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?


1 Answer 1


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.


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