After a machinist slips a hot, snugly fitting iron ring over a cold brass cylinder, the ring becomes 'locked' in position and can’t be removed even by subsequent heating. This procedure is called ’shrink fitting.' How does it occur? Can you conclude anything about the thermal expansion rates of iron and brass?
To explain: The occurrence of “shrink fitting”.
The information concluded about thermal expansion rates of brass and iron.
Answer to Problem 44A
Due to the higher thermal expansion rate of brass compared to iron, the ring gets locked on the brass cylinder.
Explanation of Solution
Introduction:
When any object is provided heat or cold, then that object expands due to thermal expansion, no matter what it is in.
When a hot iron ring is fitted over a cold brass cylinder, the area of brass that comes in the contact of iron gets hotter. Due to thermal expansion, iron expands. But, brass expands more. Thus, both get locked in a position that excess heating will make them tighter only, won’t free them.
Conclusion:
Thus, due to the higher thermal expansion rate of brass compared to iron, the ring gets locked on the brass cylinder. Even with the excess heat, the surface doesn’t free but gets tighter.
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