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Carbon fibre tube -- is round or oval stronger?

  1. Aug 8, 2017 #1
    Hi all, my question is if i have a length of carbon fibre tube, one is round the other is oval and orientated along its longest side is it more stiff than the round one? Reason for asking is i am trying to make a telescopic carbon fibre pole that is more rigid than the round ones on the market. The object on the top of the pole will keep the orientation of the pole at the correct position.
    Thanks in advance
     

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  3. Aug 8, 2017 #2

    DaveC426913

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    I am a layperson, not an engineer or mathematician.

    For a given circumference, compared to a circle, the elliptical cross section will be stronger through its semi-major axis and weaker through its semi-minor axis.

    If the primary bending load is in a predictable direction, you can align the semi-major axis that way. (i.e. if your pole can be kept oriented so its semi-majr axis is vertical, to oppose gravity).

    If the bending load is not predictable, or is not evenly distributed (i.e. if the pole can be rotated around its aixs), you'll have a problem - a column is only going to be as strong as its weakest cross-section, so it will be more prone to failure than a comparable circular cross-section.
     
  4. Aug 10, 2017 #3
    Thanks for your reply Dave
     
  5. Aug 12, 2017 #4

    DaveC426913

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    It occurs to me that some experimentation with lengthwise fluting/corrugation might help with stiffness.
    I think you'd wouldn't need them all the way around, just at the semi-minor axis, to add stiffness there.

    But I've never seen this done, and I'm just spitballing.
    fluted-pole.png
     
    Last edited: Aug 12, 2017
  6. Aug 12, 2017 #5

    DaveC426913

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    A question occurs to me though:

    What is the advantage of yours intended to be over a circular cross-section?

    A large diameter tube is stiffer than a small diameter tube, so if strength were the only consideration, there'd be no need to change it's shape. You'd just use a wider diameter tube (it wouldn't need to have thicker walls). This is not what you want, of course.

    Clearly, the goal then must be to achieve similar strength but with less weight.

    And that raises the question: just how much weight do you think you could save your potential customer?

    An elliptical cross-section with a 2:1 ratio would use only 71% of the material as that of a round tube (at a cost to strength of [ ?? ]).
    Let's assume an ideal case that you find a design where the strength is not compromised i.e. it equivelant to the same diameter circular tube.

    Would your 142g pole really outcompete their 200g pole?

    (I guess, if weight were a premium, such as ultra-light craft, sure. But I get the impression that's not your application.)
     
    Last edited: Aug 12, 2017
  7. Aug 12, 2017 #6

    Nidum

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    @Karl christophersen

    There is no simple general answer to this question . Answer depends on several different factors - any of which may or may not apply here .

    Can you tell us more about your application ?
     
  8. Aug 12, 2017 #7

    DaveC426913

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    I dunno. I think we can infer a general picture.

    Think of a selfie-stick for a camera. It is meant to be rigid against gravity. In the OP's case, for some reason, we can assume the stick stays oriented around its long axis. i.e it does not rotate about its long axis (in aeronautical terms this is called roll).

    So, an elliptical cross-section could be as strong vertically (i.e. pitch) as a circular cross-section, yet need not be as strong in the transverse direction (i.e. yaw). Thus, he could save a bit of weight.

    An ideal configuration would make it equally as strong as a circular section, but that might not be achievable in practice. Nonetheless, the OP's stick would have a higher strength-to-weight ratio.

    fpsyg-07-00142-g004.jpg
     
    Last edited: Aug 12, 2017
  9. Aug 14, 2017 at 10:37 AM #8

    Mech_Engineer

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    The best method for comparing bending strength of different cross-sectional shapes is to use the "section modulus," also sometimes called the "section area moment of inertia" (although this parameter doesn't have anything to do with inertia).

    An example list of section moduli properties can be found here: http://www.engineersedge.com/section_properties_menu.shtml. The two sections you're most interested in are the hollow circular tube (Case 19) and the elliptical hollow tube (Case 22). The hollow eccentric tube will have a higher section modulus in one direction, mirroring DaveC's previous post regarding strength along the semi-major and semi-minor axis.

    My copy of Roark's Formulas for Stress and Strain (Seventh Edition) has formulas for each case.

    For a hollow circular tube:
    moment-inetia-gyration-3_clip_image002_0009.jpg
    $$I_{x} = I_{y} = \frac{π}{4}(R^{4}-R_{i}^{4})$$

    Hollow elliptical tube (internal hole is elliptical too, so non-constant wall thickness):
    moment-inetia-gyration-3_clip_image002_0027.jpg
    $$I_{x} = \frac{π}{4}(ba^{3}-b_{i}a_{i}^{3})$$
    $$I_{y} = \frac{π}{4}(ab^{3}-a_{i}b_{i}^{3})$$
     
  10. Aug 16, 2017 at 3:52 PM #9
    I agree with Mech_Engineer's analysis regarding the pole's rigidity.

    As an aside, I like DaveC's suggestion about corrugating the pole. The problem would be manufacturing, specifically making sure that the radii are large enough that the CF can fill in the grooves and not be smaller than the minimum bend radius. Also, I think the manufacturing cost of using more complicated shapes with CF will be prohibitive since a "normal" pole is either probably filament wound or even just rolled up prepreg. If cost is no object, because it's carbon fiber you could get away with making more complicated structures, like this: http://www.isotruss.com/. Depends on the ultimate goal.
     
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