 If you remember, we're looking at what happens when a long slender, the axial dimension being the greater one on these type of members, and we're loading it in some kind of torsion or twist. And we found out the other day that the maximum shear stress then will be, of course, a function of the torsional load. This is the applied load just like the axial force was before. Also a function of c over j. Remember what those two things are, what's c stand for? It's true now, and it's going to be true for the rest of the term that we use this c in this way, so it's very useful that you recognize it. It's the maximum, well, right now it's radius, but we'll be looking at non-circular cross sections in a little bit, and so it could be the maximum distance, for example, we could look at t-beams. We're going to find what's called the neutral axis, and then c is going to be the greatest distance from that, which could be either up to that edge or down to that edge, depending upon the geometry of the cross section and where this neutral axis falls, but it does mean the maximum distance from some neutral spot, and for circular things, of course, a neutral spot at the center makes some good sense, so it is the radius for these circular cross sections. We're not actually going to do it, but if you want to just read through one of the chapter sections we're skipping, it discusses, I believe, non-circular torsional, loads on non-circular members, also on what are called thin wall members, which is any very, very thin wall tube of some kind. So that will stand for the maximum distance for the next couple weeks for lots of the things we looked through, whether they're circular or not, and remember what the j is? j. I mean, if nothing else, go by the first letter and see what it stands for, because that always works. Phil, you remember? What does it say with j? My favorite student comes to the rescue, that's the polar moment of inertia. It's just like our regular moment of inertia that we had, we looked at a lot in static slash term, only for circular shapes, circular cross sections in this case. The intermediate equivalent of that, of course, this is just the maximum load, but that's where most of the concern is. We can also look at some intermediate radius, some radius where we're inside the piece, but that's not quite the concern, certainly not on solid pieces, because the maximum stress is at the outer surface. But it could be if we have a tubular cross section, you could be worried about the stress at the inside where maybe there's cladding or adhesive or something similar to that. Alright, so we're going to add to this a little bit now, as we do just what we did in our axial loading, where first we looked at the loading and its direct effects in terms of the stress induced in the material. Then we looked at the deformation under those loads, which is really the step we took beyond what we could have done in statics. We could have done the internal stresses in statics and not gotten out of the general purview of that subject, but adding the actual deformation of the pieces. And we're going to do that now, look at the deformation of some piece when loaded by some torsional load, as would happen in a driveshaft. So if we put a couple reference lines down here, so imagine down the parallel to the axis but inscribed on the outside, maybe not a real inscription, but just for a reference point of view, and then one from there down to the radius and see what happens to that under this torsional load. As applied like that, we expect that this front face here would turn in some measure, greatly exaggerated there, and that the line inscribed on the outside would also have some angle to it. So again, the deformation caused by the loading greatly exaggerated here, just so we can see it, but it would look something like that. So let's put a little bit to this. Let's imagine this is some length dx. We have an angle here that we have already investigated. That happens to be, of course, the shear strain. Remember, that's the deviation from an original angle of 90 degrees. And if you want to look at it from some side that maybe if I draw it on the edge, this reference line that was like that, and now it forms to that because of the torsional load, this is indeed a strain because we can look at this angle here as the original 90 degree angle that's now been deformed. And the deformation of that 90 degree angle is our strain. So we've seen that before. I don't remember if we've seen it in exactly this way. And then this angle on the edge here, we'll call that some amount of deformation dd and then this arc length subtended right there, if that's the right word, is the shear strain times that distance over which it acts. That's the arc length from the axial direction and from the end-on direction. Then, of course, we've got the equivalent arc length, c, d, free there. The shear strain is some geometric constant of the geometry of the cross-section over dv dx and then we're almost ready to integrate that. Remember that we've got this material constant we call the modulus of rigidity which is the normal stress over the shear stress. So we can put that in and we get then combining those two, we get that the maximum shear stress is equal to two constants, the product of two constants independent of the load times however much angular distortion we get per unit length of material. But that maximum shear stress, we already know it would be this. So now we can combine the angular deformation with the torsional formula, the loading itself, and now we can get angular deformation as a function of the load and other geometric properties. So I think that baby there looks like it's ready to integrate. So move the dx over, integrate, everything's a constant, those two things are constant, everything else is also a constant. So we're dx, of course we're integrating over the whole length of the thing on this piece is. And we get then that the total angular deformation is, let's see, the C's cancel on either side, we get T, this integrates just to L and G comes down at the bottom. So I think that the total angular deformation is a function of, of course, the load, the greater the twist, the greater the angle, your angular deformation is going to be. Also the length of the piece, that makes pretty good sense if you've got a real short piece and you're loading it, it's not going to twist as much. When you've got a much longer piece, it's just going to twist more. There's more flexibility in those. You can, you can go home and test that with just a kind of a slender piece of wood. You can just grab it on the ends and twist it, grab it at different places. G, of course, being a material constant, that makes sense too. There are materials, different materials are going to twist different amounts, steels much different than rubber. And then, of course, J, the cross-sectional polar moment of inertia. So as simple as that, I believe, just so you know where we are to anchor it a little bit, that's equation 515 in the book. I'm not sure how many of you have both additions that we're using here, well, including the international addition, but it's around there somewhere. Just so you know where we are. So that's pretty easy to use. We're going to run a little test, test problem with it. So imagine some long slender piece, maybe a drive shaft, maybe an axle or a spindle of some kind, 54 inches in length, 1.5 inches in diameter, with a load of 250 foot-pounds, and a modulus of rigidity of 11.5 by 10 by 3. So we've got, what's that? That's about 4.5 feet long and about an inch and a half in diameter. So that's probably the kind of thing you might have in your trunk that you go take into the bar fight with you. They're going to grab the pool cues, right, Travis? You've got this. So you're more well prepared. So we can figure out that name. The amount of twist in the end now. We'll check the units as we go along to see what happens here. 250 foot-pounds, now everything else is in inches, so probably that needs to be converted to feet, sorry, inches, times the length of the piece makes sense. The longer the piece, the greater the angular deformation. The 11.5 times 10 to the third, and this is kilo-pounds per square inch. And then J, I think I have J. I don't need you to calculate it. You can confirm 0.497 inches to the fourth. So we've got pounds, and pounds are going to cancel. Once you take care of the individual, let's get length squared. That's not on the bottom. That cancels that length to the fourth. So this is unitless, meaning then the measure of these angles are, of course, ratings. So you can look at that real quick, see what we've got, the kind of thing we're talking about, how much angle. Yeah, if you don't convert this feet to inches, then it's not going to combine with the other inches to cancel out. And you go to the kips to pounds, or the pounds to kips one way or the other. It might be easy to forget. It might be easy to forget, is that what you said? No, not this group. Some other, maybe an MIT or something, those dudes forget this stuff, but not you guys. No, you guys know to watch your units. It's a hard part. Sometimes they say, yeah, it's the hardest part. I mean, this isn't difficult. You're right, it's when you get the numbers in, and we're working with some very, very big numbers and very, very small numbers, that's as hard as it is to do any of this stuff we're doing here. And don't forget, we can also find the maximum shear stress in the problem by calculating that. Remember, I gave you that little, that little bonus piece the other day once I corrected it from the board. So that's pretty easy to figure out too, and that would have units of PSI, so that's not going to take too much to work with that. Torsion has the same units as moment or torque, foot pounds, something like that. So that's going to have PSI or KSI as the units if that's more appropriate. So you run the numbers and they got them yet? Joey, look at something on your screen. No, I want to share it with us. Yeah, it's 28.3. Radiance? Yeah. Let's see, yeah. There's about six radians in one twist, so that's what, that's about four full twists around. I don't think you can even do that with one of those stringed cheese things. Get that to go full four twists, so I want to check your units here. Make sure I've got everything was right, yeah? All things to match. Somebody else got something a little different? Sounds like you may have missed the unit factor of 1,000. Phil? 0.0283. Okay, that's what I have. So you're the 283, Travis. It just sounds like you missed a little bit. Now that turns out to be about 1.6 degrees, which is more than enough for there to be the possibility of trouble with gears meshing at either end of these if that's the purpose of this as some kind of drivetrain. And the towel max, you get something like 4.5 KSI. All right, so not too difficult to put those things together, but we can certainly come up with problems that get a little bit more difficult since the equations themselves aren't terribly simple. As we said, and as Travis proved so kindly for us, we can get to be down to just a problem of watching the units. In fact, this can be very good if you think about it as a method, oh no, I don't have it on the board. This can be very useful as a tool to actually measure what G is in a material as the actual testing method for determining the modulus of rigidity. Apply load on a piece of known geometry, measure its deflection, and calculate G from that, which is not unlike what we did when we found E. Apply a load, measure the deformation, and then get E from that, which is a little bit different in the actual application. All right, another quick one for you. Now we have some tubular piece with a load. We'll actually determine the maximum torque allowable for an angle of twist of no more than two degrees. Maximum deformation to two degrees by the torsional load that can be tolerated to stay within those limits. We have to work through the same thing using now some SI numbers. What J is for a tubular cross-section, and you know which radius is C1 and C2 because that couldn't be a negative number. That's a tube, so you don't forget. Heartbreak, but that's with rolls. Is that what they were? Do you need to put in the modules of rigidity? Oh, I would, yeah. This is 77 giga-pascals. Remember what the SI prefix giga means? End of the ninth. Okay, again, watch your units. We have millimeters, and they're going to be to the fourth power. So that gets to be a pretty big small number. I appreciate it. A couple students are volunteering to fall for the small trap in this problem, but not everybody under the bus protects the others. Chris, did you find a small trap? A potential one, but you lightly stepped over it. You saw what looked ahead and it would be a quicksand. He finds so far. I keep going. Got a number yet? We're looking for the applied torque. Check your units. These should be in newton meters or kilonewton meters. Whatever is appropriate. I happen to have kilonewton meters. It doesn't really matter. That's a six-one, okay? Oh, something went wrong. Got something yet, David? Yeah. That looks good. So Chris stepped over the trap, but didn't notice there was a pit full of vipers on the other side. Is that right or did you throw that in? Let me go away. Good. Where did you fix that? Okay. Take it out of there. Get in the habit of fixing your own mistakes so you don't come back to them later and not notice that they're on. Phil? What's next, though? You don't have the same answer as he does. Most of you are catching an hour. Just two degrees cannot go in the equation like that. It must be radians. I think most of you caught it by now. Some started without catching it, but most of you caught it by now. 34.9 radians. So make sure you get that. Degrees is a unit that doesn't just disappear like radians conveniently does. They don't stay in there if you have it in there. Reconsider. They'll lose not on a keyboard. Keypad. I noticed everybody's getting that now. Something like 1.83 kilonewton meters. Units should work out. Be careful on these. What's the actual value for that? I don't know if I have it separate. Yeah, do 1.02 times 10 is minus 6. And that's meters to the fourth. This is 60. Remember, these are diameters. These are radii. I don't know why they do that. Why they can't put diameters in here and then pull whatever constants left over out to the 1.5. I'm not really sure, but it's just one of those things we like to do as students. Alright, everybody catch all the little pieces in and get something like that. Not quite yet, David. Can you do something like rock and capillary and gang? You're not getting that? No. Yeah, these are diameters you were given. These are all radii in the equation. Be very convenient if this was based on a diameter and the 1.5 to the fourth was just pulled out and added to that. But that's too nice for us to do. We're not going to do that kind of thing for students. Alright, so let's step it up a little bit. Imagine we have a long gear shaft with a couple points at which power is taken off or added by maybe some gears at certain points. And we've got a couple things going on. So we've got an applied torque and that one an opposite direction on the other two. So maybe the center one's a drive, gear, and then the other two are power takeoffs. So, and then of course the ends are going to be some kind of just a stabilizer, of course. It's all one solid shaft, 30 millimeters in diameter. Let's see what other little pieces do we have. Okay, let's label points A, B, C, D, and E. Because there's different lengths at each one, between each one. So the loads are the applied is 450 newton meters. Take off at B, 275, and at C, 175. And the distances between these, B to C is 500 millimeters and C to D is 400 millimeters. Now the trouble here is, well not the trouble, but the obvious extra complication we didn't have before is between B and C there's going to be a bit of angular deflection, deformation, and between C and D there is as well. And I want to find the maximum deflection and the total deflection. It may be different, most likely they are. Because B to C is going to twist one way, C to D is going to twist the other. So one or the other of those could be the maximum deflection, but the two together, since they'll be in opposite directions would give a different angular deflection. So let's find the maximum shear stress and the total angular deflection. You also need two other little pieces. G is 80 megapascals and the extra dimensions on either end to the bearings 120 millimeters on each end. And as you can imagine the total angular deformation will be a sum of all the other deformations along the length of the drive shaft. So we did a problem sort of like this on Monday. So for any loaded section you need to look at what the maximum shear stress is in that section. I'll give you J so you don't have to do that. Actually, you can skip it and remember to use that little piece we have there. And you don't even have to sleep in a radius. You can use just the diameters given. And of course our usual angular deformation, you will need J for that. So I'll go ahead and give it to you so you don't have to fuss with that. There's the solid polar moment of inertia in terms of diameter. So make it a little easier for you. So there it is in millimeters so you don't have to do that. What's the internal torsion in the shaft between A and B? Between A and B you have to find the internal torsion for each one of these sections so that you can investigate which one is the maximum shear stress and how much angular deformation there is in each section so that you can add up the total deflection end to end. So what's the torsional load A in the material A to B? The internal torsional load. What, Phil? You had an answer earlier. You're back and off. Yeah, there is no load. Remember for any of the pieces we've looked at it's always based on equal and opposite of either end. For the section A and B there's no torsion here it's just held by a stabilizing bearing. We assume it's a high quality one from A's hardware again. So in A and B from in the section A to B there is no torsional load meaning there won't be any shear stress and there won't be any angular deflection in that section. So it's only B to C and C to D that are loaded because D to E is also not loaded. There's no torsion in the sections D to E for the same reason. Makes sense, Joe? What's the I? The I just has to do with the fact there's going to be a couple sections you're going to have to figure out the angular deflection for each of the sections because each of the sections has their own internal load length, B to C is 500, C to E is 400 millimeters. So that's just an index, a summation index. Don't forget as you calculate these that the angular deformation in the two sections are going to be in opposite directions. So you're going to assign one as negative or just pay attention to what the difference is but the direction of the angular deflection is also important in the end. So you might do well to assign one direction as negative or as positive. That'd be even more, you may have seen wrought iron stair railings where they keep the metal and they put that real nice twist into it. Your piece would be wavy on that. It'd be more like the rubber bands on the wind up airplanes. But at least you recognize something wasn't quite right. I didn't put in a number like that. So check where the places are that things can go wrong. This is millimeters. This is meters. These are millimeters. This is megapascals. These are just straight newtons. So there's a place that things can go wrong. So we start checking some of these. We get a torsion in the section B to C. Well, if we take off everything on the other side, we'll end up at the 275. So that's all the internal load is in that one. And for C to D, we can do the same thing out the opposite end. So we've got the torsional load of means we can double check then what the maximum shear stress is. That's D cubed because that was simplification of C over J and both of them had radius terms in. Yeah, that's from beyond. Double check. Thank you for watching. You're looking for C to D. We have 51.9 megapascals. Now we don't know if that's the maximum because we also need to check the other section of C to D. The shear stress does not depend upon the length of the piece. It only depends on the internal loading and the geometry of the cross section. So if you get this one, again, 16C pi D cubed. Travis, do you have this one? The maximum shear stress? 33.1. Yeah, something maybe a little bit around. Precision of these numbers isn't terribly important because you don't design the very limit of the material's ability to withstand these loads to put in effect the safety of two, three. Remember that the twist is in different directions for these two sections. To be that way, either one of these torsions you can assign as negative and then when you add them together, they'll cancel. But make sure you understand which is the total deflection on that, Travis. Same as Chris. Now you bought our check. For a little while, Travis, you had what, 600,000 radians? Something like that? Yeah, I don't know if we can cover that with a factor of safety. For each of these. I did give you G. So you have to watch out for your units. Tom, you got something? Which proper units? This should be unit. So you have this for DC. Is it 600,000 radians? Is it less than that? Yeah. Chris, what do you have for this first section, B to C? 1, 6. 2.16 to negative 3? Yeah. And to the negative 2. 0.0216. Which do you have, Travis? That or one extra distance? Five. So you're less than 600,000 radians. Alright, well, let's see. The load. 275 newton meters. The length. Half a meter. G, 800 mega-pascals. Everything's in newtons and meters. That's 10 to the sixth square meter. That's a Pascal. So that's a mega-pascal. And J. J is in millimeters. So to get it into meters to one meter, over 1,000 millimeters, but it's millimeters to the fourth. So you have to put that whole conversion factor to the fourth. It's going to be times 10 to the minus 12th for 79 times 10 to the minus 9th. Does that sound right? David, did you get that too? Okay, so now the units are all okay. Newtons, newtons, meters, meters to the fourth. And then what's that come out to be? 21.6 radians. So 21.6 radians? Does it really? No. That's it. Let's make sure I gave you all the right numbers. Newtons, newtons, meters, millimeters. Okay, now I can't read that. Oh, yeah. Sorry. Now look at that. Good. That's the easiest one to fix. Sorry about that. Let's see. So that's kind of the ninth right here. Which is what we need, right? We need three on the bottom. Okay. Yeah, see what happens when you don't fix the little mistakes and you leave them and then find them a year later? Still is. You know, when you take this a year from now, it's the same mistake. It'll be there. So that comes out to be a little better, I hope. That comes out to be 0.0216 radians. 1, 2, 3. Yeah. Okay. And in the opposite direction? 0.0110. So maybe if we call that the positive direction. So that would make this first one negative. And then they add together to give a deflection. They cancel each other a little bit. We still get some in the negative direction. So I know that end to end there's going to be a little bit of a twist in the direction of the middle torque. 0.06 radians or about minus 6 tenths of a degree. We agree on most of those numbers. Joey, you've got those now too. And all we've got is little pieces fixed. Okay. Not difficult equations to use. The setups aren't all that difficult. We'll watch in the unit, especially if I mislead you with the wrong units from the start. What you don't know is I'm obligated to do that because the equations are so simple I have to screw you up in other ways. We talk about that at the professor meetings all the time. Okay. The next little part we need is the, well the very same type of thing we did is we were looking at the axial loadings. After we looked at the loads themselves and the stresses they induce, then we looked at the deformations they cause. Then we applied what we've learned there to statically indeterminate loaded. Meaning just what? What's a statically indeterminate problem? Using just the equations of statics, it's unsolvable because there are too many unknowns. For example, imagine we've got some kind of axial member between two rigid supports. I'm going to draw that. It might be easier just to put it on. Especially since what we have on this form is that it's not a constant cross-section. From one end to midway down. So we have a solid piece per half of it and then a tubular piece per half of it. Right down the middle is a horizontal load of 90 foot-pounds. So whichever one of the drawings you prefer labeled as A and B or the two ends sees the middle. One of the pieces cannot give you a year. I don't have the radii, but I have the polar marmalance of an inertia formula. A is C. It is 57.6 times 10 to the minus 3 inches and 42.6 times 10 to the minus 3. Now the way we get out of the static indeterminacy of this is the fact that because it's mounted at two ends, two rigid ends, then there's no overall total deflection. Using that, we'll be able then to come up with another equation. Well, in fact essentially that is the other equation and we can then solve the problem. Find the reactions at the walls. Oh, yeah, sorry. Our statics only tells us that the reactions are nothing more than equal to the total load, but we don't know how much of the load is at each one of the supports. So we need then to add to it the deflection part that the deflection between A and C must be equal and opposite to that between C and D. So the total deflection end-to-end is zero because of the two rigid supports. So that then allows us to determine just how much is at each one of the ends. Anything missing this time? I haven't told you what the material is. Remember what happened last time with the axial loading? The material constant, this G, in that case it was E, cancelled out, but the solution is the same no matter what this material is. Which I think is always very interesting. And the lengths happen to be equal, which is convenient. Well, it just makes it a little easier. When you set this up, those two will cancel. G cancels, we get just the inverse ratio of the two whole moments of inertia. Wouldn't be that simple except the lengths are the same. Could there have been two that they were different materials? That would certainly have been a possibility. But given that you can now solve that. Now we're down to just these two equations as enough to solve it. So it's a matter of fairly simple mathematics. Now it could be the reaction, so it could be the torsion at A and the torsion at B, the applied load in the middle. And they should be in the same direction, yet 51-7. Now it's just a simple ratio of the two and 38-3. Yeah, if this material is different than that material, then the Gs wouldn't cancel. But the reality is that you would still just retrieve something as having the same density throughout the electricity. Oh, density doesn't come into it. Probably the rigidity throughout the length of any given material. I mean with the safety factor, it's enough to take more. Well no, it depends on what the two materials are. If this was brass and this was rubber or something, that'd be very, very different. But it's not too big a deal. It would just be in here. It would be U-D, G-A-C here, and G-C-D. That's C-B. C-B. All right. C-B there. Okay, again we have to realize what the deformation does for us in giving us the other equation. All right, questions on this one? I'm just going to set up. We don't really have a lot of time, but I'll set up another one for us that just steps things up a little bit in terms of the static indeterminacy of the problems. Imagine we have a drive shaft that looks something like this. Just to add to it. That is hollowed out as well. On edge looks something like there's two applied loads, one here, here, and then some of the dimensions. The hollowed out section is six tenths of a meter. Oops, not quite to scale. Let's clean it up a little bit. Solid part of the big shaft is two tenths of a meter, and then four tenths of a meter for the other part. 30 millimeter OD on the end, twice that for the big section, and then the inside part four millimeter ID inside diameter. G equals 77 megapascals. And now we want to find the deflection of that free end, which we'll call A. Find the angular deflection of the end and we'll be there. And that's all the pieces. That's the length of this section. Rigidly mounted at the other end, so it's zero deflection there. Two minutes or so, so that's the angular deflection of the intermediate pieces for all the pieces. So you're going to have some angular deflection in here, some here, to determine the total angular deflection of that free end there. Double check, see if you can get it. That's like 2.3 degrees is the total deflection of that end. Even with each of those, oh, sorry, double check here. These loads are not in opposite directions. They're in the same direction. So see if you can get the 2.3 for Monday, and we'll do another statically to look at stress concentrations and drive shafts.