Answer :
Answer:
r_d= 8.1382 mm
Explanation:
In order for these two cylindrical specimen to have same deformed hardness They must be deformed by the same percentage cold worked. For the first specimen
percentage cold work = [tex]\frac{A_0-A_d}{A_0}\times100[/tex]
[tex]\frac{\pi r_0^2-\pi r_d^2}{\pi r_0^2}\times100[/tex]
[tex]\frac{17^2-11^2}{17^2}\times100[/tex]
percentage cold work= 58.13 %
For the second specimen, the deformed radius is computed using above equation and solving for radius
r_d
[tex]r_d= r_o\sqrt{1-\frac{percentage cold work}{100} }[/tex]
[tex]r_d= 14\sqrt{1-\frac{58.13}{100} }[/tex]
r_d= 8.1382 mm