We’re at it again… We couldn’t get the right Aluminum alloy, and were trying
to decide if a vertical arm could withstand the load we were going to put on
it.
We have a vertical, ~5 foot section made from 2x2 Al tubing, 1/16" thick -
Flimsy but light. We were concerned going to a weaker Aluminum alloy might
cause failure given our loading conditions.
Worst case scenario, this vertical piece would have a 180-lb axial load (in line
with the beam) and a 1090 in-lb bending moment throughout, from a grabber
which sticks out from the beam top. This setup gives us a long member in
compression, loaded eccentrically. (Simply put, the load is away from the
center of the beam.)
When beams are compressed, they will buckle or crush. Crushing is like a can
getting smashed. The yield strength is passed, the metal gives way, and
the can crumples into a little disc. Buckling is different; it occurs when part
of the beam bends out of line. You can see this if you press down on a
vertical meterstick. The middle part of the stick will bow out to the side.
So we had this 5 foot vertical section that was raising some buckling flags.
It is long, has a thin shell, despite it’s decent radius of gyration (all of the
material is far away from the center.) I looked at the beam with the other
advisors, and we found the design was adequate. I’m posting a cleaned up
version of the problem we ended up doing, so other teams may benefit.
Remember, buckling is something to consider. Watch out for:
Long, slender beams.
Heavy loads far away from the centerline of a beam.
Beams with most of the material toward the center of the beam.
Heavy loads on an unconstrained beam.
(Think antenna ball with a brick instead of a ball. Bad idea.)
cnc/waterjet triangles on all 4 sides of it, that way the load will be transferred throughout the tube and thus should make it stronger. Are you worried about side loads? or what type of deflection? if you were to make triangles along it it would essentially act like trusses on a bridge (if it were laying down sideways).
2x2 should be fine…then again, if you are still worried, get some machineable wax or some sort of light, hard plastic and make a core for the 2x2.
Trusses are very efficient means of transferring load. Weight efficient, that is. When it comes to pure buckling, total weight really doesn’t matter. Critical load, the load at which buckling occurs, equals pi^2EIa/L^2.
E is the Young’s Modulus of the column (or column/core combination)
Ia is the area moment of inertia (mm^4 or in^4)
L is the length of the beam.
This equation is based on some assumptions, but the criteria for validity is, generally, that L/r must be grater than 100. r is the radius of gyration mentioned before. r^2 = Ia/Area.
Would a bridge with solid plate sides instead of trusses hold more load? Probably, depending on geometry conditions.
Would it weigh too much? Yep.
Would it be better or worse for dynamic conditions (think Tecumseh Narrows bridge)? Much worse. Mass is a major player for dynamic performance.
Tacoma Narrows Bridge, and I’m not sure your comparison is apt, although the rest of your analysis is a great contribution to the hive mind. The Tacoma Narrows Bridge (the first one, not the one that stands today…) had a very thin bridge deck without deep stiffeners under the deck. High winds set up harmonic vibrations which caused the bridge to shake itself to pieces. It wasn’t an issue of buckling under load, it was a failure to tune the bridge for a low enough or high enough resonant frequency (and I sure hope that’s the right term – I am NOT a civil engineer) to prevent it from vibrating like a guitar string.
Did your post imply that extra mass would be worse for dynamic conditions? I am trying to understand this, if this is what you meant. Tacoma Narrows Bridge #2 is substantially “beefier” than the first one, including a much deeper bridge deck.
I knew it started with a T, but I always screw the name up (the video will NEVER leave my memory).
Every part of my message dealt with buckling, except the last sentence which mentioned dynamic conditions. I should have elaborated on what I meant by dynamic conditions, but I wanted to keep my message short. The point I was trying to make was that when dealing with buckling, mass is not a contributor to the critical load. However, when dealing with dynamic conditions (aka natural frequency, resonant frequency, etc.), mass is a big player. After re-reading the message, I can see how someone would make the buckling connection to the last sent ace. I apologize for the confusion.
Don’t forget torsional stress. Allot of load analysis formulas ignore torsional stress because they are analyzing a load bearing beam in 1 or 2 dimensions. The arm will be subjected to torsional stress. We’re using a 2" square tube with a 1" tube foamed and epoxied inside to account for these stresses. (fiberglass). Tube in tube can help also butted tubing can help.
for clarity, if L/r is greater than 100 it will buckle before it yields. You would always want to keep L/r (called the slenderness ratio) less than 100.
The equation we used per structural pipe code was kL/r had to be <120 for steel structures (E~30 Mpsi, Sy~45 ksi)), where k is dependant on the end constraints. Since worst case k=1.5 for pinned ends, it’s conservative to just use L/r <80. I ran the numbers for 6061 aluminum and it seems to me it was L/r <60 ; I have the calcs at work and I’ll post them tomorrow if I can find them.
It’s actually a pretty simple calc to derive. From Eulers equation given by Paul, setting the stress at the critical buckling load equal to the yield strength Sy, L/r = pi*(E/Sy)^1/2.
For steel (properties given above) L/r = 81
For 6061-T6 Aluminum (E~10 Mpsi, Sy~36 ksi), L/r = 52
We already figured it out. We’ve got a FOS of about 8 against buckling, and 11 for crushing. I’m posted the solved problem so the FIRST community can benefit.
Gdeaver, our arm will not be seeing significant torsional stress. We looked at the worst case scenario involving a robot trying to stack, and would be relatively motionless. The force we used would result from a piston device jamming, and we would not deploy this device if the robot was turning or driving. This formula addresses a 3-dimensional beam.
You might have been talking about a bending moment. We do have one of those, espicially since out column has an eccentric load. All that is accounted for in the problem. Given our L/r ratio is about 150, buckling needed to be addressed.
(Students - You can calculate this ratio based on your beam geometry. I didn’t actually find the L/r ratio in the problem because I correctly assumed it was bad. I did find the moment of inertia, I, of the beam cross section, and the area. rg=(I/A)^(1/2), and our L/rg comes out to 158.50.)
If a robot is not interfered with, I agree the torsional stresses are relatively low. However you can bet the other alliance is going to be hitting your robot and arm. This is going to cause major torques on the arm. IF a team has not designed for this, they could be looking at some ugly twisted metal. Last year most teams where fighting for the platform and we could cap with out interference . This year interference is part of the game and we are going to have a good amount of weight hanging out there. We’re adding reinforcements to deal with it, but we’re worried.
