Considère's Necking Criterion; Elastic Springback of a Stretched Sheet: 21-Mat-A4 Deformation Behaviour and Properties of Materials, May 2017, Q7

Question 7: Considère's Necking Criterion; Elastic Springback of a Stretched 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.

(a) Derivation of the Necking Condition $\varepsilon=n$

Given. True stress-true strain hardening law $\sigma=K\varepsilon^n$; the load is $P=\sigma A$; the material deforms at constant volume, $A_0L_0=AL$.

Find. Show that the maximum-load (necking-onset) condition occurs at $\varepsilon=n$.

Approach. Necking begins where the load $P$ reaches a maximum, $dP=0$ (beyond this point, the shrinking cross-section reduces the load faster than strain hardening can raise the stress, so deformation localises). Express $P$ in terms of true strain alone, using constant volume, and set $dP/d\varepsilon=0$.

Load in terms of true strain. Constant volume gives $A=A_0e^{-\varepsilon}$ (true strain $\varepsilon=\ln(L/L_0)$), so $$P=\sigma A=\sigma A_0e^{-\varepsilon}$$
Maximum-load condition. $$\frac{dP}{d\varepsilon}=A_0e^{-\varepsilon}\left(\frac{d\sigma}{d\varepsilon}-\sigma\right)=0\;\Rightarrow\;\boxed{\frac{d\sigma}{d\varepsilon}=\sigma}$$ (since $A_0e^{-\varepsilon}\neq0$).
Substitute the power-law hardening rule. With $\sigma=K\varepsilon^n$, $\dfrac{d\sigma}{d\varepsilon}=nK\varepsilon^{n-1}$. Setting $d\sigma/d\varepsilon=\sigma$: $$nK\varepsilon^{n-1}=K\varepsilon^n\;\Rightarrow\;n=\varepsilon\;\Rightarrow\;\boxed{\varepsilon=n}$$ Necking (plastic instability) begins exactly when the true strain equals the strain-hardening exponent — a higher-$n$ material can be drawn to a larger uniform strain before it necks.

(b) Pre-Release Length for a 6.2 m Permanent Length

Given.

Quantity	Symbol	Value
Original (undeformed) length	$L_0$	5 m
Desired final (permanent, after release) length	$L_f$	6.2 m
Elastic modulus	$E$	65 GPa
Yield strength	$\sigma_y$	200 MPa

Find. The length of the sheet while still under load, immediately before the stress is released, so that the sheet's permanent length afterward is 6.2 m.

Check

No strain-hardening data ($K$, $n$, or a full stress-strain curve) is supplied for the magnesium sheet in this question — only $E$ and $\sigma_y$. The sheet is therefore idealised as elastic–perfectly-plastic: once yielded, the flow stress stays at $\sigma_y$ for any further plastic strain, so the elastic strain recovered on unloading is fixed at $\sigma_y/E$ regardless of how far the sheet was stretched.

Approach. On release, only the elastic portion of the strain recovers instantly (Hooke's law); the elastic-perfectly-plastic idealisation fixes the recoverable strain at $\sigma_y/E$. Add that recovery (referenced to the original 5 m gauge length, consistent with ordinary engineering strain) back onto the desired 6.2 m permanent length to find the loaded length just before release.

Elastic strain recovered on unloading. $$\varepsilon_{el}=\frac{\sigma_y}{E}=\frac{200}{65{,}000}=\boxed{0.003077\ (0.308\%)}$$
Recovery length. $$\Delta L_{el}=\varepsilon_{el}\,L_0=0.003077(5)=0.01538\ \text{m}\ (15.4\ \text{mm})$$
Length required immediately before release. $$L_1=L_f+\Delta L_{el}=6.200+0.0154=\boxed{6.215\ \text{m}}$$ (check: yield strain $\sigma_y/E=0.31\%$ is far smaller than the $\approx24\%$ total engineering strain reached, confirming the sheet is deep into the plastic regime, where the perfectly-plastic idealisation is the reasonable approximation the given data supports.)

Quantity	Value
Necking condition (part a)	$\varepsilon=n$
Elastic recovery strain (part b)	0.308%
Elastic recovery length (part b)	15.4 mm
Sheet length before release (part b)	6.215 m

Concept & Theory

Topic. Considère's plastic-instability criterion; elastic springback in sheet forming.

Key relations. Necking onset $d\sigma/d\varepsilon=\sigma$; power-law hardening $\sigma=K\varepsilon^n$; elastic recovery $\Delta L_{el}=(\sigma/E)L_0$.

Why this works. At maximum load, the rate at which strain hardening raises the flow stress exactly balances the rate at which the shrinking cross-section reduces load-carrying area; beyond that point geometric softening wins and strain localises. On unloading, stress and strain retrace the elastic modulus line exactly (Hooke's law holds regardless of prior plastic history), so only the $\sigma/E$ portion of the total strain is recoverable — the rest is permanent (plastic) set.

Common pitfall. For necking, forgetting the area-reduction term and setting $d\sigma/d\varepsilon=0$ instead of $d\sigma/d\varepsilon=\sigma$; for springback, referencing the recovery strain to the stretched length $L_1$ rather than the original length $L_0$, or forgetting that with no hardening data given, the flow stress at unload must be taken as $\sigma_y$ itself rather than some other assumed value.

Source. Dieter, Mechanical Metallurgy, Ch. 3 (plastic instability, necking) and Ch. 8 (sheet-forming, elastic springback).

Question 7 of 8: Considère's Necking Criterion; Elastic Springback of a Stretched Sheet

Question 7: Considère's Necking Criterion; Elastic Springback of a Stretched Sheet (20 marks)

(a) Derivation of the Necking Condition $\varepsilon=n$

(b) Pre-Release Length for a 6.2 m Permanent Length