Roark ?

another day in paradise, or is paradise one day closer ?

Specifically, Table 24a, Case 1 Simply Supported (Neither Fixed nor Held); Uniform Pressure q Over Entire Plate in the 6th edition

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Roark will get you a more realistic estimate of the stress and deflection.

Round plates have a closed form solution (re Timoshenko), it's the rectangular ones that are difficult to analyze.

another day in paradise, or is paradise one day closer ?

Use the Rayleigh-Ritz or Galerkin method. Simple and effective for single thin plates in bending with loading that causes a deflection profile (and thus deflection trial function) that you can estimate reasonably accurately - uniform load is one of those cases. Alternatively, if you have faith in tables for slightly obscure problems, use "Roark´s formulas for stress and strain".

rb1957:
Round plates are more difficult to analyze in my opinion, since they often require use of polar coordinates to simplify calculations, and the exact solution to the round plate for even uniform load is not straightforward to derive as far as I can remember. For rectangular plates, Navier´s solution is easy enough to implement, as are the Galerkin and Rayleigh-Ritz methods in conjunction with a suitable trial function of trigonometric or polynomial type. I do not see any challenges specific to rectangular plates.

I've looked at Timoshenko, and find the formulae to depend on D (flexural rigidity), where D = EI. Seems that for a circular plate I = bd^3/12 varies greatly depending on where you measure "b". I suppose worst deflection will be at the center so b is diameter?
I don't have Roark's book, but found on engineersedge.com "circular plate uniform load edges simply supported equation". I was searching for Roark and found this.
stress at center = 1.238pr^2/t^2 where poisson's ratio is 0.3
deflection at center = 0.696pr^4/Et^3
p being load in psi
r is radius
t is plate thickness
E modulus of elasticity

My first guess was that 3/8" thick plate would be adequate. The results from Roark's formulae are stress = 15.3ksi = 0.42Fy for 36 ksi steel.
Deflection = .46" at the center = L/107 for a 49" diameter. (Deflection greater than I'd like to see).

I know for sure that this plate will never see 400 psf, but the drawings say the existing 4" thick concrete floor on composite floor deck is designed for 400 psf live load.
Increasing thickness to 1/2" brings deflection down to 0.19" = L/255.
The downside is that 1/2" steel plate 53" diameter for 2" bearing all around, weighs 313 pounds. Upside is that it won't move much once it's in place.

Thanks for the help! Please feel free to comment. Do you agree?

"Thanks for the help! Please feel free to comment. Do you agree? "
Not quite. The flexural rigidity of an isotropic linear elastic plate (such as a steel plate) is D = E*t^3 / 12*(1-v^2) , where t = thickness and v = poisson´s ratio = 0.3 (for steel) and E = 200,...,210 GPa are the parameters.

The formula you found on "engineersedge.com" seems to include the relevant parameters, but you should double-check the source of that formula and ensure that it is from a reputable source, or apply the Ritz or Galerkin method and solve it yourself. If you have access to finite element software, modelling the plate, load, boundary conditions and solving the problem is trivial and would take maybe 30 minutes at most. I cannot comment on the specific calculations you´ve done.

PS. For thin plates (Kirchoff plate) in bending, the deflection should not exceed 0.2,...,0.5 times the plate thickness, because somewhere after that (depending on boundary conditions and loading), the bending and membrane modes couple and the simple plate model no longer applies.

Sorry, centondollar. I was writing a response while your post was being sent. I'll look at Rayleigh-Ritz and Galerkin too.

For a point load applied in the middle of the plate, a trial function would be "w(x,y) = B*(y^2+x^2-R^2) in cartesian coordinates (centered on the origo of the plate), where B is the parameter to be solved (Ritz, Galerkin) and R is the radius of the plate.

Since that provides constant moment (Moment = function of second derivatives of deflection) in the plate, it is not suitable for this problem. Perhaps something like

w(x,y) = B*(y^4+x^4-R^4) + C*(y^3+x^3-R^3) + D*(y^2+x^2-R^2)

would be more appropriate. B, C and D are the constants to be solved, then input into the approximation again and used to calculate moments and shears (functions of displacement partial derivatives), normal and shear stresses, and finally the von Mises stress. This approximation is kinematically admissible (at edges, R=y or R=x and thus deflection is zero), and provides quadratic moment and linear shear.

Please note that these are my spontaneous thoughts and the result of some googling "circular plate trial function". Some book might have a better approximation than what I have presented, but the idea is this:

1. The trial function should satisfy essential boundary conditions.
2. The trial function is a sum of several linearly independent deflection interpolants multiplied by unknown parameters (w.i) to be solved: w(x,y)= SUM(i=1, n) w.i*polynomial (or parameter*trigonometric function)
3. The solution should be complete in the energy. This criterion is mathematical (exact solution minus approximation, measured in energy, is smaller than a constant), and not important for practical problems.

D is I per unit width. the loads Tim O'shenko derives are also "per unit width", so it comes out in the wash.

another day in paradise, or is paradise one day closer ?

There is no "unit width" in a circular plate, since a sector (giving a cone) of a circular plate is not equal to unit width in a rectangular plate. I find it hard to believe that Timoshenko derived a unit-width solution for a CIRCULAR simply supported plate. The rectangular plate solution is expressed as a "long beam" in his book, but that is a completely different problem setting.

In any case, this plate is finite in size and has two-way plate action (Mxx, Myy, Mxy), so unit-width formulations are not useful.