80/20

[Jeff Rodriguez]

173 is building a 2nd robot for fun. We want to use the 80/20 stuff, but i'm stuck in a dilemma.
For those who have used it before:
Would it be better to make the frame of the base out of the 1010 or the 1515-lite?
It looks like the 1010 is hlf as havy as the 1515, but the 1515 is twice as strong as the 1010. Which is more important? Is the strength that important, and have a bigger influence, or shouls i go with the 1010?

If you watch our matches, you will see that our drivers can get pretty crazy, and the bot will take alot of impact. Any advice you have would be great. Thanks

[Jnadke]

I think we're using the 1515, I'm not completely sure (Bot Pic). But anyways... I think we used that and we still manage to bend one of our protection beams on the front at MMR (Front pic: it's the thing with the tinted polycarb over it). You could see the curvature when looking at it from above, but we didn't notice it until nationals...

Of course... that was hitting the goal, full-speed (12fps), without stopping. Nothing else broke... except maybe the goal.

Other than that, the stuff has held up fine. We stand on our robot all the time and nothing has broken.

[Alavinus]

Considering the bend strength of the 1010 I think is 3600 lbs, I think the 1010 should be fine. We used it to make the entire frame of our robot,with no ill effects. (We liked to get into pushing contests too) Anyway, the only part that was more beefed up then the 1010 was the 2010 we used on our arm to make it wider. I would say that 1010 is more than adequate for the frame.

[Andy A.]

Perhaps use the 1010 for the frame, and brace it with something more beefy? It means a bit more work on the frame, but 80/20 is so easy to start with.

Also, consider adding foam padding of some sort, that should take care of most impact related bends...

-Andy A.

[patrickrd]

While a number like 3600 lbs may sound impressive, that is not necessarily the case.

What you care about here is a collision. As an example I will choose a pretty bad (you could say worst possible) collision of two FIRST robots. Each robot is traveling towards the other with a speed of 10 ft/sec and each robot has a mass of 130 lbs.

We can calculate the momentum of each robot:

P = (10 ft/sec)(130 lbm) = 1300 lb*ft/sec

Now when these two robots collide, conservation of momentum will cause them both to come to a complete stop. Thus the impulse caused by this collision is:

J = Pi - Pf = 1300 lbm*ft/sec - 0 = 1300 lbm*ft/sec

Now, I will assume you have a piece of extrusion on your robot that is simply supported on both ends. This piece is on the 36" side of your robot and the other robot hits only this piece and exactly right in the middle of the piece.

The next quantity that must be determined is the duration of the collision. To calculate this in reality would involve some really complex dynamics and study of elasticity of metals, so I am going to use some approximations. Again these are probably worst case. I am estimating a robot collision takes about 1/100 sec to occur. (this is absolute worst case! i speculate it takes longer)

If we assume the force is distributed equally over time during the collision (i.e. each robot exerts a constant force on the other over the timespan until both robots come to a stop), we can calculate the average force from the impulse:

F[avg] = J/t = (1300 lbm*ft/sec)/(0.1 sec) = 130000 lbm*ft/s^2
F[avg] = 130000 lbf

So, now we have a simplified problem where we have 130000 lbf acting upon the center of a simply-supported extrusion piece. this means that each end of the extrusion has is supporting half the weight. Thus there is a reaction force of 65000 lbf on each end. The maximum moment on the beam will also occur in the middle (1.5 feet from the end) This moment is:

M = (65000 lbf)(1.5 ft) = 97500 lbf*ft

I will now convert to inches since the rest of my calculations are in inches:

M = (97500 lbf*ft)(12 in/ft) = 1170000 lbf*in

Finally, the formula for maximum stress is:

S = M*c/I

Where S = stress, M = moment, c = maximum distance from neutral axis, and I = moment of inertia about neutral axis.

FOR 1010 EXTRUSION:
I = 0.044 in^4 (from catalog)
c = 0.5" (from catalog)

S = M*c/I = (1170000 lbf*in)(0.5 in)/(0.044 in^4)
S = 13295450 lbf/in^2

FOR 1515 EXTRUSION:
I = 0.185 in^4 (from catalog)
c = 0.75" (from catalog)

S = M*c/I = (1170000 lbf*in)(0.75 in)/(0.185 in^4)
S = 4743240 lbf/in^2

Since the yield strength of aluminum is 13,000,000 lb/in^2, the 1010 extrusion will yield (permanantly deform), but the 1515 has a factor of safety of 2.7.

However, the 1010 just meets the requirements for yielding, and that is with my worst-case assumptions for everything. If it were my decision, I wouldn't hesitate to use the 1010 since I don't think my worst case scenarios would ever be satisfied.

Hope this helps answer your question

Patrick

[Jeff Rodriguez]

That helps alot. I took physics last year, and forgot all those formulas. I'm also too lazy to figure out all that That made up my mind. Thank you