 All right. What we've looked at so far, at least in terms of the beam design we've been working on in the last... I guess we started last week right after break. We've been looking at nothing more than pure bending. Whatever type of loads we might have, whether distributed or not. However, at certain places, and it doesn't matter where, because this does happen everywhere, we've got not just the bending, but also, as we know, we've got shear across the face of that beam. Last week, we took our first looks at how to design the cross-section to best resist the bending, but we're going to look a little bit now, a little bit more at the shear. However, we're going to take an... it's not an obvious change here. I think, in fact, and this is the seventh time I've taught it, and every time I teach it and I sit down and review my notes, and I look at this, and I think, she was. You know what I'm saying? I'm just amazed where this comes from that we're looking at. We're not going to look just at this cross-sectional shear. All we need there to resist the cross-sectional shear, the transverse shear, is we just need enough area because we know that the shear stress is equal to that shear divided by an area. We don't need anything like we did with the bending where the normal stress was a factor of the size and shape of that cross-section. Remember that the C has to do with the size of it and the I had to do with the actual shape. The moment of inertia. Basically, what we are finding out from there is that whatever the cross-section, we needed as much area away from the neutral axis as was practical. And that's why I-beams are so very useful in construction purposes because they have a very, very big I because of so much of this area that's away from the neutral axis. But the weight wouldn't be any greater than a beam that would be much smaller. Those two have the same area. They have the same weight. But this one has a much greater I and thus a much greater resistance to bending stresses. Now we're going to look a little bit at these transverse stresses and the trouble they cause. However, what it turns out to be for us is it's not the transverse shear itself that we need to work with because, as I said, that's easy to handle. We just need enough area. So only if we've got enough area we can withstand the transverse shear. What we have to look at is the fact that as beams bend and our subjects some kind of transverse shear and it's very easy to see with a cantilever model, if the beam has no longitudinal resistance, imagine we had this beam made up of many, many layers. It could be like a piece of plywood but I'm thinking more like some very, very thin maybe veneer pieces of wood just stacked together and not even bonded to each other. If we apply this kind of load then there's going to be a lot of bending because each of these sheets can slip past each other and of course this is a greatly exaggerated picture as we often have to do. Whereas if we took those very, very same sheets and this is just what plywood does and bonded them together so that they actually stuck to each other and imagine we do that, we can very simply do it. Imagine we put something, some kind of, some kind of, I don't know what to call it, some kind of thing there, that's a good word, thing, thing there that keeps those, keeps the sheets from slipping past each other as happened here. When that happens we get a very much better resistance to the bending and it's quite easily demonstrated with a simple textbook. If I let the sheets slide past each other which is this kind of picture then there's almost no resistance to bending. This is sticking out just because it's stiffer but you can see if this was our beam and we allow these sheets to each slip-plack past each other it has almost no resistance to bending at all. But if I put this on here to do nothing more than hold the ends in place which is this kind of, I think I called it a thing then it does much better in resisting bending. Remember before it slumped all the way down here. The idea that you can get though is if we can even do better with these in here maybe if I move these things to the side so there's less slipping all the way around it does even better. So what we see is that those loads like these and like these that cause cross-sectional transverse shear also give us some kind of trouble that we need to address of this longitudinal shear along there. We need to address how to keep the fibers of the beam and it's especially true with a wood beam that is truly made of fibers. It's the rings of the tree growth that actually make up layers in a beam like that. If we don't address the ability of the beam to withstand those kind of shears then we're also going to have greater deformational failure. We might not have material failure but we clearly get a beam failure there that is not going to be a good design. We have to address that as well. So that's what we're going to look at now as we take a peek at these things. So we've seen at least part of it before if we look at some elemental piece of a beam in any kind of bending we know that that has the transverse shear that we've been looking at anyway we'll put it in terms of shear stress rather than in terms of shear itself because it does have to do with the area. But we're now going to look at these longitudinal shear stresses that run down the length of the beam in the axial direction. So here's our setup for this. As usual what we're going to do is look at an elemental piece and how the forces stack up along that and then expand that out if you will to the entire beam. So here's a chunk of the beam just taken out there. Axial direction, so that's the x direction. Across the interface that's also going to be our neutral axis. That's some kind of prismatic cross section. Remember what a prismatic beam is, what our term prismatic means. Bring to a symmetry along one of the axes. Symmetry with respect to the y-axis. x-axis is our longitudinal length down the length of the beam. Our y-axis is typically up from that. So just some prismatic cross section whether it's an i-beam or a circular beam like a tube or a rod would be or even just a simple square beam, it doesn't matter because there is y-axis symmetry. The cross section is mirrored across the y-axis. So the shape of that is not of material interest yet. And so we'll look at some elemental piece right there with the flush with the top and some little bit of distance down. And remember then our reference values. The distance from the neutral axis to the greatest extent of the beam in the y-direction is c. We'll call this distance here y1. And what we're interested in is the shear along this bottom face which is going to be all the way across the beam across the bottom of that little piece there. So that's our setup. And we'll call this area a. Give it a big fat script a. It's always fun to draw new symbols and get your own style with them so that's what we're going to call that one. Kind of a scripty. I think the book calls it a prime. I'm not really fond of using primes because they look too much like exponents but I think that's what the book uses. But it doesn't matter. We're going to be done with that in a little bit. So we've got this... Now if we look at it in perspective we've got this little piece here that runs all the way across the beam and that's the elemental piece we're interested in. We don't need to look at an elemental piece in the z-direction because it's symmetric in the z-direction but it's the same over here as it is over here because of the symmetry about the y-axis. So we only need to look at this little piece delta x-wide and not too concerned with the other direction because it's a square piece. So we've got this piece delta x-wide takes up an whole area dA and we're kind of mixing d's and deltas but as usual we're going to be done with those very shortly so we're not going to worry too much about it. And so let's look at the loads that we have on here. Perhaps there's some kind of distributed load across the top or it could be a point load it doesn't matter which we take care of all of those things as we look at them and in terms of the force this is w remember that's the size of the distributed load per unit length but the unit length here is delta x so that's the entire load force on that piece there. We also have of course shear so this will call v and that's also v but they're on a slightly different side so we'll call this this side we'll call this c make it the same notation so I don't drop the layer oops we'll call this cd and this is c' d' just to designate those sides and corners and the like as we need them for reference so that'll be vd the shear on the d face this is the shear on the c face and then we also have the normal stresses and remember those are linearly distributed from the neutral axis so however down far that is so we have those on one side I'll call that that's the normal stress on the d face times the area over which it's acting dA and we have the same kind of thing over here hope you started your drawing big because even mine's getting corrupted so that's on the c face and the area over which it acts and that's also down to the neutral axis and then we've got one last little piece of it and that's this longitudinal shear that we need to investigate now and we'll call that delta H God only knows where that came from but we've got that there so we'll sum the forces in the x direction and because of our static limitation they'll sum to zero so let's see we've got delta H this new piece we're looking for the longitudinal shear will you be needing longitudinal shear on top? no because we're actually at the top of the beam so there's nothing there we're looking at the internal shear and then what else is in the x direction well it's these two shear stresses so integrating over that whole area A this whole cross sectional piece there it's what? sigma c minus sigma d d dA I think we're okay and that's got to equal to zero we can change things up a little bit because this part here remember that sigma equals minus m y over I the normal stress at any place is a function of the bending load m the position across the interface and the cross sectional moment of inertia so this is for the entire cross section here that I but that's what we established last week so that's enough of new but we can put that in here so that now this all becomes we can now solve for this delta H delta H equals m d minus m c over I because those are constants they can come out this m is the moment at the d phase c is the moment at the c phase and I those are all constants they come out and we're left with just the integral over this exposed area above our plane of interest as established by y think that gives us all the pieces we need and this little part here is entirely the geometry of the cross sectional piece above our plane of interest so this is the first moment of area above the plane we're interested in as established by this y here that we're doing right there right across here and so we can look at that place that we need to depending on how far away we go whatever that cross sectional area happens to be we're going to call this capital Q and for our nice regular solids it's nothing more than y bar a above the plane now the reason the way this becomes of interest and we'll do a problem very much like this is imagine we have a an I beam we want to make out of 3 wood planks you've got a whole bunch of 2 by 6's or something and you want to make an I beam out of them so you put them like that you'd be very very concerned with how you're going to attach these to each other so that they don't come apart at those interface planes where those these cross section these beams are joining each other so maybe you'd put nails through there you might also put adhesive all the way along that face as you put these together and it's this very shear longitudinal shear here we're trying to pull the beams apart at those places if loaded that you wouldn't want the beam to do what we showed in doing with the earlier illustration you wouldn't want these planes shifting off of each other which is why you can glue them and screw them to make sure they fit make sure they stay in direct and intimate contact to maintain the integrity of the beam so we're almost done with this we almost have it put together so this becomes delta M over I where delta M is just the difference in the moment on either side times Q now we're almost done with all the pieces we need so let's see what we've got so we've got delta M over I times Q now we're going to grab something from a little bit in the past remember this delta M is the difference in moment on either side it's not in this picture because we did a force balance we're going to add to this our relationship between the moment and the shear that's from our shear moment diagrams we'll macroscopically look at it a little bit where instead of the differential values we'll use the deltas because that's what we've already got here delta M can come out and be replaced with V delta X so again V Q over I delta X where I remember delta X is how much in the X direction of this piece we're looking at and so I think so we're real close to what we need now we have the shear per unit length is the transverse shear remember this is a longitudinal shear over here is delta H this is the transverse shear that we've been calculating since all the way back into midway through our statics last night midway about a third of the way into statics last fall we came up with this transverse shear Q remember is the first moment of area of the section above our plane of interest and then I is the moment of inertia of the entire cross sectional area the entire face so that's as tricky as anything for this as those two pieces so now we can determine what the shear is per unit length the way we use that is if you were going to nail these you'd put nails every couple inches and it's easy to use that as our distance delta X so we can figure out how much shear there's going to be per nail and then we can make sure we got nails with enough thickness in area that they can withstand that so we'll do that very problem as an illustrative example of what we need to how we work with this alright so here's here's a beam you're going to build this summer for your parents deck made out of three planks nailed together dimensions 20 millimeters each planks exactly the same each planks 20 by 100 millimeters so of course this is a cross section take a second to make it look like a very realistic beam alright so there's our beam in cross section and then we want to find the shearing force in each nail expecting to put them at approximately one inch intervals we'll put them at 25 millimeter spacing between each nail so that's our delta X that we're looking at the spacing between each nail we want to find the shear force in each nail with a calculated shear usually the maximum shear let's say 500 we would find that just like we would have found it last fall we look at the beam and see how it's loaded figure out what the reactions are and then figure out what the shear moment diagrams look like see what the maximum shear is and then build the beam to resist that maximum shear because we've seen now that it determines what the what the cross sectional area is alright so our beam in cross section looks like that and we're worried about the nails withstanding the shear at that interface there where the two planks are put together we don't need to do the other side because it's symmetric about the Z axis so we don't need to figure out where the neutral axis is by symmetry we know it's right down the middle because it's symmetric not only about the Y axis but also about the Z axis but we've got all the little pieces we need alright so we've got the shear we need the Q the first moment of area with respect to the neutral axis of the part above the plane of interest which is this entire area there not just the part of area directly above the intersection of those two planes but the entire area all the way across so that's our A so we need the first moment of area of that then we also need the moment first moment of area of the entire cross section Q and remember this is of A nice simple shape we don't need to actually do the do the integral figure out what it is that's right in the middle and this is 50 millimeters up to there and then another 10 up to the middle point of the beam so this is 60 millimeters that's Y bar and then times the area of that entire piece which is 20 by 100 that's the area of A and that just comes out to be 120,000 it's not a well I guess it is a volume in a way I'm not sure what's a volume of there you go alright other piece we need is the moment of inertia of the entire cross section the central moment of each piece plus the parallel axis theorem as applied so this is 112 20 by 100 Q so that's 100 sorry 112 BHQ for the web that's what that piece is called by just the web and then the other pieces are the central moment of inertia of the flange which is 112 B which is now the 100 dimension HQ which is the 20 that's the central moment of inertia of the flange D squared which is 20 by 100 from an outer space continuing this square bracket here plus 20 by 100 that's A and then D squared is well that's again the 60 that we had before so that's AD squared for the flange except so we just multiply that square bracket by 2 there's 2 flanges so there's the moment of inertia of the web which happens to sit right on the neutral axis anyway so we don't need the parallel axis theorem on that and then the moment of inertia of the flange times the parallel axis theorem times 2 because there's 2 flanges that comes out to be 16.2 times minus 6 meters so now we have all the pieces well now we'll be able to find the shear per unit length then we'll put in the fact that we have one nail so by the way this is often given the symbol little Q also known as the shear flow if that's the amount of shear that's flowing down the cross section there as we look at it so we want to find out how much shear there is per nail so we figure out that Q the Q we've got, big Q we've got I we've now got just multiplied by delta X which is the distance between each nail you know it's Newton's times Q which is up there in millimeters but make it into meters meters cubed that's Q over I which is 16.2 times 10 to the minus 6 meters to the fourth so let's see what units do we get then we get Newton's per meter so the amount of shear longitudinal shear remember shears down the length of this space per unit meter but we only want to do it for every nail which is each nail is 25 millimeters apart so we put that in as our delta X length we get 93 Newton's per nail and now we can make sure we've got nails that are of such a material in such a cross-sectional area that they can withstand the shear stress the calculations aren't really very complicated you just have to be careful that the Q the first moment of area is faced upon this bit that's above the plane of interest and that the I use in the equation is for the entire cross-section that's really about the only places where students tend to make some mistakes remember that's the shear the longitudinal shear in that plane but to see what it is in terms of the shear stress we need to look at a fairly similar picture as we've been looking at before so here's the beam in from the side with some kind of cross-section to it it might look like doesn't really matter the idea is all the same so we're now looking at this this delta H across some piece there which will look like the part that's resisting that shear so we can find out the shear stress is this bit there right on that face there that's where the shear resides is on the bottom face of that little trapezoidal piece looks something like that this then becomes the average shear stress is the delta H over whatever that area is which is T V Q over I delta X is the average shear stress oh wait I forgot there's delta X here so there's actually two delta X's good because they cancel so we get the average shear stress which is about equal to the tau XY that we were looking at so many weeks ago across the cross-section like that just to be accurate let's let's see if we look at the beam from above so now the D axis coming out like that the Y axis right at us this distribution of this shear stress is something like that and what we're looking at is the average shear stress and so we are underestimating it a little bit because there is a maximum of down the center in the center region we're underestimating that a little bit and for rectangular cross-sections the book shows that maximum shear stress for non-rectangular it's a little different but it comes out very easy for nice rectangular cross-sections by the way that A is the cross-sectional area of the entire piece ok so let's run through it again with another example a lot of little pieces had to come together but as usual a whole bunch of little pieces were left with just one or two big working ideas let's figure out what we've got let's do a problem or start at the finish and see what we've got so we've got a beam simply supported a little bit of an end sticking out there and a distributed load up to that point kind of looked like a diving board 400 pounds per foot so we've got the entire youth group on the diving board 400 pounds per foot and one brave kid out on the end there happens to weigh 4.5 kg big kid football player 8 feet and 4 feet now are the dimensions and the cross-section is simple rectangle 3.5 inches in width but of unknown height to design for that so we've got maybe some wooden beam like that we've got 3.5 inch thick boards but we can order them to any height we need meeting these designer restrictions allowable shear stress of 0.250 KSI and allowable normal stress 1.75 we're getting more capability now of designing a good cross-section because we can deal with it not only in terms of bending but in terms of longitudinal shear as well so we need to know what the maximum shear is to avoid that trouble we need to know what the maximum moment is to avoid that trouble shear moment diagram is pretty easy to draw we can do them real quick the reaction, the first reaction there is 0.65 0.65 kips so we start from there and then we have this uniform load so we know the shear has that very same slope takes us down to a point 0.385 kip so as we might expect there's a worry about shear failure there and then the reaction at the roller is 8.35 so it jumps up by that much it jumps up to about 4.5 kip so that's our worst case so there's our shear diagram in terms of the maximum moment all we need to really worry about is what are those areas because remember that's delta m we have 0 moment here because that's pinned we have 0 moment there because that's free so whatever one of these is the greatest delta moment will be the worst case that area turns out to be minus 18 and this area turns out to be plus 18 because it comes to 0 so we know that our maximum moment is 18 and it's right at the roller exactly what the shape of the moment curve is on the other side isn't really of concern here because what we need is to make sure that the beam resist the maximum moment which we can now figure out at least in terms of h, remember h is unknown so this comes out to be let's see, we know that allowable normal stress the moment is 18 kip feet c is 1 half h remember h is unknown but we'll be able to solve for it here and the moment of inertia is 112 b which is 3.5 3.5 over 12 feet and then times hq which we don't know what it is but we do have a value for this so we can solve for an h the height of the beam we need to resist bending failure and then we'll go back and see if that's adequate for the shear failure longitudinal shear failure so that h becomes 14.6 inches so we need a beam to resist the bending failure of at least 14.6 but now we'll check the allowable shear stress and see if we're okay with that and that's 3b over 2a we've got the pressure the maximum shear is the 4.5 and the a we now have at least a design area we're checking now to see if it's adequate so we divide by the area we've now pit for design to resist the bending failure checking it now for longitudinal shear failure and this comes out to be 0.33 0.33 kips square inch and that is less than our allowable limit so that h of 14.6 is adequate it will protect in both bending failure and longitudinal shear failure which for beams when a wood beam does fail longitudinally because of whatever the loads might be it's interesting the way it fails if you look at it and it tends to fail something like that right down the planes where this longitudinal is the greater it actually does pull the planes apart and then the bending separates them even farther I think there might be a picture on the book on that so you can double check it before we do another example just to make sure that you understand what the q is that we need for the wood beam we had the wood eye beam because the concern was right at this place that plane there and so we needed the entire area above that plane if we were making this out of steel plates and we're welding them we'd run the welds down there and it's still essentially the same same area we're still concerned with the area above that plane if we're taking cast eye beams that are already one piece most of the steel eye beams are not made up of separate pieces welded together they're cast this way if we needed to strengthen that for some reason by putting some kind of channel over that which is not uncommon I guess so we want to rivet maybe through those places there as we attach this channel beam to the top now this area the plane of concern is this one here and the area beyond that is this entire area of the channel even though some of the channel actually hangs below that plane available right at the plane we have to take in the whole area to account for this longitudinal shear down the whole piece so that's not as obvious as it might be for some of these other ones and if we were saying making a a box beam a couple boards nailed together as you might do here in the Adirondacks it's nice to have wood as a big part of construction and so you might want nails through there so the planes of interest then are these planes here where the longitudinal shear could shear through the nails so the area we use for that is then the area of that piece in the middle there between those two planes so sometimes these areas are a little bit difficult to determine so we're going to we're going to avoid overly complex ones and do ones that are a little bit more obvious so we can make sure it works ok so here's one for you to do as usual I'm working way too hard so here's one for you to look at have a beam well let's not worry about the whole thing we don't need to find the reactions and all that let's just take a look at it in one place we'll look at it there we know that the reaction here is 15 kilonewtons and so we know that the shear across that face is also 15 kilonewtons and the cross section is something like this sort of a modified I beam the University of Tennessee beam maybe that's what it is 20 millimeters there 80 down the web 20 on that little bottom flange and 60 across the bottom of the T and we want to find at two places we want to find the shear there and there all those two places A and B find the shear average shear stress and that's that delta H thing we've been looking for alright so you're going to need the moment of inertia of the beam I'll give that to you and I'll give you where the neutral axis is so you don't have to find those you can work on the new part so the moment of inertia of this beam comes out to be 8.63 times 10 to the minus 6 with the neutral axis 68.3 millimeters up from the bottom normally in a problem you'd have to find those but I'll save you the trouble and just double check that you can find the average shear stress the new part of what we're working on today have I got the picture alright very shear stress add a piece where the transverse shear is 15 kilometers and we have a beam with that kind of cross section and check the two spots the two planes of interest two different calculations basically what you need to find is that q that capital Q the first moment of area of the area beyond the plane those are the two parts you need mostly the other two parts are the other parts I and the wing tier are over the same for either calculation just to make sure if nothing else you can find what q is the beam if you'd love to integrate most students will avoid it if they can and for the appropriate area that I've been crosshatching and read the area above the plane of interest beyond the plane of interest from the neutral axis got v, got i t I hope is pretty obvious it's the the width of that plane did I give that to you and specifically give it to you it's 20 as well it's been all those things and there's the time to do it and it's okay I think order may be the same thing okay now I've got it watch your units on q y bar for a is from the neutral axis up to this midpoint which is the full 120 minus 68.3 minus 10 as we come down to this 4 41.7 does that sound right and then the area of a what did you use fill what you use for the area here to find qa it's this area here 100 and should have gotten 83.4 times 10 to the minus 6 I'll watch your units I've got a bunch of them in there but in the end you've got to work them out to one spot 83.3 that right Chris you're looking shocked and amazed that we agree that amazes you straight in my review that amazes you now we can find the average because we've got all the pieces now the shear was 15 3.4 times 10 to the minus 6 meters cubed I gave you 86 8.63 times 10 to the minus 6 meters to the 4 and t would you use for t you have that Tom the 20 millimeters z dimension of the interface of interest by putting it in meters so we get kilonewton meters squared as we should for average shear stress okay and I have right now 725 kilopascals that's an order of magnitude off from what you have you have 7.25 10 to the minus 6 just multiplication whatever it comes up to be you got the same orders of magnitude on everything up there 83.4 and 8.63 oh 2 meters oh I see I see what the difference is okay I just wasn't looking yeah you're right what I had in my nose was 1.5 I just missed the decimal place on my nose so your answer is right David you got that I'm still trying to work on some of your theories because there was one I didn't get any answer as to how it was figured out I believe that was the maximum shear stress was equal to 3 to v over 2a yes remember that's for a rectangular beam only and that's where all the simple parts when you accept a rectangular beam a whole bunch of things get very much easier and a bunch of b's cancel a bunch of h's cancel the book lays that out but that's a rectangular beam only and we're not working with one of those here and it does not apply to a rectangular portion thereof the entire beam has to be rectangular to use that 3v over 2a alright so let's check the other two pieces and then I'll throw another problem for you and we'll double check it when you get back on Friday so for q b you got what number Travis? 70 times 10 to the minus 6 meters q should be sufficient and then the average shear stress then at that point you guys would have gotten like 6.08 megapascals 6.1 megapascals Chris you got that too well you did let me set up a problem for you you can look at it over Friday I know you wouldn't dare come back without having that completed alright a beam made out of it's two side boards joined to one intermediate board if you were using this as a structural as well as a decorative beam of some kind it's nice to have the full side piece and not showing any of the joint there so then with nails cut through cross section looks something like this with nails through there given dimensions 250 millimeters on the inside 150 on the side 30 of the thickness of that board same for the top board spacing between the nails 100 milliliters and an expected maximum shear of 800 kilonewtons for wood beam 800 kilonewtons is pretty high so expected shear of each nail as a diameter of 2 millimeters so find the average shear stress in each nail all pieces there alright average shear stress in each nail just like we did over there bq over it so you're going to need to find the cross sectional moment of inertia which means you're going to need to find the neutral axis obviously it's going to be somewhere up near the top but you need to find where so that you can find the inertia then you also need to find q for the appropriate area which as I showed you with the box beam this is the same just doesn't have the bottom to it is that area there between the two planes we don't use the bit of the side because that's not supporting any of the shear that's going across in those spaces where the nails are alright then with all that you can find the shear stress per average shear stress per per nail so that's the problem so you're not even dare walking the door on Friday without that being done I know you guys try not to smile on it because you're anticipating the fun of coming in with it done and shutting me out for once alright that's a wrap