 Hopefully, that's recorded, because I've got something up on the screen, so I can't actually see what we've got going on. We'll trust it. All right. We were looking at this simple gizmo yesterday, it was just a triangular support holding a simple 30 pound or 30 kilo Newton load at the end there. We were going beyond what we always used to do to get us through the statics part. All we did in statics was figure out what the forces were in the members and it was good enough at that. Now we're in a portion in this general topic where we need to determine what materials, how much of the material, what shape of the material we need to use in these structures to make sure that those loads that are in each member are supportable by that material. We finished up yesterday, I believe, we had just found the stress in the member B, C. Let's see, if I remember, that was A, B, that was C, right? That was B, is that how we labeled them? B is a B. So B is down here. So we had just found out, just determined what the whole stress was in that diagonal member B, C. No reason other than we got to start with one, so we started with that one. Do you remember how we defined this term normal stress at the end yesterday? How do we find normal stress? F over A, where that F is what force? It's that force felt down the member. It's the type of thing we looked at when we were looking at the shear moment diagrams last year. It's the force in the member itself, along the member that it's feeling. Maybe we could have labeled that then the force in B, C. We found that from a free body diagram on this point P. That was just a very simple thing to do. That way we got the load to tell us what the forces were in each of the members. What is that area? It's very important to use the right area. If you don't, you're not going to get the stress right. You're not going to get the rest of the design right. What was that area that we used in there? It's the cross-sexual area. This member B, C is a circular member, which you can tell because of the break in the diagram. It's got that kind of shape to it. That's an engineering designation that we've got an imaginary cut through a circular member. That diameter was given as, what, 20 millimeters? We, of course, then can find the area from pi r squared, where, of course, r is 10 millimeters. We came up with that, I believe. It was 314 times 10 to the minus 6 meters squared. The force in the member was, I remember, 50 kilonewtons. Then we know that the stress is simply the force that is being supported by that member itself divided by the area that is doing that supporting. Remember, this is the cross-sexual area, so that whatever force is in this member is perpendicular to that cross-sexual area. And so we finished up just coming up with 159 mega-pascals. What's a Pascal? It's what? Newton per square meter. A matter of convenience. If you don't want to use Pascal, if you want to leave it in Newton per meter squared, no sweat on a big deal. Pascal's not going to take the insult that you intend for or have by ignoring that. It's just a matter of convenience. But if you want to ignore it, you can. If we're doing this in English units, you will most likely come across PSI. And since those are English units, there's no one to honor what a special unit there is between those PSI. Also, as a possibility, remember, we're doing some pretty large numbers here. That's why we're already a mega Pascal. We might see KSI, which is what the K stands for there. In this case, remember, we're in English units. So it's, well, it actually is. It's kind of a mix of SI nomenclature with English units. This is a kilo-pound. So a KIP takes a kilo-pound, 1,000 pounds. You can never figure out why the abbreviation for pounds is LB. I have no idea where that comes from. So don't blame me for that. All right. If I remember, that's where we got to. We've just come up with this number here on yesterday's class. The end of yesterday's class, right? That's where we finished. All right. What we can do now that we couldn't do before is we can start to determine what material should we choose that can maintain that type of stress. Now that we are actually sizing the object, we can make sure that we get a material that will hold that stress. So if you have your book here looking to open the back cover, if you don't have your book here cozy up with someone who does, introduce yourself. I know there's 200 people in here today. I already got something. I can't. This is for you at least. All right. What we have in the back of the book, average mechanical properties for typical engineering materials, which are just what you might think they might be. We go down, cast iron, steel. We go all the way down into concrete and woods. The types of things you would expect us to build stuff out of. This is mostly a class on how to build structures that are strong enough to remain static, to not accelerate it any time. And so we can start to look, this first page I have happens to be in the SI unit. So if we go down to the lower page, if I can do it, then we have the very same materials just now listed in SI units. And if we go over, we start to see things like the yield strength. We'll define these specifically, shortly, but you can start to see that now we have some limits we can start looking at. We know we need to have a material that can withstand at least 160 megapascals. So we don't want to pick any material that these numbers, even when we're not sure what these numbers mean, you can tell by the wording we got to stay within here somewhere. We certainly don't want to start thinking about any materials where the sheer numbers we have here are well above the sheer numbers on this chart here. For example, way down here we have a 90 for tension. This material is in tension. So we would not want to make this out of plastic reinforced glass, fiberglass of some kind. It's just we could not make it. We'd have to use a lot more of the glass than we want to in the design here. For some reason we are limited to a 20 millimeter diameter as part of the design. So we need to pick a material where only 20 millimeters of that material will be able to support this filling 50 kilonewton load. Any other material below that's going to fail. Those are the type of decisions we're going to start making this term that we never made before. So if you take a quick look in the book, let's see, maybe steel. This object sort of looks like steel anyway. We're talking about yield strength somewhere around 200 megapascals. That's a little bit above here. When you go to design these type of things, depending upon what the use is, you might want to pick a material that's got a stress limit, maybe double of what you expect to be in there. So there's a factor of safety. That allows for assumptions you need to make that sometimes approximate answers. It allows for manufacturing irregularities that occur, whether it's the production of the steel or the production of the member itself. You want a factor of safety in there. You don't ever want to pick a material that just barely gives you what you need. You're too close to the edge there. You're too close to the chance of failure. Remember, these numbers that are in here are statistically determined. They take a thousand samples, and these might be the published averages, which means there are going to be a lot of them above that and a lot of them that are below that. Maybe you'll get a little more for your money, but maybe you'll have too little for what you paid for or you're going to be under. So you need to build in a factor of safety, and we'll use that kind of thing as we go along here, as we go through this. So that was our first calculation of the shear stress in that member. We're not done with that member yet, because as you look at it, you know that we looked at it, we looked at, and I don't see this thing supposed to show up there on the screen. Why isn't that working? Then what's this for? It's not going to work. Okay, I don't know what. All right, we're having technical difficulties. We're not going to delay with a point here. Back to the document camera. We looked at the shear stress along this cylindrical part through the center. We just took a diner across there, figured out what the area was, figured out what the stress is through that piece. There might be somewhere in this piece where there's actually less area available for the support, and we need to concern ourselves with that. If you look up at the piece up here, it's a larger piece. There's generally more structure available there, because that piece is now 40 millimeters in one direction, and only 20 millimeters in the other, but then it's got that big hole through it. So we need to look also at other places like that to determine are those places where there's actually less material available for the support of this 50 kilonewton load. So if we look at that one piece up there, we've got this sort of set up there. I'm sure you want to put technical freelance sketching and we'll be able to do amazing things like this at the board. We have the cross-sectional area here that we've already calculated what the stress is across that area. It's also true, though, that this area has to withstand that 50 kilonewton load, just as this piece down here does, because this 50 kilonewton load goes through the entire piece. We don't need to actually recalculate the stress there, but we do at least need to see what the area is. If the area is less than it is here, then the stress is going to be greater at that point. The load will be the same. If the area goes down, the stress goes up. That's a pretty easy area to recalculate. We have a piece that's... What is it? That's 40 by 20. Is that right? Is that how you read in that diagram? Bless you. But we've taken out some material for the hole. This is a 25-millimeter hole itself. So we have to subtract from that how much area is in the hole, taken out by the hole, which is 25 millimeters by 20 millimeters. Is that how you read it? You see it in the diagram. Maybe if we look at this in perspective, that's terrible. We're not going to work with that. We're going to have to wait until the end of technical freeing, sketch, and manual. Anyway, what we're looking at is just the cross-sectional area. It still needs to withstand that 50-kilometer load, just like the main member itself did. How much area is available with this portion? 14 times 10 to the minus 6 meters squared. Half of it? Well, you can do just one of those holding 25 kilonewtons, half a load, because the other half's over here, or you can do both the areas holding the whole 50 kilonewtons. It's the same number. What area I have calculated here is the total area here, so I can compare it to the total area there to see if we have more material or less material in support right there. 300 squared? 300 millimeters squared. Or, as we already know from yesterday, that's 300 times 10 to the minus 6 meters squared. We actually have a little bit less material there. Then we did down the length of the member through the center. So this becomes a greater concern because there's more normal stress in that area than we had down through the middle of the member. And you need to check that at all the other places as well. The other end of this bracket tends to look very much the same. And so you're not going to be concerned about that, but you also then have to go to each of the other members as well as the attaching points themselves, which also have to be picked out of some engineering material and designed to an appropriate area. And so this very simple problem that we dashed off in a few minutes in statics last year now becomes much more involved in engineering problem as all things would because all of these things need to be engineered just as much as any of the others. There are other concerns beyond just what the stress is. This is, if you remember, a member intention. If you look at the chart in the book, you'll notice that all of the materials have values for tension and values for compression. For many of the materials, these numbers are just the same. For example, this first number is aluminum. It's the same intention as it is in compression as are most of the engineered metals, the alloys, copper, magnesium, and steel alloys, titanium alloys. When you look down at some of the other ones, give it a second to focus, you see that some of these numbers down here are very different. For example, it's hard to get the whole thing on the chart. This is the value for wood in tension very much lower than this value here which is wood in compression. The very same thing we talked about yesterday, that wood in general is terrible in tension but excellent in compression, which is why the depth you might build this summer, you put wood posts there, it's easy to get, it's easy to work with, and it's very good in compression, so you would use that for your support posts on your deck, but you wouldn't hang the deck from an upper store using wood supports because wood is so bad in tension. One tenth of the strength in tension that it is in compression. So we also need to look at this bottom member and what kind of compressive stress is in it and then make sure it too we choose an appropriate design for it too. We already have a design area specified from this problem. What we don't have though is a particular material chosen. So what's the stress in that horizontal piece? And don't forget there's a couple possibilities. Notice that this piece is rectangular in cross-section through the main part. At this end it maintains that same cross-section but now has a hole through it again. At the other end it drives out into a y-shaped yoke of some kind so there's actually more material at this end than there is at the far end so you need to look at all of those parts all of those places to determine what the stresses are in those pieces. So just for a quick practice on your part determine the normal stress in that member AB both through the rectangular center what the stress is over here where the pin has been put through to hold the two pieces together that we're a quick but there's a different area at each of the two points we're talking about even down at the farther piece. Remember to pay attention to in these problems and make sure you document it on your work whether we're talking about tension or compression at any one spot. Load in there was 40 kilonewtons area through the center here you're fairly simple engineering drawings but more involved than we ever had to look at in statics last fall 30 by 50 millimeters squared through that rectangular center part two meters remember you have to square that conversion factor because the unit is squared itself don't give us the same units we have back here but you have that stress yet through the center part so I don't give it to me I don't see if you can agree to put duty she's fastest of all obviously there's going to be less area down here at this end A so it's going to be a point of concern what do we have for the normal stress in the center of the member AB DJ what you have you said 26.7 megapascals everybody agree with that I don't agree with that either three things I need to hear on this type of no I need of course the magnitude I agree with the 26.7 what else do I need we're already struggling with that is it megapascals or is it kilopascals that it's megapascals be careful with this you mess that up that's a factor of a thousand that your company is either going to over pay for you're going to end your engineer the object, the structure we okay that would be right that's what you guys you get the 26.7 megapascals Pat you're okay be very careful with these units we're going to be working with some very big numbers and some very small numbers in this class and they have to be right you can't be within a factor of a thousand and count that is okay pay attention because I said I need three things with these kind of numbers I only have two I have the magnitude and I have the units what's missing I need to know whether it's compression this is almost a vector this is like a one dimensional vector we need to know is this compression or tension in this subject you can either just write compression or you can put a minus sign I have trouble with that because it's just easy to forget what's minus and what's plus whereas when you write compression there's no question that that material this is a compressive stress we're calculating here if this is made of steel that's not a concern we've already seen from the tables that for the most part the stress the ultimate stress of these materials is the same tension and compression but we'd have to look at that for other types of materials let's move on to another part of this design that we need to worry about as long as we're okay with this before I clear the board it's a fairly straightforward calculation however as we're seeing there are little things that you can get in there either unit problems direction problems if you're not paying attention to whether it's tension or compression also you need to make sure that you calculate the right the correct area in any of these another type of part we have to look at for example if we look at the pin that's in up at the the corner C the pin itself we have a 25 millimeter diameter that symbol is is often used to designate diameter 25 millimeter diameter pin it's got a contact with the member CB if you remember it's that's intention so there's a 50 kilonewton force on that pin the member CB that is supporting it's attached to the wall with the bracket that must be pulling back some force itself should be pretty apparent by how much it has to pull back as well 50 kilonewton a little bit of a couple formed here but it's not significant it may be part of why at some of the other brackets for example the one down at point A is a U shaped bracket so it helps eliminate the couple that's formed by these two offset forces but that's probably not very much to be of great concern what is of greater concern is the fact that this pin if we imagine a cut across here we've got the 50 kilonewton load being exerted by the member BC there what's resisting that load at this imaginary cut but the ability of the material to withstand shear that shear 50 kilonewtons itself if it's going to withstand a 50 kilonewton shear that is in the pin this is a very very different form of stress on that pin P sorry pin C than we had inside the member CB itself that was a normal stress this is a shear stress for what the symbol is for shear stress what? Tau used to these different symbols if you want put a little C on it there because we have other places where there's shear stress we need to be concerned with so you need to make sure you're keeping all these things straight how is it defined how do we find the shear stress sure check your notes that's why we keep them it's the shear the load being resisted divided by the area doing that resistance very similar calculation of what we just had the normal stress it's just a different force goes into it but it's still that cross-sectional area that is resisting that force so we we know what the shear stress is it's 50 kilonewtons and the area is that cross-sectional area where the radius is one half of that diameter 25 millimeters so calculate that real quick we have the area so double check that 21 times 10 to the minus 6 meters squared that's good so that's all similar units to what we had before so whatever that calculation comes out to be we'll know the units go with it two megapascals is that sufficient for the answer for this problem and specifically for this material steel direction makes no difference it's an isotropic material meaning that its properties are the same in all directions that's not at all true with wood because of the grain that grows in wood the properties can be very different across the grain than they are along the grain which is why it's always very important when you're building something well you don't even think about it because when you go to buy a 2x4 for building a structure it's always with the grain running along the along the the board it'd be a little difficult to get an 8 foot board with grain running perpendicular to it but with a diameter tree there aren't many of those left so it's partly the length but as we'll see in starting in a couple weeks it's very important that for most of the structural uses of wood that the grain actually run along the length of it to do what it needs to do so we don't have a directional component on this with steel and not with the shear in this case so the answer of 102 megapascals is sufficient not so with the normal stresses we need to know whether we're in tension or compression with each of those calculations now if a member of CB that could go through a shear stress could it the member itself well not not quite true if you look at the ends there might be some some shearing here as this part is being pulled this way but this part is all trying to pull it back there might be a worry of shear failure back across that where the island itself just rips out there's even another concern we need to look at in that area what we're going to do as well we'll get to in just a second so as you're learning as you get you're only in your second year here as you get through the four years and then maybe graduate school in a couple years experience it gets more and more involved the more you go along which is why not any dope on the street can become an engineer what's called single shear there's only a single place where we're worried about the shear occurring we don't have to look at any other part of it we do have a place in fact there's more than one where we have a pin in double shear if you look down at pin A same diameter pin probably cut from the same stock material that pin C was there's always an engineering advantage a cost advantage an efficiency advantage if you've got a bunch of pins in a thing if they're all exactly the same it makes manufacture a lot easier you don't have people putting these things together and they have to select specific pins for each one they can just grab a pin and put it on and it makes mass production a lot easier so as we look at pin A it's got the same diameter as C did remember now that we have the member AB in compression we had 40 kilonewtons 40 kilonewtons coming from the member AB there which must be resisted by the contact with the bracket itself other things being equal we know that those two forces being supported by the bracket are being supported by the pin because of the bracket these must be 20 kilonewtons each so you can immediately tell that the shear in the pin A is going to be much less than was the pin in shear C the load is less actually sorry that we don't want 40 kilonewtons there do we we want 20 kilonewtons because we have the area that those forces being supported over two places so we have the same area path the load you can also look at it as the whole load being supported by twice the area and you get the same number either way so we have a center portion of pin C sorry pin A is in double shear stress is being felt by pin A you know why you're faster than me because you're not using your cell phone you've got a real calculator now it makes a difference because for one thing you stop accidentally dialing your brother you're actually doing a calculation so what do you get for the the shear stress experienced in pin A what do you have Speedy alright that's what I have 40.8 remember no directional component for this you're making all of these pins out of the same stock material you have to design for this shear because whatever supports that she was going to support this one more than enough but it may be cheaper for you to just make all the pins out of the same stuff even though you're well over the support criteria for this pin based on this concern over here but you have to look at each of the pins individually so that's an example then of double shear there's also a big concern in the design of what's going on here at pin B this one's a bit more complicated than the other two because there's a lot more going on here at pin B just to help draw in the perspective there the diameter of that pin is the same you can see it up here in this diagram that the pin A is sorry pin B is also 25mm in diameter but we have several other things going on we have the member CB coming down in the center then we have the Y shaped yoke of A B and then we also have the hanger for the weight itself all of those are exerting forces on that pin B at different places so we've got this center section where the load is coming in from the member BC maybe I'll draw it coming in like that remember BC is in tension so as the pin pulls down on BC BC pulls back up on pin B then we have this double yoke of the member A B so maybe I can draw that like that's a horizontal thing so I'll put those two there maybe always somewhat difficult to do things in three dimensions remember there is an angle here of that below the horizontal and then we have the the hanger itself holding 30 kN but it's doing so over two places so we might draw those like that maybe that's the best we can do with that sketch is that a bit of what everybody sees in there going on at pin B and so in each one of these interfaces between the loads we need to figure out what the shear is and you need to do each one of those to determine where the concern is we already had a limit of two megapascals on one of the pins we need to make sure that one of these places is not a greater concern just for illustration we'll take a look at this bottom one down here that end piece that has a 15 kN load there and we'll throw in the 20 kN load we're looking at not this interface but this third one third one here there must be a shear at face that withstands that load that's 20 kN this is 5 kN then should be able to do a fairly quick calculation to see that the shear must be 25 kN back there again this happens to form a 3, 4, 5 triangle the shear stress at that point is the 25 kN over the cross-sectional area that's withstanding that shear which was the same as it was for all the other pins 1 times 10 to the minus 6 meter square and so the shear stress expected there something like 51 megapascals well less than the 102 we found the first pin we looked at I think it was pin C we looked at first but as part of the engineering design you decide should we go with a smaller pin here save some on material or is it better to have all the pins be exactly the same material and just not have to worry about it mostly it would depend on whatever the structure happens to be and probably depends on how many you're making millions it's probably worth it that all the pins are the same and you just don't threaten that concern to worry about other things if you're only making if this is a custom job and every part of it is custom made then maybe each of the pins should be custom sized for a particular load they need to stand so we have one other quick type of thing we'll have to look at this on Friday problem and then we can start on to other parts of it one more type of stress to actually look at with this example you might want to if you don't want to carry your book every day you might want to at least photocopy the inside back cover so you have these numbers with you we're going to be dealing with those numbers a lot well we don't need them yet but you might want to later do the inside cover too