Thermal stresses and compound bars
Thermal strain versus thermal stress, fully and partially restrained bars, bars in series, and compound bars solved by equilibrium plus compatibility, with a gap problem and a rod-in-tube problem.
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Why it matters
Engines, exhaust systems, brake discs and battery packs heat up and cool down every time a vehicle runs. A part that wants to expand but is held by bolts, a press fit or a stiffer neighbour develops stress even with no external load. Bimetal parts, aluminium pistons in steel-lined bores, and steel bolts clamping aluminium heads are all compound systems whose stresses come from the same two ideas: equilibrium and compatibility.
Key ideas
Free thermal expansion. An unrestrained bar heated by ΔT simply grows by α·L·ΔT and has no stress. Thermal strain on its own causes no stress; stress appears only when the expansion is prevented.
Fully restrained bar. If a bar is held between rigid supports, the supports must push back enough to cancel the whole free expansion. The mechanical strain equals minus the thermal strain, so σ = −E·α·ΔT (compression on heating, tension on cooling). The stress does not depend on length or area.
Partial restraint (gap or yielding support). If the support allows a gap Δ before contact, or itself yields by Δ, only the expansion beyond Δ is suppressed: σ = E·(α·L·ΔT − Δ)/L, provided α·L·ΔT > Δ.
Total strain bookkeeping. In every problem write: total strain = thermal strain + mechanical strain, ε = α·ΔT + σ/E. Compatibility conditions are statements about total strain or total length change; Hooke's law applies only to the mechanical part.
Bars in series between rigid walls. Two bars end to end between fixed walls carry the same force P (equilibrium). Compatibility: the sum of their total length changes is zero (or equals any gap). Stresses differ if areas differ.
Compound bars (bars in parallel). Two materials joined so they must change length together (a rod inside a tube with end plates, a bimetal strip, a bolt through a sleeve).
- Equilibrium with no external load: the force in one equals the force in the other, opposite in sign: σ₁·A₁ = σ₂·A₂ in magnitude.
- Compatibility: both have the same total strain.
- Result on heating: the material with the larger α is held back and goes into compression; the one with smaller α is pulled along into tension. With an external axial load P as well, the load shares in proportion to axial stiffness A·E, and thermal stresses are superposed on top.
Assumptions and limits. Linear elastic behaviour, uniform temperature through each member, α and E constant over the range, and no buckling of the compressed member. Real values of α and E depend on the alloy and temperature, so use data-book values. For long rails and pipelines, thermal compression can cause buckling, which is why expansion joints, loops and bellows are used.
Formulas
δ_free = α·L·ΔT
- α: coefficient of linear expansion (1/°C or 1/K); L: length (m); ΔT: temperature change (°C or K).
σ = E·α·ΔT
- Stress in a fully restrained bar (Pa); compressive when heated, tensile when cooled.
σ = E·(α·L·ΔT − Δ) / L
- Δ: gap or support yield (m). Valid only when α·L·ΔT exceeds Δ.
ε_total = α·ΔT + σ / E
- Total strain = thermal + mechanical. Use this in every compatibility equation.
Compound bar, heating, no external load (material 1 has the larger α):
σ₁·A₁ = σ₂·A₂ (1 in compression, 2 in tension)
(α₁ − α₂)·ΔT = σ₁ / E₁ + σ₂ / E₂
- A: cross-section area (m²); E: modulus (Pa).
σ₂ = (α₁ − α₂)·ΔT·E₁·E₂·A₁ / (E₁·A₁ + E₂·A₂)
- Closed form for the tensile stress in the low-α member; σ₁ follows from equilibrium.
P₁ = P·A₁·E₁ / (A₁·E₁ + A₂·E₂)
- Share of an external axial load P carried by member 1 of a compound bar.
Worked examples
Example 1 (standard): restrained bar with a gap. Given: a 2 m steel rod (E = 200 GPa, α = 12 × 10⁻⁶ /°C) is fixed at one end and has a 0.5 mm gap to a rigid wall at the other. It is heated by 60 °C. Find the stress.
- Free expansion:
δ_free = α·L·ΔT= 12 × 10⁻⁶ × 2000 × 60 = 1.44 mm. - Expansion suppressed by the wall: 1.44 − 0.5 = 0.94 mm.
σ = E·(suppressed)/L= 200 000 × 0.94 / 2000 = 94 MPa. Answer: 94 MPa compression. Without the gap it would be E·α·ΔT = 144 MPa.
Example 2 (GATE level): compound bar. Given: an aluminium rod (A_a = 600 mm², E_a = 70 GPa, α_a = 23 × 10⁻⁶ /°C) sits inside a steel tube (A_s = 900 mm², E_s = 200 GPa, α_s = 12 × 10⁻⁶ /°C). Both are 500 mm long and joined at their ends by rigid plates. The assembly is heated by 80 °C. Find the stresses and the change in length.
- Aluminium (larger α) goes into compression, steel into tension.
- Equilibrium: σ_a·600 = σ_s·900, so σ_a = 1.5·σ_s.
- Compatibility:
(α_a − α_s)·ΔT = σ_s/E_s + σ_a/E_a, i.e. 11 × 10⁻⁶ × 80 = 8.8 × 10⁻⁴ = σ_s·(1/200 000 + 1.5/70 000). - The bracket is 5.0 × 10⁻⁶ + 2.143 × 10⁻⁵ = 2.643 × 10⁻⁵ per MPa, so σ_s = 8.8 × 10⁻⁴ / 2.643 × 10⁻⁵ = 33.3 MPa.
- σ_a = 1.5 × 33.3 = 49.9 MPa.
- Common strain from the steel: 12 × 10⁻⁶ × 80 + 33.3/200 000 = 9.6 × 10⁻⁴ + 1.665 × 10⁻⁴ = 1.1265 × 10⁻³. Check with aluminium: 1.84 × 10⁻³ − 49.9/70 000 = 1.1265 × 10⁻³.
- Change in length: 1.1265 × 10⁻³ × 500 = 0.563 mm. Answer: steel 33.3 MPa tension, aluminium 49.9 MPa compression, extension 0.563 mm.
Common mistakes
- Applying σ = E·α·ΔT to a compound bar. That formula is only for full restraint by rigid supports.
- Putting thermal strain into Hooke's law. Hooke's law applies to mechanical strain only.
- Getting the senses wrong: on heating, the higher-α material is in compression; on cooling, it is in tension.
- Forgetting that series bars carry the same force, not the same stress.
- Ignoring a gap, or applying the gap formula when the free expansion never closes the gap (then stress is zero).
- Mixing mm and m in α·L·ΔT, or using GPa with mm² without converting.
For GATE ME
Typical questions: stress in a bar between rigid walls with or without a gap, two bars in series between walls, a rod in a tube or a bolt through a sleeve heated or cooled, combined mechanical and thermal loading, and the temperature at which a gap closes. Practise writing equilibrium and compatibility as two separate lines before any numbers.
Quick check
- A steel bar (E = 200 GPa, α = 12 × 10⁻⁶ /°C) is held between rigid walls and cooled by 40 °C. Find the stress.
- In a copper–steel compound bar that is heated, which member is in tension?
- Does the stress in a fully restrained bar depend on its length?
- A bar of length 1 m and α = 20 × 10⁻⁶ /°C has a 0.4 mm gap. At what temperature rise does the gap close?
Answers: 1. 96 MPa tension. 2. Steel (smaller α). 3. No. 4. ΔT = 0.4 / (20 × 10⁻⁶ × 1000) = 20 °C.
Interview questions
All Engineering Mechanics and Strength of Materials interview questionsTry answering each one aloud before you open it.
1.What are thermal stresses and how do they occur in materials?Concept
Thermal stresses are stresses induced in a material due to changes in temperature. They occur when a material is constrained and cannot freely expand or contract with temperature changes. This constraint leads to internal forces that manifest as stresses.
2.Explain the concept of a compound bar in the context of thermal stresses.Concept
A compound bar is two or more members of different materials joined so that they must change length together, such as a rod inside a tube with end plates, a bolt through a sleeve, or a bimetal strip. When the temperature changes, each would like to expand by its own α·L·ΔT, but they are forced to a common length. The member with the larger α is held back and goes into compression throughout its length, the other is pulled into tension, and the two internal forces are equal and opposite because there is no external load.
3.How do you calculate thermal stress in a material?Concept
For a bar fully restrained between rigid supports, the whole free expansion α·L·ΔT is suppressed, so σ = E·α·ΔT, compressive on heating and tensile on cooling, independent of length and area. If there is a gap Δ or the supports yield, only the excess is suppressed: σ = E·(α·L·ΔT − Δ)/L. For compound or series members you cannot use this directly; you write total strain = α·ΔT + σ/E for each member and solve equilibrium and compatibility together.
4.Why is it important to consider thermal stresses in engineering design?Application
Thermal stresses can lead to material failure, warping, or structural damage if not properly accounted for. In engineering design, considering thermal stresses ensures the reliability and safety of structures subjected to temperature variations.
5.What happens if a compound bar is not designed to accommodate thermal stresses?Application
If a compound bar is not designed to accommodate thermal stresses, it can lead to separation at the interface, cracking, or even complete failure of the structure due to the differential expansion or contraction of the materials.
6.How can engineers mitigate the effects of thermal stresses in compound bars?Application
Engineers can mitigate thermal stresses by selecting materials with similar coefficients of thermal expansion, using expansion joints, or designing the structure to allow for some movement. Additionally, thermal insulation can be used to minimize temperature changes.
7.What is the role of the coefficient of thermal expansion in thermal stresses?Concept
The coefficient of thermal expansion (α) determines how much a material expands or contracts with temperature changes. It plays a crucial role in calculating thermal stresses, as materials with higher α will experience greater changes in dimension, leading to higher stresses if constrained.
8.A steel rod (E = 210 GPa, α = 12 × 10⁻⁶ /°C) is heated from 20°C to 100°C while being constrained. Calculate the thermal stress developed.Numerical
First, calculate the change in temperature: ΔT = 100°C - 20°C = 80°C. Then, use the formula for thermal stress: σ = E·α·ΔT = 210 × 10⁹ Pa × 12 × 10⁻⁶ /°C × 80°C = 201.6 MPa.
9.Explain why different materials in a compound bar experience different thermal stresses.Concept
In a compound bar both materials end up with the same total strain, but each would freely expand by a different amount, α·ΔT. The difference has to be made up by mechanical strain: the high-α material is compressed and the low-α material stretched. Equilibrium makes the forces equal and opposite, so the stresses depend on the areas (σ₁·A₁ = σ₂·A₂), and the stiffnesses E·A decide how the mismatch strain is shared.
10.A brass rod (E = 100 GPa, α = 19 × 10⁻⁶ /°C) and an aluminium rod (E = 70 GPa, α = 23 × 10⁻⁶ /°C) are each held separately between rigid walls. If the temperature rises by 50 °C, which rod has the higher thermal stress?Numerical
For full restraint σ = E·α·ΔT. Brass: 100 000 × 19 × 10⁻⁶ × 50 = 95 MPa. Aluminium: 70 000 × 23 × 10⁻⁶ × 50 = 80.5 MPa. Both are compressive and the brass rod has the higher stress, because the product E·α is larger for brass even though its α is smaller. Note that if the two rods were joined as a compound bar instead, the stresses would depend on their areas and would have to be found from equilibrium and compatibility.
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