Cold upsetting force with friction

Open-die forging (upsetting) reduces the height and increases the cross-section; volume is constant.

(a) Final diameter

$$\frac{\pi}{4}d_0^2 h_0 = \frac{\pi}{4}d^2 h \quad\Rightarrow\quad d = d_0\sqrt{\frac{h_0}{h}} = 40\sqrt{\frac{60}{30}} = 56.57\ \text{mm}$$

(b) True strain

$$\varepsilon = \ln\frac{h_0}{h} = \ln 2 = 0.693$$

(c) Force at end of stroke

Flow stress at the final strain:

$$Y_f = K\varepsilon^{n} = 500\,(0.693)^{0.25} = 456.2\ \text{MPa}$$

Final contact area:

$$A = \frac{\pi}{4}(56.57)^2 = 2513\ \text{mm}^2$$

Shape factor for friction:

$$K_f = 1 + \frac{0.4\,\mu\,D}{h} = 1 + \frac{0.4\times0.2\times56.57}{30} = 1.151$$

Force:

$$F = K_f\,Y_f\,A = 1.151\times 456.2\times 2513 = 1.32\times10^{6}\ \text{N}$$

Without friction the force would be $456.2\times2513 = 1.147$ MN. Friction therefore adds about 15 percent. The average die pressure is $K_f Y_f = 525$ MPa.

(d) Ideal work of deformation

Initial volume: $V = \frac{\pi}{4}(40)^2(60) = 75\,398\ \text{mm}^3$.

Average flow stress: $\bar{Y}_f = \dfrac{K\varepsilon^{n}}{1+n} = \dfrac{456.2}{1.25} = 365.0$ MPa.

$$W = \bar{Y}_f\, V\,\varepsilon = 365.0\times 75\,398\times 0.693 = 1.907\times10^{7}\ \text{N\,mm} = 19.1\ \text{kJ}$$

(Actual work is larger because of friction and redundant work.)

Answer: final diameter $= 56.6$ mm, true strain $= 0.693$, forging force $= 1.32$ MN, ideal work $= 19.1$ kJ.