Cantilever press fit shaft/dowel pin bending 1

[yacpro13]

Hi everyone,
quick question for you:

Imagine we have a 1/2" diameter dowel pin/ground shaft, 2" in length, with the first inch press fitted in a steel 'immovable plate', and with the other inch horizontally sticking out, a la cantilever beam.

Assume the dowel pin carries a certain load.

Dowel pins are usually checked for shear failure, and bearing/tear-out failures are also checked. However, how do we properly take into account the potential hole deformation due to bending?

I mean, imagine that instead of being 2" in length, the dowel pin was 6" in length (yet still only press fitted on the first inch), with the same load.
Or, alternatively, imagine that the 'immovable plate' was made of plastic instead of steel.

Certainly the hole's edges would deform (bottom on one side, and top on the other side).

This struck me today while I was doing a dowel pin calculation at work, so I figured I'd start a discussion.
Is this essentially covered by checking for bearing stress? This was not evident to me.

Thanks.

 

[desertfox]

Does the dowel pin stuck out carry any load or only the bit between the components? otherwise I don't see any problem

 

[rb1957]

i guess you could distribute the load along the protruding part of the dowel, and so get bending about the base. however, i'd expect that as the dowel yields, then the load will redistribute closer to the base, reducing the meoment. that said, i'm sure there's some bending in the dowel, and for typical dowels it's not critical (compared to shear), but for atypical dowels (very long ones) i guess you should check it.

Quando Omni Flunkus Moritati

 

[TVP]

I get the impression that then OP is concerned about local deformation of the hole, not bending stress/deformation of the dowel. My answer to this question is to evaluate the stress in the hole using an appropriate calculation or FEA. The distribution of the force about the hole will not be uniform, so the resultant stress is not purely of the "bearing stres" type.

 

[drawoh]

yacpro13,

I load dowel pins in shear, only. The easiest solution to your problem is continue loading dowels in shear, only. Press-fit pins are pushed into the hole, as opposed to directly clamped by screws. I would not regard them as all that secure in anything other than shear.

If someone was determined to load a press-fit pin in shear and I had to analyse it, I would work out the bending moment at the base. I would work out the forces at the base required to oppose the bending moment, and subtract them from the local press fit.

--
JHG

 

[yacpro13]

Thanks for all of your inputs.

As TVP pointed out, I am more curious about the local deformation of the hole caused by loading the press-fit dowel pin in bending.

Drawoh:
What do you consider "shear only"?

As mentioned, if you have a 1/2" dowel pin press fitted on the first 1" and exaggeratedly sticking out 4" and carrying a distributed load over the 4", shear failure can easily be checked. However, what about the fact that the equivalent load is effectively acting 2" from the shear surface?

This causes some bending, which could result in local deformation of the hole's edges.

Thoughts?

 

[drawoh]

yacpro13,

Shear means that the sideways force is exerted at an elevation of zero. The moment load on the dowel is not significant.

In machinery, dowels are locating devices. If you want to apply sideways force, attach screws, and tighten them down.

--
JHG

 

[rb1957]

yes, but ... the clearance on the dowel is usually less than of threaded fasteners, so it'd catch the shear before the fasteners would. we'd term it a shear spigot.

Quando Omni Flunkus Moritati

 

[drawoh]

rb1957,

If you tighten the bolts down hard, the pieces will be retained by friction contact. If you are doing crash safety analysis for an aircraft, and they do not trust friction contact, you can still assume that the bolts have to shear before everything fails.

--
JHG

 

[Tmoose]

RE:Clamping friction from fastener torque VS neighboring dowels as locators and load handlers.

Dowels have their place, but relying on them when loads are impulsive and frequent is asking too much.

Attached is an image of the flywheel flange of the crank of an aircooled 4 cylinder Porsche. From the factory the flywheel attachment is eight 8 mm hardened dowels pressed into the crank and snugly fitted in the flywheel. (Stock aircooled V-Dubs use foour dowels) The flywheel is retained by one big bolt (Porsche and VW call it a gland nut) engaging the threads in the crank, and tightened to some hundreds of lb-ft.
After a clutch job if the bolt is not properly tightened the common result is the flywheel and crank get beat up as shown in the picture.
In the 70s and 80s we did a fair amount of repair of damaged VW and Porsche flywheels and cranks, and upgrading VWs to 8 dowels. So tightening the gland nut properly apparently was not a trivial task.
A VW legend, the late great Gene Berg was finding that at some level of power and abuse even a properly assembled 8 dowel flywheel would loosen, so co-developed an attachment method the added friction via a wedging taper as a much more secure method.

http://www.geneberg.com/article.php?ArticleID=207

VW pretty much abandoned the dowel thing on the type 4 aircooled 4 cylinder.

http://www.914world.com/bbs2/uploads_offsite/lh6.googleusercontent.com-15592-1370559193.2.JPG

Conversely when Chrysler converted their 60s big block into the 426 Hemi they merely added 2 bolts to the existing 6. The holes in the flywheel are sized with clearance and the bolts are conventional without fitted shoulders. Pure clamping friction from a handful of (boo, hiss) torqued fasteners. No dowels, and NO loosenening, shuffling or fretting problems when abused by decades of hoodlums of all ages.

 

[rb1957]

yes, we don't trust friction, and yes, we size the fasteners to carry the shear; and if we've got a dedicated shear spigot, it'll react shear first (the day-to-day loads).

Quando Omni Flunkus Moritati

 

[Tmoose]

Looks like I failed to attach the picture of the crank wallowed-by-dowels.

 

[MikeHalloran]

Conversely when Chrysler converted their 60s big block into the 426 Hemi they merely added 2 bolts to the existing 6. The holes in the flywheel are sized with clearance and the bolts are conventional without fitted shoulders. Pure clamping friction from a handful of (boo, hiss) torqued fasteners. No dowels, and NO loosenening, shuffling or fretting problems when abused by decades of hoodlums of all ages.

I don't know about Chryslers, but on the few Chevys and Fords I've wrenched, the flywheel bolts are not exactly conventional. The heads are rather thin, further thinned by a deep axial relief at the shank intersection, and sort of divided into pie sections by deep forged radial grooves on the exposed head face. It appears that the heads themselves are intended to deform rather like a super-stiff Belleville spring when torqued hard, as specified.

Further, the flywheel faying surface on the crank is typically Blanchard-ground around the flywheel pilot, probably to a minimum specified rms finish so it effectively has quasi-radial micro-teeth that Brinell their way into the softer surface of the flex plate or flywheel when the magic bolts are torqued.

I don't think it's quite accurate to describe the joint as totally friction driven.



Mike Halloran
Pembroke Pines, FL, USA

 

[yacpro13]

This is all very interesting info.

I gave a bit more thought to the initial question, and I think it boils down to dowel pin engagement length.

The machinery handbook suggests that a dowel pin should be engaged in both parts by a minimum length of 1.5 to 2 times the diameter. This is pretty much how dowel pins are used usually.

However, for the purpose of this discussion, would you agree to the following statements:

-with an engagement length of 1.5 to 2 times the diameter in both parts, it can be assumed that the dowel pins are loaded in shear only.
-if the engagement length is greater than 2 times the diameter, but is the same for both parts, again, it can be assumed that the pin is loaded in shear only.
-if the engagement length is significantly different in both parts (ex. 2 times the diameter in one part, and 10 times the diameter in the other part - exaggerated for the purpose of discussion only), then it cannot be assumed that the pin is loaded in shear only and bending should be considered.
FEA would need to be performed in order to analyze the local deformation of the hole caused by the bending of the pin.


I think this covers it.
The whole reason this came up in the first place is because I saw a design proposal go across my desk, where two press fitted dowel pins were installed in a reaction plate, and the pins needed to resist a specified torque.

 

[racookpe1978]

But "hole deformation" is going to be a function of 3x things:

The relative strength of "dowel material" vs base "hole material" AND attached "hole material",
The assumption of no movement of the sliding plate away from the fixed plate,
The assumption of a very closely fitted dowel into both holes.

Only if all three material are near-identical in shear will the normal shear calculation apply: A lead dowel inserted into a tight hole between two steel plates will deform the dowel at lead shear stress under normal fitup tolerances - BUT, if the two plates are "perfectly" pressed up against each other and cannot be pushed away from each other to give room for the lead to "squish" and deform, there will still be resistance to movement by the lead.

If a steel pin is used under a cheap wooden shelf, too heavy a bookshelf load will deform the pressboard shelf AND the hole in vertical wall of the bookshelf. The steel dowel, on the other hand, will not be affected and will not bend, even if the wall were oak.

A "perfectly fitted" dowel in one hole will bend differently if the second hole is "loose" and the dowel longer (bigger cantilever effect) but if there is no room to deflect, then there will be little cantilever force.

If both holes are loose, and the dowel material is stronger than the base materials, the force will be trying to egg-shape both holes! A hard dowel into a softer material is going to change resistance as it deforms the hole: Initial pressure will be focused more than later pressure: The deformed hole will also deform in length (down the hole) as well as diameter, so contact area increases with the wear. Movement slop also increases with that wear of course!

Note that a vertical cantilever force is going to be focused on the "top" of the hole near the surface, and the "bottom" of the hole deeper into the material. Local stress variations could be important - depending on the material (wood or concrete may have local voids and hard spots, good metals will be more uniform.)