 We're through chapter three now into chapter four. And I've talked before, I don't know, chapter four strikes me as a little odd. I remember every time this spot in the year, I look through my notes and I go back and I try to figure out what it is about chapter four that's different than what we've done before. And it isn't a lot. There's a lot of what we're gonna do in chapter four is reminiscent of what we've already done. We are gonna take it a little bit farther. So we'll continue with our look at axial loading. A little bit of a review here. Just make sure we've got all the pieces. If we've got some kind of part or member of some structure or even a machine of some kind, it could be. And it's loaded actually, meaning it's loaded in the direction of its greatest dimension. You mean that axial loading is much smaller in width than it is in length. And the loading is in the direction of the greatest length. So if we have something like that, we'll even put a couple numbers to it just to run through them. So let's say there's 100 kilonewton load there and a diameter of 50 millimeters and a test section in here that will give 300 millimeters. So you can see it's a lot longer in the direction of the load than it is transverse to the load. Very shortly, not chapter five. I don't think maybe chapter six. We'll look at loads that are in the direction of the lesser of the dimensions, transverse loads, loads perpendicular, at least perpendicular to start with in terms of the greatest dimension of the part. All right, so we stress this part and we find that we get some kind of extension in the axial direction of perhaps the test results or something like 0.2 plus a little bit millimeters and a little bit of extension in the, actually be a contraction in the radial direction, transverse direction minus 0.01215 millimeters. All right, so nothing we haven't done before. It's sort of running through it again. Make sure we've got all the pieces we've meant to come through so far. Maybe I'll put a little Y on here just to represent the transverse direction. Maybe an R would do since we have a circular piece but it's not a great concern either way. From this, and this is exactly what a test for these very properties would be, we can find directly from this the elasticity of modulus and the, it's sometimes called the elastic strength or something like that. So it's a very same type of number closely related to the spring constant that you saw in physics when we played around with these blue springs. And then this blue brick later new, remember that from Friday. We'll get to each of the pieces as we go. So this elastic modulus defined as the normal stress over the normal strain. Now the load over the cross-sectional area carrying that load divided by the elongation divided by itself divided by the original length. And then from all that, well I don't know if we can clean it up a little bit. We can calculate the elastic modulus from the real quick test that could be running. And if you've looked at the videos, these tests are very quick. It probably takes longer to load the machine and actually goes to run the test. Takes only a few minutes to run the part and certainly at least through the elastic section. Remember this stress strain relation and especially this elastic modulus region. We're most interested for our design purposes in that linear region, the elastic region where if the load is relieved then the strain is also relieved. If there's any residual strain it's because we've gone past the proportional limit somewhat and we're most interested in this linear section. It's that slope that actually is the modulus of elasticity. So we can real quickly put in these numbers. You can finish it up as a double check of the units. We'll use the test length there as the original length and the elongation, nope not the minus one. The one in the axial direction and then the area, I think I had that already. Yeah, 1.96. All right, that's just of course pi R squared. So as a Monday morning warm up run through those a little bit, double check your units and give me an appropriate answer in gigapastals. So that's your Monday warm up there. Pretty good morning to warm up there in a little bit late, this is just review. Chuck those numbers, this bottom one here is just the cross-sectional area and then put that in gigapastals. We've got a lot of different sizes here, the Kila and Mila, lots of business things going on there. We've got to take care of them all. Travis, you there already? Ready to commit? Yeah. We'll see, Thomas counting over his decimal places. Let's pray. A little bit around if it's not of great concern that we have a lot of precision on these numbers remember. We're gonna build in a big factor of safety depending upon the application and what could happen in the case of failure but the factor safeties are typical from 1.5 up to 5, 6, 7, even depends upon what would happen if we had catastrophic failure. We just lose the use of the machine which is the type of failure you might have in a snow blower if a part failed or it could be a severe loss of life or even loss of life and loss of a huge building or something, we don't know what could happen. Lots of these kind of things in history. All right, a couple of people have the answer. Comes down to about 70 giga-pass counts. Pretty typical for the structural solids we're looking for. All right, then the next goal is Poisson's ratio. Remember how that's defined? If not, it's in your notes, right? Should be. Bill, you remember? Absolute value of the x over e1. Actually, e epsilon. It's the normal strain, the very same strain that we've got here where it's the axial deformation in the direction of the load. And then because the volume, as the density stays constant, if one dimension's getting greater, the other dimension has to become less to keep the volume the same. And we take absolute value just to make that a positive number. Since one of those is an expansion, one's a contraction, they're always in opposite directions to each other. So you can figure out that real quick because we've got both of those pieces. The normal or axial strain, we've already got here, didn't calculate it separately, but we've got the two pieces there. And then the lateral strain where DO can be this original diameter. And we've got all those numbers, they're just there on the board. So you can go through them real quick, get something like, this should be unitless, because certainly the strains going into it are also unitless. You can put in percentages if you wish. It just has to be the same on the top and the bottom. You can have one percentage and one not. And this number falls on fairly typical range, which I believe I mentioned on Friday for Poisson's ratio. Working Joe? Not yet. Remember what I told you for a typical ratio for Poisson, a typical value for Poisson's ratio? Yeah, this is a very typical one at about a third. Turns out it's a very important number evidently in civil engineering, in terms of what can happen with the types of materials we use for pavement, asphalt and asphalt mixtures, I guess they're called. My background in civil engineering is a bit limited. All right, from this we can find some other things we know from the pieces we've got there. I don't think I gave you this on Friday. It's developed a little bit later in our book. I'm not sure why, I don't know why they don't bring it up here and the derivation of it is not important to us. Just turns out to be a nice relationship between these quantities we've now had to generate what we call the modulus of rigidity, the shear modulus, if you will. Don't forget that that is related to the shear stress over the shear strain, just like the elastic modulus was the stress over the strain. So is this. We can find that G now for the two test values we just calculated and then from that we can calculate anticipated strains and shears depending on which one we have. And you put the pieces in there you get something like 26 gigapascals on that one. Same order of magnitude of the elastic modulus. There, the numbers are all give or take a little bit about the same. Except for this I think everything was just a run through of some of the stuff we'd already done before. But they will specifically be taking one look here at elastic loading or axial loading as we go through this. Right, look friendly and familiar or makes you feel very comfortable with that stuff. Make sure you're good at getting the units just down right. Practice paying attention especially to the SI prefixes here. Have I related to you the tradition it's not so much anymore. In fact, I don't think our book even practices it anymore. It's fairly typical that the SI prefixes and the scientific notation be kept in multiples of three. Have I told you that before? Has to do with the old practice of using slide rules. Slide rules, since slide rules can calculate the power on the 10 of any result, the user has to do that himself and it's just a lot easier to do it if everything's in a multiple of three. But with calculators now, that's not a big concern. I didn't get my first calculator until my senior year. Of course it was also uphilling through the snow to get to school. Travis, how many points do you want to lose? Should be taking that down. All right, so continuing with this axiom loading and another type of problem we looked at. Actually, our book does not introduce this until chapter four. We've already used it and when I put it up you guys didn't even blink at it. It's just such common sense. And that's this idea of different cross sections undergoing different loads. And the total response is simply the response, by response I mean deformation of the piece responding to the loads is typically, we just simply add them all together. So we did a problem kind of like this one and figured out the deformations for the individual sections and just simply added them together. And it was so common sense, you guys didn't even flinch when I put it up. So I'm not sure that the author couldn't have gotten away with just putting it in. A P4, since we've got three other forces on this piece, maybe it's a strut of some kind. Give it some dimensions. Two meters across there, one there, one and a half there. And some areas, cross sectional areas of course, where meters, yep, this one is twice that, A1, A2, A3, and A1 are the same. The two end pieces, the smaller pieces have the same cross sectional area. We did a problem very much like this where one would find out under these loads, what is the total elastic response of the piece to those loads. We did one sort of like this. It's the sum of all the individual deformations. Boy, yeah, I don't want to do that. We did that before and we'll take this to be a structural A36 steel, the whole piece. It certainly could be that the individual sections are different pieces. It could even be that a single piece is made out of two different types of materials. If so, we just break it into the smaller pieces and continue the solution from there. So for each of the individual sections, we need to know just how much deformation is undergoing there. You need to know the internal forces in that section. That's the load that goes in here. The PI is the internal load in each of the individual sections. For example, we can take the first section, put an imaginary cut midway through the piece somewhere and very quickly determine that anywhere along that first piece, A to B, there must be an internal axial load of 100 kiloohms. So we can figure out the deformation of just that section by applying that load over the entire original length of that section. That's assuming there's no intermediate forces applied, which there are, there's no other force until point B and you can figure out the individual deformation for any of the piece to go ahead and run down the rest of the structure from there. Okay, Tom, I just reset the tape for that first time. All right. Okay, so you can do that real quick. It's again just putting in the numbers, then go to the next section, put an imaginary cut somewhere down through the piece between B and C and then again figure out what that internal load is. And you have to do this anytime either the cross-section changes, the material changes, or a new load is applied, or any combination of those. You can figure out what the load is at the, through the piece B, C and just by inspection, you can see it's in compression there because of that new load applied at that shoulder between the two pieces. And then we can figure out a deformation in this part B, C, whatever load is through there times its original length, the entire original length. Remember this imaginary cut is at some place, somewhere in between. It's just to, it allows us to determine what the internal forces are along that piece. And then the area B, C. And again, you have to make these imaginary cuts anytime between changes in area because of that changes in material because of E, this one happens to be all one material doesn't have to be and anytime new loads are applied. So you can imagine some pieces you can get pretty involved as you go through all this. So let's say E is something like 200 10 to the 7th pascal. So you can chunk those numbers. Remember, you must keep very close account of which ones are extensions and which ones are compressions. We have one last section there. We have to do this. Nothing else changes at these loads. Don't expose an internal load and then try to carry that through to the next piece. And so it looks like we're gonna have a little bit of tension. Again, then you can find out all of these elastic responses of this material, that this member to the loads. And then you just add them all off and you've got the total material deformation of this structural member, whatever it might be under these loads, whatever they might be. All right, so run through those numbers if you would and let's see if we all agree. I think your piece got stacked on. Of course, this whole business of these internal forces is because we always have static equilibrium for every part and every sub part there of. But John didn't work. Got a shot of espresso there. Phil, not okay. David, all right? Sure, I got all the right numbers up there. A little late for you guys. Oh, I'm sorry, yeah. All right. This isn't 10 to the 7th, it's 10 to the 9th. You just move the decimal place. You don't have to redo anything. Just move the decimal place. So that's why I got 0.1 meters instead of 0.1 meters. A millimeter. Sorry about that. I'm like not exactly that in every class because I never have any pens with me to change those things in my notes. So that tells you if you had 0.1, using the 10 to the 7th, just move it out too and you can just do that to the rest of yours. Don't forget to get in the habit of putting a zero in front of the decimal. It's sort of an unwritten professional standard to which you should subscribe. So that's the first one agreed. Now we've got the right numbers in there. How about the second one? Phil, you have that one? DC, 3.75 times 10 to the minus 4. Phil, negative? Negative. Yeah, you can't miss that. That's compressive, of course, in there. The middle piece is in compression. 3.75, what? You had 3.75 times 10 to the minus 4th? Yeah. Meters? Meters, yeah. Okay, so 3.75 times 10 to the minus 6. Negative because that's compression, that's contraction. That's vitally important to this total deformation of the piece to sum over the whole piece. And then who has that last piece there? You have it, David? This last one? Let me see here. The delta, right? 0.001125. Is that what others got? 0.05 millimeters, I think. Is that what you've got, Travis? That's the one I had. Okay. I got 0.1211. Okay, the load, let's see. C, what do you have for CD? 150. And then over the, or times the 1.5 millimeter length, times 1.5 over the smaller area, the 0.001, then times the modulus of elasticity. So Travis, you got what? 1.125 millimeters. And that's what you had, David, too? 1.125, and you said that was millimeters. So 10 to the minus 3rd measures. No, John, something completely different? No, neither. Negative 6. Bill, what'd you have? Santa. Same thing as what's up there. Chris? Oh yeah, what's up there? Okay, got it? All right. And then the total deformation, remember, is the sum of all those little individual pieces. This is called superposition. We can supermote one solution of a small part over each other for the total. And what's the total come out to be? The total expected deformation of this piece. 1.75 millimeters. All right, everybody okay on the numbers then? Tom, all right? Ken, you okay back there with that? Sorry? Yeah, well, we've got a piece, not quite six meters, not quite, a little over five meters. So that's what, 15, 16 feet long, and it's changing length. Would you see it? And if this is some structure, some part of a structure for some reason that's between two rigid supports, it might not be able to withstand that kind of change. So we have to consider all those things. We're gonna look, starting tomorrow, at just what to do when we do have this kind of constraint on the piece itself. Because this is statically indeterminate. We cannot solve this problem with the straight statics that we have so far up here. We're gonna have to bring into it this deformation of the material itself. It makes for a pretty interesting problem when we get there. Sounds like you might need to apply linear equations. No, not on this. Well, yeah, if you're talking about it, you'll have a system of equations that need solving. Yeah, but they are linear. Okay. Now, one thing, again, that we've sort of glossed over, I mentioned it briefly, but haven't looked at it with any great detail, is this business of just what this stress distribution is under some given load at different locations across the part. So imagine we've got a part here that's of width B, which is very much less than its length L. So we've looked at lots of these kind of pieces before. In fact, this is the situation that defines our axial loading. And we've figured out the stress by simply taking that total load, dividing the cross-sectional area into that load and we get the anticipated stress. I did mention it early when we came up with this, but didn't pay much attention to it after that. This is really the average stress. What we're assuming is that anywhere across the piece, we're assuming that this stress distribution is uniform. It's not quite the case, but it's not gonna hurt us too much. If we look at a cut, again one of our imaginary cuts across the piece, and a point load being applied like this, so now we're at a depth of about B into the part. If we look at the true stress distribution, there's our average value we have been using. At this point about B into the piece, we get a real stress distribution of something a little bit like that. There is a maximum stress somewhere about the middle of the piece. And again, remember this is a point load I'm applying at the end. And that maximum load is about 103% of the average we have been using. So it's not a big error, especially if we're talking about very long pieces. The overwhelming majority of the length of the piece is at a negligibly different maximum stress than the minimum, than the average we use. The minimum stress is at about 97%. So we're not making that grade of an error by using the average value for the stresses. No, not so it's perfect, it's very easy to remember that. Once you get about a length into the piece that's equivalent to about the depth, then you essentially have a uniform profile from then on through the piece. Until you get down to the other end when it starts reversing. So that leaves a huge majority of the piece. And if you remember, all of the test sections were always in the middle here. If you have a very long piece compared to its width, it's essentially under uniform stress everywhere. We don't have to concern ourselves with it too much. But you can imagine there might be some place where we do need to concern ourselves. So let's go a little bit less deep through the piece. So again, we have this piece B wide, whatever B is. And now we're at a depth of about half B. And there's, again, no particular magic about those numbers. It's just a place where I can give you some of the results from this type of thing. So our average is about there that we've been using for, what, three weeks now? Is this over the fourth week of class? The real stress distribution is a little bit more severe. Remember, this is a point load. And now that maximum is about 139% of the average. So things are getting more severe at the end. This isn't negligible. 3% we don't have to worry about. 40% you might have to start worrying. And this is part of the reason we put in these factors of safety. The minimum is on the order of 67%. So we're very much over-estimating the minimum, severely underestimating the maximum. And it's the maximum that's gonna cause failure when there's the maximum stress somewhere could cause failure. What a happy tune. Trouble from home. A lot of some vibration. That's what we call for my generation, good vibrations. All right, we can go even a little bit farther. Let's go to 1 quarter B. Again, this is an imaginary cut through the piece. And it's no great surprise that things get even more severe. So again, for reference, the average we have been using, now it's quite severe. The actual stress distribution, much more something like that. Now, that maximum is at about 258% of the average we've been using. The minimum at about 20%. Now that's not the concern because we don't design for the minimum. We design for the maximum. So in here, we have this principle presented by some vanille. Sorry, Samantha, I know what it does to my female students when I speak French. It's dastardly the effect. Somewhere in the 1700s where he proposed, he addressed this by saying that once we're a minor length past the location of these point loads, then the stress distribution is essentially uniform and it's more than sufficient to use the average stress levels there. That you only need to concern yourself either at the ends or at places where something sudden happens, either a new load is applied or there's some drastic change in the solid itself. And so we'll look at those kind of things. Now, thanks, Tom. So we do have sudden severe changes in the piece and we have to take them into account. So imagine some piece here under some kind of axial load. We're gonna have to give it a little bit of depth here. So we're going 3D, Hollywood's doing it. So we'll do it too. I have a piece here under some kind of axial load. Again, don't forget, doesn't have to be tensile. I just have to draw something. But then imagine we've got some kind of hole in this piece. What that means is that the piece in general is gonna be much more subject to these underestimations of the maximum stress if we're using the average stress across the piece. Because once we've gotten away from the end and the piece starts to even out, we then get to a change in the piece that essentially causes us to start over. If we put an imaginary cut right where that hole is and look at the stress distribution there, we'll see something, again, not like the average uniform distribution we've been assuming, but a much more skewed distribution. And this piece, it happens to look something like that. And that would of course be mirrored on the other side. We're now, again, grossly underestimating the maximum stress by our previous habit of using the average. It turns out that just what that maximum stress is, is a function of the geometry of the part itself. So it has to do with the thickness T and the width W. No surprise there, that's the cross-sectional area, T times W. But it also has to do with then this, the diameter of the hole as well. In a second or two, I'll talk just about how those things come up. Just wanna give you another example of where this type of thing happens in a fairly typical piece that we look at. If we have some kind of piece that looks something like this, we find that very, very close to this contraction in area, of course, when we go to a much smaller area, the stress goes way up. But not only does the stress go way up, but the maximum stress goes way up. Not only have we brought down the area, but we've gone to a change in geometry that has caused this non-uniform distribution to reoccur. So the actual stress distribution of that piece looks something like this average. We have been using it way back here. We now have these maximum stresses that we'd be underestimating even if we use this minimum area, this much smaller area to calculate the average stress. The load doesn't change, but the area changes. So we use the minimum area to calculate what the stress is. But even that grossly underestimates what the maximum stress is here. So we need to take that into account. And as you can imagine, it again involves the geometry. The thickness of the piece, make sure what similar book uses just so we don't get messed up on these. Yeah, they use W. The width of the piece, the radius of these fillets, the lesser area that we're transitioning to there. So how do we take this into account? It's grand that Sanfana told us about it, that there are places where the average stress is insufficient to predict the maximum stresses. But how do we take these different geometries into account? We define what's called the stress concentration factor. Find the stress concentration factor as K defined as simply this maximum stress for some geometry divided by the average stress that we would have used coming up to this point for many different geometries. Tables and graphs exist that show us exactly that kind of thing. I don't have such a narrow picture here. There must be something, some part of this that's not doing it for me. All right, this is in your book. It's in section 4.7. Oh, by the way, I've added section 4.7 to the schedule and one homework problem doing just this. So double check the schedule to get this, the minor changes to help us with this. What we have is a chart. I don't know why the screen's not giving me the full picture. I'm wondering if there's something here. More software, more software boondoggles. I don't know, but all kinds of fun things here I can switch around. I don't know what that does. Don't you feel like you're at the optistetrician? A, or B? A, or B? Anyway, what we've got here is just these different geometric parameters. Then we can calculate, given those, given a particular R over W, R is the radius of the hole versus the entire width of the piece. And for a given R over W, whatever the values of the geometry are, you can go up to the line and then read right over and there's your stress concentration factor. So we would calculate, in the old way we have done, P over A. In this particular example, we use the reduced area of the place where the hole is. This is the minimum area in that piece. We calculate the average stress just using P over A min. We look up our stress concentration factor on the graph that would give us a maximum stress expected and we designed to avoid that stress rather than the average stress we had been using. And of course, for the other geometry I've got up there, there's the same type of thing. We just, we have multiple curves that have to do with this H over W ratio. The minimum width compared to the maximum width. And then along the bottom comes into effect the radius of this fillet at the shoulder right there. So again, you figure out your R over H value, go up to whichever curve matches your W over H, then you go over and read your stress concentration factor and then calculate your maximum stress from that. As a handy demonstration of just why this radius is important and why we see fillets in so many engineering applications like this, a nice little demonstration. You can do this yourself. I don't know if it'll show up. Well, let's see if it will if I give it some background. I've got two pieces here. One paper cut with a very sharp interior corner and then another piece where I've given it a little tiny bit of a fillet. And all I did was use a hole punch to do that, cut these lines and then put a hole punch in there. And you can do this at home. If you take the piece and pull on it, it doesn't take much for it to damage right at that corner where if you take this piece, it takes quite a lot of, did you see the veins on my neck sticking out? Wasn't that amazing? It's significantly harder to break that piece than it was to break the first piece there. So that's why we put fillets in engineering applications. If you've ever used some kind of, well, if you've ever had a tarp or something that's gotten a tear in it, you have some big tarp that gets a bit of a cut or a tear in it. If you punch a little hole at the end of each of those just outside of the tear, if we look at this greatly blown up, you can see the tear like that. And if you put a hole there, then you're much less likely for that tear to continue to increase through the piece if this tarp is under some kind of tension, which most tarps are, because you strap them down. So if you ever have any kind of tear, you can save it for a bit longer by just punching some kind of hole there by whatever means. And that's sort of like, does the very same thing that these fillets do. All right, so let's do a problem with this just to see how it works. It's not very difficult to use, but it does help with these kind of things to practice going through some pieces, some example problems. And there's lots and lots of different geometries that these charts exist for. Our book just happens to have these two only, so the only ones we'll look at. So imagine we've got some part like this or some sub part of some machine like this under some kind of load. And the dimensions are 40 millimeters across the minimum width, 60 across the maximum. Ours also happen to be all symmetric, solids are symmetric on either side of the axial direction. A radius of the fillet of eight millimeters and a thickness of the piece. All right, so you can, if you've got your book, you can pull it out. If not, I can throw it back up on the screen quickly again. But if you've got your book, this is section 4.7. I can't pull, well, I can do it enough. You've got the all the dimensions there and there's enough of the chart you can see there. Again, the bottom is R over H. The vertical axis is the K that we're looking for. Now, R is the radius of the fillet, H is this minimum dimension across the reduced piece. All right, so given an allowable stress of 165 megapascals, find the load, the largest load that this part can withstand. Okay, you've never done it before, but it's easy enough, I'll let you do it. So take a couple of seconds, get the geometry pieces out of there and then look up on your chart. If the exact value for your part is not there, you'll have to interpret between lines, just like you would do on the axes. Notice that our average stress is just what we would have calculated before using the minimum area H over T, but that assumes that it's a uniform stress distribution across the piece when we know now, close to that shoulder, that's not quite true. Figure out the average as we would have before using the minimum area, but we know that that too is going to be the maximum stress, but that's the allowable you were given over the stress concentration factor K from the table. So take a second to practice that little bit and remember that there's a- That's R over W down there. No, R over H, the radius of the fillet compared to that minimum height of the piece, I guess, because T is the width. T doesn't really come into this other than for the calculation on the average stress value that we need to work the stress concentration factor. Again, and this is in your book, section 4.7 that I added to the schedule and has one problem, and it's not too terribly much more than what you've got here, just a chance to practice it. The only difference with the problem is it's got both holes and shoulders, so you have to put them together. Figure out your W over H, figure your R over H, get K right off the chart, then you can figure out what the load is based on what you're given up there. Difficult, not particularly. You heard the stress count concentration? Some of you have, I bet. The idea is that as we have these restrictions, the stress concentrates and stress concentration factors. So we have a much greater density of stress, the stress because it's been now concentrated into a minimum area, and we also need to understand that a certain distance downstream, if you will, we have this severe non-uniformity of the stress distribution. So you can kind of look at these like stress flow lines and we'd see the same kind of thing because of a whole, we have a much greater concentration of the stress of those minimum areas because it's the same stress trying to squeeze through a lesser area. Also realize you're just estimating off this graph, so pay attention to whether you want to guesstimate in a little bit greater K or a little bit lesser K, just to build in a little bit more safety in the design. SI edition, the international edition, it's even still section 4.7 in there. And I know it is between the two editions that we use section, seventh edition and eighth edition that we use in here. It's okay to overestimate a little bit, but you don't want to overestimate too much because that just makes the part perhaps prohibitively expensive. So let's see, some of the geometric factors we need, w over h, w is the major width over the minor width. Watch the units, 1.5 and we have that curve there happens to be essentially the middle curve here. The bottom geometric factor we need, r over h is the eight millimeters over 40, 0.2. So 0.2 on the x-axis up to our 1.5 curve were a little bit over 1.7, maybe 1.73 or something like that. Could go to 1.8, it depends upon the application. Don't forget that even with these numbers, we're still going to build in a factor of safety after this. So you don't want to overestimate too severely. So we can then solve for p, p equals the sigma max, which is the allowable stress we were given over that the sigma max, where I'm solving for p, times ht, which is the minimum cross-sectional area because that's where the greatest stress is and so we're looking even at greater stress than that and then divided by k. We're at 165 megascattals. Cross-sectional area, what, 400 millimeters squared, 7.3. And you get about, forget you gotta watch all the exponents and the like. That doesn't sound right either. Don't forget, this is millimeters squared, so this is 10 to the minus three squared, not just 10 to the minus three. 36, depending on what you used for k, somewhere around there's 38. Give or take a little bit, 38 kilonewtons. Please don't. That's what I have, I have kilonewtons. 10 to the minus six here, we have 10 to the minus three squared here. I've got kilonewtons, that's what I had. Just double check everything. Megascals, millimeters, that's all I had about that. Right, David? Chris, agree? Ken, look good? Okay, all right, we got two minutes left. There's no sense starting something else. Don't forget though that this has been added, this section 4.7 has been added to the schedule, it's on angel now, and one problem has been added where you just have to go through this very same thing with a slightly different geometry. It's actually a geometry with a couple shoulders and a hole, and you have to do it for each of them to figure out which one is the worst case.