 All right, we'll roll it, roll five. All right, any questions of our little doobie, what? Just early for you. We're working over that one triangular gizmo now that we're beginning to get the idea that with the things we're learning here, we're gonna be able to start specifying sizes, dimensions, shapes, materials, all kinds of things that we could never do in statics. That's the point of what we're doing here. So we're looking at a few things in particular as we go through this. We've got two of them so far. This for a quick review brings back up to speed. We haven't been here for three or four days. Normal stress. Was there a symbol for it? A sigma, it's a lowercase sigma. Sometimes there might be a little n in there just to imply it's normal stress, but we don't really use that for anything else. It could be that we have different stresses and different directions, so we might put a sigma x on it to imply that we're talking about whatever force in the x direction and the cross-sectional area perpendicular to that. It's always the case with normal stress, hence the name normal, the word normal, that the force and the area maintaining that force are perpendicular to each other. We'll look very shortly at the possibility of a simple problem like this, that we'll look not at a plane horizontally through this piece, but an oblique plane, and in that case then we'll take the force that's there and break it apart into a normal component and we'll look at the normal stress across that, but then there'll have to be a shear component also across that piece. And so remember, these are imaginary cuts we're putting through that material. So we can do anything we need to with them to understand the entire piece. But in this case, that would be the normal component because that's the piece that's perpendicular to the area maintaining it. So keep an eye on that, that these directions are arbitrary because these cuts through this material are imaginary and so we can set up a coordinate direction however we want, but it's always true that the force is perpendicularly whatever area it is we're talking about. Notice when we do this type of thing, of course the cross-sectional area changes in size because the area of this imaginary plane is different depending upon what the angle is that's through the piece. We also had the shear stress was the symbol for that one as you get used to these different designations. It's pretty common, I haven't seen any of book that uses anything but these two. There's then the shear stress which is defined very much in the same way in that it's a force supported by some area but in this case the area and the force that we're talking about are parallel to each other. And that's exactly the type of thing that shear is. I hope you are familiar enough with it that shear is the sliding of two surfaces over each other which is just what this parallel component of the force would do. We would tend to make those two pieces if the material is gonna fail there, it would fail by those two faces sliding over each other if it was in shear failure. So we've got those two types of stress and we looked at parts of both of those in that ismo we were looking at. Here, hopefully I'm gonna be able to figure out a way to put that into the video. You guys have it, maybe our guests won't have it. Don't forget we're always worried about Ashley and how things are going for her in the Netherlands. And not that it's that happening, yes. Just in case she's running from somebody, what? She's now answering, I do not worry about this one. No, no. I told her to keep her surprised and how things are going. We do have one other type of stress that we can look at in these pieces for, well, at any one of the pin connections where we have two surfaces, two different objects, the pin and whatever it's holding are in contact with each other. For example, we'll take just any one of them, it doesn't really matter. We have one piece coming down like that held into maybe one of the brackets by a pin that runs through here. We've looked at the possible failure of the pin itself as it could be sheared through by the opposing forces. Maybe this bracket's being pulled up, the holding bracket is trying to hold it back down that's gonna cause a possible shear failure of the pin itself. But there's another type of stress in here that we need to worry about. And that's the fact that any one of these components is pressing directly on the pin and vice versa. For example, if I take a look at just this component of the structure coming down like this and I'll break it off, I'll put one of our imaginary cuts through there, leave the rest of it off. As that surface is pushing down on the pin, the pin is pushing back on that surface there where the two objects are bearing on each other. Gives us concern of what we call bearing stress. It's not a very common mode of failure for these materials, much more likely that these things will fail in either one of these other modes long before it'll be a bearing stress failure. But it's still something we need to look at at particular times. So maybe we'll give it a little B for subscript bearing defined in the same way all our other stresses have been whatever this load is, maybe we'll call it P over whatever area is that's supporting it. This might seem very much like the normal stress we've already talked about. The difference being that normal stress was interior to the material. This is at an exterior surface where two surface, two materials, two different components come together. And it's the connection between those two that we're trying to support as we construct this object. This is actually a real surface here. I've just eliminated the rest of it for the view. But this is where the material, in fact, we can put for example a side view of exactly what we're talking about and here's the hole here where the pin goes through and we're talking about this surface right there where the pin and the strut are bearing on each other. So that is a very real surface in this case. The possibility of failure here is that this material would crush somehow, not a very likely mode of failure, but a possibility could certainly be a concern if there's impurities or voids, little air bubbles as the steel was allowed to cool steel or the aluminum was allowed to cool the possibility that there were little air pockets in there makes it almost spongy. And then we have sort of a crushing failure there. This is occasionally a huge concern that women's shoe wear because women love, I'm sure, do be maybe if you bring in your favorite pair of high heels on Monday. The entire woman's weight, almost the entire woman's weight is concentrated into a very small area. And depending on what the flooring is itself, wood, marble, carpet, those aren't big concerns, but if you have a linoleum floor, I don't know if you have linoleum in your house, but there are certain linoleums that are a little bit cushiony, that they're a little bit spongy themselves. Typically it's a type of linoleum you might find in the kitchen because you drop blocks of things in the kitchen, drop heavy pots and beer bottles and other kinds of things. And if they hit the floor, they can dent the floor. Not uncommon that women's high heels would actually cause a depression in the floor. That's a failure in the floor material due to a bearing stress. So often times, well, I don't know if you've ever seen it. I barely seem to remember I have at one time seeing a sign at a place saying no, high heels allowed because of that very possibility of destroying the floor, putting little dimples across it as if a tiny guy and stilts had walked across the floor. So that's one other type of failure here. And we can look at it with this object that we've got. Just pick any one of the pins, look at the possibility of this type of failure. For instance, we've got this pin down here at A. And if we look at that bracket at A, sort of a U-shaped bracket, we're only gonna look at the part of it in concern to us. Jake, see, you can't skip out on technical freehand sketching, otherwise you won't be able to do beautiful. And that's so bad, so bad, yeah. There we go. What I've done is looked at this bracket here but cut it through to expose the bearing surface. The cut that we're imagining is across here. We've eliminated some of the material. But the bearing surface we're worried about is this bearing surface here. It's across this face that the force from the pin itself is being exerted. Yeah, that piece across the bottom AB, remember, is in compression. So the pin will actually be pushing in on the bracket just as we draw here. If we look at the pin itself, it's got a return, a reaction force from the bracket. It's very hard to draw. An action reaction paired with what's on the bracket. But it also, though, is seeing the force from the component itself all across that area, distributed over that area. So we have to see how to calculate that type of stress as well. It might seem like it's quite difficult. If I look at this hemispherical surface from above, look something like that. The force from this bearing surface is uniformly distributed across there. We're concerned with the total load that represents. And we calculated it from the other things we've done there. The trouble is, remember, that this bearing stress we're worried about, it's the normal component across this circular surface that is going to do the crushing. It's this little component there, the rest of it is sort of a shear across that. It's the little piece perpendicular to the surface that's going to do the bearing crushing that we're worried about here. So the trouble then becomes, oh man, we've got to go all the way around here, account for the fact that at every place that normal component changes. It's easy enough to figure out what the area is, just the area of a semi-cylinder, I guess. But it sounds like we're, oh, we're gonna have to integrate and all the way around that area as this normal component changes with angle. Sounds like a real pain. And it would be. You've all taken physics too, right? And I think you remember, well, if you haven't, it's quick enough to bring it to mind. If you remember, in fluid statics, you were looking at the pressure on submerged surfaces. If we've got, say, some submerged surface here, we're worried about the pressure to the water bearing on that surface. We only had to take, actually what I need to do is a curved surface because that's what we're talking about here. So we could even do, for some reason, maybe have exactly the type of thing we've got there. Again, it's the normal component that we're worried about. And that's all difficult to calculate, except that once you do the integration, if you remember from physics too, I hope, what turns out is you only have to worry about that area and the pressure at that point. And you don't have to do the integration around the whole surface itself. Sound familiar from physics too? A little bit? So all we have to worry about is not the circular area and the changing force around that area. We only have to worry about this area across here, the force on that area, which we already know because it's the force in the component itself, the connection there. So for bracket A, we already know what some of those forces are. If you remember, this is 40 kilonewtons in that horizontal member, is that right? 40 kilonewtons across the bottom. So each of these little pieces, total 20 kilonewtons, we don't have to worry about the distribution of that force, we just need the total force. So that's 20 kilonewtons if we're looking at the force inside one part of the bracket. And the area over which that force acts, this blue area in here is what, 25 millimeters, which is the diameter of the pin itself and then the width of the component, which is also 25 millimeters. So this happens to be square 25 by 25 square millimeters. This right here, this isn't the pin. This is another view of this component across here. So it goes all the way down to that U-shaped yolk at the other side. I wish I could figure out how to get this smart board to work up there, but I don't have trouble doing it somehow, I can't get the two to go the ones. So we're talking about this little surface right there and it's cross-section, the hole, but only one half of it, because that was a double-sided bracket there. And so we get whatever that bearing stress is, it comes out to be 32 megapascals. Compared to some of the other stresses, we had a shear stress of 50 and a normal stress of 160. They're about, not that those numbers are directly comparable because each one of these exerts a different type of stress in the material itself. We had the original that was either trying to pull the material apart or push it in and crush it and cause it to fail in that way. Or the shear stress was just trying to rip them sideways or the bearing was trying to just crush it at an immediate exterior surface. So the numbers of themselves are not that comparable. However, you can certainly see that there are differences in magnitude in all of these numbers. So that's the force on both the surface that the member in on the pin. Yeah, because those are an action-reaction pair. The force on the bracket is equal and opposite to the force on the pin, because they're caused by the mate of those two surfaces to each other. Any other questions? Because what I have for you now is a piece, I've got an object for you to do. It's some kind of breaking object. We've got a couple things we want to find. And I'll have to give you the limits because what we're looking at this class, remember, is the ability of certain materials to withstand these stresses. And so part of our design concerns will be staying within those limits because otherwise we're worried that the material itself could fail. So take a second and familiarize yourself with a picture and then I'll put up our design limits. All right, fairly simple gizmo. As we pull on it with force P up at the top there, cause the attachment piece AB to become under a normal stress. And we want to make sure that that piece doesn't fail. Also because of all the loads, the two loads here, plus the applied load up there, there's going to be a reaction force on the pin down at C. So we want to make sure that it can withstand the shear and bearing stresses that are down there. All right, here's our design limits. Here are the things you're given as the engineering constraints. For whatever reason in this design exercise, the allowable normal stress in the part AB is already given. This is determined by whatever the material is. For whatever reason, perhaps this is known to be a certain kind of steel or alloy or something, which has inherent in it a characteristic ability to withstand a certain normal stress that is published by the manufacturer of that material themselves. So you have to go by that limit. If we find greater stress in that, we need to bulk up the piece. Well, we need to find the diameter of this as it is anyway. So we'll make sure that the area is sufficient such that we do not exceed this stress, except that we don't want to be that close to this. So we're going to put in a factor of safety, 3.3. I'll let you think about how you're going to apply that, because there are different ways to do it. And they all come out to be the same result, if done carefully enough. The pin C has a sheer limit of 350 mega-pastels, also with a factor of safety of 3.3. This might be a company design limit. It might be some kind of national bureau of standards design limit. Some reason they're determining the need to maintain a factor of safety of 3.3. All right, so you need to determine a couple things, of course, to figure out what the load is in what the normal stress is in the component AB, you need to find out what P is. Not a big deal, I hope. That's the type of thing where you're doing in statics last fall. You also need to find out to prevent the sheer failure in the pin. You need to find out what the reaction is in the pin C. Now, typically in statics, we always found the components CX and CY. But remember, with what we're doing now, the sheer is essentially non-directional other than it's across the piece parallel to whatever surface we're concerned with maintaining that for. So what you really need to find out is the magnitude of that support of the reaction at point C because we don't really care what angle it is across the pin. We just need to know what it is so that we can then size the pin, make sure it can maintain that. So that's your journey here for a couple minutes. The thickness of that little bracket, that would be found with concerns for the bearing stress. So yeah, I do have a number for that too. So we'll just put that in. The bearing stress at pin C will want to limit to 300 megapascals, so you can say less than or equal to on all of them. All right, any troubles? Everybody able to get started on this? What would you do with it? We will, shortly, yes, but not yet. We're gonna find out is the length AB has to do with another type of failure mode that we haven't looked at yet. No, you're going to need to, because that's what you need to do to find out how big that pin needs to be to maintain this shear stress limit. You're gonna need to find out what P is, so you can figure out what this stress is, and you'll need C to also size the bracket, so that the bearing stress isn't too much. That'll kind of fix everything. Let me tell you what the diameter of that piece AB needs to be. There's a couple different ways to do it. There's a couple different ways to do it, so I just want you to think about what that is. You know what the limit needs to be on sigma A. You know that. You will hopefully already by now know what P is. A is the unknown, and then from A, of course, a factor, a straight, how do you incorporate it? Yep. That's what I want you to think about. Not because it's particularly difficult. It isn't, if you just sit and think, where can I apply this such that it'll do what I want it to do. But there are a couple different ways to look at it, and there are some, there are ways to look at it that are wrong. For example, we have, the design point we're looking for is this diameter, but in the calculation here, that's diameter is squared. So you have to worry if I put it in there, is that going to square the factor of safety? Is that what I need it to do? Is it going to put in the square root of the factor of safety, and that's not what I need it to do. So there's different ways to think about it. I'll give you a couple minutes, and then if you're still stuck, I'll spring you free. Has anybody applied that factor of safety already in some way? That's good, it's not a big trick. It's not a mysterious thing there. Just a couple ways it can be done. What'd you do with it, Pat? Multiply what? Remember, this is set by the material itself. So whatever stress is in the piece, A, B must be less than or equal to this. Which means then that what we're interested in is maybe we'll put it this way. What is our fail point? We worry about it failing at 600 or so. However, we want a factor of safety, 3.3, so we're not going to go up near, anywhere near this 600, we're going to stay below it by the factor of safety. So those numbers are all given by 182. So you can see, you lose a lot of your design margin. Factor of safety, whether it's the safety of the user or the public, or the safety of the machine itself, I don't know, go apply it here, then go apply it to the D because of the squirt factor. But that's defined industry by industry. I need to do it, but I still need some software. It'll be smaller before I send them to iTunes, but I do have an account on that. Donations will be accepted to the fray cocks. But you have a number, a diameter, 0.53. Are we all doing the same problem? What'd you get for P? Okay, good, I would hate to struggle with that one because that was static stuff. P is 40, and that gives you the area, and that's what. So you have R there, R was five. R was 2.26. Really? It's not what I got, and I never make mistakes on these things if you remember. What do you have calling? Now remember, what I'm asking for is the diameter. So don't give radius, don't give the area, meters, half a meter. I got thrown off a few units because they have a bag in there and then they have a kilo in there. Yeah, you've got to watch those in this class. Finally important, especially when you then go square because those squares can mean different things. You're squaring then that factor of 1,000 that you might have screwed up up to. Do it sometime and do it correctly. Usually it's easier to get things into the units you need right from the start. I need one of those little red lights that goes on. It says on the air. It's quiet right here. So when and where and how you apply these conversions is a matter of personal preference. However, generally it's easiest if done before you're into the equations. Because once you have the equations, even though these are simple, they're very simple equations. So our limit here is two mega-pastiles. It might be easiest to make it right here before you put it into the equation itself. You configure that we're gonna want maybe millimeters or meters or something, but that's a little bit easier conversion than all these others. What's mega, the SI prefix mega? What? Yeah, 10 to the sixth. So just take out the M and put in what it stands for right there, area first. We know the load to be the 40 kilonewtons. So the unit's sort of a step of the time as you go. Factors, safety's already in there. We don't have to worry about it again. It doesn't have units. Colin, is that okay? Is that right? Those numbers? Do we have to give just that? Just to give you a half a meter. Just a unit and a little feed, there you go. Now, of course, what if you are gonna make a billion of these things and you want to use stock material for these, and you find that 17 millimeters is not a standard material, a standard size that's manufactured. We'd like to use just off-the-shelf stuff because it's cheaper. Gotta go bigger. Maybe 16 millimeters is available, but that could take you under your factor of safety, 3.3. Maybe it doesn't matter. If I remember this number, it was actually 16.7. Maybe you're still okay. Well, no, you wouldn't be. That would push you under the 3.3. And then if that part fails, somebody gets killed, your design notebook comes out. They calculate the factor of safety you've actually used was 3.2 or 3.1. No sense going to court, and I don't settle out of court, you're not like Fred. Looks like the average factor of safety that most places use is around that number? No, that's pretty high. That's pretty high actually. But it depends upon the application. It depends upon the concern. If this is a very cheap piece to make, you don't mind going to a very big factor of safety at 3.3, because it's just not gonna cost that much. The greater the factor of safety, it does come in terms of more material handling, because not only you've got to pay for more material, but now you've got to shift, differentiate 3.3 times as much stuff as you would have to shift before if that's a concern. But it depends on what the application is. If this thing fails, perhaps it's absolutely catastrophic and people will die, or the machine will be completely destroyed, or maybe it's part of a ship and then the ship is uncontrollable or something. It could be too that to replace that part is virtually impossible, because it's so hard to get to it. So you don't ever want it to fail, because you just don't want to have to take the thing apart. You'd imagine there's parts in the space shuttle that are terrifically difficult to get to. Incredibly expensive just to swap out everything, change a 10 cent piece, and then put everything back in. So the factor of safety, even though it's a safety, it protects different things. It protects lives, it protects machinery, it protects schedules, it protects dollars. All of those things are concerns. What about the sheer pin seat? Do we have that? Remember, once again, we're doing very much the same type of thing. Hopefully, you pretty quickly have know what the, what the forces are in pin C, because it's the same as the, it's equal to an opposite, the forces on the L bracket itself. And we already know the horizontal and vertical components of that. You just have to figure out the full magnitude of P. Excuse me. But then you also have to figure out the allowable design stress point, sheer stress point in this. Again, it's a three, four, three factor of safety. Could be just coincidence. Those are the same. It could be, there's a lot of different places where those things can come from. Because again, it depends on what it is you're trying to protect. So that should be 106 megapascals. Got to resist some sheer by the size of the pin C. Pin C is in double shear. Remember we talked about that? So that any one part of it, it's got a half a, what was the total force C? Remember the magnitude of the reaction of C. 40 horizontal component and then the 15 plus the 50 vertical component. Should be able to figure out what C that is pretty straight away. Yep, 76.3. Kill a newt in this. And any one possible shear point on the pin, I guess we'll draw the whole pin because we've got the L bracket itself in the middle that has that 40 sorry 76.3 kilonewtons. In the bracket holding the pin, that's half of that load. And remember on the pin, we don't care about the direction of that force C. The pin's circular and going in any way. It could be the pin isn't circular or the material is anisotropic. Meaning it's orientation is important. But in this case, it's just a circular pin that can go in any way. And so the direction of C doesn't matter just it's magnitude. And then it's these faces here that must resist that shear. And then that's what you have to size. And then from that you get the diameter of the pin. And then we're done. So what's the diameter of the pin? What did I hear? Were you even up to venture a column? Did you have something? Yeah. Specified 22 millimeters or whatever a stock size might be. On Monday, see if you can't come up with the thickness of the brackets because of this bearing stress. Remember what we need is this projected area across that opening. Sort of, I tried to sort of shade it in there. It didn't quite come out in the top. So have that for Monday. Have a good two hours.