 So let's talk a little bit about something called distributed loads. So far, we've been talking about a concentrated force, or a point force, a force represented by a single vector applied at some specific point location. Well, that's an idealization that really simplifies reality. Force must, physical force must actually be applied over some sort of surface. Force distributed over an area like this has units of force per area, where the term per, again, means division. So for example, we have pounds, which is a force, per square foot, or pounds per square inch, or 1,000 pounds, or a kilo pound, per square inch, is another example, or Newtons per meter squared, if we're using SI units. And that is a specific unit known as the Pascal, and often abbreviated with PA. These are also units of pressure, although pressure is considered to be a scalar value acting in all directions, as opposed to a directed collection of force pieces. This directed collection of force pieces is what we call a distributed force. And we'll say that that distributed force is a type of vector that has a magnitude and direction similar to a larger concentrated force vector that might represent it, but it's distributed over a particular area. The distributed force is uniform if it has the same magnitude over the entire area. One example of that is something like atmospheric pressure on a surface. If we consider it on what we'll call the human scale, the scale we sort of normally experience. For example, if I have some sort of canister, maybe a can of soda, there is force pressing all over on the outside of the can due to the pressure of air. However, distributed forces can also vary. For example, something known as hydrostatic pressure, which is the pressure that a body feels when it's under water. You might have noticed that if you dive to the bottom of a pool, you'll feel more pressure on your chest, more pressure on your ears, and everything else, the deeper you go. Well, that's because the pressure is increasing. If we think about the entire depth of the pool and maybe the pressure that it's exerting or the amount of force that it's exerting when it pushes on the sides of the pool, this force increases the deeper you go. And that is the example of a linearly increasing distributed force. Notice it has a value of 0 at the surface. And there is more force the deeper you go. Yet another example of a varying force, a distributed force, might be a snow drift, where the different depths of the snow might result in different weight loads below different points. In two dimensions, force might be distributed over a length. For example, here's some sort of beam. And we might think of the force as being applied over the length of that beam or some portion of the length of that beam. We'll often abbreviate this with a vector q, which relates to some concentrated force total divided by, in this case, divided by the length over which the force is being applied. This has units of force per length. For example, pound per foot, pound per inch, or newtons per meter are common ones we'll see. You can think of this sort of as a distributed force applied to a width. For example, if I have an entire area with a length and a width, and that area has a distributed force throughout it, we'll call that distributed force f, small f. Well, we can think about some portion of that force as just the portion that's being applied. We're going to collapse the width in and think about everything along that, everything at the same sort of position in the x direction as being sort of concentrated in the middle. And so we're essentially applying just this segment here across the width. So our value q, which is our force per unit length, we can think about as being our overall force per unit length. Well, if our force, if that concentrated force is the same as our distributed load applied to an area, and that area is length times width, notice that the length will cancel out. And effectively, we have a relationship between our linearly distributed force and our area distributed force, which is scaled by the width of whatever you're doing. So it's important to think about your force, whether or not you're distributing it over a linear section, or you're distributing it over a three-dimensional or a two-dimensional area. An equivalent force system is one in which the concentrated force and a distributed force produce the same equilibrium conditions. This sort of depends on the placement of the concentrated force. For a uniform distribution, this is usually the center of the line or the center of the area of application. So for example, if we have a two-dimensional distributed force applied over some length, we might consider an equivalent concentrated force as being applied in the middle of that length. And its magnitude would be equal to the magnitude of the distributed force times the entire length. Similarly, if we're thinking in three dimensions and we have a force distributed over an area, an equivalent force system might include a concentrated force applied at a point, again, at the middle of that area. And the magnitude of that force would be the distributed load multiplied by the area of application. In this case, the distributed load times the length times the width of this rectangular application. When discussing distributed loads, it's important to understand what area is actually being referenced. Often, it is not the surface itself, but some sort of projection of the surface. So let me discuss this. Let's consider a trust roof. Notice this roof might be made with a series of trusses. And it might have certain things that are loading it. One of the common loads might be a snow load. Well, when snow falls on top of the roof, it falls pretty much in a vertical direction and lands on the horizontal surface. And then it might fall a little further in some places. But the snow load is usually considered to be being applied to whatever horizontal surface there is there. Let's go ahead and give some dimensions here. We're going to call this distance across the width. We'll go ahead and call this vertical distance the height. And we'll call this length of the entire thing, well, this distance all the way back, the length. If you think about the snow load, usually they discuss the snow load in terms of this projected area. In other words, it's assumed that it's projected over the length times the width. So a relationship might be that the total force due to the snow load is equal to the force, the distributed force of the snow load, times the length, times the width. Now it'd be the relationship between the total force of the snow and the distributed loading of the snow. Notice this would be different than if we were doing a load from the roofing material. If we were trying to consider how much the roof itself, the roofing material itself places on the truss, notice that would be this sort of diagonal sloped portion here. Notice we'd have one on each side. But if I just want to consider sort of this right side here, my roofing material load would actually get applied on the slope itself. So that area is going to have slightly different dimensions. In fact, let's go ahead and define an angle here, an angle theta for the relative slope of the roof. And we can find a relationship here. Let's go ahead and call this slope side. We'll call that distance s. We know that there's a relationship here between the height. The height of the roof is equal to s times the sine of theta. Whereas the width of the roof would be related actually half of the width of the roof would be related to s cosine of theta. But if we think about the roofing material, the roofing material is going to be applied on this diagonal. So the relationship between the total force due to the roofing material would be the distributed load of the roofing material times the length times this diagonal distance s. And if we were considering both sides of the roof, then we would have a factor of 2 for both sides of the roof. Another common load for something like roofing might be a wind load. Notice in this case, the wind load would be considered as being applied to only one side. And that is considered as being applied only to the vertical dimension. So the resultant force from the wind load would be equal to the distributed force from the wind load times the length, the entire length here, but then times the height. In this case, this is the height of the projection. That's the area over which wind loads are considered to have been applied when they're originally measured. So if we want to think about all the loads that are sort of being applied to this roof, we can think about, first of all, the directions that they're going in. The snow load looks like it's being applied in the vertical direction. The roofing material is also gravity, so it's also being applied in a similar vertical direction. And the wind load is being applied in a horizontal direction. So if we were trying to add up all the forces, or the different forces that were being applied to this particular roof, we would have to consider the different components, the different directions. Let's see if we can figure out what the distributed component would look like in the x direction. Well, that's simply going to be the wind load. And that distributed is going to be whatever the total load is in the x direction divided by this entire surface. Well, we know that that total load is the force of the wind. Well, we know the force of the wind is the distributed load of the wind times LH. Notice we can cancel out the L's. And if we notice that H equals S sine theta, we can also cancel out the S's and be left with a sine theta. So our total force in the x direction, our total distributed force in the x direction, would be the force of the wind, which is all in the x direction, times the sine of our angle theta. That there's a trigonometric relationship between how the wind is considered to load and how it is actually applied to our angled roof. If I do a similar thing for the forces in the y direction, I can think about the distributed force in the y direction as being some total force in the y direction applied over. Well, let's just consider one half of the roof. And that's going to be equal to the total snow force plus the total roofing material force in the y direction. And if we think about our snow force, our snow load, is going to be the distributed load of the snow times a length times, and we'll just consider half the roof. So times half of the roof there over SL, plus the distributed load of the roofing material. And again, we're only considering half of the roof, so we'll get rid of the two that's there. And again, we can cancel certain aspects here. We can see that the L's cancel out in all of these places. The S cancels out over here. And if I recognize that W over 2 is equal to S cosine of theta, I can also cancel the S's. And we get a relationship that says my distributed force in the y direction is equal to my snow load, which has a component that depends upon the angle and my roofing load, which, since it's generally measured along the actual slope of the roof, doesn't have to be scaled by the angle. If we then think about the roof as being a series of n trusses, we'll notice that the trusses on either end effectively only count for half of the roof area. And we can divide the roof into little sections held up by each truss, where the ones on the end are about half a section each. We can think about each truss as supporting some portion of the distributed force as it's applied over some length of the roof, some fractional length of the roof. In other words, my linear distributed force would be equal to my area distributed force times the length of the roof maybe divided into n minus 1 pieces. And again, my linear is measured in something like pounds per foot, which would balance out with my distributed pounds per square foot times the length measured in feet, in which case the extra foot cancels out and both sides are measured in pounds per foot. So if I can figure out that length or that segment of length, I can find out some sort of distribution along the truss itself. So now if I now look at the truss from a front view, let's consider this truss as being a pretty simple one, I now have a distribution of forces. On the right side, we have the combination of the roofing material load, also known as dead load, and the snow load, and the wind load, whereas on the left side, we don't have the wind load. But if I look at each of these pieces, let's go ahead and call this q on the right side and q on the left side. If I look at these pieces, we know that trusses assume that the loads are at the joints. So each joint needs to be associated with some linear piece of the load. So for example, I might take this joint down here and divide the member that's attached to it in half and assign that portion there a length of s over 2 to this lower node. For the joint here in the middle, I might assign half of each of those. Actually, I'm sorry, this is a distance of s over 4 for that lower part. This is 2 of those quarter pieces, which is s over 2, whereas it's the entire length. And then this upper part might also be distributed over a segment of s over 4. So I've now basically broken this into one, two, three pieces. And each of those would be the distributed force qr times the length, s over 4. Here it would be qr times s over 2. And this would be qr times s over 4 again. Similarly, on the other side, I could apply, apply and apply ql times s over 4, ql times s over 2, and ql times s over 4 again. Notice at this top node at the peak of the roof that we both have the ql, the distributed force from the left side of the roof, and qr, the distributed force from the right side of the roof acting together. So we would have to add those two vectors together. Notice for each of these qls, we have a measurement of pounds per feet. And we're multiplying by a length, likely measured in feet, which gives our answers, our forces here in units of pounds, which is what we would like for loading at each of the trust joints. Why do we need to know and work with distributed loads? Well, one of the primary reasons, distributed loads work as standards, because they scale independently and thus apply to situations regardless of size. In other words, I cannot tell you what the entire snow load is on your roof without knowing the size of your roof. However, I can tell you what the snow load is expected to be for an area of your roof, and then you can do the multiplication. I could tell you that there would be a certain weight per square foot, and then you would be able to do the calculation based on the size of your roof to figure out the entire load. And this is one of the key reasons why we need to be able to work with distributed loads and translate them into concentrated or point loads.