Derive Lame's constants

Lame's constants $\lambda$ and $\mu$ express the isotropic stress-strain law in the compact form

$$\sigma_{ij}=\lambda\,e\,\delta_{ij}+2\mu\,\varepsilon_{ij},\qquad e=\varepsilon_x+\varepsilon_y+\varepsilon_z$$

that is, $\sigma_x=\lambda e+2\mu\varepsilon_x$, $\sigma_y=\lambda e+2\mu\varepsilon_y$, $\sigma_z=\lambda e+2\mu\varepsilon_z$, $\tau_{xy}=\mu\gamma_{xy}$ etc.

Derivation

Hooke's law: $\varepsilon_x=\dfrac{1}{E}[(1+\nu)\sigma_x-\nu\Theta]$, with $\Theta=\sigma_x+\sigma_y+\sigma_z$.

Adding the three normal equations: $e=\dfrac{1-2\nu}{E}\Theta$, so $\Theta=\dfrac{E}{1-2\nu}e$.

Solve for $\sigma_x$:

$$\sigma_x=\frac{E}{1+\nu}\varepsilon_x+\frac{\nu}{1+\nu}\Theta=\frac{E}{1+\nu}\varepsilon_x+\frac{\nu E}{(1+\nu)(1-2\nu)}\,e$$

Compare with $\sigma_x=\lambda e+2\mu\varepsilon_x$:

$$\boxed{\lambda=\frac{\nu E}{(1+\nu)(1-2\nu)}},\qquad \boxed{\mu=\frac{E}{2(1+\nu)}=G}$$

For shear, $\tau_{xy}=G\gamma_{xy}$ with $G=\dfrac{E}{2(1+\nu)}$, which confirms $\mu=G$ (the shear modulus).

Inverse relations: $E=\dfrac{\mu(3\lambda+2\mu)}{\lambda+\mu}$, $\nu=\dfrac{\lambda}{2(\lambda+\mu)}$. The bulk modulus is $K=\lambda+\tfrac23\mu$.