 what the stresses were at those two points A and B. Now, you don't have to take elaborate notes on this because I've put this up on Angel so you can see it. But there's something in this business of finding out what the stresses are at A and B that we're going to need right now. For this bracket, we're trying to find the individual stresses at those two points. And then, you can just see it on the top of the screen that says it, and sketch on differential elements at these points. We're going to need that for the next couple of days with what we're going to do. What that means is at these two spots, A and B, you need to do the type of thing we've done before where you can imagine a differential element at that spot. A little piece of only delta x by delta y or dx by dy, which I guess technically are actually smaller. The whole solution is given for you there, and it's not a whole lot more than figuring out the equations that we use for the different stresses and strengths. First thing I had to do was figure out the internal forces at that phase because it's those forces that determine what the stresses are combined with the section properties, which is just the pure geometry of the situation. And then the individual stresses were found for the different points. At each one, there's a little bit of different section properties, which is the main reason there's a change for those. But for example, the top one across, we're looking for the stress, the normal stress at point A. That was partly due to this normal force. There was a bracket there with a force being applied like that in the problem. That breaks into components like that. And it's this first component right here that's causing a normal force at point A. So that's what this N over A was. If you go back to the free body diagram, that's this horizontal component N equal to the horizontal component of the force there. So it was just simply N was equal to 200, which is a horizontal component across here. And so it was a matter of doing that for the different pieces. The bending moment, the My over I is zero because the cross section was simply rectangular. The neutral axis straight down the center. And point A is right on that neutral axis. So there is no bending moment. Remember, the neutral axis is that point where there are no normal stresses due to bending. Above it's in one direction, below it's in the other direction depending on which direction the moment is. So that's why the bending stresses were zero there. So with that solution then for the normal stresses at A, 533 in tension, I can then draw that on this element. It's in tension. And that's the kind of thing we're going to be doing for the next couple days, possibly even a couple weeks, where we need to see what these particular stresses are on little elemental pieces. And then what we're going to do with those next, that's starting today, is if you ask anybody who's ever taken this course, if they remember the parts we're going to go over in the next two days, they'll all immediately say, yeah, I remember that. I couldn't do it on the fly right now, but I remember going over it. It's that common and that important of a piece of the history of this topic, if you will. Then the second little part here, at B we did the thing, thing found out the normal forces. There's again still this horizontal force, N that was causing a normal stress across the whole cross section. So that's the same for both of these. If there's a normal component to some force, remember that will exert a constant, well we take it to be constant. It's a little bit variant across the cross section, but we already looked at the fact it doesn't cause that much trouble. And then the mc over i is, it's c, because point b is at the bottom point on the cross section all the way down here, which is the farthest away point. And when we're looking at the bending stresses, remember those are, my color constant, those are across, linear across the cross section in varying, varying linearly across the cross section and zero at the neutral axis. So this is point b down the very bottom. It's got the maximum. That's why c is in there. The minus sign is in there. Oh, I guess I drew it the wrong direction, because these two are in opposition to each other. The force I called N puts this cross section in tension, but the bending due to this force here actually, and you have to figure out the effect of both of them, actually puts the bottom in compression. So that's why there's a minus sign on that equation. This is a tensile stress. This is a compressive stress, and thus the minus sign between them. And you can usually gather that from inspection. That came out because I found that the bending is in this direction at that interface doing a free body diagram of the piece itself. There's a horizontal component causing bending in one direction, a vertical component causing bending in the other direction. And it was the horizontal component bending at this point. It was the dominant. And so the total effect of M is compression at the bottom, thus the minus sign between the two. So now I have the normal stresses at b, which was minus 1,000 in compression. And so I can draw that in the element b just like that. Just as if we're doing a free body diagram on this little element, only we're drawing the stresses rather than drawing the forces on these differential elements. Then due to this transfer shear, we have to figure out what the shear stress is at each of these places. That was dependent upon what q is, and that little diagram will help. Point a was right in the middle, so its q is based on the entire area above that point away from the neutral axis. B is on the outside surface of the element at the very edge of the cross section. And remember the neutral axis is right down the center. B has no area beyond that away from the neutral axis. That was our area q. And so thus qA is calculated on the top half. qB is actually zero itself making the whole shear stress zero on element b. And then we draw in those two little pieces. The shear stress at A is 600. So we put that in. And remember we've already looked at the shear stress all the way around is the same. And it needs to be that for both the force and the moment balance on that element. And all we need to be concerned with is just what the directions are. And then there's no shear stress on B, so that element is complete. The only thing that possibly could have changed is if there was any equivalent of this n force that might have caused some kind of downward force on either of the elements. For instance, if maybe there was some kind of object sitting on top of this bracket, then there might be some kind of vertical component to these stresses. There just doesn't happen to be in this case. So what we're going to do in the next couple days, well what we're going to do today and Friday is look very specifically at these stresses on differential elements, different places in the piece. And then some of the other work we'll do is looking at these type of compound loads here that actually contribute to the stress at different places in different directions. Because this one adds both tension and transverse shear and bending. And we take account of all of those little pieces. But it's this little picture here that we need. So again I've put this up on Angel so you can see the entire solution. And you can step through it. But it's these little pictures here that we're going to need today and Friday especially to establish this little bit of the coarse strength of materials that you can find any engineer in America who will immediately recognize at least the general content of the work we're going to do today. So let's get to that. Now that you're all excited about it. Easy folks, easy. Don't get done. Maintain your composure. I know you're itching here. Alright, so let's back us to the general case of the type of things we're going to be looking at here. A couple weeks ago we established on a 3D differential element. So we've got in some direction the x-axis, the y-axis, and the z-axis. And so this is a three-dimensional differential element of dx, dy, dz on the side. We did this a couple weeks ago. And we established from there what the general state of stress was on these solids. There might be some normal stress in the x-direction. I'll draw it as tension, but there's no reason it needs to be. It's an arbitrary choice I need to make as I draw these. So I'll draw it as tension. So on that face there's some kind of some possibility of normal stress. And the problem we just did, we actually saw some normal stress in the x-direction. There might be some normal stress in the y-direction and in the z-direction. Depends on what the problems are, what the loading is. That's what we saw in the problem that was just on the board. That these stresses can even change for different places in the same piece just because the variation of bending across the cross-section. There might also be the possibility of some shear stress. And remember our convention was that we put which face it's on and then which direction it's on. So that's tau x, y there. And I happen to draw it in a positive direction, but it need not be great concern because it could be anything. So this is tau on the y-face in the x-direction. We have on the z-face in the y-direction. And then other components to those as well. You can put them all on. Remember the first letter is the face, second letter is the direction. And then don't forget all of that is on the back side too where I can't really draw. It would be a terrible mess to try to sketch that in. And then we also look at the fact that if you do a force and moment diagram on force and moment balance on any of the directions that the companion shear stress was the same for just the opposite, face opposite direction for any of those. Is that supposed to be tau y-z? Yeah, so this is supposed to be y, x. Thank you, get it? Yeah, we couldn't expect Travis to catch that. He's only halfway through his coffee. So it's also true for xz and yz and its component. So I won't bother writing all those down. That's our generalized three-dimensional state of stress. Depending on what the problem is and just which way those arrows are pointing in any problem, depending on whether we have compression or tension, etc. We, however, are going to generalize this even more. Actually, I guess it's not generalizing it more. We're going to specificize it more. If that's not a great word, I don't know what is. Good luck to my foreign viewers. We're going to keep this as a two-dimensional problem where we're only going to worry about the x and the y directions. We're going to take the z direction out of it. So our generalized state of stress is going to be nothing more than that. And we can do that by having our moments in the z direction only. Remember, that's the right-hand rule for the moment direction. We could put the thumb either in plus z or minus z, but that means bending in the xy plane alone. And only have x and y forces on the piece. There will be no z forces. To start with, if we have sufficient time, we'll add a little bit to that. So rather than draw that two-dimensional problem in the 3D representation, we make it even yet a little bit simpler because then it does allow us to view the backside stresses that I can't draw on the three-dimensional picture just to remind us that these things are always in balance. Remember to, I have to draw them one direction or the other. So I'll draw attention, but it could very easily be either compression or those stresses could be zero as we found it for the problem we just did some of the other forces were zero. And then the shear stresses across the face. In fact, as often as not, we won't even add that xy because all of those, remember these four shear stresses on the faces are all the same as they must be for a moment and force balance. What we're going to do now, and when we touched on this actually several weeks ago, we did it when we were only looking at tension. We did it with something like the third week or so where we took just the axial loads that we were looking at that time. We had no other loads in the first week or two, even three of classes. We only looked at axial loads. But we took this differential element and looked at it at other angles. And if you remember, we discovered that at 45 degrees, we're actually at the point of maximum shear across these faces and no normal forces if I remember. So what we're going to do now, we've actually touched on before, we're just going to do it in greater detail. What I want to open up is the possibility now of not looking at the stresses in the xy direction only, but look what happened when we ourselves as analysts look at different angles for this differential element. We're not changing the load in any way. Remember these stresses are caused by external loads causing the internal forces in the piece. But what about the possibility that there are greater concerns at other angles? This is an arbitrary choice of direction for this differential element to put it in a horizontal and vertical direction. We have already seen that if we do it at other angles, other stresses, they don't appear because they're there anyway. It's just, it gives us a chance to see them. So what happens when we put a differential element at an arbitrary angle, somewhat off-center in some way? And now look what are the stresses not in the x direction, but in the x prime direction, this new, and for a start this will be an arbitrary angle, but what we're going to find very soon is there are specific angles of great concern that we need to look at. So we're going to find a way to come up with these stresses at different angles for our differential element. We're not changing the loading. We're not changing the orientation of the piece itself. We're just changing the orientation of the differential element that we are analyzing. This is purely a mental exercise at this time. We're not physically doing anything to the solid itself or to the loading problem that we're looking at. We're just going to put things a little bit on an edge and see what these stresses are if we turn our, an element of analysis on a little bit of an angle. Alright, so what we'll do here to get started is we'll take a little piece of both differential elements so that we can see how they compare to each other. So we'll pull this little piece out here. It's got a little bit of the history of our original direction, and that's, that's, we're going to need that as a place to start. That's no different than the kind of things we just came up with on the last problem, but it also allows us to get our new angle in there. So if we call that theta, that angle is theta there. And then we need to do what we always do. We balance the forces and balance the moments. So let's see. We'll call that area delta A, which makes this area there the original y direction area delta A cosine theta. So we can relate the areas to each other because turning on this angle changes our direct calculation of how much area these stresses are acting on. So we need that to balance the forces. So that's just our area look there. And then we'll take our differential element and add some forces to it. We'll balance the forces. That will allow us then to solve for these stresses at different angles than just simply horizontal and vertical. So we've got this 4 or this stress, sigma x prime acting on that face. And again remember I'm arbitrarily drawing attention, but it could be anything, even 0. And that's acting over an area delta A. So that's the force on the x prime face. And we can do the same thing for the other faces. So on the x original x face it's sigma x acting on an area delta A sine theta. And as you can suspect when we go ahead and balance these we're going to lose the delta A. So the actual size of the differential element is not the effect here. And on the y face we have a form of sigma y, the normal stress. Oh, yeah, Travis you must be down past halfway in your coffee. Good catch. Yeah, sigma y in that direction, sigma x in that direction. Times the area over which it's acting. So that's the force diagram there. And then we can do the same thing for the shear stress. Tau x prime y prime acting on an area delta A x y acting on an area delta A. Cosine theta on that face. Same thing on that face as well. So we get all the pieces. This is tau x y delta A sine theta. And so we can do now a force balance on that element. And that will allow us to relate the original stress directions to our new arbitrary stress direction. That way we can see are the stresses of greater concern at some other angle than just the original horizontal and vertical. Remember for pure axial loading we found the shear stress was at a maximum of 45 degrees. Something we would not have seen if we just sat at the original horizontal and vertical angle. Alright, so I'm going to condense a lot of this for you because what's left is all algebra including some trig identities. So summing the forces using your favorite trig identities. And I know you all have them memorized. You can, and I believe our book shows you the steps of all this if you want to follow the algebra but that's not necessary. What we're trying to get down to is now a relation for the stresses in the other directions besides the horizontal and vertical directions. Then we can analyze those and see if there's other angles of greater concern than just the horizontal and vertical. So made up of several terms once we get finished that actually turned out to be the average of the two original stresses. Plus sigma x minus sigma y over 2 cosine 2 theta. Twice the arbitrary angle of new direction, the new analysis direction. I just need to fit this in plus tau x y. So it's based of course on the original stresses at the new angles. So there's the stress in the new x direction, I'm calling x prime. Remember nothing's changed with the problem itself, the original setup. We have not turned the piece we're analyzing. The problem we just did, we had that bracket. We haven't turned the bracket at some angle theta. We have not changed the loading in any way. All we've done is changed the angle at which we're making our analysis. Purely our own mental effort. So that I'll call equation one. We'll need three because we need to also sigma y prime and tau x prime y prime. So that for some reason is the next equation. Putting these in the order there in the book. Not for any great reason. So minus x sigma x minus y over 2 sine 2 theta. Make sure you get the pluses and minuses all correct because otherwise you're doing a different problem. Plus tau x y cosine 2 theta. That 2 theta of course is coming from the application of the trig identities. Minus minus plus. Okay. It's looking alright so far. And then the sigma y prime we find from sigma x plus sigma y over 2 minus sigma x minus sigma y over 2 cosine 2 theta minus tau x y sine 2 theta. Let me check those. So I'm going to get the minuses right or you're not going to. Alright that looks okay. Alright and there we've got now. Based on an original problem with ordinarily own mental stresses. The sigma x plus the sigma y's and the tau x y's. We can now look at what the stresses are at any other angle of concern. So let's just step through it and see how it works. Just so we run through it. So we get some problem. Just like the bracket problem we open fast with. We analyze it. Determine that x stresses, normal stresses are 10 megapascals. The y stresses happen to be compressive 5 megapascals and the shear stresses. Now remember our convention on these. This is the negative direction and we're going to need that. The shear stresses there. So we have to set up all these little pieces here. So just to make sure we get the minuses on the right. Tension we take to be positive. So sigma x is 10 megapascals. Sigma y is compression. So that's a minus 5 megapascals. And tau x y, that convention there is our negative direction. Down on a positive face, positive x face we take to be negative. So we can put all those in these equations. Just to speed things up a little bit because you don't know it yet but several of these terms we're going to need again. So it's worth calculating them separately and putting them in. So we're going to need the average stress as you're going to see in a little bit. It's vitally important to some of the rest of the analysis we do. It appears in two of the equations right there anyway. So you calculate it once and then you can just stick them in for that. Save yourself a little bit of trouble. And then this one I've defined myself just because I'm fantastically creative at certain moments. But we've got this sigma x minus sigma y over 2. It appears in all three equations. So let's just calculate it once, give it a little name. It doesn't play as big a part in what we're going to do from here on as sigma average does. But it just is a way to simplify things for you. It just makes these equations a little bit simpler. So go ahead and calculate that separately as well. And then you can plug those two, those into the three equations yourself. So go ahead and do that while I recast these equations in this abbreviated format. It just makes things a little bit easier. See what makes that equation just a little bit easier because we're just packaging together some of the stuff we've already done anyway. Go ahead and start plugging those things in. See what we get. Just to make sure we understand what we're getting. Oh, I'm sorry. And do this at an angle of 45 degrees. So that means x prime, y prime are like that. Remember the original loading does not change. So object does not change. We're just going to take a look at it in a different direction and see if things appear that we wouldn't have seen otherwise. We're writing those equations using this slightly abbreviated notation. That's sigma average we are going to need again as we continue our look through this bit. Don't forget your minus signs. Each of those little pieces is important. But we've got them all correct here. So it's just a matter of putting them in. That sigma difference I put there, that's purely just a matter of convenience. I invented it's not in the book. It doesn't play a big part in what follows, not like sigma average does. So figure out all those pieces. And then we'll put it all together. We're just taking a look at things at a different angle just to see if something a greater concern doesn't come up. So we'll just prepare a new direction. 45 degrees and we'll figure out what the new stresses are in those new directions. So watch your minus signs. Don't forget the two theta. We need numbers so we can instantly look at them and say which one is greater than the other. I can't do that with fractions on them. Those fractions aren't hard. But last week you had your pi's left in there. I can't tell if 3pi over 8 is different than 7pi over 15. I'm not going to teach you. We're engineers. Chris, you need to live with us. Live against us. Five weeks of school left. We're still trying to straighten you out. Okay, when you get these numbers to check with somebody else make sure you're getting the same thing. We're looking for three different values in the new directions. The loading has not changed at all. The object being loaded has not changed at all. We're just simply tilting our head sideways to look at things. All the video shots on American Idol now that tilt the camera sideways and zoom in and out. Who goes to a concert that does that themselves? Well you guys probably do your half asleep. Do you guys agree? Yes or no? For one of them. And nobody's going to check with Chris because he does his fractions. So he's no fun to play with. Will? David, who do you check with? Anybody? Jim and Joe. Hey, Joe. My name is Andy. You're so dead. You're right now in the principal's office. Did you agree with David? Did you agree with Samantha? Yeah? See, Chris, you shut yourself out of this. Well it's your own fault. You're going to work in fractions instead of real numbers like the rest of the world about it. Alright, let's see. You should have gotten, let's see, sigma x prime is, what do you have, Tommy? Negative 3.5. Don't forget, the negative is vitally important. Negative 3.5 mega Pascal's, which means now on the x-space we see compression. Remember that it's not a change in the loading in any way. We're just looking at a different direction. We now see that there's compression of 3.5 in that direction. How about y prime, sigma y prime or tau xy? No, we're not going to you, Chris. Travis, what do you have for tau x prime y prime? Negative 7.5 mega Pascal's, so that's on the negative direction. Once we get that established in all the other directions, I'm going to actually label it with the negative because that's not clear if the arrow is in the wrong direction. And then for the last one, sigma y prime, you have somebody, 8.5. So we see a normal tensile stress, 8.5 in the y direction of 45 degrees. Now, just as a quick exercise because these angles can start to get a little confusing. I want you to redo it for 120 degrees. No big deal, it's just a quick calculation. But I also want you then to sketch it yourself to show me what it looks like. Because not only do we have to come up with these numbers, we need to picture where they are and what they're doing. This is the type of thing that those people who work with carbon fiber worry about because there's a definite bias to that material because it's actually a weed fabric. And so the angle at which you lay down the carbon fiber material has a great impact on the structure's ability to absorb the stresses in different directions. It's also a great concern with wood because wood is very much a directionally biased solid because of the wood grain. We're just going to do it real quick with 120 degrees and then make the drawing that goes with it. I don't doubt you can calculate things. I want to make sure that you can actually set them in very well. I'm probably driving you nuts by saying it but just past experience of teaching this we're not changing the loading. We're not changing the object itself. We're just taking a different look at it just until NARC heads a little bit as we analyze the problem after we determine what the original direction stresses were. And you can imagine this is very easy to set up as a simple program. In fact if you put something at Google you'll come up with a whole bunch of athletes that will calculate this for you on why even some are better than others and will even make the drawing. But that's what I want to make sure you can do. Just do a couple of decimal places one or two and most of the stuff, the detail gets submerged in a factor of safety anyway. We agree? We check. Did you make your drawing? I know you can calculate these numbers. That's nothing. I want you to make the drawing. Sigma X prime 395 megapascals. Sigma Y prime 105 al-X prime Y prime is 9.5. In this direction of analysis they're all positive. So if you have a material that can't take a lot of tensile normal stresses you might want to put it at 120 degrees and that's the material that you're using for the solid. You have wood that is not very good in tension. You can angle it a little bit. It might be a little stronger in the other directions if you will put it in the weaker direction. Everybody get those numbers? Any numbers you got? Two more. A meatloaf song, two out of three ain't bad. 19.5. 9.5 is okay? Maybe you got those numbers? No way. One out of three? No. One out of three. Watch the minus signs. It's real easy. Alright we've got those numbers. Now the X prime angle is at 120 degrees so there's 90, 30 past that will be something like that. And so that then orients our solid. Remember just a differential element. Something like that. X prime direction is positive 3.95. Y direction also positive but about half that size if you want to scale your stress vectors. That's easy. Positive normal stresses are easy to draw. X prime, Y prime remember that's positive on the X face which puts it in that direction. And that's a positive 9.5 so we have I think that's the biggest shear we had on any of the directions we looked at. We started with 6, went to 7.5, now we're at 9.5. So if you have a material that's weak in shear you don't want that material oriented at the 120 degree angle. This at the sketch or the calculation. Yeah that's the X positive X prime face. We're going up on that face. Remember this was the positive X face we're going down on that. That was a negative. That's just an arbitrary convention. But arbitrary convention works extremely well for us if we're consistent with it. Be consistent with arbitrarians. Then we're allowed to be arbitrary. Remember that when you become parents. Because there's nothing more fun in the power of being arbitrary than for kids. Alright everybody get to this picture. Alright so we're going to take the next step with it and this is the piece. You can go to any bar right outside of an engineering lab on a Friday afternoon and you'll be able to talk with the guys about this. Which is just a really fun possibility in your lives. I'm sure for social contact. Most of them are crying in their beer about the fact that you can't find Pong in a bar anymore when they're just getting good at it. Alright so I'll leave those pieces up there. We'll meet those for reference. Well we're still going to use them. But here's the next step. We're going to do a little bit more mathematics with them to get to our next spot. So we've got those equations there. So I'll just write down what the steps are. Not go through the algebra. That's not the point of this course. So square one and two. Add them and then simplify. And you can imagine that those simple words are about 45 minutes of algebraic calculating and the like. But the result, what we end up with is known throughout the world. Anybody who's ever taken this class. So we get this. Sigma x prime minus sigma x plus sigma y over two which is sigma average. So we can put that in a second. That quantity squared plus tau x prime y prime squared plus sorry equals the quantity we call sigma difference squared plus tau xy. So we can simplify that with those little addition things there. Sigma x prime minus sigma average quantity squared. Square equals sigma diff squared plus tau xy squared. Make sure I got all that right. Alright, everybody got the pieces because here's what we do with this. Notice that this term sigma average, sigma diff and tau xy, those are all known from the original analysis. I give you a problem just like we open class with. These things are all known quantities before anything has been said about some other angle. Those are already known can be taken as input. These two pieces are the variables. Those are functions of the angle chosen which can be arbitrary but won't necessarily be very soon. So we'll call those variables. So I give you a problem like we open class with. You figure out the stresses in the ordinary orthogonal directions, horizontal and vertical. That sets these three values that I've marked known as constants. Then you look at some other angle and the other two terms then it becomes variables. So we're going to rename one of those terms again. So this on the right hand side that's a quantity that's known. That's a constant no matter what the angle is. That value is already determined. So we're going to call that r squared. So what I have are variables we could call this one, this variable x, call this variable y. Then you notice the equation we've got then is x minus a where a is the average squared plus y squared equals a constant that I've called r squared. That's just a very, very generic form of this equation. You recognize the form of that equation. Chris does. Even without fractions. Anybody else recognize it? David you do. What is it? Seems to be a circle that was positioned in the x-axis variables. It's a circle with the center a zero. Zero because there's nothing added on the y-term and of radius r. This is what's known as Moore's circle after the engineer who first established Otto Moore back in I believe the late 1800s. And you drop that in any bar frequented by engineers that still happens to be in business. You'll instantly make friends. So let's figure out what Moore's circle looks like. There's a step-by-step process in the book for how to draw Moore's circle. I don't like it because it leads to really ugly circles. The easiest thing I think to do is to draw the circle first and then we'll put in the axes and the known values and then we'll get a much better picture. It's much easier to draw the circle and then put the points on it. You just get a better sketch. But the circle comes from the known quantities here. Alright we're going to put right through the center in the horizontal direction the normal stress axis with positive going to the right. And we know the center to be at A and A is sigma average. So the center of the circle is that sigma average. I told you 20 minutes ago or so that that was going to play a big part in what we were doing to come up with from here on. That establishes the center of our circle on the sigma on the normal stress axis. Then in the vertical direction and just where this falls depends upon the individual values and we'll see that when we go through an example. In the normal the vertical direction we'll put the shear stress axis. However it's upside down. This is the positive shear direction. This is the negative shear direction. And I'll show you why we've done that in a couple minutes when we go through a problem. Now that circle is established by these known values. A is a known value coming into this analysis. R is also a known value coming into this analysis. Once you've done the original analysis in the horizontal and vertical directions you've established this circle. Any other angle you choose will give you another point that lies on this circle. So all we need to do now is relate the angle of analysis to where it falls on that circle. But these all have a radius R. That's a known quantity once you've done the original analysis. We could have figured that out 10 minutes in the class when I went over that first problem without knowing anything else about what we've done here. That would already be established. But that's a very important value because it is the same as the maximum shear stress possible in the problem. And if you add that quantity R to the average shear stress you get the maximum normal stress as well. That point right there would be sigma average plus R. So R is as important as sigma average is. And if you subtract R from sigma average you get the minimum stress possible. Alright, so what I want to do is go through the problem we just did because we already have all those numbers. And we can learn how to sketch out more circles. So for the problem we just did, let's draw the original state of stress, the original calculated state of stress. So that was 10 in that direction, sigma x equals 10. I think we had a sigma y is in compression so that's 6 mega passed out. But I drew it in compression so I also want to put a minus sign because then that confuses the reader. Clear the direction and the magnitude are all obvious. It was 5? Oh yeah, it was 6 in the shear stress direction. And that was negative as well. But I drew it in the negative direction so I'm not going to put the negative sign on it because then you don't know if the error was drawn in the wrong direction or not. Okay, and then from those we calculated sigma average which was 3.5, no, 2.5. And we can also figure now what r is. And that's equivalent to the maximum possible shear stress. However what you don't know yet, what we don't know yet is at what angle that occurs. And that's what we're going to use more circle to find out. So calculate that. If you would real quick, so we got it. That's the square root of sigma diff tau x by each square. And we already have sigma diff, so you have to redo anything. We got tau xy from the original analysis anyway. And you get, what do you get for r? r is 9.6. Yeah, that's what I got. Okay, now we can draw more circle and then we can figure out what the angles mean in there. So draw the circle, put the horizontal axis through it. Because that's not going to change. No matter what the problem is, the horizontal axis goes through the center there. We know the value of the center is 2.5. And so on this particular more circle, the units are mega Pascal. So we know that's a 2.5. We add r to that, which is 9.6, and we get this point here, which I believe is 12.1. All right? Subtract r from sigma average, and we get this other point on the other side. And that will allow us to put in the vertical axis then. And that value, oh I didn't write it down. What is that one? 2.5 minus 9.6. 7.1. 7.1. Negative 7.1. And now the negative sign is important because remember we're trying to place the axes. So that's 9.6. And we want the zero point between there. No. Yeah, that's 9.6. And so that's 2.5. So the vertical axis would fall right about there. And that, remember, is an upside down shear stress axis. So now we know what the maximum shear stress would be. However, we don't know at what angle that occurs yet. We haven't figured that out yet. Joe. I can just sign the vertical axis. Is that what you want? No, no, it's not arbitrary at all. It has to do with these values here. That's where if sigma equals zero, because that's where the vertical axis goes. And if that's 2.5, and that's minus 7.1, it's about there. It's precise location isn't terribly important. However, you don't want it in a completely wrong place because it's important to know that the minimum normal stress is negative. We expect the minimum stress to be compressive in this case. Alright, now, here's how we relate that to relate those maximum values to the original location and how we find out what angle these maximum stresses occur. So we take the values given, sigma x and tau xy, and we plot that point. So it's at about 10 and about 6. So that puts us right about there. And our book, I believe, calls that point A. That's the original sigma x tau xy point. Remember tau xy is positive down. See what the book tells you to do is plot that point, plot this point, and then draw your circle. You can imagine how difficult that is. It's a lot easier to draw the circle and then fit everything else to it. You get a much nicer drawing. Now that point there goes right through the center, and I believe the book calls that point g back there because it labels a whole bunch of points on these things. That's just, that point is sigma y tau xy. Remember, that was compressive, so it's a minus. That's our up direction, so this is at minus 5 and a distance tau xy. Okay. Here's the deal. That now establishes for us the direction, the angle of these maximum stresses. There's some angle, and if we go to that angle, we're going to have these maximum stresses. We could be, that's this point here. We need to know this angle between them. There's the maximum and minimum normal stresses, and we know what those are. This is 12.1 megapascals. We know this is compression of minus 7.1 megapascals. Remember, any point on the circle is a possible state of stress for our solid, and all the different angles possible. If we happen to be at this point of stress, then we are at zero shear stress. So this is a point of maximum normal stresses, zero shear stress. And the angle at which that maximum condition occurs is two times theta on the circle drawing. And so this is theta there. It's not drawn quite to scale. So now we can take the circle, take the analysis, figure out where the maximum shear stresses are. We're saying that theta is equal to two times. Yeah. On this circle, remember this circle was all established. Everything in those equations was two theta. So that's, be careful with that. But there's the reason we put positive down. Because now that and this are in the same direction. If we had positive up on the shear stress axis, then those would have been in opposite directions. And this can be confusing enough. Is it tau xy negative? No. Tau xy. Oh, for this one? Yeah, I guess so. So you could have come up with that a little earlier. Because that changes everything. But we'll go through it again. Draw our circle. Draw this normal stress axis. You know it's going to be mega pascals. The center is at 2.5. And then we're at a point of 10 and negative 6. So it should have been, that's about halfway. What's it right up there? There's point A. There's point G. And our axis was about there. So there is our axis, our angle to establish maximum, the direction of maximum stresses, normal stresses. So that's where we'd expect in the x direction positive and the y direction negative. And that's the angle in there. Where are you in your coffee there? You could have jumped on that a little bit earlier. Okay, we can figure out just what that angle is. Because let's see, tangent to theta. And by the way, this is called, this direction is called the principal direction because it leads to maximum stresses. So we call this angle theta p, the principal angle. We can analyze any other angle just by putting it on the drawing and going around the circle. This happens to be the principal direction. And that's the, remember this is tau xy in the x direction. And the y direction is sigma diff. I think that's the piece we need. And that's 2 theta. So for our problem, we can figure out what it is then. The principal planes, that's what these are called, the principal planes is one half the arc tangent of tau xy over sigma diff. And we've got both those numbers, which is why it's convenient to calculate sigma diff and sigma average ahead of time. Because you got to use them quite several times. And so for our problem that comes out to be just a little under minus 20 degrees, right? Minus 19-3. And that's the minus angle we do there. So we know if we go clockwise in our analysis we'll find the direction of maximum normal stress. There's also another angle we can figure that will give us the maximum shear stress. And that's actually just this direction. And that's called theta. This is twice that. It's called 2 theta s. It just happens to be the other direction. What happens to be if you look at the analysis, if we go 45 degrees, we've got the very same solid just turned on a different angle. So the way to find that principal plane is minus sigma diff over tau xy itself. Or you can take 2 theta p and subtract it from 90 and you'll get theta s. You'll get 2 theta s. Or you can take theta p and subtract it from 45 degrees because of the factor of 2 on the two of these. So I think that comes out to be 25.7 degrees. And at that angle for the x and there's y at 25 degrees, which is about here, we're at a maximum shear stress. And in this case it turns out to be negative on that x space. But tau max for the shear stresses, if we move to an analysis angle of theta s that would put us at the top and bottom point on our circle. At the maximum shear stresses, the normal stress is sigma average on all four faces. Which in this case is positive 2.5. Could be negative. Depends on where the circle falls for different problems. Oh, we're way over time. We're having so much fun. I'd love to pay you time and time. What? Okay. As you analyze these, some people are very comfortable with this circle. Some find it very confusing. You can do anything you need to do with the equations as well and just forget the circle. But you can google Moore's circle. You'll come out with a hundred Moore's circle calculators on the internet. So you can always check your problems.