Question

For an orthogonal cutting operation, tool material is HSS, rake angle is 15° , chip thickness is 0.5 mm, speed is 55 m/min and feed is 0.3 mm/rev. The shear plane angle (in degrees) is

Answer 1

Answer:

The shear plane angle is 34.45°.

Explanation:

Orthogonal cutting is the cutting process in which cutting direction or cutting velocity is perpendicular to the edge of the surface. For orthogonal cutting operation feed is the chip thickness.

Given:

Rake angle is 15°.

Uncut chip thickness is 0.5 mm.

Speed is 55 m/min.

Feed or the chip thickness is 0.3 mm.

Calculation:

Step1

Chip reduction ratio is calculated as follows:

$r=\frac{t_{c}}{t}$

$r=\frac{0.3}{0.5}$

r = 0.6

Step2

Shear angle is calculated as follows:

$tan\phi=\frac{rcos\alpha}{1-rsin\alpha}$

Here, $\phi$ is shear plane angle, r is chip reduction ratio and $\alpha$ is rake angle.

Substitute all the values in the above equation as follows:

$tan\phi=\frac{rcos\alpha}{1-rsin\alpha}$

$tan\phi=\frac{0.6cos15^{\circ}}{1-0.6sin15^{\circ}}$

$tan\phi=\frac{0.57955}{0.8447}$

$\phi=34.45^{\circ}$

Thus, the shear plane angle is 34.45°.