 All right. We were working on bolted and welded connections, and up to now we have worked on simple connections. A simple connection would be one that the load passes through the centroid of the weld group or through the centroid of the bolt group. It's not a simple connection. That means it's also got some vending moment applied to it. There's two ways that you can apply a vending moment on something. There's the plate to which you're going to bolt the thing, like that. And then you can put a load on it, like that. And the surfaces that are shearing are called the feying surfaces. And this load that's out here off of the centroid of the bolt group, first off we'll take it and we'll move it to the bolt group and pretend it's there, and hope maybe nobody will notice. And then we'll be able to say that the shearing force in each bolt is the total force divided by the number of bolts. But when they come over and try and put this other piece on there, they say you're going to have to move that load, it's in the way. So then we will pick the load up and we will put it over where it belongs and on top of the direct force on the bolts, it'll now cause a moment or a torque. And that force, that moment will tend to roll the bolts about the elastic centroid for a little while. After a while some of the bolts will go plastic and then this centroid really won't stay there. First thing you know the whole thing will roll about some other point but that won't be elastic. And the other thing that you can have is, here is the connection and again here are the bolts and if you load it like this, then the bolts will be subjected to some direct shear and some twisting shear. You'll notice in both cases the bolts are shearing. So you'll be able to take any forces straight up and down on the bolts plus any forces due to a moment like that bolt lives here and a moment like that would cause a force like that on the bolt. You will be able to add the direct force where you put the force at the centroid plus then you added the moment it caused this, you'll be able to add those vectorially and then you can tell me how much force is in that bolt. And the other thing is if you have this thing bolted to say the flange of a column and you put the load out here. Well now then you're not actually twisting that thing. You are direct shearing it, so that's the first component of force we'll have and then the second thing you'll have is you're tending to peel this thing off. So you're putting these three bolts on the top because the load is out here positioned like that, have no idea if you can really see what that is in two dimensions. And then you're tending to peel it off the top bolts are being placed in tension. So there's two different kind of actions that you could put on a bolt when you have a non-simple connection and you have everything in sight and you just have to do them all. That's not usual. Alright so first we'll take the one where the load is placed on the plate. There's your bolts. Put the bolt bolted to the flange of the beam and you're going to put the load out here somewhere and you're going to cause, first you're going to lie, you're going to pick it up and move it so it goes through the centroid and that's going to do this to the bolts and then you're going to put the moment that also goes on there because the load didn't really go through the centroid. It went out here P times E times the eccentricity, that'll be another shearing force so we'll be adding those two shearing forces vectorially to get the final answer for the bolt that's most highly loaded. So for an example, here's a load P in your text. This is out of an older book I can tell because I don't really have page numbers on it but the pictures are all the same they're just in a different position in your book. Here's the load P. I have arbitrarily decided that this thing is placed at a 3, 4, 5 angle so I can do number work so you see what I'm doing. I have taken the load from its true position right here, came about that close to the centroid that's what you would call the moment arm, you'd call that eccentricity, that force comes that close but it didn't go through the centroid. So the first thing I'm going to do is I'm going to pick that load P up and I'm going to move it right on top of the centroid. Here that load is right on top of the centroid of the bolt group. The centroid of this bolt group is in the center due to symmetry about two axes. If it was not a symmetric set of bolts like that, I could still solve for the centroid of the bolt group. When I put that force through the centroid of the bolt group, the total force on each bolt is the total load divided by the number of bolts. That's what they call, this is uppercase P for the total load on the connection. This is a lower case P for an individual force in just one bolt. And that is the load on a single bolt due to the loads applied at the centroid equal total force divided by total number of bolts. And the angle at which they act, 3, 4, 5. That's the angle theta at which they act. Then I'm going to come and admit that your load did not go through the centroid. Therefore I'm going to multiply P times E and cause a moment in this direction about the centroid. Here is that moment about the centroid. Here is the centroid. We assume that, and this is pretty close to true, that if you twist something about its centroid, you tell me where it lives, you draw what they call a ray or a position vector to the bolt under discussion. And that bolt would be loaded perpendicular to its position vector. Which means that if this slope right here is a 3 by 4, 3 vertical by 4 horizontal, 3 vertical by 4 horizontal, then this force right here will be 3 horizontal by 4 vertical. 3 vertical, 4 horizontal, 3 horizontal, 4 vertical. That's what makes it at right angles. And then I will probably take, because these are kind of ugly, trying to get the sum of this vector plus that vector. I'll probably go ahead and break this down into a vertical and a horizontal component. And I'll break this one into a vertical and a horizontal component. And for this bolt, I'll add his vertical with his moment vertical and his horizontal central with his horizontal moment. And then I'll take the sum of the squares to get the resultant of the vector. He shows it, you know, just flat slamming them together by using vectors. But it's not easy with all the angles that are in there. This will be the final thing we're looking for. We're going to put the force on the centroid. And for a given bolt, we'll have a lowercase p, which will be big p over n. And it'll have an x component right here. And p sub cy will have a y component. Then this one right here, if that's the one you're studying, it'll have an x component, I've showed it to the left, and it'll have a y component. We'll take that sum, that number squared plus that number squared, take a square root, find the load in the bolt. That's our long-term goal. Now it was easy getting the force in each bolt due to the centroidal loading, but for the moment loading, it's a little more to it than that. Generally, the stress in a solid circular bar that has a moment on it, you called it a torque usually, the same thing, was the stress. Stress is equal to m c over i. But that really was made for continuous media. It was meant where you had like a little d theta here and a d a scratched out. And the distance from there to there was r. That was r, that was d r, that was d s, which was r d theta. That times that was pretty near square, so you called it d a. And you went to the mathematician and you asked him what the polar moment of inertia was, it says integral of r squared d a. Go back to your strength of materials book if you've forgotten how you did that. I don't have all these little areas. I got one there, one there, one there, one there, one over there. And so what I'm doing isn't quite up to par for the theory, but it is on the conservative side. And so rather than integrating across the entire section, I'm just going to take the sum of all the r squared. That's where does this boat live? How far out does this boat live? How far out does this boat live? R squared times d a, which will be the area of one bolt. And called j, the sum as opposed to the integral. Like I said, it gives you slightly conservative results. So the stress would be, and I'm subscripting some i's here, because I think it's important that you know we're talking about a single bolt here. We're not just talking in generalities. This is the stress in shear on bolt number 17 is equal to the moment applied to the entire group of bolts times the distance where bolt 17 lives divided by the polar moment of inertia of the entire connection, sum of the r squared is d a. And you used to see an mc over i, and you used to see in tc over j. It's kind of a mixture there. Take your choice. I put tc over j, where both of them have the units. Newton meters or foot pounds or something like that. And i is anything between one and how many bolts you bought to put in the connection. We'll probably want to try and figure out which ones are the most highly stressed so we don't have to study them all. Here's the distance from the centroid of the connection. The centroid of the bolt group or the weld group to the point where the stress is being computed, namely which bolt you're interested in. J is the polar moment of inertia. Stress is perpendicular to d. We already mentioned that. Force is perpendicular to that position vector. Then the stress would also be in that direction. It says it's really only applicable to right circular cylinders, but its use here is conservative. We get somewhat larger yielding stresses that are actually in the bolt by this. And so that means we're on the safe side. It says if you use the parallel axis theorem, it says if you really want to know j for true, you ought to be adding to this term the polar moment of inertia of the bolt about its own centroid. But the truth is the diameter is 7 eighths of an inch. You raise to the fourth power times pi and then you divide it by 32. It's just a waste of time. Doesn't have anything really to give you. If you have all the same fasteners, then of course the area can come outside and there isn't any need to tag the a with the d. And it'll tell you the stress is the mc over i a sum of the d squared. Now then, breaking them down into components. Shear force in each fastener caused by the moment is as opposed to p sub c, which meant p with the force of the centroid. This is p little force unit force in one of the bolts as opposed to big p total force applied. The force in each little bolt number 18 due to moment is equal to the area of bolt number 18, all the same probably. Don't have to be, but almost always is, times the stress in shear in bolt number 18. But that's equal to the area times the stress in shear in bolt number 18. We just said was md sub i a sum d squared. A times the stress md sub i a sum d squared. So the areas of the bolts cancel out. Equal to m times where does the bolt live, how far out? Divided by the sum of everybody's bolt, all of them squared. Then from that you'd have two components. For instance, if this was bolt number one and this is where he lived, then he would have a vertical component and he would have a horizontal component, which you should be able to find knowing the slope of that line. And you should know where the slope of the line is cuz you know where the bolts are. So kind of a summary now. We're gonna break all these things into components. We already told that p sub centroidal force on a given bolt in the x direction would be p sub x component, well he didn't, we have it either. I guess that's the first time we've done that. Before now all we did is we just found the force in each bolt and we knew the angle. Now then he's gonna say that if you would like to take your p force before you even get started, break it up into an x component and a y component, put just the x component at the centroid. Divide the p sub x by the number of bolts and that'll give you little p in the x direction on each bolt. That's what he does, takes the total force in the x direction, the x component of the original force divided by how many bolts you got. And you'll have to keep track of which way it goes. Same way with a y, total force divided by a number of bolts. There's your total centroidal force. It's nothing but p, but the only reason it's got a subscript c is because you lied about where it is, I need you to know that that's at the centroid. Got an x and a y component. Here's where force one lives, what did I do this for? Yeah, here's where force one lives, d1 lives. There's a centroid of the group, that's bolt number one. Bolt number one, you should be able to tell me how far it is horizontal to the bolt and how far it is vertically to the bolt. And not only that, since you know this, you ought to be able to take that number squared plus that number squared, take a square root and get d1. So I'm gonna tell you that you have been asked back on this page right here to calculate the sum of the d squareds. In other words, bolt number one's distance from the centroid squared plus bolt number two's distance from the centroid squared. Bolt number 37, distance from where he lives to the centroid squared. I'm gonna tell you that for any bolt, the x component of where he lives squared plus a y component of where he lives squared is equal to that number squared. That says that the sum of d sub one, forget the sum, d sub one distance from the centroid to where he lives squared is equal to, forget the sum, is x1, there's your x1 squared plus there's your y1, y1 squared. You got another bolt? Fine, add it to this. This would be two of these and you put in two of these. You got 18, you got 18 of these, there's 18 of those. It says that d1 squared is equal to x1's position squared plus y1 squared and so on. Now if you ask where we're going, that's a fair question. Have no idea. However, we are gonna get there. Now, he says the x component of the force in bolt number 37, due to moment is the personal force in bolt number 37, due to moment his x component is equal to the total force due to the moment in bolt number 37 times y over d. Now let me give that a try. Here's the centroid, here's bolt number 37. For illustration purposes only, this bolt is 12 inches from the centroid of the group in the horizontal direction and 5 inches lower than the centroid of the group. And there's bolt number 37. Bolt number 37, having this position vector, will have a force in it normal to that position vector. And to make it normal, since the position vector has a x of 12 and a y of 5 and x of 12 and x of 12, it's gonna have to have a y of 12 to be normal. And since the position vector has a y of 5, he's gonna have to have a x of 5. And you can see drawn reasonably to scale. That's where the 12 goes and that's where the 5 goes. So when we call this the x location of the bolt and I'm getting ready to get you a component, I'm gonna have to say that this bolt's moment force is equal to this bolt's moment force, maybe 200 pounds, for example, and I'm gonna have to multiply it times to get the x component, I'm gonna have to multiply it times this over that. That means I'm gonna have to say it's multiplied times y over h or d. So, and that's what he did here. He says, do you want the x component? I say, yes, I do. He says, it's y over d times the whole thing. I say, no, no, no, I want the x component. He says, let me repeat. You got to multiply it times y over d. If that doesn't look strange to you, you know, you're doing something besides watching this, looks strange to me. But I see why. The position vector has an x of 12, but the force we're trying to find the components has a y of 12. It's still got the name x. But to get the x component, you multiply y over d to get the component in the vertical direction multiplied times the total p in the bulk due to the moment. We already have a nice equation for p due to the moment somewhere. Remember where? We'll find it. And by the same token, if you want to know the, let me see how this is broken up. p sub x that times that is equal to, okay, no, he's just continuing here. Okay, I was thinking he was doing something new. He says, do you agree to this? I say, yeah, I think I have to understand that. He says that's y over d times p sub m, which is md over the sumd squared. I said, no, no, no, no, no, no, where'd that come from? He says, we'll look a little higher up. p due to the moment, the whole thing not broken into components, md over sumd squared. There it is right there. There's the moment in the force due to moment in a given bolt is md over sumd squared, where that d is for that bolt. So there's your md for that bolt divided by the sum of the d squared. And that's equal to, I said, dang, you're not through yet? Oh, no, he said I'm not through yet. He says, that's y over d, do you agree? I said, yeah, well, yeah. He says, I'm just checking. You got to agree every step of the way here, you'll be lost. Times m times, where does that bolt live from the centroid of the group divided by d divided by the sum of the d squared, and the sum of the d squared is the sum of the x squared plus y squared. Back here, I already agreed to that. It's too late to change my mind now. Sum of d squared equals sum of the x squared plus y squared. There'll be a whole bunch of bolts in here. There'll be 33 bolts in here, and there'll be 33 bolts in here. And he says, and therefore the d's cancel. This was a personal d for the bolt under discussion, as was this one. So I really ought to have a little i on it. Now then they're gone. He says, you want to know the x component? Do the moment in bolt number 37. I say, I would like it. He says, go get the moment, e times e. He says, multiply times the y location of your bolt under discussion from the centroid of the group and divide by the sum of the x squared plus y squared. He says, you can do this just one time, and you'll have every bolt ready to go. And you could also take in one time and do this work, and then all you got to do is change y. Similarly, I said, OK, you're not going to do the whole thing again? He says, unless you really want to see it, I didn't raise my hand. Similarly, the y component of the force in the bolt under discussion due to the moment applied, not taking into account the p sub c force in the bolt, centroid loading, m sub x as opposed to m sub y over the same sum x squared plus y squared. It's on any single bolt. And he says the total fastener force is you found a centroidal loading x component, and you found a moment loading x component. And those two vectors are parallel and can be added just to add them up, just like a scalar. Therefore, you're going to have the sum of all of the forces in the x direction squared. You'll come up here. You have a y component due to centroidal loading, and you have a y component due to moment loading. You're going to calculate that. You're going to add it to this. You're going to take that sum, and you're going to put it right here. There's that sum squared. Take a square root. It's going to give you the force in the bolt, and you're going to find out what kind of bolt it's going to take to do the job. Now, here he is kind of belatedly showing you that if you have something like that, that you have a component there, your x and your y components. This one shows you for an angle. If it's at an angle, then the angle from the horizontal transposes into an angle from the vertical to get it to be perpendicular. Example. It wants to determine the critical force in the bracket shown in figure 8.5. Here's our problem. It says the vertical centroid of the fastener group, the vertical height, the centroid of the fastener group can be found. Well, I don't know how to find vertical centroids. I've been doing that all my life. I would take like a T beam, and I would call it area one, and I told myself that area one lived at its centroid, and I drew a reference axis at some place, and I wrote down the equation A1 times where does he live, plus A2 times where does she live, divided by A1 plus A2, and that told me the location of the elastic neutral axis. You can do exactly the same thing with a set of bolts. You draw a reference axis some place, probably there, and then you say that area times how far it is off the reference axis, zero, plus that area times how far it is off the reference axis, which maybe is five, plus that area times how far it is off the reference axis, which looks like maybe seven, times two is the total tendency to roll about that axis, is equal to sum of the bolts times the centroid location of all the bolts, just like you did on this. So here he goes. He draws an axis at the bottom. It's for reference only. Once we find the centroid, we're going to go ahead and put the centroids on this thing. Two times zero. There's two bolts right there, right on the reference axis. He didn't include them. Plus two bolts at five inches. Plus two bolts at five, six, seven, eight inches. Plus two bolts, five, six, seven, eight, nine, ten, eleven. Add eleven inches, divided by one, two, three, four, five, six, divided by eight bolts. Centroid of that bolt group is at six inches up from the lower set of, lower pair of bolts, six inches. Horizontally, dead in the middle. There is a 50-kip load sitting at a one, two, square root of five. I took that and I multiplied that times one over square root of five, 22.36, right there. And 50 kips times two over the square root of five, 44.74. The first thing I did is I picked the 50-kip. Well, I probably didn't. You know, I didn't pick the 50-kip up and put it over here I probably picked the 44.74 and put it right down that axis and divided it by one, two, three, four, five, six, seven, eight. 44.74 divided by eight. The name of that force is P due to the centroid loading in the Y direction. It's equal to 44.74 divided by eight bolts. Then I did the same thing with the Y. I just picked it up and put it in the centroid. I looked around and hoped that nobody was looking. I looked around, very few people are. Most are texting. Not everybody's texting, but... And now then I'm going to take care of the moment. I know there's a moment times an eccentricity. I know that moment's going this way around the bolt group, but I'm not going to take this thing and try and find that goofy moment arm. I'm going to take this force and multiply it times this moment arm and I'm going to subtract that force times its moment arm and that's going to be the moment about that group. So here is P sub Y, 44.72 over eight bolts. So the force in the Y direction will be the Y component of the force divided by eight bolts for this one and then the X component will be the X component of the 50 kit force. I scratched it out because I put the components on there, so that had to go. And this number comes out 17.78. I can't be right. That's not what I'm looking for. I don't know what that is. But that's not what I'm looking for because I'm looking for 44.72 divided by eight. That isn't either one of these numbers. Here it is down here. I'm not sure what that is. We'll come back and see. Here's your X component, 22.36 over eight. Here's your Y component, 44.76 over eight down. Now then he's going for the couple. The moment on that thing is... He doesn't ever calculate these things in the order that I kind of think he ought to calculate them. Moment, moment, moment, moment, moment. Well, let's just take it in his order. First place to get the force due to the moment, we are going to have to know where a bolt lives in the Y direction out of its total distance where it lives. And then we're going to have to calculate the sum of the X squared plus Y squared. So he says, look, you need all that stuff. Just go get one. All right. He's going to get this one first. Here's the sum of the X squared plus Y squared. Here is the centroid location of the group. This is the X component. This is five and a quarter. The centroid is right in the middle. So that means that this bolt has an X location of half of five and a quarter. Half of five is two and a half. Half of a half is a quarter. So that's going to be two and three-quarters inches. Here it is. It's going to be two and three-quarters inches. This guy's X, which I'll want a square. And I'll want the same number squared, same number squared, same number squared. A minus still comes out the same number squared. So he says I'm going to have eight little X positions from the centroid for the eight bolts, 2.75 squared. Now I need their Y squareds from here. This was six. So one of the numbers is going to be for this guy will be a six squared. There's one of them. There's two bolts. There's two of them. The two goes on the outside. Then this bolt is up here five and the centroid is up here six. That's a one. There's your one squared, two bolts. This one is up eight. This is six. So that's a two. There's your two squared. And the final one is up at five squared. So the sum of all those X squared, Y squared is 192 inches squared. Now your MC over I is a little weird. You say, well, that doesn't look like an inch to the fourth to me. Well, if you remember the real moment of inertia of the thing was A times the sum squared. So the A canceled out. So all we were left needed to calculate was this. The A was the other inches squared that we would have had for a real polar moment of inertia in inches to the fourth. So it shouldn't be confused by the units. Then he says piece of M, X. Wait, wait, wait. I said, no, no, no. You can't do that to me. I have no idea what you're even talking about. Piece of M, X. Let me go see what I approved. Piece of M, X. Piece of M, X. Here we go. Moment forces. Piece of M, X. Here we go. I agreed to this. Piece of M, X was the applied moment times where is it live, Y with respect to the axis divided by the thing I just calculated. All right, so I'm okay. It says piece of M, X is equal to M, Y over the sum of this. I agree. M is 480. Where the devil did the 480 come from? 477, E, C, B, D. Geez, man, we got that back on page one. The moment was equal to, I remember, remember we had a 44.7 times that dimension. We subtracted 22.36 times that dimension because it was on the corner of the plate. That was the 480 moment applied to the plate. So I agree with the 480. There's our moment at 480.7, the wide distance to some bolt he's playing with. See which bolts he's playing with. Who has a Y of 6? Here's the centroid. He has a Y of 6 and he has a Y of 6. Wonder why he's playing with these bolts. These bolts are out here at 7. They'd be further out. Well, let's check it out. Go back to basics. Number one, when you put a 50-kip load on these bolts, this bolt has a downward and to the left force. When you put a torque on this set, everybody has a ray or a position vector and the force is normal to it. So discussing this bolt, this force goes down into the left and this force goes up into the right. Those are canceling each other. I don't think that's very likely to be a big force in a bolt. How about this one? This is down into the left. Ah, look at this one. This one goes down, but it goes to the right. So the leftward component of this one is going to detract from the rightward component. So I don't think that's probably the worst bolt in the group. How about this one? This force causes down into the left and this moment causes perpendicular to that. Oh, that's a down into the left also. So now then the two downs will add and the two less are going to add. So that's a pretty good guess is maybe that's the worst bolt that we have. Why would this one not be as likely? Why would this one not be as likely? They're both to the left, but one of them's going up and one's going down. That's right. So they're fighting each other in that respect. They're canceling each other in that respect. So most of the time you can kind of figure out which bolt is probably the critical bolt. And if you really can't, maybe there's one up here. Geez, it's so far out there. There's no telling what might be going on out there. You may just have to study two of them. But you really shouldn't expect anything like that in the real world on a final. Yeah, sure, but not in the real world. Not funny at all, huh? Okay, so that's the bolt he's studying right there. Yes, sir. All right, this one. This one. Okay, you'll notice that this is going down into the left and this is going down into the left. However, do you see how this one is a lot further out? So his down into the left isn't near as big as his down into the left. That's another reason we didn't really check any of these interior bolts. They're almost always lightly loaded. Now, incidentally, by this elastic method it's very conservative because you're going to be called on to stop when the first bolt says, I've reached the limit. That's all I can take. And all of these other people are saying, pour it on, pour it on, we can take it. He's, no, no, no, no. By the elastic method, you're going to stop when the first bolt says that's it. So we were here. We had the 480. We understood why he took a Y of 6 because he's looking at this bolt right here which here's your centroid and that's 6. He has a X of 2.75 so when we're going for his Y component we put in the X, be 2.75, cranking that out, you get 14.98 kips just due to the moment in the X and just due to the Y you get 6.8 to the left and down. So then we come back and take 2.7 to the left plus 14.98 to the left 2.7, 14.9 is the total just added as a scalar not as a vector although they're added vectorially 21.78 down we have a 5.9 and a 6.8 there's your 5.9 your 6.8 12.76 total X squared plus total Y squared take a square root, total force of the bolt 21.7 kips Now then the question is so what? Well, that force is in the bolt you need to go give me a bolt that'll do 21.7 kips could be a 490 threads not to the head threads to the head could be grade A, grade B class A, class B but we got tables for that you knew that's where we were going didn't you about to work that stuff out again I need 21.7 kips and 21.7 kips by LRFD group A A325 if I take a 1 inch bolt it'll give me 31.8 but if I use a 7.8 inch bolt then it'll give me 24.3 this is of course already with the fee I never understood why this table didn't write on it and I think you sure ought to write it on it it's obviously what it is but it's fee or sub in it's got right above the column it is yeah oh well then I see why he didn't do this because he's got this other trash in here good point okay maybe I'll just kind of point to that so I won't look foolish again not likely, not likely good point now then I like to use a smaller bolt you're not going to get the 24.3 out of threads included in the plane of shear but if you'll you can still use that A325 bolt if you will keep the threads out of the plane of shear because you get the 21.7 none of these bolts down in here really going to do the job unless you go maybe to an A490 bolt then you can get the threads included A490 is going to cost you more and not only that you're ending up still with a three quarter inch bolt now three quarter inch bolt well that's no better than this one well but here you had to exclude the shear the threads from the plane of shear so this bolt might be worth the money you just have to put the pencil to it and see because that is smaller holes and so you are destroying less metal at different points where you drill these holes you just have to see and here they are excluded that's no help at all that's how you find out how much the things what size bolt that you can use on those quick and easy and what have I got an X here for I wonder kind of bolt is this yeah no good at all nobody uses them anymore you may find them in different places alright now what really happens well what really happens is when you load the bolt and they got this from a test A325 bolts three quarter inch diameter threads included one of the more common bolts that are used all over the place and they found they plotted all the data for thousands of these things and this is what they found they found that you don't get a nice yield on these bolts because of the materials they are made out of and because of their strength and everything you just kind of creeps on up there and after you just deform them so badly you can't hardly say that's hardly a bolt anymore they are still taking about 74 kips at around .34 of an inch that's a third of an inch worth of distortion includes the distortion in the bolt the distortion in the plate it was a distortion of any kind they counted it and at about that point they said that's it you can't put any more load on it some out there says no we don't want it it's just too far down the road that number I forget is something like maybe 71.6 72.6 at about 72.6 for this bolt they want you to quit the nifty thing is is back over here on the allowed stress stuff when this bolt the worst loaded bolt you had reached 21.7 kips and you picked a bolt that had that capacity actually I guess we would go on up maybe to 24.3 when it reached that number you had to stop and all the other bolts they weren't being used at all if you will take this into account that this bolt can come on up here to 71 and it really can although it's going to be pretty distorted then all of these other bolts that were all playing way down in here they'll be on up in here and they'll be taking major parts of load and they'll give you a great additional capacity of the connection that's of course why we've been doing plastic design the whole time because we always got truth and we always got more load carrying capabilities out of the beam, out of the column, out of the connection out of the plates out of the beam bearing plates out of the column what are those plates called base plate very good if that's what he said alright we'll do some more of that next time we'll start there what you got you keeping up with it okay can you figure it out from the book email me and let me get in there I can't keep up there's just too much good stuff what was it we're using the tables that we're using that we're going to get into now email me and let me back off on that because we don't have that many days left there you're probably learning a lot because you have to dig it out on your own I understand