 stress strain diagram for different materials and what was a particular interest to us was this very steep linear section very steep by I mean that these these strain numbers on the x-axis are very very small we're still not in a large deformation zone but the best part about this is of course that this is linear and it's retraceable this there's not what we call hysteresis I think you've taken some of you were in circuits or something might have looked at hysteresis hysteresis is any curve that goes one way up and comes another way down and then might retrace that shape again we do not have hysteresis in this region we go up this line relieve the load come right back down this line that means it's very very predictable and the slope itself is a characteristic of the material whether loaded or unloaded it's no different than other type of numbers we have for materials like density we call this the elasticity modulus the modulus however you want to put it and it's the stress over the strain anywhere along that elastic linear region for the material so we got our first look at that as a material property now and it includes of course the actual deformation of some of the materials this modus elasticity you can look at in a way as the load on a piece on a material because don't forget that the stress is the force being applied divided by the cross sectional area normal to the direction of that force is absorbing that load and then the strain is an actual deformation due to that load so this is kind of like a ratio of load to response for a material and it's this is the major design region in which we stay because of its predictability linearity and the like and somewhere up along there where the linear portion starts to disappear we call that then the yield stress and that too is a characteristic of the material how much stress that the material can take before it yields because beyond that point the curve starts to come over and release the load then we come down in a essentially parallel region but now there's some built-in strain even to the point where if we completely release the load we still have residual strain there's some permanent plastic deformation in the material the other number that was of some concern and depends on what kind of curve we've got as to what kind of material we're looking at but there was this peak ultimate stress that's also characteristic of the material for the most part that's not a design region you want to go anywhere near because once you're past that then the necking starts and sooner or later there's rupture of the material certainly there's substantial plastic deformation with that region because now you're way out here in the straining an awful lot of deformation has occurred by then all right so let's let's do a quick problem where we kind of use that business so imagine a beam 12 foot long beam and this is this is a good chance for us to bring in something more to it this is an IV in cross-section which believe it or not is because it looks like an eye we're going to be looking at a lot of these these are generally come in a lot of pre-determined cross-sectional dimensions and you pick the type of beam that may be of best choice for your design the beam designations are something like this so this is this is this is the beam we're using for this we'll look in a few weeks at just why an I beam is such a strong configuration for being in so common in cross-sectional materials the W just means it's a wide flange beam and that's that's just a generic descriptor of the general shape of the beam the 18 is the height of the beam give or take a little bit and the 97 if this this is an English dimension beam so this is a height of 18 inches and the 97 is it's linear weight pounds per foot of beam so if we have a 12 foot beam here then we have 12 times 97 pounds as the weight of the beam so we're going to take into account the weight of the beam in this case put a right dead center and that's 97 pounds per foot times 12 feet is about 1164 pounds take that into account also we're going to put some imagine a small shear pad there and maybe about there and these will put about eight feet apart and that allows a couple things for this beam one it absorbs any of the up or down forces just a little bit of flexibility but it also allows the beam to change shape thermally because of the great temperatures the change in temperatures from use as a beam like this might go through the year the beam can change in length and these shear pads can absorb that we've got the beam there and the shear pad there if the beam grows in length because of thermal effects then the shear pad can absorb that by shearing over a little bit and then when the thermal changes disappears being as a path and return back to normal and the beams nominally as design all right so these are about eight feet apart and the center of the pad is about four inches in from the end and we also have just to spice things up a little bit another point load of 40 kips and that's about 40 inches in oh then it's that's about 32 inches over to where we've placed the equivalent load representing the weight of the beam all right last last little bit we need is the a look at these shear pads themselves because it's not enough for us just to look at the frontal area because it's the whole area of the pad that's absorbing absorbing the the shear so the front that we can see there is eight inches by a half an inch but then the pad itself also goes back about 11 inches so this this part we see there the same as the part we see there eight and eight inches by a half inch and then the beam rests on this whole top area that's eight by eleven at the side of the sheet of paper and the last little bit I think we need just for setup is that the modules elasticity of the shear pad itself is three thousand remember what the units are he is stress over strain strain is essentially unit list of the same as the units for stress which are units of pressure since we're looking at English this would be psi all right so we want to find find the compression in the varying pads the fact that this beam is considerably is quite heavy is going to compress these pads a little bit I want to find the compression of pads and because of that the resulting angle of the beam and as a designer of the rest of the structure the angle at which this beam might be when loaded is very very important because it may make something else in the rest of the structure not work quite right okay what we need to know of course is the force that's doing the compressing of these bearing pads so if we call this and a and that and b then we have some load there that's serving to compress that pad and we want to find out what that compression is then the comparison of the two will allow us to find the angle of the beam once loaded so let's see how we're going to get there let's see the compression is part of the strain so if we knew how much strain was in it we know the original length that's being compressed that's the half inch so if we knew the strain we could find in the compression well the strain the only other place strain comes from so far for us is in this modulus of elasticity we know that for the pad so if we knew the stress then we could find the strain once we know the strain then we can find the deformation of the pad and the stress of course we can get from looking at whatever the load is and what's the area absorbing it so the load we can find that we could have done that last fall that's just straight statics to find the load then we can find the stress once we have the stress with the modulus of elasticity we can find the strain once we find the strain then we can find the deformation so a real quick statics analysis we can find then the loads on the bearing pads and then go up through this little chain of chain of solution if you will this little trail that we can take and we can solve the problem I'll leave it as an exercise for you to approve that these loads are 27.1 kip at the left side and 14.1 on the right side that's just from last fall that's just statics some of the forces some of the moments and you can solve for those two things so continuing through then with with with a so the stress in bearing pad a will just be the load at a which is 27 kips 27,000 pounds and then the area that's absorbing that load is which give me the dimensions so make sure what you're talking about 11 by 8 inches it's it's this that area across the top of the bearing pad the 8 by 11 8 inches half 8 no sorry 11 inches remember don't put inch marks in an equation itself because it looks too much like a exponent it's okay on a drawing because it's implied there and so I think that comes out to be as usual confirm my numbers I think that's 308 so now we know the stress we're given the modulus elasticity where does that come from yeah that's generally supplied by the manufacturer you may or may not want that tested depends upon your level of comfort with what the manufacturer has done if you have a history with that company you might trust them if it's a brand new company you might want to test a couple of those pads yourself but that is a characteristic a material characteristic generally supplied by the manufacturer so now we have the stress we divided by the modulus of elasticity that'll give us the strain and then we can get the deformation so the stress we know to be 308 psi modus elasticity 3000 psi make sure you watch your units and all these problems especially since we go through several of the levels of prefixes for the si notation that's a little over a 101027 1027 it's it's unitless does that make sense it does if that's the strain remember the strain is length of the distance of the deformation divided by the original length that absorbed the load and now we can find out the deformation taking that strain oh maybe I should have had little a's on each of these just since we're doing the pad at a and then the exact same calculation through for the pad at b so we know the deformation will be about 10% of the original length and that's the that's the half inch there and so that comes out to be five one three inches that's how much compression will find we we uh calculate at a now this is assuming that the compression is uniform over the bearing pad that the beam stays level which is a bit of an assumption especially since we're admitting that we expect the beam to actually be at an angle which would cause a little bit different compression on one side of the pad and on the other and we'll uh you're able to look at that once you finish the whole business uh plus there's another assumption in there that we'll get to um so you can do the same calculation for the other side of the beam and you'll get the tilt there about half of what was experienced on the left side and the tilt of the beam we can just use those two uh the difference of those two I guess the entire length between the two which is a hundred and eight inches that's the eight feet that'll give us the uh the essential give us the angle it's actually the tangent but it's a very small angle uh maybe I'll call it beta and that comes out to be about 0.29 times 10 to the minus fourth radian so it is it's very small might not be a worry but you'd need to find out after the building collapsed that maybe you should have paid some attention to it why is it a hundred inches because what we're looking at here uh if I drop a little bigger the the beam was here now it's kind of like this so it's that distance uh the difference between the two ends because one end went down that much the other one went down that much so that's uh that's this upper number here and then the original length there and so that's the tangent of that angle yeah I don't know where the eight feet came from that was a figment of my imagination that's what that was god only knows I think uh oh is that is that uh that's nine feet so I don't know that's the number I used I don't know it's my garage so I'll go measure it now assuming the uniform uh compression on the pads uh we get that angle that's about a seven percent error uh ignoring the fact that the pads really don't exaggerate this the pads really deform non-uniformly across them and we're assuming an average deformation and that the the pattern makes level but it's a very small angle not a very big error it would be well under the factor safety anyway now what we'll look at in a few weeks near the end of the course is uh further effects of this now we're going to see we're going to imagine uh that this beam dips by that amount of course greatly exaggerated here what we're also going to look at in the final weeks of the course is how the beam itself will deflect because of these loads the beam itself will deflect something like this again very very exaggerated giving a different angle here than there is over there we'll also be able to take those into account as we look at that kind of deformation as well we're assuming through this whole thing that this beam is rigid and does not itself deflect because of the transverse loads it's just simply uh we're just simply looking at the compression of the pads themselves so there's actually two angular effects and we're ignoring both of them at this stage that was you're still frown just stuck that it's your friday frown okay so you double check that you've got that that number there uh the rest is just trigonometry figuring out how the uh what the angle results in all right so i don't say well we're going to sort of revisit something we looked at before we just didn't do anything with it then um imagine we have some solid space we have some solid some material piece we'll give it axes directions because we're going to have to refer to those here so we've got some material piece there whether that's the entire beam or just a an elemental piece of it is immaterial because uh what we're worried about is the response to those loads that are applied so imagine then a axial load is applied to this piece as we've seen so far that's going to cause the piece to elongate but because the volume is the same it's also going to contract in the other directions so we might see a final heavily exaggerated piece that looks something like this under the load greatly exaggerated the piece will elongate but also contract not too bad for free hand at the board let me replace this force to illustrate it still apply at the end okay is that a helpful picture see what happens we we did a problem where we briefly looked at this in fact we'll revisit that problem in a second all right so we have an original unloaded shape and then a resulting loaded shape again greatly exaggerated you can imagine if this was a a rubber block that this kind of thing would happen uh you can just stretch a rubber band and if you look closely we'll notice that it contracts in the other direction uh orthogonal to the axial load so we actually have a couple strains in the piece here we have the the normal strain in the x direction which is how much it deforms in the x direction divided by the original dimension in that direction so for our purposes I guess it'd be twice however much you drew sticking out here or out of the front but there's also a transverse strain this is axial or normal strain there's also a transverse strain in the other direction the fact that there's a change in dimension in the y and in the z direction and however much those are of course going to depend upon the loads and the materials and a couple other factors we're going to assume that is the strain in the y direction is the same as that of the z direction since they are both transverse direction this arbitrary which one's the line which one's the z this means that we're assuming the material to be isotropic in other words it has no uh no no uh no bias in terms of the direction the piece is uh actually loaded it's the same in any direction this is not like wood wood is very different in different directions for these kind of strains because of the grain of the wood so wood has to be put down in a certain way steel is just poured into the shapes that are used as you can imagine the ratio of these strains is quite important and this is called Poisson's ratio given the greek letter new defined as the lateral strains over the axial strains we take the absolute value because the as drawn these uh transverse strains are negative the axial ones are positive for the picture I drew and so this just makes the ratio a positive number is all typical values for it typically new is less than about a half very common numbers are like 0.2 0.3 for materials and so we can actually use this calculation we had this problem I think last week we had an axially loaded piece that was originally 500 millimeters in length and 16 in diameter 16 millimeters OD OD means outside diameter there's no inside diameter on this but not atypical to do something like that and then we showed uh or I gave you numbers for the response to that loading it uh elongated by 300 micrometers here's the 500 the elongation was 300 micrometers but it also decreased in radius its original radius was bigger and there was a decrease in the radius of 0.24 and that would serve for our del y and our del z in this radial direction all right remember that problem we calculated the strains in the two directions the axial strain is the change in length the 300 micrometers over the original length 500 millimeters all I'm doing is reproducing the same numbers we found earlier so that has a strain of 600 and the radial strain the uh transverse strain the minus 2.4 micrometers over the original diameter 16 all right so that's that's just that's just brings us to the end of that problem that we finished I think a week ago all right and then we can use that then to calculate Poisson's ratio by looking at the axials sorry the transverse strains over the axial strains 1500 over 550 over 600 so about a quarter of 0.25 or we expect for every unit of axial strain there'll be about 25 percent of that will be the amount of transverse strain so simple as that in calculation simple as that to use not much more we can do with it it is also however a material property so you'd expect it to come from the manufacturer or from your own testing all right so as simple as that is we'll take another step with it we've got well let's see we've got stress normal stress and shear stress we've got strains that departure from the oh well the first strains we had were the normal strains then we had some shear strains remember was the change in a 90 degree angle whether imaginary or actual we added the modulus of elasticity which is something like load and response of the material that was the first of them that was a material property so we then added Poisson's ratio also a material property and so we'll take it a little step farther and add a little bit to this list so imagine we have some piece whether this is an elemental part of a loaded piece or an actual part itself and we see it undergoing some kind of shear we looked last week why all of these must have the same magnitude and be oriented in that direction or each one of them 180 degrees in the other direction and the response of the piece might be something like that this angle here is half the shear strain the other half being on the other side and we define one more term let's see we have the modulus of elasticity that was the axial stress to the axial strains we define the modulus of rigidity in the same way only do it for shears rather than normal loads so it's given the the letter g because there's a g in rigidity I guess and we can look at it in the same way the load versus the response the load is the shear stresses and the response is the resulting angular strain so we define the modulus of rigidity as something like that in the same way we did the modulus of elasticity as load versus response again this is also a material property so just to kind of bring all these together if you look in the back of the book you only don't have to all put it up we have well let me zoom out first show the whole thing in the back of the book and this should be any strength materials book our tables like these ours just happen to be immediately in back cover so they're very easy to get to we have anglish units on the first part of the page and then the second part of the page the very same thing in si units down the right we have different materials different structural materials aluminum copper cast iron steel alloys concrete plastic and even wood different types of wood and then across the top we have a lot of these material properties we've now been looking at specific weight is a slight density only with weight instead of mass modulus of elasticity be careful of these units all these numbers here happen to be in ksi there's the modulus of rigidity the yield strength is a number we're going to have to pay a lot of attention to notice for most of these materials it's the same intention as is an impression ultimate strength we're not going to use very often because we don't want to design out to that limit percent of elongation in a two inch specimen i'm not real sure about that because i don't know if that's at the ultimate yield strength or the uh sorry the ultimate strength of the yield strength then there's poisson's ratio for all these materials and they're all on the order of a third somewhere in that region and then coefficient of thermal expansion we've talked about the thermal effects a little bit i don't i don't know that we'll actually get to any of those in the term but they're pretty straightforward as you imagine you have a delta t a change in temperature from unloaded to loaded conditions and then you figure out how much expansion in the piece as a as a ratio just like the strain is so it's a fairly straightforward number so you got to kind of be careful watch out with the units as much as anything when you use these numbers we'll do a we'll do a quick problem then using some of these new things we just got now we've got this this whole table some of them are just the the loads but then some of them are material material responses to those loads as material properties okay so let's imagine we have a shear block the shear block is an elastic maybe a polymer of some kind uh rubber that can absorb shear loads uh on so imagine we have a rigid plate on each side of this with the shear block in between we'll assume it's rigidly mounted to some surface and then there's some kind of load here that then causes the shear block to deform and it might do something like this the that'll cause this plate to come out some length because of the the shear response of the shear block and it might do something like that if we look at it on the side as that upper plate is subject to some kind of load that causes the entire shear block to deform to absorb that load something like that the reason it's a curved shape like that is because the the bonding of the shear block to the plate would cause the block to actually have no slope to it there and then the entire center section is absorbed by the slope there absorbs the the load so let's see given a couple things here except for the block 50 millimeters length 160 height 40 all those in millimeters the modulus of elasticity and you can see how they they painted it right on the side here 600 megapascals and this maximum deformation here is 0.8 millimeters that's the the same thing I drew down here inside so want to find a couple things find average shear strain what do I mean by average what assumption do you make seeing that word average we don't worry about the fact that the strain varies with location on there because remember strain is the angle in the material and it's very different at different points so just assume then this average deformation and that's then simply the angles there 0.8 over the original 40 I believe so that's that's very straightforward I'm going to give you something else it's a little bit more involved now that we have that because that's just the response of the material I want you to find the load P that caused that response in the material it's the load P that's causing the shear it must be balanced on the other side for static equilibrium there's also obviously other things going on to resist the the moment that those caused the couple but that's the mounting of it to rigid plates that will supply that we're not looking at that in particular so the shear must be that load over the area that's absorbing it we can find that because we know the modulus of the last rigidity so that will come from the fact that we've been given G we just calculated the strain we can find the shear once we find the shear we can find the load so I'd like you to do that as much as anything the difficulty is just making sure you've got the right area in the calculation there's a lot of shapes and distances and numbers here you've got to make sure you've got the right one so that's your task for a minute or two a lot of areas in these problems we've got to make sure that we actually have the right area in the calculation a lot of areas a lot of dimensions a lot of changes lots of units make sure you're all getting the same numbers lots of large si numbers in here mega pascals killer I think we've come across giga a couple times don't carry too many sig figs remember we're one we're going to have a huge factor of safety on all this too these are all just you know just everything here is based upon a fair amount of values notice that this state was a big round number 600 so to take the answer to any considerable precision doesn't doesn't really match what you've got going in you've got pretty approximate numbers going in it would be unnecessary yeah you don't need that kind of precision coming out okay and you have to be careful about that because if you send a dimension drawn down to manufacturing in your company and you have an extreme position on it because there's so many sig figs they'll try to build it to that and they'll charge you for that effort and if you put a lot of precision in it takes a lot of machining expertise to uh to finish that and they'll build you for it so we can find the sheer stress and then we can find the load from the sheer stress as long as you have the right area in there what do you have for the sheer stress have we got that number boy got it 12 12 what we make a mega Pascal which is what we had on there so you're going to be really good with the si prefixes in this class and now we've got the sheer what area did you use to absorb that sheer Travis what do you use a hundred so that's that's this side there Chris this sheer is it's actually that's all along the top of the along the top of the piece so you have to use the 50 by 160 and then you have to watch your units so the area is this higher top area there because that's the the area that's absorbing that sheer stress and you get for that 96 what units kilonewtons you also get these are all in millimeters here so that's a total of 10 to the minus six right there yeah so it should be 96 kilonewtons problems kind of wrap up what we've done in this chapter because this pretty much ends chapter three so let's see maybe we'll do a big one and then get out of class one right too we'll use some of those things that we just had we won't use all of them but that's up for you and I saw a lot of stuff here in chapter three so again a loaded beam held by two supports different lengths this one's 220 millimeters this one's 210 10 millimeters shorter each of those supports are the same both of them are 2014 t6 aluminum which is in the book at least I believe it is let me double check real quick yeah it's there so you can look up any any of the material properties you might need the beam itself is rigid and three meters long there's an 80 kilonewton load and the design criteria the design criteria is that under this load the beam must remain horizontal which means it must be placed carefully at the right location if it's too far to the left then this support will compress too much won't be horizontal and vice versa but goes to the other side now if we did that in statics oh there's the number I was looking for I did have another both of these are 30 millimeters that would be circular cross section now we we could have uh we could have done something with this problem in statics however what we all we would have come up with in in statics maybe we'll call it the pure static solution because we're now taking we're still in a statics course but now we're looking at deformable solids the pure static solution would have done nothing more than told us that the load should be equal and the itself should be in the middle of the beam that's all we would have gotten if we done pure statics that's not going to work because if there's an equal load in the two of these one of them will deform more than the other which one if the loads in both of these struts were the same which one would deform more then the beam would not be level the longer one because remember the the strain has to do with the original length this one's longer under the same load it'll strain more so we actually have four unknowns of course the the reactions in the struts but we also don't know where to place the load yet what's the fourth unknown there's actually a fourth and a fifth but they're equal if that's a hint uh what is the compression what is the response of these struts to those loads such that the beam remains level so del a equals del b but both are unknown so of course we need four equations what are those four equations it's a static class so the sum the force is better sum to zero the moments had also better sum to zero in fact you're going to have to do that two times i'm not sure of any other way to do it but do the moment some of the moments about each end uh so that's three equations and the fourth equation is actually this one here where the deformations themselves are the same and that's something we never had in statics but now that we're looking at the deformable solids we do have that kind of thing as a way for us to solve these problems in statics this would have been an indeterminate problem more unknowns than we had equations remember we only have three equations in statics in um now that we have the deformations we have other other ways we can make these calculations as needed so so take a few minutes and come up with uh where we should place the load and what the deformations are at least set up the equations um they're actually not that difficult to solve really so of course is that then you do the sum of the moments about each end each one of those will eliminate one of the knowns unknowns and have the other and this stress of the deformation equation actually takes a slightly different form would look something like that since they both have the same cross sectional area we don't need a subscript on that but for each of the two deformations we know from our definition of e stress over strain that we can then solve for the deformation as those that'll give you four equations and four unknowns you can actually I believe yeah you can get x right from this one well no not right from it actually the uh the two moment equations are going there when you write the moment equations one of them will be in terms of f a one will be in term fb you make that substitution the f's cancel leaving just the x then you can go back to either equation and get the forces in so set up the moment equations put them in here and uh we'll leave it at that because after that it's all just algebraic make sense set the two moment equations that'll be in one in terms of f a one in terms of fb you can put those both in then actually solve for the remaining uh solve for the x from that make sense a little bit when you set up the other moment equation looks like everybody's getting the pieces of it some of the moments about a a moment caused by the load must be opposed by the moment caused by b so there's your first equation solve that for fb and put it in the deformation equation well how'd you get those how'd you get what's this 10.6 oh that's the elasticity i don't know john he's in 2014 no longer my favorite student because of what you did i'm sorry you cannot claim that status the moment equation of b oh i see what'd you do like all summer what yeah the wrong page english that's right okay a big life for you it's a flat world out there so you're going to be cognizant of all systems and cultures and the sixth uh pascal and the log of the ratio of sorry boson ratio 0.35 uh i want that because i want to find the final diameters i hadn't put that up yet but that was one of the things to look for my final diameter and the only boson's ratio for that once you find the axial load not the actual axial strain these kind of go into there that'll give you a single equation in x as the only unknown all those other quantities are known but i don't know why you want to do two moments yeah are they dependent uh are they dependent uh no no x is an unknown you can't use them to solve for either one and you can't use the two to solve for you that's the difference because this x remains an unknown this is two equations three unknowns so they're not dependent equations they're independent in this case if if uh we didn't take into account the deformations then they are dependent equations but you don't need one to solve for either of those anyway you'd use the sum of the forces and the uh sum of the uh and one of the moments maybe is there another way to do it this is use second order yeah because when you put it in here then then you have your third equation i have to make this to get on the class question i guess or come into class questions one day i actually started in the chapter though work did you do it yes i use actually algebra that's x yeah it's i i have so it's going to have to be a little bit past halfway because of the extra length on a there's not a lot of extra x but a little bit so it's going to have to be a little bit past um past the midpoint we're getting right near the end so i'll give you the answers you don't check you can get them this should all lead to x equals 1.53 meters so it's only barely past the midway you got that too chris good once you've got that you can go back into these first two equations and get the loads as 39 1 on a of course that's a total load of 80 so not a huge difference make sure you can get those and then i'll leave it to you to come back on monday with the final diameter of leg a you can do either leg but uh it's the same calculation and you'll need to push on the ratio for that