This is a rapid-fire concept trap bank for the parent topic. Each line is a question ::: answer reveal. Cover the answer, commit to a verdict and a reason, then check. If you can only say "true/false" but not why, you haven't understood it yet.
Every trap below leans on a handful of letters. Before you use a symbol you must be able to say what it means and what picture goes with it. Here they are, each earned before use.
Before the traps, pin down the two governing statements so every reveal below has something concrete to point at. Each is earned, not just asserted.
Figure s01 — Temperature vs. position along the rod. The blue line is the true steady-state profile T(x): it is perfectly straight because the slope (yellow arrows, drawn identical at three slices) never changes, which is what constant flux demands. The red dashed curve is a forbidden profile whose slope varies; it would require heat to pile up in some slices. Green dots mark the fixed-temperature ends T1 (hot, left) and T2 (cold, right).
Figure s02 — The stress sign convention. Top (red): a heated rod between two rigid walls wants to expand; the walls push inward (yellow arrows), so the rod is in compression, σ<0. Bottom (blue): a cooled rod wants to shrink; the walls hold it stretched (yellow arrows point outward), so it is in tension, σ>0. "Rigid walls" here means fixed-displacement (mechanical) ends.
A rod that heats up but is free to move at both ends develops zero thermal stress.
True — with no fixed-displacement constraint, the rod simply elongates by αLΔT and no internal force builds up. Stress needs a mechanical constraint, not just a temperature change.
If two ends of a bar are held at fixed displacement and it is heated, the stress is compressive.
True — heating makes the material want to grow, the walls push back to keep the length constant, and being squeezed inward is compression (σ=−EαΔT<0 for ΔT>0).
Cooling a bar with both ends held at fixed displacement produces tensile stress.
True — cooling (ΔT<0) makes it want to shrink; the walls resist by pulling outward, so σ=−EαΔT>0 → tension.
Thermal stress in a fully constrained rod depends on the rod's length L.
False — σ=−EαΔT contains no L; a longer rod expands more in absolute terms but the strainαΔT (and hence stress) is length-independent.
In 1D steady state with constant k, the temperature profile is a straight line.
True — constant k gives d2T/dx2=0, so the slope is constant and T=C1x+C2 is linear (see figure s01).
The heat flux q varies along a 1D steady-state rod with fixed-temperature ends.
False — in steady state dq/dx=0, so q is the same at every cross section; whatever enters one face leaves the other.
A material with higher thermal conductivity k always develops higher thermal stress.
False — k sets how fast heat flows (the temperature field), not the stress; stress −EαΔT depends on E, α and ΔT. High k can even reduce gradients and thus reduce stress.
CFRP is favoured for dimensionally stable structures partly because its α can be near zero.
True — with α≈0, the term −EαΔT collapses regardless of stiffness or temperature swing, so thermal stress and warping nearly vanish (see Composite Materials in Spacecraft).
The minus sign in Fourier's law q=−k∇T means heat can flow uphill in temperature.
False — the minus sign enforces downhill flow: it points the flux vector q opposite to the gradient arrow ∇T, so heat moves from hot to cold, obeying the second law.
"The beam heats 50 °C so it expands 0.5 mm; therefore it is stressed."
Wrong: free expansion is displacement, not stress. Stress appears only if that 0.5 mm is prevented by a fixed-displacement end. Check the mechanical boundary condition first.
"Average temperature is 25 °C, so I'll just use ΔT=25−T0 everywhere for the stress."
Wrong for a non-uniform field: each slice has its own T(x), so stress σ(x)=−Eα[T(x)−T0]varies along the bar. An average hides the tension at the cold end and compression at the hot end.
"Heat flux is q=kdT/dx."
Missing the minus sign. Without it q would point up the gradient (cold→hot), violating physics. Correct form: q=−kdT/dx.
"Since the hot end wants to expand, it must be in tension."
Backwards: wanting to expand while being held means it is being compressed by the constraint → σ=−EαΔT<0. Tension is what you get when it wants to shrink (cooling).
"The rod stress is σ=−EαΔTL."
The stray L is wrong; stress has units of pressure (Pa) and EαΔT already gives pascals. Multiplying by length gives a force-like quantity, not a stress.
"d2T/dx2=0 works even if k depends strongly on temperature."
Not generally — that clean result assumed constantk. The true equation is dxd(kdT/dx)=0; if k=k(T), you cannot pull k out, and the profile is no longer straight.
Why does steady state give dq/dx=0 rather than q=0?
Steady state means no accumulation of energy in a slice, so flux in equals flux out — the flux is constant, not necessarily zero. A nonzero constant flux is exactly a rod steadily conducting heat through.
Why does the thermal stress formula σ=−EαΔT contain E (stiffness)?
Because stress = stiffness × elastic strain, σ=Eεmech. The constraint forces εmech=−αΔT; multiplying by E converts that strain into stress.
Why does the same formula carry a minus sign?
The constraint cancels the thermal strain, so the mechanical strain is the negative of the thermal one (εmech=−αΔT). That negative sign flows straight through Hooke's law into σ=−EαΔT, encoding "heating → compression."
Why is the stress range Δσ=EαΔT (magnitude) the quantity that matters for fatigue?
Fatigue is driven by cyclic loading — the swing between max and min stress each orbit — not the absolute level. A large swing cracks a part even if peaks are modest (see Fatigue and Fracture Mechanics).
Why can pre-stressing a strut into permanent compression improve fatigue life?
If the part stays compressive (σ<0) through the whole thermal cycle, it never experiences tensile reversals that open and grow cracks; keeping the stress on one side of zero suppresses fatigue crack propagation.
Why do designers add flexible (compliant) mounts to reduce thermal stress?
A compliant mount relaxes the fixed-displacement condition. If the ends can move a little, the mechanical strain — and therefore the stress −EαΔT — drops toward the free-expansion (zero-stress) case.
Why does a spacecraft in low orbit see a thermal cycle roughly every 90 minutes?
One orbit carries it through sunlight and Earth's shadow once; entering and leaving shadow drives the hot–cold swing, so cycles track the orbital period (see Thermal Environment in Orbit).
Exactly zero — no temperature change means no thermal strain to fight, so σ=−Eα⋅0=0 regardless of how rigidly it is clamped.
What happens to the stress if α=0 (ideal zero-expansion material)?
Zero thermal stress at any temperature — with nothing wanting to expand, the constraint does nothing. This is the design goal behind near-zero-CTE composites.
What is the temperature at the midpoint when both ends sit at the same fixed temperature T1=T2?
The whole rod is at that temperature — the linear profile has zero slope, so T(x)=T1 everywhere and the heat flux is zero.
In the fixed-temperature but free-to-move (mechanically unconstrained) case, what is the thermal stress?
Zero for uniform heating — the free ends let the bar expand unimpeded, so no mechanical strain is forced and no stress develops, even though the bar physically lengthens.
If k were zero, what would the steady-state flux be?
Zero — a perfect insulator carries no heat, so q=−kdT/dx=0; the rod cannot reach the two-fixed-temperature steady state at all without heat flowing.
For the linear profile T(x)=T1+LT2−T1x, where is the thermal stress largest in magnitude?
At whichever end is furthest from the reference temperature T0, since σ(x)=−Eα[T(x)−T0] scales with that temperature difference — often one of the two boundaries.
Recall Quick self-test
Fixed-displacement rod, heated: sign of stress? ::: Compressive, σ=−EαΔT<0 — it wants to expand and can't.
Does thermal stress depend on rod length? ::: No — σ=−EαΔT has no L.
Fixed-temperature but free-to-move rod, heated: stress? ::: Zero — expansion is unopposed.
What is the full steady-state conduction equation, and what does constant k turn it into? ::: ∇⋅(k∇T)=0; with constant k it becomes d2T/dx2=0, giving a straight-line T(x).
