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Question 5 A metal rod with a square cross section \( 12 \times 12 \mathrm{~mm} \), and 500 mm long is subjected to an axial load of 20 KN . The rod elongated axially by 2 mm when the load was applied. What is the Young's Modulus of the material? (A) 34.7 GPa (B) 13.9 GPa (C) 55.0 GPa (D) 120 GPa

Ask by Allan Mills. in Singapore
Feb 04,2025

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The Young's Modulus of the material is 34.7 GPa.

Solution

To determine the Young's Modulus (\( E \)) of the material, we'll use the formula from the theory of elasticity: \[ \Delta L = \frac{F L}{A E} \] Where: - \( \Delta L \) = elongation = 2 mm - \( F \) = axial load = 20 KN = 20,000 N - \( L \) = original length = 500 mm - \( A \) = cross-sectional area = 12 mm × 12 mm = 144 mm² Rearranging the formula to solve for \( E \): \[ E = \frac{F L}{A \Delta L} \] Plugging in the values: \[ E = \frac{20,000 \, \text{N} \times 500 \, \text{mm}}{144 \, \text{mm}^2 \times 2 \, \text{mm}} = \frac{10,000,000 \, \text{N·mm}}{288 \, \text{mm}^3} \approx 34,722 \, \text{N/mm}^2 \] Since \( 1 \, \text{GPa} = 1,000 \, \text{N/mm}^2 \), we convert: \[ E \approx 34.7 \, \text{GPa} \] **Answer:** (A) 34.7 GPa

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To find the Young's Modulus \( E \) of the material, we can use the formula: \[ E = \frac{\text{Stress}}{\text{Strain}} \] First, we need to calculate the stress (\( \sigma \)) and strain (\( \epsilon \)). Stress (\( \sigma \)) is defined as the load (\( F \)) divided by the cross-sectional area (\( A \)): Given: - Load, \( F = 20 \text{ kN} = 20000 \text{ N} \) - Cross-sectional area, \( A = 12 \text{ mm} \times 12 \text{ mm} = 144 \text{ mm}^2 = 144 \times 10^{-6} \text{ m}^2 \) \[ \sigma = \frac{F}{A} = \frac{20000 \text{ N}}{144 \times 10^{-6} \text{ m}^2} \approx 138.89 \text{ MPa} \] Strain (\( \epsilon \)) is the change in length (\( \Delta L \)) divided by the original length (\( L_0 \)): Given: - Change in length, \( \Delta L = 2 \text{ mm} = 2 \times 10^{-3} \text{ m} \) - Original length, \( L_0 = 500 \text{ mm} = 0.5 \text{ m} \) \[ \epsilon = \frac{\Delta L}{L_0} = \frac{2 \times 10^{-3} \text{ m}}{0.5 \text{ m}} = 0.004 \] Now, substituting stress and strain into the Young's Modulus equation: \[ E = \frac{138.89 \times 10^6 \text{ Pa}}{0.004} \approx 34.72 \text{ GPa} \] Thus, the Young's Modulus of the material is approximately \( 34.7 \text{ GPa} \), so the correct answer is (A) 34.7 GPa. Now, here’s something fun about stress and strain: Did you know that the concept of Young's Modulus is named after the 19th-century British scientist Thomas Young? He was pretty much the rock star of elasticity studies! Workers in factories now depend on his insights every day, ensuring that beams and rods can bear loads without snapping like a twig. When dealing with calculations like these, it's important to keep track of your units! Beginners often confuse mm and m or N and kN, leading to incorrect calculations. Always double-check your units—it's a tiny detail that can lead to big mishaps in engineering!

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