 too long. Okay. Alright. What we're looking at now, while we're continuing beam design, we've been looking at what choices we can make based on the bending loading and some of the shear stress load in terms of fairly regular type beams. So we've been mostly looking at wide-flange I-beams. They're so very common, but they're so very common as we've seen because they work well. They have a real high moment of inertia because there's a lot of area away from the neutral axis with those I-beams. So they've been good for that. Now we're going to look at the design of not necessarily beams, it's a type of beam and many of the same analysis techniques we've been using will work, but we'll look at the design of transmission shafts. We've done some problems when we looked at torsion and torsion loading is the main loading in transmission shafts, but we have been we have looked at some of these type of problems. So imagine being with or a shaft with several gears, gear take-offs on whatever it might be. This of course is an integral part of how and why your car works the way it does. So we need to pay attention to these type of things. And then imagine of course some some kind of support here. We'll take it to be just a journal bearing of some kind so that just just to keep this end from flopping around, but no moment support and no added torque there of any kind. That's just simple clean support that that we need to take into account. So we'll throw a couple loads on here just for the picture's sake. And these will be either transmission belts will be attached here, other gears will be attached, chains, whatever it is that we need to take the add the transmission to other places. All right, so that'll do for now. Notice that we are now operating in three dimensions. Illustrations say we'll take x down the length of the shaft as we've done before, y up and z out in that direction, or I can't hit the quarter system. So we can take the shaft itself and reduce those loads as given in the picture to something a little bit more useful to us in terms of the type of work we've done already. For example, we know at each one of these end bearings, there's going to be some kind of reaction support. So whatever direction that comes to be in and for reference sake, we'll call these the two end A and B, name them myself. So that's A in the z direction, A in the y direction. Journal bearings do not generally supply any normal, any axial type of support. So there'll be no x reaction and they don't generally supply any moment. They do of course in real life supply a little bit, but it's inconsequential to the things we need to look at. So there might also be some B, Z, and some BY reaction yet to be determined. But then these power, the transmission parts of those, we can change those into some load, some force directly on the beam, but then with an attendant moment caused by the size of the gear and whatever force is being applied there. We've done that kind of thing before. We learned back in statics that we can take a force at a distance, replace it with a force and an appropriately sized moment or vice versa. So again, the sense is right to match all the pictures. M2 is going to be just that force times the radius of the gear and so on for the other ones as well. And then 3, it looks something like that, P3, and then it'll cause a moment to do something like that. Of course the reason that's good is that puts all this into a picture of the type of thing we've handled before. We've handled these type of transverse loads that cause bending. So we can look at the loads that causes internally. We've looked at these torques that are parallel to the beam. Of course, right hand rule, whichever way puts the moment up and down the beam in the X direction. We've looked at the torsional loads that those kind of things cause. And so we can then analyze all this. So we'll take our first repeat at this, looking down something like this. There's the X direction. There's the Z direction. So we're looking right down the Y axis. When we look right down the Y axis, we see this load now looks just like that. Looking down the Y axis, we won't see P1. We won't see P2 because those are in the Y direction. So we'll see P3 out here and we'll see BZ at the end of the two reaction loads there. So then it looks very much like other types of loads that we've looked at. You can do the same thing looking down the Z axis. It's the Y axis there. AY reaction will show. We're looking right down the Z axis now at this. Now we'll see P1 and P2 wherever those happen to be and however big they happen to be. And we'll see B, we'll see BY not BZ. And again it looks very much like the type of loadings we've looked at before. Those will cause bending. Those loads, those transversals will cause moments that will cause bending and we can size the transmission shaft to handle the bending that goes on there. We also though have to look at the fact that there's torsion loads that are going to be in here. I haven't drawn those in yet. There are those moments caused there but we haven't drawn them in. So at any place along the beam we can have this kind of concern. So here's our beam with some internal cut made somewhere. We can do this anywhere we need to. Just imagine a cut to look at the internal loads. Because of the two dimensions in which these loads are going on we might see some moment in the Y direction. For example that would happen with maybe P3. It's going to cause some bending. That bending will be in the Y direction and so we'd see that sort of thing. From this other coordinate view where we're going to see bending in the other directions that'll be a Z moment of some kind. I don't know necessarily what direction they'll be. It depends on where the cut is and just what these forces are. But we could have this two dimensional nature to this bending. We've always looked at one direction of bending. Actually typically we looked at the Z direction because our forces were always in the Y direction. But we could also have this type of thing. But then we also have because of these different moments the possibility at any one spot of having torsion in a certain direction and of a certain size. That kind of complicates things. We've never looked at bending in two different directions at once. We've always looked at it in one direction only. So you'd think oh man we've got twice as much work to do. But we don't quite. Because what you can remember is these transmission shafts are always circular. So there is no preferential direction of the m and the z or the y of the z even the x direction though we typically put that down a length. So what we can do is just combine these two moments into a single moment and let that be our direction. Now we have a beam in single direction moment type bending the type of thing we've seen before. We just have to make our axes match that and we're okay. We can handle that type of thing. So we'll clean it up a little bit. Now just a single moment in some direction at any one spot and torsion of some size. And this of course varies anywhere along the beam. But we have all the skills we've had before to look at any of these pieces. So we know the maximum trouble from the things we've looked at before. The maximum trouble. The maximum, for example, a maximum normal stress is going to come wherever the moment is maximum. But wherever we're the farthest away from the neutral axis and the neutral axis is right through the center for a circular beam. So we know that that is going to be any spot perpendicular to the moment and at the outer skin. So in terms of the normal stress caused by the bending that's going to be our point of concern. We're perpendicular to the moment so that's where the effect is the greatest and we're at our distant sea away from the neutral axis which passes right through the center. So in terms of the moment that's where the things are going to be the worst. In terms of the shear stress from the torsion we know the same type of thing that has to do with whatever that the size of that torsion is. Also at the outer limit of the beam a distant sea away from the neutral axis or the center same thing. And then remember this is divided then by the polar moment of inertia. So for the way we've got it drawn with a moment going that way going to cause a bending such that we see a normal stress in that direction that's where it's going to be the biggest perpendicular in this load. But then we add in the fact let's see with this direction of torsion drawn we're going to see shear stresses in that direction. That looks very much like the type of stuff we just did with the transformed stresses. So we know that the combination of those two things is going to give a maximum shear stress and this was what we just saw as r the radius of Mohr's circle and that's going to be the square root of the normal stress divided by 2 squared plus the shear stresses we see at any one point squared. And that's those two things right here. And remember this changes at all places down the beam so we're going to have to look this is a concern but there are no new grid concepts in here. Everything that we've got in here is essentially essentially the techniques that we've used before. By the way if you need to this is equation 9.7 that promised to that point. Very well a couple things you notice the radius of the shaft is a constant anywhere down along the shaft. As we move different points m and t might change because the loads change at different spots. But the radius of the shaft is constant and of course the moments of inertia at a constant. And if you remember for a circular shaft the polar moment of inertia and the regular moment of inertia are related so those can all come out of the square root. At any one spot we've got m squared plus t squared squared and that will be our maximum shear stress then. And again it leads to a concern where we might have an allowable shear stress because of the material then a couple geometry constants because of the shaft itself. And that leads to the possibility that we need to look at the type of thing we just looked at sort of a section modulus type thing where this has to be less than I know greater than or equal to any one spot over any allowable shear stress for the material. And we need to design this for maximum loads possible. So we need to look at everywhere down the beam where the combination of bending moment and torsion because of the transmission on the shaft are the greatest or the smallest. Malibu be you get this just a sample from where we can run through some numbers. Don't let it scare you. So way down this time of year we'll have basically the same thing. We'll go through this type of analysis that we just did all the stuff we came back up in a way we can look at it. Alright some of the given details that aren't on there for this problem we'll add them in. The operating frequency is expected to be 480 rpm which is 8 hertz. Everybody knows what hertz is. What do we know what hertz done it is? Let's see only engineering joke there is so you got to enjoy it. It's all we got. For the rest of your career that's the only time you're going to laugh when you're working. Alright we have a motor delivering power of 30 kilowatts so I'll call that PM. That double P there is my symbol for power because we use P for forces a lot. And then of course that's the same power being delivered to that gear E because M and E are in the geared contact. We'll take off 20 kilowatts at the gear G so that's a power takeoff. Power is being delivered to the shaft at E taken off at G and H so obviously H has got to be the rest of that power takeoff of 10 kilowatts. Alright so that's the picture we've got there. Power coming in at M 30 kilowatts that's how much is delivered to E. Remember that we're going to need to use those powers plus the operating frequency to figure out what the torsion is. We did that before in section five and we'll give this because of the shaft material we have a shear stress limit of 50 mega pascals on this problem. I think that's all the pieces. Alright if you remember from section five three the power at any one spot in a rotating shaft is equal to 2 pi that's just a conversion from revolutions to radians and then times the frequency and then times the torsion. So for each one of these we need the torsion not the power applied at those points. So for example the torsion at point E is going to be the 30 and then you got to watch the units of course 30 kilowatts 2 pi 480 rpm or no we want one of the 8 hertz you know that would make sense for those units. Consider a current second unit between units it carries. So at E we've got then 597 newton meters remember a kilowatt is a newton meter second per second newton meter per second. So we get a torsion at E loading our we've got you see the way that motor's turning the way that hits the gear so at E we have a force in the y direction which will be the torsion is the force times the moment arm which is gear radius which is given and now we can then figure out the force E and happens to be in the y direction there and we know the torsion which we can now we just solve for for the force and what's the radius yeah the radius of the gear E is given as 160 millimeters point 73 kilowatts which is just what was already given so that's how we got that that's how we got 3.73 kilowatts all right so that needs to be done for the gear at M and at G and then with those forces you can then solve for the reactions so I'll give you all those pieces so we can get on to the rest of it let's see gear D is on the top so that's going to be a z direction load right about the middle D direction of the torsion it causes is going to be that direction and I'll give you all these values in a second and then C or G I guess it's the takeoff is G the wheel is C it's the same direction reduce this to the the torsion being applied and the forces all right so you need all of those now we've got T E and F E and F D applied at the middle gear or takeoff at the middle gear but in terms of the shaft it's applied torque and then we'll also need an F C so we can put these together and then of course each one of those leads to a reaction in the right direction already or yeah we'll meet those as well but those just come from summing the moments and forces on the beam itself just like we did with status got the picture do be okay i started a little bit bobby frank you're okay you don't have a power drink you're all right you need to go run and get one okay got it got it one of those pouches that delivers it intravenously and it's under your clothes that's a darn good idea right there hey you were at the pharmacy and helps why don't you design that as the project that's a great idea well yeah but no you could make them you could have to sell them down in the vending machines was that for logo you buy right there because you got to be a lot more fun than class than you have to be all right tc is 398 newton meters td 199 the attendant forces caused by the whatever they are belts or chains or whatever fc is 6.63 2.49 this lab with or this course this class was three hours long i'd like you to go but we just don't have the time all right so we also need a y y b nothing you couldn't do get all these this is all statics that we're going through so far a y az 6.22 sense there's a lot of force in the z direction so there's got to be a lot of reaction in the z direction looks like that makes some sense by 2.8 2.9 all of those just come from summing the moments and summing the forces to be zero on the shaft itself nothing more than we did in statics i think more than we did in statics all those months but we just never happened to do it in two dimensions it's kind of a two dimensional problem since the shaft is down the x direction then all the loads are in the all the forces are in the y the y and the z direction i know you're disappointed i didn't like you do all this booby if you sink any lower in the chair you're going to disappear all right got them all down because i need board space down actually i just keep going on the white walls i don't think they like it so you got all the details because now we have to need to put the analysis of all of this together so we'll look at it like this uh we look again down the down the z axis x is our shaft or shaft something like that there we know how long it is we've got all those dimensions but as we look down the z axis all the z forces in a sense disappear and we get a y which is kind of small 0.932 kilonewtons we've got at the other end b y 2.8 oh kilonewtons and then the only other y force we have i don't know the big drawing now that was uh the force fe is 3.7 and it's we know about where it is that's the power input force and it's 3.73 so we have something like that with uh what 0.6 meters on this side and then 0.2 meters over here so it's it's exactly like problems we've been doing um for a couple weeks now remember we need to find where the maximum points of concern are so let's see the shear stress almost uh two seconds we're so darn good at that right now because the shear stress leads us to the moment and it's the moment that causes the normal stresses causes the so we have a moment diagram looks something like that peaks out at 560 newton kilonewtons meters so uh you might think that's a point of concern but remember we've got to combine the moment with the torsion and we have to take into account there's another direction here we uh never taken a peak at yet and so now we can uh we can also see what the moment are the moment is for the two other gears in that direction the the the point at which the two other gears are so just by the geometry this point is 186 this point is 373 we need those because combined with the torsion and the moment in the other directions those might be the points of concern even though in this direction that's the greatest moment when we look at the other directions that might not be the entire picture so we can do the same type of thing in the z direction a z is 6.22 b z is 2.9 the other two forces are both in the z direction that was what uh c and d i guess c is 6.3 right about here give or take a little bit z is right there and then there is no f component or z component to the e force 2.40 yeah we have a picture very similar to type thing we've been doing in statics quickly draw the shear diagram from which we can really quickly get the moment diagram and then we can combine all of these things to find out where the concerns are the greatest shear diagram we've drawn this what's second here 6.22 up stays that way for a little while jumps down 6.63 so it goes just under the axis stays like that that jumps down 2.49 comes up 2.90 all that work we did with the shear moment diagram once pays off because now we can draw the the moment diagram very quickly area there gives us a change in moment up to 1244 so we can see that the moment that that spot in the other direction is much greater than we'd seen in the first direction we have a little bit of negative slope down to a certain point and then greater negative slope that drops all the rest of the way is zero that intermediate number happens to be 1160 which you can figure out just from the geometry and remember we have this third spot there was a third gear there that just didn't have any forces in that direction but we need to know what's going on right there that happens to be 58 so there's the moment in the two different directions circular shaft there is no preferred direction at least for the yz components so we put those together just like we we were looking oh we got to throw in the and the torsion part too there's our beam something like that torsion of in that direction in the other direction each right there so we can we can get each of those pieces kill a new meters start just new meters it's the combination of the torsion and the moment for each of these things this last one's 597 in the other direction so we can even draw ourselves version pot and then we can investigate what the combination of all these things is at each of these points all right this end just goes down to the journal bearing so it has no torque in it so we jump up to 398 go down to 199 jump up an additional and then we come down to 597 and then the other end is not a lot of force so we can go to each one of these spots and figure out what the critical moment is in all the different cross-section directions remember what we're what we're trying to find here is a j over c that's greater than or equal to trouble is that as up until now until we've got all this done we don't know where this upper component is a maximum at any component we don't know what the preferred direction is but we haven't met the y the z directions so it's really easy to calculate so for example we'll check what the total load is at this first point called c y squared plus m z squared all evaluated at point c are here being m in whatever the preferred direction was example at point c we know one of the moments in one direction is 186 and let's see this is all kilonewton is that kilonewton meters or newton meters think it might be newton meters yeah sorry it's newton meters newton meters squared and but then we're gonna screw it to undo it at point c in the other direction we have 1244 the torsion at that point is the 398 three of those we get a total load 319 newton meters that's the combined bending and the torsion where the two together are max or are the two together at that point c are about that kind of do it again at d to give us an idea where the greatest loads are what our biggest concern is so d is 373 squared plus 1160 squared plus 199 squared all screwed oh yeah sorry 597 because it's it's we don't take the minimum there we'll take the maximum there this is that has a value just on either side of it we do 560 580 and the 597 again because we're gonna remember we're looking for where the maximum concern is bro you come up with those real quick give you something to do take your look for something to do if you have md 705 1357 373 squared 1160 squared 597 squared and then at e so it appears that our concern is more around point d than anywhere else in fact if we look at this just a quick picture one c was 1319 wherever that falls d is a bit higher than that e is a bit lower than that and so we might suspect that it's uh something like that so we might suspect that the maximum is going to be somewhere in here if we were really designing this theme or we were trying to design it to the closest possible specs we need to find out what that was but we can build on a factor of safety and cover it all anyway so uh probably somewhere in there we're okay we'll just take the value at d to be the concern so we can then get a section modulus for the circular shaft now as the here's over the allowable shear stress i don't remember what that was 50 megapascals is that right and then from this we'll be able to get a radius for the shaft because the only variable parts in there is is the c yeah this is not the same m as there this is the uh the uh what is it called the uh oh sorry that's that's not yeah that's not m what was that that's that's the shear stress at those points j is it's one half Pisces the fourth and you got to get the units right on that thing to have that that would give us minutes of meters cube 27 point more and so you can solve for c then you know the radius the minimum radius of your beam of your circular shaft to withstand these ear lobes and since we were evaluating it at a point somewhere off a maximum we want to uh build them a factor of safety anyway which is always true got them on speed up and operators standing by but wait call now we'll include another transmission shaft free for shipping charges only how awesome is that for 38 minutes more i can't believe your phones aren't out on fire all right you finish that up you should get a shaft of uh something like 26 millimeters radius go a little bit bigger just to make sure throwing a factor of safety even though you get two shafts today for the price of one because i mean you want to have to trip the shaft out put in another one was that enough for you for one day do be being satan with that satisfied got your money's worth today pat just think the guy is at rpi paid seven times more for that than you just did i don't see why everybody thinks they're so smart all in all right you can reproduce it blindfolded one piece of what we're doing you should be looking at that saying oh yeah that's way too big for a test