 All right, no questions about about the column business that we started on On Friday, all right, let's do a little bit of warm up one One I want you to size a column a Square column and we'll use wood So find the size of square column under two different possible loadings just to Get things going here a little bit Remember Depends a lot on Young's modulus We use a factor of safety of 2.5 Pretty standard 1.5 is actually cut in a little short Length of two meters about a six-foot column so about something you'd have on a deck And then allowable normal normal stress of 12 Make sure I got those right 13 oh, no, I'm sorry. Yeah, you can pass gals looks like grandpa 13 grandmas All right two loads for kilonewtons and for 200 kilonewtons, so we'll keep it a square post so the Moment of inertia is easy Minus square post because that's that's what that picture looks like. It's like where sound oh Not a red oak Frank has your foot doing Where'd you get a hat from color? I don't beat up some little kid at the mall for it All right, so we had that critical load equation Couple different ways to apply the 2.5 factor of safety, but they all work out about the same So based on those is the critical loads come up with an appropriate Calm dimension for a square almost those properties without the slenderness ratio That didn't really change anything terrifically up on the board you can have a square column That will support these two loads given those so essentially it's just this I've already given you the load You need to apply a factor of safety He you've got l you've got i over a for a square column reduces to a pretty simple Number that's the same one remember Lenderness ratio in it using the radius of gyration which you don't have to use Because that was just sort of Algebraic change in this equation Slightly different way to look at the same problem at the units. You're right. We're gonna get the a no with i in there Well, I'm sorry. Yes, that's that's actually for the Normal stress That's what that you would have cut that with the units anyway. He's given i Is unknown because that's dependent upon the dimension you can solve for a from that what you want this to be as big as possible Because that will protect more so if you multiply Factor safety by this you're protecting for a greater critical load. You'll have more Cross-sectional area doing that they're protecting for a greater load than that If you divide and I have a beam that will have a very very low critical low and won't take much to make it break you want You divide then you're going to make a beam that's only good for What 40 kilomens what you want is a beam that's greater for Well more than this low. It is tricky sometimes what? What you've done always solve for i multiply that but that would be the same thing However, you want to look at it, but if it said like this isn't the maximum allowable force then you want to provide that If I'd said this this Well I can't see how you'd say this is the maximum allowable force Without that factor safety already in that and that's what that tells you is that the factor safety says The maximum allowable load will be 100 times this I mean 2.5 times that load that's the maximum allowable load Got something Bob put your wedding reception out on that deck See you gone at four times today Now you got some Did you already use the factor safety don't use it again Use it once there are different ways to use it This is the end. It's just multiply the expected load by the two hand Design for a greater expected load Or see before right because BHQ be an age. You're the same Since it's squared things to square in here Something or you've got to check for the normal stress limit as well Lovel stress 12 megapascals is expected with this beam first 25.9. Well, you use the 98 down here It's close enough, you know, you don't want to be cutting The critical loads already in there. I mean the factor safety is already in there. So you don't want to be cutting us any closer That comes out to be just the 25 kilo Pascal's Mega pass so we're well 25 megapascals almost double the Allowable shear stress to beat that beam up. You can look at this Is that a shear stress limit are the general idea is the same There's a buckling limit that looks Something like that Where as long as you're inside of that curve You're going to be okay curtain point You have the bowel or the yield Normal stress limit So that the rear design curve then looks like that for the normal stress concerns After after a certain point after a certain slender that's ratio, even if you didn't use this calculation, there's no numbers on that Buckling is a concern on one side Normal stress failure yield stress on the other side Of some some sort of division in between Certain with these and then we're all done is that the model we initially used for this of course a bit artificial in that no one really makes a beam that's Simply pinned at both ends and more likely to make a beam that has different Methods of fastening there or up there as well. So there's other a couple possibilities so we Define an equivalent length where if you have other certain type of Support methods at the top or the bottom Then you change the equivalent use that number in the straight calculations as before The ones we've got there where it's pinned top and bottom That's the type of mounting that we use to come up with this anyway So for that case, that's our base condition of L equals one nothing will change. That's what we're assuming for this column support method anyway another possibility is That we model it as pinned at the top Embedded at the bottom if it's a metal Welded there if it's in a cement base, it could be sunk into the floor itself The point being that this kind of loading this kind of support can offer some Moment which will make things a bit stronger than the simply pinned One and in this case Then le is reduced a little bit. We can imagine an even stronger It's kind of like getting 30% beam strength for free if you will Because of that change in the number is that same kind of support at the bottom But at the top sort of the same type of thing that remember allow for for strange shrinkage due to the compression of the beam itself, but this model Moment support at both ends Reduces it even farther. It's like getting out twice as much being for free if you will then one last possibility is that of Being that's got an embedded type support at the bottom and we model it with no Lateral support at the top which is what this channel type thing was supposed to Represent the fact that the beam couldn't the beam top couldn't go side to side It could only go up and down as the beam buckled now There's the possibility that not only do you have bustling But there's no lateral support, so the the beam could essentially just I guess flop side to side So this of course is the weakest of all of the Possibilities So it only works for half as much being this too doesn't mean that you can make a beam twice as long And it's just as good. It means you can only make a beam half as long Because of that don't look all right then so we'll look at our last problem of the year We decide just what what type of Loading it represents. So we have a beam here cross section e by two oriented in that direction eight feet long with a cable support there be above the Wall anchor to support some particular weight, so you need to determine what that is make the beam out of sorry 61 t6 aluminum liable Normal stress of 37 ksi yield normal stress for the allowable find other things one of these Support what the critical load is You've got I you do not have le you have to decide from one of these possible situations Oh, I you do have I forgot what gave you the cross section So determine what the load will be When you have all the stuff on that side, but you think I'll eat it Don't pass through that point. So there must be some moment there I don't know if they need to look at it or I was Part of this part we looked at the bending moment in the beam Which is different than the moment about a we looked at the bending moment in the in the beam MC over I That's how we've got to here. That's how I got in here When we develop that last Friday The W you don't know which will make w a factor p or w That is within that critical load limit, which one you think it is Well, it's clearly not the first embedded at this. They're really not the Third moment supported that end from all I leaves the second and the fourth is Possibilities so a couple of you thought it was the second one Anybody think it was the fourth one your professional careers or some things you might actually want to say out loud Even if they're cooking around in here You want to just shrug and get a lot of them doing just jesson Well that the question and I guess is is is there any possibility of that end going? That way if we look at this from above then we have the beam and the wire Running down like that, but that's where the tension is that's causing this Load that gives us the possibility of bucking is there anything available to keep That beam in from going left or right if we consider the ideal situation this this cable has no possibility of Stretching in it If the beam happened to deflect a little bit To one side or the other and the cable Can the cable prevent that when in doubt you need to go to the more conservative choice even safer? Design so if in doubt which one of these is the more conservative number four I Don't see that for this design. There's any reason that these and couldn't go side to side In my mind that was load Support situation for We've you've probably seen Cases where those kind of beam or those kind of you know a sign or something might have Two cables Running like that now that is a situation where we have to take that the end Can't go one way or the other because one cable prevent From going the opposite way each time so I think you're right that Well, you're not a professional explainer But on that yeah, no there'd be two Yeah, that's top you So this the single cable that goes down Both are the problem the big one. No, this is a side Looking at from time Though with the cable right in the right down the center I don't see any reason why the end of it couldn't couldn't move laterally and Prevent the end from moving laterally and then put two cables again here seen from above Above pictures are we small? Of it down at the end in which case we would still use the same for too well well don't forget though that there's a Tendency of this thing to buckle this way Much more than this way And so if you're concerned with the possibility of it bucking buckling up or down Then you have to take into account the other direction on the cross-section for the moment of inertia All right, remember we looked at that on Friday Is a possibility if this was a square cross-section? Then there'd be no preference for which direction would be buckling you have to look at it in both directions Yeah, the possibility that yeah, our Y is less than rz so it's more likely to buckle with the The beam actually moving in in or out. It's more likely to buckle in the Upwards direction From the upward view would be from the side. You'd be okay. And I seem reasonable No W there cable response to that and Some load there and that's the seal load that we're going to see on the That the beam needs to prevent the shape of the triangle it should come out the peak was four-thirds W just based on The Static of that that end piece the lesser lesser direction What's I you have is I might have that it's a number and the slenderness ratio L e yet as long as you use the minimal form by using our Y So anybody take that through then I give the L over R squared. So these are these are still the same thing Just slightly different version 4,016 something like that. Did you check for possible normal stress failure? No, well, well, we'll just do a simple buckling one like we would at Friday Pinned at both ends straightforward like that So that'd be a beam design problem buckling possibility problem and What do we have super positions?