 Alright, we're going to do, we're going to change gears a little bit here now, now what we're doing is going to be actually kind of, students usually find this pretty interesting in that we're going to be doing what structural design engineers themselves would do. A whole bunch of the stuff we've looked at before, we've looked at a lot of rectangular timbers to be used as beams. We've looked at certain built up timbers, we've looked at box beams and some eye beams that somebody made by nailing a couple of boards together. Now we're going to actually do what most structural engineers would do, which is specify the limits on the problem, which is the span of the beam and the expected load on that beam and then essentially go to a catalog and pick the appropriate beam that's already manufactured, it's a lot cheaper that way, you don't want to do a bunch of one off design to try to keep the cost of structures down. So that's why I asked you to bring your book today because we're going to be using the beam tables right in the back. We'll be mostly concerned at least today with bending stresses and shear stresses, the two big ones that we've looked at for the most part in just straight standard structural type beam applications. Bending can be a problem in some ways in certain type of building applications, but for the most part we're mostly concerned with bending and shear, and it's the bending that normally dominates the design. Shear can be a concern and should always be looked at, but it's generally bending that is the big concern, and it's also the bending that dictates the design of the types of beams that we're going to be looking at. Alright, so let's go back to our bending equation where we found the normal stresses in a simple bending to the transverse load, and if you remember that was MC over I. There was a minus sign in there, but tradition dictates that we just take the minus sign out, and by inspection you pay attention to whether this is a compressive stress or tensile stress that's expected. For steel it's not a big concern. Steel is pretty much the same in compression as it is in tension, but for wooden beams it's very much a concern. There's like an eight times difference in most woods between what a wooden beam can take in compression and what it can take in tension, very, very weak in tension in general. So we're going to change up the look of this a little bit. We're going to define what we call the section modulus. I'm not sure why we didn't do this right from the start, but our author chose not to so I chose not to. We're going to define the section modulus, big capital S there, defined as the cross-section moment of area, first moment of area of the beam divided by C, which if you remember is the greatest distance in the beam from the neutral axis. That of course makes this then M over S. So if we need to reduce the stresses we need to increase the section modulus. Greater S gives us lower stresses. Notice too that that section modulus is purely a matter of the beam's cross-sectional geometry. It has nothing whatsoever to do with the load. It's the M part that's the load and the S is entirely geometry. And there's no reason we couldn't have done this right from the very start when we first came to this bending equation. We've had for some reason our author preferred that we not do that. So he's got a lot more experience teaching us an idea. So we'll follow his lead and introduce it now. So let's take a quick peek at the type of thing that can do for us just so we get familiar with it. We'll just do it real quick for two rectangular beams. A six inch by four inch beam and then one of the same area and thus the same weight, a three by eight beam. So we can figure out the section modulus real quick for either of those which is labeled for convenience. Remember I is 112 BHQ when in that configuration we've taken the neutral axis to be horizontal as we look at it. And then C of course just the distance from the neutral axis which in this case will be right down the middle to the middle part or up to the edge part. So we can calculate it real quick just a matter of putting in those numbers. In fact S actually reduces to one sixth A times H just to make things even speedier. And they have the same area and so we can compare the section moduli for these two beams just to show that even though the area is the same, the weight the same, the amount of material is the same, the section modulus is substantially different between the two. So for this first one I believe it's 24 inches cubed, has units of volume but remember it's not, it's the first moment of area actually inches to the fourth. Oh no section modulus has units of length cubed squared and then one more. And for this other beam is 32 inches cubed so if you're trying to avoid bending stresses then the much longer, much skinnier beam is the one to apply. There are other concerns when beams get loaded in transverse ones we don't look at they can twist under that load. We haven't really looked at that. It's not uncommon when using a long skinny beam like that as floor joist as you might on a deck to stabilize those beams somehow either put some cross pieces like that to help those keep those beams from twisting sideways in a way we haven't looked at or to even put some kind of board just across there just to stabilize that board. If you if you go get one of the meter sticks especially the two meter ones lay them on edge like this and push down on it you'll see that it'll twist sideways a little bit in a way we haven't looked at but for pure bending this is the superior of the choices to take even though the weight's the same and the cross sectional area is the same. We know that to mostly be a factor of increased eye because there's a lot more area a lot farther from the neutral axis. Something going on up here? That's age. Oh yeah no the area cross sectional area times age. So let's see how beam designers then apply this. It's easy enough to do it for rectangular beams but most major construction is not done with those so we'll do it with with other structural beams and if you look in your book if you brought it not you can look up here in the back of the book are the geometric properties of structural shapes This means then various eye beams, channel beams, L beams, more eye beams and we have it in both at least in most of our books at SI and in English units so what you see in this table a description this is sort of a catalog number of the beam and I'll explain a little bit better well you can see what these these numbers mean inches and pounds per foot so the second number is actually the weight of the beam the greater this number the heavier that beam is it can be that you pick a beam that's so heavy that the calculation for bending needs to be increased because now you have the weight of the beam and the problem that might be significant then you have to refer to this little diagram there what all the other columns mean for the most part the first number over here remember this is the pounds per foot this first number this 24 for example is basically the width of the beam give or take a little bit and it's fairly easy to remind yourself of that if you look here at the table you see a 24 there and you look over and you see oh I think I said the width it's the depth of the beam which is basically the first number you see for the most part they they are all in agreement there so it's the this dimension on the beam the depth of the beam alright so keep those handy we'll do we'll start a problem here and I'll show you then how we use these tables to actually decide which one of these beams is the better to use for a particular problem so we'll take a very simple setup we've done this before a cantilever beam of some kind and an expected load 15 kips on an 8 foot beam so we know from having done this type of thing before that the maximum expected shear is 15 kips and that's constant across the beam you can you can almost do the shear moment diagrams in your head for something a set of this simple and a maximum expected moment will be right here at the place where the beam meets the wall because that's where the force has the greatest moment arm and that of course then is the 15 kips out at an 8 foot moment arm so the bending start the maximum moment we expect right there at the wall which is no big surprise if if you when you remember probably as a kid when you go and stand out on the end of a tree limb it breaks back here it doesn't break where you're standing because of the greater moment arm you were probably doing that calculation as you fell to the ground I'm sure I know I was as a kid finish the calculation if you take a good bounce you get a little bit more time that way so I believe this is 1440 kip feet as the expected moment now let's say it's a steel beam with an allowable rest of 24 ksi that allows us then to calculate the section moment I started the section modulus by simply in the maximum expected moment over this allowable stress that's just the the bending equation the stress equals NC over I solved for the section modulus that now allows us to find the minimum section modulus finding the maximum will over protect the beam and quite possibly be too heavy itself and too expensive so we want to find the minimum section modulus that will allow us to protect for this beam and then there's at various problems various stages the application of factors of safety so we've got an expected moment maximum moment of the 1440 no that's kip feet it'll we'll need to fix the units yeah because this is the kids per square inch is this is that keep inches okay yeah see I don't pay attention to my notes because I'm not sure you guys are paying attention to me so it's a way first to check out there we go kip inches so the units are already okay thank you John that gives us an expected section modulus of about 60 units then we get a section modulus of inches q just like we'd expect it would be so we go to appendix B and we look at various beams that give us at least that section modulus so the section modulus here notices for for two different axes two different directions the x and the y has to do with the standard ordinal directions there and our neutral axis is in the x direction so we look at the x axis section modulus and we go down until we find a couple that are just barely above the 60 so here's 68 for but we wouldn't want the 57 6 so we will put down that beam on our list the w18 40 so it make a little list as we go down that table of those beams that look kind of young so the w18 40 go down a little bit farther we're going down this section modulus in the x direction so we got that one beam go down and look at some others here's 72 7 we don't want the 56 5 it's too small so we'll go to 17 7 we got a w16 45 look at a couple others where's our scone here so we go down a little bit farther there's the 62 7 but we don't want the 54 6 it's too small section modulus so w14 43 and this is this table in the back of the book is nothing more than the standard tables that the beam manufacturers are going to use so you can you can look down this table you can call them up and say I need I need you know 2000 w14 43s for the building I'm putting together can you give me a bit on them and so we've got we've got a couple good candidates what we see is that remember this is the linear weight of the beam the second number so all three of these that we happen to pick out you can pick as many as you want as long as they have a section modulus just above the 60 that we calculated is our minimum we can see that this one is the lightest so for our purposes in this class that would be the one to pick other considerations would be cost availability other concerns that the designers might have but for our purposes we're just looking to make sure that we maintain the integrity physical mechanical integrity of the building and so we'll just check this one last thing we need to check is then the actual beam weight this is 40 pounds per foot we need eight foot beams so we see that that's going to add her being about a hundred three hundred and twenty pounds compared to a load of fifteen thousand pounds it's not a big concern there are times where the load is increased significantly by the weight of the beam itself and so you just need to go back into the calculation we if it was a concern we now add a uniform load here that represents the 40 pounds per foot and review the calculation to find out what the maximum expected bending is the weight would be taken as a uniformly distributed load and that's it that's the game of selecting a particular I beam of interest for a particular problem so let's uh let's apply a little bit more so simply supported beam as we've looked at before with this kind of loading for 60% of it we have a uniformly distributed load and then a point load at the middle there so this is three meters and one meter on either side of that point load 20 per meter and 50 kilo Newton you don't have to draw the shear moment diagrams but we do need the maximum shear and moment to be able to calculate the appropriate section modulus because we need to protect for the worst possible situation on this beam not just some average value so we need the maximum shear and minimum maximum shear maximum moment all right just to help things a little bit I'll give you these two kip eight kip just to speed things up a little bit and then we can figure out the maximum shear and maximum moment are oh yeah killing it sorry of course I mean killing it I wouldn't screw up the others twice in one day how dare you imply that thank you guys all right we can figure out we can figure out the shear moment diagrams we know that we're going to start with a shear of 52 kilo Newton's there and come down at the rate of 20 kilo Newton's per meter so that is a slope such that it gets us down to about minus 8 then the shear is constant for a little while because there's no distributed load and then it takes a jump of 50 down and then it's constant again with no load so we go down now minus 58 there and then back up so with that we know that our maximum shear concern is out here at the very end but we have to do the same thing for a moment because that generally dominates just back us off a little bit so I have some more room for the moment diagram we draw this real quick we started zero moment because these are two simply pinned ends so we know those points exist and then we have a positive slope that decreases to zero we know that to be a parabolic shape so we hit zero right when the shear heats are hit zero remember the shear is the slope of the moment diagram then we continue off curling over a little bit and then we're constant for a bit and then even more constant for a bit so I guess get you in here looks something like that I guess but it's pretty obvious where the maximum is it's right there as would make sense because we know we have zero moment on either end and we have a closed area above on the shear diagram and a closed area below and those two have to be equal which would put the peak right there and the area of either one of those is then the change in moment between those two points which will give us the maximum moment coming back to everybody I owe and so that's a that's 67.6 kilonewtons meters and our maximum expected moment all right so with an allowable shear stress based on whatever material is expected for use for steel for a stress limit of 160 megapascals then find a section modulus and recommend a beam that would work best for it we are going to have to go to the SI tables because it's an SI problem so those are just a couple pages down you just have to watch the fact that the things are in millimeters the eighth edition that's page 804 so we need to come up with a recommended section modulus we now know the maximum moment remember this is a minimum section modulus we're looking for stress so we have these numbers now just have to make sure that in the right units and notice a power couple powers of 10 is taken off of the number in the column if we do this in millimeters it's going to be a very large number so they didn't feel any need to put all those zeros in the column over so the whole tentative of third out of the column we got a section line just make sure the units work and recommend a couple beams three four even five need to just to make sure you got the right number of beams just to make sure you've gone through the catalog pick the pick the best possible best possible we don't have a book move this around as you need or scooch over with someone and take a look at their book but you all need to know how to do this because of nothing else we're going to revisit it on the final test yeah section modulus has units of millimeters or length cubed so you have to get this ratio which are the units of length cubed you have to get this down to millimeters cubed point to the base length you know just because that's what happens to be in the table I guess you can hand it with this into spread sheet in the meters don't forget to that don't miss to notice that the weight is kilograms per meter it's actually the mass per unit length not the weight per unit length so you have to make sure those units work as well calculated section modulus no sense going into the tables we don't have that one right got something Tom in terms of ten to the third millimeters cube because that's what we need to get into this table directly so we'll need this front number times ten to the third millimeters cube so whatever you give me you may have an offer so now we can go into the table and look around those these are way over the top we need to come down quite a bit maybe we put the 632 in here we don't know what we don't could be there's a beam available below that we don't know from the table but it's free to write them down we're looking for the lightest one so go ahead and take that 632 4W4 1039 section modulus 632 what else looks yummy take this 475 at the bottom we don't know what's below that our table they may or may not be a beam that's made like that w3633 now there may be other considerations of course cost availability the depth of these beams may be of importance depends on you know what kind of things you need to put in the inner space between the joy so that you need to run pipes or cables or something so what do we look at anything anything above 425 so here's 54 7W3 1039 you coming up with the same beams as possibilities everybody's got their finger actually running down the table can what table you looking at mine up here David when I converted into millimetres cubed I actually got 0.425 well remember remember when you convert from from millimetres you also have to cube this you don't just write a cube down here did others get this okay okay so double check those days a lot of power of tens in there we just picked up the 54 7 this 535 here 250 45 say what I would have that is what match mine it was not you got to check that sometimes yes others got this one last possibility I guess 583 259 583 results on the table was saying because of the way they published I missed one well 448 oh yeah yeah we're 422 so we want the next being 46 so you've got you've gone through the beam catalog picked out a couple possibilities and chosen which one 360 as a first step choose that one because it's lighter don't forget the value given here is kilograms per meter not newtons or kill a newtons per meter so we do a quick calculation of the beam weight compared to the total load beam weight we have a 5 meter beam and a 33 kilogram per meter unit the mass per unit length and then need to multiply that by g to get the beam weight and it comes out to be what 1.6 kilo newtons which is way less than the 110 kilo newtons we've got on here barely over over what 1% 92% of the of the total load so this is probably more than sufficient this 42 5 won't go up but not me over the 475 so it's the first step for us that would be the beam to order it depends on I don't remember if the factor safety is on what no if the factor safety is already on this number which it might be if you've got a factor safety of two and a half this is way smaller than the beam can actually handle plus there are different materials used for these beams notice that this there's nothing about the actual material of these beams it's nothing more than the geometry of the cross-section that's in these tables yeah well the density would be an error but who knows what that is yeah I guess the density would be in there so there are variations on this so if if anybody's got a smartphone you can download the u.s. steel app and order these beams for us right now okay nobody's going to all right yes I guess kids are not as wired as rich reported because somebody should have done that already you know we could have these beams here by the end of class somebody jump on it for us okay so I'm the only one to you I'm working way too hard it's time for you to take over all right look for a wide flange beam that can support this situation uniform distribution of the load 100 and 1300 pounds per foot for a 16 foot beam so this section is 10 foot the overhang is six feet and the allowable gear stress is 24 ksi allowable shear stress 14.5 will just take the maximum shear and then divided by the beams cross-sectional area which is also in the tables we'll have to go back to the English tables of course but the cross-sectional area is right there on the table you don't have to figure anything out just compare the shear or to what's allowable all right so recommend a couple beams and then pick one and then order say 500 up so you'll need the maximum shear the maximum fairly straightforward setup though if you can find what the maximum shear maximum moment are without doing the shear moment diagrams you're welcome to my I can't do that as well it's just a lot of you do the shear moment diagrams I think so come on for the maximum moment the maximum shear and then recommend a section modulus because being check the air cross-sectional area of the beam figure out if the stress is a little allowable well the section modulus will guarantee the normal stresses are but you've got to check the shear stresses as well you may just need to go to a beam with greater area with attention to our answer subtle thing that happens in here with the shear which tables on there what the little bit just gives you a little more conservatism in the value maximum moment maximum share can do without the shear moment diagrams more power to you I can't but if you're going to do it by eyeball you better get it right it's your units right because the area is an inch of square the section modulus and inches q then it goes down at the slope of 1300 minus 8 8 it's a jump up of 16 6 7 it's about right but then a moment let's see I know it both ends there's no sustained moment we know right here there's zero slope on the moment diagram and a bunch of curvy shapes in between so we can figure out this area will be the change in moment here that'll give us the peak at 6.7 that area is 6.7 figure out how far down this takes us and then this should bring us back to zero gives you two points you need to check and you can check them both probably most easily by looking at the area of the shear diagram above whatever this area is which is very easy to calculate will be this change delta m down to here that'll give you the change in moment between the same two spots we know we finished with zero moment so we can back up and get the maximum moment there you know what's 6 7 there and I think this comes out to be 23 4 we don't care too much whether it's compression or bending we've got to protect for whatever the maximum is watch your units allowable stress so recommend them a section just have to get the units right now pick four or five beams again just to make sure you cover them you may find that either load is increased enough that the normal stress has become a problem or that the shear is a problem you need to go to a greater area don't make this more problem than it is like those kind of channels we're trying to be a little more modest here at times build okay so pick a couple beams that look good this year my list that looks good watch your units kips cancel we get square inches on the bottom on top we get inches cubed so that's good and so remember x x axis we're looking for something that gets us just over 11 7 without going too far over because you're just paying for being it'll be 14 9 there 10 9 is not good enough there's 11 8 just skim by good enough we'll take it up there's going for it wait you're giving away the answer already not making everybody else work for it okay look look juicy is fun everybody loves to shop that's what we're doing I want the orange what 14 W 12 by 14 second modules 14 9 might want to write down the area on this one because we're gonna have to check the shear stress so areas for one six and it's an inches squared so that beam will do us 13 the 10 15 an area of 4.41 might do us shop for a couple when you go into the gap you don't just pull on one pair of pains you pull out a couple come into the room come out let your mom look at them see if they're okay these the same beams everybody else was looking at let's go down to the six could be substantially lighter let's see 10 2 is not good enough 13 4 is good enough 6 20 586 same beans you guys picked give or take a little bit and so the lightest one is that so you can check how the weight of this well 14 pounds per foot compared to 1300 pounds per foot it's not going to be a big concern but we can check the shear stress now here's the deal with the shear stress be careful with this because this is the entire cross-sectional area of that beam we're looking at I beams but as you may or may not remember when we talked about shear flow the flanges don't really carry much shear most of the shear is carried by the web so that's really the area then that we should use for the shear stress so we check it for our beam what is that that's the TW by the D minus 2 TF if you want to be really precise about it which being really on 12 by 14 so we do the 11.9 times TW give us a pretty good picture of what is going on a bit bit of shares held by this part so just to make the calculation more straightforward we look at it as if there are no flanges when we calculate the area for the shear stress so we take the maximum expected shear over the area of the web and excluding the flanges and see if we're below the allowable shear stress and that was what the D times TW so D for the beam we picked is 1191 TW is 2.2.2 inches 2 tenths of an inch that's what it says that's pretty darn small it's less than a quarter of an inch oh well that's what it says kind of scary if you ask me and we get 0.7 kips per square inch and our allowable is 14.5 so we're okay if it wasn't we'd have to find a beam in the greater area so this thing doesn't add appreciably to the normal stress and it's adequate in shear stress and notice that this area that we're using is about half of the total area of the beam exactly there because there's so little shear held by the flanges we take out of the shear calculation and we get a pretty good picture of what being would do us if everybody agree with most of those calculations this that's the web this is the web and these are flanges yeah just just to make it if you're if you're borderline with this then you need to really look closely but don't design the borderline don't don't design to just get by with these barely gotta gotta put some margin in here so that you don't end up in court defending what you did if you're cutting things that thin don't have a whole lot of defense alright any questions about this one okay another way we can approach these kind of things that may help keep cost and weight down is for sort of a selective beam design imagine we have a problem like this simply supported beam but that's an 8-liter beam and we'll make it simple I've known the load to be a big deal of this point low right in the middle expected of 500 kilo Newton but here's the thing happened to have in stock so we want to use these you know somebody bought way too many of these so now we got a user some W690 by 125 beams now I don't believe that's actually one in the book but I'll give you the information we need for it could be that it's in the new version no we only go up to 610 so this is a bigger beam than our book happens to have so we somebody bought these you know got them on eBay real cheap now we got to use them but here's the deal area 16,000 section modulus is 3510 that's 10 to the third millimeters cube and cross sectional moment of area 1190 10 to the 6th alright so for whatever reason there's the beam we've got the trouble is it's adequate for the expected bending moment remember we've done this type of problem before so the the moment diagram well this this shear moment diagrams are very easy to draw because of the symmetry of the problem and the fact that it's only a point load the trouble with this beam that we have to use because the boss wants to get them off the inventory that maximum moment comes out to be minus a thousand or not minus but a thousand that's killing newton meters the trouble is with this beam if you calculate over an allowable normal stress just like we had before 160 mega Pascal that this beam is adequate for only a certain portion in the middle of the beam there's more moment here than this beam can support looks something like something like this so for this portion out here the beam is okay but there's greater moment than this beam can support so what we're going to do is we're going to look into inventory and we're going to weld some plates on just in the center where we need the extra protection so that'll make the cross section of the beam which might look like that we're going to weld some plate on because we also have that at inventory and it's 16 millimeters in thickness we've got lots of this plate we can make it as wide as we need to so if we need to add enough plate to protect for this worst case point right here in the center that'll over protect the rest of the beam but there's no sense going any farther to the ends because that part of the beam is already protected see the picture then so we've got we've got this tail in the beam now we're going to put on a plate that will protect the rest of the beam in fact it will over protect it for those shaded portions so your job is to find out how wide the beam should be this be and I mean the plate should be and how long it should be L which is fairly easy to do just find out where these the points are that the moment diagram goes over the exceeded allowable part that we could everybody see the picture so when you do this when you look at MC over I you're going to have to recalculate a new moment of inertia for this new this this center portion where we added these plates because we now have all of this added area with these extra 16 millimeter plates that we've welded on so we're getting close to and I'll give you the skeleton of the solution turns out this portion here is 2.24 meters so that tells us how long the plate should be but now you need to find how wide the plate should be to handle the rest to do that we need to find out remember the section modulus is I over C C is increased now because of the extra plate but so is I so we need to figure out what those are this is got to be such that handles the maximum moment expected right in the center of the beam that thousand divided by the allowable shear stress of this particular steel so we can put all those pieces in remember that C increases by 16 millimeters and so we need to increase the moment of inertia by actually we need to be greater so let's write it that way we need an eye that's greater than this MC over allowable no I mean MC MC over sigma that turns out to be 222 times 10 to the minus 3 meters to the fourth remember that's partly made up of the moment of inertia of the beam which is right out of the tables well we don't have it in our tables but I gave it to you it's just that number right there plus the fact that we've welded on these two plates top and bottom the moment of inertia those is easy to figure because it's just a simple rectangle 112 yep using the same thing H cubed where H is given it's the 16 millimeters but B is not you're looking for that so you have to solve for this plus don't forget 1 half 80 squared because these plates are often the neutral axis so that will be a function then of the unknown B and you can solve for it so we'll know how long the plate should be we got that just from the moment diagram so that we know what part we haven't protected with the regular beam how much extra we need to protect and then we have now the width of these plates you should get about 267 millimeters that was real quick just because we're at the end yeah oh sorry there it is yeah that's 678 so that's this distance 678 millimeters your boss is going to love you because not only if you use the beams that somebody bought in excess but you've also used up all that extra plate that was lying back behind the last warehouse what glue it up you might want to weld it super good all right that was quick at the end there see if you get the same things now no right duty now ever that's one you could knock that out about 12 minutes on an exam