Slip condition between two parts joined together by shrink-fitting?

AI Thread Summary
When a moment M is applied to a pipe connected to a bearing joint via shrink-fitting, the slip condition between the pipe and the bearing's inner ring is determined by the frictional force. The frictional force, calculated as F_friction = mu*A*p, must balance the applied force F = M/a to prevent slip. If F_friction is less than F, slip will occur, indicating that the inner ring will not keep pace with the pipe. The discussion confirms that the friction force is the sole force affecting the inner ring's motion in this scenario. Understanding this relationship is crucial for analyzing the behavior of the bearing under applied moments.
Arcturus82
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Consider a bearing joint together with a long pipe (with radius a) by using shrink-fitting. The grip between the pipe and the inner ring of the bearing give rise to the surface pressure p at the interface. If a moment M now is applied to the pipe, what will the slip condition between the two parts be? Is it correct to assume that if the maximal frictional force between the two parts over the contact area A, i.e. F_friction = mu*A*p (mu = friction coefficient) cannot balance the force F = M/a, we will get slip? That is that we get slip between the two parts if F_friction < F ? To be clear I am talking about the situation where no other forces are acting on the bearing, i.e. the bearing does not have contact with any other surface than the pipe.

My thought is that the friction force is the only force that will accelerate the inner ring of the bearing. So if this force is lower than the force F, then the inner ring will accelerate slower than the pipe and thus the two parts will slip.

Any help would be appreciated.

All the best.
 
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I don't know about the part about acceleration, but the slip criterion determined by this development is definitely correct.
 
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