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Newtonian Physics Question

  1. Jun 15, 2016 #1
    Hello All,

    I have a question for which I'd like an authoritative answer, if some physics genius would be so kind. I'll try to provide as much background as I can but follow up questions for clarity are more than welcome!

    So I'm working on the design for a patent application and it involves a series of custom aluminum square tubing pieces held together with either standard 5/8" hitch pins or any 5/8" in bolt. The drawing speaks volumes if you would like to refer to it as you read. Here are the particulars:

    Piece A
    • 2" square aluminum tubing (this dimension cannot change)
    • About 18" long
    • 5/8" D holes placed about 3" apart
    Piece B
    • 2.5" square aluminum tubing
    • 3/16" thick wall
    • Holes close to the edge are 2" from the edge and the inside holes are 3" from the outside holes
    Now, since the wall thickness of the piece B is 3/16", the inside is 2 1/8 x 2 1/8, which leaves a gap of 1/16" all around when piece A is fixed to piece B using the 2 hitch pins as shown in the drawing.

    The problem...because these pieces are meant to attach to each other one after the other (up to 6 or 7 in a row), and there is a 1/16" gap between each connection, the entire system sags about 5 or 6" on the end. This is not surprising, and the obvious answer long-term is to do one of the following:
    1. Reduce the gap (which means increasing the thickness of the wall of piece B)
    2. Add a second hole (done...originally these pieces had only one hold and one hitch pin holding them together)
    3. Reduce the engineering tolerance for the hole(s)
    4. Add some sort of shim at each connection point.
    I'm getting the question, I promise...the question is not how to redesign it to reduce the sag, as I have a pretty good idea how to do that. The question, which is what brings me to this forum is as follows:

    If I raise the holes off the center-line will it reduce the "slack" anymore than leaving them on center-line? For example, if all holes were moved vertically 1/8" of an inch, would there be less slack than where they are now? To make it relevant to physics...if there is a downward force on piece A, would the force on the pins be different if they were moved vertically 1/8" above center?

    Any help you could provide would be helpful!
    Thanks,
    Matt
     

    Attached Files:

  2. jcsd
  3. Jun 15, 2016 #2

    Simon Bridge

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    You look at similar situations to see how other engineers have solved the same problem ... pick the one that best suits your need.
    And example not on your list is to provide a tongue that can be screwed down or a spring that impacts the inner tube. You've seen these, you cut 3 sides of a rectangle in the side of the bigger tube and push the resulting tongue in. When the smaller tube is inserted, the tongue holds it in place against the far side of the outer tube.
     
  4. Jun 16, 2016 #3

    jack action

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    On the images, the tolerance on the holes are so slack that the tubing gap is creating all the sagging. This means that the tubing edges are taking all the vertical forces and the pins are taking the horizontal forces. As long as the fit is not tighter for the pins, this should be the case wherever the holes are.
     
  5. Jun 16, 2016 #4
    @Simon Bridge, thanks for the response but the question is not "how do I redesign this". The question is, "all else equal, if all holes (2 pins) are raised vertically 1/8" above the center-line, will the slack in the system be more, less, or the same as when the holes are all centered (as shown in the drawings)?

    @jack action, thanks for the response. In the picture with the entire system shown (2nd pic), there is only one hitch pin per connection, so you're right in that piece A pivots and the backside of piece A hits the top of piece B. However, the redesign (1st picture) shows 2 holes and 2 pins, so now the back of piece A will NOT hit the top of piece B. But, as I said above, the question is not all about how to fix the design, it's about how force on the pins changes if the holes are moved vertically 1/8". If I understand you correctly, you're saying those forces will not change? That was my original assumption but I asked the question in this forum b/c I wasn't entirely sure and it has been nearly 20 years since I took physics in University.

    Thank you both for your responses!
     
  6. Jun 16, 2016 #5

    jack action

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    If the fit of your holes are still large enough, the back of piece A will STILL hit the top of piece B.

    Assume the distance between the 2 pins is half the distance between the ends of each tube. Therefore a gap 1/32" between the pins and the holes could be enough to let the back of piece A hitting the top of piece B. Of course, if you locate them appropriately, then each fit may cancel each other, but that demands some pretty accurate machining which may make the insertion of the pins more difficult too.

    If such accurate machining is good with you, you could just as well use one pin with a very well localized hole in each tube, such that the 2 bottom faces are exactly in the same plane. In that case, the pin would take the vertical load.
     
  7. Jun 16, 2016 #6
    This is really a discussion on torque. In this case the torque can be thought of as a twist or downward force to the tubing. Mathematically, torque is defined as the cross product of the vector by which the force's application point is offset relative to the fixed suspension point (distance vector) and the force vector, which tends to produce rotation. The magnitude of torque depends on three quantities: the force applied (weight of the tubing plus guy standing on it) the length of the lever arm (changes depending on the two designs in discussion - the two pins or single pin and inside tubing edge) connecting the axis to the point of force application, and the angle between the force vector and the lever arm. If both pins are moved vertically the forces are equal - no change. if the forward pin is moved slightly upward it would change the angle (Slightly) and thus associated torque. Torque=R(Position Vector) X F(Force Vector) X Sin (Angle between force and position vectors).

    All else being relatively equal, you could "preset" the anticipated tubing gap by creating unequal holes (in sets) with the outside hole being slightly raised from the inside hole. The two pins would share the torque moment equally (small difference in force angle) but most importantly the severe drop due to the pipe gap could be minimized so that unloaded it would sit more horizontally.
     
  8. Jun 16, 2016 #7
    @Clay Perreault , thanks. That was my contention...that the force would be equal in both scenarios. But when it comes to the pieces beyond the pins, I'd really like to know if it will "sag" more or less in either scenario. Assuming neither scenario causes piece A to ever touch piece B on the top/inside, I believe both scenarios would create equal amount of sag, simply due to the engineering tolerance of the holes (again, assume this D is constant). I did think about offsetting 1 hole to create the angle, but again, I still think it's going to come down to the engineering tolerance of the holes and the sag would be the same.
     
  9. Jun 18, 2016 #8

    CWatters

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    If we come up with a solution here can you still patent it?
     
  10. Jun 18, 2016 #9
    @CWatters Not sure what you mean. The solution is already "Patent Pending". As the original post stated, I'm not looking for a solution...the question was about force. See Clay's post for the answer (the one I was looking for)...the torque, force, sag, or whatever, won't change regardless of the position of the holes (all else being equal).
     
  11. Jun 19, 2016 #10
    If I raise the holes off the center-line will it reduce the "slack" anymore than leaving them on center-line? For example, if all holes were moved vertically 1/8" of an inch, would there be less slack than where they are now? To make it relevant to physics...if there is a downward force on piece A, would the force on the pins be different if they were moved vertically 1/8" above center?
    Suppose you could raise the holes in the inner section by just less than a 1/16th all up or all down along the centre line.
    The closer to 1/16th the less sag but also less tolerance.
    You could do the opposite also by doing the outer section only.
     
  12. Jun 19, 2016 #11

    Nidum

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    You need one pin and one roller or rubbing block .
     
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