 I changed the format and then I have to upload. So, all right, Colin from, oh you gave me that homework, okay, homework for that session. All right, we were doing superposition to find the beam deflection and we're gonna use that again now to determine the reactions for over-contact over-constrained beams. What over-constrained means to us is that those are beams that are statically indeterminate, the type of beams where there's actually more support than is needed for the minimum static equilibrium we have to always have in this class. For example, this type of beam has always been statically determined. If we knew its length and we knew what this loading happened to be, then we could determine what the reactions at the wall were. It was sufficient to use our static equilibrium equations and we could figure out all reactions. Well, there's only two reactions. There's the reactions at the wall of the shear and the moment that go with an embedded support like that. However, if we have the situation like this, that beam is over-constrained. There's more support than it really needs. We never took this possibility of this beam actually bending in statics and we never did until just this last month of the course. But the reality of it of course, and especially in this course, which is occasionally called something like the strength of deformable solids or something like that, which is what really happens in life. If we didn't support it here, then this whole beam would sag down. And if that's not allowable, which it wouldn't be, for certain buildings in the life, we've got to support it there in real life. Maybe not even a roller, maybe even a pin support, or maybe not even that. It could be that you want to embed it at both ends. And that's an even more severely over-constrained situation and beyond the capability of us to find those supports based just on our static equilibrium. The sum of the forces and the sum of the moments. But if you're building something, that's probably how you do it. The situation is that if we did it for just the cantilever beam and then actually supported it that way, then it's sort of like a safety factor building. It's now gotten more support than the minimum we require back in statics. Back when you were engineering students, only this high last fall. Just a little baby engineering students. However, there are still, even though this has more support than it needs, there's still troubles with this. If this is loaded somehow, it's going to deform and that deformation, even though it's only a minor sag, could cause undue damage or stresses or even separation at those points. Even if the beam itself can withstand the bending, it doesn't mean that those supports can. So there's troubles with over-constraining that wouldn't necessarily be apparent. So we'll look at a fairly simple technique to determine what the reactions are for this over-constrained load, even though our static equilibrium equations won't quite suffice. And we'll use superposition to do it. So we'll take that beam and break it into two cases that we add together because individually, we can figure out what's going on with these. So we have the first situation that we can handle and by can handle it, I mean that we can either come up with the reactions or we can use this new beam deflection stuff we're working on to help us with these because this kind of loading is in the tables and we know that deflection will do something like that. The slope will stay zero at the wall because of the embedded nature of the support. That's why we don't pin those kinds of things there because then we have under-constrained situation. And we know that there'll be some deflection here at the end for that kind of loading. So that's our first phase of the superposition solution that we're gonna use. Then we're going to add to it another solution that we can do and that's the case of a single point load out here at the end that will restore the deflection we suffered in the first one. We have to size up P so it's big enough so that its deflection upward is the same as this deflection downward we suffered in the same one and that will return us then to this situation of an undeflected beam and we'll have then the reaction here which will be P and then once we have that, we can use the statics to determine the other reactions at the wall if we needed those. That's in the book, in the table of already done solutions, but is this one? So opening up your book, go to authentic C and double check because it's not that we can't do that, we could do that, we could figure out the moment as a function of X and then integrate back to the deflection. But it'd be nice if it's already done for us. Go to authentic C, notice that page of simply supported and then another page of cantilever beams, right? And is this solution in there? It's in there if we flip it upside down which is no big deal. We just have to turn it over, make sure we handle minus signs right and we're gonna be okay. So we'll call this one one and this one two. So for the deflection down of one, the maximum deflection, notice one of the columns given there is V max, I think. Now that I don't have my book, so Pat, you have to take over? Yeah, there's a V max, what is it? Negative P L cubed. Negative P L cubed. Over three E I. Over three E I. Remember E I is the type of thing you get if the beam is already specified, you have some in stock or you have others you know you can purchase or experiences told you that's what you want, whatever the case may be. Now hang on, that's not for this beam, is it? Or V max? Yeah, that's for the lower one. Come on here, work with me Pat. I'll have to charge you when I'm on you by the books if you're not here, just help support me there. So what is V max for the uniformly loaded cantilever beam, do we, hang on. We trust Pat to do this right this time? At least I don't. V W L over eight E I. Okay, we need that then to be equal to the maximum deflection for the second model where P is an unknown and we'll be able to set them equal and opposite and then solve for P. So what's the loading on that one? The maximum deflection on that one. That was the PO cube that were what? Three E I. Three E I. Now it's negative but we're flipping it over and needed to go up so we'll make it positive. It's negative in the book, we're flipping it over so we'll make it positive and then we want those two then of course to be equal. And then that'll be our method for solving for the reaction P. Now notice what happened in every single one of the other over constrained problems we did. We started with a couple of these type of loadings. Remember there was some vertical load on here and it was over constrained at the ends because our original, the loading we started with the class width was strict axial loading. Then we looked at the strain that caused. Then we looked at some over constrained torsional loads of some kind. When we were looking at our first opportunity to look at those things where the ends couldn't move at all but the shaft in between was being twisted somehow for those. As well as for this, notice that it's the same for any beam, any material we choose. These solutions in a sense are universal other than of course it depends upon the loading and the length of the beam that will cancel out but the EI itself cancels out, which is kind of interesting. And then you get simply three eighths WL is the reaction of this over constrained reaction that we have at that end support. What happened to the negative? Oh sorry, what we need to do is we need to add these together to equal zero and then when we solve for them, then we don't have the negative. I missed this step, v1 plus v2 equals zero or v1 equals minus v2 because one's down, one's up. So that'll take care of the negative. And what's this term WL? Put that into words, what that is. Well not just that, but what are the units on W? Oh sure. Yeah, units might be something like kilonewtons per meter. Units on L might be something like meters. That's the WL is this total load. Remember if we take all of that out and replace it with a single force, it would be equivalent to the area there. So three eighths of the total load will be at P. And then we can use the two of those to go back to figure out what the shear is at the wall. I know that there's a total of omega L there. There's three eighths, not omega WL. And then we can figure out how much shear the wall needs to supply. And then also how much moment it must supply. We can just finish those with straight stat, just summing the forces and summing the moments. In fact, I don't even think that I didn't even do it. So you can do it if you want, just not now. Actually, yeah, V's pretty obvious. The moment, place it with that, the total load, you know it'd be at the center. So you'd have V by then. And well, V is gonna be what, five eighths, WL, right? Figure out the moment at the wall. We can use the same general method to also find out the moment with another superimposed solution problem, two superimposed solutions that will allow us to find the moment. We had this simply supported uniform load. The beam would tend to deflect like that. Is an insufficient picture because we know that the real bending will be such that there's no slope at the wall because of the embedded load there. So we imagine then that what this also needs is a moment in that direction that will pull this end up to zero slope. And then those two, we can break into two separate solutions we can do and then simply add them together. That solution plus the moment at the wall. That's the picture of what we're doing, but then we use this fact that the deflection down has got equal to flexion up. What kind of boundary condition, if you will, do we need on this one to be able to put the two solutions together, the two separate solutions together that we can do such that they equal the one solution that we couldn't do separately? We need some condition that links these two to be equivalent to these two. Like we had this, we had the deflection of the ends that we needed to negate. And it has to do with what we can get out of the table for these two things. So look in the table, there's a couple things available. These two, both these loadings are in the book, right? Columns that follow and the three columns have, we're at three columns, yeah, three columns that follow. There's the picture in the first column. Then there's a column for the slope, a column for the deflection. That's what we just used on the first. And then there's the elastic curve itself. A few times when we get the deflection, that wasn't the elastic curve of the deflection that we were getting? No, that was the elastic curve we got from that. And then you find out where that's a maximum and you get that center, center column. Isn't that center column maximally deflection? Maximum deflection. Yeah, it might not be V max for all of them. It might be the deflection at a certain point. It depends on each one of the loadings. What we need is that the slope at the wall, that's the nature of a cantilever beam at the wall itself. What's the angle, the slope for this loading? It's symmetric so it doesn't matter which side, but we're working on the left side. So I think they call that theta one. M is the, over what? It's basically this. Did I switch, what did I switch? No, I didn't switch it. Oh, that did? That turned the book over. What is it? And then we want to combine you with the one above it, which is in the opposite direction. So this one is what? Minus WLQ, is that right? 24EI. Those things are all supposedly set. We want that to equal 0, but what we wanted more to equal is the angle also seen with this at the moment end and realizing that in the picture we have, our moment's going the other way, isn't it? So this beam would tend to deflect up because of the way it's drawn. And so that's the angle that we need to turn around the other way. So we've got to flip this picture over as well. And make sure you're looking at the angle that's at the moment end. What is it? Should be, this is a positive value. Should we say negative? Well, just read it right out of the book, which is? M not L. M not L over 3EI. We need that angle to be equal to this angle. Now let's see if we've got the direction right. This one's negative. We've got a negative. This one's positive. We've got a positive, we want those two to be equal. It's them both to be 0. That way we get M equals 0. Or L equals 0, or EI equals the infinite. So we want that solution. That solution possibility there. So you get a slope at the wall. Notice again, EI cancels. One of the L's cancels. And you can get a moment. Negative WL squared. What's the negative mean? That was one of our senses for negative moment. We called it moments that make the beam smile positive. Remember, this is making the beam frown. So it's negative. I think we don't use the right hand rule for that. Because we have the possibility of moments at either end that will be in opposite directions of each other, but will do the very same thing to the beam. Because there's essentially no difference between that case we have pictured there, but a case where we do the same thing on the other end, and the reaction of the beam is essentially the same in either one. It's not a symmetric loading, but the angle here will be the same as the angle there, at least in absolute value. But by right hand rule, those two moments are completely opposite, yet they do the same thing. Did we ask they don't want to take a zero? Or did we sum them up to zero? Well, we wanted the angle to actually be zero. We want both of these to be zero, which means they're equal to each other. But if we set it equal to zero and solve for that, we get a moment of zero. Well, that just means the beam's unloaded. So what we really need to do is set they're equal, they're equal to zero, but then solve that part of the equality, not be equal to zero part of the equality. That makes sense? No? We have a equals b equals zero. Now, there's we don't want to solve b equals zero. That will give us an answer, but it'll give us an answer of the moment being zero. And that just means the beam's unloaded. There is no moment. So all we're doing then is solving the unloaded beam, which of course has a slope of zero. So we want to solve this part of the equality where the two are equal to each other. They just happen to be equal to zero. We don't want to solve this one equal to zero either because then again, it's just an unloaded beam. We're not really solving any problem. So we're setting the equal to zero to get the right moment. Now, you might ask the question, do, well, we just did two solutions. Now, we did this solution of superposition to find the moment. We also did the original solution we looked at where this business equals this problem plus when we'd solved for this reaction and then could use that to go back and find out what the moment was. Or we could solve for the moment directly using this and then go back and figure out what the reaction p was. They shouldn't get the same thing. But the question might be, does this solution give the same beam deflection that that solution does? Two solutions. The one down here and the one we did first give the same beam deflection equation. Remember, that's the curve the final loading has itself. And if you take the last both of these solutions, the last column being the beam deflection curve for each of the particular loadings, add them together, you do indeed get the same beam deflection equation for both. It comes out to be wx squared. Remember, it's the deflection as a function of x times minus x squared minus plus 5 halves lx. The beam deflection of this one, the beam curve of this one, adding it to the beam curve of this one, and comparing it to the beam curve for this one, add it to the beam curve for that one, and both give the same result as you can check, and then that's the result there. And that would be then this final beam curve. So we have the reaction at both ends and the beam curve itself by adding the two solutions together where neither the book didn't have the original over constrained problem. Are they always going to be equal? They should be. I guess they're, well, if we're using those simple models to put them all together, yeah, it should be. I don't see why it wouldn't. So if it's wrong, then maybe it would be a mistake going through. Yeah. If we do a problem with this test, do we have to simplify that and factor everything out and whatnot, or? I would not believe it. I'd factor out major things like this, but then I wouldn't do anything more with that. Because you mean factor this part? Because it's a. No, I was just saying. Right? I don't know, and I was just at the home of problems. I can't do another part before, when you add all the constants together and just write it as a general form. And then I wrote underneath that a simplified version that's like that. Do we have to write that version? Like that you have on board there? Or could we just leave it in the form when it's like, you know, number plus C1, and then sub and C1 and C2? The thing is there's no standard way this has to finish up. Somebody else might leave the x squared in there, and it's still correct. So I can't say you have to factor it to a certain level, because everybody's going to see that slightly differently. However, I think the over-constrained problems are probably a little much for an exam. So we'll just do a superposition. Well, for the exam, we'll probably just do a beam deflection problem alone. I don't know if I'll do it. Well, it might do a superposition, because otherwise then you have to integrate, and that can be real time consuming, real front there. But we'll probably just do a simple superposition problem. Is that OK? Yeah, I thought it would be simple, but yeah, sounds good. Well, these are fairly simple. You have to, this one wasn't a big stretch, was it? But if I leave out over-constrained, see the over-constrained, it's hard sometimes to see just what it is you need to put together to get the back. But for a simple loaded beam that doesn't happen to be in the table, but is made up of two things that are, that's a lot easier to see, I think. The one we had yesterday, I think we had just a simply supported beam with a uniform load, and then a single load at a single spot, didn't we? Isn't that the one we did yesterday? Yeah, that's just what we did. And that was pretty easy to see what two cases make up that one. You just add together the formulas in here, whether yours is off first, load first, or not, or last thing you heard. You just add together the two of those separately. Just to really get that part where we had to split for A and for B and B, we had two different equations. Do we add them together after, or do we keep them separate? And do we keep them off the streets? Well, it's tricky to write what the deal is at the board, whereas if you're putting it in a spreadsheet, it's a lot easier to do. The trouble was with this part of the solution that, as written in the book, it's a one-sided solution. The book lists that distance is A, and this distance is B. And then the solution they give in the book is for 0 up to A, which means it only works for this side. Then so that would give you some kind of beam deflection that's accurate for this end. It would give you the right slope. It would give you the right displacement here. Then you have to turn the solution around and do it from the other side, which now this side would become A, that becomes B, and the solution's good from this side up to the load. And that gives you a different solution, but they would match anyway. When you put the two solutions together, it's very easy to do on a spreadsheet, because you can have a column for 0 to this point and then a column for 0 to this point. But then when you graph them, you graph them 0 to that end and just put the two calculated deflections together. And then they intersect just perfectly right there. The slopes match as they should, and the deflections match, because they're done from the same problem. It just flipped around from other different ends. But how do I write that on the board as a single equation? I can't really, because the, well, we could. It's, I guess, that you'd have to make the variable here, not the x that's in the book. You'd have to make that variable L minus x, which gets kind of messy, because you have x squared, which now becomes L minus x squared. You have L times x, which now becomes L times L minus x. So it gets kind of messy, but it's pretty easy to do on a spreadsheet. But it's very difficult to write up the board as a single equation. Maybe I should give you a take home, then. So you could fix babies. All right. So here's one for you to take a look at. At least set up the equations, even if the algebra needs to be left over. Set up the situation you need. What things you need to put together to get the proper solution. So again, a simply supported beam, however, an over-constrained one. So we've got three supports there, L and 1 third L, a uniform load, which could be the weight of the beam itself, or it could be maybe something like a snow load. Look at the book, decide what two solutions you want to add together, what the deflection quantity is that you want to equate. First problem, we did the deflection. The second problem, we did the angle. So you've got to decide somewhere among those kind of things for two models to find the reaction. Two cases in the appendix C that we can put together to make the over-constrained problem. And then what parameters you need to equate between the two. You need to find the reactions. So whatever it takes to do that, I guess you could do it with the deflection curves. What two models, of course, one I think is probably fairly easy, just a straight uniform load, simply supported. But then what are you going to add to it get to our original problem? A force here that represents that support. In what direction? Because that's what that reaction is going to do. The reaction is B. Once you get that, then you can get the other reactions using some of the forces and some of the moments. But then what parameters do you use that's going to deflect? We have this intermediate support where there can't be any deflection. So whatever deflection we get here, we've got to undo here. Because this beam is going to deflect something like that. So we need to have it as such that V1 is the opposite of V2. You can do that by taking the beam curve for this, adding it to the beam curve for this, which will have B as an unknown. And then setting that beam curve equal to 0 at a place at x equals 2 thirds L, I would think that would work. I think it'd be a little more difficult, probably. But I have the sense that you'd like to take that challenge on for the benefit of the class. Is that what everybody has a feeling? I think so. They're all smiling, Jake. Problem with it, deflection is the easier way to do it. Because there's not much to that, is there? Yeah, well, for the off-center load, it's a little bit, but not terrible. And for the uniform load, Bmax, well, we don't want Bmax. You're actually going to have to use the beam curve. Because Bmax, remember, is at the center where this deflection is not. You have to do V1 evaluated at x equals 2 thirds L. It's evaluated at x equals a equals 2 thirds L. Once you get all those pieces together, a whole bunch of stuff cancels beam curves. I don't know if that's up to you. I don't have a book, so I can't look at it. B2, did you take the lighter belt? Yeah, it's upside down, because this is a positive deflection. And we need that because beams can deflect positively. This beam, when loaded, will probably do something like that, where there's actually going to be some upward deflection of the beam at that far end. So well, Jake will let us know if that's just what happens when he has those beam curves together. Can I do that? For number two, how would you have to do from? Just up to the lower deflection. Yeah, and then you have to do the other one. You just add those together to get the total, or? Well, you add them together by doing it from one end and then doing it from the other end. But when you do that, they don't share the same x. Here, x goes this way up to here, a distance of 2-thirds l. And when you do it on the other side, x goes from here to here, a distance of 1-third l. How would I define on this region? That's why I said it's easy to do it on the spreadsheet. In a spreadsheet, you have this two curves. This goes from 0 to 2-thirds l that way. And put that down there. And then you can do a second column. I guess the column is 1 and 2. This one you do from 0 up to 1-third l that way. And then calculate the deflection that goes with that. And then they should meet right there. The slope should be the same and the deflection should be the same. Because then you could take a third column, just 0, down there, just for graphing purposes. You don't actually calculate anything with it. And then plot these deflections. That's how I did that plot I showed you yesterday. It's the telling presentation for the class. Brandon's all excited about that. He doesn't poke his eye out. If you come up with this beam curve plot of it, I'll take that for some extra credit. Now, since we're at the end, if you do this, you should get that v equals something like 0.6AAWL, the total is just under 69% of the whole load. Which is kind of interesting because it's at 2-thirds the distance, but it's got a little over 2-thirds the load on it. It's kind of interesting. I think it's because a longer stretch has more flex to it than does a shorter stretch. So a shorter stretch is transmitting more of the force. Anyway, double check, see if you get that. And then we'll see what Jake does with the load curve. Jake, you're not in the mood? I just don't know. Well, think about it. Say with this spreadsheet and try it.