 Okay, we'll just do a real quick one, just to remind ourselves how fast this can go, especially if we use those relations. Pay attention to what the loading is, what the constraints are, and things can go really quite quick, sometimes with some of these. So, we've got a simple one here, with a couple point loads, one distributed load, just to make things nasty. Well, we're fine, I don't think we can tell that one, nasty. Okay, there we go. So, 20 kips, kips there, and 1.5 kip of a foot, and then the distances between these six, eight, 10 and eight. Six foot, eight foot, 10 foot, and then there's 10 piece there, eight foot. All right, nothing more than we were doing last fall, nothing new here other than just a little bit of warm up, plus we're gonna speed things up a little bit as we get better at these. Of course, we can determine that these reactions are 18 and 26 respectively. Do that with your statics, double check. Remember, to do that, you need to replace the distributed load with an equivalent point load, but then, once you've done that, you're done with that equivalent point load. A lot of students will leave that there, and it's not a real load, it's just a convenience, if nothing else, that allows us to actually calculate the reactions. We only put in that equivalent point load for finding the reactions to find the shear moment diagrams then. We take it out, and we're gonna find that the shear moment, especially the moment diagram, the moment diagram gives us a very clear picture of what the beam itself is doing, as we'll see in a couple weeks. But then, to finish the shear moment diagrams, we only need to use some of our relations and go quickly through using the relations to find the points as we go through these. All right, as you remember, for a reaction, an upward reaction here, that's going to cause us to actually go up by that amount, then nothing happens to the next spot, so it's very easy to do most of the shear diagram. This is actually quite easy. We just take jumps in terms of the size of the point loads, and then nothing else happens until the next point load. So as long as we don't have a distributed load, these are really pretty easy to do. So we took a jump up at 18 for the reaction, a jump down of 20 for the point load, then we take another jump down of 12, so that takes us to minus 14 here, minus two there, 18 there, and nothing else happens until we hit the next point load. We jump up by 26, so we're at 14, up 14, up another 12, puts us about right there, and then we have the distributed load that takes us out to a free end. That free end's not going to have any residual shear, so we'll finish at zero, and we can get the shear diagram just that quickly. And remember, take that point load out that we used for the distributed load here, because it does not apply in the shear diagram. I've seen over the years, a lot of students will leave that point load in there, and then in some way try to incorporate it in the shear diagram, but that point load doesn't exist. It's not a real load. We need to get back to the real loading that we have that distributed load there. And then for the shear moment diagram, well, we can look at the loading and realize we've got this pinned end here. Since it's pinned, it's free to rotate. There won't be any moment at that point. We can sort of say the same thing for the other end. It's a free end, so there can't be any moments in the beam when we get to that point. So we know those two points exist. Then the rest of it is just a matter of putting little pieces together. We know, remember, that the shear equals a slope of the moment diagram at any place. We've got this positive 18 slope. So we have a positive 18, sorry, we've got this positive 18 shear. We've got a positive 18 slope. And the area there will give us the change in M, which is 108. So we go up as high as 108. We know it's linear down to there. Then we have a very small negative two slope, but it's easier to take the area, find the change then to the next point and then just connect them with a straight line. And so let's say we have two times a is 16. So we come down to 16, it takes us down to about a 92. And then we just connect that with a straight line. Not much easier way to do it than that. Then we have a minus 14 slope. We know that area is, one minus 14 times 10 is 140, takes us down to a minus 48. So that would be right about here. Give or take a little bit. Again, connect it with a straight line. And then anytime we have a uniform distributed load, we'll have a straight line on the shear graph and a parabola on the moment graph. So just look at the moment graph. We know we're gonna finish at zero. We've got a lot of positive slope and that positive slope decreases until it hits zero. And the parabolic shape that gives us that is just a simple curve like that. We're all done. Then remember what we're paying attention to. Now, starting today really, is what's the maximum shear and the maximum moment? And in a little bit we'll look at where those are as well. Because what we need to do is protect the beam from failure from the maximum shear, which in this case is the 18. It's only for a little section of the beam, but typically we just make the beam uniform all the way across. It's a very cheap, inexpensive way to do it. So we'll protect the beam for this section, but it'll be protected all the way across because that's the worst section. Same thing for the moment. We're gonna have to build the beam to protect it for this one spot of greatest moment, which means we're grossly overprotected at other spots, but it may be so cheap to do it that way that that's the preferred method. All the beams in this building I'm sure were just stock beams, all uniform in a cross section. Probably all match each other. It's just a very cheap way to do it. Don't forget, we don't protect for just that one little spot there. We know where that spot is, know what its value is, and then tack on a big factor of safety anyway. So we don't design for this spot. We design for that spot plus a whole much more so that there's virtually no chance of failure. You know from the history books that when buildings fail, they tend to fail catastrophically. So we're gonna try to avoid that, especially if you go into structural engineering. So now what we're gonna look at is just what do we do to make a beam that is protected for these kind of failures. So this section then, oh my goodness, that's pretty chopped. That's a new color chopped, that's very pretty. Look at her look, she's smiling, she likes that. So what we're gonna do is design for prismatic beams. To start with, we're just going to determine what is it we need to know about the cross section so that we can do this economically. And then in a little bit, a couple weeks, sometime after break, we're gonna actually look at the deformation caused by this. We know these beams bend. We're gonna use that today to establish things, but we're not gonna quantify the deformation yet. Kind of like we did in the other things we looked at. We look at the loading in the beams and then we look at their deformation. So first thing, I just need to simply define for you what prismatic means. We're going to look at beams of any cross-sectional shape. Could be circular, you'll see that we can handle virtually any shape. So we're gonna let, for the most part, that will be our axial distance, our axial dimension. We'll put x down the length of the longest dimension of the beam. Typically we'll let y go up and then using the right-handed coordinate system, z will come out from the side there. So our prismatic beams are those defined such that they are symmetric with respect to, that's what WRT is, symmetric with respect to the y-axis. So if we look at the cross-section, we'll see our, if we look right at the cross-section, look down the x-axis, there's our y-axis going up, our z-axis coming out. We're gonna look at any beam that is symmetric about the y-axis. So any rectangular beam would certainly satisfy that. We could look at triangular or circular beams, but as most of you know, the most common beams in construction are i-beams, and those are symmetric about the y-axis. So they satisfy our definition of prismatic beams. And we could look at any other shape thereof. So that's our first requirement for prismatic beams. Our second requirement is that the moments, all these bending moments in the beam are exactly as pictured are in the plus or minus z direction, which means all of the loads, the transverse loads, if that's what's causing the moment, are in the y direction themselves. Exactly what we've got here. All of these loads are in the y direction. They cause bending that is all in the z direction. Any one of the typical type of loadings we've looked at before. At any place we have bending like that, either plus or minus, depending on what it is we're talking about. What that's gonna cause for us is when the beam deforms, when this beam actually bends under these loads, and this one would probably do something like that, greatly exaggerated of course, all of the deflections are in the plus or minus y direction because of these z moments. So that's our whole setup for the design for prismatic beams, and it's now that we're gonna actually talk about these cross sections. What do we need in terms of the actual shape and dimensions of the cross sections to help us protect the beam against those moments? And the shear, we just need enough area to avoid shear failure. Here we're gonna actually need particular shape because, well, you'll see why in a little bit. All right, that's the setup. Is everybody comfortable with that? Pretty straightforward, pretty simple. We're gonna be looking at very regular geometries in this way, at least certainly for the foreseeable future. So let's start the setup and see where this takes us. That'd be a jock. Leave it ghostly apparitions. All right, so here's the deal. We have some beam in some kind of loading. I don't care what. We're gonna be doing all kinds of different things. I don't have a concern here. We can keep it a simple picture to start with. Some kind of central loading that's going to cause some kind of bending like this. I don't care what the specifics are. Just for now, to set things up, I want this kind of curvature because what I'm gonna do is now take a look at a small piece of that bending, that bent zone if you will, and really try to set it up. So we'll take a little elemental piece of the bending there and really blow that up so we can take a real good look at it. So we have that little, that sort of trapezoidal region here. Now it's got a little bit of curve on the top, a little bit of curve on the bottom. So that's our greatly exaggerated piece of the beam there. I hope you appreciate it. It's greatly exaggerated. So that's a prismatic beam here in our typical bending mode that we're gonna be looking at. Now obviously, I hope obviously, the top is being compressed because of this bending and the bottom is being stretched in tension because of the nature of that bending in a straight, unloaded state. Of course, there's no deformation, but now there's some kind of compression at the top, some kind of tension at the bottom. Well, the only way for that to occur is if there's some place in between where we go from the compression on the top to the tension on the bottom, somewhere in between where we have zero, we'll call that the neutral axis. Neutral axis. I'll just draw it through the middle there and put a little NA on it. It stands for neutral axis. And that's where the load is... Actually, I don't want to say the load. I want to say that's where the bending is zero because we could still have shear there. So the load might not be zero in its entirety, but we're only looking at the bending right now. So there's some neutral axis. I don't know where yet. I know where. You don't know where yet where the bending is zero. It's compression on the top. Somehow it varies to zero at the neutral axis and then it becomes tension on the bottom. It may be linear in between. It might be parabolic in between. It might be constant compression and also the shift to tension, but we've got some place in the neutral axis where we go from compression to tension as we go down through the beam and we need to find that spot. All right. Let's define for us a couple more things. Let's see. Let's call this entire angle here. Let's call it del theta. That's this angle that we're subtending here with our piece that's delta x in length. Remember, x was the axial dimension. So we have a little piece delta x here. That's actually the length of the neutral axis. If the neutral axis has zero bending, then it's the original delta x in length. One other thing we're going to need is at this point there's some radius of curvature and I'll call that rho. We can find some at least local center of curvature and the beam has a radius from that point of a delta r. So there's our setup with that. All right. Everybody comfortable while we now will take another look much more closely at what's happening there so we can figure out what the deal is. So again, I'll take that picture. I'll blow it up some so we can really see what's going on. Oh, by the way, these spots that were originally planings remain planar. There's no deformation of those, at least not with what we've got going here because we don't have any axial looking here. So these are still planar surfaces on the edges there. All right. So there's somewhere there is then our neutral axis. We don't know where yet and we don't know what the distribution is above and below that. But we're going to use that. That's very important to us. That's got a length delta x long because it's undeformed. It's bent but it's not compressed or intention like all of the other parts of the beam are. So we're going to measure our coordinate distance y from there. Remember, y goes vertically across the cross-section. So we're going to measure it from the neutral axis up to some other surface. That's my positive y direction. That takes me up to some other surface that I'll call delta x prime in length. It's compressed as I've drawn it. It's a little bit shorter and it's a distance y from this unknown neutral axis. All right. What else are we going to need here? A radius of curvature of rho down to the neutral axis. So down to this intermediate axis of some kind, the radius of curvature is rho minus y. And as you imagine, we're going to vary y across the entire cross-section and see where we get. So that's our geometric setup there because we take a look at all parts of it as we go across. Comfortable with that, John? Or just have a last day before spring break heading? Okay. So let's see. Let's look at the normal strain. Remember, the normal strain defined as some deformation over the original length. That's just the original definition of the normal strain. For us here, it's going to be, it's going to be, let's see, the deformation is delta x prime minus delta x over, oops, leave a little space after the equal sign. Delta x prime minus delta x over delta x, because that was the original length of my little section here. But I'm going to take the limit of that as delta theta goes to zero. And that will, what I've got is, without that limit, I don't have a rather macroscopic strain, but I don't have the instantaneous strain, if you will. All right, so let's see. Okay, delta x prime, that's the arc length of the radius times the, that's delta x prime. Agree? That's just arc length delta x, or delta theta, of course, being in radians. Minus delta x, which is rho delta theta, and that's all over rho delta theta. Cancels, we have a rho minus rho left over on the top. So rho cancels. We don't even need to limit anymore because delta theta canceled. So we get the rows canceled. We get minus y over rho. That's pretty simple. That's pretty simple. That's by golly, that's a linear distribution of the strain across the cross section. That couldn't be any much more simpler for Chris. And, of course, in terms of the normal stress, then, which is related to the strain by Young's modulus, we can then take out the strain, replace it with minus one over rho. And so we even have a distribution and we even see it's linear for the stress as well, which means if we look at some beam under some kind of loading, we know there's some neutral axis somewhere. We don't know where yet. And we know that, at least for this type of bending, that we have compression that varies linearly on the top and tension that also varies linearly on the bottom. So that's one way we can draw that distribution. Don't trouble us, we just don't know where the neutral axis is yet. But much to our relief couldn't be much simpler than linear. So for the bending shown, we have linear varying compression on the top, half of the, well, the top. I say half in quotes because I don't know where this neutral axis is yet. That's our next job. We can also say then, of course, that at the extremes, the maximum strain is at this very top surface up here. And that's not at y, but that's at c. Remember c is our convention for the maximum distance. It was always maximum distance from where our coordinate origin direction was, but now it's going to be from the neutral axis. We don't know where it is yet, but we're going to place that in a second. And, of course, we can also say the same thing for the maximum stress. It also varies linearly in the same way. So those are two pretty useful things. Combine that with the fact that it linearly varies in between. So once we determine where in the world this neutral axis is, we're going to be cooking with gas, as we say. Those deserve a box, don't they? You betcha. Okay, so we need to figure out where that neutral axis is. So that's the task our current task now. Alright, so we have to figure out what this neutral axis is. So we have some being like this with this unlocated neutral axis with this linear variation. And we need to find this neutral axis. It's a statics class, so the first thing we need to guarantee is that the forces sum to zero. Because there's compression above and tension below, we need to make sure that those two things balance each other, that we have no residual forces left over. Because if we do, we have then acceleration, not in this class. So we're going to make sure that the stress over the area will then be zero. So that's actually the sum of the forces because the stress is the force over the area and that now needs to be zero as we integrate over the area. So remember, the cross-sectional area. Okay, so let's see what can we put together now. Let's take that there. Now, this sigma x, we know it varies linearly from zero to sigma max at the extreme. So we're going to put that in for sigma x. The local strain. That's going to vary linearly between the maximum, not y, c, because we're using the maximum. Make sure I get this right. No, the y was okay. The rho is not. We want to be at c. Yeah, rho cancels out and get just c in there. Actually, rho doesn't have a part of it. We're all we're doing is setting up this linear variation. Yeah, okay, that's what we need. Now, here's the beauty of this and we're not getting to the answer. That is a constant. It doesn't vary anywhere in the cross-section because that's the maximum stress at the outer limit of the beam. That doesn't vary. Neither does c. So we can bring those out of the immigration. We're just integrating then y, da. Again, over the entire area. Yeah, that looks good. And that's got to equal 0 because that remember is the sum of the forces. Most of you should be tingling with anticipation. And what's there? Tommy? Because this should look very familiar. What is that? David? Let me think here. Something I have to do, polar moment of inertia? Well, not polar. Polar has to do with circular cross-sections. This is any kind of cross-section if we have a circular cross-section. But this is the first moment of area. First moment of pretty funny. No, it's exactly like I wrote. You come up here and try to write down here in the bottom. Okay. First moment of area. That's great because that's what we've been figuring for some time. We did that back in statics. So let's look at that a little bit and see what that means to us. If you remember we had to find this spot y-bar which was the location of the centroid of area. And that was defined as the integral of y-da exactly what we've got there over the integral d-a itself. That just came from this being over here so that we had essentially a y- integral d-a equal to the integral y-d-a. We're finding this location of the centroid essentially the balance point of the area. Now we just got that which means that this part there equals zero. That's what we just established over here when we summed the forces to zero. This all means then that our neutral axis passes through the centroid of the cross-section which we know how to find. We spent some time doing that in the fall we'll review it real quick today but that allows us to place this neutral axis and we now can establish the distribution of the stresses across the cross-section. So that's pretty useful to us. Again, deserving of a pink box because we know how to find the centroid for a lot of very regular cross-sections like just a simple rectangular beam we don't have to calculate anything we know where it is if we have a beam that's symmetric not only about the y but also about the z we immediately know where this cross-section is or sorry the centroid is so it makes things quite simple so that's our first static requirement for this class remember this is a class in statics itself the second is that the moments must also sum to zero so we're going to integrate some moment arm times the force but we've already got that x da that's df so that's a h little force times its moment arm and that's going to be equal to the moment across the abyss let's make the same jump we made before when we put this in for sigma x so this becomes y that x gets replaced by sigma max times y over c oh yeah from that the minus sign always pull out produced or lost the minus sign earlier well it's going to turn out it doesn't really matter because we know exactly what the directions are anyway so the tradition is the minus sign and the loop is even left off again minus sigma x sigma max over c is a constant comes out of the integral we're left with we're left with y squared da which again should look familiar this is the second moment of inertia or that which we typically just call capital I and we've got lots of those in the in the book there and that's with respect to the neutral axis so that's the moment of inertia as calculated from the neutral axis and we're going to have some compound shape so we're going to need all that little bit there as we put things together so where that ties these things up we get that the maximum stress is minus mc over and the local stress is minus my so because the stress varies linearly in between what we typically do now is don't even we write this down without even the minus sign being there because it's so obvious from the bending whether that's impressive stress or tensile stress because of the way these beams bend it'll be compression above and tension below for that kind of bending and if we have bending in the other direction then we have tension above and compression below with it being 0 at the neutral axis it's two different ways to draw those very same things you get the idea, you can tell from observation what kind of stresses you've got whether it's compression or tension but the biggest revelation with this is that the smaller the moment of inertia the greater the normal stresses as I goes down the stresses go up and vice-versa what that means to us and our experience is that an I beam in cross-section has a lot of area the neutral axis because of the symmetry we know it's going to go right through the center there's a lot of area for an I beam that's away from the neutral axis which means for the same cross-sectional area which we need to resist shear we have the same area but we have a much greater moment of inertia with an I beam because of this additional area away from the neutral axis as we'll remind ourselves here shortly doing another calculation of finding this this greatly increases the moment of inertia gives a greater I a lower normal stress so that's why we use I beams in construction could have done a rectangular beam but those are very heavy this is much lighter with a relatively greater moment of inertia so that's very, very useful to review I beams so real quick calculation with it we'll keep it simple for this first one in rectangular cross-section exactly what you use if you were building a deck this summer you would use 2 by 8 for the choices maybe just 2 by 6 depending and you orient them in that way so for find the maximum moment that causes yielding in other words design normal stress we're looking for is the yield stress that's the type of thing that's in the back of the book for different materials and we want to find what moment keeps us at or under that yield stress a couple of values to use just to run through the calculation we'll assume a yield stress of 36 ksi because of the symmetry we know the neutral axis goes right through the center and we'll give it a dimension of 1.25 inches that way and this dimension 8 tenths of an inch there for this real quick example what's the maximum moment we can withstand with that and we just have to calculate those just a quick run through as we do these got to watch the units of course yield stress I gave you is 36 ksi C I gave you that's 1.25 inches we just need I for a nice rectangle which is who will integrate that for me no one will nice rectangular shapes remember like this they're in the book so you just have to look them up this one happens to be 1.12 B H Q where B is the base and H is the height make sure you get that right when you look at the pictures of those in those tables in the book the beam has a width of 0.8 inches a height of not 1.25 but twice that 2.5 inches and that's Q and so we've got I we can put that in there simple as that couldn't be be a whole lot easier units okay k kips per square inch two of the inches to the fourth we've got another one on the bottom we're left with kip inches which is a good unit for a moment 30 kip inches just a quick refresher most of this well all of the shapes we're going to need are available on the tables in the book what we're going to have to do we're going to have to look at a lot of compound areas because if you look in the book for the tables of moments of inertia the I-beams not there neither are a lot of the other types of shape it's pretty common to use built up box beams especially when you're working with wood it's a really cheap way to get a lot of moment of inertia as well as a nice aesthetic look when you have these it looks like big solid beams but they're really just built up beams and cross sections it's not uncommon too you might want to run some kind of pipes or wires or something down those beams just to make the house look a little cleaner in appearance so let's let's review then how to find the neutral axis and the moment of inertia with respect to that neutral axis for some non standard shapes, some built up shapes alright so let's look at a beam like this very simple to start nice long beam with a single point load at the center 4 kilonewton meter beam so the maximum moments at the center and for whatever reason let's use a cross section that looks like this and then we have to figure out what the maximum anticipated stresses are so for whatever reason the cross section looks like that an inverted T maybe maybe they were going to put a toy railroad that runs around there I think that was it that'd be fun I live in a house that looks like that alright so in millimeters the dimensions are that little piece there is 10 millimeters wide and 150 millimeters high the bottom piece is 15 millimeters by 120 but it is symmetric about the y axis as our requirement for the built up beams alright so we need to do two things first we need to find where the neutral axis is because the moment of inertia we calculate is with respect to that and if we don't know where it is we can't calculate the moment of inertia so we need to find I with respect to that neutral axis alright so here's what we do first step we just pick some reference location from which we measure everything that will allow us to find our that defines the neutral axis and once we know where that is we pin it to the cross section and we don't need this reference position anymore so I'll pick a reference position I don't know what did I pick so I'll pick it's arbitrary I'll pick a reference position at the bottom of the beam and I'll just measure everything from there that will allow me to figure out where y bar is from our experience we know it's going to be some place like that once I found out where that is it's a part of the cross section I don't need that reference position anymore it's just a convenience for us to quickly establish where the reference position is so pick a reference line just a place from where to measure things until we find out what y bar is and then to calculate the moment of inertia with respect to that that alright one easy way to do that is to break these into pieces so I'll call this piece piece one up here and this piece two there there's different ways you can do it but they should all come out in the same way and then make a table out of that so I have two parts one and two and I'm going to need to find the if you remember that's the kind of thing we had to do to figure out where the centroid of a compound area is area of each of those pieces plus the location of the individual centroid of those individual pieces so the areas are easy area one is 10 by 150 1500 mm squared area two is 15 by 120 which is 1800 that's just looking at the chart figure out what these individual areas are right off the table and I'm going to need the sum of those areas for the denominator so what is that 33 3300 square millimeters that there as easy could be is my denominator now I need the location of the individual centroid with respect to my reference line so one is let's see it's a different it's a distance of 15 up and then half way up itself so that's 15 plus 75 is that right that's the location of y1 from my reference line which is just a matter of convenience could have picked it anywhere doesn't matter we're just looking for the neutral location so we can then be done with that reference thing so what is that that's 90 agreed everybody see that what then is y2 yes what do the others say we're looking for the centroid this is the location of the centroid of the second piece with respect to my reference line due to its regularity we know that to be right there in the center it's at 7.5 millimeters now we don't need those summed up but we need that times the area summed up so we need to figure out in the next column of the table this one's 135,000 and that's millimeters 90 times 1,500 135,000 and the other one's 1,500 and those we need added up because that's the numerator and then we can figure out where y bar is so those add up to 4,885 millimeters cubed so now we can do this 4,885 cubed over the sum of the areas 3,300 the units work out we're looking for the distance of the centroid of this compound area from our reference axis once we've got that it places it on the cross section and we don't need the reference position anymore it was just an artificial convenience to find all this part of it but then we're done with it and that is 45 millimeters and look at your solid and make sure that that makes some kind of sense we've got this piece here that piece there and by simple observation we'd expect it to be somewhere in there if we had this piece cut out of cardboard that's where we'd expect it to balance and by golly doesn't 45 look pretty darn appropriate measured from our arbitrarily chosen reference line which we don't need anymore because now we know where the neutral axis is so we know that this beam when bent by that 5 kilonewton load start 4 kilonewtons we know that the beam we know the distribution so if we look at the beam in cross section there's that bottom plate and we know that 45 millimeters up from the bottom is this neutral axis and that our stress distribution is linear about that we'll have compression along the top because of that type of bending and tension along the bottom and in fact we can even figure out how much now because let's make sure we've got them right what's the maximum moment for a beam with this kind of simple load right down the center it's actually 4 times the distance to that it's going to have a maximum moment right in the center as you'd expect if you were standing on a board like this you'd expect it to break in the middle right where you're standing so it's 4 times half the distance which is 10 kilonewton meters what's C what's C remember our convention is that's the maximum distance of beam material in the cross section away from where the neutral axis so C is that distance because that's the greater distance away from the neutral axis so there's greater stress at the top of this beam than there is at the bottom let's see what is that we've got 15 we've got 165 minus 45 so that's 120 120 millimeters as usual we've got to watch the units and then I didn't figure out the moment of inertia, did we? we got too excited here we didn't figure out the moment of inertia we're going to want to do that we've got the neutral axis but then we didn't figure out the moment of inertia with respect to that neutral axis and so right there, calculate i it didn't even read it okay it's pretty easy to do that in the same way using the same kind of table that we used before we found solid it's the sum of the moments of inertia with respect to the neutral axis so we've got make a table again it's pretty easy to do that the thing is now we're going to have to use the parallel axis theorem because none of these pieces so we'll have to figure out their individual moments of inertia then we'll need to find a di for the parallel axis theorem which you probably remember as the centroidal moment of inertia times plus 80 squared which is the parallel axis theorem being applied so we just have two rectangular solids we figure out their own centroidal moments of inertia that's the values right out of the book just using the calculations there so you can double check these 2,812,500 millimeters to the fourth very big number just because we're using millimeters remember that's right out of the book just using for these rectangular shapes 112 BH cubed this one's 33 750 now we don't need that summed up we need to sum up the moments of inertia with respect to the neutral axis so that's the only way down to there that we're working the areas we already had those on the other table but we do need to figure out what d is d is the location of the centroid for the individual piece with respect to the neutral axis so this one it's that distance nice rectangular piece has its centroid right at the center but we're trying to figure out the parallel axis there in the distance we are from that center remember we're up 45 here which puts us 30 in we need to go to 75 so that's 45 sound right? and so now I can do i plus ad squared to get the moment of inertia of that upper piece with respect to that neutral axis that we placed comes out to be 585 5 million 580,000 and this is millimeters cubed this is millimeters to the fourth oh yeah it is yeah this is this one's the fourth this was right that should be the fourth millimeters to the fourth and you can double check these numbers when you get home the location of the centroid of the second piece with respect to the neutral axis is here the centroid of the second piece is right in the middle of it that distance is d2 so we're 45 up and we come back 7.5 so that's that 37.5 don't round off too much on these we can't at the end but not done too much as you go through these so this is 2 million 531,250 and that's what we add up to get the moment of inertia of the entire piece 8,415 8 million 415,000 millimeters to the fourth and that's what we put under here the units work out in general you're going to have to watch meters and millimeters but the units work out in general two length on the top four on the bottom so we lose two of those we're going to get kilometers per meter squared when we fix it which is kill past out so it's going to be alright and so now once you watch the whole piece comes out to be 71.3 mega pascals in compression and the the bottom I think comes out to be and what is that oh that's just compression two points we don't need which if this is wood explain why they use this cross section of the beam wood is very good in compression so you don't necessarily need very much wood at the top it's not very good at tension so you're going to need a lot more material down there for the wood to resist the tension at the bottom number which I don't have to have but now we can draw in cross section our distribution of the stress we know the neutral axis to be at 45 millimeters up from the bottom just have to be where we picked our reference and the top is in compression the bottom in tension and we know the maximum stress to be 71.3 mega Pascal's compression at the top Travis you're frowning for some reason my answer is double what? double this let's make sure we got the right numbers in 10 was the that was the moment I had a 5 written down there why do I have 5? it's not going to be perfectly an actor of 2 so there's something else wrong you have 5 that's not much oh ok it is yeah it is 5 because that's the if we drew the shear moment diagram we get a shear up of 2 only half of that 2 so there's the shear and then the moment is 2 times 5 only goes up to 10 so that's why we got the half that this was wrong this should be 5 doing things on the fly and then what the addition was wrong by a lot or a little bit? well I think when you wrote the first 5 you said 5, 8, 5 I don't know 5 oh yeah this is 8, 5 sorry that better? sorry about that alright make yourself nice diagrams that are real clear don't make them too small they get crowded do these tables you don't have to I just find it's an awful lot easier to do the tables to keep all these things straight so that I get all the pieces especially since I kind of used a couple things from the first table I didn't want to redo them you get really big numbers when you do it in millimeters so you might want to go to meters before but then that makes some of the other numbers a bit smaller so personal preference just as long as you watch your units and you are consistent with them okay so I think that takes us into spring break