In a 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 m/min, depth of cut, t = 2.0 mm, feed, 𝒔_𝟎 = 0.1 mm/revolution and chip velocity, V = 20 m/min, the shear strength, 𝝉_𝒔 of the material will be ___________________ MPa (round off to two decimal places).

The correct answer is 391.89.

1. Identification of Given Parameters from the Question and Image:
Based on the problem statement and the provided diagram:

• Rake angle, _{$\alpha =0$}
• Shear force, _{$F_s=400\text{N}$}
• Cutting velocity, _{$V_c=100\text{m/min}$}
• Depth of cut, which corresponds to the width of cut in orthogonal turning: _{$b=2.0\text{mm}$}
• Feed, which corresponds to the uncut chip thickness in orthogonal turning: _{$t=0.1\text{mm/rev}$}
• Chip velocity (denoted as _{$V_f$} in the image diagram): _{$V_f=20\text{m/min}$}

2. Finding the Shear Plane Angle (_$\varphi$):
The velocity triangle shown in the image represents the relationship between the cutting velocity (V_c), chip velocity (V_f), and shear velocity (V_s), with angles 90 - α, ϕ, and 90 - (ϕ - α). From the sine rule applied to this velocity triangle:

$$\frac{V_c}{\cos (\varphi -\alpha )}=\frac{V_f}{\sin \left(\varphi \right)}$$

Substituting _{$\alpha =0$} into this equation simplifies the relation to:

$$\frac{V_c}{\cos \left(\varphi \right)}=\frac{V_f}{\sin \left(\varphi \right)}$$

Rearranging the terms to find _{$\tan \left(\varphi \right)$}:

$$\tan \left(\varphi \right)=\frac{V_f}{V_c}=\frac{20}{100}=0.2$$

Taking the inverse tangent:

$$\varphi ={\tan }^{-1}\left(0.2\right)\approx 11.3°$$

3. Calculation of Shear Strength (_{${\tau }_s$}):
The shear strength of the material is defined as the shear force per unit area of the shear plane:

$${\tau }_s=\frac{F_s}{A_s}$$

Where the shear plane area _{$A_s$} is given by:

$$A_s=\frac{b\times t}{\sin \left(\varphi \right)}$$

Substituting _{$A_s$} back into the shear strength equation yields:

$${\tau }_s=\frac{F_s}{b\times t}\sin \left(\varphi \right)$$

Substituting the values:

$${\tau }_s=\frac{400}{2.0\times 0.1}\sin (11.3°)$$

$${\tau }_s=2000\times 0.195946\approx 391.89\text{MPa}$$

Rounding to two decimal places, we get the shear strength of the material to be 391.89 MPa.