Question

During pure orthogonal turning operation of a hollow clindrical pipe, it is found that the thickness of the chip produced is $0.5mm$. The feed given to the zero degree rake angle tool is $0.2mm/rev$. The shear strain produced durign the operation is

Answer 1

$\gamma =cot\varphi +tan(\varphi -\alpha )$
Given $\alpha =0$
$t=0.2mm$
$t_c=0.5mm$
$r=\frac{t}{t_c}=\frac{0.2}{0.5}=0.4$
$tan\varphi =\frac{rcos\alpha }{1-rsin\alpha }=r$ $\because \alpha =0$
$\therefore \gamma$ (shear strain) = $cot\varphi +tan(\varphi -\alpha )$
$=\frac{1+ta{n}^2\varphi }{tan\varphi }=\frac{1+{0.4}^2}{0.4}$
$\frac{1.16}{0.4}=2.9$