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 as shown in the figure. Now the bar is heated resulting in a uniform temperature rise of \( 10^\circ {C} \). The coefficient of linear thermal expansion of the material is \( 20 \times 10^{-6} / \degree C \) and the Young’s modulus of elasticity is 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:

For thermal stress problems, calculate the thermal strain using the coefficient of thermal expansion and use the Young's modulus to find the axial stress.

Answer 1

Given:
- Length of bar, \( L = 1 \, {m} \)
- Gap before heating, \( \delta = 0.1 \, {mm} = 0.0001 \, {m} \)
- Temperature rise, \( \Delta T = 10^\circ {C} \)
- Coefficient of thermal expansion, \( \alpha = 20 \times 10^{-6} / \degree C \)
- Young's modulus of elasticity, \( E = 100 \, {GPa} = 100 \times 10^9 \, {Pa} \)
Step 1: Thermal Expansion The free expansion of the bar without any constraint is given by the formula: \[ \Delta L_{{free}} = \alpha L \Delta T \] Substituting the values: \[ \Delta L_{{free}} = 20 \times 10^{-6} \times 1 \times 10 = 0.0002 \, {m} \] Step 2: Axial Stress The gap before heating was \( 0.0001 \, {m} \), so after heating, the bar would attempt to expand by \( 0.0002 \, {m} \), but the wall restricts it. Therefore, the actual elongation is the difference: \[ \Delta L_{{actual}} = 0.0002 - 0.0001 = 0.0001 \, {m} \] Now, the axial stress \( \sigma \) is given by: \[ \sigma = \frac{E \Delta L_{{actual}}}{L} \] Substituting the values: \[ \sigma = \frac{100 \times 10^9 \times 0.0001}{1} = 10 \times 10^6 \, {Pa} = 10 \, {MPa} \] Thus, the magnitude of the resulting axial stress on the bar is \( 10 \, {MPa} \).