the 2500 figure stand for". I'll try and explain the idea of how Miners works. Think of it like this:

    The material starts off with a fixed amount of "fatigue resistance". When it's all used up, it fails. In one flight of the aircraft, you use up some of it doing 1000 cycles at S = 130. If that was the only thing going on, the life would be 9.6e6 cycles, and 1000 of those would use up 1.041e-4 of the total "resistance".

    But there's also 450 cycles as S = 190. If THAT was the only thing going on, the life would be 406000 cycles and 450 of them would use up 1.1073e-3 of the total.

    For the S = 290 the life would be 11982 cycles and 2 of them would use up 1.6692e-4 of the total.

    What Miners law says (BIG assumption - there's no deep "logical" reason why it's true, but it works pretty well in practice) is, you can add up these separate amounts of "damage" and say that in one flight of the aircraft 1.379e-3 of the total "resistance" got used up. So it's all gone in 1/1.379e-3 = 725 flights.

    Hope that helps.
    Last edited: Nov 29, 2006
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