 Let's see who took a look at that problem I left you with on Monday. Any volunteers? All right. But we're going to need the very same things that went into that anyway, so let's review it. What we were working with was our ability now to deal with composite beams. We had worked out how to find the stress in beams of homogeneous nature, all one material. We're still working with prismatic cross sections, meaning they're always symmetric about the y-axis. That's going to continue. There's a section in the book that deals with non-symmetric beam cross sections, but we're not going to get to it if you need it. You can double check it if and when you do need it. But what we have worked up is the ability to do beams of different materials. And so I left you with a beam, something like that, 200 millimeters across the top, 20 millimeters on that flange there, as well as down here on that t-section. And that, I believe, was a steel part of the beam. And then bonded to it, we had, I believe, some kind of oak timbers. Specifically, what it was is too much of a concern because I believe I gave you the needed Young's modulus. So there was our set. That looked vaguely familiar. Oh, and I had a given load of 50 kilometers. All right, so we're looking there at the cross section of the beam, this being our y-axis up that way. And it doesn't matter really where the origin is, the z-axis that way. And that's the direction of our moment using the right-hand rules. So if our thumb is in that direction, then the moment is such that it's the usual type of bending if we look down at a great length of the beam, something like that, that being the x-axis. So that's where we left it. The technique we're using, we don't know how to find, oh, by the way, we are trying to find maximum normal stress in the beam. We only know how to find that for a homogeneous beam, a beam of a single material. But this is a beam of two materials. So our bending stress equation doesn't apply. It only applies for a single material. This is a two-material beam. But we developed the technique where we would take out one of the materials and replace it with an appropriate amount of the other material. And then we did have what we consider an equivalent beam. It's very artificial in its size and shape, but it does allow us to accurately calculate the stresses in the beam there. So it doesn't matter which of the materials you take out and which you replace it with. You get the same answer either way, but we have to do one or the other. So did I actually suggest to do the one way or the other? All right, so I have to pick one or the other. I'll replace the steel with wood in the beam. So anywhere there's a steel part of the beam, I'll take it out and replace it with wood. But I can't replace it with the same amount of wood that there is steel, because steel is very, very much stiffer, you can tell, by the much, much higher Young's modulus. So we employ this factor n, and we replace the very stiff wood with n times more, sorry, replace the very stiff steel with n times more wood than we had the steel. And that extra area will then be able to absorb the same stresses, and that allows to calculate it as a single cross-section beam. So n comes out to be 16 here. So I'm going to take out the steel and replace it with 16 times more width of wood. And then we'll have an old wood beam. It's absurdly artificial in its cross-section. So I've got the steel then is going to look something like that now that we've taken out the steel and replaced it with wood, where this top piece now is 16 times the original 200 millimeters. But we're using wood instead of steel, so we have to use an awful lot more of it. And the bottom down here is 16 times the 20 millimeters. And that now, as ugly as it might be, is all wood. And it looks like wood. That's what I'm amazing sketched at. And then we leave the original wood in there, which I forgot to give you the dimensions on that, 300 by 75. So the wood's 300 tall, 75. That remains the same. So we have this additional 75 here. The whole thing was 300. That doesn't change. Now we have a beam that's all wood. Remember, it's just an artificial equivalent beam to what we originally had. But it's now of a single material, homogeneous material. And so now we can employ our cross-section, our bending equation. We're given mc, remember, is the distance from the neutral axis. The trouble is we don't know where the neutral axis is. Seems apparent that it's going to be somewhere about there, but we don't know where that. And we need that because c is measured from there. And the moment of inertia, area moment of inertia, is with respect to that neutral axis. It appears that the distance c will actually be that one because the neutral axis is going to be a little bit more near the top because of the little bit greater area of that upper flange part that was still in there. So that's all the setup, bit of a reminder from what we did on Monday. And then everything I gave you to do on Tuesday. I mean, sorry, I gave you on Monday for today. So who did it? Anybody? Thank you guys for doing work in your other classes. We're going to have to stop that. All right, we'll do it real quick because we're going to need this procedure for what we're doing with the extra part of today's work. So we need to find out where this neutral axis is. So as easy as anything, it's just picking a reference line of some kind and measuring it from there. You can pick a reference line anywhere you want. So what I happen to do for this one is midway down the bottom part, I pick my reference line. So that puts it 150 up from the bottom and 150 the interface with the top flange. No particular reason to do that. It's arbitrary wherever we pick it because what we want to do is locate this neutral axis. And once we've located it, then it sticks to the cross section and we don't have to worry about anything from there on. We don't need an arbitrary reference point because now we know where the neutral axis lays and it becomes a permanent part of the cross section. So this is a relatively easy thing to do. Maybe I'll call that piece one. And this whole bottom piece is piece two. Remember, you don't need to do these different sections because the whole bottom piece is wood. There's no reason to split it out into the different parts. So we've got those two pieces. We need the location of the centroid for each of those with respect to our arbitrary reference, the area, and then we need the product. Sue, and it's the area and the product of the area and the individual centroids that we then add up. So we have two pieces here. With respect to our arbitrary reference, it should be, let's say, 150 millimeters up and then halfway into the beam to the centroid of piece one. So that looks like it's 0.16 meters or 160 millimeters. Does that agree? Everybody understands specifically what you're looking for there in that first part of the table. It's very important when you need to take your time doing this because it's really easy to accidentally slip and look at the wrong thing or just go through it a little too quickly. So we're looking for the distance up to that point. So we have that extra 10 millimeters there. The area, 16 by 200 by 20. I don't know if I have that number separately. What? 64,000. But that's millimeters? Yeah. All right. So we just have to check the units when we go. I just didn't have that number in progress. And then the product of the two, might as well do it millimeters, either just swap back all at once. We'll just make this one millimeters because that's 160. So John, 160 times 64,000, 1,240,000. And that's millimeters down to Q. And then we do the same thing for two, 10 million. We probably would have caught that because we know about where the neutral axis would lay. And then the second piece, it's the location of its centroid, with respect to my arbitrary reference point, is 0. Because this goes right through that. What that does is make things an awful lot easier. It eliminates, or it doesn't eliminate the area. But it does eliminate the other two. Just makes the calculation a little cleaner, a little simpler. And so the area of that becomes the 2 times 75, plus 16 times 20, times 300, 141,000. We're trusting John on this. We could do worse, I'm sure. And then whatever those totals are, and we can then finish this. That one's easy, I'll leave that one. And that's a millimeter cubed. These two, what should that be? 2, 205, it's 49 millimeters. We'll call it 50. So we know that this distance right here is now 0.05 meters. Seems about right, about where you'd expect it to be. All right. What next? Now that we've located the neutral axis, we need to figure out what the moment of inertia of this artificial cross-section is with respect to that neutral axis. And remember the moment of inertia of a compound shape, which this is. It's got the long rectangle top and the fat rectangle bottom. It is the sum of the individual moments of inertia. Remember though that this one requires an adjustment for the parallel axis theorem. So the moments of inertia with respect to the centroid here is pretty easy. That's the 112 BH cubed. It's straight out of the front cover of the book. Nice regular shape. And then the A, of course, is the same area we had there. The middle column for each piece. And D is the distance the individual centroids are from our neutral axis, which for the bottom piece, we already know that happens to be y bar. So you can see some of the wisdom there making that as our original reference line. A lot of the numbers are already available. All right. Without belaboring that too much, you can double check that you get the same thing here. I got 2.19 times 10 to the minus 3 meters. And it's meters 2.19 times 10 to the minus 3. All right. You can double check that one. Again, if you prefer, it does do quite well if you make a little table out of it. Very much like we did here, where you have the individual values of the centroidal moment of the area and then the parallel axis there. That's all geometric mechanics, I guess if you will. But now we can get to the actual material mechanics by applying that bending equation we have there with the appropriate values that we will have. Remember, though, that we have two different materials and we have to figure out the stress in each. We've made this artificial beam entirely out of wood so if we use those straight numbers, that's what we're going to get is the stress in the wood. C being the greatest distance from the neutral axis. It could be up or it could be down. We're not sure it could be different for anyone of the problems. In this case, it happens to be down. And so that distance is, the whole piece is 320 across. And we're up 0.05, remember, or 50 if we're going millimeters up from the neutral axis. So C comes out to be 0.2, 0.2 meters. That's what you just said, John? Got that? That's this distance here from the neutral axis down to the very bottom divided by our moment of inertia. That's meters to the fourth. And that will give us kilonewts per square meter or kilopascals, 4.6 megapascals is what I have. Sounded about right? That's the stress we expect in the wood at the bottom because that's the farthest distance away from the neutral axis, remember. When we have a uniform material and we look at it from the side, we know that the stress is linearly, nearly varying on either side of the neutral axis. The top being a compression, the bottom being an tension. What, however, is the stress in the steel itself? That's the stress in the wood because our artificial cross section here was an all-wood beam. But now we've got to go back and figure out what the stress is in the steel, remember how we find that. You have to be a little bit careful with it. For the most part, it's the same calculation with a little bit of an adjustment. Since the steel can have so much more of the stress, it's the factor n, just make sure what you use for i. i again is the greatest distance from the neutral axis to the farthest bit of steel, which is the very same number we used before because it's this part of the steel down here that's in the greatest stress. Might be tempted to go up to the top because that's where most of the steel is, but that's not what the concern is. The concern is, where are we the farthest away from the neutral axis? So we use the same C we used before. And so in fact, it's just n times the value we had before. n times 4.6 megapascals, 73.1. And again, then for each of those, you can compare those to the yield stresses to see if we're safe, whether one or the other will fail first. Mostly a refresher, just with a new problem, because we're going to need that kind of stuff for what we're doing next. We need the very, very same ideas. So any questions with that one? Everybody comfortable with it? Phil, you're all right? Samantha, comfortable? Like that? Colors are OK? All right, colors are awesome. We all know that. All right, so we're going to take the very same idea and do another composite beam. Remember, we have two terms we use. Compound cross sections are cross sections made up of several different shapes. Composite beams are beams made up of different materials. So we're going to do another type of composite beam now, one with which you're all very familiar, maybe not in the specifics of it, but in the generalities of it. One of the most common building materials, especially for structures, is concrete. However, concrete beams offer their own advantages, but their own disadvantages as well. The advantages are that it's very easy to cast concrete beams in a wide variety of cross sections, some which may be chosen for the aesthetics. They'll just look nicer than maybe a steel beam might. You can even do a measure, sculpt concrete to look all kinds of nice ways. So if what you wanted was a T beam, if you do it in concrete, maybe you can make it so that it'll look a little bit nicer to the eye if those beams are exposed. You can also do other things very easily, cast cavities in those for IT cables or piping or any other pieces you might need. It's also rather easy, even in significantly tall skyscrapers, to deliver concrete to the site in its liquid form, pump it up, cast it in place, and then you don't have to carry in the significantly long beams that some of these buildings might require, especially as architects are trying to open up spaces, meaning there's less interior walls, longer stretches for beams, and some of these newer buildings. Concrete's also not all that expensive. It's, for the most part, dug out of the ground. However, there are things that worry us in the use of concrete beams. In any beam that's loaded and supported in our usual way, our simplest way, we look at simply supported at either end and then some kind of loading in between that is downward, whether it's a uniform load or a point load, it doesn't matter. The trouble is there's going to be, generally, a moment like this, the very type we've been looking at. And when we look at the cross section of the beam, and we'll just keep it simple and use a nice rectangular one, we know that for that kind of bending moment on a beam looked at from the side in that way, that the top is going to be in compression. By the top, I mean anything above the neutral axis. Below the neutral axis, everything will be in tension. The trouble with that is with a concrete beam is concrete is terrible in tension, which means the entire bottom of the beam is exposed to almost guaranteed failure for a simple concrete beam, a beam of nothing but concrete material throughout a homogeneous beam. You've got this entire bottom part of the beam, especially the lower part that's exposed to extreme tension, and concrete is terrible in tension. So the solution is then to not have a single material concrete beam, but have a beam that has running down it the entire length, some reinforcing bars or what we know as rebar. You've all heard of it. I know if you pay attention as you come into the building, you'll see it because the concrete is starting to break away with the age of this building, and you'll see little bits of the rebar down there. Some of that is used sort of as a network throughout the material just to sort of hold it together, and it's in all direction. But we're going to look at it at rebar, running the length of the beam that will then absorb all of the tension. And we can then use the ability to use all the advantages of concrete, but have much less of the disadvantages of concrete. So here's how we're going to approach this. Here's our idea. So here's that beam in cross-section, a couple reinforcing bars running down the length. Remember, this is the cross-section. For our analysis purposes, what we're going to do is again realize that from the neutral axis above we have concrete in compression, which is no worry. Well, it's a worry in that there are limits to that, but in general, concrete in compression is not a significant worrisome factor, certainly compared to the bottom of the concrete being in tension. So we're going to transform that beam. Remember, we cannot analyze directly beams made of two different materials. So we're going to do what we did before. Replace the rebar with an equivalent amount of concrete and then analyze our beam as a single homogeneous beam from there. So we'll leave the top alone, because that's concrete in tension, and we're not real worried about that. What we're worried about is the concrete in tension alone. So at the very same level, we're going to take out those rebar pieces and replace them with an equivalent area of concrete that is a distance N, A, S in size, where N is the usual piece it was before, the steel rebar Young's modulus compared to the concrete Young's modulus and A, S is the area of the steel rebar in the original piece, which is these three pieces right there. What that means to us is that for the most part, we are completely disregarding any of the concrete below the neutral axis. We're throwing it out, it's in tension, concrete doesn't do well in tension, so let's completely take it out of the analysis if we have it in a mode where it's going to fail in the first place. And so now here's our equivalent cross-sectional area made up entirely of concrete for purposes of analysis. And then we can do our dual analysis to figure out what distress is in both the concrete in compression at the top and the steel in tension at the bottom. Got the general picture. We've got a beam made of concrete with installed rebar. We take out all the concrete at the bottom because it's in tension, even though it's only minor tension near the neutral axis, but we're going to take it all out because concrete in tension is just not an easy thing to work with in practice. We're going to replace the steel rebar area with an equivalent amount of concrete area and then use that beam as our beam for analysis. Well, remember this is an entirely artificial replacement. We're doing just for the purpose of analysis, but we'll have, yes, this concrete will be in tension, but we've got so much of it that we're all right. Number on N, we've got it coming up here. We'll do it in a second. N is like eight, at least something like that. All right. Trouble is we don't know where the neutral axis is. We need the neutral axis on this beam, so we're going to use a slightly different way to do it than we did before, and that's by, for each of the two areas, we're going to figure out where the neutral axis is by adjusting, placing it such that Y bar A above equals Y bar A for any of the area below. It's a rather straightforward, easy way to do it. In practice, it looks something like this. We don't know where that neutral axis is, so we'll call that distance X. Again, it's arbitrary. It's the same thing as what we were doing when we picked an arbitrary reference point earlier. We'll call that distance X. Let's give some total height to the beam. We'll call that L, A with B, and so Y bar A above is, remember from the neutral axis, it's Y bar is right there, so that'll be X over two. That's this distance here. Times the area of that section above, it's B times X. Remember what we're trying to do is find out what X is, so this is artificially placing it. That's Y bar A above, and if we have a different shape of T beam or something, then we just do that for the two little pieces, and I have them together, and that's got to equal Y bar A below. Y bar, oh, let me adjust this a little bit. Sorry, this L is only down to the level of the rebar. Remember, all the rest below it is concrete, we're completely throwing that out of the picture. So from the top of the beam, which is concrete that we leave in compression down to the middle of the rebar that we've replaced now in our imaginary picture. So now Y bar for the area below is that distance, which is L minus X. And L is not considered unknown, it's the height of the beam from the top of the beam down to the rebar, so that's one of the design parameters. And then the area below is NAS. We're again, N is known because we know what material we're planning on using and what their strengths are, and then we know how much rebar we're planning of putting in. So the only unknown in there is X, and then we can just, at that time, solve for it, and then we know the location of the neutral axis on our new beam. No, this is very, very thin. All we're doing is giving this an area of NAS. We're not actually giving it any thickness. So it's just, this is all L, we take X out of it, we're left with L minus X, and that will then give us X. And then from there, we need to calculate the moment of inertia with respect to the neutral axis of our artificial cross section, and then that's what we use to calculate the bending stress, MC over I. We can do it again for both the steel, below, and tension, and the concrete above in compression. All right, so that's the picture. That's the steps. Both are some of the specifics. Make sure that they work out now. So here's a beam with cross section exposed. We've got two rebar running through there, and then the dimensions are all centered to center on the rebar is six inches, with a little piece of three inches on either side. We have two steel rebar centered and separated at six inches, with three inches on either side. Beam, and this is location of the rebar to the top of the beam is five and a half inches. Oh wait, sorry, nope, that is down to the bottom. It's four inches down to the rebar. So that's all the way down to the bottom. Four inches just down to the center of the rebar pieces. So we've got an extra inch and a half of concrete below, which we're gonna throw out in our analysis. And the rebar is five eighths in diameter. I think that's the picture. All I need to do is give you some bending load for it, 40 pit fictions. Obviously, I hope, a bending moment such that the top's in compression and the bottom's in tension. So we've got a beam, let's see, it's 12 across there, 5.5 high, two pieces of rebar running through it at four inches. So that's the cross section. The modulus, elasticity, ratio, is, steals like 20 times 10 to six PSI. Concrete's about 1 eighth of that, 3.6. It comes out to be something like 8.1. So, oh wait, let me move that out of the way. Just so I can keep it right beside this, our equivalent beam. Same height, we're gonna take out the rebar and replace it with an equivalent amount of concrete. So at some unknown spot, we have this neutral axis, where all of that material above that is in compression. I just don't know where the neutral axis is yet. That's what we have to find next. And right at the same level as the rebar was, rebar was, we replace it with a very, very thin strip of area NAS of concrete. Notice what that means is we completely ignore any of the concrete below the neutral axis. All of that is thrown out and ignored, even though it can withstand some tension, but in general it's so poor in tension we're not gonna allow it into the analysis. So that's our equivalent beam. As artificial as it looks for purposes of analysis, it works quite well. What's next? We need to find out what axis. We need to place the neutral axis and then calculate the moment of inertia from there. So y bar a above the people, y bar a below, which means 12 times x. Well, hang on, that's a. Y bar is x over two, x over two times 12x. That should be y bar a above. Agreed? Y bar a below, let's see, y bar below is the four inches minus the x. It's just the distance from our neutral axis where we don't know where it is yet down to the center of our concrete strip. So that's four inches minus x. And then the area below is NAS. We know what NAS, and we know what AS is because we've got two 5 eighths inch rebar in there. NAS is 8.1, AS is two times higher square two because there's two bars. It's what? Yeah, it's 614 square inches. That's the cross sectional area of the rebar where we're assuming all of the tension will reside. So we've got N, we've got AS now. We just solve for x, 1.45 inches, which would seem about right. Two inches is midway between the top and the rebar. And we're gonna be a little bit above that because we've thrown out so much of the bottom area. All of the bottom concrete area is gone. This is just the area of the steel. So now we know what x is. It's about where it should be. Check the teams. Now what? That was always our first step. Determine where the neutral axis is. Once we've done that, now what do we do? Do what? Find the moment of area with respect to that neutral axis. And that's, as it has been before, it's just there's a little bit of a trickier step. Not a tricky step, it actually makes things very easy. It's just not necessarily an obvious step. So if we make a little table out of it, we need to find the centroidal moment of inertia plus D and then we have plus 80 square. So for the top piece, it's just a nice rectangle. It's centroidal moment of inertia. Remember the moment of inertia about its own centroid is 112th BHQ, actually in this case BX. D is, remember that's the distance of the centroid of one from the neutral axis. Which for our purposes is x over two. Then you can actually put the numbers in. See if we get the same thing. And then that gives, oh well, I might have that number. Whatever it comes out to be. It's just plug in the pieces in now. The area is 12x. So it would be 112BXQ plus 12XX over two squared. 12XX over two squared. One half, I don't know, IC plus AD squared. I just don't have to have that separate number, so. That's just algebra. Now here's the little bit of tricky part. This strip is so skinny and narrow that we take its moment of area to be zero. No bother calculating. If we're gonna calculate it, 112th BHQ, but it doesn't have any height for all intensive purposes. All we're concerned with is its area. So it's just so skinny, we leave that out of it. D is the L minus X we had before. Cause we may not know how thick it is, but we do know where it is. And then of course its area is NAS. So this just becomes NAS times four minus X squared. That's I, which is zero, plus AD squared. You can see how easy this would be to do on a spreadsheet. All right, so that gives us an I of, when we put all the numbers in, 44.4, that can't be, I just wrote the units down wrong. I to the fourth, inches to the fourth. 44 inches to the fourth, I believe. I'll find the stress. We need to find the stress in the concrete and the stress in the steel. So you do those. We've got moment 40 kips. Kip inches, a cup inch. We've now got the moment of inertia with respect to the neutral axis. And it should be then kips per square inches, KSI. But I want to see what numbers you choose to do this. Travis, while you were out in the most important thing you probably missed, I don't think you were here for it, is when we figure out the moment of inertia with respect to the neutral axis, we do it by summing up the centroidal moments with parallel axis theorem applied for the strip that replaces the rebar. We consider it to be so very thin. We know what its area is, but we don't give it any thickness. It's so very thin that its centroidal moment of inertia is zero. It's kind of a subtle piece. I don't want you sitting there trying to figure out what the thickness of it is when we assume it doesn't really have any. And now everybody is hopefully working on the stresses and the concrete and compression at the top and the steel and tension at the bottom. Easily taken care of, but you've got to pay attention to what you put in for C to make it all along, out of the unit, so that's easy. Now for your C, what do you have? Let's see. It's the greatest distance. 40 inches. Would you use for C to find the stress in the concrete? Well, if you use this distance, what did others use? No, you're forgetting something. We don't have any concrete down here for our purposes. We've thrown it out completely. We assume that any tension on the concrete is going to cause the concrete to fail and we take it out of the calculation completely. So we're only concerned with this stress in compression. We assume that there's no concrete there to fail in tension, take it out completely. It's a conservative estimate because we've taken a lot of the material out that would withstand at least some tension, but it's so bad in tension that we take it out of the calculation. And then I, where did I put that? 44.4 inches to the fourth. So we're left with kips per square inch. Should I ask, who's my favorite student and did that right? Samantha, you did it that way, right? Yeah, Joey, you did it that way? Okay, perfect. We have, like, are you nodding that Joey did it that way or you did it that way too? All right, so one female, two male, prompments and kips. So that comes out to be 1.31 KSI. Now, what about the steel? Multiply it by n, n is 8.1, m is 40 inches, and what do you use for i? That's the four minus x, so we want just that space to the 2.55, which divides it by, is that still a greater distance? Of course it's a greater distance, but there's no sense calculating the stress down here and material that we're assuming isn't even part of the equation. We've taken it out completely. We don't need to calculate the potential stress in the concrete because we assume it's gonna fail anyway. It's there, its real purpose now is to do nothing more than hold the rebar in place. That's all the concrete does. Yeah, it's a greater distance. In fact, the distance all the way down there would be 5.5 minus 1.45. What is that on top of the top? Use what on top? 8.55 inches per cm. No, we're not figuring out the tensile stress in the concrete because we're assuming it's gonna fail. So why calculate the stress? We want to know where this concrete is actually used for the strength of the beam, which is in compression on the top. I was just backing up in other problems. It's not hard for you to do that. Yes, this is a completely different problem where we've actually thrown away part of the beam. Not replaced it with something else actually disposed of, thrown it out. And the tension on the steel is 18.5. So that's tension and that's compression. If we happen to have a concrete beam loaded in some other way, for example, if we had a beam that was maybe loaded this way, that beam we expect to bend the other way. We want the rebar on the top of that beam cross section rather than on the bottom because this is gonna have a completely different type of moment than would be the original beam that we are looking at. It can be used in any way as needed. Chris? Because you guys had a higher value for the steel. Would you, in design, use a smaller amount of rebar because you've ever done it? I haven't. Well, no, once we've got the stress in the rebar, you have to look at the yield stress of the material and then you decide, including a factor of safety. If this is approaching the yield stress of the steel, then we decide we're gonna put in more rebar. Well, that's because it's so much higher than concrete value that you've got the concrete of the kiln loop. Yeah. Well, no, we don't necessarily compare it. Again, it depends upon the yield stress of the concrete and compression, which is pretty hot. I don't have my book here, so we can look them up. I don't think I have them written down. The other thing that can be done, since I don't work in this field, and I've never worked in this field, I'm not sure what the point is, but a balanced beam is one such that the stress in the concrete is the same as the stress in the steel. A beam that's designed in that way is considered balanced, that they have equal stress in the two. I'm not sure what the point of that is. It just is there. Yeah, yeah, you'd either adjust this beam for what the maximum load might be, or you'd adjust the area such that they're balanced. Okay, one for you. Here's a concrete cast T-beam with two rebar in the very bottom there. Width dimensions, four inches there. The T-part has got a width of eight inches. This lower web is 18 inches. The bottom thickness here is six inches across. The rebar is two inches up from the bottom of the area, or let me see what that is at the rebar. One inch, I believe it's one inch diameter. Double check that, because you just have to find the area anyway, I'll just give that to you, because that I did right down. 1.57, well that's the number we need in the end anyway. That's the area of the steel. Oh, and I need to give you a load. Oh, find out our, find the maximum moment, so you're gonna have to double check which one might fail first. The maximum allowable stresses with a concrete three KSI compression for the rebar, 40 KSI intention. So find the maximum moment, which I have to do is figure out which one of these is the controlling stress. What that means is put these stresses in the bending equation, M sigma equal to MCI, put in that for the stress, and then solve for the moment, see which one's the lower, and I'll tell you which one of them. Two materials is the controlling material. All right, so first, draw yourself an equivalent cross section. You have to determine just where the neutral axis was going to be here. Suspicion, I would hope, would put it somewhere about there. A little piece there, just a compound shape, break it into two little rectangles, and then the bottom here, we put in enough concrete to simulate the steel by putting in n times lower of that area. All right, so first thing we have to do is place the neutral axis, and then find the moment of inertia with respect to that, and then calculate a maximum moment allowable by using the stress limits given, and I'll have to do that for each piece, and don't forget the factor then. All right, maybe as a hint, it might be a little bit easier if that's your unknown value X, there down from the bottom of that upper flange to wherever you might make that cut, just because the flange section here has already got a constant thickness, so that might be a little bit easier for you. Y bar A and below are equal, and the compound section, you just add those together in three sections. Because we need Y bar A, and we don't know where Y bar is for that leftover T piece, we know where it is for the two little squares, so it's always easiest to break these down into the simplest possible pieces that for which Y bar A and I are already known. But you work along and see if we don't get the same thing, and any numbers in here will be in inches. So Y bar A is equal to four, eight inches by 22. No, not you again. Y bar is the distance from our neutral axis up to the centroid of that particular piece, which is halfway through the upper flange. What is this? That's two inches, and then an additional distance X down to neutral axis. So it's X plus two. Y bar one, X plus two, or two plus X. And then A is four by 22. Eight, six, and eight. Maybe that makes sense? Is that right? Pretty much on top of these things. Quiet, so. Yeah, I was also tinkering with something of another month and wanted to see how that worked out. That's most of what I'm thinking. And then for the other piece above, it's Y bar is X over two, and its area is six by X. That's Y bar A, it's centroid right in the middle. It's height is X, so it's X over two. It's area is X by six. That's the little square here. Oh, it's two now. Yeah, it's one. So that's the area of the Y bar A product above. Below, we have from the top flange to the rebar is not 18, but 16. So that's, and then that's our distance down there. That's 16, Y bar, and the area is NAX. 16 was the whole thing. 16 minus X, right? That all right, David? Is that what you taught? Yeah, 16 is from the bottom of the top flange, but we're going from the neutral axis, so that's 16 minus X. Now NAX. Okay. That's the Y bar A, and is, whatever that ratio is there, 40 over three. Oh. 7.63. Oh wait, no, no, I'm sorry. Because that's the, we don't do the ratio of the stresses, we do the ratio of the strength. I just don't happen to have the same numbers as the one before. So different brand of concrete and the steel numbers the same, that was 2.9, that was 10 to the sixth. No, 29, that was 10 to the sixth. And the concrete on this process had to be 3.8. Concrete is mixable in all kinds of ways. Before we had 3.6, this one happens to be 3.8. So there's our N, 7.63, and the area I already gave you. All right, we're right at the end, so I'll make sure you've got the numbers before you go. X, when you solve for it, comes out to be 15.7. 15.7 is 0.1573 inches, about where we'd expect it to be. I, with respect to that neutral axis, once you find that 35, 35.7 inches to the fourth, then you can figure out the maximum moment for the two of them, whichever one is lower, is the one you'd use then. Don't forget the factor of N in there. So the moment with the concrete at its maximum would be 25.51, moment with the steel at its maximum, remember the maximum stress limits were given would be less than that, so it controls, this is 11.70. So the steel, the stress in the steel is the controlling factor, and so there's your design moment there. No, that's already got the N in it. Oh, and then yet you could apply a factor of safety to that by cutting in half, maybe for a factor of safety of two, perhaps more, depends on those. There are industry standards, there might be company standards as well. Call it legal office, they'll tell you what today's factor of safety is.