# Compute annular thermal expansion?

How do you compute annular thermal expansion?

For example, suppose you have a metal washer with an inner diameter of $$D_i$$ and an outer diameter of $$D_o$$, and the material has a thermal expansion coefficient of α, then how do the inner and outer diameters change as a function of temperature?

If the annular body has no constraint that prevents it to expand, the body undergoes an isotropic deformation (just a scaling equal for all the dimensions).

Given the formula of the thermal expansion coefficient at constant pressure (to be more precise at constant stress, that is the case of a body free to expand, without constraints introducing stresses in it)

$$\alpha = \dfrac{1}{V}\left( \dfrac{\partial V}{\partial T} \right)_p$$

For small volume changes, a first order approximation of volume expansion reads

$$V(T) \sim V(T_0)\left[ 1 + \alpha (T - T_0) \right]$$,

while for linear dimensions

$$L(T) \sim L(T_0)\left[ 1 + \lambda (T - T_0) \right]$$,

where $$\lambda = \alpha /3$$ is the linear thermal expansion coefficient.
(If you need to know where factor $$1/3$$ comes from, I'll edit the answer)

Thus, both inner and outer diameter increases for an increase of temperature expansion, following the expression for linear expansion above, i.e.

$$D_{in}(T) \sim D_{in}(T_0)\left[ 1 + \lambda (T - T_0) \right]$$
$$D_{out}(T) \sim D_{out}(T_0)\left[ 1 + \lambda (T - T_0) \right]$$

i.e. they approximately increase by $$\lambda (T - T_0)$$ percent.