Discussion
The maximum reduction in cross-sectional area per pass (\(R\)) of a cold wire drawing process is
\[ R = 1 - e^{-(n+1)}, \]
where \(n\) represents the strain hardening coefficient. For the case of a perfectly plastic material, \(R\) is
Answer: D
R is 0.632.
A perfectly plastic material does not strain-harden, so n = 0:
\[ R = 1 - e^{-(0 + 1)} = 1 - \frac{1}{e} = 1 - 0.368 = 0.632 \]
That is, at most about 63% reduction in area per pass. The larger values come from taking n greater than zero: n = 1 gives \(1 - e^{-2} = 0.865\), and n of about 0.75 and 0.5 give 0.826 and 0.777, reductions possible only for strain-hardening materials.
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