short beam stresses 3

[electricpete]

I have something resembling a short cantilevered cylindrical beam (it is actually a pin inserted into a tight fit). A predetermined load is applied a short distance (perhaps one half diameter) from the fixed boundary. (hence the term short beam).

We have used a NEMA LI-1 G-11 polyester glass laminate pin and seen failure at the location where the pin exits the tight fit. There was a “shear lip” at the final failure point, which suggests it was ductile failure rather than brittle. The NEMA G-11 has relatively tensile strength roughly 30 KSI in both directions.

Let’s say I replaced it with a steel pin that has tensile strength 90 ksi. Is it reasonable to conclude that the inrease in strength if roughly x3? Or might it be more?

One aspect I was wondering about was the effect of Young’s modulus in this problem. The G-11 has Young’s modulus around 10% of steel or even lower. I know for a tensile specimen or for long beam, Young’s modulus is completely irrelevant when determining stresses in response to a predetermined applied force. Does the same apply for short beam problem?


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[rb1957]

one thing you need to consider is the effect of the increased E of the pin will have on the material around the edge of the hole.

 

[msquared48]

For short, deep beams, the shear modulus, G, comes more into play.

Mike McCann
MMC Engineering

 

[desertfox]

Hi electricpete

I think I would be tempted to just consider the shear of the pin, normally for shear a figure of around 60% of the UTS is the failure point, so you're 3x times figure sounds reasonable.

 

[electricpete]

Thanks for those comments.

one thing you need to consider is the effect of the increased E of the pin will have on the material around the edge of the hole.

the pin fits in a steel hole, so I'm not worried about that.

I think I would be tempted to just consider the shear of the pin, normally for shear a figure of around 60% of the UTS is the failure point, so you're 3x times figure sounds reasonable.

Thanks.

For short, deep beams, the shear modulus, G, comes more into play.

That's what I'm really interested in. For a short beam with predetermined applied force: E and G affect the deformation, but do they affect the stresses?

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[desertfox]

Hi electricpete

I don't think E or G will play apart in the stresses but they will play apart in the deformation.

http://www.roymech.co.uk/Useful_Tables/Mechanics/Strain.html

If you consider a beam the stresses calculated do not rely on E, however when deflection is considered then E is important.
Similar with shear stress, shear force divided by shear area give average shear stress, now consider deflection we introduce G.

desertfox

 

[electricpete]

If you consider a beam the stresses calculated do not rely on E, however when deflection is considered then E is important.
Similar with shear stress, shear force divided by shear area give average shear stress, now consider deflection we introduce G.

Thanks. I am familiar enough with long-beam calc to know that is true.

I just want to double-check if the same holds for short beam. I'm understanding you to say that it does.



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[desertfox]

Hi epete

Yes thats exactly what I am saying.

 

[electricpete]

Thanks all for responding to my basic question.
It was helpful to verify I'm not getting too far off-track.

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[msquared48]

Indirectly, but not directly, E and G can have an effect on the stresses in a beam in at least one circumstance.

If you have a cantilever beam that is loaded at the tip with both a compressive axial force and a force transverse to the axis of the beam, then any deflection of the beam laterally due to the transverse force will be amplified further by eccentricity of the compressive force. The more deflection, the greater the stress, and E and G relate directly to the amount of deflection. This is commonly called "Sidesway" in structural engineering and is common in moment frames.

Mike McCann
MMC Engineering

 

[dhengr]

Epete:

I generally agree with Desertfox, and I think most of what he said above is just fine as far as it goes, but I’ll add a few things to his ‘it is just a shear vs. shear area problem, at about .6(UTS)’ statement, and your understanding or assumptions about long beam theory being applicable.

My first shot at the problem would be, like his, (shear load)/(shear area) vs. some allowable shear or ultimate shear stress, and that’s fairly straight forward for homogeneous materials like steel, and that’s where his .6(UTS or YTS) comes from. But, for fiber reinforces polymers (FRP’s), your “polyester glass laminate pin,” you have a slightly different animal, it is not homogeneous or isotropic. And, off the top of my head I don’t know the relationship btwn. E & G, or if their really is one; this stuff just doesn’t act the same way we think of steel, etc. But, on the plus side, if you really understand the true stress situation in your pin, you can lay up the fiber reinforcement to its best advantage to take these stresses. And, in that case the fibers take all (the majority) of the stresses as tensile loads, when oriented properly.

Next, our standard long beam theory and methods really do not apply in your situation. The stress picture that we normally think of really goes to hell on short beams and beams with applied loads within one or two times their depth or dia. (your pin) from their supports. A basic assumption of our std. long beam theory is that we can ignore this situation under normal conditions and loadings. But, your’s is that exception... can’t ignore the true stress picture. Then you would also have to consider the combined stresses, in the pin, from your press-fit of the pin, and the potential that this process might cause some stress raisers on the pin surface. One of these notches, nicks, stress raisers, along with the high combined stresses is the starter of the failure mechanism. Look in some good Strength of Materials or Theory of Elasticity texts for some of this subject matter.

I would not go to these lengths unless I was really trying to prove something, but these issues are worth keeping in mind as you address your problem. My first thought, not knowing the pin dia. or actual materials, would be to wind a few of the outer layers of fibers at 45° to the axis of the pin, and at 90° to each other (or in opposite directions) then cover that with a few final layers of the resin. This will most likely strengthen the pin against the kinds of stresses causing the failure. You are probably just using a pultruded pin with all the fibers oriented along the axis of the pin. And, this would not be particularly strong or consistent w.r.t. shear failure.

 

[electricpete]

More to think about - thanks.

The application is an anti-rotation pin on a motor sleeve bearing. It is a 5/16" diameter 1" long pin with a tight fit into the housing, and then sticks about 3/8" into a slightly larger bearing hole.

The objective is to prevent rotation of the sleeve bearing, which screws up the oil ring, which screws up the lubrication, which screws up the bearing (obviously) as we observed.

The most obvious stress on the pin should be associated with friction between shaft and bearing which tries to spin the bearing (stationary housing restrains it via pin). During start, before the oil ring gets moving, the friction is much higher.

The end of the pin and the bottom of the bearing pin hole are both rounded, so I can imagine this can translate into some compressive stress.

I don't remember seeing any obvious rounding of the hole in the housing, could be a source of stress as mentioned (depending on how tight the fit is). Maybe there are some hardness parameters of the pin that are important here as well? (imagining a sharp steel corner of the hosuing hole sticking into the low-E G11 pin).

Attached is a photo of failed pin. Does this give you any clues as the the nature of the failure or the type of stresses that caused it?





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[electricpete]

By the way, we are aware of insulating requirements for the outboard insulated bearing bearing of this non-vfd motor. We would be installing the steel pin into the inboard bearing position which is not at all insulated from the housing except for G11 pin, which does no good when remainder of bearing is uninsulated.

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[dhengr]

Epete:

That’s just a cleaving shear failure and it seems you can even see the curvature (shape) of the sleeve bearing OD and the housing ID mating surface. I’ll bet the land or lip I think I see in the lower right (5 - 6 o-clock) on the pin was actually on the outboard end direction of the sleeve bearing, where you might expect a little tensile stress rather than just cleaving shearing. Although, I don’t know where in the length of the bearing the pin is located, I am thinking of the deflection or movement of the sleeve bearing w.r.t. the housing ID, however small. You can certainly see the orientation of the fiber lay-up in (with) the length of the pin. And, that material is not particularly strong in shear across the grain, it is kinda brittle. It’s considerably stronger in tension (normal bending stress) parallel to the fibers. Are these pins actually made from rods or are they machined out of a block of G-11 material?

The obvious question, can you make the pin dia. bigger? Ask pin suppliers about best orientation of fiber reinforcement strands for this kind of shearing loading and failure. Or, best type of fiber reinf’g., or type of resin for this condition. Draw a full size end view of the sleeve bearing, housing and bored holes for the pin. I see the 3/8" dia. hole in the sleeve bearing, out at the bearing OD, acting like a cleaver on the pin, every time the motor stops, starts or changes rpm, due to a fairly small relative rotational movement btwn. the two parts. That oversized sleeve hole should be cone shaped so it bears on more of the pin length, and the corner should have a small, smooth, radius on it. The hole in the housing should have a small radius on it too, so your are not driving the pin into (trying to bend it over) a sharp edge. You might question where is the best location (least abusive) for this pin to do its job? Or, should the pin be pressed into the sleeve or into the housing to best tolerate and absorb the relative rotational motion btwn. the two parts. I don’t quite understand the insulation issue, you already have a metal sleeve bearing on an iron or steel housing, so why not a steel shear pin?

 

[electricpete]

That oversized sleeve hole should be cone shaped so it bears on more of the pin length

What does that buy us?

I have in mind from long beam theory that spreading the applied force from a point load to a uniform load-per-length of short portion of length does not change much, except to the extent it moves portion of applied force closer to the fixed end slightly, decreasing moment arm, and decreasing moment at the fixed end. Anything different for short beam? Or some other reason?

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[electricpete]

Rethinking the cone-shaped hole: it seems a cylindrical load would potentially share load along length of the pin. A cone-sized hole would apply load only at the end where the pin contacts the cone. Right?

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[dhengr]

Epete:

Reread my post with a little more detail in mind, I do try to make all of my words and thoughts mean something. I don’t have a pat answer to your problem, I’m just throwing out all the ideas I can as I think about the problem and the materials involved, so work with me, and dig a little deeper for my meaning. “ ” Draw a full size end view of the sleeve bearing, housing and bored holes for the pin. I see the 3/8" dia. hole in the sleeve bearing, out at the bearing OD, acting like a cleaver on the pin, every time the motor stops, starts or changes rpm, due to a fairly small relative rotational movement btwn. the two parts. That oversized sleeve hole should be cone shaped so it bears on more of the pin length, and the corner should have a small, smooth, radius on it. The hole in the housing should have a small radius on it too, so your are not driving the pin into (trying to bend it over) a sharp edge.” “

Take a piece of your pin material and wack it a few times with a hammer and a cold chisel, both sharp and dull chisels, and see what happens to your pin material. I’ll bet it cracks, crushes and shears. In some fashion or another the oversized hole in your sleeve bearing is doing this same thing to your pin in the housing hole. It’s acting like a chisel on your pin material, and with some fatigue and fracture mechanics in the picture too; and some constant chisel like wear on the pin material from vibration, change in rpm, start and stop, etc.

The sleeve bearing must rotate to bring the pin into play. That is 3/8" dia. straight bored hole, and 5/16" dia. pin; or {(3/8) - (5/16)}/2 = 1/32" clr. all around the pin in the hole. Knowing the OD of the sleeve brg. we can find what rotation will be required to take up this clearance, maybe a half degree or one degree or so; and this will cause the sharp edge of the 3/8" dia. hole to act like a chisel on the pin. The cone shaping of this hole would only be .3751 or .3752 dia. to .375 at 3/16ths up into the hole; to stop this chiseling action. Given the way this FRP material works, you might actually be better to make it act a little in bending, because it is certainly stronger w.r.t. bending normal stress parallel to the fibers than it is w.r.t. shear across the fibers. The way you have this detailed right now, you don’t have a short beam bending problem, it’s strictly being crushed and sheared across the fibers.

There is a Hertz bearing stress btwn. the hole of one material and the smaller pin of another material, and this plus the chiseling action of the sleeve rotation will lead to some very high and concentrated bearing stresses on the pin. And, that pin material doesn’t yield in bearing to conform and redistribute the bearing stress over a larger area, it just tends to crush and crumble. There may be some advantage to a steel shear pin, in that, it would better match the mechanical properties of the sleeve bearing; and the sleeve bearing could not abuse the pin so readily. I think I would sooner have the pin wear a compatible shaped hole in the sleeve bearing if I had my druthers, since this really isn’t a shear pin anyway.