EM 304 Exam Problem: Failure Temperature of the Space Shuttle Columbia by @mes on 3Speak
In this video, I solve Problem 1 of the EM 304 test that Dr. Judy Wood gave her students on February 6, 2003, five days after the Space Shuttle Columbia disaster. If a thermal tile is lost, how hot...
Tags: #math #science #engineering #nasa #physics #education #scifimultiverse #chessbrothers #threespeak
Problem 1 of EM or Mechanics of Materials, 304, test number 1, February 6,003. So last Saturday, February 1, 2003, the Space Shuttle Columbia disintegrated during reentry through the Earth's atmosphere. One possible cost could be the loss of thermal tiles. So here is how this setup there, there's the main gear and so on, and then if you zoom into this part right here, so they have like these thermal... tiles there to prevent overheating and so on. So, yeah, so if you zoom in there, you're going to have a little thingy like this. That is the support, it's aluminum tube. So if you go to the wing there, so wing, skin, stringer covers, web and trust supports 11 honeycomb covers, interesting. Anyway, so it looks like this. So you have these plates, these are metal plates, and now you have this aluminum tube across, that's basically holding up the... top and bottom of this wing there. So that's holding up a skin there. So it looks like that, that's this part. So now we're told, assume tiles were lost in a localized zone shown below, or shown above. Actually, assume below. So assume that there's thermal tile right here, and this is all that was removed. So basically, that foam hits it, and now this is gone. So basically, it strikes this, disintegrates it, or dustifies it, and that's gone. Yeah, so we're told assume tiles were lost in a localized zone shown below, exposing just one of the many aluminum stiffening tubes connect to both the top and bottom framework of the wing. And then assuming, yeah, assume the remaining structures completely insulated from heat and remains at zero degrees Fahrenheit. So basically, it's saying that this is gone, but still assume everything else, including this part here, is all going to be zero degrees Fahrenheit. Just assume that, yeah, it's all of zero. This is the only thing that's being heated, this tube from this zone. That's all that's being heated. These plates, assuming everything is just going to be zero. So just for simplicity purposes. Yeah, so assume the remaining structure is completely insulated from the heat and remains at 0 degrees Fahrenheit. Assume the one exposed tube is heated uniformly, so just heats completely at all throughout at once. So at 0 degrees Fahrenheit, the aluminum tube is stress-free and there are no gaps. All right, yeah, basically, there's no gaps anywhere, it's completely covered, and at zero degrees, there's no stress on it. So, basically, the only stress that's going to happen is when it's heated and expands, and then that causes the resulting stress on, basically, where, on the supports it's holding it in itself as well, because it'll be compressed as it's trying to expand. So heat shields, insulate the structure from temperature, various, as shown below, and assume the top and bottom of the framework is made of a solid plate. So these are solid plates, top and bottom, like that. and you just have thermal shield, thermal tiles, and so on. And yeah, this length L is between here. So this is L, this is our L there. And these are structural support surfaces, thermal tiles, and then we're asked, what is the temperature at failure right here? So in Fahrenheit, and it's a heat shield removed in this zone only, and there's the thermal tiles, and so on, L is constant. So let's say it's constant height there. I mean, L is just constant, it's not changing. Yeah, so if it's not changing, basically, it wants to expand, but then it's going to be being shrunk at the same time. So basically, the thermal and compression strain has to be, well, equal. All right, and now we have the zoomed -in portion. So we take this and then zoom in like that of this aluminum tube there. So we have over here, and we have three different sections to look at inside. They're basically the elements inside. The first one's part A. And this will later in the video, we'll see. we'll be making use of this. So this part here, this element right here, is at the connection point, just on this, along this. plane there. So as this is expanding upwards or wanting to expand, it's going to be putting a bunch of shear stress along this thing over here. So it's going to be sheer like that, and it's going to be, and you have to draw the corresponding diagrams like that. So it's going to be shearing, and then you have to be the equilibrium because it's not moving there. So you just put them accordingly and so on. So we'll do that later in this video. So that's part A, and then again, these are the structural support surfaces. and this is going to be a thickness of sport plate is 0.25 inches or a quarter inch, and now we have Part C is basically on this element over there, move it over there, so it's going to be like that, and this is on the tube itself. So as it's expanding, as one expands, but it's not going to be allowed to because the plates can be forcing it not expand, just due to thermal expansions, and this can be compressing it, so you can have a bunch of compressive stress along the C. yeah, along this C element is basically indicating basically all them can be in compression like that, and assume it's uniform, really heated, and so on. So that's that, and then part B is this part right here, and that's actually on the support plate, but instead of shearing like that, it's going to be like, yeah, basically within this, or actually just in here, somewhere around there, so basically in compression. So we're going to have this as it as is expanding, as it's pushing on it, now we have to compare this stress on this actual. yeah, the support plate, and that's going to be also in compression and so on, and find where it fails, and so on. And here we're told that it's an end connector, quote, rivet. And yeah, so that's useful, well, that's used in this part here, so that's the center of this cylinder, right, there's tube. So center portion, we're told, is part of end connection. So this center one is connected to this right here, so this is the internal one, it's connected to this. and we're basically told to assume it's a quote rivet and does not support connections. Basically, just there just to keep these together, but it's not supporting load. All the load is based on this tube, this outer tube there. So then when we do our compression, the calculations, we all need to consider this outer area there. Yeah, so we're also told right here, we're told the diameter inside, that's going to be, this is one inch. So this is from here to here. So we're ignoring the center there. Diameter outside is 1.2 inch. That's by 1.2 inch. And basically, the area of this yellow part, we're already given, is going to be 0 .11 pi inch squared. All right, so we have that. And now let's go further. Now we'll look at the material property. So we're given L is constant. So this is going to be constant right here, this length right here. so this L is constant, this is going to always be constant, so even though it's wanting to increase in thermal stress, it's going to be compressed, yep, from the support plates. And from there, we can obtain a formula and so on, because the, the displacement is going to be easier. So note, this is the length of the aluminum tube within the structural support and surface, yes, structural support surfaces or support plates, as shown in the earlier figure above. And also, now we're told the delta T max, or the change in temperature, equals to 3,000 degrees, and it's maximum possible. And then from Wiki, if you just, yeah, again, per degrees Fahrenheit. We'll go over that soon. But basically, zero, it's assuming it's rigid. So it's not going to be expanding due thermal expansion. And then the Yong's modulus is infinity. So it's, again, super rigid. This is sheer modulus, G is infinity, and then the Poisson's ratio is zero. So it's all basically rigid. We'll explain this in a bit. So now we're giving the maximum shear stresses 10.2 KSI, so basically 1,000 PSI, or pounds per square inch. Yeah, pounds per scrounge, 10.2,000 pounds per scringe. And then the, this is the compression, uh, is, uh, the actual compression maximum intention is 104k, and then, and in, yeah, I mean, this intention, yeah, is 104k, so compression is stronger in compression than, uh, I mean, intention than compression. Aluminum, uh, this is the type of aluminum is, 2014T6 so the coefficient of aluminum of this type of aluminum is 12.0. Yeah, coefficient of thermal expansion is 12.8, microstrain per ferrier per degree Fahrenheit. And then the Yang's model list is 10 .6 mega S.I, I mean, mega, I mean MSI, basically a million pounds per square inch. And then the sheer model list is 3 .9 million pounds per square inch. And the 0.34, and again, we'll explain this all soon. And then the sheer, the maximum shear, the shear stress, this is going to be this tau max is going to be 19kSI, and then sigma max in tension, the actual axial stress, is going to be 60 KSI, and also 60 KSI and compression and tension. So that's maximum for the aluminum. And now we're basically given the properties above. were asked, you have four things we're here, so to determine the maximum temperature of the aluminum tube without causing failure at A, B, or C. Note, failure may occur in the plate or tube. So basically, we'll find out which one of these fails first, at this point B, at this point C, or at this point right here. So at the, at the, at the, uh, this inside the plate on in shear, this one in compression or this in compression. All right. and that basically finds out which temperature is the lowest to obtain these maximum figures, right, and the strength of them. So, Part B says, where is it most likely to fail first? Choose location A, B, or C, and Y, and justify your answers with calculations, and then using your answer from Part A, C, calculate the stress state in locations A, B, and C, at the failure temperature. And then Part D, show these stresses from Part C on each of the volume elements A, B, and C provided on the right, this is, well... In the test, it was provided on the right. This is provided above. I'm just going to write that here. Stresses on the volume is provided above. All right. Representing the state of stress at each of the three locations. Note volume, put this right here. Note volumes, volume element C, B, and A, are provided on the next page, along with the workspace, are provided, yeah, let's say below. Right here, so I'll say it below. This is her exam, was like that. I'm just going to write provided below. And yeah, this part here, you may present your results for Part D on either page, show Part D on only one page only. Yeah, I'm going to just erase this part of the test itself, but yeah, we have, yeah, I'm just doing a tutorial on it on my tablet right here, or my Microsoft Surface Pro. And also going further location descriptions, we're also given point A, zone of material inside top plate. So that's the one inch here. Point B, surface of plate facing tube end. and surface of tube and facing plate. And what this means is basically this connection point in compression. Basically, they're looking at this is like the end of the tube and bottom of the, I mean the top of the plate, or the top of the plate facing this tube section there. So basically in compression between these two zones there. So that's the element there. And, yeah, and going further here, so in point C, zone of material in outer surface of the aluminum tube. Yeah, and that's just the overall compression of the tube and so. All right, and now let's finally take a look at the solution. So note that I'll be using GROC AI's assistance in the solution, as well as Dr. Judy Wood's Solution Key. There's links to the GROC one. Again, yeah, double check everything before I go through, and also a lot of terms I got from Wikipedia as well. Quick summary, so thermal expansion properties of the support plates. and aluminum tube, so in summary, the support plates are considered idealized as perfectly rigid and non-expanding, hence having a coefficient of thermal expansion of zero, and infinite yang's modulus, infinite shear modulus, and a Poisson's ratio of zero. So basically, as you're pulling it, well, it's not going to laterally transverse, these are going to be zero. So it's not going to expand in any way, just idealized, this is very simple. setup there. And all we're looking at is just the strength limits. So while the aluminum tube has a coefficient of thermal expansion of 12.8 microstrain per degrees of Fahrenheit, a young's modulus of 10 .6 MSI, or million PSI, a million pounds, million pounds per square inch, shear modulus of 30.9 MSI and Poisson's ratio of 0 .34, basically the transverse strain is 34%. of the actual strain. So if you're going to be expanding this by, let's say, 100 millimeters, the transversal is going to be, well, 34 millimeters. Yeah, pretty much like that. So basically 34% of this expansion. But, yeah, 34 would be a bit higher here than this case, but for that drawing it. It's a pretty interesting setup. And now let's take a look at the strength limits as we're given. These are just the allowable stresses. we are also given the following strength limits for the support plates and the aluminum tube, at which the material is considered to fail, either yield, or ultimate strength, depending on the design criteria. So the plates, these have a maximum shear stress of 10.2 KSI, so basically maximum allowable shear stress. And then the in compression, the Sigma Max, compression of the plates is 104kSI in tension equals maximum allowable normal stress and tension, and the compression is weaker at 70kSI, or maximum allowable normal stress in compression. And the aluminum tube, we have basically the shear max is 19, yeah, 19 KSI, and then the compressive and actual tension stress. and tension, actual stress is 60k aside, tension, Sigma Max, 60KSI, and compression. This gets uniformly heated, everything else is just zero, and so on, there's a constant length L, and we're asked the temperature at failure, and so there is the segments, there's A, there's B and C, and we're going to go over C first, this one is on the aluminum tube, and then it's going to be in compression, and then we'll go over B, and then we'll go over A, and that's just basically the order that... Dr. Judy Wood solved in her solution scheme. So, we are given that one aluminum tube is exposed and heated uniformly while the top and bottom support plates, and the rest of the structure stay at 0 degrees Fahrenheit. So, yeah, this one, remember, it's heated uniformly. Like that? And we are also told that at 0 degrees Fahrenheit, the aluminum tube is stress-free, and there are no gaps, so nothing in between. And so, basically, as it's expanding, it will always be in compression, it's being compressed, there's no gaps, you can just freely expand, and so on. But at 0 degrees Fahrenheit, yeah, yeah, so at 0 degrees Fahrenheit is going to be, there's nothing, there's no stress on it. But there's no gap, so as any temperature increase, it's going to start expanding, and it's going to be compressing at all-time, uniformly. So the plates are model as rigid, so Young's module less equals infinity, sheer modular equals infinity, the, this was the Passons ratio is equal zero, and the coefficient of... of thermal expansion is also zero. So the length of the tube is fixed, so length of this is going to be constant. And that's because, yeah, the plates cannot be moved. So the tube is always going to be constant there. So only the exposed tube wants to expand due to thermal expansion. The rigid plates prevent that expansion, producing compressive force in the tube, and equal and opposite reactions on the plates. Failure can occur in the aluminum. tube or in the plates, bearing or in shears, basically bearing stress on here, or shear on the side here, which is ripping it apart. Alright, so what do we have here? So, because L is constant, here, just copy it down and paste this segment right here, there's a plate, there's this part right here. So, because it's constant, well, this is going to be, let's say this tube right here, or this tube, it's going to be wanting to increase. So I'm going to do a, I'm going to put it like this. So it's going to be wanting to basically thermally expand, and I'm going to put this sign as delta, and then using the thermal expansion set up there. Yeah, just put it like that. This is for the aluminum, all right? So that's going to be wanting to expand, and this is, this is going to be, I'll put it like this here, thermal expansion. So this is going to be the thermal expansion. All right here, just fix that up. So this is going to be wanting to thermally expand, but this is fixed, so this is going to be equals to constant. So if it's constant, well, we're going to have a compressive strain. This is just put the delta for the amount there, but so compressive side going inward so that the length L is always going to be constant. And this one's going to be, I'm going to put this as sigma. So this is going to be a compressive strain there. Yes, this is just going to be, this is for the aluminum. So this is compressive, I'll just put compression. I'll put compressive strain right there. Or actually, instead of strain, just because we use strain for a different definition, which was that meter over meter setup, I'm just going to put compression. And it's not, yeah, just put compression there. So that's compression, this thermal expansion there. Yeah, so thus, what we have is, so thus, I'll put it like this, I'll put it here. Thus, the delta L, the change in L, is going to be, well, the total length change, is going to be total, yeah, and this is going to equal to, well, zero. So this is going to be equal to zero, and in other words, this just equals two, well, the difference now, this is going to be the expansion, thermal expansion, minus, the change due to compression. So these have to be, yeah, it has to be equal to zero, though. So then this right here, this minus that, has to go the total delta L. So delta L has to be just basically zero. Delta L is equal to zero. So in other words, delta equals zero, put this in constant there. So rearrange this, what we get is, this is going to be the delta of, in thermal expansion. equals to the delta of the compression. All right, and now note that this change in thermal expansion, so note this right here, this is the delta in thermal expansion, this just equals to, well, we have the coefficient of linear thermal expansion, so this is going to be like this. This right here is the coefficient, coefficient. And here's move things around and just pause video and added this full thing. So this is a coefficient of linear thermal expansion. The number of the units were, well, we had the strain, so it's going to be strain per Fahrenheit, so m over m, and this is going to be over Fahrenheit. I'll put this like that. Yeah, so we have that. So when we multiply by the temperature, times by the temperature increase. Then we start off at 0 degrees Fahrenheit, now it's increasing delta T like that, and that's going to be in Fahrenheit, so then this would cancel, so we're going to be left with a meter over meter, and then times it by the length L. So this is going to be, I'm just going to write this here, this is temperature increase. Yeah, this is temperature, increase again, in our temperature, in our temperature, in our case, it's when reentry, when it's, when the space shuttle is going through the atmosphere, rapid speed, and pressure, and so on. All right here, just fixed up, up, and yeah, so then this is going to be meters, so then what we'll do is cancel this, cancel this, we're just going to be left with the, well, the total thermal displacement there. And now also, for this part here, so this is the compression, so note also, also. Now we're going to look at Yonge's modulus. I'm just scrolled back up to Yonge's modulus that we covered…