Welcome to Module 10 of Applications in Engineering Mechanics. Today we're going to identify zero force members in truss structures, and we're going to explain why zero force members are used in truss structures. So, first of all, why are zero force members used in truss structures? These, these are a member that have no force in them. there's three reasons. One is they may increase stability of your structure. They could provide support if the load changes. And sometimes they're added just for, to be aesthetically pleasing, like in the roof truss of a, of a maybe a sports stadium or a bridge. You may want to add some extra members that don't carry forces just for aesthetics. So how do you identify zero force members? Well, we can do so by free body diagram, like we've done before. But you're actually going to get good enough to do this by inspection, as you'll see as we go through this module. So, zero force members by identifying quickly through inspection, it'll, it'll speed up your analysis. I want you to go ahead and look at this same truss structure in these two different situations, and I want you to find the forces and members BC and CD using the method of joints for both the left and the right truss. Once you're done, come on back. Okay, let's go ahead and look at joint C on this left structure. And I'm going to go ahead and assume the force in the te members are in tension. And you can see very quickly that summing forces in the x direction gives me FBC equals 0. And summing forces in the y direction gives me FCD equals 0. So by using the method of joints, I found very quickly that these two force members have zero forces at this point. let's go ahead and analyze the same structure now for a different loading. Again I'm going to drop the joint free body diagram of joint C. So I've got joint C, but now I have FBC to the left. FCD down. But I also have an external force F to the right. And so, once again, if I sum forces in the y direction, I find that FCD equals 0. So that's a zero Force Member. But if I sum forces in the x direction because of the loading change, I get minus FBC plus F equals 0. And so, FBC in this case ends up equaling my external load F. And so just because the loading changed in these two situations, the structure is the same. In one case, the member BC is a zero force member, and over here it's actually carrying a load. if I also had a [COUGH], if I had a external force, let's call it. If I had an external force that I might call P here, then I would also have a force in member CD as well. So that's how you analyze it, you can get really good at this and just look at your structure and see the zero force members. let's try it for some more problems here. So here is two more truss structure, let's do the first one. Actually you could try to do them on your own and then get as far as you can. If you can't do them come on back we'll do them together. first of all for member, excuse me, joint A on this body. You should of by inspection now seen that there's only a a tension force in AD in the x direction, so FAD has to be equal to 0. There is only a, a forced FAB in the y direction ,so you should very quickly be able to say, okay or just by inspection FAB and FAD equals 0. So we've got FAD equals zero, FAB equals 0, so those are two of the two force members. Are there any other two force members in this structure? think about it, see if you can find and do it. And if you come back with me here, and we'll do it together. Let's look at joint E. So kind of a special situation here. I've got two assumed tension forces, FCE and FEF. If I sum forces in the y direction, I see I've got two forces. So there's nothing I can tell from that. But if I sum forces in the x direction. You can see that I only have an x component of the force FCE. And so FCE itself must be equal to 0. And then once I find that FCE is a zero force member, I can now sum forces in the y direction. And I'm going to get minus the y component of FCE, minus FEF equals 0. But since FCE was found to be 0, FEF itself is also a zero force member, and so two more zero force members. So, when I look at a joint like this in the x direction, I see there's only one force component, FC must be equal to 0. By zeroing that out, I see FEF also has to be equal to 0. Let's do another structure. when I start with a structure like this, I may go from left to right. I look at joint c here, by now you should be looking by inspection and saying, okay, in the y direction, I'm only going to have a tension in, in BC potentially. But that's the only force I have, and so it has to be equal to 0. Let's draw it just to confirm. So here's joint C. I've got FAC, FCE, and FBC. If I sum forces in the y direction, I therefore get FBC has to be equal to 0. So there's my first zero force member. You go ahead and see if you can find a similar situation in the truss and do the next a zero force member. Okay. What you should have found is, if you look over here at joint r now, very similar, exactly the same situation as we had at joint c. If I sum forces in the y direction, FQR must be equal to 0. Alright, let's go ahead and look now, I've got FQR is equal to 0. Now I'm going to look at joint Q. But instead of having x defined to the right and y defined up, I'm going to reorient my orthogonal directions for x and y as I've shown here. I'm going to draw draw joint Q, and I've got my free body diagram. I've got FNQ, I've got FQS, and I don't have to put in FQR because it's zero. All I have is FQP. And you can see now, by summing forces in my newly defined direction for x, only the x component of FQP is in that direction. So FQP itself must be equal to 0. So if I sum forces in my newly defined x direction, I get FQP equals 0. And we can keep on going now, once I have FQP equals 0. Now I can see that joint P only has one force in the y direction, FNP, so it must be a zero force member. And I keep going, now joint N ends up being just like joint Q. Using the redefined x and y directions, so FNM is equal to 0. And finally, FLM is equal to 0. And that's, that's all the zero force members in this truss, and that'll really speed up your analysis. So, I would like you to go ahead and now use what we've learned in the last three modules to solve this problem. this is a combination of the plane truss section of the course. you need to find the forces in member CD, D and DJ. you can speed up the analysis by identifying the zero force members if that helps. You can use the method of joints or the method of section, or you can use both, come up with the answer. I've got the solutions in the module handouts, and I'll see you next time.