High School

A cylindrical metal specimen has an original diameter of 11.05 mm and a gauge length of 51.7 mm. It is pulled in tension until fracture occurs. The diameter at the point of fracture is 6.58 mm, and the fractured gauge length is 68.3 mm. Calculate the ductility in terms of:

(a) Percent reduction in area (percent RA)
(b) Percent elongation (percent EL)

Answer :

The ductility in terms of Percent Reduction in area is 64.54% and Percent Reduction in elongation is 32.11%.

A cylindrical metal specimen having an Original diameter = 11.05 mm

A cylindrical metal specimen having a Gauge length = 51.7mm

Diameter at the point of fracture = 6.58 mm

Fractured gauge length = 68.3 mm.

Percent Reduction in Area = ((do/2)² - (d1/2)²)/(do/2)²

Calculating percent reduction in area

do = 11.05 mm, d1 = 6.58 mm

So,

Percent Reduction in Area = ((11.05/2)² - (6.58/2)²)/(11.05/2)²

Percent Reduction in Area = ((5.525)² - (3.29)²)/(5.525)²

Percent Reduction in Area = (30.526 - 10.8241)/30.525

Percent Reduction in Area = 19.7019/30.525

Percent Reduction in Area = 0.6454

Percent Reduction in Area = 64.54%

Calculating percent reduction in elongation

Percent Reduction in elongation = ((do) - (d1))/(d1)

do = 68.3 mm, d1 = 51.7 mm

Percent Reduction in elongation = (68.3 - 51.7)/(51.7)

Percent Reduction in elongation = 16.6/51.7

Percent Reduction in elongation = 0.3211

Percent Reduction in elongation = 32.11%

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