In a turning process using orthogonal tool geometry, a chip length of 100 mm is obtained for an uncut chip length of 250 mm.

The cutting conditions are cutting speed = 30 m/min. rake angle = 20°.

The shear plane angle is _________ degrees (round off to one decimal place).

The correct answer is 23.5.

Step-by-Step Explanation:

In orthogonal cutting, the chip thickness ratio (also known as cutting ratio), denoted by r, can be calculated using the length of the chip and the uncut chip length. By definition:
$$r=\frac{L_c}{L}$$
where:
- $L_c$ is the length of the chip = 100 mm
- $L$ is the uncut chip length = 250 mm

Substituting these values:
$$r=\frac{100}{250}=0.4$$

The relationship between the shear plane angle ($\varphi$), the rake angle ($\alpha$), and the chip thickness ratio ($r$) is given by the following formula:
$$\tan \varphi =\frac{r\cos \alpha }{1-r\sin \alpha }$$

Given the rake angle $\alpha ={20}^{\circ }$, we substitute the values of $r$ and $\alpha$ into the formula:
$$\tan \varphi =\frac{0.4\cos \left({20}^{\circ }\right)}{1-0.4\sin \left({20}^{\circ }\right)}$$

Now, calculating the trigonometric values:
- $\cos \left({20}^{\circ }\right)\approx 0.93969$
- $\sin \left({20}^{\circ }\right)\approx 0.34202$

Substitute these values back into the equation:
$$\tan \varphi =\frac{0.4\times 0.93969}{1-0.4\times 0.34202}=\frac{0.37588}{1-0.13681}=\frac{0.37588}{0.86319}\approx 0.4354$$

To find the shear plane angle $\varphi$, we take the arctangent of 0.4354:
$$\varphi ={\tan }^{-1}\left(0.4354\right)\approx {23.53}^{\circ }$$

Rounding off to one decimal place, the shear plane angle is 23.5 degrees.