 But then other than that, the usual things apply, I guess the only thing he doesn't know about is cell phone policy in the class. Oh, no, no, no, no, no, no, no, no. I understand that you're all a wired generation and that you can't stand being out of touch. So I say leave your cell phone on because that's, you know, you guys could be, well, if you ever had a date, if you ever had a date and your cell phone went off, you'd just answer it right there. You'd say, hang on, sweetie, I got another call, hey, hey babe, that's nothing to you guys. So cell phone policy in this class is leave it on. It's just that if it rings, I get to answer it and I can say whatever I want to whoever is on the phone, which means if it is your girlfriend, she won't be by the end of the call. If it's your broker, I'm going to sell at the market and unless I think it's going to be a bad day, then we're going to buy at the market. So that's the deal. Leave it on if you want. But if it goes off, I get to answer it and that includes texts. So when that little ding goes off or a text has come in and you get all red, I'll know it's you. So you might as well just hand it to me so I can answer the text. Fair enough? All right. You've taken statics. Okay. Because I know everybody else. Where are you coming from? Clarkson. Clarkson? Ah. Every year I have a rescue effort trying to save somebody from Clarkson, repackage them, and send them back out somewhere. Good. Glad to have you. Welcome. You're on Professor Vanden, by the way. My office is just right next door. So as soon as I get around to it, I'll have office hours posted there. But most everybody's learned. My door is open and I'm in there and you need help. Ask for it. If I can help you, I will. And usually you've taken advantage of this several times. So have you. Two minutes. It's usually worth it. Don't spend two hours looking at a problem. That's a complete waste of your time. And I imagine you don't like wasting time. So don't spend two hours on a problem. All you have to do is email me, call me up. Also on the course management system, I have a discussion board. If you have a question, please call me up. All you have to do is email me, call me up. Also on the course management system, I have a discussion board. If you have a homework question posted there, that's worked a couple times. It's not, it doesn't work real fast. So don't do it at 7.30 in the morning when the homework's due at 8. But if you're a couple days ahead and you've got something that's not working, put in a question. If somebody else puts in a good answer, it could be worth some extra credit points. A couple hundred thousand points. I'm getting it all out. Any questions? Anybody having trouble getting the book? Either version, this is the older version. That's fine if that's what you come across. This is the one from which I've pulled the problem of numbers in the line. So take a second just to make sure the problem numbers are the same. It's a real good book. I've used it now. This is the sixth year in a row. I have no intention of changing this. Good book. All right. Any other questions, comments, concerns? All right. This class builds directly off of what we did in statics. This, in fact, at many schools, and maybe even Clarkson is one of them, I know HVCC is one of them, where statics and this class are actually combined together into a single class, four credit hours instead of two classes of three credit hours each, the way we do it. I think it's better this way because there's just a lot of stuff to cover. This class generally is well taken by students because it has very much to do with your everyday experience and how materials are chosen for engineering. What material is chosen for a particular application? How big it is? We're going to look into why ceiling joists and I-beams and girders are like they are. You would think that there's no great point to making a girder in the shape of an I-beam, but there very much is. And we're going to see exactly why. I'm going to help you learn how to build a good deck this summer. Then after graduation, we'll all go there and sit around for a couple hours enjoying ourselves because we know that deck will hold us and we'll know why it does. We're going to look at why concrete is reinforced like it is with rebar. All of those things that you see as you go out and about every single day, we're going to look at why those things were chosen to look like they are. So it does very much come right off of our work in statics. And if you remember, there were only two things we were concerned about in statics last fall. Only two things bothered us. Nothing bothered us. We just breezed through it. Only two things that we pay attention to. Let's see who remembers what those two things were. Now that was one of our tools to do the two things we needed to do. Jake, we were very, very much concerned with the fact that the sum of the forces on any object was zero. Actually, that's not what we were concerned with. What we were concerned with was that the acceleration of any object in this class was zero. To get zero acceleration, we had to make sure the forces sum to zero. The whole point being that what we're looking at in the most part in statics and in this class are structures and we don't want structures to accelerate. It's tragically disastrous when bridges accelerate, when buildings accelerate. So we want things to stay where they are in exactly the same way they are. So that's one of the things we were very concerned with. What was the second? Who do you guys have for a teacher? I'm going to have to talk, what? Some of the moments. The second thing was that the sum of the moments had to also be zero. What that really meant to us was that the angular acceleration was then zero. Those were the two things that carried all the way through statics and was vitally important. That's still going to be true. That's why this course follows right on the heels of statics because it goes with that and immediately comes from that. Those two things are always true at all times, are important, and sacrosanct in this class. And that will continue to be the case. To do those two things, our number one tool is Brandon, the free-body diagram. We don't see, there are very, very few problems that will come through here that we won't have at least some form of a free-body diagram involved in it. What we're really going to be concerned with a lot, and it's something we hit kind of at the end, the last half of statics I guess. The first half of statics, a lot of what we were looking at was the forces at the reactions. What do we have to do to a structure as a whole to make sure that there's no acceleration in any way? So we looked at the forces at the reactions, the moments at the pin supports, and all those type of things we were looking at. Then more in the second half, maybe second, two-thirds or so, we started looking at the forces in the individual members of a structure itself. Remember we had trusses and machines and other types of structures where we started to look at what were the forces in the materials, in the members of the structure itself. That's what we're going to be concerned with much more now, because it's those forces in a particular member that will cause that member to form. If a member deforms, it may cause the structure to accelerate. It may deform so much that it actually fails. A girder can bend so much under the weight, and it's happening around here this time of year when we get this much snow and ice that roofing supports will deform so much that they'll actually fail. And then the roof accelerates. That's not how the homeowner looks at it, because they're not there watching it accelerating, remarking, oh my gosh, look, it's accelerating. Dear, come on out here and watch the barn accelerate. They usually catch it after the fact and realize that the materials deform so much. In fact, in some places this course is called something like the physics of deformable solids or something like that. We are going to look at how much material we react to the forces that are in it. We will then choose an appropriate material to limit the deformation, limit the chance of failure. We will also physically design the material, make it look like it needs to look. And this is why we have I-beams for support girders instead of just a straight bar, which I think might be the cheapest thing to do. We'll look at what the cross-sectional area and the leg is to handle these things. Alright, to warm us up a little bit, we'll look at a simple statics problem like we did, and then we'll immediately take the next step from there to what we want to do with the class, with this class. And it'll help remind us a lot of the things we've got. So imagine we've got some solid there. Was that your cell phone? How's it going? That was my favorite from our department. Well, yeah, that does remind me. If there are times you do need to take a cell phone call, or you expect one to come, just let me know. You're perfectly primed. If it goes off, sneak out and take it, stuff does happen where you do need to take phone calls. Alright, so we've got some support member there, and let's imagine we've got another support there with it that holding a weight of some kind. And remember, unless told otherwise, we assume all these connection points are smooth pin joints, meaning that no moment can be resisted between those members. They can wobble freely about those points as they wish. This member, actually we need to correct it a little bit. We'll put that in on some kind of roller support. Remember that even though that seems like a very impractical way to do things, then there'd be nothing to hold this up. If that member had any of its own weight, unless told otherwise, we'll take all of these members to be massless themselves. It's a sufficient approximation for those times when the mass of the support structure is significantly less than the mass of the load itself. We will look at the weight of the members themselves and put that into our calculations a little bit later. But for the most part right now, we'll take these to be massless and leave it at that, keep things a little bit simple. So I'll put some dimensions on this just to size things up a little bit. It's a problem to work at. So that's exactly the type of problem we did in statics last fall. We worried about certain things like can the structure, can the supports hold? In other words, we had to find out what the reaction forces were at these support points. We needed to, well that's all we did in statics. We just figured out what those supports, what those reaction forces were. In class, we can take the next step and say well can we design that support so it will indeed hold those reaction forces. Because this little turnbuffler, whatever it is, hinge of some kind needs to be sufficiently stout that whatever the forces are there will not cause this thing to fail. Can the members of the structure themselves hold? As we figure out what the forces are in the members themselves, will that fail? You can just look at this from the experience you had in statics, realize this piece up here is going to be in tension. This piece down here is going to be in compression. Can we choose a material that will be good in tension there and in compression here? If not, the piece may fail. You know from your own experience too that certain materials are better in tension than they are in compression and vice versa. Wood is an excellent example of a material that's cheap, easy to work with, real easy to come across, can look great, but it's much better in compression. It's much better as a support post than it would be in tension. You rarely find a situation where you're hanging things from wood supports. Occasionally you will find somebody will put up a ski rack in their garage or something, they'll just run a couple boards down and then hang some skis from there. That's fine, skis don't weigh a lot, so in a garage it's no big trouble. But you never want to suspend say this floor hanging from wood supports. Wood is not good in tension. Concrete is not good in tension. It's great in compression, not good in tension. We'll come up with ways to handle that later in this class. So once we start looking at these questions, this can the members hold, can the supports hold, brings other questions to it to determine if they can hold the loads. We have to decide what materials we'll use, also we have to decide how we're going to shape those materials, what shapes work better than others. If you were building a deck this summer and you knew you could support it with wood, and you knew that wood of a certain size might work, you shouldn't be a big surprise that a 4x4 piece of wood is going to work better than a 1x16 piece of wood. They have the same area holding the load, but it shouldn't be a big surprise to you and your practical knowledge of how things work. If you don't want to support a deck with a 1x16 board, even though that has the same area as a 4x4 does, you see decks all over the place that are supported with 4x4s, maybe even 6x6s. We're going to find out why that matters. We're going to look specifically at this type of thing. Alright, so let's look a little more quickly into this particular piece itself, and then we're going to take the next step, what we need to do to make sure that we understand all, or start understanding all of what goes into these pieces. So just for convenience, I'll label those corners. And if we want to find out what the forces are in those two members, the members AB and the members BC, we can look at a free body diagram of the joint B. That will involve both the two members, and will allow us to find out what the forces are in that member itself. So we've got the load, 30 kilonewtons there. Hopefully it's capably supported by the fact that some of the force will go into the member AB, some of the force will go into the member, what I call it, CB. Hopefully it'll take you about three seconds just to make sure, or just to come up with what those two forces are. Kind of a warm-up shake, the sugar-plum cookies out of your head. You have to deal with the reality now that the peppermint molters aren't for sale at Starbucks anymore. Hang here when that time of year passes. So take about two seconds, figure out what those two forces are, the forces in those members themselves. It's about as simple as a problem can be in this class. We're using the fact that kind of went quickly by here, but these are the type of things you're going to need to remember from statics, that all of these are two-force members since we are assuming them to be massless. We're going to make some two-force members and all two-force members have forces directed right along their own length and direction. Frank, you already got it? Other students will be a little friendlier to you than they have been to other students in the past. We usually send them out in tears after about four weeks. We just had it. We put it in the off-run. We tried to get it a DJ. Brandon, you've seen a couple of Clarkson students come and go. Pat, you got an answer? Be the nicest guy. Turn back to Frank. Introduce yourself. Check your numbers. What about me? Do you guys agree? We should all agree. We're all doing the same problem. Do you guys agree? The units. Don't forget your units. You're not doing the same units. We can't even check to see if... In fact, was anybody listening last night to a furry-owned companion? Not last night, the night before. There was a joke about units on the furry-owned companion. A furry-owned companion? They had an adorondack on you. Well, never mind. It's up to myself. What's the best joke ever? What? What was a little mind? I always get the best joke. Did you get the student? Well, I'll have to see that. All right. You guys didn't agree? Did you argue it out some? We're still debating. All right. Does anybody see the... There's a simplicity to this problem that if you catch, can very, very quickly solve the problem for you. This triangle is what we call a 3, 4, 5 triangle in that this side and that side are multiples of 3 and 4 using Pythagorean's theorem that makes the hypotenuse 5, the 3, 4, 5 triangle, which means also when we form a triangle out of these three, as we do when we sum the forces and they should sum to zero. That means any time you're summing forces that should sum to zero, if you put them together, that will form a closed figure. So if we have a 30 kilonewton force there, some unknown force there, and some unknown force there, because the angles here are the same as the angles here, then this is also a 3, 4, 5 triangle. So you can immediately then finish what the forces are. 30, 40. Most of you, though, I figure found what that angle was and then just use the sign, which is fine, except that we had a little conflict here between Pat and Frank. Who was right and who was wrong? He was right. Ah, wow, Pat. Your teacher in statics must have sucked. All right, so now we know that in the horizontal member, we have 40 kilonewtons. Is it compression or tension? Just by observation you can figure it's probably compression. As this weight pulls down, it's going to pull in on this piece, but you can also tell from the drawing that we had to put. Actually, we've got some backwards on this drawing, don't we? Because to make it work, we really should have had that one going up, that one going in. Sure, everything balances. If we pick wrong, there's no sweat. We can just reverse it. As long as it all works out in the end, we're okay. If this member is pushing on this joint, then this joint is pushing back on the member, and that indeed is compression. If this member is pulling on that point, then the point is pulling back on that member, and that's indeed tension. So that'd be just like we expected. If you did that using the directions we had initially, you should have come up with minus signs here. So that's kind of self-correcting. Alright, so what we're going to look at now for the rest of this term in various guides is what's actually going on inside the material itself. So we'll kind of blow it up a little bit, look at some cross-section here that we expose, because that force of, in that case, 50 kilonewton's in tension is being transmitted through this material all over that whole face that we've imaginarily exposed. We don't really expose it because that makes the material, the member itself fail, but we know that over that area, we must be able to sustain a fully 50 kilonewton load of tension if that material, if that member can't support the 50 kilonewton's, it's going to fail and the structure will accelerate. So the things that are concerned to us now are of course not just the load itself, we're not going to draw the distributed load, we'll draw just the single load, some load in particular that that member needs to support, and it needs to be supported by a certain cross-sectional area. That should make sense to you. You know yourselves that if you have a big load on a little tiny piece of material, you've got a good chance of failing. If you put in more material, you have a better chance of keeping the material from failing. That's just common sense to you. Remember though that this is the cross-sectional area of the structural member itself that we're using there for ourselves in a way to look at both the load and the area supporting that load. You might remember this from the very tail end of physics one when we talked about it. We'll define the stress. This is the Greek lower case letter sigma. It's our symbol for stress. It is defined as the force we're supporting divided by the area, the cross-sectional area of the member that's supporting it. That's our very first marker of how the material itself is going to support this. Couple things of particular importance to us here is since we're talking about a force that is perpendicular to the area supporting it, this is specifically then the normal stress. Remember normal means to us in physics and engineering, perpendicular. This is the force perpendicular to the cross-sectional area of the material supporting it. It's not quite true that the load is evenly distributed over this piece, but we're going to assume for our purposes that it is uniformly distributed across this cross-section sort of like I drew there without any trouble of that assumption, that simplification for the most part. It is very close to a uniform distribution across there. It's also true that the distribution of the force changes as we go along the piece itself. It's a little bit more parabolic at the ends and a little bit flatter in the center, but we'll be able to assume that it's a uniform distribution all the way across there. All right, let's see. Oh, we've got to check units. On stress square meters, the usual one we'll use in SI units. It's a good chance here for us to honor a dead white male German physicist as we always do with units. We'll define a Newton per square meter as a passed out. Is that pressure of course over the area? Yeah, that's very, very much. You can look at it as a pressure on that cross-sectional area because that's just how pressure was when you looked at that in physics too. You looked at fluid statics. The force perpendicular to the area supporting it. So this is a very same unit, a very same idea. Once we have that kind of number, we can then also make sure that we choose a material that can withstand that kind of stress. It's very easy to test materials to see if the material can withstand a certain type of stress, a certain amount of stress, and then we can start selectively choosing our materials for construction of a structure so that the materials chosen adequately support a stress of that size. You can imagine that if we knew we had a force of 50 kiloohms to support, we can't go to a catalog and just look up and buy a particular, select a particular material that supports 50 kiloohms because that has nothing to do with the size of the structure itself. So they're not sold in terms of how much force they can hold. They're sold in terms of how much stress they can hold. So if we've got a certain force we need, then we can also pick an appropriate area to make sure we can support that force. So we'll see that the materials we look at, one of the things we'll look at is how much stress can that material support, and then we make sure we have enough area of that to support the particular force that we're looking at. All right. That's our first of just what we're going to do with these supporting materials. We can also do the same thing with the compressive piece, a slightly different force, and we'll be able to start looking at these in some detail. There are other kinds of stresses that we have to withstand. You can imagine that where we pin these pieces together, the pin also has to be able to withstand the forces on it. Not just the members themselves, but the actual connections themselves also have to withstand the forces on them. If we look at one of the pins, we wouldn't want to just stick in some... You don't want to bring those two things together, line up the two holes, and stick a Q-tip in there and hope it's going to hold, because it won't. There's other forces on it. We can look at it something like this. We know that this piece is in tension, so that there's certain forces in one area where it's connected, and then as the other piece comes in here and is on the other side of the pin, there might be forces like that. The example of the type of thing that can happen with the pin itself as it's connected through there. This exerts different forces on it than the type of forces we've got there. If we imagine an imaginary cut through there and look at just that piece of the pin itself where we made this imaginary cut, we've now exposed again this cross-sectional area and whatever the pin is, it still needs to support this load here, whatever that might be, maybe it's the 50 kN, maybe it's the 40 kN. But to resist that load imposed upon it, there must be the ability of the material itself to exert a force in the opposite direction across that face that we've exposed. Do you remember in statics what we called those transverse forces across a material? We call that shear. So we define then another type of stress the materials need to support. The member itself, along the bulk of its length, only needs to maintain a, to resist a normal stress. But in the pins, we have this transverse stress. So we define then shear stress, similar case tau, and that's defined as the shear stress over the area that's supporting it. I think that also makes sense. If you have a very, very small pin with a very, very small area supporting that particular load, there's going to be a lot of stress in that. If there's too much stress on the material, the pin itself will fail. So to make sure that we lower the stress, we can increase the area for a particular load. So that just makes common sense to the things you've done before. And as before, we'll assume, unless stated otherwise, which we will definitely have to do with the shear later. But for now, we'll assume it's a uniform distribution across that face. So this is actually the average shear stress across that face. But we can pick materials that will also maintain those particular stresses. All right. So what I'd like us to do now is we're going to return to this problem but in much greater detail. What we're going to look at now is this structure actually has a much more deliberately engineered piece. Same dimensions, same load. We're still assuming that these materials are essentially massless compared to the load imposed upon them. And so we want to look at some of the particular pieces. For example, let's look at BC itself, the diagonal piece across there. We know that all the way along BC, at every point along BC, we need to withstand that 50 kilonewton load intention. The trouble is, as we go along that piece BC at different places, the area changes. There's more material at certain places than there is at others just because of the shape of the piece itself that we've designed to actually be able to put it together. So up at the top at sea, we've kind of got this flattened end so we can pin those two pieces together. We can pin the top of the diagonal piece together with its support against the wall and then drive a pin through there. Then down at the bottom, we've got not only the bottom of BC coming together, that NB, that diagonal NB, that's got to fit in with whatever it is that's supporting the load itself, that attaches the load, and its connection with the cross-member piece itself. So we have a lot of different things to look at. A lot more detail we're concerned with that we weren't concerned with before because things are clearly very different at different places along this piece. So if we're interested in something like where the greatest stress is in that piece BC as we look at the stresses in that part BC, the load's 50 kilonewtons everywhere along there. That entire piece is in 50 kilonewtons' attention. That's set. So as we're looking at the stresses, that 50 kilonewtons is the same anywhere along there. Obviously the stresses will be the greatest wherever the area is minimum. If the area is at a minimum, the stress will be at a maximum. So nothing else. That should be our first point of concern in the design of this diagonal component of this very simple structure. So if we look at that piece BC, the diagonal goes across there, we can see that, let's see, up at the top at C, it's a flat end of, looks like it's 40 by 20 millimeters where it's flattened up there at the end. That's more area for supporting this load than it is anywhere along the middle where it's just a simple round member of 20 millimeters in diameter. We've got a 20 millimeter diameter somewhere in the center, and that needs to support this 50 kilonewtons load. So by quick initial inspection, that would be our first place of concern. Just because that's the smallest area supporting this 50 kilonewtons load. You need to break here and tell me what the area is for a round member like that diagonal piece is. Not all the members are round, but this one is. You can tell from the drawing that it is such. So what's the area of a round piece that's 20 millimeters in diameter? Don't forget that the units we want to get down to, of course, are just to make sure that everybody knows how to calculate the area of a round. You've already talked to him. Brandon's next. Brandon's behind you. He's not ready yet. That's Colin. Colin, you ready? He's probably got the number. He's just not ready to talk to him about it yet. It's a little early in the morning for social skills. You do? From where? Is that why you're sitting so far apart? Did you know he was going to be in here? Same year? And same prom date? No? Didn't get that bad? How well did you know each other? Went to what? Man. That sucks. So much for being a sneak into the class and disappear. Can you tell us some good stuff about Colin? The important things in life. Got it? We can start there. 3, 1, 4 meters squared. Everybody else get that? I didn't. We're sitting in millimeters squared. Let's say 10 squared is 100. So it's got to be 314 millimeters squared. We convert to meters squared. This is where we grew up. Be very, very careful with this. Because what we're going to find, we're going to find out in just seconds here, some of these numbers we're working with are very, very large and very, very small. So we're not always going to want to stay in the same specific units. We're going to want to use the scientific prefixes that we've somewhat become used to. How many millimeters in a meter? One meter is 1,000 millimeters. But we're working with millimeters squared. So you need to square that number. Which makes this area, let's see, which makes this 50,000 newtons. It was 50 kilonewtons. This area will be 314 times 10 to the minus 6th meter squared. But to really watch, you're going to need to know how to convert millimeters and meters back and forth, especially millimeters, not just squared, not just cubed, but we're going to find millimeters to the fourth, to meters to the fourth and back and forth. So we get the units just right. If you miss these units by a simple factor of 1,000 or so, you can very easily over-engineer design, costing your company a ton of money that's wasted, or under-engineer design, which means the design fails. What's that come out to be then? 59 times 10 to the sixth. It's fairly standard in engineering, not so much so now that it used to be that we try to keep a multiple of 3 on the exponent, because those are the most common prefixes, the millikilo, mega, giga, all those things. So we're expecting a stress along the major length of that diagonal member of around 160 mega-pascals. As a starter point then, we can go into a reference material of some kind, either material put out by a steel manufacturer or we've got tables of this in the back of the book, and we can pick a material that can withstand 160 mega-pascals as a starter point. We're going to have to be more careful with this than that, but as a starter point, that will be okay. Tomorrow morning we start again and we'll look at some of the other spots on this same structure in terms of the stresses involved in the pair.