Hooke’s law with thermal effects

Thermal expansion

During a change in temperature $\Delta T$, from an ambient temperature, a piece of material will change in volume — it receives a thermal strain.

A simple model for how this thermal strain $\eps^{\term}$ varies with the change in temperature $\Delta T$ is

$$\eps^{\term} = \alpha \, \Delta T$$

where $\alpha$ [1/${}^\circ C$] is a material parameter called the coefficient of thermal expansion.

During a combination of mechanical and thermal load (change in temperature), we obtain the total (observable) strain as the sum of mechanical and thermal strain.
$$
\begin{align}
\eps &= \eps^{\rm{mech}} + \eps^{\term}
= \frac{\sig{}}{E} + \alpha \Delta T \qgives \\
\sigma &= E(\eps – \alpha \Delta T)
\end{align}
$$

Note
Only mechanical strains give rize to stresses.

For example, if the material only undergoes thermal strain (the total strain), then the stress becomes zero:

$$
\eps = \eps^{\term} \qgives \sig{} = E(\eps^{\term} – \eps^{\term}) = 0
$$

Thermal deformation

The deformation of a bar with length $L$ is obtained as the integral of the strain along the bar
$$
\delta^{\term} = \int_L \eps^{\term} \mathrm{d} x = \int_L \alpha \ \Delta T \mathrm{d} x
$$

A common case is having a bar with a homogenious material and assuming a constant change in temperature along the whole bar. The thermal deformation can then be written
$$
\delta^{\term} = \alpha \ \Delta T \int_L \mathrm{d} x = \alpha \ L \ \Delta T
$$

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