 All right, if you remember, on Tuesday, we were looking at pure bending. What did that mean to us? Yeah, our moments were in one plane. We were not looking at any kind of axial loading, it was just simply easiest picture we could possibly have, this pure bending that's caused by a load, a transverse load, and or applied moments, but those moments are perpendicular to the axial direction of the beam, and perpendicular to our arbitrary, generally chosen y direction, which was up there, then that makes these moments either into or out of the board in the z direction. Pure bending of prismatic beams, and what did I mean by that? What's a prismatic beam? As you learn from the very first day in here with the cross-sectional area, and now, starting with this week, the cross-sectional area shape is vitally important to what we're doing. If we look at the cross-section, we have some cross-section that doesn't have to be rectangular, doesn't have to be circular, can be anything we want it to be, anything we need it to be, that is symmetric about that y axis. So a cross-section, an example of that is just like that. So we came up with, on Tuesday, then the normal stress in the x direction is a factor of whatever moment is being applied. I bet you make sense, that's what's doing the bending, that's causing the stress. Times the position in the y direction along that beam divided by, and this is the last little bit we got developed there, the area moment of inertia, and we'll review today how we come up with that. Remember what the minus sign was, what the deal with that was? This is the distance in the beam from the neutral axis, and that's where we put our origin. So, finally, we need to find that, that's the centroidal area, the centroid of the area of that cross-sectional, we'll review how to find that today. Then y is the distance from that to any point in the beam. The minus sign is that, if we have curvature like that, that puts the center of curvature somewhere way up here, and then our y direction is measured from the neutral axis towards that center of curvature. This minus, then, goes with the fact that that kind of bending, y moving in a positive direction with this kind of bending, we have a beam that sees compression on the top, tension on the bottom, with that point of zero stress being right on the neutral axis, the neutral axis being right through the centroid. Let me not use C because you have a different beam force. So, if we have bending in the other direction, then the center of curvature is down below. Our y now measures towards that as the positive direction, and again, then that gives us compression on the top. Positive y gives a negative stress, which is compression. Negative y gives a positive stress, which is tension. And so, that's the purpose of the minus sign in there, that's where it came from. Then we had also, of course, then our maximum stress was at whatever point was the greatest distance from the neutral axis. And you may need to pay attention a little bit to that because it's just, it's not necessarily true for any particular cross-section we have that the greatest distance from the neutral axis is up. It might be down, and you need to pay attention to which one of the directions is that point of maximum distance from the neutral axis, giving us the maximum stress through the beam at that point. So, we'll do a problem, a rather straightforward problem. Get us warmed up again, mostly, get us warmed up again to finding where the neutral axis is, where the centroid is of a non-regular shape, and then also finding the moment of inertia of that shape. So, very simple problem, about as simple a problem as we can use, so we're not focused on that. Just a mid-load, four kilonewtons, and then we don't even need to mess too terribly with what the moment is, it's almost a straightforward calculation. But let's use then a non-regular cross-section of the beam. So, this is some kind of T-shape, maybe as simple as two boards nailed together then used as a floor joist or something is the type of picture we have here. Dimensions, all of these in millimeters. So, if the top board's got a width of 10 and a height of 150 millimeters, then that bottom board is 15 by 120. Want to find a couple things. Now, we haven't messed with this before. Well, we've calculated the stress on nice regular shapes. We've never done this before with an irregular cross-sectional area. So, it's a good chance for us to use our, remember these are called the flexion formulas. So, I want to find three things. The maximum stress in the beam, whether it's tension or compression, we need to know that. If this is wood, wood is superb in compression, very bad in tension. For this kind of bending where we'd expect tension at the bottom of the beam, this sort of beam might make some sense because now there's more material down here that we'd expect in tension where wood is the worst. Just the type of thing you would do if you were building something, if you're worried about it being weak at some place, you're going to put more material in there. Just the very same thing you would do. Then I want to find also the stress at a couple spots at point K, which we'll put here a distance 40 millimeters down, and then a point here on this flange that we'll call H. All right, so we want to find the stress at those points as well. For whatever reason, if we were laminating something on there that's very, very thin, it has nothing to do with the structural integrity, but you might be worried about that lamination bundling and coming loose, and then there's an aesthetic problem with the beam because not only do beams sometimes have to be structural, they have to be pretty in time, so we got to work on all that stuff. All right, we already know how to find these stresses, whether it's the maximum or at some particular point, we'll use our fletcher formula. But what we do need is that first moment of the area, the moment, area, moment of inertia. All right, if you remember, there's a couple things we need to do to find that. First thing we need to do to find that, first thing we need to do is find the centroid itself, and then once we've found that, then we can find moment of the area moment of inertia. So it's a two-step process. The two steps are very much the same. I don't, if I remember the way our book does it, I don't like so much how they execute these steps. And I think it's very easy to get confused with the way they do it. So I'll do it, I'll do it my way. You decide which one you like, whether you like doing it with the way the book does it, or whether you like doing it this way. I don't care which. But to do this first part, to find the centroid, we have here also then two steps we need to make. First thing to do is pick some reference location. It doesn't matter where it is. It's an arbitrary location that allows you to put in all the values you need with respect to that reference location. And then once you found the centroid, you don't need this reference anymore because now the centroid is a part of the cross section itself and doesn't need any reference from there. So pick some reference location. It doesn't matter where. You can pick the top of the beam, some join in between or the bottom of the beam. It doesn't matter. So just to go with my numbers, I'll pick the bottom as a reference. And then from there, we calculate where the location of the centroid is. And that's our Y bar. A lot of these cross sections, you can look at them and get kind of a feeling where it should be. You should remember that because there's a big chunk of area down here, that that's going to bring the centroid down towards that somewhere. So we expect the centroid to be someplace in there somewhere. That'll help you because when you get through with the calculation, if you find that the number puts a way up here, you know you did something wrong. And you just have to go back and fix it. Maybe just simply something as simple as a minus sign. That's our distance Y bar that we're looking for from this reference point. Once we know what that is, then we know where the centroid is in the cross section and we don't have to worry about anything else. So to do that, we need to take the individual parts. I happen to have two parts of the beam here. So maybe just for reference, I'll call them A and B. We need to find this product for each of those. However many sub parts there are and we can divide it up any way we want. Add those together and then also divide that by the total cross-sectional area. Now what the book does, if I remember, the book then works out this equation. It just expands this equation, puts all the numbers in. I think it's a real easy way to just get something wrong when you can be a little bit more regular about it. So I like to make a table where I'll put in there the values for each of these two things. Keep them straight. It's just a lot easier way to do this without making a mistake. It's not a big deal on a two-piece beam, but we're gonna have four, five, even six-piece beams that we could do. And so it's, you know, get in practice with this one, it's easy. Then it's no sweat when it's much more difficult. So we're gonna need those two values for each of the sections. We're going to need that product for each of the sections. So it's gonna be really easy setting this up on a table. And then we need the total of that third line. That'll be the numerator. The total of the second line, that'll be the denominator. We just pull them out of that table then and survive. Also, though, pay attention to what the units are. So this one will be in millimeters. Obviously, this will be millimeters squared. This will be millimeters cubed. Obviously, it's very important we get the units right, as always. All right, for piece A, we locate its centroid with respect to our reference line. This is piece A, a simple rectangle. It's centroid, its own centroid is right in the middle. Remember, we're looking for the centroid of the whole piece, but this right here is the centroid of an individual piece. It's in the middle, which is 75 plus another 15, which is ninth. For piece B, its centroid is, with respect to our reference line, the dead center of this rectangular piece at seven and a half. The area of A, 10 times 150, 1500, the area of B, what's 15 times 120, 1800. Those two numbers, we need added up. So 3, 3, 300. Then we also need the product of each of those in the third column and then those added up. So you can do those real quick. Careful with these tables, some of these numbers get really big. And if you haven't left yourself enough room, you just have trouble writing them all in. So you do those numbers real quick. Also, make a little habit as you do this, that you line things up so you're less likely to make a mistake when you're adding numbers because for a lot of the students, the best mass students in the school, you're also the ones who might be most likely to make the easy mistakes rather than the hard ones. So that totals to I think 148.5, right? This is just to locate the centroid. That's the only thing we're trying to do here. With respect to our arbitrary reference line, then we'll know where it is and we don't need the arbitrary reference line anymore. Pull these numbers right out of the chart. That's adding up the two above it. Is that right? I didn't do that right on the wrong day. And then we just pull our numbers right out of the table, which is what makes it so easy. Notice the unit's workout. This is a distance that we're looking for here. We're gonna be left with millimeters. What's it come out to be? 45 millimeters. That locates for us now the centroid or what we now know to be the neutral axis. That's the point where once this beam is in pure bending, that's the point that we'll have zero stress. And then the stress varies linear from there. It doesn't even matter what the area is. It simply matters the y distance from the neutral axis. So we already know that if we look at our theme on the side, we're gonna have that upper part there and then this little flange part here and the neutral axis, we now know to be 45 millimeters up. And so we already know what our stress distribution is gonna look like. We don't know the values of the stress, but we do know what the distribution looks like. Everything above the neutral axis with the type of bending that we have will be in compression. Everything below the neutral axis will be in tension with a linear distribution in between. Doesn't even matter that this is much wider down here. The stress distribution is still linear throughout the entire beam. So we've located the neutral axis, which was good. That we now know goes right through the centroid. So we found that. Now we need to find the moment of area. That's the same basic type of thing that we did here to find the moment of area. To find i, then we just plug everything into here. Again, it works very well if you have a little table where in each of these, for the two parts we happen to have in this problem, we need to find the area moment of inertia of that piece itself. We need to find the area moment of inertia of piece A and that of piece B. Put that in there. And then if you remember, we have what is called the parallel axis there. And so we have to apply that because we want this moment of inertia to be with respect to the neutral axis. And all of these component parts that make up this cross-section are off of the neutral axis by a certain distance. And we have to apply that little bit. So then we can put whatever we want in the table, what was things up. Let's see, we've already got the area. So I'm just running out of space. So you can put out whatever much you want in there. And then it's that last bit here that we need to add up. We need to find the area moment of inertia, each of the pieces with respect to that neutral axis. So I see, let's see, you have your book. The front cover of the book over here are the location of the centroids with respect to particular directions for some very regular solids. And that's exactly what we've got. We've got two rectangular solids. The only thing we have to be careful of is making sure you get the right dimensions in the right direction. As drawn in the book for a rectangular solid, it looks like this. And we need the moment of inertia with respect to this x direction because that's what our neutral axis is. It's this horizontal direction. It doesn't matter which direction is rectangular. We just fill in the parts there from there. So Ix is 112 BHQ is the height. And it doesn't matter whether it's this shape or this shape, we just change what the B and the H are. So for this first one, it's 112. B is 10. That's the horizontal dimension across the bottom. A is 150, so it's 112, 150, Q. It was dimensions up to the fourth. And so you can see that's gonna be a big number. But obviously, you can't get the 10 and the 150 reversed. You gotta be real careful with this. Take the picture exactly as it lays right out of the book. For part A, B is one, is 10, H is 150. Just leave it in the orientation in which it lays. And you got that number? Remember, they get big. So you gotta give yourself some space, 2,812, 500, 500 millimeters to the fourth. You can go to meters, that'll make that number smaller, easier to put in the table. But if you go to meters, then these numbers come smaller. So it's kind of a wash. That's the moment of inertia of piece A. It's a centroidal moment of inertia with respect to its very own centroid. But its centroid is a distance between the two. This centroid is a distance D away from our centroid here. So we need to put in that D. Centroid of A is right down the center here. So that distance is 75. This distance is 45 that we just calculated. So the distance, little distance D is 75 plus 15 is 90 minus 45. Should be 45. See if you agree with that. Then you need to do A times D squared. I'll fill in these numbers. And you determine the next line for part B to make sure you're reading it right. And then we do IC plus AD squared to get the last part. So the numbers are getting pretty big. Sorry, what? D is the distance between the centroid of our entire cross-section that we found in our first step. The distance between that centroid and the true centroid of the component piece all by itself. So a nice big drawing, carefully done table can make a big difference here. So are you guys, see if we all come up with the same numbers on the component piece of the cross-section B. Obviously the fewer pieces you can make out of the cross-section and the more regular they are, the easier they are to handle. So let's see if you get the same values for the second part, the second piece B that I got. Maybe it'd be Jake and Pat and me who don't even come on regular days. You thought, oh, it's the last day for spring break. We'll have a party. Nothing like pizza day 30 in the morning. You guys are college age bachelors. You probably had cold beats for breakfast anyway. Better cap and crunch. Oh, don't have any milk. I'll use beer for the college age bachelors. Boy, did I give that life up. That sucks. It's terrible. What? Beer and cereal, old beer and cereal. That's not disgusting, it's the best. Any of it. Beer is the best. And the best part is cap and crunch and beer are about the same color. So it doesn't look disgusting. It just looks normal. Got some numbers, Jake. Pat. How'd you get the last number? The I and I. This one? Yeah. It's I plus AB squared. Because we're doing the parallel axis theorem on each of the little pieces. Each of the pieces is off of the neutral axis and we want this centroidal, we want the area moment of inertia to be with respect to the neutral axis of the entire cross section. We got some numbers, we can check. Bobby, you still going? Another 30 seconds or so and you can put in it. That on camera. Yeah, you should, I should. Do a little commercial for who is that? We're doing a commercial for? Apple. Apple. I remember two or three years ago you came in with one of the first eye touches and had some pictures on it that weren't appropriate for sharing during class time. If I remember. Yeah, I remember, I think I remember your phone went off right in the middle of the class. You started giggling. There's a text, yeah, yeah. I notice not how you even do people come to class in a moment. Which is wonderful. It wasn't that she can present all of it. All right, what do you have for the centroidal moment of area of the second piece? Which is the equation and calculation right out of the cover of the book, the 112th DHQ. What do you have? 3375. For D, right, 37.5. Remember that's the distance between the centroid of part B and the neutral axis we found in our first step. That's D. And so it's seven and a half up to the centroid. And then take that away from the 45, it's all the way up to the neutral axis. And then AD squared, well A, but A we had was the 1800. So this is just straight math here. I'm gonna have to trust you on that one. 2531, 250, 2250, is that right? And multiply well, 2565. That's the last column that we need. We need the sum of those two because the moment of the area of a component, a compound cross-section is the sum of the moments of each of the compound sections. So that's 8, 4, 1, 5, 0, 0, 0. And that's millimeters to the fourth. We're ready to do this. The moment, I think everybody had or you could have done real quick. Oh, I didn't give you a length of the beam. You can't find the moment without a length of the beam. Sorry about that. Five meters, length of the beam. So it makes the moment then five kilonewton meters. We just use our statics to find that business. And then I, we have now here, 8, 4, 1, 5, 0, 0, 0. Now you gotta watch your units. That's millimeters to the fourth is in meters. So you've gotta get that straight. And then the y location is the distance each of these points of interest are from the neutral axis. So we can figure those out straight away. That's the maximum moment by here at the center. In fact, if we look at this beam, you know by symmetry that each of these is two kilonewtons. So the shear diagram will go up two, down four, up two. And then the moment diagram, that slope, which is two is minus two. And we know the slope comes down to zero because it's just a no moment end. And then that maximum slope. As you'd expect from your own experience is standing on boards when you were a kid, they're more likely to break dead center because that's where the maximum moment is. Nothing here you haven't learned as a kid. You were doing all these calculations on the fly in your brain as a kid. Sigma max, we have all those numbers. What M we've got in a Y or sigma max. The greatest distance from the neutral axis, that's up to the top, that's 120 millimeters. Probably in this case, it's as easy to make that meters into millimeters. In the end, we generally want stress and passcals. So we've got to get to that anyway. Oh, and then the I we have. So your answer, you can do the multiplication, but the units are as important as anything. So tell me what this is in mega units is anything else because you throw these numbers off by a factor of 1,000. You just cost your company a $150 million loss when the building comes down. Don't come crying at me. Because I've got it on tape. I told you to watch your units. Yes, mega passcals. So minus, we know it's going to be compression with the type of bending we've got. What do we get? This is 71.3 minus 71.3. Now, you have to check if this is indeed wood, you need to check the bottom intention as well. Even though it's not as far away, the tent wood is about 1 tenth the strength and tension that it is in compression. So you'd have to be 1 tenth the distance away from the neutral axis to have the same type of load that we would have there. So you definitely want to check the bottom in tension to compare it, but we didn't say what the material was here. So sigma k, those two numbers the same, the m and the i are the same. What do you put in for y? Let's see, it's a whole distance was 165. That was 45, meaning that's 120, meaning this is 80, isn't it? A is on, 80 is on the same mag border, mag is on 120, so your conversion should even be pretty easy. Frame break, Wednesday, that requirement, it just makes the, the numeral itself easy to handle. What do we have for sigma k? The normal stress, axial-nominal stress at point k. 47.5 minus 47.5, the minus meaning it's in compression, which we knew anyway. So a lot of times we don't even mess with the minus sign on these because it's obvious from the problem what it's going to be, but just to be careful. What you put in for y of point h, because don't forget it's out here on the flange, so what do you put in for h? Let's see, this was 45, right? And that's 15, so we're at a point 30 below the neutral axis, but we're out here on the flange. So what do you do about that? Doesn't matter. The stress distribution is only with y has nothing to do with x because, or sorry, that's z, because we're looking at prismatic beams. So this is actually a minus 30 because we're in the direction away from the curvature, center of curvature, which is good because that tells us that this is in tension. You do those numbers, 17.8 megapascals. And for wood, ultimate strength of wood in compression is somewhere about minus 30 megapascals. So we're well over that anyway. We're not going to use wood here, which that's a pretty thin piece of wood, 10 millimeters. So you would have expected that. For tension, the ultimate strength is about a 10th of that. So good luck with this one, fellas. This is how you're building your boathouse, which is what everybody's working on in the spring break, right, working on your boathouse, getting it ready for the coming season, huh? Another question? Yep. I mean, it's probably like I'm coming in there, but I try to find the eye doing the way the book did it or like it takes, they'll take like the whole face and like the total light that's attracted by like, the part of the stuff that's out there. Do the beam not as two pieces like this, but do it as one beam like that and then you subtract out those? Yeah. Yeah, you can do that. I think that's a different part, I don't know why. Probably screw it up. Probably. The centroid is not, the neutral axis is still at the same place wherever that is. And then you've got to get the center of that, that's your D. You've got two of those. So when you add them up, you don't have to do it again. You just take one of those and add it in there twice because they're gonna have the same area, moment of inertia with respect to the neutral axis. And then you've got this hole on here. Don't forget that it's centroid right at the center. Oops, I have to put the neutral axis right through the center. But the neutral axis, as we know down here somewhere, so you have to have that D in there as well. I can just do 112 base times the height of the whole square and then subtract it by 112 base times the height of the smallest. Well, for one, don't forget, I already have it raised. It's not the base times the height, it's the base times the height cubed. Okay. No, you can't just do 112 base times the height of the whole section minus the smaller sections because what you miss is the parallel axis there in part of it. You have to do that. The only time you don't do this is when D happens to be zero and that's when one of the components has a centroid that's right on the neutral axis anyway, which is pretty easy to happen if we have, say, not an I-beam, but an H-beam. An H-beam, you know automatically, just by looking at it, the centroid of this whole area is going to be right there. The neutral axis is going to be right there and every component piece has its centroid right on the neutral axis already. So for that kind of thing, you're not going to have this column being there because D's going to be zero for everything. That is a cross-section that has not just symmetry about the y-axis, but also about the z-axis, which we did not have here. We only had y-axis symmetry. We didn't have d-axis symmetry, okay? So check your numbers. It should come out exactly the same. There's no reason you should get a different I doing it a different way because that's just an arbitrary choice by you. As the laborer, it can't affect the end mathematics, which is all this is. All right, any questions? Yeah, we've got one minute, no sense starting anything. This kind of stuff is pretty easy to do to set up on a spreadsheet and then you can put in different widths for things. So if you're interested, I can give you an extra credit problem for break that you can do fairly easily on Excel, if you're interested. If you're not even going to look at it, what? Let me explain it real quick when you're off. It's to look at a couple cross-sections, both an I beam and an H beam, but you do the same thing with them. You vary, I give you a total cross-sectional area and then you vary the dimensions, you vary the dimensions however you want. Oh, that'd be one, that'd be two maybe. You can vary the dimensions however you want so that you stay within that area. So in the extreme, H one is the same as H two and you have a beam that's like that. At the other extreme, B two is at a minimum H one and you get a beam that's like that. Somewhere in between is a beam that has the optimum characteristics of the cross-sectional area for that beam. That's the kind of thing and then you build a spreadsheet where you vary H and B, two of them I set, two of them you vary them and then you can make the graph of the stress as a function of however thick the pieces are and you get a pretty interesting graph. I think that I remember it looks something like that. So that's the kind of problem I can put up there if you want an extra credit problem to do in Bahamas. I'm not taking my backpack. You don't take your backpack, take your laptop, take grandants, yeah, can you do spreadsheets on that? Yeah, yeah. That sucks, you can't get away from it then. You got no excuse. All right.