Thermal stresses and compound bars

Why it matters

Mechatronic assemblies mix materials and warm up in service: a steel ball screw between fixed bearings on an aluminium frame, steel bolts clamping an aluminium motor housing, copper tracks and solder joints on an FR-4 circuit board. When thermal expansion is blocked, stresses appear with no external load at all, and they can buckle a screw, loosen a bolted joint or crack a solder joint. Bimetallic thermostats and thermal actuators use the same effect on purpose.

Key ideas

Free thermal strain. A bar free to move changes length by δ_T = α·L·ΔT and has no stress. Thermal strain is α·ΔT, independent of length.

Thermal stress needs restraint. Stress appears only when the free expansion is prevented, fully or partly. If a bar is held between rigid supports and heated, the supports push back with exactly the force needed to cancel the free expansion: σ = E·α·ΔT, compressive on heating and tensile on cooling. This stress is independent of the bar's length and cross-sectional area.

Partial restraint. If there is a gap Δ before the bar touches a support, only the part of the free expansion beyond the gap is suppressed: σ = E·(α·L·ΔT − Δ)/L when α·L·ΔT > Δ. If the supports themselves yield elastically (springs, a flexible frame), the suppressed expansion is shared between bar and support.

The general method: superposition plus compatibility. Every thermal-stress problem is solved the same way:

1. Let each part expand freely (thermal strain α·ΔT).
2. Apply the unknown forces needed to bring the parts back into compatibility (mechanical strain σ/E).
3. Write equilibrium for the forces and compatibility for the displacements, then solve. Total strain in each part = α·ΔT + σ/E, with σ positive in tension.

Bars in series between rigid walls. Both segments carry the same force P (equilibrium). Compatibility: the sum of the thermal extensions equals the sum of the mechanical shortenings, Σ α_i·L_i·ΔT = Σ P·L_i/(A_i·E_i).

Compound bars (parallel, rigidly joined at the ends). For example a steel bolt inside an aluminium sleeve, or a steel rod cast into a copper tube. The two parts must end up with the same length (same total strain), and since there is no external load the force in one equals the force in the other, one tensile and one compressive. The material with the higher α is in compression after heating (it wanted to expand more and is held back); the other is pulled out into tension. On cooling the signs reverse.

Combined with mechanical load. Superpose: find stresses from the external load sharing (strains equal, loads add) and from temperature separately, then add.

Practical points. α for steel is about 11 to 13 × 10⁻⁶ /°C and for aluminium about 23 × 10⁻⁶ /°C; take the exact values for your alloys from a data book. Long screws are often mounted fixed at one end and floating at the other so that they can expand. If a fully restrained slender bar is heated, it can buckle before it yields; compare the compressive force with the Euler load.

Formulas

δ_T = α·L·ΔT

• δ_T: free thermal elongation (m); α: coefficient of linear thermal expansion (1/°C or 1/K); L: length (m); ΔT: temperature change (°C or K).

σ = E·α·ΔT

• Thermal stress in a bar fully restrained between rigid supports (Pa); E: Young's modulus (Pa). Compressive on heating.

σ = E·(α·L·ΔT − Δ) / L

• Bar with initial gap Δ (m) to a rigid support; valid only when α·L·ΔT > Δ, otherwise σ = 0.

σ₁·A₁ = σ₂·A₂

• Compound bar with no external load: force in part 1 equals force in part 2 (one tensile, one compressive). A: cross-sectional area (m²).

σ₁/E₁ + σ₂/E₂ = (α₂ − α₁)·ΔT

• Compatibility for a compound bar with α₂ > α₁; σ₁ (tension in the low-α part) and σ₂ (compression in the high-α part) as magnitudes.

ε_final = α₁·ΔT + σ₁/E₁ = α₂·ΔT − σ₂/E₂

• Common final strain of the compound bar; length change = ε_final·L.

Worked examples

Example 1 (standard): ball screw between fixed bearings. Given: a steel ball screw of root diameter 25 mm and length 1 m is held between two fixed bearings on a rigid base. It warms by 30 °C in service. E = 200 GPa, α = 12 × 10⁻⁶ /°C. Find the thermal stress and axial force; then repeat if one bearing allows 0.2 mm axial float before it bottoms.

1. Full restraint: σ = E·α·ΔT = 200 000 × 12 × 10⁻⁶ × 30 = 72 MPa compression.
2. Area: A = π/4 × 25² = 490.9 mm². Force: P = 72 × 490.9 = 35 340 N.
3. With float: free expansion δ_T = α·L·ΔT = 12 × 10⁻⁶ × 1000 × 30 = 0.36 mm. The gap absorbs 0.2 mm, so 0.16 mm is suppressed.
4. σ = E·(α·L·ΔT − Δ)/L = 200 000 × 0.16/1000 = 32 MPa compression. Answer: 72 MPa and about 35.3 kN with full restraint; 32 MPa with 0.2 mm float. A 35 kN axial load on a long, thin screw can approach its buckling load, which is why one end is normally left floating.

Example 2 (GATE level): steel bolt in an aluminium sleeve. Given: a steel bolt (A_s = 200 mm², E_s = 200 GPa, α_s = 12 × 10⁻⁶ /°C) passes through an aluminium sleeve (A_a = 400 mm², E_a = 70 GPa, α_a = 23 × 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 and the force.

1. Aluminium has the larger α, so it is compressed and the bolt is stretched.
2. Equilibrium: σ_s·A_s = σ_a·A_a, so σ_a = σ_s × 200/400 = 0.5·σ_s.
3. Compatibility: σ_s/E_s + σ_a/E_a = (α_a − α_s)·ΔT = 11 × 10⁻⁶ × 60 = 6.6 × 10⁻⁴.
4. Substitute: σ_s × (1/200 000 + 0.5/70 000) = 6.6 × 10⁻⁴, that is σ_s × 1.2143 × 10⁻⁵ = 6.6 × 10⁻⁴.
5. σ_s = 54.4 MPa (tension); σ_a = 27.2 MPa (compression). Force = 54.35 × 200 = 10.9 kN.
6. Check the common strain: steel 12 × 10⁻⁶ × 60 + 54.35/200 000 = 9.92 × 10⁻⁴; aluminium 23 × 10⁻⁶ × 60 − 27.18/70 000 = 9.92 × 10⁻⁴. They agree. Answer: bolt 54.4 MPa tension, sleeve 27.2 MPa compression, force ≈ 10.9 kN. On cooling back, the stresses vanish; on cooling below assembly temperature, the joint would lose clamp force.

Common mistakes

• Writing σ = E·α·ΔT for a bar that is free at one end. No restraint means no stress.
• Using σ = E·α·ΔT for each material of a compound bar separately; the parts restrain each other, so you need equilibrium and compatibility.
• Getting the sign wrong: the higher-α material is in compression on heating, not in tension.
• Assuming stresses in a compound bar are equal; it is the forces that are equal (stresses are inversely proportional to areas).
• Ignoring an initial gap, or applying the gap formula when the expansion never closes it.
• Mixing mm and m in α·L·ΔT.

For GATE ME

Expect fully restrained bars (stress and reaction force), bars with a gap, two bars in series between walls, compound bars of two materials heated or cooled, and combinations of thermal and mechanical load. Practise the free-expansion-then-compatibility method; it handles every variant.

Quick check

1. A brass bar (E = 100 GPa, α = 19 × 10⁻⁶ /°C) is fully restrained and heated by 40 °C. What is the stress?
2. Does the thermal stress in a fully restrained bar depend on its length?
3. A 2 m steel rod (α = 12 × 10⁻⁶ /°C) is heated by 40 °C and is free. What is its stress and its elongation?
4. In a heated steel-aluminium compound bar, which part is in compression?
5. A bar's free expansion is 0.3 mm and the gap to the wall is 0.5 mm. What is the thermal stress?

Answers: 1. 76 MPa compression. 2. No. 3. Zero stress; 0.96 mm. 4. Aluminium (higher α). 5. Zero, the gap never closes.