[MUSIC] Hi and welcome back. in today's module we're going to continue talking about miner's rule and randomly varying stresses. The learning outcome for today's module is to be able to utilize miner's rule to calculate damage and life. So, let's get started. Last lesson, we started this example with a varying fluctuating stress found at the critical location on a component. The material was steel, there's a fully adjusted endurance limit of 30 ksi and ultimate strength of 120 ksi. And what it's asking is, for the accumulative damage of the part or D for Miner's rule, so we're looking for D. All right, so to tackle this problem, the first step is to start to break down the cycles, so we can do that here. We're going to call the first set of cycles going from 40 to -20, A. And so what we'll do is, it's always a good idea to make a little graph. So we're going to call Stress A sigma for Cycles A. And then we're going to have our sigma max, our sigma min. I'm sorry, this is more of a chart, and our n. Okay, so for our stress at A, our maximum is 40 and our minimum is -20. And that repeats, you can see here's one cycle, and here's the second cycle. So it repeats two times. The next situation we see is Stress Cycles B. And that goes from 80 down to -60, and it repeats just once. So we get one full cycle of that in this 10 second increment. And then our next set of stress cycles are Stress Cycles C, you can see it goes from 60 to -40. So 60 to -40 and it repeats one, two times. And then finally we're going to have Stress Cycles D. And that goes from 30 to -5, and it repeats once, okay. So once you've tabulated all of your cycles and figured out the maximum and the minimum stresses and the number of times the cycles are repeated, the next step is to try to figure out N, right? So essentially, we've figured out this little n for each stress cycle. Now we need to figure out the big N. None of these cycles are fully reversed, they're all fluctuating cycles. So we're going to have to calculate N using Goodman, and that means we have to figure out the alternating stress and the mid-range stress. So that's our next step. So for stress cycles A, our alternating stress is going to be our maximum stress, which is 40 minus our minimum stress, which is -20 divided 2. So our alternating stress ends up being 30. Our mid-range stress is equal to our max stress of 40 plus our minimum stress of -20. So our mid range stress ends up being 10. And so at that point we can start to graph the Goodman Diagram. So if we have our mid-range cycles on the x axis and we'll have our alternating cycles, alternating stresses on the Y axis. We can go ahead and draw the Goodman line by plotting the ultimate strength of the material of 120 KSI and the endurance limit of 30 KSI. And then we can go ahead and plot the mid-range stress, which is 10. And the alternating stress, which is 30. And we can see our point here lies above the Goodman Line. So we know that N is less than one and we have finite life. We could have also just calculated the factor of safety using the Goodman equation. So in order to be able to calculate the number of cycles this part can survive. Remember, we need to figure out a sigma rev equivalent that's going to cause the same damage as these cycles, but that's fully reversible. And we did this actually last module. So we used the equation, sigma a over one minus sigma m divided by s ultimate. And that is equal to 30 over 1 minus 10 divided by 120, which gives me 32.7. And that's my fully reversible stress that causes the same amount of damage as the cycles we see here at A. And so you remember what we were doing there, is we were taking a line and going from the ultimate strength through the point of interest, which was our cycles to the y intercept to give us that fully reversible stress. Okay, so now that we have the fully reversible stress, what we need to do is we need to go ahead and repeat this for each of our stress cycles. So we did it for A, I going to let you guys repeat these equations for Stress Cycles B, C, and D. And if you do that, what you'll find is you'll get a chart that looks like this. So we have our maximum stresses, our minimum stresses, our alternating stresses, our mid-range stresses for all of the cycles. We have the number of times the cycles are applied in this 10 second increment. And we also have the sigma rev equivalent for each of these cycles, which is a fully reversible stress that causes the same amount of damage as the actual stresses being applied. So now at this point, what I can do is I can go ahead and start to calculate my life for each of these stresses, the total life. So you'll remember that N is equal to, in this case it's going to be my sigma rev equivalent, divided by a to the 1 over b. I need to calculate my a and I can do that because I've been given the f, the ultimate strength, and the endurance limit. And so if I go ahead and I calculate a, I end up with, 322.8 ksi. And b = negative one-third log of 0.82, which is f, times 120, which is my ultimate strength, divided by 30. It's the log of all of that, and it equals -0.17. So then I'm going to go ahead and start calculating the number of cycles for each of these. So N in this case for Stress Cycles A is going to be my fully reversible equivalent stress. So 32.7 divided by 322.8 to the -1 divided 0.17. And I end up with 602,900 cycles. And you guys can go ahead and repeat these steps for Stress Cycles B and C. Stress Cycles D, these are different. Because what you'll notice is, we determine the fully adjusted endurance strength was given to us in the problem as 30. And in here, the fully reversible stress is 19.5, which is well below the endurance strength. If I went ahead and I calculated the factor of safety using the Goodman equation, I would find that my factor of safety is greater than one. So for the stress cycles, at stress level D, my life there is going to be infinite. Okay, so go ahead and calculate that and what you'll find is that you can fill out this entire chart and you'll end up with A life of 600,000 for Stress Cycles A. For B you'll end up with a significantly lower life, right? because the stress is a lot higher, of around 4000 cycles. And for C you end up with a life in between A and B, which makes sense at about 30,000. And in D you have an infinite life. And so now we can plug in to miner's rule. So we can say D is equal to the sum of N1, which is two cycles, and the total life would be 602,900 cycles plus 1 cycle over the total of life of 4,368 cycles plus 2 over the total life of 30,911 cycles plus 1 over infinity. And which you end up with is a D, or total damage, equal to 0.00029. So every 10 seconds this part is getting a total damage of 0.00029. Okay, so now we've figured out the total accumulative damage that 10 seconds of these stress cycles will impart on the component. Next module we're going to figure out how to calculate the life based off of the total damage. So the life in time, so how many hours can this part withstand these stress cycles that repeat over and over again. And we're also going to look at some of the limitations of miner's rule and some of the areas where you need to be really careful when you get varying fluctuating stresses. I'll see you next module. [MUSIC]