Different stresses

The total load on the 3"x3" is 90 lbs.

The total load on the 4"x4" is 160 lbs.

The 4"x4" plate has to span a longer dimension.

Are the plates simply supported on 2 sides, or 4 sides, or fixed on how many sides, .....?

Hi

I think it simpaly supported 4 sides.How would the result change if t is supported on 2 sides.

Since 4"x4" plate has to span a longer dimension is it critical for stress with higher load?

thanks

From Roark's Formulas for Stress and Strain

Simply supported all edges: width = length

σ_max = 0.2874(q)(l)²/t² (at center of plate)

Fixed all all edges: width = length

σ = 0.1386(q)(l)²/t² (at center of plate)

σ_max = -0.3078(q)(l)²/t² (At center of long edge)

Where:

q = uniform load (psi)

l = length of side (inches)

t = thickness of plate (inches)

Oh yeah -

Yes, the 4"x4" plate will be the critical case due to higher stress as can be seen from previous equations where the length is squared on top.

A little disclaimer for those equations as well - they only work for a uniform thickness plate.

You may be into the computer field, but you do not have the foggiest understanding of the problem you are attempting to solve. Why would you expect the reaction to be the same when the dimensions of the plate are increased? And why would you expect the stress to be the same when the dimensions are increased?

Before you can solve such problems on a computer, it is essential that you understand the physical aspects of the problem and that you can solve the problem, even if only approximately, by hand methods. You must realize that computers are extremely dumb. They can generate lots of output, but if the input is wrong, so is the output...or alternatively, garbage in...garbage out.

Hi,

I was thinking load is same per unit area thats why iwas thinking so.But now I understand thta

The total load on the 3"x3" is 90 lbs.

The total load on the 4"x4" is 160 lbs.

thanks