thermal expansion of tubes

[pesy]

I am trying to calculate the thermal expansion of a tube--more specifically, I am trying to calculate the stress in a tube when two concentric tubes expand and the coefficient of thermal expansion of the inner tube is greater than that of the outer tube. I know how to do this for linear problems, but I don't know the equation for radial thermal expansion.
Does anyone know this or have a good reference?

Thanks.

Pesy

 

[GregLocock]

Oddly enough Shigley discusses it. I don't have him with me, so I can't give you a better reference. Shrink fitting and related concepts is very hard to solve until you've seen it done once, then it is pretty straightforward. Unfortunately the whisky has dissolved that brain cell, in my case!



Cheers

Greg Locock

 

[ProEpro]

It is the same in the radius as it is in the axis of the tube. Use the diameter as the dimmension then multiply by the coeficent of thermal expansion and the change in temperature.
This confussed me once until I thought about the circumfrence expanding with temperature.
ProEpro

http://www.whitelightdesign.com

 

[Hush]

You know the contact pressure and the change of the circumference at the OD of the inner tube is the same as at the ID of the outer tube, plug in the equations and solve for the unknowns.

 

[prex]

A general solution of this problem may be obtained by treating your pipes as thick cylinders (formulae from Roark).
A thick cylinder subject to an external pressure p changes its outer radius a of:
-pa((a²+b²)/(a²-b²)-ni)/E
where b is the inner radius, E the elastic modulus and ni the Poisson's ratio.
Similarly if the pressure is internal, the change of inner radius is:
pb(a²(1+ni)+b²(1-2ni))/(a²-b²)/E
Now the equation to write says that the change in outer radius of the inner pipe is equal to the change in inner radius of the outer pipe (including the effect of temperature).
By using the subscript i for inner and o for outer, recalling that a_i=b_o and calling T the change in temperature and k the linear coefficient of thermal expansion, this becomes:
k_iT-p((a_i²+b_i²)/(a_i²-b_i²)-ni_i)/E_i=k_oT+p(a_o²(1+ni_o)+b_o²(1-2ni_o))/(a_o²-b_o²)/E_o
This equation is immediately solved for p (I suggest the use of a spreadsheet avoiding algebraic manipulations, if you are only interested in a numerical result!) and the stresses from Roark's formulae (no stress due to thermal expansion, of course).



prex
motori@xcalcsREMOVE.com

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[prex]

I forgot to add that the treatment in my previous post assumes that the cylinders are allowed to slide longitudinally one with respect to the other, and this seems quite unrealistic.
The longitudinal stress generated by the constrained longitudinal expansion will give rise to radial stresses too that cannot be easily calculated for thick cylinders. However, if you are concerned with this, you should be able to calculate a longitudinal stress based on the assumption of thin cylinders (just equate the longitinal expansions), whereas the effect of transverse expansion due to this longitudinal stress should be by far negligible.
Note also that the effect of Poisson's ratio in the formulae for thick cylinders becomes negligible (you can take zero for it) as soon as the ratio of thickness to mean radius is of the order of 0.1 or less.
prex
motori@xcalcsREMOVE.com

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[pardal]

On this same thread but I only want to deal with longitudinal thermal expansion.
My task to do is to generate a compression force between two plates when they rise it's temperature from ambient to 480°C
This 300 mm diameter round Sae 1010 plates 38 mm thick, will hold a stack of about 100 each Sae 1070 1mm thick to be flattened .
This, can be called sandwich, will be hold by a 40 mm round SAE 4140 threaded bar heat treated as a 12.9 grade bolt.
To build up a increasing compression , I intent to put a spacer over one plate and under the nut that will tighten all the array , this spacer shall be of a material with a different thermal expansion from the others components.
I thought on a aisi 304 hollow bar as it have a thermal exp coefficient about 17E-6 micro mm/meter/°c
and the carbon and alloy steel had about 10e-6 .
My concern is about the behavior of all this array at the 480 C temperature, it will be about 4 hour since this temp is reached.
The task to cope, is to avoid a loss on the compression, so all the 1070 plates are gathered side by side , so they will last flattened when this force is released, of course that it will happen after the eat treatement.
Any comment will be granted.
You can reach me at k281969@hotmail.com
but keep your post here for mutual benefits.
Pardal

 

[prex]

Don't really understand your goal.
Can't you use the bolt to compress the stack?
Your choice of materials sounds correct, but of course the stack will have to withstand the compressive force generated by the differential expansion, if you want it to be useful.
Another concern is how much differential expansion you'll need: this will determine the length of the spacer. With one meter you get a compression of about 3 mm at 480°C.
prex
motori@xcalcsREMOVE.com

http://www.xcalcs.com

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[pardal]

Hi Prex:

Yes I will use the bolt , but who ensures me that at the oven, when increasing the temperature all plates will stand toghether as when cold.
About the 3mm compression : I will do the calc so only about 1.00 mm is originated .
By this time I do this task by my self,so I can check all variables, but I need to do a fixture to be handle by heat treatment worker, who are not skilled for this task.

Last week I sent a mail to your adress, but it returned.
"what's the meaning of REMOVE at your domain?
To be removed??
Hope you get it clear.

Thanks for your interest for this post.

Pardal

 

[prex]

Of course for 1 mm you need some 300 mm for the spacer (take 500 if you want to be sure or make a test).
However you must be aware of the fact that both the spacer and the bolt will yield at that temperature, both because the yield stress lowers and because the stress corresponding to that elongation is quite high (over 600 MPa): a test could tell you if there is accumulation of plastic deformation at every heat up and after how many treatments you'll need to replace the bolt and the spacer.

Yes REMOVE means what it says. However it is preferable to continue the discussion on the forum, unless you have something personal.
prex
motori@xcalcsREMOVE.com

http://www.xcalcs.com

Online tools for structural design