Q.For a certain isotropic elastic solid, the bulk modulus is $K = 1.6\times10^{11}\ \text{Pa}$ and the shear modulus of rigidity is $\eta = 6.0\times10^{10}\ \text{Pa}$. Using the relation between $Y$, $K$, $\eta$ and $\sigma$ (no derivation required), calculate

Given: $K = 1.6\times10^{11}\ \text{Pa}$, $\eta = 6.0\times10^{10}\ \text{Pa}$.

1. Young's modulus, using $Y = \dfrac{9K\eta}{3K+\eta}$:

   $3K = 3 \times 1.6\times10^{11} = 4.8\times10^{11}\ \text{Pa}, \qquad 3K + \eta = 4.8\times10^{11} + 0.6\times10^{11} = 5.4\times10^{11}\ \text{Pa}$

   $9K\eta = 9 \times 1.6\times10^{11} \times 6.0\times10^{10} = 8.64\times10^{22}$

   $Y = \frac{8.64\times10^{22}}{5.4\times10^{11}} = 1.6\times10^{11}\ \text{Pa}$

2. Poisson's ratio, using $Y = 2\eta(1+\sigma)$, i.e. $\sigma = \dfrac{Y}{2\eta} - 1$: $\sigma = \frac{1.6\times10^{11}}{2 \times 6.0\times10^{10}} - 1 = \frac{1.6\times10^{11}}{1.2\times10^{11}} - 1 = 1.333 - 1 = 0.333$ …

Substitute the given $K$ and $\eta$ directly into $Y = 9K\eta/(3K+\eta)$ to find $Y$; then use $Y = 2\eta(1+\sigma)$, rearranged to $\sigma = Y/(2\eta) - 1$, to find $\sigm …

• Arithmetic slips when combining powers of ten in $9K\eta$ and $3K+\eta$ -- keep the exponents of 10 explicit at every step rather than combining mentally. …