Ratio of applied stress:
What should be the ratio of applied stress, s, to yield stress s_{Y} in a large thin plate so that plastic spread along lint of crack is equal to 1 /10 crack length?
Solution
For large plate the crack length is 2a,
Along crack length, q = 0
s_{xx} and s_{xy} vanish along crack in x-direction. For plastic deformation s_{yy} = s_{Y }= yield stress.
The plastic spread is to be upto a distance of r = 2a/10 = a/5
i.e. a stress which is only 63.25% of yield stress will be able to cause a plastic deformation equal to 1 / 10 of crack length.