Plate thickness calculation

How and at what locations is the plate supported?

The easiest way is to model it with software like Solidworks or equivalent, but you may need to find someone who has a copy and is versed in its use.

You also need to know how the plate is supported and how the load is distributed. Are you putting 50 lbs of sand on the plate or is it a rigid 50 lb box?

You also need to specify the maximum deflection in the plate that you are willing to accept under load.

Also, one would need to know if a certain deflection is the limiting factor, or simply the yield strength of the material (with an appropriate safety factor, of course).

you didn't say where the load is applied, evenly accross the whole plate or all in one point in the center?

"50 lbs distributed load"

Concentrated load at the center of the plate

Jeez 50lb distributed load is next to nothing....

Cooking foil would probably support it if carefully clamped around the edges.

1 mm would be fine...depends how much flex you can permit...

As usual...insufficient information...does not compute...

Del (whose going to to try the cooking foil eh? eh? Go on you know you want to )

Cooking foil would probably support it if carefully clamped around the edges

How many foiled folds...

Guest,

When I was working in the ship yard, had to design hatches for a special operations craft with minimal weight, due to the weight requirements that it had to be able to airlift on a Galaxy C5 A/B transport.

Requirements was the the hatch had to seal at minimum of 2 psi.

one hatch size was 24"x24", the largest hatch size was 24" x 48"

I believe I used 6061 Alum. (been over 10 years) Did a pulmonary calculation on the deflection of the plate on the larges of hatches.

Then design a proto type and hydro-tested it. Ended up using the prototype also.

But on the larger hatch I believe we used 3/16" Alum. Did not do a burst test . but we tested it until it leaked on the seals at 4+ psi (+ 4600 lbs). And there was deflection.

Do not know what type of reenforcement or geometry is on the plate.

The distributed load would be approx 0.087 psi.

don't sound like much, but I do not know what your tolerances are?

Thats my 2 cents, and I expect change back.

phoenix911

If you want to use the Classical Plate Equation, check out this site which actually gives elastic solutions to rectangular and circular plates with various edge conditions.

If you simply want to find the ultimate capacity of a plate of given thickness and you know the yield moment per unit of width, use a Yield Line Analysis. Essentially, what you are doing is equating internal and external work. Internal work is the sum of the yield moment times rotation times width. External energy is the sum of applied load times displacement.

In the case of a square plate, yield lines form on the diagonals (neglecting corner levers). Assume the center of the plate is displaced 1 unit when the yield lines are fully developed.

Internal Work = 4*m*a*β = 4*m*a*2/a = 8m

External Work = wf*a²*1/3

where m is the yield moment of the plate per inch, a is the length or width of the plate and β is the angle of rotation of each triangle between the support and the yield line (in this case 2/a).

And wf = factored distributed load per unit of area

Equating Internal and External Work:

m = wf*a²/24

This does not account for corner levers and a factor of 1/0.9 or 1.111 has been found to account for it adequately in most situations. Thus the required yield moment per inch of plate is about 1.11*wf*a²/24.

You would determine the thickness of plate by the relationship:

m = Φ*Z*Fy where

Φ is a factor to account for variability of material (probably about 0.9 for aluminum)

Z is the plastic modulus per unit width

Fy is the yield point of the material