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Exercises · 8.8

Q.A piece of copper having a rectangular cross-section of 15.2 mm×19.1 mm15.2\ \text{mm} \times 19.1\ \text{mm} is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain.

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Using Young's modulus for copper (Y=1.1×1011 PaY = 1.1 \times 10^{11}\ \text{Pa}, Table 8.1) and the relation σ=Y⋅ε\sigma = Y \cdot \varepsilon, dividing the tensile stress by YY gives the strain. The printed answer for this question is 0.127\boxed{0.127}.

Young's modulus is the material's stiffness -- it tells you how much strain (fractional change in length) a given stress produces, as long as the deformation stays elastic. For copper, Table 8.1 gives Y=1.1×1011 PaY = 1.1 \times 10^{11}\ \text{Pa}. The key relation is:

σ=Y⋅εorε=σY\sigma = Y \cdot \varepsilon \quad \text{or} \quad \varepsilon = \frac{\sigma}{Y}

where σ\sigma is tensile stress (force per unit area) and ε\varepsilon is the strain (dimensionless). So the plan is: compute the cross-sectional area, find the stress, then divide by YY.

  1. Convert dimensions to metres. The cross-section is 15.2 mm×19.1 mm15.2\ \text{mm} \times 19.1\ \text{mm}.

15.2 mm=0.0152 m,19.1 mm=0.0191 m15.2\ \text{mm} = 0.0152\ \text{m}, \qquad 19.1\ \text{mm} = 0.0191\ \text{m}

  1. Calculate the area.

A=0.0152×0.0191=2.9032×10−4 m2A = 0.0152 \times 0.0191 = 2.9032 \times 10^{-4}\ \text{m}^2

  1. Find the tensile stress. Force F=44,500 NF = 44{,}500\ \text{N}.

σ=FA=44,5002.9032×10−4≈1.53×108 Pa\sigma = \frac{F}{A} = \frac{44{,}500}{2.9032 \times 10^{-4}} \approx 1.53 \times 10^{8}\ \text{Pa}

  1. Apply Young's modulus. …

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