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I am interested in knowing how much weight a piece of cardboard can hold in proportion to it's size.

Suppose we have independent variables $x$ and $y$ corresponding to length and width and that these two vectors are in $\mathbb{R}^2$. Then I think (could be wrong) the formula would be

$$f(x,y) = \lambda_1 x + \lambda_2 y $$ where $\lambda_1 = \lambda_2$ if the piece of cardboard has square dimension.

However I am kind of stuck from here because I don't know how to get the $\lambda$ scalars to complete the formula. I tried looking online and there doesn't appear to be a general formula (surprised because cardboard is so common in our lives). Does anyone have any insight into figuring out this problem?

edit: the weight is assumed to be evenly distributed

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    $\begingroup$ Your best would be to run trials with varying $x$ and $y$ then interpolate to get a multivariable function $f$ $\endgroup$
    – user28616
    Mar 19 at 18:27
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    $\begingroup$ You still need to account for how the load is distributed in the cardboard. A 12" x 12" square could hold a 5 lbs plate no problem, but put a tack on the bottom of the plate it will cut through your cardboard. Are you evenly loading the entire surface or part of it? Then it becomes a question of shear stress. $\endgroup$
    – jko
    Mar 19 at 18:32
  • $\begingroup$ oh I forgot to mention it, the weight is assumed to be evenly distributed. I added that to my post, thanks! $\endgroup$
    – Evan Kim
    Mar 19 at 18:35
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    $\begingroup$ And how is the box supported? All around, or just on two ends (like a person carrying with two hands)? $\endgroup$
    – Wasabi
    Mar 19 at 18:52
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    $\begingroup$ Two cardboard layers with a wavy filling? Spacing? Filling angles? $\endgroup$
    – Solar Mike
    Mar 19 at 18:59
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Your question lacks many essential specifics required for a meaningful comparison. However, in the general sense, while the larger box have more volume/room, thus may hold more weight, but a smaller box will be more effective, and stronger, compared to the former. The reason is the smaller box posses the shorter span length, for everything else holding constant, it will deflect less, and has less stress.

You can do a simple experiment at home, by putting same weight in the same manner on boxes with different size, but having the same wall thickness. You will find for the boxes with same base area, the square one will be out perform the box with rectangle foot print.

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Put a modest floor layer on the bottom (another piece or two of cardboard), and the failure location (with careful handling, no bouncing resulting in concentrated load) becomes the bends at the edges, so I would guess the max load should be proportional to the perimeter.

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