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/a₁ =[0.22(s/s_Y]² = [0.22´0.6325]²

or, r = 0.0194 a₁ 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.