Elastic constants and triaxial loading

(a) Definitions and derivation

• Poisson's ratio $\nu$ is the ratio of lateral strain to axial strain in a uniaxial stress state: $\nu = -\varepsilon_{lat}/\varepsilon_{axial}$ (0.25 to 0.33 for metals).
• Bulk modulus $K$ is the ratio of hydrostatic (uniform) pressure to the volumetric strain it produces: $K = p/(-\varepsilon_v)$.

Generalized Hooke's law for a cube under three stresses:

$$\varepsilon_x = \frac{1}{E}\left[\sigma_x - \nu(\sigma_y+\sigma_z)\right]$$

and similarly for $\varepsilon_y,\varepsilon_z$. Under hydrostatic pressure $p$, $\sigma_x=\sigma_y=\sigma_z=-p$:

$$\varepsilon_x=\varepsilon_y=\varepsilon_z = \frac{-p(1-2\nu)}{E}$$ $$\varepsilon_v = \varepsilon_x+\varepsilon_y+\varepsilon_z = -\frac{3p(1-2\nu)}{E} \;\Rightarrow\; K = \frac{p}{-\varepsilon_v} = \frac{E}{3(1-2\nu)}$$ $$E = 3K(1-2\nu)$$

(b) Numerical

Strains:

$$\begin{aligned} \varepsilon_x &= \frac{100 - 0.3(-60+40)}{200000} = \frac{106}{200000} = 5.3\times10^{-4}\\ \varepsilon_y &= \frac{-60 - 0.3(100+40)}{200000} = \frac{-102}{200000} = -5.1\times10^{-4}\\ \varepsilon_z &= \frac{40 - 0.3(100-60)}{200000} = \frac{28}{200000} = 1.4\times10^{-4} \end{aligned}$$

Volumetric strain:

$$\varepsilon_v = (5.3 - 5.1 + 1.4)\times10^{-4} = 1.6\times10^{-4}$$

Check: $\varepsilon_v = \frac{1-2\nu}{E}(\sigma_x+\sigma_y+\sigma_z) = \frac{0.4\times80}{200000} = 1.6\times10^{-4}$.

$$V = 100\times80\times50 = 4\times10^5\ \text{mm}^3, \qquad \Delta V = \varepsilon_v V = 64\ \text{mm}^3\ \text{(increase)}$$

Elastic constants:

$$K = \frac{E}{3(1-2\nu)} = \frac{200}{3\times0.4} = 166.7\ \text{GPa}, \qquad G = \frac{E}{2(1+\nu)} = \frac{200}{2.6} = 76.9\ \text{GPa}$$

Answer: $\varepsilon_x = 5.3\times10^{-4}$, $\varepsilon_y = -5.1\times10^{-4}$, $\varepsilon_z = 1.4\times10^{-4}$; $\Delta V = +64\ \text{mm}^3$; $K = 166.7$ GPa; $G = 76.9$ GPa.