Square hollow aluminum beam: why is 4×4 mm adequate for my loads?

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skyturnred
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beam bending problem.jpg

I am trying to find the width of a square hollow structural member that can withstand the forces above. However, my calculations are showing me that a square hollow aluminum tube of 4 mm by 4 mm with a wall thickness of 1/16 of an inch would be adequate, which I don't believe.

Is anyone able to see where I am going wrong? I believe it may be a units issue but I can't figure out why.

Thank-you in advance
 

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1. Is the beam supposed to be accelerating? Most beam problems are computed assuming the beam is in static equilibrium.

2. Mixing SI and Imperial units is just asking for trouble. Pick one or the other, and stay consistent throughout your calculations.

3. It's unrealistic to design a structural member using a maximum allowable stress = yield stress, especially when that material is aluminum. Most allowable stresses in bending are typically limited to 0.6-0.66 * yield stress.

4. Your beam is analyzed as if it is simply supported at the ends. Is this a realistic support condition for this construction?
 
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Thank-you for the response.

1) Yes, the beam is accelerating. It will be part of a machine that will apply a maximum of 3.5 lbs-f upwards on both ends of the beam.

2) Thank-you

3) My intention was to design it to the maximum allowable stress and then simply double the dimensions of the beam to give myself a decent margin of safety. However I did not know about the 0.6*YS rule of thumb. That is very useful and I will use that in my calculations instead, and at the end will simply add on an extra 10% margin of safety.

4) the beam will be supported with 4 bolts in total (2 on each side). I will drill holes pointing in the direction of the x-axis. These holes will be on the ends of the beam. I simplified my problem because I intended to add enough of a margin of safety for it to be not as important.

I guess I should have been more clear in the goal.
The beam can be a maximum of 166 grams in weight, while providing enough strength to avoid yielding given the forces shown above.
If I use a square hollow structural tube of 1" x 1" x 1/16 ", it will weigh a total of 144 grams. I was just trying to see how much (if any) of a margin of safety a 1x1x1/16 aluminum 6063-T5 square hollow tube would provide.
 
If the ends if the beam are going to be bolted, then assuming simple support conditions is also not realistic. I would check the beam assuming

1. fixed conditions at each end.

2. adjust the moment of inertia to account for the removal of material to accommodate the bolts.
 
Where did the downward 3.5lbF point force come from? The force due to the acceleration will be a distributed load and will depend on total beam mass. If I've understood the situation correctly. That is.