Help with cantilever beam stress

[CraigM]

Tom:

Here are some pictures of the experiment.

I repeated the experiment with different setup that allowed me to measure very accurately. Caliper says that rod is .627. Interesting that the end that the bucket hangs on measures about .635 for the first half inch or so. Mushroomed from the previously mentioned use of that rod as a punch. I tried to take pictures of the caliper readings, but couldn't convince the camera to focus on the caliper. I used the caliper depth gage to measure from the top of the bar to the top of the rod. By keeping the depth rod on the caliper against the end of the bar, I got a very repeatable setup. The measurement was about .25 behind the bucket handle.

When I first set it up, the bar was lying flat, and the weight of my hand deflected it when I tried to measure. The first readings were deflection of .761, and return of only .744. This may have been due to my hand deflecting the bar, or it might have yielded. Could lso have been the rod settling into a position. The clamp on the angle is far from ideal. It took very little force to spring the bar measureably. I put the bar up on edge as in the picture and was able to repeatedly load and unload the rod. The deflection was .759 each time, and each time it came back to the starting measurement. I did not reposition the rod at any time. What happened to 7/8 deflection? Good question. Last time I measured it with a tape measure marked in 32nds. Seemed like I was getting good measurements, but maybe not. Still, 1/8" seems a lot of difference.

If the rod yielded at a calculated stress of 66ksi, I'd expect something that left no doubt. Is it yielding slightly and workhardening? The rod checks as straight as it ever did. It has a few slight wiggles, but always did.

The clamping setup is as close to cantilever as I can get. I was trying to remember shear/moment diagrams because I guess the setup could also be considered simply suported at the clamp with a balancing force at the end inside the square tube. If I remember it right, the shear/moment diagram would be the same as far as the cantilevered end goes, and would not change the analysis.

 

[Farmerford]

I worried about this same thing many years ago, and came to the conclusion that we are using the wrong neutral axis; put differently, we assume that the neutral axis is at the center of the cross section and therefore that the distance from the neutral axis to the extreme fiber is 1/2 the diameter for a round, and 1/2 the height for a square or rectangle. But the "real" neutral axis is not in the center; it is actually located a very small distance from the bottom surface.

See the attached sketch. The top example is the typical textbook formula for section modulus that assumes that the neutral axis is in the center. But the neutral axis is defined as the point where the fibers are in neither compression nor tension. This first example assumes that when the bar bends, all the fibers above the centerline are in tension and the fibers below the centerline are in compression. If that is true, then the neutral axis is through the center and the formula gives the correct stress for the extreme fiber.

But metal does not deform like example one assumes. If it did, the bottom surface of the beam would compress exactly the same amount that the top surface stretched; otherwise, the fibers in the center woud be stretched or compressed. Said another way, the example assumes that the two ends of the bar (that is, each side of the cross section) pivot around the center when it bends.

But experience teaches that metal does not bend like that, because it does not behave the same way in compression that it does in tension. Since metal is virtually incompressible, almost none of the fibers in the bar are in compression except those at the very bottom surface. All the other fibers are in tension. In effect, when it bends the two parts of the bar rotate (relative to each other) around the bottom surface (more precisely, a point just inside the bottom surface, since at least a few fibers on the lower surface must be in (very great) compression to maintain the bar in equilibrium).

Therefore, the neutral fibers, and therefore the neutral axis, are for all practical purposes on the inside surface of the bend. Thus the distance from the neutral axis to the extreme fiber is actually the height of the bar rather than one-half the height and the section modulus is thus doubled.

Since the stress is torque divided by section modulus, when the section modulus is doubled (ie, when the neutral axis is the bottom surface) the stress is halved. And naturally, the deflection is half or even less. Compare the two formulas on the attached sketch.

The enclosed example tries to demonstrate these concepts.

Thus when designing using the assumption that the neutral axis is in the center of a beam we are (perhaps unknowingly) adding a safetly factor of about two.

These concepts are clearest to me if I consider the wide flange (or "H") section (often mislabeled as an "I-beam"). Assume it is welded at one end to an immovable wall, and that we load the other end until it finally deforms and collapses. We know that it will break at the wall since the torque is greatest there. And we know from experience that when it deforms it will do so by elongating at the top flange and then down the web, and that neither the bottom half of the web nor the bottom flange will be compressed into a shorter space. That is because the neutral fibers are not in the middle of the web, but rather at the bottom outside surface of the bottom flange.

You get the same result if you consider suporting the section on a pivot in the center and loading each end equally until it breaks in the center.

Many reference books, such as Machinery's Handbook, give formulas for section modulus assuming the neutral axis is in the center (actually the center of gravity) and also assuming it is on one surface.

I typically use the formulas with the neutral axis on one surface and then incorporate a generous safety factor in the final sizing. It is easier for me to make all the calculations as accurately as I can and then apply the safety factor in one operation rather than to try to build in a small safety factor at each stage.

 

[CraigM]

Farmerford, I've never seen an I-beam fail, but I have bent steel from time to time. It does seem to bend like you describe, mostly stretching on top and not changing much over the anvil.

As far as applying that logic to designing a beam to make it strong enough, you are bucking at least 100 years of proven engineering experience. If I remember correctly, the neutral axis is the plane where the monent of inertia (I) of the material above is equal to the I of the material below. Said another way, there is as much strength above as below the neutral axis. For a round rod, that has to be the center.

The big thing with the equations in Machinerys and the like is that they only apply up until the material yields. After that, you are no longer looking at elastic deformation and all bets are off. Since the reason for doing all of this is to find out how much load a bale spike can carry before it yields, the equations should work. Granted, they don't seem to be working, but they use those equations to design airplane wings and bridges, so I must be missing something.

The classic example to demonstrate the tension and compresson in a material behaving elastically (not yielded yet) is the marked eraser. Take a big rubber eraser, Draw a line down the center (this is the neutral axis) and a bunch of crossing lines. When you flex the eraser, you can see the tension in the top by the crossing lines moving apart, and the compression below by them moving together. The continuous change from max tension on top to max compression on bottom sets up shear stresses through the material.

 

[Brad_Blazer]

Your ruler says L = 20". You had it at 30" in post #1. Looking at your c-clamp, the effective length is more like 21"

Using a modulus of 30,000,000psi which is independant of temper, I get a deflection of 0.73 - not too shabby.

I used D = 0.625. There is probably some mill scale on the rod.

 

[CraigM]

I can hardly see to type because of the bright red glow from my face. Guess I win bonehead of the week. Post #1 ended with 'can someone tell me what I'm missing? You did. I have 30 stuck in by brain as the first guess for length of the spikes on the fork.

I agree that the clamping setup probably increases the effective length of the experiment. Using the proper length in the calculations at least gets the experiment and the theory in the same ballpark. If we assume that the effective length is 21" as confirmed by your deflection calculation, that gives a stress of a bit over 46ksi. I still have a little trouble with that one, but given what I learned from your link to McMaster, and what Henrythe8th said about steel making and meeting minimum yield specs, it's possible.

I'm going to re-test the other pieces that gave me such high stress values. This setup isn't great as far as the clamp, but the old pipe vice was worse. It was just quicker than cobbling another rig together.

 

[keeney]

The clamped end of the bar is not as precisely fixed as you might think it is. Even though you have it clamped so it can't deflect vertically, it can still move and thus stretch longitudinally. The additional length of the bar extending inside of the square tubing is distributing some of the stress as that steel is undoubtedly also stretching along the top cord. More of the length of the steel is stretching than your calculations are accounting for, so the amount of strain per unit area is less than you think. By crude estimation, you probably have an error factor of 1.6 up to a theoretical max of 2x due to ignoring the portion of the bar inside the square tubing.

You will get better results modeling it as a combination of a point load at the end, another point load at the clamp, and a tapering continuous load along the inside of the square tubing offsetting the other loads.

Think of the different result you would get if the rod was tightly fitted through a press fit hole in a 1" plate and plug welded (ignoring the heating issues) on the other side to completely immobilize it.

- Rick

 

[Brad_Blazer]

Ha ha we all make "dumb" mistakes. It's a good thing you posted such good pictures.

If you clamp the bar to the top of a flat surface so it comes off a good sharp edge I think you'll make a pretty good cantilever. Your tall c-clamp will work a little better too grabbing a thicker bite. Add another c-clamp further back to control the bar.

It's not all that magic. A center point support with a "balance beam" load on each end follows the same deflection path as a cantilever beam of half the length.

With a 50# share of the bales in a load distributed over 36", the rod you are testing seems just strong enough. The bales won't bend the rods but they probably will bend when you dig them into the ground or dump too close to the ground, a pile etc. I would use 1" square solid HRS for that application. If you have a supply of the 5/8" stock cheap/free you could weld 3 together in a triangle to make a robust spike. I'm sure you'll want to use it for moving brush, etc. so you might as well over-build it.

If you go with the 5/8", at least you'll be able to straighten them when they do bend, and they'll work harden to boot.

Brad

 

[TomOfTarsus]

Craig:

You win an offical "Tom award" (guess who it's named after). Kinda like when you go to measure "exactly" with a tape and you hold it on the 1" mark and then go back and forget to knock off the 1".

Looks like you got this one. Sorry I got back to the thread late, I've been out taking a half-dozen felled mulberries apart all weekend, and I'm just setting to to design my Carry-all box.

Glad to see tha gravity & physics are still working out your way, you had me going for a while!

Best,

Tom

 

[CraigM]

My rationalization for the crudeness of my setup was the balance beam concept that Brad mentioned. As more of the rust gets knocked off my memory, I realize that the problem with this is that there would have to be as much rod inside the square tube as outside, and it would have to only touch the inside of the square tube at the clamp and at the end. Possible with a little blocking, but not what I have. I tried the experiment with a piece of A36 flat stock, and it behaved predictably and took a slight set at right around 36ksi.

A toolmaker I once worked with had a poster that read...'There comes a time in every project where you just have to shoot the engineers and get on with the project.' I think that point has arrived for my project. Thank you all for your patience and advice...and the Tom Award.

 

[MichiganIron]

When you calculate the deflections, what are you using for the Modulus of Elasticity? Are you using 29,000 for E, or a modified value.

As long as we can all calculate the Moment of Inertia properly, the only two variables are:

Modeling assumptions
Modulus of Elasticity

We calculate deflections for arena roof trusses pretty regularly. Actuals are generally pretty close, although those deflections are from Work-Energy calculations.

Sounds like it's time to step back and re-evaluate the big picture. Does the mathmatic modeling match whats being done in the lab? If it does, the modulus of elasticity can be evaluated from the measred deflections. We could then tell if we all flunked mechanics of materials 101.