Thermal stresses and statically indeterminate axial members

Why it matters

An aircraft sees temperature swings of 80 °C or more between ground and cruise, and it is built from materials that expand at different rates: aluminium skins on titanium fittings, steel bolts through composite panels. Where parts are not free to expand, the mismatch becomes stress, often large enough to matter. The same "equilibrium plus compatibility" method also solves any axial member with more supports than statics can handle.

Key ideas

Free thermal strain. An unrestrained bar heated by ΔT changes length by δ_T = α·L·ΔT with no stress at all. Thermal strain alone never causes stress.

Thermal stress appears only when the expansion is prevented. If both ends of a bar are held by rigid walls, the total strain must be zero, so the mechanical strain must cancel the thermal strain: σ/E + α·ΔT = 0, giving σ = −E·α·ΔT. Heating a restrained bar puts it in compression; cooling it puts it in tension. The stress does not depend on the length or the area of the bar; the wall reaction force does (P = σ·A).

Partial restraint. If there is a gap Δ between the bar end and the wall, the bar first expands freely. Stress develops only if α·L·ΔT > Δ, and then σ = −E·(α·L·ΔT − Δ)/L. If the supports yield elastically, the stress is lower again.

Statically indeterminate axial members. When there are more unknown forces than independent equilibrium equations, add one compatibility equation per extra unknown, written in terms of deformations, then link forces to deformations with δ = PL/(AE) + αLΔT. The steps are always:

1. Equilibrium: write the force balance with the unknowns.
2. Compatibility: write the geometric condition (total length change zero, equal deformations of parallel members, and so on).
3. Force–deformation: substitute δ = PL/(AE) (+ αLΔT if temperature changes).
4. Solve, then check the signs.

Bar fixed at both ends with an intermediate load. For a uniform bar of length L with a load P applied at distance a from end A (b from end B), compatibility gives R_A = P·b/L and R_B = P·a/L. The nearer support carries the larger share. With 60 kN applied 0.4 m from A on a 1.2 m bar, R_A = 40 kN and R_B = 20 kN.

Compound bars (members in parallel). A steel rod inside a copper tube, joined at the ends, must have the same final length. Under an axial load the two share it in proportion to their axial rigidities AE. Under a temperature rise the member with the larger α is held back (compression) and the other is pulled along (tension); the two forces are equal and opposite because there is no external load.

Assumptions: linear elastic behaviour, uniform temperature across the section, no buckling of the compressed member, rigid end connections unless stated, α and E constant over the temperature range.

Formulas

δ_T = α·L·ΔT Free thermal expansion (m); α = coefficient of thermal expansion (1/°C or 1/K), L = length (m), ΔT = temperature change (°C or K).

σ = −E·α·ΔT Stress in a bar fully restrained between rigid walls (Pa); negative = compression for a temperature rise.

σ = −E·(α·L·ΔT − Δ) / L Restrained bar with an initial gap Δ (m), valid only when α·L·ΔT > Δ.

δ = P·L/(A·E) + α·L·ΔT Total length change of a member carrying force P (tension +) and temperature change ΔT.

R_A = P·b / L, R_B = P·a / L Uniform bar fixed at both ends, axial load P at distance a from A and b from B (a + b = L).

P = (α₁ − α₂)·ΔT / [1/(A₁·E₁) + 1/(A₂·E₂)] Force in each member of a two-material compound bar under ΔT (N); member 1 (larger α) in compression, member 2 in tension.

Worked examples

Example 1 (standard). A steel rail of length 2 m is placed between two rigid walls with a gap of 0.5 mm at one end. The temperature rises by 40 °C. α = 12 × 10⁻⁶ /°C, E = 200 GPa. Find the stress (a) if there were no gap and (b) with the gap.

1. Full restraint: σ = −E·α·ΔT = −200 000 × 12 × 10⁻⁶ × 40 = −96 MPa.
2. Free expansion: δ_T = α·L·ΔT = 12 × 10⁻⁶ × 2000 × 40 = 0.96 mm.
3. The gap of 0.5 mm closes, so 0.96 − 0.5 = 0.46 mm of expansion is prevented.
4. σ = −E·(α·L·ΔT − Δ)/L = −200 000 × 0.46 / 2000 = −46 MPa.

Answer: (a) 96 MPa compression, (b) 46 MPa compression.

Example 2 (GATE level). A steel rod (A_s = 500 mm², E_s = 200 GPa, α_s = 12 × 10⁻⁶ /°C) is placed inside a copper tube (A_c = 1000 mm², E_c = 100 GPa, α_c = 17 × 10⁻⁶ /°C) of the same length, and the ends are rigidly joined. The assembly is heated by 80 °C. Find the stress in each.

1. Equilibrium: no external load, so tension P in steel = compression P in copper.
2. Compatibility: both end up with the same length, so α_s·ΔT + P/(A_s·E_s) = α_c·ΔT − P/(A_c·E_c).
3. Rearranged: P = (α_c − α_s)·ΔT / [1/(A_s·E_s) + 1/(A_c·E_c)].
4. Numerator: 5 × 10⁻⁶ × 80 = 4.0 × 10⁻⁴.
5. Denominator: 1/(500 × 200 000) + 1/(1000 × 100 000) = 1.0 × 10⁻⁸ + 1.0 × 10⁻⁸ = 2.0 × 10⁻⁸ N⁻¹.
6. P = 4.0 × 10⁻⁴ / 2.0 × 10⁻⁸ = 20 000 N.
7. σ_s = 20 000/500 = 40 MPa (tension); σ_c = 20 000/1000 = 20 MPa (compression).
8. Check: steel strain = 12 × 10⁻⁶ × 80 + 40/200 000 = 1.16 × 10⁻³; copper strain = 17 × 10⁻⁶ × 80 − 20/100 000 = 1.16 × 10⁻³. Equal, as required.

Answer: steel 40 MPa tension, copper 20 MPa compression.

Common mistakes

• Calling the stress in a heated, restrained bar tensile. Heating with restraint gives compression; cooling gives tension.
• Thinking the bar length affects the fully restrained thermal stress. It cancels out.
• Forgetting to subtract the gap, or applying the gap formula when the gap never closes.
• In compound bars, writing equal stresses instead of equal strains (or equal length changes).
• Mixing mm and m inside α·L·ΔT.

For GATE AE

Expect numericals on fully and partially restrained bars, bars with a gap, two-material compound bars under temperature change or load, and bars fixed at both ends with a load in the span. Conceptual questions test the sign of the thermal stress and its independence from length. Practise the four-step equilibrium–compatibility routine until it is automatic, and always check that the final strains satisfy compatibility.

Quick check

1. A bar between rigid walls is cooled. Is the stress tensile or compressive?
2. Does doubling the length of a fully restrained bar change its thermal stress?
3. A uniform bar fixed at both ends carries 30 kN at its mid-length. What are the reactions?
4. Compute E·α·ΔT for E = 70 GPa, α = 23 × 10⁻⁶ /°C, ΔT = 50 °C.

Answers: 1. Tensile. 2. No. 3. 15 kN each. 4. 80.5 MPa.