 Okay, up to this point we've looked primarily at axial loading. By axial we mean aligned with the major dimension of the mechanical component we're looking at, whatever it might be, whether it's tension or compression. We've looked at both that hasn't been a big difference between the two for us, but the term axial does mean in line with the greatest of the dimensions. Typically we'll put the x-axis down there from now on just sort of for reference. I don't think we'll do much else than that. Now we're going to look real quickly. This is not a big chapter. There's not a whole lot going on in this one. We're going to look at torsion. This is the twisting of a component like a dry shaft of some kind where now there's a moment applied to each end as you put a big twist on. We can draw this in two ways to signify this. I typically will do something like this because it's just one less step to try to figure out what's going on. But we can also use the right hand rule as a way to designate the effect of torsion or the loading of torsion or torsional load. So if you put your fingers in the direction that that one has to be curdling, it puts your thumb out. So we could also do that. We wouldn't do both, but we could do one or the other. So just like with our force load, we need them to be equal and opposite to have static equilibrium here. We'll typically do something like this. The straight vector can work a little bit better if we just have a simple side view of the picture. But even then, you've got to put down your pen, you've got to get out your right hand and give it a twist to figure out what's going on. It's just kind of a disconnect. So I think it's easier to do just what we have there, but either way, that's just a visual reference that we need to understand what's going on, particularly with the directions because we're going to be very concerned with multiple power takeoff driveshafts, places where there's multiple places that torque is being applied at some kind. And so we've got to get those directions right, just like we had to get the direction right when we were talking about any other type of loading. So that's our general setup. We're going to bring a lot of little pieces together, so it's going to seem rather scattered until we get down to a certain point and we've got just what we need to do to keep going. But until then, we're going to have to kind of wrestle with things a little bit as we bring all these little pieces together. So let's imagine some kind of imaginary cut. Now right somewhere in the middle of the piece, so we'll expose that. And we've got some torsion applied to it like that. If we look at individual pieces, maybe some little DA right there, some distance row from the center, because of the way the load is being applied there, that's going to cause some kind of force, maybe we'll call it DF there on that piece. This is a force being applied from the part we took off. Remember these are internal forces we're looking at to see what happens to these elastic materials under these loads. So as we expose that face, take this piece away, this is the effect of that piece on the internal face that we exposed, again an imaginary exposure of those kind of faces. And we can look at all kinds of little pieces at different rows, different distances away from the center, and so on around the piece. And then of course, we want to integrate over the entire face. So we take all those little pieces, we have some moment arm, times some force being applied, and if we integrate over the entire face, you're familiar with that symbol, integration over an area, or sometimes it's a surface area, but this will be just the cross-sectional area. And if we add all of those little pieces up, then of course that's the total torsion or torque being applied to this piece if we go over the whole face. This little bit itself right here, that's the, we can replace that with the shear stress because it's the force acting over that little elemental area we've got there. So we can put that in as well, we've got row, and then tau dA. Okay, so there's our first little piece, just to kind of set things up as we're going, as we remember that we're looking at the inside surfaces here to see what the internal forces are. So we're going to hang on to that for a little bit. Actually I'm going to need a little more room, so I'm just going to move it down here, we've got the row, okay so we're going to hang on to that for a second, we'll come back with that, that was just sort of a preliminary setup. Now we're going to take this same piece, and instead of looking at an internal surface in cross-section, we're going to look at an internal surface down the length of the piece. So that's the original outline of the piece. We're going to go to some intermediate surface and see what's going on there. So there's some intermediate surface at some radius, row. Now here's another convention we're going to use from here on out as well. The outer radius, the maximum row, we're going to designate as C. So that could be C is defined as row, max, where row is the radius, the distance from the axial direction, from the axial direction, not along the axial direction. Alright, so we've got this exposed interior piece. Let's imagine a line scribed on that surface all the way down the axial direction, and that would actually let it go through the piece would be a planar surface in the axial direction from the axis down the center out to this imaginary surface that we've now exposed. Yeah, no, the length is not arbitrary. What's arbitrary is our radius from the axial direction, but the length is still whatever it was here. In fact, we will need that length, so we'll call it L. Alright, as I put a little twist to it, imagine this end fixed, which would mean the torsion is applied by the reaction to the mounting here could be causing that same torque down there. Now, as we twist this piece, then this plane is going to deform down to here somewhere, but the far end is fixed, so we're going to have a little movement, something like this, so that point will go down to there as this piece twists, just a big long thing, got a little turn to it, and we would see exactly that type of thing if we actually had this line inscribed on the side there. Well, if we look at it with some critical eye here, we should realize that this angle is actually the shear strength, and I'll label this angle here, label that, fooey. That means that arc length, the distance this little point moves along the outer center, this arc length here, we can then say is L gamma, and it's also rho fooey, where fooey is however much twist is put into the piece relative to the other end. Or the shear strain is rho over L fooey. In other words, the shear strain along the piece is directly proportional to the angle of twist in the piece, because rho and L are just some constants of geometry, length of the piece and whatever radius we happen to be. It's nothing to take this a little step farther, actually not that one, the one below it is more useful because it actually gives us the shear strain, so I'll go right from here. We can then say the maximum shear strain is going to be, well, at the maximum radius, which is C, so we know now where the maximum shear strain is at the outer surface, remember that's the, that's the radius C, oh there I got it there, so C equals rho max, if we put maximum rho in here, we'll get maximum shear straining, if this piece is going to fail, it's going to fail at the outer surface, rather than inside somewhere, it's going to start to fail due to over shear, over torquing, it'll start to develop cracks in the outer surface and may or may not be that's enough for it to fail there, but you can imagine you don't want that kind of thing to happen either way. Alright, so there's the first little piece we're really getting at what kind of loads are in here, I'm not really sure why, but I'll give it to you, I don't remember that this is much, I mean great shakes here, if we put these two together we get that the shear stress, sorry the shear strain is rho over C times the maximum, that just says it's linear, the shear strain varies linearly with radius, from zero to the center to the maximum, so we got that piece, I don't know that that's a great shake, if you do remember though we have the modulus of rigidity, remember how that's defined, both of these modulus are material characteristics, both of them define as load over response of material, so for the elastic modulus that was stress over strain, this was the same kind of thing only done for shear, so the load there is the shear stress divided by the shear strain, so that we can use in here where we have the shear strain, we can then put it in and we get that the shear stress at any point is again linear through the material, and the shear stress is a little bit more important to us in terms of the design concerns because that's the load that was being applied to the piece, what that means is if we look at the end of the, look down the shaft and there's some radial distance rho and we graph across the piece the shear stress it's experiencing, we see that it's linear with radius and so is zero at the center and grows linearly to a maximum stress at the outer surface, so for twisting the piece like this we see this kind of shear stress that varies linearly across the piece with the maximum shear stress at the outer surface, so that would be the place where if it was going to fall, fail, that's where we'd expect it to fail, nice in one way is that that's easily inspected, if the worst case was somewhere in interior it would be much harder to inspect that, and if we happen to have a tube rather than a solid shaft and we graph the shear stress across the piece we notice that well where we don't have any material we're certainly not going to have any shear stress, but it's still linear across the piece, but we have a minimum shear stress at the inside surface and it varies linearly between the maximum shear and the minimum shear across the cross section throughout the piece, good day for drawing skills, this is a good class for it, particularly tough drawing, good circles that are concentric, right John? Ok, alright let's take it a step farther remember we're trying to pull a bunch of little stuff together to make bigger ideas that we can use more readily, we just had this integral over the surface that gives us the total torque, but now we can bring this piece into it, because we've got the shear stress right there, so we put that piece in and we get rho times rho over C tau max, and we get rho times dA, that's just bringing the part we just set up down into there, and it's an integral we can pull out the constants which are C and the maximum stress, so we can pull out those two pieces, we get the integral of rho squared dA, we just bring it up here oops, got the constants pulled out, because tau max, well that depends upon the radius of the piece and whatever the loading is, so we're not looking at dynamic loads, we're looking at static loads, so that doesn't change, and then we have this rho squared dA in there notice this rho squared dA is entirely dependent upon the cross sectional geometry of the geometry, whatever the shape is and whatever the size is, this is what we call the polar moment of inertia, J, J for polar, J for, I don't know, jazzing things up a little bit, polar moment of inertia, for circular cross sections it's half pi C squared, which happens to be Ix plus Iy, those are the moments of inertia that we remember from before with this kind of coordinate, if x is there and y is there and z comes down the piece at us, then this is essentially the moment of inertia about the z-axis, these are the moments of inertia about the x and y-axis and those two would sum to give you the polar moment of inertia, we're not going to use that, we're not going to use this part, this part of course is going to be very useful, and for a tube, so what we call this, that's for a solid shaft for a tubular shaft, which is if you, next time you drive train on the bottom of your car, it actually won't happen much anymore, since rear wheel cars are very rare, but if you ever look at the side view of a delivery truck, you know, just something I'll take and produce and like, you'll see the shaft running there to the rear wheels, that's a tube and if it breaks open and you can grab it, run out there and get it from the guy before he puts it back in and you can check this, you can see it's a tube and it's polar moment of inertia, essentially the difference between the inner and the outer where C1 and C2 are the outer and inner radii, respectively, so now we can now say that the torque, actually be the maximum torque, if we have the maximum shear stress in there at the maximum radius, we can now calculate directly, and that maximum shear stress is a material property, just like the normal shear stress, yield stress and the like would come from the manufacturer, so with this type of number, so just to put a couple numbers to it, we'll do a quick calculation, imagine some kind of tube with some kind of torque applied to it, the equal and opposite reaction at the wall would be in the other direction around the back, we'll say that's 1.5 meters in length and the inside and outside diameters 40 and 60, ID means inner diameter, OD means outer diameter, and a typical maximum shear stress for steel, it's about 120 megapascals, so we'll put that down, actually the typical maximum is greater than that, so we'll say this is the design criteria, so we need to determine what the maximum torque then, that this material can withstand, got a little pieces there, I believe so, I'll give you a little gift, I think I have a little gift here down, or actually I won't bother since this is a tube, I have a little gift that's from a solid shaft, give you a little gift, just to show you how nice it is, alright, so calculate then the polar moment of inertia for this, then we've got the other pieces that we need to put in it, we can just calculate that, and you can put it together with your correct units, and so you give me that in kilonewton meters, just to put some numbers here, get used to these things, we need to calculate J for this, the polar moment of inertia for this particular tube and cross section, C is the maximum radius where the maximum shear stress is going to be experienced and for design purposes, we're just saying this one we want to keep at 120 or below, because there's the outside diameter, so the outside radius is half that, 30, Monday morning, I think we're not going to have to kick in for like 15, 20 minutes at least, maybe Charles would bring a coffee for everybody, you guys are welcome to set up that coffee pot in the lab if you keep it plain made for you, no, it's a torque, so it has units of moment arm time and force, so kilonewton meters, so double check your units, see what you get double check that, remember J for two is one half pi C1 to the fourth minus C2 to the fourth, so that's going to meters to the fourth, there's meters squared under here, you can get meters under there, you'll be left just meters on the top once you straighten out all the units, so the polar moments of inertia can be real big numbers, they can be real small numbers depending upon just what units you're in, wait a second, that's for solids, sorry, for solids then, is it C to the fourth or C squared? C to the fourth, did I write down squared? You wrote down squared. Oh okay, I'm sorry, yeah, catch that then, just how we used to write pi r squared, I think that's what I was doing, but yeah, the polar moment of inertia for either of these, before we have the same units, is everybody here, so I'm going to have to watch the video, so I hope they see it, so make sure you get your unit conversions right, this is in millimeters to the fourth, you need to have the unit conversion from meters to millimeters and that itself has to also be to the fourth, to get the units to work, still not what I got, not that that has proven anything in the past, yeah, that's supposed to be C to the fourth, sorry, both of them are C to the fourth, this one just happened, this is nothing more than the polar, the inner polar moment of inertia is just subtracted from the outer, if you have the right textbook on them, that's on Thunder's, okay, that's more like what I had, we're not agreeing, right, not there yet, what do you have? I can't see, that's not what I had either, that's better, make sure the unit conversions right, what do you have for polar moment of inertia, you have that, David, with units, um, hold on, Travis, oh, you got it? No, David, try someone else, okay, Samantha, you got it, the polar moment of inertia for this, what units, asking me, it's your number, oh, okay, yeah, that's what I got meters to form, so you got to take your mega-pascals, remember that's newtons per square meter, and then the 30 milliliters on the bottom, and that's all the pieces, and you should get 4.08 kilo Newton meters, Travis, trouble with that one first? Yeah, the easiest one, I was using diameter and symmetry. Oh, yeah, be real careful about these, it is very typical, diameters are what are given in dimension drawings and the like, but the polar moment of inertia, if you want, you're welcome to put D over 2 in there and then carry out the constants, get the slightly different equation. Alright, what, uh, that's the, uh, maximum, what's the minimum shear stress in this situation, and why are we concerned with the 2? What's the minimum shear stress expected in this piece? Minimum shear stress, those aren't good, it's for shear stress, so we've calculated something else. Remember this comes from the fact that the shear stress is linearly distributed through the material, so I'm looking for the shear stress on that inner surface. It could be that for whatever reasons, there might need to be some coating there or cladding of some kind, and so you're worried, maybe not about failure in the material, but failure in perhaps some of the easy, there's some bonding characteristics between the two materials. So if it varies linearly, it'll just be then C2 over C1 times that's just the tau over C, and the torque for the piece is the same, and that does not vary across the cross-section, that's the load, and J is also open for the cross-section. So that should come out to something like, I think, 80 megabascals. Not where you'd expect the material itself to fail, but there might be something, some kind of coating or cladding or inner material to which it's bonded that we need to concern ourselves with. Okay, all set? Alright, another problem, we're going to take another step with this one that is very interesting in its result. So we have some kind of shaft, 8 feet in length, solid shaft with a radius of 2.24 inches, allowable shear stress is 12 ksi, you find the maximum torque, maximum torsion, or the maximum moment, any of those words that can be applied to this piece. Not greatly different than what we just did, you just find something different than what you had before and some of the numbers have changed, and give it, if you would, in inches. So we can all get down to the same place and compare our numbers using a capinch. Alright, so same kind of thing, it's just now we're given a maximum shear stress, need to find J and see, oh here's the little gift I'll give you. For solid shafts, C over J reduces to 8 over pi D cubed. How useful is that? Any other professor giving us something that cool? 8. It's a symbol for 8. Alright, you're going to have C over J right there, so that can maybe speed things up a little bit for you, maybe not. We're going to do a problem where we want to find the diameter. For some reason it's more common that these things are in diameters when designed, but the formulas have the radius in them just to screw you up a little bit. This is what they do at any of the conferences. Do you have any ways to screw the students up? Alright, in kept inches, so watch your units. Watch your unit conversions. We've got to be pretty careful with these things. So we're looking for the maximum, we can just use the equation as put, using the allowable stress in this case. And since we're looking for the maximums, we are at the outer radius, so you see. So the equations are simple. I think the unit conversions, as much as anything, you need to watch carefully. Lots of squares, cubes, quartals, whatever they're called. I don't make up words. There's a word for everything. I use it, I use it the most correctly. They've got something? Check with Travis? No, not usually. Check with Chris? You guys are usually three finished first results. Well sure, it's right without any units on it. You can always make up your own units. Like mine, I don't make stuff up. Let's see what J is. Or did I do J over C? No, I did J. 39.5 inches to the fourth. So we have kips per square inch. We have inches there. So we're still okay with units. And then, well we just have C. That will then give us kip inches directly. We don't even have to do any conversions on this one. I don't think. I'm not agreeing on numbers. Let me make sure I make something down wrong. 12 PSI, 2.24, allowable, yeah. That's the only stuff in there. What do you have? There, that's what I have. We just need to use this one. We don't even need to do anything with it. We don't want to find the allowable torsion using the allowable stress, shear stress. And that will be of concern at the outer surface. So we have a maximum radius in there. Still agree or did they talk about it? What's the difference? Did you use what? Did you check then to make sure this is right? Well, is your A based off of the length? Huh? Well, is your A non there based off of the length? This? Yeah. No, this is just simply C over J. C is one half pi C to the fourth. No, that's J over C. So we get one over one half, yeah, that's C over J, one over one half pi C cubed. Yep. So we flip all that over. That's two and C is D over two. Oh, that's C. So you cube that and get an eight. We get, maybe that should be 16. Yeah, not eight. Because we already have that one half in there. Pi, D cubed, yeah, I think it's eight. And not eight. Would that be the, doesn't make a difference? Okay. Didn't realize that D was there. Well, I said it was a G. I didn't say it was a good one. 16. Now it's 211 and KIP inches. All right. Good engineering advice there. Never trust the doddering old fools in the office. All right, we got that fixed. Good thing you haven't gotten a tattoo of that yet. Okay, same thing. Shaft, design another shaft with the same weight and material as that. Only it's a two with an outer diameter of six inches. So you're designing the drive shaft or something. There's your first design over there. Redo it. Same weight. Only now with the two of, not that much greater diameter. This is 2.24 radius. So this is three inches in radius. Not that much bigger. And redo it. So same allowable stress. Find the maximum shear, maximum torsion allowable for that design. Same weight. Same length. All that's changing is going from a solid shaft to a two tubular shaft and you want the same weight. Same material. So obviously the cross section is going to change. So J is going to change. That 16 pi over d cubed is not going to apply. That was for a solid shaft. So only slightly larger in diameter. Depending on what the application is that might be more than sufficient. The weight is going to be the same. So that's not a concern. So if it's the same weight, what does that mean is also the same. Cross sectional area. So the same length and that doesn't apply. So the same weight, same material. The density's got to be the same. It all comes down to just the cross sectional area. So area, I don't know what you could call this. Example one and this example two. So all we need is the A1 equal to A2. You're given the outside radius that will allow you to find the inside radius. Pi over c squared where we've got c over there. It's got the equal pi over times c1 squared minus c2 squared. And c1 you were given. So you have to find c2. Then you have to find j from that. Pi over c squared. We've got something there. Let's see. I think it's black or below it. Well, let me make sure it's white. I have kip inches, so that just changes the decimal place, not the digits. So I don't have those digits. Yeah, but that's still important. That's still not what I got. I don't even know what the problem is. What? Maximum radius is the same on the previous problem. Yeah. You've got to go to this maximum radius. This is now c max or c1, I guess we called it. That's actually two times. See, it's down to diameter. Nothing yet? That's not what I got. Let me check. That's your inner radius. For the inner radius? Chris got something that was... What are you getting for the inner radius? Nothing yet? Yeah. Conveniently comes out to be 2. That's part of why the 2.24 was kind of an odd number to start with, because I started this problem from a 3-inch, 2-inch tube. Okay, that looks more like it. Let's see. What's what? C is the radius from there. This just says that the cross-sectional areas must be the same. That gives them the same weight. If they have the same density and same length, all we then care about is the cross-sectional area. So this is the one over there, the c. And you're given c1, or you're given od. You can find c2, and it should be 2 inches. You got that, David? 2 inches? Well, what we're looking for is the maximum torque. The maximum torque that can be applied. So you should get c is... The inner radius is 2. Do I have a j? J I think is 102 inches of the force on about right. And then you can find then the maximum torque. Now it's just the same as here, except the numbers have changed. Keep the same allowable stress. We already know if we have j in inches to the fourth. The maximum radius where the maximum stress occurs is 3 inches. You're on the wrong side. The other way, the thing radius can do that. What do you get? Now look at that number. That's interesting. Yeah. Okay, I think most of us are there then. Finally, we're real close to it. And you get a maximum applied torque that will take us right up to the allowable shear stress. Now maybe there's a factor of safety built in there already or not. We don't know we're not that deep into the problem. And the maximum you have is 408, and that's kip inches. They're right. Same weight shaft, only a little bit thicker, and we almost double the allowable torque using a tube instead of a solid shaft. I think that's interesting. That's why it's much more common to see tube. What we could also do then is say, well, let's use a much lighter tube. If this is sufficient, let's go down to this with a lot less material. We could reduce the radius. We could keep the radius the same, increase the inner radius, whatever it might take. We could lighten the shaft and really reduce that at the weight. If this is a car and you can do this kind of stuff, or a truck for a long haul, this kind of stuff adds up. Now, what I think is also interesting is if you take this even farther, if you keep increasing the radius just to see what happens, so what do I call that C1? So if you keep increasing C1, what happens to the maximum allowable load? It's almost linear. In fact, it's one of those graphs where you've got to hold a straight line up to it to see. It just barely is concave downwards, almost imperceptible on the screen unless you go up to really high outer radii, which mathematically you can do. I took this up just for fun to a 25-inch outer radius. It's the same weight, but the inside radius was something like 24.9 inches, meaning the tube now had a thickness of about a tenth of an inch for a drive shaft. What this doesn't take into account is failure is another way. Then you have a shaft that's essentially the thickness of a piece of paper, a cardboard, and there's a whole lot of other modes of failure that just shear stress. That's just going to buckle and warp and wrinkle and all kinds of things, but it may not fail in shear, which was the concern. So mathematically, you can take this way up, but you get such an absurd number that other things could become of concern. Be careful of what you can do with things mathematically that become realistically impossible, even if the particular mathematics and the paper say it's okay. Don't be careful with that. I remember when I was first working for General Electric, I was running codes that would calculate coolant rates, and I let the computer code run where it wasn't really optical anymore and I was essentially pumping ice, solid ice through the reactor core. I thought I was brilliant. Well, I mean, it was very cool. It was when I looked at the temperatures and realized that my coolant had frozen, but of course the computer code doesn't know that, so it just keeps happily running along. I thought it was on something. Split two and two together. All right, so let's take it a little bit of a step farther. I imagine we do have some kind of drive shaft. A couple places, we have maybe a couple gears on there. We've clasped for sketching. So in a couple places, we've got some gears on here. Do what you can with it. If you'd rather, we can just draw it this way and then put our torques applied to it so we have a couple directional torques being applied here. That way on that one. Same way on this one. Some drive shaft, maybe there's some generators or other power takeoffs put on it. You can do the drawing however you want, whichever one of those works for you. Either one's that easy to do as long as you get the picture. The thing is, on this one, this particular one, we have two different shafts put together. We have a solid shaft on the outside and then a tube on the inside. Yeah, so three are in one direction, one's in the opposite direction. It's a drive shaft sometimes. Put some numbers to it. Distance between those two is 0.9 meters. Distance between these is 0.7 and distance between these is 0.5. Now that's going to be more of concern to us now or in a bit when we start actually looking at the deformation like we did with the axial loading. We looked at what the loads were first. Then we looked at the response in the material to that in terms of the deformation and the deflection. We'll do the same thing with this one. We'll look at next is just how much twist gets put into these pieces. If that starts happening, then gears don't line up. All kinds of trouble starts with that. So we'll call this one gear A. So it's got a torque of six kilonewton meters. The next one is V and then D and they are 14, 26, and 6. Of course those are all kilonewton meters. And the last little bit, the tube itself is 120 millimeters OD, 90 millimeters ID outside and inside diameter. And last little piece, an allowable stress for the two solid sections, allowable stress of 65 megapascals. We want to find then the appropriate diameter for these two solid sections. Find D in the two solid sections and then also tell the maximum normal stress, sorry, normal shear for each. Then assume of course there's some kind of bearings or something that exert no load of their own, just to hold the thing. For the solid sections then we have a given limit on the shear stress. From that, if we know what the load is on those sections, we can determine a design recommend diameter and then we can also figure out then the maximum shear stress. Actually the maximum for the solid section is already given so what we really need then is the maximum for the tube section because you're going to design to the allowable shear stress anyway so that by definition that will be the maximum. So we need the amount of torsion held by each of the sections. I'll do a small piece up there. We do what we've done before, put an imaginary cut somewhere and determine what the internal torsion must be. That's what we're using the calculation. So for example this little end piece we know at one end there's six kilonewtons, six kilonewtons of meters. We know we must statically balance these things so internally between A and B we know there must be six kilonewtons of meters. So we can use that to find D use that to find D in that first section and the dimensions are already given for the center section so you can determine the maximum shear stress in that one as well. Like with an imaginary cut in the other sections and balancing the torsional loads on each. Figure out what the internal torsion is in the material that we designed. Remember to put these imaginary cuts wherever either the cross-sectional area changes or the load changes. One either changes, then the internal loads change. Put the applied loads on and then you know what you have to balance. If you're given what it is in the solid section, the two outer sections, then you also need to find out what it would be for the inner section. And that's just for the outer sectional, outer two sections. For the solid shaft only. Why do you need this? Because if you don't know what the D is, you can't find out given the load and they allow the stress you need to find out in the geometry that can withstand that. Essentially designing the cross-sectional area knowing the load on those two sections and they allow the stress. Don't forget to ask for D, not C. Somebody down the hall calls you up and asks you for those numbers and you tell them C when you're thinking you're telling them D. Sure. We know now the allowable load. The maximum shear stress cannot exceed that and the shear stress is caused by whatever the torsion is in the section and that's what we have here now. That's the internal moment times C over J. Yeah, C over J. No, if it's in terms of the load then it's J over C if you're solving for that step. By the way, this is known as the torsion formula. It's always cooler when it has a real name. This is the 6 kilonewtons internal between A and B. That's what you're trying to find. This is a good place to use that C over J but I tried to give you it but it screwed up. So it was what, P over pi is D cubed. I think that's upside down. I give you C over J or J over C. So that was okay. Yeah, because it'll have P on the bottom. So if you put that in then it's just a little easier to solve for D there. But remember that's for a solid chaff only. What's E? Checking humans. What? Yeah. So you can solve them for T but you're going to have to put that C over J. You're going to have the J over C here. So that's the 65 megapascals. D, you're looking for actually either one of these is work. You still have to solve for D either way. And T is the internal 6 kilonewtons in meters. You just switch more to T max. No, I solved for C. It's J over C. There's C over J. Yeah. So that clock's wrong but you guys are having so much fun. Alright, this should give you and I saw a couple of you had it. So just give you the first part you can double check the second parts. 77, 8 millimeters. 78 millimeter. Then you decide is there common bar stock that we'd use and go to 80 millimeters would that work? We don't have to have any custom made for the next kind of decisions. Turns out this section has the same load so D is going to be the same for this section and you can find a maximum of 86.2 in the tubular section there. Okay, so double check those. If you don't get those we'll reconfirm them on Wednesday. Friday.