Tuesday, October 29, 2024

Mechanical Engineering Question

I had hoped to find a calculator online for this...  The wooden sheet on which my big reflector sits is beginning to deform at one of the corners where a caster goes.  I want to replace this with something more durable before I have a catastrophic failure.  So what thickness of steel do I need to support 150 pounds without too muh deformation?  that chatgpt gave me this formula during a lucid, hopefully nonhallucinatory phase:

To calculate the maximum deflection of a rectangular plate supported at the four corners, you can use the classical plate theory (Kirchhoff-Love theory). For a rectangular plate under uniform load, the maximum deflection wmaxw_{\text{max}} can be estimated using the following formula e:

wmax=qa4Dw_{\text{max}} = \frac{q \cdot a^4}{D}

where:

  • qq is the uniform load per unit area (N/m²),
  • aa is the shorter side length of the rectangular plate (m),
  • DD is the flexural rigidity of the plate, calculated as:

D=Eh312(1ν2)D = \frac{E \cdot h^3}{12(1 - \nu^2)}

In this formula:

  • EE is the modulus of elasticity of the material (Pa),
  • hh is the thickness of the plate (m),
  • ν\nu is the Poisson's ratio of the material.

Steps to Calculate Maximum Deflection:

  1. Determine Material Properties: Find the values for EE and ν\nu for the plate material.
  2. Calculate Flexural Rigidity DD: Use the thickness hh of the plate.
  3. Define Loading Conditions: Determine the uniform load qq and the dimensions of the plate.
  4. Apply the Formula: Substitute the values into the formula to find wmaxw_{\text{max}}.

The load is not centrally located but this seems a worst case assumption.  The casters at the four corners of the current 28" plywood square are not sitting at the true corners of course but close.

load = 150 pounds / 2.2 * 9.8 giving 668N/m^2 (okay the load is not the same eveywhere but this is an approximation)

a = 28 inches/39.37

thickness is .25 inches /39.37

E is 200 GPa for steel

v is Poisson's ratio, which all examples seem to regard as .3

D calculates as 4385.9

maximum deformation is .04 m or about 1.5 inches.

This seems absurdly high for 1/4" steel plate.  The plywood sheet there right now is not eforming that much.  Do you see what I am doing wrong?

UPDATE: Simpler solution: buy another piece of oak plywood the same dimensions.  That has worked for two decades just fine.  About $90 at Home Depot.  Use existing board as template for holes.

5 comments:

  1. I would believe that 1/4" steel has similar flexibility to 3/4" oak plywood.

    Remember that stiffness is proportional to the cube of the thickness.

    Try putting a 'lip' around the edge of the plywood. An extra inch or two of thickness makes a big difference ...

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    1. I suspect you are correct, but I do not want to buy a pretty heavy piece of steel and discover otherwise. Home Depot has 3/4" thick oak plywood, which has been working for 20 years okay.

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    2. Screw a 2x4 'frame' around the edge of the plywood to massively increase its stiffness ...

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  2. Civil engineer here. You are overthinking it and using the wrong formula and assumptions, if I understand the question correctly. You have a 28" square wood platform with a caster support under each corner. The wood is bending up at the corner (or down in the middle) due to the heavy load on the top of the wood, which is unsupported other than by the casters at the corners. I am assuming the caster assembly itself is not bending. Adding a piece of reinforcement only at the caster will not likely help unless it has a much larger footprint than the caster does. If the caster has a 4x4" plate, for instance, the wood is bending along that plate. If you add a thick steel 4x4" plate above the caster, the bend in the wood will be the same, because it's the wood that's bending, and in the same place. If you add a 1/4" thick steel plate that measures, say, 8x8" with one corner aligned with the outer corner of the 4x4" caster plate and rigidly bolted through the wood and the caster plate, well that will move the zone of failure (the bend) to the inner two 8" edges of the new plate. This may solve the problem because although the wood remains loaded, there is now four times as much edge there as before. In such a scenario you would not need 1/4" steel, and 1/8" would be more than enough. It is important to use sufficient screws/bolts along the entire perimeter of the new steel.

    Another possibility would be to use a piece of angle iron (a steel bedrail for instance) attached all the way around the perimeter of the wood (a 28x28" frame) and well attached to the wood and the casters all the way around. I would weld this into a rigid square frame. One leg of the angle would be attached flat horizontal to the underside of the wood base and the other leg would point up (vertical) at the edge of the wood. Pointing down would likely interfere with your casters. Because this rail would likely be only 1-2" wide and with my assumption of a 4x4 caster plate, you would need to shim the resulting gap over the caster and weld or bolt the shim in place, not so much for strength but to keep the casters and wood flat and level.

    If you have stem casters instead of plate casters, you would do essentially the same thing, but drill a hole in your steel plate for the stem.

    Jeff in MS

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    1. I am not trying to save this old piece of plywood but replace it. 3/4" thick oak plywood is available at HomeDepot. The current piece is at least twenty years old.

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