Question 8 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 — 16-Mec-B8 Engineering Materials, undated sitting (the printed footer reads 16-Mec-B8/May 2019). Three hours; 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.
Note on this sitting. Five of the eight problems restate standing 22-Mec-B8 archetypes with fresh or unchanged data — the magnesium sacrificial anode, the FRP consolidation essay (problem 5, now asking for three routes and for their applicability to primary versus secondary structure rather than four routes and their trade-offs alone), the ABS-versus-phenolic selection (problem 6), the aluminium–lithium floor-beam substitution (problem 7, restated in kilograms rather than newtons) and Considère necking (problem 8). Problems 1, 2 and 3 are new to the subject: they are Faraday's-law corrosion rate, Fick's first law applied to carburizing, and an atom-counting exercise on a silicon wafer. Every calculation has been re-worked from this paper's own numbers.
Question 8: 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 $K = 307\ \text{MPa}$ and strain-hardening exponent $n = 0.37$, both stress and strain being true (natural) measures. Plastic deformation conserves volume, so $A L = A_{0}L_{0}$ throughout, and the volume of wire to be strained is 0.07 m3.
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 0.07 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 $$\varepsilon_{u} = n = \boxed{0.37}$$ 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, $$\sigma_{u} = K n^{n} = 307(0.37)^{0.37} = 212.51\ \text{MPa}$$ 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 $$\text{UTS} = \frac{P_{max}}{A_{0}} = \sigma_{u}\frac{A_{u}}{A_{0}} = \sigma_{u}e^{-n} = 212.51\,e^{-0.37}$$ which evaluates to $$\boxed{\text{UTS} = 146.8\ \text{MPa}}$$ The area has shrunk to $e^{-0.37} = 0.6907$ of its original value, so the engineering strength is about 31 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, $w = 307(0.37)^{1.37}/1.37 = 57.39\ \text{MJ/m}^{3}$. Since one megapascal is one megajoule per cubic metre, the numbers may be read straight off the stress axis. For the requested volume, $$\boxed{W = 57.39\ \text{MJ per m}^{3} \times 0.07\ \text{m}^{3} = 4.018\ \text{MJ}}$$
It is worth checking the magnitude for plausibility. The mean flow stress over the pre-necking range is $w/\varepsilon_{u} = 57.39/0.37 = 155.1\ \text{MPa}$, which sits sensibly between zero and the 212.5 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.