Question

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?

Answer 1

Given

• Rectangular cross-section: \( 15.2 \,\text{mm} \times 19.1 \,\text{mm} \)
• Force (tension): \( F = 44{,}500 \,\text{N} \)
• Young’s modulus of copper: \( Y_{\text{Cu}} = 1.2 \times 10^{11} \,\text{N m}^{-2} \)

1. Cross-sectional area

Convert dimensions to metres:

\( 15.2 \,\text{mm} = 15.2 \times 10^{-3} \,\text{m} \) \( 19.1 \,\text{mm} = 19.1 \times 10^{-3} \,\text{m} \)

Area:

\( A = (15.2 \times 10^{-3})(19.1 \times 10^{-3}) \,\text{m}^{2} \) \( A = 290.32 \times 10^{-6} \,\text{m}^{2} \approx 2.90 \times 10^{-4} \,\text{m}^{2} \)

2. Stress

\( \sigma = \dfrac{F}{A} = \dfrac{44{,}500}{2.90 \times 10^{-4}} \,\text{Pa} \approx 1.53 \times 10^{8} \,\text{Pa} \)

3. Strain

For linear elastic deformation:

\( \text{strain} = \epsilon = \dfrac{\sigma}{Y} = \dfrac{1.53 \times 10^{8}}{1.2 \times 10^{11}} \approx 1.28 \times 10^{-3} \)

Resulting strain \( \epsilon \approx 1.3 \times 10^{-3} \) (dimensionless).