 Okay, you ought to be able to get the loads in these plates depending on how they're connected. Let's say it was pretty obvious, pretty obvious. Load P comes in, goes through the shear surface in the bolt, comes out load P, and this one in the right-hand plate. If you put load P in there because the strength of the bolt at yield here is the same as it is there. That means that half will go this way and half will go that way. Here it is failed. You'll notice that this top half, if you draw a free body of it, there's one shearing surface. So that'll be 0.6 f sub y perhaps if they're basing on yield or 0.6 f sub u. If they're basing it on ultimate times the area of the bolt. For this little piece to pull out, this little piece is subjected to a shear on the top and a shear on the bottom. The bolt is in double shear at that point. That's why that loads twice as big. These are the ones that get interesting. Total load is the load P. A third goes in that plate and two-thirds goes in that plate. Reason to be in if you draw a free body of that top plate. Let's just say that force is shear in the bolt. Then he says a third and a quarter we'll see. Whatever this force is, it's the shear in the bolt because that's one bolt that's been cut. Whatever that is to pull this little plug right here out with this plate, you're going to have to have shear bolt, shear bolt. You have to have two times the shear in the bolt. This one has two shear in the bolt and this one has two shear in the bolt and this one gets pulled loose with one shear in the bolt. So you add up 1, 2, 3, 4, 5, let's see, just on one side. 1, 2, 3, 4, that's got to be a total of 4. Since this one's going to be twice that one, then this will be a fourth of P, two fourths of P and a fourth of P. Whereas on this side, these have to add up two shear in the bolt. This one has to add up to two shear in the bolt because it's got to pull that little plug out on two surfaces. So that's a half of P and that's a half of P. You only have this many plates. These are the forces you get. So obviously, if you're asked for the bearing stress someplace, you're going to have to know how much load is in each plate so that you can get the bearing stress in that plate. If all the plates are the same thickness, then obviously either this plate or that plate would be the one you would be checking because it's going to have twice as much load in the plate. Now we were working with columns, trying to find out how strong they were. In general, we had a strength F critical. Here we go. Based on Euler and his opinion, worked nicely with a correction for the columns aren't really straight. Then past that point, we have an empirical solution, a combination of theory and practical things, namely that this is going to come in at Fy. So we had two different equations. We found a break point at which the two equations controlled to the left, to the right, different equations. We had a bunch of tables that also were worked out for us. We had two basic tables so far. One of them would give you F critical based just on the strength of the steel and the KL over R for your shape. And if you wanted to use that one, you could see if that's one of the first one that shows up here. Now they don't use that one. They're using the second table. Second table would tell you not the critical stress, but it would actually tell you the load that's in the column. So we already did one with the stress. Now we're going to do one with the load tables. He wants the lightest wide flange that can stand dead 62 live, 125, 1.2 dead plus 1.65. He wants you to put 275 kips on the column. Effective length is 24 feet. I don't know if that's a pen pen 24 feet. I don't know if that's a fixed fixed with a point six of the length is equal to 24 feet. Doesn't matter. Effective length is 24 feet. He hasn't told me how it's braced. So it's must be unbraced. And if it's unbraced, it's going to buckle about the weak axis. So by not saying anything, he's said a lot. And he wants a 992 50 ksi steel shape. So what we'll do is we look in the book and we say, what did we pay all that money for? The guy says, well, I have tables that'll give you the permitted strength, the fee of point nine times a nominal carrying capacity of a w eight by 58 of different links. Not only that, unlike that stress equation, which has no idea what the shape looks like. This is basically the allowed stress. You will still have to check it for slender elements. In my case, if it has slender elements past some point, I will appropriately calculate the true load. The nice plus if you're not going to go to graduate school. So we come down to a 24 foot column. So we've got a little note here says effective length KL with respect to the least radius of gyration. Well, since they normally buckle about the weak axis, that makes sense. That's what he is tabulating for us. If you'd like another $50, he'll include another table for effective length about the strong axis buckling. These have are sub y embedded in the numbers. We go to 24 feet. We need 275 kips on a 24 foot length. No, no, no, no, no, no. Okay, there's another page of eight buys next to this. Here's the Oh, there are no, there are no more eights. That's all the eights. Okay, an eight by won't work. Now we'll see if some 10 buys here are our lighter shapes. We're gonna start down light till we find something that works and will not go any higher. We need 275 kips on a 24 275. No, no, no, yes. That's what it's all about. Is it simple? Yeah. And again, a little uglier when you go do some other things. Yeah. Would we turn you loose with this without you understanding hopefully what's going on? No, you'd kill us all. So I'm going to tell you that a 10 by 54 is an acceptable shape. Now I'm going to go to the 12 buys. Would you remember the number 54 for me 10 by 54? I'm in the 12 buys. I'm on the 24 foot unbraced weak axis length link. I need 275. No, no, no, no. That's it. It's guaranteed. Good to go. If you don't believe it, you're welcome to go put the kale over R into the appropriate equation either to the right of the break point or to the left of the break point whatever this KL turns out to be and multiply that times the gross area and multiply that times what? Repeat. You get the critical stress and you multiply it times the gross area giving you the nominal load and then you multiply that times what? The resistance factor of how big? Yeah, well, this is a column. And here let me give you a little hint. There we go. There's a little hint. The guy says they may forget what that number is. He says I said, okay, if they forget it's in the table, right? Well, yeah, but I had this little space left at the bottom. Okay, good enough. What is this? Is this lighter than the last one? 10 by 58 is okay. This is heavier? That's what he found. You're right. It is heavier. Okay. Let's see if a 14 by is better. Looking for the same 275 on a 24 foot unbrace weak axis effective length. No, no, no. 275. Yes, even heavier. So as the column gets deeper, it's a lot better for a beam. But as it gets deeper for a column, usually the column, the squatty ones that have more moment of inertia about both axes turn out to be often will be better. But you have to check them and see. Now he says what if the shapes are not in the column load tables? Well, then about all you can do is drop back to your critical stresses or the equations and make a guess. First, he suggests this as good as any. There's a lot of different suggestions in different books. He says pick a f critical. How? Just pick one. Well, how about 83? How does that sound? 83. A 992 steel 83. We'll guess the critical buckling stress is 83. No thoughts on the matter? Like one. Okay. Well, you could start at the bottom. Yeah, you might get a number because one is possible. But how are you going to get a critical buckling stress bigger than f sub y? For any reason to start out any lower any higher than 50. That's if the column has no length. So he says my guess is probably a pretty good number is he's telling you the procedure first. Pick a f sub critical. Theoretically as maximum at f sub y. Determine how big the shape ought to be. Get an area that'll be fee f critical times gross area. We just discussed has to be actually will set it equal to but it has to of course be greater than the request. That means the gross area you come up with has to be at least this big. Then you go pick a shape that has that much area. Then you compute the truth, the real critical buckling stress for that shape. And if it's perfect, it's perfect. And if the thing is 10 times too big, well, then you're going to have to change the critical stress such as you get a more reasonable shape. And if it's unsafe, you pick another critical stress the other way. Then when you're through, you're just out of luck. You got to do your own slenderness ratio calculations and your own comparisons. And it's all it's up to you then. Hopefully everything will be in the table. Tell you now it won't be. So he says I need an 18 by shape of a 992 steel dead, live, effective length. This is about the weak axis and said otherwise. So I guess it is. So we'll be able to use those tables as long as this is about the exact the weak axis. Looking your tables, you see they go up to 14 buys. And then the 18 buys can be used as columns if that's what we're using in the area. But they're not that commonly used. And so he doesn't have them in the tables. There are no tables for them. So we pick the ultimate load. He says I'm going to go ahead and pick about two thirds of 50 ksi. I'm going to start at 33 and see what happens. And now that I have nothing new to tell you, this is identical to what you did before today. The last thing we did, you calculate the required gross area, load divided by nine tenths of 33. 20.2 square inches of steel. He's going to try W 18 by 71. He needs 20.2. He needs 20.2. Here are 18 buys needs 20.2 20.2. There's the first one W 18 by 71. Here's that what he's trying? Yeah, that's where he got the first one from because it was the lightest one that had the required 20.2 square inches of steel. Of course, it's on page one dash 20. This is in your properties of shapes. Page number should be listed. Here it is. So now that we've decided to do an 18 by 71, I tell you he's right. You can check these pages. It has, well, there it is, has 20.9 square inches, has 20.9 square inches, has a K of 26 times 12 inches in a foot, the minimum radius of gyration for a W 18 by 71. W 18 by 71 must be on the next page. Here I tabulated what I found on the next page so I didn't have to sit here and fumble. There they are. This one has 1.7. So he has effective KL over R. So 183. Less than 200. What's that all about? What's 200? That's not a break point. That's the maximum slenderness ratio that they suggest you're stupid if you go past that point because things don't have any strength hardly at all. If you say, well, in this case, I'm going to because I need to, that's okay. But they don't even have table values past that point in any of the tables. They want to lead you away from that so you don't get stuck by mistake. Here is your equation for now let's see 183.5. Here is your Euler opinion of life. I squared E divided by K over R squared pi squared E. There's your K over R squared 8.5 KSI. That's Mr. Euler's opinion. I don't know if that's going to put you to the left or to the right of the break point. I'll have to calculate the break point. That number comes up so often it's probably good for a tattoo someplace. Just put 4.71 square root of E over F sub y is equal to 113 for 50 KSI steel. Our KL over R is oh good lord 185 and the break points 113. So our KL over R is way bigger than the break point number 113. So we are way down in here. Our critical stress will be this big or this big or this big depending on where our 183.5 falls on that curve. It's easy enough 0.877 times what you just calculated F sub e of 8.5. That's just no stress at all. This could possibly work too too little critical buckling stress. So phi sub c, p sub n, phi sub c, F critical times area, stress times area, 0.9 times stress permitted times 20.9 square inches of steel. That was a bummer. So he says since this was so far off he's going to go to a stress about halfway between what he guessed and what he got. He doesn't want to go all the way to 33. He could. He can try anything he wants to. It's his time. He's going to try 20. Then they've got a new gross area with a 20 KSI in here giving you 33.3 square inches of steel necessary. Come back here and you look for 33, 33, 33, 33, 33, 33. There you go. A W18 by 119. You're 18 by 119 and he repeats just like it says on the shampoo bottle. Rinse and repeat. And you repeat till you get the right answer. This may be the right answer. He recalculated the critical stress. Breakpoint still the same. Our break point turned out to be at 116. That was our KL overall for this shape. And therefore, we are a little to the right of the breakpoint. Therefore, we're still using the Euler equation 0.877 times 21.27 f sub e for the critical buckling stress, put the critical buckling stress with a 0.9 times area 589. Cowley man, bummer. Still no good. There's very close try the next size and it works. Oh, no, no, this is true. Our made up number C was just to get us a shape. At that time, the made up number went away and is no longer discussed. At that time, you see, you say, I think this shape may work. And the first one didn't work horribly. Well, that may be the made up numbers part, but it was no part of checking if it works. No, what you did was, yeah, you can start at anything you want to. You can start at 50. I give you a much smaller shape. I give you a little tiny shape. And then when you put the little tiny shape in there, you're going to get whatever you get for f critical, but it's going to have such a horrible slenderness ratio that it's not going to work. So you're going to say, okay, 50 was a bummer. Me, I said 20 was a bummer. I forget what it was, whatever we started with was really bad. So whereas I'm going to go for 20 up to 30, you're going to be coming from 50 down to 40. Go either way. Doesn't make a bit of difference. Because every time you make a guess to guess a shape, you're going to check that shape. And if that shape works, it's good. It's got nothing to do with the guess. The guess would just lead you into getting a starting point or something to analyze. That's what all design really is in general. Make a guess at something, then you check it. Down at the bottom, he says the shape is not slender because there's no footnote when you go to the dimensions table. There's our dimensions table. He says there's no C on here, so it's not got any slender elements. Then he said, so local buckling does not have to be investigated. You just did. You just investigated local buckling by looking at this thing and saying doesn't have a C on it. Look at this one. That one needs to be investigated because you used a bad set of tables for F critical that didn't have slender and element buckling taken into consideration. You have to do that on your own. This was a loud stress design. As always at some time during the semester, I start drawing and can't stop. So here is case D all nicely drawn out. Somebody asked, Well, how could you really get such a thing? That's case D. Here's how you put the brace rod in there. So these things mean no translation at the top. Without the brace rods, it translates at the top. See the no brace. This one looks fixed. It's really looks fixed. This one looks fixed. It's pinned. The plate will bend. The footing will bend. This is pinned here. This is pinned because it's not welded nice and solid and there's not a lot of other detailing to get the load through there. But dang, the plates 18 inches thick, and he's got 43 bolts on it on a eight foot by eight foot by 30 foot footing. Why do you might call that fixed? Don't have to spend a lot of money just to make it fixed. Usually it's not worth the money. That's case E. Here again is other tables we can use. These are usually pretty good for single story buildings. They're real good for multi story buildings because as you go on up in there, these things aren't really fixed or pinned. They're attached to girders and we'll get into that later on. Just need that table. Here's the thing squats down with brace rods and horizontal loads. You notice it doesn't move enough to count for anything without the brace rods. It moves to the right. This buckles as a 0.7 theoretical practical point eight. This buckles theoretically at a two fixed and pinned. And when it's this way, I believe practicals also to upside down like this. And this is the ground. And that's the column or the girder. I believe that's 2.1. More review. There was Mr. Oilers opinion. There was point eight seven seven. There was the spot where things just went terribly awry 113 for 50 ksi steel. Here are the equations that control if you're forced to use them. You will be forced to use them for 56 ksi steel because your critical stress tables don't have anything but 50 ksi steel 36 ksi steel things they really make on quizzes. You find some very novel steals. Here's the same picture. The only thing is that I like this. The book uses this. The examples in the not the code, but the specs use this. But some people do this. Some people the break point is fixed. In other words, it's fixed at 4.71 square root of so and so. For KL of R it's fixed at 0.44 f sub y in this direction. Want to use that for the break point? Good enough. Same thing. Same thing showing some wrinkling showing basically the comparison between global buckling and local buckling. Global block buckling has a break point and two different equations. Local buckling has the same thing. Have a break point above which you have slender elements, although we don't cover how to do it in 4.46. You'd have to assure me that the shape you're going to use has no reduction of load by telling me there you are to the left of when the radius of gyration of the little element itself really gives you fits. Once you get into beams, you're going to find we couldn't live with just two regions. We got three regions. Initially, it was proposed to have five. That's what the engineer said. Yeah, fat chance. So give me give me three. All right. Here's an example that you probably should check out. Here's the same column 30 foot long. It's a weird number 23.35 feet long and 15 feet long. How much load does it carry? Here's the kale of our about the weak axis is not braced for this are 2.48 for a 992 steel. Here's your kale of our break point. Here is you are 145 is to the right of the break point. And therefore you are in the elastic region. And therefore f critical is 0.877 11.9 KSI is how much stress you can put on it. That still has to be multiplied times 0.9. And then that has to be multiplied by gross area. That would tell you the load. If you increase it to 23 feet, you redo the K. Oh my, it comes out 113 right on the money. And that's something. Use either equation. They both give you the same answer. I would certainly use the this equation because it's a lot easier than this equation. The stress permitted goes from 111.9 to 19.66 for a 15 foot column. You put in the new length our kale of our 72. That's the left of the break point. That means you're over in the goofy equation region, the in elastic region. There's your f sub e Mr. Oilers going right on through the roof to 53 54.3. The true critical buckling stress is 0.658 f y over f sub Euler times f y goes to 34. And so you ought to look at that. You ought to see what's going on. Here you were in one region. Here you were smack dab in the middle of two regions. Here you were in the in elastic region. R means radius of gyration is killing you. Lambda sub R means that's a break point number. Plano lambda with no subscripts is you. That's your. So in other words, you will be checking is lambda sub u is it less than or greater than the break point lambda when the radius of gyration really kills you. See this shape here? See how that's lambda sub R right there. Below this point things were going pretty good. Past that point, man, I mean, you are really taking a hit. The radius of gyration of column got so big that you stopped being on this nice almost f sub y curve and you're down here taking a beating. Here's your lambda sub R is 200 right there. Didn't be out there. I'm sorry. That's not your lambda sub R. That's your lambda. Lambda sub R is fixed. Lambda sub R is 113 for 50 ksi steel. Regardless of what the whether it's a wide flange or what I'm sorry. No, no, look at here. Yeah, that's not linear. So if you're saying is it a direct function of your lambda? No, flat, flat, flat, flat, but you get nothing. You get nothing. You get a little you get a tiny bit. Geez, man. Okay, now you get something. They get a lot. Now you keep a lot. It's not linear. No bore on effective length. So far, our shapes have been buckling about the weak axis, which makes sense. However, if you come in here and realize that your pinned pinned column is very weak about the weak axis, you see what the shape looks like. If you look at the top, here's the flange, there's the web, there's the flange. If you say this is terrible, I'm not getting any strength out of it. If you brace it in the middle, you will force the shape to buckle in two nodes, making the effective length half as much as it used to be. Picking up a factor of about four, as far as strength. Now then, the thing is, if you do that, now it's possible that it won't buckle about the strong weak axis, it'll buckle about the strong axis. It's going to buckle about the whoever's got the biggest KL, KL over R. That's the one that's going to buckle about the largest slenderness ratio. You want to see me make it buckle about the weak axis? I put another one of those here, I put another one of those there, and you say, well, it's still buckle about the weak axis, even though it did one, two, three, four nodes, okay, nine more there, nine more there, nine more there, nine more there. I'll get to the point where the whole thing won't buckle inside the wall. This is usually where the sheetrock is. It'll buckle in, and all the little kids will get hit with bricks. It'll come into the building instead of buckling in the wall. And here that is. Here's that brace. You see that brace you put there to hold everything? Guess what happened when it buckled? The brace just went with them. It just went over like that. You say, well, I'll fix that. I'll put a wire out here in the street, and a wire out here in the street. Now it won't buckle about any axis. Not gonna work, not worth it. Get a bigger column if you can't live with that one. This one, KL, about the X axis, he has his 26 feet. Here the KL, the effective length is 13 feet. Some good reading, mostly what I said. Here's a kind of a picture of it. That's a wire going down to the, going inside the wall. It's forcing it to buckle out of the paper and into the paper, making it a shorter effective length. How effective is the true length over L? That's true length over 2. Here it is when it buckles in the plane of the paper about the strong axis. There it is, another attempt trying to make it make sense. But the only thing I know is to go tear the sheet rock off of something and show you because it's hard to draw it. There you go, man. That took an hour. Here's a girder and a girder, and it's pinned at the top and pinned at the bottom, and here it is braced about the weak axis. It could still fail about the strong axis, but if it fails about the weak axis has to fail, fail, fail, fail, go back to zero, fail, fail, fail like that. There's a base plate, strong axis buckling. What we really need is we need like a picture of it drawn in 3D and get it on a computer and then where you could roll around and see this stuff. There it is buckling about the weak axis. So, example 4.9 has a W12 by 58, 24 feet long, pinned at base, braced in the weak direction, and it's braced at the third points. At every 8 feet, he's got a brace. He's pretty smart. He braced the weak axis. You know, he didn't brace the strong axis. Here's this brace. See that little circle thing there? There it used to be. When it buckled about the strong axis, it buckled on a 24 foot length and the little braces just went along for the ride. So that means that the kx about the strong axis is 24 feet long, times 12 inches in a foot. x, the ky about the weak axis, they're only 8 feet long, times 12 inches in a foot. The radius of gyration goes with the buckling axis. It's buckling about the x-axis, 5.28. Check it out. Look it up in the book. You shouldn't believe anything. That means that the slenderness ratio for buckling about the strong axis is 54, and the slenderness ratio, the k over ry, is only 38 for weak axis buckling. It means you have put so many braces in there, it will not first buckle about the weak axis. It'll buckle about, it will not first buckle about the weak axis, right? It's going to buckle about the longer slenderness, the strong axis. Once I know that, I don't even check this. There's no reason to put this number into the critical stress equations. There's no reason to go look it up in the tables. This is the way it's going to buckle. So then again, I go and find out where I am with respect to, oh no, he says, I was getting ready to go find the break point, find out which side I was on, and all that kind of stuff. He says, look, remember that table 422? I forgot about that. He says, go look it up. You have a k over r of 54.55. Here's where I got the r from. That's the gross area. You'll notice this particular shape has no slender elements. That one wouldn't be my first choice. Got slender column elements. It has slender elements as a column. This one down here has slender elements as a column and in flexure. We'll get into that later. All right. And there's where I got the radius of gyration strong and weak from that table. And here are my critical stress numbers. I'm going for a 50 KSI steel. I'm going for a 54.55 and 54 and 55 is between these two numbers. If I were you, I'd just, and I'm telling you, you can in this class just go ahead and take the worst one. Don't go all the way up to 36.4 because that's not true. You can interpolate, but that's not the purpose of this class is to teach you to do that, nor for you to take that amount of time. They're reasonably close. Just pick the lower of them and say the strength is 36.1 KSI. That's the F critical. So he interpolates 36.24. 36.24. There's your 36.24. There's your, uh oh, he forgot the fee. I'm sorry. It's built into built into what? Oh, it's built into the tables. It's already there. How do you know that? Well, that's not fair. Proved to me that it's built into the tables. I see you looking at them checking all my numbers at the top of the column. That's exactly right. It says fee, F critical, fee, F critical. And if you don't believe him, you know, if you're not sure what he used, then you look at the bottom and you see what fee is for use in that table. And therefore that shape is good for 616 kips. No slender elements, so it's okay. If this thing that we now chose had slender elements, we would have to know how to do check the slender elements or forego the use of that section in this class. Now, if you say boring, well, that's because you haven't seen any of this yet. I mean, you see, you know, A or B, you know, you see a grade and that's a long term goal, but boring on the way, but on a day to day basis, the stuff is great fun. It really is. Here's a yardstick. I'll leave it for you to check it out. Check the moment of inertia about the week, moment of inertia about the strong. It's braced about the week at these points. And you should be able to find out, strangely enough, the KL of R for both of them turned out to be about the same, but this would still be the choice because it's bigger. Now, he says the available strength in these load columns are based on the effective length with respect to the weak axis. That's right. He warned us about that in the tables. See if I can find one right quick. He warned us that this was only good about the weak axis. So I won't use it for strong axis. That's quite obvious. Don't use it for strong axis. He says, well, not quite that simple. You remember how you got that answer? Yeah, I used a KL of R. He says, and you divided your KL by R sub Y. And I did indeed and says, and then we used one of the appropriate equations to calculate the critical. Yes, that's true. That's what we did. And then we got the strength. He says, if I take the R sub Y back out of that calculation by dividing the true KL by R X over R Y, basically putting the R sub Y back R sub Y and divided by R sub X, he says, you get a phony KL. Yes, you would indeed. It wouldn't mean anything to me. He says, don't give up just yet. He says, if you do that, if you take the R sub Y back out and put the R sub X in to the equations, you'd get the right answer. I said, well, yeah, but that's not what you did in the tables. He says, if you will take the K X sub L and divide it by the R X R sub Y, you will get a K, which won't make any sense, but which if you enter it in the table in effect, you have taken the R sub Y back out and put in R sub X, you will get the right answer. I don't believe any of this. That guy's nuts. What's he smoking? I don't know. He says, well, give it a try. Just give it a try. See, you like it. Just give it a try. It's okay. Got a compression. Remember, shown so-and-so, pinned both ends, supported in the weak and at mid-height. Okay, this thing is going to either buckle nine-foot weak axis or 18-foot strong axis. He says, got to have a service of 400. That service is just dead plus live or whatever, just the sum of the load. That's what they call a service load. Equal dead and equal live. So half is dead, half is live. If it cares, I still want to slide the lightest shape. Great. I'm ready to go. 200 dead, 200 live, 1.2 dead, 1. live. I need 560 kips. Get out of the way. Give me my tables. I need 560 kips. He says, assume the weak axis controls and enter the column mode tables with nine feet. Okay, sounds good. Beginning with the smallest shape, nine feet need 560. Check this out. This is, those are the, that's the other table. There's a nine feet I need 560. Now, I notice I cross these out so I don't get accidentally get some loud stress. Starting with 8B, I need 560. No, no, no. Been here before. There you go. That'll work. If it buckles about the weak axis, I get 634 kips. 634 kips. He says, now don't forget to check the strong axis because you don't know which axis is going to buckle about yet. No, that's true. So I'm going to go to 8 by 58 and I just want to calculate it the long way for a W8 by 58 because he's fixing to use some phony nonsense here. W8 by 58. Here's a W58. I promise you I used all the numbers that are in the book. Buckling load, if it buckles about the strong axis is it's k is a 1. It's not nine feet long. That's 18 feet times this. And I see he is this written down. I say, what, what the devil are you doing there? He says, did I do something illegal? I said, no, but you did something stupid. You multiply it times 1. He says, well, just leave it alone. Okay. Here's my 1. There's my 18 times 12. He says r sub x over r sub y. Anything illegal? No, no, no. Times r sub y. He says, do you see how you have taken the x thing and divided it by r x over r y? I say, yeah. So he says, well, put r x over r y in there. There's r x, there's r y, 365 over 2.1. That is 1.74. Put the r sub y in there. Well, that's 2.1. I said, you just put it in. You took it back out. He says, that's what I'm telling you. You can do in the tables effectively. You can divide this by this and you get the right answer if you use that as the k l over r nonsense. He says, keep trucking. All right. So here we go. This is what I would have done in the first place. This is one times 18 times 12. That times that's 3.65. That's what I would have used 3.65. All this didn't change anything. He says, therefore, the k l over r is 59. That's 59. Now then, go to the critical buckling stress for 59. There's 59.11. The answer for 50 k s I see is somewhere between 34.6 and 34.9. If you interpolate it, it gets 34.8. And therefore, you get an answer for an 8 by 58 is 595 strong axis buckling. Oh, goodness. Sure glad I checked it because it didn't come up to the 634 we were planning on. Okay. Now, it's okay. Even the 595 is still above the required. And so I'm getting through with it here. But by the equations, we get 595 strong axis. He says, look, come up over here now and see what I was trying to get you to do. What I was trying to get you to do was to take the k l over r for the strong axis, which was 18 feet and divide it by r x over r y. You and I just calculated that r x over r y was 1.74. So okay, I'll do it divided by 1.74. That's a slenderness ratio of 10.34. Where'd you get it from? He says, you saw where I got it from. He just don't like it. I say, well, it's not nine. And it's not 18. He says, it's funny. Enter the 10.34 into the table. Okay, 10.34. Here's the table. Here's the table. Kale funny is 10.34 10.34. I need 560. No, need 560. No, need 560. No, need 560. It says a W 8 by 58 will give you in that region. What did we pick before? W 8 by 58. 8 by 58. W 8 by 58. What were we trying to get lost in this stuff? W 8. That's right. We were trying an 8 by 58. Okay, and look at what this phony thing tells us. The phony thing tells us we'll get 560 guaranteed. I mean, even if you interpolate to 10.34, the two numbers that are on the phone right around the phony number are going to give you the same number. And as a matter of fact, when you interpolate it, you know what? That dang thing works. You get 596. Now, if you're not thoroughly confused, then you're not listening. Those of you who aren't confused, you say, maybe I'll figure that out tomorrow. No, you won't. Practice with it tonight. We'll do some more of that next time. It's good stuff. It really is. Thank you. Hey. You too.