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Force due to shrinkage

Force due to shrinkage

(OP)
Hi guys,

A (hopefully) simple question but I can`t find a conclusive answer.

Let`s say I have a LDPE ring with a steel ring inside. Both are placed in a room at -30°C (coming from +20°C).

Because the shrinkage of LDPE is bigger than the shrinkage of steel, the LDPE ring will exert a force on the steel ring.

How can I calculate this force?


Can I calculate the force with the next formula: F=(E*A*ΔL)/L ? Ofcourse taking the shrinkage of the steel ring into account.

What is A? The area when I cut the ring through? Both top and bottom, so 2 x(Thickness * Width) or only 1 time?

Really hope someone can help me with this !

RE: Force due to shrinkage

There are shrink fit calculations in most machinists handbooks.
Give the very low modulus and strength of the LDPE it would have to be much thicker (20 or 30 times) than the steel to exert any real force on it.
I would guess that if you cycled this a few times between RT and -30C that the LDPE ring would be looser.

= = = = = = = = = = = = = = = = = = = =
P.E. Metallurgy, Plymouth Tube

RE: Force due to shrinkage

(OP)
Well our test data show the LDPE ring exerts 5-600 N on the steel ring, which would make it shrink an extra 0,2mm. 0,5 would be to much that`s why I need to calculate the worst case.
The LDPE is 3mm thick, the steel 2mm.

RE: Force due to shrinkage

You need to work two different equations.

If the inner pipe was solid then the force would be higher as all the contraction goes into stress in the PE. If there was no pipe there would be no force as the PE pipe would simply shrink.

The inner pipe is quite thin. If there was a high force then it compresses ( but doesn't buckle?) hence relieving some of the force.

When the two forces and two changes in circumference of the two pipes equal each other then you have the solution, providing that the two start in direct contact and you don't buckle the inner pipe/

Of course this ignores the fact that PE as a creep phenomena so although it will start to reduce circumference, over time it could easily "relax" and the circumference springs back to the original size.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

RE: Force due to shrinkage

LI has it correct (as usual)
If the interface pressure is 600N, then you have to find where this balances between stretching of the outer one and compression of the inner one.
Since this is elastic it will be driven my the modulus, with LDPE at 45,000psi and steel at 30 million psi, and the thickness being similar, I find it hard to believe that the steel will change size at all.

= = = = = = = = = = = = = = = = = = = =
P.E. Metallurgy, Plymouth Tube

RE: Force due to shrinkage

(OP)
Ok, Thank you both very much for the replies !

It would be very good if the steel is unaffected or compressed less than .5mm

RE: Force due to shrinkage

I would have said buckling is your biggest issue at those sorts of thickness.

Also don't forget the inner ring will shrink anyway. I suspect that will be 90%+ of your reduction in ID.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

RE: Force due to shrinkage

While the rings are in contact, the inner ring is always in (circumferential) compression and the outer in tension, so only the inner ring has the potential to buckle. Since the inner ring is a much stronger material, with a much higher modulus, buckling cannot occur unless the outer ring is much thicker than the inner.

Develop the deflection equations for each ring then solve the simultaneous equations knowing that the deflection (delta radius) and pressure at the contact surface is the same for both rings.

je suis charlie

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