Question

In an orthogonal machining with a tool of 9° orthogonal rake angle, the uncut chip thickness is 0.2 mm. The chip thickness fluctuates between 0.25 mm and 0.4 mm. The ratio of the maximum shear angle to the minimum shear angle during machining is

Answer 1

Given $\alpha =9{}^{\circ }$

$t=0.2mm$

$t_{cmax}=0.4mm$

$t_{cmin}=0.25mm$

$\tan {\varphi }_{max}=\frac{rmax\cos \alpha }{1-rmax\sin \alpha }=\frac{\frac{0.2}{0.25}\cos 9{}^{\circ }}{1-\frac{0.2}{0.25}\sin 9{}^{\circ }}$

$or{\varphi }_{max}=42.09{}^{\circ }$

$\tan {\varphi }_{min}=\frac{t_{min}\cos \alpha }{1-r_{min}\sin \alpha }=\frac{\frac{0.2}{0.4}\cos 9{}^{\circ }}{1-\frac{0.2}{0.4}\sin 9{}^{\circ }}$

$or{\varphi }_{min}=28.18{}^{\circ }$

$Ratio=\frac{{\varphi }_{max}}{{\varphi }_{min}}=\frac{42.09}{28.18}=1.494$