Calculating force to bend a rod?

[Kevin Sevcik]

No prob, Don. Though that's pretty low on the Young's Modulus. It should've been on the material props page, possibly listed as Elastic Modulus instead. At any rate, actual value is 70-80 GPa. But in those linear equations, it only enters into deflection anyways.

As far as introducing the students to this concept... I still occasionally have difficulties convincing ours that lightening holes should be drilled in the sides of horizontal members as opposed to the top and bottom. If I were doing this I'd love to just run the equations on a 1 x 1 rectangle with material removed from the top vs. sides and then fail them both. Admittedly, buckling can come into play if the holes are too big, but still.

[MrForbes]

Kevin--I suggest you get the students to calculate moment of inertia for several different cross sections, including a flat bar in both X and Y orientations...then let them play with the formula for stress due to bending, and see how the stress varies with the different sections and orientations.

I spent some time with a few students during build last winter, I think they learned these concepts well enough to develop a good understanding of why the lightening holes should be on the sides.

Also Ken, the modulus of elasticity is pretty much the same for all flavors of aluminum, the alloy doesn't affect it enough to worry about it for the calcuations we make.

[DonRotolo]

Well, I went and bent the rod. Instead of heating it, I decided to increase the bend radius to 1.5" instead of 0.75". I could not measure the force required, because I bent it with a torque instead of a pure bending force as originally stipulated. At the end of a 24" rod I used not less than 200 pounds, so the moment around the end was at least 400 Ft-Lb, and that's literally near the upper limit of my strength.

The photo below shows the effects of yield strain, causing obvious granularity on the outer surface, but it did not fail. A tighter bend would require a lot more force - quadruple if I remember correctly - and I am sure it would cause the surface to fail.

I did try to bend it the origianlly stipulated way, and my 5" chinese bench vise wouldn't even deflect it, even cranking the handle down with a pipe wrench to the point where I was worried about breaking the lead screw.

So maybe a bit more than 4000 pounds of force...

Don

[EDIT] Re-did the calculations with l=4.5", E as 80 GPa, c as 0.375" and I as 0.075 - deflection was 1.09 inches at 500,000 pounds of force...[/EDIT]

[Kevin Sevcik]

Awesome job on the bend, Don. Also, I think I figured the elongation of your original bend too conservatively. I was figuring based on the inner radius being at 0% elongation, but as I think about it, that's patently false. Zero strain would be at the neutral axis at or near the midplane of the rod. So instead, your just completed bend would have a theoretical 20% elongation on the most extreme fibers. Your original bend would have been a 33% elongation. Which actually probably makes more sense now that I sanity check things. (Sanity check being, at 0 bend radius, the first method would give infinite elongation) At any rate, the tighter bend still would probably have been pushing things, just not by the insane amount I originally thought.