Calculate the length of plastic deformation:
Calculate the length of plastic deformation along 45^{o} in case the applied stress σ is 63.25% of yield stress in the plate of above Example
Solution
In this case (referring to Eq. (1)) q = 45^{o}.
So that q/2= 22.5, 3q/2 = 67.5^{o} and cos q/2=0.924, sin q /2 = 0.383, cos 3q /2 = 0.383, sin 3q/2 = 0.924 .
The condition of plasticity istmax = t_{Y} = s_{Y} /2
∴ r/a_{1} =[0.22(s/s_{Y}]^{2} = [0.22´0.6325]^{2}
or, r = 0.0194 a_{1 }or 1.94% of crack length.
Compare that under a stress which is 63.25% of yield stress the plastic zone extends 10% of crack length in the direction of crack length but 1.94% of crack length at 45^{o}.