Let's see who took a look at that problem I left you with on Monday. Any volunteers? All right. But we're going to need the very same things that went into that anyway, so let's review it. What we were working with was our ability now to deal with composite beams. We had worked out how to find the stress in beams of homogeneous nature, all one material. We're still working with prismatic cross sections, meaning they're always symmetric about the y-axis. That's going to continue. There's a section in the book that deals with non-symmetric beam cross sections, but we're not going to get to it if you need it. You can double check it if and when you do need it. But what we have worked up is the ability to do beams of different materials. And so I left you with a beam, something like that, 200 millimeters across the top, 20 millimeters on that flange there, as well as down here on that t-section. And that, I believe, was a steel part of the beam. And then bonded to it, we had, I believe, some kind of oak timbers. Specifically, what it was is too much of a concern because I believe I gave you the needed Young's modulus. So there was our set. That looked vaguely familiar.  All right, so we're looking there at the cross section of the beam, this being our y-axis up that way. And it doesn't matter really where the origin is, the z-axis that way. And that's the direction of our moment using the right-hand rules. So if our thumb is in that direction, then the moment is such that it's the usual type of bending if we look down at a great length of the beam, something like that, that being the x-axis. So that's where we left it. The technique we're using, we don't know how to find, oh, by the way, we are trying to find maximum normal stress in the beam. We only know how to find that for a homogeneous beam, a beam of a single material. But this is a beam of two materials. So our bending stress equation doesn't apply. It only applies for a single material. This is a two-material beam. But we developed the technique where we would take out one of the materials and replace it with an appropriate amount of the other material.  It's very artificial in its size and shape, but it does allow us to accurately calculate the stresses in the beam there. So it doesn't matter which of the materials you take out and which you replace it with. You get the same answer either way, but we have to do one or the other. So did I actually suggest to do the one way or the other? All right, so I have to pick one or the other. I'll replace the steel with wood in the beam. So anywhere there's a steel part of the beam, I'll take it out and replace it with wood. But I can't replace it with the same amount of wood that there is steel, because steel is very, very much stiffer, you can tell, by the much, much higher Young's modulus. So we employ this factor n, and we replace the very stiff wood with n times more, sorry, replace the very stiff steel with n times more wood than we had the steel. And that extra area will then be able to absorb the same stresses, and that allows to calculate it as a single cross-section beam. So n comes out to be 16 here. So I'm going to take out the steel and replace it with 16 times more width of wood. And then we'll have an old wood beam. It's absurdly artificial in its cross-section. So I've got the steel then is going to look something like that now that we've taken out the steel and replaced it with wood, where this top piece now is 16 times the original 200 millimeters. But we're using wood instead of steel, so we have to use an awful lot more of it. And the bottom down here is 16 times the 20 millimeters. And that now, as ugly as it might be, is all wood. And it looks like wood. That's what I'm amazing sketched at. And then we leave the original wood in there, which I forgot to give you the dimensions on that, 300 by 75. So the wood's 300 tall, 75. That remains the same. So we have this additional 75 here. The whole thing was 300. That doesn't change. Now we have a beam that's all wood. Remember, it's just an artificial equivalent beam to what we originally had. But it's now of a single material, homogeneous material. And so now we can employ our cross-section, our bending equation. We're given mc, remember, is the distance from the neutral axis. The trouble is we don't know where the neutral axis is. Seems apparent that it's going to be somewhere about there, but we don't know where that. And we need that because c is measured from there. And the moment of inertia, area moment of inertia, is with respect to that neutral axis. It appears that the distance c will actually be that one because the neutral axis is going to be a little bit more near the top because of the little bit greater area of that upper flange part that was still in there. So that's all the setup, bit of a reminder from what we did on Monday. And then everything I gave you to do on Tuesday. I mean, sorry, I gave you on Monday for today. So who did it? Anybody? Thank you guys for doing work in your other classes. We're going to have to stop that. All right, we'll do it real quick because we're going to need this procedure for what we're doing with the extra part of today's work. So we need to find out where this neutral axis is. So as easy as anything, it's just picking a reference line of some kind and measuring it from there. You can pick a reference line anywhere you want. So what I happen to do for this one is midway down the bottom part, I pick my reference line. So that puts it 150 up from the bottom and 150 the interface with the top flange. No particular reason to do that. It's arbitrary wherever we pick it because what we want to do is locate this neutral axis. And once we've located it, then it sticks to the cross section and we don't have to worry about anything from there on. We don't need an arbitrary reference point because now we know where the neutral axis lays and it becomes a permanent part of the cross section. So this is a relatively easy thing to do. Maybe I'll call that piece one. And this whole bottom piece is piece two. Remember, you don't need to do these different sections because the whole bottom piece is wood. There's no reason to split it out into the different parts. So we've got those two pieces. We need the location of the centroid for each of those with respect to our arbitrary reference, the area, and then we need the product. Sue, and it's the area and the product of the area and the individual centroids that we then add up. So we have two pieces here. With respect to our arbitrary reference, it should be, let's say, 150 millimeters up and then halfway into the beam to the centroid of piece one. So that looks like it's 0.16 meters or 160 millimeters. Does that agree? Everybody understands specifically what you're looking for there in that first part of the table. It's very important when you need to take your time doing this because it's really easy to accidentally slip and look at the wrong thing or just go through it a little too quickly. So we're looking for the distance up to that point. So we have that extra 10 millimeters there. The area, 16 by 200 by 20. I don't know if I have that number separately. What? 64,000. But that's millimeters? Yeah. All right. So we just have to check the units when we go. I just didn't have that number in progress. And then the product of the two, might as well do it millimeters, either just swap back all at once. We'll just make this one millimeters because that's 160. So John, 160 times 64,000, 1,240,000. And that's millimeters down to Q. And then we do the same thing for two, 10 million. We probably would have caught that because we know about where the neutral axis would lay. And then the second piece, it's the location of its centroid, with respect to my arbitrary reference point, is 0. Because this goes right through that. What that does is make things an awful lot easier. It eliminates, or it doesn't eliminate the area. But it does eliminate the other two. Just makes the calculation a little cleaner, a little simpler. And so the area of that becomes the 2 times 75, plus 16 times 20, times 300, 141,000. We're trusting John on this. We could do worse, I'm sure. And then whatever those totals are, and we can then finish this. That one's easy, I'll leave that one. And that's a millimeter cubed. These two, what should that be? 2, 205, it's 49 millimeters. We'll call it 50. So we know that this distance right here is now 0.05 meters. Seems about right, about where you'd expect it to be. All right. What next? Now that we've located the neutral axis, we need to figure out what the moment of inertia of this artificial cross-section is with respect to that neutral axis. And remember the moment of inertia of a compound shape, which this is.  It is the sum of the individual moments of inertia. Remember though that this one requires an adjustment for the parallel axis theorem. So the moments of inertia with respect to the centroid here is pretty easy. That's the 112 BH cubed. It's straight out of the front cover of the book. Nice regular shape. And then the A, of course, is the same area we had there. The middle column for each piece. And D is the distance the individual centroids are from our neutral axis, which for the bottom piece, we already know that happens to be y bar. So you can see some of the wisdom there making that as our original reference line. A lot of the numbers are already available. All right. Without belaboring that too much, you can double check that you get the same thing here. I got 2.19 times 10 to the minus 3 meters. And it's meters 2.19 times 10 to the minus 3. All right. You can double check that one. Again, if you prefer, it does do quite well if you make a little table out of it. Very much like we did here, where you have the individual values of the centroidal moment of the area and then the parallel axis there. That's all geometric mechanics, I guess if you will. But now we can get to the actual material mechanics by applying that bending equation we have there with the appropriate values that we will have. Remember, though, that we have two different materials and we have to figure out the stress in each. We've made this artificial beam entirely out of wood so if we use those straight numbers, that's what we're going to get is the stress in the wood. C being the greatest distance from the neutral axis. It could be up or it could be down. We're not sure it could be different for anyone of the problems. In this case, it happens to be down. And so that distance is, the whole piece is 320 across. And we're up 0.05, remember, or 50 if we're going millimeters up from the neutral axis. So C comes out to be 0.2, 0.2 meters. That's what you just said, John? Got that? That's this distance here from the neutral axis down to the very bottom divided by our moment of inertia. That's meters to the fourth. And that will give us kilonewts per square meter or kilopascals, 4.6 megapascals is what I have. Sounded about right? That's the stress we expect in the wood at the bottom because that's the farthest distance away from the neutral axis, remember. When we have a uniform material and we look at it from the side, we know that the stress is linearly, nearly varying on either side of the neutral axis. The top being a compression, the bottom being an tension. What, however, is the stress in the steel itself? That's the stress in the wood because our artificial cross section here was an all-wood beam. But now we've got to go back and figure out what the stress is in the steel, remember how we find that. You have to be a little bit careful with it. For the most part, it's the same calculation with a little bit of an adjustment. Since the steel can have so much more of the stress, it's the factor n, just make sure what you use for i. i again is the greatest distance from the neutral axis to the farthest bit of steel, which is the very same number we used before because it's this part of the steel down here that's in the greatest stress. Might be tempted to go up to the top because that's where most of the steel is, but that's not what the concern is. The concern is, where are we the farthest away from the neutral axis? So we use the same C we used before. And so in fact, it's just n times the value we had before. n times 4.6 megapascals, 73.1. And again, then for each of those, you can compare those to the yield stresses to see if we're safe, whether one or the other will fail first. Mostly a refresher, just with a new problem, because we're going to need that kind of stuff for what we're doing next. We need the very, very same ideas. So any questions with that one? Everybody comfortable with it? Phil, you're all right? Samantha, comfortable? Like that? Colors are OK? All right, colors are awesome. We all know that. All right, so we're going to take the very same idea and do another composite beam. Remember, we have two terms we use. Compound cross sections are cross sections made up of several different shapes. Composite beams are beams made up of different materials. So we're going to do another type of composite beam now, one with which you're all very familiar, maybe not in the specifics of it, but in the generalities of it. One of the most common building materials, especially for structures, is concrete. However, concrete beams offer their own advantages, but their own disadvantages as well. The advantages are that it's very easy to cast concrete beams in a wide variety of cross sections, some which may be chosen for the aesthetics. They'll just look nicer than maybe a steel beam might. You can even do a measure, sculpt concrete to look all kinds of nice ways. So if what you wanted was a T beam, if you do it in concrete, maybe you can make it so that it'll look a little bit nicer to the eye if those beams are exposed. You can also do other things very easily, cast cavities in those for IT cables or piping or any other pieces you might need. It's also rather easy, even in significantly tall skyscrapers, to deliver concrete to the site in its liquid form, pump it up, cast it in place, and then you don't have to carry in the significantly long beams that some of these buildings might require, especially as architects are trying to open up spaces, meaning there's less interior walls, longer stretches for beams, and some of these newer buildings. Concrete's also not all that expensive. It's, for the most part, dug out of the ground. However, there are things that worry us in the use of concrete beams. In any beam that's loaded and supported in our usual way, our simplest way, we look at simply supported at either end and then some kind of loading in between that is downward, whether it's a uniform load or a point load, it doesn't matter. The trouble is there's going to be, generally, a moment like this, the very type we've been looking at. And when we look at the cross section of the beam, and we'll just keep it simple and use a nice rectangular one, we know that for that kind of bending moment on a beam looked at from the side in that way, that the top is going to be in compression. By the top, I mean anything above the neutral axis. Below the neutral axis, everything will be in tension. The trouble with that is with a concrete beam is concrete is terrible in tension, which means the entire bottom of the beam is exposed to almost guaranteed failure for a simple concrete beam, a beam of nothing but concrete material throughout a homogeneous beam. You've got this entire bottom part of the beam, especially the lower part that's exposed to extreme tension, and concrete is terrible in tension. So the solution is then to not have a single material concrete beam, but have a beam that has running down it the entire length, some reinforcing bars or what we know as rebar. You've all heard of it. I know if you pay attention as you come into the building, you'll see it because the concrete is starting to break away with the age of this building, and you'll see little bits of the rebar down there. Some of that is used sort of as a network throughout the material just to sort of hold it together, and it's in all direction. But we're going to look at it at rebar, running the length of the beam that will then absorb all of the tension. And we can then use the ability to use all the advantages of concrete, but have much less of the disadvantages of concrete. So here's how we're going to approach this. Here's our idea. So here's that beam in cross-section, a couple reinforcing bars running down the length. Remember, this is the cross-section. For our analysis purposes, what we're going to do is again realize that from the neutral axis above we have concrete in compression, which is no worry. Well, it's a worry in that there are limits to that, but in general, concrete in compression is not a significant worrisome factor, certainly compared to the bottom of the concrete being in tension. So we're going to transform that beam. Remember, we cannot analyze directly beams made of two different materials. So we're going to do what we did before. Replace the rebar with an equivalent amount of concrete and then analyze our beam as a single homogeneous beam from there. So we'll leave the top alone, because that's concrete in tension, and we're not real worried about that. What we're worried about is the concrete in tension alone. So at the very same level, we're going to take out those rebar pieces and replace them with an equivalent area of concrete that is a distance N, A, S in size, where N is the usual piece it was before, the steel rebar Young's modulus compared to the concrete Young's modulus and A, S is the area of the steel rebar in the original piece, which is these three pieces right there. What that means to us is that for the most part, we are completely disregarding any of the concrete below the neutral axis. We're throwing it out, it's in tension, concrete doesn't do well in tension, so let's completely take it out of the analysis if we have it in a mode where it's going to fail in the first place. And so now here's our equivalent cross-sectional area made up entirely of concrete for purposes of analysis. And then we can do our dual analysis to figure out what distress is in both the concrete in compression at the top and the steel in tension at the bottom. Got the general picture. We've got a beam made of concrete with installed rebar. We take out all the concrete at the bottom because it's in tension, even though it's only minor tension near the neutral axis, but we're going to take it all out because concrete in tension is just not an easy thing to work with in practice. We're going to replace the steel rebar area with an equivalent amount of concrete area and then use that beam as our beam for analysis. Well, remember this is an entirely artificial replacement. We're doing just for the purpose of analysis, but we'll have, yes, this concrete will be in tension, but we've got so much of it that we're all right. Number on N, we've got it coming up here. We'll do it in a second. N is like eight, at least something like that. All right. Trouble is we don't know where the neutral axis is. We need the neutral axis on this beam, so we're going to use a slightly different way to do it than we did before, and that's by, for each of the two areas, we're going to figure out where the neutral axis is by adjusting, placing it such that Y bar A above equals Y bar A for any of the area below. It's a rather straightforward, easy way to do it. In practice, it looks something like this. We don't know where that neutral axis is, so we'll call that distance X. Again, it's arbitrary. It's the same thing as what we were doing when we picked an arbitrary reference point earlier. We'll call that distance X. Let's give some total height to the beam. We'll call that L, A with B, and so Y bar A above is, remember from the neutral axis, it's Y bar is right there, so that'll be X over two. That's this distance here. Times the area of that section above, it's B times X. Remember what we're trying to do is find out what X is, so this is artificially placing it. That's Y bar A above, and if we have a different shape of T beam or something, then we just do that for the two little pieces, and I have them together, and that's got to equal Y bar A below. Y bar, oh, let me adjust this a little bit. Sorry, this L is only down to the level of the rebar. Remember, all the rest below it is concrete, we're completely throwing that out of the picture. So from the top of the beam, which is concrete that we leave in compression down to the middle of the rebar that we've replaced now in our imaginary picture. So now Y bar for the area below is that distance, which is L minus X. And L is not considered unknown, it's the height of the beam from the top of the beam down to the rebar, so that's one of the design parameters. And then the area below is NAS. We're again, N is known because we know what material we're planning on using and what their strengths are, and then we know how much rebar we're planning of putting in. So the only unknown in there is X, and then we can just, at that time, solve for it, and then we know the location of the neutral axis on our new beam. No, this is very, very thin. All we're doing is giving this an area of NAS. We're not actually giving it any thickness. So it's just, this is all L, we take X out of it, we're left with L minus X, and that will then give us X. And then from there, we need to calculate the moment of inertia with respect to the neutral axis of our artificial cross section, and then that's what we use to calculate the bending stress, MC over I. We can do it again for both the steel, below, and tension, and the concrete above in compression. All right, so that's the picture. That's the steps. Both are some of the specifics. Make sure that they work out now. So here's a beam with cross section exposed. We've got two rebar running through there, and then the dimensions are all centered to center on the rebar is six inches, with a little piece of three inches on either side. We have two steel rebar centered and separated at six inches, with three inches on either side. Beam, and this is location of the rebar to the top of the beam is five and a half inches. Oh wait, sorry, nope, that is down to the bottom. It's four inches down to the rebar. So that's all the way down to the bottom. Four inches just down to the center of the rebar pieces. So we've got an extra inch and a half of concrete below, which we're gonna throw out in our analysis. And the rebar is five eighths in diameter. I think that's the picture.  Obviously, I hope, a bending moment such that the top's in compression and the bottom's in tension. So we've got a beam, let's see, it's 12 across there, 5.5 high, two pieces of rebar running through it at four inches. So that's the cross section. The modulus, elasticity, ratio, is, steals like 20 times 10 to six PSI. Concrete's about 1 eighth of that, 3.6. It comes out to be something like 8.1. So, oh wait, let me move that out of the way. Just so I can keep it right beside this, our equivalent beam. Same height, we're gonna take out the rebar and replace it with an equivalent amount of concrete. So at some unknown spot, we have this neutral axis, where all of that material above that is in compression. I just don't know where the neutral axis is yet. That's what we have to find next. And right at the same level as the rebar was, rebar was, we replace it with a very, very thin strip of area NAS of concrete. Notice what that means is we completely ignore any of the concrete below the neutral axis. All of that is thrown out and ignored, even though it can withstand some tension, but in general it's so poor in tension we're not gonna allow it into the analysis. So that's our equivalent beam. As artificial as it looks for purposes of analysis, it works quite well. What's next? We need to find out what axis. We need to place the neutral axis and then calculate the moment of inertia from there. So y bar a above the people, y bar a below, which means 12 times x. Well, hang on, that's a. Y bar is x over two, x over two times 12x. That should be y bar a above. Agreed? Y bar a below, let's see, y bar below is the four inches minus the x. It's just the distance from our neutral axis where we don't know where it is yet down to the center of our concrete strip. So that's four inches minus x. And then the area below is NAS. We know what NAS, and we know what AS is because we've got two 5 eighths inch rebar in there. NAS is 8.1, AS is two times higher square two because there's two bars. It's what? Yeah, it's 614 square inches. That's the cross sectional area of the rebar where we're assuming all of the tension will reside. So we've got N, we've got AS now. We just solve for x, 1.45 inches, which would seem about right. Two inches is midway between the top and the rebar. And we're gonna be a little bit above that because we've thrown out so much of the bottom area. All of the bottom concrete area is gone. This is just the area of the steel. So now we know what x is. It's about where it should be. Check the teams. Now what? That was always our first step. Determine where the neutral axis is. Once we've done that, now what do we do? Do what? Find the moment of area with respect to that neutral axis. And that's, as it has been before, it's just there's a little bit of a trickier step. Not a tricky step, it actually makes things very easy. It's just not necessarily an obvious step. So if we make a little table out of it, we need to find the centroidal moment of inertia plus D and then we have plus 80 square. So for the top piece, it's just a nice rectangle. It's centroidal moment of inertia. Remember the moment of inertia about its own centroid is 112th BHQ, actually in this case BX. D is, remember that's the distance of the centroid of one from the neutral axis. Which for our purposes is x over two. Then you can actually put the numbers in. See if we get the same thing. And then that gives, oh well, I might have that number. Whatever it comes out to be. It's just plug in the pieces in now. The area is 12x. So it would be 112BXQ plus 12XX over two squared. 12XX over two squared. One half, I don't know, IC plus AD squared. I just don't have to have that separate number, so. That's just algebra. Now here's the little bit of tricky part. This strip is so skinny and narrow that we take its moment of area to be zero. No bother calculating. If we're gonna calculate it, 112th BHQ, but it doesn't have any height for all intensive purposes. All we're concerned with is its area. So it's just so skinny, we leave that out of it. D is the L minus X we had before. Cause we may not know how thick it is, but we do know where it is. And then of course its area is NAS. So this just becomes NAS times four minus X squared. That's I, which is zero, plus AD squared. You can see how easy this would be to do on a spreadsheet. All right, so that gives us an I of, when we put all the numbers in, 44.4, that can't be, I just wrote the units down wrong. I to the fourth, inches to the fourth. 44 inches to the fourth, I believe. I'll find the stress. We need to find the stress in the concrete and the stress in the steel. So you do those. We've got moment 40 kips. Kip inches, a cup inch. We've now got the moment of inertia with respect to the neutral axis. And it should be then kips per square inches, KSI. But I want to see what numbers you choose to do this. Travis, while you were out in the most important thing you probably missed, I don't think you were here for it, is when we figure out the moment of inertia with respect to the neutral axis, we do it by summing up the centroidal moments with parallel axis theorem applied for the strip that replaces the rebar. We consider it to be so very thin. We know what its area is, but we don't give it any thickness. It's so very thin that its centroidal moment of inertia is zero. It's kind of a subtle piece. I don't want you sitting there trying to figure out what the thickness of it is when we assume it doesn't really have any. And now everybody is hopefully working on the stresses and the concrete and compression at the top and the steel and tension at the bottom. Easily taken care of, but you've got to pay attention to what you put in for C to make it all along, out of the unit, so that's easy. Now for your C, what do you have? Let's see. It's the greatest distance. 40 inches. Would you use for C to find the stress in the concrete? Well, if you use this distance, what did others use? No, you're forgetting something. We don't have any concrete down here for our purposes. We've thrown it out completely. We assume that any tension on the concrete is going to cause the concrete to fail and we take it out of the calculation completely. So we're only concerned with this stress in compression. We assume that there's no concrete there to fail in tension, take it out completely. It's a conservative estimate because we've taken a lot of the material out that would withstand at least some tension, but it's so bad in tension that we take it out of the calculation. And then I, where did I put that? 44.4 inches to the fourth. So we're left with kips per square inch. Should I ask, who's my favorite student and did that right? Samantha, you did it that way, right? Yeah, Joey, you did it that way? Okay, perfect. We have, like, are you nodding that Joey did it that way or you did it that way too? All right, so one female, two male, prompments and kips. So that comes out to be 1.31 KSI. Now, what about the steel? Multiply it by n, n is 8.1, m is 40 inches, and what do you use for i? That's the four minus x, so we want just that space to the 2.55, which divides it by, is that still a greater distance? Of course it's a greater distance, but there's no sense calculating the stress down here and material that we're assuming isn't even part of the equation. We've taken it out completely. We don't need to calculate the potential stress in the concrete because we assume it's gonna fail anyway. It's there, its real purpose now is to do nothing more than hold the rebar in place. That's all the concrete does. Yeah, it's a greater distance. In fact, the distance all the way down there would be 5.5 minus 1.45. What is that on top of the top? Use what on top? 8.55 inches per cm. No, we're not figuring out the tensile stress in the concrete because we're assuming it's gonna fail. So why calculate the stress? We want to know where this concrete is actually used for the strength of the beam, which is in compression on the top. I was just backing up in other problems. It's not hard for you to do that. Yes, this is a completely different problem where we've actually thrown away part of the beam. Not replaced it with something else actually disposed of, thrown it out. And the tension on the steel is 18.5. So that's tension and that's compression. If we happen to have a concrete beam loaded in some other way, for example, if we had a beam that was maybe loaded this way, that beam we expect to bend the other way. We want the rebar on the top of that beam cross section rather than on the bottom because this is gonna have a completely different type of moment than would be the original beam that we are looking at. It can be used in any way as needed. Chris? Because you guys had a higher value for the steel. Would you, in design, use a smaller amount of rebar because you've ever done it? I haven't. Well, no, once we've got the stress in the rebar, you have to look at the yield stress of the material and then you decide, including a factor of safety. If this is approaching the yield stress of the steel, then we decide we're gonna put in more rebar. Well, that's because it's so much higher than concrete value that you've got the concrete of the kiln loop. Yeah. Well, no, we don't necessarily compare it. Again, it depends upon the yield stress of the concrete and compression, which is pretty hot. I don't have my book here, so we can look them up. I don't think I have them written down. The other thing that can be done, since I don't work in this field, and I've never worked in this field, I'm not sure what the point is, but a balanced beam is one such that the stress in the concrete is the same as the stress in the steel. A beam that's designed in that way is considered balanced, that they have equal stress in the two. I'm not sure what the point of that is. It just is there. Yeah, yeah, you'd either adjust this beam for what the maximum load might be, or you'd adjust the area such that they're balanced. Okay, one for you. Here's a concrete cast T-beam with two rebar in the very bottom there. Width dimensions, four inches there. The T-part has got a width of eight inches. This lower web is 18 inches. The bottom thickness here is six inches across. The rebar is two inches up from the bottom of the area, or let me see what that is at the rebar. One inch, I believe it's one inch diameter. Double check that, because you just have to find the area anyway, I'll just give that to you, because that I did right down. 1.57, well that's the number we need in the end anyway. That's the area of the steel. Oh, and I need to give you a load. Oh, find out our, find the maximum moment, so you're gonna have to double check which one might fail first. The maximum allowable stresses with a concrete three KSI compression for the rebar, 40 KSI intention. So find the maximum moment, which I have to do is figure out which one of these is the controlling stress. What that means is put these stresses in the bending equation, M sigma equal to MCI, put in that for the stress, and then solve for the moment, see which one's the lower, and I'll tell you which one of them. Two materials is the controlling material. All right, so first, draw yourself an equivalent cross section. You have to determine just where the neutral axis was going to be here. Suspicion, I would hope, would put it somewhere about there. A little piece there, just a compound shape, break it into two little rectangles, and then the bottom here, we put in enough concrete to simulate the steel by putting in n times lower of that area. All right, so first thing we have to do is place the neutral axis, and then find the moment of inertia with respect to that, and then calculate a maximum moment allowable by using the stress limits given, and I'll have to do that for each piece, and don't forget the factor then. All right, maybe as a hint, it might be a little bit easier if that's your unknown value X, there down from the bottom of that upper flange to wherever you might make that cut, just because the flange section here has already got a constant thickness, so that might be a little bit easier for you. Y bar A and below are equal, and the compound section, you just add those together in three sections. Because we need Y bar A, and we don't know where Y bar is for that leftover T piece, we know where it is for the two little squares, so it's always easiest to break these down into the simplest possible pieces that for which Y bar A and I are already known. But you work along and see if we don't get the same thing, and any numbers in here will be in inches. So Y bar A is equal to four, eight inches by 22. No, not you again. Y bar is the distance from our neutral axis up to the centroid of that particular piece, which is halfway through the upper flange. What is this? That's two inches, and then an additional distance X down to neutral axis. So it's X plus two. Y bar one, X plus two, or two plus X. And then A is four by 22. Eight, six, and eight. Maybe that makes sense? Is that right? Pretty much on top of these things. Quiet, so. Yeah, I was also tinkering with something of another month and wanted to see how that worked out. That's most of what I'm thinking. And then for the other piece above, it's Y bar is X over two, and its area is six by X. That's Y bar A, it's centroid right in the middle. It's height is X, so it's X over two. It's area is X by six. That's the little square here. Oh, it's two now. Yeah, it's one. So that's the area of the Y bar A product above. Below, we have from the top flange to the rebar is not 18, but 16. So that's, and then that's our distance down there. That's 16, Y bar, and the area is NAX. 16 was the whole thing. 16 minus X, right? That all right, David? Is that what you taught? Yeah, 16 is from the bottom of the top flange, but we're going from the neutral axis, so that's 16 minus X. Now NAX. Okay. That's the Y bar A, and is, whatever that ratio is there, 40 over three. Oh. 7.63. Oh wait, no, no, I'm sorry. Because that's the, we don't do the ratio of the stresses, we do the ratio of the strength. I just don't happen to have the same numbers as the one before. So different brand of concrete and the steel numbers the same, that was 2.9, that was 10 to the sixth. No, 29, that was 10 to the sixth. And the concrete on this process had to be 3.8. Concrete is mixable in all kinds of ways. Before we had 3.6, this one happens to be 3.8. So there's our N, 7.63, and the area I already gave you. All right, we're right at the end, so I'll make sure you've got the numbers before you go. X, when you solve for it, comes out to be 15.7. 15.7 is 0.1573 inches, about where we'd expect it to be. I, with respect to that neutral axis, once you find that 35, 35.7 inches to the fourth, then you can figure out the maximum moment for the two of them, whichever one is lower, is the one you'd use then. Don't forget the factor of N in there. So the moment with the concrete at its maximum would be 25.51, moment with the steel at its maximum, remember the maximum stress limits were given would be less than that, so it controls, this is 11.70. So the steel, the stress in the steel is the controlling factor, and so there's your design moment there. No, that's already got the N in it. Oh, and then yet you could apply a factor of safety to that by cutting in half, maybe for a factor of safety of two, perhaps more, depends on those. There are industry standards, there might be company standards as well. Call it legal office, they'll tell you what today's factor of safety is.


What's up, everyone? It's your girl, Kenzie here from Layton Sports Cards and I am ripping 2021 Panini National Treasures football hobby four box full case break number 110 tier random teams. Here's a look at the random results. You can always find these at LaytonSportsCards.com under the results tab. Link is also in chat for you. All right, best of luck. Here we go. Good one, Joshua. Good one. No shot. Still, I'm like in awe that Caleb is voluntarily that just that blows my mind. First up out of 99 Joe Mixon bangles and we've got two 35 Tyree Kill for the Chiefs, Century Patch Dolphins to 49 Ronnie Brown, named fabric patch Debo Samuel for the Niners, Treasure Patches for the Seahawks Russell Wilson. Seahawks. We've got Gold Treasure, Tremendous Treasure, Seven of Ten Rookie of Rondell Moore Carnals. These three color packs, they're Carnals. Rookie Auto to 35, Racy McMath Titans. Racy McMath for the Titans. We've got Crossover RPA here to 25, Tylen Wallace for the Ravens. Auto to 35, Randall Cunningham Eagles. Eagles throw back on out there. And how about this one? Ryan B with the Jack, Crossover, RPA to 99, Trevor Lawrence. Boom. Kaboom. Nice tealaw, two of 99 on the three color patch. Heck yeah, we'll bag that one for you. Very nice car. All right, starting off hot with NT. We've got to 10, Two Attack of Iloha Dolphins out of 99 Walter Payton for the Bears. Little Bears. Dual Rookie Patch Auto Vikings, Kellan Mond and Keeney and Wahoo. And I filled your combo materials. Century Patch, one of five, Jags, James Robinson. Nice low number one there, going back to the Jags with Ryan B. Got Rookie Dual Patch to 49, Diamond Brown. Diamond Brown. Got a triple here, most likely be random. Trevor Lawrence, Zach Wilson and Trey Lance. So Jags is Ryan, Jets is Eric R and Niners is Phil K. So yes, this will be random at the end. All right, we've got one of one, Laundry Tag, Rookie Auto, Laundry Tag, Rookie Auto, one of one. There you go, Harper. We'll bag that for you, man. Going back to Harper with the Cowboys to 10. How about a Tony Dorset? Dorset to 10. What is going on tonight? To 25, Prime Pairings, Rookie Autos, Trevor Lawrence and Travis Etienne Jr. Another one. Ryan B with the Jags, two T-Laws and two boxes. Mag that. And cross over RPA for the Patriots to 25, Ramon Dre Stevenson. Heck yeah. That answers my question if we are going to sing that. That bill is not working. A headache, Joshua. Honestly, all the booms and big hearts that we've been pulling, I just have a headache now and I took some Tylenol and it's not working. I can't take anything else. It's the only thing I'm really allowed to take. We've got to 35, Sam Darnold Panthers. Why do I not see myself? So, 99, Terry McLauren Washington. Marshall Faulkton, 99th century for the Colts. One out of five century patch, Tyler Higby. Rams. Looks like a St. Louis Rams patch on that one. What a five. We've got NFL Gear Trial materials to 10. Going back to the Cowboys. Dak Prescott, Zeke Elliott and C.D. Lamb. Here you go. Cowboys starting to heat up here. Here you go, Harper. 299, dual rookie patch, Devonte Smith Eagles. Oh wow. Who is that? Cut auto after cut auto today. That's so sick. That's going to be a random. It's not associated with anything. Yeah. Just like flabbergasted of tonight. Well, we have a one of one cut auto pedals at American Actor. That's so sick.  Recently passed away a few years ago, 2018. There you go. It's in the longest yard, smoky and abandoned. Is there before my time? Movies. Dukes of hazard, without a paddle. I think I've seen that one. The last movie star. That's sick. Bean in 97. There's some old movies as far as you're even born. That is really cool though. Love the cut autos. We'll bag that. That'll be a random to all at the end. Then we've got rookie jumbo, prime signatures to 99, Patriots, Ramon Dre Stevenson. Pat's getting some nice Stevenson's tonight on the four color, out of 99. We've got to 49, Jim Kelly on the bust. Oh my gosh.  I wouldn't be able to afford this car, regardless. True. Oh my gosh. Those are the best to have the logo in it. Because the logo is so boring. Like patches are usually so boring when it comes to true. How about this one, y'all? Three of five. Partial logo patch auto. Not sure. Harris, Steelers, David G on the not sure. That is so cool. Even I wouldn't bet that is not. That is sweet. What a case. Two T-laws, Naji RPA. Cut auto one on one. Last one here. Here we got a one on one printing plate. Out of 49, Tyreek Hill Chiefs to 99, Chris Godwin for the Bucks. Treasured patches to 99, James Robinson, Jax. Century patch to 10 on the gold Ozzy Newsom for the Browns. Ozzy Newsom to 99, Franchise, Treasures, Russell Wilson, Seahawks. Dual rookie patch to 49, Eli Mitchell. I'm trying to guess who's signature that is. Look like Swift for a second. Same position. You got the right position. It is a rookie. Got printing plates. Ah, one of one, Trace Sermon. Niners. Phil Kay. Kind of, D Swift kind of has that, like the D and then the little squiggle after it. There you go Niners. Phil Kay. We'll bag that for you. Very nice. Cross over RPA to 99, Trace Sermon. Niners, what? Back to back to back there. Very nice. You go Phil. We've got patch auto. You don't see this guy too much. Titans. Material Treasures, Vince Young. Vince Young to 49 for the Titans. And Stevenson's QB year crossover, RPA to 99, Mack Jones. Boom! Austin B. What a case of NT. Wow. Mag that. Give me a second to set up the random. Again, we had the triple year for the Jags, Jets and Niners. Then the one of one cut auto of Burt Reynolds. Uzi's favorite actor. Oh yeah, that's true. All right. Triple patch NT. So all the teams go into the random for the cut auto and this one was Jaguars, Jets and 49ers. We've got four and four that is eight times. Eight times on the random. I'm going to do the triple quarterback patch here first. Again, whatever teams end up on top, we'll take home each card. Triple here first eight times. One, two, three, four, five, six, seven, eight. Niners with the Tray Lance heated up at the end there. And then the Burt Reynolds. Again, all teams are in on this one. Eight times. Good luck everyone. One, one, one, two, three, four, five, six, seven, eight. Niners again. Wow. All right. Niners taking home the Burt Reynolds cut one of one. All right, let's do a recap here. We had Eli Mitchell. Niners, Russell Wilson, Seahawks, Ozzy Newsome, Browns, James Robinson, Jags, Devontae Smith, Eagles, Triple for the Cowboys, Prescott, Elliott and Lamb, Tyler Higbee, Rams, Marshall Fault, Colt, Stiamy Brown, Washington, James, Waberson, Jags, Vikings, Duo of Kellan Mon and Wong-Woo, Rondell Moore, Cardinals, Russell Wilson, Seahawks, Debo for the Niners, Ronnie Brown, Dolphins. Up next, the Vince Young, Titans, Trace Sermon, RPA Niners, Jim Kelly, Bills, Remandre Stevenson, Patriots, Tony Dorsett, Cowboys, Randall Cunningham, Evil, sorry, Tyler Wallace-Ravens, Race and Big Math Titans. We had the Remandre Stevenson, Jumbo Booklets for the Patriots, Pitted Grand Slam. Patriots did do well. We had the One of One, Trace Sermon, Ricci Otto, Niners, Semi Fajoco, One of One, Cowboys, Laundry Tag, Duo Otto, T-Law and Travis Etienne to 25 Jags, Trevor Lawrence, crossover RPA to 99 Jags, the three of five, Najee Harris RPA, gosh that's so sick, partial logo, and the Mack Jones, crossover RPA to 99. All right everyone that'll do it for NT, we've got more football coming right up.


Here's exactly how much money I make as a cardiology fellow. So first let's quickly break down exactly what a cardiology fellow means. That means I've finished my four years in medical school. I finished my three years of internal medicine residency. I even worked as a full-time internal medicine doctor and was board certified there. But now I'm doing extra training over three years to become a cardiologist. So right now I can't be an independent cardiologist and so I make nowhere close to the salary that you would make there. But over the next three years my average salary would be anywhere from 70 to $72,000. Now keep in mind that that number will vary depending on where you are. Geographically your institution as well as the cost of living in the city that you are practicing in. So for me it's about $70,000 which is enough to get by in the city that I'm in as long as I live in kind of the outskirts and not five minutes away from the hospital. Now if you enjoyed that breakdown you want more behind the scenes in my life as a full-time cardiology fellow as well as interested in more tips and pieces of advice to help you on your medical journey hit that like, subscribe, and follow for more content just like this one.


So we compute the various expectation values of each of the bi, gx, and the later things in the last class. We now want to use that to understand what the strings get. So how are we going to do that? Let's start with a simple example. Well, first I'm going to remind you of the formula we used in the game. It's very easy to find the i of yi, probability of the i, expectation value is equivalent to the world. yi in j, probability i in s in j, 2 alpha prime ai. And here is the right last statement. Okay, now let's study the three-tacton sketch. The three-tacton sketch is going to be a member of the rules, because we've got three c operators. So let's say a c operator and y1, a c operator and y2, and a c operator and y3. Times e to the power i k1 and y1, e to the power x at y1, e to the power i k2, x y2, and e to the power i k3.x at y2. Okay, now we want to understand the six-tacton sketch values. Well, the probability iteration is up, because the probability iteration is up. On season, the probability iteration is up. We can calculate the expectation value separately. And that's what we've already seen. It's y1, 2, y2, 3, y1, 3, we've computed. So that's the part we see. Now let's compute the probability axis. Okay? So the probability axis goes to this office. We've got e to the power, so we get y1, 2 to the power, y2, 3 into k1, k2. y2, 3 into 2 to the power i k2, k3, and y1, 3 into 2 to the power i k1. Now, from general principles, all three vertex operators are fixed at particular points, y1, y2, y3. Right? If we could choose these points to be any way, well, in this simple calculation, I'm not sure if they're in the procedure. Yeah, that's true. I'm not sure if they're in the procedure. I mean, it's consistent to check on our procedure that this the satisfaction should not end with the insertion point from the device. So first, it sounds crazy. I think it works. This depends on the momentum. Then you see the probability of this in a two-fold. And how? Remember, the k1.k2 is equal to the k1 plus k1. k1 plus k2 minus k1 plus k3 squared. k1 plus k2 plus k3 squared is equal to k3 squared. So each of these is given by the master value of the master value. So the master value here is k squared. The master value is minus 1 by 2. k squared is 1 by 2. So this is equal to minus 1 by 2. So yeah, 1 by 2, 1 by 2, 1 by 2, 1 by 2. OK? Sorry, I just had a 2. The problem was connected to that, with a 2. So we could choose the y1 to be raised to the power minus 1. Certainly y1 to be raised to the power minus 1, y2 to be raised to the power minus 1. And we can get the solution. There is a big truth, or a truth here, that in this case, the example is more less true than the amount of the additional factor of 2. But this is sort of the important thing, more degrading than me. And that's it. Now there's a moment in k1, k2, and k3. We place them at some point, y1, y2, y3. But it doesn't matter what this point y1, y2, y3 are, because we can choose them to be anything, by the appropriate, more based master, the appropriate, unfixed, conformable transformation that needs a lot of reaction. But, you see, any value of the change that acts in a one-to-one fashion on the power machine cannot change the relative ordering of points. Particularly on ordering. Now the way that we need to do the activation is to regard the, the version to the open string is the upper half place. So the upper half place can formally be related, it can formally transform over into the disk. And you can identify points at infinity, so they become boundary. So, the points that we put, such as y1, y2, and y3, if they're going to have a linear order, these points are going to cycle. So it's just cycling the ordering of y1, y2, y3, on the boundary of the world machine, that cannot be changed by quantity. So, if we want to sum over all the digital ways of inserting our graphics operators, we should sum over all the scientific orders. And if we're a bigger scientific order, we can choose this y1, y3, y3, and y3. Is this correct? Is this correct? Coordinate transformations and unfixed characteristics procedure that allow us to take y1, y2, y3, anything, have to be one-to-one. So, y1 cannot change the relative ordering of points on the boundary. Is this correct? Yes, that has to be summed over xx. Now, since this invisibility is a simple case, we have to do this, we do not care. It's not a question of y1, y2, y3, anything. So, if y1 were to be done in the past too, we're not doing this calculation that the length of the truth track and such. That doesn't drop in the end of the question. So, for this particular calculation, we would not care very much. That would soon become important in the end of the question. And it will also become important in this calculation when we generalize the calculation in a particular way. So, that's what we need to do in the end of the question. So, that generalization was a problem. So far, we've been talking about how open-stream is living on the strip. But it's not an amount of information that goes into the stream. Once you have boundary conditions, you've got some special boundary points.  That's not the propagation of the theory it has been in the past. That we can't change easily since it's very, very time-consuming. However, you might, once you add, you boundary use a tree. You increase the freedom that lives just at the boundary of a tree. The question of how you can add new boundary degrees of freedom in a way that preserves the formula in the end and all the other good properties of the problem. I think that is an intricate and interesting question.  All you've got to do is to make very... No, you'll do better. All you've got to do is to make very complicated charts. Okay? There are other ways to do it and that makes it more complicated because I've got boundary states, so they're there already. At some point, of course, the research will end up with this. But as a matter of fact, you've got the very trivial set of boundary degrees. The boundary degrees of freedom will only have to be given a vector space that is the final dimension. Okay? So there's a final dimension vector space during each boundary. Okay? And we work in some basis to make this space. So there's some inner problem of the vector space, the inner space, and we work in some basis, some common basis, we use the basis vectors to recall them. Okay? So the basis vectors are, you know, column one, zero, zero, zero, zero, zero, one, zero, zero, zero, and so on. And then the only difference is boundary degrees of freedom. The evolution of the evolution of the boundary degrees of freedom will be true. Okay? Which means that along any given boundary the boundary degrees of freedom don't change. Okay? Basically because, basically what I'm saying is that it's that the open strings in the picture come in different flavors. But flavors are labelled by an index, one index i, around from one to n, where there is anything. One index i, around from one to n, on the left boundary, around from one to n, on the right boundary. Our open strings on the strip, in addition to being labelled by, you know, the full Hilbert space that we obtain by quantizing these open strings is specified by a vector and the tensor product. After Hilbert space we have before, I mean the oscillator is in zero mode, and this n-square dimension Hilbert space it can have some boundary degrees freedom on the left. Okay? So you can think of it as some matrix, n for the matrix, times wherever we have the form. Okay? Think of it as n-square different species of open strings. So n-square different species of open strings. Okay? The species does not change in evolution. Right? They can change in interactions. Each species, because identically, I don't know how to think. Is this clear? I cannot see very systematically what you mean today. There is a very consistent generalization. It's one of a much larger, there's a full theory one can build up, consistent boundary states of boundary interactions one can put to a straight. I won't stop to try to build this volume. We just look at this as one example. It's something that's more or less trivial and therefore when you look at it carefully more or less clearly consistent with the generalization of the volume of this normal n-square condition. Okay? So now the important thing is that states of the open strings and then vertex operators are labeled not just by the vertex operator label links that we've got before. These are by IPX and derivatives and so on, but are also by n-square different species. That can also be a wide line, a long, long one, just not n-square different species. So the index cannot change at all. So if you have an open string and its index cannot change at all, however it comes and interacts with two output strings. So the index around here is the insert. The index around here is the insert. The index around here is the insert. Okay? Now these open strings are replaced by the vertex operator. Any computation involves the insertion to your whole vertex operator. It tells us that you have to put in addition to what you had before, you also have to put in the trace of yawnawa yawnawa yawnawa yawnawa yawnawa. No, it is actually the three cations cations that we have with you. The three cations that we have with you. Anything in the large sphere depends on these new boundary decreases. Okay? So let's do that. All we've already seen would be that the answer is one from the matter part. Okay? So first we had lambda 1, lambda 2, and lambda 3. So this would be the pi k x1 in the pi k x2 and in the pi k. k1, k2 k1, x2, x3. These are our three values. We've already seen the expectation of what orderly we choose. Okay? We get one from the matter part and matter goes up to the end. So let's do the precisely what we've already seen. So let's put in the random three or one, three, and two. So these two they should try to get all the numbers. They don't give us equal boundaries for the lambdas. Because one of them makes us place of lambda 1, lambda 2, lambda 3. And the second one we first place of lambda 1 lambda 3. So we want to do, we can write this as for instance, place of the anticognitive of lambda 1, lambda 2, lambda 3. Times 1 belongs to the three tachyons that we have to use. Okay? Where the tachyons are labeled where n cross i means matrix, but the x operator lost a point. And the answer to that is we're going to do another three-point function calculation. This time of this calculation two tachyons will be what is the physical interpretation of this calculation? If I take two tachyons of the tachyon what will help me learn whether the tachyon is charged or how it's charged under this case? Well, if we take the tachyons of the tachyons of the tachyons it's the natural gas of this gas. And the gas goes on to be non-immediate. Okay? And the tachyons will be non-immediate tachyons. Okay? But let's start seeing. So let's start seeing. So I'm going to do this to compute. Okay? So the three operators of gas, that's how the gas goes on to operate at number one. So we have e to pi k1 dot x. And then it's associated with the polymerization factor e dot x. And let's start down with this matrix signal level. Then we have e to pi k2 dot x. This is associated with the tachyons. And e to pi is e to pi k3 dot x. Again, this is a root. e to e to pi k3 dot x to tachyons, that's how it works. Or the matrix. It makes a problem. That there is a problem. Also the C3 matrix. It can be done once again. We can fix that. Okay? Now let's compute the C3 matrix. So once again we have two orders. We have the ordering let's say one, two, three. And the ordering one, three. Again, let's do the calculation first for the first. Let's do the calculation for one, two, three. Okay? First thing for the C3 that's the first one, y1, 2, y1, 3, y2, 3. Okay? And that will be common for both buildings. We'll add that on the side. Okay? Now there's also the calculation for the loudness. So the first guy gets I'll try to do the calculation more or less. The first guy will get a trace of lambda, lambda. Well, we have to distinguish between the traces. Let's say we take up traces of that. I'm not sure if you should dimension this but let's say that we would take up traces of this. That matrix is one. Okay? So, lambda1, lambda2, lambda3. And this guy of course, and that would be something other than it. This guy would have a trace of lambda1, lambda2. So we have to figure out what model class is it and what model class is it So that's what we have to do. Now we get by contacting this guy and this guy and this guy. In order to do that we need the rule that we developed last class. The rule that we developed last class helped us deal with insertions of this data. So the rule that contracts the exponential size of the data for the data and adds the additional factor of the data. Okay? So of course let's do the contractions of the data and then you are going to be using that. So let's look at the number that we got, that's y1. So we have y1 and 2 to the power over. Well, it's 2r times k1 dot k2. Now once again, k1 dot k2, 2k1 dot k2 is k3 squared minus k1 squared minus k2. k1 squared is 0 and k3 squared is equal to k2. So we get 0. 2k1 dot k2 which is k3 squared minus k1 squared minus k2 squared k1 squared is 0, k2 squared is equal to 0. And similarly we get y1 3 to the power of y2 3. For y2 3 we get 2k1 squared so we get k1 squared minus k2 squared minus k3 squared. k1 squared is 0. But k2 squared is k3 squared at each one by alpha k. So this is minus 2r times k1 dot k2 but we get the formula for 2k1 dot k2. Any reason that earlier it was minus 1 this is now. Now what do we get from the insertions? So what do we get from the insertions? So from the insertion we get if you remember from the formula it says we get 1 over there is a number where 1 over y1 minus y2 into k2 plus 1 over y1 minus y3 into k3. The general formula was check where this guy is inserted this guy is inserted into the formula. Just take yi minus yj to all other insertions and have the k of the other guy. The guy is going to be dotted with the polarization tensor. Yes? Okay. In this case, can we make an insertion? Nothing more. Now let's process the formula. This formula idea has what? We have k2 into k3. The first thing we have to find out is that k3 k1 dot epsilon is 0. k1 dot epsilon is 0. That is one of the physical state conditions the BRSP closed conditions of our experiment. Therefore by the way the conservation k2 dot epsilon is minus of k3 dot k1 plus k2 plus k3 is 0. Each of these, this k2 will be replaced by k2 minus k3 by 2. k3 can be replaced by k3 minus k2 by 2. So this is equal to epsilon dot k2 3 by 2 into 1 by y1 2 1 by y1 which is epsilon dot k2 3 by 2 into now we get y y2 3 over y1 2 y1 it's y2 minus y3 y1 2 y1 less. So now let's all work together. The first thing we have to take into consideration is that everything is going to be in the length of the each section. That part is going to work. That part is going to work because that part is going to work because now we got this times y 2 3 to 1 by 1 by y2 3 2 k2 1 by y1 3 y1 3 y2 3 and that cancels these sections of the series. What we get from the other one but notice that the other order is just obtained by replacing 3 by 2. Replacing 3 by 2 is epsilon dot k2 3 into lambda 1 into lambda 2 up to epsilon. I would like each of you to go home and make this minus sign to epsilon dot k. The method under these two order is to contribute to the relative minus k. That's the thing that's very important. Now to reach this commutator that has to be no polarization tens. The lambda has to be trivial. Where is the minus sign? k3 2 is minus 2. If lambda has to be trivial this would have just been 0. So we would have in the statement that the tacking on is uncharged under the gauge force. In the general case it's not trivial. But we also see how this thing this is the the basic structure of the two categories. It's the commutation that appears. The commutation that appears multiplied by the matrix of the gauge force. Now this is in fact the statement that the tacking on is interactive with the gauge force are exactly like a minimally coupled and joint scalar interaction matrix. What is that? That's why there's a Lagrangian minimally coupled action. The Lagrangian minimally coupled action is a stress of dv5. That's dv5. Two tacking of the magnitude that goes on interacting from comes from one of these d-news having the area. And the level of that is commutated by dv5 twice. And this one has the odd derivative. This is exactly the same as the swamp structure. You know where we put the commutator doesn't matter. It would be uncoordinated by many tools. It's the same, let's say. Where we call it one and two words. Just after we put two cyclical odd bits with a relative amount of time. And then this is exactly the scalar structure. The derivative here is the fact that the has a k.  You know, we've got a little more detail than we have here. It says that the sign of the rule of the scattering amplitude that follows from this is the factor. But the open string theory I mean, it's really seem very much like open string theory has not a billion gauge bosons there. And the datum's is charged like an adjoint scalar under least now. Look at our specific theory of datum. Okay? Every field in the problem are the gauge bosons setters. But every field in the problem enters as an cross-end. So it's plausible that every field in the problem is charged in the adjoint of this, of this gauge. So what I'm saying is that if you work out the first state sign of the state, two copies of that scattering against the datum once again you find the derivative. If you're detecting that basically all states of the theory are charged in the adjoint of this u and d2. Okay? The one reason that expression in this regard is the gauge boson itself. Expression because it should give you all the interactions and if our derivation is correct, it should give you the scattering amplitudes that follow a problem. That follow a problem the interactions of gauge bosons itself in the y-axis. So something that's interesting to do is to calculate the 3-point the 3-point spectrum of 3 gauge bosons. Okay? And then later that makes a sense. It's more of a system. Yeah, you just have to do all this stuff again and again. If one new element is set as a rule for a contraction of this stuff, the other element is you remember there were also the oscillators, the active oscillators in one contract. You just have one derivative and there's nothing going to contract. Yes, we didn't have to use that. So now we have three derivatives. So there will be an oscillation of contractions fair-wise in addition to taking the derivative that is just this piece for everything that exists. Okay? And then there will be one term when you take this piece for everything that exists which will be Q e to 1. And then there will be three other terms which will be the area. Where you take this term for one of the two delixes and the oscillators will be out. Is that clear? Do you know what I'm talking about? Do you know what I mean when I say the oscillators? Do you remember what our rule was when we contracted these delixes? There's a part that contracts that's what's going to be our oscillators. Okay, the contract therefore only adds other elements. And there's a part that has background value that will be the Q impact. The background value is a linear domain. So the term when you take all the three background values will be Q e to 1. And the term when you take one parameter, the group delixes come back to Q e to 1. There will be four terms because if the oscillators can choose there are three ways to choose it in which space they come back to Q. Okay, so there will be four terms that have to be back up. Let us go to the result of the purchase. So now together apart from these very special functions on Q e to 1. Q e to 2. Q e to 3. Plus sign. And then there's a special job which is just I'll go back to Q e to 1. Q e to 3. Q e to 2. Q e to 1. Q e to 3. The rest of lambda 1. Now as interval. The doctor tells us that interval no matrices because it's omni-ray. We were just in U1 carry. We get 0. Which is great because gate was also interacting with itself. You interact with each other. I'll simply define the rules that come out of the usual question. The three-point line of notes are done. You should check it. It's different for instance starting from the fact that it is three-dimensional in one day. And it's different also that it has an extra alpha minus. And it reflects the fact that string theory is language. String theory. For the masses, increase or decrease. Reduce the degree of radio-energy. But as an extra interaction that are not there in the middle of the problem. It's correct. The rest of this is that to introduce some I don't have a ground signal. It's proportional to alpha prime into 2 mu f mu alpha f alpha. For two of these f's and the linear is from alpha, yes? Yes, just the linear is from alpha, yes? And that is the rest of this. Why is that? Why is that? Let's see. So, this is effectively L mu A mu minus L mu and similarly L mu alpha L mu A alpha minus L mu. And why do we get this structure of interaction? Let's see. I have to say I have to check this. I'm just going to use the question scheme. As for prime, this is very much important. No, the other prime is that that could occur. And that is the other prime. We see that. We see that and you say that this thing would just be non-zero. Yeah, you're probably right. I'm sorry, I have to check this. You're probably right. Yeah, it's probably right. Okay, let me use this in exercise. Let me produce this thing. It's going to have a 3D edge. It's going to have this contract. You're almost 79. How are we going to get this structure? How are we going to get this structure? Let's see. No. You see, I don't know this. You see, you see, this is supposed to be incompatible with ui, that's stressful. We got q-ray, we got an anti-selective. And that's probably just we got q-ray. Yeah, so maybe this thing doesn't come with ui. No, no, no, okay. I think that's what I think it's going to be the same with ui. Because of that. Exactly. Because the symmetry part is symmetry in ui. Exactly. So even if you're not there, this would just happen. So it's consistent with this kind of thing. This thing will be reduced by the term of being the such a term effect in the ground. I have a check, but it's something like this, but I do the job, check it out. Please do the exact same thing. Okay? And try to convince yourself that what you get from here is exactly what you expect from the ordinary industry. And what you get from here is that the term is not inconsistent with string theory. It could be a gauge theory of learning. Because if all the derivatives are large in the ground, this term is much smaller compared to this. This has two issues there. One is the power of alpha, compared to this term. So even if the derivative is large in the limit where the length is greater than large in the U.S. ground, they are there, but they matter only for configurations of varying of length scales or of energy scales or all of the string scales. If we have a varying of length scales large compared to string scales and it's lower compared to string scales, that term is negligible compared to string scales. It's the scale. No energy to get back to string scales. There are connections around it. It's very even string scales. But we know it's very simple. It's not the only girl. This is the effect of action that comes from some. You know, some of it over all our string theory. This is one down in the effect of action. The first to effect of action. And we have independent ways of understanding that the string theory is very finite. So those issues aren't on the issue. It's just basically the same. When you're starting to encroach in theory, you have the gravity. And then we have high level of interest than the gravity but the whole theory is finite. The regularization that makes sense. We should expand it out in the lattice space and increase it. Increase it higher and it helps. The whole theory is nice in theory. And the manager can start a very good one in the string theory in theory. It should do something like e to the power minus there is squared and there is 4. That's the fact that the suppressor is high. It will go worse and worse as you're going to higher levels. I know you're very normal all the time. Yes, it's certainly not easy to make consistent theories of gravity, especially gravity by hand, doing something. So string theory manages it in a very beautiful and intricate way but I do not include it like that. But just if you want a rough analysis of it, string theory isn't all about gravity. So, I mean, it's not not local. But at least there's no string theory. There's no such thing. What do you mean by the gravity in theory? I don't know anything. It's to move on to the discussion of four points. So we're going to just move on to the discussion of the let's see how it will happen. It's actually the first formula of a string theory that is string theory. And you get to understand this beautiful formula that likes to let it pass to the world. Okay. So, it's not normal that I mean, there should be the other type of string theory that's possible in one sense when we work on it. Even if one of us has one. But there's really direction to all of us. Yeah, but they don't appear to have three points. They don't appear to have three points. Three points. Yeah. We're going to have to get the three points. We've always got to get the three points. It's exactly the two points. There's no further reason. Of course, even the three points would get the electric quantum mechanics. What's the difference? Is this the most general string I know? No, this isn't the most general string but, oh, you're asking this. Just from a word, okay, let me get this one. Yeah, I see that any little more, more Fs would necessarily come from your only hand. Yeah. They have four Fs. They have four Fs. Yeah, so maybe you get this one. Basically.  Yeah.  I mean, it's very... Okay, now, in this question when we do this, the same thing in so many terms that could appear wrong. Okay? But we understand a lot of that from the first time we've seen this. From the first time. Okay? But certainly, if you were just looking at things without any differences, that could appear wrong. Yeah, but there's obviously any string there. If you have a particular answer, although if it's a particularly particular, it's something particular. So, that's a lot of the aspects. Let me ask. Okay, and this is what I think is, whatever it is, the question is, I mean, to leave relations between this and other things. Yes, yes. That's what we do. Right? And that's the next time we're going to do it.  But let's just this one, you know, let's just see the interaction. Okay, now, let's just quickly move on to the four points. The four points that appear wrong. Now, as you remember from our first question, it is usually possible that in a delegation that is not just writing a number but in case. Right? Because on the version, you mean? Because you can fix three of the four vertices of the fourth one, is where it reaches. Right? Let's set up the aggregation to that. So, we've got let's say four vertices of the fifth in one at x, it's a lambda one. It's pi k two dot x, it's number two. That's all. I'm thinking about it that for formal reasoning, we can set those three points to be wherever we want. Let's set them to be the main interests. Let's set them to y one equal to zero, y two equal to one, y three equal to zero. So, this y i k is from C's we have. y one two, y one three, y one four, y two three, y two four, this is the number of C's. Okay. Now, anything that does not involve y four is just a number. So, the only, so the terms that involve y four are y one four, y two four, y three four. And then there are some numbers. The numbers are like y one minus y two is one. But anything equal to y three is infinity. We have infinity to the power. Okay. So, that's this is also infinity. It's infinity to the power. You know, more precisely, it's y three to the power. Three is the limit of y three. Let's see what we get from the momentum market. From the momentum contractions we get, we have e to the power k one three eight and so on. We got this product of i, right. We got e to the power two alpha prime, k i dot k g. From the point. Let's see what k i dot k g is. What this k i dot k g is is this. Sorry, y and j. Let's just add those terms that involve y three. All such terms are infinity to the power two alpha prime into k three dot k one plus k two plus k four. And we say this as well as y three is basically y three to the power. So it's infinity to the power of this. So that's the infinity to the power minus k three squared. Minus k three squared was minus one alpha, right. So that's infinity to the power minus c. That's what I'm talking about, right. With the c's, only three are fixed. Only those three c's are here. Why won't you y two three, y y one two, y one three, y two three. That's how it was formed. Here we didn't get it, guys. And what we did get was infinity. So if you take it over, that doesn't have c's. So I'm talking about that. So there's infinity to the power minus k three squared. So what's left is all the stuff that involves y four. So what do we get? The stuff that involves y four is okay, once again there's a y one two. That'd be okay. So the y four into k one dot k k four and y four minus one into k one dot k two dot k four. Okay. Now, there could be some time. Due to that decision, s is equal to k one plus k two, that would be squared minus time. And there could be four components. It's kind of non-dimensional. Alpha prime t is equal to minus k one plus k three. It's where alpha prime u is equal to minus k one plus k four.  So that object there is so our answer involves the d y four y four to the power minus u minus two y four minus one to the power and then this would be similar except we have x minus x. So that's the kind of detail that we have to do. That's the kind of detail that we have to do. If you want to solve this problem. Now, one last thing that we stopped at was this y four is integrated over what? It's integrated or minus integrated. But you see, what's really important to stop. We have y one, y two, y three except that zero one is this object and the integral here from in this way from zero to one which was the same answer as the integral here to one thing to delete just because we couldn't prioritize it. The only way to integrate is what? Order the various vertex operators inside. So instead of returning this integral as going from minus infinity to infinity we can think of three different integrals. One from each range minus infinity zero to one and one to infinity and the part that is from minus infinity to zero which has the same answer as one to zero to one so why do we integrate vertex operators in a product? Actually there are six different combinations. Each of them is proportionate or very similar in terms of integrated vertex operators. So the six different combinations are the following. Y four here or y one, y two, y three and y four here or y one, y two, y three and y four here. Y one, y two, y three because of the thickness I could do it. We could have had y one, y three, y two and y four y one, y three, y four and y two and y one, y two, y three and y four. y four go in this range in each of these paragraphs and each of them will be less than one to zero to one where the appropriate vertex operators desert desert. So what I'm making and I'm going to make that quite probably just to give you the answer is that the answer is some of six different terms. Each of which is an integral of this form from zero to one. In different terms this power is used in excess of energy. In different terms and k one dot, we have k one dot k four here and k two dot k four in these two rectangles because we were going, y four was going between one and two. So k one dot k four and k two dot k four when y one goes between two and three this would be replaced by k two dot k four and k three dot. What each of these is done is used at six to seven rectangles each of which is the same form compared to x k. Now the other thing to remember is that by the variable change y four goes to y four minus one for one minus one four we should have discriminated. So this is symmetric in the interchange of x and u on the interchange of one and two. So that's not what order this through. It's a choice of u and x. So the three times are not the invariance x, t and u okay and we get six terms two, two of which will have s and u, two of which will have x and t, two of which will have t. And what we have to do is the thing is we have to keep track of which diagram gives which dependence so that we keep track of the lambda. So that's two is one. Two is one. Let's find out. This diagram you already see four which is u and one and two which is four and two four and two which is the same as one and three. Yes, why do I have to do this? This was one three. Yeah, this was one three. You and me. What about this side? This is now going to give us two and four which is the same as one and three. And two and four and three which is the same as one s. This is going to give us four and one so that's u and four and one so that's u and four and three that's s. So that's one and two. This has y four, y one by three which is u and s. This has y four by two the same as one and three that's t and y four, y three is s and I'm here so y four, y three which is s and y four, y one which is u. I guess we've got two s's now let's do this again. y four and y one which is u y four and y three which is s this one is y four and y three is s and y four and y one which is u and this one is this is two and three. Three and two. So this one is y one, y four which is u and y four and y two which is t. We've got two s's two s's so that thing one and four is a u one and four is a u yes this one is u one and four is a s one and four is a u u t okay, we have two ut's ut ut we have two s's s t s t and two u s's okay, that's the problem okay, so now, suppose we write down this the answer, whatever this is as e of s t s t s this thing is multiplied by what? trace of, how do you look at those that give us s t s t s t lambda one, lambda two lambda four, lambda three plus lambda one, lambda four that one is lambda three lambda four lambda two. So the two diagrams give us s e of ut and then the second order is two diagrams and give us u t so u t at least two we get one, four, two, three plus one, three, two, four a of s u which is one, two, three, four plus s of one, four, three okay, we continue next time because we have to derive the differences actually, for t t, that's wrong okay, we continue next time it's amazing if we want to know how we