 A little bit of a summary of where we get going and 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 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 applying 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. And this can be done with cables and very thin members that 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 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 stresses. So that's a little bit of a summary, plus a little extra thrown in. 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. All right then. Let's refocus our attention on this pin A. 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 part of 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 back side of that pin back there. And the pin itself is pressing on this bracket as if it's 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, 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 in 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. 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 2. 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 or each bearing stress. In this case, it's F over 2 over A where A is this blue cross-sectional area which is the same as that area there which is what's 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 2? Did this kind of thing figure out 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 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 enough 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. All right, 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 stressors 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 because 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 for us 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 A0. 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 our 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 arbitrary angle 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 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 the 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 is certainly never greater than one we find that we always have a normal stress at an inclined angle less than the normal stress we have experienced at the regular axial direction as we originally had. So that tells us at off angles at 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 plane, 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 A zero 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. 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 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 the strictly cross-sectional piece this is our member on edge we're looking remember oblique angles of theta so if theta equals zero we're at our original situation we looked at when we opened things on Monday and that's just been 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 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 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 half of what it was originally so this is a we'll call this sigma one half for some other reason no other reason than just to call it something question John this is in general I assume that once you know material you're working with 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 hosted 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 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 to each other but this is at approximately 45 degrees because of the materials inability to withstand the maximum shear at that oblique angle remember our look at all of this was at an imaginary 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 a column on a deck or some kind of structural 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 naturally 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 a sheer 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 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 we'll talk about that very test probably on Monday 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 a bearing stress here we're just talking about the normal and the sheer stress and again for the most part 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 under some kind of loading 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 all 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 it looks like a Christmas ham that's what that looks like and we'll put a a cornet system on it just to help us orient things and we'll put the origin of that cornet system 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 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 don't 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 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 our piece there we've got this normal component of the stress in the X direction and 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 this shear component and go ahead and break it into the two pieces in our X sorry, 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 particularly 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 X, Y, 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 X 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 they draw something so I just happen 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 we can also do 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 stress 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 their direction is arbitrary and then there's 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 wide 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 X 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 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 we have nine stress components the nine that you see and then those are each mirror image 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? it must be maintained at all times all of the forces sum 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 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 instructed but it sure complicates the drawings so we can draw the forces on this we'll take just the X, Y plane 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 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 member 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 and 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 X Y delta A as well force balance there as well and then I have the same type of thing with the shear stresses on the Y faces so that's tau Y X 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 X delta A there as well alright 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 so I don't need to do anything 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 force balance those forces are also automatically satisfying a moment balance as well because not only are they equal 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 collinear 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 x y 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 divided 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 but 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 counter clockwise the y face couple happens to be clockwise so that's the minus sign between the two notice how this simplifies things for us specifically 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 tau xz equals tau zx tau yz equals tau zy as well makes things terrifically easier for us 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 upon intended that I'm arbitrarily drawing these as tensile forces tensile stresses we're going to see lots of situations where either or 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 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 stresses three normal stresses and the three possible shear stresses we're down now to just six stress components for an entire 3D analysis of the stress on a structural member no assumptions going in here I didn't assume that these would be the same well I did initially assume 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 what we're looking at so far is simple axial loading some load on a cross sectional area 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 normal stresses there in that situation there's no possibility of any shear stresses because I have no transverse component to the stress 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 this way now if I look at an elemental piece not oriented in this way but oriented that way in fact let's 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 all of the phases 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 with the oblique plane we open class with we saw that it off angles we get shear stresses across the faces we also know that 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 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 you know I'm sorry if I think about it 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 those 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 that we're going to see anything other 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 them okay 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 good the shear on the bottom left hand side is this supposed to be going the other way oh yeah it's too hard to draw talk see I'm in three different time periods 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