 All right, we'll cover a little bit on some thermal effects if there's time at the end and we'll just deal with any questions before the last test. This is in the eighth edition of Hibbler. This is generally section 4.6, so I don't often get it get into it, so it's kind of nice to do this a little bit. It's fairly straightforward. The idea, of course, that you've all experienced in some way is that we've assumed so far that everything put together as all the pieces were at the same temperature, well, we didn't assume it, we just passed right over it. All the pieces put together at the same temperature and that the operating conditions for the object was the same temperature as used for the actual construction. But you know from experience that generally things tend to expand a little bit given some temperature increase. So if we start with a piece that is some temperature T, at some temperature T plus delta T, just some increase, it can be a decrease, of course, but then it would be generally a shrinkage. We see this expansion, this lengthening of the piece, and we'll call that del, which is our usual symbol for these minor changes in dimension. We'll put a little T on it for change in length due to temperature increase or thermal considerations. And we have to put these in, because any of these kind of strains, this is a strain, it's a thermal strain, a thermally induced strain, but stress does generally come with it. So we define this experimental property of materials, experimentally determined only, the coefficient of thermal expansion. And it's simply a matter of taking material and monitoring how much its dimensions change as the temperatures are to change. Possibly you did something with this in Physics 2. I think the symbol used by the physics books is a beta. We, this book happens to use an alpha for the coefficient of thermal expansion. The defined as, as the amount of length of change with respect to the original length times the amount of temperature change there was from the original condition. And notice that this little part over there is actually what we've called the strain, the length, the ratio of the length to the original length of the material. So, and it's experimentally determined, it's a property of the solid. It's in the back of your book as the very last column of the thermal properties of the book. So it's very easy to look up. Tends to have units of something like 1 over degree Fahrenheit. Remember the strain has no units. So this has just the temperature on the bottom. 1 over degrees Celsius. And if you think about it for a second, it doesn't actually matter if these are temperatures in Fahrenheit or Rankine or in Celsius or Kelvin because the amount of temperature change is the same size for whether it's Fahrenheit or Rankine. Are you guys familiar with Rankine? I was about to ask you what that was. It's the equivalent of Kelvin. It's zero in degree Rankine. It's the same as zero in degrees Kelvin. And then the temperature degrees from then on are Fahrenheit size degrees and from zero at Kelvin they're Celsius size degrees. So, since all we're worried about is the change in the temperature, it doesn't matter which of the two you use just as long as you're consistent with whether it's SI or English units. But when you have the change in temperature Fahrenheit and Rankine are the same change in temperature Celsius and Kelvin are the same change in temperature as long. So we'll put a little tea on that epsilon since that's the thermal strain or thermally induced strain. And it's not very difficult to measure. Did you happen to do that lab? We have the equipment to do it. And did you do that in physics too? We have some long pipes. We just run some hot water through them and a micrometer at the end and you measure the temperature change with the length change. So it's rather easy to do. And then come up with the thermal strain that way. So the way it's going to affect us is ways we've looked at before where if we have a piece that might be constrained, we imagine first allowing it to lengthen because of temperature change. Of course, we're assuming all properties are constant throughout the pieces. An isotropic, homogeneous, constant, uniform cross-section, structural member of some kind. There is some expansion in the radial direction. The thickness gets a little bit more, but it's a minor concern. Plus it's of the same ratio if we're talking about homogeneous isotropic materials, which of course we are. So we imagine that there's some temperature change causes the piece to expand. But because of the constraints of the supports, we also know that there's got to be some force induced in the material and that's going to be the strain, sorry, the stress that we calculate. And this, maybe we'll call it del P, those two have got to be equal and opposite. And from that we'll be able then to get the stress in the material along with the observed strain, induced strain I guess is a better word, since it doesn't really strain because of the immovability of the constraints there. So we have this thermal expansion that we expect that is a function of this coefficient of thermal expansion, the original length, and the temperature change of the material, relatively straightforward calculating that. And then the amount that we need to force it back to give us the original condition of being between immovable constraints, we've been doing that for weeks and know how to find it. And from the fact that the two of these are equal to each other, then we can find the stress induced because of this thermal expansion. So given the change in temperature, the thermal expansion, the other material properties, we can figure all of this out. Then from that we get the stress. Remember when we were calculating stress it was always per original area, so the fact that this expands regularly is not of great concern. So when we put these two in there and solve, we get something like minus e, which is as always a Young's modulus, alpha, delta t. The minus sign works, well algebraically it comes into the piece anyway because of the original equation, but it does give us a compression if the temperature change is possible. And vice versa, which is exactly what we saw there. So we can handle the fact that the piece may actually get colder, delta t would be negative. If delta t is negative the two cancel each other. We then have a positive stress because we're actually going to have to put the piece in tension to bring it back out to its original position. Just this picture essentially runs backwards with a negative delta t. And in fact we'll just do a couple problems now and then be done with it. It's very straightforward, very much like some of the other stuff we've done. Especially when we were doing statically indeterminate problems, which we did, I don't know what, 10 weeks ago. It hasn't learned a lot, haven't it? Despite your efforts to the contrary sometimes. All right, so we'll do a quick problem with it. Imagine between two immovable strains. It's not necessarily that we're just putting these pieces between big walls. It's just, we have to model what's going on with just the piece itself. So it's sort of like a free body diagram business, I guess, where there certainly are other things in the problem. We're just trying to look at the one piece and then when you get to real engineering you're going to have to put all the pieces together. All right, so we've got a piece here made of two diameters, but the two pieces are each the same length. The areas, 1.2, 1.2 square inches. So that's the areas already given there. And we'll use a 36 structural steel. And just means that in the book you can get E and alpha for that material right out of there. And imagine that this was put in there with no stretch, just a tight close fit at the original temperature of 95 degrees. Sorry, 75. Wait in the term. Can't even read my one hand right now. Not that I always look at anyone. 75 degrees. But operating condition is expected to be somewhere around minus 50 degrees Fahrenheit. And so, yeah, so we've got all the pieces. So we want to find the stresses in the two pieces. The two different sides, all this on the 1 and that one, 2. So find the stresses in 1 and 2. There's actually zero strain because there's no change in dimension, but it's the very same type of thing as I just pictured a little bit before. It's just, in this case, because of the drop in temperature, we expect there to be a decrease in the length because the thermal expansion will be negative since delta T is negative. So we'll have some delta T, some delta T back in this direction as the piece changes length. And then we want to find the reaction by finding the force that would be required if we pulled it back to its original length by modeling that a change in length due to the tension we're going to have to apply. This, of course, means that in some way it's actually attached, maybe welded to those immovable supports. So as it changes lowers temperature, tries to shrink but can't, and that gives us the reaction of the support pulling it back to its original length. Alright, so we'll need some of those material properties. The alpha coefficient of thermal expansion for A36 steel, just in the back of the book, 6.6 times 10 to the minus 6 per degrees Fahrenheit. Remember that's per change in degrees Fahrenheit, which in this case we have what, minus 125. I don't see where I have the E actually written down, so somebody's got their book. 29 times 10 to the hoop. KSI. Okay, thank you. Alright, so we can figure out the change, thermal change in the pieces that we expected. This is the length they would change if they were free to do so, but they're not because of the attachment to the extreme pieces or to the support pieces. But that's the business we just looked at, alpha, L, delta T. So we've got all that. It's easy just to put it in and observe what the expected change in length would be. Notice that this does not, we don't have to split it into the two parts because of the different cross-sectional areas. That's not of an effect of it. So it's a very, very straight forward number to calculate. And because of the very small coefficient of thermal expansion, it's a very small number itself. It seems to have to be minus 19.8 times 10 to the minus 3 and that's in inches. So that's what its change in length would be if it was free to do so. But because it's attached to an immovable support, we need to pull it back out to an appropriate length in the other direction, which we know is going to have to be plus 19.8 times 10 to the minus 3 inches. And we do that just as we did before. This time we do need to figure out what are the two contributions because of the two different cross-sections of the pieces themselves. Now that P, the P1 and P2 is actually the same throughout the piece and it's just this reaction support we're looking for. So we can pull that out in front. The E1 and E2, it's the same material. So we'll just drop the subscript on that and pull that out in the front. Just a little bit of algebra. So it becomes L1 over A1 plus L2 over A2. Remember these are the original lengths and the original areas just as we've done this before. So let's say we've got all those pieces. We've got P, E, L, and A. It's all in there and we can then figure out that the P we don't have, we set it equal to this and solve for P. That's what we're looking for. We need that to find the stress. So this turns out to be 19.1 kip. The potential for the pieces experiencing in order to keep it at the same length despite this rather extreme change in temperatures. And then from that we can find the stress in the individual pieces. It's that piece that forced over the area of each of the pieces and again the original areas. So not a difficult, not a difficult concept, not difficult calculations here just a matter of keeping the numbers straight. And of course there's half the stress and the bigger piece because there's twice the area to absorb that piece, that stress. Alright, I think one of the interesting aspects of it is to determine what actually happens to this point where the two pieces come together. You might think that that will not move, the two pieces will move on either side of it, but we can figure out which by determining what the strain is. Remember that's just found from alpha delta t. That will give us the strain. And so we can figure out the strain in the two pieces. For each piece it's the thermal strain plus the mechanical strain. So for the first piece it would be just that and we can find this from our regular type calculations. That will give us the thermal strain. And you notice it's going to be the same for both of them. For both of section one and section two that comes out to be minus 825. It's 10 to the minus 6th of a strain is in units of something like inches per inch. And that's good for both of them because it has nothing to do with the area or the original length in that straight calculation. But we can get the mechanical strain for piece one remembering that Young's modulus is defined as the stress over the strain. It's been a while since we looked at this. So we know that that's going to be then the stress that we just found over Young's modulus for the material. So the stress we just found is the 31.9 and Young's modulus, oh we had it I raised it, 29 times 10 to the third KSI. So the mechanical strain in the first one is plus 275 times 10 to the minus 6th. Just to remind you we also often check out the 10 to the minus 6th and just put in the symbol micro inches per inch. And that, oh wait wait wait wait, sorry that's not the right, that's the total stress. That's the two of them put together. So let me get this one in there first. It's 1100 micros. So the stress in the first piece is the two of those added together for the first piece. So we've got that now. It's the strain due to the shrinking from the delta T and then the mechanical strain due to expanding it back, leaving an induced strain of the two of these added together and that is the 275. A pop plus 275 micro inches per inch. So that's the total inches per inch is induced in the piece due to the thermal and the mechanical strain in just that first section. Then you have to do it again for the second section. If you do so it comes out to be the opposite of that. Which makes sense because there was no change in length of the material. What's interesting, I think, is we can now find the change in length of each of the two sections due to the strain experienced in each of those. And so that's the strain times its length, the original length of the remember. And for piece one it experiences a positive 0033 inches and for piece two turns out as it should that it's the opposite of that because the total change in length was zero. Remember it was stuck between two immovable supports at either end. However if you look at this it means that that midpoint actually shifted a bit to the right. So we get a little bit of a shift of that point to the right to the tune of 3.3 mils or thousands of an inch. Which I don't know that would be obvious from the start. I think your intuition would tell you that the interface between the two different areas would probably stay at the same spot but it doesn't. It actually shifts to the right. So if that's there because some part of the machine or structure connects there that also has to be taken into account in addition to the stresses and strains in each of the different pieces. Relatively straightforward just kind of a matter trying to keep all the little pieces straight all the numbers straight in whichever spot they go. But like that interesting conclusion that the middle of the bar actually shifts to the right a little bit. Okay. So one more problem that will be done with it. Imagine a rigid bar fixed at one end and with two kinds of supports. A brass cylinder there and a threaded tie rod there. Each of them on an immovable support of some kind at the bottom. So again we're going to be concerned with two different pieces. So for the problem there the properties are. Again this is a 36 steel. So any properties we need are right out of the book. We'll do it in metrics. So that's 200 megapascals sorry gigapascals. Coefficient of thermal expansion. Actually we're not going to model any temperature change in that piece. We're going to put all the temperature change in piece 2. As if this was some part of a structure and for some reason 2 would heat up and 1 wouldn't. Which for buildings could be a concern because you'll have part of the building in sun and part of the building in shade. So you have to get these pieces to fight against each other sometimes. The diameter is 22 millimeters and the length is 0.9 meters. So that's the piece 1. Piece 2 is a solid brass cylinder. We'll use the properties of Young's Modules 101 gigapascals. This is going to change temperature. So we'll need it. Coefficient of thermal expansion. It's diameter 30 millimeters. And it's temperature, oh I need two other pieces. The distance between these is 0.45 meters and 0.3 meters there. So this all snugly fits together and that is a matter of just being able to slip the brass cylinder in there and snug down the rod so it's just snug. Not any great stresses induced there. So we want to find the stresses in the brass when the change in temperature, putting together 20 degrees, we expect that to get to 50 degrees. So that will be a plus 30 degree change in temperature. So the piece 2 is going to expand. This is a rigid bar here which means we don't take it to have any deflection of its own in this first part of the model. Otherwise we would have to take in that into account. But we looked at that a couple weeks ago so we could add that in. But we're not adding it in on this one. And so we need to find the stresses in the brass. It's not a big deal if we wanted to find it in the other piece as well. Because of this we expect a free body diagram of that rigid bar to look something like actually the whole structure. There's going to be a reaction there in the brass piece that will be up. And because that will push the bar up it's going to put this structural steel rod into tension so we can figure that out as well. And then of course there's going to be stress or reactions back here. I think I called that point B. Let's label them if we need to. A, C, B, D, X. But we're not concerned with those when we weren't asked about that. So if you were actually doing this you'd have to figure it out. And the piece is going to deflect something like that because the brass rod is in the brass cylinder as it heats up it's going to push that point up and since the bar won't bend then it pushes it up as the uniform. Similar triangles. Yes, that is missing. It's point three. So actually let's label this the distance point D moves. Point D is going to be made up of two parts to it. It's going to be partly due to the thermal expansion in D or in the brass cylinder minus the relative movement of B and the 2D which is the mechanical part. What we've called del P before. And that's due to the fact that it's going to see a reaction at that point that will cause it some additional strain. So we have to figure out the two of those together. Now the del T is not too big a deal. That's just the thermal part to it. We've got all those. And then the relative strain is going to be the reaction. It's length with two on all of those since we do have different materials. And so you can figure out all of those pieces. And then put them in. Remember though that RB at this point is unknown. We're going to need to find that because we need to find the stress in that piece. So that will give us one equation with RB unknown. You can also though do a static free body diagram to at least get the ratio between RB and RS. You get the ratios just from summing the moments. Without E you would make sense because we don't need to break the parts of E. And you get S over RB. The ratio between those two is .4. You can just do that. That's something you could have done last fall. It leaves us with an extra unknown. But then we have one last little piece that we can throw in for the whole thing that we also know as our last necessary equation that this is also .4 times del C. And that's just using similar triangles. So that should be all the pieces we need. We can then find RB from this. These two going here. RB will be an unknown. Del C is an unknown. But we know then the ratio of the two. And so we can put together all the last little pieces we need. L1 I guess it is. E1 over A. So we've got all the pieces we need to find the reactions. Those reactions are all unknown. So are the actual deflections themselves. But we have enough equations and enough unknowns to find them all. I think that's all the pieces for equations for unknowns. That's unknown. That's unknown. And that's unknown. But we have four equations. Not in any particular order. These two together. So all that's left in it then is the algebra. Put all together. I have to decide if I want you to finish all the algebra. I won't kill you. We've got the time. Chris you're not in the mood. You're going to kill me. You might kill you. On the last thing. That's exciting news. Some of it is pretty straight forward. This R is over RB. Remember it just comes from the summing of moments with respect to point E. You can't do any better than that because we don't know what EY is and we don't need it. And then some of the other pieces you can figure out completely. For example that thermal, if that brass rod was free to go, it would actually move 162 micro-inches. No, sorry, micrometers were in SI units. So I'll save you a little trouble and give you that one. What else can I give you? Yeah. These two will be in terms of RB. This will be in terms of RS. But we know the ratio between the two so we can solve for them there and with the thermal expansion. So that's all the pieces. You can do it. And a little bit. David and I are both feeling the effects of not having. Oh you have coffee today. I was going to ask for Monday actually. Well if you like cold coffee why don't you tell me. I would have given you the one nine o'clock this morning. If you're both warm. So just start putting the pieces together and the unknowns will start eliminating themselves or being found. We can check those. Okay Phil? No, it's okay. It's the last day. You guys don't have anything else going on today, do you? No, testing dynamics. What's going on? Probably going to mess you up with the chalk prints all over here. It would be very nice if you had them. David is coming together. They'll give you all the pieces you need. Yeah, I think so. I was going to do it last thing of the year except for the test next week. But that, I am worried about. Because if you don't want to have them like the area, square meters. We don't need A to A1 because we're going to ask for the stresses in that one. But it would be easy to find them because you're going to have to be able to find RS as well. But you can find RB, you know, the ratio of 2 to 2. RS is the force in the steel. So you don't want to use that for the force in the, oh, there's time to point 4. Okay. So you're getting a stress of 70? No. Do you have something for Travis? No, but if you average your 2, it will be very close. I got 37. We can check maybe some numbers as we go along. Thermal, I'll give you delta 2, delta T. The mechanical part is 4.2, 10 to the minus 9. RB is unknown. So you have that part to it. Unit's all checked. We'll give you delta C, maybe 4. That's RS. So that should give you then a chance to put those two together and then just use the two ratio parts, delta C to delta D and RB to RS. Anybody have anything left? There's RS. Okay, I don't have to have RS written down, but you can get RB from that because of the static part, RS over RB is 0.4. He has 34. Should be positive since it's tension. Yes, of course. And then you have RS, so you could find RB. Okay, that's the, you used the wrong area. Yeah. Yeah, you used 11 millimeters for the radius. You need to use 15 so that, and that will bring yours down by quite a bit. And so you might have to number that. Yeah, I think that's it. 37, yeah, 37 to 6 megapascals. And Samantha just got the same thing. Yeah, be careful and be real clear with your labels. Make sure you get the right area, the right radius into the right slot. It's really easy to mix and match these when you're getting them. So you can also figure out then, once you have the reactions, you can figure out what delta C and delta D actually are. How far those pieces actually move as the bar gets, the rigid bar gets pushed out a little bit. And that just comes from putting those pieces in those two numbers. Where's your van yet? It's still working. 0.315. What happened to have RS written down? One of you has it though. 106, 33, and that's kips. No, no. Killing it. 0.315, that's right. For delta C, I have 0.315 millimeters. That comes from just using this. Now that you have RS, the last day of class. I'm going to ruin my summer. What's that Chris? How many meters does that thing move? 12,000 meters? RS. I didn't happen to have RS, but Samantha's numbers came out. Did you get that for RS? No. Oh, maybe I just heard it wrong. Excuse me. For RB, yeah, it's just news. Oh, it equals 11, I didn't see that. Don't you have that Travis? No. Then for RB, I have 26. And if we all agree on those, then delta C and delta D, everything can be finished. 0.315. So it's not very much. These things don't move a lot, but it might be enough that things don't work quite right when they do move that much. There are certainly situations where one part can heat up and the other part doesn't. And even if they did both heat up, there's no external expansion. Coefficient. You got some of those? I'm good. Jack. And vice versa. I wrote them down right in my notes, but I always rewrite those because they're messy the first time I go through them. You got something else entirely. Oh, here's number four. So I do have those swapped then, you think? That's what I have. You get the two of those put for different. Delta D should be larger. Delta D. That's what you have? Yeah, that didn't make sense. I guess what you have over in the corner is number four. Should be delta D. Well, if it's based on the steel and piece one. Put that 11.284 down to the center of the negative nine. Times RS equals 0.126. All right, let me see. Let me see if I gave you. Okay, delta T is 1-6, that's right. I'll see. Oh, yeah, I think that might be swapped again. I think you're right. I think this is supposed to be a B. Which would then put these back to where they were. Is that right? This is really delta D. Because I have delta C, but then I have all 1's. In here and the X. Does that fix it then? Do you agree with that then? Yeah. That should be the error. Yeah, that's what I'm talking about. But you agree with it. Phil, just peg the work hominator and you're done. Oh, no, you've got it. You've got those same numbers? I don't know. That's not an idea. Just write down the answers, erase everything else. It's all just algebra though. Let's get the letter straight. What's David getting? We're close enough. A couple of people got it. We'll just put in the last bits of the algebra together. Did you get it in the end, David? Oh, I didn't. We're supposed to find the stress in the end. 37.6. Megapascal. Yeah, 26. Well, I've got it. I'm putting all these pieces together. You can put these two in there. And then you know the ratio of the two changes because the, assuming that bar is rigid. And then you can put the ratio of the two reactions, which is just the statics. Is that the whole thing there? Yeah, that's... The term B times 0.3, yeah. Over 101, the sum of the E, yeah. Times 10 to the ninth pascals. Times the area, which was 706 times 10 to the minus 6 meters. That's squared. Yeah. Yeah, 1.62 times 10 to the minus 8.4. But it's mostly all algebra. I'm going to leave a little bit if you have any last questions before our next week's test. If you don't, then stay on the algebra. I think it's 8 Thursday morning. What's this name? 8 Wednesday morning? It's on the schedule and I double checked it. It was right. Yeah, it's 8 Wednesday morning. Open books, open notes. Is that the usual? Sure. That would help. That's what we got here. Still get it wrong. That bit it wrong. Plus you've all learned how to address my answers at the board. I think you've gotten bigger pascals wrong consistently. Yeah, well, consistent. That's what's important. I don't mind if I grade wrong as long as I grade wrong on everybody's. All right. Any concerns about then next Wednesday's test? And you can do it either here or in the lab. Open book, open notes. It's over the beam design. Which was using those charts in the back. W8 by 28, those kind of things. So we don't want to do any shaft design. You do shaft design. Yeah, I've got to give you something you can get done. It's like 18. Yeah, and then the beam collection curves. That gives you by direct integration or superposition. And then the columns. So no concerns. No concerns. Just none you express at the moment. Okay. I guess we're done.