 in pure bending. We're basically looking at beams. Remember there's a couple requirements of the problem. One is that we'll set X this way, Y that way, and then Z out of the board. Our problems are subject to a couple restrictions. Remember what any of those happen to be in terms of both the bending low at the moment and in terms of the cross-sectional shape itself, which we're now getting to the point where we can size that, get the right size of cross-section, right shape even, to help us get a beam that can withstand the loads. Remember with the crisp. Symmetric, not symmetric about the X axis. Symmetric about the Y axis. So if we look at the end here, looking down the X axis, we'll see the Z and the X like that. Our cross-section must be symmetric about the Y axis. It doesn't really matter what it is. We can do all kinds of shapes as long as it's symmetric about the Y axis. That's our only requirement at the moment in terms of the, well it'd be nice if it actually looks symmetric wouldn't it? There we go. And then the only other part of the concern with that is if you remember we need to find two things with this cross-sectional. We need to find what we call the neutral axis. We'll review in a second why we need that. But it turns out it goes through the area centroid anyway. So it's just a matter of finding the center of that area. And then from there then we can find the bending. Because the stress as a function of Y, if you remember, was whatever moment is being imposed. And there is another one of our restrictions. The moments must be in the Z direction. Remember using the right-hand rule. That gives us a moment that's purely in the Z direction. And then we found that the stress was a function of that moment, the position across the cross-section and then the moment of inertia, area moment of inertia. And so if we look at some cross-section or some profile, some stress profile, we saw that for this kind of bending the top is of course in compression, the bottom is in tension and it was linear between the two. It was zero at this place that we call the neutral axis, which goes through the area centroid. So if you remember, that's pretty much where we left things off. Chris, you okay? You look like, I might have put something off there wrong. Technically there's a minus sign here because positive Y was up, but that gave us a compression for this positive moment. Remember this was our positive, our convention for positive moment. So if we had positive moment and positive Y, we needed the minus sign to tell us that that was in compression above the axis and tension below the axis. But tradition has it that the minus sign is often just left off because it's so obvious just from inspection of the problem, whether it's compression or tension. You just look at it and you're going to know what's going on with that piece. So let's do a quick reminder of how we find this centroidal axis because that's where our stresses are zero and then we know they're linear in between and that they're a maximum where we're at the maximum distance from that neutral axis. And that was the symbol that we gave the letter C. For some reason they picked that to represent the maximum distance from the neutral axis. And remember this neutral axis might not be right in the middle. So the greater distance might be below the neutral axis rather than above. It depends on where the neutral axis falls. It depends upon what the cross section looks like. So we'll remind ourselves how to find that neutral axis because we're going to need that with what we're doing next. Oh, there is one other thing. If you remember, this stress distribution actually came from the strain distribution which we established first. However, because our material is homogeneous and isotropic, isentropic which means it's the same material throughout. Same density, same material properties and everything. The stress distribution led to the strain distribution, sorry the other way around. Stress distribution which is what we found first to be linear led to the stress distribution. But that's also linear because they're related by a constant. If you remember they're related by Young's modulus. And since that's a constant, then if the strain distribution is linear, so is the stress distribution. We're going to need that in a second when we get to that. So we've got a quick problem here. I want to find the neutral axis for a cross section made up, just exactly the type of thing you might build out of three boards if you are looking to build some kind of support beam of some kind. All these measurements in millimeters. And we want to use that to find the neutral axis which just by inspection we figure is going to be someplace around there. Give or take a little bit. So when you get an answer for the neutral, where the neutral axis should be, you can look at it and check and see if that's even reasonable for what you might expect. So just to A things a little bit, I'm going to label some of these dimensions. We'll label this. This will be piece one and that is width B1 and height H1. Just, I'm going to go through this very quickly because this is just a review, just a warm up from spring waves. And I'll call this distance H2 and this distance on the second one B2. And you'll see why in a second why I did it that way as a refresher for that too. So if you remember what we need to do is find the position of the neutral axis which I believe we call a y bar which we find by looking at each of the individual pieces and calculating the product yi which is the location of its own centroid, the individual pieces. I only have two one here. And then sum that over the total area, total cross sectional area and that'll locate the y bar for it. We only need one other little piece. We have to pick some arbitrary reference line. So I'll choose it from the top but you can choose it from anywhere you want, anywhere that's convenient. If you'd like to do it along the bottom of this upper plank, no trouble. You want to do it from the bottom of being no trouble. All you're trying to do is locate the position of the neutral axis and once we find that, it's fixed to the cross section. We don't need that reference line anymore. We just need the reference line to get going a little bit as we plan through things. All right. So I think it works well if you make yourself a little table with just less chance of making some kind of error. So we have these two parts here. For each of them we're going to need to locate the centroid. For each of them we need the area and then for each of them we need the product of those two. So it works nicely if you do it in the table. The location of the centroid of the first piece, of course, is right in its own middle which is one half h1. Remember that's locating the centroid of piece one with respect to our reference line. Piece two is also right in the middle. So that's a distance from our reference line of h1 plus one half b2. And you can get h1 and b2 from the reference line from the drawing itself. I'm just trying to lay out a skeleton solution to this because it's just a reminder for us. So the area a1 is b1h1. Remember b1 is not 200. That's the inside dimension of this channel beam that you made out of some planks. b1 is going to be 20 millimeters plus another 20 on each side to get to the outside. So be careful of those kind of things. It's, excuse me, really easy to slip off with that. And of course the area of piece two is b2h2. However, there's two of them. So we can just double it or you can call it number three if you want and just add it on separately that way. It doesn't matter. It all comes out to be the same anyway. And then we can multiply those and I'll actually put those values in so you can double check them later. See, this comes out to be 75,000 millimeters cubed and this one comes out to be 600,000. And so we total that up to 675. We also total up the area because we need to divide that 675 by the area and the area is 12,000. So we're all done. We got all the pieces we need. Remember to watch your units. Be careful with it. Don't rush through this thing. Notice this is just a personal habit but I put a little space where typically in America we put commas. In lots of European countries they see commas as decimals. So it's very hazardous for us to use commas if somebody that we might be working with in Europe would see that as a decimal and have a completely different number for their design. So if you just leave a space, you get the benefit of that grouping by 3s that keeps the mistakes down but there's not a misinterpretation of what a comma or a decimal might mean. Alright, so with that we get, we divide those two get y bar 5625 millimeters. Let me divide those two. And that's right about where we'd expect. So we now know that y bar is 5625. We're down that far from the top, our arbitrary reference. That fixes the neutral axis right on our cross section and we don't need to worry about where the reference line was anymore for next. Now we know where the neutral axis is. One last thing we need to find then, we need to find the area moment of inertia of that cross section with respect to the neutral axis. To do that we sum up the individual moment of the inertia with respect to their own centroid and then we apply the parallel axis theorem because the axis we're concerned with, the neutral axis is probably not going through the centroid of any of the pieces. So again a table might help. As you do that we've got two pieces here. We need to find out their centroidal moment of inertia. That's in the front cover of your book. Just open that there. You'll find the shapes for the different regular solids and certainly rectangles are in there. For these pieces it's 112d1h1 cubed. You've got to be careful with the other one because as it's drawn in the book piece two is 90 degrees from the picture in the book. If you look at it in the x axis or just take it in the y axis direction but that becomes h2b2 cubed just because the orientation of those rectangles are 90 degrees to each other. For each one of those we need to figure out what d is because that's part of the parallel axis theorem, the ae squared. So in this case I'll just write down the numbers. You can double check where they get them. Remember this is the distance from the neutral axis that we just placed to the centroid of each piece. So that distance there is d1. This distance here is d2. As measured from the neutral axis we don't use the reference line anymore. We just use the neutral axis. By coincidence they happen to be the same in this case. It's not usual because remember it had to do with all the widths and the lengths and all kinds of things from that. It just happens to be the same to the two of them. We'll have units of oops to the fourth. Now we can add i plus a which we already have plus d squared to now get the moment of inertia of each piece with respect to the neutral axis. Then we just add those two up to get the moment of inertia for the entire cross section. So these numbers turn out to be because we're at millimeters they turn out to be pretty big. 11.8 times 10 to the sixth this is millimeters to the fourth and then two times because there's two solids two portions on either side 11.4 times 10 to the sixth. You'll have to double check it. My recollection is that maybe this first six should be a 7. So double check that number. Now we'll get it right when we take this again next year. And those add up to be 34.6 10 to the sixth millimeters to the fourth. And now that gives us the i that goes into there. The y is measured from the neutral axis and if we had something given low then we could figure out the stress at any point across the solid. Look a little familiar now we kind of flew through that in a sort of a skeletal way. Phil are you checking whether this is a 6 or 7 for me? I know that it is a 6. Okay I know that one's right so yeah I guess so if we look at that that'd be right. Okay we're gonna need that again coming up here of the finding the moment of inertia and the location of the neutral axis for a cross section of compound shapes. We're gonna need that and we're gonna need these two ideas as a reminder for what we step into next. Now with all that remember one characteristic of the beams that's held true throughout all of this bending that we've been doing is that it's all the beams are all made of one material no matter what the shape of all these different pieces are that are put together they're all the same material. There's certainly the possibility that we might want to do something other than that. If we have a beam in simple bending like we've looked at before we know that the top part is in compression the bottom part is in tension. You also know that wood is very good in compression and terrible in tension so you might say well why don't we make the top part of the beam wood because it's cheaper it's easier to work with it's very attractive in a structural setting especially a home it's great in compression maybe we'll make the top part wood and the bottom part maybe steel or aluminum or some other material and the beam could also be lighter that way so there's several reasons we might want to do this any one of those might be worth it to you if you ever become a structural engineer. The trouble is when we look at the loading then in the piece remember we first established that the strain the normal strain in the x direction is linear then because those two materials were the same we also then established that the stress was linear in the same way because of the constant Young's modulus that related those two this is only if the beam is homogeneous meaning the same material throughout which is not the case if we start looking at a two material beam so the trouble comes now that the strain will be indeed linear that doesn't change that was only dependent upon the cross sectional area this came from the fact that stress and strain are related by Young's modulus if we have a different Young's modulus for the top that we do for the bottom we can now expect some kind of stress distribution that may look like like this wherever the neutral axis is we know that it's linear through there let's see here's here's our second material across there so we know it's linear through there but now we have a different material that would have some different type of stress and have its own distribution that would look something like that this by the way this this little sketch is assuming that e2 is greater than e1 if it was less than e2 is less than e1 then the stress distribution of the top would be greater however we can't just calculate that by applying the new the different Young's modulus because the trouble is we don't know where the neutral axis is for this distribution it might not be at the same place it was for the uniform cross sectional beam so we need to do something to establish where this is then when we've got that we can apply this linear distribution depending upon what e is on either side these are called composite beams because they're made of two different materials we're assuming that the whatever is holding the two materials together is perfect and will not fail itself we'll look into that a little bit couple weeks so whatever is bonding these two together is as most things are in my classes perfect so we've got to figure out some way now to handle this so that we can figure out what the stress distribution is thing is we just don't know where the neutral axis is for that distribution because this is a completely different beam so here's what we're going to do a design of composite remember prismatic beam prismatic is that deal where it's a it's symmetric about the y axis so we have this beam we'll just start with this simple shape like this one where at some location not necessarily right in the middle that's just too convenient I'll put in we'll change the beam into a made of two different materials the top one with some young's modulus and the bottom one with some young's modulus because remember that's what determines what the stress is the strain is still linear the stress is no longer quite linear well it's linear but it's disjointed leaf linear I guess if you will so here's what we're going to do everything we've done so far has been for beams of homogeneous material the same material throughout so what we're going to do is going to take that composite beam turn it back into a beam of homogeneous material and then analyze it like we would before so we've got we've got this beam here let's see we'll call it a width B and we'll keep that on the top then we'll take out the material at the bottom that's it's one or the other that are giving us trouble I'll assume in this case that the young's modulus for the lower part is greater so I'm going to take out that stiffer material and replace it with the original material that's not as stiff because it's not as stiff I'm going to need a lot more of it so I'm going to put in a whole bunch more of this material and now it's entirely homogeneous and essentially equivalent to the beam we're trying to design of two different materials so now this beam is entirely of the material with stiffness A1 analyze that we've been doing for for a week or two now depending on what you did over break which I'm sure was homework trouble is how big do we make this how much wider must this be then it was originally so that we have enough softer material in here now to replace the stiffer material we took out turns out that it's a factor that we call n times wider where n is simply the ratio of those two so if the bottom was twice as stiff we take it out use the less stiff material and use twice as much of it to get the equivalent ability to withstand the stress you can if you'd rather go the other way leave the stiff material in replace the stiff the softer material with the stiffer material and no trouble you just use less of it so this would be then be over n actually I don't want to hatch that pink because that was my one material so this is my two material you can do it either way it doesn't matter it's going to come out with the same results and you can double check that if you want now we take this new cross section I don't want to be two at once so I'll erase that one we take this new cross section and we now find out where the neutral axis is for this cross section from that neutral axis we now find out the moment of inertia and then once we've done that then we can find what the stress is as a distribution as a function of y position across the beam with respect to the neutral axis which is a problem we we started the week before break and we just reviewed again today no it does matter because it depends upon what the original shape was it's still fundamentally rectangular and it depends upon what the materials were because that tells us how wide the new part is we've taken out this stiff material here on the bottom so that the entire beam is made of the softer material so we needed a whole bunch more that softer material in the bottom to represent the stiffer material for which we're designing we'll do a problem we'll step towards that sometimes it's your way to actually go through this all right so let's imagine we have a beam that's got an aluminum plate on the top and wood on the bottom now I'm depending on what loaded carries that may or may not make a lot of sense but it might be aesthetically pleasing to have this wood on the bottom so we've got the quality of the aluminum across the top but the aesthetic view the beauty of wood if you were a structural designer for a great Adirondack camp your customers are not going to be real happy if the beams are aluminum they're going to be a lot happier if the beams are wood so we'll design for that kind of wood that kind of beam so let's see the Young's modulus for wood is 10,200 ksi for the aluminum it's 1700 so significantly wait I got those wrong that doesn't make any sense sorry reverse 10,200 for the aluminum 1700 for the wood that makes more sense and the dimensions here the aluminum is a quarter an inch the wood is eight inches the width three inches okay so that's our pure cross section looks like that wood on the bottom aluminum on the top we're going to take out one or the other and replace it with an appropriate amount of material we can take out the aluminum replace it with wood and we're going to need a whole bunch more of that wood or we can take out the wood replace it with aluminum we won't need very much of that because the aluminum is so much stiffer doesn't matter which so I'll take out the aluminum and replace it with an appropriate width of wood so now this cross section is all wood now remember this isn't the final design isn't going to be what it looks like but this is what we need to do to to make it equivalent for our design purposes so the bottom is the wood we left that in it's not quite the scale it was supposed to be that white maybe I'll redraw that just so it's not confusing that remains the same we replace the top with an equivalent piece of wood that's wide enough to maintain to withstand the same stress of the aluminum wood how much wider must this be then it's going to be n times 3 inches where n is the ratio of these two you can do it either way it doesn't matter just make sure that what you're doing makes sense in the end this comes out to be 6 so the width up here will be 18 inches that we assume to be an equivalent being it's not a very attractive one it's not what you're going to install for your customer but it is one for which we can figure out where the neutral axis is we can figure out what the moment of inertia with respect to that neutral axis is and we can figure out what the stress distribution is all right so we we need to to figure out where the neutral axis is all again pick a reference line from the top from where I'll calculate everything now all the calculations are for this new cross section first thing we need to do is figure out where the neutral axis is then once we've established where that is we know it's going to be somewhere up near the top here because of that extra area of the top and keep my numbers straight I'll call that piece one this piece two we can do like we have before figure out where y is the area is two some the area and some that product for piece one it's centroid with respect to our arbitrary reference line is halfway across the quarter inch thickness so it's an eighth of an inch point one two five that's in inches its area is what eighteen by one quarter which is fifty six twenty five I think all right done no I looked at the wrong one four and a half it's the product it's fifty six twenty five all right piece two it's centroid right dead center which is four inches down through the piece but we're already a quarter of an inch down so this is four point two five its area is the eight by three that's twenty four so we can add up the area twenty five and y mark or y times a for that individual piece is a hundred and two we know this to be them hundred and two five six two five the neutral axis then is those two things divided we figure it's got to be two or three inches down here that looks like what it's going to be actually turns out to be three point five doesn't hurt to carry a little bit extra figures fill you love these problems carry through some of the significant figures until the end so that's why bar we know to be just under three point six inches from our arbitrary reference line now what now we figure out the moment there chef with respect to that arbitrary reference line which is the sum of the individual ones with respect to their own centroids plus the parallel axis there for each of the individual pieces we need to find its centroidal moment of inertia that multi moment of inertia with respect to its own centroid then since we're using the parallel axis there we need to then move it over a little bit and then that's the last piece we sum up there let's see if I have all those numbers they're all rectangular so it's the 112th BH cube just make sure the orientations right this first one it's point oh two three four four and that's inches you can check these we're going through these kind of quick because these are just geometric calculations that you can make this one comes out to be 128 D that's the location of the centroid of the individual piece with respect to the neutral axis that we've now placed and that is for this first one three point three point four seven three four remember that's that's down three point five nine nine back up a quarter of an inch three I said yeah I just like four then they don't do that on my paycheck three point four seven you guys are awake after break and the second one is point six five one three now it doesn't matter if D is above or below the neutral axis we don't care about a minus sign being on that because in the end we use D squared anyway so it doesn't matter whether we're above or below one of the few times we don't have to worry in the least about minus signs okay so ad squared a we've already got from the first table d squared this becomes 51.3 by the way that's inches this is still inches so 51.3 on that last one 10.2 oh wait sorry that's just a that's just a d squared not i plus ad squared oh it doesn't change my 51.32 for that one and 138.2 you can check all these for testing on this kind of problem it'll probably be a take-home one because there's so many places you can make little tiny mistakes that it's virtually guaranteed in the test the time of a test and the pressure of a test you're just going to make some little mistake that really doesn't indicate you're you're not getting the idea anyway so I'll give you take home see if I can ruin your weekend 189.2 that's the moment of inertia with respect to the neutral axis now before we finish this problem because that was that was the easy part I mean you could have done that the week before break what we need to do now is calculate the stress distribution in these pieces so let me erase this part we'll apply some load to this and then figure out what the stress distribution is remember the straining distribution remains linear because that only had to do with the cross-sectional shape didn't have to any new material the stress that comes from that has to do with the material so let's imagine a moment was applied of 6,000 pound inches in our usual sense 60,000 yeah 60,000 and then from that then we can figure out what the stress distribution is so here's our beam now we've got all the things we need to remember the stress is a factor of that moment of inertia but here's the little piece we need to be careful with we now know the neutral axis to be right about here so the strain distribution is linear just like it was before it's just now we know where the neutral axis is and it's not where we would have placed it with just that simple rectangular beam so this the strain remains the same still linear still zero at the neutral axis it's just now we know where the neutral axis is of our equivalent beam the stress distribution is a little bit different the stress distribution well the bottom is just the same remember we had wood on the bottom and we made the beam entirely of wood so that's still going to be the same that's still going to be just like it would have been if we calculated it before those very same numbers that we've got there so we've got a applied moment of 60,000 pounds inch see now remember is for the wood is all the way down to the bottom from the neutral axis that number comes out to be I don't have it actually written out you're gonna use it for me eight and a quarter because that's the entire beam minus that neutral axis 3.599 they we had 8.25 minus 3.599 and that's inches that's that's how far the the farthest away piece of wood is from the neutral axis and divided by the 189.5 inches to the fourth and we're left with good units PSI and it's 1473 PSI in the wood we know from the problem is compression I guess I didn't put any direction on the moment so evidently it's being bent the opposite way as usual all right so that now gives us the distribution of the stress in the wood now the distribution in the aluminum is a little bit different we do the same basic calculation doing the aluminum looking for the maximum we know it's linear in between it's the same moment times C for the aluminum which is that little bit left over in fact that is Y bar well we already figured for Y bar over I but we need to multiply that by that factor and that we had because if we don't we'll just get the same distribution we had there but the aluminum can withstand a much greater strain than can they look so we have to put back into the calculation that that actually was aluminum so n was 6 m is 60,000 we already know what the units are going to be we're okay with that Y bar is all it goes in here 3.599 and I is the 189.5 and we already know what the units are going to work out to because the hen is unitless and we get 68.37 PSI which is six times greater than it would have been if that was just a wood piece there and that's now linear all the way back to the neutral axis that's a little hard to draw much greater compression ability of the aluminum which is part of the reason it may have been in there might have been that for that low the wood would not have been sufficiently strong so you put in a little bit of plate across the top and absorb a lot greater stress and then put the wood on the rest on the bottom for perhaps aesthetic purposes or for weight purposes and we've got the picture there now all right it's this little step there the students often forget you have to apply that factor event and remember you can do this the other way if you want n to be the lighter one over the stiffer one you can do that too you're gonna find the y'all you still all come out with the same calculation it's you know either way you just have to be consistent with it all right any questions for you do your own let's see if you're gonna be a hard one or an easy one actually they're they're about the same get to linear let's start with a fairly straightforward only it's a bit different than what we just had all right we've got a beam here not to scale as usual that's made up of two materials on the outside is brass and the inside is steel again might do that for aesthetic purposes brasses a lot of times a much more attractive metal than steel might be so the brass is point four inches thick the steel point seven five symmetric the other brass pieces the same and all of three inches high all right the load is 40 kip inches the young's my dose 29 6 6 yes I for the steel 15 times 10 to 6 all right find the stress the maximum stress in in either now it's not a calculation we can do straight away because our we can only analyze a beam of uniform cross homogeneous material so we have to take this one and replace it with an equivalent beam entirely new cross section figure out the moment of inertia from that we need the moment of inertia to figure out what the stress is