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,

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

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,

$\begin{array}{cc}K& =\lambda +\frac{2}{3}\left(\frac{E-3\lambda +r}{4}\right)\\ & =\frac{6\lambda +E-3\lambda +r}{6}\\ K& =\frac{E+3\lambda +r}{6}\end{array}$