## Materials Science and Engineering: An Introduction

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%