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Post in hole problem

  1. Nov 7, 2013 #1
    Supposing there is a metal cylinder that fits into a perfectly machined hole in a block of metal with the cylinder protruding some length from the block. If you place a force on the edge of the part of the cylinder that is protruding in a direction perpendicular to the hole, is there a way/method to characterize the forces at the interface of the metal cylinder and the block?
    Intuitively, there would be maximum compression at the neck of the cylinder where it meets the block opposite where the force is being applied, but is there a mathematical way of showing this? ANd what goes on at the end of the cylinder that is embedded in the block? Would there be compression at the edge on the same side as where the force is being applied?
    Would appreciate any pointers on how to solve this.
     
  2. jcsd
  3. Nov 8, 2013 #2

    Simon Bridge

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    Start with a sketch and follow up by deciding on a model.

    Useful to consider how it would be different if the hole were not "perfectly machined".
    Maybe have a little bit of imperfection - say the hole is a micron wider than the cylinder.
     
  4. Nov 8, 2013 #3
    If the hole is a little wider then the post will incline as it is pressed and have two contacts with the hole one at the neck where the post emerges from the hole and one at the end of the post. Then if the post is relatively thin and inflexible you can use the law of levers to approximate the forces at each point. But how do you get a mathematical expression for the forces at each point on the surface in the "perfectly machined" case?
     
  5. Nov 8, 2013 #4

    nvn

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    dimitri151: Could you give us the dimensions (mm)? And the alloys?
     
    Last edited: Nov 8, 2013
  6. Nov 8, 2013 #5
    The application I'm interested in is in dental implants. So the material is titanium grade 4 and 5. The actual form of the implant is conical and it is threaded. The dimensions are on the order of 3-6 mm diameter with 8-13 mm embedded in bone and 6-10 mm above the bone. That's clearly a much more complicated system, so I thought about starting with as simple a case as possible and drawing what conclusions could be drawn from that.
     
  7. Nov 8, 2013 #6

    nvn

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    dimitri151: In post 1, in your simplified representation, is your cylinder a hollow, cylindrical tube? Or is it a solid, circular (round) rod, above and below the embedment?
     
    Last edited: Nov 8, 2013
  8. Nov 8, 2013 #7
    I suppose the solid case would be the easiest.
     
  9. Nov 8, 2013 #8

    nvn

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    dimitri151: OK, you have a vertical, tip-loaded cantilever, consisting of a solid, round rod, of diameter d1. The surface of the metal block is point A. The cantilever tip is point B. The horizontal load applied at point B is force P. The resultant of the embedment horizontal, compressive, reaction force (force F1) just below the surface of the metal block is located at point C. The resultant of the embedment second horizontal, compressive, reaction force (force F2) is located at point D. The tip of the embedded portion of the rod is point E.

    The cantilever length (distance AB) is L. The distance from point A to point C is a1. The distance from point A to point D is a2. The distance from point A to point E is L2.

    Summing moments about point D, and solving for F1, gives, F1 = L*P/(a2 - a1). Summing horizontal forces, and solving for F2, gives, F2 = F1 - P.

    E.g., let's say L2 = 10.5 mm. Let's say you arbitrarily assume a center of rotation, within the cantilever embedment, at 3.5 mm from point A. Let P = 400 N, L = 8 mm, a1 = 1.2 mm, and a2 = 8.2 mm. Therefore, we have, F1 = 8*400/(8.2 - 1.2) = 457.1 N. And F2 = 457.1 - 400 = 57.1 N.

    You can adjust dimensions a1 and a2 slightly, to see what effect it has on reaction forces F1 and F2.
     
  10. Nov 8, 2013 #9
    Thanks nvn! Is your setup like this?

    *******B <----P
    *******
    *******
    *******
    *******
    *******A_________
    F1-> C*******
    *******
    *******
    *******D<---F2
    *******
    *******E

    How can you justify that your second reactive force is at D and not at E? I have an intuitive feeling it would be at the exact end of the rod? Similarly, why is not C precisely at the level A or very close to A? How can you not assume where the center of rotation is, that is, can you not somehow mathematically determine where the center of rotation will be? I agree there will be rotation, or something like an infinitesimal rotational displacement. If you applied a force at thend of the cantiliver you couldn't somehow get a purely translational motion, but how to demonstrate it mathematically?
     
  11. Nov 8, 2013 #10

    nvn

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    dimitri151: Yes, your diagram is correct. Excellent work. You could put your diagram inside [noparse]
    Code (Text):
     
    [/noparse] tags, to preserve the spacing.

    The loading is slightly distributed along the embedded length. Reaction forces F1 and F2 are the resultant forces of the distributed loading. Therefore, they are slightly away from the ends. But you can put them at the exact ends, if you wish, to see the effect. You generally will not be able to compute the exact center of rotation; that would be extremely complicated to attempt, and inconclusive. It is better to approximate (or guess) a center of rotation.
     
    Last edited: Nov 8, 2013
  12. Nov 8, 2013 #11
    I think finite element analysis is used typically to find the surface force distributions but I thought there might be some method for finding these things without a computer. It seems you should be able to integrate something along the surface to find it ( although I know not what).
    I'm glad you think there is a point about which the cylinder will rotate. It just intuitively feels that way. That point (again without any proof ) seems like it will be below the level of the surface of the block. Then on the side opposite point A,B at the coronal part there would be compression of the surface, then moving down that side the compression would decrease and become zero around the level of the point of rotation. Below that point there would be tension increasing to the apex. And something like the opposite of that would be the case on the other side, tension at the coronal part, a null point at the middle, and then compression in the apical part ( on the same side as P that is).
     
  13. Nov 8, 2013 #12

    nvn

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    True.

    Perhaps there are research papers dedicated to the study of the force or stress distribution on the embedment of a round rod cantilever (?), although I do not know of any. And I have not searched for any.

    Your second paragraph in post 11 is correct, except I do not think there would be tension. Only compression.
     
    Last edited: Nov 8, 2013
  14. Nov 8, 2013 #13
    I was confusing models. Yes, in the metal case there would be no tension since the parts are not adherent, but in a biological model if there is some adhesion on all sides of the implant in its neutral state then I think it would be as I described. Maybe like a metal rod embedded in a hard rubber block that is adherent to the rod over the surface in common.
     
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