 All right, so far we've been looking at the stresses that a material is subject to based upon some kind of loading. These are internal stresses we're talking about. For the most part, we're assuming that they're constant or we're dealing with the average values. This is not necessarily true, but it certainly suffices for the problems we're doing. Part of why it works is because even if we look at the true distribution of stress internal of some kind of solid, we treat it as an average, which means in certain places we're over estimating the stress, certain places we're underestimating the stress. But then when these kind of things go to design anyway, a factor of safety is applied, which is generally calculated as something like the maximum anticipated load, and this can be on the forces themselves, it could be on the stresses that result from those forces. Remember that even though the forces through a member may not change, the cross-sectional area may change, which causes the stress to change. The maximum load over the design load, in other words, once you've calculated what the maximum loads are, and again we may take this as an average value, which means we're a little bit worried that we're underestimating the stress in certain places. We apply this factor of safety, which takes us way over the top. So if we're calculating for these average stresses, we apply a factor of safety, which can be somewhere on the order of 2, 3, 4, 5, depending upon what the application is, where you figure out what the maximum load is, and then you simply design for twice that. So you're well over the maximum that we're missing because we're averaging, so we eliminate the concern. It also eliminates uncertainties, there's uncertainties in tests, uncertainties in manufacture, it's certainly possible you get some corrupted material from the manufacturer, it doesn't have quite the strength it is advertised to have. Very high factor of safety when the possibility of failure could be more catastrophic than some other situation, which may mean greater loss of life, but it could also mean that whatever machine part does fail could be more catastrophic to the rest of the machine than some other less crucial part where you could apply a lower factor of safety. Because generally the higher the factor of safety, the more the cost, because you're going to have to put in more material to increase the area, or you're going to have to put in more members to decrease the force in each of the members for the area of which you design. We'll work with a, we'll do a couple problems though, if it hasn't already come up, I'm not sure, in the problems as we go through these, I don't want mine to go off. Alright, so we've come up with some stresses, we've got two types here so far. The normal stresses due to axial loads and the shear stresses due to transverse loads, and by axial and transverse, I mean the direction of the forces causing the stress, the orientation of those forces to the area that's absorbing that force. Normal stresses are always perpendicular to the cross-sectional area used to calculate the stress, shear stresses are parallel along the area that are absorbing that stress. Alright, so that's nothing we couldn't have done in statics. There's nothing there that has anything really to do with the strength of material. We very easily could have calculated all that in statics. But now we're going to add to it a little bit and start looking at the actual result of these forces on these members. And we'll see that as what we call strain. And again there's going to be two types of strain. We'll have a normal strain and we'll have a shear strain. And I'll work up both of those now. As a material, as a member is loaded, we know now that of concern is not just the load on it, but also its cross-sectional area. The greater the cross-sectional area, the less the stress. And without even knowing anything in the particulars of what happens to materials, the lower the stress, the better. So we know that for some given cross-sectional area, and one of the typical ways we do it in engineering is we in the member itself draw what the cross-sectional area looks like as if we've cut through there and then turn that area on its side so we can see it. That's one of the ways that we demonstrate what the cross-sectional area looks like. If it was a square member, I might draw that if it was a triangular member, they'd all look the same on the side, but they'd look very different when we just draw the little sketch that shows what the cross-sectional area is. Often that's entirely sufficient to then actually calculate that area as you need. So whatever the cross-sectional area is, as of yet we are not worried about what the shape is. We're just worried about what the area is. We'll find out very shortly how the shape of that area plays into it. We can have beams with the very same area and they're going to act very, very differently under loads as we'll see. So we know that if we've got some member with some cross-sectional area under some load that we can figure out the stresses in that piece. What we haven't looked at yet is the response of the member to those loads, other than to say if the loads are too great or the area is too little or the combination of the two in ratio, the stresses are too great, the piece could fail. What the piece will do before it fails, however, is actually elongate by some measure. Let's say by an amount dealt. The normal strain in an elastic solid and though it may not necessarily seem like it, structural pieces made of iron, steel, wood, all of these pieces are somewhat elastic. They will deform a certain amount. If you don't deform them past a certain limit then when you release the load they will return to their original size. Of course you can imagine with steel and the like, this deformation is very, very small but with certain other structural materials like rubber which can be used in certain structural situations as we'll see, this deformation is large enough to even be visible. So our symbol is eta, the lower script, sorry, that may be sigma, I'm not sure, epsilon, there we go, epsilon, yeah that sounds better, thank you, you speak Greek. Our definition of the normal strain epsilon is simply that amount of elastic deformation divided by the original length of the piece. Simple as that. Load a piece, measure how much it deformed, you can calculate then the normal strain on the piece. What are the units? The units come in about 12 different flavors, whatever you want because it could be meters per meter, millimeters per millimeter, or you can look at the units as if they simply cancel and there are no units at all. The strain could be just simply a pure number with no units. This deformation is generally very, very, very small compared to the original length. So it's quite likely that the strain values will be on the order of something like 10 to the minus sixth meters per meter of original length. And of course same for inches, of course the units do disappear if you so wish. We can also call this deformation of 10 to the minus sixth units per original unit length, we might call a micro. Sometimes just left it like that, remember that scientific notation, sorry the SI prefix notation of this is 10 to the minus sixth. So sometimes it's as micro, actually they wouldn't do an M in there because that implies that there are still units. It's not uncommon and I believe our book does, it actually puts it in, it calls it a radian. Remember radians are sort of mystical units that come and go unneeded and so that's one place where they're just thrown in. Remember the radian came from a ratio of lengths itself. It's also possible that you could list it as a percentage where it's the percent of deformation to original length. So in there you'd have to of course multiply the calculated value by 100. Was that compression going to be a negative? Sorry? If it was a compression going to be a negative? Exactly, right, yes and that's important because strain in one direction is very different from strain in another direction. Some materials are very, very different in their reaction to compression than they aren't attention. So that's precisely true. It's also true that if we have a solid that's twice as long with the same area and the same original load that the deformation will be twice what it was in the first one I showed. By the way, let's put the load on there. So this now would be a deformation of twice what it was before. And you can think of these elastic solids very much like springs, especially a linear spring in a certain region this response is linear. We'll see surely more specifically what happens if we have a back to the original length now with twice the load we can expect twice the deformation compared to that original picture. So we have those possibilities plus any other combinations thereof. Just what this deformation comes out to be of course depends upon what the solid is itself as well, what material this is. Material is very different than this piece we were talking about was rubber of some kind. So we're going to relate the material properties itself to this response in the next day or two of my hit and fray. Alright, to exercise that a little bit, let's try a problem. I mean we have a simply pinned beam of some kind supported by two rods about here and these are also pinned and then one a little bit farther out and a little bit longer. Alright, the dimensions, all of these dimensions are in meters. So two there, 1.5 there, 2.5 meters across there, 2.7 and then another half meter on the end because it's the end where we're going to apply the load and some load P is there at the end. Now for our purposes here, so far we're going to assume that this beam, the horizontal beam there is what we call a rigid beam meaning it itself is not going to deflect under any loads. It shouldn't be too much a stretch of your imagination to imagine if we loaded a beam in that way that it's going to deflect due to those loads maybe something like that. We're going to assume with the rigid beam assumption that there is no deflection like that in this beam. It's not accurate of course because the beam will deflect like that but that's something we could much near the end of the class and the last month of the semester we'll look at the deflection sideways of beams to transverse loads like this. Rigid beam no deflection under the load applied. However with what we're doing so far what will happen as this rigid beam is loaded these two members will absorb that load and because of the natural strain the normal strain in those members they will elongate. So the beam may not deflect itself but it will rotate a little bit about that one pinned end due to this stretch if you will in the supporting members. Alright so we want to figure out something about what that is and get used to the numbers given. Alright we'll call this support one this is support two and we're given that the strain in two the normal strain in two is something like 800 micrograds. Now right now we don't have the capability of calculating that beforehand as we would in a design situation where we would imagine we're faced with this scenario and we predict based upon what the materials are what this strain will be but it's something that's very easily measured because it's nothing more than the geometry of the situation. If we built this up loaded it we'd simply measure the increase in distance from the top of support two to where it now currently is. So what we're given is the information of what this deformation is in piece two. Just simply measure it. So our task then is to find the strain in that first piece. What does it suggest? Well one thing we can always do is start with the definition. We know that the strain is going to come from however much the piece one deflects deforms compared to its original length. Piece two or piece one sorry is back here somewhere so we know it's going to deform an amount like that. If we can determine what this amount is we can use similar triangles to determine what that amount is and then simply calculate the strain. How does this help us find that which will give us that which will give us that. Very very geometric problem of course is the deformation of two divided by its original length. So take a second or two calculate those up even though with a strain the units do cancel you still have to get the relative size of these numbers correct. A collection of 800 micro meters per meter of original length is almost a millimeter for original length. Original meter length. Is that right? One two three for the decimal place and then one two three for the milling. So that piece is originally almost three meters long it's been deflected over two millimeters which may be enough to make this whatever this piece really is not working it's it's an in ready on intended purpose. If we were storing bowling balls on this they might not stay. If you go into the bowling ball storage business they may have that deflection. Yeah that's what I have. I thought it was I thought at the moment there I thought it was the strain but I forgot. That's just the deformation. That's a measurement of mine. Which came from the strain. Yeah that's that's purely a geometric measurement if this was actually a test piece you would simply measure that. Well maybe not simply. You can do a micrometer stand of some kind. We have one in the physics lab. From that by similar triangles you can figure out what the deformation of one is and then from that you can find out what the strain is. So strain is a very very geometric problem as it will remain so when we look now at after this a shear shear strain exactly the way that springs act linear springs act and we'll we'll put that together in exactly that form possibly on Friday maybe not tell me What do we have for the deformation of the first support member? Phil you have that? 0.96 millimeters. Agree? Yeah by similar triangles. That again is faced upon the assumption that the beam itself is undergoing no sideways deflection. There's no axial deformation in the beam just merely a rotational displacement about the pinned end. And then we include the fact that the original length was 1.5 millimeters and so how did you express your answer then? Remember there were several choices for ways in which we'd express this strain if being generally a very small number. A first strain was expressed in micro radians. Generally is a good idea you put it back into the units that started the problem if there's not a request to the contrary that works as well as any other. And so how many then micro radians is the strain in piece 1? 640. 2. It's what? 640. 640. How about as a percent and you can express it however you wish. Generally on tests I'll put it in, I'll put the way I wanted to express just makes grading easier if I don't have to convert your answer to some other set. This expressed not as a micro radian or micro 640 micros but as a percent. Just make sure just a little practice making sure we're going to have a lot of things we have to do where we very carefully have to watch how many powers of 10 we have in these problems which is the same thing as where is the decimal point. As we get to other numbers these are very small numbers some of the stresses are very very big numbers when we put the two together we get even greater experience. Tom, Joe either one of you have this as a percent perhaps you can pass your coffee forward please just grounds he'll take it. Who's got it for me? Of course David, Chris you do too. Did you check with each other? Travis, Travis, Phil you've got it as a percent. A millimeter is 10 to the minus 3 meters that's going to divide to 0.64 times 10 to the minus 3 and how many percent is that? Somebody save me here. David confirm that Chris very very small very very small numbers and we're going to find that with what we when we put strain and stress together that's actually going to cause a little bit of trouble for us as we actually go to visually demonstrate the response of these materials to the load and the stress that's in the piece itself. All right same thing here however what if as this thing is put together there's going to be manufacturing tolerances what if as these things are put together there's a one millimeter gap in the pin connection on piece two is that going to change things then what is the strain in the piece one is that going to change things in any way assume that it still has the same strain in it though. Yeah there's there's a little bit of slot in the pin connection in this such that as this load causes the beam to move to rotate a little bit it'll move a millimeter at this connection before there's even contact full contact in the piece just manufacturing tolerances or possibly a certain size pin was planned for but not actually available for a combination of both how does that change things so that content the connection at two is going to drop by that amount but what is that amount now such that it still has the same strain that's the strain that should lead us to the actual elongation of that piece what is that elongation if that's the strain which was the same strain we found before what now is this actual elongation of piece two this comes directly from the strain the strain's the same as it was before so should this be the same as it was before it sure should because it only comes from one place this number hasn't changed that number hasn't changed then that number's not going to change either so the actual elongation of the piece two is the same what however is the deflection of the connection point one millimeter more because it's going to move to take up the slot then it's going to start straining this piece up to the strain we found in the first place so maybe we'll call this capital del which will be del two so this will be del two there capital the delta two equal to del two plus one millimeter and so that changes things back here so you're going to have to find a delta one but again now a similar triangles as it was before this is del one plus some unknown amount just simply due to the greater deflection down here well actually that's not quite right is it there is no slop in this piece down here so so this will be del one itself since there's no slop in that connection by similar triangles but there probably will be slop in both of them just by the nature of physical connections you know the more precisely things are built the more expensive they are to build so depending on what this application is it may actually have quite a bit of slop in what's this then the deflection here we know this to be 3.16 millimeters so this is it was 0.96 is it going to be greater than that yeah because it's a it's a larger triangle of nothing else so I believe it's 1.4 millimeters sound about right and now the strain in this piece is that 1.4 millimeters over I can remember the original length of the piece unstrained that's just that let's get that number we'll take a quick break to reset the tapers David you ready depending on just how much round off you took through this problem something like that 930 940 remember there's a big factor of safety in these things anyway so the last significant figure or two is not going to be a great concern you're going to design way over that limit anyway all right stop take for a bit all right again we're getting our our first chance here to look at the actual response the physical response of these materials these elastic medium to the loads being applied to them we just looked at the normal strain defined as the amount it deflects under load divided by the original length purely a geometric problem so we'll add to that now the shear the response of the materials into shear also as a physical deformation of the solid so if we have some either some structural solid or some elemental piece there of subject to some kind of shear load and we looked at how on a cubicle piece like this these shears must be the same magnitude all around and oriented in somewhat that fashion we looked at that on Friday what we didn't look on Friday is what the response of the material is to those type of load again I've arbitrarily chosen which direction these all lie in I could have easily take the other way but piece under that kind of load will start out uh as a square shape and then given what I show here will deflect to something like that that in there we might call uh uh del again that was fairly useful um if we had coordinate direction on this which we often do in these pieces we would call this then del x possibly just to just to highlight the fact that there are going to be other strains in other directions and that as shouldn't be a surprise is also some function of the original length um we find then the shear strain as the change in angle of some original angle in this case it happened to be originally 90 degrees it now undergoes a change to uh whatever that angle is dependent upon the geometry and we'll label this gamma x y x y meaning this is the piece with our view of this piece is in the x y direction x y plane so in this case then the shear strain in the x y plane for that little picture is defined as the original angle 90 degrees or pi over 2 because this is generally done in radians minus whatever the current angle is so piece that that corner angle started out at 90 degrees it's now some deformed value theta prime I'm not exactly sure why it's got to be theta prime why couldn't we just say that but that's uh some of the tradition of this this area and so we get now a change in the angle of that much as shown this is a positive shear strain if the piece was such that the angle group would be greater than 90 degrees as this one over here happened to do that would be a negative shear strain and again it's a purely geometric problem that's by the way the average shear strain uh locally we can't get different shear strains if you imagine a rigid plate on top of a deformable solid like a piece of rubber and this by the way when we have this type of thing that absorbs some kind of transverse loads in it by deforming elastically is called a shear block these are very common at the ends of bridges to absorb the thermal expansion of the bridge itself with temperature change you can imagine in in an area like New York for example if you go down here exit 14 and putting a new bridge across the north way there they're doing it in relatively cold weather in august it could well be over 100 degrees that bridge is going to get substantially larger by substantially I mean we could see the change it would be it'd be visible to the eye and this rubber block can absorb that the thing is it will probably deflect more like that rather than like this which means that the shear strain is a little bit different at different places along here that we actually had a grid drawn on here we see that that grid if it was made up of squares would have different shapes at different places in that case then the local shear strain the non-average shear strain sort of like instantaneous deflection would be the local deflection divided by the local distance there in other words just the slope of the shape of the piece itself now is the whole block rubber just the top block or these are rigid plates as what the reason I was the reason I want these to be rigid plates is because I don't want them to elongate which adds a different dimension here to the strain on this side would be different than the strain on this side is the plate itself change length which would be a normal strain in real life that indeed happens however again for these kind of applications generally the deflection in the support plates is substantially less than the deflection in the elastomer the rubber like solid that's absorbing that movement all right so if we have a deflection due to shear loads like that we might actually get a response of the piece to become something like that highly exaggerated and not to scale that's supposed to be a parallel graph why it doesn't look like one this is this is in all the things I have to draw especially in this class you're learning already I hope this is a very graphical class it depends a lot on your ability to draw piece this is one of the hardest things to draw because those faces don't stay parallel to their original location is that better that's an awful lot better anyway half of the strain is there half of the strain is over there they may not be the same depending upon what those what the piece is itself this is a positive strain and if instead of getting narrower they've got now we're deflecting that way that's an example of negative negative strain yeah look look at your book for for drawings done by the graphic artists that they hire by the way that's part of why these books get to be very very expensive all these drawings done not by the author they're done by graphic artists they hire to do each one of them in considerable detail all right so let's uh let's oh this uh this again not only is it a entirely a geometric response a deformation of the the piece determined purely by the geometry before and after but it can be actually measured in that way you can go to a piece and scribe on it a square and then look at the deformation of that square due to the loading it's got to be enough that it's measurable it might not be um but it's it's purely a geometric response to the solid so as an example let's start with a simple plate 720 millimeters on one side 480 on the other in original unstressed unstrained dimensions and again this could be a particular part of the whatever it is a machine of some kind or a support plate of some kind or it could be an actual inscription on the piece itself let's say it deflects to something like this again not necessarily the scale and that deflection here is half a millimeter the deflection here is a quarter of a millimeter find that the strains at the two places so we'll call uh that one one that strain deflection one and that two because the total sorry wrong gamma one and gamma two because the total of those is those two things added together because remember the original 90 degree corner here the total change in that angle is the strength the shear strength the two of those added together is the total shear strength original 90 degree angle down here and the change in that angle is the shear strength the total shear strength which is the two of those added together again uh rather simple calculation uh purely geometric but has to do with some numbers of rather different sizes this deflection of one end is much smaller than the original as than the length of that side transfers to that plus don't forget you have to get the signs right if any how'd you find gamma one anybody have it this is by the way Greek letter gamma it can be an either but the straight calculation if you use these numbers here it will be radians automatically you can convert it to degrees but typically it's given as a as a in radians or unit less as well it can it can have the same type type of numbers same type of presentation that did the the normal strain how'd you find gamma one there's this shear strain in the in the first location actually um no we just take the original the original numbers um the the these are very very small angles right so that's why we don't need to take the tangent because the tangent of a very very small number is the same as the tan as the number itself and that's also why you didn't use that method when you showed us the differential means yeah yeah you can if you see it better that way take the tangent but and then this is over the original pieces itself those tapes comes out to be give it to me and uh radians micro radians and percent and for some of these since they're very very small you have to be you have to carry enough uh significant figures through the problem and then round it off at the end generally how many radians is that David you ready 0.001215 001215 how many micro radians is that that stands for 10 to the minus 6 so move it over six places 1215 micro radians how much percent is that this one's um millimeters per meter already just move it over two so it could be less than just over a tenth of a percent reflection so again very small numbers but uh numbers that may be enough of concern that they can't be ignored depends again on the on the situation all right any questions for it clear up uh this as much of any as anything is just simply a matter of uh trying to weigh through the geometry of the problem seeing what uh what is deformed and what has to find a 90 degree angle that's deformed find uh what it's changed to all right so a wrap up problem uh some kind of structural member under some load 12 columns original length 100 millimeters original diameter 16 millimeters due to the load this piece will get substantially longer well i don't know substantially that's a relative term but because the density is the same and the volume is the same then not only will it get longer but it will get narrower as well not all the way back to here it'll it'll spread out to the original connection uh distance but remember we're looking at average strains here so let's say it increased in length by 300 micro meters not very much barely visible and now the uh and the change in diameter i'll call that del d del remembers our symbol for change in length due to this loading is 2.4 millimeters so if that's our x direction that's our y direction that's our z we need to find the normal strain in the x direction and the normal strain in the y and z directions since it's circular those two will be the same anyway maybe we can just call it the uh strain in the radial direction all right we'll double check that on friday because that's in the class