Consider a turning operation shown in Figure 1 in which a flat form tool with angle $\alpha$ and clearance angle $\delta$ is used to cut a V-groove on a circular job with a groove angle 2$\lambda$.

• As shown in Fig. 7.3.2 the tool rake face will have the same form on the intersection of rake plane XY with the circular V-groove with the job.

• If the width of the groove at the outer periphery is 2b and its depth equal to h.

$tan \ \lambda = \frac{b}{h}$

• In the tool rake plane XY, point Y is at the outer radius of the job and point X is at the root of the groove.

Let, OY = R = Outer radius of the job.

OX = r = Radius at the groove bottom = R – h

$\angle XOY = \theta_1$

$ \angle XYO = \theta_2$

From triangle OXY,

$\frac{R}{sin \ (180 - \alpha)} = \frac{XY}{sin \ \theta_1} = \frac{OX}{sin \ \theta_2}$

Or

$XY = \frac{R \ sin \ \theta_1}{sin \ (180 - \alpha)}$

$= \frac{R \sin \ \theta_1}{sin \ \alpha}$

Again

$sin \ \theta_2 \ = \frac{OX \ sin \ \alpha}{R} = \frac{r}{R} \ sin \ \alpha$

Angle $\alpha = \theta_1 + \theta_2$

The angle between plane perpendicular to XZ and rake face plane is equal to $(\alpha + \delta)$

Therefore distance $d = XY \ cos (\alpha + \delta)$

Again

$tan \ \psi = \frac{b}{d}$

$= \frac{b}{XY \ cos \ (\alpha + \delta)}$

$= \frac{b \ sin \ \alpha}{R \ sin \ \theta_1 \ cos \ (\alpha + \delta)}$

Example: A groove of 90° angle, having a depth of 8 mm is to be turned on a shaft of 150 mm diameter with the help of a form tool. If the tool has a rake angle of 15° and clearance angle 10°, determine the angle ground with tool in a plane perpendicular to the end flank edge.

Solution:

R = 75 mm

$\alpha$ = 15°

$\delta$ = 10°

$2 \lambda $ = 90°

$tan \ \lambda = \ tan \ 45° \ = 1 \ = \ \frac{b}{h}$

h = 8 mm

$\therefore$ b = 8 mm

r = 75 – 8 = 67 mm

If $\psi$ is the semi angle ground on the tool in a plane perpendicular to the end edge of the tool.

$tan \ \psi = \frac{b \ sin \ \alpha}{R \ sin \ \theta_1 \ cos (\alpha + \delta)} $

$= \frac{8 \ sin 15°}{75 \ sin \ \theta_1 \ cos \ (15 + 10)}$

$sin \ \theta_2 = \frac{r}{R} \ sin \ \alpha$

$= \frac{67}{75} \ sin \ 15°$

= 0.2312

$\theta_2 = 13.37°$

$\theta_1 = \alpha - \theta_2 = 15 – 12.37°$

= 1.63°

$tan \ \psi = \frac{8 \times sin \ 15°}{75 \ sin \ 1.63° \ cos \ 25°}$

= 1.071

$\psi = 46.96°$

Angle ground = 2$\psi$ = 93.92°