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.