 All right. Huh? That's a good idea. What? That's Friday there. What about Friday? That's Friday. This Friday? That's Friday. Go on. Okay, probably be the same kind of thing. Maybe two class, two problems in class, one for take home. Take two a little more time maybe. All right. Okay, we're, if you remember, we were looking at bending and beams. We were looking at pure bending, taking it that for the most part, our beams are prismatic. Remember what that means? We're looking for the most part of pure bending and prismatic beams. Remember prismatic beams? You don't really have to worry about it too much, I guess, because we're not gonna look at any other type of beams. Yeah, I don't have it on there. We didn't, yeah, it was section eight, we skipped. Prismatic beams, if you remember, are those with symmetry about the y-axis where we typically take x to be axial, axial down the beam itself. And then our bending is either in the plus or the minus z direction using the right hand rule to find applied moment of some kind. So we've been looking at that. We also have been looking at cross sections that might be a little bit irregular in that they're not always circular, not always rectangular. They're always prismatic, but they've also been isotropic. They've also always been a single material in the beam. Now we're going to look at composite. They'll still be prismatic in shape, but now we're gonna allow for the possibility that part of the beam will be made of a different material. So maybe this upper part of the beam is one material and the lower part of the beam is another material. It's not uncommon to do that kind of thing, especially with structural wood materials to have plywood is certainly a composite beam. There are also lots of things where, lots of times where we might have a steel beam for a lot of strength and then maybe some wood decoration applied to it and we'll be able to look at that now to see will the wood itself be able to take any of the load or are we putting that on merely for decorative looks and the like. So we'll start out real simple with a beam just like that in cross section, we'll let it be a simple rectangular beam. We know automatically where that neutral axis is. Do you remember, I hope, what the importance to us is of the neutral axis? Sorry? It's not touching or impression. Yeah, that's the transition point between, for example, in that type of bending, that's the transition point between the compression we feel on the top for the type of bending I showed and the tension we feel on the bottom. So there'd be compression on the top and then tension on the bottom and that transition from one to the other is through the neutral axis. Remember also that neutral axis coincides with the area centroid of that cross section. That was especially important for us to know how to find that when we're looking at more irregular, though prismatic shapes. All right, so we're gonna add the possibility that there's two different materials. One material on the top, one material on the bottom with the interface between the two that can be anywhere. Need not be at the neutral axis. So we're gonna have that type of situation now and we need to look at what the deal with this is. Now, things change a bit with this different material. This can be the strain diagram. If you remember, both the strain and the stress are linear across the interface with this neutral axis being the intercept, if you will, between the two. Things are a little bit different though with the stress. The stress diagram on this is gonna be a little bit different because of the different material we have at the two points. The bulk of the material for this example will stay exactly the same. We'll have compression at one place and tension at the other. That's not gonna change. That's a direct response of the strain. But because these two materials have different modulus of elasticity, remember that's the E, then the stress diagram will be a little bit different. However, it's still linear, still intercepts through this neutral axis, but for example, if the upper material is a little bit softer, a little bit lower modulus of elasticity, then the stress will be still linear, but the upper material will only be able to hold less stress by a factor of the difference in the modulus of elasticity. We'll set that up now. So for reference purposes, I'll call the top material one and the bottom material two, that the stress is directly related to the strain. By the modulus of elasticity, so we put a one on each of those. Actually, I don't necessarily need a subscript on the strain. The strain's both the same. That's a mechanical response. And then for the lower one, the same type of thing. Oops, that should be a two there. And for both of these, we remember that it's linear. All right, so let's set up and see if we can figure just what this difference is and then how we're gonna be able to calculate what these strains and stresses are in these composite beams. All right, so here's our cross section with some place being this transition from one material to another. Beam of cross sectional area with B and height H. And we'll look at some little differential piece of area as we've done before. And we'll do that anywhere across the intersection and anywhere across the face. In this upper part here, D, F, one, the force on that little bit of interface there, of course, is the stress on that face over the area of it. No different than what we've seen before. And we know what that stress is if we remember. And then you can see that we're gonna integrate over this entire face. We'll be able to pick that up. Same thing for face two. So just everything we've got above just change any subscripts to a two. And then we'll be able to determine what the forces are across the whole face. And then of course those forces have got to balance to zero because this is a static situation. So when we do that, we get that D, F, one, which will equal one over N, D, F, two, where N is defined as the ratio of the modulus of elasticity because everything else drops out. Fairly straightforward, just a simple static setup here because the forces of force must sum to zero. I guess in this case the differential forces but it doesn't matter. It all shakes out to be the same. So our method of analysis then is to do this. Here's the original cross section, neutral axis. In this case, pleasantly enough, it's right down the center because of the regular shape of the cross section. Though it need not be. There's our transition material. Now I have to make a pick here. So I'll say for this example of N greater than one. Doesn't need to be. It depends on whatever the materials are but in this case we'll say that because what I have to do next uses that assumption but it doesn't matter. The same kind of thing can be done in the opposite direction. So in other words, E two is greater than E one here. E two, this bottom material being the stouter, the stiffer, the more resistant to bending if you will. We can't analyze that type of beam with the methods we've got. For pure bending and a prismatic beam, our methods will not handle this situation where we have a difference in material. So we're going to transition this beam to something a little bit different. If the upper material here is a little bit softer, that you see the trouble is all of our methods can only handle a single material. We can't handle this composite beam with our methods. So we're going to make an equivalent beam out of a single material that has the same characteristics as this beam. So we'll take out this stouter, stifter material that I made with that initial assumption there. If I take out this material that has a greater modulus of elasticity and replace it with just this material, I'll need a lot more of that material down here. I'll need to make the beam a lot fat. That's taking out this E2, replacing it with a lot more E1 because E1's less. Now this beam is an entire single isotropic material which our methods can't handle. Remember, this beam was a width B. So the question becomes how much of this second material or how much of the second material must I replace with the first so that I have a beam that can handle the same amount of stresses? And the answer is that beam must now be a width NB where N is that ratio of the modulus of elasticity. When we do that, a couple things happen. One is our stress now is exactly as we could calculate it before. Where the top is in compression, the bottom is in tension through the neutral axis. Remember before when I had this other material, I had that little jag I had to put in because of the different material. So the neutral axis after we change the sheet, we do have to, as we calculate, if you remember our stress is calculated from MC over I. In this case, if I'm using C, it's the maximum. And we have the understanding of the minus sign in there, but just by inspection, we know where it's compression and where it's tension. The thing is the I here is of the new cross-section. The neutral axis is where it was before, but we do have to use the new cross-section to determine what the first moment of area is. It will allow us to find out what the maximum stress is and the upper material and the maximum is in the lower material using the methods we've had before to only work with isotropic solids, solids of a single material. All right, so that's the setup. Let's give it a try. So we'll always know, and if we just know what the material is about, right? Yeah, yeah, yeah. That's exactly what we're gonna have right now. So imagine we have a steel core, 0.75 inches there and three inches high and then has on it a brass cladding. Brass is kind of pink looking, so it's a perfect choice. So there's this brass cladding here now. Maybe that's cosmetic. Maybe it's for erosion purposes or something brass does not rust as readily as steel might. And we'll let that thickness be 0.4 inches on both sides. Either way, we can replace the stiffer material with much more of the softer material or we can replace the softer material with much less of the stiffer material but the calculations all come out to be the same. So it doesn't matter which you do just as long as you do it in the right direction. But I'll pick it such that we'll take out the steel and put in the brass. So we look up E for steel, we look up E for brass and they're different from about a factor of two. Give it a take a little bit. Is that to be 1.93? So I arbitrarily choose to replace the steel with a higher modulus of elasticity with the softer brass. I'm gonna need more of the brass. So now our cross section looks something like, well we still got that 0.4 inches on this side but now I'm gonna take out the stiffer steel and put in more brass to replace it because the brass can't hold as much stress on its own so I'm gonna need more of it and then I'll have the original 0.4 inches over here. So that this is now a single material in cross section. How much do I need to put in this inside? N times the 0.75 which is what? 1.2, 1.448 I believe. That makes the entire beam now conveniently because calculations always work out perfectly. Oh, not 2.5, 2.25 inches if I recall or if I did it right. And now we do the very same calculation with that one which we've been able to do before. All right, give it some kind of load. Let's say it's 40 kip inches. Also known as a kip inch. Brandon, how was spring break? Brandon, he's busy. Come back. Ended a little early, real busy doing what? Not bumming around like some of the rest of you. You did what you do. As hard as Brandon did, because Brandon barely made it in here today. You came bouncing in, smiling your face. Let's go. All right, so we've got some things that we need to calculate here. For example, we need to find the modulus of the, sorry, the first moment of area, the moment of inertia, which is actually the second moment of area, of the new cross section, which is a too big a deal because it's still rectangular, so that's nice and easy. We've got all those numbers. It now has a base of two and a quarter inches and the height was still three inches, right? Yep. And so, straight calculation of what that comes out to be. 5.063, I believe. Remember, this is all of the new cross section. Maybe new should even be in quotes. It's not the true cross section. We're not actually replacing the beam with this solid brass beam. Just doing it so for our calculations so that now we can figure out the maximum stress. Remember, this is all brass, so this will be the stress in the brass, which is different than the stress in the steel because they have different modulus of elasticity. Luckily, the moment of, or the neutral axis is the same for both. We don't need to, for this example, we don't need to change that. We will in a second. Oh, yeah, that was the 5.063. Oh, sorry, inches to the fourth. You're gonna skip per square inch, which is 119. Now, here's where we need to be careful because that's not the same as the stress that's in the steel. The strain is the same in the two because they're mechanically connected, but their response to that depends upon, their response to that depends upon their individual modulus of elasticity. And so we will find the stress in the steel by simply multiplying it by that factor n, which is about double, but not quite expected because the greater stiffness of the steel, it can take almost twice the stress. With those two numbers, then we can decide are those dimensions appropriate for what we need it to withstand in life? Straight forward. Gets a little bit more complicated when we have non-rectangular cross sections. So if you've absorbed that simple start, we'll do one of those. If we do a clean start, we gotta crawl before we can walk, walk before we can run, run before we get hobbled and have to use a cane and we start all over again. All right, so imagine this beam. We have a T beam. Of course, symmetric about the Y axis. So this is steel, 200 millimeters across the top, 20 there, 300 down, and also 20 there. This flange is 300, but I'm gonna wait for a second before I put in my dimension on that. Now, what we're gonna do is perhaps for decorative purposes, most people like wood more than they like. Like a couple pieces of wood, doesn't it? Should've brought brown chalk for today. That just wasn't thinking. Most people in their homes prefer to look at wood over steel, so maybe we're putting some wood in there for decorative purposes, but also expect to take some of the load. And that load, oh, and then down here, 300 millimeters, 75 cross here, and of course, symmetric on both sides, all the dimensions in millimeters. So we do have a little bit of the steel beam hanging out at the top there, sticking out. For steel's 200 gigapaxi scowls, for the wood, I think I picked oak, so it looks nice, 12.5, quite a bit less. We need to find them 50 kilometers. And remember, that's what we were doing in the shear moment diagrams last, well, two weeks ago now. Find the maximum stresses. We need to know at both the steel and the wood. If we make it a steel over the wood, I think is 16. So if we take out the brass, which is much greater in its modulus of elasticity, if we take out that steel, we're gonna have to put in a lot of wood to replace it. So we can do either, but typically, if you just do the same thing as what we've set up, then it's less likely to make a mistake. So we're gonna have to put in a whole bunch of steel, take out the steel, put in a whole bunch of wood to replace it, where now this across the top is 16 times 200 millimeters. That was the original dimension, the 200, the bottom is now a width of 16 times 20 millimeters. Then we've got the original 75 millimeters of wood, or maybe that looks about like that. That's our new beam, and that's why it's so big because the wood, we just need a lot more wood to replace the steel we took out to get the equivalent response. All right, which means, of course, we're gonna have a new neutral axis that we have to determine because we need to calculate our eye with response to that. Remember how to find the neutral axis from, it's with respect to that, that we need to calculate this new moment of inertia. Pick some reference point, and then from that reference point, calculate that quantity, that is the reference point, and from there. It's not too difficult, it's kind of two simple composite shapes. And we need to figure out where the individual parts are. Probably just as easy to do that, that's part one, part two. It'd be much simpler than that. Probably the easiest way to do it tends to go pretty straightforward. You don't have to, but anything else, just extra calculation. Piece one, with respect to our arbitrarily chosen reference point. Yep, of course. Halfway down to the second one, let's see, that's 20, and then half of the 300 is 170 by 16 by 200. Remember, we're doing this with respect to the new area. We have to find out the new neutral axis, not because that's where the stresses are zero. We'll actually need that for the old axis, but because that's how we calculate our moment of inertia with respect to that area. What, two times 75 plus 16 times 20 times 300, in fact, to this arbitrary reference, and then otherwise, we know it's right in the center, because we're doing prismatic things, don't we? With respect to that neutral axis, it might do well to make another table for that as well. That's all with respect to this new neutral axis, this new century. 120 millimeters, not meters, new cross-axis, remember, with respect to that neutral axis. Yeah, you got that? Yeah. It's up to you. Like a table art book happens to just write this out, but it's up to you. You'll need the centroid of each of those with respect to that neutral axis, and then the d is the distance from that neutral axis. I see, remember, it's just the 112th ph cubed straight out of the book for rectangular cross-section, and then ad squared is the parallel axis there, and part of it. It helps, if not, you don't have to. If you want another spring break, so you can do it the right way. 120, and then you get i with respect to that. You have this upper little piece. D is the distance from the new neutral axis, which we just located as the 120, you found distance up to that. This other one has its neutral axis, or down there, so that would be d2. They're not drawn in the center, but they would be, because these are prismatic beams. Those are symmetric about the y-axis. D6, y of mine in meters would be different. Times 10 to the left, 6th. Yeah, still doesn't sound quite right. What'd you have, Bobby? Hey there, anybody else? What do we get? Little numbers to go over, so he's just going. That's this, checks of the neutral axis of each individual piece. That's what the NA stands for, with respect to the neutral axis that we calculated, which is why we had to find it. Times 10 to the minus 4th, and minus 3rd meters to the 4th. Do we confirm that, Bobby got that? Yeah, TJ got it, okay. So do 80 squared, even the square? No, yeah, but very easy to forget those squares. Throughout this maximum stress, 50. The wood is the greatest in stress, the greatest distance away from that, which is the 200 millimeters. That's the new neutral axis, and then over the move, I hope you was killing the Newton's per square meter. Remember, that's in the wood, because for this part, we're imagining that's all we've got. Then we've got to back up and get the stress in the steel, which will be a lot greater because it can hold so much more. 0.6 megapascals, the steel will be n times that. That's at the same location, because that's still the farthest distance, that's where the steel will also be in the greatest stress. And so we gotta make sure that both the steel and the wood can stand that. We don't need to go back and redo this, taking out the wood and putting in the steel, which we could, we'd have the original T-beam with little tiny pieces of steel, or replace the wood with steel, the 116th, the original with gear. But these numbers should still come out to be the same, so there's no reason to do it again other than as a check. I'll take home, so you don't have to worry about, you can easily double check all these little numbers you're calculating. What's the whole test about? Chapter four, five, and six. So we have steel loading, corrosion, and then