Here's a quote from a website I found. Sorry that I didn't copy the URL, but it was one of the first of several that popped up when I did a web search on the subject.
"Question - When water expands as it freeze, how much force (ie. psi) does it exert.
Application: if a say a crack in a foundation wall freezes with water in it, How much
force would be exerted in that crack.
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It depends on the temperature. At 0 degrees Centigrade, a very small pressure increase
will melt ice. As the temperature drops, the pressure required to melt ice increases
rapidly, by approximately 145 atmospheres per degree Centigrade. At -22 degrees
Centigrade, ice can exert 2,700 atmospheres (~40,000 lbs. per square inch) of pressure
without melting. (Data are from the web page
http://www.benbest.com/cryonics/pressure.html)
Beyond this temperature and pressure, things get complicated because there are many
different forms of ice."
This web site provided some useful information, but the other did little to provide the type of information I was looking for, so I did the following analysis.
The elongation required to accommodate the expansion that takes place when water freezes isn't that dramatic. It appears the properties of elongation may decrease with a drop in temperature. The elongation appears to be on the order of 3 to 5% if the change in volume (water to ice) is about 10%. The elongation of carbon steel should be on the order of 25% at room temperature.
The following is my solution to the problem. Assume you have a sphere with a volume of water equal to 10 cubic units. Upon freezing the volume would increase to about 11 cubic units.
The volume of the sphere would be V=4/3 X Pi X r cubed (I wish there was a way to write formulas easily)
Circumference is = 2 X r X Pi
Elongation is = ((C2-C1)/C1) X 100%
The radius would increase from 1.33 units to 1.38 units upon freezing (assuming there is no restraint). All units are rounded off
The circumference would increase from 8.4 to 8.7 units
Elongation would only be 3.2% to accommodate the increase in volume of the sphere.
Initial Volume 10 = 1.33 3.14 r^3
2.394521335 = r^3
Radius1 1.337807408 = r
Circumference 8.40143052
Final Volume 11 = 1.33 3.14 r^3
2.633973469 = r^3
Radius2 1.380987598 = r
Circumference 8.672602114
Elongation 3.22768359
Instead of a sphere, if you use the area of a circle, that is a section of pipe with a length infinitely small and consider it to be a constant, , the elongation increases, but not close to what you would expect the elongation of steel to be. It worked out to be something less than 5% if I remember correctly.
If the thought process is correct and if I didn't make any blunders, it would appear the ductility of the carbon steel drops considerably as the temperature decreases. It would be interesting to see if it coincides with the data for the transition temperature (ductile to brittle transition) obtained by the Charpy Impact text.
Best regards - Al