 Today I'm going to talk about beam reactions and distributed loads. In order to understand this topic you have to know about translation and rotation equilibrium. So in case you're not so sure about your knowledge about translation and rotation equilibrium yet I would suggest you go to my website. In the mechanics part you have a couple of presentations as well as video to refresh your memory about those two subjects and then you can come back and hopefully you will understand what I'm talking about here when I'm talking about beams and their reaction to distributed loads. What is a beam? So beam is a horizontal structural element that usually holds vertical loads. So it could be just a beam between two supports and then some additional forces on it as we have here or sometimes you see it as a support for the roof as we have here in the second picture. The second new item are distributed loads. So far we only have talked about loads that act on one single point. Now distributed loads will act over a certain area. These are actually more realistic loads as everything you might put on a beam usually has a surface area. We're going to consider two main types of distributed loads. One will be uniform distributed loads and the other one is triangular distributed loads. We want to do calculations with these distributed loads. We first need to figure out what are their equivalent point loads. Now in the uniform case this is quite simple. The equivalent point load will be the surface area times the low presurface area and it will act exactly in the middle of the distributed load. In the triangular case it's a bit more complicated. The equivalent load will be one half the surface times the load, the maximum load on the right side. Basically what this is is the surface area of the triangle here and the central point where this equivalent load is acting is one third from the bigger side of the triangle and two thirds along the way from the smaller side of the triangle. Now the beam reactions we already have calculated them in many cases are simply the forces that come from the supports of the beam. Here I have one on the left on the right side but you could very well have an additional support if you put a column let's say here in the middle. How do we calculate these reaction forces? Well first we're going to replace the distributed loads by their equivalent loads and then we draw the free body diagram, solve the rotation equilibrium and the translation equilibrium. Nothing here should be new except that we have now these three distributed loads. Everything else you should have been able to solve before. Now let's have a look at an example. Here we have a 300 kilogram beam that is loaded with 500 Newton per meters as shown and you have the other dimensions down here. Two meters is where the load is distributed and the whole beam is five meters. Now what was the first step? The first step is calculating the distributed loads or finding their equivalent point load. Now in this case this is fairly simple. We had 500 Newton per meter over a length of two meters so we get a total of 1000 Newton. Also the load is acting in the middle of the distributed area so exactly one meter from the right end. Next step is drawing the free body diagram. I already found my equivalent load here. Of course we have gravity as you're still on the planet we have a load on the left, we have a load on the right and here I chose to put my pivot on the right side. Note that once again you can actually put the pivot wherever you want as long as we're in static equilibrium it has to be in static equilibrium around any point on the object. Once we have the free body diagram we're going to solve for the rotation equilibrium. Now this you should be able to do. So pause the video while I'm getting a coffee you're going to do this on your own and then I'm going to come back in a few minutes. So pause the video and solve it by yourself. The answer that you should get for the support A force is 1700 Newton. So pause it, give it a try and then we're going to come back later. So did you solve it? Let's have a look. So we have what forces are causing us torques here. So we have the torque of left support. We have the torque due to the gravity, the weight and we have the torque due to the load must be zero. The torque through to the gravity as through to A is clockwise therefore minus and at five meters distance fA. The one due to gravity is counter clockwise so positive. The thing was 300 kilograms so 3000 Newtons and we are at the distance of 2.5 meters. Gravity always acts in the mill times 3000 and then the load is at the distance of one meter so plus plus one times and we calculate its 1000 Newtons is zero therefore my fA must be 2.5 times 3000 plus 1000 divided by five which gives us the 1700 Newtons you have seen in this solution before. Now what's the next step? Exactly solve the translation equilibrium. So for the translation equilibrium in x direction we have nothing to solve the only thing to solve is in y direction. So as before pause the video and give it a try by yourself. Okay that was easy right? All you have to do is look what forces you have in y direction. So we had force A, we had force G, we had the load and then we have force B and the total of that if the beam is not starting to move or anything should be zero. So we had plus the 1700 from the rotation equilibrium that we already solved and then minus 3000 Newtons for the weight minus 1000 Newtons for the load plus space again plus fB is zero therefore my fB must be 3000 plus 1000 minus 1700 plus 1000 minus 1700 gives me exactly the 2300 Newtons as the solution suggests. That's it for today.