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{align}

\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{align}$$ Plugging these into the equation for $$\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{align}

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{align}$$