In wire-drawing operation, the maximum reduction per pass for perfectly plastic material in ideal condition is

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ESE Mechanical 2014 Official Paper - 2
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  1. 68%
  2. 63%
  3. 58%
  4. 50%

Answer (Detailed Solution Below)

Option 2 : 63%
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Detailed Solution

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Concept:

The maximum reduction in cross-sectional area per pass (R) of a cold wire drawing process is

\(R = 1 - {e^{ - \left( {n + 1} \right)}}\)

where n represents the strain hardening coefficient.

For perfectly plastic material, n = 0;

⇒ R = 1 – e-1

\(= 1 - \frac{1}{{2.718}} = 0.632\)

⇒ R = 63.2 %

Alternative Method:

For wire drawing drawing stress (σd) is given by

\({\sigma _d} = {\sigma _o}\ln \left( {\frac{{{A_o}}}{{{A_t}}}} \right)\)

Ao = initial area, At = final area

For maximum reduction

d = σo) where ‘σo’ is the permissible stress

\( ⇒ 1 = \ln \left( {\frac{{{A_0}}}{{{A_t}}}} \right)\)

\(⇒ \left[ {\frac{{{A_o}}}{{{A_f}}} = e} \right];\left[ {R = \frac{{{A_o} - {A_t}}}{{{A_o}}}} \right]\)

\(R ⇒ 1 - \frac{{{A_f}}}{{{A_o}}} = 1 - \frac{1}{e} = {\rm{\;}}0.632\)

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