 sort of messed up the statics part of it. So we'll talk about this problem and then do another one that has to do with it and then see what we need for it for what's coming up. So we had this support structure of some kind and this was a 0.16 inch, right? And there were a couple loads. The top load was 75 down, 75 kip. We had some kind of load up there which was what 45 and then 30 it's a bottom in those directions, right? And you were to figure out the displacement of this bottom point I think I called it D. And the part that we hadn't gotten to that was in chapter four and really went for fair to expect for this exam. However, when you see it, I hope that kind of will kind of make sense anyway. Most of you were approaching it anyway. It's the sum of any individual displacements that there might be. The fact that these different sections are under different forces. Some parts are under compression, some are under tension. And it's the addition of all those different displacements across the thing that's going to give us the total displacement. And we figured out in the same way we would have figured out any individual one, we just then have to add it all together. So individual section, you just need to figure out what's the force in that section, what's the length of that. And we do have an area change. However, it was all the same material. Yeah, we even said it was 836 steel. So that was this simple concept, which is kind of a no-brainer and I think everybody was skirting on it anyway, is technically in our book a chapter four topic, rather than a chapter one, two, or three topic. Chapter four, yeah, chapter four topic. So that's the part that really should not have been on the exam and I had no business putting it on. So for that, I apologize for that. We're talking about a possible solution. We've already voted unanimously not to press charges with the administration, I thank you for that. And we did do that formal vote. We got the tape on, right? Now it is gone. Bob, you got a hand up? TJ, did you have a hand up that had a green plaid shirt on it? I swear. All right, but the part that where most everybody messed up anyway, and I don't know that it's very a messed up there, was simply the statics. What are the forces in each of these pieces? Because there's three different sections that have different loads in them. And it's those three sections that you're going to sum. So you need to know the force in each one of those sections. The length, that's pretty simple, that's sort of straight off of there. The area too, I just gave you that. I didn't even give you the radius and make you calculate the area, which after grading the exam, I found out I would have tricked a couple of you, because we're still working on the area of the circle. So we need to find out the loads in each one of those pieces. So to do it, let's imagine a cut here somewhere just above this section, somewhere up in here somewhere. So here's our dividing line. This is section two, because this is going to give us whatever load is in that section that I have to do a labeled one. And that will pertain all the way to the tippy top where you would see the reaction force, because it's going to have to be the same force all the way in there, because anywhere along here, none of the load below that changes. So we have the 75 down, the 45 up, 30 down. So p1 must be 75 minus 45 plus 30, which is 60. And that's the same as would be the reaction at the very top. But we don't necessarily want the reaction. We want the force in this piece to determine is it being stretched, is it being compressed, and by how much. So now we have p1, 60 kip, its length is the 12 inches. I think only one person put 12 with a little tick mark. Be careful of that kind of stuff, because then that looks like 12 to the 11th inside an equation. So don't set traps like that for yourself, or for other people who might have to look at your work, which will happen in businesses. Your work is not a secret that you keep locked up necessarily, some of you sometimes have to share. And then e, well actually e comes out of the whole summation, so we'll just do that once, which was what, 29? Yeah, 29 times 20 is 6. That's a pound per square inch. So we've got to watch the, we have chips there. But, okay, so that's, that with this e, the change in length of that first section. As we know it to be intention, we know that's going to be positive, it's going to be in the elongation. Then you have to add to that what happens here in this second section. But to do that we need to know what the force is in the second section. So now we make a cut, there's some force there, we don't know what in that second section due to this 45 load clarity down there. And those are all chips. So p2 must be, yeah I got it clearly now that we see those two things, I've got it in the wrong direction. But that's okay because we need to know what it is. And in fact in this equation then we know that's in compression. So we can either just put the minus sign in there or leave it as plus and put the minus on the force. It doesn't matter, it's going to be the same thing anyway. 15 kip, this length is also 12 inches and its area is the same, 0.09. We're sorry, 0.9, the e underneath the fill. That piece, del 2. And it's a minus because that section's in compression. And we'll then shrink a little bit. So this piece will grow a little bit, this piece will shrink a little bit, those two will sort of counteract each other. Then the last piece, imaginary cut in there to figure out the load in that last piece. And that's clearly in tension, a tension of 30. So even plus 30, its length was 16, its area is less, what was it, 0.3? Yeah, 0.3 square inches. The only thing we have to do is watch out because we have kips here, inches here, 1 kip is a thousand pounds. Kind of a mess on the board here, squeezing it all in. I think it's fair to expect you to have gotten those. And I don't know that anybody really did. This, even though it's kind of obvious when you see it, isn't it? I think, I hope, this problem is kind of half a clinker. So for those of you that wandered to end, we're trying to decide what to do about that. One possibility is do that problem again. Either give you a take home, can you do it, or we take class time to do it. Another possibility was change the points. These were 20 point questions, two 20 point questions. So for example, if you got a 35 out of 40, that's about an 88%. I can make them into two 30 point questions and leave everything else the same. Then it becomes a 55 out of 60, which is about a 92%, if I remember the numbers. So I could do that. Just 20 points on everybody's ratio, top and bottom. Drive the mat, teachers, nuts. Because what more fun is that than let it do that? That is so much fun. Notice that's off-screen. So now everybody wherever watches that video is going to go, oh, what did he do? I'd love to know. See, it's fun to write stuff like it. I mean, you go to my math 108 class, I teach once a while, they'll look at that and go, yeah, what's wrong? You guys are skim to crawl online. You can tell the absurd nature of math, but that's hard to do. So that's also one possibility. Then I don't know, maybe there's other possibilities you can come up with. And we don't have to decide right now. You can think about it. You can email each other over the weekend if you want. Say, I don't know, let's go ahead and press charges. One time in our life we have a case, we've got a nail, and I'm not going to explain the whole story to Brandon. Huh? What up? Press charges? Yeah, but we already have on tape that you're not going to. So anyway, you want to just stew over it for the weekend? Is it possible to have a take home question and then take the higher of the two? Sure. So if by chance you don't know? Sure, that's certainly a possibility. Okay, so what I can do is, is that the general consensus? I don't want to go on a couple nodding heads when others are going, uh, I'm in there. Like the higher of the two, like you'll take this one or the 20? No, no, the first problem stands because there's nothing wrong with that. Right. First problem is a legitimate problem and your score on it's legitimate. Second problem we're talking about. So I can give you another second problem that'll have to do with deformation and you can do this take home and then, uh, I'll take whichever of those is the higher. Yeah, yeah, it's gotta, it's gotta be something I can actually do. No, I do kind of like this. I feel like we're doing advanced work here this way by stealthily zipping into the other chapters. No, I'll try to keep it to chapter three. Did we get it right? No, not all the way right. And there's, there were two parts to it. There was the static part and then the mechanics and materials part and everybody messed up the static part I think. So I'm not totally without grounds. All right, so I take home and I replay and I take the higher of problem two. 20 point take home, the higher of either two. So should we, should we vote? You want to put your heads down on your desk to vote? Do we not? Something else? All right, the general, Colin, you okay with that? Okay. Uh, and what I'll do is, uh, is I'll put that in an angel over the weekend and tonight or tomorrow morning or something will come up with a problem. I'll post it there and then you guys can do it. And no reason not for Monday, I guess, because it was a, shouldn't be more than a half an hour legitimately because it was a half hour problem. Maybe a little bit more could you have take home? What are you going to post it? As soon as I get it this afternoon or tomorrow morning, but I'll send you an email that it's there. Okay, everybody can live with that. And I will need these back because I don't know what you got on the second one. So you got to give me these ones back as well. And you can either do it now or on Monday, but I will need those ones back. So glad we're still friends. All right. So let's, uh, let's see if you can actually do this. No, I don't have to instruct on it because I already did. We will need this for the rest of the business that's coming up that'll, that'll take us through the most of the rest of chapter four. Chapter four is a quick one. Not too much to it. In fact, we'll finish chapter four on Monday. But we do need this, this idea. So I'll give you another one. You can figure it out. Two meter piece there in length area, 0.001 square meter. Yep. Some other piece to it. One meter in length, but twice the area, a meter and a half of the original piece could be some kind of, uh, some kind of axial bearing shaft or some kind. So that's the piece. You've got there. And it's loaded pollutants there, engine length from, from A to D all the way from the start down to the end. So very much the same problem we just had in this class. Things always, always are statically in static equilibrium. So yeah, it's fixed at N D somehow. And what you're just doing, give or take a little bit, but one more smarter, P L over A D E. And notice we could, we could make them of different, uh, different materials, particular sections of different material wouldn't be any problem. It's all the same material though. Then, then E just comes out of that summation. Let's see the answer. The, the dynamics exam. Now it's, that one is legitimate. Double check that one. Uh, if you want to start earlier, you can. It just, I did not tell 930 how to go serve for a sentence in the math lab. I'll be back at 930 and then you can start if you want. Where's your math lab? 15 minutes. It takes me seven minutes to walk over and seven minutes to walk back for 15 minutes over there. And when I get there, there's two other tutors already on duty. So, but it's a, it's a contractual obligation I have to put in 15 hours of contact on a, a term. I do more than that in the fall. It doesn't carry over. We're the same problem. So I want to check as you go along, you're getting the same loads in each piece. Load in each of the three sections. Take the center piece. All by self. But then it's not in static equilibrium. The forces aren't balanced then. Forces are always balanced in this class. Remember, it's, it's been true. Several times last fall I told you that that's, we needed to keep that for what we're doing in this class. That everything is always in static equilibrium in this class. Several schools run the two together statics and mechanics materials in a single four-hour class. Which used to do those two three-hour classes. Gives us more time with the mistakes. It applies to the whole as well as to any part thereof. For the materials, remember there are internal forces. Those are the things we're worried about in this class because those are the things that cause the deformation. Now, the loads in the material itself, we're not, the external loads are important because they determine what the internal loads are in the material. The thing to do is make a couple cuts, a couple imaginary cuts anywhere, one at a time through those three pieces to expose those internal loads. We need to know not only how big they are but whether they're tension or compression. The size of that imaginary cut used because every part thereof is in static equilibrium. Check with each other what the loads are in these three pieces. And remember anytime there's an external load applied that's considered a new section because the internal forces are going to change on either side of that. Well, it's okay. Yeah, I got 50,000 kilonewtons there because it's lengths. It's 1.5 meters. You only have one meter now there. Yeah, but the middle piece doesn't have the 50,000 kilonewtons you just put forcing this piece to be 50,000 kilonewtons or newtons. Okay. All right, I see. All right. Now don't just label these as the, for example, don't just say del C because that doesn't make any sense. What happens to point C depends upon what's happening over here and here and here. So put it as a region. You can either do gel one. Don't just label it as a single point. And unless you're taking into account everything that's happened because all comes into play. Well, maybe not since we do consider this fixed and immovable. Then I guess by definition del D is zero because it's fixed to the rigid support. Jeff, first, at least make sure you got the internal forces the same. No sense going farther than that. Those are all wrong. We've got P1, P2 and P3 that come up with or PAB, VC and CD if you'd rather. Internal loads. Yeah, well, they're taken to account of whether it's in compression or in tension. Internal load in piece one is pretty straightforward because it's only got that 100 kilonewton load on it. Its length is two meters, a E for A36 steel. We just had that, right? Yeah, we had it in English units. We needed an SI unit, so be careful that 200 times 10 is a seven. Am I reading that number right? No, 10 to the ninth square meter. So the units work out. So it'll be a deformation in meters. It'll be positive because it gets longer. That section gets longer actually. Bobby, you got that? Then go until something changes, at least in terms of the load. We don't have any change in load until we get here to B, so we'll go into the next section. Imaginary internal cut so that we can find out what the internal load is. Must be in that direction, change in length of piece two, or if you'd rather BC, whatever notation you choose, is fine. That's the load in two, the length in two, the area of two, and they all have the same material. Though notice they don't need to. We could change that there if we wanted to. If this was a different type of material in there. We have all those numbers, as long as you're careful with the right ones. This piece is in compression, so probably convention wasn't then that's negative. All the same units. Get the right length, though, one meter. Get the right area. It checks that the units are the same as up here. The right everything in there's okay. So the center section, we have a negative in here for compression. Center section actually decreases by 375 micrometers. Like, okay, you got that? Did it all, oh, didn't add it all these up at once. And then the next piece, static equilibrium. We get for P3, 50,000. Putting all the numbers in. You've gotten, I believe, just the opposite of that, didn't you? Wait, wait, the total definition is up there. Just coincidence. No guarantee that kind of thing would happen, just the way the numbers shook out, because that's on the area of the load and the length. Since calling all right, internal forces are, and whatever these deformations are, we can use these now of what are called statically indeterminate problems. In fact, we've already looked at one, statically indeterminate problems. We've already looked at one, and we've been sort of hinting around them even since all last fall. If you remember, every time we did some kind of beam problem, simply supported beam, always put some kind of roller thing under one end. Reason being, if we didn't, we can solve that problem, do that. If we did it more like you'd imagine we actually do these type of things, and of course there's got to be some kind of load on here, whatever it is. If we did that type of thing, we have, and I guess we'd need some kind of component of a horizontal load here just to make the problem actually work out in its entirety. If we knew what these loads were, and the lengths and all those other types of things we need, we can have those rollers in there. We have a problem of how many unknowns? Four unknowns here because we've got a reaction, a full reaction at each side. We only have three equations available. That's why we had to put the roller in here because otherwise we have too many unknowns for the number of equations available, and those problems are statically indeterminate. Meaning, we cannot, using our normal statics, determine what the reactions are. But we can solve statically indeterminate problems now using the deformation. In fact, we've already done it once. We had a problem last week, I believe, where we had this beam with two different length legs, and once loaded, it had to remain horizontal. To solve that then, we used the fact that the deformation of the two legs had to be the same. So we had to take into account the area of those, the length of those, the material of those, and adjust the load accordingly so that each side deformed by the same amount. Originally, this problem was statically indeterminate. We were able to solve it using now the deformation ideas that we didn't have last fall. I'll give you a problem, a bit of a warm-up, and then we'll continue with it on. I think we've got some kind of support piece here that is actually made up of two materials inside the other. So if we look at it on end, it's only two materials. One an outer tube, and one an inner. They don't necessarily have to touch, it's just easier to draw it that way. Not considering any of the adhesives or bondings or any of those type of things now. And then we imagine it loaded with maybe there's some kind of cap here then upon which we impart a load. That cap we're considering kind of like our supports. It's rigid, undeformable, plays no piece in the problem. We're having all that, does forces, distributes the load evenly over the two things. So this problem we do in, well we do it the way we do anything else. I guess we look at free body diagram of the two objects. There's the tube and there's the rod and have the cap there because everything's got to be in static equilibrium. Creativity we'll call it P. It gets distributed partly that piece and partly onto this piece. We don't normally draw the loads as distributed, but we kind of have to just because of that tube thing there. But I guess we could reduce it to looking at the cap in most simple terms. The cap is going to be pushing on the tube and the rod and those are going to then be pushing back. So there's some load in that tube and some load taken by the rod. As the cap pushes on them they push back on the cap and we need to find out what those are. So that's PR for load on the rod is PT. In statics we only have one equation plus PR. The total load must be partly taken by the tube and partly taken by the rod. And last fall that's where we would have been stuck in that the deformation of the tube must equal that of the rod. However much the rod gets compressed the tube's got to get compressed the same amount. Otherwise some kind of gap would form in here between the cap and the tube and the rod themselves and that doesn't make any sense. Couldn't push on that and have one of them shrink farther away until it's no longer even in contact. Yep. Has to, yeah. Because otherwise if we press on this a little bit if that del wasn't the same say that the del of the rod was a little bit more. Now the rod's down here no longer in contact. Well how would that happen? There's nothing to force it down to there because so they can only they can only go to the same length. Otherwise one of them loses contact with the forces that are causing that. All right so we we have a second equation then now. So we have two equations now, two unknowns. Those equations don't involve anything more than the geometry. The area, the length, the material involves forces so we're not really introducing any new unknowns. All right so that's that's one way of looking at static interminacy and we'll do another method. Leave with me these exams if you want I need them for reference when you do the new problem. I'll post that and send an email when it's ready. Also the dynamics guys if you want to take start the test at 930 you're welcome to. Still get an hour 20 but if you want to start early you can.