Trusses: 3. Truss Zero Members Simple Joints

Let's look at this very, very, very, very ugly truss that is loaded with all kinds of loads. There are difficult questions and there are very simple questions. And that's the tactic of the ARE. They'll give you a terrorism question that looks so difficult, but then they'll ask a simple question to see if you understand. And that's what I hope this page will do. Very good. There's two kinds of joints, simple joints and nasty joints. So an L, such as this one, or a T, such as this one, are the simple joints. Anything with diagonals is going to be a nasty. So avoid the joints that have diagonals. And when you look at a truss like this, I want you to think of the simple joints. analyzing these, it's important to recall that compression is when a member is pointing towards a joint or pushing on a joint, and tension is when it's pulling on a joint or away from the joint is tension, towards the joint is compression. So let's look at this ugly truss and let's look for the simple joints. Where I see one. This one has a diagonal. Forget it, that's a nasty. So this is an easy joint. This is an easy. This one. This one. These are the simple joints without diagonals. So if the question is around a simple joint, maybe the answer is simple. It could be complicated, but maybe it's easy. But if the question is around a complicated joint, I don't want to look at the complicated joint. So looking at this truss, and if the question calls for find the force in member Fn, this one. I'm using green in order not to imply tension or compression. So how much is the force in member Fn? We look at the member Fn. It's surrounded by two joints, F and N. F is a nasty. I don't look at F. I look at N. And what do I see at joint N? Let me zoom in and let me not see anything except joint N. Now looking at this joint specifically, it has a vertical, it has two horizontals. For every single joint, the sum of vertical forces should equal to zero, the sum of horizontal forces should equal to zero, which says these two are equal and opposite. This one has no partner. It has to be zero because the sum of verticals is zero. There's only one vertical. Got to be zero. You might be thinking, where's the 6 skip going? I don't care because I don't see a 6 skip when I zoom in here. The joint on its own must be in equilibrium vertically and horizontally. I don't know where the 6 skip goes. It goes somewhere else. It cannot come to end. Okay. So in other words, the member Fn is zero. It carries no load. the sixth skip has to go somewhere else. Excellent. When we look at member DQ, we see the joint D. That's a complicated joint. We look at Q and we see one vertical, two horizontals. Must be this one is zero. Now that joint is not loaded. When you have a simple joint that is loaded, such as E, I have five kips at E, I have one vertical, I have two horizontals. These two horizontals must be equal and opposite. And vertically, these two must be equal and opposite. I have five kips down. I need five kips up. Now, is this five kip in member DE and member, what is it called? EP. Is the five kip in member EP towards the joint E or away from the joint E? Let's recall. This is compression. This is tension. It looks like this one is towards the joint. So the force in member EP is 5 kips of compression. So that 5 kip that started at E comes down here, and it's to be triangulated at joint P. Which way was the direction of the 5P at E? It was towards the joint. Well, it must do towards the joint at P. So there's the 5 kip. Joint P is nasty. I'm not going there. Now on the ARE, if they ask me about member DP, that is nasty, that is nastier. I'm not going to answer that question. It'll take me half an hour to answer that one. Look for the simple joints. Excellent. Let's look at member CR. And please look at the joints individually. CR, you do not look at the C and the R at the same time. You don't see the three kips and the 11 kips at R. No, I don't want to talk about R. I'm just going to zoom in to C. I see 3 kips. I need 3 kips in compression. That's it. End of story. Ask me about this one. It's too much work. I cannot answer how much the force in this one is or that one without extreme work. I'm not getting into that for the ARE. So the 3 kip is coming down here as compression. Excellent. What happens with member GM? GM, let's not look at M. That's a nasty joint. Let's look at G. I see one vertical. There is no load here. If there were a load here, that would be different. There is no load. That one is zero. Looking at the member HL, I see 7 kips at H. That's a nasty joint with all these members framing into it. I don't want to deal with this one. So let me zoom into L. Looks like there's 8 kips pulling down there. It looks like we need to balance it with 8 kips. Now, is the 8 kip up or down? It's up. Is up away from the joint or towards the joint? It's away from the joint, so this is 8 pips of tension. I don't know anything about these. They're too much work. So did we do all the easy joints? Not yet. There's two joints remaining that are easy, joint J and joint A. Let's do joint A first, and I hope you won't struggle with this one. Okay, the roller is equal to one reaction. Is that correct? Don't ask me to calculate that reaction. I'll spend the rest of the video and another three videos trying to find that answer. That's not the way to go. If I would like to calculate the force in member AR, let's talk about AB. AB is equal to the reaction. Whatever that is, that is equal and opposite, but I'm not going to spend the time on that one. I have to get the reactions. I can't. If they give it to me, if they give me the reaction, then yes, AB is equal to that reaction and compression. It's a vertical over a support. Okay. Now, AR. Let's think of AR. The roller is only one vertical reaction, which means horizontally there's only this member. So the force in member AR must be zero. Take it out. No one's going to miss it. Mind you, if this were the pin instead of the roller, then it would have been two reactions. And this reaction would have had to have been 1 kip, because sum of horizontal equals 0. The 1 kip is the only horizontal load on the strata. Okay, well now this one is not 0 anymore. It has to be 1 kip. And it looks like it's going away from the joint. That's 1 kip of tension. Okay, but we don't have a pin here. All we have is a roller. So this member here is doing nothing. This one. because there is nothing opposite to it. Let's look at this joint J. There's one horizontal only, it's zero. There's one vertical only, it's zero. When you put a vertical load at J, this one is no longer zero. If you put a wind load or something horizontal at J, then this one is no longer zero. But in the absence of load, both of those members are zero. Very good. So let me go back to this sheet now. and confirm that this joint is one of the simple joints. It has one vertical, got to be zero. This one, this one, they're all zero members because there's only one vertical. And this is a pin here, so that reaction is equal to this one. But it looks like there's no horizontal. Maybe this is zero and so is this one. This one is definitely zero. This reaction is equal to whatever this load is, because that's a simple joint. Looking at the Pratt truss, whatever this load is, this is equal and opposite in compression. In the How truss, looks like this one is zero. In the Fink truss, they didn't put the zero members, you don't need them. In this How truss, this one is zero, this one is zero. In the bow string, I don't see any zeros. They don't have them in. Those would have been zeros. Looking at the overhang, that's a simple joint. That's zero. That's zero because there's no load here. It looks like this one is a zero member. It looks like at this joint here, it looks like this has to balance the load above. It looks like this one is zero. Okay. Whatever this load is, this is equal and opposite in compression. Looking at the cantilever, I see a roller here. A roller is one reaction, so this is equal to that in compression. There's nobody in the vertical direction. That one must be zero. Let's look for some more of these simple joints and try to answer some questions as we go. Calculate all the members you can without doing too much math on this page. So the idea is there's different configuration trusses, but all these trusses are loaded with 3 kips, 6 kips, and 9 kips at the top. How many members can we find without even calculating the reactions? Let's see what we can do. I want to identify the simple joints and see what I can do. So looking at cross number 1, I see simple joints at B, at F, and at D. It looks like this one is a zero member. There's no horizontal. It looks like this one is 3 kips into the joint or compression. Looks like at F, there's nothing going on in CF. That member is zero. It doesn't take any of the 6 kips. Looks like CD is the only horizontal at joint D. looks like that is zero. DE is doing nine kips opposite the nine kips and it's towards the joint that's compression. That's it. All the remaining joints are nasty joints. I'm moving on. I just want to do the simple joints. Looking at truss number two, I see CF is zero. It doesn't take any of the 6 kips. I see AB is 3 kips into the joint compression. I see BC is 0. I see CD is 0. I see that ED is 9 into the joint, that's compression. The 9 comes down to here as compression, and the 3 comes down to here as compression. Into the joint is compression. I didn't flip the arrow, I just said compression at B, 3 kips, and compression into the joint at A. Looking at truss number 3, joints A, C, E are nasty. Joint F is easy. This one is 0. So you could get a truss that looks like this. Look for the simple joints. Looking at truss number 4, it looks like C, F has to do 6 kips towards the joint. that's compression. The 6 skip comes down to F to be triangulated. Very good. Looks like AF is 0 because there's no horizontal loading. Looks like EF is 0. Looks like this one, DE is equal to the reaction. I don't have the reaction. Looks like AB is equal to the reaction at A. Fine, move on. Looking at truss number 5, I see 6 kips of compression in the member CF because of the joint C. Every other joint on this truss is nasty. Move on. Looking at truss number 8, I see 0. I see 3 kips compression. I see 4 kips because that's the only horizontal at joint B. And this 4 looks like it's away from joint D. don't look at joint C. So 4 kips of tension in BC, okay, well it pulls on C with 4 kips. Looking at joint D, the other simple joint has one horizontal, it's 0, has a vertical equal and opposite to 9, so it's 9 in compression, comes down to E for triangulation, and the 3 kip comes down to A for triangulation, and it's pushing into the joint A, and the 9-kip is pushing into the joint E. I'm done with the simple joints, move on. Looking at truss number 7, I see two zero members. I see CF is zero, therefore not much going on here. AB, looking at B, I see a zero member, and I see four. That's it for the easy joints. Let's move on. Looking at this deep truss, I see joint F with one vertical, zero. I see three kips here, compression. I see nine kips of compression here that transfer down to triangulation for triangulation. That's it for this truss. No, there's one more zero and another zero. Okay.