Question

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

• A. 0.865
• B. 0.826
• C. 0.777
• D. 0.632

Answer 1

The correct option is D 0.632

Given:

Maximum reduction in cross sectional area per pass,

$R=1-e^{-(n+1)}$

For perfectly plastic material, strain hardening coefficient n will be zero.

So, $R=1-e^{-(0+1)}$

$=1-e^{-1}=0.632$

Maximum reduction per pass,

R = 0.632