Engineer's Notebook §2.1

Thermal expansion

Most materials grow when heated. Over a moderate temperature range the change in length is proportional to the original length and to the temperature change:

ΔL = αL ΔT
ΔL
change in length
m
α
coefficient of linear expansion
1/K
L
original length
m
ΔT
temperature change
K
A
area
m²
V
volume
m³
E
Young's modulus
Pa
σ
stress
Pa
A bar resting on a roller, drawn at its original length with a dashed end mark, and at its heated length. ΔL = 4.8 mm L = 10 m change drawn ×100
Fig. 2.1-1Change in length exaggerated; scale noted under the bar.
α ΔT, strain4.8 × 10⁻⁴
Eα ΔT if restrained−96 MPa
ΔL = αL ΔT4.8 mm

Worked example

Uses the values in Fig. 2.1-1. Change them and the working updates.

Given
  • αsteel12 × 10⁻⁶ /K
  • Loriginal length10 m
  • ΔTtemperature change40 K
Find

The change in length ΔL.

Solution
ΔL = αL ΔT = 12 × 10⁻⁶ /K × 10 m × 40 K = 4.8 × 10⁻³ m
Answer
ΔL = 4.8 mm

If both ends were fixed, the stress would be −EαΔT = −96 MPa.

Related cases

  • Area

    ΔA ≈ 2αA ΔT

  • Volume

    ΔV ≈ 3αV ΔT

  • Bar fixed at both ends

    σ = −Eα ΔT

Assumptions
  • α is constant over the temperature range. In reality it varies with temperature.
  • The whole bar is at one uniform temperature.
  • The material is isotropic and free to expand, with nothing restraining it.
  • Values are typical. Concrete varies with its aggregate, roughly 7–12 × 10⁻⁶/K.

Derivation
  1. The coefficient is defined as the fractional change in length per kelvin: α = (1/L) dL/dT.
  2. For small changes, ΔL ≈ αL ΔT.
  3. If the ends are fixed, the thermal strain α ΔT must be cancelled by an equal and opposite elastic strain σ/E.
  4. So a restrained bar carries σ = −Eα ΔT: compression when heated, tension when cooled.

This restrained stress is why rails, bridges and pipework need expansion joints: it does not depend on the bar's length.