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 )}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/376e7d0e28413323dfad0319d7d46b4197a59190)
Verify:
![{\displaystyle \mu ={\frac {E-3\lambda +r}{4}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9c73aa15507141dac0b7c07ca38e800fa182d3ef)
![{\displaystyle K={\frac {E+3\lambda +r}{6}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a610207c5250eafe42dc5300c3beb1ef4ed8e438)
where
.
From the second equation that has been given
![{\displaystyle E\lambda +E\mu =3\mu \lambda +2\mu ^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a98af20cfd53679322fdca52f5725dc107c7b71a)
or,
![{\displaystyle 2\mu ^{2}-(E-3\lambda )\mu -E\lambda =0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dc7ee8963a0a66d496d9ccf3a1559de794dc9b5c)
Therefore,
![{\displaystyle \mu ={\frac {(E-3\lambda )\pm {\sqrt {(E-3\lambda )^{2}+8E\lambda }}}{4}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/424631e7005f4796e0a6ab6a62af7d3a1e68755f)
or,
![{\displaystyle \mu ={\frac {(E-3\lambda )\pm {\sqrt {E^{2}+9\lambda ^{2}+2E\lambda }}}{4}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/73330f60480ce4040743c29f8f43f1f197ff71a0)
or,
![{\displaystyle \mu ={\frac {E-3\lambda \pm r}{4}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/24739ff5cd51aed1e465e9bc65ae258e81840c5e)
To find out whether the plus or the minus sign should be placed before
in the above equation, we put everything in terms of
and
. 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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8dd2f02d337b093dc69f54d2779c3c9f8042cb66)
Plugging these into the equation for
, multiplying both sides by
and dividing by
, 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 )}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/459315464b2ab1c3d9f671406fe7756aa97e1160)
The limiting value of
is 0.5. Plugging this value into the above equation, we get,
![{\displaystyle 0=-1.5\pm 1.5}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2be2a4002cfa1e06533ae0076113f0abbb3777cc)
The above can be true only if the sign is positive. Therefore, the
correct relation is
![{\displaystyle \mu ={\frac {E-3\lambda +r}{4}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9c73aa15507141dac0b7c07ca38e800fa182d3ef)
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/48ea0296d79418bc410119134e2f78e6053e0e5b)