## Q. 6.7

Nanoindentation of MgO
Figure 6-20 shows the results of an indentation into single crystal (001) MgO using a diamond Berkovich indenter. The unloading stiffness at a maximum indentation depth of 1.45 μm is 1.8 × $10^6$ N/m. A calibration on fused silica indicates that the projected contact area at the corresponding contact depth is 41 μm². The Poisson’s ratio of MgO is 0.17. Calculate the elastic modulus and hardness of MgO.

## Verified Solution

The projected contact area $A_{c}=41 \ \mu m^2 \times \frac{(1 \ m)^2}{(10^6 \ \mu m)^2}=4.1 \times 10^{-11} \ m^2$

The reduced modulus is given by

$E_{r}=\frac{\sqrt{\pi } }{2\beta }\frac{S}{\sqrt{A_{c}} }=\frac{\sqrt{\pi }(1.8 \times 10^6 \ N/m) }{2(1.034)\sqrt{(4.1\times 10^{-11} \ m^2)} }=241\times 10^9 \ N/m^2 =241 \ GPa$

Substituting for the Poisson’s ratio of MgO and the elastic constants of diamond and solving for E,

$\frac{1}{E_{r}}=\frac{1 \ – \ 0.17^2}{E}+\frac{1 \ – \ 0.07^2}{1.141 \times 10^{12} \ Pa}=\frac{1}{241\times 10^9 \ Pa}$
$\frac{0.9711}{E}=\frac{1}{241 \times 10^9 \ Pa} – \frac{1 \ – \ 0.07^2}{1.141\times 10^{12} \ Pa}$
$E=296 \ GPa$

From Figure 6-20, the load at the indentation depth of 1.45 μm is 380 mN (380 × $10^{-3}$ N). Thus, the hardness is

$H = \frac{P_{max}}{A_{c}}=\frac{380\times 10^{-3} \ N}{4.1 \times 10^{-11} \ m^2}=9.3 \times 10^9 \ Pa=9.3 \ GPa$