Question 7 of 8: Necking of a ductile wire — Considère's criterion
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, December 2017 — 16-Mec-B8 Engineering Materials. Three hours, open book; any non-communicating calculator permitted. Eight problems, all of equal value; any FIVE constitute a complete paper, so each problem is worth 20 marks. Candidates are urged to submit a clear statement of any assumptions made. All eight problems are solved below, because the set as a whole is the study resource.
Groover, Fundamentals of Modern Manufacturing, 7th ed. — shaping and consolidation of polymer-matrix composites.
Note on this sitting. Two of them are worth flagging because the fresh data changes the answer, not merely the arithmetic. Question 8 raises the copper content of the incumbent alloy, which makes it denser, and as a result both candidate alloys now meet the customer’s weight request so the selection turns on strength parity rather than on who clears the bar. Question 4 changes the aeroplane from a twelve-seat business jet to a hundred-seat commercial transport and the stretch from 25 to 30 per cent, which sharpens rather than softens the damage-tolerance objection. Every number below has been re-worked from this paper's own data.
Question 7: Necking of a ductile wire — Considère's criterion (20 marks)
Given. A uniform ductile wire in simple tension whose plastic flow curve is the Hollomon power law $\sigma = K\varepsilon^{n}$ with strength coefficient $K = 338\ \text{MPa}$ and strain-hardening exponent $n = 0.43$, 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 0.75 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.
Part (a) — 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 that 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.
Part (b) — 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 $$\varepsilon_{u} = n = \boxed{0.43}$$ 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, $$\sigma_{u} = K n^{n} = 338(0.43)^{0.43} = 235.13\ \text{MPa}$$ 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 $$\text{UTS} = \frac{P_{max}}{A_{0}} = \sigma_{u}\frac{A_{u}}{A_{0}} = \sigma_{u}e^{-n} = 235.13\,e^{-0.43}$$ which evaluates to $$\boxed{\text{UTS} = 152.95\ \text{MPa}}$$ The area has shrunk to $e^{-0.43} = 0.6505$ of its original value, so the engineering strength is fully 35 per cent below the true stress at the same instant — a bigger gap than for a lower-$n$ metal, because the specimen has drawn further before becoming unstable.
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, $w = 338(0.43)^{1.43}/1.43 = 70.70\ \text{MJ/m}^{3}$. Since one megapascal is one megajoule per cubic metre, the numbers may be read straight off the stress axis. For the requested volume, $$\boxed{W = 70.70\ \text{MJ per m}^{3} \times 0.75\ \text{m}^{3} = 53.03\ \text{MJ}}$$
It is worth checking the magnitude for plausibility. The mean flow stress over the pre-necking range is $w/\varepsilon_{u} = 70.70/0.43 = 164.4\ \text{MPa}$, which sits sensibly between zero and the 235.1 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.