Trusses: 5. Trusses Method Of Sections Math

So let's look at the method of sections for solving the force in a truss member, just one member, or two maybe, maximum three. Okay, this page, and rather this problem, is based on an ARE problem from ARE 3.1. The method of sections is very sharp, it's smart, It's a very interesting, structurally sexy concept that you just have to understand. Once you understand it, the math is real easy. So try to follow, please, just the concept and the strategy of how to do this thing. So let's see. The method of section says take a section, a section cut, from one end of a truss all the way to the other end, not passing through more than three members, including the member you're looking for. You will get two segments when you cut the truss with a section cut from one end to another. You get two segments. Either segment is in equilibrium, horizontally, vertically, rotationally. The segment, even though the truss is not complete, but the segment of the truss is in equilibrium, doesn't move horizontally, doesn't move vertically, does not rotate clockwise or counter. Very simple. So let's try to understand the strategy of taking a section cut. So, if I am interested in finding the force in member CL only, all I want is member CL, which is this one. Take a section cut passing through no more than three members. Include CL in your section cut. Very good. I cannot take a section cut here. I cut through way too many members. If I take a section cut here, I didn't get the green member. So, how about I take a section cut through here. I pass through one, two, three members, and I get one segment and a smaller segment. Either segment is in equilibrium. Very good. So, we're going to practice it right now. I'm just interested in focusing on how to take the section cut to include the member you want and not more than three members. Very good. So let's try another one. If I'm interested in finding the force in member EK, where is EK? Let's see, this one. Take a section cut from one end of the truss to the other, passing through no more than three members, include EK. Looks like that. 1, 2, 3 There is a right side that you can think of That segment is in equilibrium Up equal down, left equal right Clockwise equal counterclockwise Or if you want, you can think of that segment It too is in equilibrium Very good So that's how we take that section cut Let's try one more time and then we'll do the math I'm interested in the member FH this one. To find the force in this member, take a section cut, passing through no more than three, include FH in your section. I think that is a very nice section. I can work with the left side. It's way too many unknowns. Or else I can just look at that small segment and force it to be in equilibrium. I find my unknowns. Okay, so there's the logic for you. Excellent. Let's do the following, What is given? A bunch of loads on A, B, C, D, E, F, G. And we have the reactions. They gave it to us. This was a problem from way back when. Excellent. So, to find CL, let's leave CL alone. No, we'll do CL. Do I have room on the page? Yes, I do. Okay, so to find CL, we decided that we would take a section cut like that And force the left section, because it's easier to work with, into equilibrium. So let me see if I can do this live, because it's not, I didn't pre-prepare it. So I'm going to do that, and I'm going to say, take a screenshot, and put it onto copy. Sorry, I didn't prepare this ahead of time. I have all the other section cuts. Okay, paste. Excellent. So there's my section cut, and now what I would like to do is the following. Let me do this. So we're going to put this bubble, take a screenshot, and we're going to put it right here. Copy, paste. Very good. So now, how do I do this? I can't do it. I need to rotate it. Let me rotate it. Very good. I'm sorry. I'm learning this software as I go. Okay. Ooh. Very good. So that's where I would like to be. Now, I look at one side of the truss or the other. Right now, whatever this force is, it's not on my section. Only from the section cut to the left is what I'm interested in. Excellent. And we wanted the force in member CL. This was C. And this member is CL. And I cut through it. Let me assume it's round. I'm interested in the force in member CL. There's a force in there. It's vertical. So, there's whatever members you cut through, there used to be forces, internal forces in them, that are now external. Which is these three? I cut through these three members. Whatever internal force was inside the member is now external. I have three equations. Some of horizontal equals zero. Some of vertical equals zero. sum of rotations about any point you want is equal to 0. I have three equations. Don't cut through more than three members because all you have is three equations. That's where that number came from. So I have an unknown called BC, this one. I have an unknown called CL. And I have an unknown called KL. Very good. So now let me explain one thing. There is, how many unknowns do I have? I have three unknowns. Two of them are horizontal, and only one is vertical. The blue one is vertical. But the sum of vertical forces should equal to zero, as well as the sum of horizontals. But I have two unknowns in the horizontal in that tan highlighter, so that's too many. Verticals, the sum of verticals should equal to zero. Here's all my verticals on this section cut. Here's all my verticals. So the sum of all these four forces should equal to zero. Well, let me make that happen. Twelve up. Twelve up. I have one down. I have two down. And I have Cl. The sum of these should be equal. So these should add up to twelve down to balance twelve up. Well, then Cl must be nine plus two plus one. So this is nine. 9 plus 2 plus 1 is equal to 12. That's how easy this is. Now, which way is the arrowhead of 9? It has to be down. Here it is. Down. Is down pushing into where you cut the member or pulling? Here we have to stop and do a little convention. If the force is pushing into the member, now we're not talking about joints. We're talking about members. This is compression. If the force is pulling on the member, then it's tension. What happens here, we did the method of joints earlier. If the joint arrowhead was away from the joint that was tension, well, it pulls on the member and then it ends up pulling on the other joint. Likewise, if the force was pushing on the joint in compression, we looked at the other end, we said it's pushing into the other joint. it passes through the member and compresses it. So essentially what I'm trying to say is pushing into your section cut where you cut it is compression, pulling on that section cut is tension. So look at your section cut, is the arrowhead pulling, running away from the section, or is it towards, or pushing into the section. So now I'm looking at this nine kips, And I'm seeing it pushing. I'm seeing this situation. I'm seeing a member. It has a section. And I'm seeing pushing into it. It's pushing into it. So that 9 is compression. I talked a lot. This takes 10 seconds to solve. If we know where to take the section cut and make sure we don't flip our arrowheads, we're good to go. Let's try another one. So the answer to member CL is 9 kips compression. Very good. Let's try another one. Let's find the force. Let's do the easy one first. Let's take FH, this member here. To find the force in member FH, we said take a section cut here. And there's a left side, there's a right side. I cut through three members. Let me look at the right side. It's easier. This one is pre-drawn. It looks like this. Here's the right side of that section cut. I have a section cut. And I cut through three members. One, top chord. Then the member I'm interested in, it's called, what is it called? FH. And then there's this member. And each one of these has an unknown force. Again, I look at my unknowns. This one is horizontal. This one is horizontal. This one is vertical. Well, we have three things. Sum of horizontal is not going to take me anywhere because I have two unknowns. Sum of moments has never been on the ARE before, so don't worry about it. I do it in my structures class, beyond the scope. Fine. So it must be sum of verticals equals zero. Okay. Let me highlight them. Verticals. This one is vertical. This one is vertical. This one is vertical. 24 minus 5 is 19. That's it. You have to be 19. This force must be 19 to help the 5 make the 24. To help the 5 means I have to be down. 19 plus 5 down equals 24 up. Is this pushing on the tip of the member or pulling? Looks like it's pushing. Again, one more time. Here's the member. If you have that, you have compression. If you have pulling, you have tension. So this one is 19 compression. Easy. Let's do another easy one. Find me the force in member BM. Take a section cut here, and let's see what we come up with. Here's the drawing. It's already pre-drawn. So we cut through one, two, three members. Don't look at this side. It's too much work. Let's look at the left side. Excellent. So here it is. You have to put all the forces and reactions on there. I had an external 1 kip at A. I had a 12 kip reaction. B is not on my section cut, so I don't even look at that. I cut through three members, the top chord, the bottom chord, and that vertical. So again, I have one vertical plus one plus is equal to 12. This guy has to be 11. It has to be down to help the one equal the 12. Down is towards the member. This is 11 compression. Good. Now let's try one of the uglier. Let me erase a little bit. Very good. Delete. So, help me find the force in member EK. This one. Take a section cut right here. 1, 2, 3. The right side has 3 forces. The left side has 4. I'll take the easier one. Let me look at the right-hand section and redraw it. That's what it looks like, and we're interested in this member. Excellent. So I have a vertical and a horizontal this time. Because it's a diagonal, I can replace it with two components. Do I have a slope? I have a slope, one to one. So those two are equal, V and H are equal. Then I have this unknown top chord, I have this unknown bottom chord. So how many unknowns do I have? One, two, the diagonal is three. Yes, it has two components, but once you know one, you know the other. Very good. Again, let me look at how many horizontals and how many verticals I have. In the horizontal direction, I have 1, 2, 3. Way too many. In the vertical direction, I only have 1. Here's the 1 vertical plus the vertical plus the vertical plus the vertical should equal to 0. I have 25 down, I have 24 up, looks like I need 1 kip to help the 24 in the same direction. So this guy needs to be 1 kip, let me erase that V, it needs to be 1 kip in that direction, up. So the other component is to the right, because they are both components of this guy. This guy was doing that. And its two components are V and H. If the vertical is 1 and the slope is 1 to 1, then the horizontal is also 1. And the diagonal is that much. But the point is it's pushing compression. So that's it. That's the force in that member. Let's try one more. Let's do this one on the other side. Find the force member BL. Take a section cut, passing through no more than 3, include BL in your section. So let's do that. And let's take this side, and let's redraw it with the 2 kip, the 1 kip, the 12 kip. And 1, 2, 3 unknowns. It's right here. So I have 1 kip. I have 2 kip. That's 3. I have 12. I need a 9. I am horizontal. I can't help you with vertical equilibrium. I am horizontal. I can't help you. I am horizontal and vertical. Let's not look at the horizontals. There's too many of them. Vertically, it looks like you need to be a 9. To help the 2 and the 1, 9 plus 2 plus 1 is 12. Let me erase this guy. And let me remember that the slope of this truss was 1 to 1, which means the rise equals the run, the rise is 9 kips, the run is 9 kips. The arrowhead is going down to help the 2 and the 1, but in my book, as far as the section goes, I needed 9 down for equilibrium, but now I'm looking at the member and I'm looking at this force and seeing it running away from that section cut. So must be, this is all tension. So the answer is 9 square plus 9 square, whatever that math gives, and tension. Because it's running away, let's please remember, the convention is important here. That's tension, and that is compression. Pushing or pulling. Very good. Let's do another one diagonal, which is this one. Let's find the force in member CK. So take a section cut, passing through no more than 3, include CK. I want this one. Oops. So we're going to take a section cut here. We're going to look at the left. And we're going to see this segment of a truss that is in equilibrium. And we cut through three members. 1, 2, 3. So I redrew it down here. And we're interested in the force in the diagonal number. I have 12. I have 1 plus 2 plus 2. I have 12 up. I have, if I add these up, I have 5 down. Looks like I need a 7 down. You are horizontal and vertical. Your vertical must be 7, because that's the only vertical in the picture. So if the vertical is 7, then the horizontal by slope is also 7. These are doing that kind of arrowhead, which is also tension. It's pulling on that member from the section cut. So the answer is 7 square plus 7 square, square root, and tension. Okay.