Elastic Modulii and Poisson Ratio

What are the $Lam\acute{e}$ constants $\mu$ and $\lambda$ ?

First, we’ll define the Bulk Modulus “ $k$ ” by considering a block oriented along the principal strains:

Initially, $V_{0}=l_{0}w_{0}h_{0}$

After deformation, $w=w_{0}[1+\lambda_{1}]$ ; $l=l_{0}[1+\lambda_{2}]$ ; $h=h_{0}[1+\lambda_{3}]$

$V=lwh$

$\frac{V-V_{0}}{V}=\frac{\Delta V}{V}=\frac{w_0(1+\lambda_1)l_0(1+\lambda_2)h_0(1+\lambda_3)-l_0w_0h_0}{l_0w_0h_0}=(1+\lambda_1)(1+\lambda_2)(1+\lambda_3)-1$
$\approx \lambda_1 + \lambda_2 + \lambda_3 = \epsilon_{kk}$ where $O(\lambda^2)$ terms have been eliminated.

Now, from eq. 4 in Section 8: Linear infinitesimal Elasticity, we know that $\sigma_{ii}$ can be expressed as follows:

$\sigma_{ii}=\lambda \delta_{ii} \epsilon_{kk} + 2 \mu \epsilon_{ii}$
Summing both sides on $i$ (and on $k$ , as always) and noting that $\overbrace{\frac{\sigma_{ii}}{3}}^{\text{summed}}=-P$ (pressure – e.x. hydrostatic)

$\longrightarrow \frac{\sigma_{ii}}{3}=\underbrace{(\lambda + \frac{2}{3} \mu)}_{k}\epsilon_{kk}$

$P=-k\frac{\Delta V}{V}$ , where $k$ = “bulk modulus”

note: $\frac{1}{k}$ is sometimes called the “compressibility”, since “ $k$ ” relates hydrostatic pressure to volumetric strain. Really though, compressibility is determined from $\frac{k}{G}$ , where “ $G$ ” is the “shear modulus”. $\frac{k}{G} >> 1 (\rightarrow \infty) \longrightarrow \nu \rightarrow \frac{1}{2}$ , where “ $\nu$ ” is the Poisson Ratio. “ $\nu$ ” = $\frac{1}{2}$ $\longrightarrow$ incompressible material.

Next, we’ll define the Young’s Modulus, “ $E$ ”, the Poisson Ratio, “ $\nu$ ”, and the shear modulus, “ $G$ ”. Any two modulii are needed to define an isotropic material ( $E, \nu$ or $k, G$ , etc.)

We need $\boldsymbol{\epsilon}$ in terms of $\boldsymbol{\sigma}$ ;

We’ll start with eq. 4 in Section 8: Linear infinitesimal Elasticity, which is re-written, below:

$\sigma_{ij}=\lambda \epsilon_{kk} \delta_{ij} + 2 \mu \epsilon_{ij}$ (1)

Take the trace:
$\sigma_{11}=\lambda(\epsilon_{11} + \epsilon_{22} + \epsilon_{33}) + 2 \mu \epsilon_{11}$
$\sigma_{22}=\lambda (\epsilon_{11}+ \epsilon_{22} + \epsilon_{33}) + 2 \mu \epsilon_{22}$
$\sigma_{33}=\lambda (\epsilon_{11}+ \epsilon_{22} + \epsilon_{33}) + 2 \mu \epsilon_{33}$

$\sigma_{11}+\sigma_{22}+\sigma_{33}= 3 \lambda \epsilon_{11} + 3 \lambda \epsilon_{22} + 3 \lambda \epsilon_{33} + 2 \mu \epsilon_{11} + 2 \mu \epsilon_{22} + 2 \mu \epsilon_{33}$
$\longrightarrow \sigma_{kk}=3 \lambda \epsilon_{kk} + 2 \mu \epsilon_{kk} = (3 \lambda + 2 \mu)\epsilon_{kk}$

$\longrightarrow \epsilon_{kk}=\frac{\sigma_{kk}}{3 \lambda + 2 \mu}$ (2)

(2) $\longrightarrow$ (1) and invert:

(1) $\begin{equation*} \epsilon_{ij}=\frac{1}{2\mu}(\sigma_{ij} - \frac{\lambda}{3\lambda + 2 \mu}\sigma_{kk}\delta_{ij}) \end{equation*}$

Now we’re ready to find $E$ , $\nu$ , $G$ :

Consider the case of uniaxial tension: $\boldsymbol{\sigma}=\begin{bmatrix}\sigma_{11} & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix}$
$\epsilon_{11}=\frac{\sigma_{11}}{2 \mu} (1-\frac{\lambda}{3\lambda+2\mu}) = \underbrace{\frac{\cancel{2}(\lambda + \mu)}{\cancel{2} \mu (3\lambda+2\mu)}}_{*} \sigma_{11}$

* – must be $\frac{1}{E}$ since those of us that have done laboratory testing of linear, isotropic, materials, know that $\sigma=E\epsilon$ for a simple uniaxial test

(2) $\begin{equation*} E=\frac{\mu(3\lambda+2\mu)}{\lambda+\mu} \end{equation*}$

Now, $\epsilon_{22}=\frac{1}{2\mu}(\underbrace{\sigma_{22}}_{=0}-\frac{\lambda}{3\lambda + 2\mu}(\sigma_{11}+\underbrace{\sigma_{22}}_{=0}+\underbrace{\sigma_{33}}_{=0}))=-\frac{\lambda}{2\mu(3\lambda+2\mu)}\sigma_{11}$ ,

where $\epsilon_{22}$ is the lateral strain that occurs from the uniaxial (longitudinal) stress.

By definition (i.e. as defined in undergraduate mechanics of materials),

$\nu=-\frac{\epsilon_{22}}{\epsilon_{11}}=\frac{\frac{\lambda}{2\cancel{\mu}\cancel{(3\lambda+2\mu)}}\cancel{\sigma_{11}}}{\frac{\lambda+\mu}{\cancel{\mu}\cancel{(3\lambda+2\mu)}}\cancel{\sigma_{11}}}=\frac{\lambda}{2(\lambda+\mu)}$