Wire drawing operation is performed on a perfectly plastic metal without any strain hardening. Assuming no friction and no redundant work, the maximum possible percentage reduction in area in a single pass is closest to

The correct option is 63.2.

To understand why this is the correct answer, let us derive the expression for the maximum possible reduction in area during a wire drawing operation under ideal conditions (no friction and no redundant work).

For a perfectly plastic material with no strain hardening, the flow stress remains constant at a value $\sigma$₀.

In ideal wire drawing (also known as homogeneous deformation), the drawing stress $\sigma$_d required to draw the wire is given by the formula:
${\sigma }_d={\sigma }_0\ln \left(\frac{A_0}{A_f}\right)$
where:
• $A_0$ is the initial cross-sectional area of the wire.
• $A_f$ is the final cross-sectional area of the wire after drawing.

Physically, the maximum drawing stress ${\sigma }_d$ that can be applied to the drawn wire without causing it to yield and fracture at the exit is equal to the yield strength of the material, ${\sigma }_0$.
Therefore, the limit for a successful drawing operation is:
${\sigma }_d\le {\sigma }_0$

To find the maximum possible reduction, we set the drawing stress equal to this limiting value:
${\sigma }_0\ln \left(\frac{A_0}{A_f}\right)={\sigma }_0$

Dividing both sides by ${\sigma }_0$ gives:
$\ln \left(\frac{A_0}{A_f}\right)=1$

Taking the exponential on both sides:
$\frac{A_0}{A_f}=e$
which can also be written as:
$\frac{A_f}{A_0}=\frac{1}{e}\approx \frac{1}{2.718}\approx 0.368$

The percentage reduction in area (r) is defined as:
$r=\left(\frac{A_0-A_f}{A_0}\right)\times 100$

Rewriting this expression:
$r=\left(1-\frac{A_f}{A_0}\right)\times 100$

Substituting the value of $\frac{A_f}{A_0}$:
$r=\left(1-0.368\right)\times 100=0.632\times 100=63.2\%$

Thus, the maximum possible percentage reduction in area under these conditions is exactly 63.2%.