[MUSIC] Hi, and welcome back. In today's module, we're going to continue talking about fluctuating stresses and the modified Goodman diagram. The learning outcome for today's module is to be able to calculate the factor of safety using the modified Goodman criteria. So let's get started. Last time we discussed how you can plot a fatigue diagram with the midrange stresses on the x-axis and the alternating stresses on the y-axis, and how the modified Goodman line goes from the ultimate strength to the endurance strength. The region below the modified Goodman line If you fall into that region, with your alternating and mid-range stresses, you'll have infinite life. But if your mid-range and alternating stresses are above the modified Goodman line, then you'll have finite life. So let's take a look at how these equations can apply to a real life example. So here we have a 1045 hot rolled steel rod, and it's undergoing cyclic axial loading. They've already calculated the nominal stresses at point A which is right here in the rod. And the rod has a hole in the center of it so these nominal stresses are accounting for the fact that there's less material in the center of the rod. They're at a minimum stress of 2 ksi and a maximum stress of 10 ksi. And then they also give you a stress concentration factor for the rod with the circle in it of 1.7. There's an ultimate strength of 82 ksi and they give you a full adjusted endurance strength of 20 ksi. So the example problem would like you to use the modified Goodman criteria to determine the factor of safety. So, to do that the first thing we need to do is look at the minimum stress which they have calculated as, the minimum is 2 ksi. And the maximum is 10 ksi. So the first thing we need to do with this information is find the alternating and mid-range stresses. So, the alternating stress is, and this is going to be the nominal because we're going to add in the stress concentration factor after, is 10 ksi the maximum minus the minimum, divided by 2, which is 4 ksi. And the midrange stress is the maximum plus the minimum divided by 2 which is 6 ksi. So if we were to draw the diagram showing the fluctuating stresses with this is our zero and here's our stress and here's our time. Our max stress is right here at 10 and our min stress is here right here at 2. So we have these lines going out. And then our midrange stress is right in the middle at 6. And you can see that the stress is going to fluctuate up by 4, and then down by 4, and continue like this. So that is what the part is experiencing. It's being pulled at 10 ksi and then released down to 2 ksi and then pulled again to 10 ksi. Because there's a hole here and this is a nominal stress, they haven't accounted for the stress concentration factor. We're going to assume that there's no plastic yielding at the hole. And therefore we're going to apply the stress concentration factor to both the alternating and the nominal stress. So our stress concentration factor we need to calculate first. That's going to be our kf, which is 1 + q (kt -1), and kf will be 1 plus 0.9. They gave us the q and the k t factor times 1.7 minus 1, which is 1.63. All right, so then what we can do is we can apply the kf to our midrange and our alternating stresses. So our alternating stress is going to be 1.63 times 4, which is 6.5 ksi. And our mid-range stress is going to be 1.63 times 6, which is 9.8 ksi. Okay, so if we went ahead and we looked at the graph, here's our modified Goodman line. And you can see it's going from our ultimate strength, which was 82 ksi right here to our endurance strength, which they gave us to be 20 ksi. And then if we go ahead and plot our alternating stress on the y-axis, it's right here at 6.5ksi. Our midrange stress is on the x-axis at 9.5ksi. So our point is here, it's clearly below the modified Goodman line. So that's indicating that we're going to have infinite life. But we can check that using these equations. So to check for the modified Goodman line, we say n is equal to 1 divided by our alternating stress divided by our endurance strength, our endurance limit plus our mid-range stress divided by our ultimate strength. And that is equal to 1 divided by 6.5 divided by 20 plus 9.8 divided by 82 and we get an n of 2.24 which is greater than 1, indicating we have infinite life. The other thing you need to check is that you're not going to get yield on the first cycle, and you do that using the Langer yield criteria. Which is that, let's see, n in this case is equal to the yield strength divided by your alternating stress plus your midrange stress, which in this case would be 45 divided by 6.5 plus 9.8. And that gives you 2.76 so you're not going to yield on the first cycle. Great. Okay, so the next thing we need to figure out is, it says if infinite life is not predicted, estimate the number of cycles to failure. In this case we did figure out that infinite life was predicted. But let's say it's a little more complicated. So what you can do, before the next module, is go ahead and calculate out this example. and instead of using a minimum stress of 2 ksi and a maximum stress of 10 ksi. You can use a minimum stress of negative 10 ksi and a maximum stress of 15 ksi. So go ahead and try this at home, and then we'll work through the next module, and we'll see how to estimate the number of cycles to failure if infinite life isn't predicted. I'll see you next module. [MUSIC]