Question

A cylindrical metal specimen having an original diameter of 11.53 mm and gauge length of 51.4 mm is pulled in tension until fracture occurs. The diameter at the point of fracture is 7.80 mm, and the fractured gauge length is 68.3 mm. Calculate the ductility in terms of (a) percent reduction in area (percent RA), and (b) percent elongation (percent EL).

Answer 1

Answer:

a)Percent reduction in area=54.23 %

b)Percent elongation=32.87 %

Explanation:

Given that

Di= 11.53 mm

Df= 7.80 mm

Li= 51.4 mm

Lf= 68.3 mm

A= π/4 D²

a)

Percent reduction in area ,RA

$RA=\dfrac{A_i-A_f}{A_i}\times 100$

$RA=\dfrac{D_i^2-D_f^2}{D_i^2}\times 100$

$RA=\dfrac{11.53^2-7.8^2}{11.53^2}\times 100$

RA= 54.23 %

Percent reduction in area=54.23 %

b)

Percent elongation EL

$EL=\dfrac{L_f-L_i}{L_i}\times 100$

$EL=\dfrac{68.3-51.4}{51.4}\times 100$

EL=32.87 %

Percent elongation=32.87 %