Question

The following observations were made during orthogonal cutting of brass tube on a lathe

width of cut = 0.45 cm

cutting speed = 7.5 m/min

Rake angle $={20}^o$

uncut chip thickness = 0.21 mm

chip thickness = 0.60 mm


Then the induced shear strain is ___

• A. 2.685

Answer 1

The correct option is A 2.685
$\varphi -\beta -\alpha ={40}^o$

$r=\frac{t}{t_c}=\frac{0.21}{0.60}=0.35$

$\varphi =ta{n}^{-1}\left(\frac{rcos\alpha }{1-rsin\alpha }\right)=ta{n}^{-1}\left(\frac{0.35cos{20}^o}{1-0.35sin{20}^o}\right)$

$\varphi ={20.48}^o$

Shear strain $\gamma =cot\varphi +tan(\varphi -\alpha )$

$=cot{20.48}^o+tan({20.48}^o-{20}^o)$

$\gamma =2.6858$