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Machine Problems in Trusses and Frames · Acedra
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Machine Problems in Trusses and Frames
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0:06
JEFF HANSON: Hey, gang, welcome back. Today we're talking about machine problems. What's a machine problem? Why is it in the Trusses and Frames chapter? Well, a machine problem is very similar to a frame problem, in that I don't know that I've ever worked one, a machine problem, that ever had any global equilibrium. It's not hooked to the world. If you're going to get a machine problem, this is kind of the quintessential machine problem--
0:30
a good, old-fashioned pair of pliers or something like that. Right? What is a machine problem? A machine problem has moving parts. So that would be something like a pair of pliers or the bucket on the front of a front-end loader or a tractor or a lift or something-- something that has moving parts. It's similar to trusses, in that it might contain two-force members. It's similar to frames, in that it might also contain multiforce members. So the good thing is--
1:04
frame problems and machine problems, exactly the same. You can follow the same recipe, the same steps, and it's going to get you the same answer. So how do I do this? What I have here is a pair of pliers that look like some nice locking pliers, here-- better known as "vise grips," right? How many pieces is this made up of? Well, one, this bottom handle. Two, this top jaw. Three, this little guy right here. And then four is this bottom jaw. Right? So there's four pieces that make this thing up.
1:38
So it says here, find the forces on all members of the pliers. So at every pin-- and including over here at point F, where it's holding what looks like a nut over there-- find every force there. So basically, they're asking us, find everything. OK? So step 1-- let me write my recipe back over here. OK? Step 1. Here's my favorite. OK? Look for two-force members.
2:14
Look for two-force members. Why do I need to find two-force members? Because, number one, you don't want to draw a free-body diagram of a two-force member. You draw a free-body diagram of a two-force member; what do you find out? Well, you find out that F equals F. Right? Super not helpful! Yeah, I know, right? Number two is that two-force members take me from two unknowns to only one unknown. So that's a really nice thing, because if you're solving a problem and you're like, man,
2:44
there's too many unknowns in this problem-- I can't solve it-- you probably didn't spot one of the two-force members. OK? Do you see two-force members in this problem? Step 1-- do you see any? I bet you you don't. All right? Is there one? Is there two? Is there zero? What's a two-force member? Pin connected at both ends-- no forces or moments in the middle. OK right? Yeah. Here's one right here-- this member FAB. So-- FAB-- two-force. Right?
3:22
And actually there's one more that you probably didn't get, and that is the nut, right over here, right? That little nut is a two-force member. What is he experiencing? Compression. Right? Remember, two-force members exist in tension or compression. So this is a two-force member, over here, as well.
3:46
All right. So, next, step number 2. We've identified it. So we don't want to draw a free-body like that guy. Yeah? So step 2, draw FBD of each part.
4:06
So we want to draw a free-body diagram. Now, if we eliminated the two-force members, that leaves us with one, two, three parts. Right? So we have three free bodies. And we want to fully label them-- all the arrows on them. And remember, if we don't really understand which direction a force goes, we can just guess. We'll just get a negative if we guess wrong. And then step 3 is to solve-- just solve it. That's kind of a high-look recipe, there, for doing these problems.
4:42
So what I want to do next is, I want to sketch this free body, that free body, that free body, and let's see-- let's see if you can do it. So here's the top jaw. I'm just gonna-- not as good a drawing, here, but whatever. There's that top jaw, let's say. And then I've got the bottom jaw, which is kind of like that, maybe. And then I've got that bottom jaw over here--
5:19
something like that. So there's that guy. Now, let me put some dots on here, so you'll know what's going on. On the top jaw, what do we have? We have this. We'll call that 50 pounds. That's given. Here's point A, maybe-- point A. Point C is down here-- point C. And then over here is point F. And then, same thing over here. Here is point F-- this guy, point C-- and then, down here, point D. And then on this guy, I've got D. And then, up here, I've got B. And then, over here-- boom-- 50 pounds.
6:09
OK? So there's our three free bodies. Why don't you put some arrows on it? Push Pause, put some arrows on it, and come back and see if you get the same one I do. Ready? Go. Did you do it? All right, let's go. Here's what I know. Now, I know that that little nut is in compression. So if you squeeze me, what's my reaction to you? Oh! I'm going to push back on you, right? So here's that nut. I'm going to do this. I'm going to put-- this I'm just going to call this FY.
6:42
And then over here, the other side of the nut is pushing down-- FY. OK? What else do I have? Well, on this one I have point C-- pin connection, right? Pin, pin, CY, CX, A. Ooh! If I squeeze this together, now, that guy's a two-force member, right? You think he's in compression or tension? Compression right? So what's he going to do? Compression members do what to me? You squeeze on me; I push on you. Right? So here's what it is. And we'll call this guy FAB. Now, that's what I said, right?
7:22
It takes you from two unknowns, which would have been, like, a pin connection, to just one unknown, because I know the direction that that two-force member acts. So it eliminates one of my unknowns. Now, I'm going to have to find an angle, but I think that that's pretty easy. We can do that, can't we? As a matter of fact, the angle is this triangle right here, which, this side of the triangle is 1. And what is this side, over here? Well, that's-- 4-- 5.2, and this side is 2.4. So 2.4 off of 5.2.
8:01
How much is that? OK? That's 2.8, isn't it? So this side of the triangle's 2.8. So this angle right there, which is the same as this angle up there, we can do tan. 1 divided by 2.8 equals-- inverse tan-- answer-- equals 19.65 degrees. So this is 19.65 degrees. There you go. OK, what else do we have? On this free body, do I know about CX and CY? Oh, since FAB has a component going this way, then CX has to go that way, doesn't it?
8:52
OK? And let's see. If I put my finger here, and I take the moment, that guy rotates me what-- counter. This guy rotates me counter. This guy rotates me counter. So that guy has to rotate me clock. Right? Now, I could have guessed that, but now I know that CY goes down. If I solve for CY and I get a negative, I'm going to really scratch my head because I think that should be absolutely correct. OK? How about these guys? Oh, look here. C, over here; C, over there.
9:30
So what do we know-- if I have C on this free body, what do I know about that free body? It has to be the same thing except in opposite directions, doesn't it? And so CX, instead of going to the right, goes to the left. And instead of going down, it goes up. OK? Now what do I have left? Well, I have D. So that has an uh and an uh. Let's put a little DY and a DX.
10:05
CX goes to the left, so DX needs to go to the right. I know that. Right? I can go up here-- DX DY. That means I know that this guy has to go this direction. Right? Pin connection at B is B? No! B is the other side of my two-force member. So here we go. There he is. That's the other side of that. At an angle of-- what did we say? 19.65 degrees. And that guy is FAB. OK? Let's see. Let's see, here. Hmm. If I take my moment, here, this guy rotates me uh clock. That guy rotates me uh clock.
10:55
So this guy has to rotate me counterclockwise. So he's got to go uphill. And then that means this guy has to go downhill. There you go. Did you get all the same arrows that I did? I know for a fact that, when I solve for each one of those things, I better get a positive answer. Right? OK? So we're ready to rock and roll, here, folks. We got our free-body diagram all drawn up. Where do you want to start? We want to start somewhere where we know something. Right?
11:27
Maybe that top guy up there is a good place to start. Maybe we take the moment at D-- I like for you to see the strategy. Right? Take the moment at D. Then I automatically find that guy. And then I can come back and get those two guys. Right? And once I have those two guys, then I already know that guy. I know D and these guys. Yeah I think I can-- that works good for me. So let's start with this top guy, here. OK? And I'll go right over here. We're going to take the moment at D--
12:00
summing the moments at point D. This being positive. So we're taking the moment, here. What do we have? And don't forget, we're going to break this guy into two components, here-- one there, one there. And let's put the B down here, so we can see him-- B. OK, so this x component is what? FAB times the sine of 19.65, and this guy, FAB cosine 19.65. So let's see if we can do it. So let's do this guy first-- the 19.65. Put my finger at D. This rotates me--
12:50
oh, that's counterclockwise, so that's positive. So FAB sine 19.65-- FAB sine 19.65 degrees, times how far away? How far is it from D over to B? From D to B-- oh, there it is, right there. Right? It's 1 inch. We'll put a 1 there. OK? What else do we have? We have this guy, which rotates me about D-- ooh negative huh? So that's minus FAB cosine 19.65, times-- what's the distance from D to B? From D to B is 0.2. Isn't it? 0.2 inches. Anything else? Oh, yeah, the 50! Which rotates me negative--
13:43
-50 times, how far away? How far is it from D all the way over to there? Well, it's 1 plus 4 is 5. OK? So we can solve that, can't we? Let's do it. Where'd my calculator go? Here in my pocket. So let's see. The sine-- uh-- Clear. The sine of 19.65 is 0.336. So this whole thing is going to be 0.336FAB minus the cosine of 19.65 times 0.2, right? 0.188. 0.188FAB equals 5 times 50-- 250FAB is equal to 0.336 minus 0.188 equals 0.148. And then 250 divided by 0.148 equals-- woo!
15:03
--a lot-- 1,689.2 pounds. OK? What does that mean? Man, you put 50 pounds of force on these pliers, the mechanical advantage there really starts adding up. Doesn't it? So this thing is going to put 1,600 pounds on that little member, there. That's a lot. So, can we find DX, now? Is that hard to find? Well no. DX has got to be equal-- right? Sum of the force in the x. That guy has to be equal to that guy, doesn't he? So DX has to be equal to 1,689.2 times cosine of 19.65-- times cosine 19.65--
16:00
1590.8. There's that guy. OK? And then what about DY? Well, sum of the forces in the y equals what-- 50 plus DY minus FAB sine 19.65-- so, 1,689.2 times to the sine of 19.65. Right? So that means that DY is equal to-- here we go-- 1,689.2 times the sine of 19.65 equals-- And then, what? Move that to the other side and then subtract 50 off of it. Right? Minus 50-- 518.03. OK? There's another one. So now I know you, I know you, I know you. Where are we going now?
17:16
Well, we could do a couple of things. We can go here. Let's go here. Yeah, let's go here! FAB, do we know you? Yeah, we found you, right? 1,689.2. OK? And again, FAB, one of those things-- I see a vector at an angle, man. I want to break it into components, don't I? There you go. OK? So this guy is 1,689.2 sine 19.65, 1,689.2 cosine 19.65. OK? Here's what I'm thinking. On this free body-- bam-- we take the moment at C, and then we find FY. We find out how much force is on the bolt, don't we?
18:11
Let's do that. Sum of the moments about point C. What do we have? Well, we get the 50. That rotates us-- what is that-- positive. So 50. Now, how far is it from C over to the 50? From C over to the 50 is 4-- it's 5.2. OK? And then I got the 1,689, which rotates me about point C. That's negative-- minus 1,689.2 sine of 19.65 times, how far away-- from C to A-- from C to-- it's 4. OK? And then I got this guy, which is above point C, so he rotates me-- what is that? Counter? That's positive.
19:15
So, plus 1,689.2 cosine 19.65 times, how far away? How far is it from C up to A? Oh, that's too hard. From C up to A-- OK, let's see. I've got to go down 0.7, down 0.7-- no-- down 0.5. Right? And then go up an inch. So the distance from here to there is how much? It's also 0.5, isn't it? OK? 0.5. Anything left? What else is left? Oh! This guy over here, right, who rotates me positive. So, plus FY times, how far away from C to FY-- duh to tuh-- is-- uh-oh-- on, there it is-- 1.2 right?
20:17
Boom-- long equation, but that's no big deal. OK, so how much is FY? FY Is equal to-- oh, I've got to add all this stuff up. Right? And I'm going to change all these signs and move it over to the other side. Can I do that, right quick? Can I do this? Will you allow me? If I put n equals right there-- whoop, whoop! And I just change all of these signs-- minus, plus, minus-- let me get rid of all this, now. OK? There. I just moved FY to the other side. Right? So here we go. 1,689-- whoop-- Clear.
20:56
1,689.2 times the sine of 19.65 time 4 is 2,272-- minus 50 times 5.2-- times 5.2-- equals-- minus one more-- minus that guy-- 1,689.2 times cosine of 19.65 times 0.5 equals 1,216.7. And then divide that-- divide it by 1.2-- leaves me with 1,013.93. 1,013.93 L-B-S-- pounds. What? You put 50 pounds of force over here, and it puts 1,000 pounds of clamp on your nut over there? That's a really good force multi-- plier! [LAUGH] OK. Maybe that's where that came from, eh? What do you think?
22:19
I don't know. OK, so who else do we have to find, now? CX. CX. This shouldn't be too hard, should it? CX has just got to be equal to that guy, doesn't it? So, 1,689.2 cos 19.65-- boom.
22:41
1,590.8 1,590.8 L-B-S. OK? That guy has to equal that guy. That's it! And then one more equation-- sum of the forces in the y. What do I have, here? I have this guy, going uphill, which is-- how much is that? 1,689.2-- I'm trying to-- I put this in my calculator 14 times-- sine 19.65-- sine 19.65-- equals 568. 568. Right? That's this guy, right here. Who else is going the y direction? FY is going uphill, plus 1,013.93. So I got that one and that one. CY is was going downhill-- minus CY.
23:35
And who else? Oh, yeah-- minus 50. OK? So move the CY to the other side. And what does CY equal? CY is equal to-- here we go-- 568 plus 1,013.93. minus 50 is 1,531-- 1,531.9 L-B-S. Now, do I know everything? Do I know-- oh! We haven't even done this free body halfway. Oh my-- wait. I know you. I know you. I know you. I know you! I know you! Oh, come on. We know everything, don't we? So what are we doing? We find a little bit of information on one free body, we take it to the next free body,
24:35
and we just work our way across. Piece of cake, right? I hope this helps, and I'll see you on the next video.
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