# Introduction to Elasticity/Constitutive example 6

## Example 6

Given:

For an isotropic material

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

1. $\mu ={\frac {E-3\lambda +r}{4}}$ 2. $K={\frac {E+3\lambda +r}{6}}$ where $r={\sqrt {E^{2}+9\lambda ^{2}+2E\lambda }}$ .

### Solution

From the second equation that has been given

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

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

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

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

$\mu ={\frac {E-3\lambda \pm r}{4}}$ To find out whether the plus or the minus sign should be placed before $r$ in the above equation, we put everything in terms of $\nu$ and $E$ . Thus,

{\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 $\mu$ , multiplying both sides by $(1+\nu )(1-2\nu )$ and dividing by $E$ , we get

$2(1-2\nu )=(1-2\nu ^{2}-4\nu )\pm {\sqrt {(1-2\nu ^{2}-4\nu )^{2}+8\nu (1+\nu )(1-2\nu )}}$ The limiting value of $\nu$ is 0.5. Plugging this value into the above equation, we get,

$0=-1.5\pm 1.5$ The above can be true only if the sign is positive. Therefore, the correct relation is

$\mu ={\frac {E-3\lambda +r}{4}}$ Plugging this relation into the first of the given equations, we have,

{\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}} 