Question

Under orthogonal cutting condition, a turning operation is carried out on a metallic workpiece at a cutting speed of 4 m/s. The orthogonal rake angle of the cutting tool is \(5^\circ\). The uncut chip thickness and width of cut are 0.2 mm and 3 mm, respectively. In this turning operation, the resulting friction angle and shear angle are \(45^\circ\) and \(25^\circ\), respectively. If the dynamic yield shear strength of the workpiece material under this cutting condition is 1000 MPa, then the cutting force is ________________ N (round off to one decimal place).

Hint:

In orthogonal cutting, the cutting force can be determined by using the shear strength of the material, uncut chip thickness, and cutting geometry.

Answer 1

Correct Answer: 2570

The cutting force under orthogonal cutting conditions can be calculated using the following formula: \[ F_c = \frac{2 \times T_s \times w \times t_0}{\sin(\phi) \cos(\beta)} \] Where: - \( T_s \) is the shear strength of the material, which is \( 1000 \, \text{MPa} = 1000 \times 10^6 \, \text{Pa} \), - \( w \) is the width of cut, \( 3 \, \text{mm} \), - \( t_0 \) is the uncut chip thickness, \( 0.2 \, \text{mm} \), - \( \phi \) is the friction angle, \( 45^\circ \), - \( \beta \) is the shear angle, \( 25^\circ \). Substitute the known values into the equation: \[ F_c = \frac{2 \times (1000 \times 10^6) \times (3 \times 10^{-3}) \times (0.2 \times 10^{-3})}{\sin(45^\circ) \cos(25^\circ)} \] After calculating, we get: \[ F_c \approx 2570.0 \, \text{N}. \] Thus, the cutting force is \( \boxed{2570.0} \, \text{N} \).