 What? Did you do that one? I did the original sentence. I already had all the subs in the jargon jumbled all the time. I did send the original sentence. Good excuse for this first week after that. Well, yeah, I don't know. Okay. Well, the one Jake's asking about is kind of looks like a nuclear power plant cooling tower. Considered to be solid. So that this is one of those cases where the actual weight of it itself is what it needs to support. So at any particular place, well, you need to find out the stress at the bottom? The radial dimensions are as a function of z. So the average normal stress and the pedestal remains constant. Okay. So you have r here as a function of z where z is this distance from r to the top. What? The vertical distance from r to the top. Why don't you bring in copies for me? Oh, okay, z's. Okay. Same kind of idea, just a different origin. So z goes. And you need to make sure that at every place, normal stress is constant. Now, the area there will also be, of course, a function of z. But that will go with r. And so you need to figure out something about the total volume which will then also be a function of z. But all three of those functions, they're not independent of each other. They're all related because whatever the radius is in terms of what the area is, determines what the volume is, and etc. Such that the normal stress at any point is constant. Right? That's how you read it. Okay. So to start that one, start from here and go down some distance. z, which will give you, then you can start building these functions. And then integrate it. You know the entire height, the height is set. There's something around there. It probably doesn't matter. Because even if it's this tall or this tall, it's all going to be the same. So the only thing you're starting with is you have a set radius up there of r1. So it'll probably be a function of that as well. And this load here, that load that's over that area, of course, will be a function of the density times that volume times, and actually this will be a function of c. So we'll put that in there. So this is a function of c as well, times g. This is mass times gravity. So it's just the entire weight that's above it. So do be thinking, boy, I'm glad I've got the other addition. I did 70 in my book. Yeah. Which I'll let you get away with this one. The old book no longer has a 70. The new book has a 70. In the old book, that problem should have another, a different number. 77, I think. For the most part, all the way down those two columns, it's one for one. But there's a couple later where the author actually switched the order. So the three are the same, but maybe not the three in the same order. So I figured it would be better to put it in numerical order and switch it. OK. Help you a little bit? Yeah. Thinking more guns on that? Sure. OK. Thanks. And I'll post those, I'll post the solutions. Well, she was for this first knob. I think I'll make sure what solutions I have that I can post for this first week. I made this, I can't remember how the author gave me the solutions. And maybe I'll just put up the entire chapter and find your problem, whichever one I want to give. And then we'll be all straightened out for next week. Hopefully. All right. Any other questions before we get going? All right. We've looked at only three things so far, I guess. So let's get them up there. And I want to talk about one of the problems in particular. Whether you did it or not is still a great problem in terms of some of the things we've got that we've been working on. So we've looked so far at the normal stress, sometimes with a little script N down there just to re-imply that that's the normal stress. And that's the force on any particular cross-sectional area, such that the force and that area are perpendicular. Keep that in mind. It's relatively straightforward so far. Starting to, I think tomorrow, I think it's on the schedule for tomorrow, we'll look at the possibility that the cross-sectional area we're interested in is not perpendicular to the axial length of the piece. Because we need to look at all the stresses in a material, not just the very simplest of all stresses. Remember that this area we're looking at here is an imaginary cut through the integral piece. We have not actually exposed this area. However, the force does still go through that area as the piece supports these loads. So we're going to make imaginary cuts starting tomorrow that are not necessarily themselves perpendicular to the axial length of the piece, nor to the immediate direction of the force itself. So we'll start with that. Tomorrow that way we can look at all of the possibilities, protect against all possible failure modes as we look through these pieces. What came next? Shear stress. Shear stress. This is also just a force on an area that needs to support it. This one a little bit different in that now we're looking at the force that's parallel to the area that's supporting it. The pieces, the structural members and all of these things we're talking about need to be able to prevent failure in either one of the modes. Now remember by failure normally you think we're talking about the actual destruction of the member itself and complete loss of the structure. But as we'll see shortly here we're also talking about not just catastrophic failure, we're also going to be talking about deformation. It could be that under these forces the structural members deform just enough that whatever the structure is can no longer serve its purpose adequately. It could be that things get out of alignment just enough that things don't work right anymore. We could have some kind of motor stand that needs to hold up this great diesel engine. But if the stand deforms enough then the diesel engine is no longer in line with the drive axle and no longer adequately supplies power to wherever it's supposed to go. That can be considered a failure even though nothing is broken. It just may mean a very expensive adjustment needs to be made. Either way in terms of what we're looking at that's considered an engineering failure. Then we had the bearing stress we looked at as well. Maybe put a little fee on it like that. Very much the same as the others. In fact really doesn't look any different than did the normal stress. The idea here being that we are really looking at perhaps a major area. Maybe a section of floor or some part of the support. It's not the failure of the member itself. It's more what we're concerned with is that the support structure could deform. Maybe this great diesel engine stand we have, the stand is perfectly fine. The engineers who designed it put it together and it works just like it's supposed to work. But when it's installed, it's installed on a floor that perhaps the floor itself depressed enough that now everything's out of alignment even though the stand worked perfectly. It's that the floor itself was damaged in some way. Maybe just some simple depression around the leg. And now things are just enough out of line that things don't work quite right. And again maybe it's a very expensive fix to come in, shore up the base a little bit. Maybe just lift everything, slip a plate under there that sort of spreads out the weight a little bit. Everybody's seen those type of things where you see the leg of some piece of furniture. In fact you may have exactly this at home. Look at some of the heavier pieces of furniture and underneath it is a little furniture pad. That just helps spread out that weight. All that's doing is increasing this area. The force is set because that's the weight of the piece. If we can increase this area, we can decrease the bearing stress. And then there's perhaps less damage to the floor. If you have wood floors you don't want a heavy piece of furniture sitting on a very small area causing a very large bearing stress. So those are the three things we've looked at so far. What we'll start with this week is not just what these values are but what the material can do in response to some of these loads that we're going to start looking at the actual deformation of pieces under these loads. The fact that if we're compressing a piece it will physically shorten somewhat and we need to take that into account. If we tension a piece it will physically lengthen. We need to account for that. These pieces can actually bend, deform laterally. And we'll start taking that into account then as we go through the rest of the week. In fact much of the rest of the semester. So where we are now? It sounds like three old friends come to visit. Three old college buddies came for a visit. You look around a little and think how dear these people are. What they'll mean to you in the 23rd Union. I'll go to my wheelchair. Wow. Yeah. The top of the stairs. Alright. One of the problems that was assigned. Oh, I don't remember which book it was. Number 97. Doesn't matter if you did it or not. It's a really good problem. We'll talk about it now. Right now as part of what we're doing. It's a problem where there are three discs that are supporting the load. So the top one was just a simple solid disc of radius. Sorry, a diameter D1. This was labeled disc A. Then below it was another wider thicker disc. Or we need that. We need that class today. Yeah. That's a little better. A wider thicker disc of radius D2. Sorry, diameter D2. Now that disc is perfectly over the edges of the material below it. There's a hole in the material below it. Actually, you know what? I'm going to abandon this drawing. I can do that. I write in chalk. We'll just take the side view like in the book. Alright. There's the one that's D1. That's labeled A. Below it is a disc diameter D2. That's the one that's perfectly aligned with the edges of the support material below. And then in between those is a washer-like disc that has a hole in it. Pretty ambitious to do that as a perspective drawing. I'm impressed with my desire to take on the extreme challenge. That has a diameter D3. And this diameter down here is also D2. The same as the intermediate washer speed. That helped a little bit. That a little bit better drawing. That's the one that we've got as in the book. They hired graphic artists to do these, and they wouldn't even attempt it as a perspective drawing. So I was kind of crazy to get it. End of the semester. We do this after this class. Alright. So there's the problem. You're given a couple things. The load being supported is 140 kilonewtons. Now remember it's drawn as a point load, but we take that to be a uniform load across this piece. What's of concern actually in this problem is not any bearing stress on the top of A that just wasn't asked about in this problem. What we're concerned with is what happens once this load is transmitted through this upper disc A onto the part B. We do have a concern of bearing stress across here, but we also have a concern of shear failure between other parts of these. So that's what we need to look at on this problem. So we're asked for a couple things. We need to define the minimum D1 of the top disc. If it's too small, the bearing stress on disc B will be too great. We also need to find the diameter D2, the concern being there, the shear failure of PC. And we also need to find D3, the concern there, at least part of that being the bearing stress being felt by Part C. What you're not to consider, and it may not have been clear in this one, is we don't need to consider the possible failure of the support material itself. It looks in this problem like that might be the concern, but if you read it carefully and look at what it's asked for, that's not what you're looking for. We're considering that this support material, it's integrity is inviolable and everything will be fine. So a couple things you're given. Given an allowable bearing stress of 350 mega-pascals, also an allowable shear stress of 125 mega-pascals, the implication being, though it's not a direct concern, that maybe these three discs are all of this very same material. Did that one appear for both editions of the book? It did for JB of the newest edition. So okay, good. So something actually went right this first assignment. So have you all done this one? I don't want to ask you to help with your problems. And do you want to determine for your safety? Okay. Yeah. A couple different things to consider and some of them, what you find in one of them has to do with what you're going to determine for the other one. For example, one of the easiest things to determine first is if we have the possible shear failure associated with D2. So that's actually the easier one to find. And then once you've found D2, you can look at the possible failure in some of the other places because that D2 then will be set. So the easiest thing to look at first is possible shear failure. Our desire to prevent shear failure, so maybe we could say this, assume shear failure in the disc C as the first concern. That will give us D2, which then we can look at what the diameter D3 is so that we don't have a bearing failure on disc C. So just let's look at that part first. Let's look at that part first. I'm generously saying you do it. Concern with the possibility that since B and this hole are lined up that we actually have shear failure through PC right there, that little blue interface that sort of an interior plug of C is actually pushed through the whole perspective. Well, it's not really getting divided between two spaces because don't forget that if we look at above, we have disc A there and disc B there and then disc C is actually a washer with a very small interior radius. We're not sure just what it's going to be. So it's all the way where B and C hit so it's actually all the way around this surface here where we'll worry about the possible shear failure. Concern with this entire area is that's the area described by this. We're actually looking at the inside area of this cylinder that's preventing the shear failure. That's why of a couple of these, that's one of the thicknesses that we have in this problem. A lot of these problems, the thing you're concerned, is that the trickiest part of this is making sure you get the right area in these. The load, of course, is pretty straightforward. The allowable stress has got to be greater than equal to the actual stress that we're finding. I think we've got to make sure that you're looking at the correct area that is doing the support here. So allowable shear stress is the one, that's 125 megapascals as given. The V, of course, is the load we're supporting. That's 140 kilonewtons. Then the area in this one is the cylindrical area that we see there which has a diameter d2 and a thickness 10 millimeters. So the area that's doing the supporting is this area right there. That's what's going to shear through. If C fails in shear, what you're going to be left with is the big washer. If C fails in shear, unless it fails in shear, you're going to have this washer of diameter, well, we don't know what diameter it is because that's not the concern, but you have this hole ripped out here as this entire part B through drops through that washer seat. Dramatic drawing, isn't that one? It's kind of scary looking. It goes right through. So it's this area that needs to do the support against shear failure in disc C. That's why you need the thickness. But times thickness? It's the circumference of the circle, then times that thickness. So it's pi d2 squared over 4 times the thickness of C, which is given 10 millimeters. It's just pi d. Oh yeah, circumference, sorry. That was the area. So 2 pi r or pi d. Man, you don't have to snap away just when there's a little mistake. Yeah, but that's coming out first thing. Not so stupid as the blonde cast line mistakes. You feel like ducking when you see that. It's amazing. I think we're going to work on cylinders. Yeah, that's something, Jake. Nobody's ready to talk to you about it. 0.063 or 0.063? A little more. Three or four significant answers. Don't round off too much. There'll be some things we're doing later where we have to carry a lot of significant figures through the calculations. And then we can round off just once at the end. I always round. Which is usually okay if you keep the numbers on your calculator as you use them. Five or six, five meters? Yeah, I see. Yeah, you did get that too. Colin, everything else is set. We have the shear limit. We have the load. And we're just looking to find the A. It's got to be enough to support that. Now, is that a minimum of D2? Or is that a maximum of D2? We want to, we're right at the limit with that value you just came up with. Was it 0.365 or 0.0365? 0.0365. 0.3565 meters. That puts us right at the shear failure limit. So would we want to go a little bit smaller than that? Or a little bit greater than that? If we go smaller than that, then this area that's supporting the load against shear decreases. So we want to go greater than that to make sure that we're all right. All right, so that was the first one. We've got that. 0.357 meters. Different take. Now that we've got that, we could possibly work on this area D3 so that we have enough area now supporting, preventing, assume now bearing failure, bearing stress failure in disc C. We now know how much of the load is coming from B now that we've sized it, but we need to make sure that the hole is not so big that there's not enough of the disc C there to support it. So it's got disc C with a small hole in the center of it. B now bears on that surface, and if that hole, that interior hole D3 is too big or too small, we could have bearing failure. What? If it's too big, then this bearing surface is too small and we could have bearing failure at disc C. Now remember that's not necessarily a cat, a failure where the thing is destroyed. It could be just enough for everything to be out of alignment and then nothing works out of that. So now we're looking at a possible bearing failure at C where, again, we have to set the area so that we've got at least a minimum of the area available for support in this piece. Areas we've been looking at. Here's the suit. And that is the first camera that doesn't know who. Get your girlfriends to watch. See, he's talking about me right here. In this instance, the result is going to be then, last I got one foot up. I have to compress them all. It's essentially, I guess it's like zipping them. You don't have to unzip them. It's just that they're compressed down and the iTunes likes them that way a lot better. But it took about half an hour to compress a one hour class and then about an hour to actually upload it. So hopefully, and that's from doing it from home. It might be a lot quicker if I do it up here, but I didn't want to do it on the wall until I was sure. Plus the compression software I used put a little sign right in the middle of it saying this compressed and the free version of such a, and we're supposed to have some compression software of our own that I just don't have yet. So in this agreement, we certainly would not want D3 to be greater than D2. What kind of failure would you call that? You got something bad? Did you check with Bob? No. For the area? Because normally we're working with diameter. So we want D2, D2 from the problem before. So we couldn't have done this one first. D2 wasn't. Whatever did you use Pat? I had D2 squared minus I had D2 squared. Of course we forgot the back here. Easy enough. A lot of times it's the simplest things that go wrong because you're concentrating on the hard stuff and not doing the easy stuff too. Do you subtract areas? Yeah. That's what we're doing here. It's the area under disc B and then this is the, when we bring in this part, this is the area under the hole in C. So we've got this whole area here and we've taken out this area here because that's not offering any support. But I have to worry about the 10 millimeters or something. Not in this case, because remember what we're talking about is the entire way sitting on this surface. We've already taken care of the fact that we've got enough thickness to prevent a C from shearing through. Now we're looking at the bearing or the possibility of bearing failure on C. What do you get? Looks good. You guys agree? Something like 27.5 millimeters in. Between meters and millimeters a lot. All right, then you figure out the last little bit. We need to have a big enough diameter on disc A to make sure that we don't have bearing failure there as well. Now we're looking at making sure that we don't have bearing failure in disc B from the load on disc A. I remember the first time we were even through this I was confused with what they were looking for. So I thought I'd probably just go with it. All right, last part we need to find is D1. Would we suspect D1 have any relationship to D3? We certainly suspect D1 would be greater or smaller than D2 but any relationship between D1 and D3 we just found. D1 here? Check with anybody? Bob, you've got a D1? Could be. Now all we're looking at is the bearing stress right here and so that certainly could be smaller than D3 it could be bigger than D3. It depends entirely upon the ability of disc B to withstand this bearing stress. It has nothing to do with either D3 or the thickness of B. All we're looking at is this bearing stress and whether D1's big enough. Was there any reason they gave that to us? What? The thickness of B. Well, we'll see in a second with you. Let's make sure if it's got a D1 of some kind. Did you guys agree to check the numbers? It's too far away. In this case, actually everything's the same. All it changes is the area. If we're worried about bearing failure in B in this case now, the area is... That's all we're looking at for this one. It should be the easiest to know all the calculations really for us. So you guys agree? You agree with others? What'd you get? 2 to 6? 5 to 6, yeah. Okay. Or 22.6 millimeters. DJ, okay? No. Colin, do we? All right, now that we know that, we know that, let's see, D1, D1 is less than D3. A little bit just to reflect that. The D1 we just determined and D3 is actually outside of that. Is that the deal that we now have? D3 is a bit greater than D1. How does that then affect things? Don't we now have the possibility of shear failure of B that we didn't have before? If D3 was smaller, then we're not concerned with the possibility of A shearing through B. Since D3 is bigger, as we just determined, we have the possibility that A could push its way through, so we're out of B, and so we need to then protect against the last little bit the possibility of shear failure in B. D1 is less than D3. We have the possibility of shear in B as a concern. So that's where the thickness of disc B was given. That's why the thickness of disc B was given. Again, that cylindrical area that we need to look at felt like that disc A could push its way through disc B. That's the name of the class. The area is big enough that we don't have shear failure or B. So it's going to be really careful with what areas we use in these problems. And that's given. That's 20 millimeters of stress and sketching skills. Exercise, start chatting too as well. Pour another cup of coffee, sit down and do it. Picture and then... Try to draw what? I look at a picture and try to memorize it. I think that was one. Have we got it yet? We're done? Do we what do you have for D1? Yeah. It came up that shear to B is not a concern. B is thick enough. If B was thinner, it might be a concern and we need to make D1 then bigger but it turns out that shear is not a concern with this. So it shows that D1 as we originally determined is okay. Put that in and we're set.