Question

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

Answer 1

Answer:

% reduction in area==PR=0.734=73.4%

% elongation=EL=0.42=42%

Explanation:

given do=12.8 mm

df=6.60

Lf=72.4 mm

Lo=50.8 mm

% reduction in area=(($\pi$*(do/2)^2)-($\pi$*(df/2)^2)))/$\pi$*(do/2)^2

substitute values

% reduction in area=73.4%

% elongation=EL=((Lf-Lo)/Lo))*100

% elongation=((72.4-50. 8)/50.8)*100=42%