Linear, Areal, and Volumetric Expansion
The linear expansion coefficient is alpha = (1/L)(dL/dT); for small changes a free isotropic solid obeys Delta L = alpha L Delta T, with areal and volumetric coefficients approximately 2 alpha and 3 alpha.
Why this shows up in the exam
Expansion gaps in rails and bridges · Heating rings and holes for fitting · Dimensional correction of sheets, cubes, and containers
Learn the idea
Small unconstrained thermal expansion scales with original size, temperature change, and the proper expansion coefficient. Heating increases typical interatomic spacing, so every free dimension grows; for an isotropic solid the first-order area and volume changes collect two and three identical linear contributions.
🧠 Memory hook: A hole expands as if it were made of the same material.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- Delta L = alpha L₀ Delta T — small free linear expansion when alpha is effectively constant
- Delta A = 2 alpha A₀ Delta T — first-order area expansion of an isotropic solid
- Delta V = 3 alpha V₀ Delta T — first-order volume expansion of an isotropic solid
- gamma = alpha_x + alpha_y + alpha_z — first-order volume coefficient for anisotropic principal directions
How to approach it
- 1Identify whether length, area, or volume is requested
- 2Choose isotropic or directional coefficients
- 3Use a ratio when comparing bodies and test the sign
Common slip-ups that cost marks
- •Applying 3 alpha to an anisotropic body
- •Using final rather than reference dimensions inconsistently
- •Keeping second-order terms in a stated small-change problem
🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.
Original chapter practice
Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.
A wire 1 m long and cross-sectional area 2 mm^2 extends by 1 mm under a 200 N load. Find Young modulus.
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