Calculating force to bend a rod?

Technical Technical Discussion
1

I’m an electrical engineer, so I’m not as familiar with this as perhaps I could be. It’s also been over 25 years since I’ve had Machine Design…

I have an aluminum rod, 6061-T6, 0.750 inches in diameter. I will mount it onto a fixture which is a piece of 3” angle iron held at 45 degrees to the horizontal, and push it down with another piece of angle iron (the ‘point’ will be relieved a bit, r=3/4" or so) until the rod is bent 90 degrees. See the thumbnail for the ‘unbent’ condition.

How many pounds of force (maximum) will I need to bend this?
Will the rod remain structurally sound, or will the outside edge become so stretched that most of the strength will be lost?

If someone could remind me of the way to do this - I used to know - I’d be grateful.

Don

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2

Don,

If you’re looking for a one-off, eFunda is the place.

Properties for your cross section are here.

Material properties are here and here.

Finally, the calculator for a simply supported beam is here.

I think you’d plug in a load until you exceeded the yield strength by a good bit. Problem being, you’re specifically trying to yield it and that gets into the non-linear range of things that aren’t as pretty.

The actual equations to figure the same thing aren’t very bad, but it’s 1am and I don’t know if I’d need to cover everything from torques and moments up to the bending beam stuff or if I could skip ahead a few chapters.

Incidentally, first run the equation thing until you hit yield, then run it until you hit the ultimate tensile strength. UTS would be your worst case load.

Though now that I look at the numbers, I’m skeptical that this will work as you describe it. I think a bend of that radius will be stretching your 6061-T6 too much. Even if the bar only ends up about half as thick at the middle of the bend, with a radius of about 1.125 on the outside, you’d be stretching the outside by 50%, and the elongation at failure is only 12%. So I think you’d likely crack the outside of your rod before you’re done. I suspect you’ll need to increase your bend radius, heat the bend area, or both.

EDIT: Quick calc puts the answer around 800 lbs for yielding T6. Probably a bit more than half a ton to actually bend it. O stock, meanwhile, would be around 200-300 lbs and less likely to crack on you. Of course, you’d have to heat treat it after you bent it. On the whole, cutting a 45 bevel and welding the joint and strengthening brace sounds… well… about a half ton easier.

3

looks like you’re just trying to bend a piece of aluminum rod, using a press, so it really doesn’t matter how much force it takes to bend it, as long as the press is strong enough and your jig doesn’t bend.

But you really need to figure out if the part will be strong enough when you’re done bending it…and we have no clue what you’re planning to use the part for, so we can’t help you figure that out.

I suggest you do an experiment, if you have some extra rod available…bend it and see what happens. If you can use a pipe instead of a piece of angle to bend the inside radius, it will probably result in a useable piece of aluminum rod (assuming your design can work with a larger bend radius)

4

Well, it’s more of a feasibility thought experiment, rather than a specific part to be fabricated. And yes, of course, I’d really end up just pushing on it until it bent - I cannot imagine that a 20 ton press wouldn’t handle it.

What i was looking for was some ‘order of magnitude’ estimates, and if anyone with some experience with this sort of thing had noticed something important that I’d missed. Additionally, i want to learn how to calculate it, rather than an empirical approach: The team already uses far too much ‘cut and try’ methods, I want to kick it up a notch and show that it can be calculated, at least for a first iteration. Give me a circuit, no problem, but a dumb metal bar, I’m strugging.

So, there are some things to consider. Like elongation being 12% to failure, but I’d be putting it through >50%. That sounds like a deal-breaker right there. Or at least, I’d better warm up the torch :smiley: .

OK, now that I have some direction for the equations, let me fool with it for a while.

Thanks!

Don

PS: Ah, what is it for… Think of a vertical shaft, and a 90 at the bottom to serve as an axle. Non-FIRST application. Vertical shaft can rotate a’la swerve drive, wheelbarrow wheel on axle driven via chain by window-lift motor attached to vertical shaft.

PPS: I agree that cut & weld is the better solution - but my intent is to learn the calculations, not make the part.

5

This doesn’t answer your question but your question sparked my question. What advantages do a single bent piece have over two pieces cut and welded to make the part?

6

Sounds like the steel parts that hold the front wheels on my riding lawnmower, and I would try to use steel for a part like that. Look at the yield strength vs. ultimate strength for different materials and different alloys. A soft material will allow you to bend it successfully…6061 is not very bend friendly.

7

I’m an electrical engineer, too, so consider the following with healthy skepticism:

You might consider 1/16th inch wall steel tubing as an alternative to solid aluminum. At 3/4" o.d. this wall thickness would give about the same weight per foot. Steel has two advantages: (1) compared with aluminum, there is more margin between the yield and ultimate strengths of steel, so you run less risk of breaking the stock while you try to bend it, and (2) the finished hollow steel piece will be stronger than the solid aluminum. The force required to bend the steel will be about 50% higher than what you’d need to bend the solid aluminum.

I have heard that thin wall steel tubing can be bent to smaller radii without too much risk of buckling if you first fill the section to be bent with Bismuth Alloy – the alloy will melt at a fairly low temperature (maybe 120 to 180 deg F?) and after you form the bend you can melt the alloy out again using hot water.

8

Simply put, the primary advantage is that you don’t have to weld it. Welding is wonderful for joining two pieces of metal, but most all welds are weaker than a solid piece of material for a few reasons. First, your weld has to be perfect. Any voids or cracks in the weld compromise its integrity in the long run. Second, welding is HOT. When you weld, you’re basically melting your two parts together and adding extra material to make up for any gaps and to provide some extra material to reduce sharp corners. This means that the material around the weld is getting very hot. So all around the weld is an area of material you’ve just applied an impromptu and fairly uncontrolled heat treatment to. This is the Heat Affected Zone (HAZ) which you’ve basically just air quenched and softened by a good bit.

Finally, sharp corners are bad. Any sharp corner anywhere inevitably concentrates stress at that point and weakens the joint. So a large radius bend will hold up much better to load than a weld with a small fillet on it.

As with everything there’s plenty of other things to be concerned about with welds like hydrogen sulfide cracking and other corrosion issues and… well. lots of stuff that doesn’t come up much in FIRST bots.

9

I think Richard is right on with the steel tubing recommendation, since as the previous poster mentioned, the 6061 is going to fracture on the bottom side due to too much elongation.

You should use “chrome-moly” steel tubing, which is typically 4130 alloy. Pretty readily available stuff from many sources. Instead of using bismuth, you could also fill it with sand or small ball bearings - the only caveat with these “loose” materials as opposed to the bismuth is that you need to plug the ends of the tube before bending to make sure the loose material is constrained.

hth,
Ken

10

Ken, the ductility problem with 6061 applies to 4130 as well…if the part is to be bent, it ought to be made of relatively low strength material.

Bending 3/8" od 1/32" wall soft steel tubing with a bender is tricky enough, I can’t imagine trying to bend 3/4" high strength 1/16" wall steel tubing without a bender without having it kink.

But I can imagine lopping off a piece of that tubing at a 45 degree angle and welding it back together with a 90 degree bend in it.

11

And third, I don’t have a welder…:frowning:

Probably. I’m going to try to bend the aluminum with some heat and see what happens, which will certainly mess up the calculations, but I was hoping to use aluminum 'cause it’s an outdoor application. (I suppose the 'bot won’t last 2 years anyway, the steel should last longer…)

Ay, there’s the rub - needs to be strong, but machine well enough, and bendable. Three competing requirements. Not having any Bismuth handy (have you priced out that stuff?), maybe tin-lead solder, or sand, might be the way to go. (Or galvanized plumbing pipe with an elbow :smiley: :ahh: ) Maybe I need to bolt the joint together.

Anyway, again the idea is to learn about the calculations, so when we get into the season in a month I’ll be able to maintain my “know-it-all” persona…:cool:

Thanks all again, this is great.

Don

12

Actually, both 6061-T6 0.049" wall and 4130 0.035"-0.068" wall can be bent. We have been doing it on our robots for years, especially the 6061. Its funny - the first time you ask someone to bend it, they say you can’t do it, and they sometimes show you a bent wrinkled up cracked mess. But we knew a guy (recently retired UAW technician) who could make some pretty nice tight bends in 6061 tubing. He used sand. I’ll try and dig up some pictures.

Another possibility if its an outdoor application would be stainless steel. If anyone remembers the “wings” we had on our 2003 bot, they were made from tightly bent 3/4" OD stainless tubing.

Ken

13

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.

14

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

15

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.

16

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.

17

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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bentrod.jpg
5d4bdf9053bbcea870169f1ba2624e61538e3f55.jpeg474×677 33 KB

18

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.