PHYSICS / CHAPTER 35

Thermal expansion: geometry, measurement and restraint

Follow what can move, what the ruler measures and when heating produces stress.

8 sections8 worked examples7 checks to tryRead at your own pace

A ring warmed to release an insert and a rod heated against a rigid stop behave differently for the same reason: the material wants to change size. The useful question is which parts can actually move, including the scale used to measure them.

We will start with free expansion, then compare holes, liquid vessels, warm rulers and constrained rods. Each example keeps the reference temperature and approximation visible.

What you will learn to do

  • Use signed temperature intervals in a first-order expansion model.
  • Explain why a freely expanding hole enlarges.
  • Relate area, volume and density changes without changing mass.
  • Separate liquid expansion from vessel capacity and ruler readings from actual length.
  • Find contact thresholds before calculating thermal stress.
  • Use compatibility for composite rods and integrate a varying coefficient.
Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Thermal expansion: geometry, measurement and restraint
Learning goals:
- Use signed temperature intervals in a first-order expansion model.
- Explain why a freely expanding hole enlarges.
- Relate area, volume and density changes without changing mass.
- Separate liquid expansion from vessel capacity and ruler readings from actual length.
- Find contact thresholds before calculating thermal stress.
- Use compatibility for composite rods and integrate a varying coefficient.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

BEFORE YOU READ

Bring back one idea

Try a sentence or a sketch before opening the comparison. If this is new, use the explanation to get started.

How does a fractional change differ from an absolute change in length?

Compare your reasoning

A fractional change is ΔL/L₀ and has no units. Two rods can have the same fractional expansion but different extensions because their initial lengths differ.

Revisit the explanation →

01 / Thermal expansion: geometry, measurement and restraint

Start with a temperature change and a free length

A warm metal ring may release a tight insert, while a warm rod trapped between rigid supports can push harder on them. Heating is common to both situations; freedom to change shape is not. Before calculating a length, decide whether the object can actually reach its unconstrained size.

For a small temperature change, a freely expanding uniform rod changes length in proportion to its starting length and the temperature interval. The coefficient α describes the material in the stated range. It is not a speed: heating slowly or quickly to the same uniform final temperature gives the same free expansion in this model.

A rise from 20°C to 70°C is a 50 K interval. Adding 273.15 to both endpoints cancels in the subtraction. Absolute temperatures matter in many other thermal laws, but the expansion formula uses a difference.

At the microscopic level, increased vibration alone is not a sufficient explanation: a perfectly symmetric harmonic potential would not shift the mean spacing. The asymmetry of real interatomic interactions allows the mean spacing to change. The sign and size remain material properties; expansion on heating is common, not universal.

ΔL≃αL0ΔTL≃L0(1+αΔT)\begin{aligned}\Delta L&\simeq\alpha L_0\Delta T\\ L&\simeq L_0(1+\alpha\Delta T)\end{aligned}

This is a first-order free-expansion model about the initial temperature.

Explain this equation

Read it in words

The fractional length change is approximately the expansion coefficient times the temperature interval.

Symbols and units

L0L_0
Length at the initial uniform temperature.Metres (m).
LL
Unconstrained length at the final uniform temperature.Metres (m).
ΔL\Delta L
Final length minus initial length.Metres (m).
α\alpha
Linear thermal expansion coefficient over the stated interval.K⁻¹.
ΔT\Delta T
Final temperature minus initial temperature.Kelvin interval (K); the numerical Celsius interval is identical.

Signs and operations

A positive α gives extension when ΔT is positive and contraction when it is negative. The product αΔT is dimensionless; ≃ signals an approximation.

When it applies

Uniform temperature, unconstrained expansion, a coefficient appropriate to the material and interval, and |αΔT| much smaller than one. Do not apply a free length directly to a constrained rod.

Make it concrete

A 2.0 m rod with α = 15 × 10⁻⁶ K⁻¹ heated by 40 K grows by 1.2 mm.

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Equation (LaTeX): \begin{aligned}\Delta L&\simeq\alpha L_0\Delta T\\ L&\simeq L_0(1+\alpha\Delta T)\end{aligned}
Read it as: The fractional length change is approximately the expansion coefficient times the temperature interval.
Symbols:
L_0: Length at the initial uniform temperature.; Metres (m).
L: Unconstrained length at the final uniform temperature.; Metres (m).
\Delta L: Final length minus initial length.; Metres (m).
\alpha: Linear thermal expansion coefficient over the stated interval.; K⁻¹.
\Delta T: Final temperature minus initial temperature.; Kelvin interval (K); the numerical Celsius interval is identical.
Signs and operations: A positive α gives extension when ΔT is positive and contraction when it is negative. The product αΔT is dimensionless; ≃ signals an approximation.
Conditions: Uniform temperature, unconstrained expansion, a coefficient appropriate to the material and interval, and |αΔT| much smaller than one. Do not apply a free length directly to a constrained rod.

Help me explain the equation in words, predict how changing a quantity affects the result, and apply it to a small numerical example.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

WORKED EXAMPLE

A contraction still uses the original reference length

A freely supported 1.50 m rod has α = 16 × 10⁻⁶ K⁻¹. It cools uniformly from 75°C to 25°C. Find its length change and final length to first order.

Choose an approach

Subtract final minus initial temperature before multiplying.

  1. The interval is negative because the final state is colder.

    ΔT=25−75=−50 K\Delta T=25-75=-50\ \mathrm K
  2. Multiply the small fractional change by the reference length.

    ΔL=1.50(16×10−6)(−50)=−0.00120 m\Delta L=1.50(16\times10^{-6})(-50)=-0.00120\ \mathrm m
  3. Add the signed change to the original length.

    L=1.50−0.00120=1.49880 mL=1.50-0.00120=1.49880\ \mathrm m

What this gives usThe rod contracts by 1.20 mm. The final model length is 1.49880 m; the sign belongs to the change, not to the length.

Check the result another way

The fractional contraction is 0.0008, much smaller than one. Repeating the temperature subtraction in kelvins gives the same −50 K.

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Worked-example topic: A contraction still uses the original reference length
Problem: A freely supported 1.50 m rod has α = 16 × 10⁻⁶ K⁻¹. It cools uniformly from 75°C to 25°C. Find its length change and final length to first order.

Guide me through reconstructing the argument myself. Ask why each step is justified before moving to the next.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

STOP & THINK

A freely expanding material has a positive α. Which change makes its length smaller?

Choose an answer
Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Thermal expansion: geometry, measurement and restraint / Start with a temperature change and a free length
Problem: A freely expanding material has a positive α. Which change makes its length smaller?
Choices:
1. A positive Celsius temperature
2. A negative temperature interval
3. A slow heating rate

Help me choose a model and explain my reasoning before checking my answer.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.
Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Thermal expansion: geometry, measurement and restraint
Focus: Start with a temperature change and a free length
Learning goals:
- Use signed temperature intervals in a first-order expansion model.
- Explain why a freely expanding hole enlarges.
- Relate area, volume and density changes without changing mass.
- Separate liquid expansion from vessel capacity and ruler readings from actual length.
- Find contact thresholds before calculating thermal stress.
- Use compatibility for composite rods and integrate a varying coefficient.
Equations in this section (LaTeX):
\begin{aligned}\Delta L&\simeq\alpha L_0\Delta T\\ L&\simeq L_0(1+\alpha\Delta T)\end{aligned}
Meaning: This is a first-order free-expansion model about the initial temperature.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.
Back to chapter contents ↑

02 / Thermal expansion: geometry, measurement and restraint

A hole follows the same geometric scale factor

Imagine marking a grid on a thin plate before cutting out a circle. Under uniform, free, isotropic expansion, every separation in the grid is multiplied by the same factor. Removing the central piece does not make the remaining boundary move inward: the hole grows with the grid.

The important conditions are uniform temperature and freedom to expand. If a cold insert constrains the hole, or heating is strongly uneven, stresses and gradients enter. The simple geometric result no longer describes the whole assembly.

Area contains two lengths and volume contains three. Squaring or cubing the length scale factor produces first-order fractional changes 2αΔT and 3αΔT. Products such as (αΔT)² are discarded at the same approximation level.

ΔAA0≃2αΔTΔVV0≃3αΔT\begin{aligned}\frac{\Delta A}{A_0}&\simeq2\alpha\Delta T\\ \frac{\Delta V}{V_0}&\simeq3\alpha\Delta T\end{aligned}

These factors assume isotropic free expansion; anisotropic materials can have different coefficients along different axes.

Both boundaries enlargeHoleDashed: initial · solid: expandedFree, uniform heating · exaggerated scale
Original schematic: an unconstrained plate and its circular hole enlarge together. Dashed outlines show the initial geometry; solid outlines show an exaggerated common enlargement, not the physical size of a typical thermal strain.

WORKED EXAMPLE

Do not subtract expansion from an empty space

A free isotropic plate has a 50.0 mm circular hole. With α = 20 × 10⁻⁶ K⁻¹, it is heated uniformly by 80 K. Find the hole diameter increase and its approximate fractional area increase.

Choose an approach

Apply the linear scale factor to the hole diameter, then count two dimensions for area.

  1. The common fractional length increase is small.

    ϵ=αΔT=0.00160\epsilon=\alpha\Delta T=0.00160
  2. The diameter follows the same scale as any other separation.

    Δd=(50.0 mm)(0.00160)=0.0800 mm\Delta d=(50.0\ \mathrm{mm})(0.00160)=0.0800\ \mathrm{mm}
  3. The first-order area increase has twice the fractional length increase.

    ΔA/A0≃0.00320=0.320%\Delta A/A_0\simeq0.00320=0.320\%

What this gives usThe hole becomes wider by 0.0800 mm, and its area increases by about 0.320%. Empty space is not a material with an opposite coefficient.

Check the result another way

Directly squaring 1.00160 gives an area factor 1.00320256. The extra 0.00000256 is the second-order term omitted in the first-order answer.

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Worked-example topic: Do not subtract expansion from an empty space
Problem: A free isotropic plate has a 50.0 mm circular hole. With α = 20 × 10⁻⁶ K⁻¹, it is heated uniformly by 80 K. Find the hole diameter increase and its approximate fractional area increase.

Guide me through reconstructing the argument myself. Ask why each step is justified before moving to the next.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

STOP & THINK

A free, uniformly heated isotropic ring expands. What happens to its inner and outer diameters for positive α?

Choose an answer
Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Thermal expansion: geometry, measurement and restraint / A hole follows the same geometric scale factor
Problem: A free, uniformly heated isotropic ring expands. What happens to its inner and outer diameters for positive α?
Choices:
1. Both increase by the same fraction
2. The outer grows but the inner shrinks
3. Both increase by the same number of millimetres

Help me choose a model and explain my reasoning before checking my answer.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.
Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Thermal expansion: geometry, measurement and restraint
Focus: A hole follows the same geometric scale factor
Learning goals:
- Use signed temperature intervals in a first-order expansion model.
- Explain why a freely expanding hole enlarges.
- Relate area, volume and density changes without changing mass.
- Separate liquid expansion from vessel capacity and ruler readings from actual length.
- Find contact thresholds before calculating thermal stress.
- Use compatibility for composite rods and integrate a varying coefficient.
Equations in this section (LaTeX):
\begin{aligned}\frac{\Delta A}{A_0}&\simeq2\alpha\Delta T\\ \frac{\Delta V}{V_0}&\simeq3\alpha\Delta T\end{aligned}
Meaning: These factors assume isotropic free expansion; anisotropic materials can have different coefficients along different axes.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.
Back to chapter contents ↑

03 / Thermal expansion: geometry, measurement and restraint

Mass stays fixed while volume changes

A sealed solid sample does not gain mass merely because it warms. If its volume increases, its density decreases. Start with ρ = m/V, keep the mass fixed, and put the expanded volume in the denominator.

You can either retain the denominator from the first-order volume model or expand its reciprocal consistently to first order. The numerical difference is second order, so carrying many digits from the denominator does not improve the underlying material model.

ρ≃ρ01+3αΔT≃ρ0(1−3αΔT)\rho\simeq\frac{\rho_0}{1+3\alpha\Delta T}\simeq\rho_0(1-3\alpha\Delta T)

The second expression expands the reciprocal to first order as well. Both assume isotropy, fixed mass and free expansion.

WORKED EXAMPLE

Check density through conserved mass

A free isotropic sample initially has density 8000 kg/m³. Take α = 16 × 10⁻⁶ K⁻¹ over a 100 K rise. Estimate its final density.

Choose an approach

Use the volume coefficient 3α, not the linear coefficient α.

  1. Compute the fractional volume change.

    3αΔT=0.00483\alpha\Delta T=0.0048
  2. Keep the mass unchanged when dividing by the larger volume.

    ρ≃8000/1.0048≈7961.8 kg/m3\rho\simeq8000/1.0048\approx7961.8\ \mathrm{kg/m^3}
  3. A completely first-order expansion gives nearly the same estimate.

    ρ≃8000(1−0.0048)=7961.6 kg/m3\rho\simeq8000(1-0.0048)=7961.6\ \mathrm{kg/m^3}

What this gives usThe model predicts approximately 7.962 × 10³ kg/m³. The density decreases; the small difference between the two estimates is beyond first-order precision.

Check the result another way

Multiplying the retained-denominator density by 1.0048 times the initial volume restores the initial mass exactly within that volume model.

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Worked-example topic: Check density through conserved mass
Problem: A free isotropic sample initially has density 8000 kg/m³. Take α = 16 × 10⁻⁶ K⁻¹ over a 100 K rise. Estimate its final density.

Guide me through reconstructing the argument myself. Ask why each step is justified before moving to the next.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.
Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Thermal expansion: geometry, measurement and restraint
Focus: Mass stays fixed while volume changes
Learning goals:
- Use signed temperature intervals in a first-order expansion model.
- Explain why a freely expanding hole enlarges.
- Relate area, volume and density changes without changing mass.
- Separate liquid expansion from vessel capacity and ruler readings from actual length.
- Find contact thresholds before calculating thermal stress.
- Use compatibility for composite rods and integrate a varying coefficient.
Equations in this section (LaTeX):
\rho\simeq\frac{\rho_0}{1+3\alpha\Delta T}\simeq\rho_0(1-3\alpha\Delta T)
Meaning: The second expression expands the reciprocal to first order as well. Both assume isotropy, fixed mass and free expansion.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.
Back to chapter contents ↑

04 / Thermal expansion: geometry, measurement and restraint

A liquid level compares two expanding objects

A liquid-in-vessel experiment does not observe the liquid volume in isolation. The cavity holding it also changes volume. For a vessel filled exactly to its brim at the starting temperature, first compare the liquid’s free volume increase with the increase in vessel capacity.

Use a volumetric coefficient for the liquid. For an isotropic vessel, its cavity’s volumetric coefficient is approximately three times the solid’s linear coefficient. Subtracting the coefficients gives the apparent expansion to first order.

The formula predicts overflow only when its result is positive and the initial fill is at the brim. If the capacity grows faster, the liquid falls below the brim instead. Water near its density maximum and materials with changing coefficients require data for the actual interval; a single positive coefficient is not universal.

ΔVexcess≃V0(βℓ−3αv)ΔT\Delta V_{\mathrm{excess}}\simeq V_0(\beta_\ell-3\alpha_v)\Delta T

Compare the free liquid volume and the vessel capacity at the same final temperature.

Explain this equation

Read it in words

The excess volume is the initial filled volume times the difference in volume-expansion coefficients times the temperature rise.

Symbols and units

ΔVexcess\Delta V_{\mathrm{excess}}
Liquid volume minus vessel capacity after heating; a positive value can spill from an initially brim-full vessel.m³.
V0V_0
Common initial liquid volume and vessel capacity.m³.
βℓ\beta_\ell
Volumetric expansion coefficient of the liquid.K⁻¹.
αv\alpha_v
Linear coefficient of the isotropic vessel material.K⁻¹.
ΔT\Delta T
Common uniform temperature change.K.

Signs and operations

The minus sign accounts for the growing container capacity. The subscript ℓ identifies the liquid and v identifies the vessel; 3 converts the vessel’s linear coefficient to its first-order volume coefficient.

When it applies

Small changes, initially brim-full vessel, free isotropic vessel expansion, uniform final temperature, negligible evaporation and coefficients valid over the interval. Spilled liquid is evaluated at the final temperature to this order.

Make it concrete

For 300 mL, βℓ = 600 × 10⁻⁶ K⁻¹, αv = 10 × 10⁻⁶ K⁻¹ and ΔT = 40 K, the excess is 6.84 mL.

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Equation (LaTeX): \Delta V_{\mathrm{excess}}\simeq V_0(\beta_\ell-3\alpha_v)\Delta T
Read it as: The excess volume is the initial filled volume times the difference in volume-expansion coefficients times the temperature rise.
Symbols:
\Delta V_{\mathrm{excess}}: Liquid volume minus vessel capacity after heating; a positive value can spill from an initially brim-full vessel.; m³.
V_0: Common initial liquid volume and vessel capacity.; m³.
\beta_\ell: Volumetric expansion coefficient of the liquid.; K⁻¹.
\alpha_v: Linear coefficient of the isotropic vessel material.; K⁻¹.
\Delta T: Common uniform temperature change.; K.
Signs and operations: The minus sign accounts for the growing container capacity. The subscript ℓ identifies the liquid and v identifies the vessel; 3 converts the vessel’s linear coefficient to its first-order volume coefficient.
Conditions: Small changes, initially brim-full vessel, free isotropic vessel expansion, uniform final temperature, negligible evaporation and coefficients valid over the interval. Spilled liquid is evaluated at the final temperature to this order.

Help me explain the equation in words, predict how changing a quantity affects the result, and apply it to a small numerical example.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

WORKED EXAMPLE

Subtract the expanding capacity

An open vessel is exactly filled with 300 mL of a liquid. The liquid has β = 6.0 × 10⁻⁴ K⁻¹ and the isotropic vessel has α = 1.0 × 10⁻⁵ K⁻¹. Both warm by 40 K. Estimate overflow, neglecting evaporation.

Choose an approach

Compute two volume increases before taking their difference.

  1. The free liquid would occupy an additional volume.

    ΔVℓ=300(6.0×10−4)(40)=7.20 mL\Delta V_\ell=300(6.0\times10^{-4})(40)=7.20\ \mathrm{mL}
  2. The vessel cavity also becomes larger.

    ΔVv=300(3.0×10−5)(40)=0.360 mL\Delta V_v=300(3.0\times10^{-5})(40)=0.360\ \mathrm{mL}
  3. Only the excess over the new capacity leaves the brim.

    Vspill≃7.20−0.360=6.84 mLV_{\mathrm{spill}}\simeq7.20-0.360=6.84\ \mathrm{mL}

What this gives usApproximately 6.84 mL spills. Treating the vessel as unchanging would overestimate it.

Check the result another way

The final free liquid volume is 307.20 mL, while the vessel can hold 300.36 mL. Their difference reproduces the spill.

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Worked-example topic: Subtract the expanding capacity
Problem: An open vessel is exactly filled with 300 mL of a liquid. The liquid has β = 6.0 × 10⁻⁴ K⁻¹ and the isotropic vessel has α = 1.0 × 10⁻⁵ K⁻¹. Both warm by 40 K. Estimate overflow, neglecting evaporation.

Guide me through reconstructing the argument myself. Ask why each step is justified before moving to the next.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.
Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Thermal expansion: geometry, measurement and restraint
Focus: A liquid level compares two expanding objects
Learning goals:
- Use signed temperature intervals in a first-order expansion model.
- Explain why a freely expanding hole enlarges.
- Relate area, volume and density changes without changing mass.
- Separate liquid expansion from vessel capacity and ruler readings from actual length.
- Find contact thresholds before calculating thermal stress.
- Use compatibility for composite rods and integrate a varying coefficient.
Equations in this section (LaTeX):
\Delta V_{\mathrm{excess}}\simeq V_0(\beta_\ell-3\alpha_v)\Delta T
Meaning: Compare the free liquid volume and the vessel capacity at the same final temperature.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.
Back to chapter contents ↑

05 / Thermal expansion: geometry, measurement and restraint

The measuring scale can expand too

A ruler measures by counting its own intervals. If those intervals become longer, fewer of them fit along an unchanged object. A smaller reading can therefore mean that the ruler has expanded, not that the object has shrunk.

Separate the actual length from the displayed reading. If one nominal millimetre becomes (1 + αΔT) actual millimetres, multiply the number of displayed millimetres by that factor. If the specimen also expands, compare its fractional expansion with the ruler’s instead of treating either as an absolute standard.

WORKED EXAMPLE

Recover length from enlarged ruler divisions

A ruler calibrated at 20°C has α = 20 × 10⁻⁶ K⁻¹. At 70°C it reads an unchanged specimen as 500.0 mm. What actual length does that reading represent in the reference unit?

Choose an approach

Find the actual length represented by one warm ruler division.

  1. The ruler interval has increased by a factor of 1.001.

    f=1+(20×10−6)(50)=1.001f=1+(20\times10^{-6})(50)=1.001
  2. The specimen spans 500.0 of these larger intervals.

    L=(500.0 mm)(1.001)=500.5 mmL=(500.0\ \mathrm{mm})(1.001)=500.5\ \mathrm{mm}
  3. Dividing an actual length by the enlarged interval returns the reading.

    500.5/1.001=500.0500.5/1.001=500.0

What this gives usThe actual length is 500.5 mm in the reference unit. A stretched scale under-reads an unchanged specimen.

Check the result another way

A 500.0 mm unchanged specimen would instead read about 499.5 mm. The direction is consistent in both versions of the question.

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Worked-example topic: Recover length from enlarged ruler divisions
Problem: A ruler calibrated at 20°C has α = 20 × 10⁻⁶ K⁻¹. At 70°C it reads an unchanged specimen as 500.0 mm. What actual length does that reading represent in the reference unit?

Guide me through reconstructing the argument myself. Ask why each step is justified before moving to the next.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.
Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Thermal expansion: geometry, measurement and restraint
Focus: The measuring scale can expand too
Learning goals:
- Use signed temperature intervals in a first-order expansion model.
- Explain why a freely expanding hole enlarges.
- Relate area, volume and density changes without changing mass.
- Separate liquid expansion from vessel capacity and ruler readings from actual length.
- Find contact thresholds before calculating thermal stress.
- Use compatibility for composite rods and integrate a varying coefficient.
Equations in this section (LaTeX):
Build the physical reasoning before choosing an equation.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.
Back to chapter contents ↑

06 / Thermal expansion: geometry, measurement and restraint

Use free expansion until a constraint becomes active

A gap can absorb free expansion before a support starts exerting a thermal force. Compare αL₀ΔT with the available gap first. Below the contact temperature there is no axial thermal stress in the ideal freely sliding arrangement.

After contact, thermal strain still tends to lengthen the rod, but an elastic compressive strain offsets part of it. With tensile stress defined positive, total axial strain is αΔT + σ/Y. A completely fixed length has zero total strain; a rod allowed to fill a gap g has total strain g/L₀.

This superposition is a small-strain elastic model. It assumes straight axial deformation, rigid supports and no slip after contact. Long slender rods may buckle; materials may yield; supports may move. Those are changes to the mechanical model, not corrections to the sign of α.

ϵtotal≃αΔT+σYσ≃Y(gL0−αΔT)\begin{aligned}\epsilon_{\mathrm{total}}&\simeq\alpha\Delta T+\frac{\sigma}{Y}\\ \sigma&\simeq Y\left(\frac{g}{L_0}-\alpha\Delta T\right)\end{aligned}

The second relation applies only after the free expansion has closed the axial gap.

Explain this equation

Read it in words

The permitted total strain is the sum of the thermal strain and the elastic strain needed to satisfy the supports.

Symbols and units

ϵtotal\epsilon_{\mathrm{total}}
Actual length change divided by original length.Dimensionless.
αΔT\alpha\Delta T
Unconstrained thermal strain.Dimensionless.
σ\sigma
Axial stress, positive in tension and negative in compression.Pa.
YY
Young’s modulus in the assumed linear elastic range.Pa.
gg
Initial free axial gap that can be filled before contact.m.
L0L_0
Initial rod length.m.

Signs and operations

Compression has a negative σ under this convention. The symbol g here is a geometric gap, not gravitational acceleration. Before contact use σ = 0 rather than the second equation.

When it applies

Small uniform thermal and elastic strains, rigid end separation, one freely sliding end until contact, no frictional restraint, no buckling or yielding and uniform temperature.

Make it concrete

For L₀ = 1 m, g = 0.36 mm, α = 12 × 10⁻⁶ K⁻¹, Y = 200 GPa and ΔT = 80 K, contact began at 30 K and the final stress is −120 MPa.

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Equation (LaTeX): \begin{aligned}\epsilon_{\mathrm{total}}&\simeq\alpha\Delta T+\frac{\sigma}{Y}\\ \sigma&\simeq Y\left(\frac{g}{L_0}-\alpha\Delta T\right)\end{aligned}
Read it as: The permitted total strain is the sum of the thermal strain and the elastic strain needed to satisfy the supports.
Symbols:
\epsilon_{\mathrm{total}}: Actual length change divided by original length.; Dimensionless.
\alpha\Delta T: Unconstrained thermal strain.; Dimensionless.
\sigma: Axial stress, positive in tension and negative in compression.; Pa.
Y: Young’s modulus in the assumed linear elastic range.; Pa.
g: Initial free axial gap that can be filled before contact.; m.
L_0: Initial rod length.; m.
Signs and operations: Compression has a negative σ under this convention. The symbol g here is a geometric gap, not gravitational acceleration. Before contact use σ = 0 rather than the second equation.
Conditions: Small uniform thermal and elastic strains, rigid end separation, one freely sliding end until contact, no frictional restraint, no buckling or yielding and uniform temperature.

Help me explain the equation in words, predict how changing a quantity affects the result, and apply it to a small numerical example.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

WORKED EXAMPLE

A gap delays the onset of compression

A straight 1.00 m rod can lengthen freely into a 0.36 mm gap before meeting a rigid stop. Use α = 12 × 10⁻⁶ K⁻¹ and Y = 200 GPa. Find the contact temperature rise and the stress after an 80 K rise. Exclude buckling and yielding.

Choose an approach

Solve the free-expansion threshold, then balance thermal and elastic strain after contact.

  1. Closing the gap requires a specific temperature rise.

    ΔTc=0.000361.00(12×10−6)=30 K\Delta T_c=\frac{0.00036}{1.00(12\times10^{-6})}=30\ \mathrm K
  2. The permitted strain is less than the final free thermal strain.

    ϵelastic=0.00036−0.00096=−0.00060\epsilon_{\mathrm{elastic}}=0.00036-0.00096=-0.00060
  3. Convert the elastic strain to signed stress.

    σ=(200×109)(−0.00060)=−120 MPa\sigma=(200\times10^9)(-0.00060)=-120\ \mathrm{MPa}

What this gives usContact occurs after 30 K; the final stress is 120 MPa in compression. Counting all 80 K as fully restrained would miss the initial free motion.

Check the result another way

Only the extra 50 K after contact produces compression: Yα(50 K) = 120 MPa. Both routes give zero stress exactly at contact.

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Worked-example topic: A gap delays the onset of compression
Problem: A straight 1.00 m rod can lengthen freely into a 0.36 mm gap before meeting a rigid stop. Use α = 12 × 10⁻⁶ K⁻¹ and Y = 200 GPa. Find the contact temperature rise and the stress after an 80 K rise. Exclude buckling and yielding.

Guide me through reconstructing the argument myself. Ask why each step is justified before moving to the next.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

STOP & THINK

The same rod is heated only 20 K, less than its 30 K contact threshold. What is the axial thermal stress in the stated model?

Choose an answer
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Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Thermal expansion: geometry, measurement and restraint / Use free expansion until a constraint becomes active
Problem: The same rod is heated only 20 K, less than its 30 K contact threshold. What is the axial thermal stress in the stated model?
Choices:
1. Compression Yα(20 K)
2. Tension from the remaining gap
3. Zero

Help me choose a model and explain my reasoning before checking my answer.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.
Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Thermal expansion: geometry, measurement and restraint
Focus: Use free expansion until a constraint becomes active
Learning goals:
- Use signed temperature intervals in a first-order expansion model.
- Explain why a freely expanding hole enlarges.
- Relate area, volume and density changes without changing mass.
- Separate liquid expansion from vessel capacity and ruler readings from actual length.
- Find contact thresholds before calculating thermal stress.
- Use compatibility for composite rods and integrate a varying coefficient.
Equations in this section (LaTeX):
\begin{aligned}\epsilon_{\mathrm{total}}&\simeq\alpha\Delta T+\frac{\sigma}{Y}\\ \sigma&\simeq Y\left(\frac{g}{L_0}-\alpha\Delta T\right)\end{aligned}
Meaning: The second relation applies only after the free expansion has closed the axial gap.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.
Back to chapter contents ↑

07 / Thermal expansion: geometry, measurement and restraint

Compatibility links unequal rods in series

Two rods in series between rigid supports share an axial force, not necessarily a stress. Equal cross-sectional areas make their stresses equal; unequal areas require σᵢ = F/Aᵢ. Their individual length changes need not vanish even when the combined length is fixed.

Add the unconstrained thermal extensions of both rods. The common compressive force must produce a total elastic shortening of exactly that amount. Each rod contributes compliance Lᵢ/(YᵢAᵢ). This is why simply averaging expansion coefficients or moduli usually loses the constraint information.

Fcompression=∑iαiLiΔT∑iLi/(YiAi)F_{\mathrm{compression}}=\frac{\sum_i\alpha_i L_i\Delta T}{\sum_i L_i/(Y_iA_i)}

Use the positive magnitude of compression here; the signed tensile force is its negative. The assembly is initially stress-free and straight, with a common temperature rise and rigid end separation.

WORKED EXAMPLE

Fixed total length does not fix each rod’s length

Two equal-area rods in series are initially stress-free between rigid supports. Rod A: L = 1.0 m, α = 10 × 10⁻⁶ K⁻¹, Y = 100 GPa. Rod B: L = 2.0 m, α = 20 × 10⁻⁶ K⁻¹, Y = 200 GPa. Both warm by 30 K. Find the common stress magnitude and each length change; exclude buckling and yielding.

Choose an approach

Use equal stress for equal areas, but sum the individual strains times their lengths.

  1. The total free extension would be 1.5 mm.

    ΔLfree=0.00030+0.00120=0.00150 m\Delta L_{\mathrm{free}}=0.00030+0.00120=0.00150\ \mathrm m
  2. Elastic shortening per unit stress adds in series.

    ∣σ∣=0.001501/(100×109)+2/(200×109)=75 MPa\begin{aligned}|\sigma|&=\frac{0.00150}{1/(100\times10^9)+2/(200\times10^9)}\\ &=75\ \mathrm{MPa}\end{aligned}
  3. Rod A shortens overall while rod B lengthens by the same amount.

    ΔLA=0.00030−0.00075=−0.45 mmΔLB=0.00120−0.00075=+0.45 mm\begin{aligned}\Delta L_A&=0.00030-0.00075=-0.45\ \mathrm{mm}\\ \Delta L_B&=0.00120-0.00075=+0.45\ \mathrm{mm}\end{aligned}

What this gives usThe stress is 75 MPa in compression. The interface moves by 0.45 mm even though the end-to-end distance remains fixed.

Check the result another way

The two changes sum to zero. Equal stress does not imply equal elastic strain because the Young’s moduli differ.

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Worked-example topic: Fixed total length does not fix each rod’s length
Problem: Two equal-area rods in series are initially stress-free between rigid supports. Rod A: L = 1.0 m, α = 10 × 10⁻⁶ K⁻¹, Y = 100 GPa. Rod B: L = 2.0 m, α = 20 × 10⁻⁶ K⁻¹, Y = 200 GPa. Both warm by 30 K. Find the common stress magnitude and each length change; exclude buckling and yielding.

Guide me through reconstructing the argument myself. Ask why each step is justified before moving to the next.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.
Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Thermal expansion: geometry, measurement and restraint
Focus: Compatibility links unequal rods in series
Learning goals:
- Use signed temperature intervals in a first-order expansion model.
- Explain why a freely expanding hole enlarges.
- Relate area, volume and density changes without changing mass.
- Separate liquid expansion from vessel capacity and ruler readings from actual length.
- Find contact thresholds before calculating thermal stress.
- Use compatibility for composite rods and integrate a varying coefficient.
Equations in this section (LaTeX):
F_{\mathrm{compression}}=\frac{\sum_i\alpha_i L_i\Delta T}{\sum_i L_i/(Y_iA_i)}
Meaning: Use the positive magnitude of compression here; the signed tensile force is its negative. The assembly is initially stress-free and straight, with a common temperature rise and rigid end separation.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.
Back to chapter contents ↑

08 / Thermal expansion: geometry, measurement and restraint

A coefficient belongs to a temperature range

If α changes noticeably during heating, a single endpoint value does not represent the whole interval. For small total expansion, add the contributions α(T)dT over the interval using the initial length as the reference.

More precisely, the instantaneous definition is α = (1/L)dL/dT. Integrating this definition gives ln(L/L₀) = ∫α(T)dT. The usual first-order result follows by expanding the exponential when that integral is small. A coefficient fitted using the initial length over a finite interval is an average coefficient; do not confuse its definition with an exact instantaneous constant.

A bonded bimetal strip is another reminder to state the constraint. The higher-expansion layer tends to become longer on heating, so the strip bends with that layer on the outside of the curve in the usual thin-strip model. A numerical curvature additionally needs thicknesses and elastic properties; the coefficient difference alone is not enough.

ΔLL0≃∫T0T1α(T) dT\frac{\Delta L}{L_0}\simeq\int_{T_0}^{T_1}\alpha(T)\,dT

Use this first-order integral when the total fractional expansion is small, with a freely expanding, uniformly heated specimen.

WORKED EXAMPLE

Use the interval average, not the final coefficient

For a free 2.0 m rod, α(θ) = (8.0 + 0.040θ) × 10⁻⁶ K⁻¹, where θ is the numerical temperature rise in kelvins from the initial state. Estimate the extension after a 50 K rise to first order.

Choose an approach

Integrate the coefficient across the interval or use its mean, since this α is linear in temperature.

  1. The endpoint coefficients are 8 and 10 in units of 10⁻⁶ K⁻¹.

    αˉ=(8+10)10−6/2=9.0×10−6 K−1\bar\alpha=(8+10)10^{-6}/2=9.0\times10^{-6}\ \mathrm{K^{-1}}
  2. The integrated fractional increase is small.

    ∫050α dθ=(9.0×10−6)(50)=0.00045\int_0^{50}\alpha\,d\theta=(9.0\times10^{-6})(50)=0.00045
  3. Multiply by the initial length only after finding the fractional change.

    ΔL≃2.0(0.00045)=0.00090 m\Delta L\simeq2.0(0.00045)=0.00090\ \mathrm m

What this gives usThe first-order extension is 0.90 mm. Using the final coefficient throughout would give 1.00 mm and overestimate the expansion.

Check the result another way

Direct integration gives [8θ + 0.020θ²] × 10⁻⁶ from 0 to 50, also 0.00045. The exact instantaneous-coefficient model would give L₀(exp(0.00045) − 1), differing only beyond first order.

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Worked-example topic: Use the interval average, not the final coefficient
Problem: For a free 2.0 m rod, α(θ) = (8.0 + 0.040θ) × 10⁻⁶ K⁻¹, where θ is the numerical temperature rise in kelvins from the initial state. Estimate the extension after a 50 K rise to first order.

Guide me through reconstructing the argument myself. Ask why each step is justified before moving to the next.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.
Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Thermal expansion: geometry, measurement and restraint
Focus: A coefficient belongs to a temperature range
Learning goals:
- Use signed temperature intervals in a first-order expansion model.
- Explain why a freely expanding hole enlarges.
- Relate area, volume and density changes without changing mass.
- Separate liquid expansion from vessel capacity and ruler readings from actual length.
- Find contact thresholds before calculating thermal stress.
- Use compatibility for composite rods and integrate a varying coefficient.
Equations in this section (LaTeX):
\frac{\Delta L}{L_0}\simeq\int_{T_0}^{T_1}\alpha(T)\,dT
Meaning: Use this first-order integral when the total fractional expansion is small, with a freely expanding, uniformly heated specimen.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.
Back to chapter contents ↑

BRING THE IDEAS TOGETHER

What should stay with you

  • Identify the reference state and which motion is permitted before using an expansion formula.
  • Uniform free isotropic expansion scales holes as well as outer dimensions.
  • Area and volume factors count dimensions; density changes inversely with volume at fixed mass.
  • Measurements compare a specimen with a scale or a liquid with its container.
  • Thermal strain and elastic strain must satisfy the actual supports.
  • Check small strain, material coefficients, uniform temperature and elastic stability.

RETRIEVE, THEN EXPLAIN

Try it without the examples

Work these out before checking. If an answer surprises you, identify which assumption or step needs another look. A correct choice is most useful when you can explain why the alternatives fail.

SELF CHECK 1

A free isotropic sheet has area 0.50 m² and α = 10 × 10⁻⁶ K⁻¹. It warms by 40 K. What is its first-order area increase in cm²?

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Thermal expansion: geometry, measurement and restraint
Problem: A free isotropic sheet has area 0.50 m² and α = 10 × 10⁻⁶ K⁻¹. It warms by 40 K. What is its first-order area increase in cm²?
Requested answer unit: cm²

Help me choose a model and explain my reasoning before checking my answer.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

SELF CHECK 2

A straight rod is fully restrained from its initially stress-free state. Y = 100 GPa, α = 15 × 10⁻⁶ K⁻¹ and ΔT = 20 K. What is the compression magnitude in MPa? Exclude buckling and yielding.

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Thermal expansion: geometry, measurement and restraint
Problem: A straight rod is fully restrained from its initially stress-free state. Y = 100 GPa, α = 15 × 10⁻⁶ K⁻¹ and ΔT = 20 K. What is the compression magnitude in MPa? Exclude buckling and yielding.
Requested answer unit: MPa

Help me choose a model and explain my reasoning before checking my answer.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

SELF CHECK 3

A specimen and ruler have the same α and the same uniform temperature rise. What happens to their length reading in the common free-expansion model?

Choose an answer
Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Thermal expansion: geometry, measurement and restraint
Problem: A specimen and ruler have the same α and the same uniform temperature rise. What happens to their length reading in the common free-expansion model?
Choices:
1. It increases
2. It stays unchanged
3. It decreases

Help me choose a model and explain my reasoning before checking my answer.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

SELF CHECK 4

An initially brim-full vessel has a larger volume expansion coefficient than its liquid. For a small common temperature rise, what does the model predict?

Choose an answer
Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Thermal expansion: geometry, measurement and restraint
Problem: An initially brim-full vessel has a larger volume expansion coefficient than its liquid. For a small common temperature rise, what does the model predict?
Choices:
1. Overflow equal to the liquid expansion
2. The liquid falls below the brim
3. The liquid mass increases

Help me choose a model and explain my reasoning before checking my answer.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

APPLY THE CHAPTER

Take the ideas into a problem

Try a complete problem next. Commit to a method before revealing a hint, then compare your reasoning with the worked solution.

The ruler changes along with the specimenIntermediate · Free sample · Hints and guided reasoningCount the three directional changesAdvanced · Hints and guided reasoningCompare liquid growth with new capacityIntermediate · Hints and guided reasoningOnly the heating after contact creates stressAdvanced · Hints and guided reasoningSeries rods share force, not stressAdvanced · Hints and guided reasoningAdd expansion over the temperature intervalAdvanced · Hints and guided reasoningA long rod warms upIntermediate · Hints and guided reasoning

COME BACK TO THE IDEA

What can you explain later?

Try these from memory at your next study session, then again a few days later. Say your reasoning aloud or sketch it before looking. Adjust the gap to what you can explain.

Next study session

A warm specimen gives the same reading on a warm ruler as before heating. Must its actual length be unchanged?

Compare your reasoning

No. If the specimen and the ruler expand by the same fractional amount, the specimen spans the same number of enlarged divisions. The reading is a ratio of two physical lengths.

Revisit the explanation →

A few days later: change the situation

A rod has a small free gap before a rigid stop. Why is YαΔT not the compression magnitude for every temperature rise?

Compare your reasoning

Initially the rod expands freely and has no axial stress. Only after the gap closes does additional heating require elastic compression. First find the contact threshold and then impose the support constraint.

Revisit the explanation →

Compare the physical reason, the assumptions and the conclusion. Your wording can differ. If one part is missing, revisit that section and try explaining it again after a break.

Back to chapter contents ↑