Answer
$r_{d}$ = 8.8 mm
Work Step by Step
Given:
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 15 and 12 mm, respectively. The second specimen, with an initial radius of 11 mm, must have the same deformed hardness as the first specimen.
Required:
compute the second specimen’s radius after deformation
Solution:
To have the same deformed hardness, the two specimens must have deformed to the same percent as a result of being cold worked. Thus:
%$CW = \frac{A_{0}-A_{d}}{A_{0}} \times 100 = \frac{πr_{0}^{2} - πr_{d}^{2}}{πr_{0}^{2}} \times 100 = \frac{π(15 mm)^{2} - π(12 mm)^{2}}{π(15 mm)^{2}} \times 100$ = 36% CW
For the second specimen, the deformed radius is computed using the above equation:
$r_{d} = r_{0} \sqrt a- \frac{CW}{100} = (11 mm)\sqrt 1 - \frac{36 CW}{100}$ = 8.8 mm
