Q) During orthogonal machining with rake angle 10° and uncut thickness 0.125 mm, the values of Fτ and F_c are found to be 217 N and 517 N respectively. Average chip thickness = 0.43 mm, width of chip = 2 mm, cutting speed = 120 m/min. Compute chip thickness ratio, shear angle, resultant force, shear force, normal compressive force, frictional force, merchant’s constant, cutting power, consumption, shearing power, consumption and frictional power consumption.

Solution:

$\gamma=10^{\circ}, \quad t=0.125 \mathrm{mm}, \quad t_{c}=0.43 \mathrm{mm}, \quad V=\frac{120 \mathrm{m}}{\min }=2 \mathrm{m} / \mathrm{sec}$ $b=2 \mathrm{mm},$

For $r_{c}, \varnothing, R, F_{a}, N_{a}, F, C,$ power consumption

$r_{c}=\frac{t}{t_{c}}=\frac{0.125^{\prime}}{0.43}=0.2907^{\prime}$

$\tan \emptyset=\frac{r_{c} \cos \gamma}{1-r_{c} \sin \gamma}=\frac{0.2907 \cos 10}{1-0.2907 \sin 10}=\frac{0.5908}{0.8958}=0.3015$

$\emptyset=16.78^{\circ}$

$R=\sqrt{F_{p}^{2}+F_{q}^{2}}$

$R=\sqrt{517^{2}+217^{2}}=60.69 N$

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

$=517 \cos 16.78-217 \sin 16.78$

$=357.02 N$

$F_{G}=F_{p} \sin \gamma-F_{q} \cos \gamma$

$=517 \sin 10-217 \cos 10$

$=471.5 N$

$V_{c}=2 m / s e c$

$V_{s}=\frac{V \cos \gamma}{a(\emptyset-\gamma)}=\frac{2 \times \cos 10}{\cos 6.78}=1.983 \mathrm{m} / \mathrm{sec}$

$V_{t}=v \cdot r_{c}=2 \times 0.2907=0.5814 \mathrm{m} / \mathrm{m}$

Power consumption for cuts = $\frac{F_{p} V}{1000}=\frac{517 \times 2}{1000}=1.034 k W$

Power for friction = $\frac{F V_{t}}{1000}=\frac{303.48 \times 0.5814}{1000}=0.176 k W$

Power of shearing = $\frac{F_{s} V_{s}}{1000}=\frac{432.34 \times 1.983}{1000}=0.857 k W$

$\mu=\frac{F_{p}+F_{q} \tan \gamma}{F_{p}-F_{q} \tan \gamma}=\frac{217+517 \tan 10^{\circ}}{517-217 \tan 10^{\circ}}=\frac{308.161}{478.737}=0.5804$

$=0.6437$

$=\tan \beta$

$\beta=32.77^{\circ}$

Merchants$=2 \emptyset+\beta-\gamma$

$=2 \times 16.78+32.77-10$

$=56.33$