Pin Design

[ayoung802]

I found most of my answers on here, but one remains. When talking about shear pins .577 of the Ultimate Stress is used for the failure point. I'm specing a pin for a hydraulic cylinder, but I don't want it to fail or yield. Does that mean I should use .577 of the yield strength?

Thank you in advance.

 

[BobM3]

For steel, the shear yield stress is about .58 of the normal yield stress. However if there is fatigue involved there may be other things to look at.

 

[TVP]

Take a look at these previous threads:

thread404-155277

thread404-148616

 

[ayoung802]

Thanks for the responses and I did see one of the posts already. Looking back over it let me correct my earlier statement to .577 of Yield strength.

However I guess what I'm asking, is there a Yield Shear Strength or do pins not yield in shear? The cylinder clevis is only supported on one side and I don't want it to get bent under load.

Thanks again hopefully this helps clerify my question

 

[TVP]

Yes, there is a shear yield strength, and it is theoretically 0.577*TYS where TYS is the tensile yield strength. If you don't want the pin to bend, definitely keep the shear stresses below ~ 1/2 of the tensile yield strength. Keep in mind that shear applications can still have bending present, which complicates things.

 

[ayoung802]

Thanks TVP I just wanted to make sure that is the correct way to deal with the shear yield. I have calculated the Shear Force and the Bending Moment.