Thermal analysis — conduction in structures, thermal stress

Overview

Spacecraft experience extreme temperature gradients in orbit — from +120°C in sunlight to -150°C in shadow. Thermal conduction through structural members creates temperature distributions that induce thermal stress, which can warp solar panels, misalign antennas, or even crack materials. This note derives the governing equations from first principles and shows how to predict both temperature fields and stress fields in spacecraft structures.

Core Concepts

Derivation: 1D Steady-State Conduction

Goal: Find temperature distribution $T(x)$ in a rod of length $L$ , with fixed temperatures $T_1$ at $x=0$ and $T_2$ at $x=L$ .

Step 1: Conservation of Energy

Consider a thin slice of thickness $dx$ at position $x$ . In steady state (no time variation), energy in = energy out:

$q(x) \cdot A = q(x+dx) \cdot A$

where $A$ is cross-sectional area. Expanding $q(x+dx) \approx q(x) + \frac{dq}{dx}dx$ :

$q(x) = q(x) + \frac{dq}{dx}dx \implies \frac{dq}{dx} = 0$

Why this step? Steady state means no energy accumulation in the slice. The derivative of flux must be zero.

Step 2: Apply Fourier's Law

In1D, $q = -k \frac{dT}{dx}$ . Substitute:

$\frac{d}{dx}\left(-k \frac{dT}{dx}\right) = 0$

For constant $k$ :

$\frac{d^2 T}{dx^2} = 0$

Why? The second derivative being zero means $T(x)$ is linear in $x$ .

Step 3: Integrate

$\frac{dT}{dx} = C_1 \implies T(x) = C_1 x + C_2$

Step 4: Boundary Conditions

At $x=0$ : $T(0) = T_1 \implies C_2 = T_1$
At $x=L$ : $T(L) = T_2 \implies C_1 L + T_1 = T_2 \implies C_1 = \frac{T_2 - T_1}{L}$

Derivation: Thermal Stress in a Constrained Rod

Setup: A rod of length $L$ , heated uniformly from $T_0$ to $T_0 + \Delta T$ , with both ends fixed (can't move).

Step 1: Free Thermal Expansion

If unconstrained, the rod would elongate by:

$\Delta L = \alpha L \Delta T$

Why? Most materials expand linearly with temperature: $\frac{\Delta L}{L} = \alpha \Delta T$ (definition of $\alpha$ ).

Step 2: Constraint Force

The rod is held at original length $L$ . To prevent expansion $\Delta L$ , we must apply a compressive force $F$ that causes elastic strain:

$\varepsilon_{\text{elastic}} = \frac{\Delta L}{L} = \alpha \Delta T$

But elastic strain is $\varepsilon = \frac{\sigma}{E}$ (Hooke's Law). So:

$\frac{\sigma}{E} = \alpha \Delta T$

Step 3: Non-Uniform Temperature

If $T = T(x)$ , each slice has different expansion. Local thermal strain:

$\varepsilon_{\text{thermal}}(x) = \alpha [T(x) - T_0]$

If the rod is constrained, the total strain is $\varepsilon_{\text{total}} = \varepsilon_{\text{elastic}} + \varepsilon_{\text{thermal}}$ . In a statically determinate structure:

$\varepsilon_{\text{total}} = 0 \implies \varepsilon_{\text{elastic}} = -\varepsilon_{\text{thermal}}$

$\sigma(x) = E \varepsilon_{\text{elastic}} = -E \alpha [T(x) - T_0]$

For the linear temperature profile $T(x) = T_1 + (T_2 - T_1)x/L$ :

$\sigma(x) = -E \alpha \left[\frac{(T_2 - T_1)x}{L} + (T_1 - T_0)\right]$

Linear stress distribution from $\sigma(0) = -E\alpha(T_1 - T_0)$ to $\sigma(L) = -E\alpha(T_2 - T_0)$ .

Worked Examples

Worked example Example 1: Aluminum Beam in Sun-Shadow Transition

Problem: An aluminum beam ( $k=200$ W/m·K, $\alpha=23\times 10^{-6}$ /K, $E=70$ GPa) of length $L=1$ m connects a sunlit panel ( $T_1=100°C$ ) to a shadowed radiator ( $T_2=-50°C$ ). Find:

Temperature at midpoint
Heat flux
Maximum thermal stress (both ends fixed) Solution:
Midpoint temperature: Using $T(x) = T_1 + \frac{T_2-T_1}{L}x$ : $T(0.5) = 100 + \frac{-50-100}{1}(0.5) = 100 - 75 = 25°C$ Why this step? Linear profile → midpoint is the average of endpoints.
Heat flux: $q = -k\frac{T_2-T_1}{L} = -200 \frac{-50-100}{1} = 200 \times 150 = 30,000 \text{ W/m}^2 = 30 \text{ kW/m}^2$ Why positive? Heat flows from hot (x=0) to cold (x=L), in +x direction.
Thermal stress: At $x=0$ (hotest), if we take $T_0 = 25°C$ (reference): $\sigma(0) = -E\alpha(T_1-T_0) = -70 \times 10^9 \times 23 \times 10^{-6} \times (100-25)$ $= -1.61 \times 10^6 75 = -121\text{ MPa (compression)}$

At $x=L$ (coldest): $\sigma(L) = -70 \times 10^9 \times 23 \times 10^{-6} \times (-50-25) = +121 \text{ MPa (tension)}$

Why? Hot end tries to expand → compression. Cold end tries to contract → tension.

Worked example Example 2: Spacecraft Strut Thermal Cycle

Problem: A titanium strut ( $\alpha=8.6 \times 10^{-6}$ /K, $E=110$ GPa, length0.5 m) cycles between -100°C and +50°C every 90 minutes. If fully constrained, what is the stress range?

Solution:

Stress range $\Delta\sigma = E\alpha \Delta T$ : $\Delta T = 50 - (-100) = 150 \text{ K}$ $\Delta\sigma = 110 \times 10^9 \times 8.6 \times 10^{-6} \times 150 = 142 \text{ MPa}$

Why this matters? Stress range drives fatigue life. Aluminum alloys typically fail after $10^6$ cycles at ~100 MPa range. This strut experiences one thermal cycle per orbit. With a 90-minute orbit: $\frac{1440 \text{ min/day}}{90 \text{ min/cycle}} = 16$ cycles/day → $16 \times 365 \approx 5{,}840$ cycles/year. For a 10-year mission: $\approx 5.8 \times 10^4$ cycles → still a significant fatigue concern. Designer must:

Use low-CTE materials (CFRP: $\alpha \approx 0$ )
Add compliance (flexible mounts)
Pre-stress to keep stress always compressive (no tension reversals)

Common Mistakes

Active Recall Questions

#flashcards/physics

What is Fourier's Law of heat conduction? :: $\vec{q} = -k \nabla T$ , where $\vec{q}$ is heat flux, $k$ is thermal conductivity, and $\nabla T$ is temperature gradient. Heat flows from hot to cold.

Why does thermal stress arise in a heated, constrained rod?

The rod wants to expand by

\alpha L \Delta T

, but constraints prevent this expansion. To suppress the expansion, internal forces develop, creating stress

\sigma = E \alpha \Delta T

For a rod with ends at

T_1

and

T_2

, what is the steady-state temperature distribution?

T(x) = T_1 + \frac{T_2-T_1}{L}x

(linear). Follows from

\frac{d^2T}{dx^2}=0

in steady state with constant thermal conductivity.

If a titanium strut (

\alpha=8.6 \times 10^{-6}

/K,

E=110

GPa) is heated by 100K while fully constrained, what is the thermal stress?

\sigma = E\alpha\Delta T = 110 \times 10^9 \times 8.6 \times 10^{-6} \times 100 = 94.6

MPa (compression).

How does thermal stress scale with material properties?

\sigma \propto E

(stiffness),

\sigma \propto \alpha

(CTE),

\sigma \propto \Delta T

(temperature change). To reduce stress: use low-

E

materials (flexible), low-

\alpha

materials (CFRP, Invar), or reduce

\Delta T

(insulation, coatings).

What is the difference between steady-state and transient thermal conduction?

Steady-state:

\frac{\partial T}{\partial t}=0

\nabla^2 T=0

(Laplace). Transient:

\frac{\partial T}{\partial t} \neq 0

\frac{\partial T}{\partial t} = \frac{k}{\rho c_p}\nabla^2 T

(diffusion equation). Spacecraft thermal cycling is transient.

Why is thermal stress a fatigue concern in spacecraft?

Spacecraft experience thermal cycles every orbit (~90 min). Stress range

\Delta\sigma = E\alpha\Delta T

can be100+ MPa. Over tens of thousands of cycles, this causes fatigue crack growth and eventual failure.

Memory Aids

Recall Explain to a 12-Year-Old

Imagine a metal ruler lying on a table. You heat one end with a candle and cool the other end with ice. The hot end wants to get longer (atoms vibrate more, push apart), the cold end wants to get shorter. But the ruler is one piece — it can't split in half!

So the hot end pushes against the cold end, and the cold end pulls back. These pushes and pulls are thermal stress. If you heat the ruler too much, it might bend or even crack from these internal forces.

In a spacecraft, imagine metal beam connecting a solar panel (super hot in sunlight) to a radiator (super cold in shadow). Same thing happens: the beam is trying to expand and contract at the same time, creating stress. Engineers have to make sure the beam is strong enough, or they use special materials that don't expand much (like carbon fiber), so the stress stays low.

The key: Heat makes things expand. If you don't let them expand freely, they get stressed out — literally!

Connections Thermal Environment in Orbit — solar flux, albedo, Earth IR define boundary conditions for $T_1, T_2$

Material Selection for Spacecraft — $\alpha$ , $k$ , $E$ drive thermal-structural performance
Finite Element Analysis — real structures have 3D geometry; FEA solves $\nabla \cdot (k \nabla T) = 0$ and coupled thermal-stress
Thermal Control Subsystems — heaters, radiators, MLI used to manage $\Delta T$ and reduce thermal stress
Structural Dynamics — thermal stress can shift natural frequencies, affecting vibration response
Fatigue and Fracture Mechanics — thermal cycling causes high-cycle fatigue; predict crack growth
Composite Materials in Spacecraft — CFRP has near-zero or negative $\alpha$ → "athermal" structures
Deployable Structures — hinges and boms experience high thermal gradients; stress can jam mechanisms

Summary

Thermal conduction in spacecraft structures follows Fourier's Law, producing temperature gradients that drive thermal stress $\sigma = E\alpha\Delta T$ when expansion is constrained. For 1D steady-state, temperature is linear between boundaries. Stress is maximum where $\Delta T$ is largest. Key design levers: material CTE ( $\alpha$ ), stiffness ( $E$ ), thermal isolation (reduce $\Delta T$ ), and compliance (allow expansion). Thermal cycling causes fatigue — critical for long-duration missions. Always check constraint conditions: free structures expand without stress, fixed structures develop high stress.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, is note ka core intuition bilkul simple hai — jab spacecraft orbit mein ghoomta hai, to ek side Sun ke saamne hoti hai (+120°C tak garam) aur doosri side deep space ki taraf (-150°C tak thandi). Ab jab kisi metal beam ka ek end garam hota hai aur doosra thanda, to garam wala hissa phailna (expand) chahta hai aur thanda wala hissa sikudna (contract). Lekin agar beam dono ends se bolted ya fixed hai — matlab constrained hai — to woh freely move nahi kar sakti. Isi wajah se andar internal forces build up hoti hain, jinhe hum thermal stress kehte hain. Yaad rakho ek clean chain: heat flow (conduction) temperature distribution set karta hai, temperature distribution expansion drive karti hai, aur constrained expansion se stress paida hota hai.

Iske peeche do main equations hain jo tumhe zaroor samajhni chahiye. Pehli hai Fourier's Law, $\vec{q} = -k\nabla T$ , jo batati hai ki heat hamesha hot se cold ki taraf behta hai (isiliye minus sign) aur kitni tezi se behega woh material ki thermal conductivity $k$ pe depend karta hai. Steady state mein jab hum energy conservation lagate hain, to nikalta hai ki temperature rod ke along linearly change hota hai — bilkul straight line. Doosri equation hai thermal stress ki: $\sigma = -E\alpha\Delta T$ , jahan $E$ Young's modulus hai aur $\alpha$ expansion coefficient. Yahan minus sign ka matlab hai ki agar rod garam ho rahi hai (expand karna chahti hai par nahi kar sakti), to woh compression mein aa jaati hai.

Ab yeh matter kyun karta hai? Kyunki spacecraft har 90 minute mein ek orbit complete karta hai, matlab hazaaron thermal cycles jhelta hai apni life mein. Yeh stresses telescope ke optical bench ko micrometers tak bend kar sakti hain jisse pointing errors aate hain, antennas misalign ho jaate hain, aur joints thak ke (fatigue) crack ho sakte hain. Isiliye engineers ko design stage pe hi predict karna padta hai ki temperature field kaisa hoga aur usse kitna stress aayega — taaki structures ko sahi size aur material ke saath banaya ja sake. Basically, yeh analysis spacecraft ko orbit ke temperature swings mein tootne se bachati hai.