Hypothetical question: What would happen if I were to cool down a steel cylinder to make it contract, and then slide it all the way into a cylindrical hole in a room-temperature solid steel box? The hole's diameter is very slightly bigger that's the diameter of the cylinder when cold, but meaningfully smaller than the diameter of the cylinder when in room temperature.
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9$\begingroup$ Why Hypothetical? This is a real world solution used in bushings, bearing races, iron-shod wagon wheels, and both positive and negative temperature-differentials are a go-to for a frustrated mechanic struggling with recalcitrant assemblies. $\endgroup$– CriggieMar 27 at 2:21
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4$\begingroup$ This technique was (and still is) used to fit steel tyres onto cast-iron railway wheels c. two hundred years ago, but was probably 'invented' long before that. $\endgroup$– MikeBMar 27 at 8:33
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3$\begingroup$ @Criggie I enjoyed helping our workshop staff do this for a 20+kg vibration damper for a vacuum line. An undersize hole in the main body, a bucket of liquid nitrogen to cool the inlet and outlet pipes, and a big soft mallet were all it took to make a vacuum seal $\endgroup$– Chris HMar 27 at 12:26
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3 Answers
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Assuming your hypotethical question as what it would be for a difference of 200 degrees celsius and assuming the hole as rigid.
The volumetric expansion index of steel is
$$ \text{ $9*10^{-6}$ - per 1°C – for regular steels; $16*10^{− 6} $ per 1°C – for low-alloyed steels.}$$ Let the length and radius of the steel rod be
- $L$ = length
- $R$ = radius
The volume of the rod will try to increase by the amount of $$\Delta V=9*10^{-6}*200=0.0002$$ Because the rod can expand only axially the length $L$ has to increase.
$$L_\text{final}=L_\text{initial}*1.0002$$ And it will undergo compressive lateral stress proportional to strain it would experience if free to expand radially!
In reality the rod will expand the cylinder proprtional to their bulk modulus of elasticty. The bar will elongate a bit less and will expand laterally a bit.
If we close the top of the hole immediately after inserting the rod, and assuming a rigid confined hole, the rod will undergo complex 3d stresses related to its bulk modulus of elasticity, which for steel is 156-165 GPa.
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The cylinder would expand until constrained by the cylindrical hole. After that it depends on the relative strengths. If the block has thin walls it may be stretched by the expanding cylinder and might even crack. If not the combination would behave like a solid block.
What you have described is a standard industrial technique for interference fit. See Liquid Nitrogen Shrinking of a Shaft for a YouTube example.
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$\begingroup$ The box is completely solid besides the hole. How could the combination behave like a solid block? The metal cilynder will reach room temperature, and it'll have a diameter that's smaller that the diameter that it's supposed to have in that temperature. It'll just... Be squashed? Can steel be squashed like that? $\endgroup$ Mar 26 at 14:43
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$\begingroup$ Yes. It can be compressed. It is, for example, in ultrasonic horns used for ultrasonic welding. Continuously welded railway tracks are normally in tension but on very hot days will be in compression. They're just giant springs! $\endgroup$ Mar 26 at 14:45
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1$\begingroup$ @RamRachum "completely solid" that is your misunderstanding. Is glass a solid? Is water incompressible? Check out the use of liquid nitrogen, make sure you know about the temperatures for nitrogen... $\endgroup$ Mar 26 at 15:27
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7$\begingroup$ The rod can stay at the same diameter after it's warmed to room temperature. It will simply exert enormous force pushing outward. It may plastically deform itself or the block into which it was inserted, and in doing so relieve some of that pressure. $\endgroup$– DrewMar 26 at 16:22
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Every material can stretch or compress, even steel. The equations used to predict how tightly the rod would be squeezed in the hole are the same equations that would apply if the hulk stuck his fingers in the hole, stretched it out like a rubber band and placed it around the rod. Or, more realistically, if there was a taper and you hammered the rod in.
All of these scenarios are called interference fits, since if the steel were to relax into it's natrual shape, the parts would interfere with each other.
