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 of 10 s$^{-1$. The flow stress of the alloy at the extrusion temperature is given by $\sigma = 10 \, (\dot{\epsilon})^{0.3}$ 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 ............ × 10$^6$ J m$^{-3}$ (rounded off to one decimal place).}

Hint:

In extrusion, the ideal work per unit volume is obtained as flow stress multiplied by true strain $\ln (A_0/A_f)$.

Answer 1

Step 1: Flow stress at given strain rate.
Given $\dot{\epsilon} = 10 \, s^{-1}$. \[ \sigma = 10 (\dot{\epsilon})^{0.3} = 10 (10)^{0.3} \] \[ \sigma = 10 \times 1.995 = 19.95 \, MPa \] Step 2: True strain in extrusion.
True strain in extrusion = $\ln \dfrac{A_0}{A_f}$, where $A_0$ and $A_f$ are initial and final cross-sectional areas. Initial diameter $D_0 = 300$ mm $\Rightarrow A_0 = \pi (D_0/2)^2 = \pi (150)^2 = 70685.8 \, mm^2$. Final diameter $D_f = 75$ mm $\Rightarrow A_f = \pi (D_f/2)^2 = \pi (37.5)^2 = 4417.9 \, mm^2$. \[ \epsilon = \ln \dfrac{70685.8}{4417.9} = \ln (16) = 2.773 \] Step 3: Ideal plastic work per unit volume.
Work per unit volume = flow stress × true strain. \[ W = \sigma \, \epsilon = 19.95 \times 10^6 \times 2.773 \] \[ W = 55.35 \times 10^6 \, J m^{-3} \] Final Answer: \[ \boxed{55.4 \times 10^6 \, J m^{-3}} \]