Question

At room temperature of 25°C, a gap of 1 mm exists between the ends of the rods 1 and 2 as shown. Given the cross section area \( A \) of the rods is \( 1500 \, \text{mm}^2 \), Young's modulus \( E = 75 \, \text{GPa} \), and the coefficient of thermal expansion \( \alpha = 23 \times 10^{-6} \, \text{°C}^{-1} \). When the temperature has reached 150°C, the magnitude of normal stress in each of the rods (in MPa, rounded off to two decimal places) is \(\underline{\hspace{2cm}}\).
\includegraphics[width=0.5\linewidth]{image19.png}

Hint:

To find the thermal stress, calculate the strain using \( \epsilon = \alpha \Delta T \) and then use \( \sigma = E \times \epsilon \) to find the stress.

Answer 1

Correct Answer: 90.6

The thermal strain is given by:
\[ \epsilon = \alpha \Delta T \] where \( \Delta T = 150 - 25 = 125^\circ C \).
Thus, the strain is:
\[ \epsilon = 23 \times 10^{-6} \times 125 = 2.875 \times 10^{-3}. \] The stress \( \sigma \) is given by:
\[ \sigma = E \times \epsilon = 75 \times 10^3 \times 2.875 \times 10^{-3} = 90.625 \, \text{MPa}. \] Thus, the magnitude of normal stress is \( \boxed{90.63} \, \text{MPa}. \)