Necking instability, ultimate tensile strength and plastic work: 22-Mec-B8 Engineering Materials, December 2013, Q3

Question 3: Necking instability, ultimate tensile strength and plastic work (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 a Hollomon power-law flow curve, deforming at constant volume:

Given data
Quantity	Symbol	Value
Strength coefficient	K	310 MPa
Strain-hardening exponent	n	0.40
Volume of wire to be deformed	V	0.1 m³
Volume constancy	AL	A₀L₀ (constant)

Find. (a) the differential condition on σ and ε that marks the onset of necking, and (b) the ultimate tensile strength of the metal together with the plastic work needed to strain 0.1 m³ of wire up to that point.

Flow curve and load curve for the wire. The load P/A₀ = σe^−ε peaks exactly where the tangent to the flow curve has slope equal to the stress itself, at ε = n = 0.40. The shaded area under the flow curve up to that strain is the plastic work per unit volume.

Approach. Necking begins at the load maximum, so set the differential of the load to zero, eliminate the area using volume constancy to obtain the Considère condition, apply it to the power law to find the uniform strain, then convert the true stress at that strain into an engineering stress and integrate the flow curve for the work.

Express the load in terms of true stress and current area. At any instant the tensile load carried by the wire is $$P = \sigma A$$ where σ is the true stress and A the current cross-section, not the original one. Necking is a geometric instability, so it is the behaviour of P, and not of σ, that decides when it starts.
Impose volume constancy to relate area to strain. Plastic deformation conserves volume, so $$AL = A_0L_0 \quad\Rightarrow\quad \frac{A_0}{A} = \frac{L}{L_0} = e^{\varepsilon}, \qquad A = A_0e^{-\varepsilon}$$ using the definition of true strain ε = ln(L/L₀). Differentiating the constancy condition also gives the differential form $$\frac{dA}{A} = -\frac{dL}{L} = -d\varepsilon$$
Set the load to a maximum — the Considère criterion. Necking begins when the load stops rising, dP = 0. Differentiating P = σA, $$dP = \sigma\,dA + A\,d\sigma = 0 \quad\Rightarrow\quad \frac{d\sigma}{\sigma} = -\frac{dA}{A} = d\varepsilon$$ and therefore $$\boxed{\ \frac{d\sigma}{d\varepsilon} = \sigma\ }$$ This is the answer to part (a): at the onset of necking the rate of strain hardening exactly equals the flow stress. Physically, hardening can no longer compensate for the loss of section, so any local thinning grows rather than heals.
Apply the criterion to the given power law. With σ = Kεⁿ the derivative is nKεⁿ⁻¹, so the condition becomes $$nK\varepsilon^{n-1} = K\varepsilon^{n} \quad\Rightarrow\quad \varepsilon_u = n$$ $$\boxed{\ \varepsilon_u = n = 0.40\ }$$ For a Hollomon material the uniform strain is numerically equal to the strain-hardening exponent, which is why n is quoted as a formability index.
Evaluate the true stress at the instability. Substituting the uniform strain back into the flow curve, $$\sigma_u = K\varepsilon_u^{\,n} = 310(0.40)^{0.40} = 310(0.6931) = 214.9\ \text{MPa}$$ This is a true stress, referred to the necked-down area, and is not yet the answer to part (b).
Convert to the ultimate tensile strength. The UTS is by definition the maximum load divided by the original area, so using the area relation from Step 2, $$\mathrm{UTS} = \frac{P_{max}}{A_0} = \frac{\sigma_u A}{A_0} = \sigma_u e^{-\varepsilon_u} = 214.9\,e^{-0.40} = 214.9(0.6703)$$ $$\boxed{\ \mathrm{UTS} = 144.0\ \text{MPa}\ }$$ The section has shrunk by 33.0 % at this point, which is exactly why the engineering strength is so much lower than the true stress.
Integrate the flow curve for the plastic work per unit volume. The work absorbed per unit volume is the area under the true stress–true strain curve, $$w = \int_0^{\varepsilon_u}\sigma\,d\varepsilon = \int_0^{0.40}K\varepsilon^{n}\,d\varepsilon = \frac{K\varepsilon_u^{\,n+1}}{n+1} = \frac{310(0.40)^{1.40}}{1.40} = \frac{85.95}{1.40}$$ $$\boxed{\ w = 61.4\ \text{MJ/m}^3\ }$$ The units deserve a comment: 1 MPa is 1 MN/m², which is 1 MJ/m³, so the numerical value of an area under a stress–strain curve in MPa is an energy density in MJ/m³ with no conversion factor at all.
Scale to the stated volume. For the 0.1 m³ of wire specified, $$W = wV = 61.4 \times 0.1$$ $$\boxed{\ W = 6.14\ \text{MJ}\ }$$ Because the deformation is uniform right up to necking, every element of the wire has absorbed the same energy density, so this simple product is legitimate; past necking it would not be, since the deformation localises into the neck.

Results for the necking wire
Quantity	Symbol	Value
(a) Instability condition at necking	dσ/dε	= σ (Considère)
Uniform (necking) true strain	ε_u	0.40 (= n)
True stress at necking	σ_u	214.9 MPa
Area reduction at necking	1 − A/A₀	33.0 %
(b) Ultimate tensile strength	UTS	144.0 MPa
Plastic work per unit volume	w	61.4 MJ/m³
(b) Work for 0.1 m³ of wire	W	6.14 MJ

Check: the work calculated is the plastic work only. The Hollomon law is fitted to the plastic part of the curve and is unbounded as ε → 0, so it cannot represent the elastic leg; the elastic energy stored at yield is of order σ²/2E ≈ 0.02 MJ/m³, some three thousand times smaller than the plastic work, and is rightly neglected. The calculation also assumes room-temperature, quasi-static deformation, so the work appears almost entirely as heat rather than as stored energy.

Topic. The Considère criterion for tensile instability, and the relationship between true and engineering measures of strength.

Key relations. True strain ε = ln(L/L₀) and true stress σ = P/A; volume constancy A₀L₀ = AL; Hollomon flow curve σ = Kεⁿ; instability dσ/dε = σ, whence ε_u = n and UTS = Ke⁻ⁿnⁿ; plastic work per unit volume w = Kεⁿ⁺¹/(n+1).

Why this works. A tensile test is a competition between two effects. Straining the metal hardens it, which raises the load it can carry; straining it also thins it, which lowers the load. Early in the test hardening wins and the load rises, so any incipient neck is self-arresting: the thinner region is harder and stops deforming. The criterion dσ/dε = σ is precisely the point at which the two effects balance. Beyond it, thinning wins, deformation localises, and the load falls even though the material in the neck is still hardening. Everything the tensile test reports after that instant — the falling engineering curve, the fracture strain — describes a neck, not a material.

Common pitfall. Reporting the true stress 214.9 MPa as the UTS. The UTS is an engineering stress and must be referred to the original area, which requires the factor e^−εu; omitting it overstates the strength by 49 % here. A second trap is integrating the flow curve to the fracture strain rather than to ε_u: the question asks for the work “to necking”, which is exactly where the uniform-deformation assumption expires.

Source. Dieter, Mechanical Metallurgy, Ch. 8 (the tension test: instability in tension, Considère construction); Callister & Rethwisch, Ch. 6.

Question 3 of 8: Necking instability, ultimate tensile strength and plastic work

Question 3: Necking instability, ultimate tensile strength and plastic work (20 marks)