A cylindrical metal specimen having an original diameter of \(12.8 \mathrm{~mm}(0.505\) in.) and gauge length of \(50.80 \mathrm{~mm}(2.000 \mathrm{in} .)\) is pulled in tension until fracture occurs. The diameter at the point of fracture is \(6.60 \mathrm{~mm}(0.260 \mathrm{in} .)\), and the fractured gauge length is \(72.14 \mathrm{~mm}\) (2.840 in.). Calculate the ductility in terms of percent reduction in area and percent elongation.

Question: Calculate the ductility of a cylindrical metal specimen in terms of percent reduction in area and percent elongation, given the original diameter is 12.8 mm, fractured diameter is 6.6 mm, initial gauge length is 50.80 mm, and fractured gauge length is 72.14 mm. Answer: To calculate the ductility in terms of percent reduction in area and percent elongation, follow these steps: 1. Calculate the initial and fractured cross-sectional areas using the formula \(A = \pi(\frac{D}{2})^2\): Initial area (\(A_i\)) = 128.3 mm² Fractured area (\(A_f\)) = 34.2 mm² 2. Calculate the percent reduction in area: % reduction in area = \(\frac{(A_i - A_f)}{A_i} \times 100\) = 73.3% 3. Calculate the percent elongation: % elongation = \(\frac{(L_f - L_i)}{L_i} \times 100\) = 42.0% Thus, the ductility of the cylindrical metal specimen is 73.3% in terms of percent reduction in area and 42.0% in terms of percent elongation.