Discussion
Details pertaining to an orthogonal metal cutting process are given below.
| Chip thickness ratio | 0.4 |
| Undeformed thickness | 0.6 mm |
| Rake angle | +10° |
| Cutting speed | 2.5 m/s |
| Mean thickness of primary shear zone | 25 microns |
The shear strain rate in s-1 during the process is
Answer: C
The shear strain rate is about \(1.0104\times10^{5}\) s-1.
First the shear angle, from the chip thickness ratio r = 0.4 and rake angle 10°:
\[ \begin{aligned} \tan\phi &= \frac{r\cos\alpha}{1-r\sin\alpha} = \frac{0.3939}{0.9305} \\ &= 0.4233,\ \phi = 22.94^{\circ} \end{aligned} \]
The shear velocity along the shear plane is:
\[ \begin{aligned} V_{s} &= \frac{V\cos\alpha}{\cos(\phi-\alpha)} = \frac{2.5(0.9848)}{\cos 12.94^{\circ}} \\ &= 2.526\ \text{m/s} \end{aligned} \]
Dividing by the shear zone thickness of 25 microns gives the strain rate:
\[ \begin{aligned} \dot{\gamma} &= \frac{2.526}{25\times10^{-6}} \\ &= 1.0104\times10^{5}\ \text{s}^{-1} \end{aligned} \]
Dividing the cutting speed itself by the zone thickness gives exactly \(1.0\times10^{5}\), near but not equal to the keyed value; the velocity that slides across the shear zone is the shear velocity, not the cutting speed. The other three options do not follow from these data.
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