This is module seven of an App, of Applications in Engineering Mechanics. Today we're going to make use of the method of joints. Or sometimes called the method of the pins to analyze truss structures. Here's a truss that we're going to look at. And we're going to find the forces in all of the members of this truss. And we're going to use the method of joints, which I talked about last time. And this is the rules for cutting through trusses. So we're going to isolate one joint. I'm going to go ahead and start off by isolating this joint at B. I'm going to cut through each member only once. And I don't cut through the joint itself. I continue to cut to isolate the entire portion within the circle and I am going to assume tension in each of the cut members, so here is a free-body diagram of joint B. So I've got an external reaction here B Y from the roller and I'm going to assume tension in F BC and tension in F BE. And from geometry I see that this is a four, I'm sorry, a three a four on three slope. There you go. Four high, three vertical, horizontally. Okay, so that's my free body diagram. Since I do have an external reaction in that free body diagram, I'm going to have to solve for it, so I'll do that by solving the entire structure's free body diagram. So let's do that together, I've got F BY. I've got a pin over here, so I'm going to have AX and AY. For the force that's in the joint B, I only need BY, but let's go ahead and solve for AX and AY as well, since I'm going to eventually need those as well. So let's begin by using the equation of summing the moments about A, setting it equal to 0, I'll call counter-clockwise positive. And I'm going to have minus 1 times its moment arm, which is 3, minus 3 times it's moment arm, which is 6, minus 2 times its moment arm which is 4, plus BY times its moment arm which is 12 and that's going to equal 0. And so what that gives me is BY ends up equaling 2.42, and it's positive, so B y is 2.42 kips up on the truss. As I said, that is all I would need to go back and solve for joint B's free body diagram, but let's go ahead and solve for A x and A y, because we'll use those later. So I'm going to sum forces in the x direction, to find A x. I'll choose to the right positive. I've got A x and the only other x component force I have is 2 kips, so it's plus 2 equals 0. So A x equals minus 2. So instead of to the right, A x is actually to the left. And then finally, I have to solve for A y. I'll sum forces in the y direction. Choose up positive and I get let's see, Ay minus 1 minus 3 plus 2.42 equals 0. Or AY equals 1.58. So AY is 1.58 Kips up, and we'll use that latter on. Okay, so now we can go back to our Free Body Diagram of Joint B. I put in the By of 2.42 and we can now use the equations of equilibrium to solve for our first two members FCB and FEB. So let's sum forces in the y direction. I'm doing a lot of this together with you but I'm pretty soon here, I'm going to let you take off and do it on your own, since it just gets to be the same procedure over and over again. So we've got the 4, 5ths component, of FCB plus 2.42 for the BY equals 0. And so I get FCB equaling minus 3.025. Which, that means that instead of tension, I assume them all the members in tension, but instead of tension, this member is in compression. So FCB is equal to 3.025 inn compression, and I'll just put the symbol c in parenthesis there for compression. So that's one of my forces in one of my members. Keep on going, let's do some of the forces in the x direction, set it equal to zero. And I've got minus FEB minus 3 5ths of FCB equals 0. Substituting we found that FCB was equal to minus 3.025. And so that gives us FCB equals 1 point, well FBC, FCB, same thing, 1.0, sorry 1.815 and it's positive, so that means that member is in tension, FCB is equal to 1.815 kips in tension. So now we've got two of the members. So we're just going to keep on going now. We've taken care of this member, and this member. let's go ahead now and isolate another joint. I'll go ahead and choose to isolate joint C. We're going to need to draw a free body diagram there. We know that the force in member BC or CB is 3.1, 025 kps in compression, so I'm going to draw that like that. And we don't know these other two members, so we're going to assume I'm in tension so this is going be FCD and FCE. That's a good free body diagram, and we can go on from there. So, here we go, there is joint C. What I'd like you to do now is to sum forces in the X and Y direction and come up with the force in members EC and DC. There is some forces in the Y, that's the answer that you should have gotten for F EC. Now I'd like you to go ahead and sum forces in the X to get the forces in DC. There is the answer for the forces in DC, you should go through it step by step on your own. So now, we have this member and this member and this member and this member done. We've got to go to another joint. We'll keep going. Method of joints. One at a time. So let's go through here. And for this one we just found out that from the, the Isolation of joint B that member FBE or EB is 1.815 in tension. We just found on this free body diagram that FCE is 3.025 in in tension. We have three kips down at point E, we don't know what these two forces are, so we're going to assume them in tension. And we're going to go ahead and analyze that joint, so please go ahead and do that on your own. So here's joint E with the sum of the forces in the x and the sum of the forces in the y direction. and so now let's total up, we've got this member, this member, this member, this member. So we going do that joint D next. And so for joint D, one kip force down at the pin, or at the joint itself, D. We just found out from mem, analyzing member, or joint C that member CD is in compression. And it was 1.63 kips in compression. then we have FDE here it is, 0.0725 in tension. This is unknown, so we'll just call that FAD. That's a free body diagram of joint D. And, here is joint D. And oh, I forgot to put in, no, I, I put them all in. And now I sum forces in the Y direction. I can find FAD. You actually only have to do one of these equations. I've double checked by doing some of the forces in X direction and find that that also gives me the same answer for FAD, so FAD is 1.975 in compression. So now we've actually gotten all the members. we can do one more check by isolating joint A and make sure that the values that we found for FAD and FAE are correct. So if you'll do that on your own, double check, I'll just show you the answers on the next slide. So there you go! I've double checked joint A. I found out the external reaction, if you go all the way back to the free body diagram of the entire structure was 2 kips to the left. Here's my other members, if I sum forces in the x and the y direction I get 0 equals 0 and so the structure checks. So we've taken care of the force in every one of the members of this truss. So let's go ahead and have you do one on your own. Here's a truss where you can analyze using the method of joints. And I've got the solution in the module handouts. And we'll see you next time.