 If we get a chance to go into a bar and say, hey, I know about Mohr's circle. All right, well, that's what we're going to continue with today. If you remember, we were taking loading configurations that we found on a particular elemental solid, but we started with it oriented quite naturally in the horizontal and vertical direction. Typically, this is just some elemental piece of some greater object, like a beam under some kind of loading. And we're concerned with looking at one little place. We want to see what the stresses are on that little piece right there. And that's all we're doing. So the actual stresses that we see on these elements are wholly and solely a function of the loading that started the problem. So we might have some kind of normal stresses in the x and y directions. And we might also have shear stresses. In any type of problem we look at, any one of those could be pointing the other direction or might even be zero in a problem. It's very easy to come up with problems that don't have any of those in particular. Well, I guess we wouldn't want to come up with a problem that didn't have any of those, because that's a pretty boring problem in this class. But what we're looking at starting with Mohr's Circle, the whole idea of Mohr's Circle is just one way to look at everything, is if we take exactly the same situation, the loading doesn't change, the object doesn't change, nothing changes over here. All we're doing is looking at it at a slightly different angle. Because, well, it's no different than in physics one, we get a particular force in a problem, and then we pick a direction for the coordinate axes, and then we break the force into its component directions. This is sort of the same type of thing. Once we turn our coordinate system to a slightly different direction, then the forces have different sizes just because of these coordinate directions that changed just nothing more than changing the way we're looking at things. I have to stress, I can't stress strongly enough that the loading has not changed. All we've changed is the way in which we're looking at these things, all the way in which we're analyzing the problem. So some stresses can change direction because of that. It depends on the angle and what the magnitudes of all these were. So I'm just arbitrary picking some directions and magnitudes as I sketch these in. And we have to do that because we need to find out are there directions where the stresses get so big that perhaps our material cannot withstand those stresses. And we might not see that if we just keep it in the horizontal and vertical direction. We need to take a different look at these things and find out just where these greater stresses lie. And once we had that idea, we set up the equations. There's a couple little parts that go into each of the equations. And a couple of these actually have greater meaning later. The average shear stress is simply the arithmetic average of the two shear stresses on the two faces. Remember the prime designates our new angle at which we're looking at things, x and y is the original horizontal angle. So the average stress actually comes into things in two ways. We also need this quantity. It recurs in several of the things that we come at looking at. This you won't find in any book. I made this up just because I'm lazy and I don't want to keep writing out that factor. It's just a lot easier to write this shorter part. So then we laid out how we have the three equation that give us these transformed stresses. That some different angle of analysis blessedly did not bother with the algebra that got us to this point. But we did then get a chance to practice them a little bit and we'll do that again today as we go through these. So all I'm doing is rewriting these so that we've got them as sort of a summary. Remember that sigma diff, it carries some meaning for us as we go through these as we'll remind ourselves again in a second. But you won't find that in any book. It's just a way to simplify things that I invented a couple of years ago just because I'm lazy. But it recurs over and over so why not save ourselves a little bit of trouble? That's all we're doing. Those are the three equations of the transformed stresses that we came up with. We also then took the next step of combining two of them and simplifying and realizing that when we did that we got the equation of a circle that's the very famous and well-known Mohr circle. Every young engineering student is expected to be familiar with. I'm not sure Mohr circle is all that helpful. I sometimes find it even more confusing to use because anything we need to know we actually could find with the equations straight away. In fact, one year I didn't even teach Mohr circle but then I got nervous that I would be putting out students that never even heard of it and our transfer schools wouldn't be happy with that. So I've gone back to at least showing it. It's not too difficult to step through and it does carry some of the information. What it gives us, let's see it's a circle based on center sigma average and we use the x-axis as the normal stress axis. The radius of this circle is actually the maximum expected shear stress because we put the y-axis as the shear stress axis with positive going down and we'll remind ourselves in a second just why we need that. And so we've got these couple of points here, we find that there will be a maximum shear stress. Remember what this circle is, it's the locus of all of the values that we can see as we go through all of the possible different angles of analysis. So whatever those angles are, let's make it on the directional for now. We see that in that locus there's some angle at which we're going to have a maximum shear stresses on the faces and average normal stresses on the faces. There's also some angle where we'll have the maximum normal stress and no shear stress because that's right on the x-axis and there's even some face with minimum shear stresses and any points in between are possible. Anywhere on that circle is the points that are possible for these angles and notice that some of the points repeat themselves are just 90 degrees images of each other which is this square moved at 90 degrees and we end up with the very same square anyway. So we don't need to do every angle, we only really need to do 0 to 90. After that we start repeating ourselves. We also found that these original stresses can be plotted right on here as points. If we do so, of course they fall right on the circle as any points at any angle would and once we do that that defines a diameter, I think the book lists those points as A and G. And when we've done that then we have this idea of these particular angles. We've defined a diameter with our original stresses in the untransformed state just merely lined up with whatever the orientation of the solid was. But there's some angle that brings us back to this line here to bring us back to that point where we have the maximum normal stresses and zero shear stress and that angle was called theta p. And on the unmoor circle it happens to be 2 theta p just because every angle in the transformed equations is a 2 theta angle. And those are called the principal angles and the stresses found there are called the principal stresses and the directions that we get from those are called the principal planes. And we find those angles by just using some of the transformed stresses that we've got actually using the original stresses. And so if we happen to be at those angles, actually we can put these three drawings, well we have two drawings now, the original and now the theta p direction and we'll have a third and a second. But we can put all those together so that maybe that helps see it and we'll do that with a couple sample drawings. So we have our original analysis of the loading, whatever it might be, can be positive, can be negative. And now it's some transformed angle direction in the same direction as that, this is our original x and y direction. And now it's some angle theta p in that same direction so we can draw it like this. We now have an analysis that tells us, that brings us up to this point right here on the circle where we're at maximum normal stress on one face, minimum, and with the picture I have here it just happens to be negative so I'll draw it that way, no shear stresses in that direction. Remember that puts us at this operating point on the x axis with no shear stress. There's also another angle called theta s which if you look at this carefully do it a couple times is just 45 degrees away from theta p, since it uses the same values just in inverse to each other. And that angle is opposite direction there, 45 degrees away so that's 90, that's about 45. And we have another analysis direction, remember the loading hasn't changed, we're just looking at it at a different angle. That puts us at this operating condition where the normal stresses are the average values and the shear stresses are the maximum. So we have average shear stresses on both faces, actually all faces. Why is it at that angle? Because of the way the circle I happen to draw is actually a negative. Remember once you've drawn this circle and all these points are set by whatever the original stresses are, once you've got all the original, so we'll go through a couple of problems. This axis could fall anywhere, any of these are possibilities, the shear stress axis could fall there, it could fall over here, all of these are set by the values mostly sigma average that puts the center and r or tau max that fits the radius. So that shear axis could fall anywhere and depending on where it falls I just happen to draw such that the minimum shear is a negative. I guess another possibility is it could fall like this too. Now this is not the maximum shear because it's actually quite a bit smaller. It's this shear, I mean not shear, normal stress that would be the maximum one. And then these are maximum shear stresses at this second of the principal angles. So maybe we could label that one p and that one s just to remind us which of the angles we're at. Okay we'll do a couple problems, just walk through together a couple of problems. Okay including calculating the angles and what the principal stresses are drawing the circle as we go through each of these. And we'll end up with this type of drawing here. Okay so we come upon a problem with a loading such that we calculate on an elemental piece somewhere in the solid that we expect these kind of orthogonal stresses. We expect in the x direction that there'll be 50 megapascals in tension, in the y direction 10 in compression. Well I hope it's obvious, maybe it needs stating anyway, that the units on these have to be the same if you actually want it to be a circle. You can't have one direction in megapascals and the other direction in kilopascals. It's not typically a problem in any of ours that we look at but still needs stating just in case. And we have 40 megapascals as the ordinal shear stress. And so we want to find out the principal stresses and the principal directions for these particular shear stresses. Alright we need a couple pieces. We need shear average, I'm sorry normal stress average because that gives us then the center of the circle. Whether you want to draw the circle or not, we still need those because it's part of the principal stresses. So let's see that's 50 minus 10 because that's in compression which is 40 megapascals divided by 2 is 20 megapascals. We'll need that value sigma diff which is the x stress minus the y stress. So that's a 50 minus a minus 10 which is a 60 divided by 2 is 30 megapascals. And then r which is the radius is sigma diff squared plus tau xy squared and then screwed it. So we've got the sigma diff that's 30 squared plus 40 squared added square root of 60 should be 50. So now we can sketch out the circle. The book says plot the points that you need and then sketch the circle. It's very difficult to get a decent looking circle. I think it's a lot easier just to simply draw the circle and lay the points out from there now that you've got them. Now put down your axes where they belong. We know that this axis goes right through the center and that that point is 20 megapascals. So both axes in megapascals center at 20. We know the radius is 50 so that makes this far right point then 20 plus 50 is 70. And far left point 20 minus 50 is minus 30. And now when you've got those you can place your y axis to the shear stress axis. So that's a distance 50 or about 20 so that'd be right about there. Give or take a little bit. It's just a sketch and remember the positive direction is down. Remember why we put the positive shear stress direction down? When we get these angles drawn on here the angle in the circle is the same direction as the angle on the drawing. If we had positive shear stress up then clockwise on the circle would be counterclockwise on the circle. Just extra pain. We don't need extra pain. Alright so let's see if we can plot point A. There's different ways to find out what these shear stress or these principal angles are. One of the ways is to plot point A which is sigma x tau xy which is 50. And is this positive shear stress or negative? You have to pay attention to the directions. On the positive x face it's in the positive y direction that's a positive shear stress. Now it's up on the x face and then we plot that point down. So we're at about 50 down about 40. If that doesn't plot right on the circle something's wrong because this circle represents all the possible stresses at any angle. So it always should plot right on the circle itself. And so there's our angle 2 theta p pictured right there. And we can calculate that directly. So that's plus 40 over sigma equals 30 theta p. Now notice if you remember the calculators don't do it but this actually represents two angles between 0 and 180. If you used a chart of tangents you would be able to pull the two angles out of there at 266 and 116. But those are 90 degrees apart which just means our element has been rotated 90 degrees and we get the same picture anyway. So one angle is not more important than the other. So now we can plot our first of the principal transform stresses by actually putting in some of these values. At 26.6 degrees which is about what that looks like we know that the maximum stress, 70 megapascals, the minimum will be the minus 30. And at those points there's no expected shear stress because we're right on the x axis. And the other principal direction, tangent 2 theta s is minus sigma diff over tau xy which is 45 degrees off of this anyway. Minus 18.4 degrees. This is theta s minus, the minus indicates the clockwise direction of the x axis. And we know at that angle all faces will have the normal stress of 20 megapascals. We know its intention because the center happens to be a positive 20. It might not always be. We'll do a couple problems where things fall in other places. And we know the maximum shear stress at that angle, the maximum shear stress at that angle is whatever the radius was which is 50. If you have a material that's very, that might be oriented at that angle, that's some bias in the material like there is with wood. There's very clearly a directional bias with wood because of the grain. If it's weak in shear you wouldn't want that grain oriented in this direction. You'd want it oriented in this direction where there is no expected shear stress. And so you can selectively place the material such that it withstands these stresses the best. Definitely the case with things like carbon fiber or the fabric, the carbon fiber fabric itself is just a woven fabric. Very definitely has a directional strength bias to it. And then they carefully design the direction in which they lay down the, I think it's called the warp and the weave of the material. Alright, that's one example. It just happened to look very much like the other sketches we'd had where the vertical tau axis, shear stress axis fell just inside the left hand side of the circle. Just coincidence, just to allow us to comfortably get into some of these problems. But now we know what the principal angles are. These are the principal planes, those directions, and the principal stresses on those. Questions on that? Before I close the board and we do another one, let's do this one. A more of a designing oriented one that shows that we don't always concern ourselves with these principal angles. Sometimes we are more concerned with other angles. So imagine steel pipe that we expect to load in the axial direction down the length of it. And it looks something like that. So give it a 12 inch outer diameter, 1 quarter inch wall thickness, and we're going to load it axially. So put a little cap over it, give it some good solid base. So once loaded it's going to look like that and that will be 40 kips. Now the deal is this pipe was made by starting with a triangular piece of steel, rolling it up and welding the seams together. And you can do this kind of thing with a piece of paper maybe you have before. Take a long triangular piece like that and then roll it up and you'll get a tube with a seam that goes around the piece spiraling. Exactly the way the tubes inside of paper towel rolls are. If you've ever taken those out and chased your little brother around the house. What's along? You'll notice that after you've done that for a while and the thing starts to come apart, you see that it was made out of a single piece of paper that was spiraling wound together. So that puts this, what will be a welded seam on there, have to weld the two pieces together all the way around at a particular angle. So let's assume that that weld which goes around the piece is at an angle of 22.5 degrees. This should actually be the angle on this original piece. So we need to calculate what the stresses are going to be and then look at them at this 22 degree direction to make sure the weld is going to be strong enough to hold. Oh wait, sorry, there's more load to it. There's also a torsional load. A torsional load of 80 kip inches. So what we want to do is look at this loading, look at some elemental piece in the ordinal directions because that lines up with the solid itself and the loads, and then take a look at the stresses at an angle of 22.5 degrees. We'll call that the y direction and this the x direction. So we have a normal stress in the y direction because of this compressive load over the area that's supporting it, which is the cross-sectional area of the pipe itself. So we'll need that value, but that's pi times the outside radius minus the inside radius squared. That's 9.23 square inches. That's this area. That's the area that's supporting that load. The load itself is with 40 kip for the 9.23 inches. So that's easy. We could have done that in the first week, 4.33 ksi. So that's back. That's stuff from a long time ago. That's easy. Now there's a shear stress due to this torsional load. Whatever this load is, is trying to twist the piece that way. That's going to tend to make our little elemental solid can't to the right, I guess, if you will. And we know how to figure out what that is. That's tau c over j, where j is the polar moment of inertia. And hopefully you remember how to calculate that. It's in the book. That comes out to be like 387 inches to the fourth. That's like 1 half pi times c0 to the fourth minus ci to the fourth for our inner radii. And so we've got all those pieces now. So what we do with these three inches is the torsional load. What's c? Six inches. If we look at the pipe on n, we're looking here at an elemental piece on the outer surface because that's the farthest piece away. Remember, that's where the shear stresses are the greatest. This is 6.0 inches and j is 387 inches to the fourth. Gives us kips per square inch ksi, just like what we want. And that's 151 ksi. And so now we can draw our little elemental piece. We're originally lining it up in the direction of the loads and the piece itself. We know it's going to have a compressive normal stress of 4.33 ksi in the y direction. That's because the load is in the y direction. And then because of this torsion, the torsion on the piece there is going to try to twist the piece in that direction. If this was made of rubber, you can describe that little circle on there, I mean that square on there, and then twist in this thing, you'd find it sheared in just that direction. So that gives us then the direction to go with the magnitude that we've got there. And once we've got that, we've got the other faces. And so that's a expected shear in the ordinal direction. The direction is lined up with the loads and the piece itself of the 1.5 ksi. What we want to do though is look at what those become at this 22.5 degree angle where the weld exists. So the probably easiest way to do it is to just pull out those three equations and put in theta equals 22.5 degrees. So we need a couple pieces to go in there. The average normal stress is you calculated for me. What do you get? Got something? Remember, it's sigma x plus sigma y over 2. Sigma x is zero. We have no load in the x direction. And so it's the 4.33 divided by 2. That 4.33 though, remember, is impressive. So this is a minus 2.17 ksi. Sigma diff is sigma x minus sigma y. So we have a minus, a minus. It's plus 2.17 ksi. And then you can fill in all the little pieces and calculate everything from the equations. Let's go. What's square? What's square? Is that the base? What's square? This square? The ksi. This square? Yeah. That square is coming because we need to know what's happening at the weld. So that's where we place our elemental solid. Remember this idea of an elemental piece is an academic one. It's one we do in our minds. We don't actually have a piece that we pull out of there. It's a piece that we look at of size delta x, very small dimensions, and to see what the shear stresses are on that piece. No, it's not down at the base because we're looking at the weld. It's this little square here. So you can put those pieces in the equations. Sigma x prime, we're now looking at an angle of 22.5 degrees to see now what the stresses are right lined up perfectly with the weld, and that's going to be the sigma x prime there that we come up with. Sigma average, I'm sigma diff cosine 2 theta, where theta is the 22.5 degrees plus tau xy sine theta 2 theta. And you can put all those values in because we've got them all now. We're going to go ahead and calculate it. You should get 0.423 psi. So very, very small expected stress in the x direction, and it's positive so we know that there's a slight tension there. Draw it small. We don't know what the other ones are going to be if we want to guess these to scale, but we've got to figure if we're working with 4.3 and 1.5 that 0.43 is going to be pretty small in the diagram. So we put it that way. Sigma y prime, so that remember is a normal stress along the length of the weld, which is not really the direction you worry about a weld failing. You're more worried about it failing because a normal stresses transverse to the weld or shear stresses in the direction of the weld. So this one we didn't expect to be very important. It's not the direction we expect a weld to fail anyway. So we need to calculate the other pieces. So that's a minus. So just taking the equations as we had before, we have another set of equations that you can save a little time in your life if you want to have a tattoo or somewhere. We don't have to keep you walking them up. We've got all those values. Remember theta is 22.5 degrees. We put them in and calculate it. We get minus 4.77 ksi. We can draw that on the side. So about 10 times bigger than in the x direction. 7 ksi. And compress it. Those stresses actually hold the weld together. And then the shear stress that we expect at this weld angle, we've got all those pieces as well. Minus sigma diff. Minus sigma diff sine 2 theta cosine 2 theta. That's just a straight, sorry, plus, that's just a straight component, trigonomic components there. So we get minus 0.467 ksi. So pretty small shear stress. But it's up to the technicians, I guess, to determine that that actually is something this weld can withstand because that would depend upon the type of weld, the material, the quality of the weld. And it's a negative. So we know which direction it goes. Sorry? So we know that that shear stress is going to be 467 ksi. And probably as part of this, you would have found out what the principal's angles are and decided let's not put the weld on that angle because things do get worse there. 22.5 degrees is a very easy angle, make half of 45. So it's easy to measure, easy to cut pieces at that angle. And now we know what stresses we'd expect in that direction, in the direction of the weld itself. Got it? Okay? John, when you return, would you pull that door shut? It's getting a little noisy out there. Thanks. Phil, all right, you're squinting at it. You okay with that? Yeah. Okay. No reason on this one to draw more circle. There's no great value in it. But you can if you want. It'll just show you the same thing we've got here. We'll show you where this point was. And then we've moved 22.5 degrees off of that, which is not the same as going from the point A that we've plotted before with these values to the x-axis, the normal stress axis. That's the principal angle. That's not what this is. It's just an angle that was of concern because of the way the piece was constructed. All right. Comfortable with that then? Matt, all right. Pretty straightforward. Some of the problems look a lot harder. I know the circle can sometimes be a little confusing. But as you see, when we get to some real life problems, we don't always need it. All right. So let's do another problem that's got that starts from the load like this one did and we'll take it all the way through the principal angles and find out what the maximum expected stresses are and like. All right. So imagine we've got some component of a piece here. It's like some bracket. We'll put the, we'll line the coordinate directions up with that. Oh, we don't want to order it up there. We'll put it in that direction. We'll put the x-direction right down there, lined up with that piece of bracket arm. Y will go up and z will be in that direction. So it helps to line things up with the piece and the load if you can. And since we're looking at sort of end-to-factory problems, it's exactly what we'll do here. So in the z-direction, but applied at the end up here, there's a load of 150 pounds. So that's going to try to twist that piece. It's going to bend that piece over and it's even going to try to shear that piece off depending on what the piece is good at resisting. Hopefully it'll resist all of it. And that's in the, actually in the minus z-direction. That's just so it's clear. Minus k hat. So it's in the minus z-direction. All right. Let me put some dimensions on this piece so we know what kind of loads that's applied. This arm is 18 inches long. For whatever reason, we're going to concern ourselves with a spot right here. And that's up four inches from the bottom. That little spot's up four inches from the bottom and down 10 inches from the top. So the whole piece is 18 inches. I'm sorry, 14 inches in that y-direction. And that's for whatever reason. That's the point where we want to do our analysis. We want to look at that point. We'll label it point H. Find principal stresses and angle. Point H such that it's completely in the x-y direction. So if we took that picture and blew it up, here's our shaft. It just happens to be right there on the front of it. And that'll be in the x-y direction itself. So it's not on the back side that gives a little z component to it. It's right there on the front that if you were looking right down the z-axis, you'd be looking right onto that little square H. For whatever reason, that's where we want to make our analysis. So we need to know then for that piece what the normal stresses are in the x- and the y-direction. So let's say the normal stresses in the x-direction depend upon what forces are in the x-direction and the area. I guess I had to put area x because it has to be whatever area is in the x-direction that's resisting that load. And remember, this is all function of this outside load here. Let's see. Well, we don't have any forces in the x-direction. We have a force in the z-direction, but not in the x-direction. So that's easy to do, that one's zero. So we have no x-direction, normal stress on that element, H, down the main shaft. And there's no bending in that x-direction either. In the y-direction, it'd be whatever force is in the y-direction and whatever area is resisting it. We don't have any y-force either. However, we do have some bending because this 150-pound load is going to try to bend that upper piece over, which is going to put... Well, by observation, we see it's going to put this intention, this point at the front intention, because it's going to try to push the back of this bracket over, which is going to stretch the front of the bracket and compress the back of the bracket because of this bending. The moment is the 150 applied 10 inches away at a moment arm of 10 inches from that piece. Oh, I forgot to give you one thing. I forgot to give you the diameter of this bracket. That's 1.2 inches. It's a solid piece of round stock bent at a 90-degree angle. So everybody see that moment that's being caused at this point H. It's the 150-pound load trying to bend the top of the bracket with a moment arm of 10 inches because it's 10 inches above our point H and the back will be in compression. The back will be in compression because it's trying to bend the bracket over that way. What's C? Diameter is 1.5. C then is the radius. It's the distance from the center because of the symmetry that's the neutral axis goes right through the center where distance 0.6 inches away from that half of the 1.2. Simply the radius. And I is the moment of inertia of the cross-section which is the same there as it is down in H which is made just from a piece of round stock. So we'll put that here for our review 1 quarter pi R in the fourth and we've got all that in there 0.1018 inches to the fourth. All right. No wide erection load to do any normal stress just purely bending stress there. We've got all that number and we know it's going to be intention so we know this is positive. We have 8.84 KSI. Remember when we first got this bending equation it actually had a minus sign in it you have to look at it by inspection to know whether it's going to be positive or negative. We can tell by inspection that at the front of the piece where H is that'll be intention. So we know this is a positive stress and we can now draw that in 8.84 in the wide erection. Is that KSI? Yeah, that'll be KSI. Okay. Then we also have the fact that this 150 pounds out at 18 inches it's trying to twist that whole piece it's trying to twist the piece in that direction. We know that that's going to cause a sheer stress like that on the elemental piece because of that twisting motion because of this torque there in that direction to go with the moment that we had which I guess would be in the z-direction which I can't draw because we're looking down the z-direction in that picture. But we know how to figure that sheer stress due to some torsional load. The torsion is the 150 at 18 inches. So the force is 150 the moment arm is 18 so we know what that load is. C is the same thing because our element H is out at the very edge of the material and J we squeeze it in here is actually 2 times I if you remember and so it's twice that which is 208 which is 2036 and so we can calculate that now that sheer stress it's 7.96 You figure out the principal stresses the principal angles and sketch more circle and sketch the the two principal direction elements that goes with this the P element, the S element that's your task and we'll make that a get out of class question for the weekend. The principal stresses and their angles sometimes find the principal planes sketch more circle sketch the two elements P and S at the principal angles that's your task should you decide to accept it it's an admission impossible so you have to accept it if you get it right if you get it early you can go and not then we'll put it together we'll need a nice circle center, I know that's a good spot get that freehand circle on camera wouldn't it that's a beauty alright so you have to figure out where the center lies where the axes are of course we know where the X axis is it's always right through the center the units will be KSI must be in both directions where they're going to fit on the circle otherwise it's going to be an oval I'm Joe nice circle maybe you're not a visual guy who I am I just don't have any paper today run on the lab and get some or ask tomorrow some or get a smartphone and take a picture of the board as it goes by there's always paper available somewhere great circles if that's your own nothing else I'm going to be able to say that ACC students can draw a circle I'm a good one much better than last year, huh yeah, you're practicing home nope just grows on ya got some values you can figure out I'll make ourselves a little bit of space here the calculational part just put the answer that's the road work principal angles principal stresses sketch of the other two elements no no, it's done from the original the original, these original values yeah plot the point A just to double check that your angle makes sense this one's a little bit different than some of the other ones we've looked at the circle looks fundamentally the same as the others our operating point, if you will it's a little bit different you can find the principal angle without more circle just use the equations it's just nice if these things all confirm each other if you're doing alright, it makes it tough if you're busy, it's wonderful guys on the streets, out of jail kick you out of the program you can even go work on the circle there aren't you going to do a free hand circle for us? everybody else did it looks like everybody's getting some of the easy values I'll go ahead with you too then so let's see, we're going to need sigma average we don't have an x direction but we do have a y now there are normal stresses in the z direction but because of where h is we're not concerned with those so it's not like those things aren't there we're just not concerned with them so that's easy 4.42 ksi sigma diff is x minus y so it's minus 4.42 maximum everybody's getting these same numbers? this is the easy part 9.1 so we can place some of those points we know now where the center is 4.42 and we know that the maximum is 9.1 beyond that 13.52 and the minimum is minus 4.63 is that right? and so our y axis our tau axis is right there just about down the middle of it too just by coincidence it'd have to come out about the same as all the others it looked could have fallen anywhere but for the values we have we know it fell there now point A sigma x tau xy sigma x is zero tau xy is just under 8 and well since it's zero and it's on the circle we know that point A falls right there and so we can draw the diameter point G is just what things look like 180 degrees from that but that's still the same piece it's just this piece 190 degrees on it remember 180 on the circle is 90 on the picture so we can then find the principle find the principle angles and we've got that 796 sigma diff is minus 4.24 so this is minus 1.8 we take the r tangent of that divided by 2 because we have this angle 2 theta well this is 2 theta p and you get 30 or minus 30 minus 30 which means on the picture it's twice that minus 60 and that's then actually this angle in that direction it's just the shorter angle we could go the other way like we did in all the other pictures but that's 180 degrees opposite it's just the very same picture 90 degrees on it's edge so you're just getting the same thing it's redundant to go more than 90 degrees around alright so we know at that point that what the stresses will be and so we can draw that in we're almost done let's see where it's going to draw I guess put the original original one there at minus 30 degrees about like that we have our p direction and we know there that the principle so that'll give us the x direction so that's a minus 4.63 the maximum is on the other side that's in tension 1352 all of those in KSI and in the same way we can figure out what theta s is it's just 45 degrees off of that so that's what 1514 149 is that right I didn't even add it down yeah 14 5 degrees and that's our s direction and we know there we have the average normal stress on each base which is a positive 4.2 so we sketch that in and at that point also we have the maximum shear stress which is 9.1 and so we can sketch all those in there this no that's sigma it's either sigma it's just the normal stress as plotted when we plot point A we use sigma x which for this problem happened to be 0 which that's why we're right on this axis oh that's right because of the totals this point uses sigma y sigma y minus tau xy which is the mirror image at that point across the circle okay good luck in the the bar in Warnsburg where all the engineers go talk about more circle maybe we have a more circle video game we could play now the tong is common alright with that okay that's a wrap then for the weekend bye