. 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 pu

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. 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 pu

lled 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

Engineering

1 answer:

Mrrafil [318] · Answer 1 · 2026-04-11T12:25:44+03:00

% area reduction==PR=0.734=73.4%. % elongation=EL=0.42=42%. Explanation: Given do=12.8 mm, df=6.60 mm, Lf=72.4 mm, Lo=50.8 mm. The % area reduction is calculated using the formula % reduction in area = ((π*(do/2)²)-(π*(df/2)²))/(π*(do/2)²). Substituting the values yields % area reduction=73.4%. For % elongation, use the formula EL=((Lf-Lo)/Lo)*100, resulting in % elongation=((72.4-50.8)/50.8)*100=42%.