# Computing the required thickness of an aluminum sheet to support a given load

I want to fabricate an aluminum shelf for my media center. 22 1/16” x 37 5/8” with a rectangular cutout 2 1/2” deep on one of the long edges of the sheet with 6” of material on either side of the cutout. (CAD model)

The shelf will be supported by four shelf pins on the short edges near the corners.

The question I need to answer is how thick the aluminum needs to be to support my intended load (100 lbs of equipment, more-or-less in the center).

I've been playing around with SimScale and (assuming I'm doing it right) I've determined that a tenth of an inch of aluminum is sufficient, but I'd like a better methodology for computing the minimum required thickness than "guess and check".

What is the right way to compute the required thickness of the sheet of aluminum? Modelling the problem as a rectangle secured on the short edges with a 100 lb load in the center seems like the first step, but coming from a CS/EE background I'm lacking the statics knowledge to enable me to properly set up the problem.

An adequate simplification would be to treat it as a simply supported beam (there is not much added benefit to treat it a plate).

For a simply supported beam with:

• Length L = 37 5/8''
• width w = 22 1/16'' - 2 1/2'' = 19 9/16'' (the width is at its thinness point)

$$M=P\cdot \frac{L}{2} = 1881.25 [lbf.in]$$ where:

• P is the load (100 lbf)

The maximum stress that will be developed is:

$$\sigma = \frac{M}{I}\frac{t}{2}$$

where:

• I is the second moment of area of the plate $$I = \frac{w\cdot t^3}{12}$$
• t is the thickness

the maximum stress needs to be less than the allowable stress:

$$\sigma_{all} \ge \sigma = \frac{12 M}{w\cdot t^3}\frac{t}{2}$$

Therefore $$t \ge \sqrt{\frac{6 M}{w\cdot \sigma_{all} }} = \sqrt{\frac{2 P\cdot L}{w\cdot \sigma_{all} }}$$

Assuming you use 6061 which has a yield stress of 40000 psi with a safety factor of 1, then the minimum resulting thickness is 0.1201 inches (which is consisted with your calculations).

For comparison purposes, assuming you applied a safety factor of N=2 (therefore the $$\sigma_{all}=\frac{40000}{N}=20000 [psi]$$, the minimum thickness would be 1.7 inches.

## additional check with deflection.

Its best to check with deflection.

The max deflection for a simply supported beam with a concentrated load is:

$$d_{max} = \frac{P L^3}{48 EI}$$

where:

• $$d_{max}$$ is the max deflection. Usually, its defined as a fraction of the beam span length to avoid visible sagging (e.g. 1/250 L =0.1505'' ).
• $$E$$ is the elasticity modulus of aluminium = 10000 ksi

So again (after some algebra):

$$t = \sqrt[3]{\frac{12 P L^3}{48 E\cdot d_{max}\cdot w}}$$

If you apply the values, then the minimum thickness results in 0.3564 inches. As you can see its double the minimum thickness when you calculate based on stresses.

So, if you use 0.17 inches the shelf will not break, but it will have a noticeable by eye sag, which will not be pleasant.

• Note to OP: when you get much thicker, it may make sense to construct the shelf out of multiple differently shaped pieces (like an I-beam or T-beam), instead of just a single sheet, to save on the weight and cost of the sheet Sep 6 at 8:44
• Or you can just add a stiffner at the edge (both sides of the long edges). It offers a significant improvement. Sep 6 at 9:17
• yes, that also counts as "multiple differently shaped pieces" and is probably a more sensible shape for a shelf. Sep 6 at 9:19
• So do rolled edges or a fold… that will change the deflection. Sep 6 at 9:24