Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Statically indeterminate tube in Torsion

  1. Jul 9, 2015 #1
    Hey guys.

    It's been a long time since I've done anything beyond a simple shear stress calculation, so have what will hopefully be a basic question.
    I've found similar examples but not ones I was confident I could apply to mine.

    As per the dodgey paint sketch I've made below, I have this scenario where I have a hollow cyclinder a-c (door shaft)which is driven by Fd which is acting on distance Lad.
    Lets assume that all horizontal arms a-d, b-e, and c-f are rigid, so basically I really only need to look at Ta Tb and Tc

    The hollow cylinder is supported by a thrust bearing at 'a' in all directions, x,y,z, and by a roller at 'c' which supports x and y.

    Lets ignore the weight of this single welded frame.

    Now points 'e' and 'f' can either hit a fixed object at the same time, or independently.

    1) In the case that only 'e' or 'f' hit a fixed object, that solution is statically determinate, and Fd = Fe, or Fd = Ff.

    2) In the case that 'e' and 'f' are both limited by a fixed object, such as a wall, I need to determine the distribution of forces, or the Torque, at points 'b' and 'c'. I believe I need to work with the polar moment of inertia to solve this?

    Thanks a million in advance,


    Attached Files:

  2. jcsd
  3. Jul 9, 2015 #2
    1) Yes.
    2) Yes as well as all the engineering properties of the arms.
  4. Jul 10, 2015 #3
    Hmm ok but it doesn't really help me, i'm having trouble relating it to any existing examples.
    Most examples of statically indeterminate beams in torsion have one or both ends fixed, whereas this is not the case.
    Also, I think I agree I need to consider the properties of all arms (even though they are relatively very rigid compared to the shaft) because if i assume the three horizontal arms are rigid, then i think that means all forces and moments would then theoretically stop at 'b' because the joints then may become "fixed" constraints. Though unsure.

    So can anyone help here?

    I tried modelling up a frame simulation to try solve it using Inventor Professional, however my results suggest I am not setting up the frame and/or constraints correctly.
    I'd prefer to at least look at this as simply as possible on paper before I start using F.E.A, so any help is welcomed.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook