Thermal expansion and stress in constrained members explains how structural and machine elements change length when temperature changes. This cheat sheet helps students connect free expansion, restraint, compatibility, and internal axial force in bars, rods, bolts, and composite members. These ideas are essential for designing bridges, frames, piping, rails, and mechanical assemblies that must survive temperature variation.
It is especially useful when solving mechanics of materials problems involving supports or materials with different coefficients of thermal expansion.
The central idea is that a free member expands by delta_T length = alpha L delta_T, but a restrained member develops stress because that expansion is prevented. For a fully restrained uniform bar, thermal stress is sigma = -E alpha delta_T, where compression occurs for heating and tension occurs for cooling. In indeterminate systems, equilibrium must be combined with deformation compatibility, such as total change in length being zero or equal to a known gap.
Composite members require each material's modulus, area, length, and coefficient of thermal expansion to be included in the compatibility equations.
Key Facts
- Free thermal elongation is delta_L = alpha L delta_T, where alpha is the coefficient of thermal expansion.
- Thermal strain for an unconstrained member is epsilon_T = alpha delta_T.
- Mechanical strain in an axially loaded member is epsilon_m = sigma / E = P / (A E).
- Total axial strain is epsilon_total = epsilon_m + epsilon_T when both load and temperature change act in the same direction convention.
- For a fully restrained uniform member with no change in length, sigma = -E alpha delta_T.
- A positive temperature change in a fully restrained member usually creates compressive stress, while cooling usually creates tensile stress.
- For a prismatic axial member, mechanical deformation is delta_m = P L / (A E).
- Statically indeterminate thermal problems require both equilibrium, such as sum F = 0, and compatibility, such as sum delta = 0.
Vocabulary
- Coefficient of thermal expansion
- The material property alpha that gives the thermal strain produced per unit temperature change.
- Thermal strain
- The strain caused only by a temperature change, calculated as epsilon_T = alpha delta_T.
- Constrained member
- A structural member whose free thermal expansion or contraction is partly or fully prevented by supports or connected parts.
- Compatibility
- The deformation condition that connected parts must satisfy, such as equal displacement or zero net change in length.
- Thermal stress
- The stress produced when a temperature change is restrained and the member cannot freely expand or contract.
- Composite member
- A member made from two or more materials that deform together but may have different E, A, L, and alpha values.
Common Mistakes to Avoid
- Using delta_L = alpha delta_T without multiplying by length is wrong because thermal elongation depends on the original length L.
- Assigning the wrong sign to thermal stress is wrong because heating a fully restrained bar creates compression, not tension, under the usual tension-positive convention.
- Treating every thermal problem as fully restrained is wrong because partial restraint, gaps, springs, or flexible supports change the compatibility condition.
- Forgetting mechanical strain in compatibility is wrong because the actual length change includes both P L / (A E) and alpha L delta_T.
- Using one material property for a composite member is wrong because each material may have different E, A, and alpha values and therefore different force sharing.
Practice Questions
- 1 A steel bar has L = 2.0 m, alpha = 12 x 10^-6 /C, and is heated by 50 C while free to expand. What is its change in length?
- 2 A fully restrained aluminum rod has E = 70 GPa, alpha = 23 x 10^-6 /C, and is heated by 40 C. Find the thermal stress and state whether it is tensile or compressive.
- 3 A brass bar with A = 600 mm^2, E = 100 GPa, L = 1.5 m, and alpha = 19 x 10^-6 /C is fixed between rigid walls and cooled by 30 C. What axial force develops in the bar?
- 4 Why can a two-material composite bar develop internal stress during heating even when no external axial load is applied?
Understanding Thermal Expansion and Stress in Constrained Members
Temperature stress depends on the reference condition. A bar is not naturally stress free at every temperature. It is stress free only at the temperature and shape at which it was assembled or installed.
Engineers call this the stress free reference state. A rod fitted exactly between rigid walls on a cool day can be compressed later when the building warms up. If the rod was installed with a gap, it expands freely until that gap closes.
Stress begins only after contact occurs. This is why installation temperature, initial clearances, bolt tightening, and support movement matter as much as the final operating temperature.
Real supports are rarely perfectly rigid. A wall, bracket, bolt, or neighboring member can stretch under force. That flexibility reduces the thermal force because part of the attempted expansion is allowed by movement of the support system.
In a calculation, the deformation of every connected part must be included. A long slender bar is more flexible than a short thick bar made from the same material. Its axial stiffness is controlled by area, modulus, and length.
The member with lower stiffness takes more of the deformation needed to satisfy compatibility. This is a useful way to check results before doing detailed algebra.
Composite members show why material properties cannot be treated separately. Consider two strips firmly bonded along their length. If heated, each strip tries to expand by a different amount when their expansion coefficients differ.
Bonding forces them to end at the same length, so internal forces develop even when no outside load is applied. One material may be placed in tension while the other is in compression. The force balance requires equal and opposite internal forces across a cut.
The shared length change supplies the compatibility condition. This behavior appears in bimetallic thermostat strips, laminated panels, electronic circuit boards, concrete reinforced with steel, and pipes with protective layers.
Students should choose a sign convention before writing equations and keep it throughout the work. Treat elongation as positive and compression as negative, or use another consistent choice. Separate the thermal part of deformation from the force caused part, then add them for each member.
Draw a simple displacement sketch showing which points can move and which points must end together. For nonuniform temperature, split the member into regions because each region has its own temperature change and free expansion. Check units carefully.
Expansion coefficients are usually given per degree Celsius or per kelvin, while modulus may be in gigapascals. A final physical check is essential. Heating against restraint should not produce a tensile result unless another load or geometry effect changes the situation.