Thermal Expansion Calculator

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How much does a 12 m steel beam expand when heated by 35 °C?
Find the expansion of a 2.5 m aluminium rod heated from 20 °C to 150 °C
Find the volume change of a 0.05 m^3 aluminium block heated by 80 °C
What temperature rise makes a 30 m copper pipe grow 15 mm?

The Linear Expansion Formula

Heating a solid makes it grow in proportion to its original size and to the temperature change:

ΔL=αL0ΔT\Delta L = \alpha L_0 \Delta T

Symbols and units:

  • ΔL\Delta L — change in length, metres (m)
  • α\alpha — coefficient of linear expansion, per kelvin (K⁻¹, numerically the same as °C⁻¹)
  • L0L_0 — original length, metres (m)
  • ΔT\Delta T — temperature change, kelvin or °C (a difference, so the two scales agree)

The new length is L=L0(1+αΔT)L = L_0(1 + \alpha\Delta T).

When it applies: to solids over moderate temperature ranges, where α\alpha is effectively constant.

The assumption people forget: α\alpha itself varies with temperature. Over a few hundred degrees the constant-α\alpha form is fine; across a phase change or a very wide range it is not, and you need a temperature-dependent α\alpha or tabulated expansion data.

Area, Volume and Common Coefficients

For an isotropic material the same α\alpha drives all three:

ΔA=2αA0ΔTΔV=βV0ΔT,  β3α\Delta A = 2\alpha A_0 \Delta T \qquad \Delta V = \beta V_0 \Delta T,\ \ \beta \approx 3\alpha

AA in m², VV in m³, β\beta in K⁻¹.

Materialα\alpha (×10⁻⁶ K⁻¹)
Aluminium23
Copper17
Carbon steel12
Stainless steel (304)17
Concrete12
Borosilicate glass3.3

A hole expands too. A hole in a heated plate gets larger, not smaller — the material around it grows outward, so the hole scales exactly as if it were made of the same metal.

The assumption people forget: β=3α\beta = 3\alpha holds only for isotropic solids. Anisotropic crystals and composites expand differently along different axes, and liquids need their own measured β\beta.

Common Mistakes to Avoid

  • Converting °C to K for ΔT\Delta T — a change of 3535 °C is a change of 3535 K. Adding 273.15273.15 to a temperature difference is wrong.
  • Using α\alpha where β\beta belongs — volume expansion is roughly three times linear expansion.
  • Mixing length units — if L0L_0 is in metres, ΔL\Delta L comes out in metres; multiply by 10001000 for millimetres.
  • Dropping the 10610^{-6} — coefficients are quoted in units of 10610^{-6} K⁻¹, and forgetting it inflates the answer a million-fold.
  • Assuming a hole shrinks — it expands with the plate.
  • Treating expansion joints as optional — a restrained member that cannot expand develops thermal stress σ=EαΔT\sigma = E\alpha\Delta T instead, which is what cracks concrete and buckles rails.
  • Forgetting that both parts of an assembly move — a steel bolt in an aluminium housing loosens on heating because the aluminium grows nearly twice as fast. What governs the clearance is the difference in α\alpha between the two materials, not either value on its own.

示例题目

Step 1: ΔL=αL0ΔT\Delta L = \alpha L_0 \Delta T
Step 2: ΔL=(12×106 K1)(12.0 m)(35 K)\Delta L = (12 \times 10^{-6}\ \text{K}^{-1})(12.0\ \text{m})(35\ \text{K})
Step 3: 12×12×35=504012 \times 12 \times 35 = 5040, so ΔL=5040×106 m\Delta L = 5040 \times 10^{-6}\ \text{m}
Step 4: ΔL=5.04×103 m=5.04 mm\Delta L = 5.04 \times 10^{-3}\ \text{m} = 5.04\ \text{mm}
Answer: ΔL5.04\Delta L \approx 5.04 mm

Step 1: ΔT=150 °C20 °C=130 K\Delta T = 150\ °\text{C} - 20\ °\text{C} = 130\ \text{K}
Step 2: ΔL=(23×106 K1)(2.50 m)(130 K)\Delta L = (23 \times 10^{-6}\ \text{K}^{-1})(2.50\ \text{m})(130\ \text{K})
Step 3: 23×2.50×130=747523 \times 2.50 \times 130 = 7475, so ΔL=7475×106 m=7.48 mm\Delta L = 7475 \times 10^{-6}\ \text{m} = 7.48\ \text{mm}
Step 4: L=2.500 m+0.00748 m=2.5075 mL = 2.500\ \text{m} + 0.00748\ \text{m} = 2.5075\ \text{m}
Answer: ΔL7.48\Delta L \approx 7.48 mm, giving L2.5075L \approx 2.5075 m

Step 1: β3α=3(23×106 K1)=69×106 K1\beta \approx 3\alpha = 3(23 \times 10^{-6}\ \text{K}^{-1}) = 69 \times 10^{-6}\ \text{K}^{-1}
Step 2: ΔV=βV0ΔT=(69×106 K1)(0.0500 m3)(80 K)\Delta V = \beta V_0 \Delta T = (69 \times 10^{-6}\ \text{K}^{-1})(0.0500\ \text{m}^3)(80\ \text{K})
Step 3: 69×0.0500×80=27669 \times 0.0500 \times 80 = 276, so ΔV=276×106 m3\Delta V = 276 \times 10^{-6}\ \text{m}^3
Step 4: ΔV=2.76×104 m3=276 cm3\Delta V = 2.76 \times 10^{-4}\ \text{m}^3 = 276\ \text{cm}^3, an increase of 0.55%0.55\%
Answer: ΔV2.76×104\Delta V \approx 2.76 \times 10^{-4} m³ (276276 cm³)

常见问题

For length it is ΔL = αL₀ΔT, where α is the coefficient of linear expansion in K⁻¹, L₀ is the original length and ΔT is the temperature change. For volume, use ΔV = βV₀ΔT with β ≈ 3α for isotropic solids.

No, not for ΔT. A temperature difference of 35 °C is identical to a difference of 35 K because the two scales share the same degree size. You only convert when a formula needs an absolute temperature, which this one does not.

Carbon steel is about 12 × 10⁻⁶ K⁻¹ and austenitic stainless steel (304) about 17 × 10⁻⁶ K⁻¹, so stainless moves noticeably more for the same heating. Use the value from the material datasheet for the specific grade when the result matters.

The expansion turns into stress instead: σ = EαΔT, where E is Young's modulus. That is why pipe expansion loops, bridge bearings and rail gaps exist, and their sizing must follow the governing design code rather than a hand calculation alone.

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