Thermal analysis — conduction in structures, thermal stress: Worked examples

This page is a workout. The parent note gave you two engines:

Conduction: $T (x) = T_{1} + \frac{T _{2} - T _{1}}{L} x$ (linear temperature between two ends), and heat flux $q = - k \frac{T _{2} - T _{1}}{L}$ .
Thermal stress: $σ = - E α Δ T$ when expansion is blocked (compression is negative).

Here we hit every kind of input those engines can receive — every sign, the zero cases, the "one end is free" case, the limiting cases, a real word problem, and an exam twist. Before each solution, Forecast: cover the answer and guess the sign and rough size. That guess is where learning happens.

Recall What do the symbols mean again? (open if rusty)

$k$ ::: thermal conductivity, W/(m·K) — how easily heat flows through the material
$α$ ::: coefficient of thermal expansion, 1/K — how much a 1 m rod stretches per 1 K of warming
$E$ ::: Young's modulus, Pa — stiffness; stress needed to strain the material by 100%
$Δ T$ ::: temperature change from a stress-free reference $T_{0}$
$σ$ ::: stress, Pa. Positive = tension (pulled apart), negative = compression (squashed)

The scenario matrix

Every problem in this topic is one row of this table. Our job is to hit all of them.

Cell	What makes it different	Example
A. Heating, both ends fixed	$Δ T > 0$ → wants to grow → blocked → compression	Ex 1
B. Cooling, both ends fixed	$Δ T < 0$ → wants to shrink → blocked → tension	Ex 2
C. Non-uniform $T (x)$ , bar constrained	one average $Δ T$ sets a single constant $σ$ (equilibrium)	Ex 3
D. Zero / degenerate input	$Δ T = 0$ , or one end free, or $α = 0$ (CFRP) → stress = 0	Ex 4
E. Fixed–free (partial constraint)	expansion partly allowed → stress between 0 and full value	Ex 5
F. Limiting behaviour	make $E$ , $α$ , or $L$ huge / tiny — what dominates?	Ex 6
G. Real-world word problem	conduction + stress chained, units, kW/m²	Ex 7
H. Exam twist: fatigue cycle	it's the range $Δ σ$ over a cycle that matters, not one value	Ex 8

Related deep topics: Fatigue and Fracture Mechanics, Material Selection for Spacecraft, Composite Materials in Spacecraft, Thermal Environment in Orbit.

Example 1 — Cell A: heating, fixed–fixed

Step 1 — Find $Δ T$ . $Δ T = 70 - 20 = 50 K$ Why this step? Stress only cares about the change from the stress-free reference $T_{0}$ , not the absolute temperature.

Step 2 — Apply the fixed–fixed formula. $σ = - E α Δ T = - (70 \times 1 0^{9}) (23 \times 1 0^{- 6}) (50)$ $σ = - 80.5 \times 1 0^{6} Pa = - 80.5 MPa$ Why this step? Both ends held → total strain forced to zero → the elastic strain must exactly cancel the thermal strain, giving $σ = - E α Δ T$ .

Verify: sign is negative → compression, matching the forecast. Units: $Pa \cdot \frac{1}{K} \cdot K = Pa$ . ✓

Example 2 — Cell B: cooling, fixed–fixed

Step 1 — Find $Δ T$ . $Δ T = - 100 - 20 = - 120 K$ Why this step? Cooling gives a negative $Δ T$ ; carrying that sign correctly is the whole game.

Step 2 — Apply the formula. $σ = - E α Δ T = - (70 \times 1 0^{9}) (23 \times 1 0^{- 6}) (- 120) = + 193.2 MPa$ Why this step? The two minus signs (formula sign and cooling sign) multiply to a plus → tension, exactly as the physics says.

Verify: positive → tension. ✓ Bigger than Ex 1 because $∣Δ T ∣$ is bigger (120 vs 50). A cold bracket is often the dangerous case — tension opens cracks. See Fatigue and Fracture Mechanics.

Example 3 — Cell C: non-uniform temperature, one constant stress

Step 1 — Write $T (x)$ . $T (x) = T_{1} + \frac{T _{2} - T _{1}}{L} x = 100 - 150 x$ Why this step? Steady-state conduction gives a straight line (parent note, $d^{2} T / d x^{2} = 0$ ). We need it to compute the average.

Step 2 — Equilibrium forces a constant axial force. Cut the bar anywhere. The axial force $F = σ A$ pulling on the two cut faces must balance (no other axial loads act). With $A$ constant, $σ$ is the same everywhere — it cannot track $T (x)$ slice by slice. Why this step? This is the correction to the naive "each slice independently blocked" idea: individual slices are not independent; they are welded together and must share one force.

Step 3 — Compatibility with the average temperature. Total elongation must be zero (ends fixed). Free thermal elongation uses the average rise: $⟨ Δ T ⟩ = \frac{1}{L} \int_{0}^{L} [T (x) - T_{0}] d x = \frac{T _{1} + T _{2}}{2} - T_{0} = \frac{100 + ( - 50 )}{2} - 25 = 0 K$ $ε_{total} = \frac{σ}{E} + α ⟨ Δ T ⟩ = 0 ⟹ σ = - E α ⟨ Δ T ⟩$ Why this step? The bar's net length change is what the fixed ends veto — and net length change depends on the average temperature, not any single point.

Step 4 — Evaluate. $σ = - (70 \times 1 0^{9}) (23 \times 1 0^{- 6}) (0) = 0 MPa$

Verify: because $T_{0} = 25$ is exactly the average of $100$ and $- 50$ , the average rise is zero, so the net axial stress is zero. Look at the figure below. ✓

Figure — Thermal analysis — conduction in structures, thermal stress

Figure — why axial stress is one number, not a curve. The dashed lavender line is $T (x)$ falling from $+ 10 0^{\circ}$ C to $- 5 0^{\circ}$ C; its average (the mint level) sits at $+ 2 5^{\circ}$ C, equal to $T_{0}$ . Because the average rise is zero, the uniform axial stress (coral band) is a single flat value at $0$ MPa — not the tempting slice-by-slice curve. The take-away: for a uniform constrained bar, only the average temperature enters the axial stress. (Point-to-point bending stresses from the temperature shape are a separate story handled by Finite Element Analysis.)

Example 4 — Cell D: zero and degenerate inputs

Step 1 — Case (a): free end. $Δ L = α L Δ T \neq = 0, σ = 0$ Why? Nothing blocks the growth, so the strain is all thermal, none elastic. Displacement without stress — the classic Mistake 1 trap from the parent note.

Step 2 — Case (b): no temperature change. $σ = - E α (0) = 0$ Why? No $Δ T$ , no thermal strain to fight. Trivially zero.

Step 3 — Case (c): near-zero CTE. $σ = - E α Δ T = - (E) (0) (200) = 0$ Why? Even with the ends fully fixed and a huge $Δ T$ , if the material barely expands there is nothing to constrain. This is why Composite Materials in Spacecraft and low-CTE alloys are chosen — see Material Selection for Spacecraft.

Verify: all three give $0$ , but only (c) tells the design story: kill $α$ and you kill thermal stress regardless of temperature swing. ✓

Example 5 — Cell E: fixed–free (partial constraint)

Step 1 — Full fixed–fixed stress (the ceiling). $σ_{full} = - E α Δ T = - (110 \times 1 0^{9}) (8.6 \times 1 0^{- 6}) (150) = - 141.9 MPa$ Why this step? This is the maximum possible — the case where zero expansion is allowed. Heating gives compression (negative), consistent with our convention. Everything else is a fraction of it.

Step 2 — Allowed strain reduces the elastic (stress-producing) strain. Free thermal strain: $ε_{th} = α Δ T$ . If 60 % is released, only 40 % is elastic: $ε_{elastic} = - 0.40 α Δ T$ Why this step? Stress comes only from the strain the constraint forbids. Release some expansion → less forbidden strain → less stress. The minus keeps compression negative.

Step 3 — Stress. $σ = E ε_{elastic} = - 0.40 E α Δ T = 0.40 \times (- 141.9) = - 56.76 MPa (compression)$ Why this step? Multiply the forbidden (elastic) strain by stiffness $E$ to convert strain into stress — the same $σ = E ε$ we used for the ceiling, just with the reduced strain.

Verify: $∣ - 141.9 ∣ \geq ∣ - 56.76 ∣ \geq 0$ — the magnitude sits between the free (0) and fully-fixed (141.9) bounds, exactly as a partial constraint must be, and the sign stays compressive. ✓ This is why designers add "compliance" (flexible mounts) — see Deployable Structures and Structural Dynamics.

Example 6 — Cell F: limiting behaviour

Step 1 — Aluminium. $σ_{a} = - E α Δ T = - (70 \times 1 0^{9}) (23 \times 1 0^{- 6}) (100) = - 161 MPa, ∣ σ_{a} ∣ = 161 MPa$

Step 2 — Invar. $σ_{i} = - (140 \times 1 0^{9}) (1.2 \times 1 0^{- 6}) (100) = - 16.8 MPa, ∣ σ_{i} ∣ = 16.8 MPa$ Why this step? Stress magnitude scales with the product $E α$ . Invar's $E α = 1.68 \times 1 0^{5}$ vs aluminium's $1.61 \times 1 0^{6}$ — a factor of ~9.6 smaller.

Step 3 — Limit $α \to 0$ . $lim_{α \to 0} (- E α Δ T) = 0$ Why this step? $E$ is finite and $Δ T$ is fixed, so the product vanishes with $α$ . Stiffness cannot save you if there is no expansion — and it cannot doom you either.

Verify: $∣ σ_{i} ∣/∣ σ_{a} ∣ = 16.8/161 = 0.104$ , i.e. Invar carries about 10 % of aluminium's stress; both stay compressive (negative). Low $α$ beats high $E$ . ✓

Example 7 — Cell G: real-world chained problem

Step 1 — Temperature at $x = 1.5$ m. $T (x) = 120 + \frac{- 40 - 120}{2} x = 120 - 80 x, T (1.5) = 120 - 120 = 0^{\circ} C$ Why this step? Linear conduction profile; just plug in.

Step 2 — Heat flux and power. $q = - k \frac{T _{2} - T _{1}}{L} = - 200 \cdot \frac{- 40 - 120}{2} = - 200 \cdot (- 80) = 16, 000 W/m^{2} = 16 kW/m^{2}$ $P = q A = 16, 000 \times 4 \times 1 0^{- 4} = 6.4 W$ Why this step? Flux is per unit area; multiply by $A$ for the actual watts flowing down the boom.

Step 3 — Uniform axial stress from the average rise. $⟨ Δ T ⟩ = \frac{T _{1} + T _{2}}{2} - T_{0} = \frac{120 + ( - 40 )}{2} - 40 = 0 K$ $σ = - E α ⟨ Δ T ⟩ = - (70 \times 1 0^{9}) (23 \times 1 0^{- 6}) (0) = 0 MPa$ Why this step? Same equilibrium logic as Example 3: a uniform constrained bar carries one constant axial stress, set by the average temperature rise — here exactly zero.

Verify: flux positive ⇒ heat flows in $+ x$ from hot to cold. ✓ $T (1.5) = 0^{\circ}$ C sits between the two ends. ✓ Average rise zero ⇒ net axial stress zero (though the temperature shape still drives local bending stresses handled by Finite Element Analysis). ✓ For the whole thermal picture see Thermal Control Subsystems and Thermal Environment in Orbit.

Example 8 — Cell H: exam twist (fatigue cycle)

Step 1 — Temperature swing. $Δ T_{swing} = 50 - (- 100) = 150 K$ Why this step? The cycle spans this full range; the stress swings across it.

Step 2 — Stress range. $Δ σ = E α Δ T_{swing} = (110 \times 1 0^{9}) (8.6 \times 1 0^{- 6}) (150) = 141.9 MPa$ Why this step? From peak tension (cold, $σ = + E α ∣Δ T_{cold} ∣$ ) to peak compression (hot, $σ = - E α Δ T_{hot}$ ), the difference between the two extreme stresses is $E α Δ T_{swing}$ — a positive range, which is what fatigue analysis needs.

Step 3 — Cycle count over 10 years. $\frac{1440 min/day}{90 min/cycle} = 16 cycles/day$ $16 \times 365 \times 10 = 58, 400 cycles$ Why this step? Each orbit is one full thermal cycle; multiply out the mission.

Verify: $Δ σ = 141.9$ MPa matches the parent note's Example 2. ✓ Nearly 6×10⁴ cycles at ~142 MPa is a real Fatigue and Fracture Mechanics driver — mitigate with low-CTE materials or pre-load. Detailed life prediction uses Finite Element Analysis. ✓

Recall Self-test

Cooling a fixed–fixed rod gives which sign of stress? ::: Positive — tension. A rod with one free end and $Δ T = 80$ K has what stress? ::: Zero — it expands freely. Two materials, same $Δ T$ : which has less thermal stress, low $α$ or low $E$ ? ::: Whichever gives the smaller product $E α$ ; low $α$ usually wins. For a uniform bar with both ends fixed and a varying $T (x)$ , what sets the axial stress? ::: The average temperature rise $⟨ Δ T ⟩$ , giving one constant $σ = - E α ⟨ Δ T ⟩$ .