Question

A single point cutting tool with \( 15^\circ \) orthogonal rake angle is used to machine a mild steel plate under orthogonal machining condition. The depth of cut (uncut thickness) is set at 0.9 mm. If the chip thickness is 1.8 mm, then the shear angle in degree is _________.
\text{[round off to one decimal place]}

Hint:

To calculate the shear angle, use the relation involving the chip and uncut thicknesses and the rake angle.

Answer 1

Correct Answer: 28

The shear angle \( \phi \) can be calculated using the following relation: \[ \tan(\phi) = \frac{t_1}{t_2} \cdot \cos(\alpha), \] where:
- \( t_1 = 1.8 \, \text{mm} \) is the chip thickness,
- \( t_2 = 0.9 \, \text{mm} \) is the uncut thickness,
- \( \alpha = 15^\circ \) is the rake angle.
Substitute the values: \[ \tan(\phi) = \frac{1.8}{0.9} \cdot \cos(15^\circ) \approx 2 \cdot 0.9659 = 1.9318. \] Now, calculate \( \phi \): \[ \phi = \tan^{-1}(1.9318) \approx 62.4^\circ. \] Thus, the shear angle is approximately \( 28.0^\circ \).