Clevis strength??Same old thing but differant??

[heloman78]

I'm working on a control rod that has two identical clevis type rod ends screwed in to each end of a tube. What I'm trying to work out is the ultimate tensile stress the clevis ends can take before the bolt (pin) tears out. This is a little different than typical clevis/double shear calculations because the high strength steel bolt (pin) is going to tear out of the aluminum clevis before it ever gets close to (double) shearing bolt that runs through the clevis. So I’m wondering how do I go about this, is this simply a pin tearing through a boss, times two (two "ears" of the clevis)? Is it shear strength of the material (KSI) times the cross section area (inches) of both ears of the clevis? This rod is mounted in a push-pull configuration so the load is in perfect tension with the control rod's major axis and normal to the clevis pin as in typical double shear calculations. Or in other words, picture the typical clevis/pin double shear type problem in reverse; the pin will not fail before it tears out of the clevis. I hope I explained this clearly; can anyone point me to an example?

 

[rb1957]

check Bruhn, section D ... shear tear-out of the lug. his method doesn't expressly cover your concern, but he shows that one solution covers all lug failures.
Niu almost certainly covers it too.

 

[heloman78]

Thanks RB, I just don't think I'm looking at this right.

 

[rb1957]

a clevis is "just" two lugs, each reacts 1/2 of the applied load. most people will assume a 60:40 split between the lugs (accounting for some variation of loading of geometry in service) and so design the lug for 60% load.