Question

A cylindrical metal specimen having an original diameter of 11.77 mm and gauge length of 46.1 mm is pulled in tension until fracture occurs. The diameter at the point of fracture is 7.96 mm, and the fractured gauge length is 66.0 mm. Calculate the ductility in terms of (a) percent reduction in area (percent RA), and (b) percent elongation (percent EL).

Answer 1

Answer:

% reduction in area = 54.26 %

percentage elongation = 43.16 %

Explanation:

a) percentage reduction in area $= \frac{(A_1 - A_2)}{A_1 } * 100$

$A_1 =\pi r^2 = \pi * 5.88^2=108.8 mm2$

$A_2=\pi * 3.98^2= 49.97 mm2$

% reduction in area =$\frac{(108.8 - 49.97)}{108.8} * 100 = 54.26 %$

b)percentage elongation $= \frac{(66 - 46.1)}{46.1} *100 = 43.16 %$