#### Answer

%RA = 59.66%
%EL = 46%

#### Work Step by Step

Given:
cylindrical metal specimen with original diameter of 12.8 mm and gauge length of 50.80 mm
diameter at point of fracture is 8.13 mm
fractured gauge length is 74.17 mm
Required:
ductility in terms of percent reduction in area and in terms of elongation
Solution:
Using Equation 6.12, it follows:
%$RA = \frac{π (\frac{d_{0}}{2})^{2} - π (\frac{d_{f}}{2})^{2} }{π (\frac{d_{0}}{2})^{2}} \times 100 = \frac{π (\frac{12.8 mm}{2})^{2} - π (\frac{8.13 mm}{2})^{2} }{π (\frac{12.8 mm}{2})^{2}} \times 100 $= 59.66 %
Using Equation 6.11:
%$EL = (\frac{l_{f}-l_{0}}{l_{0}}) \times 100 = \frac{(74.17 mm - 50.80 mm)}{50.80 mm} \times 100 $ = 46%