Chapter 6
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.
Step-by-Step
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 \ GPaSubstituting 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