Help with cantilever beam stress

CraigM · Mar 25, 2009

I'm designing a hay fork for my B2150. It will lift 4 bales at a time. The bales are stacked 2 wide x 2 high to make a 200 pound cube about 36 on a side. I envision 4-8 round spikes, like a tine bucket

A quick test of a possible spike using a 50# bag of feed in a 5 gallon bucket says that 5/8" diameter rod clamped in a pipe vice will hold 53# at a distance of 30" from the support, and spring back when the weight is removed.

With that as a starting point, I got out the books to 'experiment' with different rod diameters and lengths. When I calculated the stress in the 5/8" test, I got 66ksi. Interesting for mild steel that should only be able to withstand 30-36ksi. I went back to the shop and tried again with other rods and was able to get calculated stresses as high as 100ksi without bending the rods.

The equations I'm using are:
Stress in a cantilever beam = (My)/I. M is bending moment, y is distance from neutral axis...the radius of the rod in this case, and I is (Pi/4)r^4 where r is the radius of the rod. Pi is that magic 3.14 but I'll have to write it as Pi since I don't know how to make the symbol.

When I do the calculations with 5/8" dia, 53#, and 30" I get M = 1590 in#, I = .0075 in^4, y=.3125 in. This gives a stress of 66ksi.

It's been way too long since school and engineering lab. Can someone tell me what I'm missing?

Huskerplowboy · Mar 25, 2009

Your math all sounds right, I got the same answer. Google for "Beamboy". nice freeware beam stress program, even has some channels, I-beams, etc preloaded in there for I, etc.

Huskerplowboy · Mar 25, 2009

Did you check with a straightedge or something to see if the bar sprung all the way back? Could be that it took a very slight set that was hard to notice.

Also, what grade is the steel you have, and is it hot rolled or cold rolled? Cold rolled bars have higher strength than their hot rolled siblings due to work hardening during the cold rolling process. Might account for the difference, but not up to 100ksi, that's getting up in the heat treated alloy steel range there!

MichiganIron · Mar 25, 2009

Here's what we got:

5/8" Diameter Rod - S (section modulus) = .0239685 in3

Momemt = 1.590 in-kips

fb= M/S
fb= 66.33 ksi

To still be in the elastic range on the stress/strain curve, your steel would have a tensile strength of 100ksi. Possible, depending on the grade of steel, especially in rounds. Round bars come in tensile stregths well over 150ksi.

Where did you get the stuff?

TomOfTarsus · Mar 26, 2009

Like a starved Chihuahua on a pork chop, I can't stay away from this one, mainly because it gives me an excuse to use Mathcad!

(hopeless geek engineer). I checked the math as well (it's spot on), and went on to calculate the deflection and slope at the end, which should be nearly 2-3/16" and 6.2ｰ (or .108 in/in) at the bucket handle. That's pretty noticeable, did you see anything like that? It seems to me that if you had the rod clamped horizontally in a vice, the bucket would want to slide off the end.

I'm not doubting your word - your tech skills are obviously strong enough. Actually, your attitude seems like mine - you want the numbers to reflect reality.

So like Michigan Iron says, you must have some mighty good stuff there, maybe heat treated 4140 or something. Unless, like Huskerboy says, it took a set and you just couldn't tell (easily).

Curiously yours,

Tom
(calc's attached as graphic)

Egon · Mar 26, 2009

Now add in the bending forces when you run it into the ground.

Brad_Blazer · Mar 26, 2009

What kind of feed was it? I just measured a 50# bag of sweet feed and it was roughly 26" x 16" x 4.25" = 1768in^3 = 7.65gal

It would never fit in a 5 gal bucket.

Mild steel can have a pretty wide range for yield strength.
McMaster-Carr

CraigM · Mar 26, 2009

We all get the same textbook stress, but it still doesn't add up. I repeated the experiment today, but measured the deflection of the rod at the bucket. Actually 1/4" from the bucket, but that should be close enough. The rod deflected 13/16" (.813) and sprung back exactly to the starting point, as best as I could measure. Tom, I didn't calculate the deflection, but your equation is the same one I would use, so we should get the same answer. Naturally, the experiment and the math don't agree.

Michigan and Husker, the steel came from the hot roll rack at the local supplier. They also sell A36 steel, which yields at 36ksi, and stressproof steel, which I believe is equivalent to 1040/1045. That has a yield of 42-45ksi. If one of these had found its way into the rack, it still should have bent as the calculated stress was almost 1.5x the yield. The ends are mushroomed a bit from using it as a long punch to drive a stuck shaft out of the tractor, so I doubt it's anything special or heat treated. And remember the other tests Iv'e done that calculate out to a little over 100ksi, but the rod didn't bend. That was a 1/2" rod that I'm sure is nothing special.

Brad, actually it was oyster shell for the hens. Sort of a feed. I figured that calling it feed would save questions about oyster shells. That page from McMaster is very interesting. I had no idea that you could get mild steel to that kind of strength. I assume that's by heat treatment. It still doesn't answer the problem entirely since heat treatable steel normally comes from the supplier in a soft state. And it wouldn't explain why I get less than half the deflection that the math predicts. Still, it is a great suggestion. I will call the steel place tomorrow and get some numbers.

So, we all get the same calculated stress, which is good news. My brain hasn't entirely turned to rust. But we can't explain the differences between the theory and reality. That part makes me nervous, especially with something so apparently simple. I don't believe that we are all doing the wrong calculations, so the only answer is that the steel is something fancy, but I doubt that too. This is going to bug me until I figure it out. As Tom said 'hopeless geek engineer'.

Thanks Everyone

henerythe8th · Mar 26, 2009

Here is the answer, mostly, to your question...

If you buy 'regular old' 36ksi structural steel keep in mind that the 36 ksi is the minimum yield strength. Making steel is an art, more than a science and making a batch that doesn't meet the strength requirements is expensive.

It is a lot easier and more cost effective to make a batch of steel that will exceed the requirements.
Years ago we were tearing up an implement on a potato farm, I can't even remember what it was. Calculations showed that a 5/8 -inch grade 5 bolt would fail and protect the implement. We ended up with 'no grade' 1/4-20 bolts with the threads in the shear plane, by experiment...

Now, you may have a different grade of steel than A36 but expect it to have a higher strength than 'advertised' as well.

TomOfTarsus · Mar 27, 2009

Good point and true, henerythe8th, but what is bothering me (and Craig, at least) is the fact that the deflections aren't anywhere near where they should be, and that is based on an entirely linear calculation (no "yield" or permanent bending).

With Craig holding his bar in a vice, it's hard to believe he'd measure less deflection than a built-in cantilever calculation, which of course assumes an infinitely stiff "wall" that the beam is attached to. That's difficult to duplicate in real life.

So, yeah, you can heat-treat plain ol' steel up to some pretty good yields, but for steel the elastic modulus is well established, so the deflections shouldn't be that far off, much less in the "too low" direction.

Are we to the point that "this thread is worthless without pics?"

This starved Chihuahua is still perplexed. Is gravity not working too well out your way, Craig?

Are you able to get the OD accurately (i.e not eyeballing from a tape measure)? I've missed diameters significantly that way, and this case is pretty sensitive to it - a 3/4" rod calc's at only 1.04" deflection and to match your .813 would probably be a 20mm rod (I hate metric). Just for giggles, wrap a piece of duct tape or paper around it, cut it with a razor at the meeting, unwrap it, measure and divide by pi. Unless, of course, you have calipers or a micrometer handy. & unless you are alrady sure of the OD, then we're back in the Twilight Zone.

Tom