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I've recently designed what is essentially an aluminium box. I've done the entire design process of CAD software SolidWorks and then done some simulations of loading of the box.

Through iterations I've figured out the minimum thickness of the aluminium (AL6061 T6) box material to be quite small - 0.25 mm. This seemed almost too small to me so I looked up kitchen 'tin' foil for comparison. The Wikipedia page puts tin foil thickness at between 0.2 mm and 0.006 mm. That means that the upper end of the tin foil is easily comparable with the thickness of my box.

I'm aware that tin foil is probably not AL6061 and it's probably not tempered at all but this has thrown me into quite a bit of confusion as to whether or not I can trust my simulation results (I've not given any specific details of my simulation since I don't want to turn this into a homework style question - but I'm happy to share if it helps).

So my question is; can I trust the results at AL6061 T6 is significantly stronger than aluminium foil (it would help if I can find what aluminium the foil is made of)?

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  • $\begingroup$ Showing all your calculations would definitely help. $\endgroup$ – Paul Nov 14 '15 at 23:07
  • $\begingroup$ @Paul show all my CAD calculations? I'm not sure what your asking for here. The FEA matrix is produced by the CAD software and solved by the CAD software. I'm not asking for specific answer to numerical problems I'm asking if/where my intuition is flawed $\endgroup$ – ThePlanMan Nov 14 '15 at 23:09
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    $\begingroup$ It's just that, as-is, we don't have enough information to judge the plausibility of that thickness. Is it a 1x1x1 cm box? Is it a 1x1x1 m box? What sort of load is it supposed to support? Just its own self-weight? An elephant? $\endgroup$ – Wasabi Nov 14 '15 at 23:23
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    $\begingroup$ Always be wary of 'black box' solutions - answers produced by a 'magic box numbers'. I've been using software packages for 3 decades and I've seen people suggest ridiculous solutions based on blind acceptance of an answer from a computer package. There's an old adage "if it looks right on paper, but doesn't feel right in you guts it's probably wrong". Computers & their software are tools to assist you in getting a solution. $\endgroup$ – Fred Nov 15 '15 at 0:05
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    $\begingroup$ With everything, "garbage in - garbage out". If you use software to simulate / model something, you need to make sure that all information you enter applies. I am sure that solid works employs finite element methodology. This would mean that you need to make sure that the programe has / produces enough shells to allow for deflections. Also your boundary conditions need to be correct to get a plausible answer. I have never used Solid Works however. $\endgroup$ – NamSandStorm Nov 18 '15 at 11:22
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Most soda cans in the US are on the order of 0.1 mm thick according to the sources I've seen. So 0.25 mm seems plausible to me, assuming that you made no errors in your calculations.

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The Wikipedia page puts tin foil thickness at between 0.2 mm and 0.006 mm. That means that the upper end of the tin foil is easily comparable with the thickness of my box.

Ooook. Let's look at this, from an engineering perspective.

First, scales. The smallest measurement from Wikipedia vs the largest is a 33x factor difference. This is perhaps hard to conceptualize in mm, but imagine a 1m stick. Now imagine 33 of them. This is the difference between those measurements.

This means you need to be very careful looking at those ranges and saying "oh that means my measurement is about tin foil." Understanding orders of magnitude is a very important engineering principle.

So my question is; can I trust the results at AL6061 T6 is significantly stronger than aluminium foil (it would help if I can find what aluminium the foil is made of)?

It is good to question unintuitive software results. But be careful to not combine different things.

You may (as in this case) attempt to compare things which are not good to compare. Experience helps with this.

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  • $\begingroup$ Additionally, if concerned about buckling stiffness, it goes with the cube of the thickness, so that 33 turns into 35,937 times stiffer. Even the "small" difference between .2 and .25 turns into twice the stiffness. $\endgroup$ – Rick Nov 17 '15 at 14:13

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