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}

or

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)}

(3)   \begin{equation*} \nu=\frac{\lambda}{2(\lambda+\mu)} \quad 0<\nu<1/2 \end{equation*}

Lastly, consider pure shear: \boldsymbol{\sigma}=\begin{bmatrix}0 & \sigma_{12} & 0\\ \sigma_{12} & 0 & 0\\ 0 & 0 & 0 \end{bmatrix}

\sigma_{12}=\lambda \epsilon_{kk} \delta_{12} + 2 \mu \epsilon_{12}=2\mu\epsilon_{12}

Since \gamma =2 \epsilon_{12} and \sigma_{12}=G\gamma (by definition),

(4)   \begin{equation*} G=\mu \end{equation*}

It can also be easily shown that G=\frac{E}{2(1+\nu)}