 Some we got something done on the homework, so we're all right for one week Who knows who did what problems you pretty much do any problems you want a little bit of both? That's why I think what I would have done You know I can't remember I told you I think as I sat one through the problem sets I Remember last year thinking I don't remember assigning that problem. I think the actual problem set I had there was not from The addition before this but from the addition even before that and I just never taught it So it drives me nuts, you know if you sat and look through those The pictures are the same the layoffs the same the text is the same they change a few sections around a little bit But then most of the rest of was just jumbling Problems and some of them was a very same problem that turn at 90 degrees and put in new values It's such a rack and I apologize to students for this it is it's it's It's a racket and then you know they They have no trouble whatsoever sending instructors as many copies as we want and then You know they collect and some some professors donate theirs some some just keep them all some I Just sell mine on half comm because I pretty well underpaid anyway, so I don't squirm about that too much anyway All right, let's go a little bit. We We've been looking at this simple Idea Objects under Some kind of load so we'll take the the various simplest of the objects and the very simplest of the loads a Strict axiolo, I haven't draw attention could have been compression doesn't matter all of the argument comes from there's going to be very very much the same what we've done so far is look at what happens in an intermediate imaginary cut and of course that load must go through any part of that material as a whole if any part of that material any time is supposed to be under Static equilibrium And from that knowing that this has some cross-sectional area Whether it's circular or square doesn't matter we from that got an idea of this normal Normal stress that the material is under going What I'll do for now is I'm going to call this a Subzero because that's going to be a reference area I could put an n I guess because it is the normal area not being normal to axial length of The piece if you will But what we're going to do now is do this very same thing Only instead of making a A normal cut we're going to make a An oblique cut still true the load P must be transmitted through that cross-sectional area this Oblique cut is at an angle of theta from the normal and then for us Exposes a couple different things going on two things two things we need to look at One is that now this area That's supporting this load a theta is a little bit bigger Than was a zero because we've tilted back our OB oblique plane a little bit because of that we've now exposed some extra area and Without without too much of a flavor month. I guess we can Relate those two that this will be a zero over cosine theta that right I think that's okay that because cosine theta is less than one So a theta is always greater than a zero and If they that happened to be zero the cosine is one so we're okay. I think I got that right All right, so that that's one thing we can do with that The other thing we can do with that and I just want to take this drawing and Bring it up there. So it's a little bit bigger. I've got a little bit more room for it We've got this load P Axial to the piece and remember it could be compression to none of this argument would change I just had to draw one or the other we can break that into two components One component Perpendicular to that face and one component up to that face. So we'll take the load Now reference that load to our To our our oblique direction but that does for us to show us that there's this Normal force, maybe I'll call it F And this this normal let's see that's up. That's what that's That's P. Let's see. That's P over cosine theta as well. Isn't it P times cosine there It's always good to take notes in chalk. It's so easy to correct. That's P cosine theta And that is normal to the force Sorry to the face which leads us to a normal stress on this oblique plane of Well, whatever load is normal to that face, which is our new F Over the area that's holding it which was our a theta and if we put in what we know about a theta and What we know about F? We get then that this is P cosine Theta Over a zero all I'm doing is getting back to our our our original Load our original cross-sectional area and there's an extra cosine theta under there So this becomes P cosine squared theta Over a zero and notice that that P over a zero Was our original Normal stress maybe a better change this let's make this Not a and let's make this a theta at our reference plane That way we don't have two sigma n's on the page so this then becomes Sigma and our original normal stress Times cosine squared theta Stress a maximum. Yeah when theta equals zero or in other words our original situation It's it's at the the greatest normal stress is When we're looking at a plane straight across the piece That's our that's our biggest concern theta equals zero degrees No big deal. I guess Because as we go off angle We reduce the normal force but increase the area so of course the stress is going to drop so if we have a Failure due to normal exceeding normal stress. We would expect that failure To occur essentially perpendicular to across the piece We can take that into design considerations if we need to as we start looking at more technical Materials where we can design them for certain strengths in certain directions. We can handle that Now we look at the shear across the face We have a pure axial load But now we see that in certain directions there are actually shear concerns that arise From a purely axial load, so we have to take those into consideration Let's see. This is this is p sine theta shear stress across that face Remember is the load over the area Supporting it which in this case is our a theta so if we put in the shear we know with respect to the original load and whatever angle we're looking at and We put in this a theta That gives us p sine theta. Oh, sorry. Sorry over a zero. This actually has as magnitude of Our original normal stress even though it's a shear stress This is just that gives us a magnitude sine theta cosine theta check That would be like Like when it's flat, right? Yeah, that's our original orthogonal Cross-sectional area actually if you if you if this is a very regular solid then this is Actually the minimum possible cross-sectional area Any other angle is just going to increase the area and remember these are not Real cuts we make through here. These are imaginary cuts to so we can look at what's going on Internal to the material angle is this a maximum? That's that's a little bit harder to see At at theta equals zero The sine is zero so that's certainly we wouldn't expect to be a maximum at theta equals 90 Well, that's when we're laid all the way over and we are Cross-sectional area goes all the way along the edge of the piece that ain't make any sense But again, this disappears because then cosine is zero But at do we would you say? At 45 degrees This is a maximum That means is under pure axial load We are worried about shear failure at an angle of 45 degrees and I believe in your book There is a picture of a piece of material that was under a pure axial load and Then failed actually ruptured if it's not in the book or if they removed it from the new book I'll get a copy and put it up on angel for us, but the failure looks like This you'll see you'll see kind of a conical piece at roughly 45 degrees as This this part pops apart because of the shear failure at the 45 degrees Roughly roughly in here 45 degrees and you see this kind of conical Piece that just popped out of this stuff. It was the shear that actually made it fail At this 45 degrees it was greater than the normal stress And so we had that that kind of conical 45 degree shape failure to it there All right, we're gonna have to keep that in mind a little bit as we go on to some other things So okay before a clean board over there, I'll start erasing over here Take my time So here's where we're going to take that next Let's let's look at a more general situation so here's a Here's some engineering thing we've got some component Some some piece of a bridge piece of a machine piece of some weapon you carry around. I don't know what you guys do and References reference it in x y and z directions and this could be under some kind of load Who knows what there's one force there another force over here another force Out the back going that way who knows what kind of loads could be on this thing Could be loads all over the place of course what we're worried about is not each Individual force necessarily, but what we're interested in of course is the net load on any of those Or due to all of those because of course that must sum to zero in the end We need to have static equilibrium on this at all times So we'll sum all the forces together. It could be one single force You know who knows where it's going to come out what it's going to do Maybe it looks something like that and then of course we know That we must have it in static equilibrium, so we have that that out the other side We're looking for a general case, so so individually what's going on is not our big concern So what we're going to do is what we've done before is we're going to take our engineering object whatever it is But we're going to imagine Exposure of one face Make an imaginary Internal planar cut through the material to see what the loads are on that face Now to make it easy and since it's arbitrary anyway, we want this face to be Perpendicular to the x direction Just just for reference purposes and that's no big deal Because we can point the coordinate system anywhere we want So we want that face to be perpendicular normal to the X-direction that face is parallel to the YZ plane. Does it look like that near John? But it does jakes does because he takes technical free-hand sketching. This is no big deal for you. You're laughing at this stuff all right and Because of this whatever this net force is And just like we had here this could give rise to both a Being perpendicular to that face. That's easy to do that's very easy to set up Because our cut through the material Was perpendicular to the x-direction so automatically we have this normal stress That that will exhibit itself a little bit more problematic is Whatever the other component of this resultant is This sheer component because it could lie in any direction Maybe it lies in that direction Just because of what all the other forces were and what this resultant is determines then over that area That exposed area then what these stresses are Now well, that's not a big deal because no matter what direction this sheer is We can break it into components Sorry, what I should have is not not the sheer itself. I should match it That's the stress sheer stress That I'm exhibiting there Because it's over expect sheer be over that area It doesn't really matter what direction that is because we'll break it into component directions That are parallel to the y and z directions So we'll take it out make a drop like this well Let's leave it like that You're done Bob. All right. I'm gonna raise this guy down We would there's progress here. You cannot stand in the way of progress. All right, so we'll we'll take our engineering material with its exposed face That's perpendicular to the normal direction the x-direction So we have this Normal stress there. We're going to break this sheer stress Into components parallel to the y and parallel to the z directions So that'll be one component there call that towel Because it's on an x-direction face Remember this cut imaginary cut through the material is perpendicular to the x-direction That means it's what we call an x-direction face But it itself points in the y direction. So I'll call that tau xy That's that's the vertical component of this sheer and now I'll do the z component And that will be tau Because it's a sheer stress It's on an x-space But it's in the z direction. So I took this this stressed material Put an imaginary cut through the material perpendicular to the x-direction and Then that allowed me to expose these stresses Interior to the material based on arbitrarily chosen coordinate directions Yeah Now it's parallel to the y-axis Now it's parallel to the z-axis. Everybody okay with that? Challenge Imagining that about the challenge sketching it. Here's what happens next with it We made that imaginary cut that first imaginary cut Perpendicular to the x-axis parallel to the y-z plane We're now going to do that In the other directions and there's five more directions we need to do Because we need to make a cut on a plane That's perpendicular to the y-direction and one that's perpendicular to the z-direction So that's that's three total We need to make three more cuts because we also have Minus x minus y and minus z directions to cut When we do that what we're left with then of course is a Cube that's what we're left of our engineering material after mate We've made cuts across all of the faces all of the possible directions Three coordinate directions But each one has a plus and a minus face on it and on each face now as exposed these various same kind of things in the coming out from that face we have a normal stress That's perpendicular To that face itself like that Jake you got about it about it y'all jazzed about this is exactly what we're doing and Technical freehand sketching yesterday on that and we have the two shear stresses that we've exposed So that goes right up the face in the extra sorry y-direction That's tau xy Across the face. That's tau xz remember the first letter tells us what face we're on The second letter tells us in what direction it lays itself So that's just a Retrint of what we've already done here, but we've also done the other directions So coming straight out of this face straight up in the y-direction We have sigma y remember those could also be in compression I happen to draw attention because I have to draw something, but they could just as easily view the other way and be in compressions and that's also perpendicular to that upper face and we also have a Shear laying across the face be tau. What would its subscripts be? it's on Y face In the x-direction, so it's tau yx and we might have a shear in this direction too, which would be tau Y z it's on the y-face and it's in the z-direction How's everybody doing with these drawings? Yeah, you don't want to draw too small or you're gonna be hurting and make these kind of big good everything fit in there Pat not too bad Jake do I need to call professor hamster over to check this? Got stepped on do be beautiful Awesome, that's a big all right, then we have this one more exposed face here so we've got a Sigma in the z-direction there remember that's a normal That's normal to that face And I just arbitrarily chose drawing in the positive direction But it could be compression as well any one of these could lay in the opposite direction. I have to draw something Then we've also got a shear across that face Which I'll call tau Sub what sub what tau It's on a z-direction face It's in the x-direction itself. So it's tau zx and then there's one more Possible there I happen to draw in the positive direction. I don't need to that's tau z Why exactly the same type of thing on the three back sides that aren't exposed But for equilibrium Everything on the back side. It's got to have the opposite direction of any of these So Sigma x coming out this side It's got to come out the other side on the back because this piece is always going to be an equilibrium and These values come right from the forces exerted in those directions themselves All right, so let's let's do this Little bit will need this. Well, let's see remember the forces must always sum to zero The forces are all of the form For instance the force in the x-direction is the stress in the x-direction times the The area over which it's acting, but it's a cube. So all these areas will be the same Remember that's that for this that particular one it's that front exposed face all the forces take that General form the stress times the area over which they're acting So we could sum all those up and we'll we'll do so rather quickly in a second Also, don't forget though All of the moments must sum to zero two That's more of a concern. Well, that's not a concern with the forces that are causing these normal stresses because those are all Colinear with the whatever forces on the opposite face. So those forces are automatically zero but all these shears are caused by forces that are Equal in magnitude opposite in direction and Separated by some distance whatever the size of our cube is here So we're going to have to balance all those moments what we'll do is We'll Take one look at one face We'll look at the look right down the z-axis now We're looking right up here up along the z-axis So our face is a square like that. We'll give it each side a length A it's a cube so it's a cube of a by a by a just so we have that as reference You can imagine I hope that that is not going to be material to the solution in the end And if we draw the forces on these faces Let's see we have the force here is Sigma x times the area over which it's acting Which we can either just call the whole area delta a or we can call it little a squared It doesn't matter. They're all the same for all of them. So we'll just call it delta a Shouldn't be a big stretch of your imagination here that we're going to be done with those very shortly We have a shear going across this face of Magnitude tau It's on an x-space it's in the y direction and it's operating across the very same area So now I'm drawing the forces Stresses I had over here in blue forces. I have over there pink. So it's a slightly different drawing Just because I'm drawing the forces so it's a sheer stress acting over an area So I'm just calling one face of this cube Delta a I Could write a squared it doesn't matter Because as you can Probably start to anticipate now that's going to cancel out in the end. We'll be done with it Because this this whole study can't depend upon what the size of the cube is We'd have to redo it for every single little piece and we wouldn't we'd never get any general general idea. What's going on? So if I draw the faces, I'm sorry the force out of this face That's tau. Sorry sigma y Also acting on the very same area because it's a cube. I have this shear stress tau this is y-ax Operated on the air. That's the shear across that face. I have the same thing going out the back sides Sigma x Delta a it's in the opposite direction Sigma y Delta a just the area of this side one side of the cube also the shear going across tau x y Delta a Tau y x ready to see things in multiple dimensions. All right The sum of this is remember now a free body diagram. These are the forces the sum of the forces is Identically zero. We don't even need to mess with that every force on here has an equal and opposite one Which it always must have had Because this piece has always been an equilibrium right from the start So we don't even need to worry about that one But let's look a little more closely at the fact that the moments must sum to zero This normal stress forces we don't need to worry about. There's no moment arm there It's these shear stresses. We need to worry about So we can take this one tau x y Delta a that's the force The moment is due to the couple of these two on opposite faces equal in office go in opposite directions separated by a moment arm of a These two Shear stresses here that cause a moment Now we've got another moment in the opposite direction on the opposite face. So it's minus tau y x Delta a That's the force the shear force The couple is due to that moment arm between the two Some to zero because the moments must sum to zero And we can do the same thing on the Opposite we could look down the y-axis and look down the x-axis do the very same drawings Let's see since this all equals to zero and the area of the the Dimension of the side that was the moment arm of the couple Cancels the size of the face Cancels as well leaving us with this very useful fact that tau x y Equals tau y x that's very useful because it sure sure simplifies all these drawings Now we don't need to do anything more than Sigma x in that direction remember those could be compression. I just happen to draw attention Sigma y in that direction and then all of the shear stresses are Just will just call them tau It's got to be tau X y or y x. They're the same thing. It's got to be that because we're looking at an x y drawing So we're not even going to mess with a subscript on that because if we go around in all the other directions all those shear stresses must be the same So that's all we need Describe the whole piece the only thing that would be missing is whatever sigma z is itself All right. He yeah, the only other thing. Well, we don't need to add that to it Okay, everybody everybody you feeling stressed yourself see see you thought you thought stress was stuff like oh My gosh, where's my cell phone? Is it charged? You thought that was stress. This is stress man you got to understand the forces and The stress they cause and you have to go the stress of trying to draw three-dimensional pictures Under pressure because we're moving kind of fast through all these pictures That's real stress. Not this Do I have a date for Saturday night? not yet prospects We're we're we're hoping for you. Don't forget. There's a there's a what chemistry calm All right any questions on this we're going to come back to this in just minutes I Well, actually we may have to wait until Until Friday to come back to these these shear stresses on these spaces and the kind of thing that they can cause So everybody all right I can erase the whole board if I want to let Professor and Professor Hampshire see this drawing on Jake this this this this video All right So that's that's our first look at all of these Stresses Plus a little peek at some of what can happen if the stresses are too great But there's a there are other concerns as We apply these lows and that has to do with how the pieces themselves Actually deform these deformations. There's failure. That's certainly one deformation But we may have deformations that are are are more elastic in nature So we're going to look at those type of things because these are these are Things that have to be very very closely understood by engineering designers Because any time a piece undergoes load It's going to deform If it deforms Elastically, then if that load is removed it will return to its normal length and everything's fine If it deforms in elastically You may have a design failure of some kind and the piece may need to be redone may need to be repaired may need to be redesigned But we're going to look at elastic deformations. This is what happens every time a truck drives over a bridge It loads all the members in different ways But once that truck's gone the bridge returns to its unloaded state ready for whatever next truck drives by So we're going to look at all of those type of deformation The first one and the simplest one Any time in this class I draw this kind of thing. This means it's an unmovable Anchor whatever I attach to it That thing itself will not move. We're only concerned with what's going to happen to the part that is attached to it So I have a simple piece here Which I'll happen to draw in tension because of that tension Well, we already know that there's going to be a normal stress there. We've looked at that some We know that at certain angles. There's additional shear stressors would be concerned with What I want to concern ourselves with now is the fact that once loaded like this And I'm exaggerating it that piece will tend to In this case elongate because I happen to draw attention if I drew it in compression It would tend to shrink so we have this exaggerated elongation of Some distance will call del greatly exaggerated here Well, it wouldn't be I guess because this piece could be rubber or something. What we'll do with that is we'll define The normal strain we've looked at the stress Now we're going to look at what we call the strain normal still means the same thing it did before the load is Perpendicular to the area supporting it a little bit more of what we mean here is that the Elongation the deformation is perpendicular to the load or to the face. So we'll define the normal strain Use the symbol that's a lowercase sigma Actually, no, it's a epsilon. Sorry Okay, so on Defined as the deformation we see under load and that could be negative divided by the original length of the piece itself So we'll call that L for simplicity simple as that put it under load See how much it stretches compare that to the original length And that's the normal strength But we don't have in there The size of the force causing it nor the area supporting it All we're looking at is the response of the material to that load you Maybe your experience With how things stretch whether it's a rubber band or a spring indicates that the original length does have something to do with it You have a short spring. You can't pull it very far where if you have a long spring and you can really pull it far Simple as that. What are the units on it? What now? That doesn't matter. I'll watch the tape See what you said there Bob Is that I now have a directional mic pointed right at you. You always sit in the same spot We're just gonna get an extra camera Put it right on you never know What are the units units might be? Meters per meter Inches per inch they might be nothing because they cancel There are also other options of it that we do with this this deformation as you can imagine especially with engineering engineering materials like steel and and Aluminum and iron and those type of things this deformation is not very big You could you couldn't get some piece of metal and hang from it and actually notice any stretch going on There's very very small for a very large piece like on a bridge Where the pieces are quite long it could become noticeable Especially if there's a whole bunch of pieces and each one of them has a deformation of its own and those tend to add up over the length of the structure So with bridges this kind of thing has to be very closely monitored It tends to be more like on the order of 10 to the minus 6 meters per in meter or 10 to the minus 6 inches per inch So we also then might say it's a micrometer per meter or a micro inch per inch Or might eliminate that completely and just call it a micro because the units don't matter so cancel out the meters it leaves just the SI prefix micro and so we can call it that Yeah, but but that's not sufficient We want to throw more possibilities in here because the more possibilities the more Opportunity for the funnelmen of students the more excited we are about that. So we might also call this a Micro rap for radians is kind of a magic unit itself So I think that's where it comes from Or we can look at this elongation as a percentage of the original length So whatever number we get there Multiply it by a hundred percent and call it that So it might also come as a percentage So you you choose any one of those you like All right, let's let's start a setup of a type of problem. We might see doing this It's a very fairly simple idea just you you load something and it's gonna stretch some some amount So imagine we have here beam of some kind pinned at one end But supported at two places So supported there by a cable and supported there by a cable this distance Three meters that distance four meters But I'm going to split it in the middle with some kind of load That's on this beam So it's two meters on either side of that. Let's label some parts here. We'll call that point a E Just for reference. We find that due to that load the piece itself displaces By ten millimeters on that end right one more thing to me the length of each of these two cables We'll call four meters. Ah, so clearly it's not not to scale Because that four meters looks pretty darn close to that ten millimeters so for your benefit It's just a cartoon All right, so the the support cable DB Stretches by a certain amount as does the support cable EC Find the strain in each of those so find the strain in BD and the strain in C E as well Just as a as a practice step for us to get used to the units and sizes and the simplicity really of this calculation Remember, we have the strain as the deformation By the divided by the original length itself. So both most of those in fact This this one over here Really couldn't be a whole lot more straightforward Just really much pretty much comes right off the pace there. Just remember you've got to get the right lengths in there Watch your units a little bit. I'll leave it to you to express it in Three different sets of units your favorite from that list as much as anything these problems tend to be problems in geometry While it's fairly straightforward seeing what the elongation of the cable CE is It's a little bit different how much elongation the cable BD itself Underwent just a little bit of a problem of geometry and we have this one yet in your favorite set of units the elongation For this one right off the picture 10 millimeters Original length four meters did you pick? because we can we can reduce this to There's something at least simpler to write What so what percent? All right, fair enough One way to write it point two five percent Anybody do it in in micros? Okay, and had what meters over meters or Could just leave that off and just say it's point zero zero two five Pat took it multiplied by a hundred called a percent Another possibility, and then we've done 25,000 I started twenty five hundred microbes All right, your ticket for regid mission on Friday is To have that if you don't already if you already have that then you can stay until Friday