Dose adapter sleeve hinder the axial displacement of inner ring of spherical roller bearings?

Question

This is supposed to be the locating bearing in my design, the outer ring is supported by the housing and a taconite seal (taconite seal is hidden). So it is fixed in place and can't move at all.

There is no significant axial forces working on the bearing system. I want to know if the inner ring can move to the left side due to thermal expansion of the shaft while it is not supported axially by anything else ? (same direction as the red arrow)

I think the conical shape of the sleeves adapter prevents any movement of inner ring anyway am i right? The length of the shaft is 50 inches and the diameter is 4.5 inches

Answer 1

We would be able to better estimate the lateral expansion if you edit your question and add the length and radius of the shaft. But it seems it will be okay at a first estimate, say your shaft is 10 inches. The expected variance in its length will be +/- 10* 0.0005= 0.005 inches or 0.125 mm.

Answer 2

Expansion coefficient of steel is 7.2* 10^-6 for each Fahrenheit degree of change.

Calling your shaft's length L and assuming the range of design temperature is 100, we get the expansion: delta L =

$$ \delta L = L \times 100\times 7.2^{-6} =L\times 7.2^{-4} $$ Or very roughly you have your shaft elongated by to L* 1.001.

If your bearing can not tolerate this much play it has to be able to contain the shaft back with a force of

F = E*e =29*10^6(0.001)psi= approx. 29000 psi.

This is just a first estimate but gives some idea about the order of forces involved.

Edit Ok now having the length of 50 inches we assuming the block is of the same family of steel with more or less same expansion coefficient there is no need to be that concerned about differential expansion unless we are talking extreme work temperature changes which may cause contortion and warping in the block. Otherwise assuming a difference in temperature between the block and shaft of +/- 50 degrees we will have thermal expansion of the shaft, E = 50 * 0.0005 *25mm/inch = .625 mm elongation or shrinkage. Now you need to check with your bearing manufacturer to see if the bearing can take this much displacement. Another thing I recommend to check is systems natural frequency and its spectra of vibration to check for destructive resonance modes.

One cautionary note: your shaft area of 1/4* 4.5^2*pi = 15.9 sq. inch can produce a considerable compressive load, need to check for that. All the temps are preliminary hypothetical numbers, you want to use correct data in your estimates.