Designing arms against buckling

[Paul Copioli]

Greg,

With regards to your first paragraph, the 2 x 2 without cutouts is more resistant to buckling than if you cut trusses in the 2x2.

Your second paragraph is the ticket. A lightweight core (even balsa or styrofoam) will definitely help the buckling scenario.

-Paul

[Greg Perkins]

are you sure paul? because wouldnt the the trusses distribute the load more evenly over the surface?? eh, anyways, im batting 50% cant beat that!

[Paul Copioli]

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^2*E*Ia/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.

[Rick TYler]

Originally Posted by Paul Copioli

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.

- Rick TYler
(An old guy with a sticky shift KEy)

[Paul Copioli]

Rick,

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.

-Paul

[Gdeaver]

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.

[Gary Dillard]

Originally Posted by Paul Copioli

This equation is based on some assumptions, but the criteria for validity is, generally, that L/r must be greater than 100.

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