Question

Details pertaining to an orthogonal metal cutting process are given below:
Chip thickness ratio $0.4$
Undeformed thickness $0.6mm$
Rake angle $+{10}^o$
Cutting speed $2.5m/s$
Mean thickenss of primary shear zone $25$ microns
The shear strain rate in $s^{-1}$ during the process is

• A. $0.1781\times {10}^5$
• B. $0.7754\times {10}^5$
• C. $1.0104\times {10}^5$
• D. $4.397\times {10}^5$

Answer 1

The correct option is C $1.0104\times {10}^5$

$r=0.4,t_1=0.6,\alpha -10,V_C=2.5m/s$
$\varphi =$ Shear angle
$=ta{n}^{-1}\left(\frac{rcos\alpha }{1-rsin\alpha }\right)$
$=ta{n}^{-1}\left(\frac{0.4cos{10}^o}{1-0.4sin{10}^o}\right)={22.94}^o$
Shear strain rate
$=\left(\frac{cos\alpha }{cos(\varphi -\alpha )}\frac{V_C}{\Delta Y}\right)$
$=\left(\frac{cos10}{cos(22.94-10)}\frac{2500}{0.025}\right)$
$=101046.8=1.10104\times {10}^5/s$