Question 5 of 13beginner🔧 ApplyNumerical3 marks

A piece of copper having a rectangular cross-section of 15.2 mm × 19.1 mm is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain?

Correct Answer

The resulting strain is approximately 1.39×101.39 \times 10⁻³ or 0.14%.

Exercise: EXERCISES | Q: 8.8
For More Understanding

Explanation

This question requires calculating strain from stress using Young's modulus. From the context, we can derive Young's modulus of copper: the text states that for a 0.1% strain (0.001) in a wire of 0.1 cm² cross-section, copper requires 1100 N. This gives Y = Stress/Strain = (1100 N / 10⁻⁵ m²) / 0.001 = 1.1×101.1 \times 10¹¹ N/m². The solution follows the same methodology as Example 8.1 in the textbook.

Solution Steps

  1. Step 1: Calculate the cross-sectional area A = 15.2 mm × 19.1 mm A = (15.2×1015.2 \times 10⁻³ m) × (19.1×1019.1 \times 10⁻³ m) A = 290.32×10290.32 \times 10⁻⁶ m² = 2.90×102.90 \times 10⁻⁴ m²

  2. Step 2: Calculate the stress Stress = Force / Area Stress = 44,500 N / (2.90×102.90 \times 10⁻⁴ m²) Stress = 1.53×101.53 \times 10⁸ N m⁻²

  3. Step 3: Determine Young's modulus of copper from context From the context: For 0.1% strain in copper wire of 0.1 cm² area, force = 1100 N Stress = 1100 N / 10⁻⁵ m² = 1.1×101.1 \times 10⁸ N m⁻² Young's modulus = Stress / Strain = (1.1×101.1 \times 10⁸) / 0.001 = 1.1×101.1 \times 10¹¹ N m⁻²

  4. Step 4: Calculate strain using the relation Strain = Stress / Young's modulus Strain = (1.53×101.53 \times 10⁸ N m⁻²) / (1.1×101.1 \times 10¹¹ N m⁻²) Strain = 1.39×101.39 \times 10⁻³ ≈\approx 0.14%

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