Thermal Expansion: Common Mistakes and Fixes (3)

hard • 4 min read
Tags Thermal Expansion

Question

A steel rod of length L0=2L_0 = 2 m at 20°20°C is heated to 120°120°C. The coefficient of linear expansion of steel is α=1.2×10−5\alpha = 1.2 \times 10^{-5} /°C. Find: (a) the increase in length, (b) the new length, (c) the strain produced if the rod is clamped between two rigid walls so it cannot expand.

Young’s modulus of steel Y=2×1011Y = 2 \times 10^{11} N/m2^2.

Solution — Step by Step

ΔT=120−20=100°C\Delta T = 120 - 20 = 100°\text{C}

ΔL=L0αΔT=2×(1.2×10−5)×100\Delta L = L_0 \alpha \Delta T = 2 \times (1.2 \times 10^{-5}) \times 100

ΔL=2.4×10−3 m=2.4 mm\Delta L = 2.4 \times 10^{-3} \text{ m} = 2.4 \text{ mm}

L=L0+ΔL=2+0.0024=2.0024 mL = L_0 + \Delta L = 2 + 0.0024 = 2.0024 \text{ m}

If the rod is clamped, it cannot physically expand — but it would have wanted to. The compressive strain induced is:

strain=ΔLL0=αΔT=(1.2×10−5)(100)=1.2×10−3\text{strain} = \frac{\Delta L}{L_0} = \alpha \Delta T = (1.2 \times 10^{-5})(100) = 1.2 \times 10^{-3}

The thermal stress is:

σ=Y⋅strain=(2×1011)(1.2×10−3)=2.4×108 N/m2\sigma = Y \cdot \text{strain} = (2 \times 10^{11})(1.2 \times 10^{-3}) = 2.4 \times 10^8 \text{ N/m}^2

Final answers: ΔL=2.4\Delta L = 2.4 mm, new length =2.0024= 2.0024 m, thermal strain =1.2×10−3= 1.2 \times 10^{-3}, stress =2.4×108= 2.4 \times 10^8 Pa.

Why This Works

A rod heated freely expands by ΔL=L0αΔT\Delta L = L_0 \alpha \Delta T. If you prevent that expansion (rigid walls), the rod is effectively “compressed” by an amount equal to what it tried to expand by. That compression produces internal stress.

Strain is dimensionless (ΔL/L\Delta L/L). Stress = Y×Y \times strain. So thermal stress σ=YαΔT\sigma = Y\alpha\Delta T — independent of length, surprisingly.

Linear: ΔL=L0αΔT\Delta L = L_0 \alpha \Delta T

Area: ΔA=A0(2α)ΔT\Delta A = A_0 (2\alpha) \Delta T (factor 2 because area scales as L2L^2)

Volume: ΔV=V0(3α)ΔT=V0γΔT\Delta V = V_0 (3\alpha) \Delta T = V_0 \gamma \Delta T, where γ=3α\gamma = 3\alpha

Thermal stress (clamped): σ=YαΔT\sigma = Y \alpha \Delta T

Alternative Method

Combine in one shot:

σ=Y⋅ΔLfreeL0=YαΔT=(2×1011)(1.2×10−5)(100)=2.4×108 Pa\sigma = Y \cdot \frac{\Delta L_{free}}{L_0} = Y \alpha \Delta T = (2\times 10^{11})(1.2\times 10^{-5})(100) = 2.4 \times 10^8 \text{ Pa}

Skips intermediate steps.

Three thermal expansion traps that cost students marks:

  1. Using °F or K mixed up. ΔT\Delta T in °°C and ΔT\Delta T in K are equal (both are intervals), but α\alpha is given per °C, not per K. Conventionally same; just don’t switch to Fahrenheit.
  2. Using β=2α\beta = 2\alpha for area always. This works for isotropic materials (same α\alpha in all directions). For anisotropic crystals, area expansion = sum of two perpendicular linear expansions.
  3. For volume of a hole in a metal plate, students think the hole shrinks. No — the hole expands at the same rate as the metal. Heat a ring with a hole in the centre, and the hole gets larger.

When two rods of different materials are joined end-to-end and heated, total expansion = sum of individual expansions. When they are clamped at both ends, calculate the thermal stress in each from the constraint that total length is fixed.

Common Mistake

Students often forget that a hole expands along with the material. JEE 2021 had a question about a steel ring fitted around a wheel — the question hinges on whether the hole gets bigger or smaller on heating. The hole gets bigger. Imagine the metal as a “scaled-up” version of itself: every point moves outward, including the boundary of the hole.

Want to master this topic?

Read the complete guide with more examples and exam tips.

Go to full topic guide →

Try These Next

A steel rod 1m at 20 degrees C — find length at 100 degrees C

medium

Thermal Expansion: Conceptual Doubts Cleared (2)

medium

Thermal Expansion: Numerical Problems Set (1)

easy

Why are gaps left between railway tracks — explain with thermal expansion

hard