 And the questions, well let's say we're doing, oh yeah, we haven't had a class since the last test. So there could be questions, are there? Yeah, tell me the questions. We did. Remember, we were talking about statically indeterminate problems. Nobody corrected the spelling on the board, so you can go back to the video and see if you can catch it. You can catch it until I look to the video. But that's what we were looking at. We're looking at those questions where the static equilibrium is incomplete in terms of solving our problems. As I told you all last fall, all the statics we were doing last fall was to lean to stuff we needed to do here to, it was very much a part of what we're going to do here. Typically in our two-dimensional problems, this is three equations. And most of our problems are such. For a full three-dimensional problems, that's a full six equations then. But for most of what we're doing is three equations. So statically indeterminate problems are those that have more than three unknowns because usually because there are more supports than are necessary. A simple example that we've looked at before is that type of problem where that beam has more reactions than we could solve for with the three equations. Another one, and we'll look at it now, is the type of problem where we have some kind of support structure between rigid immovable supports. Remember that's what we take this type of thing to be. So if we load this somewhere with some load, this to us is statically indeterminate because we can certainly draw a free body diagram of it. This is the reaction down at one end. This is the reaction of the other due to the presence of that load. We can't solve that with our statics equations. In fact, the only equation we can come up with is that at least as drawn, we can't do any more than that statically. So we still have to satisfy static equilibrium, but we also have to come up with some other things that will help us solve that problem. Oh, by the way, I didn't see any questions over the weekend on the take home, so is that going okay or we just started it? No, absolutely not, it's about five o'clock, right? Yeah. So that's the values of 10 to the sixth, or did that involve the ninth, seventh, and third? Did that change the second and third? Well, I didn't, I just said it was steel and aluminum. I didn't say what kind. So I double checked, you know, make sure because anyone that has typos, so that those values use those values as there. Jake, you say that every time. Yeah, but quite the story of my life, is that, okay, no, that makes it easy. All right, so the way this simple type of problem works, well, you probably already have an idea of how we do this one. Same picture, just moved over a little bit, so I have a little room. We'll even allow the possibility that the load is not necessarily centered. The guiding or overriding equations, of course, are the static equilibrium ones. We just showed in what they might be, that the reactions must equal the load, that the total deformation of the material, remember that's the new thing we brought in this term, that we didn't have all the other, all through fall. This is the fact that these materials are elastic and nature intended to form. This allows us the extra equation we need and we can then solve it. What should the deformation be for a problem like this one? Finish with here. What's for this type of problem, this type of structure? This could be a building structure with a floor right here that's loaded and then we look at it as this type of thing. The total deformation allowable in this support structure is what? Well, even more specific than that. Huh? These are both rigid and immovable. Therefore, the total deformation must be zero. A little bit of stretch there, a little bit of compression there because of the type of force that I've shown there. But total allowable between two rigid supports is zero. And then that allows us to finish the problem since we'll have two deformations. There'll be some load in the first one. Typically, we take these to have the same material and we'll have the same type of thing in the second one. This drawing looks like they're in the same area. So it makes the two of those very straight forward in terms of how to solve those. Well, we've already looked at that type of thing. Using the constraints on the deformation itself as our extra equation. So we'll look at a new, another way to do this, another possibility for this type of problem to be solved and that's called superposition. Superposition is those times when we take two solutions, neither one of which completely solves the problem, lay them over each other, superimpose them on each other and the two solutions that were incomplete individually give us a complete solution themselves. So we'll do this within the structure of a problem very much like the ones we've seen before. Some kind of structural support of various means, different areas, take that to be 150 millimeters. So is the bottom, but midway through we put, midway through we have a change in area, midway through the lower part we'll have a load. Actually, we do in the upper part too. No way to second your, oh wait, wait, I'm sorry, it's drawing is different. That's why you should take notes and chalk for equal sections all 150 millimeters. You take notes and chalk, it's easy to fix. So each of these midpoints we have a load of 250 to 400. This is statically indeterminate because of the two rigid supports on either side. There's no way using our static equilibrium that we can find what the reactions are, find the reactions. There's two, there's one at the upper and one at the lower part we've got to find the two. So the superposition we're going to use is like this. We're going to take this to be the equivalent of two separate problems. Imagine that the bottom end is free and subject to a force there, loads being imposed. Hopefully you recognize me, it's still statically indeterminate but it allows us to make the first step in terms of the solution we want to find. Again, the free end, free of this support but free of the, so we'll just look at the two loads on this piece, pause it to deform. Sum amount, that's an incomplete solution of course because it's missing this reaction that would keep it from deforming but it's a partial solution that we can solve. That we can solve because we solved that weeks ago when we first learned about strain. We're going to add to it with our magic plus sign of superposition another partial solution that we can complete and the two together will give us this reaction and be equivalent to the original statically indeterminate solution we were looking for. This time we imagine given this, what force would be required its original length. Obviously, del L equals zero so that they cancel each other out when we add those two together. We've got the two loads. We've got no change in length of the strut and we've got the reaction force in there and so we do have this solution with two partial solutions that we can manage. That superposition, this is the superposition move right here that gives us a solution that we couldn't have found normally. All right, so let's work out the details. We've got some numbers here. So first thing we're going to find is what is this in the first part of the solution and then what is that in the second part of the solution. Probably done before. We've got an axially loaded material and because of the area changes, and by the way we're assuming this is all one material, but there's two different loads, two different areas, which breaks this whole thing into four pieces. Remember any time there's a change in load or a change in area, we need to reconsider that as a subpiece to figure out what are the loads in each of the subsections. Remember we're doing this middle solution right now. We can figure out the loads in the different sections by making imaginary cuts through the material. For example, there's a cut through section four. No loads on it. So we know that in this bottom section, there's neither tension nor compression. It's unloaded. Remember we're doing this second solution here that we can do with the reaction force from the bottom removed. Then we go up, make an imaginary cut through section three. We know that at this midpoint, there's a downward force. What was it? 600 going to do this? Is that the one there? Just because that allows me to expose the interior force and force in the material itself. And we can tell that section three then is in a load of, so you take a second, figure out the load in sections two and in one. Now it goes, you learn to get rid of another student. You're right. Nobody's going to RFI in the fall or a couple of you guys are going back to Clarkson maybe. Yeah, I guess we're not going to award a single one of the scholarships this year. Good. I don't have to go over the words. It's too bad. You guys are going to Clarkson fresh $10,000 scholarship. Even if you weren't any good. One year we didn't have anybody. We thought it was any good. They called us all something to give it to them. Anyway, somebody's got to have it. All right. You're looking for the other pieces by making again, imaginary cuts through the material with the loads as imposed. Six hundred kilobitons each other. Since it's Monday morning, you're not in the mood of talking to anybody, not even the mood of talking to me. Barely in the mood of listening. I mean, myself talking to me. P2, 600 kilobitons obviously. I hope it's obvious when you say that sometimes, but that's the only load it needs to oppose. Should we then just take two and three is as one section and not worry about it. Different areas. Different areas mean different responses to those loads. That's one of the things we've been looking at as we go along. And then obviously again, this has got to be 900 kilobyte. All right. So we can use the business we've done before. Remember, we're looking for del l. That's going to be the sum of the individual dels for each of the four sections. Actually, only each of the three sections because for this intermediate incomplete solution, this bottom section four is unloaded minus then for each of the three loaded pick them all to be the same material. So think you're that out real quick. Remember, each of these sections is 150 milliliters, a millimetres, 150 milliliters. And leave it in terms of E because I'm not going to tell you yet what the material is. Just that it's all one material. It's all, sorry for that. So when you do this, you should have some number over here. So I need that sum number. And what you guys do this weekend, they have been partying community college. We don't have fraternities so it can't go out being stupid. You have to go out and be stupid on your own. If you were, just get to pass on the way in. You could have gotten a nice cup of coffee underneath me. It's not difficult. Some of the places, I know Cumberland's farms. I think they even have super jolt coffee and super high cat and got to drink it within a few certain amount of time, otherwise it'll burn through your mug out here. So get in your stomach, let it burn through that first. Do you guys have fraternities up at the garden? RBI? For the total deformation of this intermediate solution as if the lower restraint is lifted, a solution we can do, then we'll superimpose it over a solution with the reactions put back in. We're all doing the same problem. So when you get something, check it. Watch your units. We've got kilonewtons, millimeters, and millimeters squared all mixed in there. Units of course should come out to be something like meters since this is a distance that we're looking for. It's the amount of deformation of the material itself. What I got, let's see, double check your numbers. Watch your powers of e. I'm being kind of mysterious by not telling you what the material is, so there's no value for e yet. So just all of this will accumulate to some number over e. And should have units, I'll cut by, should have units of meters all told together, including the units of Pascal's or mega Pascal's on the bottom. That's not the norm. Like that. Right digits. Pat, dang. Yeah, we've got each of the p's. Got all the p's, the l's, and the a's. Make sure you get the right ones for the right spots. Yeah, we have area. I think one of them was 400 square millimeters. A millimeter's on the bottom, meters on the top, but we're trying to convert square millimeters, not millimeters, so we need to square the unit for the conversion factor as well. Then we can cancel the millimeters squared over millimeters squared, and I left with meters squared. So I think that was what the 3 and 4 was the big one. That helped Pat. Come on, you guys have to get it. I don't have any room on the board to do it. I'm stranded. Easy, one of the easiest things to do is get them in the base units as you go. For example, p1 would be 900,000 newtons. So you make the conversion like that on the fly. L1 is 150 times 10 to the minus 3 meters. We had it in millimeters. A, we can make the conversion there. That's going to be what? 400 times 10 to the minus 6, because we had 10 to the minus 3 squared on the bottom. So just put that in. That was 250, though, not 400. Just put it in like that. So we have newtons per square meter. E will be in newtons per square meter, but it's on the bottom, so those will cancel, leaving just meter. So there's the first of the factors with i equal to 1. And across that's where it comes just from the little piece. No, the length is the length of each section. This force, for example, p1, where we made some imaginary cut somewhere through 1 here, that imaginary cut could come anywhere in that section. So that whole section, 1, has a load of 900 kilonewtons in tension. Its length is 150 millimeters, and its area was the 250 millimeter square. That's why I numbered these sections this way, because nothing changes in this first 150 millimeter. Nothing changes in the second one. There's a change between the two, but once we're in this section, it's all the same all the way down. Then we get an area change. Then we get a load change. No, Jay? One by that? Well, yeah, that doesn't matter. I know, but when I have sections for 3, I figured I'd take the whole length from 3 to 4. No, there's no load on section 4. So don't you just ignore the length? Yeah. Remember when we were looking at the strain over the original length, but it's the loaded length? There's no load on this bottom section. Why should it deform? Because it's attached to the... So what? Cut it off, and there's no difference. We cut it off and throw it away, and there'd be no difference in terms of the results. The load is up here. Down here, there's no load. There's not going to be any elongation. This doesn't come into it at all. So if that was running all the way through, it would shake the load up? If this load was down here, yeah. But that would have made this little drawing here different. In fact, then we would have had just one section 3 to 4. Anytime there's a change in area or a change in load, a change by section count. Now what do you get, Frank? There, that's the number. You got good power down now? Yeah, that looks good. What do you have calling it? It should be overeat, and we're eating ease on the bottom there. So section 2, this should become 600. The length stays the same. The area stays the same. Section 3, the load stays at the 602 hat. The length stays the same, but now the area goes up to 400. And then section 4, we don't have to do it all because it's unloaded. Pat, what do you have? TJ, he's got it. Some of the unit's out. We've got 10 to the third here, and 10 to the minus third there. That'll help a little bit. Make sure that you have us work on one of them, but then be careful you get the right numbers in the right place later. Got a duty? Same number to everybody else? I think most of us are there. What are we looking at? 1.125 times 10 to the ninth. Get that number. It's got to be just a matter of calculator or something as long as you wrote each of the pieces down in the right place. Be careful it's really easy to get going too fast here and overwrite some of this stuff. Tubi, you've got it. Colin, you're okay. Jaycee, you got that? Now that I wrote it down, you got it? TJ, you got it? Pat, you got it? Frank, you okay? Finally, you got it? Bobby? Okay. So that's the number we need because remember, it's that number that we need to recover with the second part of the solution. Now we take off those two loads, find out what force would be required to push the piece back to its original length to recover that to do this business here. Then we've already done that type of problem. Easier said than done. How do we do that? We do it the same way. Again, break it into two sections where the doesn't change and no load is added or subtracted and that simply breaks it into two sections in and we're looking for whatever change there is in those two sections is equal to the del R that we're trying to recover which we already know because it's got to be equal and opposite to del L. I'm going to put it on that. Or do you need to come up with what P2R? Sorry, P1 and P2R. Just add them together. We set that equal to this. We can solve for Rb then. A1, L2L, A2 pretty straightforward, I hope. But what's P1 and P2? Remember, we're doing a different solution. It's not the same P1 and P2 that we had with this solution. We're now adding the second part to it. Now to A2, they're known. You need to come up with P1 and P2. Put those in there, add them together. We'll have over here a function of E. Remember, I haven't told you what that is. Some function of Rb. So we need to come up with, again, just that number that goes up there. Come up with is what's P1 and P2 that go in there. There'll be functions of Rb of what we're trying to find. There'll be a reaction because of the immovable support at the bottom. Notice in this drawing, I do not have these loads. It's simply what load is required, what reaction is required to recover what those loads originally caused, unloaded other than this reaction. Dex numbers than we had over here. We have a different problem we're doing now, but we're going to add the two of them together. If we kept those loads on and added it to here, then we have twice those loads. Got something to be thinking about it and not. If we want to help 2a to be right off the picture, what are P1 and P2 for this particular problem? What? There are Rb. Rb is the only load on there. So internally, that's going to be the only load on either one of the sections. So Rb, we don't know what that is yet, but that's why this is going to be, this answer is going to be a function of Rb because then we can solve for it. All I need for me is that number that's up there. We do know it's going to be a contraction. Rb, we do it at Rb. We don't know what that is yet. So the second number is going to be a function of Rb over E. I mean whatever number is left there. Got it? I think that's it. Yeah. You have an extra power of 10 from what I did. Couple of people are coming up with it. A one. Yeah? Oh, that's so exciting. Oh, yeah, yeah. On American Idol, when they drag out the announcement so they can play a poignant music. 1950, minus 1950. What? That's nothing more than L1, A1, L2, A2. Let's see. Oh, I have a minus in here, so I have a minus, a minus. Oh, no, we'll have a minus on there. Notice, except what you need. What does this tell you in terms of you being an engineer on this project? It doesn't matter what the material is. The reaction is going to be the same when we have a problem. I don't have it up there anymore between two rigid supports. Any material is going to react in the same way in this problem. Now you can solve for RB. The reaction was at the top as well. So do that and we're done. It's your place now. Somewhere, I was going to kill a Newton's probably, because that's all the loads are. Forgetting anything way out of that. Write down less numbers. Colin, got something? Well, let's start with RB, because if you don't have RB you don't want to get in RA. Do you agree? TJ's got the same, TJ's got the same. Frank's got the same, 177 kilo Newton's. Free body on the whole piece as is the other end. We take this problem. We now know this to be in kilo Newtons. This is 600 in the opposite direction. That's 300 in the opposite direction. And you know all those forces must sum to zero. So you can find RA. What's it mean that RA is up? That's a different type of attachment. If RA was down, you could just set the strut against the immovable support and that's all you need. But in this case, RA needs to pull on the piece. So you might have to thread it in or bolt it in or weld it. What if that immovable support wasn't there and everything else was the same? What would the piece do? Yeah, but to form how? Because there's more load going down, it would shrink. So the purpose of the piece RA is to hold it up there, which I guess is kind of how you'd expect it if this was a roof and this was a basement and then this was some intermediate floor. I guess it'd be the two spots there. Okay, take away this support and leave that one. Take away this one and leave the one down here. Yeah. Or if we had RB down, like if we had RB down, then we'd have a negative sign in here. So these forces would be up? Yeah, you have to turn all of it over or none of it.