Friction in Shear Failure 1

[Tigerdawg]

I am modeling a system of two pieces bolted and pinned together which are subjected to a shearing force. How would the pins and the friction forces interact with each other in failure?

I am assuming that there is 'slop' in the bolt holes so that they are initially only contributing to the friction force through tightening. The pins are a press fit so there is no movement before they fail.

Would you need a force sufficient to overcome both the shear strength of the pins and static friction before failure?

Or would the initial failure force be solely the static friction? If this, would the system still fail if the shear strength of the pins and the kinetic friction were greater than the shearing force?

 

[vonlueke]

> Would you need a force sufficient to overcome both the shear strength of pins and static friction before failure? Theoretically speaking, yes, for strictly static loading; however structures intended to be essentially static are rarely if ever analyzed, in static analysis, using static friction.

> Would the initial failure force be solely the static friction? No.

For static (quasi-static) loading,
(1) if possible, the most preferable approach is to conservatively assume the strength of the interface is equal to the shear strength of the pins alone;

(2) the next, less preferable method is to assume the interface strength is equal to shear strength of pins plus kinetic friction force, assuming a conservative value for kinetic coefficient of friction.

Item 2 is undesirable if there's any dynamic loading.

 

[ivymike]

For engines, the opposite seems to be true most of the time. In most designs that I've seen, pins are for locating and friction is for transmitting a load. The friction force is usually considerably higher than what the pins will take in shear, and the contribution of the pins to the strength of the joint is ignored (joint is considered to fail if the load overcomes the friction). In some special cases, where the dynamic loading has a large constant portion with a smaller oscilating portion, a slightly lower cover factor is allowed with the following rationale - if the joint slips, it will only slip in one direction, and it will only slip until it loads the pins, after which it will cease to slip. Thus slip need not be prevented entirely through friction.

Friction coefficients in joints like the above are often assumed to be 0.2 or so.

 

[ivymike]

I suppose it's worthwhile to note that if the pins will be transmitting a load, they will have to deform slightly before they will transmit that load (even with a press fit). In other words, there will be some (perhaps non-negligible) distortion of the joint before the pins contribute their fair share.

 

[Tigerdawg]

Thanks for the help. You've confirmed the worst. I hate having to assume a friction coefficient to calculate system failure. There are just too many variables to get a 'warm fuzzy' about the design.

 

[Lcubed]

Tigerdawg,

My favorite "Rule-of-Thumb" regarding friction is "Don't count on it if it would help, but assume that it will be there when you least want it to be.” I don’t think any co-worker has ever disputed this in my 40 years or so of aerospace stress analysis.

This is not just blind conservatism. You can’t know exactly what pressure exists between the bolted components when they are freshly assembled; after a few cycles, or a few months, the load which would overcome the friction between the parts can’t be rationally estimated, so it is best not to expect any benefit from friction.

As to the shear load on each pin, I suggest you calculate the torque required to twist the shaft through the deflection which you stated that you can calculate, divide that torque by the radius of the pin circle, and divide again by the number of pins. I’d also suggest a healthy factor of safety; suddenly applied loads can be tricky.

Also, you may want to compare the stresses resulting from your loads to the yield strength of the pins instead of the ultimate shear strength; if the pins are seriously deformed, substantial load may be transferred to the bolts, and if the bolts are likewise permanently deformed, they may be extremely difficult to remove from your assembly if removal eventually becomes necessary.

That’s about all that comes to mind based on your description and some assumptions on my part. Hope it helps.

Lin Lawson