Material thickness required

The thickness of what?

The thickness of the chosen material. I would like to see your calculation approach and kind of curious what equation you use to for the calculation.

Have fun!

I don't want to sound like a ninny, but the dimensions of the platform pale in comparison to the shape allowed. There are about a thousand galvnaized alloys that you could use as well.

Is this a point load, or evenly distributed? There are so many unanswered parts of the equation...

Look up "beam boy" calculator. You'll have fun with that.

Making the basic assumptions:

Assuming 6'W x 12'L span simply supported, P is point load at center Mmax = (PL)/4 = 7500ftlbs = 90000inlbs

Bending Stress δ = (My)/I for plate = 12My/bd³

d³/y = (12M)/δb y= 0.5d for δmax therefore d = √((6M)/δb)

Substituting value of 36000psi gives thickness of 0.4564" check this by entering values back into max bending equation.

This neglects the mass of the plate and any S.F.

Solving for the added plate mass again making assumption the mass acts as point load in centre span; steel density = 0.2848lb/in³:

Mass (Weight lbs) = density x volume = 2952.8xThickness d = 2952.8d

Therefore max bending moment with = (P + M)L)/4 = ((2500 + 2952.8d)144)/4

= 90000 + 106300.8d

Apply this for M in d² = (6M)/δb

Giving d² = ((90000*6)/(36000*72)) + ((106300.8d*6)/(36000*72))

d² = 0.208 + 0.246d

0 = d² - 0.246d -0.208

Applying quadratic equation d = -0.35 or 0.6

Take the 0.6" thickness to be the minimum

Check by entering into δ = (My)/I = giving around 35597psi (rounding errors)

Again neglecting any S.F's. (you can apply those) and forgiving me for any lack of bracket discipline or explanations.

Thanks, Andy

How is the platform supported? on 2 op. sodes? on all 4 sides?

The computations you got are not for a platform but for a "beam". The differences are very important! As a plate you could obtain a lower value as considerd as a beam.

The other question is how load is applied. The calculus is for a concentrated central force but in fact considered as uniformly distributed over the width whis can be a false assumption. If you want a good answer you should give the right input.

The definition of your problem should be:

a plate axb supported on z sides is loaded with c lbs in the position - -

material Sy =

Which is the thickness?

Although seaming simple the problem of plates is not trivial at all.

You also need to know if the plate is just resting on it's supports or is clamped and the maximum centre deflection allowed.

All assumptions were stated in my calculations, Simply supported, point loads, no limits to deflection, no safety factors were applied as none were asked for. All of which are a case of extending the equations or applying new ones for deflection which remain separate to the stress models. But any stress problem should always start off with a basic model. Changing conditions just changes the variables the theory is solid.

Thanks, Andy

max.y= -{0.203Wb²(m²)}/m²Et³(1+0.462α⁴)

max . s_{b(at centre)} =-3W/2∏mt²[(m+1)log b/2r₀ +m(1+k)], where k=( 0.914/1+1.6α⁵)-0.6

If edges are all supported

And Load is uniformly distributed over small concentric circular area of radius r₀

_{This is supposed to be Approx formula by S. Timoshenko}

_{All above from "Formulas for Stress and Strain--R.J.Roark}

_{Library of Congress Card 53-5173.}

_{If you do not care to calculate- the harder way will be to place a 6mm plate and go on loading till you reached your net goal.}

_{Do not ride yourself YET!}