In an ideal orthogonal cutting experiment (see figure), the cutting speed V is 1 m/s, the rake angle of the tool α = 5°, and the shear angle, 𝜙, is known to be 45°.

Applying the ideal orthogonal cutting model, consider two shear planes PQ and RS close to each other. As they approach the thin shear zone (shown as a thick line in the figure), plane RS gets sheared with respect to PQ (point R1 shears to R2, and S1 shears to S2).

Assuming that the perpendicular distance between PQ and RS is 𝛿 = 25 μm, what is the value of shear strain rate (in s^-1 ) that the material undergoes at the shear zone?

The correct answer is the second option: 5.20 × 10⁴.

1. Identify the given parameters from the problem description and the figure:
• Cutting speed ($V$) = $1\text{m/s}$
• Tool rake angle ($\alpha$) = $5^{\circ }$
• Shear angle ($\varphi$) = ${45}^{\circ }$
• Perpendicular distance between adjacent shear planes PQ and RS, representing the shear zone thickness ($\delta$) = $25\text{μm}=25\times {10}^{-6}\text{m}$

2. Formulate the relationship for Shear Strain Rate:
In the ideal orthogonal cutting model, the shear strain rate ($\stackrel{·}{\gamma }$) in the shear zone is defined as the ratio of the shear velocity ($V_s$) to the thickness of the shear zone ($\delta$):

$$\stackrel{·}{\gamma }=\frac{V_s}{\delta }$$

3. Calculate the Shear Velocity ($V_s$):
Using the velocity relations from orthogonal cutting geometry:

$$V_s=\frac{V\cos \alpha }{\cos (\varphi -\alpha )}$$

Substitute the given values into the equation:

$$V_s=\frac{1\times \cos \left(5^{\circ }\right)}{\cos ({45}^{\circ }-5^{\circ })}=\frac{\cos \left(5^{\circ }\right)}{\cos \left({40}^{\circ }\right)}$$

Using the trigonometric values:
• $\cos \left(5^{\circ }\right)\approx 0.9962$
• $\cos \left({40}^{\circ }\right)\approx 0.7660$

$$V_s\approx \frac{0.9962}{0.7660}\approx 1.3004\text{m/s}$$

4. Compute the Shear Strain Rate ($\stackrel{·}{\gamma }$):

$$\stackrel{·}{\gamma }=\frac{1.3004}{25\times {10}^{-6}}\approx 5.20\times {10}^4{\text{s}}^{-1}$$