 Okay. On the last problem we were working, we added the weight of the beam almost as an afterthought. I forget how many kips of load it was carrying, four or five kips per foot. The weight of the beam was only about another 30 pounds per foot. Couldn't hardly make a difference. Sometimes it can, just the tiniest bit. If you've got a really good, efficient beam that you haven't included the weight yet because you didn't know what the weight should be, it's possible it'll put it over the limit and make you move on. Yes, sir. It doesn't matter. Let me just say this. We, no and I mean really, you know, you'll have to go back and I understand it. We're starting over basically. But we designed a beam and we're going to design another one right now. We designed a beam to handle lateral torsional buckling using those graphs. Those people that used to look like this. Of which each page in the book gives you a little tiny window on to pick a beam. That's correct. We picked a beam not knowing its weight because we didn't know what the beam was. After we found the weight of the beam it added a mere 30 pounds a foot to it. Someone asked, shouldn't you recalculate the C sub B with the new loads? You really never need to. In that case, you and I had a 192-kip-foot moment straight line, horizontal line because there was no uniform load. There were just two concentrated loads and it dropped down. I went ahead and added the weight of the beam and the 192 here went to 196.8. We took that into account when we checked to see if the beam still worked. But we didn't redo the C sub B. The C sub B now rather than being 192, 192, 192, 192, 192. If you remember that C sub B was 12.5 times M max divided by 2.5 M max plus 3 M at the quarter point plus 4 M at the half point plus 3 M at the quarter point, that was our correction factor because we were being nicer to our beam than Timoshenko's graphs were being. He was taking the worst case. So when you put in the new numbers, don't put in 192, I don't remember what this is, is 194, 195.1, 196.8. These are your new M max and your new M quarter and half and quarter points. And I just stuck them in. You can take my word for it or work it out. The added beam, of course, C sub B has always added strength. And so we got a little added strength. We got about a percent more strength coming to us than we took. We already picked the beam was okay with the added moment. You really didn't need any added strength to the beam. And so it wasn't really necessary. The point being that you can either take your C sub B on the moment diagram and excluding the weight of the beam or including it, it's not normally going to make any difference at all between the C sub B that you get to use. Here's another one, same idea. This is example 512 in your book. It's an A992 steel roll shape for this figure. There's a nine-kip service live load. And there is a three-kip per foot uniform dead load, 30% of which is dead and 70% of which is live. So three-tenths of the three you see is service dead and 70% of the three is service live. So there's two live loads on there and one dead load. He says I'm going to again neglect the beam's weight and I'll check it later. As a dead load uniform, 30% of the load shown, or .9, and 70% of that same uniform load is live. So this is your live load. Then to get your W sub U, your request, you're going to take 1.2 times anybody who's dead. So there's a guy who's dead there. Plus 1.6 times live is 2.1. 1.6 times the 2.1 gives you 4.44 so far for the uniform load. Remember the uniform load, I'm not talking this guy, the uniform load was 30% dead and 70% live of the three shown. Then to get the concentrated piece of U load, it's all live load. So it all gets the 4.1.6 treatment times the 9. It's 14.4 kips and we're ready to draw a new load diagram. So with these numbers, here's your new load diagram. You have a total of 4.44 dead plus live uniform. Here's your 4.44 dead plus live uniform. And here is the only concentrated load was live, hence the 1.6, 14.4, 14.4. So we're trying for a real load and then a shared moment diagram so we can design the beam. But we've got to take the loads as the boss gave them to us, split them up, factor them, add them all together, and then stick them back on the beam. Now you can get the moments you need for your C sub B first place. This is a uniform load, so this is pretty much parabolic. This concentrated load in the middle is going to cause a kink in the diagram there. So it really ought to be shown probably a little lower. There ought to be a noticeable kink right there. And then it comes on down. It's not that that's going to matter much as long as you get the right quarter point, half point, three quarter point moments. It's braced at three places on this one. I don't believe the other one, it was braced in the middle. So you're going to be analyzing either this beam or this beam. They're both 12 feet, so I don't care. Due to symmetry, I don't care. Just design me a beam that'll hold up this load between brace points and it'll be okay. And then don't forget that this M max is the one that'll have to resist a plastic moment as well. Yes, sir. I will show you how to do that. If M max is outside of the range, then it will not influence your C sub B, but I got one just answering that question in a minute. Just so happens. Well, no, somebody last semester says, what happens? And I say, I don't know. Let's see. So I don't care how you get these moments from your 221 class or from your 305 class. In this case, he just took a free body. Don't guess I drew one. He took a free body. Here's the reaction. 60.48 kips. He came out an arbitrary distance, X. I thought, sure, I had that picture somewhere. He put his uniform load on there. 4.44 kip per foot. He put on the beam a positive shear and a positive moment, and he solved for the equation for a moment at any point, X. 60.48 times X minus the height, 4.44 times the base, X times the distance to the centroid, X over 2. They got a nice equation so that he can find the moment at the quarter point, moment at the half point, moment at the 3 quarter point, and the moment at the end point, M max. Here's where he cranked those numbers out. So he has M sub A, M sub B, M sub C, and M max. He then drops those into his C sub B equation. He cannot get C sub B from page 3-18. I wonder what 3-18. It seemed like 3-18 was a page with nothing but C sub Bs on it. It's the chart. Right, he's got about 4 or 5 C sub Bs on it. And you'll notice that he doesn't have one like this, does he? He's got one with nothing but this on it. But there's no way to combine the two to get the C sub B for this case. So you have to work out this equation on your own. And I tell myself, I've got a picture of this on page 210B if you still have those notes in the notebook. Sometimes I'll put one in front and it'll tell me I've got one to show you. So you put those numbers in and you get a 30% increase over what Timoshenko gives you on the graphs. It's yours. It truly is. Because you did not stress these fibers nearly as much as you could have, nor these, nor these. You don't have near as much tendency to laterally, torsionally buckle as the graphs indicate. Those graphs are for and say C sub B equals 1 and nothing else. So he says enter the unbraced length on the charts with an unbraced length of 12 feet. That's correct, brace, brace, brace, 12 feet. And he says, don't go in there with M max of 406. He says, since the charts aren't any good for C sub B not equal 1, take your moment and divide it by 1.36 on the way in. And I say, okay, I'm not sure why that would work. He says, if you go back to your equations, I don't know if I've got them here. You just can't carry all this stuff. Build the battery. I don't know how that happened. The equations, goodness, they've been way back there, though. Yeah, I don't have anything with shear equations. You got equations for M nominal according to Timoshenko. And those equations were used in plotting these graphs. I'll leave it for you to go back and dig them out. And you had a L sub P. Yes, sir. I'm sorry. Might as well. Probably didn't really kill the battery. Probably was just going dead. You have a long old equation, square root, square root, square root, those kind of things. It has a C sub B on the outside. Anybody happen to have a page number on that? Doesn't matter. I'll take it. I'll leave it for you to dig it out. But I will tell you that every one of these numbers on this thing right here, except this one, has a C sub B, such that if you want to pick a curve, you can come in with the moment, and you could come in with this one, and that thing would really be C sub B greater. Now you've got to be careful. You can multiply this number times 1.1, and you can get this, and that's okay. And you can, if C sub B is 1.2, then maybe that's okay. But if C sub B comes out 1.6, and you say great, I picked a beam with C sub B of 1, my gift is a 1.6 above Timoshenko, that number is correct. The beam will not buckle until that load is reached. But the beam will not reach that level. It'll only reach up to C sub B and sub plastic. So that's the only thing you have to worry about. The C sub Bs, you can take that present on your way into these graphs. So I'll show you what it does. You need a moment that'll go 4.6.1. I'm going to let you go into my graphs, into my charts with a 4.6.1 over 1.36, to make sure that you only need 299 kip foot of resistance of moment. Here's that graph. There's a bunch of them, of course. But the first thing I did is I looked over on the left-hand side and found somebody who had 299. And I went across the bottom where it said 12. And then I said, okay, a W18 by 50 won't work. I said a W12 by 58 will work. But there's lighter ones ahead above because this is dashed. And then a W10 by 68 will work. And then I found a solid line as a W21 by 48. A W21 by 48 will hold 311. You say, well, now, wait a minute. I'm just going to need 4.6.1. And you're telling me you're going to hand me something with only 311 capability in lateral torsionally buckling. I'm going to say I admit that. However, when you come back out of the room with your W21 by 48 beam, which only has the ability to handle 311 according to lateral torsionally buckling, at that time, you'll get to apply your Christmas present to it. You'll get to multiply 311 times 1.36. And that's what we're going to do. Here's your 311, which the beam will really carry before it buckles, except you didn't bend it as bad as those charts think, times 1.36. It will not lateral torsionally buckle before 423. Now, the only thing is, any time you apply this CCB, that's the only time you ever might go over in plastic. So you have to check this beam, which is perfectly great for lateral torsional buckling, to see if you already killed him with a plastic moment. Now, let's go back and sometimes you can do this. You can follow the W12 by 48 up, and you can find his plastic moment right there. You can also get that moment where else. Then the Z-tables, that's where you can get it. It's in there, too. And that's a good place, for instance, to get this guy's plastic moment, because when you try and trace it up there and see what its plastic moment is, it's not on this page. You could track it to the next page, then you could track it to the next page, and then it'll finally, they all do, they all finally crater out at a plastic moment. But in that case, for one of these that doesn't just come right up here and die, giving you a piece of B in plastic, then it's a good idea to be able to use the Z-tables. So I think this beam's probably going to be great. Number one, it has 311, which I really get more than 311, because I got a present coming. I really get 423, and 423 is more than I need. 423 I only need 406. And now then, let's check its plastic moment. Here's its plastic moment, it's 398. That's including the fee. So here's 398. Oh, bummer! See how I don't get as much as I need? The beam fails at 398. It fails just shy of what I need with a plastic moment. So now then I got to say, bummer, it's no good. Required, it's 406. Available, m sub p, a fee sub b, m sub p. Available, there's only 398. It's no good. Enough to fix. If that one doesn't work, a 21 by 48, we simply scoot up here until we find a W18 by 55. W18 by 55 is a number a lot lower than the required 406, but I'm in the no c sub b office. Then as I walk into the no, out of the no c sub b office, I get my 1.36 applied to this. And of course I know it's going to work, because I already picked a number which, when multiplied by 136, is going to work. For the next trial shape, move up the charts to the next solid curve and try a W18 by 55. For 12 feet, the design chart is 335. 335 is correct. When you bring it out of the Timoshenko room, you get an extra 1.36. There's a 1.36 and 335. That's bigger than I needed. I needed 406 bigger than I need. And he says this is also greater than FISA b m sub p. Well, that's not good. That's not bad. That's okay. In other words, it won't laterally torsionally buckle until you get to 456, but it may hit a plastic moment before that. His plastic moment is back to the 18 by 55. We can fortunately track it down. It's 420. So your thought that that beam would go up to 456, it won't. It's going to die at 420. But fortunately, 420 is still bigger than you needed. And so both criteria are acceptable. Number one, it won't laterally torsionally buckle. And number two, you have not driven the beam past its plastic moment. Now, adding the beam's weight. You know, we're getting pretty close to everything here. In this case, the beam's weight could hurt us. I mean, it won't hurt us. It'll just be another 10 minutes. We'll pick another beam. Your 406.1 plus 1 eighth. The beam weighs 55 pounds a foot. 1.2 times 55 pounds a foot. 24 WL squared over 8. So 411 kit feet is necessary. That's still less than his plastic moment. And it's obviously less than his lateral torsional buckling moment. With the Christmas present, beam is okay. Now you say, how detailed? That's true. Cutting lots up into pieces. I used to do that with my dad. Closer minus a foot, close enough, don't care. You know, out in the middle of nowhere on the salt grass in Galveston. Of course they don't do that anymore, an inch. And you're talking a million dollars now for oil royalties and things. So don't do that anymore. This stuff here is detailed, yeah. Maximum shear. This is the load on, this was the reaction we had before. Here's where we're going to add the dead load. 1.2, 55 pounds a foot, 24 foot, half goes on each end. You need 61.3 kips on that reaction. That's your ultimate requested shear. You can go to the Z-tables. Here are your Z-tables. Here's your beam, 18 by 55. You got 212 kips of shear. Therefore you're way over and you usually always are. And so the beam's good to go. Now do you see all that? Probably. Do you understand all that? Kinda. You able, be able to do that on an exam? If you don't practice with that? I don't know that. I kind of doubt it. You'll drive around the block with this new car two or three times. It's a stick shift. I think by the time you get back you will be ready to go out in traffic. Take a quiz. Here is the answer to your question. What if the maximum moment in the beam does not fall within the region that we're designing? The beam has to hold the maximum moment, period. There's no way around that. When you first look at that you say, well this is a highly loaded beam. And this is a lowly loaded beam. And so that puts the preference on this one. You back up and you say, well except that this one is 20 feet long and this one is a lot less subject to lateral torsion. Because it's only 12 feet long. I do not know which will win. You would have to design two beams. Number one, you'll have a C-sub-B in this 12 foot region. You'll get, here's your loads. Here's your loads. They're already factored. Moment of the quarter, moment of the half. M max. M at the quarter. M on the ends. Not being used here. So you need, here the numbers are 142.81 and so on. Here's C-sub-B, 12.5 M max, blah, blah, blah. I worked out C-sub-B for the left-hand short region. Here it is. I got a 25% increase over Timoshenko's graphs. And I designed the beam flat out from start to finish. When I finally picked the beam, I made sure that it had this much plastic moment capacity. And once I had all those done, including the shear and everything else, then I was through with this section. And I proposed, I didn't really carry it out that far, but I proposed the beam. Then I went over to see if this beam controlled. Because even though M max is much lower and the other moments are much, much lower. I don't know if they're much lower, but they're lower. I don't know if they're even lower. Maybe they're the same. But whatever they were, I worked them out. And I found that C-sub-B for the long section, even though it was long, was a 1.667. Much more Christmas present in Timoshenko's world. And then I did the same thing. I designed a 12-foot beam with an M max of 328 with a 1.667. And then I designed a 20-foot long beam with an M max of 262.3 for my C-sub-B. But then I had to also 3 make sure beam B-C will carry 328. Because even though this is what you call M max when you're working at your C-sub-B, this is M max for the beam you're proposing. You're proposing to run that beam all the way across there. And if it doesn't carry that moment right there, you're gone. Does that kind of answer what you were asking? All of this stuff is stuff we've already covered. Every now and then you will find that deflections will control. We already talked about deflections controlling too much vibration or the sensation that the structure is not safe. There's a typical deflection, PL cubed over 3EI. It seems like the deflection on it. That's a deflection on a cantilever beam. I don't know. But there of that form, you can find on page 328 a moment of inertia table which will help you if you've got deflection problems. I'll show you that here. He discusses 345 stuff in case you didn't take our past 345. We'll assume that's not the case. He discusses things like here's a beam and here's a beam and here's a floor beam and here's a floor beam and here's a floor beam and here's the load and that goes to that beam and halfway should be your fair share goes to that beam. Remember all of that I hope? He discusses it. He shows you what they call a tributary area. He shows you here a concrete slab on some beams. He shows you how much is that beam's fair share. Gives you an example problem. He's told me a lot right here. He has told me that this beam will or will not laterally torsionally buckle. Sir, why not? Because your I-beams are up in the concrete. That's correct. And the way they actually do that, here's your I-beam wide flange and they'll put boards here or metal pans. They'll hitch them on to the web or to the flange. Then they'll put concrete on the top and if they need more tying this piece of steel to this thing here then they'll weld in these little studs about every six or eight inches running down the beam. That's not to support the flange. This is plenty to support the flange. That would be if you want to make a composite section, one where the concrete and the steel work together. So here's where he calculates the weights. I'll leave that for you. Slab petition, live loads, dead loads. My first interest in this is 1.2 times the total W pounds per foot plus 1.6 times the total live load per foot. 350 pounds a foot, 700 pounds per foot. 1.5 kip per foot is the 345 number I'm looking for. He says these beams that we are looking at, when they come in here and frame into a girder, they have almost no resistance to bending on their ends. The reason is the way they're hitched up. Here's the beam framing into the girder. They can weld a little steel tab on here and then drill holes and bolt it or they can bolt an angle to the web and bolt this to the angle or a T. But this little piece right here just has no ability to handle much moment. Therefore, this beam is freely pinned on that end the way it would be considered. So those are simply supported beams. They're simply supported beams with a uniform load on them and they have this kind of a moment diagram and what is cease to be for that kind of a moment diagram. 1. Not really. Good guess. Cease to be is only if you've got full blast, full blast, full blast, full blast, moment diagram, full blast, full blast all the way. That's how he derived it. He could have derived it in this one and then made other corrections, but that's not what he gave to us. 1.13, where did I get that from? I could have done the equations, but you know I'm too lazy to do that. I got it out of a table. Anybody happen to have that table tabbed? The book you brought with you? The cease to be table. Christmas presents for bending. I see all these. I can't see it from here. It looks right. What page is that on? 3-18. 3-18. Thank you very much. Was it 1.13? Well, that's pretty close. Yeah, okay. I thought it was 1.13. Thank you, sir. Okay. Well, here we go. I can tell you how we're going to find out. Well, he's not going to use a cease to be on this one, is he? Why is he not going to use a cease to be when he walks through Timoshenko's door? Why? These are important questions. They really are. They test your understanding. They say, give us a break. We can't understand yet. Why? Because we ain't going through Timoshenko's door. We only go through Timoshenko's door if they beat us with a stick into the lateral torsional buckling room. If you say this is not going to lateral torsional buckling, I'm not going in there. I never did like that place. And therefore, we don't need a cease to be. All right. Total factored load. Moment off of page 3-213. W squared over 8. 173 kit feet to have continuous lateral support so we can use the Z-tables. I brought the Z-tables with me. Surely. There we go. I need a moment of 173. 173. Okay, here we go. 173. 173. Here's one that works. It's a W14 by 30. There are others up here that'll give you 173 feet of BMSPX. However, they'll be heavier. You'll notice they run in little groups. This is the lightest one for these below. This is the lightest one, the next lightest one. This is the next lightest one. This is the next lightest one. You can check that out. 31, 35, 48. So we would try a 14 by 30. 14 by 30 gives us 177. We only need 173. So we're okay. Now we add in the beam weight. Woo, that's getting really close. Adding in the beam weight gives us 177, but it's still okay. And the old days, if this said the beam weight threw it up to 178, we'd put it down in a heartbeat. We wouldn't think twice. You couldn't possibly ask for that much accuracy. But the courts do. So I don't care if a Boeing 707 crashes into the building. They will go through your stuff, and they will ask you, did this beam meet AISC? That's correct. And you'll say, well, first I need to explain to you about accuracy and engineering. And he'll turn to the judge and say, Your Honor, the witness is hostile. Please have him answer yes or no. And he'll look at you and he'll nod. And he knows he's got you. And so does the other lawyer. And you'll say, no. Thank you. No more questions. And you can try and explain that when your lawyer questions you. There's no explaining it. It says 178. Go get one that's got 178. Maximum shear, half of the total load on the beam. Again, the Z-tables are also a good source for finding the capacity and shear for that beam. The capacity and shear for that beam is 112. Usually always quite a bit higher. And you only needed 23. So the beam is OK. Now, oh, yeah. On my notes here, it says C-next page. Well, I got a whole bunch more calculations on it. But that's OK. We were right to here where the 112 shear was bigger than what you asked. Now then, he also has a maximum deflection requirement listed back there on the previous page. He didn't want over L over 360. L was 30 times 12 inches of the foot divided by 360. Don't want it to deflect over an inch due to the live load. Now, the live load is the only one I care about because the dead load, I'll camber it out before it gets delivered to the job. I'll camber out the dead load. And I don't care. You don't have to factor the live load. You only have to make sure the service load doesn't cause this because I'm not talking about when the worst thing could possibly happen to you. The worst thing that could possibly happen to you is a 1.6 times this live load. Then it's going to deflect more than an inch. And people are already terrified. The worst thing in the world just happened to you, so that's not going to change much. And we don't care. But we do care if, as the people are on it day by day, you get more than an inch of deflection, then that's not acceptable. Therefore, I will tell you that the delta L out of page 213, 3-213 for a uniform load on a beam is 5384 WL4 over EI. There's 5 over 384. Here is your nothing but your live load only unfactored. We have an overkid per foot divided by 12 inches in a foot. That's what the slash 12 is for. Here's your length in inches. Here's your E in kips per square inch. And this is in kips per foot. And 291 is the moment of inertia of the beam that you recommended you did too. I heard you do it. About an inch and a half. It's not acceptable. Good beam will not fail. It won't break. But when the people go in it and they feel it shaking around, they'll go to somebody else's building. What I'm going to do is I'm just going to solve for I out of this equation to see what I need. No sense in picking another beam, finding it doesn't work. Pick another beam, finding it doesn't work. Pick another beam, no reason. Let's just go straight for the I. Solving for I is 5 WL to the fourth divided by 384 E. The I comes up here. The delta comes down there. Delta required. I think that should really say delta allowed because it's allowed to be an inch, not required. If your beam doesn't have 440 inches to the fourth, it's not going to do the job. And he has a table for plastic stuff. He also has a table for moments of inertia because engineers can do this all the time. I need a 440, 440, 440, 440. I don't know what that is. Somebody else has been using my page here. 440, 440. There's the first 440 I see. And here is the lightest one I see is a W18 by 35. Just because it's got more moment of inertia doesn't mean it'll still work for lateral torsional buckling. Not a problem in this case. Or if it's still got enough plastic moment because it's deeper doesn't mean that it might not be a problem with it. So you do need to check it. Trying a W18 by 35 has more moment of inertia than we need, so our deflection is okay. Delta is okay. Plastic moment. I'll get that right off the same Z-table I had a minute ago. It's good for 249. I only needed 177. 177. Now, that's after the load, after the dead load was applied. After the weight of the beam was applied. You don't see the 177. But it's bound to be on there somewhere. There we go. There's with the dead load of the beam's weight applied. 177. It's greater than that. And the shear is also greater than we need. It seems good to go. So notice your checks now. This is back on an earlier page. Used to be you had that many checks just practically right out of the box. Number one, you had to make sure the plastic moment was okay. After you went into and out of Timoshenko's room, you had to check lateral torsional buckling while you were in Timoshenko's room, including any Christmas presents you got coming. Cease of B. You had to check flange local buckling. You do that easily by seeing if there's a little F on the beam you're going to pick. And you didn't have to check local buckling. You probably should state WLB never a case for this class. Now you've got to check delta max, and you've got to check the shear max on the ends. It does not. I don't think so. It's suggested usually in other codes. AISC has one. Some of the other real codes as opposed to specifications have them. We just kind of draw them from there. If you look in Segui, back where we were talking about those limits, he had various and sundry kinds that are commonly used. You will get the exact specification to be used when you go to the city and say, can I bid on this building and say, here are the specs. He says, dang, you don't like this thing to reflect more than a half inch. He says, that's what we like. He says, do I have to abide by that? He takes it back from me. He says, next, you don't get to bid. All right, so these are the things. This is another way you could have found out that 14 by 30. You could have just gone to the zero foot length and run up until you find somebody who comes out here bigger than 173, and the first one would be a 14 by 30. That's just another way you can do it. It's easy to find this page with these numbers on it, and so it's really easy to find the lightest shape. You just go to the 173, and you look for the lightest shape above that. It turned out it didn't work anyway because it wasn't strong enough once you walked back out the door. The plastic moment wasn't enough. Actually, that would have told you the plastic moment. We could have found it right there. There was a moment of inertia table. We've already done all of that. We've already done holes in beams. Let me just get a start on that. First place, here is a flange that somebody has drilled some holes in the flange to attach it to a column, or perhaps it's a splice plate to attach to another beam. We will have to provide such that we get as much moment as we need, sometimes the full plastic moment. On the compression side of the beam, the bolts are in these holes, and the loads will go right on here, and if things get really bad, the stresses in this region right here can go ahead and build up a little bit, and the hole will close up on the bolt, and the load will pass right on through. If you don't have holes any place, but on the compression side of the beam, there are no holes. If on the other hand, you're having to develop the moment, then obviously you need a plate on the top and a plate on the bottom. That plate's intention, that side has holes. Whether or not you have to consider them or not depends on the following thought. Number one, before you went drilling holes in things, that flange was able to carry this much force in the flange. The web, of course, is something different, and it's got no holes in it, so we would not worry about it. But the flange intention was supposed to carry the gross area of the flange. That would be B sub F times T sub W. B sub F being the width of the flange, T sub F being the thickness of the flange. I think that B is just B, it's not B sub F. When you multiply that area times the yield stress that you admitted you were going to put in there, it had a force. And you told me that force would be there and I'd like that force. Say, well, you know, the truth is the flange area is going to, a little of it's lost. Not a lot, just a little. I say, don't give me that. How much of it is lost? Well, I'll have to find the diameter of the hole times the thickness of the flange and I'll have to take off two of them and so the flange's gross area will now reduce down to the flange's net area. And I say, okay, do you still have the same force in there? You say, well, obviously not. I don't have as much area. I say, well, let me go run a bunch of tests on this and let's just see what we can do. Now come back with the good news. The good news, if you haven't taken too much area out you can run the yield stress right on up to the ultimate stress and things still work just fine. It's not like this is a weak link. It's more like a discontinuity. Now you could drill so many holes in there that when you multiply the net area after taking out the holes times f sub u you still don't have maybe half of how much I was expecting before you drilled holes in it. And I don't care how much less you have if you can't take your net area times the ultimate strength, the ultimate stress. If it's not at least equal to how much I had planned with the gross area under yield you must include the holes in the flange. And we'll take it up there next time. Think about that.