Poisson's ratio for perfectly plastic, constant-volume deformation: 22-Mec-B8 Engineering Materials, December 2016, Q7

Question 7: Poisson's ratio for perfectly plastic, constant-volume deformation (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 homogeneous, isotropic prismatic bar of length $L$ and rectangular section $b \times t$, stretched axially by a small amount $\Delta L$, with the lateral dimensions contracting by $\Delta b$ and $\Delta t$. The deformation is perfectly plastic, so the volume is identical before and after.

Given data
Quantity	Symbol	Description
Initial dimensions	L, b, t	Length, width, thickness of the undeformed bar
Axial extension	ΔL	Small, positive
Lateral contractions	Δb, Δt	Small, positive (dimensions are reduced)
Volume condition	ΔV = 0	Perfect plasticity: deformation is isochoric

Find. The numerical value of Poisson's ratio ν implied by the constant-volume condition.

The bar before and after a constant-volume tensile stretch. The rectangle lengthens and thins by exactly the amounts that keep its area — and, in three dimensions, its volume — unchanged, which is the geometric content of ν = 0.5.

Approach. Write the volume as the product of the three dimensions, express the deformed volume in terms of the three strains, expand it to first order in the small strains, and impose the condition that the volume change vanishes. Poisson's ratio enters through its definition as the ratio of lateral to axial strain, and the algebra returns a single value for it.

Define the three strains from the stated changes. Taking extension as positive, the axial strain and the two lateral strains are $$\varepsilon_{L} = \frac{\Delta L}{L}, \qquad \varepsilon_{b} = -\frac{\Delta b}{b}, \qquad \varepsilon_{t} = -\frac{\Delta t}{t}$$ The minus signs are not decoration: the question states that the cross-sectional dimensions are reduced by Δb and Δt, so those two strains are negative while ε_L is positive.
Introduce Poisson's ratio through its definition. Poisson's ratio is defined as minus the ratio of lateral to axial strain, $$\nu = -\frac{\varepsilon_{\text{lat}}}{\varepsilon_{L}}$$ and for an isotropic material the two lateral directions must behave identically, so $$\varepsilon_{b} = \varepsilon_{t} = -\nu\,\varepsilon_{L}$$ The whole question is now a matter of finding what value of ν is consistent with the volume staying fixed.
Write the deformed volume. The bar remains prismatic, so its volume is the product of the three current dimensions: $$V' = (L + \Delta L)(b - \Delta b)(t - \Delta t) = L\,b\,t\,(1 + \varepsilon_{L})(1 - \nu\varepsilon_{L})^{2}$$ Substituting into the original volume $V = Lbt$ gives the volume ratio directly.
Expand to first order in the strain. Multiplying out and discarding products of two or more small quantities, $$\frac{V'}{V} = (1 + \varepsilon_{L})(1 - 2\nu\varepsilon_{L} + \nu^{2}\varepsilon_{L}^{2}) \approx 1 + \varepsilon_{L} - 2\nu\varepsilon_{L}$$ so the fractional volume change — the dilatation — is $$\frac{\Delta V}{V} = \varepsilon_{L} + \varepsilon_{b} + \varepsilon_{t} = \varepsilon_{L}\left(1 - 2\nu\right)$$ This intermediate result is worth remembering in its own right: for small strains the fractional change in volume is simply the sum of the three direct strains.
Impose the constant-volume condition and solve. Perfect plasticity means ΔV = 0, and the bar really is being stretched, so ε_L ≠ 0. The only way for the product to vanish is for the bracket to vanish: $$\varepsilon_{L}\left(1 - 2\nu\right) = 0 \quad\Longrightarrow\quad 1 - 2\nu = 0$$ $$\boxed{\ \nu = \tfrac{1}{2} = 0.5\ }$$ which is the answer to the question.
Confirm the result without the small-strain approximation. The first-order expansion is not actually necessary. Working in true (logarithmic) strains, $\varepsilon^{\text{true}} = \ln(1 + \varepsilon)$, the volume ratio is exact: $$\ln\frac{V'}{V} = \varepsilon^{\text{true}}_{L} + \varepsilon^{\text{true}}_{b} + \varepsilon^{\text{true}}_{t}$$ Setting the left side to zero gives $\varepsilon^{\text{true}}_{b} = \varepsilon^{\text{true}}_{t} = -\tfrac{1}{2}\varepsilon^{\text{true}}_{L}$ for arbitrarily large strains, so a plastically flowing metal has an effective Poisson's ratio of exactly one half however far it is drawn. That is why plasticity theory is written in terms of deviatoric stress alone.

The number is worth a sentence of interpretation, because 0.5 is not an accident of this problem but a hard upper bound. The elastic bulk modulus is $K = E/[3(1-2\nu)]$, so as ν approaches one half the bulk modulus diverges and the material becomes incompressible; a value above 0.5 would give a negative bulk modulus, meaning that hydrostatic compression would expand the material, which is thermodynamically impossible. Real metals in the elastic range have ν between about 0.27 and 0.35 — 0.33 for aluminium, 0.29 for steel — so roughly a third of the axial strain shows up as a genuine volume change: at ν = 0.33 the dilatation is $(1-2\nu) = 0.34$ of the axial strain. Once the metal yields, however, dislocation glide moves atoms past one another without changing the number of atoms per unit volume, so the plastic part of the strain is isochoric and the effective ratio jumps to 0.5. A tensile bar therefore shows a Poisson's ratio that rises from about 0.3 to about 0.5 as the plastic strain comes to dominate the elastic strain.

Results
Quantity	Relation	Value
Dilatation for small strains	ΔV/V = ε_L + ε_b + ε_t	ε_L(1 − 2ν)
Constant-volume condition	ΔV/V = 0, ε_L ≠ 0	1 − 2ν = 0
Poisson's ratio for perfect plasticity	ν = 1/2	0.5
Lateral strains at that value	ε_b = ε_t	−0.5 ε_L
Elastic bulk modulus at ν = 0.5	K = E/[3(1 − 2ν)]	Infinite (incompressible)
Typical elastic ν for a metal, for contrast	—	0.27 – 0.35

Check: the step from ε_b = ε_t to a single Poisson's ratio assumes the material is isotropic in the plane of the cross-section. If it is not — a heavily rolled sheet, for instance, which has a different plastic strain ratio through the thickness than across the width — the constant-volume condition still holds but gives only the weaker statement ν_Lb + ν_Lt = 1, so the two lateral contractions must share the total but need not be equal. The single value ν = 0.5 follows only when they are. The question says the bar is homogeneous and treats Δb and Δt symmetrically, so isotropy is the intended reading and is stated here rather than assumed silently.

Topic. Poisson's ratio, dilatation, and the incompressibility of plastic flow.

Key relations. Definition ν = −ε_lat/ε_axial; small-strain dilatation ΔV/V = ε_x + ε_y + ε_z = ε_axial(1 − 2ν) in uniaxial loading; the elastic moduli relations K = E/[3(1 − 2ν)] and G = E/[2(1 + ν)], which together bound ν between −1 and +0.5 for a stable isotropic solid; and for large deformations the exact statement ε₁ + ε₂ + ε₃ = 0 in true strains.

Why this works. Elastic strain stretches interatomic bonds, and stretching bonds along one axis does not fully compensate for the contraction across the other two, so an elastically loaded solid genuinely changes volume. Plastic strain works by an entirely different mechanism: dislocations glide on close-packed planes, shuffling atoms past one another one lattice step at a time. No atoms are created, destroyed or moved further apart on average, so the density is unchanged and the deformation is volume-preserving by construction. Setting the dilatation to zero is therefore not an idealisation imposed for convenience — it is a statement about the physical mechanism, and it is the reason that the whole apparatus of plasticity theory (the von Mises and Tresca yield criteria, the Levy–Mises flow rule) is built on the deviatoric part of the stress tensor and is indifferent to hydrostatic pressure.

Common pitfall. Losing a sign: writing ε_b = +Δb/b and then concluding ν = −0.5. The question already tells you the lateral dimensions are reduced, so Δb is a positive number describing a negative strain. A second pitfall is keeping the second-order term ν²ε² and solving a quadratic in ν; for small strains that term is negligible, and for large strains the true-strain form of step 6 is exact and much cleaner. A third is to report ν = 0.5 for a metal generally, when it is the value for the plastic part of the strain only — the elastic response of that same metal has ν near 0.3.

Source. Dieter, Mechanical Metallurgy, 3rd ed., §2-6 and §3-2 (elastic constants, volume constancy in plastic deformation); Askeland & Wright, Ch. 6 §6.2; Callister & Rethwisch, Ch. 6 §6.5.

Question 7 of 8: Poisson's ratio for perfectly plastic, constant-volume deformation

Question 7: Poisson's ratio for perfectly plastic, constant-volume deformation (20 marks)