## K

2nr 2

On the plane 0 = 0 the principal stresses a, and <x2 are equal and act in X and Y directions: ay is a principal stress. For plane stress, <r3=0 and *max = i<7i- Hence, the plastic zone size for 0 = 0, derived in section 4.1, is the correct size for the plane stress case with the Tresca as well as the Von Mises criterion.

The boundary of the plastic zone as a function of 0 follows from substitution of eqs (4.18) into (4.17), thus:

Plane strain: -— [§ sin20+(l —2v)2(l +cos 0)] = 2a2 2nr

Plane stress: -— (1 +1 sin2 0 + cos 0) = 2a 2S. The extent of the plastic zone as a function of 0 can be given as:

Plane strain: rJ0) = -—r [| sin20 + (l-2v)2(l +cos 0)1 4ii<j

4 nai

Note that eq (4.1) is indeed obtained by putting 0 = 0 in the equation for plane stress.

The boundary of the plastic zone as predicted by eqs (4.20) is plotted non-dimensionally as rp/(K,/nffys)2 in figure 4.5. The plastic zone in plane strain is appreciably smaller than the plane stress plastic zone. For 0 = 0 and v = j it follows from eqs (4.20) that the difference is a factor nine.

plane stress

Figure 4.5. Plastic zone shapes according to Von Mises and Tresca yield criteria a. Von Mises criterion; b. Tresca criterion

Therefore the plastic zone correction factor derived as eq (4.1) is not suitable in plane strain (see sect. 4.5).

If the Tresca yield criterion is used, the plastic zone shape turns out to be slightly different. From the Mohr's circles it is found that the maximum shear stress in plane stress equals Tmax = ^(j1, and in plane strain Tmax = i(cri ~°3) or Tmax = i(°'i — whichever is the largest. By using eqs (4.18) the Tresca yield zone is found as:

0 0