Box Aluminum Extrusion Strength
 

I'm unable to find anything expressing the ability of box extruded aluminum except the brinell hardness (40) which doesn't make sense to me (explain if it's what I'm looking for:).

We would like to use extrusions in the range of 1-2" outside width and we're wondering how thin the wall can be. We want to use it for the main portion of an arm, so the forces will be fairly high and it will be a few feet long.

Re: Box Aluminum Extrusion Strength
 

Rather than looking for numbers concerning box aluminum, look for numbers concerning the sheer strength of whatever aluminum you are using, and then find the equations that deal with it as a function of area.

Good luck this year.

Re: Box Aluminum Extrusion Strength
 

Eeek! lost my post...

I don't think sheer strength is the material property you need. You will need either the Young's Modulus (aka Modulus of Elasticity) or the Yield or Ultimate Tensile Strength. This, in combination with the proper moment of inertia and description of the load on the beam will let you calculate the stiffness or strength of the beam. There are plenty of good websites out there (I can't seem to find my favorite one and the bookmark isn't with me)

If you haven't done this stuff before and can't get anyone to help you, it's a bad idea to try to pick your extrusion based on what you come up with. It would be better to look at similar parts on other FIRST robots and make your selection based on that.

Re: Box Aluminum Extrusion Strength
 

Yes, my mistake. In any case, it should be a fairly simple statics equation.

Re: Box Aluminum Extrusion Strength
 

One way to figure this out is to tell us specifically what you are trying to do/make and tell us specifically what material you have in mind and we'll tell you if it is a good choice or not. :)

Re: Box Aluminum Extrusion Strength
 

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.

Thanks!

Re: Box Aluminum Extrusion Strength
 

I'd say for a SF of 3 or thereabouts you need about 1.4 to 1.5 square inches of cross section. This is the cross sectional area of a 4x2" piece of extrusion with 1/8" walls. Remember that aluminium is a maliable metal, and so any points of rotation need to be reinforced lest the holes widen themselves.

I'll stop by if I can, this week or so.

Re: Box Aluminum Extrusion Strength
 

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

Re: Box Aluminum Extrusion Strength
 

That is far, far from the truth. You should be designing for loads several times greater than that one. Perhaps if you were the only robot on the field, and there were no goals, that might be true, but with all the stuff that may be colliding into your arm from pretty much every possible direction, you need to deal with much greater loads. Just imagine a 130lb robot traveling at 8 miles an hour colliding with the side of your arm. Either you stop the robot, your own robot turns, or your arm breaks.

Re: Box Aluminum Extrusion Strength
 

That is ONE of the reasons I use a safety factor of 4 for FIRST robots. I also typically assume a 1/2g side load, but I thought that was a little too much to add for one post. I also typically use FEA for our arm designs, so adding the side load is pretty trivial once the model is built.