Answer:
0.49 mm
Explanation:
Given:
Crack length, 2a = 16 mm
or
a = 8 mm = 0.008 m
Applied stress, σ = 350 MPa
Yield stress, [tex]\sigma_y[/tex] = 1000 MPa
Now,
The stress intensity factor (K) is given as:
K = σ√(πa)
on substituting the values, we get
K = 350 × √(π × 0.008)
or
K = 55.48
also,
from Irwin's plastic zone correction factor
we have the formula
[tex]r_p=\frac{1}{2\pi}(\frac{K}{\sigma_y})^2[/tex]
where,
[tex]r_p[/tex] is the radius of the plastic zone
on substituting the respective values, we get
[tex]r_p=\frac{1}{2\pi}(\frac{55.48}{1000})^2[/tex]
or
[tex]r_p=[/tex] 0.00049 m
or
[tex]r_p=[/tex]0.49 mm
