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Stolb23 [73]
4 years ago
11

A cylindrical metal specimen having an original diameter of 11.53 mm and gauge length of 51.4 mm is pulled in tension until frac

ture 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).
Engineering
1 answer:
Verdich [7]4 years ago
3 0

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 %

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ALL CAREFULLY ANSWERED CORRECTLY

Explanation:

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Answer:

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T2/T1 = P2/P1

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5 0
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Learn more:

New technology brainly.com/question/5768621

#learnwithBrainly

3 0
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