Bulk Deformation Process

Hot working is plastic deformation above the recrystallisation temperature (about $0.6T_m$ in kelvin), and cold working is deformation below it (usually near room temperature), so the metal strain hardens.

Differences

Effect of temperature

Flow stress falls as temperature rises because thermal softening and recrystallisation occur. Above the recrystallisation temperature the strain-hardening exponent is nearly zero, and forming load reduces by a large factor. Too high a temperature causes burning or grain growth.

Effect of strain rate

Flow stress depends on strain rate $\dot{\varepsilon}$ as

$$\sigma = C\,\dot{\varepsilon}^{\,m}$$

where $m$ is the strain-rate sensitivity exponent: about 0.01 or less for cold working, 0.05-0.4 for hot working. In hot working a faster press or hammer therefore needs a higher load. Strain rate in forging $=v/h$ (velocity of ram divided by instantaneous height).

Effect of friction

• Friction between work and die opposes the flow at the interface; it raises the forming force and power.
• In upsetting, it causes barrelling and a "friction hill" of pressure, highest at the centre.
• In rolling, friction pulls the metal into the roll gap, but too much friction raises power and roll wear. The condition for entry is $\mu \ge \tan\alpha$.
• It is controlled by lubricants (graphite, oil, glass) and by die finish.

Friction therefore has most influence when the area of contact is large compared with the thickness, as in flat rolling of thin sheets and forging of thin discs.

Given: $t_0 = 25$ mm, $t_f = 20$ mm, $w = 300$ mm, $R = 300$ mm, $N = 100$ rpm, $\mu = 0.15$, $\bar{Y}_f = 200$ MPa.

Draft actually applied: $d = t_0 - t_f = 5$ mm.

(a) Maximum possible draft

The rolls can bite the strip if $\mu \ge \tan\alpha$, where $\alpha$ is the angle of bite. The maximum draft is

$$d_{max} = \mu^2 R = (0.15)^2\times 300 = 6.75\ \text{mm}$$

The applied draft (5 mm) is less than 6.75 mm, so the pass is possible.

(b) True strain

$$\varepsilon = \ln\frac{t_0}{t_f} = \ln\frac{25}{20} = 0.223$$

(c) Contact length

$$L = \sqrt{R\,(t_0 - t_f)} = \sqrt{300\times 5} = 38.73\ \text{mm}$$

(d) Roll force

The force is the average flow stress times the contact area:

$$F = \bar{Y}_f\, w\, L = 200\times 300\times 38.73 = 2.324\times10^{6}\ \text{N} = 2.32\ \text{MN}$$

(The effect of friction on the force is neglected; it would raise it by a few percent.)

(e) Torque and power

The resultant force acts at the middle of the contact length, so the arm is $L/2$:

$$T = F\,\frac{L}{2} = 2.324\times10^{6}\times 0.01937 = 45.0\ \text{kN\,m per roll}$$

For two rolls the power is

$$P = 2\times T\,\omega = 2\times 45.0\times10^{3}\times \frac{2\pi\times100}{60} = 9.42\times10^{5}\ \text{W}$$

Answer: $d_{max} = 6.75$ mm, $\varepsilon = 0.223$, $L = 38.7$ mm, $F = 2.32$ MN, $T = 45$ kN m per roll, $P = 942$ kW.

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.

In drawing, the wire is pulled through a die; the drawing force is limited by the strength of the wire after it leaves the die.

(a) Area reduction and true strain

$$r = 1 - \left(\frac{8}{10}\right)^2 = 0.36 \;(36\%),\qquad \varepsilon = \ln\frac{A_0}{A_f} = \ln\left(\frac{10}{8}\right)^2 = 0.4463$$

(b) Drawing stress

Average flow stress:

$$\bar{Y}_f = \frac{K\varepsilon^n}{1+n} = \frac{700\,(0.4463)^{0.2}}{1.2} = \frac{596.0}{1.2} = 496.4\ \text{MPa}$$

Die contact length and mean diameter:

$$L_c = \frac{D_0 - D_f}{2\sin\alpha} = \frac{10-8}{2\sin6^\circ} = 9.567\ \text{mm},\qquad D = \frac{10+8}{2} = 9\ \text{mm}$$

Shape factor for redundant work:

$$\phi = 0.88 + 0.12\,\frac{9}{9.567} = 0.993$$

Friction term: $1 + \dfrac{0.1}{\tan 6^\circ} = 1 + \dfrac{0.1}{0.1051} = 1.9515$

$$\sigma_d = 496.4\times1.9515\times0.993\times0.4463 = 429.3\ \text{MPa}$$

(c) Drawing force

Area of the drawn wire: $A_f = \dfrac{\pi}{4}(8)^2 = 50.27\ \text{mm}^2$

$$F = \sigma_d A_f = 429.3\times 50.27 = 21\,580\ \text{N} = 21.6\ \text{kN}$$

(d) Power

$$P = F\,v = 21.58\times 1.5 = 32.4\ \text{kW}$$

Check of fracture

The wire leaving the die has been strain hardened to a flow stress of $K\varepsilon^n = 596$ MPa. Since the drawing stress (429 MPa) is lower than 596 MPa, the drawn wire will not yield or break. The ratio $429/596 = 0.72$ shows that further reduction in this pass is possible but with less margin.

Answer: area reduction $= 36\%$, drawing stress $= 429$ MPa, drawing force $= 21.6$ kN, power $= 32.4$ kW; the wire is safe.