Question

In a turning operation with orthogonal machining the spindle rotates at 270 rpm and the tool is fed at a rate of 0.3 mm per revolution. The depth of cut is 0.5 mm. The rake angle is ${10}^{\circ }$ and the coefficient of friction at the tool chip interface to be 0.5317. What is the value of chip thickness?

• A. 0.30 mm
• B. 0.40 mm
• C. 0.46 mm
• D. 0.50 mm

Answer 1

The correct option is C 0.46 mm

In Orthogonal machining $\lambda ={90}^{\circ }$
Depth of cut $d=b=0.5mm$
$t=f\left(feed\right)=0.3mm/rev$
$\alpha ={10}^{\circ }$

$\beta =ta{n}^{-1}\left(\mu \right)=ta{n}^{-1}\left(0.5317\right)={28}^{\circ }$

From merchant theory,
$\Rightarrow$ $2\varphi +\beta -\alpha ={90}^{\circ }$
$\varphi ={36}^{\circ }$

$tan\varphi =\frac{rcos\alpha }{1-rsin\alpha }$

$tan{36}^{\circ }=\frac{rcos{10}^{\circ }}{1-rsin{10}^{\circ }}$

r = 0.654

$r=\frac{t}{t_c}=\frac{0.3}{t_c}$

$t_c=\frac{0.3}{0.654}$

$t_c=0.46\text{mm}$