 Hello and welcome to this video where I'm going to be doing an example of truss analysis with a method of joints on a simple truss spectrum. Let's start at the beginning. Trusses are everywhere. You can find them, for example, on bridges and roof structures. Here I have the example of the Patula Bridge in Surrey, especially the one in the front is a classical example of a truss. The one in the back uses trusses, but then something a bit different, which is a truss arc bridge in the middle section, which might be at some point the topic of another video. For now, let's concentrate on the trusses. Trusses are also used for roof structures. Here are the trusses lying on the floor. Here actually they're cheating a little bit because in a real truss there should have been a joint in this corner and no continuous section, but it is close enough so that we can consider this a truss and do our calculations with trusses. First of all, what is a truss? I already have made a little short video on this, so I will just play this video for you. This is a truss. This is a very simple truss. Trusses are rigid structures made of members connected at joints. Each member is made to hold only actual loads, torques, or lateral forces are not allowed. Therefore, all the loads applied to a truss have to be applied to its truss. How do we analyze a truss? The first step is to find the support forces. Here in this simple case, as it's symmetric it's easy, each support force holds half of the load. Then we explode the parts of the truss and draw a free body diagram for each of them. Finally, we look at each joint, solve the translation equilibrium, and by doing so find the forces acting on each of the members in the truss. Now to our example, the truss bridge with our ligament. So let's assume that this ligament has a weight of 10 Newtons and is putting all its load in the middle of the truss. The first step in all truss analysis is to do a free body diagram of the entire structure and calculate the support forces. So here is the free body diagram of the entire structure, the dimensions are just approximate assumptions here. So let's assume the bridge has a span of 12 cm, a height of 5 x 2 cm, and angles in all the triangles of 60 degrees. Apart from the free body diagram of the entire structure in truss analysis we need the free body diagram of each part of the structure, especially of the joints. So you might see that I just removed my connecting members between joints A and B and then replaced its effect on joint A over here and the effect on joint B over here. I did this for the entire structure but actually will turn out we don't need all of this so it looks more dramatic than it actually is. Note that I did assume in my free body diagram of the joints over here that all these forces are tension forces in the members so that the members are pulling the joints together. We can easily see from one of these free body diagrams that one of these arrows was pointing in the wrong direction. So what we will do, we will simply calculate assuming these are tensions with the directions as drawn and if mass then tells us that we get a negative answer that means that our error was actually wrong and that the member actually was under compression. What do we do next? Well first we calculate the support forces. For that I use the left free body diagram of the entire structure. My torque around the set axis has to be zero. I put my pivot on point A and I can calculate that minus Fc times the distance of six centimeters because I'm going clockwise and plus the force applied on E times the 12 centimeters must give me zero. I can solve this for my force on the support in E and I get five newtons. Then I do the translation equilibrium in Y direction. I get that plus my force in A minus my force in C plus my force in E must be zero and therefore the force in A is five newtons. We actually see that here we had a symmetric case so we didn't need to do all of this because we could have immediately said that the two forces on A and E have to hold each half of the load. However, if you do have a non-symmetrical case then you need to do this and if you have on top of the vertical forces also horizontal forces then you probably also need to do the sum of all forces in X direction. Next we looked at the choice and I'm going to start with joint A. I'm going to do the sum of all forces in joint A has to be zero therefore plus my force A plus my force AB sign of the angle which was 60 must be zero. I solve this for the force AB and I get minus 5.8 newtons. Note that I have a minus here remember minus to say that means compression that means this member here actually is being compressed not stretched and the arrows I drew here are technically wrong however we're not going to fix it because we already made the convention if ever I get a negative answer that means opposite as it was drawn. Also from Newton's third law of motion I know immediately that the force on A into the direction of B must be equal to the force on B in the direction of A so once I have this one down here in the corner I immediately know the one up here then I do my sum of all forces in X equals zero where I have plus force AC plus force AB this time cosine of the angle of 60 and I solve for my force AC as I had minus 5.8 for force AB I get minus minus so I end up at plus 5.9 which means I have a tangent in the member connecting A to C. Now I go to my next joint which is joint B again I will do the sum of all forces in Y equals zero so up here I get minus the force BA times sine of the angle 60 I'm not closing the side minus the force BC times sine 60 must be equal to zero and I can solve for the force on B by C which I get plus 5.8 Newton's according to our convention a plus number means tension as well as I had drawn in every bi-diagram. Next I do the sum of all forces in X so I get minus the force BA cosine 60 plus the force BC cosine 60 and then plus the force BD here in the top. I solve this for the force on B by D and I get minus 5.8 Newton. What does the minus mean again? Yes exactly we're under compression not tension as I assumed in the top right diagrams. So here is what we have so far we have solved about half of our trust structure and we could just continue in the same way and go to joint C, joint D and then join E however we already had figured out that this trust is actually completely symmetric so we can use the fact that it's symmetric and simply copy over the numbers and the tension and the compression for the rest of the trust. Now before we go away and hand this in let's think for a moment does this actually make sense? If you look at my bridge here I get that I had a compression around the top and I had tension around here. Tension means I could actually take the member out and replace it by a string. Well what do I get if I replace these blue members by a string? Well correctly an arc suspension bridge so I think actually the answer here makes really sense because it kind of shows us that we could have replaced this with an arc suspension bridge. Given that my analysis led me to a different type of bridge that actually does exist I'm quite confident that the results of my calculations are correct therefore I'm going to hand this video in and go to the next poll.