Learning topic
Thermal Deformations and Stresses
Calculate free thermal expansion and thermal stress in restrained bars using αΔT and EαΔT, with compatibility equations and a worked example.
A temperature change causes thermal expansion or contraction of a material. If a bar is free to change length, the temperature change produces deformation without mechanical stress. If movement is restrained by supports or other structural members, thermal stresses develop.
Free thermal strain
For a uniform bar subjected to a uniform temperature change and free thermal expansion:
$$\varepsilon_T=\alpha\Delta T,\qquad \Delta l_T=\alpha\Delta T\,L.$$
- $\varepsilon_T$ — thermal strain;
- $\alpha$ — coefficient of linear thermal expansion, 1/°C or 1/K;
- $\Delta T$ — temperature change;
- $L$ — initial length;
- $\Delta l_T$ — free thermal change in length.
For heating, $\Delta T>0$ and the bar tends to elongate; for cooling, $\Delta T<0$ and it tends to contract. If free movement is permitted, thermal strain alone does not create mechanical stress.
The coefficient $\alpha$ depends on the material and temperature range. In elementary calculations it is commonly treated as constant over the specified temperature change.
Fully restrained bar
Consider a bar whose ends cannot move axially. During heating, it would tend to elongate, but the restraints prevent this motion. The support reactions create a mechanical strain opposite to the free thermal strain.
If a uniform linearly elastic bar cannot change its length during a uniform temperature change, the total axial strain is zero:
$$\varepsilon=\varepsilon_\sigma+\varepsilon_T=\frac{\sigma}{E}+\alpha\Delta T=0.$$
Therefore:
$$\sigma_T=-E\alpha\Delta T.$$
- $\sigma_T$ — thermal normal stress;
- $E$ — Young's modulus;
- $\alpha$ — coefficient of linear thermal expansion;
- $\Delta T$ — temperature change.
With tension taken as positive, heating of a fully restrained bar produces compression, while cooling produces tension.
Partial restraint
In the general case, displacement need not be zero. Thermal and mechanical components are then considered together. For a bar segment in a linear-elastic model:
$$\Delta l=\frac{NL}{EA}+\alpha\Delta T\,L.$$
The signs of the terms follow the selected convention: tension and elongation are positive, while compression and shortening are negative.
Example
A steel bar with $E=200\ \text{GPa}$ and $\alpha=12\cdot10^{-6}\ \text{°C}^{-1}$ is fully restrained and heated by $\Delta T=40\ \text{°C}$. Assuming linear-elastic behavior:
$$\sigma_T=-200000\cdot12\cdot10^{-6}\cdot40=-96\ \text{MPa}.$$
The resulting thermal stress is therefore $96\ \text{MPa}$ in compression.
Engineering significance
Thermal stresses are important in pipelines, rails, long metal structures, machine components, and assemblies made from materials with different thermal expansion coefficients. Expansion joints, compensators, and details that permit controlled movement are used to reduce them.
Calculation procedure
First determine the free thermal deformation, then identify the kinematic restraints. For a statically indeterminate system, write deformation-compatibility equations and solve them together with equilibrium equations to obtain reactions and internal forces.
About this topic
Temperature changes cause thermal expansion or contraction in structural materials. In statically indeterminate systems, constrained expansion induces severe thermal stresses. This section presents thermal stress evaluation formulas and code compliance procedures.