GATE 2022 ME (ME2) – Question 55
In an orthogonal machining operation, the cutting and thrust forces are equal in magnitude. The uncut chip thickness is 0.5 mm and the shear angle is 15°. The orthogonal rake angle of the tool is 0° and the width of cut is 2 mm. The workpiece material is perfectly plastic and its yield shear strength is 500 MPa. The cutting force is ______ N (round off to the nearest integer).
Practise this question in The GATE Grind →
Show answer and explanation
Correct answer: 2732
Explanation
**Orthogonal cutting with zero rake.** The resultant force $R$ is resolved into the cutting force $F_c$ (along the velocity) and the thrust $F_t$. At the shear plane (shear angle $\phi=15^\circ$) the shear force $F_s$ is
$$F_s=F_c\cos\phi-F_t\sin\phi .$$
**Step 1: use $F_t=F_c$.**
$$F_s=F_c(\cos15^\circ-\sin15^\circ).$$
**Step 2: shear force from the shear strength.** The shear plane area is $A_s=\dfrac{t_0\,w}{\sin\phi}$:
$$F_s=\tau_s\frac{t_0w}{\sin\phi}=500\times\frac{0.5\times2}{\sin15^\circ}=\frac{500}{0.2588}=1931.9\text{ N}.$$
**Step 3: cutting force.**
$$F_c=\frac{1931.9}{\cos15^\circ-\sin15^\circ}=\frac{1931.9}{0.9659-0.2588}=\frac{1931.9}{0.7071}=\mathbf{2732\ N}.$$