 I wanted to walk through an example problem for coil springs. So in the diagram I have shown here in this example, we have two springs. So an inner spring and an outer spring, which I'm sure you can tell by my array of dots on the screen, array of little circles. And we're going to assume they're both made out of steel. They're both the same length. One is just nested inside the other, so the smaller one is nested inside the larger. And we have the criteria given there that the outer diameter, major diameter is 50 millimeters, wire diameter of the outer spring is 9 millimeters, and there are five turns of the spring. For the inner spring, its major diameter is 30 millimeters, wire diameter is 5 millimeters, and it has 10 turns in this case. So we're going to assume that, again, they're the same length, both made out of steel. The end plates on them that the support would be set on and that rest on the table or whatever it is, are fully in contact, so there's no weird stresses going on. And we want to figure out how much deflection we're going to get by setting a weight on this pair of springs together. And that weight that we're going to apply up here is going to be three kilonewtons. And how much stress we could expect to find in the springs when we do that. And we can't necessarily assume that those are the same. The same. So first step might be to go ahead and find the spring constant for each of these. So we have an equation for the spring constant that looks something like this, d to the fourth g over 8 d to the third n. And if we take g for just a generic steel, we can look up in the book or we can look up online, but a good number that we might use would be 79 giga pass scales, which would be 79 times 10 to the ninth pass scales. And then we can calculate our spring constant. Now, because we have the diameters in here and the number of turns, we can see that the spring constant isn't going to be the same for each one. So we need to calculate the spring constant or spring rate for them individually. So starting with the outer, we have nine millimeters to the fourth times, let's see, units we need to. So I've got this one in millimeters. We can do giga pass scales in millimeters if we want to, just recognizing that basically one mega pass scale is a Newton per millimeter squared. So converting that to giga pass scales, we'd have 79,000 Newtons per millimeter squared. Let's see, divided by 8 times 50 millimeters to the third times five turns. And if we calculate this out, we get 103.66 Newtons per millimeter. Great. So that effectively means that to compress this spring by one millimeter, it would take 103 and two thirds Newtons to do that. And we can do the same thing to the inner spring. So these are basically treating the springs independently. Sorry, I'm writing my units a little sloppy here that squiggly line is millimeters. And I get a spring rate of 22.86 Newtons per millimeter. So our smaller inner spring, we can see has a lower spring rate. We would probably expect that, right? The larger size, larger wire diameter of the outer spring would lead us to expect that the outer larger spring is going to be stiffer. All right. So so far, we know four supplied and we know spring rate. And that kind of leads us to use the calculation, which we know f equals k delta, which is to say that given a spring rate of Newtons per millimeter times some deflection millimeters, we get force, which of course, we could rearrange this since we know k and we know f. We rearrange this to solve for delta as f over k. Now, what do we do with the fact that we have two different values of k? Well, we need to kind of think about what's happening, right? We have these two springs acting together, right? And if I apply a load uniformly to the two of them, I don't have a way to represent two springs, but if I apply a load uniformly to the two of them, they're both going to push back, right, with some force in opposition to the force we apply. And the amount at which they push back is going to be proportional to their spring rate. So we can assume that they will both contribute to that force pushing back, such that we get this force of 3000 Newtons divided by two forces pushing back due to that spring rate. So 103.66 plus 22.86 since they're both contributing to that. And that's in Newtons per millimeter. Newtons over Newtons cancel. One over millimeters becomes millimeters or one over one over millimeters becomes just millimeters. And we get a deflection of 23.71 millimeters. Great. So we know, you know, applying this load of three kilonewtons, we're going to compress this by 23.7 millimeters in theory. Now, this is an overall force applied to the whole thing. And something that might be a little confusing, but if you think about, let me kind of backtrack a little bit here. If I pull this, this, I just drew this top plate as a line, right? But let's pretend it's, you know, more of an actual plate. And I have on one side, my three kilonewton load. On the other side, I have, and you know, I'd normally apply my spring loads right in the middle, assuming uniformity. But let's say I have one fo and an fi both applied at the same, more or less the same point. And both of those forces are opposing this opposite this 3000 kilonewtons. So they're not necessarily equal, right? They don't have to be equal. And we actually wouldn't expect them to be equal, right? Because we know our deflection. And we can see that the spring rates are different for each spring. And therefore, we'd expect this load to be taken up proportional to their stiffness, right? So what that means is we can calculate fo and fi now using this same equation, but now saying ko, delta, and ki, delta. So where before, we were putting the k's together into a cumulative stiffness of the two springs together. Now we're separating them out and saying, well, how much force does each spring contribute? So if I start with ko, I have 103.66 newtons per millimeter times 23.71 millimeters equals 2458 newtons. And for the inner spring, I have 22.86 and I get 542. And as we would expect, these two add together to balance that 3000 kilonewtons, right? It's just a proportionality. How much of that 3000 newton applied load needs to be taken up by each spring. Great. So scroll down so we can keep moving. We've got these forces. And now the next thing we want to know is, well, what kind of stress can we expect from this? And we had an equation for stress. And that is tau equals 8 fd over pi d cubed and times some k value, different k than what we were just talking about. This is the correction factor because of the torsional stress. And we're using the static form for that that we talked about before, rather than the dynamic because we're assuming the load is just applied statically. And therefore, we can substitute what we know into that equation in order to solve. Now, first we need to grab the equation for ks. If I go up here, I can see ks equals 1 plus 0.5 over c and c equals big d over little d. So for the inner and the outer, where did my pen go? There we are. c outer equals 50 over 9, which equals 5.56. c inner equals 30 over 5, which equals 6. So that's our spring index. And then we have from that k outer equals 1 plus 0.5 over 5.56, which equals 1.09. And k, oops, this is s outer ks inner equals 1 plus 0.5 over 6, which equals roughly 1.08. Great. So I've got those values. Now plug everything in. tau outer equals 8 times the force taken up by that spring times the diameter pi times 9 cubed times 1.09 equals 467.94 megapascals. And do the same thing for the inner. And I get 8 times 542 times 30 over pi times 5 cubed times 1.08, 357.75 megapascals. So I get stresses in each of those springs. And great. That's what the problem looked for. If we were to keep going with this, what we would do is take those stresses and compare them against our stress limits. Typically, we might compare them against our yield strength or our ultimate strength, allowing for a clash limit. So trying to avoid, you know, spring solid situation and what that would, the failure that that would cause. But this is where we're taking this problem is basically how much stress we expect to find. So I'll stop there. Thanks.