Thermal Assembly (Two-Segment Bar Between Rigid Walls)

stress

Take two rods of different metals, weld them end to end, and wedge the pair between two walls stiff enough not to move. Now change the temperature uniformly by $\Delta T$. Each segment wants to grow (or shrink) by its own free thermal strain $\varepsilon_{i,\mathrm{th}} = \alpha_i \Delta T$ — but the walls forbid it. Something has to give, and since the length cannot, a force does: an internal axial force builds up with no external load at all. A rail pinned between fixed abutments on a hot afternoon, a bridge girder without its expansion joint, a bolt clamping parts of mixed metals through a temperature swing — all live here.

This is the temperature-driven cousin of the composite bar: there two materials shared a load; here they share a temperature change, and the “load” is one they manufacture themselves. Like the propped cantilever it is statically indeterminate — a bar built in at both ends has a redundant reaction, and statics alone cannot say how large the force is. The missing equation, again, is geometry.

Free expansion, then the force that cancels it

Equilibrium is almost trivial: the two segments are in series with nothing pushing at their junction, so the same internal force runs through both, $F_1 = F_2$ (call it $F$). That is one equation for one unknown magnitude — but it is the compatibility condition that fixes the value. The rigid walls hold the total length fixed, so the net elongation is zero: whatever each segment expands thermally must be taken straight back out elastically by the force. Segment by segment,

$\alpha_1 L_1 \Delta T + \alpha_2 L_2 \Delta T = \frac{F L_1}{A_1 E_1} + \frac{F L_2}{A_2 E_2}$

Solve the pair together and the force is the total free expansion divided by the total flexibility:

$F = \frac{(\alpha_1 L_1 + \alpha_2 L_2)\,\Delta T}{\dfrac{L_1}{A_1 E_1} + \dfrac{L_2}{A_2 E_2}}$

The build does not ask you to trust this. It certifies that equilibrium and compatibility form a system linear in the unknowns $\{F_1, F_2\}$, runs an exact $2\times2$ solve at build time (the same solveLinear capability the propped cantilever introduced), and checks the result back through every relation. The flexibility sum in the denominator is the system determinant, guarded non-zero. What the machine proves and what still rests on a textbook is on the verification page.

The material really matters here — through the $E\,\alpha$ product

Set both segments to the same material and area and the whole thing collapses to the classic result every mechanics text carries,

$\sigma = E\,\alpha\,\Delta T$

and it holds a surprise. The stress depends on stiffness times expansivity, not expansivity alone. Aluminium expands about twice as much as steel ($\alpha \approx 23$ vs $12 \times 10^{-6}/\mathrm{K}$), so intuition says aluminium should stress more — but steel’s modulus is nearly three times larger, and $E\,\alpha$ wins: a fully restrained steel bar develops higher thermal stress than an aluminium one for the same $\Delta T$. This is the temperature-world echo of the composite bar’s “stiffer carries more,” and of the flywheel’s “titanium deflects more than steel.” Swap either segment’s metal in the widget and watch $F$ and both stresses move:

• Young’s modulus $E$ and expansion coefficient $\alpha$ enter together as the product $E\alpha$ that drives the force; a high-$\alpha$ but compliant metal (aluminium) and a low-$\alpha$ but stiff metal (steel) can push comparably hard.
• Yield strength $\sigma_y$ sets each segment’s own margin and is independent of $E\alpha$: the segment that stresses most is not always the one that yields first.
• Density plays no part — this is a force-and-stiffness story, not a mass one, so $\rho$ is not bound here.

Sign convention and scope

$F$ is taken positive in compression. A temperature rise ($\Delta T > 0$) makes both segments want to lengthen against the walls, so the bar goes into compression: $F > 0$ and the stresses $\sigma_i = F/A_i$ are compressive (reported positive). A temperature drop flips every sign — the bar is held stretched, $F < 0$, the stresses are tensile. $\Delta T = 0$ recovers the unstressed state exactly, $F = \sigma_1 = \sigma_2 = 0$ (the widget’s zero check). The slimmer segment carries the higher stress because the force is common while $\sigma = F/A$. The model is linear-elastic and assumes truly rigid walls, a uniform temperature through both segments, and a bar braced against sideways buckling; the moment either segment reaches its yield stress the linear numbers stop being the truth, and the widget says so.

Try it

Governing relations

Derivation

Steps marked modeling step are where physics enters by citation — every other line is machine-proven to follow from them, and the cited models are independently re-derived in the test pipeline where possible. See what is and isn't machine-verified.

How it fails

The widget guards first yield of each segment separately, warning the moment $|\sigma_i|$ reaches that segment’s yield strength. Because thermal stress scales with $E\,\alpha\,\Delta T$, it climbs fast: a fully restrained steel bar passes $250$ MPa after only about a $100\,\mathrm{K}$ rise. But yield is rarely the first thing that goes wrong here.

• Buckling gets there first, on heating. A temperature rise puts the bar in compression, and a bar in compression between two walls is a column. A slender segment will buckle sideways long before it yields — the failure the published worked examples quietly assume away with the phrase “suitably braced against buckling.” The critical temperature rise for buckling scales with $(r/L)^2$ and is independent of $\alpha$‘s partner effects; a long, thin restrained rail bows in the sun for exactly this reason. That is a stability limit governed by geometry and $E$ (see the Euler column), not the axial stress shown here, and this model does not see it. The widget’s numbers are the stress if the bar stays straight.

• Past yield, the elastic force is a lie. Every number here assumes both segments stay linear-elastic. Once one yields it can carry no more force as the temperature climbs further; it flows plastically, the force plateaus, and on cooling back down the bar is left with a residual stress of the opposite sign. Repeated heating and cooling past yield ratchets — each cycle adds a little permanent strain until the part cracks (thermal fatigue). The $|\sigma| \ge \sigma_y$ warning marks where the linear model stops telling the truth; it says nothing about the cycles that follow.

• Cooling loads it in tension — where brittle materials fail. Drop the temperature and the bar is stretched: $F < 0$, both stresses tensile. A brittle segment (gray cast iron, a casting) has no yield plateau to blunt a tensile overload and can simply crack; rate it against ultimate strength with a generous margin, not the yield-based warning here. A welded or bonded junction between the two metals sees this full tension too, and an interface is often the weakest link.

• The walls are never truly rigid. The entire force follows from zero net elongation. Give the supports even a little flexibility — a real abutment settles, a bolted joint draws down, a designed expansion gap of width $g$ opens — and the compatibility equation changes: the bar first expands freely to close the gap, and only the blocked remainder generates force. This is precisely why bridges have expansion joints and pipelines have loops: a few millimetres of designed give can drop the thermal force from dangerous to negligible. Model the supports as springs, not walls, and every number here shrinks.

• The properties themselves drift with temperature. This page treats $E$ and $\alpha$ as constants read at a stated temperature — the mean CTE over a modest range near room temperature, the room-temperature modulus. Over a wide swing both change: $\alpha$ generally rises with temperature and $E$ falls, so a large $\Delta T$ computed from a single room-temperature $\alpha$ is only a first estimate. A faithful analysis integrates $\alpha(T)\,dT$ and uses $E(T)$ — outside this model’s scope, and flagged on the verification page as a declared modeling choice.

• Uniform temperature is an assumption too. The whole bar is taken to reach one temperature. A real transient heats the surface and the ends first; the resulting gradients drive their own stresses (and, in a bar of two conductivities, an uneven split) that a single lumped $\Delta T$ cannot represent. Time-dependent heat flow is a separate engine this THING does not carry.

Sources

• Gere, J. M., & Goodno, B. J., Mechanics of Materials, 9th ed., Cengage, 2018 — ch. 2 (Axially Loaded Members), §2.5 "Thermal Effects, Misfits, and Prestrains": free thermal strain α ΔT, and statically indeterminate bars with temperature changes solved by equilibrium + compatibility.
• Hibbeler, R. C., Mechanics of Materials, 10th ed., Pearson, 2017 — ch. 4 (Axial Load), thermal stress: a fully restrained bar heated by ΔT carries σ = E α ΔT; a compound restrained bar is solved by superposition (free expansion, then the restoring force that returns the length).