 No, they won't give you back pay and they won't let you just take off. You have to have a hangnail All right Leon Leon Leon's not here today Mitchell Mitchell I'm gonna be talking to you specifically on this stuff. Okay, okay Bronn Bronn actually didn't show up Reinhardt Reinhardt if he can't answer the question, you know if he looks if his eyes roll up in his head, then you must answer for him Okay. All right. This is a short review There's a question of when you should be using single story out of that table that shows single story columns only Those are far if you really have a single story frame. They're good enough. They work nicely They're certainly much quicker than using the G's if you really had a single story thing You could still use the G's on it for a pin. You'd use a 10 in the G You wouldn't say this is fixed because you'd have two beams coming in and a single column coming in So it would have a G you can use that But for single story things you use that table where we just have the single story stuff Do you remember the table number on that? Then they they probably know What what page is that on 16.1 dash 511 that's that table But for multi-story things you can't just you can't use those K's You got to use G's on everything from then on now We think maybe there's a homework problem where it's Just a single story and they use G's anyway. Well, that's acceptable. It's just more work So they don't usually do that This particular one here was if the end is fixed you get a one I know you could theoretically get a zero, but they just don't like that idea and they don't believe you They do have a thing and there's is if you have Theoretical proof that you really have a smaller number you can use it, but it goes in with your other calculations and If it's pinned you ought to be using an infinite But the truth is a better number is a ten for you and it is permitted And it's really almost impossible to get something really pinned so it's safe There's the where those notes come from The one we were working with had a couple of had a member column AB had some Supporting girders good guys supporting girders good guys some bad One bad column coming in at the top two bad columns coming in at the bottom We wrote down the moments of inertia out of the tables and here were the length so we could quickly calculate G Now this is not the same frame, but it's typical the solid ones are the The members in the plane of the paper there are some other ones back here They have some bracing on the sides and no bracing as you look through the building in this direction And so dent Lee this thing says Members are oriented so that bending is about the strong axis and the reason why is the architect Wanted wind is in here and So he didn't want any bracing We told him we'll have to weld all this stuff up the members will have to be quite a bit stronger Because now then your wind loads are gonna have to get down to the ground through these moments. They said they said it's worth the price Yes, sir, sorry, I can turn this on but kind of hated to I Don't know if I can holler any louder. Well, I have to do that a click at a time There we go Hello. Hello. Thank you for telling me. Hello. Is that okay? Does that make a difference? Okay The reason why is when you put the wind loads on the side all the forces have to go through these members So this is a top view of the way that column is positioned They're all positioned so that they get bent about the strong axis So here is a similar case. You'll notice there is the member If you looked at it from the top Here's the member if you look at it from the top. Here's the top flange. There's the bottom flange There's the web if you look at it from the top All of them in the front and the back are all situated like that Now well, you may say If you load it from this side, why don't we just do the same thing just get the strength About the minor axis this really difficult to get the strength you'll need for when loading about the minor axis So whether they like it or not somehow or another they're probably going to have to put bracing To stop member D1 from moving back into the board and they do that with the bracing if it wants to move back It throws this member in compression that member in compression puts this member in tension then the load gets back down to the ground and Member a point D1 doesn't move significantly to the back. Yes, sir Tubing is fine for columns. It's not popular. It's not easy to connect to You say, why don't we just weld it? Well, yeah, you can weld it, but the walls are pretty thin When you're talking about a wide flange, here's the top flange. Here's the bottom flange Here's the member coming in from the side It's pretty easy to weld it on to there And then if you need to you can throw another plate and weld it to the web to get that force on through to the next one Tube's gonna be a problem because you're gonna have to get a little pity guy to go up in there and weld a plate inside that tube So they're not as convenient Okay This is a picture of what this column a kind of looks like, you know I don't have the ones going in the back, but basically somebody's coming in here It's welded nicely coming here welded nicely coming here probably just goes all the way across the top Or it could be welded nicely There can be plates in here and very likely may have to be To get that force in the flange on through into the web rather than bending the flange Here's one down here where they did they just said look it's gonna be pins not gonna be fixed It's not gonna be a moment-resisting connection You'll notice it's just sitting on to a bolted angle So it's just pinned on the ends Here's joint D3 is seen from the top. Here's your column You notice I told you that it was situated that way the flange is on the top flange on the bottom the web This one is a moment-resisting connection So it was welded nicely this one there was no real strength about this axis anyway So they just put an angle in there bolted the angle to the web and bolted the The beam to the Angle so it won't fall off Now this is what it looks like weak axis buckling In other words, this is the same column The weak axis is about this axis So I'm going to show you that this column went back into the paper I know it it may look like this is right left right, but it's not if you look closer. It's back Forward back Say there's no difference. Well, you got to see the difference buckling about the weak axis This piece of the column is coming out of the plane of the paper This piece of the column is moving into the plane of the paper and this is basically pinned pin So it'll basically be a 26 foot column. This will be a 20 foot column. This will be a 14 foot column That's KL about the weak axis About the strong axis I don't know I got to have all my G's and get my act together on that but here is strong axis buckling in this case because you didn't brace it this move to the right didn't move 22 feet or anything, you know greatly exaggerated move a couple of inches and Then this isn't really perfectly fixed because there's a G at the top and a G at the bottom and The columns coming in and the supporting girders coming in influence the G and the same way with this one But this column a 0 a 1 has moved to the right in the plane of the paper I don't know where to write that down and Then this one buckled and you see how that's kind of buckled like that. This is a nice pin and But this supporting pair of girders kind of brought it back up vertical, but not quite And I need a G at the top and a G at the bottom to tell me effectively how long this column is And then I guess that's the same picture So our solution to the original problem before we got off on that tangent we had supporting and Wanting to buckle people Supporting not wanting to buckle versus people wanting to buckle with their eyes Here where there here's this here it is right here summarized Here are your eyes and your links out of the table 171 inches to the fourth 12 feet long 171 inch to the fourth 12 feet long this one was 20 feet long and 18 feet long so Bad guy coming in 171 over 12. It's only one of them Two good guys trying to help you out 88.6 over 20 plus 88.6 over 18 Gives you 1.52 for G on either end you can enter this GB or GA doesn't matter Then for the other end you have two 171 bad people coming in causing you two times grief Then you have a hundred ninety nine over 20 and a hundred ninety nine over 18 for a 1.36 for the other ends G Then the alignment charts Here's the alignment chart. What is this? This is an unbraised frame That doesn't say unbraised, you know, I don't know which one this is which one is this unbraised or braised How do you know so quick? Says uninhibited side sway that is right, but I don't remember exactly unhinged uninhibited what that means or not Could you tell from the picture? See how this moved over to the side that's an unbraised frame Not one of these that's braced So we enter the two G's that we get we find out what the K is 1.45 We multiply the true length of the column times 1.45 and that's its effective length That's if it's elastic now once we get in to pick in columns We're gonna have to see whether or not the column is elastic or inelastic once you take a first shot at what the column ought to be If it's inelastic we we get some relief From this grief about how long the column is it's really a little shorter So Here he gets this he gets the 1.45 that we mentioned a second ago So our lambda is k L over R our K out of the nomograph 12 foot long column 12 inches in a foot the rays of gyration for the column that we tried See he is proposing a 10 by 33 column Got 199 inches to the fourth Well, I'm looking more for the length here the length is 12 feet So there's his 12 feet times 12 inches in a foot divided by the radius of gyration about the strong buckling axis 49.83 we have a personal break point on There's a familiar number 50 ksi steel of a hundred and thirty three hundred and thirteen So we are well to the left of 40 of a hundred and thirteen So we are in the inelastic region. So we have yielded fibers And so those fibers are not as potent as they say when this column wants to buckle and He starts feeling popping out to the side just a little bit He told us that he was a beast. He was going to roll that thing whether or not we liked it or not Since it is so highly compressed before he really gets around to even testing buckling He has a bunch of fibers that are yielded the minute He just starts to roll those outside fibers yield and he no longer is as Effective at buckling as he says he is now. You don't have to take that into account If you don't mind spending a lot more and getting a lesser grade of course but if you have a Slenderness ratio down in this inelastic region you get a towel Now here we go for the Factor load I'll take his word for it that that was his dead in his life We find that the No, the stress in there the piece of you how much stress would really be in there Would be the load over the area of the shape is twenty seven point seventy nine And you can go through the equations nothing wrong with it except we have a set of tables for towel Tells us what this stress will do to a piece of trying to see where the stress is a 990 just be a 50 ksi steel and It says the towel is twenty seven point seven nine twenty seven point eight if you go list that in the towel He'll tell you exactly the appropriate towel that you should be using Twenty seven point seven nine you can interpolate or you can take the worst One the one that doesn't let you shorten the column as much Turns out if you interpolate it you get point nine eight eight. You're only going to pick up one point two percent Shortening on the length anyway, it's hardly worth anything Here's where things get good if that column was a little shorter and holding up a lot of load Or if it was a little heavier column where the stresses in there were pretty low because it was a bigger column Then you might not you could multiply those G's by point five nine Multiplying those G's times point five nine would bring you Way on down down below a one point four Or whatever we got I guess that's the right picture that goes with what we're doing So for point nine eight eight. He says you get point ninety eight seventy seven. I agree therefore the elastic G which is what the tables with the Equations are good for these equations are only good. You'll notice e tangential and e normal or not in there To turn the elastic numbers out of your Tables times tau will give you the inelastic G And so instead of having a one point five two you get in or a one point five I don't think you can hardly put it in the table that close and this one is one point three six Reduces down to that and he says from the alignment charts you go from a one point Four five to a one point three four. No one point four three My guess is he didn't really get that out of the table My guess is he got it out of the equation. Remember we had an equation for tau. It's just the table the picture was convenient But you do get some reduction Now then he says very short because the support conditions normal to the frame Case of y can be taken as a one Whereas we talked about this case right here Where we are now studying buckling about the strong axis that's where there's our G values our G values Talking about what happens due to this load being applied to this column about the weak axis also has to be checked that was weak axis buckling and weak axis buckling was Back into the page out of the page back into the page. See how that's pen pen pen pen pen See how strong axis buckling Double did that one go see how strong axis buckling that's not pen pen. It's not even fixed fixed It's something in between G kind of guy Now that's as far as he takes it. He Make sure that things to work He could go ahead and find out how much load the thing will take Check it out, but he that's as far as he carries it. All right Torsional flexural buckling we don't cover so that's everything we got to say about columns or beams You're gonna need to know your 221 stuff where you first learned to share in moment diagrams so you can get the moment diagrams in these structures Then I have to also know your 305 work where you reviewed share moment diagrams So you can still get the maximum moments and also things like MC over I for bending stresses In 305 we limited you so that no fiber got stressed above The yield stress because our equations fell apart including the MC over I equation It was it required that the fibers stay within f sub y or the equation failed or it came apart Bending stress used to use this symbol will be using this symbol now it's a little hard to get used to and Sometimes you look at this and forget what it is But it's the symbol for the actual bending stress in the beam when you apply a moment to it You go to M over I oversee see what M over section modulus we use the section modulus a lot All of the notes you ought to read through here This is what it does is what the chapter is far here where you'll find that information and the specifications Here is the same equation that you and I wrote down earlier Except you wrote down p sub u has to be less than or equal to P sub n and then because you couldn't guarantee me that was the number. I was gonna get You said okay, well, I'll drop it down an appropriate amount so that I cover and can guarantee you You will get this this was called the design load This was called the ultimate request for load. I believe the specifications wrote that down p sub r required Because they wanted it to go for a loud stress people and for ultimate stress people and But we never used it because we don't ever do a loud stress This is what p sub r means to us means p sub u Then they had a different equation for there I think they used some kind of a thing looks like that for their factors Here is a typical beam You can either have them rolled out of steel They roll them for you if you can't get one big enough. You can have one made These things will be subjected to Moments the moments may do bad things to you among other things the whole shape May buckle but instead of buckling like this They buckle like this You probably felt that happen You take a yard stick you support it on both ends you put a load on it and you just can't In other words when the thing is standing up you can't get the thing to fail Every time you try and push down on it and try and break it it pops over to the side That's because the top half is in compression. It thinks it's a little column And so the little column goes over the side and the bottom piece is in tension He doesn't let it go too far. He says what is your problem? He says I don't know. I just wanted to pop over to the side Also, the flanges themselves may in in our case we we run them right up to the limit The flange may buckle not I showed you flanges buckling in columns. They can also buckle in beams You will be wanting to know how far this thing sticks out That will be B sub F divided by two well on each side and You will want to know how thick it is because that's what keeps it from buckling thickness of the flange So you'll need numbers like that out of the table so that you can tell the Tendency for a flange to buckle or for the whole column to the whole beam to buckle bending stresses and the plastic moments What we do is is we will have a beam here it is from the side You will put a little bit of moment on it see that little bit of moment that will cause some compression And it will cause some tension This is f sub y right here. You'll put some more moment on it and You will do this to it. That will be what we call first yield Then you put some more moment on it and this little fiber here says I've had it. I'm through I can't take any more. I won't dump what I've got like a glass beam wood, but I can't take any more First thing, you know, the stresses will look like this Then the stresses will look like this and then all of the fibers will go into the yield range And then that's the limit Really the limit because as you know Here are all your little fibers now you got some more But was there's nobody willing to take that part We're gonna have all the fibers reach f sub y and then we're gonna give it up If you go in this region you get so much deformation the beam just isn't hardly useful anyway Although it's there if you need it bad enough Beam loads find the moment Stress at this point if you wanted it stress at the maximum point was where the fibers the furthest from the neutral axis Rending stress equation Here he's showing you what I was just talking about got a concentrated load with a moment on it at a hundred kips Here's what the shape looks like and the stresses are this is 50 ksi the stresses are at 30 ksi and Some of those little fibers are up here at 30 Some of them down here or down here at 15 some of them here or down at zero run it up to 167 kips and you get the yield stress on one fiber only that's called first yield Put some more load on there the stresses start Evening out at f sub y because as you put more load you cause more moment These fibers said I'm not taking any more moment or stress So they had to dump it off to the fibers below them. This is where we're gonna work We're gonna have every little fiber working for us to the full extent possible Same picture Here C is f sub y. Here's fiber abc. Here's fiber a's current stress B and C As you reach first yield fiber a reached 50 ksi Fiber B is at two-thirds at 50 C is at one-third of 50 Put some more load on here Now then fiber a is reached yield Fiber B has also reached yield Fiber C is trying to escape. He sees it coming, but he can't get away Finally a b and c are all on this flat part and I have everybody at f sub y full plastic moment You can't do that too many times the truth is we don't ever expect it to happen It probably won't happen. It's never happened to this building been here. I don't know hard years never happened Probably won't happen now because it'll probably happen to the building next door first. It's in the way so tornado comes that's their problem But if it did happen too many times it'll it'll fail or if it even happened once There was a building in Dallas got hit by a tornado and it didn't fall down But it was really twisted. It was really bent out of shape. I don't know what they did with it yield moment is the moment at first yield plastic moment is equal to the moment at all fibers Equals f sub y kind of what happens when you look at the beam This is true This is next load kind of true This is incorrect because there's there's no way that you really get these stresses. You can't get beyond m sub y What happens is as you put more load on it a longer portion of the beam starts getting yield stresses As opposed to the earlier loads So what happens is when you put a little load on it It just deflects like you and I are used to seeing in our 305 problems Once you start getting up where this yield moment is has occurred you get a little bit kind of a kink not much First thing you know when you really reach a significant portion of this thing is reach the yield Then you get a big old kink and you get very excessive deformations You're supposed to know how to get an elastic neutral axis I'm going to show you a beam where It has fully yielded Here is a beam that is fully yielded it started out with an applied moment the top thing was 25 and 25 ksi and linear down It's got to 50 and 50 and linear up and down Then this one stayed at 50 and this one went up to 50 And this was half a 50 and 0 then this one went up to 50 50 50 and this was half a 50 And this was 0 and finally this is 50 50 50 50 50 50 50 50 50 Now you never get this little piece here really, but it's not worth worrying about it's so close to the neutral axis I'm not counting on much moment from it Here's your stress diagram for the way you and I are going to do business F sub y on the top area in compression F sub y on the bottom area in tension Looking at it from the side Now it looks kind of like they got the area in the top equal to the area on the bottom, but I don't care If your shape really looks like this Then you tell me how much is in compression. I'm going to yield it in compression You tell me how much you got I'm a yielded in compression You tell me how much is down here, and I'm going to yield it in tension And I'm going to tell you that stress in compression f sub y in compression times that area Plus that area times f sub y in compression Minus this area times f sub y in tension equals how much? Zero because otherwise it goes down the road It's not an equilibrium if you don't have this compression force in the top Whatever that area in the top is Equal to this tension force on here then your then the beam is not an equilibrium pretty picture Happens to be a wide flange Got a compressive stress all on the top got a tensile stress on the bottom The distance between the centroid of this shape whatever it is I don't care you should be able to find it and the centroid of this one wherever it is That's where the compressive load is centered. Here's where the tension load is centered I don't care what this is. We should be able to solve for it. Whatever that distance is Either one of these times the moment arm between them. This is a couple. This has to equal this That either one of them times a would be the plastic moment Now since the compression force minus the tension force has to be zero so the two forces have to be equal Since the compression force is the area in compression. I don't know what it is. I don't care I think I can find out times yield is equal to I don't know what's left But that's the area in tension times yield that says the area in compression has to equal the area in tension That didn't used to be the case Used to be the case when you solve for the elastic neutral axis for this thing here You didn't it didn't have 20 times 6 equal 20 times 6 and The neutral axis was there It was somewhere down in here And I'll show you an example But since the area in compression is equal to air in tension if you say my beam is a 20 by 6 and the web is a 20 by 6 that beam once we're doing business where all the fibers are plastic That neutral axis has to be here so that the area in top on top times the yield in Compression equals the area on the bottom times the Area on the bottom times the yield stress income in tension Without that being there again, it's not an equilibrium Here's a case see I knew where the plastic neutral axis was on that one because 20 by 6 and 20 by 6 That's easy to do Here's one is 20 by 6 and 20 by 2 Now I don't know where the plastic neutral axis is but I see you took a lot of this area off So I know I'm gonna have to go on up in there and pick up some area So that this area that's lightly shaded is equal to this area that's heavily shaded down in here Because the two areas have to be equal Here's how you should remember how to solve for an elastic neutral axis Draw a reference axis at the base How far it is to the elastic neutral axis Was a1 y1 plus a2 y2 divided by a1 plus a2 Here are your calculations if you have forgotten Here's the plastic neutral axis Since this is a 12 by 2 and a 12 by 2 The plastic neutral axis is 12 inches moved up from nine and a half inches As you were yielding the fiber the the neutral axis was moving Final position right there is you're back up ready to go Is he okay? Is he sleeping? No, all right. I can't see from there. He's got that hat on and he's behind There he is Okay Okay For the built-up shape determine the elastic section modulus s and the first yield moment And the plastic section modulus. That's what we call the plastic section modulus The elastic section modulus is listed as s in your 305 book The plastic section modulus. This is yield. This is listed the plastic section modulus is listed as z First off here is the elastic neutral axis. It's in the middle due to symmetry Here's the plastic neutral axis for a wide flange. It's in the same place But not because of y1 d1 plus y2 a1 y1 plus a2 y2 plus a y3 y all that stuff This is in the middle because this area is equal to this area Yes, none Because you and I are going to crank at rascal right up to plastic no matter what I don't care where it started It's ending up where a1 a top a compression is equal to a tension I mean he may not like it. He'll scream and holler, but I'm gonna pour it on Here are the calculations for the moment of inertia from your 305 work You should of course be able to get that you will have to get that sometimes Sometimes you'll need to know The elastic neutral axis and the elastic moment of inertia The reason being is sometimes the The requirements in the specs call for this number not to be some number not to be exceeded. You have to still be able to do it The elastic section modulus as I oversee Check it out And the first yield moment the moment at first yield would be f sub y times the elastic section modulus 446 kip feet if you're in a last if you're in a loud stress person That's probably where you're going to stop with us some factor of safety associated with it That's the first fiber that screamed i'm hurt. I'm hurt and everybody brings a little ambulance and say well don't hurt anybody else We don't do that Answer is elastic section modulus was 107 305 446 kip feet B because it's symmetric A on the top is equal to the area on the bottom so we don't have to do much work there What we do have to do is we have to Make every fiber go into compression To f sub y and every fiber on the tension side go Into tension at the uh below the plastic neutral axis Now he solves for the centroid of the shape And then solves for the force He takes eight times one multiplies it times the stress Then he takes this one times the stress And then he adds those two forces together and gives it a moment arm of how far it is from there to that centroid That's really a lot of work. I mean it may show you something. I'm not sure Well, let me just go on down here and let's talk about this He says he calculates the distance between the centroid of a t And of course the same centroid on the bottom t He calculates where the centroid is And he tells you how much area there is and he multiplies the area in compression Is the total area the whole shape over two that's the area in compression times a That gives you that force times this moment arm And then he's got another one A This is a over two times a that's right. That's the distance between them It comes out with what they call a plastic section modulus. Now. That's the first time you've seen that calculated And just let me show you which makes a lot more sense to me Let me make sure this is the same one. Now. This is a different one Well, that's okay. I'm going to refer you to him to calculate z He calculates the centroid Of a t calculates the centroid of a t It's already got it from this one and then he multiplies the area Over two times the distance between them Same calculation It says five one is that five one turn to find out this is his five one He's got an eight. Let me see what five one looks like so I can That is five one. That's eight by one Oh, this is a different one. Okay, that's okay. So I do have the calculations for him There's his shape. There's my calculations to get z all z really is is He's taken this total area times the distance to the combined centroid That's the same thing is just taken this area times the distance to his centroid Plus this area times the distance to his centroid That's how you found the combined in the first place So this is what z is z is equal to eight times one times Six and a half That takes care of the z for the top flange plus Six times a half times Halfway up is three And this is I want to add multiply times 50, but that's z for the top half You want both halves you need another one on the bottom If you multiply times 50 you're going to get uh The 50 ksi in there at this time, you know, you still have to double it when you get through Here is your z over two a1 y1 plus a2 y2 Here's f sub y for the top half 61 inches cube times 50 ksi I guess he was calculating the moment And then if you want both halves you got to have 61 times 50. There's one on the top and one on the bottom Gives you a z for the entire mess 61 times two Times 50 gives you the moment. That is what he's calculating That's the m plastic m plastic is z for the shape Your plastic section modulus times f sub y See if I got a Little piece of paper here Here's a new shape This is two inches by six inches This is 12 inches by one inch Where's the plastic neutral axis? Back up back up back up back up Well, no, where is it you tell me where to put the line for the plastic neutral axis That's correct. Why is it located 12 inches from the bottom two inches from the top? Because here on the top is 12 times 12 and the here on the bottom is 12 This is z for that shape you can do his way you can find centroids of things Z is equal to six inches times two inches Times one inch That's for the top Now the bottom doesn't look like the top, but it's the same idea It's how far it is off of the plastic neutral axis it's One inch wide it's 12 inches deep and how far is it centroid from the plastic neutral axis Six inches that's z if this is a piece of 36 ksi steel Then m plastic for that shape Six times two is 12 What is this all add up to 72 total 84. Thank you 84 inches cubed times 36 kips per square inch And do you have your calculator still going there? Thank you 264 this should be bigger than that shouldn't it 2024 3024 Okay, that's inch kips divide that by 12 for me if you would To well, maybe that's what he already told me because he knew that's what I was going to be looking for kip feet Just that easy now you can do those centroid things like he's doing and I think he thinks you understand it better that way But the truth is this is a force in compression on the top Which is this area times the yield stress This is a Area on the bottom if you wanted to find centroid you could multiply six times two Times 36 that gives you this force Then you multiply this area times 36 gives you that force The distance between them is seven inches. That's what he's doing And you'll get the same answer But I think this is just a heck of a lot quicker That's all you and I are allowed. There is more available because the stresses still could go up to f ultimate But we just that's beyond our safety. We just don't feel safe with that All right. See you next time Yes, sir To calculate the design strength of a column that's right. We had a break Thank you I think you're talking about members that are in tension because all the columns have the same fee So now you're talking about whether or not you were talking about gross section yield Which is pretty easy to predict and pretty reliable and doesn't have a lot of variation 0.9 Or if you're talking around holes drilled in things causes a lot of variation In which case the fee was 0.75 F sub y No times the area gross Now you used f sub y times a gross area And you were permitted to use the ultimate On the area net or area effective Because it was such a short little piece. It didn't have a lot of deformation in it So we didn't didn't care too much about that. We had to make sure it didn't break Breaking was such a serious thing That we made you use a 0.75 on it Okay, but again No, we we'll talk about that later But we haven't started yet the truth of the matter is there are some places we'll let you run certain things up to f sub ultimate Probably around maybe the ends where they're in shear or something like that The equations that we're looking at now are good are good for inside the beam But what happens at the connection point where you have now connections are designed separately Totally different items Right chapter eight. Maybe Okay, take a look in the book See where they see where they're at Oh there Okay Before you get to use it. Yeah, this is going to be nominal Sure No, I you can but I'm not gonna I've got to get this put together