Orthogonal cutting forces and power numerical

Given: $t_0 = 0.25$ mm, $t_c = 0.75$ mm, $b = 3$ mm, $\alpha = 10^\circ$, $F_c = 900$ N, $F_t = 450$ N, $V = 90$ m/min $= 1.5$ m/s.

Shear angle

Chip thickness ratio $r = t_0/t_c = 0.25/0.75 = 0.3333$.

$$\tan\phi = \frac{r\cos\alpha}{1 - r\sin\alpha} = \frac{0.3333\times0.9848}{1 - 0.3333\times0.1736} = \frac{0.3283}{0.9421} = 0.3485$$ $$\phi = 19.2^\circ$$

Forces on the tool face

$$F = F_c\sin\alpha + F_t\cos\alpha = 900(0.1736) + 450(0.9848) = 156.3 + 443.2 = 599.4\ \text{N}$$ $$N = F_c\cos\alpha - F_t\sin\alpha = 900(0.9848) - 450(0.1736) = 886.3 - 78.1 = 808.2\ \text{N}$$ $$\mu = \frac{F}{N} = \frac{599.4}{808.2} = 0.742,\qquad \beta = \tan^{-1}0.742 = 36.6^\circ$$

Shear force and shear stress

$$F_s = F_c\cos\phi - F_t\sin\phi = 900(0.9444) - 450(0.3289) = 850.0 - 148.0 = 702\ \text{N}$$

Shear plane area:

$$A_s = \frac{b\,t_0}{\sin\phi} = \frac{3\times0.25}{0.3289} = 2.28\ \text{mm}^2$$ $$\tau = \frac{F_s}{A_s} = \frac{702}{2.28} = 308\ \text{MPa}$$

Power and specific energy

$$P_c = F_c V = 900\times1.5 = 1350\ \text{W} = 1.35\ \text{kW}$$

Metal removal rate $= b\,t_0\,V = 3\times0.25\times1500 = 1125\ \text{mm}^3/\text{s}$.

$$u = \frac{P_c}{MRR} = \frac{1350}{1125} = 1.2\ \text{J/mm}^3$$

Answer: $\phi = 19.2^\circ$, $F = 599$ N, $N = 808$ N, $\mu = 0.74$, $F_s = 702$ N, $\tau = 308$ MPa, $P = 1.35$ kW, $u = 1.2$ J/mm$^3$.