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:


Here's some pictures of my current coping and stones that would be in one lid tray: https://imgur.com/a/4LIbrST

Here's an idea of possible solution: https://imgur.com/a/38edhS4

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:


  • $\begingroup$ 1/4" is too thin IMO mainly because it lacks sides. The folded sides make a difference. If you bolted square tubes beneath it to increase rigidity then it would be fine. $\endgroup$
    – DKNguyen
    Oct 21, 2021 at 1:20
  • $\begingroup$ Both edges? A plate has 4. $\endgroup$
    – Solar Mike
    Oct 21, 2021 at 5:00
  • $\begingroup$ Updated the description for both edges, I meant both sides have the bracket along it's length $\endgroup$
    – Steve
    Oct 21, 2021 at 14:31
  • $\begingroup$ Don't understand the 3-4" wide between brackets. Is it 3" - 4", or 3'-4" (40")? But the long side of your plate is only 24". Let's be specific, is the plate supported on two sides, with a span length of 24" (between brackets)? $\endgroup$
    – r13
    Oct 21, 2021 at 15:07
  • $\begingroup$ The brackets are 3-4” wide, so 2” will be holding the sides and 24” long. It’s actually 20” long and 4” length at the back will be sitting on concrete $\endgroup$
    – Steve
    Oct 21, 2021 at 15:46

2 Answers 2


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$


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

  • $\begingroup$ 2700 or 27000? Which? $\endgroup$
    – Solar Mike
    Oct 21, 2021 at 6:55
  • $\begingroup$ Updated description, I'm looking at 5052H32 0.25" aluminum, does that affect your calculations? What would a safe margin be? $\endgroup$
    – Steve
    Oct 21, 2021 at 14:36
  • $\begingroup$ @Steve, 5052H32 aluminum is a bit stronger. Fy=28000psi. so it does not affect our calculations. It is an improvement. $\endgroup$
    – kamran
    Oct 21, 2021 at 14:48
  • $\begingroup$ See updated description. Will 1.5” folds on each side help much? $\endgroup$
    – Steve
    Oct 21, 2021 at 16:09
  • $\begingroup$ @Steve, yes they do. effectively they add 3 inches to the plate width, approximately 0.15 times more strength, and the fold ads stiffness too. but you want to bend the fold gently not stress it to cause cracks. $\endgroup$
    – kamran
    Oct 21, 2021 at 16:19

enter image description here

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.

enter image description here

  • $\begingroup$ Thank you for the insight. Ok I looked up these angles, makes sense. Where would make the most sense to put it? The angles welded to the plate to look like that |____________| ? $\endgroup$
    – Steve
    Oct 23, 2021 at 0:38
  • $\begingroup$ Yes, attached at the free edge, preferably by welding, but can be bolted too. $\endgroup$
    – r13
    Oct 23, 2021 at 0:54
  • $\begingroup$ Figures added. You can do it either way. The one to the left is easier to assemble, the one to the right is much more stronger but needs to be continuously welded. $\endgroup$
    – r13
    Oct 23, 2021 at 1:01
  • $\begingroup$ I had another idea, updated description. The coping paver is a different height than the regular paver. What if we filled in 13" long 0.55" height space, perhaps something that would add stiffness/structure to it but minimizing cost. Here's an example: imgur.com/RVD97OW could we add something like a couple small channels in there or a couple plates or a big plate to add depth to the paver and also add structure without requiring the angles? Something cost effective, if I do another 13" x 0.55" full plate it's going to be too pricey $\endgroup$
    – Steve
    Oct 23, 2021 at 1:25
  • $\begingroup$ I don't think I understand your new idea. It seems to me you wanted to eliminate the angles but uses 0.5" thick plates at the edges instead. Well, it will work, if you weld a 6"x0.5" plate on each free edge below the plate. Then, it seems odd as the overall plate width is only 20", that will result in 20-2*6 = 8" remains in between the added plates) $\endgroup$
    – r13
    Oct 23, 2021 at 1:42

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