13 Replies Latest reply on Oct 9, 2008 6:05 PM by Matthew Johnson

    Simple Torsion Stress in Tubes

    Matthew Johnson
      I'm a new user to CosmoWorks and FEA in general. I'm working on a project that involves torsion in a hollow tube. I'm making some general assumptions based off calculations for torsion in tubes as outlined in "Mechanical Engineering Design" by Shigley and Mischke. When I initially started the project, the calculations and the FEA were off by a factor. So in order to enhance my understanding of Cosmosworks, I picked two sample problems from the book. They are of torsion in a round tube and a rectangular tube. In the round tube problem, I took the properties for 1018 steel as seen in the attached pic and assumed a 3000 lbf-in input about the centerline axis. The book indicates that I should be getting 44,427 psi shear stress whereas Cosmos is showing 78,480 psi as seen in the attached pictures. In the rectangular problem, I used the material properties defined in the problem and used a 23,728 lbf-in input, from the problem, with an axis generated from the center of the rectangle. The book indicates that I should be getting 11,500 psi shear stress whereas Cosmos is showing 130,800 psi as seen in the attached pictures. Both tubes are constrained by their bottom face and have an applied torque about the top face. Is there something I'm not addressing here?

      EDIT: I finally managed to work that counterintuitive attachment java app. Here are the attachmets. The problems are in the 7th edition pages 144-145. Thanks for the input.
        • Simple Torsion Stress in Tubes
          Steven Dinsdale
          Hi Matt,

          I do not see any pictures. There are many factors that could be influencing your results from how the study was set up to mesh size and everything in between. What edition of Shigley are you looking at? I have the 7th Ed; if you tell me what page the problem is I might be able to try and run it myself.

          Steven
            • Simple Torsion Stress in Tubes
              Vince Adams
              Hi Matt, another common mistake in torsional problems is the calculation used for J. Roark has a table of equations for the correct polar moments of inertia.

              If you can upload those pix and more details about the geometry, even the equations you used, I'm sure we can get to the bottom of this. I've been thru enough of these that I am sure that for some reason you're not comparing apples to apples.

              On a side note, this is a great exercise to help you learn COSMOSWorks! I wish more new users had (or took) the time. When asked what the best example problems are to learn FEA, I point them to Shigley and Roark.

              -- Vince
                • Simple Torsion Stress in Tubes
                  Matthew Johnson
                  EDIT: Added part pictures.
                    • Simple Torsion Stress in Tubes
                      Bill McEachern
                      I did not spend too much time looking at your pictures but here is couple of things:

                      -You need to plot the right stress components with the right reference (shear stess vs. Von Mises).
                      -you need to understnad that the formula is giving you the tube nominal maximum. If you look at the color of the bulk of the sq. tube it lines up with the values you have calculated by hand. The hand calc is not going to give much on the stress risers due to the restraint.
                        • Simple Torsion Stress in Tubes
                          Matthew Johnson
                          1) Ok, I revamped the results to show stress in the XZ. I would assume that this is the plane of shear that I need. Please see the attached pics. I'm still missing something between the hand calcs and this.

                          2) I'm not completely clear on your nominal statement. So the stresses at the risers would not be as indicative of the overall failure of the member, but is merely a function of the restraint?

                          The project I'm working on is to reduce the wall thickness of a round tube. I'm trying to target a factor of safety of 4 and manage some sort of fatigue life in twist. I have failed samples and would like to have the FEA and samples align in results before proceding further with the design. Any articles you would recommend over standard text would be appreciated.

                            • Simple Torsion Stress in Tubes
                              Bill McEachern
                              Hi Matt,

                              I hear your pain on the attachment app....they might as well get rid of he attach button as it doesn't do anything that I can tell.

                              here are some hand calcs and the cosmos calc. Von Mises is plotted but given it is a a pure state of shear it amounts to the same thing. 592 psi vs. 599. Close enough for me. Hope this helps.
                                • Simple Torsion Stress in Tubes
                                  Steven Dinsdale
                                  I just found some time to do the second problem on page 145. TauXY = 14.845; not too shabby compared to 14.809 in the book. I took T = 1. I'll try to get some pictures if I can.

                                  -Steven
                                    • Simple Torsion Stress in Tubes
                                      Matthew Johnson
                                      Ok, I see that we're all agreeing on hand calculations.

                                      So, for whatever reason, I'm setting the model up incorrectly, but don't know what. I have attached pics of two versions. The first is th 10in/9in tube with 40000 lbf/in input. The second is 1in/0.9in with 1 lbf/in input. I still don't get the same VonMis as you guys and my shear doesn't correlate either.

                                      Thanks.
                                        • Simple Torsion Stress in Tubes
                                          Steven Dinsdale
                                          I think I see your problem.

                                          The ID in the book is .9" and the OD is 1", not 9" and 10".

                                          -Steven
                                          • Simple Torsion Stress in Tubes
                                            Bill McEachern
                                            Oooooppppssss.....

                                            Through a calamity of mistakes my previous post had errors...

                                            My Cosmos model had a thicker wall such that the Von Mises just happen to produce the right number......and as I was lazy .....Von Mises does not correlate to pure shear with out a modification......

                                            Anyway here is the correct hand calc with the conversion to Von Mises for a state of pure shear...from the Cosmos help file

                                            In short your answer in Cosmos was right and mine was --well not right for the case I did the hand calc for but some other case.....(ie 8" ID).

                                            Oh and it pays to be careful with the free help.....

                                            the other thing is that the shear stress component doesn't plot the right way...maybe Vince can comment...does not appear to plot in cylindrical co-ordinate system when the temp axis is used as a reference for shear stress.
                                              • Simple Torsion Stress in Tubes
                                                Matthew Johnson
                                                BINGO! I was setting the problem right, but I had misinterpreted the Von Mises for telling me the same thing as shear. I was always coming up with VM that was 1.73 more than what I had hand calculated. By what you listed from the help (which would have clued me in earlier had I looked, RTFH), the VM=3^(1/2)*Shear, and therefore was what I was missing. So if I take any VM result and divide it by 1.73, the hand calcs work.

                                                Now a followup question as this is probably theory based now. Why does shear stress for the plane that runs the length of the tube show the correct shear value? I'm not brushed up as well on my SOM, but I would have thought the shear would have been from a plane parallel to the end surfaces.

                                                Thanks guys.