# Introduction to Elasticity/Constitutive example 6

## Example 6

Given:

For an isotropic material

${\displaystyle K=\lambda +{\frac {2}{3}}\mu ~,~~~E={\frac {\mu (3\lambda +2\mu )}{\lambda +\mu }}~,~~~\nu ={\frac {\lambda }{2(\lambda +\mu )}}}$

Verify:

1. ${\displaystyle \mu ={\frac {E-3\lambda +r}{4}}}$
2. ${\displaystyle K={\frac {E+3\lambda +r}{6}}}$

where ${\displaystyle r={\sqrt {E^{2}+9\lambda ^{2}+2E\lambda }}}$.

### Solution

From the second equation that has been given

${\displaystyle E\lambda +E\mu =3\mu \lambda +2\mu ^{2}}$

or,

${\displaystyle 2\mu ^{2}-(E-3\lambda )\mu -E\lambda =0}$

Therefore,

${\displaystyle \mu ={\frac {(E-3\lambda )\pm {\sqrt {(E-3\lambda )^{2}+8E\lambda }}}{4}}}$

or,

${\displaystyle \mu ={\frac {(E-3\lambda )\pm {\sqrt {E^{2}+9\lambda ^{2}+2E\lambda }}}{4}}}$

or,

${\displaystyle \mu ={\frac {E-3\lambda \pm r}{4}}}$

To find out whether the plus or the minus sign should be placed before ${\displaystyle r}$ in the above equation, we put everything in terms of ${\displaystyle \nu }$ and ${\displaystyle E}$. Thus,

{\displaystyle {\begin{aligned}\mu &={\frac {E}{2(1+\nu )}}\\\lambda &={\frac {\nu E}{(1+\nu )(1-2\nu )}}\\E-3\lambda &={\frac {E(1-2\nu ^{2}-4\nu )}{(1+\nu )(1-2\nu )}}\\8E\lambda &={\frac {8\nu E^{2}}{(1+\nu )(1-2\nu )}}\end{aligned}}}

Plugging these into the equation for ${\displaystyle \mu }$, multiplying both sides by ${\displaystyle (1+\nu )(1-2\nu )}$ and dividing by ${\displaystyle E}$, we get

${\displaystyle 2(1-2\nu )=(1-2\nu ^{2}-4\nu )\pm {\sqrt {(1-2\nu ^{2}-4\nu )^{2}+8\nu (1+\nu )(1-2\nu )}}}$

The limiting value of ${\displaystyle \nu }$ is 0.5. Plugging this value into the above equation, we get,

${\displaystyle 0=-1.5\pm 1.5}$

The above can be true only if the sign is positive. Therefore, the correct relation is

${\displaystyle \mu ={\frac {E-3\lambda +r}{4}}}$

Plugging this relation into the first of the given equations, we have,

{\displaystyle {\begin{aligned}K&=\lambda +{\cfrac {2}{3}}\left({\cfrac {E-3\lambda +r}{4}}\right)\\&={\cfrac {6\lambda +E-3\lambda +r}{6}}\\K&={\cfrac {E+3\lambda +r}{6}}\end{aligned}}}