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GATE 2023 ME – Question 44

Machining and Machine Tool Operations · Mechanics of machining · 2 marks · Multiple choice

In an ideal orthogonal cutting experiment (see figure), the cutting speed V is 1 m/s, the rake angle of the tool $\alpha = 5°$, and the shear angle, $\phi$, 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 $\delta = 25$ µm, what is the value of shear strain rate (in s$^{-1}$) that the material undergoes at the shear zone?

a stationary tool cutting a workpiece that moves at speed V, with the shear plane at the angle $\phi$ and an enlarged view of the shear zone where the plane RS is displaced from PQ by the distance $\delta$.
  1. $1.84 \times 10^4$
  2. $5.20 \times 10^4$
  3. $0.71 \times 10^4$
  4. $1.30 \times 10^4$

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Correct answer: (B) $5.20 \times 10^4$

Explanation

In orthogonal cutting the velocity of shear (the sliding speed along the shear plane) is $V_s = \frac{V\cos\alpha}{\cos(\phi - \alpha)} = \frac{1 \times \cos 5°}{\cos 40°} = \frac{0.9962}{0.7660} = 1.30$ m/s. This sliding speed shears the thin zone of thickness $\delta$, so the shear strain rate is $\dot{\gamma} = \frac{V_s}{\delta} = \frac{1.30}{25 \times 10^{-6}} = 5.2 \times 10^4$ s$^{-1}$.