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Originally Posted by generalbrando
Good thinking  What we'd like to do is create a few portions of an arm. We haven't decided specifics yet, but the worst loads would be that it would be a box aluminum tube of 5ft pivoting on one end and lifting a tetra on the other. The pivot would be connected to a stationary piece of similar aluminum, attached to the chassis. This stationary piece would likely go under the greatest stess. Assuming interference with other objects (e.g. robots), we're wondering what our options are - is this stuff even strong enough that we can use it to a lesser degree?
Being that we don't have the engineers or engineering skills to calculate things properly, I was hoping someone would come forward who uses this stuff on their bot and say "oh, yeah, 1/16 is all the strength you need" or something like that. But if you guys have a mathematical way to do it, by all means. I'll gladly teach my team what you come up with.
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The equation you are looking for is Stress= Mc/I where Stress is the bending stress in psi, M is the bending moment, c is the distance from the neutral axis of the section to the outside and I is the moment of inertia for the section.
The moment is easy enough to calculate. M=weight of the tetra (approx 8lbs) x the distance from the pivot (5ft) = 40 ft lbs=480in lbs.
Assume you are using a 1x1x1/8 square tube. Since the tube is symmetrical, the neutral axis is at the center, so c=0.5in. I is a little harder, for a rectangular section I = (bh^3)/12. Where b= the base of the section and h=the height. Other cross-sections, like round tubes, use different equations. In this case b=1 in and h is also equal to 1 in. So you get 0.0833 in^4. But now you need to subtract out the part that isn't tube. That part is (0.75in*(0.75in)^3)/12 = 0.0264. So I = 0.0833-0.0264 = 0.057. Notice that I increases much faster with height than with base, so if your stresses are too high, add height not base.
The whole thing becomes Stress = (480 in lbs)*(0.5in)/(0.057in^4) = 4213psi. Notice that up to this point we have disregarded the material used. Stress level depends on loads and geometry, not material, unless you are dealing in the non-linear world. For FIRST, we don't generally want to go there. What will change with the material is the deflection under a given load, but that was not part of the original question. We also neglected the weight of the beam. This was to get a Rough Order of Magnitude (ROM) estimate of the stress to see what materials might work. To add the stresses due to the weight of the bar, just multiply the weight of the bar to the distance of the center from the pivot point. Add the resulting moment to the moment due to the load and re-calculate. Since the allowable for aluminum is around 10,000 psi, the 1x1x1/8 is probably adequate. We'll assume it is for now. The weight of the bar is (1in^2-(0.75in^2))*60in*0.1lbs/in^3 = 2.6lbs and the distance is 30in for a moment contribution of 79 in lbs. This is pretty minor compared to the tetra load, so we were right to neglect it at first.
The margin in this case (10000psi allowable/4300psi load) is 2.3. Many people would consider this adequate, but personally I like a margin of around 4. These calculations are for static loading but actual competition loads are variable and dynamic. The extra margin helps account for this without the long and ugly calculations required for dynamic modeling.
Hope you found this helpful. Feel free to do your own investigation with other sections, lengths, add a gripper, etc. I picked the 1x1 because it was easy, not because I thought it would work.
ChrisH