 Forester, where's Forester? Forester, what do you know how to do, Forester? Yeah, I know, you can't hear me. They'll settle down in a day or two, maybe. What do you know how to do? Yeah, what do you know how to do? Yeah, I know, you know how to put on your pants. I'm going to talk about that kind of stuff. I mean, in this class, never mind. Just say here. OK, good enough. Carrington. Carrington, what do you know how to do? That sucks. Tidwell, what do you know how to do? OK, how might that be? Laddle towards no buckling, could. What else? It could plastically rupture. I don't want to go to somebody else. I don't think anybody else knows. You're my only hope. What else? Yeah, well, but that's rupturing. That's plastic. That's kind of figured the end point. What about the flanges and the webs? Yeah, they can't do that. They can't. They can't. That's what they can do. Yeah, that's what they can do. Thank you. Appreciate your coming today. OK, back when we were talking about lateral torsional buckling, mostly, we had. Links below which you were able to develop the full plastic moment in a graph really meant to help you in design. You'll always have a fee plugged in there for you. Then you had below this LCP, you had the full plastic moment. Now these don't tell you whether or not you're going to have web buckling, but you're not, of course, or flange buckling, which it's possible. There are a few that have that. So you'd still have to check that. So that's why he says here, compact shapes. Then you had a length beyond which the radius of gyration is really killing you. And Timoshenko's equation took over. You had a straight line interpolation between the two points, which fit the data pretty well. Your LCP was somewhere, could be in here, could be in there, could be in there. Now, bending strength of non-compact shapes, that means where the flange or the web might have a tendency to buckle, the web will never have, for any of our 50KSI roll steel shapes, or anything of lesser strength steel, only if you've got some, I don't know, some 80KSI steel, which is so strong that the flanges only need to be very thin. They might be subject to this, but only the flanges of a few of these shapes would possibly buckle. Now here is, we find the equation. Here's the equation where you determined, in this region it was plastic, in this region it was straight line, in this region it was plastic, in this region it was straight line. There's your starting point, here's the height of the drop, there's the length of the drop, and you're here minus you started at LP, which was right here. I just want to show you how similar the web behaves. As a matter of fact, it's the identical equation, except you don't get any correction for the web buckling, because they're never going to be long enough such that you may be being cheated by Tim Schenkel's equation. And here is that equation. You'll notice it doesn't have the c sub b. As the m sub p minus the starting point minus, this was your m sub r. This was the moment that it dropped to. This was m sub r. Same equation. Same equation. And the only difference is instead of having a l sub brace minus l sub plastic over l when the radius of gyration starts killing you, minus l up to where the plastic moments no longer are true, you have the lambdas instead. It behaves in exactly the same fashion. Now the only time this is ever going to really happen is if you really stress the flanges. I mean, these effects of the non-compact shapes hurting you, they'll hurt you in here, because you're asking to run the stress in the flange all the way up to the plastic limit f sub y. And they might hurt you in here. But down in here where the lateral torsional buckling is significant, there's no way that you're going to even be given equations to check out in there for buckling of the flange. It's only going to happen when you really are highly stressing the flange. Equations are the same, like I say. So here's the old equation with the l's in there. Here's your new equation. Same, same, same m sub p. Only difference is you put in the flange parameters instead of the length parameters for the beam that you're choosing. You'll notice for the flanges buckling, the l sub b is not a factor. Secondly, equation f 3 1, which is the new one, it's almost identical to the old one f 2 dash 2 on these pages, except there's no c sub b, since there isn't any l sub b. And there's no corrections that'll ever be given to you because Timoshenko's lying to you. Timoshenko doesn't even know you till you get way out in here. And then lambda sub f replaces lambda sub b. Your break points, of course, are different, because your break points are designed to tell you when the flange is buckling locally rather than when the entire beam is buckling globally. Don't remember what these are. Well, it's probably got one here. There we go. Absolutely. This is compression in the flange. The tension flange will never have this problem, because its intention and all it does is tend to get straighter and straighter. Well, the load is coming in. Here's the piece of the beam. You just extract it from anywhere that you're worried about it having problems. Here's the bending moment that's applied. Here's the force in the top of the flange. Here's the force. Here's the stress distribution. Looks like that. Then it looks like that. Then it looks all plastic. And sometimes before you can get everybody here all to plastic, this guy probably already is all plastic, but these people are still trying to get up to plastic. That means more strain in this one, and more strain. All of a sudden, this guy goes kind of like that. So he buckles out of the way due to the bending moment. He can also do that because of applied axial loads plus bending. The thing that causes this thing to really get yielded or pass yielded is a candidate for this problem. Here were those things I was looking for. There are the flexure numbers. The flexure numbers are 0.38 and 1.0, 0.38 and 1.0 for the flanges. Here are for the webs, but they just never control. In our case, these are just about going to have to be made up of steel plate and welded together, trying to get a beam that's stronger than anything that they roll. Here were the corresponding numbers. Turned out the thing printed on both sides. Here were the corresponding things for lambda subar for non-slender and slender for columns. And then, of course, the L subar things. They didn't have any local buckling stuff like this. No, not really. Not if the whole beam is subject to this. If this was a constant moment and it was 20 feet long, I don't know where it will buckle. But wherever it just happens to be the worst little bit out of shape, that's probably where it's going to buckle. So this is for the beam local buckling. This is for the flange. Excuse me. This is for the beam global. I guess maybe that's what I said. Lateral torsional buckling. This is for the flange sticking out. This is a local buckling. So very similar. Here are the break points of interest to us. Langes. Possibly, but not until you get in graduate school and start putting together your own beams by welding plates together. Got an example. Got a simply supported beam 45 feet long. This particular beam is W14 by 90. He wants to know whether or not the shape is compact, non-compact, or slender. And he's willing to do a bunch of work to find out. Here's a W14 by 90. It's not compact. Got a little flexure problem mark right there. But, you know, here's where he does it. He says the B sub F over 2 T sub F for that shape is listed on this page. That's where I got it listed, copied it. The lambda P and lambda sub R are the 0.38 and the 1. Those are the break points. That was 9.15 and 24.1. Yours is 10.2. So sadly, yes, you're down in the region where it could locally buckle. It is non-compact. And then he says, check the capacity based on flange local buckling. Well, you see the equation is identical, except for this is lambda, this is L braced minus L plastic over L sub R minus L sub P, same numbers. So there's nothing to do except stick new numbers in for the flange as opposed to for your beam. This is a property of the beam itself. These are in the tables. Far as L sub B, that's your business. Then L sub P and L sub R, those are properties of the beam and they are in the tables. This, you brace it as close together as you would like to. Note that most of the time when they work these things out, especially in a text, they don't put the fees into the very end. But they're gonna find out how strong it is due to flange local buckling. Then they're gonna find out how nominally strong it is due to local global buckling, not local global buckling, lateral torsional buckling. And then they'll take the lower of the two and then they'll stick the fee on there. Say, well, I don't have to do that book once. The reason I mention that is because if you look in the book for almost any of these numbers, you'll find M sub N, but it doesn't seem to be this one. The reason is, oh, here we go. Here's a W14 by 90, it's got flange problems. And here is your M sub PX. And see, it's already got the fee on it. So notice that whenever you look in the tables, they're gonna put the fees on there for your convenience during design. Whereas if you use Tim Schenkel's equation, he doesn't have the fee on there in the equation. You gotta add it later. So there's your M plastic. Wonder where he got that from, M plastic. You think he got it from this thing right here? I doubt it, 574, because this has already got the fee on it. And he's working it in inch kips, instead of kip feet, and he doesn't have the fee on M plastic. Now there's where he got it right here. So the plastic moment is equal to F sub Y times the plastic section modulus, 78.50. You wanna divide that by 12 and multiply it times 0.9, you'll get the number out of the table. Right here, fee sub B M sub plastic about the X axis for this shape. So we went ahead and worked out the nominal strength. This is the plastic strength minus, since he had some problems with the flange buckling, he went on and applied this equation. 78.50 is the starting point. Your drop, it drops down to get to M sub R. Seven tenths of F sub Y times the elastic section modulus, do you remember what the 0.7 was for? In other words, normally F sub Y S sub X is the elastic moment at first yield. Remember why we had a 0.7 of that number? Many words have been thrown out in here, I know. Because of what? Well, it's actually because of residual stresses in this beam when it was rolled. And so that when they really go out and test and see how much this thing has got at first yield, a lot of the fibers had already yield. So it didn't come up to this number and they said, what's wrong here? The guy says, if you go take the beam and you heat it back till it's red hot after it's been rolled and you cool it very slowly so that you don't have the flanges cool quickly. And the intersection of the flange and the web cool later on and lock residual stresses in there. You get this number, but they don't do that. They don't have time for that. They don't have the money for it. Therefore, the 0.7 is in there. That's what that's for. So the 0.7 times 50 KSI, you look this up, you'll find it on one of these pages, the elastic section modulus, to get the elastic first yield number, including residual stresses, times our lambda is 10.2. The lambda for the, see here? He didn't, he worked these numbers out. He could have found them on page 16.1-17, listed. Let's see if I've got that. There we go. There's the W by 90, B sub R over 2T sub F. So what we're looking for? No, he's looking for the 0.38. Yeah, you'd have to work that out because it depends on F sub Y. But E over F sub Y is 29 over 50, and 0.38 is the first break point, one was the other one. Here are those break points again, are right here. Due to flexure, there's the 0.38, there's the one on page 16.1-17. So now that we know all the numbers are going in here, we just crunch it out, we get 637. Needs a fee on there someday. Now then, he's gonna check the capacity based on the limit state of lateral torsional buckling. He goes and does lateral torsional buckling. We've already done a bunch of that. I'll tell you what, the flanges are not going to locally buckle. Look, this sucker's all the way out at 45 feet long. L sub P is less than 15. You could have some problems in this region. L sub R is 42 feet. I've got a curve like here. Man, I mean, this guy's out in here. He's nowhere near where these flanges are going to be buckling plastically. He's not even barely in the elasto-plastic region. So that's not gonna control. But continuing, it's just a uniformly loaded beam, simply supported fast, so-and-so. Subject load, yeah. Dead load, live load, uniformly loaded. Just wondering why he was going after this C sub B number, 1.14 correction. Out of the manual, you get everything that you need. Good, heavens, he's gonna work this out of the equations from the word go. The new addition changed this number slightly. You'll find several shapes that changed a little bit. I'm not exactly sure why. Somebody measured something a little different or they went and took the average of a bunch of people who make them. But you'll see slight differences like this. Shouldn't bother you. R sub tough stuff. H zero J warping constant. There's your 1.14 that we just got off of a uniformly loaded beam with no interior braces. Old stuff, except you don't do this. It's a lot easier to probably get this stuff off of these tables, this graph here. Go track down where a W14 by 90 comes in at zero length. Track it down on the next page, track it down on the next page, track it down on the next page, track it down on the next page, track it down on the next page, track it down on the next page. You see the difference in the dots. The dots are solid, then they're that. Then track it down to the next page and that's about what you're looking for right there. Or you can work it out. This is the critical buckling stress. 37 KSI. And the nominal moment is that critical buckling stress times the elastic section modulus. 53-20, that's what I figured. Well, less than 78.50 for M plastic and it's smaller than we got for flange buckling. We got 76.50, we're down to 5,300. So if he wanted to show us how the web could hurt us, he didn't pick a good case. He could have picked a much smaller braced length. This one has F, he's warning you this one can have problems, but I'll tell you now it can only have problems here or here or here or here or here or here or here or maybe here, no way down in here. The reason being, the only reason these things buckle is because you put a full yield stress in them and probably beyond. In other words, you were, here was your beam, here were what the stresses looked like for a while, here were what the stresses looked like for a while, that's F sub y, then here's what the stresses looked like for a while. And you're still trying to get this last little plastic moment out of the web and when you did that you put so much strain in this thing that it didn't, even when it got to this point it didn't buckle. But when you put some more moment on it, trying to get some more strain and stress and pick up some more plastic moment out of the web, this poor guy says, okay, that's it, I've had it. That's maybe here, maybe here, maybe here, but it's no way. It's easier just to do the number work rather than say, I say no way, and Larry says, ha, gotcha, in that case it did, minus 10, you know, and there's no sense in taking a chance in the real world either. All right, so this is the same one, this is the one where I've got committed all the notes and everything. Now, let me show you what they do with these graphs. This is, well I can't really give you a page, number, because this is paid something, this is paid something, this is paid something, there's all kinds of pages there. But I have pulled the number forward, W14 by 90, with a superscript F on it. And if you will, and I'm gonna put the FISA B in there too, because the tables and the graphs all have that in there. The FISA B, MSP listed in the book, got that one, I might as well just get it right here. FISA B, MSP listed for a W14 by 90 with flange problems is 574, 574. And you can calculate FISA B, MSP by multiplying fee times the yield stress times the plastic section modulus, 0.9, 50 KSI, the Z sub X is 157. So let me say this is on page three dash 24. Page three dash 24, there we go, three dash 24. It's also on page 212 C. And that's that times that times that, divided by 12 inches and a foot. So that's interesting, that's 589. So the real plastic moment is 589. You know, I swear, look at this thing here, this thing says 574. And you say, what's that all about? And not only that, if you go look on this graph, you'll find this is drawn as 574. And so here's what somebody said. We got a problem with these thin flanges. I said, what's your problem? He says, well, this is the true plastic moment. And this is the true L sub plastic, 13 feet. And then it truly drops at this rate and it gets to L sub R of something. And then Timoshenko's equation takes over. I said, and your problem is, he says, well, my guys go in there and they come over here and they pick a beam to give them that much moment. Or they take this much moment and this length and they go find the best beam. I say, great, that's why you charge them $200 for the book. He says, yeah, but look at this. They come down here when you're less than L sub P on these guys that have an F superscript and they come in here and they think they get this much moment, 589. And I say, no, they don't. They know they may not get 589 because they see the little F as a flange problem. They'll immediately go to that equation for flange local buckling and they'll see it's not 589. So nobody's gonna die because my guys know what they're doing. So wouldn't it be simpler if it could just go to the graph and pick out the number without having to check for flange local buckling? I said, well, yeah, I would, but how you plan on doing that? He said, we're gonna lie. We're gonna list 574 as the real plastic moment for anybody who's got an F. Gonna put the lateral torsional, I'm sorry, the local flange buckling number here. And we're gonna draw that as a solid and we're gonna erase that one. We're just gonna throw it away. And I say, well, then you got a problem because when they take this number and this number to go down that straight line, you're gonna tell them that this number is this number and they're gonna have a straight line drawn from here to there. He says, we figured that out too. We're not only gonna lie about the moment, we're gonna lie about L plastic. I think in our texts, they call it L plastic star. Like be careful, this is a weird thing. And I said, okay, and where do you propose to put this dot? They say, we're gonna put it right there. And so this is what they'll see in the book. They will have listed for them L plastic is 15.1 feet, not the number given by the equation for L sub P. So let me see, I'm not sure, I believe that. W14 by 90, L plastic turn. It is, it's what they did. So that means you're starting at the right height, 15.1, and you're gonna go from 15.1 at 574 and you're gonna drop at the proper slope to find out what's going on in that region. Point being for these that have the little F-sum, there's some hanky-pank going in on there. That is because you're never gonna reach that plastic moment. That is because you're never gonna reach that plastic moment. You're only going that high for these flanges. Why is that because? Why is that because? Why is that because? Why is that because? Have a little roll of textures here, man. Tsk, tsk, tsk, tsk, tsk, tsk, tsk, tsk. He'll never notice, tsk, tsk, tsk, tsk, tsk, tsk, tsk, tsk. But that's okay, I need the money. Next year, same class? Same time? No? Oh, well, I don't know about that. All right, so watch out for that. There are some numbers that are put in there that aren't quite true for your convenience. Summary of moment strength. Basically, I would suggest you write down M plastic, tell lateral torsional buckling, flange local buckling, web local buckling, and you start out, you know, go get the plastic moment, see what you can have because of that, see if it's gonna laterally torsionally buckle, if it's just continuously supported on the top flange, you can mark that off the list, flange local buckling, see if it's got an F, if it's got an F on it, then that means it could have, but only when the beams are relatively short and highly stressed. And of course, the equations to do it are here, and the numbers are in the tables for almost everything. Now, next way a beam can fail, it can fail in shear. These are very highly loaded, especially if they're relatively short. The stresses in the beam due to bending, they are distributed linearly, starting at the neutral axis at zero. Then of course, the stresses build up, this reaches F sub y, then this one reaches F sub y, and finally they all reach F sub y. In the shear stress, as opposed to the bending stresses, those stresses are distributed parabolically. You learned that in 305. You learned that the equation for the shear stress in a beam was VQ over IT, and they told you that for a roll-steel shape, it was pretty close to the shear divided by the area of the web. Now, I forget about how much, some of them are off maybe 10, 12%, something like that, which didn't seem like just a little bit, but the truth is everybody knew where we were going. We were going where the stresses aren't distributed parabolically. We're gonna have all of those little fibers in there up to the yield in shear, where this is really very true. So shear force to be applied divided by the area of the web that would be the depth of the beam times the thickness of the web. This information's on this page, 16.1-67, in the specs. Here is a typical pair of curves for the tensile test that I did the other day. Here's the shear stress test that I did the other day. The slope of that line was about 29,000. That's the modulus of elasticity for steel. The slope of this line is about 11.2. You'll notice it's much less stiff in shear than it is in tension. Not only that, it's not nearly as strong in shear as it is in tension. This number right here is right at six-tenths of that number. Now, this is F sub yield in tension. You've never seen that symbol and I only made it up just now to emphasize the fact that that is the tensile yield. And you'll never see this again for a stress yield in shear because we don't ever use it. What we use is six-tenths of the yield stress in tension because this number really is right at six-tenths of that number for all the steels that we use. Rather than having an F yield in tension and an F yield in shear, we just use, if we ever want that, we just write that down times 0.6. You see a 0.6 just about anywhere in equation that's usually working with shear and you'll see the corresponding tensile F sub y number associated with it. That works on the yield. That also works on the ultimate, both. So shearing stress is load over area and that stress can run up to 0.6 times the tensile stress and it's okay. It of course has more capacity than F sub y, could go on up to F sub u. There's some things that won't like that. For example, buckling of the web might occur and they'd never be able to reach F sub u. There's another reason we don't go up that high. Generally speaking, you're gonna tell me that the ultimate request for shear on the beam, where are you gonna get the shear from? From the load combination. The load combination can be put on the beam and where are you gonna get the shear from? What are you gonna do next to give me the worst shear in the beam? The louder whoever mumbled it was right, maybe not, maybe they were saying, shee, man, how would I know? I thought you were saying shear diagram. And a shear diagram. You draw a shear diagram, you pull the biggest shears off the diagram. That's what they're for. That's why they pay you to get them. You'll get a shear diagram. You'll tell me the request. Then you will have to make it less than. The ultimate request is gonna have to be less than what? Less than what? There, that's right. Somebody back there, their hand knows, they just hadn't got into the brain yet because they were going like this. See, and I stick it out on the end. I don't care where it goes. What I had before was just the nominal strength is this much. Then I was thinking, okay, well, let me remind them that the ultimate has to be less than, has to be less than phi v sub n. So there's your yield in shear. There's your area of web, stress times area gives you force. You can't guarantee me that force for all of the pieces you tested times an appropriate phi factor. Could be 0.9, could be 0.75. I don't know yet, we'll have to check the code. Could vary. You saw that in 305 where they said, okay, that's close enough, we'll just call it uniform. And the size of the beam, the area of the web is considered top to bottom times thickness of the web. The part really subject to buckling is after it comes out of these radiuses, and then it may have a tendency to buckle. If you remember for a beam that is bent where the tension side of the beam is in tension and doesn't buckle, just this piece up on the top might have a tendency to buckle back into the paper. Now then we're talking about is gonna buckle between where it's fixed on the top, fixed on the bottom. So the numbers are obviously gonna be different. This buckling is going to be left and right. This buckling is gonna be back into the paper. An old pop quiz, imputed in plastic for that given beam. Solutions there. Student did a nice job on it. Here we go, AISC specs for shear. First off, the ultimate has to be less than an appropriate resistance factor for shear alone times the nominal strength in shear. As we'll see, the resistance factor and the factor of safety, and the safety factor of the factor, that's some other people that don't play that game. The resistance factor depends on the width, webs with the thickness ratio, as you might imagine, the thinner it is, the worse life gets. For the, he says the basic strength equation for all the stuff we're gonna do now is that the nominal strength is six tenths of F sub Y times the area of the web, and then to take care of the different thickness ratios on the web, we're just gonna put on a number called C sub V. A sub G, whether you know what it is, depth times thickness of the web. D is the overall depth of the web, very top to very bottom, and C sub V is what they call the ratio of critical web stress to the yield, shear yield stress. It's the ratio of when it's gonna buckle to the yield stress. And it depends on whether the limit state is three ranges, ooh, I've been down that road before, web yielding, web elastic buckling, or something in between, web inelastic buckling. You've been down that road too, there's yielding, there's inelastic, and there's elastic. Depends on where on there your H over T sub W occurs. So we go first for case one. It's a hot rolled eye shape. If your H over T sub W is less than, there's a break point, you know where we're getting these break points. Little bit of theory, a lot of bit of practical, a lot of testing, there's your break point. I wrote down the number 2.24 times square root of 29 over 50 for 50 KSI steel, number works out to be 54. If your H, here's your H, that's D minus K design, minus K design. It's listed in the book. If your H over the thickness of the web, tending to buckle, is less than 54, then we have found that C sub V is a one, and you get to use no variation at all. I'm telling you, if your web is that fat and short, you'll get every little fiber out of it every time. No.9 necessary, we don't find enough variation to even count it. What is this, gray? You know, I do this in front of you guys, it's not like I'm home drunk or something, gray. Oh, I know what gray means. What does gray mean in your book? Commentary, thank you. I'm gonna write that down, commentary. There's a user note on that page, there is indeed. It's in the commentary, it tells you why we're doing this and how to do it, and how you expect you to behave. Correct. There's most shapes with 50 or less KSI fall in this region, but you know, it seems like they never do on an exam. Case two, for all other singly and doubly symmetric shapes, we want you to use a .9, because we start to find some variation. And that'll include some other things down in here, several other things. And C sub V, you always need a phi sub V and a C sub V, because here's your equation. Here you have a V nominal, so you need a phi sub V to go with it, and there's your C sub V to get the final answer. You're gonna need a phi sub V and a C sub V. It says I want you at this point, for anybody beyond this limit, the limit that I just told you right here, 54 for 50 KSI steel, I want you to drop your ability to use a one and a one. I want that one to drop to .9, because we found some variation in those more slender webs. And C sub V is to be determined as follows. If the H over T sub W is less than 1.1 times square root of K sub V E over F sub Y, duh. Okay, I understand this. K sub V, K sub V, K sub V. Here we go, where K sub V is a five. Is this value is for unstiffened webs. Well, one way you can get some more strength out of this thing is to put stiffeners in there and force the web to buckle in a little closed cell. And, but we're not doing those. Those get to be very interesting and you can get a lot of strength out of them if you can afford to weld all these things in there. But for unstiffened wells, K sub V is a five. If your H over T sub W is less than 260. He says now, the section doesn't say what to do past that. It's not an upper limit, but someplace else he puts a limit, proportioning limits for I shaped members. And he says it shall not exceed 260. So he says that's what we all consider as the limit. And for, now that I know this is 1.1 times square root of five times 29 divided by 50, that number turns out 59.2. If you're a ratio, what was the old ratio? The old limit was 54. Now then the new limit is 59.2. If it does not exceed 59.2, please drop your fee from 1 to 0.9. And, oh hey, you still get the same C sub V? Yeah, he says you still get the same C sub V. He says this is the only thing that changed. Okay, so from a 54 to 59, I'll drop the fee to this region and I still use C sub V is equal to one. Does it corresponds to shear yielding? Says now then, if you go from 59.2 up to 1.37 square root of the same number before, this now turns out to be 73.8, you're no longer in the plastic region, you're in the inelastic web buckling region. Okay, what does that mean? What do you want me to do? He says, well I still want 0.9. This is for everything from here on. And C sub V changes. It's no longer one is 1.1 times the square root of five times 29 over 50 divided by your H over T sub W. Well, I know this much of it. I know that I can work out the 1.1, it's 59.2. So the only new thing is C sub V now changes to this. He says, and if you go past this number with your H over T sub W, you should do one for bigger than this number, this number is 73.8, square root of five times Z, five times 29,000 over 50 times 1.37, says if you're in that region above that region, then you get a new C sub V. Bummer, man, are you sure? He says, yeah, I'm sure. You got nine break points on me here. He says, you don't have to do it that way if you don't want to. It's just you won't ever win any bids. I say, okay, I think I see what to do. Seriously, gotta go get my H over T sub W, find out where I fall so that I know what to do. Here's a graph of what you do. First off, your nominal shear or fee nominal depends on whether you put the fee in there. Up to this value, H over T sub W, up to this break point, you use a fee V as a one and a CV as a one and it's constant. Up to beyond this, they insist you drop the fee to 0.9 and C sub V is one. Then from the 1.1, which turned out to be 2.46 or something, I don't remember. And to this one, that's a straight line. Here is your equations, G2.4. And out in this region, here is your equation and you've got a one, a 0.9, a 0.9, a 0.9. You've got a CV of one, CV of one, CV based on H over T sub W, CV based on H over T sub W squared. Just what you gotta do. That's how strong these webs are when you put load on them. See you next time. I don't have any idea. I was planning on going through chapter nine. Yeah, chapter nine. Well, but we'll finish that by Wednesday. Or you could read the syllabus and it'll say exactly where, because otherwise everybody in their brother would be asking me when, what is it, cover? I know, I know. If somebody didn't, I would be disappointed. Yes. Oh, good. Is there anything that's given in the tables that we can find here? You do not. If you have a table that, right, I could very easily ask, see if you can do that by say, verify the number in the table. And you would, I did have one person say the number in the table is this and this is the number you asked me to check and this is equal to this, so I verified it. You know, you'd have to use the equations. Right, okay. I wanted to make sure. Full access. So we can use the table. Full access, yes. You would have to say which table. You would have to tell me what page, right. Because a lot of times you'll write down C sub V is equal to 1.63. And I look and it's not 1.63. And I don't know where you got it from. So did you make a serious mistake, you know? Or did you just look on the wrong beam? So I need to know exactly where it came from. Okay. That would be fine. Okay. That would tell me you know what you're doing. Okay. You can do that. You've said. Well, you do need to know how to use it. And if I really was worried about you not knowing how, I would ask you to verify a number in there. Sure thing. Where did, oh, there it is.