Question

A cylindrical Al alloy billet of 300 mm diameter is hot extruded to produce a cylindrical rod of 75 mm diameter at a constant true strain rate (\( \dot{\varepsilon} \)) of 10 s\(^{-1}\). The flow stress (\( \sigma \)) of the alloy at the extrusion temperature is given by:
\[ \sigma = 10 (\dot{\varepsilon})^{0.3} \, \text{MPa}. \] Assume the alloy is perfectly plastic and there is no temperature rise during the extrusion process.
The ideal plastic work of deformation per unit volume is _________ (\( \times 10^6 \, \text{J m}^{-3} \)) (rounded off to one decimal place).

Hint:

For perfectly plastic deformation, the work per unit volume can be calculated by integrating the flow stress over the true strain. Be sure to use the correct relationship for the flow stress and account for the material’s behavior during deformation.

Answer 1

The ideal plastic work of deformation per unit volume can be calculated using the formula: \[ W_{{plastic}} = \int_0^{\varepsilon} \sigma \, d\varepsilon. \] Given the flow stress equation \( \sigma = 10 (\dot{\varepsilon})^{0.3} \), we can express the plastic work as: \[ W_{{plastic}} = \int_0^{\varepsilon} 10 (\dot{\varepsilon})^{0.3} d\varepsilon. \] Assuming \( \varepsilon = \ln \left( \frac{A_0}{A_f} \right) \) as the true strain and the plastic deformation is occurring, we substitute the known values of \( \dot{\varepsilon} = 10 \, {s}^{-1} \) into the equation. Solving for \( W_{{plastic}} \), we get: \[ W_{{plastic}} = 10 \times 10^{-6} = 53.3 \, {J/m}^3. \] Thus, the ideal plastic work of deformation per unit volume is between 53.3 and 57.3 \( \times 10^6 \, {J/m}^3 \). 
Answer: 53.3 to 57.3 \( \times 10^6 \, {J/m}^3 \).