Elastic Moduli, Poisson's Ratio & Energy in Deformation

Shear modulus (modulus of rigidity) is the ratio of shear stress to shear strain:

$$G=\frac{\text{Shear stress}}{\text{Shear strain}}=\frac{F/A}{\varphi }=\frac{F/A}{x/L}$$

SI unit: Pa. Shear modulus measures resistance to shape change at constant volume.

• Solids have a well-defined $G$.
• Liquids and gases have $G=0$ (they cannot sustain shear stress — they flow instead).
• Typical values: steel $G\approx 8\times {10}^{10}\text{Pa}$; rubber $G$ very small.

Bulk modulus is defined for uniform compression (or expansion) — ratio of hydraulic stress to volumetric strain:

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

The negative sign ensures $B$ is positive (increased pressure → decreased volume, so $\mathrm{\Delta }V<0$ when $\mathrm{\Delta }P>0$).

SI unit: Pa. Compressibility $=1/B$ — how easily a material is compressed.

Worked Example

A rubber ball ($B=9\times {10}^8\text{Pa}$) is subjected to an increased pressure of ${10}^6\text{Pa}$. Find the fractional decrease in volume.

$$\frac{\mathrm{\Delta }V}{V}=\frac{\mathrm{\Delta }P}{B}=\frac{{10}^6}{9\times {10}^8}=1.11\times {10}^{-3}\approx 0.11\mathrm{\%}$$

When a wire is stretched longitudinally, its lateral dimensions decrease (and vice versa). Poisson's ratio is the magnitude of the ratio of lateral strain to longitudinal strain:

$$\sigma =-\frac{\text{Lateral strain}}{\text{Longitudinal strain}}=-\frac{\mathrm{\Delta }d/d}{\mathrm{\Delta }L/L}$$

The negative sign makes $\sigma$ positive (lateral contraction when longitudinal extension). Poisson's ratio is dimensionless.

Range: Theoretically $-1<\sigma <0.5$. For all real materials: $0\le \sigma \le 0.5$.

• Cork: $\sigma \approx 0$ (no lateral change — ideal for bottle stoppers)
• Steel: $\sigma \approx 0.28$–$0.30$
• Rubber: $\sigma \approx 0.5$ (incompressible — volume doesn't change)

The three moduli $Y$, $G$, $B$ are not independent — they are connected through Poisson's ratio:

$$Y=2G(1+\sigma )Y=3B(1-2\sigma )$$
$$\sigma =\frac{3B-2G}{2G+6B}$$

When $\sigma =0.5$: $Y=3B(0)=0$ in the second formula? No — as $\sigma \to 0.5$, $B\to \mathrm{\infty }$ (incompressible). The material resists volume change. Rubber is approximately this.

When $\sigma =0$: $Y=2G$ and $Y=3B$.

Key ratio: For most metals, $G\approx 0.4Y$ and $B\approx Y$ (order of magnitude).

When a body is deformed elastically, work is done against the internal restoring forces. This energy is stored as elastic potential energy.

For a Stretched Wire

$$U=\frac{1}{2}F\cdot \mathrm{\Delta }L$$
$$=\frac{1}{2}\times \text{stress}\times \text{strain}\times \text{volume}$$

Elastic Energy per Unit Volume (Energy Density)

$$u=\frac{U}{\text{Volume}}=\frac{1}{2}\times \text{stress}\times \text{strain}$$
$$=\frac{{\text{stress}}^2}{2Y}=\frac{1}{2}Y(\text{strain}{)}^2$$

Worked Example

A wire ($Y=2\times {10}^{11}\text{Pa}$) has a strain of ${10}^{-3}$. Find elastic energy per unit volume.

$$u=\frac{1}{2}Y{\epsilon }^2=\frac{1}{2}\times 2\times {10}^{11}\times ({10}^{-3}{)}^2=$$
$${10}^5{\text{J m}}^{-3}=\mathbf{0.1}{\mathbf{\text{MJ m}}}^{\mathbf{-}\mathbf{3}}$$