Confussed
Confussed
(OP)
We are inserting a dowel pin thru a hole into a plate. The pin to the hole has an interference fit. We have noticed that at times the pin will split the plate during insertion.
While trying to model the assembly to redesign the plate I have come across something that I can explain.
If you look at the compressive stress of dowel pin during insertion if should fail but it does not. The pin is short so buckling does not enter into the picture.
So how do you explain a stress 6 times greater than the ultimate stress of the material and does not fracture?
While trying to model the assembly to redesign the plate I have come across something that I can explain.
If you look at the compressive stress of dowel pin during insertion if should fail but it does not. The pin is short so buckling does not enter into the picture.
So how do you explain a stress 6 times greater than the ultimate stress of the material and does not fracture?
Talk To Other Members
RE: Confussed
1) What element types are you using for the pin and what are you using for the plate?
2) What material models have you specified for each?
3) If you are using a material model requiring some type of defined coefficients, what are they?
4) Is the pin entering the hole? Penetrating some distance? Exiting the back?
5) What failure method are you using to suggest a stress level of 6x the yield?
6) When you say the pin "splits", do you mean as though it shears in half (splits down the middle)?
7) What materials are the pin and plate made of?
8) Is the pin cut along a particular direction of the grain boundary? Or is it random?
Any chance you can post an archive file on engineering.com?
Garland
RE: Confussed
I am just concerned with the compressive stress in the dowel pin alone right now. We are using 12723lbf (1800psi on a 3" bore cylinder) to insert the pin. The pin never fails but if you look at the stress it would suggest that it just "blow" apart.
It is a standard heat treated 1/8 diameter dowel. So roughly an applied load of 12723 lbf over .012^2 inches = 1060250 psi about 4 times the tensile for a dowel at 58 Rc.
I have not modeled anything as of yet just trying to understand my hand calculations first.
RE: Confussed
You are trying to apply a tensile calculation to a non-tensile situation if I understand your scenario correctly. You are pushing a pin in to a hole, but you aren't pushing it against an immovable stopper. The tensile calucation that you are trying to use is for beams that are fixed on one end and loaded along the axis, but your pin is allowed to move...only restricted by friction.
The circumferential load on the pin is based on the load caused by the interference fit. The hole will relax some and the pin will be compressed some. The curcumferential pressure due to the interference will load the pin.
Not sure if any of this makes sense, but you may be able to find more information by looking for stress in rivets. They behave in much the same manner.
By the way, with a pin, I thought you were typically more concerned with shear stress. You typically select a pin material that is much stronger/harder than the material that the hole is in and you are usually trying to increase shear strength between two plates. Just an observation.
RE: Confussed
Mark's Handbook also has good information on it.
If you want to run some quick hand calcs, Hicks has an equation on it under the topic of Force Fits.
--Scott
http://wertel.eng.pro
RE: Confussed
I tried looking up "pin" in both the Standard Handbook of Machine Design by Shigley and Mischke, and couldn't find anything, but section 19-4 covers "Inteference Fit - Stresses".