Question 6 of 8: Necking of a ductile wire — Considère’s criterion

Question 6: Necking of a ductile wire — Considère’s criterion (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.

Given. A ductile wire of uniform section obeying the Hollomon power law in true stress and true strain, with strength coefficient K = 340 MPa and strain-hardening exponent n = 0.45; plastic deformation at constant volume; the wire is loaded to the onset of necking.

Find. (a) the differential equation that holds between true stress and true strain at the instant necking begins, and (b) the ultimate tensile strength and the plastic work needed to strain 1 m³ of the wire to that point.

Approach. Necking is a geometric instability, not a material failure: it begins at the load maximum, where the rate at which the material hardens can no longer keep pace with the rate at which the section thins. Setting dP = 0 with P = σA and imposing constant volume gives the governing differential equation directly; substituting the power law then fixes the strain at necking, from which the ultimate tensile strength and the plastic work follow.

1. Write the load and differentiate it. The load carried by the wire is the true stress times the current area, P = σA. At the load maximum — which is the onset of necking — the differential of the load vanishes:$$dP = \sigma\,dA + A\,d\sigma = 0 \quad\Longrightarrow\quad \frac{d\sigma}{\sigma} = -\frac{dA}{A}$$
2. Impose conservation of volume. Plastic deformation is isochoric, so for the uniform gauge length AL = constant, whence dA/A + dL/L = 0. Since the true strain is defined by dε = dL/L, this gives$$-\frac{dA}{A} = \frac{dL}{L} = d\varepsilon$$
3. Combine to obtain the necking condition. Substituting the second result into the first,$$\boxed{\ \frac{d\sigma}{d\varepsilon} = \sigma\ }$$which is Considère’s criterion and is the differential equation the question asks for: necking begins at the strain where the slope of the true stress–true strain curve first equals the true stress itself.
4. Apply the criterion to the given power law. Differentiating σ = Kεⁿ gives dσ/dε = nKεⁿ⁻¹. Setting this equal to σ = Kεⁿ,$$nK\varepsilon^{\,n-1} = K\varepsilon^{\,n} \quad\Longrightarrow\quad \boxed{\ \varepsilon_u = n = 0.45\ }$$so for a Hollomon material the uniform elongation in true strain is numerically equal to the strain-hardening exponent. This is the single most useful result in the whole topic: n is not merely a curve-fitting constant, it is the ductility available before the deformation localises.
5. Evaluate the true stress at the onset of necking. Substituting ε_u = 0.45 back into the flow curve,$$\sigma_u = K\varepsilon_u^{\,n} = 340(0.45)^{0.45} = 340(0.6981) = 237.4\ \text{MPa}$$
6. Convert to the ultimate tensile strength. The UTS is an engineering stress, referred to the original area A₀, whereas σ_u is referred to the current area A. Constant volume gives A₀/A = L/L₀ = e^ε, so A = A₀e^−ε and$$\mathrm{UTS} = \frac{P_{\max}}{A_0} = \sigma_u e^{-\varepsilon_u} = 237.4\,e^{-0.45} = 237.4(0.6376)$$$$\boxed{\ \mathrm{UTS} = 151.3\ \text{MPa}\ }$$The engineering value is markedly lower than the true stress, as it must be: by the time the load peaks the wire has already thinned by some 36 % in area.
7. Integrate the flow curve for the plastic work. The work done per unit volume in straining the material to ε_u is the area under the true stress–true strain curve:$$w = \int_0^{\varepsilon_u}\sigma\,d\varepsilon = \int_0^{n} K\varepsilon^{\,n}\,d\varepsilon = \frac{K\,\varepsilon_u^{\,n+1}}{n+1}$$Substituting the numbers,$$w = \frac{340(0.45)^{1.45}}{1.45} = \frac{340(0.3142)}{1.45} = 73.7\ \text{MJ}\,\text{m}^{-3}$$since 1 MPa is identically 1 MJ m⁻³. For the 1 m³ of wire the question specifies,$$\boxed{\ W = 73.7\ \text{MJ}\ }$$