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Unread 03-08-2007, 23:06
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Re: Calculating force to bend a rod?

Ken was planning on bending solid 6061 rod, not tubing. We bent some 5052 sheet for the robot this year, the old fellow who runs the fab shop suggested stacking two pieces and then bending it, so the lower piece has a much larger bend radius than the corner of the brake....worked great! without the extra piece the relatively mild 5052 cracked noticeably on the outside edge, and it was only .080" thick. We tried bending some 16 gage 6061, the results were not pretty.

Another idea for the solid aluminum rod would be to drill and tap the end of the horizontal piece for a 3/8" bolt, and drill a hole thru the vertical piece and file it flat, so the horizontal piece can seat against it.
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Unread 04-08-2007, 21:26
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Re: Calculating force to bend a rod?

The calculations show that I'd need about a ton (2000 lbf) to make the bend, but the outer edge would elongate way past failure - in sort, this bend can't be made as specified.

Of course, the equations given are for small deflections, and this is well into plastic deformation, so the results are only as good as an order of magnitude: 200 lbf is not enough, 20000 is too much, the required force is between those extremes. Close enough for government work....

IF I was to actually build such a thing, I might first try heating the rod with a propane torch to around 750 C, that will help with the yield to failure, but the piece would require heat-treatment again.

More likely i would take Squirril's advice, and make it a bolted joint, with a steel bolt passing through the vertical member into a hole tapped into the axle. A flat would be milled or filed onto the vertical piece. Instead of 3/8, I would venture to 5/16 to provide enough 'meat' for the vertical piece to avoid fracture, as a hole does seriously weaken it, and the flat on one side doesn't help.

Or just some steel tube filled with sand...

In conclusion, I wish I'd retained enough of my machine design course (I did get an A, causing several MechEs to dislike me for acing this course outside my specialty) to now go ahread and calculate the relative strengths of the vertical-with-hole, bolt, axle, etc. - remember, the idea is to show the students that it can be done, by hand, with a slide rule and some tables. No need for a pentium of simulation program, kinda like a time warp back 60 years.

I think I'll pick something simpler, like a beam in deflection - calculate the failure point, then run a few beams through the wringer and compare theory with real life. Put in a few holes, notches, and other stress concentrators to demonstrate that these matter.

Or maybe stick to the sparky stuff.

Should be a bit of fun this September.

My particular thanks to Kevin Sevcik for the links to the equations. I still don't know what Young's Modulus for 6061 T6 might be, so I just used Yield strength (275 MPa) in the formula, wonder how valid that might be...

Don
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Last edited by DonRotolo : 04-08-2007 at 21:28.
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Unread 04-08-2007, 22:00
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Re: Calculating force to bend a rod?

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.
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Unread 05-08-2007, 00:00
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Re: Calculating force to bend a rod?

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.
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Unread 06-08-2007, 20:35
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Re: Calculating force to bend a rod?

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]
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Last edited by DonRotolo : 06-08-2007 at 20:39. Reason: Maybe a LOT more than 2 tons...
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Unread 07-08-2007, 00:19
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Re: Calculating force to bend a rod?

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.
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