During orthogonal turning by using single Al, cutting tool, F = 0.24 mm/rev, uncut chip = 0.23 mm, ϕ = 20°, tangential force component = 800 N, radial force = 150 N. The value of Shear force is _______ N.

The correct answer is 700.

Step-by-Step Explanation:

In orthogonal turning, the shear force ($F_s$) can be calculated from the cutting force components using Merchant's force circle relationship. The formula for the shear force is:

$$F_s=F_c\cos \varphi -F_t\sin \varphi$$

Where:
• $F_c$ is the tangential cutting force component = 800 N
• $F_t$ is the thrust force component (represented by the radial force in this setup) = 150 N
• $\varphi$ is the shear angle = 20°

Substituting the given values into the shear force equation:

$$F_s=800\times \cos (20°)-150\times \sin (20°)$$

Using the values of the trigonometric functions:
• $\cos (20°)\approx 0.9397$
• $\sin (20°)\approx 0.3420$

Now, calculate each component:

$$800\times 0.9397=751.76\text{N}$$

$$150\times 0.3420=51.30\text{N}$$

Subtracting the thrust component from the cutting component to obtain the shear force:

$$F_s=751.76-51.30=700.46\text{N}$$

Rounding to the nearest integer, we find:

$$F_s\approx 700\text{N}$$