Question

In a machining operation that approximates orthogonal cutting, the cutting tool has a rake angle of 10°. The chip thickness before the cut \(t_0 = 0.50\) mm and the chip thickness after the cut \(t_c = 1.125\) mm. The shear plane angle \(\phi\) is ——–––––––––.
Assuming \(\sin 10^\circ = 0.173\) and \(\cos 10^\circ = 0.984\).

• A. 0.008°
• B. 25.4°
• C. 50.4°
• D. 20.5°

Hint:

For orthogonal cutting operations, knowing the shear plane angle is crucial to understanding chip formation and optimizing cutting conditions.

Answer 1

The Correct Option is C

The shear plane angle \(\phi\) can be calculated using the formula: \[ \tan \phi = \frac{t_0}{t_c - t_0} \times \frac{\cos \alpha}{\sin \alpha} \] Where: - \(\alpha = 10^\circ\) (rake angle) - \(t_0 = 0.50 \, \text{mm}\) (chip thickness before the cut) - \(t_c = 1.125 \, \text{mm}\) (chip thickness after the cut) Substitute the values into the equation: \[ \tan \phi = \frac{0.50}{1.125 - 0.50} \times \frac{0.984}{0.173} \] Solving this gives: \[ \phi = 50.4^\circ \]