 Alright, we were doing the actual beam deflections from last Friday. This is where given some of our standard loads we were able to figure out just not just standard loads but also standard supports. We looked at a couple but whatever those loads happened to be and we're going to show today that we can handle just about anything, whatever those loads happen to be we're going to be able to figure out what the deflection is and that's actually finding out this deflection, the amount the beam actually moves due to the elastic response of the material itself. We were able to, for some simple loadings, figure out just what that equation was, remember that was called the elastic curve. Elastic meaning a couple things to us, one is the beam will, if not pushed too far will return to its unloaded state if the loading is removed but elastic also had to do with the fact that the curve was, this curve is continuous. Remember that term from calculus, well actually pre-calc I guess you would even do continuous where there's of course no step change in the position of the beam. The only way that could happen is of course we had beam failure and we're not stressing any of these beams to the point of failure but it also meant that the slope was also continuous meaning there was no abrupt change in the slope of the elastic curve itself. This meant that the beam had no sharp corners in it, even if the beam itself was continuous it can't maintain any sharp corners. There is a way we could do that, I can't remember off hand if our book has any of these type of problems but there were some in several other books where there are actually two beams joined together by a smooth pin joint, the very type we'd see in statics last fall and these beams can, well depending on whatever the load is, these beams can have a rather abrupt change in the slope of the elastic curve itself. I don't know if it does something like that but it's the deal where there's a continuity in the beam itself but a discontinuity in the slope of the beam because of this hinge joint but like I said I can't remember off hand if our book had any of those. So we worked through this and it was usually a matter of figuring out the moment as a function of x and then integrating backwards to this curve using the shear moment, the load shear moment, slope and displacement curve all related by their integration. Now all of these, not all of these, some very simple loads of these are all done in this appendix C here, slopes and deflections of beams and you can see some of the ones we actually even did in class and what's available here is the slope usually as a function of the maximum slope, the maximum angle made by the beam under that particular load or possibly the two different slopes of the different ends depending upon what the load is. What you have to do though is be very, very careful. Don't use one of these solutions just because the problem kind of looks like the picture. It has to look exactly like the picture. It has to be precisely this case. The last column is usually the beam curve itself, this curve that we were actually working on coming up with last Friday. Also available though are the deflection at a certain point. This is not the point of maximum deflection, it's just the deflection where the load happens to be. That may be what you need. If this was a floor beam and this was a wall on that beam, you'd want to know the deflection right there because it could deflect enough that the wall separates from its own ceiling above. You don't necessarily care what the maximum deflection is when it's more critical what the deflection is actually under that beam, under that particular point load. If you did need the maximum deflection, you can take this curve here, differentiate it, set it equal to zero and find out where the maximum deflection is. There's some other possibilities. All the ones on this first page are those with simple supports, the pinned end and the roller end. Again, be careful that the loading is exactly as shown here, otherwise you can't use this. And all this is a matter of the author doing for these problems exactly what we did on Friday and then just giving you the results. You remember every time we need a moment of inertia, there's some very standard shapes available in the tables, pre-done for us, that's exactly what this is. Some standard loadings, pre-done for us with the answers right here. On the second page were several examples of cantilever supports with different types of loadings on them. Again, you need exactly the loadings shown in the picture to use this solution. Like a test problem right there. Oh, it gets even better because there are lots of problems available that are not on here but are relatively easy problems to do and we'll talk about how we do those right now. So imagine we've got a simply supported beam with a uniform load and remember that could be nothing more than the weight of the beam itself. That's exactly how the weight of the beam would show itself in one of these problems as a uniform distribution of a load per unit length of beam. But then imagine we've also got a point load at about a quarter. So this is not replacing the distributed load with an equivalent single load. This is a separate load to that shown by the uniform distribution. So it's an 8 meter beam and that load is one quarter of the way along. You'll remember one of the terms that appears very often in these and you can see it all over the place on the chart here, this EI. That's the modulus of elasticity, whatever the material of the beam is determines that and then I is the cross sectional moment of inertia of the beam itself. So it's often worth it in these problems that it's either just given or it's something if you have the separate parts to it that you go ahead and calculate it separately so that it's available for all the other times you need it. Sometimes this is termed the flexural rigidity of the beam, these two together. So we've got a beam with that already established. The first thing you should do is look in the tables and see if this solution's already available and then you can determine what it is. We want to find the slope and the deflection at certain places. None of those are available. We're kind of close, but kind of close isn't good enough. If we tried to solve it with just this one, it wouldn't work. If we tried to solve it with just this one, it wouldn't work either. However, we can use superposition. We'll do an equivalent situation of two solutions superimposed over each other. That being the solution of just the single load added to the solution of the uniform load. Both of these are available separately in the tables. We'll get a solution of the two of them together by combining these two solutions as we go through these. So we can figure out the deflection due to the first loading, add it directly to the deflection of the second loading. We'll have the deflection of the two together. So we'll look at these. This one is the second one in table C. In fact, it's the one we did on Friday. We actually did a beam where the point load was one-quarter of the way along the beam. We actually did that on Friday. So we don't need to redo it because it's either right there available in your notes. The numbers might change slightly or we can just take it right out of the table. So we're looking for the general beam deflection. I'm going to make sure I slip over to the right one. Be careful that you pay attention to what A and B stand for. So we have V equals minus P BX over 6 EIL where L is the length of the beam and P is the magnitude of the point load itself. And then, which one am I looking at? L squared minus V squared minus X squared. And so we can put all of those values in where for our particular setup we have A equals 2 meters, B equals 6 meters, and L equals 8 meters. Oh, and P equals 150 kilonewtons. So you put all those values in. We even have the EI. We can put everything in and this all reduces down to minus 187.5 times 10 to the minus 6. 28 minus X squared. That's just reducing everything. Putting all the values in and collecting terms and simplifying. It all comes down to that. And that's the particular solution for just this sub part of the problem. And if we put in, let's pick a value for example, right at X equals A. I think that's the value I put, right at where P is. This comes out to be minus 9 millimeters. So we would know from this particular loading that that distance there is minus 9 millimeters. And if you remember, we do need them to be very careful with the minus signs because there are loadings where the beam could actually go above the neutral position rather than below. For the second loading, I'll have to bring it over here. Just, I don't have a lot of room there. The second loading is also in the book. It's the fourth one down in our book as long as they didn't change the order between the versions. And so if we go over to this particular solution here, and then again, we just fill in the pieces. W, X, remember X is the position, but W is that uniformly distributed load. The value of the linear part of the distributed load over 384. Let me make sure I'm looking at the right one. Oh, I had actually slipped down. 24EI. It's hard when you can't see the whole picture on the screen. X cubed minus 2LX squared plus L cubed. And of course, we just set our values to what we were given as a problem. W is the 20 kilonewtons per meter. EI, we've got. L, of course, is 8. For that one, that's all we need to know. And then that, actually, you don't have it. Well, yeah, I do have it reduced just for simplification here. Not a bad idea. Oftentimes in problems, we need to look at a couple different points of these. And so it's nice to have it in a very simplified form. X cubed minus 16X squared plus 12. That's just all of the values we've got put in. Simplified, reduced, cleaned up a little bit. And then at the very same place where X equals A, right under the load P, the value A doesn't have a specific importance in this loading we're doing now, but it relates to the original problem so it's somewhat useful. So at X equals A, again the 2 meters in, we get a value of minus 7.6 millimeters. And so as this beam deflects at a point 2 millimeters in, we get a deflection of minus 7.66 for that separate problem. We add the 2 together. So V is the minus 9, minus the 7.6 millimeters. It is minus 16.6 millimeters for that entire problem. It's just simple superposition. So we know this beam's going to deflect in some way. And at a particular point now, we know that the deflection will be minus 16.6. And we can also figure it out at any other place along there. If it was also important to figure out what these angles are, then we simply figure out the 2 angles on the separate sub-problems and then add them together. So rather straightforward, rather quick, as long as the sub-problems you need are available on the table, then you can save yourself a lot of trouble. If they're not, then you have to go through and figure out what the moment is as a function of X, integrate it to the deflection and finish the problem that way just like we did on Friday when we first set these up. And if you go back to Friday's problem, you'll see that this specific solution is just like we came up with. Sorry? Here? Oh, what's missing is I had an X pulled out. And I didn't see it. I had it in there. So yeah, that would be there. It would be 28X minus X cubed. Good point. You've got to be really careful with these subscripts, superscripts, and all the different factors and variables as part of it. Just for your own interest, you can notice that you'll get this very same 9 millimeters if you do it from the other end of the beam where A now is 6, B is 2, L is still 8. You'll still get the minus 9 millimeters now at a point 6 meters from 0 just to show you that you can use that from either end, but you have to be very careful which end and that you make sure that the values you put in match the pictures there precisely. Okay, so we'll do a couple more. These work well with some examples. Here's a model of a hairbrush design. So it's a long plastic beam that the users are going to hold on one end with a fist and then pull down using yellow. There's your hair. Pulling the hairbrush down through the blonde here. We're going to make a model out of that as a beam like this where we'll model that end as cantilever as it would be as the users holding it then with a uniform load and we'll take it to about halfway. So that'll be our model and then we'll put some values to it. Let me check out. Okay, so the entire length is 120 millimeters. This is half of it. It goes half the length. So this is 60 millimeters across there and the load is something like... I'm going to stop doing these things late at night. The load is 50 newtons per meter with a proposed cross-section of the hairbrush. Make it out of plastic, some kind of unreinforced plastic so we'll have an E for it. And I said 10 millimeters by 25 millimeters. Plastic has a modulus of elasticity of 2 gigapascals. So E times I is something like... Oh, I didn't do it separately. So you can do it separately. Remember to watch your units. I for a rectangular shape is just 112 BHQ. So we know we're going to need that. I just didn't have it on my paper. The first thing to do is check to see if the solution is in the tables. If not, we've got to come up with a solution that represents it. So looking at the cantilever solutions... Well there's a uniform load. There's a uniform load halfway across the beam but it's at the cantilevered end, not at the free end. So we can't use that solution. So we don't have a available solution in there but we can make an equivalent picture of it that will suffice. So here's our original model out of it of the one that is in the picture which is this one here that is in the table and the uniformly distributed load goes to the halfway point which is sort of like what we've got and then we'll add to it the uniform distribution that's also in the tables of the same magnitude going in the opposite direction. That way this first portion here will cancel that portion there to leave the first part of the beam unloaded which is like in our model for the hairbrush and then we'll have a load here it'll be in the opposite direction but we'll just have to be careful remember the hairbrush it doesn't matter where you're going you have to do it both ways we just have to make sure we get it so it makes sense for the picture we've got here so I guess let's make it so it is better so it does a better match we should put this one upside down that makes more sense right David? I can see it frowning we'll make this one then like that that'll match our model precisely again these two ends will negate each other leaving it unloaded then this adds to the unloaded end and gives us just exactly what our model is so you can do that let's step through it find the deflection and the maximum angle we know it's going to do something like that the maximum we expect to be down there and we also expect that's where the maximum deflection will be as well so practice opening up if you have your book that'll help if not I'll put up the relevant ones right here yeah I can just barely get them both on the screen okay so we'll leave this one and then the first and the third here on the pages that we've got so you set those up, see what we get with them the third you just make positive sorry? yeah well for this one positive yeah you'll have to make it well W remember is positive one down put in W as negative or just change these to positive and put in the magnitudes either way all the same same with the deflection these are all assuming downward loads so all of these deflections are negative and the angles are the negative too as shown alright so see what you can do with putting those two together just so I don't get sidetracked from my notes we'll number this one two and that one one as we refer to this visually perhaps V1 and V2 as symbols and we can also figure out the elastic curve as well so there's the two solutions we need in the table let's practice and see if we all get the same thing and if the same ones looks like you're the international nation too I'll have to remember that set up the curves with the appropriate values and then just add them together for the two superimposed solutions we've done this superposition thing before and you can imagine some of them can evolve three four more of the simple models to add it together if it looks okay put some hands on the airbrush point the thumb now the thumb is along the backside there okay like that is that okay? are you happy with that? much better now now you can finish the problem I guess we can argue whether the cantilever should have started right back here but we didn't have a solution that would work for that we need to simplify a little bit but would you decide? what I've called number two here remember we're looking for the deflection at the far end they're going to be different there's different equations there because there's different loadings there so we're looking for the loading at the far end so v2 if we're looking for the maximum deflection out here at the free end that's the second of these two equations it's inverted so we take off the minus sign w over lq we said minus w over lq 192 eI 4x minus l over 2 I believe we've got all those values what do you get for eI then? I don't have an eI out separately I do have I what do you say? not 26 even 0.04 0.04 22 times 10 to the 6th Newton's per meter squared and for 112th bh q I've got 1.28 and it's 10 to the minus 8th meters to the fourth 10 to the 6th 9 to the 9 you have gpn oh yeah yeah I was doing megapascals the last time 10 to the 9th so whatever that comes up eI just don't have it calculated separately then the maximum deflection again take out the minus lq and remember all these units should work out did the units work out? I hope so since you didn't track them you better do them and then for the second one we can use the loading just as shown minus wx squared 24 eI x squared minus 4lx plus 6l squared and the maximum expected deflection the angle the slope again at the end and again the model and the solution match each other so we can use just the x picture x is no x is wherever along the beam you're interested in this so if we want to know what the maximum deflection is you'd want to x at l if you want to know it anywhere else in between reduce everything check your units make sure everything's in meters if that's what you're doing all those distances in you've got some numbers you've got some numbers the units the units is always right you too no units how can I tell if your number is right so what? radius intermediate values written down with x equal to l on all of these a lot of it simplifies you can really get it reduced down to a single piece there once you get all the numbers put in so see if we all agree you can imagine if hairbrush was too flexible it would be a little flexible yeah a little bit flexible especially if you're worried about losing hair which you are yet but you will be I'll see you guys at the 50th reunion yeah I'll be laughing at how bald fat we are you'll be laughing unless you're laughing at this already it's just going to get worse alright everybody's coming along who's got some values if you think you got them ask somebody else who does you can prepare them so we get this in radius yeah remember this is all based upon the angles being very small the tangent theta equals theta we had to use f from the very first almost the very first fold when we started going on we forgot to go on right there so yeah these are not typically very big deflections plus this is not a very large object anyway so it doesn't have a whole lot of length the flex both the angle and the absolute expansion are going to be small get your plus and minus signs right because these two deflections are in opposite directions the minus sign in there has to do with a positive load which we have always taken to be a downward load as positive small is that meters millimeters he's got 0.02 millimeters but that's minus yeah so one was down two was even downer yeah something something didn't quite work there we expect there to be a negative displacement as pictured Newton meter squared over Newton meter squared which is no units but remember that's when Newton's applies yeah well that's got to be 10 to 6 Newton's Tom got anything yet both positive plus you because that's a positive deflection this is a negative deflection and on a model we'd expect it to be oh I only have one okay yeah number two we expect to have should be positive one should be negative hopefully if they work out the total deflection is negative is that negative you want to be positive I guess brush me to pull down we'll give it an upward load on our original model but this is not Australia negative 0.0 oh I get what you mean for what? yeah she's pulling it down load would be up Phil what are you at negative 0.77 millimeters but it is negative for the total and you got 0.77 millimeters 0.012 0.004 0.004 oh that's closer to what I had I had 0.004 I had 0.04 that's meters millimeters 0.004 you have the individual number centimeters no I don't have that in the individual numbers one or the others may have slipped a power of 10 what do you have David? 10 times probably almost exactly 10 times what you had but it was negative negative yes so did you have the same thing then the traffic had but not negative all right well well that's details the physics is there just perhaps the sixth grade algebra is for either one of us I was kind of rushing through this last night did this last night? yeah which is the tail end of the long weekend easy all right so we'll double check those numbers when you've got a couple minutes to down but the basic idea is there I want to have time to get on this to the other application of this as we've seen before as long as your physics are there there's a search that a simple superposition works very well in lots of different applications all right we also have used this type of thing for those problems that are statically indeterminate we did this for torsion we did it for the strain simple elongation and we can do it now so imagine we have an over constrained beam what that means is a beam for the simple statics equations we've used before are insufficient if we have a beam like this our typical static solution from last fall will not suffice there's just too many problems for the number of equations we've got but we've now got this idea of superposition to add on as another possible or actually counts as another equation another equation actually comes into it so we can make this into two models this one of course is not on the solution so we can take it as two models that will superimpose and use that as a way for us to then get to a solution that we do have this solution is available it's the one we just looked at that will cause a deflection of something like that we expect the beam to deflect something like that remember a cantilever support the slope is zero and then there's zero displacement over here so we'd expect something as a displacement like that so we can't clearly get it with the solution we've got there but we can add to it another solution that we do have available and that's then a simple load at the end that brings the beam back up to where it would have been such that those two together we would expect to give us the solution we have there and that will give us the reaction at both walls so maybe we'd say something like this we have a v1 down and then with the second model the same magnitude deflection only would be up such that the two added together means no deflection at the simply supported end the zero slope for both will guarantee zero slope for the other solution and we'll then be able to find the full solution so let's put that together and from appendix c we have vmax as minus wl fourth 8 over ei so that's going to be our v1 and the minus is appropriate because our model matches that in the book so we know that we expect a maximum deflection at the end of that particular model of minus wl adi and we don't have to have values for these we're just doing them in absolute variables and then we add to it the inverse of that top logic so again we're using the first and the third but we've got to flip the first one over because it's in the opposite direction that we're using it here so that will take out the minus sign pl cubed over 3i that's this deflection here the maximum deflection of a single point load at the end of a can the free end of a cantilever beam and then we're saying that maybe I'll call this a 2 here would make more sense we're saying that v1 plus v2 equals 0 that as simple as it is is the extra equation we need to solve the original problem and so we can then use that to solve for p which is equivalent to the reaction at that point which is 3 eighths wl as simple as that when you put v1 plus v2 equals 0 put those two together the unknown is p that we're using to represent the reaction at that end you're not done because we don't have the reaction at the wall we also need that so that is going to take a separate model to find the reaction at the wall so we'll split that into two models as well one of them will be a simply supported beam actually we're going to have to to model this twice so a simply supported beam however we know that this one would have a slope at the left end which we don't want so we'll have to find what moment is there that will bring that back up so that's actually itself two models both of which we can do to get the original anticipated deflection not in any particular order we can do a uniform low distribution that will give us a deflection like that add to it simple moment as calculated by that and then we can use the two together to give us zero angle at the wall such that theta one and theta two are equal and opposite we need those two to find out what the reaction at the wall is for this for any of our static solutions when we find the reactions it's not just at the simple at the wall supports but also at the wall support so you know P on the far right so why do you need two separate models there because we don't have any simple model we can use to illustrate that one well I guess we could we could have made this as another solution we could have done it as this well actually it's the same two we've got there well no it wouldn't be because yeah what needed if we did this we're not going to get the moment at the wall so this isn't going to suffice we had to bring the moment into it so we we had to remodel it and then split that into this superposition so now we're back on the other page we're using this load here plus this load here only done from a different end with the moment running in a different direction so we again have to be very careful with this we're using these two middle loads and we're using the angle ones so there's the ones we're using there so we have to be a little bit careful we'll call that one that two just to stay consistent with my notes so the first one we have to be careful with the subscripts we're doing the angle over at the end where the moment is so we want this particular picture m I have listed as m a m a l over 3 e i and as pictured that will be positive so if the moment is positive then this will be in the correct direction because the sides are swapped and the moment is swapped that's like two negatives so we can use the same equation we don't want this as theta max theta just theta 2 and theta 1 it doesn't matter which end we use because it's symmetric so it's minus l cubed over 25 24 e i minus wl cubed over 24 e i minus wl cubed yeah yeah I want to change these numbers are right so I want one here two there one here two there so one's on the bottom it's arbitrary the way I have to do it in notes and so we put that together we should get m a equal to minus wl squared over 8 remember every time we did this superposition thing that do statically indeterminate problems the material itself didn't matter so we ended up cancelling out and that again is the case when we put these two things together theta 1 plus theta 2 equals 0 solve for m a the not only the material but the cross section of the beam cancels out theta 1 plus theta 2 yeah that's a negative no no that's our negative convention so does it not come out in the algebra we got m a l over 3 e i minus because I have a minus here wl cubed over 24 e i and those two have got to add to 0 okay oh this yeah you're right because we don't need this in the negative since we've already got the proper sense to it so this would be a positive then all we need is the magnitude we already know the direction we already know the direction of a moment at a cantilever wall now the question might be we use two different models here we use this model first where we have the uniform load with a point load back up on a cantilever beam then we use this model a simply supported uniform load and a simply supported beam with a moment the question might be we use those for two separate purposes we use this for the right end reaction we use that for the left end reaction the question might be do two models give the same elastic curve we have an elastic curve that we'll get from this solution we'll have an elastic curve we'll get from this solution the question is do they each give the same elastic curve if they don't then it might not be appropriate to choose two separate models so let's let's let u3 find the elastic curve for this model and u4 find the elastic curve for this model and we'll see if they're equal if they're not then you either have to decide is one the better model than the other or perhaps neither model is good and it may not be obvious which is which so do the same thing just use p and w for the loads that's w that's p actually we just need w find the elastic curve for the two which is a matter of for the one that's pictured up there which is the second one we did you've got this elastic curve over here you just need to add those two together taking into account one of them that's upside down and then for the other one we use those two cantilever solutions there those elastic curves we use this in some place either group needs to see the elastic curves people doing the left hand one your elastic curves are up there don't forget that we've got the simply the simple load at the end of a cantilever beam upside down we can go for your two beams that's these two right here you've got this load with this elastic curve and then this load with that elastic curve be careful that this is at the opposite end from which x is this one three people where the moment is actually in there use the use the one we solved that M equals WL squared over 8 because we now know that we have to pick out the minus sign yep the first and the third write out the equation for the elastic curve and put them together put them together minus x squared minus is okay, we're in the same way 24EI the EI is not going to cancel in this case adding these together but they don't add together to equal zero they add together to equal whatever they equal x squared minus 4L x plus 6L squared is the first part yep, that looks good and when we add to it this one where we now know the load we're just going to have to narrow it down yeah we know the load P as well, that's got P in it but we know what P is we know that it's three 8's one of the three 8's so we can put that in but it's upside down where P equals what we found before I don't have it in there it's three 8's WL so that will give you a low curve I'm sorry, an elastic curve for one of them and then for the other one we have to go to this one the second one where this one is at this the uniform one here it's in a proper orientation x cubed minus 2Lx squared plus L cubed and then V2 the second one with the moment this is upside down so we have to turn this over we know the magnitude of M zero I had an MA on the board so you can figure out what that is as well using the moment that we now know no minus sign there because that one is upside down and those should give the same elastic curve do you guys have yours to compare? is it real? this is a good short little but your variables have a slider function possibly and you can imagine there are other solutions for these for example we've got this L over 2 but what if it's just L as some variable of x so other books have other solutions so hopefully this works out now add those together add those two together to give a complete elastic curve and that should equal to these the ones these two are adding together this group is adding together for this elastic curve it should give the same elastic curve assuming you've got the same you've got to hurry, we're running an overtime I'll help you now, it's 100 bucks an hour alright we're at the end there so I'll put it up as a function of x it should look like Wx squared over 24EI depending on which one you put first and what you do with your algebra 5 halves Lx minus 3 halves L squared plus no, minus x squared did you guys get that? make sure I'm reading the right one Wx squared over 24EI 5 halves Lx minus 3 halves L squared minus x squared and I got that the two did give the same model it doesn't agree, it's most likely just algebra physics is nothing more than writing down the two equations and adding them together that's all the physics is did any of you agree? you got some of the terms double check that you've got them all but they do give the same elastic curves so they're both legitimate models luckily are they close to each other? what? yeah, I don't know why they wouldn't as long as you do it right? as you said you've done well