 I left you with a problem before we we broke on Friday Make a chance to look at that one. I think I left you with the answers No That was one where we had an I beam that was supported in The middle against buckling at least in that direction, but it could have buckled in the other direction So you had to test it in two directions with two different Possibilities if you remember it was something like this we we don't really expect to To support a beam in that way a column in that way That's just a way for us to illustrate the fact that this point can go up and down, but just can't go side to side So that's the only purpose of that you're not really going to Build the deck in that way. However, as we'll see in a few minutes that What's going on up there at the top and down there at the bottom greatly affects how much load of beam come support So we had this Problem where at midpoint the beam was supported against buckling in that direction But it could buckle in full length perpendicular to the board and gave you also I beam in question that I think was just one right out of the book But we had to use the particular directions and the associated Moment of inertia because it affects how things are going to go to play out So if we if we look right down the y-axis, we'd see they the web going As a hidden line that way if we look right down the length of the beam we'd see those two supports that are drawn there in that way, so there were two possible failure modes if it buckled as We could see from this plane It might do so like that if you remember that's our second of the second of the critical Modes of failure. This is an n equals two Failure where if it failed in the other direction, we have to come to the side and look at it that way In that case, it's not supported in that direction. So it's fairly low would be like that And then that's an n equals one failure mode. So we we looked at some of this stuff on Friday One thing you may not realize is that the length you use For testing this failure mode is this length here You have to put in L over 2 In there because then it then it becomes appropriate for for this failure mode Well, let's see. No, if you put n equals 2 in there You can use just regular L if you don't have the n in there then you use L over 2 And it's going to give the same result either way So watch for that The failure mode That turned out to be critical One this is the this is the failure mode that is critical meaning that it's Sturdier in the other direction because of the much greater eye that's in there. All right, so Anybody get a chance to work through that one if you did you'd find that the allowable Low here and that's after applying the factor of safety Which I believe was 2.5 on this problem. You should get an allowable load of something like three 93 kips That's on the first failure mode being the critical one the second one was not as likely because of the much greater Moment of inertia in that direction All right, so take a look through that one if you still want to There are Consequences of different ways of supporting this these beams we have here the very simple The simplest of all where the ends are pinned Which means the beam is free that the column is free to buckle such that There is an angle at both ends Generally, we take it to be a symmetric situation and in that case we use for the critical length here just simply the Full length of the beam there are other ways other modes to support these beams in That can improve things for example, for example, we can embed the beam at the bottom and Use a pin support at the top In that case the failure mode might look something like This where it'll have an angle Possible at the top is the pin support there, but because of the embedded support at the bottom it's going to tend to to be perpendicular to wall at that place just like bending modes on a cantilever beam would be in this case use 0.7 l because that's a sturdier less likely failure mode so the smaller l on the bottom means an allow a greater allowable load on the top in fact, I think the book calls these An equivalent length L sub e So if we develop the first one in class on Friday, then these are these are just modifications to those possible loadings to others that are possible It's possible that both ends Have kind of a cantilever support to it This one now if it does buckle will tend to do so with no angle at either end because of the The embedded support it could be something just as simple as a weld Down at the bottom and just weld it to the wall However to test the floor This has an equivalent length of 0.5 l and then the last of our possible failure modes is One that's embedded at the bottom, but is not Supported in any way at the top So if there's some load applied to it now it will tend to fail by Just displacing itself. So it's not a very sturdy sturdy load sturdy column support and in fact It is half as strong as what we originally established on Friday Two ML on here squared becomes four So it's one fourth as sturdy as our simplest of modes that we started with on Friday So we have those other possibilities This one's not pinned at the bottom because then if it displaced at the top there'd be nothing to hold it up at all So that's not even a Possible support mode Okay, so let's test drive a couple of those Possibilities Just want to look very much like an old little stacks problem so a cantilever beam well not quite cantilever beam because it is supported with a wire of some kind but Supported there such that it can't Can't deflect in any way at that That load point Will take six feet that dimension on the beam is three inches it's Cross-section though, however is three by two We can see that the three inch and then two inch depth there And we've looked at what that has to do with it a little bit Yesterday so being fairly obvious as we load this with some kind of Load on it that it'll buckle it'll tend to buckle where the beam deflects into or out of the board Okay, so we're looking. We need a couple other things The beam is eight feet long and made of 60 61 t6 Aluminum and we want to find the greatest possible weight that we can hang on there without any kind of buckling failure however also want to check it against exceeding the Stress limit the normal stress limit because remember this is going to be axially loaded Which was the type of thing we looked at in first place So many weeks ago now, okay, so there's a picture Of course what we need to find I hope you realize that this is this is loaded just like a column would be even though it's used as a Being all we need to figure out is What is the critical load there that could cause buckling based upon the pieces that were given? So we'll divvy this up a reference If you remember we had to look at the radius of gyration those parts just a different way to put the critical load equation where if we use the the Radius of gyration Now we also understand that we put a critical. I mean an equivalent length down there so the smaller of The radii of gyration for these two directions is going to be the one that's critical for that so you need to look at Both of these radii of gyration To see whichever one is the smaller. That's the one that we use in here Because that would make the critical load a lower limit. We want to design for the lower limit so take a take a little bit of time and work through that one realizing that This load here Will be a function of w and you could put that in there and then solve for the critical load based on Whatever the weight is that comes directly from a Free body diagram of the very end of the beam Yeah, that's the equivalent length That I just showed you for the four possible support conditions over the radius of gyration which is Directly related to the moment of inertia and The area as in right here Just like we are using mass moments of inertia and mass radii gyration in Dynamics these are area moments obviously that one of the tension must be They all it must be equal to our P there and then that's the P we put in here and then everything else will be known from the situation and you can solve for a Critical load w and you've got all the pieces I think Don't forget that this is a cantilever support, so it's a Equivalent length is at point seven free at one end pinned at one end and Freedom and cantilever to the other solve for this the same as that Make those equivalent and we know everything on this side You'll know this is a function of w you can then solve for w Relikes to this is a 3-4-5 Just the last Monday of the term blues So you can use either one of those you don't have to necessarily find the radius of gyration Because these two are equivalent You have a new state permit here in red pencil and do more damage with the red pencil Just Should have here that P equals 4 3rds w whatever w is the force on the beam itself And we can treat as a column will be 4 3rds of that Everything's given except W. Let me consent all of you have your books. I'll give you e for the 6061 1066 PSI It's not my Chris is that what you just found. Well, you don't want to You may need and what we want is Pete. We need to know what the load is on This being the axial load on the beam that could cause buckling Once you're all done also double check it to make sure it's not subject to just simple compressive stress failure L is 8 feet because of the cantilever support at one end Particularly this point seven now remember the shorter the column the greater the load it can hold so this is This cantilever support at one end is actually considered as an improvement In that it can it increases the load if this was a simple pin to end Then ellie would just be the eight feet Like the coefficient there'd be one and a little less Load in the can by having a cantilever support at one end Sorry No, this is of the of the ones that picture. This is the The second one we're assuming this end isn't free No, look at the top. See here. It's pinned at the top. That's the load That the support that we're looking at this it's cantilever at both ends. See if there's no angle under buckling here There's an angle, but none here. So the strongest of all is the one where there's No buckling at the wall. It's only the center Contacts getting Yeah Kids can't wait these days The numbers they could fell Just check some numbers as we're going along the radius gyration of the two directions That's the right directions. Okay. Yep. All those parts are known. How many were you stuck? You've got last Monday of the semester Could be worse. We could have been starting that 740 or more. Chris, then we'd be almost done. Chris, you have something? where you stopped or Are the flu is catching on everybody 360 kips I Have 32 for the load Kids really have anything on that yet. Yeah, 30 33 kips. What did I just say 32? Yeah, because Smaller one on the bottom of the bottom is the same as it being on the top which makes this number smaller And he designed for just smaller critical load Because if you designed for the greater critical load that'll fail at the smaller one And the book sometimes gives a factor safety and my safe factor safely on oiler buckling That means we expect us to fail in this buckling mode rather than Rather than just a compressive stress for le you're using a I mean for le you're using a Because it's it's a cantilever to one end So he's a point seven hell For le if it was pinned at that end And it have to be a ball joint I guess because I'd have to deflect in either direction Then you use just eight feet then then You can test it against the compressibility limit Compressive stress limit just to make sure that that's just something out of the P over a Joe, where are you? Okay? Well, we know the critical was gonna be Four-thirds of W. That's just from a force balance on the end of the beam Then you put that in here Everything over here is no we know what the material is we know the beams area, we know the equivalent length and R you use the smaller one of these Because then that gets you to the earlier two critical limits Sorry Where'd that come from? comes from Hey, so there's the R squared There's the A and there's the I So if you if you don't want to figure that you're welcome to go back here But you still need to use this lower of the moments of inertia either those two These two forms here are equivalent with simply the algebraic substitution of radius of generation types of supports Why I don't See why it should but What made I haven't looked at carefully is This trying to calculate it myself What they're doing on that? David is saying We have our base that we established on Monday now if we can't a lever One end in the other and now the failure is something like that and they're saying that if we Cut it off here Then this just looks like a smaller version of that and this is 0.7 of that tested against the Compressive stress limit, that's good because that's what you've got there is is Design based on buckling failure It could be that this won't fail on buckling. It'll fail because of compression. It's unlikely For these very long slender type of things are much more likely to buckle But you have to double check that anyway Phil Did you check that compressive limit? Now now that we know what the possible load might be before thirds that will you have it there? then over that area Cannot exceed that given limit that little y on there means yield The yield stress is he pedaling those themselves? No, that's p. Isn't it 32 kips? Yeah, I got a w of 32 kips Phil got that as well. I think Chris has it. Oh, that's right. This is speed and so then you take p and make sure that it doesn't exceed Compressive stress Which it it won't these long slender beams tend to buckle much sooner than Do the pressing failure? That's part of why they're often supported mid-span No, no Phil's got me this over now all the way back to January 23rd. Yeah, I just reset the whole thing So your question? Before you interrupt it yourself It's still Yeah, still That's W. That's what I got for W for p critical Most people are getting it so you should get a compressive stress something like 7.1 ksi which is well below the compressive stress limit as we expect so Your choice is the designers to Could you just flip the the beam just turn it sideways and use it that way just turn it Would that work take this and tell what just turn the beam 90 degrees Then it'd be you more likely to fail Because the weight of the beam would tend to Automatically make it sag or anything like that. It's going to some anyway, but not as great in that orientation So if you just flip the beam over 90 degrees, it's even more likely to buckle Because now the weight is Doing considerable more menning everybody back home on this one Joe Okay, let's bring up another one. This design problem is viewing platform of some kind with a column that you expect to Build in like that. So we'll take this top of this just setting on that column. It's as if it's pinned there and In some kind of concrete footing down there the details being that the column is three and a quarter meters outside diameters a hundred millimeters factor of safety of Three on buckling, which means We apply it there on the critical load rather than on the compressive Limit we have to check the stress, but it's not expected to be trouble and the material is 2014 T6 aluminum and the expected load each beam Each column is a hundred kilometers. So we're going to design for that. I think Like that's all the pieces you need Different kind of a little looking around on eBay came up with whole new type of aluminum to use No, that's scar grail for the people so they don't fall off and there's another guy And here he's depressed. He's thinking of jumping was 3.25 meters by 10 feet He's depressed because he weighs a hundred kilometers. Actually even more if he's in a bit On the column So his weight's calling that causing that he's not even out at the end yet. Oh, I didn't tell you what you're supposed to find I guess I will find the Find the wall thickness of this round aluminum pipe different shapes of columns Can be important to if there's fear of wind loads, which would be like Because the type of bending we look at a couple weeks ago transverse loads Perpendicular to the axial length Which would contribute to the possibility of buckling failure at the top. They're just set on the top of the beams Well, probably not that often but it could be you can set this one off then realize there's some Awesome algebra That's what's done Looster dimensions red for load putting that together I'm not there yet. You're in the game Chris I'm just late in the term. I'm just starting to think of your awesome summer Well, what Cornell and don't you work there? Who presented What Did you go late in the ceremony? Oh, yeah My understanding is she schedules the presenters and the order in which they send in their information for the program So I was very quick this year. Usually it's very late Every other guns pretty late By the time we go through everybody else's exact same speech for the exact same person every time Okay, what are you looking at? Good start we're giving the load We're giving the load But we're expecting it on a Factor safety on that so you can put it in right there which means that Actually designed for 300 kilonewtons because that's going to reduce that Other values put in So that so if you Increase this Then That'll increase I And it's I that you're really designing for you to shoot for a particular value of I so if you Don't design for the hundred kilonewtons, but the factor safety of three is under the 300 kilonewtons Then you'll have an eye that's 300 times. I start three times better than what you needed for the expected load So And when we say the factor safety on buckling that means we put it in this equation rather than the compressive compressive stress So this then Will give you an eye and I will be a function of the wall thickness So then you can solve for wall thickness What is what again for I do you can do it as a con on body? It's just a big circle with a little circle cut out of it And then that little circle as a function of T and so then you can solve for T equal to that for a circle Is one fourth R squared from the polar moment of inertia Which was one half You can do it that way or you can put it in as as the Outer radius minus T to the fourth and solve for the T Either way, it's about six or one half dozen of another We know this we know that we know this why you don't know is I and I depends upon the wall thickness so you're there you're trying to find So you can There's a couple with different way to do the algebra you can put the eye right in there and solve for T It'll be the only unknown or you can get what the value of I needs to be and then solve for T Either way, it's just a Algebra after that got all the points everything known is known then except the T Check your units as you go. We'll tell you if your algebra is right all the pieces solve for the minimum Possible I Effective length is point seven times the 68.9 gigapascals, so that'll give you a I that is It's actually the minimum I For three times the expected load, which means for the expected load it will be More than sufficient and then that I Is equal to this R zero, you know because the outside diameter is given try your dad I think that is what it is Right it up to that. Is that what you were waiting for or is there something else about it? Okay, got you good. It shouldn't be made it Oh Here Be lightweight Yeah, let's just do it. It's a one-layer roll Yeah, you get the thickness of aluminum foil you might wonder Units work on this. Let's say we get oh wait, I didn't square got a square of the Works work yeah, we'll get meters to the fourth on that one You guys there's a slightly different 73.1 that'll change it's a little bit Oh, is it Okay, so what's this one 73 Actually would have made it more even more conservative with lower E. What is it 73.9? Huh I Well, I had the I was off by a little bit on that, but yeah, I'm around seven millimeters Are you guessing? Oh seven meters Seven meter wall thickness Say if I have this intermediate I number I don't have that written down The answer is less than seven millimeters five seven at some point two millimeters with the wrong All right, so in terms of a millimeter That's pretty good, that's like a Stack of So the cans Yeah, you don't have to do that stuff if you want it don't want it, but generally if you break it into steps It's just easier to check as we go along Then you know which step is wrong Let's make sure we got 3103 that's right point seven of course 73.1 R squared This is this is diameter everybody got that into account. This is diameter here. This is radius here You got that? No This is diameter that is radius Do that I don't know typically you buy pipes on outer and inner diameters But the I Calculations we have are always based on radius don't blame me. I didn't make up this deal. I just So I pass along with all of that traps and trips who be Hitler's Think of this is very funny A lot of textbooks But I am put in diameters radius So that our zero we need over there is 50 millimeters Watch your units though And going in the last week it's a little tough for anybody else to pass you did you get something 57 Tom you haven't had this number just I write it down 2.28 times 10 of the minus six meters to the fourth and then that Was equal to this with the only unknown as the T And it's fairly easy to solve for it knows a Quadratic of a quadratic or what are you called for or take for any by one of these fights yet? That's why I say let's do the one I like to say hypercube That's that's to the fourth is hypercube Yes What's cute to the cube though? Depends on what a nine-dimensional cube would be Poor guy being gonna fail or he has a chance of size. He wants to jump We're gonna be there there's just gonna tilt down. He's just gonna roll off. It's gonna be really pitiful We might go back there anymore It's great There's a song lyric from What a shot you would be if you could shoot with me The angry guys Full murderous intent Can't go back there. Can't go check on Phil's words. I have to go past her Bill is that coming out? With this pin support we expect the failure mode Just gonna be something like that if it does Oh, that was my wrong I had the wrong I had to be for the the 60 16 alone instead of 2014 This yeah, but it's it's gonna The 68 9 down here is gonna give you even a more conservative answer So it makes that your safety 3 maybe 3.1 No, because it's It's got no directional bias to it. It's just it probably be because Either it's not perfectly vertical and installed or there's imperfections somewhere in the building of it itself Could be whether this is an extruded pipe Which they they push it out through a hole like the noodles or it could be a Seen pipe where they roll it and weld it which isn't common with the loom Become extruded There's other factors Consider maybe Maybe something's not quite right there. Maybe that wasn't cut right at 45 degrees or there's a little gap there or something So all of those kind of things factor in But Chris you guys around seven millimeters a little bit less I Yeah, I just put this whole thing equal to that number and then that goes over that goes over then you Unforth it Solver team that That takes us to the end