 Alright, this course generally, especially for those of you who are really interested in mechanical engineering, or civil engineering, this will lead directly into a course called structures. This course actually goes by several names, in fact enough names that I'm not even sure what we call it in the catalog, but it goes by several names, one of which is mechanics of materials. We're going to look at the physical properties of a material and its response to load. So it's also called strength of materials, it's also called things like deformation of elastic solids. If you think about it, that's exactly what we want structural materials to do. We want them to be able to flex under loads, and when those loads are released they return back to their original condition, ready to take the next load that comes in. You can do something more than what you do when you come into class, and the building, whether you know it or not, actually flexes under the load of us being in here when we weren't in here half an hour ago. And when we leave the building, the building will return to its original unloaded state. Actually it's sort of preloaded because the structural members of this building are holding everything above them, which includes floor, walls, and ceiling, as well as a lot of general mechanical stuff, things up there. So this course goes by a lot of names, but it does come directly off of the statics we did last fall. So everything we did there is going to lead directly into what we're going to need to do here. The very first thing we're going to have to determine is what the external loads are, and then from those external loads we're going to do exactly what we did part way through statics where we determine the internal loads in the material. That's the static response of the material itself such that there is internal shear, internal moments. We're going to also add some other types of loads to this, and it's those loads, the material responding to the external loads that is reflected in the deflection or the deformation of the structural materials that we're going to look at in this class. So that's the basic idea of what we've got going there. These two, the statics and the strength of the materials are so closely related to each other a lot of schools put these together as a single four-hour class where we have them separate as two three-hour classes, but they're that closely linked. So our number one tools and of course will be that the forces on all our objects will sum to zero, and of course the moments will sum to zero, and that's all forces taken into account in any way whether they're internal or external to the structural piece itself. It's still going to have to be a static situation for everything that we do. So let's warm up with a real quick little statics problem and then we'll make the next step as we look into the response of the material and start to get a touch on how things actually deform under these loads. To start with, we won't look at the actual deformation. We're a couple days away from that, but we will very quickly get to that. So imagine a very simply structured object of some kind where we have two simply pinned straight members holding some kind of load there. Nothing more than we did in the static. In fact, we were doing this kind of thing pretty much in the first week of statics. We went a little bit farther with it as we went along, but pretty much we weren't doing a whole bunch more than things like this in the very first weeks of static. So it's just a little bit of warm-up, but it also allows us to make the next step as we look at very different, we take a very different viewpoint of this structure than we did in statics last term. So let's say we've got a 30 kilo Newton mode there and first thing we need to know is what are the external forces on each of these members. That's going to then allow us to look at the members themselves and see how they're going to respond to those loads. What we did in the statics was we used those external loads to figure out what the internal loading was, the response of the material to those external loads. Now we're going to look at the actual physical response of these members as they react to these loads. So it shouldn't take us too long, we realize, I hope, recognize that this is a 3, 4, 5 triangle and that means that all of the ratios of all of the forces in this are also of that same ratio. So we can figure out then based on a force balance at point B here, because that will then give us the forces in the two members, since both the members meet right there, that we can very quickly determine what the external forces are and then we can make our next step and then into this new subject matter we've got here. So that's 30 kilonewtons there, obviously, and then let's see, let's label these something, we'll call that FAD and FBC. They also must have the same ratios as that 3, 4, 5 triangle because, well, why? Let me ask you why. Why must the forces be in the same ratio of 3, 4, 5 that the geometry itself is? It has to do with the fact that they're in the same directions. Well, why are they in the same directions? They're two force members. Both of these, the two structural members, AB and BC, are both two force members. Therefore, we know immediately that the forces lie in the same direction as the members themselves. Well, more specifically in the line connecting the two pins at either end. It doesn't matter what the member does in between. The forces are directed through those two pins. So then we can figure out then what the forces are without having to do too much quick calculation because we've got the same ratio in the forces that we have in the geometry itself. That's nothing more than anything we would have done in statics the first or the second we were doing that kind of stuff. What we're going to do now is take a look at the internal forces that are caused by those loads. So if we bring this little piece over here, kind of blow it up there so that you can see it, we have this, in that case the 50 kilonewton load is acting throughout this entire piece. And in fact, it's spread over that entire cross-sectional area. We'll only draw it for the most part as a single point load. But that's just for a matter of expediency on our part. But there is a bit of a distribution of that force over that area. That should make some sense to you because in general a thicker piece of material can hold a greater load than can a much smaller piece of material. Area most certainly does have something to do with it. That's not great news to you. You've got that kind of experience anyway. If you're building a deck, a dock, a wood shed or something, you don't want to do it with toothpicks. It's not going to hold. Well, you may want to do it with toothpicks. But you're not going to. There's just not enough material there to support the kind of load you're seeing. And this has to do with this cross-sectional area. A very quick word about that force distribution over that area. It's not uniform. If we look at it in different places through the piece, it tends to have sort of a distribution that's like that, where most of the force is directed through the center of the piece. It tends to drop towards the edges. We're going to treat it even more simply for the most part as uniformly distributed, but not even that. We're just going to say we have this force acting on that cross-sectional area. And we're not going to worry very much about the precise distribution of the force itself. In more advanced studies of this material, yeah, you're going to have to. But for what we're doing here and our start on this, we don't need to worry about too much more than just that there's some force in some area as we're concerned with it. So those two quantities, the internal force in a material in a structural member and its cross-sectional area are right now the two most important things that we have to concern ourselves with. Now as we look at the material and its response to this, and we're finally now bringing into account more than just the length of the member, which we looked at a lot in statics last term. We're now looking at more of what the cross-sectional area is. It should follow fairly with good fair sense that the ratio of F over A would have some import to it. It makes sense, I think, as the force goes up, things get worse for the material. As the area goes down, things get worse for the material. So the ratio of the two makes some good sense because it's their movement in opposite directions. As one goes up and the other goes down, both of those contribute to a worse case, a less advantageous situation for the material itself. What we're worried about happening in this class is things going to the point where the structure itself fails, whether the material actually breaks catastrophically or maybe it just deflects enough that the structure no longer works like it should. It's quite possible that the load could go so low even though none of the members break that the structure doesn't do what it's supposed to do in the way it's supposed to do it. So this ratio is very important to us through this class. And we call it the stress and give it the symbol sigma. Great lower case sigma. And it's the force in a member divided by the cross-sectional area that is supporting that force. More specifically, this is known as the normal stress. Remember the meaning of the word normal to us? Perpendicular. Because this is the situation where the force we're concerned with is perpendicular to the area we're concerned with. And that's indeed the case that we have here where the force is perpendicular to the area. What that should mean to you is that we can take other areas. They don't have to be normal to the axial direction of the member itself. We're going to look at areas that are taken more on an angle and see what the forces are there. We're going to find out that's very, very important because when some of these things fail, they fail not at a direction perpendicular to their own axis, but at an angle off that axis. And you've seen some of this kind of thing before. When pieces fail, they don't fail in a nice cut across the piece. They fail at some angle across the piece itself. So you've seen this kind of thing before. The units. Take a rest. David, you're working too hard here already. Chris isn't ready to work at all. The units. Sort of. Two things. One is you remember our long tradition in the physical sciences of naming recurring sets of units off dead white male German physicists. So this is known as a Pascal. One Newton spread over an area of one meter squared is known as a Pascal. It's a unit of pressure. It's the very same unit of pressure that most of you have taken physics to. So you did some fluid statics. You were looked at pressure. This is the very same unit of pressure that was used there. It's exactly what it is. This is internal pressure, if you will, in a loaded member. We typically are going to be working with very, very large forces. So it's much more likely that we'll look at kilo Pascals and mega Pascals, even giga Pascals, rather than just Pascals themselves. But that's just a matter of us taking care of the units as we work through these. Since we are American academics, we also need to look at American units. Typically, we look at them in pounds per square inch. Again, since we're talking about rather large forces for the most part and some of these structures, we'll be looking at more likely kilo Pascals. Kilo pounds, or KIP, per square inch, or also known as a KSI, which is a KIP per square inch, and KIP is 1,000 pounds, a kilo pound. So we're going to have to get used to some slightly different units because these numbers typically can get rather large. Alright, so we're looking here at the stress in a material, the force it's maintaining, divided by the area of the material that's maintaining that force, that's resisting that force, if you will. We'll typically, or the main loading we'll look at now for the first couple steps through this is either a tensile force or a compressive force. For several materials, their response to tensile loads is very much the same as their response to compressive loads. Structural steel is very much that kind of material its ability to withstand stresses is about the same whether it's intention or compression. That's not true with very other, several other common structural materials, for example wood. Wood is very good in compression. It's terrible in tension. It's a factor of somewhere around 8 to 13 difference. If you're building something, you typically don't want to hang it with wood structural members. If you're going to hang it from something, if you're going to build a shelving system that hangs from the ceiling of your garage, you're going to want to have it hang from chains or cables rather than from wood straps themselves. Concrete is also that way. Very, very good in compression, not very good in tension. Sorry, other way around. No, that's right. Very good in compression, not very good in tension. We're going to have to deal with that later in the term when we look at concrete structural members because they're very commonly used in buildings, especially a lot of modern skyscrapers. It's a very easy material to work with. It can be transported in its mixed form and then poured into the shape needed for the structure right on site. That's very, very useful to an architect or a structural engineer in putting things together. It's an ability to add an aesthetic component to your structure to make it look good as well as have it behave well under these structural loads. So we've got all those kind of things that we're going to be looking at as we go through the term. Another type of simple loading, and we've looked at it some in the statics we had before, is that a structural member can have not loads that are actually directed but are transverse to the material. Now for the simplest statics we can do on this, those two forces have to be equal for the sum of the forces to be zero. Remember it's also true of the moments must be zero. What we're talking about though when we talk about shear forces is the situation where the distance between these two is very, very small so that we're not worried about the moment that that couple is causing. We're only concerned with the shear that it's causing because if we take a look at a piece, an imaginary cut through the piece, we know that that causes internal shear forces and that can cause a different type of damage and it calls for a different response from the material itself. But it's still the ratio of the two, the force over the area on which it's acting that is our concern. This case, the force is not normal to the area but it's parallel to the area. This is known as shear stress. And since we're assuming a uniform distribution of the force across that area, we're specifically talking about average shear stress. This is the type of thing that we'll see in the pins that we're using to hold these pieces together. We never look specifically at the pin joints themselves. We just assume that whatever pin was in there was strong enough to hold it. But if we look at two members that we've pinned together, we're going to do it by probably putting a rivet or a bolt or something through the piece to hold those two structural members together. That pin itself then is under these type of forces that we have here, these shear forces here where the area of concern is right there where the maximum amount of the shear is occurring. So we're concerned with the cross-sectional area of the piece and then we're concerned with the forces that that area is supporting. I have to draw something up here. So if I draw attention, it doesn't mean that compression is not a concern. As far as the pin itself is concerned, there's no difference between a piece being in compression and a piece being in tension attached to it. The pin itself still needs to resist that internal shear no matter what the piece is, what loads it's undergoing. All right, so we're going to go to the screen here. If it comes up, good. All right, I'll give you each a copy of this. We're going to now look at that very same structural object we had before, that simple triangular piece maintaining a load of 30 kW. But now we're going to look at the pieces themselves and figure out what some of the stresses are in those pieces because the places where the stress is the greatest, assuming that a piece is made out of the same material throughout, wherever the stress happens to be, the greatest is going to be our place of greatest concern. So there's a lot of little structural pieces here and we now need to look at all of them. So we'll pick one to start with. Let's look, let's see, at the member BC, the diagonal member across the top of the piece. Now in the center of the piece, we have this cross-sectional area, a little exposed part there, for example. We have that cross-sectional area that needs to withstand the forces on there. And if I remember, that was 50 kN, was it not? And that cross-sectional area itself is, well, the piece has a diameter, you notice, of about 20 millimeters. So it has a cross-sectional area of, how do you figure out the area? We're assuming that's a circular piece. You can tell that from the type of shape that the person, the draftsman, put in. That indicates a circular piece. Same kind of thing I did here on the ends of my piece. That's a common engineering notation to indicate it is a circular cross-section. How do we figure out the area of a circle with a diameter of 20 millimeters? Five times what? Pi R squared, or Pi D squared over 4. And I've got that for you. 314 times 10 to the minus 6th meters squared. So we can figure out the stress in the piece at that cross-section by simply determining the ratio of the force to the area that's withstanding that. 50 kilonewtons in the minus 6th meter squared. What units? It's easy enough to divide those numbers. I'll give it to you. It's 159, and this is intention, so we'll tend to call that a plus. Do you figure out what units? Let everybody think. David? Joey? You don't want to think yet? It'll be 8 o'clock on the first day. 3 times 10 to the third newtons, that's a kilonewton, over, we'll call it about 3, times 10 to the minus 6th meters squared. So 50 over the 314 doesn't quite give us that. 50 over, what, 0.3, right? So that would change this to 10 to the minus, the 50 over the 0.3 will give us the 159. Is that right? So this will be now 3.14 times 10 to the minus 3. 10 to the minus 3. So we have 10 to the third over 10 to the minus 3, that's 10 to the third over 10 to the minus 3. They cancel? Of course not, they're not the same. They have to be the same to cancel. So it's 10 to the 6th. That's what's just in the newtons per meter squared or megapascals. In engineering, it has long been the tradition to keep the exponents on the scientific notation in multiples of 3. Anybody know why that's long been the tradition? Because it's certainly not necessary. We have these prefixes for just about any number. No reason we have to stick to multiples of 3. Anybody know why? The tradition is long been stick to multiples of 3 if possible. You can't possibly know. So that's why you're looking at me so puzzled. When I was a baby engineer, everything we did was done on a slide rule. A slide rule can do division of two numbers very easily, but it cannot take into account where the decimal point itself goes. That had to be done. We had to do that ourselves by doing this very type of thing, writing it down. We could do the division on a slide rule, but then we had to look at the scientific notation itself and figure out what the result was then. It's just a lot easier to do that in multiples of 3 rather than get 7s and 4s and all kinds of different things thrown in there. It was just easier to keep everything as a multiple of 3 to make life a little bit easier on the slide rule. Not a concern anymore. So it's not nearly often it's done that we keep these in multiples of 3 because you just enter the whole number on the calculator and do the division straight away and take whatever is written down. But I can't guarantee you're not going to end up working at some place where the tradition is maintained very, very closely. It could well be. It also does simplify things and it keeps the number of prefixes that we need to remember down. Killa, Mega and Giga are our main ones. We'll use the micro a little bit in a step coming up here soon. Just a little bit of history for you. I guess I should ask, does everybody know what a slide rule is? Joey, you know what? In my office I had one of those 6-foot yellow ones. That's the one I used to carry around as a friend. I used to use it instead of the calculator. Those were our calculators. In fact, one very nice thing about a slide rule is they automatically maintain the right number of significant figures in an answer because you simply cannot read a slide rule to any greater accuracy than 2 or 3 places and that's automatically maintained through the piece. One way they used to get greater accuracy, more significant figures on a slide rule is they simply built them bigger. I have my grandfather's slide rule. It's this long and it's got a magnifying glass over the slider hairline so you can read even more closely. It can be more finely divided. I imagine it was a very expensive piece. It comes in a leather embossed case with his name and gold letters on it. What? It's instrument 01. Yeah, point 01. It was a big deal and if you go on eBay, there's people that are trading in slide rules as now almost like archeological pieces. Nobody, very, very few people actually still use them. Alright, so there's other places then we need to now concern ourselves. For example, up here at the end, C, if you look at the drawing there and see what we've got, we've got the piece of NC actually flattened out a little bit so that it will then mate with the other piece there and there's a hole put through it. I'm looking at the upper end of the piece of BC, the diagonal piece that goes across here. It's actually the same thing as the end down at B. That's a concern for us because now the cross-sectional area that's maintaining the force, internal force, has changed. If we look right across there, that area still needs to withstand the 50 kilonewton load. If we look at it in cross-section, it looks something like this. There's the hole and now there's the cross-sectional area that's trying to withstand that 50 kilonewton load because the 50 kilonewton load hasn't changed anywhere along the piece but now the area has changed that's withstand that load. And so that piece is now, let's see, I think this is 40 millimeters across there, is that right? And the diameter of the pin, because that's also the diameter of the hole through that piece, make sure you know how to read these. It's that pin through there, D, that's making this hole, that's a 25 millimeter diameter hole, which means the area maintaining that 50 kilonewton load has changed. So let's see how it changes the stress in the material at that point. The force is still the 50 kilonewtons. Yeah, there's a little bit of friction between the wall attachment and that piece, but we're not concerning that ourselves. Remember, we're looking at these unless otherwise said that are freely pinned pieces. So the friction there is very, very small. So this is 40 by, what's that distance? 40 minus 25 is... Yeah, so that's 17 and a half, but there's two of them on either side so we can concern ourselves with just the full area. 40 by 15 millimeter squared, and that's 60, 600 millimeter squared. Is that right? So what's the stress in that piece? That way we can compare it to the stress in this piece. Again, this is still a tensile load, so we'll call it a plus. So the stress in that piece now, who's got the calculator out? David, you do. Did I do it right? What? The stress in the cross-section? Yeah, this is the cross-section. The 40 by 15 is right, isn't it? Isn't that the attached area? Because remember, the whole area has been removed. So it's not there to withstand and it is still at 50 kilonewtons. So the 50 kilonewton force is now only through those two areas. The pin is notwithstanding any of this tensile force. It's just holding the two pieces together. We'll have to look at the pin separately because the pin is in shear, but the member itself is in normal stress. David, what's that come out to be? Something wrong with that. That's not the number I had. What did I do? Something's wrong there. I put in 50 kilonewtons. Oh, it's the millimeters. No, no, this isn't right at all. Because the area, we also have to take into account the thickness of it, which is 20 millimeters. Then it's time for the 40 minus the 15 millimeter squared. So that didn't take into account that thickness, which is 20. So 40 minus 15 is what? 25 times 20 is 500. That's still not right. 40 minus 23. 300. No, okay, that's right. Now we're okay. Yeah, 40 minus 25. 15 is the result. It's my first day, too. Is that better? Is that now the crosshatched area that's resisting that force? Okay, that's a little bit better. So it should be about 167 megapascals. Is that about right now that we got the numbers worked out? So notice that in the center of the piece we only had just under 160. Now we've got someone over 190 as a designer your concern is more now with the end piece withstanding that load because it's under greater stress than is the center of the piece itself. Not by a lot. Typically when you design structural materials you would throw in the factor of safety of at least two anyway. But it does highlight the fact that the concern is not in the middle of the piece but at the end where the shape of the piece has been changed and is now pinned through. All right, let's see. Any questions? What else can we look at? Let's look at the piece AB now. AB you'll notice is a rectangular piece. So it's interior cross section somewhere in the center is something like that. Imagine where this cut is. It's in compression remember? I remember by 40 kilonewtons. So what now is the compressive stress in this piece through the center of the piece AB? It's got to maintain the 40 kilonewtons but over what area? It's this cross sectional area here so it's the thickness divided by the depth. Looks like it's 50 and then this is millimeters. How deep is the piece? Everybody agree with that 30? You actually have to look down at this piece here. Here's AB here and you can see then it's thickness So the area is 600 square millimeters 1500 Looks a little better. So the compressive stress in that one will give it a minus sign to indicate compression if we need to. Since we're not talking about what specifically this material is we're not directly concerned with whether it's compression or tension but we certainly will be when we specify what the material is we'll have to look at the material response in compression and in tension. So what do we get for a stress there? 26.7 One thing we're going to look at later in the term is there's a different type of failure in a piece like that that's under compression where it can deflect sideways just like a column would when it's loaded we're going to have to look at how it resists the tendency of it to bow out laterally and in what direction it'll do that. Then we have that so we can look then at at this other end here where it's now this fourth part AB now comes into a U connection and it's now the cross-sectional area on point B it's now this cross-sectional area that needs to withstand the same 40 kilonewtons that was in the rest of the piece as a whole. I think by inspection you can see that there's more area there so that's going to be less concerned in terms of failure than would the the original piece in the center of the piece. Any questions so far? Pretty straightforward. It's not too terribly different than what you might have put together yourselves anyway this idea that we're concerned with the force being exerted divided by the area that's withstanding that force makes a lot more sense to do it that way than it does the product of the two as we see. The next point of concern maybe let's look at let's look at the pin at point A we've got a U-shaped connection at the wall that's this U bracket that's mounted right on the wall and then the structural member AB fits in there and it's all pinned together now that pin A is now under shear stresses such that there's a 40 kilonewton force there and two 20 kilonewton loads there just by symmetry of it I know that it's the two 20 kilonewton loads over that pin A so if we look at this little piece here we look at that little piece there it's got a 40 kilonewton load that way from the member AB itself and we must have shear moves across that face there this by the way is called double shear hopefully obvious reasons there's shear at two places of equal size on the piece is this piece in equilibrium yes or no fill your nod yes David that piece is in equilibrium because whatever couple is caused by the 20 kilonewton is defeated by the 40 kilonewton itself so the shear shear stress then how much shear is being withstood by how much area do I put in 20 kilonewtons here or do I put in 40 what you're feeling it depends it depends on what area is put in if I put in the 20 then I put in just one of these areas withstanding that 20 if I put in the full 40 I've got two areas withstanding that that either way the ratio comes out to be the same but it's the force being resisted divided by the area resisting it in this case we have a 25 millimeter area so just one of those areas is withstanding the 20 kilonewtons if you put in the full 40 you double the area and you get the same ratio anyway what's that come out to be you got it David 491 maybe with a little bit of round off here or so I want the shear stress 20 from 25 over 2 then squared 40.7 40.7 what is that a question mark I'm not sure if they use Pascals either one it's not strictly a pressure but we still use the same 40.7 without any scientific part to it any scientific rotation part to it is mega Pascals depending on what material that is you need to look at the ability of the material to withstand the shear stress certain materials it's very very much a concern of how they react to shear stresses for example if you look at a piece of wood as a pin which is very common if any of you especially in the Adirondacks we have a lot of those timber peg homes where the girders and the joists all the structural members are held with wood pins actually driven into the holes if the grain of the wood runs this way then the wood pin is very good in shear if they cut those pins such that the wood grain ran this way then it's terrible in shear and that wouldn't last very long if you even go buy a simple simple dowel at Home Depot or Loaves all of them are with the grain this way you can't even find a piece nobody would do it it just wouldn't hold wouldn't suffice any questions there's other places that are could be of concern and we need to look if you were actually designing this you need to look at every single one of the ends as well as the places themselves for example if we look at pin B pin B is a rather interesting piece itself because because it's got several loads coming in there they all come in certain different directions for example the 30 kN which is straight down is actually over two areas that are supported by this I don't know what you call it this little bracket piece so we actually have one, two, three, four five areas of pin B that need to maintain some of the load the two skinny outside areas where that bracket is and then the two or three areas I guess that are about the same size that need to maintain where the structural pieces come in so if we try to draw these in some direction let's see on the two outside slender pieces we're drawing even the outside part of the pin that's not in contact with any area because there's no load on that there's no load on it there's no sheer concern there on those parts of the faces by symmetry we assume half of the 30 kN load is on each one of those and then the part B AV comes in this U bracket at the end of that so that's the two pieces and that's in the piece itself is in compression what was the load in that? 40 kN so there's 20 kN on each of these and then the diagonal piece itself comes down B that's just the center piece now and it's in the center piece's intention so it's pulling on the pin there and so we have those loads little difficult to draw as three dimensional things always are but that's the type of thing that's happening at pin B if we turn it on the side we'd see the two 15 loads coming down like that the two 20 kN loads doing that so the concern with this pin the pin in B is which one of those faces we have these four different faces where there's four different shears going on which one of those faces is subject to the greatest shear stress because you have to design for the greatest load in order to maintain all of the loads the area is the same let's label these these faces with some reference letters E, F, G and H what's the shear at face E actually I gotta draw that one in a little bit more on that piece there acting on that section so if I look at just the face E the first face where we're seeing shear is the face between this U-brack the load bracket and the end of the piece AB what loads are on what shear load is on that face and is that the face of concern that's maintaining just one of the 15 kilonewtons loads the same thing is going on at face H just flipped around the mirror image of it so we don't need to look at that that puts an internal shear of 15 kilonewtons at the face E that again is the load bracket pulling down on the pin and so the shear stress at face E is 15 kilonewtons and what's the area of pin B that's withstanding that 15 kilonewtons pin B is how big here's pin B 25 millimeters and so what's the shear there 30.5 looks good enough any units? alright so you get out a class question what's the shear at face F or G? they'd be the same either one what's the shear stress at face F and G it's a little bit probably easier to look at it on end because it's got the two loads like that it's only half of the loads from the members because there's two sides to maintaining it's both F and G so we're only going to put half of the loads there what's the shear and the area is still the same it's the same pin it's certainly possible I guess that we can have a pin that changes an area but we don't in this case just a simple cylindrical pin put in and then tell them got it? you ready? okay first get out a class question there's no trouble, same pin, same area but what's the shear there? once we know what the shear is we just put it in a divide it's real easy what is the shear there? is it 35? remember we don't add the magnitudes of vectors we add the vectors themselves so those two vectors added together got to be resisted by a shear across the face 25 in fact that's our 3, 4, 5 triangle again and again if you look at the two areas F and G together they're maintaining 50 kilonewms in the building area, you get the same ratio anyway and you get 50.9 make