Thermal Expansion of a Ring 2

[Creigbm]

I am trying to figure out the resulting interal pressure if two rings (one inside of the other) are heated up to a pre determined temperature. I was initially using the linear thermal expansion equation:

strain = alpha * delta T

however, I seem to remember that this was not valid for rings/spheres/etc. Any suggestions?

 

[desertfox]

Hi Creigbm

Can you supply more information like the tube sizes and materials, temperature rise and whether there is any interference between the tubes prior to being heated.
If we assume no interference between the tubes but one just slides inside the other, with the outer diameter of the inner tube touching the inner diameter of the outer tube and
in addition the inner tube is made from copper and the outer one from steel proceed as follows:-

1/ work out the increase in circumference for each tube
using the linear expansion equation assuming they are
both free to expand.

2/ calculate the difference in radii between the two tubes
by subtracting the values from step 1 and dividing by
2*pi(this difference will be the interference)

3/ If the two tubes are in contact then the copper will want
to expand more than the steel however the copper tube
cannot expand to its free state because its restricted
by the steel tube which will be forced to expand slightly
more than its free state. So at the copper / steel
interface there is a pressure that is tending to increase
the radius of the steel tube and decrease the radius of
the radius of the copper tube and this increase and
decrease of the respective radii must equal the
interference calculated in step 2.

4/ therefore:-[p*(C)^2/(E1*b)]+[p*(D)^2/(E2*t)]=interference

where p = internal pressure your trying to find

C = mean circumference of the outer tube

E1= modulus elasticity of outer tube

b = to the wall thickness of outer tube


D = mean circumference of the inner tube

E2= modulus elasticity of inner tube

t = to the wall thickness of inner tube

I will leave you to transpose the formula to find p.

Note if there is an interference fit between the tubes prior to heating then this stress will need calculating first and adding algebraically to the resulting p above for
both tubes. Finally I have assumed there is no longitudinal
stresses.

regards desertfox

 

[GregLocock]

Shigley gives a worked example for a railwheel and steel tire, which may be helpful in this context



Cheers

Greg Locock