Irodov 1.299: The Ceiling on Poisson's Ratio – Amit & Deepali Sinha Foundation

Problem Irodov 1.299

A bar of an isotropic elastic material, with Young’s modulus $E$ and Poisson’s ratio $\mu$ , is immersed in a fluid at pressure $p$ , so every face of the bar is pushed inward by the same pressure $p$ . Find:

the fractional decrease in the bar’s volume, $\Delta V/V$
the compressibility $\beta$ (the fractional volume decrease per unit pressure) expressed through $E$ and $\mu$ , and, from that relation, an upper bound on $\mu$

A small element of the bar, pressed inward by the same pressure p on every face: the stress state is the same compression p along all three principal directions at once. — A small element of the bar, pressed inward by the same pressure $p$ on every face: the stress state is the same compression $p$ along all three principal directions at once.

Given

$E$ , $\mu$ known
hydrostatic pressure $p$ on every face

Find

(a)

\Delta V/V

, (b)

\beta(E,\mu)

, and the bound on

\mu

Solution

Step 1The strain along each principal direction

Every face carries the same compressive stress $p$ , so the stress state is $\sigma_x=\sigma_y=\sigma_z=-p$ . The generalized Hooke’s law gives the strain along one axis from the stress on that axis together with the stresses on the other two.

\varepsilon_x = \frac{1}{E}\bigl[\sigma_x-\mu(\sigma_y+\sigma_z)\bigr] = \frac{1}{E}\bigl[-p-\mu(-p-p)\bigr] = -\frac{(1-2\mu)p}{E},

and by the symmetry of the loading, $\varepsilon_y=\varepsilon_z=\varepsilon_x$ .

Step 2Part (a): the fractional volume change

To first order the fractional volume change is the sum of the three principal strains, which are all equal here.

\frac{\Delta V}{V} = \varepsilon_x+\varepsilon_y+\varepsilon_z = 3\varepsilon_x

\frac{\Delta V}{V} = -\frac{3(1-2\mu)p}{E}

(1)

Step 3Part (b): the compressibility

The compressibility is defined as the fractional volume decrease per unit pressure. Substituting (1) into this definition gives $\beta$ directly in terms of the elastic constants.

\beta \equiv -\frac{1}{p}\frac{\Delta V}{V}

\beta = \frac{3(1-2\mu)}{E}

(2)

Step 4Part (b), continued: the bound on Poisson’s ratio

A real solid contracts, never expands, under an all-around squeeze, so the compressibility above must be positive. Since $E>0$ for any elastic material, this requirement bounds $\mu$ .

\beta = \frac{3(1-2\mu)}{E} > 0 \quad\text{and}\quad E>0 \;\;\Longrightarrow\;\; 1-2\mu>0 \;\;\Longrightarrow\;\; \mu < \frac{1}{2}.

\beta>0

forces

\mu<1/2

for every real solid.

Check

At $\mu=0$ (no lateral coupling at all), $\beta=3/E$ , the sum of three independent uniaxial compliances acting together, as it should. As $\mu\to1/2$ , both $\Delta V/V$ and $\beta$ fall to zero, the incompressible limit, consistent with the same $(1-2\mu)$ factor vanishing there in Problems 1.297 and 1.298.

Remark

For the same stress magnitude, hydrostatic pressure changes a solid’s volume exactly three times as fast as a single uniaxial stress does: Problems 1.297 and 1.298 each found a uniaxial volume response of $(1-2\mu)\sigma/E$ , and here all three principal directions carry that same stress at once, so the three identical contributions simply add. The bound $\mu<1/2$ is really a statement that the bulk modulus $K=E/[3(1-2\mu)]=1/\beta$ , the stiffness against exactly this all-around squeeze, must stay finite and positive.