 Okay hang on to your hats stuff isn't hard but it takes some thought it really does. I mentioned earlier that you could have a failure due to block shear. Sorry about that. One of us was destined to dump something on the floor so here is a plate bolted to a gusset plate. Here is if the gusset plate is weaker than the plate tears the whole block out of the gusset plate. Here is a beam with a coped end on it where they coped this end off so something would fit. If they were just going to bolt it to a column as you see here then they probably wouldn't have bothered coping it and you wouldn't have that little block shear problem. That usually happens when you're bolting the incoming beam into something like this into a girder. Thank you sir. I'm not used to holding it that direction but I'll... there's for an angle. The main reason I even brought it we've already had a bunch of pictures of those. Here you see they were testing something. This particular in fact these two specimens failed on the ends and they were still busy failing on the sides in shear. Rupture on net tension area. Here's a typical bolt after some severe loading. Here's the top plate and the bottom plate you can tell it was in double shear because of this part's all crushed and then this side is crushed. A lot of deformation for a bolt and it's still holding on nicely. It still has area of the bolt times f sub y in shear. 0.6 f sub y. Okay those are tension members. Now compression members have one more aspect to them. They can still fail just due to pure old compression if they're nice and short but once they start getting long then you have problems with the section buckling. We already did some buckling in 0.305. We determined that Mr. Euler was a whiz at predicting how much strength you get out of a compression member in a derivation for it and that's what we're going to use in this class also. You can read all of this. There are structural elements subjected to only an axial compressive force. Actually sometimes the column will be subjected to bending also and then they're called beam columns so you'll have to design them to resist both the column action and the bending action. Usually it's a vertical member primary function to support loads from the top. Read all this. You should know what columns are all about. Three methods have been developed over the years. They each have their own adherents. People who just think it's the best way to do business. The AISC agrees they're all acceptable. Some may be more economical under certain situations but when you get where you're going they're going to probably use one of these methods. We'll talk about them later. One's called the direct analysis. One's called the effective link. That's mostly the one we'll use. I think one of the homework problems has one of the other ones and the last one's called a first order analysis method. They're straightforward. It's just a matter of what you're planning on using. A lot of times you'll have a computer program that will already be set up to give you what you need. These are actually the loads on the structure. In this chapter we're just going to find out how strong the members are. Then we'll talk about taking the loads applied to the floors and things like that using one of these methods and getting the loads in your particular girder in your particular column. Column theory. You've been there. You've done that. You had a beam kicked out to the side. You put loads on it. You had a second order differential equation. You set up the equation for how far this thing deflected to the side depending on the load and all of that stuff. Pull it out of your 305 book or look at it here. We'll be fine. Square y dx squared is equal to m over e i. That's right out of your 305 book. There's your differential equation. There's your assumed solution. You set your boundary conditions equal to appropriate quantities. For example, on a pinned, pinned column you would tell the differential equation. You're going to integrate it a couple of times so that you can find m and delta. You're going to tell them at this point you know the moment. Since it's pinned there's none. At this end it's pinned and therefore there's no horizontal deflection. There is some here. There is some here. Down on this end you tell them that. And it comes back and he tells you pi critical is equal to pi squared n squared pi squared e i over l squared. Now in 305 I don't know if we even mentioned we might have just so you wouldn't be confused about it now. We just had pi squared e i over l squared for the critical buckling load according to Euler. Truth is the end was in there when you set one of these terms equal to zero when x is equal to zero, y is equal to zero. You're trying to get that term equal to zero. Sign of that was zero for zero but that gave you no load so you weren't interested in that solution. It is a solution or there could have been pi which is the pi that shows up in the equation. And that was the lowest number but it could have also been two pi which nobody really cared because with an n of two then you'd have four times as much load. And we figured we needed the first time the thing buckled and so that's probably what you worked with the entire time. The fact that n could be a two and still get a zero is possible though. For example here is for n is equal to one this is what the shape would look like. This is what it looks like if n is equal to two and you can force that to happen by putting a brace in the middle of the member. In other words across the top here you may have a floor and it is pinned to the floor and down here is a footing and you have it pinned to a steel plate on the footing. And if in between here you put some braces across the mid height of the column. So here's your structure that's pinned at the bottom. I don't care if you want to fix it but it's pinned in the one we have shown here. And if you come in here and you put a member across here it doesn't have to be very big. Just has to be enough to prevent this from wanting to move and buckle in a two node state. Then you're going to have to put some braces in here otherwise the whole thing they'll all just buckle like that they'll all go together. So you're going to have to put some braces in there to make sure that that doesn't happen. And if you're willing to pay that money you can make the column four times as strong. You can do it as many times as you think it's advantageous to you to do so. The solution was the lower load was you probably don't remember but there's a defined quantity called the radius of gyration. It was a square root of the moment of inertia of the shape divided by its cross-sectional area. Somebody made you learn it. Somebody told you you make money with it someday. Somebody told you that its use will become quite useful to you but never showed you why nor here it is. In our equation for pi squared E i over l squared if you substitute i is equal to square both sides. A r squared into this equation then you get p critical is pi squared E i over l squared is pi squared E i a r squared divided by l squared is equal to this. You say so what's the difference? I can look up a and the radius of gyration in the book just like I can look up i in the book. Yeah but if you're going to start talking about stresses critical buckling stresses then you take the load divided by a that's this term divided by a and the a drops out now then you have an equation for critical buckling stress and a lot of times that's what we work with instead of loads. Take your choice. You'd want to remember that E for steel is 29,000 this is the slope in this straight line portion. If you get out into this section out in here then E is not 29. E is 20. I'm sorry E is 10. I'm sorry E is 6. I'm sorry E is zero. You're going to load it out in that rate region then when I tell you the buckling critical buckling load is pi squared E anything you got the wrong E. You put in a 29 and when you gave me the stress at which it buckled that stress came out here. So you've been using the wrong E and I'm going to insist that you go find out what E is for the answer you gave me and don't put 29, 20, 10. You're going to put in the right E. Once you put in the right E then you put that back in this equation and he'll tell you what the true buckling stress is except when you get that true buckling stress and I say well okay at least you admit it's not here now you say it's here say okay I feel this coming on again yeah that's right you got the E wrong again and it takes an iterative solution to really home in on the answer. Now for this steel you know you may not care in other words if it goes above the yield you say okay forget that it doesn't go above yield anyway even if you think it's buckling it's not buckling all the little fibers are shot but when you get some nice high strength steels like we use then the stress strain diagram looks kind of like that and so you were right 29 KSI was right for a while but then while you're really out here still working and maybe before you even get to what they call f so why that tangential modulus of elasticity has changed so now it's a serious problem now you're going to have to go in and change it and redo the load and get the stress and redo the E and once you do that you got to redo it's an iterative solution as you might imagine most engineers aren't going to like that and I say look you got something a little better that's close enough yeah yeah but I mean this is where this is coming from so here's an example we got a 12 by 50 wide flange the villa the villa how much is this thing weigh 50 pounds a foot that's right my point that the 12 didn't influence you at all did you hill big uh what color is it painted don't they have a color don't know the the specification what is that for I thought that was a number 12 paint now what is that not the paint color he says it is the height it's approximate height it's in the 12 by series it's around 12 inches deep some of these things I don't mind maybe even in the 12 by's the suckers will get 13 14 inches deep because they were started off as 12 by's and they just got rolled thicker and thicker and thicker but it's about the depth supposed to hold a load of 145 kips that load there is somebody's already put a 1.4 on a dead and a 1.2 dead 1.6 live plus 0.5 snow plus plus plus plus they've already done all that this is what they want you to hold up it's got a name of peace of you 20 feet long ends are pinned he says without regard to load factors because he says he's already put them in here and without regard to the fact that you're getting ready to tell me a number that we can't use you're getting ready to give me the nominal strength he says don't worry about the resistance factor right now is this thing going to be stable what is in effect the average of a thousand tests without taking into account that a few of them were not so good he says you don't need to know the grade of steel because grade is steel 50 ksi no 90 ksi no well it's not in the equation so nobody cares uh it's a pen pen column so k is one in other words you're on node node one one node uh you know pi you know e you know i about the weak axis you know the length you should be able to find out how much the thing will carry for a w12 by 50 on page 126 that's where we get the area if we need it there are our equations for critical stress two versions of critical load one with a moment of inertia one with a r squared in it here is where we get the moment of inertia or the radius of gyration whichever version of the equation you select to use to get load or stress or whatever it has a strong axis 391 inches to the fourth resistance to bending what's the resistance about the weak axis 56.3 oh that's good you revived it just the right time there yeah we're going to hit your head follow them on your desk there and it has radius of gyration that's uh elicitous big one little one we're going to be using this one because if you have a pen to pen column nobody said anything about bracing it in the middle the brace the weak axis obviously then when it fails it's going to fail in a single node for that he says the minimum strength radius of gyration is the weak axis number 1.96 1.96 that's true i'm sorry not always let me show you one where you don't where you have braced it there and braced it there and braced it there with little braces across like that then it may pop out about the strong axis it's hard for you to see but i'm raising the pencil and i'm bringing it back down it'll buckle about its weak axis because you braced it like crazy about the small axis you know what you're really going to be looking for is you're going to be looking for a l over r for this little guy or an l over r for this one so i can't tell you that for sure but we will you would always brace the weak axis it would be pointless to brace the strong axis the strong axis in this case already isn't buckling all right so we're taking the minimum number 1.96 the biggest l over r we find is 20 feet times 12 inches and a foot you'd be surprised how many people forget that it's easy to do if the guy doesn't write his units down here that's 20 feet long divided by 1.96 inches if you write the units you'll find out there's a problem there in a hurry eat a 12 inches per foot gives you a this is called the slenderness ratio as you know and you will be looking for the maximum slenderness ratio depending on how it's braced and the properties of the column and then you plug it in the equation for load pi squared a divided by the slenderness ratio squared pi e incidentally you better be within that straight line part now times area divided by slenderness ratio squared 278.9 kips he says the applied load is only 145 and you've got available 278.9 so it's stable and it has a factor of safety of how much it could take divided by what you requested around two what is the real stress well the real stress in your column is 145 well let's let's don't i'm not going to use 145 i'm going to use the 278 because you told me 278 was how much it could take so i'm going to use the 278.9 so i'm going to got a calculator divided by was it 14.1 6 thank you square inches what is the real stress in that column 19. something okay well that's well below yield for all the steels and therefore i am happy that you are in the straight line portion that's less than f sub y for anything i know so that would be an appropriate number to pick now there's a discussion that i in effect have already discussed he says they found out a long time ago once they put the real loads on there the things buckled much earlier than they thought then they realized the reason why was you were sticking 29 in there but you were trying to pick up this much load or this much stress meaning that the e that you input into this i into this equation was incorrect and so it turned into an iterative solution the real data shows that mr euler's equation is 100 correct 100 correct 100 correct because mr euler is not saying the failure would occur at this point he's saying it will not buckle until you put this much load on it so here i have a column it's one inch tall it's on a concrete thing and when i put the load on it mr orver says that will not fail in buckling i'm not talking about anything else he says it will not fail in buckling until you get two million psi on it okay well i'll tell him i have no interest in that you've exceeded f sub y he says what i said you have exceeded f sub y he says what i don't know what you're talking about i'm a buckling guy i'm telling you when that thing will buckle okay i can see this guy is a buckling guy he's not interested in the fact that the fibers have already turned into jello somebody go do some testing and tell me what to put on this column for the real loan test test test test test orders right orders right orders right uh orders really wrong here and the guy says well that shouldn't surprise you you know the number is not going over f sub y no matter what this thing if it didn't come in at f sub y then it can't be right past some break point below a break point you're going to have to use a different equation and if you can't come up with a different equation that's reasonable to use then you're going to have to just fit a curve in all that data and we'll just go according to that equation so no i got an equation he says uh it really sucks we've got to do a trial and error to get the right curve so these are called empirical solutions where we take the data and we take the data we'll take the derivations the theory and we piece them all together and we fit a curve in there that seems to do nicely and that's what we've done for things when these things buckle inelastically where some of the fibers are already going into the yield usually they're going into the yield because they're always a little crooked so there's always a little moment and so this fiber on this side is yielding a little bit and that fiber is yielding a little these are nice and still elastic but because some of these are failing you're in a range they call inelastic buckling out in this region if the if the column buckles and you can get the load off of it before it collapses if you'll get the load off it'll go back to perfectly straight and the inelastic region if it buckles and you don't put any more load on it on a testing machine that's the way you put load you do it you know by deforming it if you take the load off it's not going to go back to straight because you have yielded some of the fibers now effective length we could just redo the entire derivation you say well we didn't do the derivation in the first place yes we did you and i in 305 did that derivation and all we would do is we would pick our new case our old case was no deflection no moment no deflection no moment if you have a fixed fixed case we would he'd say do you know the numbers anywhere so yes i do no deflection and no theta see how it didn't rotate and right here there's no deflection and there's no theta see how it didn't rotate i didn't tell you no moment because i don't know what the moment is so that's not a boundary condition you plug those boundary conditions into the equation he'll come back and he'll tell you exactly how the thing fails and how much load you can put on it this particular one that he's describing here see what he's doing here this is i think he would say what it is i know what it is from the formula it's pinned and it's fixed other inconditions can be in general many moments our non-hemorrhage is a boundary condition be different original one formulas is so and so for consider here we go a member pinned at one end and fixed against rotation and translation on the other so this is what he's talking about he's talking about a buckling situation where it looks like that if you put that in this is what you'll come up with 2.05 pi squared i over l squared and also an n in there ours didn't have the 2.05 now if you'll divide if you put a 1 over number like a 0.7 0.7 squared 1 over 0.7 squared that's 2.05 somebody noticed that and so i said look it's a little more convenient to put it down here and i say why what difference you make it's there there says well you'll notice that this 0.7 here looks like 0.7 times the original length that the thing was when it came in the box so he says that's where the pin is that's where this thing stops being bent this way and starts being bent that way and therefore the length between pins there's a pin a lot of moment not a pin not a lot of moment not a pin right there is 0.7 0.7 times the true length where this is the true length so okay so what you're saying is if I ever see one of these just measure it as it comes out of the box before we install it get l true and if i intend to pin it and fix it then in my Euler's equation i put the effective length which is k times l say do you happen to have a k for this one he says i do says you happen to have a k for this one he says i do because i'm going to just count it pin here's the load on top of it moment moment deflection deflection okay here's the other half go dig up any column any flagpole on earth you'll find we always put another half under the ground maximum moment not so maximum a little bit of moment no moment no moment pin pin effectively a column a freestanding column is how long that's not the right question effectively a freestanding column is how long how far is it from here to here not well that's just l didn't get enough number it's l white l out of the box l true that's correct how long is this mirror so how long is the column effectively two that's right here is the column he was discussing the true length out of the box was 30 feet or something when you installed it with those in condition you force this pin to come down to this point and it turns out from the derivation it's seven tenths of l true that's called the l effective the effective length yes sir if i ever lied to you go get a shovel you'll see every flagpole has another flagpole in the ground underneath it when you got me there don't you okay well there's not really another half underneath the ground whatever's in a box is all you get they steal half of it before they send it to you so l true this is 30 feet but your effective length on the column is how long it was 60 feet to l that's correct as long as you do that i don't care how what mental process you use all right and he talks about more k l is the effective length k is the effective length factor he says we've got some others in table ca 7.1 there are theoretical and recommended values for design you can use the theoretical values under really strange conditions and under very going to really be expensive because the truth is it's very difficult to really fix something in many instances not hard to make a pin out of it that that's easily done let me see let me get that table right quick before we go any further on page 16.1511 here is a fixed fixed column theoretically we just found well we didn't just find but this thing has got an axis of symmetry like this and then there's another one going the other way and then there's another one going like this these are at the quarter points and therefore the effective length of this column would be half of its boxed length half its true length here you see the theoretical value was 0.7 for a pinned fixed here's one where the column was allowed to translate but not rotate here's the upper half here's the lower half it is effectively pinned maximum moment pinned it is effectively L long that's what he says right here the effective length is one times the true length this is the one you and I did in 305 and for which he has the full derivation for here's the flagpole 2.0 theoretical and here is an upside down flagpole fixed at the top and pinned at the bottom it has a effective length of two times the box length the truth is though when you try and really make this thing not rotate it has a tendency to rotate and it's really difficult to stop it the plate will bend the connection will bend the whole footing will roll over a little bit it doesn't take very much and it really pretty much ruins this thing it doesn't run it all the way down to no good at all but they don't want you using 0.5 unless you've got some really extraordinary circumstances and can show it they want you using a 0.65 same way on this one this is very easily done in the field this is not easily done in the field 0.7 they want you to drop to 0.8 meaning this seven tenths uh 10 foot column is a seven foot column they want you to say it's an eight foot column i'm gonna ask you to make it a 10 foot column because you do have some effective change in the behavior but they only want it to be shortened from a 10 down to an eight foot column this case they want you to go with a 1.2 instead of a 1 this case a 1 is fine this case it's not easy to really fix that footing on the ground they want you to go from a 2 to a 2.1 this is a fixed pinned column this is a fixed pinned column 2 that's interesting this case they make you go to a 2.1 but when you turn it upside down it's a 2 i don't make any sense at all interestingly it really does everything these people do makes sense you got a column that's fixed on a footing in the ground that footing can rotate two to a 2.1 please you got a footing that's pinned to you got a column that's pinned to a footing but is welded to a girder now that's a different matter and that that's that's kind of doable you can weld it to another piece of steel you're not just bolting it to a piece of concrete in the soil and in that case they say good enough you can use a 2 all right back to this this is what the code this is what the specifications called for by the codes will insist that you follow number one go find me the critical buckling strength multiply times the gross area that's the nominal load you will have to multiply that times an appropriate resistance factor to bring your p nominal down to something that i feel is safe under all conditions even if you bring out the worst one that anybody's made in 10 years and you will make sure that your nominal load times that 0.9 will be greater than the requested piece of you piece of use 1.2 dead and so on resistance factor is 0.9 so specified on page 16.1 days 31 in the specs and piece of you less than that of course that's the whole purpose that's the whole purpose they're going to pay you now here are kind of pictures of these things case a case a is fixed fixed all right this one might you might be able to get away with fixed fixed there it's fixed because it's welded to a steel beam there it's fixed because it's welded to the first floor beam i don't think you're going to get away with that on the next floor down where it's bolted to a concrete footing you might get away with that there you'll notice that this is fixed fixed with no translation because in the frame they put brace rods so when the wind came and pushed on it this rod got tight and this point did not move with respect to the support beneath it here is a fixed pin here they welded it like crazy and here they put angles in there and they bolted it these little angles i mean you could beat somebody to death with them in a hurry but they're not really very strong they're just stronger than the guy at the baller's head when this thing wants to roll these things don't have a chance they just open up and this one just collapses not a lot but it only takes a little bit and this is a pin so this would be case b case b pinned at the top fixed at the bottom or upside down there isn't one of those so this one you can just turn it upside down here it's fixed at the top and pinned at the bottom braced no translation here's a frame where the guy forgot to put the braces in there and so when you put the load on it would it wanted to do it wanted to fall over to the side so that's case c case c that's how you got that translation it is fixed yes it's welded nicely to a nice big girder and it's fixed on the bottom so that's case c fixed fixed case e is pinned fixed case e is pinned fixed truth is i think that this case would also be applicable to this one because you got one end pinned and you had the other end reasonably fixed on the bottom flanges you probably get away with using the 2.0 as opposed to the 2.1 there's case c and e there's case d pinned on top pinned on bottom that's the pin pin column you'll have to put braces in here or you'll have to stop this thing from translating somehow or otherwise it'll just fall on the ground case f fixed at the top pinned at the bottom you'll notice i used angles so i can say pinned put them angles in there it's a pin connection okay first we're going to define pi squared e a divided by k over r squared as the Euler buckling load so we'll subscript it with an e got such a little e i couldn't see it the critical buckling load according to Euler the stress is p over a so it's pi squared e divided by k of r squared a cancels a we were talking about times when Euler's load gave us a good number Euler's load gives us a very good number as long as you are above to the right of this break point that break point is where these two curves seriously diverge this region is called the elastic region if you get the load off the thing will come back straight around in here the fibers are starting to yield this is the inelastic region the break point divides when you have an elastic versus elastic behavior now because Euler's perfect but Euler assumes that the load is right down the center not really possible and the member is perfectly straight that's the worst part of it all it just can't be a few of them are but some are bent to the right some are bent to the left they put in a factor to take it take that into account and they spent a fortune finding out what that number is by taking a bunch of roll shapes that came out of a rolling mill and testing it 0.877 this is an experimentally derived number this is a is the equation e 33 on page 16.1-33 if on the other hand you are to the left of the break point in the inelastic region this is the equation that best fits the data that we have found you'll notice that when the yield stress is 36 and Euler says I don't know why it quit at 36 because this number here ought to be a million then 36 over a million is zero and anything raised to the zero is one and the critical buckling stress is the yield stress so it's coming in right where we want it to not only that it has the properties that the two curves aren't perfectly tangent but they're you know there's a little kink in the curve but not much and they are both at the same height at the break point here it is again this is what region inelastic this one is and there's your break point where in the world do they get that from I don't know one million two million bucks who knows but it was very cheap rather than having a bunch of things fall down or putting columns that were too big when you didn't need them 4.71 times the square root of the modulus of elasticity divided by the yield stress of the metal they got that from taking 36 ksi yield metal 50 ksi yield metal they didn't have to worry that much that's that was 29 for everything and that's where the break point occurs that's where mr. Euler's formula really diverges from what we see in the field so if you are to the left of this point you ought to be on that weird equation you ought to be on this inelastic 0.658 raised to the fy over fe power multiplied times f sub y if you're to the right of it f sub e is perfect except the things aren't perfectly straight the fact that they're not perfectly straight is already included in this number also 0.877 times f sub e once you find that critical buckling stress you multiply it times the gross area of the shape and you got how much load that column will carry we'll tell you that beams see columns only have two regions beams are going to have three regions so you're going to have to have two break points to the left of this break point you'll be using one idea in here you'll be using a straight line idea and in here you'll be using a timoshenko idea so what I say hang on to your belts uh all he's doing here is repeating what he did the only difference is rather than giving you a break point in terms of k l over r you got to calculate that anyway so I don't really see any reason to to do it another way but there's also if there's a break point there then there's a break point here and that break point is 2.25 the ratio of f sub y over f sub e so I just scratched it out it's up to you you can use either one this one right here would say uh if your yield stress is less than f sub e over f sub e is less than 2.25 you're either on one side of the other side of the curve as long as you can remember one of them you're okay here's where it is chapter e 16.1 dash 31 here's the first thing he talks about is your resistance factor for compression members talks about the effective length now yes you told me once mother global buckling global buckling means the whole thing is buckling as opposed to a little piece of the column buckling we'll get into that later here is your equation nominal strength is critical times gross here is your critical there's your inelastic equation and here is your elastic equation you'll have to check out your personal k l overall for your personal column and see if it is to the right of this magic number or to the left of this magic number for your steel uh and he tells you how to get f sub e that's just always buckling stress this is for local buckling we'll cover it next time and I just put these pages with the global buckling just so they would be available to us all steels in elastic elastic for a yield of 50 ksi which is usually what we use for columns this number comes out 4.7 y one times the square root of 29000 divided by 50 it's 113 it's a number that gets imprinted in your brain before too long and you go find the k l overall if it's above 113 you're using the uh elastic equation this one to the left you're using this one for the critical buckling stress uh they also suggest that you don't use k l overalls over 200 they don't care if you do but if you got a k l overall over 200 it's going to be so inefficient it's just not going to carry any load at all and so nobody in their right mind goes past that number here's an example it's got this wide flange made out of 50 ksi steel 20 feet long pin pinned he wants to compute the compressive strength he gets the k l overall 96 he checks the break point 4.7 129 over 50 there's that 113 that 50 came from page 2-48 96 says to the left of the break point we're going to be using break points like crazy in here since you're to the left of the break points you ought to be using that uh goofy equation the goofy equation needs a 0.658 i got it it needs an f sub y i got it it needs an f sub b i don't got it it says okay here it is 5 squared e to about about a kilowatt squared 30.56 do the math 25 ksi here's where you get the areas where you get the radio gyration that's where i worked it out by hand multiply the critical buckling stress times the cross sectional area get 549 kips you only get nine tenths of it and nine tenths of it is 495 kips that's how much you can put on it see you next time yes sir yes the book you're in my 305 book has the same values as they use and they have a recommended you know like uh like factors which is one line underneath we do you use are you going to really design for the real world then you need the practical value absolutely in this class and the uh well this is now in the 305 class you weren't told there were two values because we'd have to explain why you know and that takes some maturity that you now have but back then we had nothing but theory thank you sure let me let me shut this down i saw you leaving