Question

In pure orthogonal turning by a zero rake angle single point carbide cutting tool, the shear force has been computed to be 400 N. The cutting velocity, \( V_c = 100 \ \text{m/min} \), depth of cut, \( t = 2.0 \ \text{mm} \), feed, \( s_0 = 0.1 \ \text{mm/revolution} \) and chip velocity, \( V_f = 20 \ \text{m/min} \), the shear strength \( \tau_s \) of the material will be \(\underline{\hspace{2cm}}\) MPa (round off to two decimal places).

Hint:

Shear force in turning operations is directly proportional to the cutting area and the shear strength of the material.

Answer 1

Correct Answer: 388 - 400

Step 1: Calculate the Chip Thickness Ratio

The chip thickness ratio is:

$$r = \frac{t}{t_c}$$

where $t_c$ is the chip thickness.

Using the velocity relationship:

$$r = \frac{V_f}{V_c} = \frac{20}{100} = 0.2$$

Step 2: Calculate the Chip Thickness

$$t_c = \frac{t}{r} = \frac{2.0}{0.2} = 10 \text{ mm}$$

Step 3: Calculate the Shear Angle

For zero rake angle ($\alpha = 0°$), using the relationship:

$$\tan \phi = \frac{r \cos \alpha}{1 - r \sin \alpha} = \frac{r}{1} = 0.2$$

$$\phi = \tan^{-1}(0.2) = 11.31°$$

Step 4: Calculate the Shear Area

The shear area is:

$$A_s = \frac{t \times s_0}{\sin \phi}$$

$$A_s = \frac{2.0 \times 0.1}{\sin(11.31°)}$$

$$A_s = \frac{0.2}{0.196} = 1.020 \text{ mm}^2$$

Step 5: Calculate the Shear Strength

The shear strength is:

$$\tau_s = \frac{F_s}{A_s}$$

$$\tau_s = \frac{400}{1.020} = 392.16 \text{ N/mm}^2$$

$$\tau_s = 392.16 \text{ MPa}$$

Answer: The shear strength of the material is 392.16 MPa.