 What the big thing that this is giving to us in this class so far, and we'll really emphasize it today, is there are real problems that come up that statics can't solve because it just doesn't have enough equations. There's just too little known. That's why whenever we had statics problem, we always had a roller at one end of those beams. Because if we didn't have a roller there, then there's too many unknowns, not enough equations. There's nothing we could do. This class is in a sense bringing up more equations for us, allowing us to use more equations to do more problems. For example, if we have maybe some kind of support piece or something that's made up of two different materials. One is a tube made up of one material, so it would have its own modules of elasticity. I happen to call that material two, so I'll go ahead and label this one two. And inside there's another material, maybe not as a support piece of any kind, but some kind of piece under some of these structural loads that we see in a lot of these problems. And let's additionally say that the bond between the two, there's either a small gap so they're not really in contact or there's no adhesive in any kind. That way as one changes length the other can change freely depending on the loads itself. There's not extra forces, shear forces available at the interface between the two materials. You probably wouldn't want to do that anyway for thermal expansion problems, so that as one heats up and expands differently than the other one, they won't bind on each other, which would cause tremendous residual shear forces. So we'll just say that the bond between the two is slippery. There's no adhesive, they're not pressed bit, they're not fit while the outer one's warm and then allowed to cool and shrink down into a tight fit or any of those type of things. It just simply slips in and is very loose. So if we look at this then in cross section and then at some kind of loading, we'll put a rigid plate there and a rigid support of some kind there and then apply some kind of load to that and we want to figure out what the reactions are going to be, what are the stresses in the materials themselves going to be. There's the inside material. Okay, so like a, I don't know what, a host is dinged on? There's some chocolate cookie thing that's kind of like this. It's a chocolate thing with cream filling. Swiss roll? Did you know when I first asked that question I was looking right at you anticipating you'd know the answer? So we have a chocolate cake with a cream filling as our support, as our structural member here and we want to figure out. That's assuming of course we know the amniascent scientist, the wood cream, which I bet is on the internet. So if we break this apart into our little pieces, we've got this load here on this plate and then resisting that are two loads, two reactions. One from the inside piece allowing some kind of support so I'll call that P1 maybe and then one from the outside piece maybe I'll draw that kind of like this and those two things together are P2. Just to remind us that that's a two but that force is resisted all the way around and of course as we already know has to do with the area of these two pieces as well as a factor especially when we're looking at the stress. So that P2 is actually the force all the way around the contact ring of the two. But I got to draw it some way so I'll draw it like that for you. And then we can look at the individual pieces. Here's the two in cross section and it's of course withstanding that much and we know that that's also going to be the reaction down here and then the inner part itself is, that's P1 right? Okay so there's the setup but all we can do with it in statics is really this piece right here. All we can say is that P must equal P1 plus P2. However much force is the total load is split somehow between the tube and the filling. And that's statics. We couldn't do any moment equation. Why couldn't we do a moment balance on this? You know, presumably we knew the radius of the tube so we know where those are. Why couldn't we do a moment balance and have it help? Because that's always, when we did, you know, there's our force balance. We can't do two force balances because we don't have any forces in the x direction so it wouldn't help if there's some forces in the x direction. Why can't we do a moment balance though? For a, for a, we need, we got two unknowns, we only need one more equation. Why don't we just do a moment balance? Well there's, there's, this force is applying a moment about that point and those forces about any other point on there, there's moments being applied. But because of two things, well they're kind of interconnected, because of the symmetry then all of those moments are going to be identically balanced anyway. Plus all of the forces are parallel so remember the kind of thing when we were looking at a three force member in statics, we had to have all those forces cross at some point and then know there's moment about that point. If they're all parallel there is no such point. So we can't do a force balance, we need some other equation, two unknowns, one equation, we're just one short equation. So we use the things that we've got now in this term, the deformation of the pieces. We have to assume that the two pieces would deform by the same amount. You can't have that force squishing the two pieces and have one of them squish farther than the other one would. What would their, what is this? No that's the little delt force. Well yeah but that's not the case, you can't change the problem to your own liking, that's not fair. You can't come in and say it's not a Swiss roll, it's a, I don't know, something else. Twinkies, twinkies can withstand anything. So we assume that, well we don't assume it just couldn't be any other way. It couldn't possibly be that the tube would deflect down to here but the inner part would keep going and vice versa, they have to go the same amount. So whatever that deflection is, and of course greatly exaggerating my little picture, it's got to be the same for the two of them. And that allows us more equations then. Del equals del one equals del two, where again the one and the two are for the two different parts of the structure. One the inner, two the tube part. And then those we can just put into there actually the lengths of each is the same. So I don't need a subject on that. Whatever the two material properties are, both the geometry of the part, the L and the A and the material property, the E. And that's just setting the deflection of the two, the deformation of the two to be equal. And with that and this then we have two equations, two unknowns. We can solve the problem. I'll just give it to you. We're going to go through the algebra of this. That's not the point of this class. It's the, oh no, that's two. No, no, it's got to, it's P itself. Yeah. P itself. The original load over E1A1 plus E2A2. And then the P2 looks just the same only these top two indices are changed to a two. Yeah, we have no minus signs in there. And then once those are known, just put those over the individual areas. And we also know the stress. And then we can design it to avoid maximum stress in life. But the whole thing comes down to this additional piece of information now that we can apply from this course that we didn't have in statics. So we can now solve problems that last fall would have been statically indeterminate. Meaning there's more unknowns than equations. Now we have more equations to bring into the problem. And those are the things we'll use for, at least for this section of problems, statically indeterminate problems that we'll use to solve them. To solve problems we couldn't have done last fall. Now this thing could be, so now this method could be applied to say something like reinforce count three perhaps? Yeah, any time you have two dissimilar materials that must undergo the same deflection you could, yeah. We're going to, and don't remember that we very often look at reinforced concrete problems in compression. We look at them in bending where we have reinforced concrete beams with transverse loads. We're still looking at axial loads here. But no reason that this couldn't have been cement with a rebar filling. It would have tasted very good, Travis had eaten it anyway. Alright, so let's try a couple more problems and other types of solutions. This is simply, I don't know, there's not a real particular name for this other than just using the fact that the two pieces must have the same deflection. But there's other types of problems that we can solve with this. So imagine a one material beam or support structure of some kind, whatever it might be, between rigid supports. And at some point in between is an intermediate load. We'll say that's at some distance down from the top, rest of it we'll just call L2. So some intermediate place there's some kind of load. This could be a column that goes through more than one floor of a building. And this load is one of the floors that's tied to that column. It's not uncommon for support beams, support posts to go through several parts. Alright, just for reference sake, we'll call the top there A, the bottom B, and the intermediate point C. And so what we're looking for in this are the reactions at the outside part. At some load P we have some kind of reaction up at A, some kind of reaction down at B. May or may not be both in compression, doesn't matter because we're going to find them anyway and we'll get those appropriate signs through the solution. So all we can do with our statics part, in fact, let me turn it around just so it makes a little bit more sense. I guess we would expect this one to be that kind of reaction. All we can say here is that P equals RA minus RB, I guess we'd say since they're actually in the same direction. That's why I made it better because then I can have a plus sign there. So that's our full statics analysis. We can't do it anymore because there is no moments because it's a purely axial problem. We just don't have enough going for us here to add anymore. So what are we to do? Well now we pull in again the possibility of some kind of deformation. Loaded as shown, we'd expect maybe L1 would get greater because of the orientation of P being down. We'd expect L1 to stretch a little bit and then of course L2 must compress by in fact the same amount because remember I said these are rigid supports. When we say rigid supports, that means that the distance between them does not change. The kind of thing where if this was a support column driven down to bedrock, we'd expect that the support at the bedrock would not move. We would certainly want to look at the bearing stress if you remember that but it would not move. We'd expect that L1 however much the upper part might stretch would be equal and opposite to L2 where 1 is the upper part of the beam above the load, 2 is the lower part below the load. So we'd expect that those two then would be our equations as long as we don't bring in too many more unknowns here we're going to be okay. So we'll take the second equation and make it let's see P1L1 over, I'll just say EA because there's no difference in the material itself, the same cross-sectional area and made of the same material anyway must equal P2L2 over again EA. Well without having a solution yet, one thing that's interesting is whatever the solution we get it's the same no matter what the material or the cross-sectional area that those two do not factor into it. And so now we can, well we're a little bit troubled this P1 remember is the force in the interior of the material P2 is the force in the lower part so we sort of have to do something to figure out what those things are so what we'll do is we'll again make one of our classic imaginary cuts but we'll make it just shy of, this is one of our imaginary cuts so I tend to make those a little wiggly so we realize it's not really a structural cut. To do that exposes this P1 that's doing the stretching and we see well that's no trouble it must be equal to the reaction so we haven't really introduced a new unknown we just hadn't made sure what it was yet. So P1 is really RA and P2 is really RB and they themselves then differ by the amount of P from the force balance and now we can solve the rest of the problem. Again it's just algebra left now so we'll put it out for you. P L2, interesting that the reaction at A depends upon the length at 2 over the original length and RB is very much the same only with the indices changed P L1 over L where L remember is the entire length of the piece, P is the intermediate load whatever that might be and then L2 and L1 themselves actually locate the load oh we don't need to do the timer today I think we fixed the software so you're fired. We thank you, good service, pick up your gold watch on the way out. We'll see what, whether this taping's fixed or whether it's fixed. I'm sure you thought that it's been fixed before. We checked it, we ran a test Monday. I let the whole class tape that on Monday afternoon. Alright so again the whole secret, the whole help here that gets us beyond what we could have done in the fall is the fact that we can use the deformation themselves somehow. And again I think it's interesting that the material itself even with the cross sectional area the piece doesn't make any difference it doesn't even show up in the solution. So this is valid for some support beam or for whatever might be on Travis's dinner plate would all be the same solution. So Travis will test that for us at dinner tonight. So again we have to look at the geometry of the deformation bring that in to get our additional equations. Okay our next step with this, this was convenient in that the cross sectional areas were the same but that might not necessarily be so. But there's another technique we can use again to find the reactions called superposition. Superposition is where we take the problem, break it into two problems we can solve that when added together give us the solution to the problem we can't solve. So again imagine between two rigid supports we want to find the reactions there. We have something like this, some kind of support structure where midway down the cross sectional area changes and we'll actually put some numbers to this one. So it's right in the middle where we have this area change. Two intermediate loads, oh I'm sorry, that's not 150, we have two intermediate loads those are at 150 millimetres. So something like this all dimensions in millimetres locates the middle of each section where we have two intermediate loads. Yeah that's the picture we've got, something like this we'll put 300 kilonewtons there and 600 kilonewtons there. So it could be again, well it's a little short for some kind of building structure but something that requires a support piece with two different cross sectional areas and two intermediate loads and we want to find the reactions at the different spots. Middle piece C there, that's B and some areas, 250 square millimetres. Don't confuse the area A for the index location A. I don't want to make this too easy, 400. The superposition solution is going to look something like this. It's the problem we want to solve but can't. We have this loaded piece and we want to find the reactions at the rigid supports. Oh actually, oh yeah, this one's RV, we're going to believe this. We can do the same problem twice, we can't find both reactions at once. Well, once we find the bottom reaction then we can find the top reaction. But what we're going to do is leave it like this. So we're going to break this into two other solutions. So this solution that we need to find here will be equivalent to two other parts that we can do. This we can't do because it's indeterminate but we can do this problem where the top doesn't move at all because it's a rigid support there. We have two intermediate loads and we remove this, we imaginarily remove that and we can use that, we can determine what the deformation of the part would be without that bottom rigid support. That's not the solution we're looking for but if we take that solution this we can find and if we add to it this solution that we can also find, we can also find this solution where we find what force would be required to cause the piece to deform back to its original. Those two solutions superimposed over each other give us this original solution, the original situation we're looking for that we could not solve. This is the superposition, superimposing. Two solutions we can do over each other to get a solution we couldn't do with the original situation. Once we then find RB from this last step then we can use statics to find out what RA is, the reaction of the upper part. So we imagine taking away the lower rigid support allowing the piece to deform, calculate that deformation, then remove the loads and calculate what force would be required to return the deformed solid to its original length. In other words, I guess, LL plus LR equals zero. See the picture? Travis, you're frowning. You're just thinking of chocolate, whatever they are. Alright, because this we've done. We did this, I think Monday we did one of these type of problems. I can't remember, may have been on its side, may not have, but that type of problem we can do. We need to find out how the different sections each deform. We have four sections. Anywhere we have a change in area and or an added load, we consider that to be a different section. So we have four sections, one down to the load, then down to the change in area, three in section four. And on this piece then, down to four deformations, the four different sections, how much they deform, all added up. So by whatever means we get all those sections. For example, if we look at this bottom piece, piece four, for this intermediate solution now, that's the one we're working on, where we've removed the support and imagine what the deformation would be. We see piece four here, once we imagine the support removed is unloaded. The next section, actually up into the piece a little bit in our imaginary cut, because we've got the 600 kilonewton load. Now we've gone a little bit past it into section three. I did start from the bottom because that was the easy part. There's no load there. That was the easy one to do. So we know that this upper part must be in tension. Wait a second. That drawing didn't work. One line too many. Here's the load there. So this will be the force in section three. We know to be 600 kilonewtons. And we expect section three to elongate. So del four is zero. Remember we're doing this artificial intermediate solution. There's no load on the bottom all the way through section four. Del three is the load there, which we now know to be the 600 kilonewtons, times the length of three over E and A. We're given that's all one material. And A is this cross-sectional area here, this 400 millimeters. So we've got all those pieces. 600 kilonewtons. The length is 150 millimeters. E. I don't know if I had one for whatever material. Oh, I'm going to just leave that. Then we can just put it in once. 400 millimeters squared for this area. That's how much in this artificial intermediate solution here, we would expect section three to stretch with these two kinds of loads. And no load at the bottom. Remember we're going to put that back in to recover this total deformation. All right, I think, no, naturally I don't have that piece written down. I've got just the final answer. So we'll go to the next section and put an imaginary cut here in section two. This 600 kilonewton load in the bottom part, we haven't quite gotten up to this load of 300 yet, so it's not there. So this part is also in tension at 600 kilonewtons. So now we can figure out how much that section will stretch. That's two del two p two l two e over a two, which are most of the same numbers except the area has changed to what? Two fifths. Leaving off the units there because we've already got them. We're already going to have to check them. We've got all the pieces. And so that's how much deformation we'd expect in this second section here. And then for the third section, we make an imaginary cut. I guess this is the first section now. Put an imaginary cut. We've now got all the loads back in. You know that this is p one. The force in the first, the top section has got to hold these two loads, 900. And so we know then that del one is a one, which is 900 kilonewtons, stretching 150 millimeters over e and the cross-sectional areas to two fifths. All the units are the same, so they're going to only have to check them once. All of those added up are going to give us this total deformation of the piece in this imaginary intermediate loading. Then we're going to find out what force is required to recover that back to the undeformed state that was required because of the rigid supports. And that will be our reaction at the bottom support. So all of these added up come out to be 1.1. With all the units checked, 1.129 times 10 to the ninth meters divided by e, whatever the material is. That's all of those pieces. No deformation in the bottom piece that was unloaded. That will stretch a bit, that will stretch a bit, that will stretch a bit, and we add them all up. So we know that this is the amount that the piece would stretch if the bottom support was removed. That's just like a couple problems we did. I think Monday we did one, and we did one the week before as well. I need some board space because now we need to imagine that the loads are removed and all we want to find is what force would be required to return us to our undeformed state as required by the original rigid supports. Alright, so I need more boards. Alright, everybody okay with this? If we got this down, Bill? Okay, let me at least put up that one little piece we've just calculated. I'll put it right here with this. One, two, nine meters. Actually that's the units on the whole thing is meters. That's already got the units of E in there. The whole thing is meters. Alright, now we need to do this little piece here, the superimposed piece. I guess we need again imaginary cuts because the area changed. So we have this rB we can't find yet. We're looking for it. That must be how much is in that section. So we just simply number those one and two. Del one will be rBL1. The entire length of that section now which is 300 millimeters over E, A, A1. Or again these numbers refer, maybe that's not helpful. Let's do this. We have four sections here. Let's call this then sections five and six. So we don't have any confusion with our indices. Confusion with these indices we head over on these other ones. Oh, so that's del five. Alright, again we don't know what rB is but when we combine these together and compare these two artificial deformations we're imagining rB will be the outcome of it. Pieces for that put together. Oh, not quite. And then imagine a cut somewhere in section six as I called it. And then we just see that section in tension under rB. Huh? That's supposed to go down, correct? Oh, yeah, yeah. So we used to draw in things in tension. Thank you, David. And that then becomes that piece to form to the tune of, again there's an rB in there, six over eA6, this whole top section is a different cross-sectional area. And those two added together is this total deformation del r2 which we know is the opposite of this. So del r is those two things added together. Del 5 is del 6. And with all the numbers and the geometry and everything else I can put in there for you we get a minus 1.95 times 10 to the third rB over e and it all has units of meters. That includes the units already on rB and e. Everything's been checked to that point. So we'll put that together with our imaginary superposition solution that's required. We know that these must be equal and opposite. So del 1 equals minus del r. Remember this is the imaginary elongation from the first part. This is the imaginary compression from the second part. We put the two over each other and we're going to get our full solution with no deformation because of the rigid supports and we'll know the reaction at that bottom support. So the opposite of that thing we lose the minus sign. We have rB and units all been checked. That's just this thing rewritten up there. Notice what interesting thing happens. Once again this solution is not at all dependent upon the material that cancels out. And so we get a final notion that the only unknown now is that rB because the e canceled even though I had given it to you, 577 kilonewtons. Now that that's known then it's a trivial matter from what we'd have done in the fall. Well, actually physics one that could have done it. You can figure out what rA is and it's just got to be the difference between those and the loads that are opposed so it becomes 323 kilonewtons. So superposition. We have a problem that was statically indeterminate because of these rigid supports. There's no way we could have used statics last fall to figure out what the loads were, the reactions were at these two supports. So we imagine removing one of the supports allowing the piece to deform because of the loads and then figure out what force would be required to return that to its undeformed state which would then give us the solution we had and we can solve the whole problem that way that we couldn't have done last fall because last fall we didn't know anything about the material responses to these loads. We just assumed that everything was rigid and did not deform. Questions on that? We can also use it for a similar type of problem. Imagine we had exactly the same piece that we understand now is in tension throughout the whole piece. This part was positive, we had to bring it back. What if we build in a pre-gap before loaded so that once it is loaded then it will deform down to the rigid solid but the reaction there will be less because some of the force is going to actually truly elongate the solid in this sense or in this case but not all of it. So the rv would be reduced, rA of course would have to go up a little bit but it might balance the two a little bit more. So you should be able to find imagine that this pre-gap is say 4.5 millimeters and this still holds. It's just the intermediate part here this will be greater than the 4.5 millimeters so you'll have to recover a little bit of it but with a much smaller rv and in this case then you get a reaction at the bottom of 115 kilonewtons significantly less than it was before which may or may not be the purpose of this gap it could be just the gap is necessarily to even install the piece and then rA increases because we still have the same support load requiring increases to about 785 in this case. So it may be that's purposely put in maybe it's required just to get the piece in there and then have to fit these things together before they load them. So you can go through the superposition problem and see if you come up with that same thing the first part here shouldn't change any. Then the second part you have del R but you don't have to recover the entire elongation from the first part all the 4.5 millimeters really. So there's something to keep you awake on the weekend. In other words, delta L plus delta R is 4.5 millimeters. Yeah, let's try it with another type of problem. What's that? That was you. Do you have a question? So is the second scenario something we're not supposed to do? No, you wish. We're not going to do it now. You should be able to do yourself using what we had before. It's just there's already a pre-gap in there that you don't have to recover with your reaction load. That's why the reaction load is so much smaller. Instead of bringing it all the way back you only have to bring a part way back because 4.5 millimeters is left in there. That of course means there's some residual strain in there that may not have been before, but not necessarily a problem. Alright, so here's a different type of problem. This is not a superposition problem. Imagine we have a beam we're supporting with 3 struts. A, B, and C evenly spaced all the same length and C, rigid support at the top and a load on the bottom bar 220 kilogits. The difference being here that the bars are not of the same material. So bars A and C made with something with an elastic modulus of 70 gigapascals and a cross-sectional area of 550 millimeters squared. But you want to know what it is. It might be could be a baked whipped cream or if you go get something out of Travis's trunk that's been in there for a while, God knows what it is now. R, C, 200 gigapascals. May or may not be that material. It just happens to have the same modulus of elasticity. And bar C has a cross-sectional area of 900 square millimeters. Is the top one A and B? No, bars A and C have that. Oh, sorry. That's B, the middle bar. Bar B, sounds like a cattle brand. She's the only one that understands my cowboy references. God bless her. Okay, we want to find the supports, find the stress in the supports, find the stresses, and find the new position of the beam. So I guess we can call it bell of the entire beam. Let's drop a little bit assuming that it remains level. So those are the things we need to find. Because if we do know this material, we know the elastic modulus, we're also going to know the yield stress and we want to make sure we stay below that. So we do need to calculate the stresses. So let's, as we have on the other ones, at least begin with the statics we know as best we can. So we've got PA there, resisting this load of 220 kilonewtons. We have PC and PB doing that for us. Yeah, maybe in your alphabet. Not in my new world alphabet. Okay, can't even spell ABC. All right, there we go. All fixed. Thank you, Travis. It must be Swiss rolls with dinkel below bayonet. All right, so the statics tells us, let's see, oh, by symmetry we know PA equals PC. Comfortable with that? That would actually come from some of the moments about point B, the middle. So then we can also say PA, B, and C must equal 220 kilonewtons. So that's some help there. Then let's sum the moments about C because in this problem we expect there to be some moments. Maybe that can help. So if we sum the moments about C, we get PA. Unknown, we're looking for that. Once we find it, we can find the stresses. So we need the forces. PA times a meter and a half plus PB is 0.75 meters must equal 220 times 0.75 meters. If I look okay for the moments about C, we've got two equations, really only two unknowns because two of the forces are the same anyway. I can even rewrite this as two PA since PA and PC are the same plus PB. That's just putting this one into that one. What's the trouble with these two equations we have here? We've got two unknowns, PA and PB. Why can't we just solve them? If we divide this equation by 0.75, we get this equation. So they're not independent. It was a noble effort. You were more than happy to write it down without protest but we've got to do something else. Any suggestions? Not independent. So we still need another equation. David? If the bar remains level, then each one of them must elongate by the same amount. So del A equal del B equal del C as required. That will give us then actually the del A and del C equations are going to be the same but that will give us another equation and no extra unknowns. PA, they all have the same length so I'll just call that L over EAA. That will be the deformation of P's A and the deformation of P's C is going to be exactly the same so it won't be an extra equation but what will be is the deformation of P's B. That will be an independent part so there's an independent equation independent of the first two and we can solve the pieces and there are so what's left. Just the algebra that you love. Notice even though we're talking about the deformations the length of these pieces is immaterial. This will be the same solution no matter what because of this constraint that the bar remain horizontal. So solving for those last little bits it's all algebra from here on out this is where you hire your little brother to do it. We get force in A and we know to be equal to the force in C because of symmetry is 33 kilonewtons and since those two together handle 66 then we know what's left over for B it's 154. So again we needed our elastic deformation of the solids to give us enough equations to solve this problem. Now we can figure out what the stresses are because we know what the forces are we know what the areas are we can figure out the stress in each of the pieces I think I have those you can double check them 59.9 megapascals and the stress in the middle one it's got a different area a different load so it's going to have a different stress it's 171 most of the most of the troubles in there even though it has a greater area has a lot more of the load one thing we're not looking at yet that we will shortly is the fact that this beam itself would deflect because these are elastic solids there's no way that can be avoided but we're not taking that into account yet we will look at the actual deformation the beam itself as well as its displacement to get a more accurate solution it can certainly be that the solution is level but to have the same deflection at A and C that you do at B is not a realistic possibility but is approximation that we're making still at this point we'll handle that in April we'll get to the actual bending of the beam itself by cross loads transverse loading like that and the last little bit you need to find the full deflection how much the beam displaces and that looks like 1.71 millimeters that's just answering these pieces right there find any one of them and you'll know the deflection of the hole the beam itself guarantees that the beam stays level alright questions on that piece? 33 solving for PA yeah oh I don't know what those became F's they should have been P's but that's just from solving the two equations we have two equations two unknowns PA and PB are the unknowns PA and PB are still the only two unknowns so you solve these two equations and you get these and we already know PA with PC so the algebra on this one isn't even all that complex and then you take those forces of course and divide by the respective areas to get the stresses questions? a lot of stress in this middle piece even though it's got the greater modulus of elasticity and a much greater area ok so can start a problem for you I've been earning my pay but you guys have been slacking off so you need something to do we'll start a problem here then last fall we couldn't have done it so here's a beam supported by two struts evenly spaced each at about a third one of them a lot longer than the other at a rigid support and all to supply support for some load P here we'll put at the end evenly spaced struts at 50 inches apart the longer one is 80 inches and the shorter one 50 itself that's one that's two so L1 is 80 inches L2 is 50 A1 is one inch squared A2 is half an inch squared they're both made of the same material ten times ten to the third KSI everybody know what a KSI is? kill a pound per square inch and for all of this the stress in those pieces they're all the same material must be kept below 30 KSI okay that brings us right up to the end so there's something for you to do over the weekend you don't have homework in dynamics and you've got in a winter supply of Swiss rolls you don't need to go out you find out the pieces now the trouble is that this maximum stress might come in one or the other that makes one of them the limiting case you might find that a large P one of them will cause the other one to fail so you have to look at both of them and again you just have to use the very same type of things we've used before the beam will deflect displace something like that and you can use our deformation informations to figure out what those two would be proportional to each other based upon the triangle that's made as they displace so the statics would not allow you to solve this but the strength of the material will allow you to solve it because you can figure out what the displacement is let me make sure I've got all the pieces for you there yeah everything we've got everything you need to solve that with the combination of the statics and the strength of the material