Question

In an ideal orthogonal cutting experiment (see figure), the cutting speed V is 1 m/s, the rake angle of the tool \(\alpha = 5^\circ\), and the shear angle, \(\phi\), is known to be \(45^\circ\). 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 \(\delta = 25 \, \mu\text{m}\), what is the value of shear strain rate (in \(s^{-1}\)) that the material undergoes at the shear zone? 

 

• A. \(1.84 \times 10^4\)
• B. \(5.20 \times 10^4\)
• C. \(0.71 \times 10^4\)
• D. \(1.30 \times 10^4\)

Hint:

For metal cutting problems, it is crucial to remember the velocity relationships derived from the velocity triangle: - Shear velocity: \(V_s = V \frac{\cos \alpha}{\cos(\phi-\alpha)}\) - Chip velocity: \(V_c = V \frac{\sin \phi}{\cos(\phi-\alpha)}\) The shear strain rate is then simply \(\dot{\gamma} = V_s / t_s\), where \(t_s\) is the thickness of the shear zone.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The shear strain rate (\(\dot{\gamma}\)) in metal cutting is defined as the rate at which shear deformation occurs in the shear zone. It is given by the ratio of the shear velocity (\(V_s\)) to the thickness of the shear zone (\(t_s\)).
Step 2: Key Formula or Approach:
1. The shear strain rate is given by \(\dot{\gamma} = \frac{V_s}{t_s}\).
2. The thickness of the shear zone, \(t_s\), can be related to the perpendicular distance \(\delta\) given in the problem. From the geometry of the shear zone, \(t_s = \delta / \sin \phi\). However, the problem states "perpendicular distance between PQ and RS is \(\delta\)", which is often interpreted as the shear zone thickness itself, i.e., \(t_s = \delta\). Let's use a more standard formula.
3. The standard formula for shear strain rate in orthogonal cutting is: \[ \dot{\gamma} = \frac{V_s}{\text{thickness of shear zone}} = \frac{\cos \alpha}{\cos(\phi - \alpha)} \frac{V}{\delta \cdot \csc \phi} = \frac{V_s}{t_s} \] A more common and direct formula derived from the velocity triangle is: \[ \dot{\gamma} = \frac{V_s}{t_s} = \frac{V \frac{\cos \alpha}{\cos(\phi - \alpha)}}{t_s} \] From fundamental principles, shear strain \(\gamma = \cot\phi + \tan(\phi-\alpha)\). The time taken to cross the shear zone is \(t = t_s / V_s\), where \(V_s\) is shear velocity.
\(\dot{\gamma} = \gamma/t\) is not correct.
The widely accepted formula for shear strain rate is: \[ \dot{\gamma} = \frac{V_s}{t_s} \] where \(V_s\) is the shear velocity and \(t_s\) is the thickness of the primary shear zone. From the velocity triangle in orthogonal cutting: \[ \frac{V}{\sin(90 - (\phi-\alpha))} = \frac{V_s}{\sin(90-\alpha)} \implies V_s = V \frac{\cos\alpha}{\cos(\phi-\alpha)} \] The problem gives the perpendicular distance \(\delta\) between the planes, which represents the thickness of the shear zone, \(t_s\). So, \(t_s = \delta\). Step 3: Detailed Calculation:
Given:
- Cutting speed, \(V = 1\) m/s
- Rake angle, \(\alpha = 5^\circ\)
- Shear angle, \(\phi = 45^\circ\)
- Shear zone thickness, \(t_s = \delta = 25 \, \mu\text{m} = 25 \times 10^{-6}\) m
1. Calculate the Shear Velocity (\(V_s\)): \[ V_s = V \frac{\cos\alpha}{\cos(\phi-\alpha)} = 1 \times \frac{\cos(5^\circ)}{\cos(45^\circ - 5^\circ)} = \frac{\cos(5^\circ)}{\cos(40^\circ)} \] \[ V_s = \frac{0.9962}{0.7660} \approx 1.3005 \text{ m/s} \] 2. Calculate the Shear Strain Rate (\(\dot{\gamma}\)): \[ \dot{\gamma} = \frac{V_s}{t_s} = \frac{1.3005 \text{ m/s}}{25 \times 10^{-6} \text{ m}} \] \[ \dot{\gamma} = \frac{1.3005}{25} \times 10^6 \text{ s}^{-1} = 0.05202 \times 10^6 \text{ s}^{-1} = 5.202 \times 10^4 \text{ s}^{-1} \] Step 4: Final Answer:
The shear strain rate is \(5.20 \times 10^4\) s⁻¹.
Step 5: Why This is Correct:
The solution uses the standard and well-established formulas for shear velocity and shear strain rate in the ideal orthogonal cutting model. The values are plugged in correctly, and the calculation yields a result that matches one of the options.