deformation Young modulus and thermal expansion

Deformation is a change in the shape or size of a body caused by external forces or by heating. For a bar of initial length $L_0$ and cross-sectional area $A$ we define the stress $\sigma = F/A$ and the strain $\varepsilon = \Delta L/L_0$; in the elastic region they are linked by Hooke's law $\sigma = E\varepsilon$, where $E$ is the Young's modulus (measured in pascals). Young's modulus is a property of the material: the larger it is, the harder it is to stretch a bar by the same fraction of its length. On heating by $\Delta T$ the length changes as $\Delta L = \alpha L_0 \Delta T$, where $\alpha$ is the coefficient of linear thermal expansion (in $1/\text{K}$), and the volume grows as $\Delta V \approx 3\alpha V_0 \Delta T$.

If a heated bar is prevented from expanding, a thermal stress $\sigma = E\alpha\Delta T$ appears — the same relation $\sigma = E\varepsilon$, with $\varepsilon$ fixed by thermal expansion instead of an external force. Beyond the elastic limit Hooke's law fails, and a permanent (plastic) deformation remains after the load is removed.

A steel wire of length $L_0 = 2.00$ m and cross-sectional area $A = 1.0\ \text{mm}^2 = 1.0\cdot 10^{-6}\ \text{m}^2$ is stretched by a force $F = 200$ N. The Young's modulus of steel is $E = 2.0\cdot 10^{11}$ Pa. Find the stress $\sigma$, the strain $\varepsilon$, and the elongation $\Delta L$.

An aluminium rod of length $L_0 = 5.00$ m at $20\ ^\circ\text{C}$ is heated to $120\ ^\circ\text{C}$. The coefficient of linear expansion of aluminium is $\alpha = 23\cdot 10^{-6}\ \text{K}^{-1}$. Find the change in length and the final length. By how much does the volume change if the initial volume is $V_0 = 2.0\cdot 10^{-3}\ \text{m}^3$?

A steel rail of length $L_0 = 25.0$ m is laid with a gap $d = 6.0$ mm. $\alpha = 12\cdot 10^{-6}\ \text{K}^{-1}$, $E = 200\ \text{GPa}$, elastic limit $250$ MPa. On a hot day the rail is heated by $\Delta T = 30$ K. (a) How much would it lengthen if it were free? (b) What stress would appear in the rail after the gap closes? (c) Is this heating dangerous?

Self-study plan with an interactive simulation (no heating and no hanging of real weights).

1. Open an elastic-deformation simulation (for example a Hooke's-law style one), attach the virtual wire, and set its length $L_0$ and area $A$.
2. Increase the force $F$ in small steps, recording $F$ and the elongation $\Delta L$ in a table. Compute $\sigma = F/A$ and $\varepsilon = \Delta L/L_0$.
3. Plot $\sigma$ against $\varepsilon$. The linear part has slope $E$; find $E$ from the slope and compare it with the tabulated value for the wire's metal. Mark the point where linearity breaks down — the elastic limit.
4. In the second part, switch on the thermal-expansion model: set $L_0$, $\alpha$, and vary the temperature. Record $\Delta T$ and $\Delta L$, plot $\Delta L$ against $\Delta T$, and check that the slope equals $\alpha L_0$.
5. Finally, switch on rigid clamping of both ends in the simulation and confirm that heating produces a stress $\sigma = E\alpha\Delta T$ while the length stays fixed.

Safety. Heating metal rods, open flames, hot water, and hanging heavy weights at home are hazardous: do all observations in the simulation. Real experiments belong in a laboratory under a teacher's supervision.

To show two key facts: (1) the slope of the linear part of the $\sigma(\varepsilon)$ graph equals Young's modulus $E$, and linearity is lost beyond the elastic limit; (2) on heating the length grows by $\Delta L = \alpha L_0 \Delta T$, while with rigid clamping a stress $E\alpha\Delta T$ appears instead of an elongation.

Two-panel sketch.

• Panel A — graph. Horizontal axis $\varepsilon$, vertical axis $\sigma$. A straight line from the origin with slope $E$; point P marks the elastic limit; beyond it the curve flattens (plastic region). Label nearby: $E = \sigma/\varepsilon$, $[E] = \text{Pa}$.
• Panel B — sketch of a bar. Above, a bar of length $L_0$ at temperature $T_0$; below, the same bar at $T_0 + \Delta T$ with the shaded elongation $\Delta L = \alpha L_0 \Delta T$. On the right, the "rigid clamping" case: the same supports, inward arrows, and the label $\sigma = E\alpha\Delta T$.

A static rope of length $L_0 = 30\ \text{m}$ and diameter $10\ \text{mm}$ has a cross-sectional area $A = \pi r^2 = \pi (0.005\ \text{m})^2 \approx 7.9\times10^{-5}\ \text{m}^2$. A climber of mass $m = 80\ \text{kg}$ hangs motionless on it. The Young's modulus of nylon is $E \approx 3\times10^{9}\ \text{Pa}$. The rope tension is $F = mg \approx 80 \cdot 9.8 \approx 784\ \text{N}$, and the elongation is $\Delta L = \frac{F L_0}{E A} \approx \frac{784 \cdot 30}{3\times10^{9} \cdot 7.9\times10^{-5}} \approx 0.10\ \text{m} \approx 10\ \text{cm}$.

The scenario tests Hooke's law in the form $\sigma = E\varepsilon$, where $\sigma = F/A$ is the mechanical stress and $\varepsilon = \Delta L/L_0$ is the relative strain. Here $\sigma \approx 9.9\times10^{6}\ \text{Pa}$ and $\varepsilon \approx 3.3\times10^{-3}$; their ratio is exactly the Young's modulus $E = \sigma/\varepsilon = 3\times10^{9}\ \text{Pa}$. The strain of about $0.3\%$ stays elastic: when the load is removed the rope returns to its original length.

A rail of length $L_0 = 25\ \text{m}$ is laid at $15\ ^\circ\text{C}$ and on a sunny summer day heats up to $55\ ^\circ\text{C}$, that is by $\Delta T = 40\ \text{K}$. The linear expansion coefficient of steel is $\alpha = 1.2\times10^{-5}\ \text{K}^{-1}$ and its Young's modulus is $E = 2.1\times10^{11}\ \text{Pa}$. The rail elongation is $\Delta L = \alpha L_0 \Delta T = 1.2\times10^{-5} \cdot 25 \cdot 40 = 0.012\ \text{m} = 12\ \text{mm}$.

The scenario uses the linear thermal expansion law $\Delta L = \alpha L_0 \Delta T$ and its consequence for the case when expansion is prevented: $\sigma = E\alpha\Delta T$. With $\alpha = 1.2\times10^{-5}\ \text{K}^{-1}$, $\Delta T = 40\ \text{K}$ and $E = 2.1\times10^{11}\ \text{Pa}$ the compressive stress reaches $1.0\times10^{8}\ \text{Pa} = 100\ \text{MPa}$ — this is exactly why gaps of about $12\ \text{mm}$ per $25\ \text{m}$ are left between rail sections.

An aluminium lid of diameter $d_0 = 6.0\ \text{cm}$ is screwed tightly onto a glass jar. The lid is doused with hot water and its temperature rises by $\Delta T = 60\ \text{K}$. The linear expansion coefficients are: aluminium $\alpha_{Al} = 2.3\times10^{-5}\ \text{K}^{-1}$, glass $\alpha_{glass} = 0.9\times10^{-5}\ \text{K}^{-1}$. The lid diameter grows by $\Delta d = \alpha_{Al} d_0 \Delta T \approx 0.083\ \text{mm}$, while the jar neck diameter grows only by $\approx 0.032\ \text{mm}$.

The scenario applies the same linear expansion formula $\Delta L = \alpha L_0 \Delta T$ to two different materials and shows that the difference of the coefficients $\alpha$ is what matters: a diameter difference of order $0.05\ \text{mm}$ relieves the tension in the thread, friction drops — and the lid unscrews. The same physics explains why glassware must not be cooled abruptly: different $\alpha$ values together with the Young's modulus create internal stresses $\sigma = E\alpha\Delta T$ that crack the glass.