Hoop Stress due to change of state 2

[DM1848]

This is my first post so please be gentle if this is obvious. I've seen several posts regarding thermal expansion of pipes and hoop stress but mine is more related to the stress and behaviour of the material inside the pipe.

A thin walled pipe is filled with a liquid that is just above its solidification temperature (pipe at the same temperature), the temperature then drops and the liquid solidifies at which time it expands volumetrically by 1%. At the moment, I consider the pipe to be long so I can treat this as a 2D problem. I've also neglected any changes in wall thickness in the pipe to keep it simple.

If the material in the pipe is rigid on solidification, then the problem is simple, the 1% increase is related back to the change in the circumference in the pipe and the stress strain curve gives the hoop stress.

But, if the material is not rigid, and has a lower Elastic Modulus and yield strength than the pipe, what happens to this solid cylindrical plug in the pipe?

I keep thinking there has to be a closed form solution based around the hoop stress equation and strain compatibility but I seem to be missing something.

Any thoughts are much appreciated...

 

[Compositepro]

Pipes can freeze without bursting. It depends on how they freeze. It is primarily the pressure of the unfrozen water that actually stresses the pipe. The expansion of freezing water will pressurize the remaining water if it has nowhere to go. So the specific pattern of freezing determines the ultimate pressure.

 

[DM1848]

It's not a freezing pipe problem, it's putting a eutectic alloy in a steel tube.

 

[SnTMan]

I'd think that the shrink fit / interference fit type calculations would do the job. I haven't checked it out though...

Regards,

Mike

 

[Compositepro]

The same logic applies as with freezing water. At the time of freezing and becoming a solid, the material has very little strength and will not stress the container. Pressure will only be generated by the liquid that has no place to go. It doesn't sound like the liquid will be trapped in your case so there will be little stress on the pipe. Additionally, the pipe will be thermally expanding as it warms and the solid eutechntic will shrink as it cools.

 

[zekeman]

As a practical matter, the compressive deformation of the frozen material is at least an order of magnitude less than the the volume deformation caused by the thin wall.
So you assume NO deformation of the material and the hoop stress would then be be 1/3 of the 1% x the modulus for the hoop material.

 

[BillBirch]

I would try to consider it in terms of strain energy as you imply. Does the contained material completely fill the container, i.e fully vented? If not, pressure would be produced by the compression of the vapour space. If so, I would be trying to determine the equilibrium between the "compression" of the frozen contents and the "stretching" of the container.

 

[prex]

Assuming, as you stated, a simple 2D problem, it should be treated as follows.
The relative change in radius of the contained solid material is
[Δ]R/R=0.01-p(1-[ν]_s)/E_s
The same for the containing pipe is (assuming a relatively thin one)
[Δ]R/R=pR/tE_w
By equating the two it is possible to calculate the pressure p exerted by the pipe on the solid and from that the deformations of both.
All this assumes elastic behavior, of course. If the Young's modulus of the contained solid is higher than some 1/5th of that of the wall, the latter will undergo a plastic deformation.
In this case, assuming a perfect elastic-plastic behavior, the pressure will be p=Yt/R.

prex

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[rb1957]

the upper limit to the problem is to assume the pipe exands to suit the new volume. A volume change of 1%, is equivalent to a change in radius of 0.5% (approx), which is a strain in the pipe of 0.5%, which is a stress of 0.005*E in the pipe. steel pipe, E = 30E6psi; stress = 15ksi.

 

[zekeman]

The linear strain for the problem is 1/3%, not 1% or 0.5% as suggested by the previous posts. It is NOT a 2D problem but a 3D problem.