Questions: Thermal Expansion: Linear and Volumetric

Question 1

An engineer calculates the volumetric expansion of a single crystal (an anisotropic material) by multiplying its linear expansion coefficient α by 3. What is wrong with this approach?

• A. Nothing — β = 3α is a universal law for all solids
• B. β = 3α only holds for isotropic materials; anisotropic materials require β = αx + αy + αz, which need not equal 3α
• C. The formula should be β = 2α for anisotropic materials
• D. Volumetric expansion cannot be calculated from linear coefficients under any circumstances

Answer: (B)

The β ≈ 3α relationship comes from the binomial approximation (1+αΔT)³ ≈ 1+3αΔT, which assumes the same linear coefficient applies in all three dimensions — the definition of an isotropic material. For anisotropic crystals, expansion differs along different crystallographic axes, so the true volumetric coefficient is β = αx + αy + αz. Using 3α for an anisotropic material can yield large errors.

Question 2

A steel pipe is installed while cold and then welded rigidly between two fixed supports. When the temperature rises by 80°C, what happens?

• A. The pipe expands freely along its length as normal
• B. The pipe shortens to compensate for the temperature rise
• C. Thermal stress builds up inside the pipe because the mechanical constraint prevents free expansion
• D. No change occurs because steel has a small expansion coefficient

Answer: (C)

When free expansion is mechanically constrained, the material cannot lengthen, so internal compressive thermal stress builds up instead: σ = EαΔT. For steel (E ≈ 200 GPa, α ≈ 12×10⁻⁶ K⁻¹), an 80°C rise would produce roughly 192 MPa of compressive stress — close to the yield strength of mild steel. This is why pipelines and structural frameworks require expansion joints.

Question 3

For isotropic materials, the volumetric thermal expansion coefficient is approximately three times the linear expansion coefficient.

• T. True
• F. False

Answer: True

This follows from geometry: if a cube of side L expands by ΔL = αL₀ΔT in each dimension, its new volume is (L₀+ΔL)³ ≈ L₀³(1+3αΔT) for small αΔT. The fractional change in volume is therefore 3αΔT, giving β ≈ 3α. This approximation is valid when αΔT ≪ 1, which holds for virtually all engineering applications.

Question 4

A material with a larger linear expansion coefficient α will also have a larger thermal stress when constrained, regardless of its stiffness.

• T. True
• F. False

Answer: False

Thermal stress is σ = EαΔT — it depends on both α and Young's modulus E. A material with large α but low E (like a rubber or polymer) can expand significantly yet develop little stress when constrained. Conversely, a stiff material with moderate α (like steel) generates large stress. Both material properties matter; using α alone to compare thermal stresses is incorrect.

Question 5

Why does thermal expansion occur at the atomic level? What property of the interatomic potential causes atoms to sit farther apart on average as temperature increases?

Model answer: The interatomic potential is asymmetric: the repulsive wall at short distances is steeper than the attractive tail at long distances. As temperature increases, atoms vibrate with greater amplitude. Due to the asymmetry, the average position during each oscillation is shifted toward the shallower (long-distance) side of the potential well. This asymmetric averaging moves the mean atomic separation outward with increasing temperature, producing macroscopic expansion. If the potential were perfectly symmetric, vibrating atoms would have the same average position at all temperatures and no net expansion would occur.

This is the microscopic origin of thermal expansion — not just 'atoms move more' but specifically that the anharmonic (asymmetric) shape of the potential shifts the equilibrium. Diamond and invar alloys have unusually stiff, nearly symmetric potentials, giving them exceptionally small α. Materials with soft, shallow potentials (polymers, metals with weak bonding) have large α.