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 option is 5.20 × 10⁴.

Analysis of the Given Data and Diagram:
Based on the provided cutting model diagram, we can identify several key parameters of the orthogonal metal cutting process:
1. The stationary tool has a rake angle, $\alpha =5^{\circ }$.
2. The work material moves with a cutting velocity, $V=1{\text{m/s}}^{}$.
3. The shear angle (angle of the shear plane PQ with respect to the horizontal direction of cutting velocity) is $\varphi ={45}^{\circ }$.
4. The perpendicular distance between the shear planes PQ and RS (representing the thickness of the thin shear zone) is shown in the zoomed inset diagram as $\delta =25 \mu \text{m}=25\times {10}^{-6}\text{m}$.

Step-by-Step Derivation:

Step 1: Calculate the Shear Velocity ($V_s$)
The shear velocity is the velocity of the chip relative to the work material along the shear plane. Using the velocity triangle relation from Merchant's orthogonal cutting model, the shear velocity is given by:

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

Substituting the given values into the equation:

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

Calculating the trigonometric values:

$$\cos \left(5^{\circ }\right)\approx 0.9962$$

$$\cos \left({40}^{\circ }\right)\approx 0.7660$$

Substituting these back to find the shear velocity:

$$V_s\approx \frac{0.9962}{0.7660}\approx 1.30{\text{m/s}}^{}$$

Step 2: Calculate the Shear Strain Rate ($\dot{\gamma }$)
The shear strain rate is defined as the ratio of the shear velocity to the thickness of the shear zone ($\delta$):

$$\dot{\gamma }=\frac{V_s}{\delta }$$

Substituting the values of $V_s$ and $\delta$:

$$\dot{\gamma }=\frac{1.30}{25\times {10}^{-6}}$$

$$\dot{\gamma }=0.0520\times {10}^6=5.20\times {10}^4{\text{s}}^{-1}$$

Thus, the shear strain rate that the material undergoes at the shear zone is $5.20\times {10}^4{\text{s}}^{-1}$.