 I procrastinate too sometimes, you know. I get away with it on the box. All right, any questions before we start something new? Okay, some more technical freehand sketching stuff for today. Yeah, we're gonna test our hand drawing skills here. Cause we're gonna look at a different type of loading. We've been looking at axial loading. We've been looking at shear. Now we're gonna look at torsion. Torsion is the type of thing that happens in a drive shaft of some kind where one end is twisting it one way. That's our torsional load. And the other end maybe is twisting the other way or maybe it's just fixed. It doesn't matter to us, what matters to us is that the entire structural member, whatever it is, is being twisted as if you're grabbing one end with each hand and twisting those in opposite direction. We also have a different way we might show that. If we use the right hand rule, put our fingers in the direction of the torsional twist, our thumb gives us the equivalent direction if we wanna do that. So we can do either of those designations for torsion, but not both. I prefer the circular twisty one just because it's more visual that that's torsion. But it also allows us to have torsion and axial loading, which we might do because we are going to be combining loads in a little bit. So remember that's right, it does that for us if we wanna do that. Some books even put a double arrowhead on it, but then that's just one more thing you gotta remember. I don't know, maybe you've gotta bring lots of real men in and you can remember all kinds of little things like that. I can't, so I like to keep it more straightforward. So I'll use this picture more often of the torsional loading that looks just like what it is, just like what's going on. All right, so as usual, we're gonna take a little look at what's going on inside the material. So we'll make an imaginary cut there to see what's going on on that face that we've now exposed, where we've got this torsion going on. You can imagine that, well, in fact, imagine you've got maybe a stack of Oreo cookies, which would be just this kind of cylindrical shape, and you twist the top cookie. It drags the cookie below it around a little bit, which drags the cookie below it around a little bit. And as you go down through this stack of Oreo cookies that you're twisting, that effect diminishes until it gets down to the bottom cookie that you're holding still and it doesn't move at all. Each one of those cookies moving over the cookie below it is a shear force. The layer right in front of this one is being twisted and it's trying to drag sideways, trying to drag in a circular way to this face that we've just exposed. So we'll look at a little elemental piece of that, some little area, DA, that's a distance, we'll say, row from the center. You can see what's gonna happen. We're gonna look at all of these little elemental areas. We're gonna integrate over the whole face and get the entire, entire deal of what's going on. So as we're trying to drag one layer over the other, that exerts a force on it, maybe something like that, where this lies right in that face. Maybe it'd help if we drew another one of these little elemental areas some other distance away and it's being dragged that way, little areas and each one of these lies in the face and is perpendicular to the radial direction located in. And then we've got these little things all over that face, all of these making these radial, these are these shear forces. If we look right down the face, rather than at a bleak angle, let's look right down the face, all these shear forces, if that's our little elemental area that's now on edge because we're looking right down to space, the shear force on that, well, heck, I can't even draw it. It lies right in the face and oh, that's not gonna help, I need all of it. So we're gonna have to edit that thing out of the video, that stinks. Hard to draw as most three-dimensional things are on a two-dimensional surface, but remember that all of these forces are lying right in this face that we've exposed. Maybe if we looked at it on end, that would help a little bit better. So we've got this little elemental face, some distance row away, and now there's this force here, maybe that helps a little bit. Does that help a little bit better? And then some place over here, we've got this other little elemental area and it's got a force perpendicular to it. Maybe that helps a little bit more, huh? When you take it next to your, maybe I'll have this drawing perfect. Anyway, we need to add all those up because all of those forces and the moment arms at which they act, all of those forces that we add up over that whole face, let's see, those are all little torques. We have a force acting at some moment arm row and we add all of those up over that entire face. Each little force, whatever size it is, whatever radius it acts, that's just a torque, all of those added up is the total load that we've got on this piece. Remember that's down here somewhere. That's the total load we've got acting on that piece. But each one of those forces is a shear force on that, little elemental force working on some elemental area. Take that, put it in there, and we get that the torque will equal radius where these things are times the shear stress around, across and over that face times the area dA and then we can integrate that over the entire area of that face and we'll be able to relate the load to the shear stress that the materials feel. Remember this is one of our imaginary cuts inside the material so we can see what's going on in the material. Now we can start to see what the shear stresses are that are trying to rip this material apart as this torsional load is exposed, or it is exposed to this torsional load. One of the characteristics of this type of deformation that it's undergoing, because it's going to cause this shaft to twist, there's gonna be an angular deformation. One of the characteristics of this deformation is that these cross-sections remain planar. They themselves do not distort. So if we expose this face, or imagine it to be exposed, and then twist this shaft, that face that we've exposed stays in a planar shape. In other words, if we were twisting those cookies, well the cookies remain cookies, they remain planar, but the different cookies, the different layers are all dragging over each other, causing this shear stress that we need to now look at. So what we're gonna do is now that we've sort of looked at things over the plane, remember we're gonna knead this in a little bit so we'll kind of set it aside. We're gonna come back to that. Now that we've looked at this face to see what's happened, we're going to take another look in a different direction. So here's the original shaft. There's the original shaft, the full thing. And remember we took a cut perpendicular to the axial direction to expose that. Now we're gonna look at a distance row and look down the shaft and see what's happening in that way. Again, an imaginary cut inside the material. I draw this, that little radius out to a point there. It's still the same radius. Just will be a better reference spot for us if I put it there. Now I'll imagine from that point down the shaft, I describe a reference line. That's just a line from where this radial point comes out that I imagine that describes it all the way down this interior surface that we've exposed. Because when I put a torsional load on this, give it a little bit of twist, what's gonna happen is that line is gonna displace as this piece twists. This original line will move to maybe here that line will move to something like that. We get this angular displacement as that piece displaces to there, that point displaces because of this torsional load. Well we've seen this type of thing before. This angle we can call gamma, the shear strain. This angle will be a slightly different one down at this end and I'll call that angle, I'll call that angle tweet. I think it's tweet, it doesn't matter because we don't actually measure that angle but it's how we're gonna be able to shake out of this the whole shear stress thing that we're looking at so we can start to see just what this deformation is as a material and just what shear stresses it's undergoing. So we can relate those two angles to each other. Maybe I'll call this point A and that point A prime. That's the one that's displaced. So that's an arc length that equals rho times phi also equals gamma times, gamma times, well whatever the whole length of is this, that's the arc length. So I'm gonna call that L something simple. That's the length of this shaft undergoing some torsion. And remember that's for small angles. A was the original point here and then I torqued this shaft and it made A shift up to A prime. It's the arc length that that little point displaced. So now I've got a direct relationship between this lengthwise strain, an interior displacement if you will. So what we can say is well obviously then, it's always nice when that happens, that this maximum angular displacement is going to come when rho itself is at a maximum. So we'll call that, let's see, we gotta give this radius this maximum possible radius some designation we'll call that C, max radius. Well that's obviously the radius of the shaft itself. So we can't go any bigger than that. There's no material, no material beyond that. So that's a rho, max, but we'll just call that. That'll be our symbol for maximum radius, which is the outer radius of the material itself. We can put that on here. So rho is any radius where we happen to be, that's less, that's still inside the material where when we get to C, we're at the edge of the material. All right, so if we put those two things together, we get then at the shear displacement, linearly dependent upon where we are in the material and it will be the greatest at the outside of the material. So if we tend to twist these shafts, they're more likely to fail on the outside than they are on the inside. There's just less going on on the inside, more going on on the outside. And we just need to keep that below the yield strength and we're okay. So let's relate that then to the yield strength. Remember that we have this definition, this material characteristic called the modulus of rigidity, that's what the G stands for. Rigidity, I guess, because there's a G in there. And that's the shear stress experience. Remember, sort of like the load over the response of the material to that load. In this case, it'd be the strain. It's very much like what we did with Young's modulus or the modulus elasticity, which was the load over the response only in an axial way. This is in an angular way. So we can use that, relate these two here. Remembering that G is a constant, so we can put that in there for gamma and we can also put it where we have shear max over gamma max, put that in there for that. And we get then that the shear stress is equal to rho C over the maximum shear stress. And we just need to keep that below the yield strength. There's another piece of the puzzle for us. We got to start tying some things together. Seems a little disjointed at times here, but we're getting close. The deal is with that, if we look at that now, it isn't immediately obvious, I bet, it wasn't to me, what that relationship actually looks like. Rho is simply the location. If we look on N, rho is just some distance in the radial direction. Now we're looking down the length of the shaft up to a maximum rho of what we call C. That's the outer radius of the material. C itself is a constant. What this means is that the shear stress is linear with radius, where at the outer edge we have T max and anywhere interior to that, we have a linear diminishing shear stress to the center. And at the center itself, there is no shear stress. That's pretty useful. It makes the analysis easy. We know where the failure point is. We know that it's linear in between. So we can even relate this then to tubular shafts, where we have an interior radius C1 and an exterior of C2. So this is a tube. We'll see the maximum shear stress at the outside surface. It drops linearly with that, so we can then predict what the shear stress will be at the inside of that tube. That's important in case we have to have two different materials bonded to each other. And then we have a bind material tube. We need to know what the shear stresses are, where those two materials are bound together and we need to figure out what stresses we'd be trying to pull those things apart. All right, so we're getting desperately close to something we can actually work with here. So let's go back to this first thing we came up with, where that shear stress is integrated over the entire surface that we exposed. And we know that that's going to be the total load. But we now have that this tau, this shear stress there is rho C times tau max. We put that in there. We get now that we have the integral of rho squared. We got a rho here and a rho here. C is a constant. Tau max is a constant. Once loaded, the shear stress, maximum shear stress is determined. That just has to do with the location. So we have integral rho squared dA, which is pure geometry. Little beast here is called the polar moment of inertia. And it should look very much like the original moments of inertia we define typically in the X and the Y direction back in statics. You're like me, you don't want to redo that integral every time we come up to some circular shaft. So it's done for us, by the way we call it J. So J for a solid shaft like that integrates. We only need to do it once. No reason anybody needs to do that integration again. One half pi times C to the fourth for a solid shaft. A circular solid shaft, by the way, but it's a square solid shaft to be different here though. I don't know what kind of shaft you want to use for torsion in a square like that, but let's add that just so it's clear. For a circular tube with two different radii like that, it integrates to I, two to the fourth, the outer one minus the inner one. So obviously it's a positive number, which is a very straightforward calculation to do. Once you do that then, you can figure out what this maximum stress is. Then the imposed load is the maximum shear stress times the outer radius times J. If we need to figure out what it is at some point in between, actually it's more useful in here. I guess if we do it this way. Actually solve for that shear stress since we often know what the load's going to be. So we just solve for that T C over J or any intermediate spot in between, we just go to that intermediate radius but the other things are a constant. These are the elastic torsion, more or less, wait, the formulas. Okay, took a little bit to get there, a little bit disjointed, seemed like we had to dance around a little bit to a couple different places, but now we've got the working formula we need. So let's imagine some tubular shaft here of some kind that we're loading, not looking at torsion plus axial loading. We've looked at axial loading the last couple of weeks. We'll bring it back inside diameter, 60 millimeter outside diameter, steel with a maximum shear stress of 120 megapascals. So we've got to stay below that with our load. So do this, now we can find the torsional load, the maximum load that we can impose. Anything above that will exceed this shear stress and would expect the shaft to rupture. We can also find the minimum shear stress since it's a tubular shaft that'll be on the inside surface. Then well, we know what the maximum load we can take is. It's from our original form of that elastic torsion formula. So let's put it this way because J over C is pure geometry. That has nothing to do with the loading itself. That's nothing but the geometry of the shaft itself still sitting in the box at the loading dock, unloaded. We know that limit on the maximum and J at C is obvious from the picture I hope and J is the polar moment of inertia. So calculate those real quick, double check, make sure that the units are working. Get used to some of these numbers, remember that spending some time getting used to real big and real small numbers here and there. Just make sure you get the right numbers in the right place and watch your units. C is the maximum radius. Careful with these, here's the outside diameter. You need the outside radius for C. Too much stuff on the page and you can't even see where you're going. But the units on this, you know what the units are on torque, on torque are something like, maybe we want a new kilonewton meters, mega newton, I don't know, you're gonna have to see where these numbers are taken out. And the English system, foot pounds or even inch pounds or pound inches tends to be said that way more often just because the English system tries to get as knuckleheaded as it possibly can at all times. Watch all these. What'd you get for J? Let's see if we agree on that. Oh, sorry, I did the same thing. Jake, you did ask me about that on the take-off question. I did, I didn't screw it up. But you just asked too late to change it, wasn't it? Okay, follow your voice. You did, yeah, so anyway. Doesn't matter, it doesn't matter, it was just as you're turning in, it's too late to fix it there, so you had to leave it as it was printed. It just changed the values to a little more absurd. What'd you get for J? Who, by hope, got what? It doesn't hurt to do these type of things separately so that if you make a mistake on any one of them, it's a lot easier to pick it out and fix it because if you don't get the same answer as this, 10 to the minus 6, it's really easy to fix that. That's J of Newton meter squared. And when we divide it by meters, we'll have Newton meters units for torque. And so divide by C, which is the radius. For example, what'd you get? Other part you were to find, the minimum shear stress. Maximum shear stress, well we're limiting that by the material or we have some design limit on it there. Where that number came from isn't necessarily part of this problem, it can be policy, can be government guidelines, could be something from the manufacturer. What's the minimum shear stress? It drops linearly through the material to the inside so you just have to reduce this by moving from the outer radius to the inner radius. And you got at the outer surface, remember on this two, and then we have the maximum shear stress out there with this load that we now calculated and it drops linearly for the center. So you need to figure out what that interior one is, but it just drops linearly towards the center. So what's about the big T and then? What, no, what would load would be zero? Just don't load it at all, leave it in the box. This is the minimum shear stress. The maximum shear stress under this load that we just calculated is 120. In fact, we use that to calculate the load, but that shear stress drops linearly through the material to the inside surface. So out here is the maximum shear stress that was set by this limit and in here is the minimum shear stress. And I wanted you to find that just to drive home that it drops linearly through the material. So at any place we could figure out what the shear stress is in your relation. That wasn't a two, it was just a solid circle, would that be zero? If this was a solid shaft, solid circular shaft, then the minimum shear stress would be at the center of the shaft, it would be zero. It doesn't drop that far because there's no material that far. So that's just, that's like seventh grade math or something, so we just need to drop it by the change in the radius. We go to two-thirds of the radius, so we go to two-thirds of that maximum. And this then is, hey, right? Megapascals, we have a ratio of the radius. Same type of problem, only we're gonna make a quick adjustment to it. Let's do this for a solid circular shaft now real quick. Normal Iowa shear stress is 12 KSI. Its radius is two-four inches. Of the shaft, a few seconds I hope to find to stay under this allowable shear stress. So very same kind of calculation that we just did, only we're using this solid circular polar moment of inertia, and then we're gonna do something a little bit different with it after that. There's nothing great that we didn't just do, only we have a solid shaft and English units instead of a circular shaft, a tubular shaft and SI units. As to double check, it's the English system. Basically, it's with the units you do whatever you feel like. Just name them after yourself if you want them. Whether you're not a dead white male German citizen. Get radius, sprinkle those all over the place. They're kind of like sesame seeds. They don't serve any purpose, but they're always there. If the torsion's in the opposite direction, is it right here, or is it not there? It would matter if we had multiple torsions, which we will look when we look at drive shafts, where we've got one gear trying to turn the thing one way, another trying to turn it the other way and then we do need to take account of direction. But in this case, we have just a single torsion and so it's kind of like a one dimensional problem. Don't need to worry about it. Okay, everybody got it? What'd you get for J? 39.5. What units? If you left it just like that, it's inches to the fourth and so you've got a maximum allowable load of 211 kip inches. All right, here's what I want you to look at now. Same type of problem. Only now, instead of a solid circular shaft, have a tubular shaft exactly the same amount of material, same weight, but two shafts weigh exactly the same. So outside radius on this one, I gotta limit it for you. Outside radius, three inches. So it's a slightly bigger shaft, but now it's tubular and it weighs the same as the solid shaft. Now find some maximum allowable load with that. That's way the same. So there's not gonna be any concern about the delivery weights or anything like that. Same cost other than the cost of manufacturing a tube over a solid shaft. It doesn't make any difference to the load capacity. Does the shear allowable load? Yep, same material. Same length. Same length, length didn't come into it. Kind of a red herring. Same length felt here because you need to find the inner radius that will give you the same amount of material as if we took this, melted it down, formed it into a three-inch tube rather than a two and a quarter inch solid shaft. So first thing you need to do is find this inside radius so that you have the same amount of material. Sure be nice if they would, I don't think it's obvious that they would be. They might be, but if I were you, I'd recalculate it. So let's see, since the same weight, same material's gotta have the same volume since they have the same length, then that doesn't matter. All they need to do is have the same cross-sectional area. So that'll help make the calculation a little bit simpler, the cross-sectional area. So you can then solve for the inner radius. Think before I call the outer C2, yeah. So that'll be C2. You need to find C1 so they have the same weight. Then you need to calculate the maximum torsional load. And you have two minutes to do it. Your boss is coming down all the way. Here's footsteps. Find C1, then double-check with somebody nearby if you're talking to anybody. See if they got the same C1. Then calculate J, maybe double-check, calculate the maximum load that this shaft can hold. Got it, got no points in there? D for five bucks and a little bottle, they can damn well better work. Guess we won't know that that's on tape, because we're not gonna ask for that in response to this class. Do you see the J values? I didn't say whether the J values were or weren't the same. It would be hopeful that they are. It'd be probably a wonderful coincidence if they weren't just that a coincidence. So you're gonna have to recalculate it. In fact, you should get a substantially larger J, don't you? What'd you get for the inside radius? Yeah, two inches. Is that marvelous how numbers always come out so nice and round? It's just like that in your entire career you watch. J, I think, was substantially larger this time. 102 inches to the fourth. Same amount of material, same cross-sectional area, substantially larger J, and J is directly proportional to the maximum torsion that this can hold. So if we put everything in, what'd you put in for C? C one or C two? It's the outer radius, so it's the three inches, and you should get 408. So the shaft is not that much bigger in outer radius, which always helps in terms of the load if we push out the radius. But with the same amount of material, we have double the load it can carry. It's much more mechanically efficient to go from a solid transmission shaft to a tubular one. In fact, the transmission shaft that runs under your car well doesn't happen as much anymore since we don't have rear-drive cars, but pickup trucks are rear-drive, so they have these big long shafts. Those are tubular shafts for this very reason. Same weight, same amount of material, little bit extra at cost in manufacturing, but you get double the load that you can carry. Double the factor of safety if that's what you need it. Also part of why bicycle frames are made out of tubes rather than just rods. There's a lot of twisting that goes on in a bicycle frame. By the way, if you ever wondered just how tough bicycle racers are, go search under bicycle racer world championships or something and see what happened to a guy over the weekend in the world championship track racing. You'll be convinced that bicycle racers are pretty tough. If you can't find that, let me know. I'll share it with you.