Thermal analysis — conduction in structures, thermal stress: Exercises

This is a self-testing page. Each problem gives you a clean statement; the worked solution hides inside a collapsible callout — try first, then reveal. Problems climb from L1 Recognition (spot the right formula) through to L5 Mastery (design-level synthesis).

Everything here builds on the parent note. Before starting, make sure you have the four tools it built:

Recall The four formulas you will use

Fourier's law (1D): $q = - k \frac{d T}{d x}$ — heat flux is conductivity times the steepness of the temperature graph, with a minus sign because heat runs downhill from hot to cold. Linear temperature profile: $T (x) = T_{1} + \frac{T _{2} - T _{1}}{L} x$ — a straight line between the two end temperatures. Thermal stress, full constraint: $σ = - E α Δ T$ — the stress a rod feels when it wants to expand by $α Δ T$ but is not allowed to. Free expansion: $Δ L = α L Δ T$ — how much length a free rod gains per degree.

Symbols: $k$ = thermal conductivity (W/m·K), $E$ = Young's modulus (Pa, "stiffness"), $α$ = coefficient of thermal expansion (per K, "how eager to grow with heat"), $Δ T$ = temperature change (K), $L$ = length (m), $A$ = cross-section area (m²).

A quick sign compass to keep at hand:

Figure — Thermal analysis — conduction in structures, thermal stress

Figure s01 — Sign compass. Left panel: a red rod is clamped between two black hatched walls and heated; the black arrows inside the rod point outward (it wants to grow) while the walls block it, so the rod ends up in compression (σ < 0). Right panel: the same clamped red rod is cooled; the inside arrows point inward (it wants to shrink) while the walls hold it out, so the rod is in tension (σ > 0). Take-away: heating a held rod → negative stress, cooling a held rod → positive stress.

Before any flux problem, fix the coordinate convention with this sketch:

Figure s03 — Flux sign convention. A horizontal rod runs left (x = 0, hot, red dot) to right (x = L, cold). The black $x$ -axis points rightward (+x from hot end to cold end). A red arrow labelled $q > 0$ points in the same +x direction, showing heat physically flowing from the hot end toward the cold end. The temperature graph above falls from left to right, so its slope $d T / d x$ is negative; feeding that into $q = - k d T / d x$ turns the negative slope into a positive $q$ . This is why " $q$ positive" and "heat flows in +x (hot → cold)" mean the same thing.

Level 1 — Recognition

L1·Q1 — Read the flux sign

A rod has $T_{1} = 8 0^{\circ} C$ at $x = 0$ and $T_{2} = 2 0^{\circ} C$ at $x = L$ . Without any numbers for $k$ or $L$ , which direction does heat flow, and is the flux $q$ positive or negative (with $+ x$ pointing from end 1 to end 2)?

Recall Solution

Heat always flows from hot to cold. Hot end is $x = 0$ , cold end is $x = L$ , so heat flows in the $+ x$ direction → flux is positive (compare figure s03). Check with the formula: the graph of $T$ falls as $x$ grows, so $\frac{d T}{d x} < 0$ . Then $q = - k \frac{d T}{d x} = - k \cdot (negative) = positive$ . ✓

L1·Q2 — Pick the formula

You are told a strut is bolted rigidly at both ends and heated uniformly. You must find the stress. Which single formula applies, and what does the answer's sign tell you?

Recall Solution

Both ends fixed + uniform heating = full constraint, so use $σ = - E α Δ T$ . Uniform heating means $Δ T > 0$ , so $σ < 0$ → compression. The rod is being squeezed because it cannot expand (left panel of figure s01).

Level 2 — Application

L2·Q1 — Midpoint temperature

A copper bracket of length $L = 0.4$ m has $T_{1} = 6 0^{\circ} C$ and $T_{2} = - 4 0^{\circ} C$ . Find the temperature at $x = 0.1$ m.

Recall Solution

Use the linear profile $T (x) = T_{1} + \frac{T _{2} - T _{1}}{L} x$ . Slope $= \frac{- 40 - 60}{0.4} = \frac{- 100}{0.4} = - 250 K/m$ . $T (0.1) = 60 + (- 250) (0.1) = 60 - 25 = 3 5^{\circ} C$ . What it looks like: a straight line dropping steadily; at one quarter of the way along we have dropped one quarter of the total $10 0^{\circ}$ swing.

L2·Q2 — Heat flux magnitude

Same copper bracket ( $k = 400$ W/m·K, $L = 0.4$ m, $T_{1} = 6 0^{\circ} C$ , $T_{2} = - 4 0^{\circ} C$ , cross-section $A = 2 \times 1 0^{- 4}$ m²). Find the heat flux $q$ and the total heat power $P = q A$ flowing through it.

Recall Solution

Start from Fourier's law $q = - k \frac{T _{2} - T _{1}}{L}$ and build it up one piece at a time so no step is hidden. Step 1 — the temperature difference. $T_{2} - T_{1} = - 40 - 60 = - 100$ K. Why negative? End 2 is colder than end 1, so going in $+ x$ the temperature drops by $100$ K. Step 2 — the slope (gradient). $\frac{T _{2} - T _{1}}{L} = \frac{- 100}{0.4} = - 250$ K/m. Why? Spread that $- 100$ K drop over $0.4$ m of length; each metre would drop $250$ K, and the sign stays negative because temperature falls with $x$ . Step 3 — multiply by $- k$ . $q = - 400 \times (- 250)$ . Why the two minus signs? The $- k$ is Fourier's downhill rule (heat runs down the gradient); the $- 250$ is the falling slope. A negative times a negative is positive: $- 400 \times (- 250) = + 100, 000$ . So $q = 100, 000 W/m^{2} = 100 kW/m^{2}$ . Positive → heat flows in $+ x$ (hot to cold), matching figure s03. ✓ Step 4 — total power. $P = q A = 100, 000 \times 2 \times 1 0^{- 4} = 20 W$ . Why multiply by area? Flux is power per square metre; multiplying by the cross-section $A$ gives the actual watts passing through the whole rod.

L2·Q3 — Free expansion length

A steel rail-like fitting, $L = 1.2$ m, $α = 12 \times 1 0^{- 6}$ /K, warms from $- 3 0^{\circ} C$ to $+ 9 0^{\circ} C$ while free at both ends. How much longer does it get, and what is the stress?

Recall Solution

$Δ T = 90 - (- 30) = 120$ K. $Δ L = α L Δ T = 12 \times 1 0^{- 6} \times 1.2 \times 120 = 1.728 \times 1 0^{- 3}$ m $= 1.728$ mm. Because it is free, nothing resists the growth → $σ = 0$ . Displacement, not stress.

Level 3 — Analysis

L3·Q1 — Stress at both ends of a gradient

An aluminum beam ( $E = 70$ GPa, $α = 23 \times 1 0^{- 6}$ /K), fully constrained, carries a linear profile from $T_{1} = 10 0^{\circ} C$ to $T_{2} = - 5 0^{\circ} C$ . Using reference temperature $T_{0} = 2 5^{\circ} C$ (the stress-free assembly temperature), find $σ (0)$ and $σ (L)$ .

Figure s02 — Temperature and stress along the beam. The black straight line is $T (x)$ , falling from $+ 10 0^{\circ}$ C at $x = 0$ to $- 5 0^{\circ}$ C at $x = L$ . The black dashed horizontal line marks the stress-free reference $T_{0} = 2 5^{\circ}$ C. The red dot sits where the temperature line crosses $T_{0}$ , at $x = 0.5 L$ — this is exactly where the thermal stress is zero. Left of the red dot the material is hotter than $T_{0}$ (wants to expand → compression); right of it the material is colder than $T_{0}$ (wants to shrink → tension). The stress therefore runs linearly from $- 121$ MPa at the hot end to $+ 121$ MPa at the cold end, flipping sign at the red dot.

Recall Solution

Local stress for a constrained slice: $σ (x) = - E α [T (x) - T_{0}]$ . Hot end $x = 0$ : $T - T_{0} = 100 - 25 = 75$ K. $σ (0) = - 70 \times 1 0^{9} \times 23 \times 1 0^{- 6} \times 75 = - 120.75 \times 1 0^{6} \approx - 121$ MPa → compression (hot end wants to grow, is squeezed). Cold end $x = L$ : $T - T_{0} = - 50 - 25 = - 75$ K. $σ (L) = - 70 \times 1 0^{9} \times 23 \times 1 0^{- 6} \times (- 75) = + 120.75 \times 1 0^{6} \approx + 121$ MPa → tension (cold end wants to shrink, is stretched). What it looks like (figure s02): stress is a straight line crossing zero exactly where $T (x) = T_{0} = 2 5^{\circ} C$ , i.e. the midpoint here.

L3·Q2 — Where is the stress zero?

For the beam in L3·Q1, at what position $x$ (and temperature) is the thermal stress exactly zero?

Recall Solution

$σ (x) = 0$ requires $T (x) = T_{0} = 2 5^{\circ} C$ . Profile: $T (x) = 100 + \frac{- 50 - 100}{L} x = 100 - \frac{150}{L} x$ . Set $= 25$ : $100 - \frac{150}{L} x = 25 \Rightarrow \frac{150}{L} x = 75 \Rightarrow x = \frac{75}{150} L = 0.5 L$ . The neutral point sits at the midpoint, halfway along, because $2 5^{\circ}$ is exactly halfway between $10 0^{\circ}$ and $- 5 0^{\circ}$ . To one side the material is compressed, to the other it is stretched (red dot in figure s02).

L3·Q3 — Which end is more dangerous?

A material fails in tension at $150$ MPa but tolerates $400$ MPa in compression (a common ceramic-like asymmetry). Using the L3·Q1 stresses, which end is closer to failure, and by what safety margin?

Recall Solution

Compression end: $∣ σ ∣ = 121$ MPa against a $400$ MPa limit → margin $= 400/121 \approx 3.3$ . Tension end: $σ = + 121$ MPa against a $150$ MPa limit → margin $= 150/121 \approx 1.24$ . The cold (tension) end is far more dangerous: its margin is only $1.24$ versus $3.3$ . Brittle spacecraft materials crack from tensile thermal stress at the cold end, not the hot end — counter to first intuition.

Level 4 — Synthesis

L4·Q1 — Fatigue cycle count

A titanium strut ( $E = 110$ GPa, $α = 8.6 \times 1 0^{- 6}$ /K), fully constrained, cycles between $- 10 0^{\circ} C$ and $+ 5 0^{\circ} C$ once per $90$ -minute orbit. Find (a) the stress range $Δ σ$ , and (b) the number of cycles in a $10$ -year mission.

Recall Solution

(a) The stress range is the peak-to-peak swing, so it depends only on the temperature swing: $Δ T = 50 - (- 100) = 150$ K. $Δ σ = E α Δ T = 110 \times 1 0^{9} \times 8.6 \times 1 0^{- 6} \times 150 = 141.9 \times 1 0^{6} \approx 142$ MPa. (This is the magnitude of the swing from most-compressive to most-tensile; per the definition above, a range is a positive distance, so the minus sign of $σ = - E α Δ T$ drops out.) (b) Cycles per day: $\frac{1440 min/day}{90 min/cycle} = 16$ . Per year: $16 \times 365 = 5840$ . Ten years: $5840 \times 10 = 58, 400 \approx 5.8 \times 1 0^{4}$ cycles. This is a real fatigue driver — see Fatigue and Fracture Mechanics.

L4·Q2 — Material trade for stress reduction

The strut in L4·Q1 experiences $142$ MPa range. A designer swaps titanium for a CFRP tube with $α_{CFRP} = 0.5 \times 1 0^{- 6}$ /K and $E_{CFRP} = 130$ GPa. What is the new stress range, and by what factor did it drop?

Recall Solution

$Δ σ_{CFRP} = 130 \times 1 0^{9} \times 0.5 \times 1 0^{- 6} \times 150 = 9.75 \times 1 0^{6} \approx 9.75$ MPa. Reduction factor $= \frac{142}{9.75} \approx 14.6 \times$ . Even though CFRP is stiffer (higher $E$ ), its near-zero $α$ dominates the product $E α$ . This is why Composite Materials in Spacecraft and thoughtful Material Selection for Spacecraft win the thermal-stress battle — you attack the $α$ factor, not the $E$ factor.

L4·Q3 — Partial constraint (fixed–spring)

A rod ( $E$ , $α$ , area $A$ , length $L$ ) is fixed at $x = 0$ but at $x = L$ is held by a soft spring of stiffness $k_{s}$ (N/m) instead of a rigid wall. On uniform heating $Δ T$ , only part of the free expansion $Δ L = α L Δ T$ is allowed. Show that the rod force is $F = \frac{E A α Δ T}{1 + \frac{E A}{k _{s} L}},$ and check the two limits $k_{s} \to \infty$ and $k_{s} \to 0$ .

Recall Solution

Let the far end actually move outward by $δ$ (less than the free $Δ L$ ). The rod is compressed by the blocked amount $(Δ L - δ)$ , producing a compressive force $F = \frac{E A}{L} (Δ L - δ) .$ That same force pushes the spring out by $δ$ , so $F = k_{s} δ \Rightarrow δ = F / k_{s}$ . Substitute: $F = \frac{E A}{L} (α L Δ T - \frac{F}{k _{s}}) = E A α Δ T - \frac{E A}{k _{s} L} F .$ Collect $F$ : $F (1 + \frac{E A}{k _{s} L}) = E A α Δ T$ , giving $F = \frac{E A α Δ T}{1 + \frac{E A}{k _{s} L}} .$ Limit $k_{s} \to \infty$ (rigid wall): denominator $\to 1$ , so $F \to E A α Δ T$ , i.e. stress $σ = F / A = E α Δ T$ — the full-constraint result. ✓ Limit $k_{s} \to 0$ (free end): denominator $\to \infty$ , so $F \to 0$ — a free rod carries no stress. ✓ The soft spring interpolates smoothly between the two textbook cases — this is the real spacecraft situation, where mounts are stiff but never perfectly rigid.

Intuition Edge cases you must be able to state

$Δ T = 0$ (no temperature change): every formula collapses — $Δ L = 0$ , $σ = - E α \cdot 0 = 0$ , and $Δ σ = 0$ . No thermal loading at all, regardless of how stiff the constraint is. Thermal stress is entirely driven by temperature change; kill $Δ T$ and you kill the stress. $k_{s} \to 0$ under cycling (perfectly compliant mounts): from L4·Q3, $F \to 0$ , so the stress range $Δ σ \to 0$ as well. A perfectly floppy mount transmits no force even as the rod cycles hot-cold-hot, so it experiences no fatigue loading — the rod simply expands and contracts freely. This is the limiting ideal behind flexure mounts and thermal isolators: trade a little positional stiffness for a huge drop in cyclic stress. The catch is that a truly floppy mount also fails to hold the structure in place, so real designs pick a finite $k_{s}$ that balances pointing stiffness against fatigue.

Level 5 — Mastery

L5·Q1 — Full design decision

An optical bench truss uses struts of length $L = 0.5$ m, fully constrained, over an orbital swing $Δ T = 150$ K. Requirement: stress range $Δ σ \leq 40$ MPa for $1 0^{7}$ -cycle fatigue life. Three candidate materials:

Material	$E$ (GPa)	$α$ ( $1 0^{- 6}$ /K)
Aluminum 6061	69	23.6
Titanium 6Al-4V	114	8.6
CFRP (quasi-iso)	70	2.0

(a) Compute $Δ σ = E α Δ T$ for each. (b) Which pass the $40$ MPa gate? (c) For any that fail, what compliance factor $R = \frac{1}{1 + E A / ( k _{s} L )}$ from L4·Q3 would be needed to bring them under the gate?

Recall Solution

(a) Using $Δ σ = E α Δ T$ with $Δ T = 150$ K:

Aluminum: $69 \times 1 0^{9} \times 23.6 \times 1 0^{- 6} \times 150 = 244.3 \times 1 0^{6} \approx 244$ MPa.
Titanium: $114 \times 1 0^{9} \times 8.6 \times 1 0^{- 6} \times 150 = 147.1 \times 1 0^{6} \approx 147$ MPa.
CFRP: $70 \times 1 0^{9} \times 2.0 \times 1 0^{- 6} \times 150 = 21.0 \times 1 0^{6} = 21$ MPa. (b) Gate is $40$ MPa. Only CFRP passes ( $21 < 40$ ). Aluminum and titanium fail. (c) Compliance multiplies the constraint stress by $R$ (from L4·Q3, $σ_{eff} = R \cdot E α Δ T$ ). We need $R \leq 40/Δ σ_{rigid}$ :
Aluminum: $R \leq 40/244 = 0.164$ — the mount must let through only $\sim 16%$ of the rigid-constraint force (very soft mounts).
Titanium: $R \leq 40/147 = 0.272$ — the mount must let through only $\sim 27%$ . Design call: CFRP meets the gate outright with no special mounting, so it is the lowest-risk choice: no floppy flexures to design, no positional-stiffness penalty, and margin to spare ( $21$ vs $40$ MPa). Aluminum and titanium would each need deliberately compliant mounts (flexures / thermal isolators) sized to those $R$ values, adding mass and complexity while eroding pointing stiffness. This is a textbook win for low- $α$ Composite Materials in Spacecraft in precision structures; see also Deployable Structures where thermal snap-through from this same stress is a failure mode.

L5·Q2 — Coupled conduction + stress budget

An aluminum strut ( $k = 200$ W/m·K, $E = 70$ GPa, $α = 23 \times 1 0^{- 6}$ /K, $L = 0.8$ m, $A = 1 \times 1 0^{- 4}$ m²) bridges a $T_{1} = 12 0^{\circ} C$ fitting and a $T_{2} = - 14 0^{\circ} C$ radiator, fully constrained, stress-free at $T_{0} = 2 0^{\circ} C$ . Find (a) heat power conducted, (b) tensile stress at the cold end, and (c) whether it survives a $200$ MPa tensile allowable.

Recall Solution

(a) Flux $q = - k \frac{T _{2} - T _{1}}{L} = - 200 \cdot \frac{- 140 - 120}{0.8} = - 200 \cdot (- 325) = 65, 000 W/m^{2}$ . Power $P = q A = 65, 000 \times 1 \times 1 0^{- 4} = 6.5$ W flowing hot→cold. (b) Cold-end stress: $σ (L) = - E α [T_{2} - T_{0}] = - 70 \times 1 0^{9} \times 23 \times 1 0^{- 6} \times (- 140 - 20)$ $= - 70 \times 1 0^{9} \times 23 \times 1 0^{- 6} \times (- 160) = + 257.6 \times 1 0^{6} \approx + 258$ MPa (tension). (c) $258$ MPa $> 200$ MPa allowable → it fails at the cold end. The conduction is fine ( $6.5$ W is easily radiated), but the tensile thermal stress at the frigid end exceeds the limit. Remedy: swap to low- $α$ material, add compliance, or thermally isolate the fitting via Thermal Control Subsystems to shrink the effective $T - T_{0}$ .

Recall Self-test recall

Direction of heat flow for hot-at-x=0, cold-at-x=L, with +x from 0 to L ::: +x direction, flux positive. Sign of thermal stress in a held, uniformly heated rod ::: negative (compression). Where along a linear gradient is thermal stress zero ::: where $T (x)$ equals the stress-free reference $T_{0}$ . Which product must you shrink to cut thermal stress, and which factor usually dominates ::: shrink $E α$ ; the $α$ (expansion) factor usually dominates the choice. Full-constraint stress from the fixed–spring formula as $k_{s} \to \infty$ ::: $σ \to E α Δ T$ . What happens to thermal stress when $Δ T = 0$ ::: it vanishes entirely; no temperature change means no thermal loading. What happens to the fatigue stress range as mounts become perfectly compliant ( $k_{s} \to 0$ ) ::: the range $Δ σ \to 0$ ; a floppy mount transmits no cyclic force.