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Mechanical Properties of Solids

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Question : 8 of 21
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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. Shear modulus of elasticity of copper is 42×109 Nm−2.2 \times 10^{9} \ \mathrm{N m^{-2}}.
Solution:  
Here, A=15.2×19.2×10−6 m2;A=15.2\times19.2 \times 10^{-6} \ \mathrm{m}^{2} ;
F=44,500 N;η=42×109 N m−2F=44,500 \ \mathrm{N} ; \eta=42 \times 10^{9} \ \mathrm{N} \ \mathrm{m}^{-2}
Strain = stress  modulus of elasticity = \frac{\text{ stress }}{\text{ modulus of elasticity }}
=FAη= \frac{\frac{F}{A}}{\eta}
=4450015.2×19.2×10−6×42×109= \frac{44500}{15.2 \times 19.2 \times 10^{-6} \times 42 \times 10^{9}}
=3.65×10−3=3.65 \times 10^{-3}
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