Relation between thermal expansion and stress

[raju_pro]

I have been trying to find an answer for this for quite some time and found this forum which could help me.

Does an initial stress (elastic strain induced in a metallic component) change its thermal expansion coefficient? if yes then what is the relationship?

In simpler terms, does 2 identical parts one under load (elastically strained) and other free state would expand through the same length when the surrounding temprature is increased?

 

[Robot B9]

No, it does not.

Imagine this example. You have two identical springs. They are both extension springs with loops on each end, they are hanging from pegs. One of the springs is under tension because there is a weight hung from it. This weight is not enough to plastically deform the spring. The other spring has no additional load, it just hangs by its own weight. One spring, the one with the weight, is under greater stress.

So you take this set up and put it in an oven and increase its temperature by 100 degrees. Both springs will, if they are perfect ideal springs, expand equally due to the normal thermal expansion effects.

The explanation is just common supper-position. Both effects just are independent and ignore each other.

 

[raju_pro]

Thanks for your response. I had the same thought earlier, but then came across this Journal Paper from MIT-Cambridge (although it is quite an old reference)

http://ieeexplore.ieee.org/xpl/freeabs_all.jsp?arnumber=5121477

Appreciate your comments on this. Thanks

 

[Mapes]

It is often assumed that the coefficient of thermal expansion is stress-independent, but some manipulation of partial derivatives shows that that's not strictly true. Consider a long rod (with length L and spring constant k=dF/dL) that can be axially loaded with force F; we'd like to know whether the linear thermal expansion coefficient

\alpha_\mathrm{L}=\frac{1}{L_0}\left(\frac{\partial L}{\partial T}\right)_F

is dependent on the load. This dependence can be written as

\frac{\partial}{\partial F}(\alpha_\mathrm{L})=\frac{\partial}{\partial F}\left[\frac{1}{L_0}\left(\frac{\partial L}{\partial T}\right)_F\right]_T=\frac{1}{L_0}\frac{\partial}{\partial F}\left[\left(\frac{\partial L}{\partial T}\right)_F\right]_T=\frac{1}{L_0}\frac{\partial}{\partial T}\left[\left(\frac{\partial L}{\partial F}\right)_T\right]_F=-\frac{1}{L_0k^2}\left(\frac{\partial k}{\partial T}\right)_F

which shows that the temperature dependence of stiffness is somehow involved!

The authors did essentially the same calculations for the more general 3-D case of stress and strain, then confirmed their predictions by experiment. Does this answer your question?

 

[raju_pro]

Mapes, thanks for your reply. So in this case if I had assumed that the Young's Modulus does not vary with temperature i.e., \frac{\partial\bf{E}}{\partial\bf{T}} = 0, then I do not have to worry about the variation of \alpha wrt to temperature ?

 

[Mapes]

If you mean \alpha with respect to stress or force, correct; the two effects are linked. (But \alpha will still vary with temperature in a way that's material-specific.)