 Okay, last and last, we're going to go back a little bit. What we've been looking at before was what kind of normal stresses were caused in beams by simple transverse loading on prismatic beams. And we're going to still keep that in a sense, as we've been looking at just some kind of simple loads, whatever they did. For the last week or two, we haven't been too concerned with just what that meant, what the loadings themselves were other than we needed to know, of course, where the maximum moment was. We'll even come back to that in some more detail in a little bit. But what we were concerned with was the response of that beam because of internal moments, causing, in the usual case, something like this, causing a lot of compression in the top of the beam and tension in the bottom of the beam. The interchange between the place, between that, had very much to do with the shape of the prismatic beam and we found that that was right at the normal axis. What we neglected in there is the fact that there are still shear stresses on these beams. We only looked at the moment. Now we're going to look at the internal shear stresses, what those cause in the prismatic beam. So we're going to have to pay attention to the fact that there's more things going on in these beams than just the simple compression and the tension that we had with the simple bending. The easiest way to explain it or to illustrate it is to imagine we have a beam made up of different layers like that, as if you just took a couple boards, stacked them together and that was your beam. If you just leave it like that and apply some load to it, some simple load, you can get a very different response from the beam if you didn't glue them together. If you just simply stacked them and then applied your load, you'll get a very different response to that than if you had glued them all together or capped the end or whatever it is you wanted to do with it. Easiest thing to draw is if we capped that end to hold it tight there. You can see this very type of thing. If we imagine this book as a beam, it doesn't have much structural integrity when I leave it free like this. There's just not that much. It's almost the individual strength of each paper itself that offers the integrity. But if I clip the ends, then it's a much stiffer beam. The comparison between the two isn't even close. You want to pass it around so you can see it? It is pretty remarkable in how easily it now bends. So we need to look at the effect then of these shear stresses here. We're going to see that that has a great deal to do with whatever shear stresses are along those different planes. Because it's that this bonding that either prevents this shear stressing or the free picture of it that allows it. DJ, you didn't like that? Did you take a good, you did not? I took it and that was the gap. Okay. Like a regular gap. All right. So if you remember, we did look at this sum. If we pull out a little piece of that element there. We did look at the shear stresses on those elements. We found it to be because of the different types of static equilibrium restrictions we had. Both moment and forces had to be in balance. If you remember, we came up with this type of shear stress where all those shear stresses all the way around that little element were the same magnitude. You might remember that from, I think, second week or something we did then. Okay. So we need to do a little bit more of an in-depth look now. This type of thing only in better detail as we look at these shear stresses in these beams that we're going to look at. All right. So imagine some loading of some kind. What it is isn't exactly important other than the fact that we know what it is. And at any point we can find out what the shear stresses are in the beam. Then we can analyze just what's going on here. So here's a greater view of the beam in side view. This is not an end view here. We're going to have to be careful we know which view we're looking at. So we'll take a little section in the beam of length delta x. Remember this generally x is our axial direction. y we tend to take up and then c in and out of the board. So we'll continue with that as before. And whatever the cross-section of this beam is, let's make it prismatic but not necessarily rectangular. So we have the neutral axis from that shape in our experience. We know that the neutral axis will be then a bit lower, lower down, something like that. And so our little piece of interest here that we have is delta x, y. In the cross-section is going to look something like this. We're going to take it as going all the way across the beam itself. We're going to look that entire piece in some detail. So we need some other points of interest. Remember the maximum distance from the neutral axis to the outer bit of the beam that we call distance c, any intermediate distance in there we call y. And this part of the cross-section that's shaded, we're going to call kind of a big script a. I don't know why they picked that for area, I guess. That's the area of the cross-section above a distance y from the neutral axis all the way to the edge of the beam. So we're going to now change our view once again, just because we're having so much fun with these drawings, j kind of, dj. So here's the beam in some perspective. Here's our little piece running across like that. So it's real important you get this whole picture in good view so you know what we've got here. So to make it bigger now, here's the side look. It's the same picture here now expanded. So this is delta x across there. It has a whole area, dA, mixing our infinite decimal and the 5th decimal, finite elements. But we're just looking for a picture here of it to see that we know what we're getting. Alright, so it's got a load. Depending on where we are, if we were right under this uniform part it would look just like that. We want to account for all of it though. So that's whatever that load curve was times that delta x. That delta x is small enough so that no matter what the load is it's essentially uniform across that piece. We're not going to try to make this more difficult than it is. Other things, let's see. We looked just the last couple weeks at what the normal stress was. It is linear down to the neutral axis. So we can draw that in. Looks like that. Just to help us a little bit, we'll call... Oh, why does the book do this kind of stuff? I never thought of that. We'll call this point C, this side C, this side D. I don't know why they use a C for something like that when we've already got a C. So this would be the stresses on the D side times the area that's acting on. So that'll be the force on that side. We've got the same kind of thing on the other side. We'll call that total force. There's sigma C, the normal stress on the C face times DA. What else were we missing? Oh, and then we have the shear stress on the D face. We haven't looked at that for the last two weeks or so. So that's what we're looking at now. That's the shear stress on the C face. The point we've been trying to get to is this that we call delta H. The shear stress along a horizontal plane. That's why that H. And that's the difference in that beam I showed of taking that book and putting the clips on it and whether or not we hold all these fibers going down this face together or not. So it's the shear face along each of those fibers or planes or however you want to call it. All the forces in the X direction, they must be zero. So let's see what we've got. We've got delta H. That's the shear stress. Actually, it's not the shear stress. The shear stress, the shear force actually, since it was measured by or multiplied by the area on which it's acting. So sorry, that's the shear force along there. This is what we're trying to determine. How much shearing is going on along the plane of the beam. So we've got that. Plus we have C going in one direction, D going in the other. So if we integrate over that whole area, the shaded area, that's the DA in the two of these. So that would be sigma C minus sigma D because they're in opposite direction, DA. And then integrate over that whole area here because that's where those normal stresses are acting. Two little things here, remember. We already know. Sorry, MY over I. That's what we've been working on for the last week. We get delta H, this shearing force all the way along the underside plane of this in the horizontal direction there. Then equals, let's see, M is constant. That's the loading on the beam. I, this is the entire cross section with respect to the neutral axis. So those are constant as well. So that comes out. We get solving this M over I integral of, wait, no, I want different places. So this is going to be M D minus MC over I. The moment, remember, can change with X, but over the face it's constant. So that then leaves just the YDA behind I's over the entire cross section. That piece there is the first moment of area of this shaded portion. Area moment of what we're calling that A there. It's not the first moment of area of the entire cross section. Just that little portion here with respect to the neutral axis. So that's the values of Y that we're putting in there for the integration. All right, we're going to speed that up a little bit in that we're going to call this Q, big capital Q. This big capital Q is defined as Y bar area is simply the distance from the neutral axis to the area centroid of this special shape, the big script A times the area of it. Now I'm going to use the fact that we know how to relate the moments to the shear themselves and this is shear we're trying to figure out what we've got going here. Or delta M, delta X. Because the delta M is what we've got right there. So it's pretty useful to now pull all this out and we get then delta H, that horizontal shearing force internal, horizontal, and axially directed down the length of the beam. All the pieces are there. So I've got V, Q, so that goes V delta X. The big Q remember is that first moment of area and then we've got the delta X. We've got almost all the pieces. The only thing we could do now to make it a little bit easier is this is the delta H on a little piece of a delta X. So if we just divide through by that we get the horizontal shearing force per unit length of beam, per unit length, which is useful because it can tell us just how much either adhesive or actually a mechanical fastener like a nail we'd have to put in to actually fasten this. So we'll do just that with an example. So I introduce to you my brown colored chalk. Does that show up okay? It's realistic. No wait, I'm not even done yet. That's just a brown colored outline. Wait until I draw in the grain of the wood. It just pops. You see it? You'd think with this kind of realism you would think you were at Disney World. You don't even know those stupid 3D glasses they give you. I mean just practically jumps out at you. All right, we were making a wooden I-beam like this. Our concern is for these surfaces here, right there and right there because if those aren't fastened well enough when the beam is loaded then we get that type of thing that we saw where the different layers just split apart, there's no support from that. So our concern then is do we either glue this or would we put some nails in? And if we put nails in, how many should we put in? That top plank to the bot to the intermediate thing and you want to put in enough nails that it'll actually do some good that this shear force won't be so great it actually shears those nails because it's that shear force right through the nail that's giving us this concern now that we're looking at this. So you want to put in enough that it'll actually do some good but you don't want to put in so many that it's a big waste of money, time, and effort to do so. But we can look at that now. We now know the shear stress per unit. Let's just put some numbers on this. We'll say this is three boards, each of them 10 by 100, sorry, 20 by 100. So that makes that web height 100 as well. You want to find the belt of age in a single nail. How much shear force is going through per nail? Because you have to see if a nail can withstand that. We need to know that the shear strength of a nail depends on what they're made of and of course their own cross-sectional area. We'll also say that this distance here between nails will say either that's the type of thing we need to calculate or check. So let's say we're just going to check and we have to know it's 25 millimeters. So that is then our delta X in this problem. There's one nail per 25 millimeters. So delta H over delta X will be then how much is in one nail. Alright, so we have to do our usual analysis of this and we'll say for the purpose of the problem that we anticipate a 500 Newton maximum transverse shear someplace. That's V. Q remember is, well let's see if you do remember. What is Q? The first moment of area of the beam, part of the beam beyond the plane of interest. Remember in the first picture, well there it is right there. You can sort of see it in ghostly fashion there. It was this upper area here. That's our A and our Q is the first moment of area of that. It's that entire piece there because that's the amount of the beam. That's the area, cross-sectional area of the beam that's beyond the plane of interest. Plane of interest is that junction between the two planks there where we need to determine what the horizontal shear is. Okay, so Q is pretty straightforward to find. It's for nice regular shapes like this, just Y bar A. With respect to the neutral axis, remember, with a nice beam like this we already know that's got to be right down the center because of the symmetry of it. So Y bar is the distance from the neutral axis up to the area centroid of that hatched area A above the plane of interest. What's that distance? 60. We have 50 of the web and then halfway through the upper plank is 10 so that's 60. And then this shaded area is 20 and that number is, what, 120,000. So we have V, we have Q, we have delta X we just need I as the last little piece. Remember I is of the entire cross-section. One of those and two of those moment of inertia for each O for that piece I labeled one. It's moment of inertia, area moment of inertia 112th BH cubed. So that's 112, 20 millimeters times 100 millimeters cubed. That's piece one. Piece two, 112th BH cubed but it's turned sideways so it's B is the 100, it's H is the 20, so that's Q. That's its centroidal moment of area. Oops, that's not an equals here, that's a plus. Then remember since it's off the neutral axis we have to do the parallel axis there in AD squared. A is 20 by 100 and D is 60 and that's squared. All of these units, millimeters, so that they match together. Let's see, that's AC, that's AD. Okay so far what about this part of the beam down here? Due to symmetry it's got the same moment of inertia as the piece above so we just primarily multiply by two because there's two beams. Two times, no. The D I guess technically is negative but it's squared anyway. There is no strict Y bar piece in here. This is just the moment of inertia that I want to shape. And so that number comes out to be 16.2 times 10 to the minus 6. All the numbers I put in were millimeters and then transferred that to meters cubed. I don't know why I just did. I guess the number's kind of a little bit better later. So now we have V, Q, I and delta X. 500 newtons. What was Q? Did I write that down? Oh, there it is right there. From the millimeters into the meters cubed fourth. So that's V, Q over I all in consistent units. I think everything's okay. Ameters cubed. See that will give us newtons per meter. How much horizontal shear we have for unit length of the beam. But then we have a nail every quarter of a meter. And so we put that in. That was a delta X part. And now we've got that. So we know that there's 93 newtons of shear in each one of the nails. So we know we need to get nails big enough to withstand 93 newtons. If we want to put them 25 millimeters apart we can put them less close together but use bigger nails. Typically, of course, gluing will help. He sees it especially if you get gorilla glue because it's awesome and amazing. If you've ever tried it, you're not getting anything apart when you put it together with gorilla glue. That's for sure. All right, so that calculation is fairly straightforward. The biggest places to make mistakes it's pretty straightforward what V is in these problems. It's the same shear we've been using, the same transverse shear we've been using ever since the first or second week. Delta X is usually fairly obvious because either it's given as the distance between the nails or if you're just going to use an adhesive all the way down there or even welding, then you're more concerned with delta H over delta X. So the biggest places to make mistakes are on the Q. So just as a reminder, that's the first area moment that's shaded part of it, beyond the plane of interest. And remember, it could go either direction. And then I is the moment of inertia of the entire cross-section. The two places are most likely to make errors in this because the V and the delta I are generally pretty straightforward. If we're concerned with a long beam, don't forget this could be V max, which means you could protect it for the maximum shear and then it's over protected for everywhere else. So it might be economical to concern yourself with a different nail spacing along the length of the beam or something, however it shakes out with that problem. All right, that's the shear force, determining the shear stress. If we look at the beam on side, take a cross-section delta X. Here's whatever plane of interest we might have had in that problem. For instance, and this last problem was just where the planks joined together. Look at the prismatic cross-section of the beam. That place, then that delta X part is there and that then is part of the beam right there. We need that because this shear stress is spread along an area that's delta X by T, where T is the thickness of the beam at that point. For our beams, it's constant all the way down. So then we can calculate the average shear stress, which is that horizontal shear, some delta A, which we've had, let's see, delta H is VQ over I delta X. Same thing we had there. The area over which that's acting is T delta X. It's the bottom face there because that's where that shear is acting. The delta X is cancelled and then we get VQ over I T as the average shear stress on that face. It's not perfectly uniform, but it does give us the average. We do have to be careful that the maximum is significantly larger than is the shear stress distribution on that down the face might look something like that. So we have maximum shear stresses that might be significantly greater than the average shear stress. It shakes out that the maximum shear stress, not the average, but the maximum shear stress for a rectangular cross-section is just 3V to A, area of the entire cross-section. So we've got the shear per unit length. We've got the average and in some cases the maximum shear stress. So we can look at a couple problems there. So given a beam here, beam cross-section, simply rectangular, of unknown height. You're to determine that with these allowable stresses, 1.7525 KSI. Remember that this allowable normal stress is wherever there's the maximum moment, it's not uniform down the length of the beam as we know from our experience, times C over I. And then the allowable shear stress, well, we just came up with that. In fact, it's a rectangular beam so we can use the 3V. So you need to find out where the maximum moment is, where the maximum shear is, and then make sure that you have an H that protects for either one of those. The easiest thing to do if any is find out what H is from here, check it in there to make sure that you're still below the allowable shear stress. Speed jump a little bit. I'll give you the reaction forces here. Remember you'll need it. In fact, we can very quickly do the shear stress diagram, find out where the maximum shear and the maximum moment are. 65 kip. This one's up. With that, we can pretty quickly do the shear moment diagrams. This reaction's 0.65 down so we know that we come down 65. Then the slope of the shear is the slope of the load diagram minus the slope, minus 400 pounds per foot up by 8.35. By the way, that takes us down to 5 kip. Then we jump up by 8.35 kip. A little out of scale, but we know that our maximum shear is now the 4.5 in the second or third of the beam. We need the moment diagram. We know that there's zero moment at the two ends, and all we care about then is the change. Maintain no moment, and that's free end. The pictures in the book sense to you it's a little easier for those graphic artists to do them than it is for us to do them at the board. Cool beam again? We're satisfied.