 Howdy, howdy, howdy, howdy. Campbell. No, it's further back from you. It's a guy in a green shirt. The one who's got his, just picking his nose. There you go. But he's still talking. Oh, he's eating it now. Oh, log. Jolt can turn around and look at him if you want to, and he'll say, I'm not kidding you. You'll have to watch the tape to see what we're talking about you. Guy in a green shirt, yeah. All right, we were working on bolts. Bolts that would twist and load a set of bolts, these you can do now because you can just use the tables. They give you an ultimate set of results as opposed to an elastic set of results. These on the other hand, these bolts will be loaded by 1.2 dead plus 1.6 live shear only on each bolt. That total divided by a number of bolts, one, two, three, four times two is eight, plus a moment about the centroid of the bolt group, 1.2 dead plus 1.6 live times 2.75 inches. Be the difference between that shear plane in the center line of the bolts. The moment for this design would be 1.2 dead plus 1.6 live coming up on this shear plane times 2.75 about that point. Actually, when you pull it out of the table, it'll have an instantaneous center of rotation as where we'll be getting the answer, but the moment is applied about the centroid of the set. So on this one, the only difference is rather than the bolt being in shear because it's loaded down plus bending because you put a bending moment on it, which again caused shear. And shear plus shear, we can get reasonably easy. This set's going to have shear plus the top bolts are going to pretend to go into compression. The truth is, the compression will go through the bottom of the connection. And then the tension will come into these bolts. And now we know how to do it, but it was obviously more challenging because this bolt now has shear and tension. So we used the interaction equation. First, you found out how much shear you used up, which was originally available to you. And if you hadn't used it all up, then you probably had some tension coming to you, which you would find using the interaction equation. So going through one of those, not much reason to worry too much about the bearing and things like that. We'll hit it a little bit, but no doubt about it. There is a shear load going through here. It's going through a 0.560 inch thick flange of a 10 by 49 made out of the same steel. Everything here is the T that they have cut to use as a connector, has a flange thickness of 0.615. So when you're checking bearing in here, you're obviously going to use a smaller number. You have tables to get a lot of this for you. First, how much load you can put on each bolt due to shear comes right out of a table, page 493C, table 7-22. See if I've got that. 493C, you ought to be down the road a little bit. 3C, there's 493C, table 7-1. They're using 3.25 inch Group A bolts. Threads included in the planar shear, single shear. Check the problem spec. And you get the strength of the bolt in shear, 17.9 permitted. Final, including fee allowed for you, 17.9 per bolt. But we put that much allowed per bolt. Then you're going to have some bearing strengths on these inner bolts. They'll have reasonably long plugs to pull out. And you'll have strength of these in bearing. They'll have little shorter plugs, but we will assume that these little links will be made long enough such that bearing will control in all of those cases. Crushing, it's bearing no matter what. So you have tables for the inner bolts, how strong they are with a spacing of three inches. Bolt bearing is equal to, see where we got the 87.8. This is pretty old. I have that table in here somewhere, surely. There we go. Available bearing strength, that bolt holds based on the spacing between bolts. That obviously is not the table for allowed bearing strength at bolt holds based on in-distance or something like that. Bound to be another table for that. Who knows? Probably I just got one of them. Ours was three-quarter. PR Subin for a standard hole with three-inch spacing. Our steel, 65-KSI steel, 87.8 kips of strength in bearing per inch of plate. Got to give me an inch of plate before I'm going to give you the full 87.8. You didn't give me an inch of plate, so I'm only going to let you have a piece of the 87.8. 87.8 times the thickness, 0.56, means there's 49 kips of strength on each bolt in bearing, well over the 17.9. So the 17.9 shearing the bolt still controls. That's plug and crushing interior bolt strength. He has given us some equations, so I went ahead and pulled the specs out now. We already discussed this. When you have an eccentric load piece of you, you're going to cost a moment about this. There's two ways that they suggest work nicely, one of which is you're willing to go try and find the real neutral axis for this item. On the next page, he shows you how. He says draw you a neutral axis. He suggests that you take 1.6 of the depth to begin with and see if it works for you. You go that high. You then tell me how wide this is. How wide he suggests you call that, be effective, is about eight times the thickness of the flange, but no wider than be of the flange, because that's all the width that's even there. Question, if you've got some engineering reason to change it, that's your business. If you make it smaller, he's certainly not going to complain about it. Knowing that area, and knowing this height is a six of the depth, you're going to find where the sum of the area of the bolts times where they are is equal to the area of the rectangle times where it is, d over 2. Set those equal to each other and see if they balance. If they don't balance, you're going to have to move this up and down a little till it does. When you find where they balance, then that's a pretty good estimation of where the neutral axis is. Knowing that is the neutral axis, you should be able to calculate the moment of inertia of a rectangle, b h cubed over 3. That sound familiar? b h cubed over 12 is about the centroid. Then you will get the right answer, because you will be adding a d over 2 squared, which happens to come out, b d cubed over 3 is the moment of inertia of a rectangle about its base. For years, I told everybody you'll never use it, forget it. Well, it turns out here you actually can use it directly if you happen to remember it. It's a moment of inertia of a rectangle about its base. Then you'll have the moment of inertia of these two guys. There will be 2 times the area of the bolt times where do they live off of your neutral axis, plus 2 times the area of the bolt times y2 squared, y3 squared. That will be the moment of inertia. I mean, we're not going to do this. We don't do this. It's not even listed in the book, but it's just so that when your boss does this, you can do anything, because you understand how these things work. You just go look around at his examples, and you say, OK, I'm ready. Once you get that moment of inertia, you will bear his equations. You'll find that r sub u total, you're going to go ahead and assume it's either straight line. And r sub u total is going to be your ultimate load times the eccentricity. You'll have to tell me what you are ultimately requesting times c, which will be the distance from your neutral axis to the outside bolt. You don't care what these stresses are. There's no way they're too big. That's steel on steel. But these bolts in tension, what we're looking for, mc over i, that's the stress times the area of the bolt. That's how much ultimate force we would agree is pretty close to the force that's in that bolt. Knowing the shear in the bolt, shear in the bolt, load over area, total load divided by the number of bolts, this will tell you how much of your shear capacity you extracted before we started playing this game. Then you've got to go to your interaction equation and tell me how much is left for you in tension and see if that matches what you need. So it's OK. It's as good as anything else. Sugui and most people just go ahead and take it as a centroid in the middle. Go ahead and pretend this isn't in compression between the plates, that the compression is taken by the bolts. And that the top bolts are in tension so that you can find out what the tension is. You're not going to use these anyway. These bolts will be in shear only. These bolts will be in shear and tension. Taking that as the neutral axis. Assuming case two, neutral axis is at the center of gravity. There's your load in shear, load over number of bolts. Here's your load due to the moment. P sub u times e is your moment. N prime is the number of bolts above the neutral axis. He would never ask you to do something he didn't define. D sub m is the distance from the neutral axis to the center of gravity of the bolts. And he's only going to take one set of bolts. He's going to use d. That d is the distance between the center of gravity. If you want to take one set of bolts times half the distance and then take two of them, that's fine. His equation just shows you one distance between the couple caused by those bolts. The whole purpose being, again, just to get that r sub ut. So that you can tell me if the thing is OK. First, again, find out how much shear force is in the bolt. Admit that you took some of the strength out in that fashion. Apply your interaction equation to see if you still have at least as much left. If you do, you're good to go. Still got to check bearing stresses and things like that to make sure that the plates themselves don't fail. Now, are you going to really do that? Well, you might. More than likely, you would use the tables, because the tables have all those bolts and information on how strong they are. Only thing this page, as far as just for the last part of this top stuff, this is for hollow structural sections. There's a table we use to get the strength of the bolts. There's a table we use to get the strength of the plate that's in bearing because of the bolts. Here's where he completes out this problem so that you can see all the numbers. We had so far found 17 point something kips of available force through the shear. We found, I don't remember, 40 or something like that, kips of force available in bearing. Here's for the edge bolts. Use table 75. Evidently, I don't have that one, but it's right there next to table 74. It says, I'm going to go ahead and assume we'll use an inch and a quarter, since that's a number in the table. He'll pull the number right out of the table. 49.4 kips per inch of thickness. Times how thick is your plate. Got 27 kips. 27 is bigger than, I think the other one was 49. Both of them are massively bigger than how strong the bolts are. There's the strength of the bolts in shear. Then to get towards the interaction equation, I'm going to have piece of you, 1.2 dead, 1.6 live, for 88 kips of load. So your shear load per bolt is 11 kips. I think you're pretty good off here. Even if this came out 17.9, there'll still be a little tension left for you. There won't be lots, but there would be some available, because if you remember that curve, looked like this, looked like this, and even if you used the full 17.9 in shear, it was still a little tension that the specs would allow you to use and which is really available for you. But as it is, you only used up 11 out of 17.9, so I'd say you got quite a bit left for tension. I don't know if you got enough, but we would see. Then you compute the tensile force per bolt. I don't know, I find this weird. Here's the free body. We had three inch spacing on all the bolts, three inches to the center, it was inch and a half, inch and a half and three. These were R sub tension, these are the tension forces. Well, I would say that some of the moments about that point is there are two bolts there an inch and a half out and there are two bolts that are four and a half inches out and there's a pair on the top and a pair on the bottom. Those multiplied times of force and tension ought to be how much moment these forces cause about that centroid. He's got the numbers here, it comes out the same thing I got, so I can't complain. Then the applied moment is load times distance. You told me you were gonna put 88 kips of load at a 2.75 inch distance. Check the 2.75 right quick, I think that was correct. It's correct, 2.75. In other words, they're doing that, that's the way they're gonna construct it. They're probably gonna leave maybe three quarter to a half inch gap, something like in there, three quarters to an inch in there and then they're gonna get back here far enough so that these things don't tear that little block out. That's not gonna tear the little block out cause they didn't cope the top out. But they gotta have some in distance here, minimum would be required and that's gonna take the, where the load comes through the bolts and this shear plane 2.75 inches. There's your force times your distance, 242 kips. Here's the force in the bolts, how much moment they can take. The force in them times 24, set them equal, 10.08 kips of tension on each bolt. And then from then on is the same old thing. Go right on down into the interaction equation, admit that you took out some of the shear times these constants out of the tables, tells you how much stress you have left. First it hadn't exceeded 90 so it's okay. It still hasn't been checked that it's okay but that's how much tension stress you can put in it. And then you, there's that dang table, a loud bearing strength that end holds. There's that 49-4. It's kind of in a weird place, right page but there's where they were checking how strong the bolts were in shear. 54 ksi times area of the bolt. Still gotta have a fee in there and how much design strength you get out of them. You get 20.4 with that stress that we just got. Stress times area times 0.75. It's a greater load than you're requesting so they were okay. That part's old stuff. The only new stuff here is how in the dickens do you get the tension force in the bolts? Now wells work the same way. They have the same two methods of being solved. One's an elastic and one is a plastic analysis. We're not about to do either one of them by hand, obviously. Doing the bolts by hand, you probably could do by the bolts by hand but there would be a lot of calculations. You could be given a connection and do enough of the bolts by hand to show that you know how to do them but the wells are even more onerous. First, there's an elastic analysis. Again, it has the limitation of whenever a weld anywhere is overstressed, he says quit and everybody else has to quit and they're still within the elastic range. Go to a plastic analysis, then that well starts to go non-linear but he doesn't say quit. He goes on up and goes very high, letting all of his buddies come on up into their plastic range to some extent and really be there if needed under unusual conditions. The elastic is done just exactly as the elastic is done in the bolts. You're assuming everything is elastic. You would locate the centroid of this set of lines and you would calculate the TC over J or MC over I, whatever you wanna call it. You have to calculate a moment of inertia about this point, the way you'd find the centroid is you would say this length times that distance plus this length times zero plus this length times where he lives divided by that length plus that length plus that length. That would tell you where that centroid was and elastically if everybody stays in their elastic range that is the point at which it would rotate. You would find a stress MC over I, TC over J, whatever version you're gonna pretend to use this for. It wasn't really TC over J or MC over I wouldn't either one really made for this so it gives conservative results. You'd find a force in the X direction, a force in the Y direction. It doesn't matter which way they're loaded generally so you could add that X component squared, that Y component squared, take a square root and tell me how much stress is there and when it reaches the peak you'd have to quit. The plastic works, here's where he actually works one elastically. You'll notice he's finding an X, he's finding a Y due to the, he's gonna have to find, put the load right down the centroid to get some stress due to the group loaded centrally in the Y, there is no X. Then due to the bending P times E you're gonna have a X component and a Y component of that stress. You'll add the two Ys together, square it, you'll add the X squared, take a square root that'll be how much stress is in there and keep that within how much stress is permitted. We're really gonna do in the elastic, you know, we'd go into it in more detail but you don't remember how to get the centroid of a group of lines, it's in the FE reference manual. What he says right here, take the sum of the centroid. Centroid of a line is the sum of the length of the line times where it lives divided by the sum of all the lines. Same way with a Y. And if it's in a third dimension you got a Z. So for example, here's where they're gonna rotate about. You'd have two components on that weld, you'd have two components on this weld, the components would be added, the PY causing some centroidal, then the X if the force really goes two directions then you'd have some moment, you'd have the components, you'd add them and square them all day long. Got the solution. Now this is still the elastic method. Yeah, that's still LRFD. Just because it says LRFD doesn't mean he's using a plastic method. Time to go through all that? No, reason to go through all that in very little. Design of the weld turned out to be a half inch elastically. Horrible. Causes, man it's so conservative, it's just ridiculous. Ultimate strength analysis, now we're talking. Now then we'll have an instantaneous center of rotation, you can assume it's wherever you want to, if it's really symmetric and if the load is really straight down it will be somewhere on that axis. But who knows? How does the weld behave? Well, it behaves very much like a bolt does. The problem is, is while we're at it, there's no reason, we know we're not gonna get in here by hand and do this. We're gonna come in here every inch and break the things up, find out where they are, find the X and Y components and work it out by hand. But, if we're gonna have a computer do it for us anyway, we might as well take into account the fact that this resisting force is not right down the axis of the weld. And therefore it is bigger than the strength right down the axis of the weld. How big did we say a weld was if you loaded it across its throat rather than along its throat? How big? Thought somebody mumbled the right answer. Maybe it was just a burp? How big? Point seven or eight? Well, no, that wouldn't point seven or eight. You mean like one point seven or eight? Best I remember it was one point five. It was 50% stronger if you loaded across the throat rather than down the throat. You didn't have to take that into account. You were permitted to not bother with it if you didn't want to. But if we're here and we're gonna go to all this work and use a computer to do the numbers for us anyway, we might as well do that. It turns out that the deformation of something at a given angle is given by this equation and they're gonna take into account the fact that the weld is not just the 100% of its strength that it's higher because you're loading it at an angle. And that angle, of course, continuously changes all the way around. Do we have, do I have a computer program that does it? I don't. I'm sure you can buy them. I'm sure there's not much to them. And he goes through an example. And he goes through an example. You wanna see one, there's one. Because the truth of the matter is you and I are gonna do what? How? What? No, we are too. We gotta have a way to get some kind of answer. We're gonna do the same thing everybody else does. We're gonna buy a computer program that does it for us, or we're gonna use the tables. And the tables work very nicely. They look complex, but they're not as long as you've been through it once. It's not much to them. And there's the table right there. And that's everything. That's all the extra strength because some of the welds are not loaded right along the throat. They have some kind of an instantaneous center of rotation. And that weld right there is loaded at an angle. It takes everything into account. And the table gives you coefficients, C, and the strength of your connection, the entire nominal strength of the connection will be the C from your table, times a factor that you will apply depending on the strength of the weld. Remember how we had E60s, E70 rods, E90s, E110s? These work for all of those. All you have to do is have an appropriate coefficient to change from changing from an E60 rod to an E70 rod. I assume these numbers are probably for E70 rods because that's 90% of what's used. I'm sorry. C1's electrode strength from table eight. I don't know if your book says that or not. Maybe I got an older printing, but that's the wrong table number. Yeah, it says it's one for E70 rods. So that makes sense because then the people, 90% of the time, never have to worry about the rod strength. They don't have to change the number given right out of the table. So the strength is C2.96, if you're on the right place in the table, times C1. So one, if it's an E70 rod, times D, what is D? And if I know, D, D. Where did you get that from? Oh my, right there in the little thing right there. See, he never asked you to do something. Doesn't tell you what it is. D is the number of 16ths of an inch in the fillet weld size. Times, length, the characteristic length. They gotta call something L and then they gotta call that a piece of L. And they gotta call this L and they gotta call that some multiple of L. Some place they gotta have a number on here so you can get the length in there. So what they do, they set up the entire set of tables calling this L, you tell them L. And then you tell him how big is K? Well, if this is 12 inches and that's six inches, then L is 12 and KL is six. And since KL is six, K times 12, K must be half. So that gives you the shape of this particular one. They got a table where this is at 20 degrees, 40 degrees, 45, all of these 65 degrees, things like that. They have tables where these are boxes. They have tables where there's nothing but two straight lines. They have tables where there's just two straight lines and the load is over to the right of the two straight lines. And then they have tables where this load is out in space. It has to go through the middle. They got all kinds of tables for your enjoyment and profit. So, take a look at an example. First off, to use even these tables here, you're gonna have to know where the CG is because everything's gonna be measured with respect to that. Well, I say everything, namely how far the load is off of that. And he could have taken it from the back, probably and worked the tables out, but he didn't. So, that's your problem. Probably, I don't know, maybe it wouldn't fit in two dimensions like this, I'm not sure. But I can find the CG of a set of lines. My lines are eight by 12 by eight. He said he wanted to solve example 8.5. It's on page eight dash 90 in Segui. So, I just copied it down here so I didn't have to go back and look. The centroid right there is, I would draw a reference axis right through that weld. I would say that the centroid is the length of that line times where does he live? That's eight times four. The length of that line times where does he live? That's 12 times zero. The length of that line times where does she live? That's eight times four. That's the sum of the moments about some reference axis. And that's equal to the sum of all the lines times where do they in common live? Which would be eight plus 12 plus eight times X bar. There's the calculation. X bar, eight times four plus eight times four plus 12 times zero divided by eight plus eight plus eight, 2.286 inches. Now the tables actually will also tell you where that is. I think they're more trouble to use than this. It's up to you, but you do have to know that this dimension right here is 2.286 inches. He says, determine the size of weld required for that example based on ultimate strength considerations. Use the tables in part eight. The weld of example 85, same one shown in table 88. That's on page eight dash 90. That's page eight dash 90. That looks like ours. He says, the following constants are need for entry into the table. First, you gotta know K. Gotta know where to enter the table there. Well, we already said that K was some percentage of this, can be longer than that. You know, that K goes on all the way out to two. But in our case, KL is eight over 12. So eight over 12 is 0.67. That's where we should be entering down that line somewhere. The other number I need out of the table is A. Whatever A is, according to him, A is the S-intricity divided by the characteristic length. Well, this weld is characterized as a 12-inch weld. And E is 2.286. So here's A is, where did that come from? That's E, excuse me. That's E. I mean, that's X bar. X bar is 2.286. Here's E. E goes from there to there. In our weld, our weld, that load was 10 inches from the edge of the weld. The weld was eight inches long, and you just went too far. I was getting ready to say E is 18 inches. You gotta back off 2.286. So E sub X is 10 plus eight minus 2.286. That gives you E sub X is 15.7 inches. You know, it's nothing but geometry. How far is it from here to there? Minus how far is it from here to there? So A is equal to A times L over L. I couldn't hardly complain with that. A times L is E sub X. Is that what he has in the picture? He does indeed say A times L is E sub X. That's our 15.7 inches. E sub X is 15.7 divided by the characteristic length. There's your value for entry in the vertical dimension. There's your entry for the K, the horizontal dimension. 1.3 and 0.67. 0.67, 1.3. Now that's gonna require interpolation in two directions. It's just the way life is. Nine times out of 10, nobody's gonna, well you can do it by hand obviously, but you're gonna probably set up a little computer program where you just enter four numbers and let them do the interpolation for you based on how far from here to here you wanna be. You wanna be 0.7 of the way from here to here. And here from there to there, you wanna be 0.31 down to get that number for you. That number will be C. Your strength will be somewhere around, I would say 1.6, somewhere around 1.6. Let's see what he really gets. Here it's solved, C. K L got the C, got the K. This is where he'll find X for you down here. Look at all that work. I don't think it's worth it. I'd rather get it by statics. So there's your A. A and get C from the tables. Well we got for C. Your A in the middle of any of them. Yeah, gotta be in the middle. You know, if it's not shown there, it has to be. Well I was the 0.67, that's from up here. See that's six to seven, so we're entering K there. So we're gonna go 70% of the way from here to here and then we gotta go 11 out of 0.11 out of 0.2 down to get to that point. So let me see. We know where we're going. All I gotta do is see the number. There's our coefficients. We're using a one. Here are the numbers for the other rods. There we go. He says by interpolation, C is 1.52 and I put a check on it, so that must be what I got. Oh no, no, no, that's interpolation in the, for the 1.3. That's one of the interpolations for K of 0.6. It's not interpolation. That's what it is for 0.7. Interpolating between the two value gives you 1.67 for the C out of the table. Then you can tell me how strong it is. How strong it is according to our table says nominal, C, C1, D times L. Not gonna do us a lot of good without the fee, so you might as well put the fee on there. That gives you the available strength. Is fee times C times C1 times D times L. You now know fee 0.75. You now know no C interpolating between the two values. That's about right. 1.67 with his high number here and the low number here. I'd say that's right. Times D is the number of 16th inch of weld. You know, he never told us the number of 16th inch of weld, did he? Never told us the weld size. What's he asking? What's he asking? That's probably what he's asking. How do we know that? This is the only thing he didn't tell us. You're right. Because now then here we have another known number of the characteristic weld length, 12 inches. So we solve for the number of 16ths of an inch that we need in this weld. D is equal to or has to be greater than piece of you move fee over to the denominator, move C over to the denominator, move C1 over, move L over. There's your 60 kips is what we were asked to give him. So on page 463 is given. There's your fee. There's your 1.67 double interpolated C. There's because you're using E70 rods and it was 12 inches long. You need that size weld. So he says it's about a quarter inch. So it's like half of an elastic design because none of the little people that are down towards the middle have a chance to go plastic and give you some strength. So the tables work fine. You see what's going on. And other than that, you just about have to have a computer program to do the work for you. Here's another typical table. This is one where it's the same thing as we had. It's exactly the same idea as this. The only difference instead of having a channel looking thing on it, you have two wells on the sides of something. That would be, for example, if you were welding a plate to something. And rather than welding it around there like that, you're welding it on the sides. It would be something, for example, if you had a column and you wanted something to put a beam on like that and you put the beam on there like that and you wanted to weld it down the sides, the front, the near side and the back side. That would be this case right here. You have the two wells down the side of the angle and you have the load at some eccentricity, A times L off of the plane right there. And here's where we would probably assume the load was applied in the center of this bearing area. Then this table would be perfect. All you gotta do is know how to go get all of these numbers. Now they also work in other directions. The strength is this. B times R sub N is your design strength. So that's, you're still gonna have to have a fee someday in here as you're right here. The nominal strength is gonna be CC1DL. You can solve for anything out of there. You can instead, if you know the load and you know how many sixteenths of an inch you want to use in your weld and you know the characteristic length of the wells here, as you see that cell, and you know fee, and you know your electrode factor, probably a one, then I can tell you the minimum C you have to use in the table. So when you come in here and you say, well, okay, from my case K is 0.4 and I don't know something else, but then I'll tell you that you go down and you can find the minimum C that's gonna work. You're not gonna be able to use a number smaller than this one. R, say that you already have C out of the table for a given K and a given A, and you'd like to know the minimum well size, the minimum number of sixteenths. That's basically what we did a minute ago. R, you say, look, I know the well size and I, that's a mistake. In the world, let that get through. What is that thing doing in our house? You know, it is amazing sometimes the typos they have on these things, they don't catch before they publish them. Has anybody got their, look on page E-8-90, see if it still says that. 8-90, that's where I got this from. It does not, okay, it's been fixed in your printing. We still have the same book, but it's a different printing. I would tell you the minimum length that you'd have to put on the angle. All right, and he's got, that's his solution. Sometimes you will have angles that the centroid does not line up with the centroid of the weld that's permitted by the specifications. Unless you have repetitive loading like on a bridge or something, and then they will say, look, we want the center of the weld group to match up with the centroid of the angle. In which case, the only way to do that is to make this one shorter and make this one longer. It's very briefly, you can some moments about this point. You'll have to know how much load you're gonna put in the angle, of course. Take 50 or whatever it is. You can some moments about this to get rid of this unknown force. You can take how much force you're gonna have on the end of it. You'll know how much, you'll know how long the angle is. You'll know the weld you're planning on using. You'll be able to say how much load you have here, perhaps 30 kips comes out. Out of that, then you still need for this to bring out some amount of force. Using that, you can solve for the force in this one and therefore how long it is. Then you just, some force is horizontal to find out how long this one has to be. Well, what that does is that lines up the centroid of the weld group with the centroid of the angle that's being connected. Do I expect you to be able to do that? No, do I expect you to kind of remember when you go to graduate school and they say, do you know how to put the centroid of the weld group along with the angle? They say, I don't know what I think I can find out. On page 506. All right, did we cover everything there is? Are you 80 years old? Well, then we didn't cover everything there is. See you in the final. Questions? I'm sorry. Is there something to do today? Not do today? Not do today. Okay. Did you work it? That was a smart move. Are you talking to me? If you're talking to me, I can't hear a word you're saying. Oh, thank you. That's what he's asking everybody else says there wasn't anything to do today. The last one's optional. This one. I can't read that. Let me come over here. I wouldn't, I wouldn't trust that anyway.