TITLE: Relation Between $Y, K, \eta$ and $\sigma$
QUESTION [0 upvotes]: How did we get the following relations between $Y, K, \eta$ and $\sigma$, where $Y$ is Young's modulus, $K$ is Bulk modulus, $\eta$ is Shear modulus, and $\sigma$ is Poisson's ratio? (Some books and Wikipedia use $Y$, $B$, $G$, and $\nu$ instead.)
(i) $Y=3 K(1-2 \sigma)$ or $Y=3 B(1-2 \nu)$
(ii) $Y=2 \eta(1+\sigma)$ or $Y=2 G(1+\nu)$
(iii) $\sigma=\frac{3 K-2 \eta}{2 \eta+6 K}$ or $\nu=\frac{3 B-2 G}{6 B+2 G}$
(iv) $\frac{9}{Y}=\frac{1}{K}+\frac{3}{\eta}$ or $Y=\frac{9 K \eta}{\eta+3 K}$ or $Y=\frac{9 B G}{3 B+G}$

REPLY [1 votes]: First note (iii) and (iv) are simple consequences of (i) and (ii). They are obtained by solving the linear system for $\nu$ and $E$ (Young's modulus), respectively.
To derive (i), consider a cube with perpendicular stresses $\sigma$ on all faces. The cube strains on each direction by $\epsilon$. We'll calculate this in two ways.
By normal stress, they stretch according to $E$ and $\nu$ as (the direct result of the normal, minus the squeezing due to Poisson's ratio)
$$\epsilon=\frac{\sigma}{E}-\nu\frac{\sigma}{E}-\nu\frac{\sigma}{E}=(1-2\nu)\frac{\sigma}{E}.$$
But by volumetric stress, it stretches according to $\kappa$ (bulk modulus) by (cubing the displacements)
$$3\epsilon=\frac{\sigma}{\kappa}.$$
Putting these together, we get (i)
$$E=3\kappa(1-2\nu).$$
Now, deriving (ii) is a bit more complicated. Consider a cube with one of its faces fixed to the floor. From the side it's going to look like a square, $ABCD$, with $AB$ touching the ground  Apply a shear stress $\sigma$ along $DC$. This displaces $D\rightarrow D'$ and $C\rightarrow C'$, with strain computing with shear modulus $\mu$ we get
$$\epsilon=\sigma/\mu=DD'/AD=CC'/BC.$$
However, we're effetively applying (decomposing the stress) $\sigma$ along both $DB$, as a compression, and $AC$, as a tension. Let's look at it from this perspective and write $\epsilon$ using $E,\nu$. Again, direct stress along $AC$ minus ratio'd along $DB$:
$$\frac{AC'-AC}{AC}=\frac{CC'/\sqrt2}{BC\sqrt2}=\frac{\epsilon}{2}=\frac{\sigma}{E}-\nu\left(-\frac{\sigma}{E}\right),$$
$$\epsilon=(1+\nu)\frac{\sigma}{E}.$$
We combine these two equations to get (ii):
$$E=2\mu(1+\nu).$$
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