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A cylindrical metal specimen having an original diameter of 10.33 mm and gauge length of 52.8 mm is pulled in tension until fracture occurs. The diameter at the point of fracture is 6.38 mm, and the fractured gauge length is 73.9 mm. Calculate the ductility in terms of (a) percent reduction in area (percent RA), and (b) percent elongation (percent EL).

User Mnille
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1 Answer

5 votes

Answer

a) 62 percent

b) 40 percent

Step-by-step explanation:

Original diameter ( d
_(i) ) = 10.33 mm

Original Gauge length ( L
_(i) ) = 52.8 mm

diameter at point of fracture ( d
_(f) ) = 6.38 mm

New gauge length ( L
_(f) ) = 73.9 mm

Calculate ductility in terms of

a) percent reduction in area

percentage reduction = [ (A
_(i) - A
_(f) ) / A
_(i) ] * 100

A
_(i) ( initial area ) = π /4 di^2

= π /4 * ( 10.33 )^2 = 83.81 mm^2

A
_(f) ( final area ) = π /4 df^2

= π /4 ( 6.38)^2 = 31.97 mm^2

hence : %reduction = ( 83.81 - 31.97 ) / 83.81

= 0.62 = 62 percent

b ) percent elongation

percentage elongation = ( L
_(f) - L
_(i) ) / L
_(i)

= ( 73.9 - 52.8 ) / 52.8 = 0.40 = 40 percent

User Skytz
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