Topic: Machining and Machine Tool Operations
During orthogonal machining with a HSS tool, the rake angle is $5^{\circ}$ un-deformed chip thickness is 0.25 mm and width of cut is 4 mm. Assuming shear strength of work material to be 350 $N/mm²$ and coefficient of friction as 0.5, determine cutting and thrust force.
- A. 1029.4 N, 406.8 N
- B. 1000 N, 500 N
- C. 1110.5 N, 425.8 N
- D. 1025.4 N, 410.8 N
Answer: A. 1029.4 N, 406.8 N
Explanation: Rake angle $\alpha = 5^{\circ}$, $\mu = 0.5$. Friction angle $\beta = \tan^{-1}(0.5) = 26.57^{\circ}$. Using Merchant's theory, the shear angle is $\phi = 45^{\circ} + \frac{\alpha}{2} - \frac{\beta}{2} = 45^{\circ} + 2.5^{\circ} - 13.285^{\circ} = 34.22^{\circ}$. The shear force $F_s = \tau_s \times \frac{b t_c}{\sin\phi} = 350 \times \frac{4 \times 0.25}{\sin(34.22^{\circ})} = 622.37$ N. The cutting force is $F_c = \frac{F_s \cos(\beta-\alpha)}{\cos(\phi+\beta-\alpha)} = \frac{622.37 \cos(26.57^{\circ}-5^{\circ})}{\cos(34.22^{\circ}+26.57^{\circ}-5^{\circ})} = 1029.45$ N. The thrust force is $F_T = F_c \tan(\beta-\alpha) = 1029.45 \tan(26.57^{\circ}-5^{\circ}) = 406.96$ N. Hence, option (a) is correct.