Shafting Interference pressure; both thin walled with Thermal loads

[Zachimus]

All, I'm a Nav Arch working a PE review course question that MechE's probably know off the bat. A thin-walled steel shaft has a thin-walled inner brass bearing. At initial temperature, the interference is zero. The bearing and shaft are heated. No other forces are applied to the system. The system is free to movements in the axial direction.

First question: Individually, the shaft and bearing are both thin-walled cylinders, but together they could be considered in the thick walled category (t/D>0.1). Is it more appropriate to treat the heated interference fit as a combination of thin-walled cylinders or as a thick-walled cylinder?

So far, I have worked the problem using thin-walled strain formulations. I would appreciate some confirmation if I am applying the correct strain equations and compatibilities.

I established a strain equation for each cylinder. Strain for the outer cylinder was equal to thermal strain plus the circumferential strain (circumferential stress (p*r/t) due to the interference pressure divided by the modulus of elasticity). The strain equation for the inner cylinder was the same, except that the sign for the interference pressure was negative. Material properties were used in each equation to suit.

For compatibility, I said that the change in diameter of the inner bearing must be equal to the change in diameter of the outer shaft.

How say you??

 

[desertfox]

Hi Zachimus

Firstly you can use thick walled vessel theory on thin walled vessels because they actually have hoop stress variation just like thick walled one's, its just that the hoop stress in thin walled vessels doesn't vary that much so a simplification of assuming equal hoop stress throughout was valid.

The interface pressure between the outer diameter of the brass and the inner diameter of the steel sleeve must be the same after they are heated.
If you consider heating the steel first, it as nothing to restrain it and therefore can expand freely, now when the brass is heated because it wants the expand freely and to a greater extent than the steel but can't, because the steel restrains it, it therefore pushes the inner diameter of the steel further than it would expand normally under the heat.
The steel however stil prevents the brass from expanding freely so they reach a compromise this can be stated as follows:-


compression of the steel + compression of the brass = difference
in free
"lengths"


desertfox