Bulk Modulus, Shear Modulus, and Poisson's Ratio

Young's modulus describes one of three independent elastic responses of an isotropic solid. The other two are the bulk modulus (volumetric response) and the shear modulus (response to tangential forces). The dimensionless Poisson's ratio links lateral to longitudinal strain.

Concept

Bulk modulus $B$

When a body is subjected to a uniform pressure $P$ (e.g., a sphere immersed in a fluid), its volume decreases by $\Delta V$. The bulk modulus is:

$B = -\frac{P}{\Delta V/V}$

The minus sign ensures $B > 0$ since $\Delta V < 0$ for compression. The reciprocal $\kappa = 1/B$ is the compressibility.

Typical values. Water: $B \approx 2.2\times 10^9$ Pa. Steel: $B \approx 1.6\times 10^{11}$ Pa. Gases have very low $B$ (highly compressible).

Shear modulus (modulus of rigidity) $G$

When a tangential force $F$ acts on a top face of area $A$ of a block fixed at the bottom, the block distorts by an angle $\theta$. The shear modulus is:

$G = \frac{\sigma_s}{\varepsilon_s} = \frac{F/A}{\theta}$

Only solids exhibit a non-zero $G$; ideal fluids have $G = 0$.

Poisson's ratio $\nu$ (or $\sigma$)

When a wire is stretched longitudinally, it becomes thinner laterally. Poisson's ratio is the ratio of lateral to longitudinal strain (taken with a minus sign so $\nu > 0$ for most materials):

$\nu = -\frac{\varepsilon_{\text{lateral}}}{\varepsilon_{\text{longitudinal}}}$

For most materials, $0 \le \nu \le 0.5$. Cork has $\nu \approx 0$; rubber has $\nu \approx 0.5$.

The three moduli of an isotropic solid are related by:

$Y = 2G(1+\nu) = 3B(1-2\nu)$

Derivation

Bulk modulus from $PV$ definition. Consider a body of volume $V$ at pressure $P_0$. Increase pressure by $dP$, causing a change $dV$:

Step 1 — From definition: $B = -V\,(dP/dV)$.

Step 2 — Rearrange: $dP = -B\,dV/V$.

Step 3 — Integrate for small changes: $\Delta P = -B\,\Delta V / V$, i.e. $\Delta V = -\Delta P \cdot V / B$.

Relation $Y = 2G(1+\nu)$ (sketch). A pure tension along the $x$-axis produces longitudinal strain $\varepsilon$ and lateral contractions $-\nu\varepsilon$. By Mohr's circle in strain, the maximum shear strain on planes at 45° is $\varepsilon_s = \varepsilon(1+\nu)$, while the shear stress on those planes is $\sigma/2$. Hence $G = (\sigma/2)/(\varepsilon(1+\nu)) = Y/(2(1+\nu))$.

Worked Example

(a) A solid sphere of volume $1\,\text{L}$ is taken to a depth where pressure is $10^7\,\text{Pa}$ greater than at the surface. If $B = 2\times 10^{11}\,\text{Pa}$ (steel), find $\Delta V$.

$\Delta V = -V\,\Delta P/B = -10^{-3} \cdot 10^7 / 2\times 10^{11} = -5\times 10^{-8}\,\text{m}^3 = -0.05\,\text{cm}^3$.

(b) A cube of side $10\,\text{cm}$ has its top face pushed horizontally by $200\,\text{N}$ while its bottom face is fixed. Top face displaces by $0.1\,\text{mm}$. Find the shear modulus.

Area $A = 10^{-2}\,\text{m}^2$. Shear stress $\sigma_s = 200/0.01 = 2\times 10^4\,\text{Pa}$.

Shear strain $\theta = 0.0001/0.1 = 10^{-3}$.

$G = \sigma_s/\theta = 2\times 10^7\,\text{Pa} = 20\,\text{MPa}$.

Common Confusions

• Fluids cannot support shear in equilibrium — for an ideal fluid $G = 0$.
• Bulk modulus is positive despite $\Delta V$ being negative on compression; the negative sign in the definition takes care of this.
• Poisson's ratio is dimensionless and cannot exceed 0.5 for stable, isotropic, conventional materials (above 0.5 the material would expand in volume on stretching, which is impossible for ordinary solids).
• Auxetic materials have $\nu < 0$ — unusual but real (some foams, certain crystals).

Key Takeaways

• Bulk modulus $B = -P / (\Delta V/V)$ — resistance to uniform compression.
• Shear modulus $G$ — resistance to tangential distortion; zero for fluids.
• Poisson's ratio $\nu$ — magnitude of lateral contraction per unit longitudinal extension; dimensionless, $0 \le \nu \le 0.5$ for stable solids.
• $Y$, $G$, $B$, and $\nu$ are linked: only two are independent.