Do you have to consider bending in pin design?

[Andrewstructure]

I am designing a pin in double shear. I have (2) 3/4" plates with (1) 1 1/2" plate in between them with 1/16"-1/8" gaps in between the plates.

It would seem that all of the force is transferred in shear and bending in the pin would not need to be considered. Another engineer says that I have to take the moment at PL/4.

 

[slickdeals]

In reality you would have something closer to wl^2/12 and wl^2/24 and not PL/4. I don't think traditional flexural analysis would be appropriate, the behavior would be analogous to a deep beam.

 

[CELinOttawa]

Bending is a question of geometry: Is there opening or slack to permit bending? If so, it may occur and must be considered as a possible failure mode. If the holes which apply the double shear are all tight, bending cannot occur and you don't need to consider it.

 

[rb1957]

just i would include bolt bending. for my industry it more a concern when repair doublers are used, increasing the length of the bolt.

PL/4 is very conservative if L is the overall length clamped up (ie 3" in your case). i would usually say that the P in the central plate is 2 forces P/2 at 1/4 (and 3/4) thickness and the outer plates react P/2 at mid thickness. now the moment is P/2*(L/8+L/8) = PL/8 .... "L/8" for the outer plate 3/4" thick = L/4, so mid thickness = L/8 and this is conservative, you can model the shear across the thickness as a triangle and get a smaller moment arm (1/3 instead of 1/2); and this is real, as the load increases, as the bolt gets closer to yielding, the shear concentrates at the interface (ultimately the moment might be zero (!?).

Quando Omni Flunkus Moritati

 

[SAIL3]

I strugled with this for many years....on one hand, theoretically, there could be bending in the pin, but on the other hand, I have a difficult time visualizing it....so, in the end I addressed it by tightening up on the tolerances..as Cel pointed out, keeping the hole size @ 1/32" greater than the pin size and the gap between the plates @ 1/16"...in reality, one would be shocked at what happens in the field where any available shackle may be used if it fits so I try and control what shackle , clevis and pin they use by keeping the tolerances tight....

 

[rb1957]

does this help ?

Quando Omni Flunkus Moritati

 

[bkal]

Suggestion by rb1957 is same as in the Eurocode 3 (EN 1993-1-8), Cl. 3.13.2. Bending capacity (resistance) of the pin depends on wheteher it is intended to be replaceable or not.

 

[Andrewstructure]

I ended up using a load distribution similar to what rb1957 posted calculating the moment as M = P/2 * gap + (t_centerlug/2*P/2)/2.

This gave a realistic bending moment.

 

[rb1957]

i think that's a little light ... P/2 is the force creating the moment, so the moment arm you've got is gap + center thk/4 ...
i'd add outer thk/2 ... the arm from the middle of the outer plate to the 1/4 thk of the center plate ...

clear as mud ?

Quando Omni Flunkus Moritati

 

[dhengr]

Andrewstructure:
Normally you don’t apply any bending calcs. or bending stresses to the pin design. But, at the same time, they can’t always be completely ignored either. We need much more info. here, to draw the full picture of your detail. What’s the pin dia. and mat’l., and the hole dias. and the pl. mat’l? The loading? Side view of the pls. with dimensions would be helpful too. Is the inside clear dimension btwn. the .75" plates, 1.5" plus 1/16", 1/8" or 1/4" (2 * 1/8")? Your description would allow any of these, and you want the max. possible gap to be as small as possible, to minimize the potential for bending. You also want the pin dia. and the hole dias. to be as close as practical; this tends to minimize potential for pin deflection due to bending, or limits bending; and the bearing stresses btwn. the pin and the plates is very much dependant upon this snug fit. Look up the Hertz bearing (contact) stress problem, and see how Dpin vs. Dhole affect the bearing stresses. I’d look at the pin as though it was a cantilever semi-fixed at the middle of the 1.5" pl. But, almost immediately it is supported by a triangular bearing load which becomes max. at the gap; then it is loaded by a triangular loading (bearing force/ unit length) which is max. at the gap and decreases as you move into the side pl. That’s the general bending picture, putting a number on how much bending actually takes place involves some engineering judgement.