I need to construct a table that will hold a water tank. The filled tank will weigh up to 1300 pounds. I have the table leg frames worked out to hold 2500 pounds so no worries there. (Here is a link to details.)

enter image description here

They are just a 28" x 34" rectangle. The tank itself has a footprint of 30" x 38" and is made of molded polyethylene, so don't count on its rigidity for support. The table top will be 36" x 48". (It is bigger than the tank solely because I can buy it off the shelf at that size without any fabrication cost.) There will be no motion to account for nor any rapid changes in the weight. So my question is how thick does the stainless steel table top need to be to safely bear 1300 pounds?

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    What have you considered as the support for the top? This Q & A may give you some thoughts: engineering.stackexchange.com/q/21054/10902 – Solar Mike Nov 11 at 5:21
  • I described the support for the top in explicit detail in my explanation above. Is it missing any required details? I do not need to improve the quality of the support. My question is not about the support however, just the top. – Kenneth Vogt Nov 12 at 6:18
  • So, what joins the legs together? is it the same material as the legs? Are there any pieces supporting the table surface between the legs? Is there any diagonal bracing of the table surface? Is there any diagonal bracing of the legs? Your "explicit detail" is lacking, do a diagram... – Solar Mike Nov 12 at 6:21
  • Here is the base support with pictures and details: benchdepot.com/cleveland-series-workbenches/1553/159/… – Kenneth Vogt Nov 12 at 6:24
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    Also I strongly disagree with the two close votes: Sizing a sheet of metal for a given load is IMO a good question for this site (which I sadly can't answer). The question needs a bit clarification and close reading but is otherwise a great fit. – mart Nov 12 at 9:34
up vote 2 down vote accepted

The Stress formula for a rectangular plate is:source,

Rectangular plate, uniform load, simply supported (Empirical) equations. Since comers tend to rise off the supports, vertical movement must be prevented without restricting rotation. Symbols used:

  • $a$ = major length of rectangular plate, (in)
  • $b$ = minor length of rectangular plate, (in)
  • $p$ = uniform pressure loading, ( Ibs/in')
  • $\nu$ = Poisson's ratio
  • $E$ = Young's modulus, ( lbs/in2)= 29,000,000 psi for a 36 steel, with 3 as factor of safety we get 18000 psi for allowable stress.
  • $t$ = plate thickness, (in)
  • $S_m$ = maximum stress, (lbs/in2)
  • $y_m$ = maximum deflection, (m, in)

$$\begin{align} \sigma_{max} &= \frac {0.75 pb^2}{t^2[1.61(b/a)^3 + 1 ] } \\ y_m &= \frac {0.142pb^4}{Et^3[2.21(b/a )^3+1]} \end{align}$$

This is for edge simply supported, but you have cantilever edges. to make the task simpler we just reduce the p by a factor of 0.75, so 0.75(1300)/ (36)(48) = 0.564 psi

$$\begin{align} t^2 &= \frac{(0.564(28^2)0.75)}{ 18000[1.61(28/34)^3 +1]} \\ &= .0987 \\ \therefore t &= 0.1\text{ in} \end{align}$$

Say pick 1/8 thick sheet for practicality.

Check my arithmetic please.

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    I would be cautios with this answer: The OP states 5.11 kN/m² load, for similar loadings but smaller spans (750mm but only two supported sides) I've seen 5 or 6 mm SS sheets. safety factor may be greater in "my" cases because the load is a person. – mart Nov 12 at 12:17
  • You are going to need support struts due to bending moments at centre where force will be greatest under highest load. 2 cross struts would reduce the need for thicker steel sheet and variation in water level will produce strain hardening of welded joints into plastic range of steel. You do not want plastic deformation: it's permanent and possibly catastrophic. – Rhodie Nov 20 at 19:55

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