Question

Calculate the maximum strength for the following two cases, using the Griffith equation. Given: (i) mineral platelets have thickness of 1 mm and diameter of 10 nm;

(ii) mineral platelets have thickness of 1 nm and diameter of 50 nm.

Assume γsurf = 1 J/m2 ; EHAP = 100 GPa.

Answer 1

Solution;

Fracture strength of mineral platelets by Griffith criteria is given by;

$\sigma_m^f=\alpha E_m\Psi$

In which;

$\Psi=\sqrt{\frac{\gamma}{E_mh}}$

$\alpha\approx\sqrtπ$

$\gamma$ is the surface energy;

h is diameter of the mineral platelet.

(i)

$\sigma_m^f$ =$\sqrtπ$×100×10⁹×$\sqrt{\frac{1}{100×10^9×10×10^{-9}}}$

$\sigma_m^f=5.605GPa$

(ii)

$\sigma_m^f=\sqrtπ$ ×100×10⁹×$\sqrt{\frac{1}{50×100}}$

$\sigma_m^f=2.507GPa$