Q) In an orthogonal cutting following data have been observed:

Uncut chip thickness = 0.15mm

Width of cut = 0.5 mm

Cutting speed = 120 m/min

Flake angle = 10°

Cutting force = 600 N

Thrust force = 220 N

Chip thickness =0.25 mm

Determine shear angle, friction angle, shear stress along the shear plane and power for the cutting operation, Also find the chip velocity, shear straight in chip.

Solution:

$t=0.15 m m, \quad t_{c}=0.5 m m, \quad V=\frac{120 m}{m i n}=2 m / s e c$ $\gamma=10^{\circ}, \quad F_{p}=600 N, \quad F_{q}=220 N, \quad t_{c}=0.25$ For $\emptyset, \beta, \tau,$ power, $V, \in$

$r_{c}=\frac{t}{t_{c}}=\frac{0.15^{\prime}}{0.25}=0.6^{\prime}$

$\tan \emptyset=\frac{r_{c} \cos \gamma}{1-r_{c} \sin \gamma}=\frac{0.6 \cos 10}{1-0.6 \sin 10}=\frac{0.5908}{0.8958}=0.6595$

$\emptyset=33.49$

$\mu=\frac{F_{q}+F_{p} \tan \gamma}{F_{p}-F_{q} \tan \gamma}$

$\mu=\frac{200+600 \tan 10^{\circ}}{600-220 \tan 10^{\circ}}=\frac{325.796}{561.208}=0.5804$

$B=\tan ^{-1} \mu=30.13^{\circ}$

$\tau=\frac{\left(F_{p} \cos \emptyset-F_{q} \sin \emptyset\right) \sin \emptyset}{b t}$

$=\frac{(600 \cos 33.4-220 \sin 33.4) \sin 33.4}{0.5 \times 0.15}$

$=\frac{(500.908-121.1058) 0.55048}{0.5 \times 0.15}$

$=2787.65 N / \mathrm{mm}^{2}$

Power $=\frac{F_{p} V}{1000}=\frac{600 \times 2}{1000}=1.2 k W$

$V_{t}=V_{r c}=2 \times 0.6$

$=\frac{1.2 m}{s e c}=72 m / m i n$

$\epsilon=\frac{\cos \gamma}{\sin \emptyset \cos (\emptyset-\gamma)}=\frac{\cos 10}{\sin 33 \cdot 4 \cdot \cos (33.4-10)}$

$=\frac{\cos 10}{\sin 33 \cdot 4 \cdot \cos 24 \cdot 4}=1.95$