Question

Following data correspond to an orthogonal Turning of a 100 mm diameter rod on a lathe. Rake angle: +15°; Uncut chip thickness 0.5 mm, nominal chip thickness after the cut 1.25 mm. The shear angle (in degrees) for this process is (correct to two decimal places).

Answer 1

$\alpha =15°$
t = 0.5 mm
$t_c=1.25mm$

$r=\frac{t}{t_c}=\frac{0.5}{1.25}=0.4$

$\tan \varphi =\frac{r\cos \alpha }{1-r\sin \alpha }=\frac{0.4\cos 15{}^{\circ }}{1-0.4\sin 15{}^{\circ }}$

$\varphi =23.3155{}^{\circ }\approx 23.32{}^{\circ }$