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Zero-Force Members in Trusses · Acedra
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Zero-Force Members in Trusses
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0:00
[MUSIC PLAYING]
0:07
JEFF HANSON: Hey, friends, I'm coming at you today with a video for trusses, another truss kind of video, kind of a little secrets of the pros video, on identifying zero-force members and how to solve a big old truss like this. Now, this might be scary to you. You look at this truss, and you're like, oh, my gosh. How many members are in this truss? This many. That's 21 members. And we're going to use the method of joints to solve it. I know you're thinking, we're going to be here forever.
0:38
What are you thinking? So let's look at this. Find every force in the truss and state whether it's in tension or compression, so the same thing we've been doing. But first, well, there's a couple things to identify this as a little bit simpler problem right off the bat. Number one, things you should be looking for. Number one, this is a symmetric truss. The left side is the same as the right side. Also, in addition to that, it's a symmetric load-- 10 20 10.
1:10
So if you have a symmetric truss and a symmetric load-- you have to have both of those-- but if you do, then you get this little bonus. Well, guess what? You get half, and I'll get half. So global equilibrium is super easy. Because this is 20 plus 20 is 40. So you get 20 kilonewtons, and you get 20. And there is step one. You'll remember this little recipe over here from our discussion on the method of joints. Find global equilibrium. Well, guess what? Done. Step one done.
1:47
Now, it says pick a joint here. But really, there ought to be a substep in here that says, look-- those are eyeballs-- look for zero-force members.
2:06
Well, what's a zero-force member? You've got to do some little mental free-body diagrams in your head. But here you go. Any time that you have two forces that are on the same line of action and you have a third force that goes like that, so let's say, and just think about this-- something to the left in the x, something to the right in the x. This guy has a y component. Who else has a y component? Nobody. So guess what? This guy has to be 0-- has to be.
2:40
And it doesn't matter if he's like that or if he's like this. Because some people say, oh, I've got something going up. I've got something going down. I got something going left. I got something going right. I don't know if he's 0 or not. Well, just take that guy and spin it-- zhoop. It's this. So any time you have two forces that are on the same line of action and you have a third one, the third one is always 0, always. The easiest ones to identify are the ones that look like a T.
3:10
If you have something to the left, something to the right, something up, but nothing down, then this guy has got to be equal to 0-- just has to be. So zero-force members are pretty easy to identify in that if you have two on the same line of action, that third one is 0, always. There's one more case, and that would be like this. Let's say you had a joint like that. Well, look here. I got something to the left, nothing to the right. This guy is 0. I got something down but nothing up.
3:39
That guy is 0. So any time you have a free-body diagram with only two members on it that are not on the same line of action, they're both 0. They've got to be. So that's how we identify zero-force members. And identifying them will make our life easier because it'll simplify our truss. So let's see if we can simplify this truss. Because that right there is scaring me. That's 21 members in this truss-- 21. Well, but number one is that, since this is symmetric, that already cut my work in half.
4:14
Because whatever's going on in member HG is going to be going on in AL. So if I find something on this side, then I automatically find it on that side. So right there, boom, I just cut our list in half. So don't be panicking anymore. Let's see if we can find any zero-force members in here. Do you see any? What do you see? Tell me. OK, all right. I heard you say BL. If you look at this free-body diagram right here-- BL-- I got something to the left, something
4:47
to the right, something up, but nothing down. So BL is a zero-force member. We can take him away. Take him away. So BL, BL. Where's BL. (SINGING) BL. Where is BL? There he is. He's LB. He's 0. Oh, and so is HF-- HF, 0, way up there at the top. Ooh, ooh, look what it did. Now look. I got two members on the same line of action and another one if I do a free-body diagram in my head of that joint. So HE is 0-- HE 0. Also LC 0. LC-- where are you, LC? There he is-- 0.
5:41
So right there, I have made my life a little bit easier. Now, there's a couple of other things that are easy. If I look at this one right here, if I look at that joint, I have something like this. Well, I got 10 going down. So what does that guy have to be? Bam. 10 going up-- gotta be, right? So guess what, CK is 10, and EI is 10. CK-- CK, where are you, CK? There you are-- 10 kilonewtons. And what else did I say? IE. That's down at the very bottom.
6:21
And that guy is going away from the joint. So what is he? Tension. So this guy's tension. That guy's tension. Wow, look how many. We already got a bunch done, don't we? Now, here's another thing that makes this easy. If you drew a free-body diagram of joint L now, if I draw a diagram of joint L, I've got this, which tells me that AL and LK, they've got to be equal. So does AB and BC and CD. So that is equal to that is equal to that. They're all the same.
7:00
Because the free-body diagram's like this, of the joint. So whatever's on this side, it's got to be the same on that side. So this is going to be OK. We're going to simplify this quite a bit here. Let's just do a couple of joints, and we're going to be there. Let's do joint A. So I'm going to do a free-body diagram of joint A. Joint A looks like that.
7:23
Let's see. This is going up. Uh-oh, this one's got to go down, doesn't it? And if this one goes to the left, then this one has to go to the right. So this is AB. And this guy is AL. And the angle is 2, 4, 6 by 2, 4, 6. Oh, it's 45 degrees.
7:47
And this is 20 kilonewtons down here. So quickly, I can do the sum of the forces in the y. And I get 20 going uphill minus AL sine 45.
8:06
So therefore, what is AL equal to? OK On Clear. 20 divided by 0.707 is 28.289. So we'll call it 28.3. So AL equals 28.3 kilonewtons. And what? It's pushing on the joint, so that guy is compression. And then I can do the sum of the forces in the x. And I get AB minus 28.3 cosine 45. And what is that? 23 times 0.707 equals-- well, that's 20. AB is 20.
8:58
And AB is pulling on the joint, so he is in tension.
9:05
So let's put that over there. AL-- 28.3. (SINGING) AL, where are you? There you are-- 28.3 in compression. And then AB is 20 kilonewtons. AB is way up here-- 20 kilonewtons in tension. And that means that BC is 20 kilonewtons in tension.
9:36
And that means that CD right here-- here's BC, here's BC-- and then here's CD. So guess what? He's got to be the same. So CD is 20 kilonewtons in tension.
9:51
And also, what we had over here that we just found. Oh, yeah, we can do LK. LK has to be the same as AL. Where's LK? LK-- right there. So that one is the same as AL, which is right there. This one's 28.3 kilonewtons compression. And then these are in tension over here. So GF is 20.
10:27
And so is EF and DE. So all these are 20.
10:38
Man, look how much stuff we just found. So we got all of this. We got EI down there. And it's kind of good to also put check marks on the ones we know. So we know you. We know you. We know you. We know you. We know you, and we know you. I know you and you and you and you and you. So where do you want to go next? Where do we go next? OK, what we could do is go here, joint K. Since I know this guy and that guy, I only have to cut through two unknowns. So that's going to get me-- here's
11:19
joint K. It's got one there, one here, one here, and one there.
11:28
And let's see. KL-- KL-- where'd KL go? Right there-- 28.3 in compression. So here he is, 28.3. And then I know this guy, which is KC, was 10 in tension.
11:53
Now, can I tell about these other ones here? Can I tell which one's up or down or left or right? Mm, not really. I kind of think this one's like this, and this one, I'm not really sure. So I'll just guess him in tension. I don't really know. I think he's probably in compression, but I don't know. So here's KD. And this one up here is KJ.
12:20
And all we need is some angles here. So we know this. We know that's 45 degrees, and that's 45 degrees. I don't know that angle in there. But this triangle right here is 2 by 4. So this angle right here, which is one I have here that I need right there, that angle is opposite over adjacent. So 4 divided by 2 is 2. And so inverse tan of 2 equals 63.44 degrees. OK, I'm ready to write my two equations. Sum of the forces in the x, sum of the forces in the y.
13:16
In the x, I've got-- what do I have here?
13:24
In the x, I've got 28.3 cos 45. I've got minus KJ cos 45. And then I have plus KD cos 63.44. So two unknowns in that equation. That's no help. Next equation. I think I'm going to have to use my equation solver to do this. The y equation, I have 28.3 sine of 45 going up. I've got minus 10 going down. I've got minus KD sine 63.44. And I've got minus KJ sine of 45.
14:23
So there's my two equations. I'm going to rewrite those where they're a little bit simpler. And so 28.3 times cosine 45 is 20. So I have 20 is equal to 0.707KJ plus the cosine of 63.44 is 0.447. So there's equation number 1. That's from my x equation. That's from here. Now I'm going to simplify this y equation. Oh oh oh oh. I moved that guy to the other side. He became a positive. That guy should be a negative, shouldn't he? There. Whew, that was close. OK, equation number 2.
15:15
Be careful, algebra. 28.3-- well, it's the same thing, right? That's 20 is equal to-- oh, that's 20 minus 10. 20 minus 10 is just 10, isn't it?
15:34
So 10 is equal to-- and we got these two guys. I'm going to do KJ, which is a 0.707KJ. And then this guy is going to be a plus sine 63.44-- 894KD.
15:56
All right, let's throw that in our system solver, and let's see what we get for that. So second system solved, 2 by 2, Enter, 0.707, Enter, minus 0.447, Enter, 20, Enter, 0.707, Enter, Enter, and then 0.894, Enter. And then the last one is 10, Enter. If y'all haven't got one of these calculators with system solver on it, dude, you gotta get one. Enter. All right, here we go. So here we go. Uh-oh, we got a negative. What does that mean? It tells me here that KJ is equal to 23.57.
16:48
And KD is equal to negative 7.46. So what does that mean? KD-- I didn't know. I guessed KD. I guessed him in tension. So what does that mean? He's really in compression. So that guy is in compression. KJ, going to the joint, that guy's in compression. So we just found four more things, didn't we? KJ-- KJ-- 23.6 kilonewtons compression. And KD-- KD-- 7.46 kilonewtons compression. And that means that if I found those two guys, I also found these two guys, so JI and ID. So let's see.
17:44
Let's don't get this wrong. JI is the same as KJ. There's JI. There's KJ. So this is 23.6 kilonewtons compression. And ID is the same as KD. So there's KD. And there's ID. So 7.46 kilonewtons compression. So what are we left with here? What are we left with? GH-- where is GH? A B C D-- oh, I know GH, don't I? GH was the same as AL. So GH and AL are the same. Where is AL? AL-- right there-- 28.3 compression-- compression. And then HI, where is HI? HI-- oh, that's the same thing, isn't it?
18:42
HI also 28.3 kilonewtons compression. And who does that leave? It leaves one dude right there-- JD. We could come back with this last guy right there. And we get-- KJ was in compression. So on the other end, he's going to be in compression. So here's joint J. I've got this. And that guy, KJ, was-- how big was KJ? 23.57. And then there's the other one-- 23.57. And these are 45 degrees, 45 degrees. And then there's JD right here-- up up. JD's got to go down, doesn't he-- JD.
19:27
Sum of the forces in the y is equal to 0 is equal to 2 times 23.57 cosine-- well really sine. But it doesn't matter-- same thing. Sine 45-- that's these two guys going uphill, they're components-- minus JD. So JD is going to be in tension. And it's going to be 2 times-- Clear, Clear-- 2 times 23.57 equals times 0.707 equals 33.32. So JD equals 33.32 kilonewtons in tension. That's the last one-- 33.32 kilonewtons tension. Bam. Look at that. It was daunting, wasn't it? But it wasn't that bad.
20:31
Dude, we are so smart. Identifying those zero-force members helps us. Now, here's the secret. If you miss it-- you're like, I didn't see it, I didn't say the zero-force member-- well, when you do this, when you do your little method of joints, you'll still get 0. It just saves you a little bit of time if you can spot them. That's all it does. You'll still get 0 when you do the math. So don't worry about that. But it might save you a little bit of time if you can eliminate those members.
20:57
All right, gang, we'll see you next time. And there'll be more helpful tips and tricks.
21:04
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