Comparing Rod Elongation Under Tensile Stress

This question compares the elongation of two identical circular rods made of mild steel and aluminum when subjected to the same axial tensile stress ($\sigma$). Both rods have the same dimensions (length L, diameter) and experience stress within their proportional limits.

Key Material Properties and Formulas

We need to compare the elongation ($\Delta L$) based on the materials' properties. The key relationship involves:

• Stress ($\sigma$): Force per unit area. Given as the same for both rods.
• Young's Modulus (E): A measure of material stiffness. It relates stress and strain. Given as $E_{steel} = 200$ GPa and $E_{Al} = 70$ GPa.
• Strain ($\epsilon$): Deformation per unit length. Defined as $\epsilon = \frac{\Delta L}{L}$.

Hooke's Law states that within the proportional limit, stress is directly proportional to strain: $\sigma = E \epsilon$.

Calculating Rod Elongation

We can express elongation ($\Delta L$) using the given parameters. From Hooke's Law, strain is $\epsilon = \frac{\sigma}{E}$.

Substituting the definition of strain, $\frac{\Delta L}{L} = \frac{\sigma}{E}$.

Rearranging to find the elongation, we get: $ \Delta L = \frac{\sigma L}{E} $

Comparing Elongation of Steel and Aluminum Rods

For both rods:

• The length (L) is the same.
• The axial stress ($\sigma$) is the same.

Therefore, the elongation ($\Delta L$) depends inversely on the Young's Modulus (E): $\Delta L \propto \frac{1}{E}$.

Given values:

• $E_{steel} = 200$ GPa
• $E_{Al} = 70$ GPa

Since $E_{Al} < E_{steel}$ (70 GPa < 200 GPa), the term $\frac{1}{E_{Al}}$ will be greater than $\frac{1}{E_{steel}}$.

This means the elongation for the aluminum rod ($\Delta L_{Al}$) will be greater than the elongation for the steel rod ($\Delta L_{steel}$).

Conclusion

The Aluminum rod elongates more than the mild steel rod because it has a lower Young's Modulus, indicating it is less stiff and deforms more under the same stress.