Thermal Stress and Deformation

Understanding thermal stress and deformation is critical for engineers because all materials change size with temperature, but real-world structures often prevent that change. From buckling railway tracks on a hot day to cracked engine blocks from rapid cooling, the forces generated when thermal expansion is constrained can cause catastrophic failure if not properly designed for. This analysis provides the foundational principles and problem-solving framework to predict and manage these stresses in engineering systems.

Free Thermal Expansion

When a material experiences a temperature change without any external restraints, it undergoes free thermal expansion (or contraction). This dimensional change is proportional to the original size of the object, the magnitude of the temperature change, and a material property known as the coefficient of thermal expansion ($\alpha$). For a one-dimensional member like a rod or beam, the change in length (${\delta }_T$) is given by the fundamental equation:

$${\delta }_T=\alpha \Delta T{L}_0$$

Here, $\alpha$ is the coefficient of thermal expansion (units of $1{/}^{\circ }\text{C}$ or $1{/}^{\circ }\text{F}$), $\Delta T$ is the change in temperature (final temperature minus initial temperature), and $L_0$ is the original length. A positive $\Delta T$ leads to expansion, while a negative $\Delta T$ causes contraction. This is a straightforward strain-displacement relationship where the thermal strain (${ϵ}_T$) is simply $\alpha \Delta T$. For example, a 10-meter steel rail ($\alpha =12\times 10^{-6}{/}^{\circ }\text{C}$) heated by $30^{\circ }\text{C}$ would freely expand by ${\delta }_T=(12\times 10^{-6})(30)(10)=0.0036\text{m}$, or 3.6 mm.

Constrained Expansion and the Development of Thermal Stress

In practice, components are rarely free to expand or contract. They are connected to other parts, embedded in rigid foundations, or otherwise constrained. If a member that wishes to expand (or contract) by ${\delta }_T$ is completely prevented from doing so, a force and corresponding stress will develop within it.

Consider a metal rod firmly clamped between two rigid walls. Upon heating, the rod "wants" to elongate by ${\delta }_T$. Since the walls prevent this, they exert a compressive force on the rod, shortening it by an amount ${\delta }_F$ due to mechanical load. For the rod to fit between the unchanged wall positions, the net displacement must be zero. This gives the compatibility condition: ${\delta }_T+{\delta }_F=0$. The force-related displacement is given by mechanics: ${\delta }_F=\frac{PL}{AE}$, where $P$ is the axial force, $A$ is the cross-sectional area, and $E$ is the modulus of elasticity. Substituting and solving reveals the thermal stress ($\sigma$) induced:

$$\sigma =\frac{P}{A}=-E\alpha \Delta T$$

The negative sign indicates the stress is compressive for a positive $\Delta T$ (heating). If the rod were cooled ($\Delta T<0$), the stress would be tensile. Crucially, this stress is independent of the member's length or cross-sectional area; it depends only on material properties ($E$, $\alpha$) and the temperature change. This is a key insight: in a fully axially restrained member, the thermal stress is the same whether the part is short or long, thick or thin.

Solving Statically Indeterminate Thermal Stress Problems

Many real systems are statically indeterminate, meaning the forces in the members cannot be found by equilibrium equations alone. Thermal loads are a classic source of self-equilibrating stresses in such systems. Solving these problems requires a methodical three-step approach combining equilibrium, compatibility, and force-displacement relationships.

Step 1: Equilibrium. Draw a free-body diagram and write the relevant static equilibrium equations (e.g., $\sum F_x=0$, $\sum M=0$). You will find that these equations are insufficient to solve for all unknown reaction forces.

Step 2: Compatibility. Examine the geometry of the deformed system. The key is to express how the displacements of various members are related based on the constraints. For instance, if two different materials are bonded together, their total elongations must be equal. This creates an additional equation.

Step 3: Force-Temperature-Displacement Relations. For each member, express its total deformation as the sum of its thermal deformation and its mechanical deformation due to axial force. That is, ${\delta }_{total}={\delta }_T+{\delta }_F=\alpha L\Delta T+\frac{PL}{AE}$.

You then substitute these displacement expressions into the compatibility equation from Step 2. This yields a new equation in terms of the unknown forces and the known $\Delta T$. Finally, you solve this equation simultaneously with the equilibrium equations from Step 1 to find all internal forces and stresses.

Worked Example: Composite Bar

Imagine a steel rod ($A_s$, $E_s$, ${\alpha }_s$) inside a copper tube ($A_c$, $E_c$, ${\alpha }_c$) of the same length, rigidly connected at the ends to plates. The assembly is heated by $\Delta T$. Equilibrium: The force in the steel ($P_s$) and copper ($P_c$) are equal and opposite: $P_s=-P_c=P$. Compatibility: As the end plates are rigid, the total elongation of the steel must equal the total elongation of the copper: ${\delta }_s={\delta }_c$. Relations: $${\delta }_s={\alpha }_{s}L\Delta T+\frac{PL}{A_s{E}_s}$$ $${\delta }_c={\alpha }_{c}L\Delta T+\frac{(-P)L}{A_c{E}_c}$$ Setting ${\delta }_s={\delta }_c$ and solving for $P$ gives the internal force. The stresses follow as ${\sigma }_s=P/A_s$ and ${\sigma }_c=-P/A_c$.

Common Pitfalls

1. Assuming Zero Stress in "Unaltered" Systems: A common error is assuming that if a structure is not externally loaded, its stresses are zero. A temperature change in a constrained, indeterminate system can generate significant internal stresses even with no applied loads. Always consider thermal effects as a potential load case.
2. Ignoring the Nature of Constraints: Not all constraints are fully rigid. A partially yielding support or an elastic connection will change the compatibility condition and reduce the magnitude of thermal stress. Carefully analyze the boundary conditions to determine if displacement is truly zero or merely related to a reaction force via a spring constant.
3. Mismatching Units and Sign Conventions: Inconsistent use of units for $\alpha$, $\Delta T$, and $E$ is a frequent source of numerical error. Similarly, maintaining a consistent sign convention for $\Delta T$ (positive for increase) and deformation (positive for elongation) is crucial for the compatibility equation to be correct. A sign error here can flip tensile stress to compressive.
4. Overlooking Stress Relief Mechanisms: In analysis, we often assume perfectly brittle or linear-elastic behavior. In reality, if thermal stresses exceed the yield strength, plastic deformation will occur, relieving stress. For ductile materials under cyclic thermal loading, this can lead to thermal fatigue, a failure mode not predicted by a simple elastic analysis.

Summary

• Free thermal expansion is calculated as ${\delta }_T=\alpha \Delta T{L}_0$. When this expansion or contraction is fully or partially prevented, thermal stresses develop, which can be tensile or compressive depending on the temperature change and constraint.
• In a perfectly rigid, axially constrained member, the induced thermal stress is $\sigma =-E\alpha \Delta T$, independent of the member's dimensions.
• Solving statically indeterminate thermal stress problems requires supplementing equilibrium equations with a compatibility condition derived from the geometry of deformation, and then using the force-temperature-displacement relation (${\delta }_{total}=\alpha L\Delta T+PL/AE$) for each member.
• Accurate analysis demands careful attention to boundary conditions, consistent units and sign conventions, and an awareness of real-world material behaviors like yielding and fatigue that can modify purely elastic predictions.