by Cre8ivdsgn » Thu May 10, 2012 8:15 pm
Bart: is that moment at the neutral axis or at Y=0?
At neutral axis I = (b*h^3)/12
At y=0, I=(b*h^3)/3
Or am I screwing up the concept of moment of inertia? I'm beginning to feel like an idiot... again! I'm just hoping some clever soul can explain this in simple terms that I can grasp!
BTW, the simplification we are using for the deflection calculation Yd = (P*L^3)/48E*I assumes the beam is supported so that you get a simple arc curve. Mount any beam rigidly (and assuming the mounts themselves are ideally rigid), and you are now dealing with three arcs and something slightly more detailed would be required to figure things out. Certainly a rigidly mounted beam would deflect a good bit less (1/2 roughly?) than what would be implied by the calc for Yd.
Bart, here would be something interesting: would you draw a beam that is 1.5" tall (call it 40 mm but do let me know which is used) and 0.125" (3mm) and whatever long? What does ProE return as the moment of inertia? It should return the moment at the neutral axis or I = (.125*1.5^3)/12 = 0.035 and in the metric case (3 x 40^3)/12 = 16000. I would not be surprised if it was returning 4x that amount or 64,000.
If I treated (in response to mattrsh) the beam as a simple rectangle and ignored the V roller surfaces I would have .78" and 1.675" (for b and h) and so in the imperial world I get 0.305 and in the metric world I get (19.812mm x 42.545mm) or 1.27 10^5 mm^4. OK. Am I being stupid here? Wait, we are at cross purposes! I have been looking at the beam as an X axis for an ORD-type platform, making the long axis verticle (h) and the short axis horizontal (b). When I swap them, I get 2.75 x 10^4 which is close enough to your numbers! Yes! I am not crazy!!!
Given the hi resolution folks seem to be achieving with the hardware these days, I can't help but think these answers matter!