What thickness aluminum plate will support 350lbs?

Question

I need to purchase an aluminum plate that is 19.625" x 24", that can support 350lbs.

This will hold pavers + possibly a person, next to a pool on top of an automatic cover. The automatic cover company sells lid/trays that are a 0.25" thick aluminum plate, but I can purchase one for 1/3 the cost. The lid they sell has sides, but I think a plain rectangle plate will do better and let us fit pavers closer. Do I need 0.25" thick or what size will do? Will 3003 0.25" suffice? Also I'm looking at mainly at 5052H32 0.25" aluminum now

I am also considering adding 1.5” folds on the two sides to add strength (similar to pictures), will that help?

Edit: Another idea is I need something to fill in 13" long 0.55" height space, perhaps something that would add stiffness/structure to it but minimizing cost. Here's an example:

https://i.sstatic.net/VPUe0.jpg

Here's some pictures of my current coping and stones that would be in one lid tray: https://i.sstatic.net/GA90A.jpg

Here's an idea of possible solution: https://i.sstatic.net/q9heO.jpg

I should add that it will be supported by a bracket at both sides for the width of the plate (the 19.625" edge -- 19.625" between brackets that are 3-4" wide).

Here's a picture of the lid from the automatic cover company to show brackets with the set spanning:

Answer 1

Let's assume the 350lbs force is imparted at the center of your plate to maximize its effect. We use yield strength of Fy=27000psi and allowable stress at 0.6*27000. We call the required thickness of the plate, h.

$M=\dfrac{350lbs*24"}{4}=2100 lbs.in\quad s= \frac{m}{\sigma}=\frac{2100}{27000*0.60}=0.13 in^2$

$s=bh^2/6=24*h^2/6=4h^2=0.13 in^2$

$ h^2=\dfrac{0.13 in^2}{4}=0.032 in^2$

$h=0.18"$

So it seems your plate is at near 0.7 capacity, which is marginally safe.

Answer 2

Estimate required stiffness parameters $I$ & $I$ to satisfy the design criteria:

$M = \dfrac{340*24}{4} = 2040$ lbs-in

$S_{req} = M/\sigma_a = 2040/16800 = 0.121$ in^3 < $S = 0.208$ in^3, ok.

$I_{req} = \dfrac{PL^3}{48E\Delta_a} = \dfrac{340*24^3}{48*10x10^6*0.067} = 0.146$ in^4 > $I = 0.026$ in^4, NG.

The plate needs to be stiffened to satisfy the deflection criteria. Since the plate is quite flexible, let's conservatively ignore it and assume only the bent plates at the edges are effective in restricting the deflection.

$b_{bp} = 0.25"$, $h$ = ?

$2I_{bp} = 2*\dfrac{b_{bp}h^3}{12}$ > $0.146$ in^4

Solving the equation, $h = 1.52"$.

Theoretically, adding two 1.5" deep bent plates on the edges should satisfy both criteria; however, for practical concerns over the flexibility of the thin plate, and loss of rigidity due to plate bending (to form the bent leg), also the fact it is a poolside application, therefore, it is recommended to add two 1.5"x1.5"x1/4" angles instead.