Question

A brass of length 1m is heated throught a temperature rise of $50^\circ$C. Find thermal stress developed in the rod
$(\alpha = 2 \times 10^{-5} \ ^\circ\text{c}^{-1}, Y = 1 \times 10^{11} \text{ N/m}^2)$

Hint:

Notice that thermal stress ($Y\alpha\Delta T$) is independent of the initial length or cross-sectional area of the rod. It only depends on the material properties ($Y, \alpha$) and the temperature change.

Answer 1

Step 1: Understanding the Concept:
When a solid is heated, it tends to expand. If this expansion is prevented by rigid supports at its ends, internal forces develop to resist the expansion. The force per unit area generated is called thermal stress.

Step 2: Key Formula or Approach:
Thermal strain is given by $\Delta L / L = \alpha \Delta T$.
According to Hooke's Law, Stress = Young's Modulus ($Y$) $\times$ Strain.
Therefore, Thermal Stress ($\sigma$) is:
\[ \sigma = Y \cdot \alpha \cdot \Delta T \]

Step 3: Detailed Explanation:
Given values:
Length $L$ = 1 m (not needed for the stress calculation, but useful to know it's a rod)
Change in temperature, $\Delta T = 50^\circ$C
Coefficient of linear expansion, $\alpha = 2 \times 10^{-5} \ ^\circ\text{C}^{-1}$
Young's Modulus, $Y = 1 \times 10^{11}$ N/m$^2$
Substitute these values into the formula:
\[ \sigma = (1 \times 10^{11} \text{ N/m}^2) \times (2 \times 10^{-5} \ ^\circ\text{C}^{-1}) \times (50^\circ\text{C}) \]
Group the terms for easier calculation:
\[ \sigma = (1 \times 2 \times 50) \times (10^{11} \times 10^{-5}) \]
\[ \sigma = 100 \times 10^{(11 - 5)} \]
\[ \sigma = 100 \times 10^6 \]
\[ \sigma = 10^2 \times 10^6 = 10^8 \text{ N/m}^2 \]

Step 4: Final Answer:
The thermal stress developed is $10^8$ N/m$^2$.