Thermal expansion of a washer

[NuclearVW]

I am working on site installing a ball mill (28 foot diameter, 44 feet long). We are trying to heat the end of it to insert the trunnion and I am trying to figure out how the thermal expansion of the head works.

The head of this mill is basically a washer shape, 28' on the outside with a 15' hole in the middle. Its slightly conical, but I am ignoring that.

I can't seem to find any info on the equations for a disk expanding in the radial direction, let alone a washer.

Basically I am trying to heat it to insert the trunnion, and I want to know how much to heat it to get the desired increase in diameter of the hole.

 

[corus]

The expansion is alpha.R.T where alpha is the coefficient of thermal expansion and T is the increase in temperature from cold.

 

[racookpe1978]

times diameter.

Thermal expansion coef x change in temperature x distance (length or diameter) that is heated.

If you have an unrestrained ring that is uniformly heated, the OD of the ring increases. AND the ID of the ring increases.

If you have a restrained ring that is uniformly heated, the OD "tries" to expand but it cannot. The ID is forced "smaller" by the retraining force around the OD.

If you have a solid round bar that is uniformly heated, the inside of course cannot "go anywhere" so it tries to expand (pushes against the OD). The OD is not restrained, so it expands.

Thermal interference fits are done by cooling the shaft AND heating the ring so there is a workable gap between the two. Very carefully positioning the two BEFORE heating the ring and cooling the shaft so that sliding fit can be made up SAFELY, IMMEDIATELY, SMOOTHLY, and CONTINUOUSLY in ONE MOTION with as little touching of the two parts as possible.
Then, when the mating surfaces are in the right spot, the two are allowed to equalize - or shrink fit together. Once the two parts are sliding across one another, you only have 2 to three minutes before binding begins for parts weighing 500 to 800 lbs. 10 to 20 seconds for parts under 10 lbs.

Usually, the shrink fit interference fit is 0.001 - 0.003 on the diameter of small parts (1.000 to 4.000 inches. The interference increases as the part diameters increase: Might be as much 0.015 or 0.020 or more at your 15 foot.


Take a 10 foot long bar and only heat the middle 10 inches of the bar?
The middle of the heated bar expands, but

 

[corus]

I forgot to say that R is the Radius in the expression for calculating the radial thermal expansion.
Depending on the size you may have to consider the time taken to achieve an approximately uniform temperature, and a time for cooling once it leaves the oven, or wherever you heat it up. That'll mean you'll have to heat the washer up higher than necessary to allow for the cooling before it's fitted. There may be a rule of thumb around for that. Failing that it can be calculated depending on the material properties, dimensions etc.

 

[NuclearVW]

I guess to be clear, this is not an interference fit, we just wanted to heat it to be sure there was ample room for the insertion. But alas, this is a moot point anyway because the contractor failed to heat it and it fit like a glove anyway. Thanks for the responses though