Q) During orthogonal machining with rake angle 10° rake tool, with depth of cut = 2 mm and feed rate of 0.20 mm/rev. The cutting speed is 200 m/min. The chip thickness ratio is 0.31. The vertical cutting force is 1200 N and horizontal cutting force is 650 N. Calculate from the Merchant’s theory, the various work done in metal cutting and shear stress.

Solution:

$\begin{array}{ll}{r=10^{\circ},} & {t=0.2 \mathrm{mm}, \quad r_{c}=0.3, \quad V=\frac{200 \mathrm{m}}{\min }=3.33 \mathrm{m} / \mathrm{sec}} \\ {b=2 \mathrm{mm},} & {F_{p}=1200 \mathrm{N}, \quad F_{q}=650 \mathrm{N}}\end{array}$

$\tan \emptyset=\frac{r_{c} \cos \gamma}{1-r_{c} \sin \gamma}=\frac{0.3 \cos 10}{1-0.3 \sin 10}=\frac{1.2954}{0.9479}=0.3116$

$\emptyset=17.31^{\circ}$

$F_{s}=F_{p} \cos \emptyset-F_{q} \sin \emptyset$

$=1200 \cos 17.31-650 \sin 17.11$

$=1952.25 N$

$F_{[?]}=F_{p} \sin \gamma+F_{q} \cos \gamma$

$=1200 \sin 10+650 \cos 10$

$=848.5 N$

$V_{c}=3.33 \mathrm{m} / \mathrm{sec}$

$V_{s}=\frac{V \cos \gamma}{\cos (\emptyset-\gamma)}=\frac{3.33 \cos 10}{\cos 7.31}=3.306 \mathrm{m} / \mathrm{sec}$

$V_{f}=V r_{c}=3.33 \times 0.3=1 \mathrm{m} / \mathrm{sec}$

Workdone in cutting $=F_{p} V=1200 \times 200=4000 \mathrm{Nm} / \mathrm{sec}$

Workdone in shearing $=F_{l} V_{s}=952.25 \times 3.306=3148 \mathrm{Nm} / \mathrm{sec}$

Workdone in cutting $=F V_{f}=848.5 \times 1=848.5 \mathrm{Nm} / \mathrm{sec}$