Question

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Supplemental Problems 10.16 A 12-in.-diameter structural nickel steel specimen was subjected to a tension test. After rupture it was determined that the 2-in. standard-gage length had stretched to 2.42 in. The minimum diameter at the fracture was measured to be 0.422 in. Compute the percent elongation and percent reduction in area.

Answer 1

( 10.16 )Percent elongation is given as ratio of final extension at rupture to original length expressed as percentage

So percentage elongation = ( L' - L / L ) x 100

where L = Original Length i.e 2.42 in

L' = Length at rupture i.e 2 in

So substituting the values given in the problem, we get

Percentage elongation = ( L' - L / L ) x 100

= (2.42-2) / 2) x 100

= ( 0.42 / 2) x 100

= 0.21 x 100

So Percentage elongation = 21%

Percentage reduction in area is given as ratio of maximum changes in the cross sectional area to original cross sectional area expressed as percentage

So percentage reduction in area = ( A - A' / A ) x 100

where A = Original cross sectional area, original diameter is given as 12 in in problem, so area is (3.142 x 12 x 12 /4) = 113.112 sq in.

A' = Minimum cross sectional area i.e minimum diameter is given as 0.422 in in problem,

so area is (3.142 x 0.422 x 0.422) / 4 = 0.140 sq in

So substituting the values given in the problem, we get

Percentage reduction in area = ( A - A' / A ) x 100

= (113.112- 0.140) / 113.112) x 100

= ( 112.972 / 113.112) x 100

= 99.876 x 100

So Percentage reduction in area = 99.876%