Question 7 of 8: Plastic Instability (Considère's Criterion); Stretch-Forming Spring-Back of a Magnesium Sheet

Question 7: Plastic Instability (Considère's Criterion); Stretch-Forming Spring-Back of a Magnesium Sheet (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

7.1 — (a) Deriving the Considère necking condition ε = n

Given. A material work-hardens according to $\sigma = K\epsilon^n$ in true stress and true strain, deformed at constant volume in uniaxial tension.

Find. Show that plastic instability (necking) begins at the true strain $\epsilon = n$.

Approach. Necking begins where the load-carrying capacity of the bar reaches a maximum, since beyond that point any further stretch reduces the cross-section faster than the material can strain-harden to compensate, so deformation localises. That maximum-load condition is Considère's criterion, and combining it with the given hardening law fixes the strain at which it occurs.

1. Write the load and find its stationary point. The axial load is $P = \sigma A$, where $A$ is the current cross-sectional area. Instability (maximum load, $dP = 0$) occurs where $$dP = A\,d\sigma + \sigma\,dA = 0 \quad\Longrightarrow\quad \frac{d\sigma}{\sigma} = -\frac{dA}{A}.$$
2. Bring in constant-volume flow. Plastic deformation conserves volume, $A L = A_0 L_0 = \text{const}$, so $dA/A = -dL/L = -d\epsilon$ (true strain $\epsilon = \ln(L/L_0)$, $d\epsilon = dL/L$). Substituting, $$\frac{d\sigma}{\sigma} = -(-d\epsilon) = d\epsilon \quad\Longrightarrow\quad \boxed{\dfrac{d\sigma}{d\epsilon} = \sigma} \quad \text{(Consid\`ere's criterion).}$$ Necking begins exactly where the slope of the true stress–true strain curve equals the current true stress — geometrically, where a line from the origin at strain $\epsilon=-1$ is tangent to the curve.
3. Apply it to the given power law. With $\sigma = K\epsilon^n$, the hardening rate is $d\sigma/d\epsilon = Kn\epsilon^{n-1}$. Setting this equal to $\sigma$ itself, $$Kn\epsilon^{n-1} = K\epsilon^{n} \quad\Longrightarrow\quad n = \epsilon.$$

Physically, $n$ is therefore not just a curve-fitting exponent: it is the uniform (pre-necking) true strain the material can sustain in a tension test, and a metal with a higher $n$ can be drawn or stretch-formed further before it localises — the property exploited deliberately in part (b), and referenced again when comparing metal and polymer deformation in Question 6(a).

7.2 — (b) Over-stretch required to reach 6.2 m after spring-back

Given. Magnesium sheet, 1.5 mm thick $\times$ 80 mm wide, original length $L_0 = 5$ m, target length after the stress is released $L_f = 6.2$ m, $E = 65$ GPa, $\sigma_y = 200$ MPa. No hardening data is supplied, so the sheet is modelled as elastic–perfectly-plastic: once yielding begins the stress plateaus at $\sigma_y$.

Find. The length $L_{\text{before}}$ the sheet must be pulled to, under load, so that after the load is released it settles at the target 6.2 m.

1. Separate the deformation into plastic and elastic parts. While the sheet is held under load at the plateau stress $\sigma_y$, it has been stretched plastically and is also carrying an elastic strain. On unloading only the elastic part recovers (spring-back); the plastic part is permanent and sets the released length. The elastic strain is carried by the sheet at its current, loaded length, so the elastic shortening on release is $$\Delta L_{\text{el}}=\epsilon_{\text{elastic}}\,L_{\text{before}}, \qquad L_f = L_{\text{before}}-\Delta L_{\text{el}} = L_{\text{before}}\left(1-\epsilon_{\text{elastic}}\right).$$
2. Quantify the recoverable (elastic) strain. At the moment of release the stress in the sheet is $\sigma_y$ (perfectly plastic plateau, no further hardening to release from), so Hooke's law gives the strain that snaps back: $$\epsilon_{\text{elastic}} = \frac{\sigma_y}{E} = \frac{200\ \text{MPa}}{65{,}000\ \text{MPa}} = 3.077\times10^{-3}.$$ The sheet's $1.5\ \text{mm}\times80\ \text{mm}$ cross-section plays no role here — it fixes the force needed ($\sigma_y\times$ area) but not the strain recovered, which depends only on stress and modulus.
3. Combine to get the pre-release length. $$L_{\text{before}} = \frac{L_f}{1-\sigma_y/E} = \frac{6.2\ \text{m}}{1-3.077\times10^{-3}} = 6.2191\ \text{m},$$ i.e. a spring-back of $L_{\text{before}}-L_f=(3.077\times10^{-3})(6.219\ \text{m})=0.0191$ m. $$\boxed{L_{\text{before}} \approx 6.219\ \text{m} = 6219\ \text{mm}}$$