 No, I put it over there in that can over there. I didn't think you really wanted to see me after class. Okay, here we go. Well, here some of us go. Not everybody's here. If you weren't here last time, I've got your quiz. Not sure I'll give it to you, but I have your quiz. See me after class. Had a couple of people come and discuss their exam, that's what that's perfectly good. Anytime this afternoon will be good, all day tomorrow will be good, Friday won't be good because I'm going to be out of town Friday afternoon teaching all morning. So if you wait till past thing, you'll have to see me next week sometime. Consistency of grading, I'm hoping for the best. Try. You can see this is one problem. Every time I take off some points, I write it down. I grade his paper and I say minus 20 and I put minus 20 right there and I see your paper and made the same mistake, minus 20, so I check it to make sure how many points I see her paper, minus 20, but I notice you all three have the same answer. So then I go and change my numbers to the right numbers and I go back and scratch out the minus 20 on all of those papers. The only thing you may see is, you know, this is, this obviously has some thought in it. Somehow you got the wrong number there and if your paper says minus six and Joe's paper says minus three, there's really a reason because I got the minus six down there. You're more than welcome to bring yours and Joe's paper and I'll explain to you, yeah, but he, see up here he knew this. You didn't even write that down, that's how come you lost the full six and I guarantee I won't change Joe's grade. I might change yours, you know, but that's what you're there for if I've rooked you. One person brought in, you can't add, Mary was right, you know, doing good to do anything after these are graded, so I'll be glad to re-add your points. We're going into now, beams bent about more than one axis and how you handle them. You can get biaxial bending because someone has a load at an angle or it's possible that they have people load dead loads and wind loads that cause bending about both the strong and the weak axis. Or he handled bending about the strong axis due to loads in this direction, but these loads coming from the side, we really haven't studied what to do with the beam that's bent about its weak axis, it behaves somewhat different. Number one, it does not, it cannot laterally, torsionally buckle. So if it's bent about the weak axis, that's one thing that you can mark off the list as not necessary to study. And are you speaking if you have the loaded at an angle rather than giving the two loads, you take the angle, the load and the angle, put a vertical force on it, p cosine theta, put a horizontal force, px would be p sine of theta. This is a top view, so this is looking at it from the load coming in from the side. This one you can't see that side load. Tends to bend the beam in two directions as opposed to just straight up and down. Now, when you put a load on a beam, you cause shear in the beam, as I'm sure you remember from 305, put a load down on a wide flange, you get a VQ over IT shearing stress, you get a VQ over IT shearing stress, you get a VQ over IT shearing stress, if you cut off that little corner, there's a shearing stress because it has a Q, has a base and a height and a distance from the neutral axis to the centroid, so it has a Q. And what really happens is when you put the load down, the shearing stress is inside the beam resist going vertically and then they flow on out to the corners. And since this one can't just start at nothing, they actually flow in and come up to the top. And if you can find a point on the beam where the sum of all the, not the stresses, but the forces times their moment arms are equal to zero, that's called the shear center, and you should put the load through the shear center. Since this stress times area and this stress times that area balance each other and these two do the same, if you, some moments about this point, these two cancel, these have no moment arms, these two cancel, that's the shear center for a wide flange. You should put the load down through the centroid of that shape. So the centroid of a wide flange is in the center. Interestingly for a channel, the shear center is not at the centroid. The reason being when you push down with 265B page numbers, you get some shear stress up which flows on to the right. This doesn't start at zero, it comes in and flows like that. So if you put the force, the shear force, or the reaction at the centroid of the shape, the shape is going to roll on you. Now it won't roll much where it's supported, you know, but the top of the flange will probably move over where it's bolted to a reaction, but out in the middle you've just made lateral torsional buckling a real problem, have torsion stresses in effect inside the channel. So what we do to counteract that clockwise moment is we move the load to the left, we bolt on an angle, you put the load out here, and then those shear stresses go and clockwise are balanced by the counterclockwise moment of the shear force that you applied. Now this one only has a vertical load. This area is where I had intended to put the shear load right here. And then when I did that, these stresses came out. And because, let's just say I put it, well I was going to put this shear force at the centroid. So let me take that one off as having been shown already right there. You know, I don't know how we're going to get it there, we're going to have to put it probably right about here somewhere. Then when I saw moments about this point, the shear has no moment arms, so there's no moment due to this, but all of these people are going clockwise and so the whole shape rolls on me. So what I'm going to do is I'm going to put the shear force over here somewhere until when you saw moments about any point, I don't care where you saw moments about, you can saw moments about this point or that point or anything else, but this is out there far enough such that the total moment about the centroid would be zero. It's the most common place to saw moments about. Well see, this guy right here is probably going to be replaced by picking him up and moving him here and causing this problem and then putting an external applied moment because that's what this does and its moment arm is how far it is from the centroid to the back of the channel and then this is called, it's listed in the book, I think it's E but I don't remember exactly what they call it, so that when you multiply, so you put a moment on the counterclockwise direction, magnitude V times that distance, these people will come up with a clockwise moment, which is whatever it is and they're equal, then it won't twist on you. You still have this going clockwise, but about this point now you have this going counterclockwise of the same magnitude. If you saw moments about this point, then what you get is these, this guy going clockwise, this guy going clockwise, but this guy going counterclockwise, so they again all balance. All this shear center and you can kind of tell where the shear center is on most things just by doing that, here's a Z shape, incidentally when you do put that load on the channel at the, even when you put it at the shear center, the thing is obviously going to bend down, it's just not going, and it's not going to rotate, which is what you're trying to prevent. Here's a Z shape, put load down on it, you get shearing stresses up, trying to resist but they don't stop, they go this way and they don't just start at zero, they come in from here, zero, fifty, five hundred, two thousand, four thousand PSI shear, then it turns the corner, it comes up parabolically, and then it turns the corner. Now then if you put the load here, this is multiplied times that, counterclockwise, and this is multiplied times that, clockwise, so it doesn't twist. So this is the shear center, however what a Z shape has a tendency to do when you load it, is it moves to the left, because these two are going to the left and you don't have any balancing force, so if you don't mind it moving to the left, a lot of times you don't mind things deflecting, but it's going to have a tendency to do that. Centroid of a T, downward force, upward shear stress, it's just like water flowing down a pipe, flows to the right, flows to the left, some moments about this point get no moment left over anywhere, that's the centroid of the T. Here's the centroid of an angle. Why? Because when you push down, you get shear stress is flowing up, they don't just start that, if you remember you had like a neutral axis, if you were asked for the shear stress in the middle of that thing, you cut it there, VQ was that times that times that, that was Q, the shear stress right here was VQ over I T. So it comes through like that, through that point, you can some moments about that point with no resulting moments, that must be the shear center for an angle. Here's what channels do, if you don't move the load off, they roll and go down, this one just goes down. Here's somebody has put a moment to the left of the web and obviously they put a moment on it, it's not that it just came because of the shape, actually put a moment on it and the shape rolled and moved down, here they put it in the right place, it just moved down. Here's an angle when you put a load on it, it's going to move down because you put a load down on it, but because of this set of stresses to the left, the angle has a tendency to move to the left, these all balance so the T just moves down, the Z as we said moves to the left. So number one, it'd be good if you always do case one, put the load through the shear center, if you don't we can handle it, but it's got a little more fault behind it. If it acts through the center, centroid, shear center, then not the centroid, excuse me, the shear center, then it's simply being bent about two axes and all I have to do is make sure I know how to design something that's bent about two axes. Now here's how they get a shear center, I don't know why but I picked a channel, I just thought it'd be as good as anything. A channel is particularly bad at trying to get the shear center, it's not if you get into all the math and that kind of stuff, but when you just say, well I guess it's 15 inches deep, they list the thicknesses 0.65 inches, they list the flanges having that height and I'll just break it up into rectangles, I'll solve for the shearing stresses at various points, I'll multiply the average shear stress times area over which it acts, I'll multiply it times the moment arm and I'll find where to put the load, 200 kips, arbitrary number, so that I know where the shear center is. So, to find the shear stress here, you put a wiggly line there, you cross hatch everything outside of the line where you want to know the answer, there's no area cross hatched, so the shear stress is zero. Come back in the middle, half of 3.72 and find how much shear stress there is there, you get this number right here. Come right here, don't come to 3.72, but come to 3.72 minus the web thickness, find the shear stress, you get 10.67 KSI. Then I don't care what it is in the web, there is shear stress in there, I hesitate to even show it, the reason being is I'm gonna some moments about this point so I don't have to do all those shear stresses, not only that they're parabolically distributed, so I'm gonna have to integrate to get the force in the web and it's not causing any out of balance, so flew on that. Then this one, these go this way, these come this way, this one's the same due to symmetry. So my whole problem is, is these stresses times these areas times this moment arm clockwise is what's causing a problem I ought to be able to solve for E, the eccentricity from the center of the web. Here's your VQ over IT calculation, same thing you saw in 305. So on some moments about the center of the, of the channel web, I get V times E to the center of the web minus the force in the flange times its centroid, force in the flange is 10 points, this is linear, because every time you come halfway you cut off half the area, when you go all the way you cut off all the area, so that's a straight line. So the average stress is this over two times the base times the height times and you can check it out, I checked it out very carefully how far it is from there to the outside minus a half of a flange, this number here, 200 times his moment arm is equal to, excuse me, this whole thing is 15 inches deep. So what I did is I multiplied one force times the distance between the two forces as a couple. You want to take one of these and multiply it times the moment arm, then add a second one times seven inches, you get obviously the same answer. So I solve for E, but that's not what's posted in the book, they don't post the distance from there to there, they post the distance from the back of the channel to there. So I take my answer minus half a web, that's 0.389 and I look in the book, 0.583. So I say well okay, let me go find out where I made my mistake because I don't want to show the students that and have them all say that's not 0.65, there isn't any mistake. After doing it about three times, I realized channels aren't rectangles, have a nice sloping face in here, means that the stress in here is not uniform, the T isn't even uniform, here the T is probably about 0.65 and here it's quite a bit more and the moment arms are going to be a beast and I'm going to have to probably take a rectangle and a triangle except I don't know and then you got this radius and you got that radius, but this is how it works, way that you would do it mathematically, only thing is you got to pick all, you got to pick the right numbers and a channel that's particular difficult to just do it like this. By far the easiest way is just to write an equation in a computer program to get the stress integrated from some place times DA to find the force, find the centroid to force the whole nine yards, that's what they've done and of course they also go get a channel and they stick it out there and they put the load right down the centroid and it rolls like crazy. So then they put an angle on the back of it and they start moving the load out and they measure E, push down on it at some distance it doesn't roll clockwise, it doesn't roll counterclockwise, when they move the force to the left some more it rolls counterclockwise and that's what they have for you. Alright now this is getting back to your book, like I say all of my figures are being dragged along through the ages. Here's how we're going to handle bending about two axis. Two axis bending, we will use a thing called an interaction formula. What it does is it says, do you have some strength about the XX axis? You say, yeah, I said where did you get it? Well I got it from M plastic, that was one of the things I had to check. I got it from lateral torsional buckling, that was one of the things I had to check. I got it from web local buckling, flange local buckling and I got it from a whole lot of things like that. I said okay but you have a strength. You got a number in mind that it's good if I put that much load on it. They say yeah. I say do you have a request for bending about the XX axis? They say yeah we went and did 1.2 dead plus 1.6 live plus a little snow plus a little of that and we put that on a surface. We multiplied the pressure on the surface between the beams so that each beam took his fair share and we know the loads, we know the required. What do you call those? It says well those are like p sub u or w sub u ultimate request. So what do you got for me for strength? So we have a nominal. I said let me see the nominal. I said no, no, no you can't see the nominal. I got to do something to the nominal before I let you see it. What does he got to do to it? You're even awake and listening man. What do you got to do to it? Just right. One of those reduction kind of factors, a resistance factor on there to take the nominal number down to what the design people get to design against. So in other words you're telling me that for moments or for p sub u and p sub n and all that kind of stuff. Basically the equation is m sub u is less than p sub e m sub nominal. So then I'm going to tell you that m sub u divided by phi m sub n had to be less than had to be less than or equal to one. And you're going to say, well, yeah, all you're doing is you're getting the ratio here. You're just telling me that you can use up to 100% of your capacity and no one will complain. And I say, I like it. Then if you're going to bend something about two axes, then I would like you to go find out how much they are requesting for you m sub u xx divided by how much is really available, including all those things you said and put it in ladle, torsional buckling, plastic moment, you name it. And then add as a percentage what they asked you to give them about the y axis and divide by how much is available about the y axis and make the sum of those two less than or equal to one. It's like you're using up some percentage about the xx and then you're using up some percentage about the yy and you shouldn't have any more than 100% use. If you look on page 16.1 day 73, basically you'll see that equation. It's equation h11b. Got a lot of other goodies that go with it because he didn't figure there was any reason to just talk about people who use up some x bending and some y bending and you're getting ready to put some axial loads on it. I'll bet you and there's going to be another chunk of percentages removed and so he puts them all in there at one time. Have one of those. This is it right here on page 16.173. Oh my goodness. Look at this. When your ultimate request over your capacity is greater than two tenths, they think this is the right equation less. This is the right equation. That sounds a lot like we're getting into this. Is this a small plate or a big plate? Is it heavily loaded or is it lightly loaded? And that's kind of where we're at. In our case, we this is when you have a lot of axial requests. For instance, if you requested half of how much you have available, incidentally, piece of R is a generic term so that the allowed stress people can play our game. All he means is for you and me should read piece of you. I liked it in the old book where they did their own thing. So now then they got to use a different symbol. You and I got to use a different symbol. That's piece of you. That's what it is. piece of C is the capacity of the column in axial load, piece of C piece of nominal. If you've used up half, this is the equation you're supposed to use. If on the other hand you have no axial load or very lightly loaded axial, this is the equation we will be forcing you to use. And since piece of R over piece of C for you and me today is there is no axial load in there. I don't want to try and get all three of these things at the same time. Here's what we said. Capacity about the x-axis ultimate, I'm sorry, ultimate request about the x-axis divided by b subending m sub n x. There's that ugly term. Plus the request for ultimate load about the y-axis divided by the true capacity, including piece of B bending about the y-axis. That's the equation you would use even though you don't know what all this other stuff is. That's when there's no axial load. So that's the one we're going to be working with for right now. Here are the references for all that. Now one thing you don't know how to do, although you do know how to do it, is bending capacity about the weak axis. You say, well, how's it going to be any different from the strong? It's just a piece of steel. Well, that's true. But when you bend a beam about the strong axis, the top goes into serious compression. The bottom goes into serious tension. And all of these little compression people, they really want to think they're columns and they laterally torsionally buckle. They don't do it real soon because these tension people are down here just tight as a string saying, we want to stay straight. But they pretty well after a while overpower them and they laterally torsionally buckles. With a plate laid down like this, I have no idea what that little stick is, and I don't know what that stick is either. Somebody asked some question that made a lot of sense at the time. So just ignore the stick. Oh, I guess that's, was that what it is? Think it's glued there with a force on it? Causing a moment? I don't know. If the plates laid down flat, all of these are in compression, but look how little their moment arm is. And all these down here, they're in tension, but they're tight as a string. And they're so close to this, that guy cannot buckle laterally. You want proof? Go get a yardstick. Take the yardstick you laid down flat. You can push it and push it and it'll bend and bend until it breaks. Turn it up. Not this way, but turn it up like this and put the load down on it. Support it at 36 inches. You can hardly keep it from lateral torsion ball. You just continuously wants to pop out to the side. So number one, weak axis bending LTB is not a consideration. Something else, though, is consideration. On the plates, we said that you could have f sub y, z sub y, that's your yield stress, that's your plastic section modulus. That was the plastic moment about the weak axis. But in beams, they're so long that if you get this beam bent about its weak axis, you still got a lot to go, but it deflects just horribly. And that's just not acceptable. But we'll let you have the full plastic moment unless it exceeds 1.6 times the first yield elastic moment. Once you're into that region, you have caused so much deflection that we're not going to let you go any higher. So this is a limit on the plastic moment in a beam. A plate, I'm not worried about it. They don't hang out that far, and they don't bend that much. And they stick out only what, four inches, five inches from the column face. These rascals are what, 20 feet long. Can't live with that. So there's a limit due to deflections. Now, I don't know why he's been teaching this longer than I have and written more books. But he says, would you like to know if this is a problem? He says, well, all you got to do is take z sub y over s sub y. And if it's less than 1.6, it's not a problem. I think that's ridiculous. Because then I got to remember which one's on top and I got to remember the 1.6. And if it is less than 1.6, I quit. I only did one calculation. And if it's greater, I still got to go calculate this. If I don't have that there, and he uses it a couple of times in the examples, that's why I mention it. If you just go calculate this, which you have to do anyway, if you just go ahead and calculate this number, and if this is smaller than that, good. And if this is not smaller than that, you already have the answer, one calculation. Well, my suggestion is if you have two limit states, it's usually not much reason to find out which one controls, just go write them both down and take the smaller of the two. Now then, bending about the weak axis. Here's a wide flange that has some yield stresses in that flange. And because it has yield stresses in that flange, and because it is sticking out, there's a possibility that it will buckle. And therefore I need you to take these compressive stresses and determine whether or not this little stick out part has a tendency to laterally, torsionally buckle. You already did it. In other words, if you bend this about the weak axis, everything on the top goes into compression up to f sub y, and this is what it looks like. Last week, you took a beam and you bent it about a strong axis, and all of these little people were subjected to a full f sub y, and you determined some beautiful break points beyond which this little piece of steel sticking out would buckle. I don't care if you caused those stresses because you bent it about the strong axis, or if you caused those stresses because you bent it about the weak axis, there's no change. If the shape was compact when you were discussing strong axis bending, is compact bending about the weak axis bending? Now, if it laterally torsionally buckles, that will influence your strength about the weak axis, just like it does in the strong axis. But you don't need any more theory on any more break points. Break points are your break points. You'll notice it says subject to flexure about the weak or strong axis right there. It's just flexure. Now here's how much you'll have to reduce it. Little note, same identical break points as for XX bending. There's the page number table B four dash one B gives you all your break points, and I got it on that page. If the shape is non-compact because the flange with the thickness, the strength will be given by the equation is identical with the equation for strong axis bending. The only sub-group change is x change to a y, y, y. That was already a y. That wasn't a y. It is now. And here's your lambda p for the flange. There's lambda p for the flange. There's lambda rigidation for the flange. And you're lambda. Exactly. The reduction in your bending strength about the weak axis will be reduced by this. It'll be here it is right here. It'll be a constant number up to a certain point after which it will reduce in a straight line fashion to some point based on this right here. And then that's all they go. They don't have one where it gets out there, mainly because none of them are slender where your past lambda plastic could happen where you're past lambda sub radius of gyration. There are your break points. Here is what the code, the specifications say code says this by reference to this specification. This is the piece of it we're going to use. We will add column extraction of strength a little later on. The definition of terms for 174, nothing that we ever use stuck in there just for continuity might get into that. And there's your equation right there that we showed about the 1.6 limit. That's your nominal moment. Flange local buckling considerations. There's your flange local buckling reduction in strength. They give you one for slender flanges. Here we go with critical buckling stresses. Here's your critical buckling stress blah, blah, blah. We don't get into it. None of ours are slender. So for example, I have been asked to determine if a W21 by 68 used as a simply supported beam 12 feet long lateral support of the compression flange only at the ends. That means this is not only L, this is L sub B. We embrace points x, x. He's already done all the work for me. There is a dead load x moment of 48 and a live load x of 144. Those are service load moments. They still have to be factored service load moments about the y axis six dead 18 live gonna use a 992 steel. I'm gonna have to go look up the flange thickness of a 21 by 68 to see what the strength is. I already did turned out to be 50. You can take my word for order. You can go check. Does it satisfy the specs? Since he said, assume that the moments are uniform over the length of the beam, I'm not going to get a C sub B a Christmas present over my tables or my graphs. C sub B will be one. Had it been a uniform load on it as opposed to a uniform moment, then the moment diagram would look like this and I would have a C sub B value which would help me out. If it had a concentrated load causing these moment, I'd have a C sub B which would help me out. I don't get one here. First he says we got a bunch of things to consider. This is one thing that I think he needs to add really. There's so many words here. It's kind of hard to tell what he's doing. What we're doing first is we are going to check the supply about the xx axis. We get all that done. We're going to see what's the supply about the weak axis. Then we're going to go find the demand about the xx and the demand about the yy and then we're going to stick it into like it's going to take a while. Then we're going to stick it into our interaction equation and see if it comes out less than one. Here is how much capacity we have. First, here's a typical curve for W21 by 68. It has a length below which it will not laterally, torsionally buckle called L sub B. It has an L sub R above which it goes into that Timoshenko kind of buckling equation. L sub B is 6.36 feet. L sub R is 18.7 feet. Z sub X is 160. The last exception mod is 140. About the x, about the y, 15.7. Plastic section, about the y, 24. Where from? Right out of the book. W21 by 68. Here's a W21 by 68. It is compact. There's no F. Any numbers you need offer there? Let me know. I don't see any. On the next page, there's your plastic. That's your elastic section modulus x. There's your elastic section modulus y. There are your plastic section modulus y. Here's a z-table. For a 21 by 68, there's plastic section modulus. Here's fee-bending him some plastic about x that give you about y. There's your, remember we were talking about LP and L sub R. There they are. Here's a chart graph you can use to get the same thing we're getting ready to use. He just wants to see you practice with the equations. Your equation says your nominal strength is equal to start off with the plastic strength. Now notice this is M sub P without the fee. Before it's over, we're going to have to put a fee on it. And if you use the table, it's got a fee in it. M sub plastic x, M sub plastic x minus 7 tenths F sub y S sub x. What is your brace length? What's the plastic? What's the razor gyration length? What's the plastic length? And then it has to be less than M plastic. D times F sub y. Plug in the numbers. First off, C sub B is a 1. M sub P x. We're to calculate that. Oh, here we go. M sub P x is your plastic moment. That's where you started at. That's this right here. It's equal to yield stress times plastic section modulus, 8,000 minus 8,000 times 6.7. F sub y was 50. S sub x, there it is right there. Pulled it right out of the table 140. Your brace length 12 feet. L sub P. L sub P. L sub P. Did I not write it down? We just saw it. There it is. He wrote it down. L sub P. The total distance across from here to here so that you can get the slope. L sub R minus L sub P. L sub R minus L sub P. It's this strong in inch kips. It's this strong in kip feet without the fee. He's going to bother putting the fee on there until he finds out if this is the thing that controls. Now we're going to have a supply about the yy axis. Shape is compact. There's no flange local buckling. Same thing he said up here. And because there's no F on the tables. So you get the full plastic moment unless it causes too much deflection. F sub y, z sub y, z sub y, 4.4. So that's strong in inch kips. That's strong in foot kips. Except that may cause so much deflection. They won't allow it. You're going to have to go multiply 1.6 times F sub y yield elastic section modulus. That's your 1.6 times your first yield elastic strength. 1.6 times 50. S sub y, S sub y. 15.7. It's a strength since that's bigger than this strength. That other one go. 1,200. I've lost it. There it is. There it is. 1,220. Yours is 1,256. And therefore you have not reached that limit. You can use this number. 1,220. So I know how strong it is about the y-axis and the x-axis. This is in inch kips. This is in this is in inch kips. What is this? Therefore okay. We've done too much stuff. M plastic y, 1220 divided by 12. That's not right. They're being foot kips. Okay. Help, help, help. Oh, look at that. How did that go so many years and nobody noticed that typo? I doubt seriously it's in your book. Thank you. Well, it's obviously 1220 divided by that's about what that is because that's 12 over 12 was a 1, but I couldn't figure out, you know, how that turned in, why it didn't turn into feet. Just a typo. How then how strong are our people? Well, about the x, I'm sorry, how much request demand do we have for our people? 1.2 times dead plus 1.6 times live only about the x. There were the numbers listed. That's the request about the x-axis factored about the y-axis 1.2 dead 1.6 live there were the numbers that must have come off of a computer program. There's the demand request having that all that well to be able to see if it's okay. There is your request, your demand. Here's your supply was the 548. Now you stuck the point nine on there. Here was your request about the weak axis. Here was the supply and it's okay. It says note that you could have got the x number m sub nominal x just this one right here. There isn't a table for this one. You could have got it from the beam design charts. And sometimes it's worth doing rather than doing all that stuff about slopes and things. For instance, when you were at the z table, your 21 by 68 had a fee sub the m sub px of 600. And you can come to these charts. You come in at 600 and you'll find him entering right here. There's your W 12 21 by 68. So now if you have to follow this too many pages, sometimes it gets more trouble than it's worth. But if we're looking for this number, our length was 12 feet. Since you've found this W 21 by 68, you can pick it right off of there that easier than this calculation. Yeah, I say it's a little easier. But if you have to track it track it down four or five or six pages, sometimes they get kind of hard to track down. And this of course is this is this came from that equation. So it's the same thing. And if you have loads not applied through the sheer center, then you already put a moment on there right on top of everything else. For example, if you put your load at an angle off of the sheer center, you're causing a moment. Then what you're really doing is you put in a vertical component that causes no moment, but you're causing a P cosine theta times that distance. So we're going to have to get you to put that on there and somehow resist that moment. I don't mind resisting this in the normal fashion. I don't mind you moving this to this axis in the normal fashion and finding it's bending about the weak axis. But the one thing when you move this to the sheer center, then you have to put this moment on that's loading case one, loading case two. You put the load not at an angle, but it still didn't go on the sheer center. Then move it back to the sheer center and give me the regular old bending about the strong axis only. But then you'll have to put a torque on there about the centroid P times Z. That's loading case two. Here's how you analyze case one. You just go ahead and put the design the beam or analyze the beam with the force on there. And then you put this force on the top. That's conservative. And you just pretend that it's a T shape and make the T itself handle this horizontal load, this bending. You'll go get the Z sub Y for a wide flange, and you will cut it in half, meaning only half will support the load. And that's really pretty conservative. They got ways to do this without that mess. You put it off center, you simply put the load on the center. And the moment you take out by P times the over H zero, that causes half the moment this causes the other half. And you make a T handled this, then that T will handle that half. You do want to if you have questions about your exam that you think I missed something, I don't say I'd be possible to grade 50 of them and not miss something. They become seeming no later than the end of next week. Yes, sir. I do not have a mother, but you can talk with him. He knows where he tell him where I put him. All right, let me close up shop. If y'all want to walk on down there, you can. I'll be there in a minute.