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Our question is what size and alloy aluminum tube can be used to replace steel tube in a vertical load application. Our current spec is 1.5” OD STD steel tubing 7 feet long. We will use 6 poles attached to steel footings to support vertical stacking of containers that each weigh 1500 lbs. the bottom container will see 4500 lbs. from vertical stacking. the containers are 45” wide and 87” long. There is one pole in each corner and 2 in the center on each side. Six tubes of steel 1.5” x 7’ steel is to heavy. Can we use aluminum tube round hex or square and what alloy ?

Thank you

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    $\begingroup$ And the steel tube wall thickness? Then you can do a comparison. $\endgroup$
    – Solar Mike
    Commented Aug 17, 2019 at 12:48
  • $\begingroup$ The compressive load capacity ability of your structure has very little to do with material strength ; It depends on the geometry. I could not readily find a compressive strength for aluminum but pine is roughly 13,000 psi so design value of 5,000 psi should work for pine. I think it is safe to say aluminum is as strong as pine ; so, one square inch of aluminum will easily support 5,000 lb vertical load. The selection of construction material depends on cost, availability, ease of fabrication and less on strength.. $\endgroup$
    – blacksmith37
    Commented Aug 17, 2019 at 20:16
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    $\begingroup$ The buckling load is more important for these situations. Even if the tubes do not go plastic, they can buckle at relatively low loads. See en.wikipedia.org/wiki/Buckling for more. $\endgroup$
    – Biswajit Banerjee
    Commented Aug 17, 2019 at 21:16

1 Answer 1

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Aluminum columns usually fail under buckling when their slenderness ratio is greater than 99. using the formula K= L/r, where L is the free factored length and r is the radius of gyration. $r=(I/A)^{0.5}$ In your case K may be smaller or bigger than one, needs more detail on the type of connections. We assume K=1.

$r= 1/4\sqrt(D^2+d^2)= 1/4\sqrt(1.562+1/64)=0.505$ Assuming 1/8 in tube thickness.

In your case K=84in/0.505in= 165 , 165<99.

$P_{critical}=\pi^2\frac{EI}{(Kl)^2}$

I let you take over from here, but just eyeballing it, it looks ok.

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