 Okay, we were working on tension members. We were trying to find out, is that too loud? The gana backs, yes, that's too loud. We were trying to find out how efficient the joints were because under certain connection conditions you don't connect everything as nicely as you could and therefore some of the stresses don't know where to go and they run around in the joint and go places that you really aren't planning on them going and it means that when you put the load on it it won't carry as much as you might anticipate. It has less than 100% efficiency to it. These are your factors that tell you how much you lose when you connect certain things in certain ways. He's got cases 1, 2, 3, 4, 5, 6, and so on. I'll have marked it case 4a, b, and c so we can discuss it. There's also a case d. You say, well, there isn't one listed. When case d is not listed he means you can't do it. I like to see it there. You can't do this. So all tension members where the tension load is transmitted directly to everything in sight, for example a plate on a plate, or example an angle on an angle where you bolt it and bolt it and bolt it. That'd be like 100% because there are no unconnected outstanding elements. He says you should use a u is equal to 1, transmitted directly to each of the fasteners, except in cases 4, 5, and 6. In cases 4, 5, and 6 you have connected that with these longitudinal wells to this, but it's kind of a special case. Here you have connected everything in sight, but it's kind of a special case. It's a tube with a plate inside of it. These are hollow structural shapes in the same way with this tube or pipe. These three are exceptions. Next, all tension members accept plates. Okay, accept plates. And yes, sir. That's okay. In this class? Okay. I think he's the one. I think those are the ones that looked up, saw me here and ran the back door. I just brought it out of the room and dumped it on you. I didn't know who it belonged to. Thank you. We try and get all this stuff back to you, but we don't know exactly how. All tension members accept plates in each hollow structural shape where the tension loads transmitted to some, but not all. In this case, this is our general equation. In other words, you take 100% efficiency, but you subtract how far is it from the connecting plate that you've bolted that T onto to the centroid of that T. That would be what they call x-bar. That's listed in the books. Here's an angle. There's a wide flange. Believe it or not, sometimes on the wide flanges, and we'll show you some better pictures of these shortly. You put a plate on the top. You put a plate on the bottom. There is an unconnected element, and that's the web. Same way if you put a plate on the web and you don't bolt down the flanges, then there are some outstanding elements that you have to account for. All tension members where the load is transmitted only by transverse wells to some, but not all. I'll show you one of those. When we number the cases three, it doesn't have a picture. I'm not sure why they don't actually show you one, but I got one back here. Plates where the tension load is transmitted by longitudinal wells only. The real problem here, and I'll show you in more detail, is you've welded it nicely. You've welded it nicely, but this little stress down here comes down a road, down a road, down a road, and I know what to do, because he sees down a road there's nothing but air. So he's got to hook it over to this weld, and that causes problems. That causes a loss of efficiency. There's a little dead zone right in here where there are no stresses, no loads being transmitted. So you'll have less efficiency. Here's the equation. If the length of the connection is greater than two times the width, in other words, if the plate looks like this and you've got it welded longer than two times this number, then all of these people, even these little last ones, have got plenty of time to get home, and they don't have to go out that little indoor where there's nothing but air. So if L is greater than 2W, you get 100% efficiency. If your L falls within 2W in one and a half, they have found experimentally that you're losing about 17% of your joint carrying capacity. If your length falls between one and a half and W, in other words, if your length is equal to W, then here's the width of the plate. Here's what you connected it to. You see how this no welds. You see how this length is the same as this length. Then some of these people are going to have a hard time getting out, and therefore you lose 25% of the efficiency. If you try and weld it like this, not permitted, just nobody can get home like that. They just all stack up on each other and they're burned to death trying to get out the door. And that's not listed, and that's why it's not listed. But the truth is, if W is greater than L, U is zero. Questions? Let's see those looks right now. What? What? What do you say? Okay. Round holla structural sections, like a tube. I'll show you a better picture of this one. If the length of the connection, that'll be how far it's welded back down this line. It's greater than 1.3 times the diameter of the tube. The diameter will be listed in the book. You say inside or outside. I say yes. Big D. If it says ID on the book, then that's what you use. I think that's the OD. Then you get 100%. In this region, you get the standard equation, 1 minus X bar over L, where X bar is D over pi. And very interesting, I think, is if you will find out where the centroid of a solid piece is times its area, minus where the centroid is of a missing piece times its area, you find that the centroid of a circular segment is D over pi. And therefore, that centroid, off of this thing right here, that's X bar. I admit it looks like it's in the Y direction, but that's what they call X bar. You can derive that. You can derive this one here also, where the centroid of a bent piece of metal like this, whereas it's centroid, that's right here, B squared plus 2BH. I've done it. I didn't believe it. It's right on the money. A1Y1 plus A2Y2 divided by A1 plus A2. Remember that? Where to get a centroid? Now then, these are for analysis. These are when you say, look, I've already got a pretty good guess what this thing is going to be. It's going to be 2 by 3 by 70 inch angle, or whatever it's going to be. If you know what it is, you're just going to do this. Basically, it's all 1 minus X bar over L, except in a case like this, where it was such a mess, we had to go out and test millions, thousands of them, and write down the right number for you. Now then, if you say, look, I haven't even got a start on the thing. I know I'm going to use some angles, but I don't know the size of the angles yet because you won't tell me what the efficiency is. I say, well, I can't tell you what the efficiency is until you tell me X bar and L. And he says, well, I can't tell you what the, and then we go back and forth. And the first thing he knows is a big fight breaks out, rolling around on the ground. We've got to have a way to get started somehow. But here's how you get started. If you're designing a W, miscellaneous, American standard, or an HP shape, or if you're studying a T cut out of those shapes, that's all we got to say, with the flange connected to a plate with three or more fasteners in the line of the direction of loading. For instance, here would be the flange. Kind of hard to draw and make sense here. I think I got a picture of it later on. Here's a channel with three or more, one, two, three, four, five, six, seven, in the line of loading. So we're talking about connecting the flange with at least three or more fasteners. Then we need to know the ratio of how thick the flange is compared to its depth. We'll let you have 90% of the connection strength. You lose 10%. If, on the other hand, it is smaller than that, then you're going to have to drop off 15%, and I'll show you pictures of those. Now then he says, now then once you get something that you can really put your hands on and you want to go back to case two, which is all tension members except so and so and so and so, and you want to use this equation, you're welcome to go back and use that equation. And if you also calculated out of case two and you say, gee, I wish I hadn't calculated case two after I got through doing this approximation, when I went and found out what they think is a better number, a better way to do it, I got a lower number. Very often they say this and they'd never say it if it wasn't true. The larger value is permitted to be used. All this stuff is somewhat approximate but does a pretty good job. What they've really found is is that if you follow these rules, those are okay, period. I don't care what you finally designed. And if you follow these rules, they won't tell you anything but something better or something worse. And if you follow this or this, you're okay. So almost always we say how much load I always pick the bigger, how much resistance I always pick the smaller. He's telling you now, you don't have to, but he's telling you now that if you do the design based on this and then you check it with this or vice versa, you are permitted and it will prove to be the true case. You can take the larger of those two loads. Most unusual, but 100% true. Now then if you're not connecting the flange to some plates, if you're connecting the web to some plates, then you're going to need at least four or more fasteners in the line of direction of the loading and you will drop out 30% of your strength. You'll have to multiply your value. Your R sub n, what does R sub n stand for? What? Nominal resistance, that is correct. Multiply the nominal resistance times 0.7 and once you multiply the nominal resistance times 0.7, what is that called? The design strength, that's correct. Talking now instead about angles, double angles or single angles, it doesn't matter. Again, he says if you calculate you by this method to get a preliminary design and then you go upstairs there and you use the one minus X bar over L and get a bigger number, you can use the bigger number. He says if you use this just to get a design, because otherwise you can't get started, and you go up there and the one minus X bar over L gives you a smaller number, he says we stand behind you 100%, it's a fact. You can use the larger of those two numbers. Here's an example with four more fasteners, one, two, three, four, five, six. See these outstanding legs standing up here on the angle? See the angle attached? It's got more than seven bolts on it. That's this case, therefore you get 80% of the strength. Here's one where you were cheap and lazy and only put three bolts. You said that's all I could get in there. Well then you're going to have to... That's interesting. With four or more you get 0.8. Oh no, I thought I said 0.8 again. It's 0.6, that makes sense because the connection is not as long. And the one minus X bar over L, that's L right there, that's L right there, doesn't include this, doesn't include this 30 feet down here, just the distance between the ends of the bolts. Said if you have fewer than three, fully on you, go up and use case two, one minus X bar over L. You're welcome to use this to get some guesses to what your number is, then pick a bigger angle, then take that angle with your measly two bolts over L and see what that person has to say and he'll tell you the truth. What if you only have one bolt? The right answer. She's shaking her head, no, you can't do that. Okay, so that's a quick rundown of where we're going and what we're going to do. This is case three. Case three said all tension members where the tension load is transmitted only by across the member, longitudinal will be along the member, transverse means across the member, transverse welds to some, but not all of the elements, you get 100% efficiency, but you must call the area nominal, just that piece that's welded to the elements. Here is an angle, it is welded to a plate with a transverse as opposed to a longitudinal weld, and you get 100% efficiency, but your nominal area, you don't get to pick that at all. Things are so bad you don't even get to pick that. You just pretend the load is coming through this bottom angle leg, straight on into the plate like it was a plate and you just got that for some other reason. Maybe you need that for bending strength later on. Maybe you need it for some compressive strength later on for some buckling resistance. Your area, this is a five by five by half, the area you must pick five by a half, connected area only. Here's a further description of case seven. A seven is these approximations as to what you can do. They're going to check B sub T in comparison with two thirds of the depth. Here is B sub F. That's B sub F. That's the width of the flange. There's the depth of the beam. If this number in your table is greater than two thirds of this number, they'll let you have 0.9. That's a rather squatty shape. Here's a shape that's pretty tall. There's your B sub F. There's your D. If B sub F is smaller than two thirds of this number, then you lose 15%. And you can kind of see why. Here your plate is connected to the top. This is a rather squatty shape. There's a lot of width here. So this centroid isn't too far off of this shearing plane. This one on the other hand, look at this mess. Here you still have a plate on the top. This plate happened to be narrower than the flange. It didn't matter. Here it's wider. It doesn't matter. Now then, though, the centroid of this piece, in other words, if I cut that into pretend it's a T, it'd be exactly what's going on. The top plate is looking for the centroid. The bottom plate is looking for the centroid of a T. That's exactly where that T would come from, from cutting a wide flange in half. Same way on this one. This is a T cut out of a wide flange. If you say, well, it's not a T, it's a wide flange. Well, I'll just go get another T cut out of a wide flange, and I'll just put them together. I don't even have to weld them because there's no shear stress in there. And then I'm telling you that you are trying to find out how far this centroid here is. Actually, we'll let you start from here and go to the centroid of that T. There's an angle. There's the joint length, the connection length. Here's this outstanding unconnected piece. That's X-bar. Now, I don't care that in the book they call that, maybe the angle looks like that. They may call it Y-bar. You're the one that has to decide whether you need Y-bar out of the tables or X-bar out of the tables. Call it X-bar no matter what. The centroid of the angle. Here's the centroid of a T. Basically, these are the unconnected, I'm sorry, of a wide flange. This is like a channel, and they're going to give you this number here, X-bar. Here, if you bolt or weld a plate to the top, you pretend it is two T's so that you can find out how far it is from the shear plane to the centroid of the unconnected. See, that one's connected. See, that one's connected. How far it is from the shear plane to the centroid of the unconnected element. How far is it from the centroid to the unconnected element? Here's a channel. You go from the back of the channel to the centroid of the channel, whether there's one or two. Here's a better view of that tube-looking thing. What they do is they cut a slot in the top and the bottom. I was thinking there was two lines there, but you only see one. And they weld a plate in here. The load comes down the plate, drops through that weld, drops through that weld, drops through that weld. But sadly, although around in here, these stresses are uniformly distributed, there's a dead zone right in here. And so at some time in its life, you're asking all the load, see this load's going up in there, all the load to go through this little piece of steel right here, and not take advantage of this entire dead zone because there's nothing but air. And therefore, you get a reduction in the strength. If the length of the connection, this is a side view. The length of the connection is greater than 1.3 dog diameter, diameter, outside to outside. You get 100%. So all you got to do is just drill that hole or groove, make that groove bigger, make L bigger than 1.3D, and you get it all. But if you say, look, I can't get down in there that far, if it's less than 1.3D, you got to go back to this equation right here, where x bar is D over pi. And if L is less than D, see how he didn't tell you, if L is less than D, you can't build it, not by these specs. For a hollow rectangular shape, here's the plate, same idea. They're looking for the centroid of this unconnected right-hand side, how far it is from there to the centroid. Here is where the centroid is. If L is greater than H, you get to use this equation. If L is less than H, you don't get anything, nothing for you. And here's where they have the plates welded on the outside. Here is we would analyze half of it. The half we would analyze would find where is the centroid of that shaded piece. You'll find the centroid of that shaded piece from the back is B squared over 4 onto B plus H. I mean, I don't believe anything. I checked it. I don't know, I don't think I got the calculations here. But check it out, it's correct. That's where the centroid is. If L is greater than H, you get to use the equation. Here's your X bar for use in the equation. If it's less greater than H, if it's less than H, you don't get anything. Now, this is back to, this was still Sugui's pictures and book. Here's the length of the connection. Here's the length of the connection if the bolts are what they call staggered. Sometimes you need five or six bolts in there and they won't quite fit. They're in a leg that's kind of small. If you stagger them like that, they'll fit. They'll probably fit anyway. You just can't get a guy with a wrench or a lady with a wrench on the thing to install the bolts. You have an angle that is welded all the way around. That's the length. Sometimes because more load is going through this part along with this part, it makes sense to make that weld longer. We'll talk about that later in this one shorter. In that case, L is equal to L1 plus L2 over 2. You take the average of those two lengths. Here are plates. I think we already did plates. But here's the problem. See the narrow plate, long plate. See how all the loads have plenty of time to get out of the building. There's a little dead zone that's not even used. See what happens when you try and make everybody get out this little short weld? When the plate is really wide, there's a tremendous dead zone in here that you're not using. Here in this one, U is a 1. Here in this one, it's so bad. That's a 5-inch weld on a 6-inch plate. You get nothing. There's a 4-inch wide plate and a 5-inch long. You take L, and you find out whether it's in this region or in this region or in this region, and he'll tell you exactly how much you can have. Evidently, 4 is less than 5. L is 5. Less than 1 1⁄2 wc. 1 1⁄2 w is 1 1⁄2 x 6. So that's 6. Well, the length is 5. 5 is less than 6. The lower end is w. So 4 is less than 5. It's less than 6. So we're in that region. That's why we're using that number. Just to make sure you know, these are longitudinal welds. These are lateral welds. For plates, since there are no holes, the gross area is equal to the net area. The effective area is then a sub n times u. And the alternatives add case 7. Oh, here it is. This case 8 first. Case 8 says if you do not know how long the connection is going to be because you don't even know for sure what angle you're going to use yet, then just tell me you're going to use an angle. If you're going to use an angle, then if you're going to put four or more fasteners, you can use this under any and all conditions. Even if you later on use a 1-x over L on it and you get a lower number, you can still do this. We've proven that this one's still good. For three or more fasteners, 0.6. Generally speaking, you're not going to find that the 1-x over L is more restrictive than these. These are pretty restrictive because, you know, they have to cover a whole lot of cases and they're pretty much picking the worst case scenario. So they're probably pretty bad. For four fasteners, for three fasteners, unlisted, but it's listed in the table there. For two fasteners, you've got to go to the 1-x over L according to this page right here. Alternatives for wide flanges and other things. If you satisfy the following conditions, we already discussed. In the flange, three or more with the width, two-thirds of the depth. In the flange with three or more fasteners with a width of two-thirds of the depth, around in that range. Then secondly, not through the flange, but connected to the web. You've got to have at least four or more fasteners to get 0.7. Then there are a bunch of special conditions that you'll run across. For instance, for bolted splice plates, a splice plate is where you're splicing two plates together with some cover plates. The effective area has to be, the nominal area has to be less than 0.85 area gross. It's given a user note. You'll find these little user notes in places where they've had some problems with something and they can't seem to make it work with the general table. So they'll just put a little subscript down there. For this kind of plate, you need to do this. Then he's got an example. I want you to determine the effective net area for the tension member shown in figure 312. Heavens only knows where that was. I guess it was on page 142 or 10 pages ago. Basically, though, it was an angle. It was used in 5-8-inch bolts and it had evidently six bolts in two rows. Only one leg of the cross-section was connected and so you're going to have a U-value. He went to the properties in part one of the manual. He went to the dimensions on the angle and he looked up a six-by-six-by-half-inch angle and he says the X-bar. This is what is shown in the book. That's X-bar. You say that's not X-bar. That's Y-bar. I say, no, it's not. He said, look in the book. It says Y-bar. Not for you and me. It's not. Since this is our plate and this is the shear plane, I need to know how far it is from here, strictly speaking. We might have should have done that, but these plates are relatively thin and we don't know how thick the plate is just yet anyway. Good enough distance from there to that centroid. If it's listed as Y-bar, in this equation it's X-bar. Is it in this equation? X-bar. I just thought you were nodding off there. Just a wee bit. So, oh, okay. Now then, what is that? That's a six-by-six, so it doesn't matter, does it? If we're going with a six-by-six angle, I don't really care if you use Y-bar or X-bar. They'll be the same. But I didn't circle Y-bar because X-bar is in there. There's X-bar for a six-by-six-by-half-inch angle. It's about 1.67 inches. 54, 55. Thank you. Just as an aside, since page 55 came up now, here's three bolts and an angle. It's case 8B and it's on this page. X-bar is known and L is unknown. If that's the case, then we're going to have to use one of our approximate numbers. See, no calculation for X-bar over L because we don't know L. Here's a single and double angle take your choice. There may be another one behind here where we don't know L still, but we've got to use somehow because we have one, two, three, four bolts. Here we only had three bolts. This was case 7B. It was a given wide flange. X-bar was known, but again L is not known. Therefore, we're going to have to test B sub F to D for this shape right here. And B sub F over D was this number. It was less than two-thirds. That was B sub F was less than two-thirds of D. He just went ahead and put it together. And X-bar was 1.28. He would find that from going to the T-shapes. Find out how far it is from here to the centroid of the T cut from that wide flange. Although he knows that. He's still doesn't know L, so it didn't really help him. And he's got a .85. So this is different cases. These are different cases where he's been working with the approximate values. Case 8A, case 8B, case 7A, 7B, 7A, 7C. This is attached to the flange. This is attached to the web. Definitions. Thank you because I thought I'd lost that problem. There's the rest of this problem. Oh, there's the picture. 6 by 6 by 1 quarter. There's the holes. Here's the 1.67 you and I just found. Got L is equal to 10. I don't know where I would have gotten L equals 10 from. Maybe that was just checking things out because L on this one is certainly 6 inches. Length of the connection is 6 inches. Therefore, U-bar is equal to 1 minus we-know-X-bar. We know L for this case. .7217. Therefore, we have a net area of .502 square inches. Where do the .502 square inches come from? Here's our net area. Gross area out of the table. Minus. How many holes does this thing run into? 1, 2, 3, 4, 5, 6. How many holes does this load run into? Ah, you wouldn't be fooled by that, would you? You say 2. That's correct. The load comes down. The load comes down, and it tears across two holes. Therefore, from the gross area, you subtracted the thickness of the angle times the effective diameter of the hole. Bolt size plus 1 sixteenths fit. Plus 1 sixteenths damage times two holes. Leaving you 5.02 square inches of net area. Which still has to be effect-sized. Because, unfortunately, you didn't bolt this leg down. It's an outstanding, unconnected element. There is your 5.02 net area. Times your U value. That's how much steel you get to pretend is yours. The rest of it was lost due to various practical considerations. He says you could also use the alternate. If you wanted to, it has three bolts in a row direction. You could have taken the U as 0.6. You see, if you had taken this 0.6, you would have taken a pretty fair hit. Because you got a U of 72% remaining for you. He was only going to give you 60% if you weren't going to really go figure it out. What you probably did is you did another angle. You came up with the other angle and you found it had 72% strength. And you say, well, gee, I might be able to get the next angle down. So you went to the next angle down. And it also worked once you plugged it in this more accurate equation. You got attention. Remember, welded is shown. Determine the effective area. It's the same angle as before. As an example, 3, 4 only part of the section is connected. So you got to use a reduced area. See if we can see what this thing looks like. So this thing is welded all the way around. Looks like it's all welded down to me. What does he mean? Only parts connected. What's not connected? The little half inch is sticking up. Yeah, this thing right here. That whole leg is outstanding and unconnected to that plate. Therefore, there's a dead zone. I'm not sure I'm going to draw it, but you've seen it before. Here's the angle. It's not easy to draw until you get them drawn. But sadly, that one's going the wrong way. It's bolted and it's bolted. This whole thing right here is just worthless to you. Therefore, your load is going to have to be reduced because of that inefficiency that you connected it. Where's his answer for that one? 1 minus x bar of rail. That's the same x bar we had a minute ago. Five and a half is the weld length. It is indeed. That's the length of the connection. You do not add, if that's maybe a quarter inch weld, you do not add any more to it. It's the distance from the end of the angle to the end of the weld. Now, staggered fasteners. Every now and then, to try and get more bolts in this thing, you'll stagger the connection. And the load will come down and all of a sudden it'll hit a point of weakness and it's possible that it will fail straight across the plate. In which case your net area is your gross area minus the thickness of that plate times the effective diameter of that hole times one. But sometimes you'll notice that it didn't do that. Sometimes he's always searching around. He says, I wonder if it's a little weaker over on this side. He'll reach out and he'll grab this hole so that he gets to pull two holes out of your area. But on the other side, of course, it's longer from here to here than it would be from here to here. And since it's longer, he says, well, I don't know. But since it's weaker through the hole, he says, well, yeah, I think. He picks the way he goes. And of course, I can tell you which way he's going to go. You check them both. You check and assume that the net area is equal to straight across the plate with only one hole in there. Are you taking into account the fact that he tore through two holes? I don't care the fact that here's the hole. On the top hole, the thing's going straight across so it's that long. On this hole, it's tearing across here. There's not enough difference in those two numbers to matter. You just use a diameter of the hole and you pick up the extra length of this line. Except we find out when we tested them, that didn't work very well. And the reason is, is what is this? See, we're using the... I don't remember. What are we using in this calculation? F sub u. Very good. When do we use yield? Where on this angle do we use yield? Gross area around in the middle of the thing where it's got a long length and it can be yielding and it's a serious problem. We don't care if these things yield. We're looking at tearing strength here. How did I get off on that? Oh, I remember why because I need to know which one of these to use. What happens is if you test a piece of steel across here like that, you find it has some tension stresses and those rascals can run on up to F sub u. But unfortunately, steel is kind of weak in shear. As a matter of fact, almost all of these numbers are 0.6 F sub u. 0.6 F sub y. And it is so common. We don't even have a V sub y. We don't even have one listed anywhere. We don't find it anywhere. If anybody wants to know what it is, it's 0.6 F sub y. That's how strong a piece of steel is in shear. In yielding. And we don't have a V sub u because it's always 0.6 times F sub u. Or we don't even have tables for it. If you want to know something's in shear, how much will it take to break it? You go get that number to break it in tension. That's how much strength it has in ultimate shear. But the final problem of this is how did this thing fail? Did it fail in tension? Yes. And did it fail in shear? Yes. And so it's kind of a combination of those two things put together. And so what we do to account for the fact that the stresses in here are some tension, some yield, some shear, some this, some that. Is we do not add the true added length of this hypotenuse. Some guy did a whole lot of testing. I forget back in the 60s, I think. It's been working great ever since. He said, you know how long you ought to count that line? I say, well, that squared plus that squared take a square root. He says, nope, too long. That won't work right. I say, why not? And he goes through all this explanation. I said, oh, OK, OK, I heard that already. How long should you make that line? He says, you ought to say that that line is s squared over 4. Shut up. Gee. I say, OK. You want to tell me what all those things mean? He says, yes. He says, you know what a train is, right? I say, yes. He says, you ever have a model train? I said, yes. He said, do you remember it was a 20 gauge? Man, it must really be important. He says, the tracks run along the gauge, right? I say, well, yeah, along the gauge. That's the way the track goes. He says, that's g. I said, OK, so what is s? He says, well, s is what we call the spacing of the bolts. So that's s. That's the spacing of these bolts. They don't have to be the same, but they usually are. You should say that this length of this line right here is s squared, whatever that is, divided by 4g. You put that length in there and you'll get the right strength. Take it into account, there's some shear in there and there's some tension in there. You'll get the right length. So basically, there's some old notes. Good stuff. Let me finish this s squared over 4g here. This is how long this line here, right here is. First off, you study this failure plane right there. You take this length minus this length, multiply it times the thickness of the plate. Then to find out if it's going to fail across this line, you take this length minus one hole minus two holes and the length here is... I'm going to take this length and the length here is s squared over 4g. You're going to add, every time you get one of these things, you're going to add an s squared over 4g. Skip all of that nifty stuff for a minute. Here we go. Now incidentally, some people like to take the total distance across or something and subtract things. That's not the way it's done in the manual. And I suggest that it's confusing. I suggest you do right what he did here. For line A, B, C, D, E, for line A, B, C, D, E, you take the total width here of 16 inches, you subtract three holes, and every time you have a slanted line, it picks up an extra s squared over 4 times g. And two of those lines did that. That's the added length that you had above and beyond going straight across. You had two slanted lines, which added to how far this thing had to go. That's your net width. How far it is effectively across the plate. Then you'll multiply the length of that line times the thickness of the plate, and that will give you the net area for that calculation. Here's an angle that has been flattened out into a flat plate. Incidentally, the width of this angle flattened out would be, here it is, 5 minus a half of a thickness plus 3 minus a half of a thickness. That's the length of the midline. That's actually how wide the plate is. They make these things out of. So when you flatten them out, that's how long they are. Here this guy has got a load coming in. He's going to check across here, take off two holes. Then he's going to take off three holes, and he's going to add an s squared over 4g for that one. If he thinks it might go back, then he'll add an s squared over 4g for this slant line. That'll be how long the line is, and you'll multiply times the thickness of the angle to get the net area. All right, just more of that next time. Thank you. They love that. They think it's already graded. How many things have I signed today? You don't have to sign that. That's just because I'm supposed to give it to you. Okay. The only thing that's different is they change that bit up to that. Wow. Okay. Okay. I don't have any idea in the world what to do. Okay, well, I hope, does it say what to do here? I've never seen such an instruction. Why don't you take that? Oh, it's for you. I have one for every other teacher. Good. You throw it away. I don't need it. That'll be nice. That'll be good. You're a civil engineer.