Question

A piece of copper having a rectangular cross-section of 15.2 mm x 19.1 mm is pulled in tension with 44,500 N force, producing only elastic deformation. (Shear modulus of elasticity of copper = 42 G pa)

• A. Shearing stress produced in the copper is $1.53\times {10}^{8}N/m^2$
• B. Shearing stress produced in the copper is $3.6\times {10}^{-3}$
• C. Shear stress in the copper is $1.35\times {10}^{8}N/m^2$
• D. Shearing stress produced in the copper is $6.3\times {10}^{-3}$

Answer 1

The correct option is A Shearing stress produced in the copper is $1.53\times {10}^{8}N/m^2$

$\begin{array}{c}\text{Given dimensions}=15.2mm\times 19.1mm\\ \text{Force = 44,500N}\\ \text{shear stress}=\frac{44,500}{\text{Area}}\end{array}$

$\begin{array}{cc}\text{Area}& =15.2\times 19.1\times {10}^{-6}{m}^2\\ & =2.9\times {10}^{-4}\\ \text{Stress}& =\frac{44,500}{2.9\times {10}^{-4}}\\ & =1.53\times {10}^{8}N/m^2\end{array}$