Necking of a ductile wire — Considère's criterion: 22-Mec-B8 Engineering Materials, December 2014, Q1

Question 1: 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 uniform ductile wire in simple tension whose plastic flow curve is the Hollomon power law $\sigma = K\varepsilon^{n}$ with strength coefficient $K = 420\ \text{MPa}$ and strain-hardening exponent $n = 0.33$, both stress and strain being true (natural) measures. Plastic deformation conserves volume, so $A L = A_{0}L_{0}$ throughout.

Find. (a) the differential equation that the true stress and true strain must satisfy at the onset of necking, and (b) from it, the ultimate tensile strength of the metal and the plastic work needed to strain one cubic metre of the wire up to necking.

The flow curve and the load it produces. The true stress (solid) rises indefinitely, but the load carried by the specimen, proportional to the engineering stress P/A(0) (dashed), passes through a maximum at the strain where the Considère tangent touches the flow curve. That maximum load, divided by the original area, is the ultimate tensile strength; the shaded area under the flow curve up to the same strain is the plastic work per unit volume.

Approach. Write the load as the product of true stress and current area, set its differential to zero for the maximum-load (instability) point, eliminate the area using constancy of volume, and then evaluate the resulting condition for the given power law before integrating the flow curve for the work.

Express the load in terms of true stress and current area. True stress is defined on the instantaneous cross-section, so the tensile load carried by the wire is $$P = \sigma A$$ where $A$ is the current area. Necking begins at the instant the load stops rising: the specimen becomes unstable when a small further extension no longer requires a larger force.
Impose the maximum-load condition. Differentiating the product and setting $dP = 0$ at the maximum gives $$dP = \sigma\,dA + A\,d\sigma = 0 \quad\Longrightarrow\quad \frac{d\sigma}{\sigma} = -\frac{dA}{A}$$ This is a purely mechanical statement so far: it says the fractional gain in stress must exactly offset the fractional loss of area.
Use constancy of volume to remove the area. With $AL = A_{0}L_{0}$ constant, taking logarithms and differentiating gives $\dfrac{dA}{A} + \dfrac{dL}{L} = 0$, and since true strain is defined by $d\varepsilon = dL/L$, $$-\frac{dA}{A} = \frac{dL}{L} = d\varepsilon$$ Substituting this into the previous result eliminates the geometry entirely and leaves a relation between the two material variables alone: $$\boxed{\frac{d\sigma}{d\varepsilon} = \sigma}$$ This is Considère's criterion, the answer to part (a). Geometrically it says that necking starts where the slope of the flow curve has fallen to the value of the stress itself, that is, where the tangent to the curve has a subtangent of one strain unit.
Apply the criterion to the given power law. Differentiating $\sigma = K\varepsilon^{n}$ gives $d\sigma/d\varepsilon = nK\varepsilon^{n-1}$, and setting that equal to $\sigma = K\varepsilon^{n}$ leaves $nK\varepsilon^{n-1} = K\varepsilon^{n}$, so $$\varepsilon_{u} = n = \boxed{0.33}$$ For a Hollomon material the uniform (pre-necking) true strain is numerically equal to the strain-hardening exponent — the single most useful result in this whole topic.
Evaluate the true stress at that strain. Substituting back into the flow curve, $$\sigma_{u} = K n^{n} = 420(0.33)^{0.33} = 291.3\ \text{MPa}$$ This is the stress on the actual cross-section at the instant of instability, not the tensile strength quoted on a datasheet.
Convert to the engineering ultimate tensile strength. The UTS is the maximum load divided by the original area. From $\varepsilon = \ln(L/L_{0}) = \ln(A_{0}/A)$ we get $A_{u} = A_{0}e^{-\varepsilon_{u}}$, so $$\text{UTS} = \frac{P_{max}}{A_{0}} = \sigma_{u}\frac{A_{u}}{A_{0}} = \sigma_{u}e^{-n} = 291.3\,e^{-0.33}$$ which evaluates to $$\boxed{\text{UTS} = 209.4\ \text{MPa}}$$ The area has shrunk to $e^{-0.33} = 0.719$ of its original value, so the engineering strength is about 28 per cent below the true stress at the same instant.
Integrate the flow curve for the plastic work. The plastic work per unit volume is the area under the true-stress/true-strain curve, $$w = \int_{0}^{\varepsilon_{u}}\sigma\,d\varepsilon = \int_{0}^{n}K\varepsilon^{n}\,d\varepsilon = \frac{K\,n^{\,n+1}}{n+1}$$ Substituting the data, $w = 420(0.33)^{1.33}/1.33 = 72.28\ \text{MJ/m}^{3}$. Since one megapascal is one megajoule per cubic metre, the numbers may be read straight off the stress axis. For the requested volume, $$\boxed{W = 72.28\ \text{MJ per m}^{3} \times 1\ \text{m}^{3} = 72.3\ \text{MJ}}$$

It is worth checking the magnitude for plausibility. The mean flow stress over the pre-necking range is $w/\varepsilon_{u} = 72.28/0.33 = 219\ \text{MPa}$, which sits sensibly between zero and the 291 MPa reached at necking, as it must for a curve that is concave downwards. Note also that this is the work to necking only; the specimen absorbs a good deal more before it finally separates, but that further work is concentrated in the neck and is not covered by the uniform-deformation analysis above.

Results
Quantity	Relation	Value
Necking criterion	dσ/dε = σ	—
Uniform (necking) true strain	ε_u = n	0.33
True stress at necking	σ_u = Knⁿ	291.3 MPa
Ultimate tensile strength	UTS = σ_ue⁻ⁿ	209.4 MPa
Plastic work to necking	w = Knⁿ⁺¹/(n+1)	72.28 MJ/m³
Work for 1 m³ of wire	W = w × V	72.3 MJ

Topic. Plastic instability in uniaxial tension: Considère's criterion, the Hollomon flow law, and the relation between true and engineering measures of strength.

Key relations. True stress $\sigma = P/A$ and true strain $\varepsilon = \ln(L/L_{0}) = \ln(A_{0}/A)$; conversion from engineering measures $\sigma = s(1+e)$ and $\varepsilon = \ln(1+e)$; constancy of plastic volume $AL = A_{0}L_{0}$; the instability condition $d\sigma/d\varepsilon = \sigma$; for $\sigma = K\varepsilon^{n}$ the consequences $\varepsilon_{u} = n$, $\sigma_{u} = Kn^{n}$, $\text{UTS} = Kn^{n}e^{-n}$ and $w = Kn^{n+1}/(n+1)$.

Why this works. A tensile specimen is a competition between two effects. Straining the metal makes it harder, which tends to spread deformation uniformly along the gauge length; straining it also thins it, which concentrates deformation wherever the section happens to be smallest. Early on, hardening wins and the load rises. Because the slope of a power-law curve falls continuously while the stress itself keeps rising, the two effects must cross, and at that crossing the load is stationary. Everything after it is unstable: a slightly thinner element now needs less force than its neighbours, so it strains preferentially and a neck forms. That is why the criterion is written as a comparison of the hardening rate with the stress, and why a material with a high $n$ — annealed copper or austenitic stainless steel, say — can be drawn much further before it necks than one with a low $n$.

Common pitfall. Quoting $\sigma_{u} = Kn^{n}$ as the ultimate tensile strength. It is the true stress on the necked section; the UTS that a datasheet reports is referred to the original area and is smaller by the factor $e^{-n}$. The second pitfall is integrating the engineering curve for the work, or integrating the true curve past $\varepsilon_{u}$; only the true curve gives work per unit volume, and only the uniform range is genuinely one-dimensional.

Source. Dieter, Mechanical Metallurgy, 3rd ed., §8-3 to §8-5 (true stress–strain, instability in tension, Considère's construction); Askeland & Wright, Ch. 6; Callister & Rethwisch, Ch. 6.

Question 1 of 8: Necking of a ductile wire — Considère's criterion

Question 1: Necking of a ductile wire — Considère's criterion (20 marks)