Physics · Thermal expansion

Linear thermal expansion

Q: What does linear thermal expansion describe physically?

The relation $\Delta L\simeq\alpha L_0\Delta T$ describes free length change from a reference length at starting temperature T₀. Positive α makes heating lengthen the body, while cooling gives a negative change. The coefficient belongs to material, direction and temperature range. A rod fixed at one end but otherwise free extends at its free end; complete restraint can instead create stress. For $L_0=10\,\mathrm m$, $\alpha=12\cdot10^{-6}\,\mathrm{K^{-1}}$ and $\Delta T=30\,\mathrm K$, $\Delta L=3.6\,\mathrm{mm}$, not a final length of 3.6 mm. Final L is 10.0036 m. Temperature gradients and material transformations require an extended model.

Q: How can I check a calculation involving linear thermal expansion?

For a rail example with $L_0=10\,\mathrm m$, $\alpha=12\cdot10^{-6}\,\mathrm{K^{-1}}$ and $\Delta T=30\,\mathrm K$, $\Delta L=0.0036\,\mathrm m=3.6\,\mathrm{mm}$. Reversing $\frac{\Delta L}{L_0\Delta T}$ returns the assumed coefficient. Also check that you use a temperature difference and length at the initial temperature. Dimensionless $\alpha\Delta T=0.00036$ is small, explaining the linear approximation. A constant differential coefficient gives the more precise model $L=L_0e^{\alpha\Delta T}$, whereas a measured mean applies directly to its interval. This mathematical check does not prove that the actual rail is free to expand; supports can change its mechanical response.

Free linear length change is approximately $\Delta L\simeq\alpha L_0\Delta T$ when material direction, temperature interval and coefficient match. L₀ is reference length at T₀.

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Formula

Calculate

Δ L = α L_{0} Δ T

ΔL and L₀ in m; α in K⁻¹; ΔT in K or °C as a temperature difference.

Variables & units

Variables & units
Symbol	Meaning	Unit
$ΔL$	Change in length	m
$α$	Material’s linear expansion coefficient	K⁻¹
$L₀$	Initial length	m
$ΔT$	Temperature change	K

LaTeX

\Delta L = \alpha L_0 \Delta T

QUANTA STUDYquanta-study.de

Diametral clearance g in millimetres versus temperature change: initially minus 0.040 millimetre, zero at 76.994 kelvin and finally plus 0.011952 millimetre at 100 kelvin. — Original free-fit calculation: both separate parts undergo the same temperature change. Diametral clearance reaches zero near 77 K and is about 11.95 µm at 100 K. The small difference is shown magnified.

Linear thermal expansion calculator

Find the change in length, final length, initial length, expansion coefficient or temperature change of a rod that can expand freely, using ΔL = α · L₀ · ΔT.

Find

Initial length L₀

Length at the starting temperature, e.g. 2 m for a steel bar.

Linear expansion coefficient ᾱ

Linear expansion coefficient of the material, e.g. 12 · 10⁻⁶ /K for steel, 23 · 10⁻⁶ /K for aluminium.

Temperature change ΔT

Temperature change, end minus start, e.g. 50 K from 20 °C to 70 °C; negative when cooling.

Filled with example values. Type your own and the result updates straight away.

Result: Change in length ΔL

ΔL = 1.2 mm

With L₀ = 2 m, ᾱ = 12 10⁻⁶/K and ΔT = 50 K, ΔL = 1.2 mm.

Working · rearrange for ΔL, substitute, evaluate

1. Rearrange for ΔL
$\Delta L=\bar\alpha L_0\Delta T$
2. Substitute values (in SI units)
- ᾱ = 12 10⁻⁶/K = 1.2 · 10⁻⁵ K⁻¹
$\Delta L=0.000012\cdot 2\cdot 50$
3. Result
ΔL = 1.2 mm

ΔT in °C is the same difference as in K; no 273.15 offset is added. Final length must remain positive. A negative ᾱ is meaningful only for the matching material and interval.

Quick tips

Estimate: with α = 10 · 10⁻⁶ /K every metre grows by 1 mm per 100 K of heating. Steel at 12 · 10⁻⁶ /K is just above that.
For ΔT kelvin and degrees Celsius are the same: a difference of 50 °C is 50 K. But never insert the final temperature instead of the difference.

Assumptions and limits

Homogeneous unloaded rod, uniform temperature and free expansion along the selected direction. L₀ > 0 is length at the start of the interval. ᾱ explicitly means the interval-average coefficient ΔL/(L₀ΔT) for this material and interval, not a universal table value. With this mean, ΔL = ᾱL₀ΔT is the interval relation. A local differential coefficient requires a justified linear approximation. At ΔT = 0 there is no length change; this zero case does not determine ᾱ. No constraint stress, fit calculation or phase transition.

Derivation & background

The differential coefficient is $\alpha(T)=\frac1L\frac{dL}{dT}$ . Integration gives $\frac L{L_0}=\exp\!\left(\int_{T_0}^T\alpha(\vartheta)\,d\vartheta\right)$ . A constant coefficient and small relative strain give the linear approximation. Measured interval mean $\frac{\Delta L}{L_0\Delta T}$ instead directly applies to its interval. The formula describes free uniform change; restraint can generate thermal stress. Under free isotropic expansion a hole grows too.

Exam blueprint

Validity range

Free uniform thermal length change in the specified material direction, without phase or material transformation. For constant differential α, the short linear form approximates small $|\alpha\Delta T|$ ; a measured reference-length mean applies to its stated interval. Constrained expansion additionally requires a mechanical balance.

Derivation steps

Integrate differential relative length change and linearise for small strain.

1 $L_0$ is free reference length at $T_0$ . Define $\Delta T=T-T_0$ and $\Delta L=L-L_0$ .
2Differential coefficient is $\alpha(T)=\frac1L\frac{dL}{dT}$ for the chosen direction.
3Separate variables: $\frac{dL}{L}=\alpha(T)\,dT$ .
4Integration gives $\ln\frac L{L_0}=\int_{T_0}^T\alpha(\vartheta)\,d\vartheta$ .
5Constant differential coefficient gives $\frac L{L_0}=e^{\alpha\Delta T}$ .
6For $|\alpha\Delta T|\ll1$ , $e^{\alpha\Delta T}\simeq1+\alpha\Delta T$ , giving $\Delta L_{\mathrm{lin}}=\alpha L_0\Delta T$ .
7Measured interval mean $\overline\alpha=\frac{\Delta L}{L_0\Delta T}$ directly describes its interval and is not the differential coefficient.

Rearrangements

Material coefficient from measurement

\overline\alpha=\frac{\Delta L}{L_0\Delta T}

This is the reference-length mean coefficient for the measurement interval. The differential coefficient is instead $\alpha(T)=\frac1L\frac{dL}{dT}$ . Temperature differences have the same numerical value in K and °C. This inverse requires L₀ > 0 and ΔT ≠ 0. If ΔL = ΔT = 0, the mean coefficient is undetermined.

Task variant

A free 2.0 m rod with $\alpha=20\cdot10^{-6}\,\mathrm{K^{-1}}$ warms by 50 K. Find its length change.

$\Delta L=\alpha L_0\Delta T=20\cdot10^{-6}\cdot2.0\cdot50\,\mathrm m=0.0020\,\mathrm m=2.0\,\mathrm{mm}$ .

Common mistakes

Using absolute temperature instead of ΔT or assuming free expansion when the body is constrained.

Use only the temperature change; constraints additionally create thermal stress.

Exam context

Bridge joints, rail lengths, brass-steel comparisons and dimensional change under heating.

Formula cluster

Temperature and material properties

Links heat transfer to measurable changes in geometry.

Q = m·c·ΔT

Heat Capacity: Q = m·c·ΔT

ΔU = Q − W

First Law of Thermodynamics

ρ = m/V

Density

Worked example

A freely expanding steel rail has $L_0=10\,\mathrm m$ and $\alpha=12\cdot10^{-6}\,\mathrm{K^{-1}}$ . Warming by 30 K gives $\Delta L=\alpha L_0\Delta T=0.0036\,\mathrm m=3.6\,\mathrm{mm}$ . Check: relative strain is 0.00036 and final length is 10.0036 m.

Applications

Expansion joints, material measurement, temperature sensors and thermally loaded components.

How do you calculate with Linear thermal expansion?

Here is how to work through a typical Linear thermal expansion (ΔL = αL₀ΔT) task step by step:

1
Task
A free 2.0 m rod with $\alpha=20\cdot10^{-6}\,\mathrm{K^{-1}}$ warms by 50 K. Find its length change.
Solution path
$\Delta L=\alpha L_0\Delta T=20\cdot10^{-6}\cdot2.0\cdot50\,\mathrm m=0.0020\,\mathrm m=2.0\,\mathrm{mm}$ .

Understand and apply the equation

A heated rod tends to become longer. Whether this produces movement, a looser fit or compressive stress depends on its surroundings. First calculate free length change using $\Delta L\simeq\alpha L_0\Delta T$ . Then check whether the body can actually move freely and whether its mating component also warms.

What L₀ and ΔL actually measure

Start with the reference state: $L_0$ is unloaded length at $T_0$ . After uniform free heating, $L=L_0+\Delta L$ . For positive $\alpha$ , positive $\Delta T$ means extension, whereas cooling means contraction. With only one end fixed, the free end moves by $\Delta L$ . If both ends can move, displacement is distributed according to the boundary conditions; the length change itself is unchanged.

Coefficient $\alpha$ is a measured material property, not an arbitrary constant. Temperature, composition and crystal direction can change it. The simple model uses a value appropriate to the direction and temperature interval considered. The same scalar value applies in every direction only for isotropic expansion.[1]

First understand relative change

A relative length change of $0.001$ means one millimetre per metre of original length. For $\alpha=20\cdot10^{-6}\,\mathrm{K^{-1}}$ and $\Delta T=50\,\mathrm K$ , this is the expected relative change in the linear approximation. A rod twice as long gains twice as much length while the relative fraction stays unchanged. This checks the scale before numerical substitution.[1]

\begin{aligned}\varepsilon_{\mathrm{th}}&\simeq\alpha\Delta T\\\Delta L&=L_0\varepsilon_{\mathrm{th}}\end{aligned}

Relative change first, absolute length change afterwards.

Explore further: differential coefficient and interval mean

\begin{aligned}&\alpha(T)=\frac1{L(T)}\frac{dL}{dT}\\&\ln\frac{L(T)}{L_0}=\int_{T_0}^{T}\alpha(\vartheta)\,d\vartheta\end{aligned}

The definition uses current length. Free thermal length change is considered under appropriate mechanical boundary conditions.[1]

The differential definition describes a very small change relative to the current length. Separate $\frac{dL}{L}=\alpha(T)\,dT$ and integrate from the reference state. A constant differential coefficient gives an exponential function. Expanding it for small $|\alpha\Delta T|$ gives the familiar linear relation to first order. Thus check two conditions separately: small relative strain and an appropriate, nearly constant coefficient over the interval.

\begin{aligned}&L=L_0e^{\alpha\Delta T}\\&\Delta L\simeq\alpha L_0\Delta T\end{aligned}

The exponential form is exact only for a constant differential model coefficient, not an arbitrary tabulated interval mean.

Tables may instead give a measured interval mean $\overline\alpha=\frac{\Delta L}{L_0\Delta T}$ . With that value the finite linear relation holds by definition for that particular interval. It must not simultaneously be treated as a constant differential coefficient in the exponential equation. The definition determines what a supposedly more accurate calculation actually means.

Rearrangements and relative strain

\begin{aligned}&\varepsilon_{\mathrm{th}}=\frac{\Delta L}{L_0}\simeq\alpha\Delta T\\&\alpha\simeq\frac{\Delta L}{L_0\Delta T}\\&\Delta T\simeq\frac{\Delta L}{\alpha L_0}\end{aligned}

Rearrangements need nonzero denominators. α has unit K⁻¹; thermal strain is dimensionless.

A change of 60 °C equals a change of 60 K. The Celsius offset cancels on subtraction; absolute final temperature does not replace $\Delta T$ . A coefficient of $17\cdot10^{-6}\,\mathrm{K^{-1}}$ means 17 millionths relative change per kelvin. Multiplying by length and temperature change gives a length again. Do not confuse the usually small addition $\Delta L$ with total final length.[2]

Heat capacity relates supplied heat to a temperature change. The expansion formula takes that change as given; it determines neither required heating energy nor heating time.[3]

Example: a free rod and controlled approximation error

Uniformly heating a copper rod

Given: $L_0=1.80\,\mathrm m$ at 20 °C, final temperature 80 °C; specified constant differential model $\alpha=17\cdot10^{-6}\,\mathrm{K^{-1}}$ , unconstrained.
Find: Find ΔL and L using the linear equation and check against the exponential model.

Approach: Use the linear approximation $\Delta L=\alpha L_0\Delta T$ ; constant differential coefficient gives $L=L_0e^{\alpha\Delta T}$ as a check. Form temperature change and relative strain.
$\begin{aligned}\Delta T&=60\,\mathrm K\\{} z&=\alpha\Delta T=0.001020\end{aligned}$
Multiply by reference length.
$\begin{aligned}\Delta L_{\mathrm{lin}}&=1.80\cdot0.001020=0.001836\,\mathrm m\\{} L_{\mathrm{lin}}&=1.801836\,\mathrm m\end{aligned}$
Check the constant differential model without linearisation.
$\Delta L_{\mathrm{exp}}=1.80(e^{0.001020}-1)=0.00183693668\,\mathrm m$

Result: Linear $\Delta L=1.836\,\mathrm{mm}$ ; the exponential model gives 1.83694 mm, about 0.000937 mm more.

Check: Using $\Delta L_{\mathrm{lin}}$ , $\frac{\Delta L_{\mathrm{lin}}}{L_0\Delta T}$ returns the assumed value $17\cdot10^{-6}\,\mathrm{K^{-1}}$ . The larger exponential length change instead gives a slightly larger interval mean referred to L₀. Leading absolute truncation error $\frac{L_0z^2}{2}\approx0.936\,\mathrm{\mu m}$ explains the difference; actual variation of α may exceed this linearisation error.

Holes grow during free isotropic heating

Why does a hole grow too? Imagine the free isotropic body initially without the hole: all distances scale equally during uniform heating. Removing the imagined insert afterwards leaves a larger distance across the rim. Hole diameter therefore obeys $\Delta D\simeq\alpha D_0\Delta T$ . For a fit, this is only half the calculation. If the shaft also heats up, its diameter changes too.[1]

g(T)=D_{h,0}(1+\alpha_h\Delta T)-D_{s,0}(1+\alpha_s\Delta T)

g is diametral clearance here for equal temperature change of both components; g < 0 means interference.

Example: two fit components heated together

Accounting separately for hole and shaft

Given: At 20 °C: aluminium hole $D_{h,0}=40.00\,\mathrm{mm}$ and steel shaft $D_{s,0}=40.04\,\mathrm{mm}$ . Specified model values $\alpha_h=25\cdot10^{-6}\,\mathrm{K^{-1}}$ and $\alpha_s=12\cdot10^{-6}\,\mathrm{K^{-1}}$ . Both parts separately heated freely by the same ΔT; not yet assembled.
Find: Find temperature change for zero interference and clearance at $\Delta T=100\,\mathrm K$ .

Approach: Diametral clearance is $g=D_{h,0}(1+\alpha_h\Delta T)-D_{s,0}(1+\alpha_s\Delta T)$ ; both parts expand freely. Find initial difference and rate of change.
$\begin{aligned}g_0&=-0.0400\,\mathrm{mm}\\{} \frac{\Delta g}{\Delta T}&=40.00(25\times10^{-6})-40.04(12\times10^{-6})=0.00051952\,\mathrm{mm/K}\end{aligned}$
Solve $g=0$ for ΔT.
$\Delta T_{\mathrm{contact}}=\frac{0.0400}{0.00051952}\approx76.9941\,\mathrm K$
Calculate both diameters at 100 K.
$\begin{aligned}D_h&=40.1000\,\mathrm{mm}\\{} D_s&=40.088048\,\mathrm{mm}\\{} g&=0.011952\,\mathrm{mm}\end{aligned}$

Result: Diameters just match at about 97.0 °C; at 120 °C the model gives about 11.95 µm diametral clearance.

Check: Substitution into $g=(-0.0400+0.00051952\,\mathrm{K^{-1}}\Delta T)\,\mathrm{mm}$ confirms both values. If the shaft stayed at 20 °C, matching diameters would require only 40 K; that is a different condition. Assembly additionally needs appropriate tolerances and clearance, not merely $g=0$ .

View diagram: fit clearance

When does interference become clearance?

QUANTA STUDYquanta-study.de

Original linear model for the separately and freely heated parts in the fit example. Below zero is interference; above zero is clearance. It does not describe a press-fitted assembly.[1]

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Prevented expansion creates stress

\begin{aligned}\varepsilon_{\mathrm{total}}&=\varepsilon_{\mathrm{mech}}+\alpha\Delta T\\{} \sigma&=E\varepsilon_{\mathrm{mech}}\quad\Longrightarrow\quad\sigma=-E\alpha\Delta T\quad(\varepsilon_{\mathrm{total}}=0)\end{aligned}

Complete axial constraint, linear elasticity and positive tensile stress; heating with $\alpha>0$ produces compression.[1]

Free extension and actual extension are different quantities. Under complete axial constraint, elastic mechanical compression must cancel thermal strain. A compliant support permits some movement and produces a different stress. Yielding, buckling and strong temperature gradients lie outside this simple model. For a bimetal, bonding, stiffnesses and geometry determine curvature; two free length changes alone are insufficient.

Mean normal stress under axial force connects the support force with cross-sectional area. The compressive stress calculated below results from restrained free expansion; it is not an additional thermal length change.[1]

A constraint loads the rod

Compressive force instead of free extension

Given: Initially unloaded rod: length 1.50 m, cross-section 80.0 mm², constant model values 200 GPa and 12.0·10⁻⁶ K⁻¹. Uniform warming by 25.0 K. Both ends are axially rigidly held. Small strain, linear elasticity and no buckling or yielding.
Find: Find free extension, actual extension, stress and axial force.

Approach: total strain is zero. Mechanical strain therefore cancels free thermal strain; tension is positive.
$\begin{aligned}\Delta L_{\mathrm{free}}&=L_0\alpha\Delta T\\\varepsilon_{\mathrm{mech}}&=-\alpha\Delta T\\\sigma&=E\varepsilon_{\mathrm{mech}}\\F&=\sigma A\end{aligned}$
First calculate free strain and extension.
$\begin{aligned}\alpha\Delta T&=12.0\cdot10^{-6}\cdot25.0=0.000300\\\Delta L_{\mathrm{free}}&=1.50\cdot0.000300\,\mathrm m=0.450\,\mathrm{mm}\end{aligned}$
Convert prevented strain into stress and force.
$\begin{aligned}\sigma&=-200\cdot10^9\cdot0.000300\,\mathrm{Pa}=-60.0\,\mathrm{MPa}\\A&=80.0\cdot10^{-6}\,\mathrm{m^2}\\F&=-60.0\cdot10^6\cdot80.0\cdot10^{-6}\,\mathrm N=-4800\,\mathrm N\end{aligned}$

Result: Freely, the rod would lengthen by 0.450 mm. Its actual length stays unchanged; the model gives compressive stress 60.0 MPa and compressive force 4.80 kN.

Check: $\varepsilon_{\mathrm{mech}}+\varepsilon_{\mathrm{th}}=0$ . Also, $\frac{|F|}{A}=60.0\,\mathrm{MPa}$ . A compliant support or buckling would remove complete axial constraint; this force would then no longer apply.

Area, volume and approximation limits

\begin{aligned}&\frac{\Delta A}{A_0}\simeq2\alpha\Delta T\\&\frac{\Delta V}{V_0}\simeq3\alpha\Delta T\end{aligned}

First order for free isotropic expansion in two and three dimensions respectively.[1]

Factors two and three result from multiplying two or three linear factors and discarding higher powers. For isotropic expansion, the differential volume coefficient is exactly $3\alpha$ . Rounded tabulated means may differ. Liquids and gases need an appropriate volume model rather than a rod model. Phase or material transformations, anisotropic properties and nonuniform temperature require more than one scalar linear formula.

Five solved exercises

Calculate · Foundation

$L_0=2.0\,\mathrm m$ , $\alpha=20\cdot10^{-6}\,\mathrm{K^{-1}}$ and $\Delta T=-50\,\mathrm K$ . Find ΔL.

Show hint

Cooling gives negative ΔT.

Learning goal: Calculate negative length change on cooling.

Show solution and calculation

Approach: free linear change $\Delta L=\alpha L_0\Delta T$ , then $L=L_0+\Delta L$ .
$\Delta L=\alpha L_0\Delta T=-0.0020\,\mathrm m=-2.0\,\mathrm{mm}$ .
Final length is 1.998 m; the model predicts contraction.

\Delta L=-2.0\,\mathrm{mm}

Rearrange · Application

A free 0.800 m specimen grows 0.384 mm over 40 K. Find mean α.

Show hint

Put lengths in matching units.

Learning goal: Find a mean expansion coefficient from measurements.

Show solution and calculation

Approach: the measured reference-length mean is $\overline\alpha=\frac{\Delta L}{L_0\Delta T}$ .
$\begin{aligned}\overline\alpha&=\frac{\Delta L}{L_0\Delta T}\\{}&=\frac{0.000384}{0.800\cdot40}\,\mathrm{K^{-1}}\\{}&=12\cdot10^{-6}\,\mathrm{K^{-1}}\text{.}\end{aligned}$ Substitution returns 0.384 mm; this is an interval mean.

\overline\alpha=12\times10^{-6}\,\mathrm{K^{-1}}

Transfer · Challenge

A fully axially constrained elastic rod has model $E=200\,\mathrm{GPa}$ , $\alpha=12\cdot10^{-6}\,\mathrm{K^{-1}}$ and $\Delta T=30\,\mathrm K$ . Find stress.

Show hint

Mechanical strain cancels free thermal strain.

Learning goal: Translate thermal strain into elastic stress under complete axial constraint.

Show solution and calculation

Approach: complete axial restraint requires $0=\frac\sigma E+\alpha\Delta T$ , hence $\sigma=-E\alpha\Delta T$ .
Thermal strain is 0.00036 and mechanical strain −0.00036.
$\sigma=-E\alpha\Delta T=-72\,\mathrm{MPa}$ , compression.
This assumes elastic behaviour without buckling.

\sigma=-72\,\mathrm{MPa}

Interpret · Application

A homogeneous isotropic metal plate has a circular hole 30.0 mm in diameter. It is heated freely and uniformly by 80 K; $\alpha=15\cdot10^{-6}\,\mathrm{K^{-1}}$ is constant and the linear approximation applies. Does the hole grow or shrink? Find the diameter increase and explain why expansion of surrounding material does not close it.

Show hint

Under free uniform expansion, all distances in the plate scale equally. Imagine temporarily filling the hole with the same material.

Learning goal: Explain free thermal expansion of a hole diameter geometrically.

Show solution and calculation

Approach: treat hole diameter as the distance between two material points on its edge.
It scales by the same positive strain factor as the plate.
The hole grows by 0.036 mm to 30.036 mm.

Check: an imagined insert of the same material would also grow before removal, matching the enlarged hole. Constraints or nonuniform temperature could change this free-expansion model.

\begin{aligned}\varepsilon&=\alpha\Delta T=(15\cdot10^{-6})(80)=0.0012\\\Delta D&=D_0\alpha\Delta T=(30.0\,\mathrm{mm})(0.0012)=0.036\,\mathrm{mm}\\D_f&=D_0+\Delta D=30.036\,\mathrm{mm}\end{aligned}

[1]

Error analysis · Foundation

A freely expanding rod with $L_0=1.0\,\mathrm m$ and $\alpha=10\cdot10^{-6}\,\mathrm{K^{-1}}$ warms from 20 °C to 50 °C. A solution adds 273.15 to $\Delta T=30\,{}^\circ\mathrm C$ and inserts 303.15 K into $\Delta L=L_0\alpha\Delta T$ . Correct the temperature difference and length increase. Why is the Celsius offset wrong here?

Show hint

The offset is added to each absolute temperature. It cancels when subtracting the two temperatures.

Learning goal: Use Celsius and kelvin temperature differences without an absolute offset.

Show solution and calculation

Approach: either convert both temperatures to kelvin and subtract or use the Celsius difference as the equal-sized kelvin difference directly.
Correct values are $\Delta T=30\,\mathrm K$ and $\Delta L=0.30\,\mathrm{mm}$ .

Check: without a temperature change, expansion must vanish; adding 273.15 to every difference would predict unphysical expansion even then.

\begin{aligned}\Delta T&=(50+273.15)\,\mathrm K-(20+273.15)\,\mathrm K=30\,\mathrm K\\\Delta L&=(1.0\,\mathrm m)(10\cdot10^{-6}\,\mathrm{K^{-1}})(30\,\mathrm K)\\&=0.00030\,\mathrm m=0.30\,\mathrm{mm}\end{aligned}

[1][2]

In brief

Specify reference length, temperature change and an appropriate coefficient. Calculate free strain first, then check the support. Holes grow during free isotropic warming; a fit additionally requires shaft expansion. Kelvin and Celsius give the same temperature difference. The coefficient definition and material-model limits determine whether the linear calculation is sufficient.

Sources for this explanation

Quanta · Updated October 2, 2026

Editorial responsibility: Quanta Study editorial team

Review scope

This explanation, 3 worked examples and 5 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.

An internal human subject review is documented for an earlier release of this page. That confirmation does not apply automatically to the revised explanation and new exercises.

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

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Change history

October 2, 2026Linked heating energy and axial normal stress in context; preserved interval coefficients, fits, constraint stress, exercises and diagram.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.
October 1, 2026Reworked the introduction and explanations; added a third complete example, an original quantitative diagram and a short summary. Retained existing exercises, models and sources; clarified mathematical typography and model conditions.

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Quanta. Linear thermal expansion. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/lineare-waermeausdehnung

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Frequently asked questions about Linear thermal expansion

What does linear thermal expansion describe physically?+

The relation $\Delta L\simeq\alpha L_0\Delta T$ describes free length change from a reference length at starting temperature T₀. Positive α makes heating lengthen the body, while cooling gives a negative change. The coefficient belongs to material, direction and temperature range. A rod fixed at one end but otherwise free extends at its free end; complete restraint can instead create stress. For $L_0=10\,\mathrm m$ , $\alpha=12\cdot10^{-6}\,\mathrm{K^{-1}}$ and $\Delta T=30\,\mathrm K$ , $\Delta L=3.6\,\mathrm{mm}$ , not a final length of 3.6 mm. Final L is 10.0036 m. Temperature gradients and material transformations require an extended model.

How can I check a calculation involving linear thermal expansion?+

For a rail example with $L_0=10\,\mathrm m$ , $\alpha=12\cdot10^{-6}\,\mathrm{K^{-1}}$ and $\Delta T=30\,\mathrm K$ , $\Delta L=0.0036\,\mathrm m=3.6\,\mathrm{mm}$ . Reversing $\frac{\Delta L}{L_0\Delta T}$ returns the assumed coefficient. Also check that you use a temperature difference and length at the initial temperature. Dimensionless $\alpha\Delta T=0.00036$ is small, explaining the linear approximation. A constant differential coefficient gives the more precise model $L=L_0e^{\alpha\Delta T}$ , whereas a measured mean applies directly to its interval. This mathematical check does not prove that the actual rail is free to expand; supports can change its mechanical response.

What assumptions and limits apply to linear thermal expansion?+

The simple free-length calculation requires uniform temperature and a suitable coefficient in the chosen direction. Constant differential α additionally needs linearisation for small $|\alpha\Delta T|$ ; across wider ranges α itself may vary. Isotropy allows the same coefficient in all directions but is not universal for crystals. A completely restrained length does not keep growing freely: the elastic axial model gives $\sigma=-E\alpha\Delta T$ . Yielding, buckling or compliant supports change that result. A hole grows under free isotropic heating, whereas an assembled interference fit has additional contact forces requiring a separate balance. Distinguish geometric free expansion from the actual mechanically constrained response.

How do I check units and scale for linear thermal expansion?+

α in K⁻¹ multiplied by ΔT in kelvins gives dimensionless strain. Multiplying by L₀ then gives a length. A change of 60 °C equals 60 K; do not insert absolute final temperature. For 17×10⁻⁶ K⁻¹, 1.80 m and 60 K, $\Delta L=0.001836\,\mathrm m=1.836\,\mathrm{mm}$ . Convert metres and millimetres consistently. Relative strain is 0.001020, or 0.102%. Also check the sign: positive α produces contraction for negative ΔT. ΔL is the addition and must be added to L₀ for final length. A missing factor of one thousand in conversion can otherwise overwhelm the small physical change.

When should I combine linear thermal expansion with another relation?+

For a fit, calculate hole and shaft separately. Equal heating gives diametral clearance $g=D_{h,0}(1+\alpha_h\Delta T)-D_{s,0}(1+\alpha_s\Delta T)$ , not merely hole growth. Free isotropic area and volume changes have first-order factors 2α and 3α. Axial constraint links mechanical and thermal strain through elasticity; complete restraint gives $\sigma=-E\alpha\Delta T$ . Required heating energy can be calculated with suitable heat capacities, but does not give elapsed time without power or a heat-transfer model. Each additional relation needs its own boundary conditions, particularly free motion versus constraint. A bonded bimetal also needs geometry and stiffness rather than just two added free length changes.

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Subject references and further reading

This list names subject references and further reading. A page or section number appears when it is part of the reference; these are not verbatim quotations.

[1]OpenStax, University Physics Vol. 2, §1.3: Thermal Expansion

More formulas: Heat and thermodynamics

\eta_{Carnot} = 1 - \frac{T_{kalt}}{T_{warm}}

Carnot Efficiency

P = \sigma \cdot A \cdot T^4

Stefan-Boltzmann Law

Q = m L

Latent heat during a phase change

\Delta S=\int_{i}^{f}\frac{\delta Q_{rev}}{T},\qquad T=\mathrm{const.}:\ \Delta S=\frac{Q_{rev}}{T}

Entropy change along a reversible path

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Linear thermal expansion

Free linear length change is approximately

\Delta L\simeq\alpha L_0\Delta T

when material direction, temperature interval and coefficient match. L₀ is reference length at T₀.

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Symbol

Meaning

Unit

ΔL

Change in length

α

Material’s linear expansion coefficient

K⁻¹

L₀

Initial length

ΔT

Temperature change

Derivation & background

Exam blueprint

Validity range

Derivation steps

Integrate differential relative length change and linearise for small strain.

1 $L_0$ is free reference length at $T_0$ . Define $\Delta T=T-T_0$ and $\Delta L=L-L_0$ .
2Differential coefficient is $\alpha(T)=\frac1L\frac{dL}{dT}$ for the chosen direction.
3Separate variables: $\frac{dL}{L}=\alpha(T)\,dT$ .
4Integration gives $\ln\frac L{L_0}=\int_{T_0}^T\alpha(\vartheta)\,d\vartheta$ .
5Constant differential coefficient gives $\frac L{L_0}=e^{\alpha\Delta T}$ .
6For $|\alpha\Delta T|\ll1$ , $e^{\alpha\Delta T}\simeq1+\alpha\Delta T$ , giving $\Delta L_{\mathrm{lin}}=\alpha L_0\Delta T$ .
7Measured interval mean $\overline\alpha=\frac{\Delta L}{L_0\Delta T}$ directly describes its interval and is not the differential coefficient.

How do you calculate with Linear thermal expansion?

Here is how to work through a typical Linear thermal expansion (ΔL = αL₀ΔT) task step by step:

Task

A free 2.0 m rod with $\alpha=20\cdot10^{-6}\,\mathrm{K^{-1}}$ warms by 50 K. Find its length change.

Solution path

$\Delta L=\alpha L_0\Delta T=20\cdot10^{-6}\cdot2.0\cdot50\,\mathrm m=0.0020\,\mathrm m=2.0\,\mathrm{mm}$ .

Understand and apply the equation

A heated rod tends to become longer. Whether this produces movement, a looser fit or compressive stress depends on its surroundings. First calculate free length change using

\Delta L\simeq\alpha L_0\Delta T

. Then check whether the body can actually move freely and whether its mating component also warms.

What L₀ and ΔL actually measure

First understand relative change

\begin{aligned}\varepsilon_{\mathrm{th}}&\simeq\alpha\Delta T\\\Delta L&=L_0\varepsilon_{\mathrm{th}}\end{aligned}

Relative change first, absolute length change afterwards.

Explore further: differential coefficient and interval mean

\begin{aligned}&\alpha(T)=\frac1{L(T)}\frac{dL}{dT}\\&\ln\frac{L(T)}{L_0}=\int_{T_0}^{T}\alpha(\vartheta)\,d\vartheta\end{aligned}

The definition uses current length. Free thermal length change is considered under appropriate mechanical boundary conditions.[1]

\begin{aligned}&L=L_0e^{\alpha\Delta T}\\&\Delta L\simeq\alpha L_0\Delta T\end{aligned}

The exponential form is exact only for a constant differential model coefficient, not an arbitrary tabulated interval mean.

Rearrangements and relative strain

\begin{aligned}&\varepsilon_{\mathrm{th}}=\frac{\Delta L}{L_0}\simeq\alpha\Delta T\\&\alpha\simeq\frac{\Delta L}{L_0\Delta T}\\&\Delta T\simeq\frac{\Delta L}{\alpha L_0}\end{aligned}

Rearrangements need nonzero denominators. α has unit K⁻¹; thermal strain is dimensionless.

Heat capacity relates supplied heat to a temperature change. The expansion formula takes that change as given; it determines neither required heating energy nor heating time.[3]

Example: a free rod and controlled approximation error

Uniformly heating a copper rod

Given: $L_0=1.80\,\mathrm m$ at 20 °C, final temperature 80 °C; specified constant differential model $\alpha=17\cdot10^{-6}\,\mathrm{K^{-1}}$ , unconstrained.
Find: Find ΔL and L using the linear equation and check against the exponential model.

Approach: Use the linear approximation $\Delta L=\alpha L_0\Delta T$ ; constant differential coefficient gives $L=L_0e^{\alpha\Delta T}$ as a check. Form temperature change and relative strain.
$\begin{aligned}\Delta T&=60\,\mathrm K\\{} z&=\alpha\Delta T=0.001020\end{aligned}$
Multiply by reference length.
$\begin{aligned}\Delta L_{\mathrm{lin}}&=1.80\cdot0.001020=0.001836\,\mathrm m\\{} L_{\mathrm{lin}}&=1.801836\,\mathrm m\end{aligned}$
Check the constant differential model without linearisation.
$\Delta L_{\mathrm{exp}}=1.80(e^{0.001020}-1)=0.00183693668\,\mathrm m$

Result: Linear $\Delta L=1.836\,\mathrm{mm}$ ; the exponential model gives 1.83694 mm, about 0.000937 mm more.

Holes grow during free isotropic heating

g(T)=D_{h,0}(1+\alpha_h\Delta T)-D_{s,0}(1+\alpha_s\Delta T)

g is diametral clearance here for equal temperature change of both components; g < 0 means interference.

Example: two fit components heated together

Accounting separately for hole and shaft

Given: At 20 °C: aluminium hole $D_{h,0}=40.00\,\mathrm{mm}$ and steel shaft $D_{s,0}=40.04\,\mathrm{mm}$ . Specified model values $\alpha_h=25\cdot10^{-6}\,\mathrm{K^{-1}}$ and $\alpha_s=12\cdot10^{-6}\,\mathrm{K^{-1}}$ . Both parts separately heated freely by the same ΔT; not yet assembled.
Find: Find temperature change for zero interference and clearance at $\Delta T=100\,\mathrm K$ .

Approach: Diametral clearance is $g=D_{h,0}(1+\alpha_h\Delta T)-D_{s,0}(1+\alpha_s\Delta T)$ ; both parts expand freely. Find initial difference and rate of change.
$\begin{aligned}g_0&=-0.0400\,\mathrm{mm}\\{} \frac{\Delta g}{\Delta T}&=40.00(25\times10^{-6})-40.04(12\times10^{-6})=0.00051952\,\mathrm{mm/K}\end{aligned}$
Solve $g=0$ for ΔT.
$\Delta T_{\mathrm{contact}}=\frac{0.0400}{0.00051952}\approx76.9941\,\mathrm K$
Calculate both diameters at 100 K.
$\begin{aligned}D_h&=40.1000\,\mathrm{mm}\\{} D_s&=40.088048\,\mathrm{mm}\\{} g&=0.011952\,\mathrm{mm}\end{aligned}$

Result: Diameters just match at about 97.0 °C; at 120 °C the model gives about 11.95 µm diametral clearance.

View diagram: fit clearance

When does interference become clearance?

QUANTA STUDYquanta-study.de

Original linear model for the separately and freely heated parts in the fit example. Below zero is interference; above zero is clearance. It does not describe a press-fitted assembly.[1]

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Prevented expansion creates stress

\begin{aligned}\varepsilon_{\mathrm{total}}&=\varepsilon_{\mathrm{mech}}+\alpha\Delta T\\{} \sigma&=E\varepsilon_{\mathrm{mech}}\quad\Longrightarrow\quad\sigma=-E\alpha\Delta T\quad(\varepsilon_{\mathrm{total}}=0)\end{aligned}

Complete axial constraint, linear elasticity and positive tensile stress; heating with $\alpha>0$ produces compression.[1]

A constraint loads the rod

Compressive force instead of free extension

Given: Initially unloaded rod: length 1.50 m, cross-section 80.0 mm², constant model values 200 GPa and 12.0·10⁻⁶ K⁻¹. Uniform warming by 25.0 K. Both ends are axially rigidly held. Small strain, linear elasticity and no buckling or yielding.
Find: Find free extension, actual extension, stress and axial force.

Approach: total strain is zero. Mechanical strain therefore cancels free thermal strain; tension is positive.
$\begin{aligned}\Delta L_{\mathrm{free}}&=L_0\alpha\Delta T\\\varepsilon_{\mathrm{mech}}&=-\alpha\Delta T\\\sigma&=E\varepsilon_{\mathrm{mech}}\\F&=\sigma A\end{aligned}$
First calculate free strain and extension.
$\begin{aligned}\alpha\Delta T&=12.0\cdot10^{-6}\cdot25.0=0.000300\\\Delta L_{\mathrm{free}}&=1.50\cdot0.000300\,\mathrm m=0.450\,\mathrm{mm}\end{aligned}$
Convert prevented strain into stress and force.
$\begin{aligned}\sigma&=-200\cdot10^9\cdot0.000300\,\mathrm{Pa}=-60.0\,\mathrm{MPa}\\A&=80.0\cdot10^{-6}\,\mathrm{m^2}\\F&=-60.0\cdot10^6\cdot80.0\cdot10^{-6}\,\mathrm N=-4800\,\mathrm N\end{aligned}$

Result: Freely, the rod would lengthen by 0.450 mm. Its actual length stays unchanged; the model gives compressive stress 60.0 MPa and compressive force 4.80 kN.

Area, volume and approximation limits

\begin{aligned}&\frac{\Delta A}{A_0}\simeq2\alpha\Delta T\\&\frac{\Delta V}{V_0}\simeq3\alpha\Delta T\end{aligned}

First order for free isotropic expansion in two and three dimensions respectively.[1]

Five solved exercises

Calculate · Foundation

$L_0=2.0\,\mathrm m$ , $\alpha=20\cdot10^{-6}\,\mathrm{K^{-1}}$ and $\Delta T=-50\,\mathrm K$ . Find ΔL.

Show hint

Cooling gives negative ΔT.

Learning goal: Calculate negative length change on cooling.

Show solution and calculation

Approach: free linear change $\Delta L=\alpha L_0\Delta T$ , then $L=L_0+\Delta L$ .
$\Delta L=\alpha L_0\Delta T=-0.0020\,\mathrm m=-2.0\,\mathrm{mm}$ .
Final length is 1.998 m; the model predicts contraction.

\Delta L=-2.0\,\mathrm{mm}

Rearrange · Application

A free 0.800 m specimen grows 0.384 mm over 40 K. Find mean α.

Show hint

Put lengths in matching units.

Learning goal: Find a mean expansion coefficient from measurements.

Show solution and calculation

Approach: the measured reference-length mean is $\overline\alpha=\frac{\Delta L}{L_0\Delta T}$ .
$\begin{aligned}\overline\alpha&=\frac{\Delta L}{L_0\Delta T}\\{}&=\frac{0.000384}{0.800\cdot40}\,\mathrm{K^{-1}}\\{}&=12\cdot10^{-6}\,\mathrm{K^{-1}}\text{.}\end{aligned}$ Substitution returns 0.384 mm; this is an interval mean.

\overline\alpha=12\times10^{-6}\,\mathrm{K^{-1}}

Transfer · Challenge

A fully axially constrained elastic rod has model $E=200\,\mathrm{GPa}$ , $\alpha=12\cdot10^{-6}\,\mathrm{K^{-1}}$ and $\Delta T=30\,\mathrm K$ . Find stress.

Show hint

Mechanical strain cancels free thermal strain.

Learning goal: Translate thermal strain into elastic stress under complete axial constraint.

Show solution and calculation

Approach: complete axial restraint requires $0=\frac\sigma E+\alpha\Delta T$ , hence $\sigma=-E\alpha\Delta T$ .
Thermal strain is 0.00036 and mechanical strain −0.00036.
$\sigma=-E\alpha\Delta T=-72\,\mathrm{MPa}$ , compression.
This assumes elastic behaviour without buckling.

\sigma=-72\,\mathrm{MPa}

Interpret · Application

A homogeneous isotropic metal plate has a circular hole 30.0 mm in diameter. It is heated freely and uniformly by 80 K; $\alpha=15\cdot10^{-6}\,\mathrm{K^{-1}}$ is constant and the linear approximation applies. Does the hole grow or shrink? Find the diameter increase and explain why expansion of surrounding material does not close it.

Show hint

Under free uniform expansion, all distances in the plate scale equally. Imagine temporarily filling the hole with the same material.

Learning goal: Explain free thermal expansion of a hole diameter geometrically.

Show solution and calculation

Approach: treat hole diameter as the distance between two material points on its edge.
It scales by the same positive strain factor as the plate.
The hole grows by 0.036 mm to 30.036 mm.

Check: an imagined insert of the same material would also grow before removal, matching the enlarged hole. Constraints or nonuniform temperature could change this free-expansion model.

\begin{aligned}\varepsilon&=\alpha\Delta T=(15\cdot10^{-6})(80)=0.0012\\\Delta D&=D_0\alpha\Delta T=(30.0\,\mathrm{mm})(0.0012)=0.036\,\mathrm{mm}\\D_f&=D_0+\Delta D=30.036\,\mathrm{mm}\end{aligned}

[1]

Error analysis · Foundation

A freely expanding rod with $L_0=1.0\,\mathrm m$ and $\alpha=10\cdot10^{-6}\,\mathrm{K^{-1}}$ warms from 20 °C to 50 °C. A solution adds 273.15 to $\Delta T=30\,{}^\circ\mathrm C$ and inserts 303.15 K into $\Delta L=L_0\alpha\Delta T$ . Correct the temperature difference and length increase. Why is the Celsius offset wrong here?

Show hint

The offset is added to each absolute temperature. It cancels when subtracting the two temperatures.

Learning goal: Use Celsius and kelvin temperature differences without an absolute offset.

Show solution and calculation

Approach: either convert both temperatures to kelvin and subtract or use the Celsius difference as the equal-sized kelvin difference directly.
Correct values are $\Delta T=30\,\mathrm K$ and $\Delta L=0.30\,\mathrm{mm}$ .

Check: without a temperature change, expansion must vanish; adding 273.15 to every difference would predict unphysical expansion even then.

\begin{aligned}\Delta T&=(50+273.15)\,\mathrm K-(20+273.15)\,\mathrm K=30\,\mathrm K\\\Delta L&=(1.0\,\mathrm m)(10\cdot10^{-6}\,\mathrm{K^{-1}})(30\,\mathrm K)\\&=0.00030\,\mathrm m=0.30\,\mathrm{mm}\end{aligned}

[1][2]

In brief

Sources for this explanation

Quanta · Updated October 2, 2026

Editorial responsibility: Quanta Study editorial team

Review scope

An internal human subject review is documented for an earlier release of this page. That confirmation does not apply automatically to the revised explanation and new exercises.

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

Editorial responsibility, sources and methods

Change history

October 2, 2026Linked heating energy and axial normal stress in context; preserved interval coefficients, fits, constraint stress, exercises and diagram.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.
October 1, 2026Reworked the introduction and explanations; added a third complete example, an original quantitative diagram and a short summary. Retained existing exercises, models and sources; clarified mathematical typography and model conditions.

Cite this reference page

Quanta. Linear thermal expansion. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/lineare-waermeausdehnung

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Frequently asked questions about Linear thermal expansion

What does linear thermal expansion describe physically?+

How can I check a calculation involving linear thermal expansion?+

What assumptions and limits apply to linear thermal expansion?+

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Linear thermal expansion

Variables & units

Linear thermal expansion calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Temperature and material properties

Worked example

Applications

How do you calculate with Linear thermal expansion?

What L0 and ΔL actually measure

First understand relative change

Explore further: differential coefficient and interval mean

Rearrangements and relative strain

Example: a free rod and controlled approximation error

Uniformly heating a copper rod

Holes grow during free isotropic heating

Example: two fit components heated together

Accounting separately for hole and shaft

When does interference become clearance?

Prevented expansion creates stress

A constraint loads the rod

Compressive force instead of free extension

Area, volume and approximation limits

Five solved exercises

L0=2.0 mL_0=2.0\,\mathrm mL0​=2.0m, α=20⋅10−6 K−1\alpha=20\cdot10^{-6}\,\mathrm{K^{-1}}α=20⋅10−6K−1 and ΔT=−50 K\Delta T=-50\,\mathrm KΔT=−50K. Find ΔL.

A free 0.800 m specimen grows 0.384 mm over 40 K. Find mean α.

A fully axially constrained elastic rod has model E=200 GPaE=200\,\mathrm{GPa}E=200GPa, α=12⋅10−6 K−1\alpha=12\cdot10^{-6}\,\mathrm{K^{-1}}α=12⋅10−6K−1 and ΔT=30 K\Delta T=30\,\mathrm KΔT=30K. Find stress.

In brief

Sources for this explanation

Frequently asked questions about Linear thermal expansion

Subject references and further reading

More formulas: Heat and thermodynamics

Retain Linear thermal expansion for exams

Linear thermal expansion

Variables & units

Linear thermal expansion calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Temperature and material properties

Worked example

Applications

How do you calculate with Linear thermal expansion?

What L0 and ΔL actually measure

First understand relative change

Explore further: differential coefficient and interval mean

Rearrangements and relative strain

Example: a free rod and controlled approximation error

Uniformly heating a copper rod

Holes grow during free isotropic heating

Example: two fit components heated together

Accounting separately for hole and shaft

When does interference become clearance?

Prevented expansion creates stress

A constraint loads the rod

Compressive force instead of free extension

Area, volume and approximation limits

Five solved exercises

L0=2.0 mL_0=2.0\,\mathrm mL0​=2.0m, α=20⋅10−6 K−1\alpha=20\cdot10^{-6}\,\mathrm{K^{-1}}α=20⋅10−6K−1 and ΔT=−50 K\Delta T=-50\,\mathrm KΔT=−50K. Find ΔL.

A free 0.800 m specimen grows 0.384 mm over 40 K. Find mean α.

A fully axially constrained elastic rod has model E=200 GPaE=200\,\mathrm{GPa}E=200GPa, α=12⋅10−6 K−1\alpha=12\cdot10^{-6}\,\mathrm{K^{-1}}α=12⋅10−6K−1 and ΔT=30 K\Delta T=30\,\mathrm KΔT=30K. Find stress.

In brief

Sources for this explanation

Frequently asked questions about Linear thermal expansion

Subject references and further reading

More formulas: Heat and thermodynamics

Retain Linear thermal expansion for exams

What L₀ and ΔL actually measure

$L_0=2.0\,\mathrm m$ , $\alpha=20\cdot10^{-6}\,\mathrm{K^{-1}}$ and $\Delta T=-50\,\mathrm K$ . Find ΔL.

A fully axially constrained elastic rod has model $E=200\,\mathrm{GPa}$ , $\alpha=12\cdot10^{-6}\,\mathrm{K^{-1}}$ and $\Delta T=30\,\mathrm K$ . Find stress.

What L₀ and ΔL actually measure

$L_0=2.0\,\mathrm m$ , $\alpha=20\cdot10^{-6}\,\mathrm{K^{-1}}$ and $\Delta T=-50\,\mathrm K$ . Find ΔL.

A fully axially constrained elastic rod has model $E=200\,\mathrm{GPa}$ , $\alpha=12\cdot10^{-6}\,\mathrm{K^{-1}}$ and $\Delta T=30\,\mathrm K$ . Find stress.