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Problem with the fixed collar on a smooth rod

  1. Jun 17, 2005 #1
    I gota problem with the fixed collar on a smooth rod. The collar is like a rigid metal T shapped pipe, and the top of the t is where the shaft of the rod goes through. (Its in any basic statics book). It says this type of set up has a normal force perpendicular to the axis of the rod, (OK thats fine I get it), and a moment. (OK, I get that too, its restraining the rotation since its RIGID and not pin.) But the problem is HOW!? does it prevent moment!

    To me, the only way to prevent moment is to have a counter moment produced where the collar meets the rod, in the top horizontal part of the T. This could be acomplished by a symmetrical distributive loading on opposite ends. ( See pic.) But the problem is that damm normal force. If you have a distributive loading like that, which kind of makes sense, because if its a very GOOD, or better yet IDEAL rod/collar, there is almost no slop between the two, and no friction. So this means that the area in contact on the top and bottom of opposing sides should be equal and opposite, so that a couple is produced. Also, one would expect the force to be maxium on the outer edges, and deminish as you approach the center. But then what about the normal force!? It is pointing up! This is distrubing becuase if a couple is acting on that part of the collar, you would expect to have zero force at its center, and most of the force along the edges. I dont know anymore.........sigh....
    Last edited: Dec 26, 2005
  2. jcsd
  3. Jun 20, 2005 #2
    Does anyone have an anwser at all? :-(
  4. Jun 20, 2005 #3


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    You don't give the actual loading of the member or any directions as reference. However, most of the ones I remember were pinned. Is this in reference to a problem of somekind?
  5. Jun 20, 2005 #4
    This is more about the theory than any particular problem. See there is a pinned collar, and there is a FIXED collar. The FIXED collar does not make any logical sense to me. If it is fixed, then it MUST prevent a moment and a force normal to the direction of the collar. BUT lets assume there is some slop in the collar, then it would rotate until opposite ends of the collar are touching the rod, and the normal forces would be at the ends of the rod. Of course they would be equal and opposite, and create a pure torque. But now there is only two points of contact, so that means the restriction of movement perpinduclar to the rod has to conincide with one of these normal forces. As a conseqence, it will create an additional torque, and you will get a different moment produced. Do you see what I mean?

    ( see pic for clarification )

    This is a gross exageration of course. But the book says the left is what happens when you have a rigid collar. I say no, the right is what REALLY happens. but you can clearly see it has to restrict both rotation and translation. Since the forces act on the two opposite ends, then one of these ends has to have a larger magnitude force so that it can also prevent translation. But this in and of itself creates additional moment. So the total couple moment needed is goign to be different when calculated using the methodology of the right side and the left side.
    Last edited: Dec 26, 2005
  6. Jun 21, 2005 #5
    Common guys, I CHALLENGE you to find a solution or error in my work :-p
  7. Jun 21, 2005 #6


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    Cyrus, I'm a bit confused as to what you're trying to understand, but I'll try and help you out. It looks like your statics book is pointing out certain types of geometry for mechanisms and this particular type is a "T" shaped thing that can slide on a rod. The book you have then points out the various possible forces this type of geometry can resist. Obviously some of the loads and moments on this T can't be resisted very well such as a rotation around the rod this T slides on. Imagine the rod rotating around it's axis -> the T can't resist this moment very well, though due to friction there could be a slight resistance.

    One thing this T can resist is the moment. I think you have that one down. If you have a load as you show in the first or last pictures, there's a load up and an equal and opposite load down. This can also be viewed as a moment right at the connection. So although it's shown in two different ways in the attachment to post # 4, they are both the same. The moment, a curly arrow shown over the intersection, is the same as a load up and a load down on the tips of the T where it contacts the tube. There's no difference between using the curly arrow or the arrow up and arrow down, it's just the curly arrow is the conventional way of showing a moment at a point.

    Another force this T can resist quite easily is a load perpendicular to the rod that it slides on. This isn't the same as the moment, it's different. This is a verticle load which doesn't produce any moment on the T at the intersecting joint. The reason I mention this is because you said:

    Well, in the way we would normally indicate a rotational moment and also a verticle load, is exactly the way it's shown on the left. The thing on the right shows the reaction loads between the T and the rod when you put a moment on it, and it's perfectly correct. But it doesn't show a verticle load. The one on the right only shows how a moment is reacted by the rod that this T slides on.

    Note also that if there is a moment or a verticle load on the geometry as given by the left hand picture, there can also be some frictional load in the horizontal direction (ie: the T does not simply slide frictionlessly along the rod, it requires a force equal to the coefficient of friction times the normal loads to move it, though if this were a linear ball bearing for example, that frictional load might be negligable.) Not sure if that's important for you or not, just figured I should point that out.
  8. Jun 21, 2005 #7
    I get that part, its not what im asking though sorry. See here is the thing I AGREE with the pic on the left in principle. But think about it, if there is a moment produced, it can be distributed (in the general case), or in the special case where there is alot of slop, so the only points of contact are the ends, then it comes down to a set of two normal forces. See pay very close attention to this part. In the extreme case where there IS slop, then the normal forces ONLY act on the ends, right? These two normal forces produce the couple. So we now know HOW the couple is being restricted by the collar. The only remaining question is how does the collar resist TRANSLATION as well. But because of the couple, the resistance to translation has to act at one of these two points of normal contact due to the couple. I mean hey, nothing else is in contact anymore!!!!! See the dilema?! The couple screwed things up! So this means that one of the normal forces on the end has to be larger than the other. The result is STILL the same couple, but you also get the added restriction of translation. BUT NOW WHAT!? Because that normal force is larger, it creates ANOTHER MOMENT! AYE!!!!!!!! :-( Hehe, do you see what Im saying now?
  9. Jun 22, 2005 #8


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    I'd agree that you can look at it either way and either way is correct. Whether the force is distributed a lot or basically just a point contact depends on how flexible the T is, how much force, ie: the strain induced in the T, etc... You can also reduce any distributed force such as this to an equal, single reation force. The distributed load is generally in units of force per linear distance, so integrating over the distance gives you the total force. The center of that reaction can also be found by integrating.

    The resistance to translation (ie: sliding along the rod) is due to friction between the T and rod which is the normal force times coefficient of friction. Perhaps the question is, given there's a moment on this T along with some translational force, what does the diagram now look like, assuming friction prevents motion?

    Let's use the second illustration you have in post #4 of the T that has some slop in it, but let's ignore the arrows you show*. If there's a force at the top of the T that is pushing to the left, the T will want to rotate as you show in the diagram with the left side coming down on the rod, and the right end coming up. Note that this isn't statically balanced yet. We have 3 loads, two verticle and only one horizontal. We need at least one more horizontal load which is the frictional load between the T and rod. That's actually in 2 places, one arrow on each point of contact. So now you have 5 arrows. There are 2 verticle ones as shown in the diagram (but reversed as noted below), and three horizontal ones. With these 5 forces acting on the T, you can now do a static balance of all the forces. Start by summing moments around one of the points, preferably a point where a force is applied that hasn't been determined yet. Hope that helps.

    *Note: The arrows you show in this figure are the forces exerted BY the T ON the rod, not the forces on the T exerted by the rod. If you want to do a force balance on the T, these arrows should be reversed.
  10. Jun 23, 2005 #9
    IM sorry but thats wrong. Im not talking about moving along the direction parallel to the rod, I said normal to it. My whole point of this thread is that you cannot ignore the arrows on my picture for the one with slop. I dont think your understanding my question, sorry. It is possible for the rod to have two normal points of contact while being perfectly rigid as well, thats another one of my points. When you put a moment on a T, the t has some slop and so the T MOVES until the normal forces create a counter moment. If you look at my picture again, it should be clear as to what im saying. If not ill try to make a better picture for you and retype the problem with a written example with bogus numbers :-)
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