 They take this first very steep part, give it a different scale, and then they finish up with the second part and a different scale. There's two things I would do if I were them that they don't do if you look at the book. One is, if I remember, this line is kind of a light blue when they do it, and then the scale is kind of a light blue, and then this line is kind of a slightly darker blue, but then the line that goes with it is a black and they don't match. This way the black goes with the black, the red goes with the red, and it's pretty easy to tell what's going on. The trouble with that is, though, that when you photocopy it, or I just sent it out on my printer, same thing as a photocopy would do, now you don't know what's going on. So that's hazardous. I also, if I were them, would add this little dotted line just to emphasize that it's meant to be a continuous line that's connected that way, and it's just easy for the eye what to do with that. So if I were them and could do a book in full color like these books are, which is hard why they're so expensive, I would do it the way I first had it with the red and the black or something like that that just stands out a little bit more, I think that's a mistake they've made with their graphs in the book. But the way I think I myself would do this if I was presenting this data is this way, where the real steep part is on an expanded scale, then there's a break, I'd worry about this and work on it a little more, I had some time to deal with that kind of stuff, but then I'd show that we've gone now to a different scale for the rest of it. Now this doesn't depend on color, it doesn't depend upon, it doesn't have the danger of any misinterpretation of which line goes with which scale. It's obvious this way which line goes with which scale. I might also maybe put up here where there's a little bit of room, put a single graph that shows the thing to actual scale for the whole thing just to kind of also illustrate why I'm doing this break out so that this kind of stuff becomes more apparent. This comes from a couple problems, the problem 3-3 in the book, it's just a data table and they said to now graph this in a way such that you think might work. So I think there's a couple different ways to look at those things. And it's part of what you need to learn. When you do work like this that needs to be shared with someone, which most work does, you need to consider carefully how it's going to look, how you're going to present what you've got. So it's the most used to your reader. I forgot about that, I didn't get that light fixed. Alright, so we were looking at a problem, didn't quite finish it. Let's bring that back up, see where we got, see what we can do to finish it. We had a beam hung from a cable at one end and simply pinned at the other. And then about a partway between an unknown weight that was hung there. Because of the presence of that weight it caused the beam to deflect position down to there. So we had a couple numbers that went with this all, 3 feet, 2 feet. So this was 2 feet from the end where the cable hung, another 3 feet of beam there. I think the deflection, oh the deflection was a point B. Did I say that yesterday I put the deflection here? Because that's what it was meant to be. Downward deflection of the beam of 0.025 inches. Huh? I did have it at the end. Too bad, what it does is it just changes the number some a little bit. So we didn't get all that far with it though, did we? But you worked on it some at home? Yeah, I got it on. But should we continue through with it some? Sure. Okay. Did anybody go farther with it with these numbers? Because if I put the deflection here that will match my numbers. But we can also re-figure it matching your numbers either way. Depending on how much you got done. I don't think so. Okay. Well we might put it here because it really doesn't change things too terribly much I don't think. The two cables were both A36 structural steel. That kind of thing was never a concern before. We never cared what this stuff was made of. We will be from now on because we're going to need material properties for us to decide some of the engineering dimensions and designs and stuff that we're going to be relying on here as we go a little bit farther. Had a cross sectional area of 0.02 inches squared. That's both cables here. Oh by the way that one was 4 feet. And at this time we're considering the fact that the beam as a real structural member would tend to bend a little bit itself in the center because of this weight hanging there. We're not concerned with that yet so we're going to assume that there's no lengthwise deflection in that beam at all. That it remains straight. We will later add into the fact that the beam will bend and then we'll have the deflection of the point and the deflection of the beam to take into consideration in some of the design problems we're going to be doing a little bit later. Alright so with those numbers we needed to find a couple things. We needed to find the strain in the cable ED which is really just a problem of geometry. We know it's original length. We can use the little bit of geometry and find out how much longer it is now and then that's the strain. It's the deflection in ED over the original length of ED. Now with yesterday if I accidentally gave you that number then it's a real quick calculation you just plug those things in and they're done. With the deflection moved back to where it should have been it's not a big deal. We just have similar triangles and so we're at a distance we could call five-thirds. This is going to be five-thirds that deflection because we're five-thirds the farther distance along so it's just a geometry problem there. And hopefully isn't too big a piece to it. And so using this drop diagram with that deflection in the right location we have a strain in that cable ED of something like that. Oh yeah sorry that's still I looked at the wrong one and that gives a deflection, sorry a strain of 0.115. That's a geometry problem that part but then the other things we were to find need to find the strain in both cables. So we still need to find the strain in the other cable that one was labeled BC. To find the strain in BC we'll need, well we need the same type of thing that we had up there. We need to know the deflection in the, sorry not ED, BC over the original length of BC. Well the original length of BC that's easy, that's given, that's the four feet right here. The deflection of BC that's not this value given here, that's the deflection of point B. Not the elongation of the entire cable BC. So that's a number we don't happen to have yet. Do we have this? That's what we were supposed to find. That's one of the things we were supposed to find the strain in both cables. We already found one, I'm supposed to find the second. We can't find this unless we have that. So then what? We do have this now appearing in another place that we got yesterday. We didn't have before. You could use the elasticity of the E equals PL over the del A. We have the fact that it's A36 steel which means we can pull out the book, remember what this E stands for? Elasticity or elastic modulus or? Young's modulus. Young's modulus. Any one of those. Some books actually impact and think in physics one we used a capital Y for that value. Same thing, just a different choice of terms there. So we pull the book open, look at the U.S. units. Structural A36 steel and we've got a modulus elasticity of 29 times 10 to the third KSI. Remember what a KSI is? Kips per square inch, kilo pounds, a thousand pounds per square inch. So this is a 29,000,000 pounds per square inch. So a million pounds per, 29 million pounds per square inch. And we know that to be... Well how did we find this Young's modulus and how did we define it? This is what we just got yesterday. It's experimentally determined. It's the only way to find this out. You've got to run the material through the test machine. Do you want to stress over the strain slope? Well, that's a little redundant. Stress over the strain is the slope. Remember we have the test data. Something like that, what we're concerned about is this linear portion, this elastic portion. And that slope is what we use for the elastic modulus. Again, experimentally determined. Probably something published by the manufacturer of that material or those structural members. It's generally not something that would be determined on site. You have the stuff delivered. You already know you long since know what the elastic modulus is because the design depended upon knowing that. You might need to test it just to make sure that the batch you ordered was as promised. But you wouldn't necessarily retest that at every step. You've got to trust the manufacturers to some measure, I guess. So we could use the stress over this elastic modulus to find the strain. Do we know enough stuff to do that? E we've got, we just looked it up. But what about the stress? We could find this divided by the elastic modulus. Then we know the strain in the cable BC. How do we know the stress? What is stress? Besides being a college student trying to drive through the snow all the other things that are pestering your lives cause them stress. You're not stressed. This is stress, man. What? In this case, it would be for this cable, it's W over A. A being the cross-sectional area of that material supporting that load, which is this cable, the two cables are the same. Well, we don't know W, though. If W was zero, then this mean wouldn't do this. We wouldn't have any problem. Good try. It's always tempting to say everything's zero and we're done. You just cross out the whole page. Sometimes you have to assume values. Sometimes you'll assume zero and you're done. Can we find W? If we can, everything else is known and we're done. If we look at the beam, we have this unknown W here we're looking for. It's the thing that causes everything to happen anyway. It's being opposed by some force there in the cable ED. We're looking for some support reactions here at A at point A. That's not area, that's just our point A. If we knew this force, we could find that W because we could just sum the moments about the point A. That way we don't even care what the support reactions are at A. They weren't asked for. They're not a concern at least in this part of the problem. We could sum the moments about A. We know they'll sum to zero. Remember this class is always done in static equilibrium just like statics was last fall. If we could do that, that would give us ED. That would give us W and we could finish the problem. How do we find ED though? We know the strain in ED. The strain in ED is just like it was in BC. It's going to be that load ED over that area divided by the elastic modulus of the material. We do know the strain. We do know the area. We do know the elastic modulus. We can get ED, the force in ED directly from here. That should equal, anybody have it? 6, 7 to find the weight that keeps this in equilibrium with that then the strain in the cable BC. With the strain in the cable BC, we could find the elongation of BC and it could be that the deflection of the beam plus the stretch of the cable BC then makes this gadget unusable. It could be that this weight is just barely above the floor somewhere and once all this stuff is put together, the fact that the cable ED stretches and the cable BC stretches and the weight now gets the floor and is out of design specs for whatever reason, for whatever purpose this thing would serve. That's why we need to look at those deflections now. The deflections, the elongations, all of those things are going together as we put this together. Anybody have any of the missing numbers still? We're still looking for the strain in BC. We need this force ED, the force in the cable ED to find that. We need the W. That was one of the requirements to find. Anybody have some of those pieces yet? A pretty fun class in that we've been here for almost three weeks and we're still just dealing with nothing more than a few simple ratios. This is the easiest class you'll ever take. It just gets easier. Bob, how you feeling? Amazing how long it can take this dust in. Shake out. We got some numbers. I should have done those. I should have taped it out of focus so it would have been like you were really here. I taped it out of focus and kept the camera going and then rub the camera over the trash cans and back. I put it up. It takes about an hour and a half to get an hour of class up to iTunes. In fact, I found it's not even worth doing at night because the whole network's a little bit slower because everybody's updating Facebook and let us know what happened with Billy Biffin's scooter that day. Any numbers to share? We've got a brain trust of five over here against three over there. I had the wrong dimension. It's the same process, right? Just slightly different numbers as all. Fair about that. So you can check them since you don't seem particularly in the mood to drive it forward too much right now. For this setup, I had a final weight, a 112 kit put on there and a strain in BC 193.112. Oh yeah, sorry. If I'd actually written down pounds, it would be .112 kit. Either way's fine. Things we can do now we couldn't do before is just make sure that once under load, all of these things are still within specs. A simple problem like hanging a weight from something might mean the weight no longer hangs like it's supposed to, whatever it is. If this was a cattle wrestler that were hanging from a tree limb, you don't want his feet touching the ground. He's not going to learn his lesson. The other cow folks are going to get the message and don't steal our cattle. So they don't show it in the movies and hang them high with Clint Eastwood. They don't show the calculation they went through first. Do you know Clint Eastwood at least? I may not have seen him hang them high, but yeah. Any other questions on this one? P.J., you got it? You think? Enough? Can I start erasing a little bit? Some of this. In case you're still going. Alright. So let's try another problem. Say we have a structural member and a nuclear reactor. Very simple piece of strut of some kind. Made of zirconium. Three feet in length. Subject to a four-kip load. I need you to find that load is safely supported. You certainly don't want structural members and nuclear reactors failing. Use a factor of safety of three with respect to yield. That means we've used a factor of safety before. Had that a couple times last week, that thing. But I don't believe at that time the term with respect to yield. Exactly what yield is. If you remember yesterday when we talked about the stress strain diagram, we had this linear portion that was of most interest to us. Because that's where the load is most predictable. The response of the material to the load is most predictable. And it's repeatable. If we load a piece and go up this curve and unload it, we come back down the curve. There are curves in engineering where you take a different path up than you do down, which can complicate things. This kind of thing may have come across in some of the work with Professor Wulmer. Possibly, I don't know. This is called hysteresis. A path that is not the same in one direction as it is in the other. That's not the case that we have here. We follow identically the same path up or down as we increase the load going up or decrease the load coming back down. At some point we lose that linearity and the curve does whatever the rest of the curve does. That takes us out of this elastic region, the slope of this line that gives us Young's modulus. It's the elastic limit we define as the yield stress. That's one of the numbers you see in the back of the book for those various materials. So a factor of safety with respect to yield means that the allowable stress for this particular problem will be one-third of the yield stress of the material, which is a big reduction. That's the kind of thing you need to do when you're working with something that a failure would be absolutely catastrophic. Certainly you know the kind of thing that happened with Chernobyl. I don't remember that it was a structural failure. I don't think it was. I think it was in trouble with their control rods. But it was devastating that Chernobyl is no more for the most part as a city. And it's been 25 years this April. How do I know that, you wonder? How do I know so quickly it's 25 years? Because that's my wedding day. Same day as Chernobyl. Not only that, it's the same day Arnold Schwarzenegger and Murray Shriver were married. So how do you think that makes me feel? My wedding day and we're on the third page of The New York Times to this day. All right. So know the load. You need to find the area that will support that load, which means you need to find the load and what stress, what area will keep us under the allowable stress with the appropriate area that will also come from Young's Modulus. I told you. Oh, did you? No, I told you. You're not listening. Zinc? Do I have to run the... Zinc? No? No? Zirconium alloy. Zirconium alloy. What, it's not in there? Well, maybe they used wood instead. No, I'll give you those numbers. Those, that's not available, so. It's the yield strength for Zirconium alloy, 57.5 KSI Modulus of 14 times 10 to the third. Those are the values of interest for the... For a second, we actually could have done this problem last week, but there's got to be more to it. If this paper is something you need to refer to months from now or your photocopy at a couple of times, it may not be clear. Keep the zero out in front of that. And then even if that dot disappears, you look at that, say, you know there's a decimal place there. So you kind of protect yourself with what she got. So it can't be both of you, right? It's also quite different than what Frank got, but I know you guys aren't on speaking terms. It's different than what T.J. got. Brandon, what'd you do? We got numbers all over the page. This... A zero radius is that 4-kip load will float in space all by itself. This is easy stuff. This is all from yesterday. Once we get this allowable, we know that it's going... or not even yesterday, last week. We know that stress is going to be P over A, where the P is given, and then A needs to be the right size to support that, to keep us under this allowable stress. Maybe the hang-up is that then have to remember what the area of the circle is. No, that wasn't it. Something else got you screwed up. Because we got some people saying something like a quarter inch, and others saying five inches. So you don't want to go to your boss and say, man, you know, somewhere between a quarter inch and five inches. You know, it's only a zirconium alloy. Are you missing some decimal places somewhere? Allowable comes out to be 19.2 KSI, right there. So we know the load P. So the area should come out to be 209, 207, I think. Something like that. Remember, we've already got a factor of three built in there anyway. So the last decimal place may not be of a particular concern. Too much. Jay, you found your mistake? Yeah. Would you like to share what you did with us, or is it algebra? Algebra. Algebra. Man. Still learning. And so you get a radius of two five eight inches. Get in the habit of keeping this zero out in front when you write numbers down. Because you're going to work on stuff and put it away for months. Photo copy of 12 times. And that dot, if there's zero in there and that dot disappears, or if there isn't one there and it appears, that zero will clear things up for you. Even if it happens to disappear in photocopying, you'll still be able to tell. And that's got to be a decimal place in there. It just wouldn't make sense any other way. But you've got to be in the habit of doing that. Good practice. You won't see the book without a leading zero on decimals. The design decision is finished. The only other thing you'd have to look at is that the radius you'd specify, or would you have to see if maybe this comes in some stock radius that would be a lot less expensive if you ordered that rather than have it custom made to that radius? What if it's available at a quarter of an inch? We've got a factor of three in it, but if you go down to a quarter of an inch, you no longer have a factor of three, and that could leave you liable if there's some kind of failure. Even if you say, well, come on, I had a factor saved you 2.9, and they said I'm sorry, the design specs, which may be legally determined by a national agency. They certainly are with other things we're going to look at later. We're going to look at pressure and the stress in the walls of a pressure tank that's very strictly regulated in terms of what kind of pressure is allowable and boiler tanks and the like. And if you were supposed to design to a factor of safety of three and you hit 2.9, that may not be good enough, but you'll be protected with that factor of safety. So you go to a greater one if you're not going to do that 2.58. All right, let's change it a little bit, figure out what load is allowable, load such that the change in length of that structural member itself is no greater than we hit that amount of, in this case, elongation than the structured self no longer works card. Maybe gaps form, you don't want gaps forming in a nuclear reactor. Well, yes, the four kips is out. I mean, if we keep the four kips, then what's there to do? You can take the four kip and check this. Well, is that radius still going to be relevant? Yeah, we now have a radius. It could be that even with the radius, we can support this load, but maybe we cannot allow this much change in length of the structure itself, and so we're going to have to modify that four kip load somehow or modify the structure itself. Put in more of these zirconium doodads. That's the technical term in nuclear reactors. In your experience with designing nuclear reactors, you mean engineers down the hall, with whom you might still be on speaking terms with any. Recurring number, four kips on there. We're going to exceed that. Maybe you checked that first, maybe you didn't. It's very easy to start with this strain because we've got the original length. When we've got the strain and Young's modulus, then we can go to the stress. Once we have the stress and the area we had before, we can then find out what the allowable load is. Basically what you did? We could have a new radius, or not because we want to. Well, I don't know, it depends. These problems are much narrower than a real design problem. Because you might need to say that, well, that's too small. What do you get, 1.62 kip? Something like that, 1.6 kip of some kind. You might find out that that's too small. We need to support four somehow. So you've got to decide what do we need to do so we can get this back up to four. Well, yeah, we may need to increase the area. We can increase the area by possibly just putting in extra structural members. That takes up volume. Volume in a nuclear reactor can be a critical issue, especially if it's in the nuclear navy in the submarine. Space is an absolute premium in submarines. So these are all the systems you need to iterate around and around until you get to a number that satisfies all the available criteria. And sometimes it does mean you've got to go back somewhere, maybe re-investigate this. Maybe this is too strict. It could be that whatever number led to this four kip, we've got to back off on that, which maybe has something to do with what the reactor's doing when the vessel's under full power or something. Lots of a million things. That's why we've got to have you here for four years before you can even pretend you're going to work in anybody's office. Yeah, long way to go on. What about what? Do you use the different material that they're going to modulate? That's a possibility, but there's other concerns. It depends on just where this might be in the nuclear reactor. There's a big concern in nuclear reactors of something called hydrogen embrittlement, where the material is subject to bombardment from atomic particles itself with all the radioactivity that's going on, and the material properties change. So sometimes some very specific, even expensive materials are required because others will not last in the nuclear reactor. A million design decisions you're going to have to make, which is part of the joy of engineering and part of why the schmo's out on the street can't do it. There's a reason that the most popular major in a college is often business or... What's the other one? When the guy becomes a corrections officer. What? Criminal justice. Criminal justice, yeah. Criminal justice. God made criminals so that we can have the criminal justice major so that we can have a place to put football players. Yeah, the tape was run, that's good. I want that one on record. So as we're looking at these things, we're looking now in a lot more detail at the material response of these items as we've got here, whatever they might be. Jake, how awesome is that? We're looking at the material response to these structures under some kind of load. So we start here with an item in some kind of initial unloaded form. So there's its initial shape. Apply some load to it and we'll make it a simple axial load. The material is going to respond. Mechanically respond to that load. Response is a concern to us. So far for the most part, except for one little time we looked at a slightly different problem, we've only been looking at the possibility that this material will elongate, at least under a tensile load like I happen to draw. If it was in compression, it would compress. But in this case, it will elongate. But since the density of the material will stay the same, if it elongates, it's also going to have to narrow down. So maybe the response of the material will be not only elongate, such that the density stays constant. Oh man, this is so much fun drawing this stuff. Jake, you've got to be all excited about the possibility of doing this drawing. This is so much like technical freehand sketching here. Not too bad. That'll do for a cartoon at the board, I guess. That looks a little bit better on the screen. A wobble on the top side to it there. But let's see if anybody's doing it better. Well yeah, it's always good when you think it's so tiny. You've got to take it. You've got to be courageous. Go big or go home. You're not working for the post office designing stamps here, Frank. Look at that, awesome job. Kind of small. Do I have to show that Professor Hampshire? Do you want me to do that? Not how'd you do? Not bad. It doesn't have to be great, as long as you can tell what it is you've got there. As long as you can tell what it is, it's going to be okay. Alright, just to help us, we'll put some reference directions here. We'll call that the y-axis, that the x, and the z coming out here. Because what we've got now is different strains in different directions. For example, we have an axial strain, which is just like what we've looked at before, which is in this case the elongation in the x direction over the original length in the x direction. In this example, that would have to be positive because the material got longer. But we also have a lateral response. We have a lateral strain, which I guess we could call in the y direction, which is whatever the change in dimension is in the y direction over the original length in the y direction. Well, for this particular drawing, we also have a response in the z direction, but we'll assume that our materials are what we call isotropic and that there's no difference in the y or the z direction, but there may be in the x direction. Wood is very much like that. Wood is very different axially than it is in a transverse direction, but once you look at the transverse directions, it doesn't matter in what orientation the wood is. It's pretty much the same across the piece, but it may be very different down the piece. So this assumption here is that the material is what we call isotropic. The material properties are the same in all directions. Well, this here is under a completely different load. It's the part that's in attention. Then the material responds transverse to that, which is a different load. But if we could take this material and put it in any orientation, and it didn't matter which one is x and which one is y, and we get exactly the same response, then it's isotropic in all directions. So we have one last thing to throw in here, and then we'll be done. It's, again, a material property in just how the material responds in an axial and a lateral direction at the same time. We have what we call Poisson's ratio. That's the Greek letter nu, I believe, and it's defined as the absolute value of the lateral strain. The absolute value simply because one of these is going to have an opposite sign of the other, because as it gets longer in one direction, it's got to get shorter in the other and vice versa. So this way we keep it always as a positive value. In this case, it'd be minus the lateral strain over the axial strain. And if you open up the back of the book, it's the second to last column in there, and you'll notice everything's on the order of 0.3. Very typical value of around 0.3 for Poisson's ratio. We looked at this briefly with a problem we had where we took a piece, elongated it, put it under tensile load, and we looked at the fact that it also narrowed down on a stress. That's one of the problems we looked at on last week sometime. Alright, so we'll start up again then with Poisson's ratio on Friday.