Question

In an orthogonal metal cutting process are given below

Chip thickness = 0.8 mm
Undeformed thickness = 0.6 mm
Rake angle = ${15}^o$
Cutting speed = 2.5 m/s

Mean thickness of primary shear zone 25 microns then the shear strain rate is $\left(s^{-1}\right)$ during the process is

• A. $1.0636\times {10}^5$
• B. $1.0736\times {10}^5$
• C. $1.0836\times {10}^5$
• D. $1.0936\times {10}^5$

Answer 1

The correct option is C $1.0836\times {10}^5$

Given data:
$r=\frac{0.6}{0.8}=0.75,\alpha ={15}^o$

$\varphi =ta{n}^{-1}\left(\frac{0.75cos{15}^o}{1-0.75sin{15}^o}\right)$

$\varphi ={41.95}^o$

$\frac{V}{cos(\varphi -\alpha )}=\frac{V_s}{cos\alpha }$

$V_c=\frac{2.5\times cos{15}^o}{cos(41.95-{15}^o)}=2.709m/s$

Shear strain rate $\left(\dot{\epsilon }\right)$

$\dot{\epsilon }=\frac{V_s}{t_{(}s)}$

$=\frac{2.709m/s}{25\times {10}^{-6}m}=1.0836\times {10}^{5}se{c}^{-1}$