Answer
$T_{f}= 41.82 °C (107.28 °F) $
Work Step by Step
Given:
railroad tracks made of 1025 steel
$T_{0}$= 4°C (40°F)
joint space of 5.4 mm(0.210 in.) is allowed in between rails of 11.9 m (39 ft) long
Required:
highest possible temperature $T_{f}$ without introduction of thermal stress
Solution:
Using Equation 19.3a:
$\frac{l_{f}-l_{0}}{l_{0}} = α_{l} (T_{f}-T_{0})$
Substituting values for $α_{l} = 12.0 \times 10^{-6} (°C)^{-1}$ from Table 19.1:
$\frac{5.4 \times 10^{-3} m}{11.9 m} = (12.0 \times 10^{-6} (°C)^{-1})(T_{f} - 4°C)$
$T_{f}= 41.82 °C (107.28 °F) $
