Question

A bar of length \(L = 1 \, m\) is fixed at one end. Before heating, its free end has a gap of \(\delta = 0.1 \, mm\) from a rigid wall. The bar is then heated, resulting in a uniform temperature rise of \(10^\circ C\). The coefficient of linear thermal expansion of the material is \(\alpha = 20 \times 10^{-6}/^\circ C\), and the Young’s modulus of elasticity is \(E = 100 \, GPa\).

Assume that the material properties do not change with temperature.
The magnitude of the resulting axial stress on the bar is ............ MPa (in integer).

Hint:

Thermal stress arises only when expansion is restrained. Always check: - Free expansion \( \Delta L_{free} \). - Gap available. - Blocked expansion = \(\Delta L_{free} - \delta\).

Answer 1

Correct Answer: 10

Step 1: Free thermal expansion of bar.
\[ \Delta L_{free} = \alpha \, L \, \Delta T \] \[ = (20 \times 10^{-6})(1000 \, mm)(10) = 0.2 \, mm \]

Step 2: Compare expansion with gap.
Gap = 0.1 mm. Free expansion = 0.2 mm. Excess expansion blocked by wall = \(0.2 - 0.1 = 0.1 \, mm\).

Step 3: Mechanical strain induced.
\[ \varepsilon = \frac{\Delta L_{blocked}}{L} = \frac{0.1}{1000} = 1 \times 10^{-4} \]

Step 4: Stress using Hooke’s law.
\[ \sigma = E \cdot \varepsilon = (100 \times 10^3 \, MPa)(1 \times 10^{-4}) \] \[ \sigma = 10 \, MPa \]



Step 5: Careful re-check with units.
Convert \(E = 100 \, GPa = 100 \times 10^3 \, MPa\). Thus stress = \(100000 \times 10^{-4} = 10 \, MPa\). (If small rounding adjustments considered, answer remains \( \approx 10 \, MPa\)). Final Answer:
\[ \boxed{10 \, MPa} \]