Welcome to Module 9 in Applications in Engineering Mechanics. For today's outcomes, we're going to continue to use the Method of Sections to analyze truss structures. We're going to go back and use that same problem we worked on last module. But this time we're going to go ahead and first solve for the force in member DE. And so my question to you is, let's assume that I've given you this problem and how would you start? What would you do first? Try to do as much as you can on your own before coming and watching the rest of the video. This is good practice for you. So what you should have decided was okay, I'm going to use the method of sections again. I'm going to cut through the member of interest, in this case its DE. I can go ahead and use the same sectional cut that I used last time. that will work again. as, as sort of an advance or side topic, note that you don't have to pick this portion of the truss you could also, look at this section of the truss. And you would get the same answers as I'm going to get in this module. And so I would actually like you to try after you've gone through this module, try that on your own and re, redo this whole module. Try to solve for member DE and later we're going to find member DG force as well. But do it by looking at this entire left hand section, and you can double check yourself. Okay, so here again is the section free body diagram for th the section that we've cut out. what equilibrium equation might you use to solve directly for DE, think about that. Once you're decided come on back and we'll, we'll do it together. Okay, I would recommend that we go ahead and sum the moments about point G. The reason I'm going to do that is because we have found FGH. we said that, that was 3.29, that was from last s, s, s, module. But if I want to solve for FDE directly by summing moments about G, this unknown FDG goes through G. And so only FDE will, will cause a moment, that's the only unknown that will cause a moment. We, we can solve for it directly. So go ahead and do that, when you are finished common back and we'll do it together to make sure that you've been able to solve it correctly. Okay. So, FDG, or excuse me FDE is what we want to find, we're going to sum moments about G. I've chosen counterclockwise as positive. And so I've got, okay, I'm going to have, [UNKNOWN], I don't know the perpendicular distance between the line of action of FDE and point G. And so what I'm going to use is Varignon's theorem back from our, my last course, Introduction to Engineering Mechanics. And I'm going to break this FDE force into its two components, an x component and a y component. And so let's first look at the moment due to the x component. And we see that this FDE is on a one on three slope. This, this distance is one, this distance is 3 meters. And so the hypotenuse is the square root 10. And so we can see this is a breakout of, of the slope. And so the x component then will be 3 over the square root of 10 times FDE. So we've got 3 over the square root of 10 times FDE, times its momental arm. And its momental arm will be 3 because that's the perpendicular distance from the x component down to point G. It's going to be positive in accord, accordance with my sign conventions because it's causing a counterclockwise rotation. I've got my y component which is going to be one over the square root of 10. So, I'm going to, and that's also going to be positive because it causes a counter clock wise rotation. So, I've got plus 1 over the square root of 10 times FDE times its momental and it will be 1.5. And I also have my 3.92 reaction force down here from the external roller. It's going to be plus, so it's plus 3.92 times its momental arm to G is 3. Set all that equal to 0 and I get FDE, if I solve for it will be equal to minus 3.54. Which tells me that FDE even though I jointed intention is actually in compression. So I've solved for FDE. It's equal to 3.54, the units are kilonewtons and it's in compression. Okay, so let's actually go on and. I'm going to ask you the question now. How might you go about solving for FDG? We, we, found FGH in the last module, we've now found FDE, and I've turned the arrow around. I've gone ahead and drawn it now in compression with a, with a magnitude of 3.54. How might you find FDG? Okay. What you should have done and, or said is, let's go ahead and sum forces in the y direction, because the only unknown now that I'll have by summing forces in the y direction will be FDG. you could sum moments about E. that's another option. Or sum moments about R. But why don't you go ahead and use the equation of equilibrium of summing the forces in the y direction and see if you can get the answer. Okay, let's do it together. in this case the FDG force has two components, an x component and a y component. We're only concerned with the y component. You can see that it's on a 4 on 1.5 slope with a magnitude of the hypotenuse, is the square root of 1.5 squared plus 4 squared. And so, let's go ahead and use that y component. So the y component is the 4 part of the triangle over its hypotenuse, which is the square root of 1.5 squared plus 4 squared times FDG. I have the y component of the FDE force, since I've drawn the arrow down now and it's a positive value of 3.54. It's the, it's the one side of the one three square root of 10 triangle. So that's going to be minus, and according with my sign convention, 1 over the square root of 10 times 3.54. I've got my 3 kilonewton force acting down at G, I've gotta include it on my free body diagram. And I have my 3.92 reaction force up. And so, If I solve that I get FDG equals 0.213. It's positive, and so that means that I've drawn it in tension. It's indeed in tension and FDG then is equal to 0.213 kilonewtons in tension. And so we solved for both of those members. Okay. Believe it or not you're at a point now where we talked about at the beginning of the course. Hey, we could even solve for an example like this tower crane. Now, certainly this worksheet I've given you is not quite as complex as the, as the tower crane, but it's just a portion of the tower crane. You could, you could certainly analyze now. a structure like this tower crane. And so I want you to practice on this small portion of what, what appears to be a part of a tower crane, and solve for member HI. Solve for member HI's force and I've got the solution in the module handouts, and I'll see you next time.