# Solving Fatigue Law Problem: Calculate # Flights

• cabellos
In summary, the fatigue life of a component fabricated from an aluminium alloy is predicted to be 9600 flights, but this answer comes out to be too large at 16. Using Miners law, it is discovered that in one flight cycle 1.379e-3 of the total "fatigue resistance" is used up.
cabellos
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?

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.

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?

Please could you tell me what answer you get for the total number of flights so I can check?

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.

ok...im working it out now

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725 flights in total? as my sum was 1.379x10^-3....

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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.

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## 1. What is the purpose of solving fatigue law problems?

The purpose of solving fatigue law problems is to determine the number of flights that an aircraft can safely make before experiencing fatigue failure. This helps to ensure the safety and reliability of the aircraft during its operation.

## 2. How do you calculate the number of flights in a fatigue law problem?

To calculate the number of flights, you will need to use the Basquin's Law equation, which takes into account the stress amplitude, mean stress, and fatigue strength coefficient of the material. By plugging in the appropriate values, you can obtain the number of flights before fatigue failure occurs.

## 3. What factors can influence the number of flights in a fatigue law problem?

The number of flights in a fatigue law problem can be influenced by several factors, such as the material used for the aircraft, the design and construction of the aircraft, the type of loading the aircraft experiences during flight, and the maintenance and inspection practices of the aircraft.

## 4. Can the number of flights be accurately predicted in a fatigue law problem?

While the number of flights can be estimated using the Basquin's Law equation and other fatigue analysis methods, it is important to note that it is not an exact science. There are many variables and uncertainties that can affect the accuracy of the prediction. It is important to regularly monitor and inspect the aircraft to ensure its continued safety and reliability.

## 5. How can solving fatigue law problems benefit the aviation industry?

Solving fatigue law problems is crucial for maintaining safety and efficiency in the aviation industry. By accurately predicting the number of flights an aircraft can safely make, it helps to prevent potential accidents and costly repairs. It also allows for more informed decision making in terms of aircraft design, maintenance, and retirement, ultimately leading to safer and more reliable flights for passengers and crew.

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