A cylindrical metal specimen having an original diameter of 11.00 mm and gauge l

Question

A cylindrical metal specimen having an original diameter of 11.00 mm and gauge length of 48.8 mm is pulled in tension until fracture occurs. The diameter at the point of fracture is 7.70 mm, and the fractured gauge length is 73.5 mm. Calculate the ductility in terms of (a) percent reduction in area (percent RA), and (b) percent elongation (percent EL).

i get:
(a) 26.1988 percent RA
(b) 50.6148 percent EL


It says my a is wrong
I know that %RA=((Ao-Af)/Ao)*100
and Ao=pi(r^2)h
I get Ao=4637.62
Af=3422.62

substitution in the above equation for %RA gives 26.1988 % but wiley says it is wrong. I tried Entering it negative but that doesnt work either. Could you please help? THANKS!

Explanation / Answer

Area = D2L/4

Apply ln on both sides

ln(A) = ln(/4) + 2ln(D) +ln(L)

Now differentiate the above equation

=> dA/A = 2(dD/D) + (dL/l)

dD = 7.70-11.0 = -3.3

D = 11.0

dL = 73.5-48.8 = 24.7

L = 48.8

dA/A = 2*(-3.3)/11 + 24.7/48.8 = -0.0938 => percentage change = (dA/A)*100 = -9.38%

=> percent reduction in area = 9.38%

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