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  1. Dec 4, 2009 #1
    I've been trying to determine which is stronger, square tubing or Round tubing. I searched the forum but came up with some other ideas but not the answer.

    Using a Square tubing that's 2"x2" AND .25" thick vs. a Round Tubing that's 2" outside diameter and 1.5" inside Diameter and both are 5' long, as my example. My Force is 4,000 lbs.

    I calculated the Maximum Stress in a beam with uniform support at the ends, and my Force in the middle of the beam. My numbers are M.S. for Square tubing is 13,713.66 psi and M.S. for Round tubing is 23,325.24 psi.

    Does this mean that the Square tubing is stronger then the Round tubing? Or am I misunderstanding the definition of Maximum Stress. Also why is it I hear Round tubing is Stronger then Square tubing?
     
  2. jcsd
  3. Dec 4, 2009 #2

    minger

    User Avatar
    Science Advisor

    One nice thing about circular tubing is that the stress is equal independent of the load direction.

    Let's walk through it though. The strength of a beam is a function of its area moment of inertia
    http://en.wikipedia.org/wiki/Second_moment_of_area

    The moment of inertia for a (let's assume solid for the moment) circle is:
    [tex]
    I_0 = \frac{\pi}{4}r^4
    [/tex]
    And the moment of inertia for a rectangle about its centerline (not diagonal) is:
    [tex]
    I_x = \frac{bh^3}{12}
    [/tex]
    Let's now assume that for comparison, the two sections have the same area, and thus the same weight. For ease of comparison, let's also assume that the rectangular tubing is square. Then to find the length of one side, a, of the sqaure:
    [tex]
    \begin{equation}
    \begin{split}
    A = &\pi r^2 \\
    &\pi r^2 = a^2 \\
    a = \sqrt{\pi r^2}
    \end{split}
    \end{equation}
    [/tex]
    b and h both equal, so the moment of inertia of the square section is:
    [tex]
    \begin{equation}
    \begin{split}
    I_x &= \frac{\sqrt{\pi r^2}(\sqrt{\pi r^2})^3}{12} \\
    I_x &= \frac{ (\sqrt{\pi r^2})^4}{12} \\
    I_x &= \frac{ (\pi r^2)^2}{12} \\
    I_x &= \frac{ \pi^2 r^4}{12} \\
    I_x &= \frac{\pi^2}{12}r^4
    \end{split}
    \end{equation}
    [/tex]
    One can now compare the two coefficients, and see which is larger:
    [tex]
    \frac{\pi^2}{12} = 0.822 > \frac{\pi}{4} = 0.785
    [/tex]

    So, based on this, a square tubing is 4.6% stronger for the same weight than circular. However, as the loading starts to go off axis, the square tubing will get much weaker. I'll let you determine the moment of inertia at a full 45° off axis.

    p.s. Might want to check my work before taking it as gospel though
     
  4. Dec 4, 2009 #3
    Thanks, I'll work on it and get back with you on my findings.
     
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