Question

An orthogonal cutting operation is performed using a single point cutting tool with a rake angle of 12° on a lathe. During turning, the cutting force and the friction force are 1000 N and 600 N, respectively. If the chip thickness and the uncut chip thickness during turning are 1.5 mm and 0.75 mm, respectively, then the shear force is ______ N (round off to one decimal place).

Hint:

In orthogonal cutting, the shear force can be calculated using the relation between cutting force and chip thicknesses. The shear force depends on the geometry of the cutting process.

Answer 1

Correct Answer: 625

The shear force \( F_s \) in orthogonal cutting is given by the relation: \[ F_s = \frac{F_t \cdot t_c}{t_u}, \] where:
- \( F_t = 1000 \, \text{N} \) is the cutting force,
- \( t_c = 1.5 \, \text{mm} \) is the chip thickness,
- \( t_u = 0.75 \, \text{mm} \) is the uncut chip thickness.
Substituting the values: \[ F_s = \frac{1000 \times 1.5}{0.75} = 2000 \, \text{N}. \] Thus, the shear force is: \[ \boxed{625 \, \text{to} \, 750 \, \text{N}}. \]