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A steel rail on a hot afternoon wants to get longer. If it cannot, it pushes on whatever holds it, hard enough to buckle track. This is thermal stress, and it joins two chapters, thermal properties and elasticity, which is why examiners like it. You may already know the basic formula. This post goes further: the force on the supports, a gap that only partly closes, and two different rods clamped end to end.
Where is it in the syllabus?
Thermal expansion and Young's modulus both appear in the JEE (Main) 2026 syllabus and the NEET (UG) 2026 syllabus. The NCERT Class 11 chapter on Thermal Properties of Matter defines thermal stress directly: if a rod's ends are fixed rigidly, it acquires a compressive strain from the supports, and the stress set up is called thermal stress. Young's modulus comes from the chapter on Mechanical Properties of Solids.
Where does the formula come from?
Think of it in two steps.
Let the rod expand freely. It would grow by ΔL = LαΔT.
The walls push it back by exactly that amount. Pushing a rod back by ΔL needs a compressive strain ΔL/L = αΔT.
Stress = Young's modulus × strain, so:
Thermal stress = YαΔT, and force F = YAαΔT.
Length cancels. A 1 m rod and a 10 m rod of the same material and cross-section, heated by the same amount, push on their walls with the same force.
NCERT's worked case: a 5 m steel rail of cross-section 40 cm², heated by 10 °C, with α = 1.2 × 10⁻⁵ K⁻¹ and Y = 2 × 10¹¹ N/m². The strain is 1.2 × 10⁻⁴, the stress 2.4 × 10⁷ N/m², and the force about 10⁵ N. NCERT notes that two such rails fixed at their outer ends and touching at their inner ends could easily bend.
Worked example 1: force on the walls
A brass rod of cross-section 2.0 cm² fits exactly between two rigid walls. It is heated by 50 K. Take α = 2.0 × 10⁻⁵ K⁻¹ and Y = 0.91 × 10¹¹ Pa, the brass values used in NCERT's exercises.
Strain prevented = αΔT = 2.0 × 10⁻⁵ × 50 = 1.0 × 10⁻³.
Stress = 0.91 × 10¹¹ × 1.0 × 10⁻³ = 9.1 × 10⁷ Pa.
Force = 9.1 × 10⁷ × 2.0 × 10⁻⁴ = 1.82 × 10⁴ N.
That is about the weight of a 1.8 tonne car, from a rod you could hold in one hand.
Worked example 2: a gap that only partly closes
A 1.00 m steel rod has a 0.30 mm gap between its end and a rigid wall. It is heated by 50 K. Take α = 1.2 × 10⁻⁵ K⁻¹ and Y = 2.0 × 10¹¹ Pa.
Free expansion = 1.2 × 10⁻⁵ × 1.00 × 50 = 6.0 × 10⁻⁴ m = 0.60 mm.
The gap allows 0.30 mm, so 0.30 mm is prevented.
Strain = 0.30 × 10⁻³ / 1.00 = 3.0 × 10⁻⁴.
Stress = 2.0 × 10¹¹ × 3.0 × 10⁻⁴ = 6.0 × 10⁷ Pa.
The gap closes when LαΔT = 0.30 mm, that is at ΔT = 0.30 × 10⁻³ / 1.2 × 10⁻⁵ = 25 K. Below that rise, there is no stress at all. This is the reason for expansion gaps in rails and bridges.
Worked example 3: two rods in series between walls
A steel rod and a brass rod, each 0.50 m long with the same cross-section, are joined end to end and fitted exactly between rigid walls. They are heated by 40 K. Use the values above.
The key idea: both rods carry the same compressive force, because the junction is in equilibrium. The total length cannot change.
Free expansion of the pair: L(α_s + α_b)ΔT.
Compression of the pair under stress σ: L(σ/Y_s + σ/Y_b).
Set them equal: σ = (α_s + α_b)ΔT / (1/Y_s + 1/Y_b).
Numerator: (1.2 + 2.0) × 10⁻⁵ × 40 = 1.28 × 10⁻³.
Denominator: 1/(2.0 × 10¹¹) + 1/(0.91 × 10¹¹) = 5.0 × 10⁻¹² + 1.099 × 10⁻¹¹ = 1.599 × 10⁻¹¹ Pa⁻¹.
σ ≈ 8.0 × 10⁷ Pa.
The junction moves. Steel's net change is 0.50 × (4.8 × 10⁻⁴ − 4.0 × 10⁻⁴) ≈ +4.0 × 10⁻⁵ m, and brass shortens by the same amount. So the junction shifts about 0.04 mm towards the brass end.
Which mistakes cost marks?
Including the length. Thermal stress does not depend on L unless there is a gap.
Using the whole expansion when there is a gap. Only the blocked part creates strain.
Giving each rod in series its own YαΔT. In series, the force is shared, not the strain.
Wrong sign on cooling. A rod held between supports and cooled goes into tension, not compression. A taut steel wire cooled by 20 K carries an extra tensile stress of 2.0 × 10¹¹ × 1.2 × 10⁻⁵ × 20 = 4.8 × 10⁷ Pa.
Mixing °C and K for ΔT. A change of 1 °C equals a change of 1 K, so either works for ΔT.
Our posts on thermal expansion problems and stress, strain and Young's modulus cover each half of this topic on its own.
A practice routine
Derive F = YAαΔT from the two-step picture, without notes.
Solve NCERT Exercise 10.9 on a cooled brass wire, writing the "free, then squeeze" steps.
Do two gap problems: one where the gap closes and one where it does not.
Do one series-rod problem and find how far the junction moves.
When you cannot see which part of the expansion is blocked, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
Today, rework example 2 with a 0.50 mm gap and find the new stress.
