Short Column Buckling With Eccentric Loading

Hello CMNKY 26

Haven't looked at this sort of problem for a few years so it's interesting!

I calculate bending stress (from the secant formula) ~ 6 x direct compressive stress for this column. Secant formula for bending stress P.d.sec(μ.L)/Z where P = load, d = offset (2"), L = height, Z = section modulus and μ = √(P/(E.I)), E = elastic modulus, I = 2nd moment of area. (E for steel, 3.10⁷psi, assumed)

You can then calculate load to give total stress = maximum design stress for the material. E.g. for load 2000lbf, I make direct compressive stress 2909psi, bending stress 16922psi, total 19831psi. Assuming it's steel, and depending on grade, this could be getting near the limit. Horizontal deflection = P.d.L²/(2.E.I) = 0.005in - well within normal limits.

Taking effective length L_eff = 2.L, Euler buckling load E.I.(pi/L_eff)² ~ 10⁶ lbf so clearly this is not limiting.

Hope this helps.....Codey

When all the math is done, it is time to consider "real world" testing.

I have seen too many mathematically correct designs fail before their stated values when subjected to Destructive Testing.

About 30% of the tested designs failed before reaching their projected failure point!

Your design is small enough to test in-house if you were so inclined.

I for one, never want to see one of my designs on the TV program,"Engineering Disasters". Or the subject of a law suit.

haha, no i doubt this problem will make it to the engineering disasters show. The guys who built it, didnt build to print, so now my boss wants to know if this widget fails if some guy stands on it. Hence me to coming up with some numbers...

But, like Codemaster, i havnt done this in a few years either. *dusting off the ol textbooks* Thnx for the quick run thru

Prbarry, I would love to break err test this thing, but sadly i doubt i could convince somone to pay for it.