A metal specimen with an original diameter of 0.500 in. and a gage length of 2.000 in. is tested in tension until fracture occurs. At the point of fracture, the diameter of the specimen is 0.260 in. and the fractured gage length is 3.08 in. Calculate the ductility in terms of percent elongation and

Question

A metal specimen with an original diameter of 0.500 in. and a gage length of 2.000 in. is tested in tension until fracture occurs. At the point of fracture, the diameter of the specimen is 0.260 in. and the fractured gage length is 3.08 in. Calculate the ductility in terms of percent elongation and percent reduction in area.

Accepted Answer

Percent elongation is simply the longitudinal strain at fracture:

[latex]\varepsilon=\frac{\delta}{L}=\frac{(3.08 in. -2.000 in .)}{2.000 ext { in. }}=\frac{1.08 in .}{2.000 in .}=0.54 in . / in.[/latex]

[latex] herefore ext { percent elongation }=54 \%[/latex]

The initial cross-sectional area of the specimen is

[latex]A_{0}=\frac{\pi}{4}(0.500 in .)^{2}=0.196350 in .^{2}[/latex]

The final cross-sectional area at the fracture location is

[latex]A_{f}=\frac{\pi}{4}(0.260 in .)^{2}=0.053093 in .^{2}[/latex]

The percent reduction in area is

percent reduction of area [latex]=\frac{A_{0}-A_{f}}{A_{0}}(100 \%)[/latex]

[latex]=\frac{\left(0.196350 in \cdot^{2}-0.053093 in .^{2} ight)}{0.196350 in .{ }^{2}}(100 \%)=73.0 \%[/latex]

Question 3.8: A metal specimen with an original diameter of 0.500 in. and ...

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