Flow curve constants K and n

The flow curve is the plastic part of the true stress-true strain curve: $\sigma = K\varepsilon^{n}$, where K is the strength coefficient (MPa) and n the strain-hardening exponent.

Finding n and K

Divide the two readings:

$$\frac{450}{300} = \left(\frac{0.20}{0.05}\right)^n \quad\Rightarrow\quad 1.5 = 4^n$$ $$n = \frac{\ln 1.5}{\ln 4} = \frac{0.4055}{1.3863} = 0.2925$$ $$K = \frac{300}{(0.05)^{0.2925}} = \frac{300}{0.4164} = 720.5\ \text{MPa}$$

Flow stress at $\varepsilon = 0.40$

$$\sigma = 720.5\,(0.40)^{0.2925} = 720.5\times 0.7649 = 551.1\ \text{MPa}$$

Average flow stress from 0 to 0.40

$$\bar{Y}_f = \frac{K\varepsilon_{max}^{\,n}}{1+n} = \frac{551.1}{1.2925} = 426.4\ \text{MPa}$$

This average value is used to compute forming forces and work in rolling, drawing and forging, because the metal strain hardens during the process.

Meaning of n

• n is the slope of the flow curve on a log-log plot. It lies between 0 (perfectly plastic) and 1 (perfectly elastic).
• A large n (about 0.2-0.5, e.g. annealed copper, stainless steel) means strong strain hardening; the metal stretches uniformly and is good for stretch forming and deep drawing.
• A small n (about 0.05-0.15, e.g. cold-worked steel) means little hardening and early necking.
• In a tensile test, the true strain at the onset of necking equals n.

Answer: $n = 0.2925$, $K = 720.5$ MPa, flow stress at 0.40 strain $= 551$ MPa, average flow stress $= 426$ MPa.