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.
