Discussion
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?
Answer: B
The shear strain rate is \(5.20 \times 10^{4}\ \text{s}^{-1}\).
The strain rate is the shear velocity along the shear plane divided by the thickness of the shear zone. From the velocity triangle of orthogonal cutting,
\[ \begin{aligned} V_{s} &= \frac{V\cos\alpha}{\cos(\phi - \alpha)} \\ &= \frac{\cos 5^{\circ}}{\cos 40^{\circ}} \\ &= \frac{0.9962}{0.7660} \\ & = 1.300\ \text{m/s} \end{aligned} \]
\[ \begin{aligned} \dot{\gamma} &= \frac{V_{s}}{\delta} \\ &= \frac{1.300}{25 \times 10^{-6}} \\ &= 5.20 \times 10^{4}\ \text{s}^{-1} \end{aligned} \]
The distractors reuse numbers from the same calculation. 1.84 is the shear strain, \(\cot\phi + \tan(\phi - \alpha) = 1 + 0.839\), a dimensionless quantity, not a rate. 1.30 is the shear velocity in m/s; dividing it by 100 μm instead of 25 μm would give \(1.30 \times 10^{4}\). 0.71 is \(\cos 45^{\circ}\), which does not give the strain rate on its own.
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