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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.stack.imgur.com/VPUe0.jpg

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

Here's an idea of possible solution: https://i.stack.imgur.com/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:

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Steve
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  • 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. – DKNguyen Oct 21 '21 at 01:20
  • Both edges? A plate has 4. – Solar Mike Oct 21 '21 at 05:00
  • Updated the description for both edges, I meant both sides have the bracket along it's length – Steve Oct 21 '21 at 14:31
  • 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)? – r13 Oct 21 '21 at 15:07
  • 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 – Steve Oct 21 '21 at 15:46
  • If I add 1.5” folds to each side, will that add a lot of strength? – Steve Oct 21 '21 at 15:47
  • Do you mean 1.5"x1.5" angle? Yes, it will work. The minimum will be 1"x1" bar on each side of the opening. Also, suggest using the checker plate instead of a smooth plate. – r13 Oct 21 '21 at 16:46
  • I mean 1.5" at a 90 degree angle on the two sides that hold brackets, so like this: 1.5" |_______19.625"______| 1.5" – Steve Oct 21 '21 at 17:06
  • The minimum depth of the leg should be 1.75", even so, I still suggest welding 1.5"x1.5 angle (L) to the bottom of the plate, so a person or two won't feel too bumpy. – r13 Oct 22 '21 at 00:49
  • Stiffness. You're after stiffness rather than strength. – DKNguyen Oct 23 '21 at 01:39

2 Answers2

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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.

r13
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kamran
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  • 2700 or 27000? Which? – Solar Mike Oct 21 '21 at 06:55
  • Updated description, I'm looking at 5052H32 0.25" aluminum, does that affect your calculations? What would a safe margin be? – Steve Oct 21 '21 at 14:36
  • @Steve, 5052H32 aluminum is a bit stronger. Fy=28000psi. so it does not affect our calculations. It is an improvement. – kamran Oct 21 '21 at 14:48
  • See updated description. Will 1.5” folds on each side help much? – Steve Oct 21 '21 at 16:09
  • @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. – kamran Oct 21 '21 at 16:19
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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

r13
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  • 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 |____________| ? – Steve Oct 23 '21 at 00:38
  • Yes, attached at the free edge, preferably by welding, but can be bolted too. – r13 Oct 23 '21 at 00:54
  • 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. – r13 Oct 23 '21 at 01:01
  • 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:

    https://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

     – Steve Oct 23 '21 at 01:25
  • 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) – r13 Oct 23 '21 at 01:42
  • Edited to add a picture of the current coping bricks that need to go into a 'tray'.. https://imgur.com/a/4LIbrST The back bricks are a different height, so need a 0.55" boost of some kind to match the front coping bricks. I was thinking maybe we could add something to fill in 0.55" on the back half that will make the bricks level and also give some stiffness. Any ideas? I guess the biggest one would be to do a full 0.55" aluminum piece 13" x 19.5" in that space. I guess welded together? Would this give me the best option? Or could I use 0.55" U channels or something else? – Steve Oct 23 '21 at 19:08
  • Edited to add an example of two plates together https://imgur.com/a/38edhS4 – Steve Oct 23 '21 at 19:27
  • The free edge without the additional plate beneath is prone to bend and deflect, I wouldn't do it. – r13 Oct 23 '21 at 19:52