Thermal stresses and compound bars

Why it matters

Rails, pipelines, bridges, engine parts, bimetallic thermostats and reinforced-concrete members all change temperature in service. A part that is prevented from expanding can develop stresses as large as those from mechanical loads, without any external force. Compound bars (a bolt inside a sleeve, a steel core in an aluminium conductor, steel bars in concrete) share loads and temperature effects between materials, and the same compatibility idea solves both.

Key ideas

Free thermal expansion. A bar of length L heated by ΔT and free to move grows by δ_t = α L ΔT with no stress at all. Thermal strain is ε_t = α ΔT. Heating a free bar does not cause stress.

Restrained bar. If rigid supports prevent the expansion completely, the supports must push the bar back by the full δ_t. The mechanical strain is −α ΔT, so the stress is σ = −E α ΔT: compressive on heating, tensile on cooling. The stress does not depend on length or area. If the support yields by a gap or a known amount Δ, only the part of the expansion that is prevented produces stress.

Superposition approach. Total strain = thermal strain + mechanical strain: ε = α ΔT + σ/E. Then apply the actual geometric constraint (total change in length is zero, equals the gap, or equals that of another member).

Compound bars (parallel bars). Two or more bars of different materials are fixed together at their ends so they must have the same final length. Two conditions solve them:

1. Equilibrium: with no external load, the tensile force in one material equals the compressive force in the other, σ1 A1 = σ2 A2. With an external load P, σ1 A1 + σ2 A2 = P.
2. Compatibility: the final strains are equal, α1 ΔT + σ1/E1 = α2 ΔT + σ2/E2 (stresses signed, tension positive). On heating, the material with the larger α is held back (goes into compression) and the one with the smaller α is pulled out (goes into tension).

Mechanical load on a compound bar. For an axial load alone, equal strain gives σ1/E1 = σ2/E2, so the stiffer material carries more stress: σ1 = P E1 / (A1 E1 + A2 E2). Combine with thermal effects by superposition.

Bars in series (end to end) between rigid walls. Here the force is the same in each part and the total change in length is zero (or equals the gap). Stresses differ only through the areas.

Limits. These formulas assume linear elastic behaviour, uniform temperature through each bar, no buckling of the compressed bar, and constant α and E over the temperature range (take values from a data book).

Formulas

δ_t = α L ΔT

• α = coefficient of linear thermal expansion (1/°C or 1/K), L = length (m), ΔT = temperature change (°C or K).

σ = E α ΔT (magnitude, bar fully restrained; compressive on heating)

σ = E (α L ΔT − Δ) / L (support yields or gap Δ; applies only if α L ΔT > Δ)

σ1 A1 = σ2 A2 (compound bar, no external load; magnitudes)

σ1/E1 + σ2/E2 = (α2 − α1) ΔT (compound bar, σ1 tensile in the low-α material 1, σ2 compressive in the high-α material 2; magnitudes)

σ1 = (α2 − α1) ΔT · E1 / (1 + A1 E1 / (A2 E2))

σ1 = P E1 / (A1 E1 + A2 E2) (axial load P shared by parallel bars)

Worked examples

Example 1 (standard). Steel rails 12 m long are laid with a gap of 3 mm at 20 °C. E = 200 GPa, α = 12 × 10⁻⁶ /°C. Find the stress at 60 °C. What would it be with no gap?

1. ΔT = 40 °C. Free expansion: δ_t = α L ΔT = 12 × 10⁻⁶ × 12 000 × 40 = 5.76 mm.
2. Gap = 3 mm, so expansion prevented = 5.76 − 3 = 2.76 mm.
3. Stress: σ = E × 2.76 / L = 200 000 × 2.76 / 12 000 = 46.0 MPa (compressive).
4. With no gap: σ = E α ΔT = 200 000 × 12 × 10⁻⁶ × 40 = 96.0 MPa (compressive).

Answer: 46 MPa compression with the gap; 96 MPa compression without it.

Example 2 (GATE level, compound bar). A steel bolt (area 400 mm², E = 200 GPa, α = 12 × 10⁻⁶ /°C) passes through a copper tube (area 600 mm², E = 100 GPa, α = 17 × 10⁻⁶ /°C) of the same length; the nut is just snug at room temperature. The assembly is heated by 60 °C. Find the stresses.

1. Copper (higher α) is held back → compression σc; steel is pulled → tension σs.
2. Equilibrium: σs × 400 = σc × 600, so σc = (2/3) σs.
3. Compatibility: σs/Es + σc/Ec = (αc − αs) ΔT = 5 × 10⁻⁶ × 60 = 3 × 10⁻⁴.
4. Substitute: σs/200 000 + (2/3)σs/100 000 = 3 × 10⁻⁴, i.e. σs (5 × 10⁻⁶ + 6.667 × 10⁻⁶) = 3 × 10⁻⁴, so σs = 25.71 MPa.
5. σc = (2/3) × 25.71 = 17.14 MPa.
6. Check equal final strain: steel 12 × 10⁻⁶ × 60 + 25.71/200 000 = 8.486 × 10⁻⁴; copper 17 × 10⁻⁶ × 60 − 17.14/100 000 = 8.486 × 10⁻⁴. ✓

Answer: steel 25.7 MPa tension, copper 17.1 MPa compression.

Common mistakes

• Saying heating causes tensile stress in a restrained bar; it causes compression.
• Calculating thermal stress for a bar that is free to expand (it is zero).
• Using E α ΔT separately for each part of a compound bar; the two parts are coupled by equilibrium and compatibility.
• Forgetting the area ratio in the equilibrium equation.
• Ignoring a gap or yielding support.
• Mixing GPa and MPa: use N and mm with E in N/mm² (200 GPa = 200 000 N/mm²).

For GATE PI

Typical questions: stress in a restrained bar or rail with a gap, compound bars (bolt and sleeve, two parallel rods) heated or loaded, bars in series between walls, and the reaction force at walls. Practise writing the equilibrium and compatibility equations before substituting numbers, and check the answer by comparing the final strains.

Quick check

1. A free steel bar is heated by 50 °C. What stress develops?
2. E = 200 GPa, α = 12 × 10⁻⁶ /°C, ΔT = 50 °C, fully restrained. What is the stress?
3. In a heated compound bar of steel and aluminium, which goes into compression?
4. Does the restrained thermal stress depend on the bar's length?

Answers: 1. None (zero). 2. 120 MPa compressive. 3. Aluminium (larger α). 4. No; only on E, α and ΔT.