Answer:
0.16 micron per day
Explanation:
Given:
The initial crack length, a₁ = 0.1 micron = 0.1 × 10⁻⁶ m
Initial tensile stress, σ₁ = 120 MPa
Final stress = 30 MPa
now from Griffith's equation, we have
[tex]\sigma=[\frac{G_cE}{\pi\ a}]^\frac{1}{2}[/tex]
where,
Gc and E are the material constants
now,
for the initial stage
[tex]120=[\frac{G_cE}{\pi\ (0.1\times10^{-6}}]^\frac{1}{2}[/tex] ........{1}
and for the final case
[tex]30=[\frac{G_cE}{\pi\ a_2}]^\frac{1}{2}[/tex] ............{2}
on dividing 1 by 2, we get
[tex]\frac{120}{30}=[\frac{a_2}{0.1\times10^{-6}}]^\frac{1}{2}[/tex]
or
a₂ = 4² × 0.1 × 10⁻⁶ m
or
a₂ = 1.6 micron
Now,
the change from 0.1 micron to 1.6 micron took place in 10 days
therefore, the rate at which the crack is growing = [tex]\frac{1.6-0.1}{10}[/tex]
or
average rate of change of crack = 0.16 micron per day
