Which deflection formula is correct for simply supported hollow aluminum tube with a uniformly distributed load (I've read conflicting answers online):

  1. δ = (5 x w x L^3) / (384 x E x I)
  2. δ = (5 x w x L^4) / (384 x E x I)
  3. δ = (w x L^4) / (384 x E x I)
  4. δ = (w x L^4) / (8 x E x I)
  5. Other?

In particular, I'm looking to calculate the mid-span deflection of a simply supported aluminum 6061 tube 3.25" diameter, 0.125" thick, 200" long, with both the weight of the hollow tube itself plus 30 LBs of total uniformly distribute load along the whole length of the hollow tube. Would be great if someone could show me the calculations.

Also, note, when calculating moment of inertia, I'm using I = p/32 x (Do^4 - Di^4).

  • 1
    $\begingroup$ Where did you get each of these? And, more importantly, what assumptions were made for each of them? That would probably provide the answer to your question... $\endgroup$
    – Solar Mike
    Commented Sep 29, 2023 at 6:07

1 Answer 1


$\delta \sim wl^3$ is dimensionally incorrect.

$5wl^4/348EI$ is correct.

$wl^4/384EI$ would be for fixed-fixed end condition.


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