 All right, say here, if I called your name, Savoy, not here. Let me repeat the instructions. Anybody got a spare eraser? Raymond? Barry? Odds? Odds? Odds? Never mind. He's here. Curtis? Mr. Curtis, what are we doing? What is all this stuff? What is this thing, this one minus piece of U over piece of E1? You remember what it was? No? Okay. Toborga? You remember, what were we doing here? What was that for? What was its purpose? I'm sorry, any less por favor? That is correct. It is for calculating bending moments. This thing right here, though, this amplification thing. Why was it amplifying? What was it? I mean, it was amplifying moments. Calzado? Not here? Casey? Okay. Why was it amplifying them? I mean, I don't know what happened. All of a sudden, it made us amplify the moments that we get out of 305. Clark? Clark? Okay. That's the direct answer. Casey? Yep. Oh, it already got you? Never mind. I forgot to check you off. Okay. Hardy? Here. That's all I'm going to get from him. Lahog? Yeah. Come on. Lahog? Why did we have to amplify those moments? Why? Yesterday. Well, we did it two days ago. Maybe that's the problem. Well, the problem was when we got it in a box, it wasn't straight. And so when we put the column load down it, we had taken into account the fact that it wasn't straight for light loads, and we put a .866. But when you really put some heavy loads on it, the fact that it wasn't straight had a little p delta, which that .877 took care of. But then that load caused the column to kick out to the side even more delta. And that more delta caused you to have a little more p delta. And so it's accounting for the fact that the column is starting to buckle just as soon as you put the load on it rather than buckling all at one time. It's a correction. It's an amplification factor for column loading. And rather than a lot of more time, the c sub m, you could have had it bent under the worst situation possible. Which is kind of what we did if we put, if you put two moments of the same magnitude on the ends to cause that bending out, that's above and beyond the original little delta that it came with in the box, then that was about as bad as it got. We turned that into a sine wave, you can use just about anything you want to and derived it on that basis, but that was your worst case. And you told me in this case, that's not, I'm not doing it that bad. I said, well, what are you doing? You say, well, one of my moment goes this way and this one goes the same way. And so rather than them both kicking it out to one side, one of them kicks it out a lot, but this one brings it back. And that's not as bad as you're saying you should amplify one over one minus p sub u over p sub e, my moments that I was planning on designing for. So okay, you know what the one over one minus p sub u over p sub e one, if you're just don't care and not willing to work, you could always make c sub m equals to one, but if you did that, you're probably not going to get the lightest section possible. Unless you really have equal moments on each end, it's just about as bad as you can get in. You're not going to get any coefficient or Christmas present or consideration or you're not going to be able to knock the moments down. This number is going to be big and it's going to be big times the regular old 305 moments and that'll tell you how much you must request. Now the c sub m has been determined more, a lot of theory, but a lot of experimental stuff too. So they find out how much amplification is knocked off when you do less than the worst case. What they find is if you have first off no transverse loads, moments at the end of this column only. Those very often come up, for instance here's a frame, it's loaded down here, it cranks the joint clockwise, it's loaded over here that pushes this down, rotates this joint clockwise, pushes it up, rotates this joint counterclockwise, there you have moments on both ends and no loads in the middle. Now if you have wind loads, someone has made sure that they go with the joints only so that the column doesn't have any lateral loads on it in the middle, which we'd have to take care of in some other fashion. But a lot of times if you just bend a structure, you know, move it to the right, move the joints to the right, there'll be moments existing inside of those columns. And if they're pretty bad, you're still, if you put M2 on this end and M2 on that end, you're going to pay the full price because that's as bad as it gets. But if you say, whoa, but I don't, I got 300 kit feet on this end from my analysis and I only got 100 on that end, well then you deserve some relief. And the relief they have found that you should be able to correct one over one minus PCV over PCV is six-tenths minus four-tenths of the absolute value of the small moment on either end divided by the absolute value of the big moment on the other end. And then they say, but if both your moments cause single curvature, that's not as bad, you should, you should get this number. But if you say, well not only is it not as bad as it could be, it's not even that bad because these things cause reverse curvature, then you deserve even more consideration because you are making the deltas smaller. What they do is, and what they found is, if M1 and M2 cause single curvature, that's pretty bad. So they say M1 over M2 is ratio of bending moments to the end, M1 is the end moment that's smaller, M2 is the other one, and the ratio is positive for members bent nicely causing reverse curvature. So it says here the ratio is positive, so you come up with your ratio and it's positive and your correction is six-tenths minus four-tenths of that ratio. See how that knocks C sub M down? And that gives you some really nice corrections to your amplification factor. However, and negative, if the moments cause single curvature bending, single curvature bending means they both go the way that kicks the beam out the furthest. In that case, that ratio of M1 over M2 has to be entered as negative. So it's six-tenths minus four-tenths times a negative number and a negative number times a negative number adds to this, and so you go from 0.6 and you have a plus this number rather than being 0.6 minus maybe 0.2. Now it's 0.6 plus 0.2 up to 0.8, and as this thing gets bigger, then you have less correction. If you don't want any correction at all, it's a one. If you have reverse curvature, maybe it's 0.6 minus 0.2 for example, and if it's same curvature, it's a 0.6 plus perhaps a 0.2. So that's how you handle the correction factor that you really have coming to you. We stand behind you when you take it if you have moments at the ends only. So that would not apply to this case, and the reason is this is an unbraced frame and these corrections are only for braced conditions. We'll get into unbraced. That's rather than p little delta. These are p times big delta and have different corrections for those. Then for transversely loaded members, he says, okay, it's not quite as easy. You can just take it as one. In other words, you can just take nothing. Well, you can always just take nothing, but that's sometimes not cheap. There's a more refined procedure we have found is provided in the commentary to appendix 8. You'll find that on page 16.1-525, and the correction factor is 1 plus psi times the ratio of piece. I don't know what r is. What is this for us? P sub u, thank you very much, because the piece of r is the generic term for a loud stress people and LRFD people. It's P sub u divided by, and we all use that symbol, piece of oil or one. Pi squared and that kind of stuff. And he says, I'll be glad to tell you what psi is for different cases. And you can fit almost anything into these cases, and if you can't, then you're just going to have to go ahead and let it be one, or do some good engineering reason why it's one of these. It's on page 16.1-325. Here's your psi. I don't know why they bother with psi, because all of this right here, here's C sub m. This is C sub m with psi plugged into this equation. Incidentally, what is this? One. That's right. When you see that anywhere in our class, day one through day 43, you ought to say one. And there it is for a fixed pin column with axial loads on it. Here's for a pin pin column with axial loads on it. C sub m is one. Here's your alpha, because he's still got alpha in there for everybody. Piece of r means piece of u, and piece of b1 is that oiler load. Fixed fix, darn if it's not the same. That's interesting. I guess we could ask why. I'm sure somebody has a reference on why, because somebody had to figure out it was the same. Here's a pin pin with a concentrated lateral load. These were uniform. Here's a fixed pin, but the load has to be in the middle. Here the load's in the middle by default. I don't know why I had to bother putting that over, too. Don't know, because I'd put the load in the middle here also. And here's one with fixed on each end. And you have different size, so you have different correction factors that are available to you if you need them, or if you have them coming to you. So this one does not apply, because C sub m is only for the braced condition. Here's a braced frame. This one would be the 0.6 minus 0.4 m over m number. And this one would have lateral loads on it. So this one would be this C sub m with a psi in it. And this is what I showed you on the previous page. Here he has about the worst case possible. And this, of course, is much less severe. So for an example, this example came from page 306 in Segui. Case two, Appendix 7.2, is the way he analyzed it. I think here he says, right here, the effective length method. And that was case two on this page. This wouldn't be a AISC manual. You'd have to be in Appendix. It's going to have to have one of those higher numbers, something like that. Oh, there it is, probably 16.1-509. That's probably where Appendix 7.2 is. So the guy used the elastic of the effective length method to get the loads. Now, you're constrained. Of course, you work with him, so that's not a problem. Or her, that means you know what they use to get the loads from the computer program. And so you use that to analyze the stuff. First, you are not permitted to reduce the EI if you want to use the effective length method. You're going to use a 572-grade 50 steel. He wants to know if it's adequate. Case of X and case of Y for this case. He's telling you what the effective length was for this one, 1.0. Has dead and live service loads. And he has dead and live service moments. And on this end, he has dead and live service moments. They're both about, this is W12 by 65, they're both about, it doesn't have like it's a X or a Y. What is your opinion? Is this about the X-axis or about the Y-axis? Certainly about the X-axis. If you have somebody bend in your column, that's the first thing you're going to be doing, trying to get it oriented, so it's being bent about its strong axis. So when you put these rotten moments on the end, you're going to make a kick after the side of Delta. And Mr. Euler's going to have an opinion about the buckling load. And you're going to ask him, please give me the buckling load about the X-axis. That's right. Because that's the axis about which you caused a Delta problem, a P Delta problem. And so when you get into the equation and it asks you for Euler's opinion, be sure you put in XX numbers. Now if somebody said, look, I've got moments about both axes, then when you do the load, you'll be finding moments about the X-axis and you'll be asking Euler's opinion. And when it bends about the Y-axis, hopefully you've at least got the big moments about X and the small moments about Y, then you'll be putting in all the Y numbers for his opinion about the Euler buckling load. P sub u, 1.2 times 70, 1.6 times 210, we're really going to ask for 420 kips unless that has to be amplified. M sub u, X, 1.2 times 11, 1.6 times 36, he's going to ask for 70.8 kip feet unless that has to be amplified. No bending about the Y, moment about the X on the other end, 82.4, and none about the Y. So he says here are the factored loads. And as the factor load amplified, this one has to be amplified just like everybody else, is equal to the P, no translation case, plus B2 times P with lateral translation in place. What you'll have to do is if you actually had a frame that could move to the side, was not braced, you'd have to go brace it, just put little pins all over the place around the joint so it doesn't translate and run an analysis. That would be what you'd get here. You wouldn't put pins, you'd put rollers because you want the columns to be able to displace, so you'd put a little roller there, you'd put a little roller there, and you'd analyze it, and that would give you all of your loads with no translation. Then you'd come in and you would move it over to the side and you would allow it to translate and then you would find the axial loads due to that. In our case, we got 420, yeah, I see the 420, there's the 420, and we have B2 piece of LT where we have no LT, because our frame doesn't need those little pins, he said it's a braced frame. It's part of a braced frame. So we have none of that, no LT. Check it out, this is equation A82, it's an appendix 8, a section where we're doing this kind of analysis, and it's on page 16.1-237. So as a quick summary, we're getting ready now to amplify the moments. We're going to take our requested moments, they're called M sub u about the x-axis, but now then since they may be amplified, I'm going to use the M sub u as amplified for my request, so I'm just not going to have anything here, just M sub no translation about the x, plus B2 times the moment with lateral translation about the x, again this will be 0, and I need to know B1. So here the P sub u's are recalculated, here's the P sub u calculated for with lateral translation, which we had none. 1.2 times 11, this is a recalculation of this moment about the x was requested, but as of yet unamplified, here was the moment about the x on the bottom as yet unamplified. It's bent about the x-axis, but that doesn't stop the fact that it's buckling about the y-axis. That's usually very confusing when you first start doing it. You say, but I heard just down the page, I saw you do it, you asked Mr. Euler about a buckling strength about the x-axis, because I'm trying to find that what your moments did to my column in the x-axis direction caused more grief with the x-axis moment. So I need his opinion about the x-axis strength. As far as how much load it will really take or how strong it is, that's strictly, that's not working on requests. That's strictly a matter of how much it takes to buckle this column. Whether it be y-y, generally is, if it's just pin-pinned, or even if it's braced in the middle, x-x, those things aren't amplified in any way. They're just requests. That'll become less unclear once you start looking through these problems. The factored loads, load two combinations shown in figure 611, he's got them down here. There's your 420, your 70, your 82, and your 420, those are as yet unamplified. Factored, but not yet amplified. He says it is from the column load tables for a 14-foot column with a K of 1, axiocompressive strength of a W12 by 65. I got some column load tables here. I bought them a couple of years ago. He says for a W12 by 65, 14 feet long, the answer is 685. That's buckling about what axis? The weak axis. That's correct. Now then, if it really was buckling about the x, that would be fine. I could do it. Could I use this table, which is not made for weak axis buckling? Yeah, you simply use KL phony, like you always did before, and you can use it. It wasn't made for it, but you can use it for the strength of phony 17-foot axis, the length column about the strong axis. So we've got a 694. Where do we put it? Is that what it was, 694? What was it? 685. So there we go. There's our strength. It's on page 4 to x19. YY capacity, 315B. I have no idea. Oh, that's right. That's where I go to see the number. Now then, the whole reason for me getting this number first is so that I can get P sub u divided by P sub c, P sub n, so that I can decide whether or not I ought to be using a lightly loaded column equation or whether I ought to be using an interaction. Heavily loaded, axially loaded equation. P sub u was 420. That was our request right here. Over 685, that was our capacity right here. Oh, wow. Well, that's a column. Therefore, you're going to have to admit to all of the P sub u over phi P sub n. You will be given a slight gift. You'll only have to include eight nights of the moment stuff. I don't see that just probably on the next page. What are we doing? That's one thing I wish he would, I don't know, then I wouldn't have anything to tell you. But what are we doing? This is the buckling capacity about the YY axis. What are we doing? We're getting ready to go find the strong axis requests because we're nowhere near the interaction formula yet. We've got a lot of things to do. And not only that, we've got to get the request for bending strength about xx. And guess on the next page, we've got to get this request for the YY axis. He's working on this part of that equation. Number one, he goes to Mr. Euler and he says, I need to know how strong this thing is about the xx axis. He says, I thought you just asked me something about that. I mean, you were talking about, the strong axis, I saw you over in my YY table. I say, just get it. He says, okay. It's pi squared EI sub x over KL squared about xx. EI, this is one way you know this is the effective length method is because somebody didn't take advantage of reducing the EI. I bet you if he could have reduced the EI, he'd have reduced this. Now, pi squared, I got that. EI got that. X axis, 533, I got that. Effective length, K is one. 14 times 12 squared gives you 54.05. Mr. Euler thinks it will buckle at that number. Since, what is this all about? Since the effective length was used in the analysis, okay, EI is not reduced. He didn't. He says, well, I didn't. I said, well, I'm just noticing that. Now, what's this all about? This is a summary of what B1x is. Once we get bending about two axes, we're going to have an x and a y. B1x is equal to 1 over 1 minus p sub u over p sub E1xx if you're a weenie. Correctable perhaps. Correction for a moment about the x axis. Correction factors, amplifications can't be amplification factors unless they're bigger than the original number. So this has to come out bigger than one. Even if this person right here is aggressively optimistic and brings this ratio down to less than one. No way you can play tricks and get less moment on the beam than the computer program first spit out. These of you, you'll notice it has no extra y subscript. It just buckles about whichever axis you tell me it buckles about. It buckles about the one that has the greatest KL over R. All right, so here we go. What in the world is he doing here? He's got a 70.8 on that end. I agree. He's got an 82.4 on that end. I agree. What is he getting these quarter points for? I don't know. Let's continue on. He has some twists which are out of balance and the twist being out of balance gives him some reactions on the beam. Looks like he didn't really use them, but some of the moments about this point, this thing here overpowered this one and so he's got a little reaction on the ends of the column. The moments both go the same way. I'm sure I wish you could have figured out a way not to make them both go the same way. Single curvature. Moment, moment at the quarter point, moment at the half point, moment at the three quarter point, moment at the max, okay. What does it sound like he's getting ready to calculate here? C sub m. That's correct. A correction for the lateral torsional buckling strength. Why greater than one? We already discussed that. All right. Don't see where that table is. No, no, we just used that table. So I didn't see where that was going to be used just right at this time. Continuing. We've got a 21 by 65. We say that the correction factor is no lateral loads, no loads coming in from the side. 0.6 minus 0.4 times the ratio of the end moments, little moment, big moment, moments. Either to be entered as a positive or a negative, whether or not you took the good case or the bad case, you took the bad case, and since you took the bad case, you're going to make this 0.6 less attractive by adding a number. If you had a better case, this would be a minus. This would be a plus, and then a minus times a plus would be 0.6 and lower than 0.6, which would be an even better correction factor for this M. Incidentally, I don't know what I said back here. I seem like I said C sub M. Oh yeah, because it's not C sub M, is it? That's the symbol you and I use for C sub B. This guy here's got to have his own name. Correction of column moments. This is correction of lateral torsional buckling strength. So, holler if you say, I don't see that number. Alright, so 0.6 minus 0.4 times the ratio. 0.6 minus 0.4, the ratio has to be entered minus, so that the minus times a minus gives you less correction as is only fair for single curvature bending. 0.9 to 0.437. Single curvature equals negative. This will always calculate less than one. Well, that's good, because it would never be, it wouldn't be a correction factor if it wasn't, and we wouldn't take it. Then the amplification is, this is the generic version. In our case, it's C sub M. We just got C sub M for X. That'll be 0.94. 1 minus, dang those ASD people, times P ultimate over P over 1. Our P ultimate request was 420. Our Mr. Euler prediction was 5405. That means that you have to amplify your moments by about 2% higher than you were given from the computer. And it's bigger than one, so it's okay as is. Sometimes this thing will be 0.6 minus maybe even nothing. 0.6 is really a big correction. And 0.6 over this would give you something like probably 0.6. You cannot have an amplification of 0.6. You've got to be at least one, or you, at your moment amplification factor, Xx. No doubt about this. This isn't complex, but this takes a while to get used to about what to do. So now I'm going for the bending Xx capacity. Because one of the numbers in my interaction formula is, how much capacity do I have bending Xx? From the beam design charts with C sub B is equal to 1 and L is equal to this, the moment, well how the devil did he ever find this thing in the first place, man? There's 43 pages where this may be on there. He just says go to the design charts. How do you think he found 14 by 65 curve in all that massive data? Very good. Start with where it comes in. And somebody else now, where does it come in? Where does it come in? At the plastic moment it comes in, at C sub B, M sub P. And where do you quickly get somebody else? Where do you quickly get C sub B, M sub P? Now see, you're not somebody else. You say, yeah, but I'd like to go home sometime today at the Z tables, that's exactly what he did. So he went to the Z tables and quickly found a W at 12 by 65 and quickly went over and found C sub B M some plastic, his 356, and so he quickly went to a page that had some 356's on it and he says, yes indeed, there it comes in right there. So he came over with a W at 12 by 65 and he followed it down and he says, you know, that's not the lightest beam. Well, we're not using it as a beam. We're using a piece of it as a beam, but it's a beam column, so the fact it doesn't show being lightest is not my problem. And I stop when the curve hits the 14 foot length and I quickly get the bending strength of C sub B M sub N. I also, I think I got a Christmas present coming on a lot of torsional buckling, but you'll notice you better not expect too much of a C sub B correction, C sub, yeah, B correction, because the plastic moment's just right here, just right very close. And if you say, well, I can't tell because it was on another page, well then you'd have to go, well, you already did. You'd have to not forget this number and not exceed that plastic moment. Continuing, he says we get a lot of torsional buckling strength of 345, which we did. Now we're going for our C sub B. He says C sub B is equal to 12.5 M max times two and a half, three, four, three quarter point numbers. If you remember the quarter point numbers are right here. Quarter point numbers are 82.5, so on, so on, so on. So you get a 6% increase, better than the chart value. However, 1.06 times the chart value is 366 and you only had capacity plastic of 356. The 366 is bigger than 356. So you got to stop at fees of B and plastic. Since it's non-compact, maybe we shouldn't be using this number. I think we can use that number anyway even though it's non-compact. Yes, it is flexure. It's flexurally non-compact. It's got really thin flanges on it. What about these tables here? What about these tables here? They're smarter than you. They're smarter than me. They take that into account. They wouldn't put it down there and then later on say, well, it failed because you didn't take into account flanged local buckling. No, it takes that into account. You don't have to compute any of that. So now we've got a W12x65 non-compact 50KSI. That's the design strength. That's already based on flanged local buckling. So that's taken care of. Then the factored moments are, the factored moments are 82.4. We've done a lot of work here. Where the devil did that come from? The factored moments were 82.4. 82.4. So that's right. This is just factored. This isn't amplified. You've got to watch his loads carefully. His words carefully. The factored load moments are the moment is that you're going to request. There's a smaller one on the other end, but you're not going to request it. You've got to request the one on the big end. It's 82.4 kip feet. Then amplified now. You had an amplification factor of 1.023. You've got to go back a day and a half to find out where in the heck you had that number. Multiplied times your original request. And that gives you 84.3. And now you're ready for your interaction equation. Interaction equation says piece of u of phi, piece of n, because you have a column. But only eight-ninths of these ratios, these were zero. You remember you already had this number written down. This ratio was 60%. That's what drove you over 20% in the first place. Plus eight-ninths of the 84.3 divided by the 356. 356. The number we pulled off the charts. And that's 0.824. Satisfactory. And I don't care, satisfactory or not satisfactory. I was doing crazy on an exam making it not satisfactory. It's easier on me because I don't have to work the problem first. I just throw one out there to see the satisfactory or not satisfactory. But it was not. That shouldn't surprise you. That shouldn't say, oh, I better work problem one again until it works and not worry about problem two and three. Move on. Lot of work. Yeah. Lot of money. Yeah. All right. A summary. First off, amplification is not appropriate for this. Why? These are not requests. Whether requests, because you didn't do some things like consider P delta, have to be amplified. These are strengths. You will be using this number unamplified because it's not appropriate to determine whether to use, I had written equations down and then people said, what's that for? Just some words. Use this ratio to determine whether to use this equation because P sub u over P sub n was greater than 0.2 to use that. So many erasures on these pages because that was greater than 2 tenths. You use this equation and you maintain that equation 100% extraction of load capacity. Or if it is greater than 0.2, less than 0.2, greater than 0.2 means it's really a column, no gift for you, little gift for you. Or if it's not really hardly even a column, then you only have to admit to confess to half of the real numbers, but you have to account for all of the ratios, the ratios totaling up less than 1. Then if you're using one of these equations, m sub ux may have to be amplified due to P delta effects. If so, you have an amplification factor times the moments that come out of the computer about the x, plus an amplification factor for no translation, plus the amplification factor because of lateral translation times a different kind of amplification factor, b1 and b2. It could have just as easily and should have, but it's the same if it's been about y or if it's been about both. You'll have one of these and you'll have a muy, b1y, bubba y, y, y, y. And then b1 is determined from this equation. If you're a weenie, it's 1 over 1 minus, dog on you, p sub u over p sub e1 about xx or yy. If you'd like to save a little money, you can correct that if it's coming to you with c sub mx. Where c sub mx is on the next page. Important note, if we know the column will buckle about the weak axis, why are we computing c sub mx and b1 sub x? If we know it'll buckle about the y axis, why are we computing c sub mx and b sub 1x, which are strong axis numbers? Because those numbers are amplification factors which are only used to amplify the requests. And the other numbers, those are capacities. So the requests about x will still have a number in it buckling strength weak. Then, continuing with the two page summary, if there are no transverse loads, this is the equation or table that you ought to use. Do they have a table for this thing? I don't remember seeing a table for c sub m. It's on this page, and I got it on this page, a copy of it, or table. Yeah, but that's really not fair, see, because the table should be for this c sub m. So I'm not sure exactly, there may be some table rounds got something in it that I meant to talk to you or show you, but since I forgot to write down the page, so r table, and that was on what page? The one you just held up for me. So this table is not for this equation. Oh, that table! Okay, thank you, thank you. Okay, equation r table. When I wrote it down, it made a lot of sense. It's got to make sense to the new person watching it too, so it no longer makes sense to me, I'm like a new person. Alright, then if you have both of these going the same way, you enter it as a, thank you, I understand, negative for single curvature, or that for reverse curvature. Then if you have lateral loads of a certain kind, then you can get the correction factor straight off of the table, so on and so. Alright, write this down just enough so that I can go ahead and put a curve up there. I'm going to provide you with a, I don't know why, I never noticed that, that's called a table. With table 210, I'll provide it, so you don't have to write that down. All you got to write down is a W14 by 132. It is 18 feet long. It's a simply supported beam. It is uniformly loaded. And it has a c sub b is equal to 1.32. And I just want you to tell me what the capacity of that beam is. Write off of what you've been doing before. All you really need is, it's a W14 by 132. It's 18 feet long. It's uniformly loaded. It has c sub b of 1.32. I want you to tell me right out of the table. Look at that. You tell me what the strength, p sub b m sub p. P sub b m sub p is on this side. The things that are grayed out, allowed stress. Here's the guy I'm talking about. I'll go ahead and mark him so he's a little easier to find. You can work together if you don't understand the problem. If nobody within three people understand the problem, well then let me know. I have to back off a little bit to make it fit. 14 by 132 is right here. 18 feet long. What is p sub b m sub u or m sub n? I guess that's nominal strength. Well, I don't know m-plasty. I don't know what those are. I just need a number. And write it down, put your name on it, and hand it in. This is the table you would use if we were finding the capacity of a column. And it asks that interaction equation says, well, how strong is it? This is where you'd get some of the numbers. If you're having a hard time because you are further away from it than I am, this thing comes up right about there and hits the curve. And I don't need an exact number. I just need right around whatever you think. If you're all by yourself and you want to move over somewhere and you're welcome to, it may take three of you to get an answer, so... No, if I knew the load on that this thing would carry, I'd put a box around it and I'd do it myself. That's why I'm asking you. That's wrong. No, that's right. What grade steel? There's 80 KSI steel where the table isn't applicable. Come on. No, of course I'm kidding. It's got to be 50 or the table's no good. Good. Help my friend there, would you? Nope, you need help, too. The question is, how strong is the beam? How much capacity does this beam have? Good. I wish you'd put units on it. Good. Nope, help my friend here. He's ready to go home. Good. What is this? Where is he? Oh, okay, okay. Let's make sure he's really here. Good. That's kind of low. Where'd you get that number from? Yeah, where'd you get that 840 from? Show me. Okay. Is that 840? Okay. Did you... Show me where you... Okay, so what is your answer? Put a box around your answer. All right, what's 840? Uh-huh. Good. Nope. Help my friend. Here you go. Yeah. Good. No. Hey, wait, wait. Right here. He'll help you. Good. Wait a minute. Where's all the other calculations that go with that? You can't just see 1.32 times something. I want to see it written down. Good. All right, where's the other guy? All right, okay. Okay. Okay. Does that make sense? Good. Now, do you see? Good. I want you to see. Uh, wait a minute. Who's the other guy? Bring me your paper up here. Where's the other guy? Who are you? Wiley? Wiley, where you at? Wiley? Is Wiley here? Okay, bring me one paper only. Good. Close enough. I can live with that. No problem. I like it. Fair enough. Perfect. Excellent. Not too shabby. Yeah, I can live with that. Good. That makes me sad. Gotcha. Anybody else? Still helping somebody? Hang on, hang on a second. I need to find all my stuff. What do I do? Good. Trying to find my notes from today. You're not sure? Talk to him. Got time to talk to him? Good. You understand? Yes, sir. Good. Good. You understand? See what's going on? Okay. All right. You have to be able to do that for both quiz A and... I know I guess for quiz B. No? No? Okay. So you don't need to do any of this for whatever reason? That table... That's what that table graphs. That's what the table is for. So you don't have to do that work. 40 times that is too big. What is that? That's in plastic. Yeah. And you could find that also on the table. That's correct. So, you want to turn it in? Oh, I might request. No. Well, don't turn it in. No, he's not figured that out at all. What is this?