Example Data Table
Typical coefficients vary with alloy, temperature, and processing. Always verify with your material datasheet when accuracy matters.
| Material | α (×10⁻⁶ /K) | Sample L₀ | Sample ΔT | Expected ΔL |
|---|---|---|---|---|
| Steel (carbon) | 12.0 | 2.000 m | 50 °C | 1.200 mm |
| Aluminum | 23.0 | 1.500 m | 30 °C | 1.035 mm |
| Invar | 1.2 | 1.000 m | 80 °C | 0.096 mm |
| Borosilicate glass | 3.3 | 0.800 m | 60 °C | 0.1584 mm |
Formula Used
The calculator uses the small-expansion approximation. For a base dimension L₀ and temperature change ΔT:
- Linear: ΔL = α · L₀ · ΔT
- Area: ΔA ≈ 2α · A₀ · ΔT (implemented as a factor 2α)
- Volume: ΔV ≈ 3α · V₀ · ΔT (implemented as a factor 3α)
These relations assume α is constant over the temperature range and strains remain small. For large ΔT or precision metrology, use temperature-dependent α(T) data.
How to Use This Calculator
- Select the model: linear, area, or volume growth.
- Choose what you want to compute (Δ, final, α, ΔT, or L₀).
- Pick a material preset or enter a custom coefficient.
- Enter L₀, ΔT, and α in your chosen units.
- For inverse modes, provide a measured change value Δ.
- Optional: add 1σ uncertainties to estimate result confidence.
- Press Calculate to see results above the form instantly.
- Use the export buttons to download CSV or PDF summaries.
Measure carefully, verify units, and document your assumptions always.
Why linear expansion matters in practice
Thermal strain changes clearances, alignment, and stress in machines and structures. A 10 m rail that warms 35 °C with α = 12×10⁻⁶ /K lengthens about 4.2 mm, enough to bind joints or buckle if restrained. Design intent is keeping predictable movement paths. Expansion joints, sliding bearings, and bellows are common mitigation features in long runs. Ignoring movement can shift loads into anchors and create cracking.
Choosing a coefficient you can trust
Published α values vary by alloy, heat treatment, and temperature. Aluminum is often near 23×10⁻⁶ /K, carbon steel near 12×10⁻⁶ /K, and borosilicate glass near 3.3×10⁻⁶ /K. When tolerances are tight, use the supplier datasheet at your operating range. For polymers and composites, α can be anisotropic and strongly temperature dependent. Treat tabulated room-temperature numbers as a starting point, not a guarantee.
Working with temperature difference correctly
Expansion depends on ΔT, not absolute temperature. A change from 10 °C to 60 °C is the same ΔT as 290 K to 340 K. For Fahrenheit, only differences convert: ΔK = (5/9)ΔF. Always confirm sensors measure the same reference points.
Interpreting the three model options
Linear mode applies to bars, pipes, and one-dimensional lengths. Area and volume modes use small-strain approximations β ≈ 2α and γ ≈ 3α, useful for plates and bulk materials. If your geometry expands mainly in one direction, stick with linear to avoid overestimating.
Reading the result panel like an engineer
The change output is the movement you must accommodate with gaps, slots, or flexible couplers. The final value adds Δ to the starting dimension. Use the same unit system across drawings; exporting the CSV or PDF helps keep a traceable calculation record for reviews.
Using inverse solving for verification
Field measurements can validate assumptions. If a rod lengthens 1.2 mm over 2.0 m for ΔT = 50 °C, the implied α is 12×10⁻⁶ /K, matching typical steel. Solving for ΔT is useful when temperature logs are incomplete.
Adding uncertainty for realistic tolerances
Input uncertainty propagates into output uncertainty. Small σ in α, L0, or ΔT can dominate when Δ is small. For example, with L0 = 1.5 m, α = 23×10⁻⁶ /K, ΔT = 30 °C, and σ(ΔT)=0.5 °C, the ΔL uncertainty is about 0.017 mm.
Good workflow for documentation and QA
Record material grade, α source, measurement tools, and the temperature interval. Confirm units, sign conventions, and whether restraints create thermal stress beyond simple expansion. Archive the exported report with your project files so calculations remain reproducible months later. Include brief notes and approval dates.
Q1. What does α represent?
α is the linear coefficient of thermal expansion: fractional length change per kelvin. If α = 12×10⁻⁶ /K, a 1 m bar changes by 12 µm for each 1 K temperature change, assuming α is constant.
Q2. Can ΔT be negative?
Yes. Cooling makes ΔT negative, so the change becomes negative and the final length decreases. The magnitude still follows the same formula, so unit conversion and sign convention are the main things to check.
Q3. Why do area and volume options multiply α by 2 or 3?
For small strains, area expansion is approximately twice the linear strain and volume expansion is about three times. That is why β ≈ 2α and γ ≈ 3α. The approximation is best for isotropic materials and modest ΔT.
Q4. Which α unit should I use?
Datasheets may list α as 1/K or as µm/m·K. They are equivalent, just scaled. For example, 12 µm/m·K equals 12×10⁻⁶ /K. Choose the unit that matches your source to avoid conversion mistakes.
Q5. How accurate are the preset material values?
They are typical reference values for quick estimates. Real α depends on alloy grade, temperature, and processing. For engineering tolerances, use your exact specification and, when possible, temperature-dependent α(T) data.
Q6. What does the uncertainty section do?
If you enter 1σ uncertainties for L₀, α, and ΔT, the calculator estimates output uncertainty using standard propagation for independent inputs. This helps you compare measurement quality and understand whether a computed change is significant.
Q7. When should I solve for α or ΔT?
Use inverse solving when you measured expansion but want to infer α or the effective temperature change. It is useful for verification tests, calibration checks, or diagnosing mismatches between expected and observed movement in the field.
Accurate inputs make expansion predictions reliable for real projects.