[kramarczyk]

Originally Posted by squirrel

For the sake of making the discussion even more interesting, how about making several different shapes and sizes of beam, and applying the same (relatively small) load to them and showing them all? For example, you might have rectangular tubing with the load applied to the narrow side, and to the wide side, and round and square tubing, and an I beam, and a channel.

Also it would be nice if you'd give a quick explanation of how to set up the constraints in Inventor so others can do this, as it's not obvious to the novice Inventor user.

I'll see what time I can get to run a series of these examples, first as the base sections, then we can send them to Jenny Craig. I'll also throw the kitbot rail in as a reference point. I used the left vertical edges as my constraints and fixed them only in two dimensions... on the axis parallel to the load, and on the axis that is both perpendicular to both the load and the axis of the beam. They were left free on the axis that would allow them to get closer to each other. A picture would help, I'll add that to the list. <edit> The part used was a 38" long piece of 2x2 inch aluminum with at 1/8" wall </edit>

Originally Posted by M. Krass

Inventor exaggerates the deformation considerably, as you're meant to use the scale displayed on screen to determine the appropriate values rather than a visual representation.

I have yet to see a system that can't scale the displayed deformation; deflections of a few thousands of a inch aren't terribly visible otherwise. You're right though, go with the numbers.

Originally Posted by sanddrag

The question on holes got me curious. Here is a summary of my test and results

1"x1"x1/16" wall 6061 Aluminum Box Tube, 18" long. 0.625" holes down opposite sides, spaced at 1" centers
End faces fixed constraint
500 psi pressure applied to a 2" long patch, on the top side center of the tube.

For the given loading and fixed constraints, the part exhibited a 29% higher maximum stress with the holes than without. The part with holes had approximately 15% less mass than the part with holes.

Got any screen shots? With the end faces held fixed I would expect to see some hot spots out near the end.

Originally Posted by sanddrag

Moment of inertia includes mass to the 1st power, and distance to the 2nd power. So, roughly speaking, when the "meat" of your beam is twice as far away from the central axis, then it gets 4 times stronger.

The applicable moment of inertia(actually the second moment of area, but commonly referred to as the Moment of Inertia, I. Don't ask me why, I have know idea why we confuse people like that.) formula for rectangular tubes is ((W*H^3)-(w*h^3))/12 Where W & H are the outer rectangular width and height respectively and w & h are the inner rectangular width and height respectively. Point being that mass is not generally involved in beam bending calculations unless it contributes a significant amount to the actual load. Additionally, the strength of the beam increases at the cube of the height of the beam, so doubling the beam depth increases the bending resistance by 8x.

http://en.wikipedia.org/wiki/List_of...nts_of_inertia
http://web.mst.edu/~mecmovie/index.html