 a little bit of a summary before we get going on just where we've gotten so far. We were looking at objects under simple axial tension or compression. I have to draw one or the other. I can't really draw both. So as often as not I just happen to draw something under under tension. But what we're looking at now is at the response internally of that material as if we again put one of our imaginary cuts through this material. And then as we look at what's happening inside, we came to appreciate the other day the fact that these internal forces are spread over some cross-sectional area that for the normal stress that we talk about is that cross-sectional area that's normal to that force that's over that face over that area. Occasionally we'll put a little end there. If we don't, we're generally implying the normal stress. But we had that normal stress, the force divided by the area that's withstanding it, which kind of makes sense. As the force goes up, the stress is going to go up. As the area goes down, the stress is again going to go up. So it makes sense that we do this ratio. And the general units on this, typically kilopascals, maybe even mega, occasionally even as far as gigapascals. But we're talking about generally structural solids that can maintain an awful lot of force over sometimes very small areas, as can be done with cables and very thin members that can withstand an awful lot of weight. So we had that as our first look at the possible type of loading and the stresses associated with that internally to the material. It's the material's ability to withstand these stresses that tells us whether or not we're going to have some kind of catastrophic failure. We also then had the other possibility of not axial loading of some kind, but transverse loading could be, as we find with the distributed loads we saw in statics, low those many months ago. But again, as we look at the internal response of the material to those loads, and it doesn't matter if it's a distributed load or not, what matters is the internal response of the material to that load. In this case, we have a force that's now parallel to the cross-sectional area rather than normal to it, and then we call the average shear stress. So we had the normal stress, and if we don't say anything before the word stress, that's the stress we're typically talking about. And then this being the shear stress. And in this case, for both of these quantities, these are average values. They are not constant over the cross-section, but we're not going to worry, at least not at this time, about the fact that these force distributions across these areas aren't uniform. We're going to treat them as if they're uniform, and so these are average values, unless we say otherwise. One other little piece we hadn't quite gotten to, and we will look at it as we wrap up this piece, the problem we were looking at on the board, and that's that sometimes one member is simply bearing on the surface of another member at that cross, at that interface where they connect, which in this case is this base piece here, at that interface where they connect, there's a bearing force as one member presses on the other. We know that again that this is a load F of some kind, this is a total load F. It's also force acting on the area that's withstanding that force. In this case it's one object bearing on another, so this is an average bearing stress. Same units, same general size as the other, but can lead to a very different type of failure. I don't know about you, you can go downstairs in my basement at home where I have the columns in the basement supporting the floor above and you'll see little minor cracking in the concrete below that column, and that's bearing stress starting to cause the concrete to break apart. Well the builder who happens to live next door swears that it's nothing to worry about, no trouble. I personally think that snakes are coming up through those cracks, but he says they aren't, so I'll have to trust him. But that's a bearing stress. It's not a failure, it's just a response to the material to that stress, but that's the type of thing that we talk about bearing stress. So that's a little bit of a summary, plus a little extra thrown in of the type of thing we started with on Monday. So any questions with that as you started to look at some of the problems? We'll finish up, we'll look at a little bit of the bearing stress on one of the parts of this member and then we'll move on from it. Alright then, let's refocus our attention on this pin A. Remember this member AB is in compression, so that member is pushing on the pin A, then the pin A itself is pushing on the bracket at A. So if we take a cross sectional view of that bracket, this course is wonderful for your drawing skills, so I hope they're up to speed. So there's a little bit of the cross sectional view of the bracket, then here is the wall to which it's attached, and then the pin itself, pin A, is in the hole that I have represented here, and then the member AB is in compression, so pressing on this pin A. So it's going to be something like the member AB is pressing on the pin there, so we have a bearing stress from member AB on the middle of the pin, and then the bracket is pushing back, it's hard to draw it on the backside of that pin back there, and the pin itself is pressing on this bracket as it fits in the piece there. Now again, that's just the bearing stress is the force divided by the area that's withstanding that force. For this bracket, things are a little bit different. I'm going to change the view some. We're going to look straight down on the bracket there. It's a little bit easier to draw. Remember, this is a cross section of the bracket, where I'm looking at just the bearing stress of the pin on the interior surface of that bracket where it meets. It's not a uniform force distribution, not in magnitude or direction as it presses on the interior surface here. It changes a little bit with position. We have sort of a circular force distribution, where that total force is, of course, this actually it's f over 2, because there's two surfaces where this bearing is going on. This is the force coming from the member, so f over 2 is at the two different bearing surfaces. So this is really f over 2. It's still a resultant force, just that direction, because the little side components on the left have an equal and opposite side component on the right that cancels each other. So all the sideways components all have equal and opposite opponents to them on the other side. So it gives us a net of this f over 2 straight on. But how do we calculate, how do we figure what the area is? It's actually quite simple. And you'll remember this very same type of thing from your statics working in physics too. The area we use because of this circular distribution is just simply this cross sectional area of the opening itself. And that's the area we use to calculate the bearing stress, average bearing stress. In this case it's f over 2 over a, where a is the, this blue cross sectional area which is the same as that area there, which is what the diameter of the pin 25 by the thickness of the bracket which is another 25 in this case. And so you can take those known values and fairly simply figure out what the bearing stress is along a curved surface even without a uniform force distribution across it, either in magnitude or direction. It all integrates to be just the total force through the, it's not really the cross sectional area, it's sort of an exposed area sometimes. Same thing you did if you took, did you do fluid statics in physics too? And you did this kind of thing, you know the force on an inclined surface, the pressure force on an inclined surface? Sure. Well it's in your book so if you didn't cover it it's in there. But we're not going to be doing very much with it. The thing is, the reason we can sort of gloss over this as we need to is generally these stresses are not the sources of the catastrophic failure we're worried about. It's much more likely that these objects, these structures are going to fail either in normal stress failure or shear stress failure or some combination of the two. So we're not as worried about the bearing stress. Often anyway it's not as difficult to see because it's a little bit more straightforward but this was one where it wasn't quite as straightforward. But the calculation of it does come out to be quite simple in the end. Harder to visualize than calculate actually. Alright so that will wrap things up for that little diagram. We can put the projector to bed for a bit here. Unless there are any more questions on that structure there's a lot of pieces in there and if you were designing that you'd have to take care of every single little part of it to make sure none of it failed because any one of those pieces failed anywhere. If any of the members failed or any of the connections failed or any of the brackets failed the structure itself is going to fail. One thing we will have to concern ourselves with in a little bit probably on Monday we'll start with it is that the failure doesn't necessarily mean that the piece breaks apart and can no longer maintain any forces. It could be that the structure elastically deforms enough so it no longer serves its purpose. It could be that under load some of these force some of these members deform enough that other parts in the structure don't match quite and then the piece won't quite work. Same kind of thing if you're putting in a new door and you don't mount it tightly enough that under the load of the door it deforms itself that it doesn't even open or close properly. So it's not necessary that things need to catastrophically fail. It might be enough that they just deform until the entire thing itself fails in some way. Alright so we're going to change gears a little bit here as we look at not just simply what's going on internally but take another bit of a view towards it as we look what happens if we're not looking at the stresses on a normal cross section but on an oblique cross section where we have a cross section taken at some angle to the normal direction. So I've drawn such an oblique angle there for us to take a look at. That's our imaginary cut plane through the piece and notice it's set back at some arbitrary angle theta and then this piece is loaded with some force P. And we'll pull away this front part, look at just the back part here because that will expose force then that oblique phase now set back at some arbitrary angle theta. Of course it's still true because of the force balance that we need on anything here that it's still force P but that force P is now acting over a slightly different area. We'll call this a theta where a zero is the original phase from which we would have calculated the original stress. So if we look at the original normal stress and by original I just mean what we've been looking at since the first day. It's the load and it's supported by a normal cross sectional area A zero. Alright let's look at this one then a little bit differently. Let's look at this oblique phase now. We have this force on it say a force P but we can break that into two components. The normal component of that force acting normally to the surface we might call F and then a transverse component that acts parallel to the surface and we can call that V because that is a shear stress so that's rather common for us to use that as the designator for a shear stress or a transverse force. So I'll cross out that force P because I've broken it into two components. I don't have three forces there. I have either the original force P or the two component forces F and V. I don't have all three L. This is V and this is again acting over this area A theta because this is the inclined phase. Imaginary cut through the piece inclined at an angle theta. So this angle as well is that angle theta. So we know that this force F is P cosine theta and it's acting on an area A theta. A theta the inclined area exposed remember imaginarily exposed by that area is the original area A0 also divided by cosine theta. So if we put those two together to see what the internal normal stress is at an angle arbitrary angle A theta theta. So I'll call this sigma theta it's F over A theta which is P cosine theta that's F divided now oh sorry this is this is A theta there not A0 divided by the area A theta which is the original area divided by the cosine so I get P cosine squared theta over the original area. No this is A0 because now I have cosine squared from here if I put these two together I get a cosine squared on top. What's nice about that is that it allows us to compare the normal stresses at an arbitrarily inclined angle interior to the piece compare it to the original normal stress that we were observing and because cosine is never greater than one cosine squared certainly never greater than one we find that we always have a normal stress at an inclined angle less than the normal stress we had experienced at the regular axial direction as we originally had. So that tells us at off angles and oblique angles we don't expect there to be greater normal stresses than we would have found in the original look at the normal stress anyway. Let's see if that same kind of thing holds true for the shear stress. So the shear stress let's see that's the load times sine theta so the shear stress the average shear stress at some imaginary plane at some arbitrary angle theta is V over the area over which it acts which is V over A theta but V is P sine theta divided by A theta which is A0 over cosine theta and so I get a new then shear stress internal on some arbitrary oblique angle that is sine theta over cosine theta. Alright so what we need to do now is look at what those are in terms of theta to see if there are other places there might be trouble that we would not have seen before if we had looked at some arbitrary angle some arbitrary angle theta internal to the piece. So if we just look at theta over maybe what I'll call the shear ratio which is I'll look at the ratio of these two sigma theta to sigma zero is cosine squared theta just to normalize it so I can compare the two. At an angle of zero degrees which was our original look at these strictly cross sectional piece this is our member on edge we're looking remember oblique angles of theta so theta equals zero we're at our original situation we looked at when we opened things on Monday and that's just then simply a shear ratio of one and from there to a maximum of 90 degrees which means we're longitudinally cutting all the way down the piece which is in its own right a bit absurd. We get a relationship between the shear ratio and the imaginary angle of our oblique plane to be something like that. We see that the worst possible situation is the original one that we've looked at this angle is, well that's not it, can't even drop because it's zero. That's where we're going to find the maximum normal stress. Any other angle from there the normal stress drops and remember this is the normal stress normal at all times to whatever face we're looking at. It's not that there's more forces it's just that as we go to oblique angles the original force has different components to it that we have to account for. When we look at the shear force however across that face and do the same kind of thing with it we did previously so t theta, I'm sorry tau theta over tau zero is sine theta cosine theta that has a bit of a different response to it does something like this where at 45 degrees it reaches a maximum which also happens to be the place where it crosses the other shear stress or shear ratio curve that being the normal response this being the shear stress response. We see it does reach a maximum at a place where at 45 degrees. So at an oblique angle of 45 degrees we see we have the maximum shear response and we're still at a significant normal of about a half, at least a half of what it was originally. So this is a, they will call this sigma one half for some other reason no other reason than just to call it something. Question John? Just this is in general I assume that once you know material you're working with it. Oh yeah what we have not talked about at all is what is the material limits on these stresses that we're finding. There's no point looking at any of the materials and how they can withstand these stresses until we know how to figure out what the stresses are themselves. So this actually turns out to be the critical situation. This is an angle of 45 degrees and you will find, and in fact I've posted two videos where they actually do these type of tests and you can see what the failure looks like. If you take a test piece and pull it apart when it fails, it doesn't fail straight across the piece in a nice clean break, it tends to fail something like, something very much more like this. When the two pieces pop apart you'll see that there's a semi-irregular surface that's now been exposed where the piece broke. It's only semi-irregular because it tends to look sort of like a 45 degree cone that formed on one end as it popped out of the other. Kind of like a socket into each other. But this is at approximately 45 degrees because of the material's inability to withstand the maximum shear at that oblique angle. Remember our look at all of this was at an imaginary angle, we determined that the maximum shear response was at an oblique angle of about 45 degrees and that's indeed where the material tends to fail. It's also true and maybe you've seen this if you've ever overloaded a column on a deck or some kind of structure or piece that was meant to carry more weight than it could, it tends to fail something like that and if you, I've posted two videos, one that shows a tensile test to failure and you'll see sort of this type of surface forming and I've also got a piece of wood in compression to failure and you'll see that it splits in very much this type of way and this is fairly close to that 45 degrees. Of course this assumes a completely homogeneous material which is not the case nationally. We have, especially with wood, there's differences in the material throughout it's not a homogeneous material perfectly which is why it's not exactly 45 degrees but that is where we predicted the maximum shear failure and that's just what's seen when the piece does go to failure. So on Angel go to check out those two videos they're not even a minute long I don't think. You have to watch them closely because what you can't tell is that as soon as the video starts the test has already begun and there's no sound that indicates that there's actually stress of the stress test undergoing and if anybody happens to have an extra $50,000 to buy us one of these test machines we could have done this ourselves. So John if you stop wasting your money on little toys on your computer you could buy us a stress test machine and we'll talk about that very test probably on Monday. Alright let's take a break while I re-step the taper. Okay also Matt there? By the way I don't think I did introduce Dale at the second class I didn't introduce assistant Linda. I assume what you're gearing towards is maybe one of you will take one class and one the other. Exactly. Once Linda gets up to speed. So we need to get a little into this. Hi everyone. Alright where's that? We are. Alright so we just had a chance to look at the internal stress at arbitrary angles now we're going to take a more general look at what we might call generalized stresses and again we're really only looking at the two stresses that we've talked about so far. Well we talked about three but we're not going to talk about bearing stress here. We're just talking about the normal and the shear stress and again for the most part these are average values. We're assuming the part to be homogeneous by that I mean a uniform throughout it's not like a piece of wood is where it's very different in one place than it is in another. We're just assuming it's a perfect piece of material and the force is uniform over the surface. So we have some object we're concerned of and I'll even keep its shape general under some kind of ring and again just to illustrate how very arbitrary any of this can be so that we do indeed get a very generalized situation. We have some object we don't care what its size, shape or purpose is. It's under some kind of load and we're not even concerned with what that load is. But what we'll do again is take an imaginary cut through the piece and expose the interior of it so maybe it'll look something like that. Looks like a Christmas ham that's what that looks like. And we'll put a cornet system on it just to help us orient things and we'll put the origin of that cornet system dead center just to make things nice and easy for us. So we'll call that the X, that the Y and that the Z direction. So now that we've done that and taken all of these forces, figured out what the resultant is, we know that there will be some component of that resultant force right down the X axis normal to the face. That's just the nature of no matter what the resultant comes out to be, we could resolve one component of it to be in the X direction. So we'll also imagine that to be acting on a little piece of area we'll call delta A. And then the other component of the resultant force, no matter where it points, it's going to have some component of shear. Well let me draw it in even a more arbitrary direction because we can't possibly know where it's going to be. So often some direction parallel to that face is some shear component and it's the two of those together that came from the resultant force of all of these original forces acting on it. So since they're both acting on the same area, we can make the components, not the force components themselves, but we'll make them the shear components because they're both divided by the same area. So we have that normal component of stress, whatever it is, whatever magnitude, we do know its direction. We purposely resolve the resultant into an X component. But the Y component, or not the Y component, the shear component could lie in any direction along that face. So what we'll do is make things very simple on ourselves. We'll take this exposed face with this coordinate system conveniently located on the center of our piece there. We've got this normal component of the stress in the X direction. Remember there could be compression. It depends on what the forces were originally and where I made this imaginary cut, but I have to draw something so I drew tension. And then we'll take the shear component and go ahead and break it into the two pieces in our Y and Z directions, and then we are able to figure out all pieces we had before. So I need sort of a system of writing this thing down. So I'm going to use this system here. For the shear stresses, I'm going to have a two letter subscript to it. The first letter is the face that the shear is on. And by that I mean a cut like this that is perpendicular to the X direction we call an X face. So this is tau sub X. And then the second letter in the subscript is going to be the direction, the coordinate direction that that shear component lies in. So this would be a tau XY shear, a shear stress on an X face in the Y direction. So this piece down here is a shear stress. And then what would the two letters be that describe what shear stress that is? It's still on the X face. That's the only face I've exposed right now is a face in the X direction. It's a cut that's normal to the X direction. And that one happens to be the component that's in the Z direction. So I have a piece that looks just like that. So that no matter what my loading is, no matter where I actually have this cut, we can resolve the resultant force, calculate the resulting stresses, both normal and shear, and draw them in something like that kind of designation. Remember the actual directions of these is arbitrary. I had to draw something, so I just happened to draw them all in that direction. We can then do that in the five other directions. We can do it on the back of the face in what we call a negative X face. We can also do a Y, two Y cuts and two Z cuts. And come down to all those cuts together, expose just a big cube, again with the coordinate directions conveniently located at the center of the cube. And that exposes the shear stresses, the internal shear stresses, sorry, the internal stresses, both normal and shear on all of the faces. So the ones I've already drawn there, I have a sigma X, we'll call it a normal stress in the X direction. And then I've got those two shear stresses. Remember their direction is arbitrary. And then there's an exact copy of that kind of thing on the back face that I'm not going to draw. It's just too difficult to draw behind the board. And then on this Y face, I've got the same kind of thing. It might be tension, it might be compression, but there's still going to be some stress along that face. I'll just arbitrary choose to draw all of these in the positive direction. There's a shear stress that will be exposed along that face. What should I call it? Tau, it's on the Y face on a Y face. We have two Y faces. There's one on the bottom as well. Tau, Y face, but it's in the X direction. And then I might have a component in that direction as well. It's on the Y face in the Z direction. And again, the very same type of thing on the bottom of this cube that we've imaginarily exposed with our cuts through the material. And then there's also a normal stress to the Z faces, a shear stress on these Z faces, one in the X direction and one in the Y direction. And then again, of course, on the back side, on the back Z face that I'm just not going to bother drawing. In total then, we've got nine stress components. In a generalized case, generalized meaning for any arbitrary loading and any arbitrary orthogonal directions, we can break down any problem to this kind of thing. We have nine stress components. The nine that you see and then those are each near-imaged on the back, but not to the point where we have 18 stress components because on the back, for example, the normal stress in the X direction, we have to have the exact same normal stress in the X direction on the back side as well. Why is that? Static equilibrium must be maintained at all times. All of the forces summed to zero, these stresses are called, these X stresses are caused by X forces. So for every X force in that direction, we have an equal and opposite one in the back direction. So those aren't new components. We only have then the nine stress components exposed. And don't forget that the moments must also sum to zero. So we're going to generalize things even a little bit farther, make things a little bit easier to us. We're going to actually look at the force balance on these things. So let's imagine our cube has a length a on the side. We're only going to look at it in two dimensions. We've done this in so many of our other classes where we look at two dimensions rather than all three because what happens in the third dimension is not necessarily any more instructive but it sure complicates the drawings. So we can draw the forces on this. We'll take just the X, Y plane of view of this type of thing. We'll ignore the Z direction. It comes out of the board at us. And so on this X face now, which is the very original first face we drew, I'll draw not the stresses but the forces on that piece because I need to do a force balance on this. So the force on that face is sigma X acting over an area delta A. Remember that was the size of the element that I first exposed. And of course on the back side, we know there must be an equal and opposite load on the back side. And again, arbitrarily drawn as tension. And the sigma Y forces sigma Y delta A and an equal and opposite load on the back side. And remember to that, sigma Y delta A. So automatically by virtue of that, we've already got the force balance satisfied at least for the normal stresses. For the shear stresses, I'll draw them something like that to remind us this is across the face. Remember we're looking down the Z direction. We're looking along this X face with this drawing we're just doing there. So that's tau XY acting on an area delta A. Remember we're doing a force balance here so I want the forces not the stresses. It's not going to be a big concern. All these delta A's are the same anyway. And this is on the back face. So this is tau also on the next face, also on the Y direction, also acting on an area delta A. So that's tau XY delta A as well. So we have an out and make, we have a force balance there. And then I have the same type of thing with the shear stresses on the Y faces. So that's tau YX delta A. And down here in the opposite direction as it must be because I need to balance that force on the top. I have tau YX delta A there as well. All right, we automatically have our force balance satisfied. So it gets a big happy check mark. Because I drew it that way, we know it must be that way so it's not like I'm assuming anything. We know that we have static equilibrium on all these pieces. So I just drew it that way right from the start. We don't need to do any more with the force balance there. That especially works for all the normal stresses. All the stresses perpendicular, normal to the faces. Especially because not only are those automatically satisfied with force balance, those forces are also automatically satisfying a moment balance as well. Because not only are they equal and opposite, but they're co-linear so there's no moment caused by any of those. So at least for the normal forces we also have a moment balance satisfied. That's not true of the shear forces though. Because they're equal and opposite, but they're not co-linear. They're separated by a distance A. So we need to satisfy a moment balance here with the shear stresses. So we'll take this couple first, tau XY delta A. That's the magnitude of the forces in this couple. This is the couple of these two shear forces on the X faces. But what's the magnitude of the couple? This is just the forces in the couple. How do we calculate the magnitude of a couple formed by two forces? We have two equal and opposite forces separated by a distance that causes a moment. Causes a couple. The magnitude of that couple is the magnitude of the forces times the distance between the two. Remember this was a cube of sine A. So the magnitude of the couple caused by the shear stresses on the X faces is that. So that's the X face. There's also a couple caused by the forces on the Y faces as well. And we know that those must be equal and opposite so that they sum to zero. So I'll assume it's equal and opposite. Turns out I drew it that way anyway, so no great trouble with that. Sorry, this is tau YX, not XY. I'm working with this couple now caused by these shear stresses. The forces is the shear stress times the area over which it acts times the distance between the two forces that make up the couple. So that's the Y face couple. And I know they must be equal and opposite that they sum to zero. And I drew the minus sign because I actually have them drawn in the opposite direction each other. So just taking that into account. The X face couple happens to be counterclockwise. The Y face couple happens to be clockwise. So that's the minus sign between the two. Notice how this simplifies things for us terrifically. Because the A's cancel, the delta A's cancel. Not only is delta XY equal to delta YX, but I can do this on the other faces as well. And I get delta XZ equals, sorry, tau XZ equals tau ZX. And tau YZ equals tau ZY as well. So that makes things terrifically easier for us. Now we're down to not nine stress components. We're now down to six stress components. We have the two stress components in our X and Y direction. Again, I stress pun intended that I'm arbitrarily drawing these as tensile forces. Tensile stresses. We're going to see lots of situations where either or are compression. I have those two. I have a third in the Z direction. And I have shear stresses that must be drawn in that relationship to each other. Or we don't have a force balance. Two on adjacent sides must either point towards each other or point away from each other depending on which corner you're looking at. And it's not even important anymore what I designate them. I'll just call them the shear stress tau. Because they're the same on all of the faces because I can do this in any directions. I can do this all the way around the cube. All of the shear stresses must be the same. And they must orient themselves in this way. We could have the possibility that any or all of them are in the opposite direction as drawn. But then the relationship between the shear stresses must still be like this. That they tend to point to opposite corners or they tend to go away from opposite corners depending on what the actual loading is. And remember it's the external loading that determines the magnitude and the directions of any of these things. We'll look at problems that do exactly that very shortly. So now we're down to six stress components. The sigma, the normal stresses in those directions, and then the three different parts of the stresses in those directions. If I did need to concern myself with the z direction I might further the designation tau xy. But all four of these have the very same magnitude. They must or we don't satisfy both the force balance and the moment balance. If we happen to look at any of the other two directions we just take into account the other two possibilities for the shear stress. So that's six components, three normal stresses, and three possible shear stresses. We're down now to just six stress components for an entire 3D analysis of the stresses on a structural member. No assumptions going in here. I didn't assume that these would be the same. Well I did initially assume that these would be in the directions shown. We'd have the opposite directions of each. But it turned out that that's the way it indeed had to be. So we get down to this not with any assumptions. It's just the reality of the nature of the three dimensional stresses on a solid. Okay so let's put this together with something we just did earlier. Again all we're looking at so far is simple axial loading. Some load on a cross sectional area. And if we look at some small elemental piece in the member itself, remember we of course do not actually expose any of these areas because we can't do that in a structural piece. For a simple axial load all we're going to have is the normal stresses there. In that situation there's no possibility of any shear stresses because I have no transverse component to the force so I'm not going to have a transverse component to the stresses if we take the very same situation. This is not a different situation. It's the exact same one we're just going to look at in a slightly different way. Now if I look at an elemental piece not oriented in this way but oriented that way. In fact let's make it nice and regular. Make it at 45 degrees. I know from the oblique plane study we did to open class that I have shear normal stresses on all of the faces now before I only had them on one. I only had an axial force on the member. I'm only going to have axial forces in resultant. But when we took a side look at these we got extra components on the other faces. Plus I'm going to have shear stresses now exposed on those faces. When you go back and look at the oblique plane we opened class with we saw that it off angles we get shear stresses across the faces. We also know that though those will be at their maximum in this piece. We also know though now from what we just looked at our generalized case that they must be oriented in that fashion in relation to each other. So we've been able to combine our strictly orthogonal place view with an oblique orthogonal view where we know now from what we looked at earlier that things are at a maximum at 45 degrees. No I'm sorry if I think about it I've actually happened to draw the shear stresses in the wrong direction for this tensile loading because their components are going to be in that direction not as I drew here. So let's fix that. And these sigmas remember are not necessarily the same. So maybe I'll call those x prime and y prime just to indicate we're at an oblique angle but we do know if that angle happens to be 45 degrees that that is the maximum shear stress that we're going to see. Anything other than that any other angle, arbitrary angle than that we're going to see less shear stress. Remember what the magnitude of these shears were, the normal shears were to the original normal shear it was half of the original. In fact those two would be the same. Sigma original divided by 2. I guess if they're going to be the same we don't need any kind of subscript on it. Any questions? As we go through these things remember that the shear stresses must be oriented in relation to each other in this way. They might be in opposite directions but if one's in an opposite direction they all are. And no matter what the angle all four of those are the same magnitude or we don't have a static equilibrium on an element which we must for any element, any parts, any total structure that we might look at. Questions? Phil, questions? Good. The shear, is this supposed to be going the other way? Oh, yep. It's too hard to draw, talk. See I'm in three different time periods. I'm drawing one thing, I'm saying one thing and I'm thinking what's coming up next. Good catch, thank you Phil. They must always be there. They're either pulling the piece this way or they're pulling the piece that way. In fact we'll look at that specific deformation caused by those next week.