AcademyTemperature and Heat
Questions

Academy

Thermal Expansion

Level 1 - Physics topic page in Temperature and Heat.

Principle

Most materials change size when their temperature changes, and the change is approximately proportional to the original size and the temperature interval.

If expansion is prevented, the missing expansion appears as mechanical strain and produces thermal stress.

Notation

\(L_0\)
initial length
\(\mathrm{m}\)
\(V_0\)
initial volume
\(\mathrm{m^{3}}\)
\(\Delta T\)
temperature change
\(\mathrm{K}\)
\(\Delta L\)
change in length
\(\mathrm{m}\)
\(\Delta V\)
change in volume
\(\mathrm{m^{3}}\)
\(\alpha\)
coefficient of linear expansion
\(\mathrm{K^{-1}}\)
\(\beta\)
coefficient of volume expansion
\(\mathrm{K^{-1}}\)
\(Y\)
Young modulus
\(\mathrm{Pa}\)

Method

For modest temperature intervals, the first-order model keeps only the term proportional to \(\\Delta T\). That gives a linear rule for each dimension.

Free linear strain
\[\frac{\Delta L}{L_0}=\alpha\Delta T\]
Length change
\[\Delta L=\alpha L_0\Delta T\]
Final length
\[L=L_0(1+\alpha\Delta T)\]

For an isotropic solid, every linear dimension expands by the same fraction. A cube with side \(L\) has volume \(V=L^3\), so the first-order volume expansion is about three times the linear expansion.

Expanded cube
\[V=V_0(1+\alpha\Delta T)^3\]
First-order expansion
\[V\approx V_0(1+3\alpha\Delta T)\]
Solid volume coefficient
\[\beta\approx3\alpha\]

If a rod is clamped so its length cannot change, the thermal strain and mechanical strain must cancel.

Total strain fixed
\[\frac{\Delta L}{L_0}=\alpha\Delta T+\frac{F/A}{Y}=0\]
Thermal stress
\[\frac{F}{A}=-Y\alpha\Delta T\]

Rules

These are the compact thermal-expansion relations.

Linear expansion
\[\Delta L=\alpha L_0\Delta T\]
Final length
\[L=L_0(1+\alpha\Delta T)\]
Volume expansion
\[\Delta V=\beta V_0\Delta T\]
Isotropic solid
\[\beta\approx3\alpha\]
Constrained stress
\[\frac{F}{A}=-Y\alpha\Delta T\]

Examples

Question
A steel rail has length
\[18.0\,\mathrm{m}\]
at
\[10\,\mathrm{^\circ C}\]
Estimate its length increase at
\[40\,\mathrm{^\circ C}\]
using
\[\alpha=1.2\times10^{-5}\,\mathrm{K^{-1}}\]
Answer
\[\Delta T=30\,\mathrm{K}\]
\[\Delta L=\alpha L_0\Delta T=(1.2\times10^{-5})(18.0)(30)=6.48\times10^{-3}\,\mathrm{m}\]
The increase is about
\[6.5\,\mathrm{mm}\]

Checks

  • Use a temperature interval in kelvins or Celsius degrees; the interval size is the same.
  • Holes in a freely expanding solid expand with the surrounding material.
  • The approximation assumes \(|\\alpha\\Delta T|\\ll1\).
  • Constrained expansion can produce large stresses even when the free length change is small.

Previous

Absolute Temp.

Next

Heat Transfer