 We're going to solve this simple problem here, which consists of a beam with a complicated distributed load. Now the question asks us to calculate the forces F, A, and F, B, which are applied here at A and B, required to keep this beam in static equilibrium. Now earlier we looked at how we would actually calculate or deal with distributed loads, and what we would normally do is take that distributed load and replace it with a resultant force, which we'll call R, which will have a location, X bar, that will be statically equivalent to the distributed load. And we did this using the equation R is equal to the integral over the length of the beam of the distributed load as a function of X dx, where X would be the distance along the beam. And then we would find the location X bar as the integral over the length of X times the distributed load as a function of X dx divided by the total resultant load, which is the integral over the length of W of X dx. Now we have a slight problem with this problem that our distributed load is actually a discontinuous function. So formulating this expression of W of X is a little bit of a pain. However, there is actually an easier way to deal with this problem, and that is by just dividing it up into sub-areas that we know the solution. So let's take a look at that. If we take a look at our loading here, what we would actually like to do is come up with a resultant force for that distributed load. But instead of coming up with one resultant force, what I'm going to do is take that distributed load and divide it up into smaller known areas that we can deal with. So the first area I will define is a triangle, which I'll denote with a Roman numeral 1, and that area will have a resultant force happening at a particular X bar position. I can then define a second area, and it will have its own resultant and position, and then a third area and a fourth area. So what we can do is use the known solutions for the resultant and the position of the resultant for a rectangular and a triangular area in order to set up this problem. So if I look first at area 1, what we need to do is consider this triangular area and also consider the points at which it's acting. So not so much the fact that W is 6 there, but just the height W is 6 here, but we want to deal with the height of this triangle, which is 2. So that's why we have this 2 here, and it starts from X equals 0 to X equals 4. Now we know that the resultant force is going to occur, we derived it earlier in the class, that this is 1 third the base, which I'll call A, so if I call the base of that triangle A, it's 1 third from the side of the triangle, or we could also say it's 2 thirds from this point. So if we use that result, we can see in our case that A would be 4 minus 0, and we will get that X bar 1 is 2 thirds times 4, which is 8 over 3 meters. Now we also know that the resultant is the area of this triangle, which will just be 1 half the base times the height, which will give us 1 half the height, which is 2 times the base, which is 4, which gives us 4 kN. Now if we move on to area 2, it is also a triangle, but now the height of our triangle is 4 and it starts at X equals 4 and goes to X equals 10. So when we define X1, we have to take the base here, which is 10 minus 4, which is 6, and we are 1 third of 6 from that edge, but that edge starts at 4. So we get that X bar 2 is 4 plus 1 third 6, which is 6 meters. Now for the resultant, we get 1 half the base times the height, so 1 half 6 times 4, which will give us 12 kN. Moving on to the third area, we now have our rectangle here has a height of 4 and a base of 4. This is 4 and this is 4, and it starts at X equals 0. So our resultant force will be in the middle of this rectangle, which will be 1 half of 4. So we get that X bar 3 is 2 meters, and the reaction force itself is going to be the base times the height, so 4 times 4, which gives us 16 kN. And going just a little bit more quickly for area 4, there's a typo on here, let's correct that. We now have starting at X equals 4 to X equals 10, so our base is 6 and our height is 2. So we get that our X bar is 4 plus 1 half the base, which is 7 meters, and our resultant is just the base times the height, which is 12 kN. So now we have the magnitudes and positions for all of this here, and we can translate that onto our free body diagram and perform our equilibrium. For the first part of our equilibrium, we will do some of the forces in the Y, taking upwards as our positive reference. Now if we see that, we see that FA and FB are acting upwards, while all of the resultant forces are acting downwards. So we'll get FA plus FB is equal, or minus 16, minus 4, minus 12, minus 12 is equal to 0, and if we rearrange that, we will get that FA is equal to 44 minus FB. And just as a reminder, that value is in kN, all of our forces we have been calculating are in kN. Now we can do some of the moments, and I will pick point A as my reference point. And if I do that, the force at A passes through that point so it has no moment arm, where the force at B will create, has a moment arm of 4 plus 6, and when I cross that with FB, I'm going to get a counterclockwise moment. So I will get FB times 10, and that is positive because I've used counterclockwise as my positive reference. Now all of the other forces are acting downwards and will create a clockwise moment, so it will be negative. So first we get 16 times the moment arm of 2, minus 4 times its moment arm of 8 over 3, minus 12 times 6, and then minus 12 times 7. Simplifying that, I will get that FB is 198.6 repeating, and then I divide that by 10 because it was FB times 10. I can take that result which is in kN again and complete it and simplify it and stay close to reasonable significant digits, and I get 19.9 kN. Now if I sub this value back into the equation for force equilibrium, I can solve that FA has to be 24.1 kN. So this becomes our result that we calculate for this problem. Now I always recommend when you've calculated the results, ask yourself, do these results make any sense? Usually you can analyze the results a little bit and see if they sort of came out as expected. So here we can see that FA is larger than FB. So does that make sense? Well if we look at our original diagram and the loading, the loading is much more concentrated. If I kind of draw a line down the middle, the loading is concentrated on the left hand side more so than on the right. So we would expect the force at A to be larger. So yes, this answer makes sense. Okay, I hope that solution helps you understand this problem a little bit better. Thanks.