Two previously undeformed cylindrical specimens of an alloy are to be strain hardened by reducing their cross-sectional areas (while maintaining their circular cross sections). For one specimen, the initial and deformed radii are 16 mm and 13 mm, respectively. The second specimen, with an initial radius of 11 mm, must have the same deformed hardness as the first specimen; compute the second specimen's radius, in mm, after deformation.

[tex]\left(\%(W))_{1}\right.[/tex] = [tex]\frac{\left(A_{0}\right)_{1}-\left(A_{f}\right)_{1}}{\left(A_{0}\right)_{1}} \times 100[/tex] = [tex]\frac{\left(\pi r_{0}^{2}\right)_{1}-\left(\pi r_{1}^{2}\right)_{1}}{\left(\pi\left(r_{0}\right)^{2}\right)_{1}} \times 100[/tex]

[tex]\left(\%(W))_{1}\right.[/tex] =[tex]\frac{\pi(16)^{2}-\pi(13)^{2}}{\pi(16)^{2}} \times 100[/tex]

[tex]\left(\%(W))_{1}\right.[/tex] = 33.984

To calculate the percentage of cold work for the specimen 2:

[tex]\left(\%,(w))_{2}\right.[/tex] = [tex]\frac{\left(A_{0}\right)_{2}-\left(A_{f}\right) 2}{\left(A_{0}\right)_{2}} \times 100[/tex] = [tex]\frac{\left(\pi r_{0}^{2}\right)_{2}-\left(\pi r_{1}^{2}\right)_{2}}{\left(\pi r_{0}^{2}\right)_{2}} \times 100[/tex]

[tex]\left(\%,(w))_{2}\right.[/tex] =[tex]\frac{\pi(13)^{2}-\pi r_{f}^{2}}{\pi(13)^{2}} \times 100[/tex]

The deformed radius is calculated by

[tex](\% \ W)_ 2=(\%\ W)_1[/tex]

[tex]\frac{169-r_{f}^{2}}{169} \times (100)[/tex] = 33.984

[tex]r_{f}[/tex] = 10.562 mm