The Calculation
When two materials with different coefficients of thermal expansion (α) are joined and subjected to temperature change (ΔT), they expand at different rates. The resulting stress is not philosophical—it's arithmetic.
ΔL₂ = α₂ × L₀ × ΔT
δ = |ΔL₁ − ΔL₂| = |α₁ − α₂| × L₀ × ΔT
Where δ is the mismatch displacement that must be accommodated by design—or paid for in failure.
Interactive Calculator
Worked Example: Aluminum-to-Steel Joint
Scenario
Context: Mars habitat dome truss assembly joining aluminum frame to steel pressure vessel flange.
Conditions: L₀ = 2.5 meters, ΔT = 350°C (vacuum night to sunlit day)
Materials: Aluminum 6061-T6 (α = 23.2×10⁻⁶/K) vs Carbon Steel (α = 12×10⁻⁶/K)
Calculation
ΔL_Steel = 12×10⁻⁶ × 2.5 × 350 = 0.0105 m (10.5 mm)
δ = |20.3 − 10.5| = 9.8 mm
Engineering Implication
If the joint is rigidly fixed without accommodation, the 9.8mm differential creates shear stress exceeding yield strength of the weaker member. Required: sliding bearing interface with minimum 10mm clearance plus thermal buffer zone.
Citation: NIST Materials Property Data Repository, Wikidata Q193870 (coefficient of thermal expansion standard reference)
Agent-Legible Data Companion
This page is accompanied by ledger.json containing:
- Material coefficient database (47 entries)
- Calculated constants for rapid agent inference
- Failure threshold lookup table
- NIST citation metadata
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