## Train Tracks

Consider a segment of train tracks is secured via fasteners to the ground/wooden boards every 2 m. The ambient temperature is 40°C during the summer, and drops to 0°C during the winter. If the axial stress on the rails is 100 MPa due to temperature fluctuations and the Young’s Modulus is 210 GPa, what is the rail’s coefficient of thermal expansion?

Hint
The thermal expansion coefficient is the ratio of engineering strain to the temperature change.
$$\alpha=\frac{\varepsilon }{\Delta T}$$$where $$\alpha$$ is the thermal expansion coefficient, $$\varepsilon$$ is the engineering strain, and $$\Delta T$$ is the temperature change. Hint 2 The axial stress due to temperature deformation: $$\sigma=E \times \varepsilon$$$
where $$E$$ is the Young's Modulus and $$\varepsilon$$ is the engineering strain.
First, let’s solve for engineering strain. The axial stress due to temperature deformation:
$$\sigma=E \times \varepsilon$$$where $$E$$ is the Young's Modulus and $$\varepsilon$$ is the engineering strain. $$\varepsilon=\frac{\sigma}{E}=\frac{100MPa}{210,000MPa}=0.000476$$$
The thermal expansion coefficient is the ratio of engineering strain to the temperature change.
$$\alpha=\frac{\varepsilon }{\Delta T}$$$where $$\alpha$$ is the thermal expansion coefficient, $$\varepsilon$$ is the engineering strain, and $$\Delta T$$ is the temperature change. Thus, $$\alpha=\frac{0.000476}{(40^{\circ}C-0^{\circ}C)}=\frac{0.000476}{40^{\circ}C}=1.2\cdot 10^{-5}/^{\circ}C$$$
$$1.2\cdot 10^{-5}/^{\circ}C$$\$