Wire drawing stress, force and power

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 $429/596 = 0.72$ leaves a reasonable safety margin, so a somewhat larger reduction in this pass would still be possible.

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