# Stresses in Plastic Zone

In view of the fact that in plane stress, ац = 0, а22 = aY at the blunted crack tip and ац + а22 must be harmonic, choose

ац + а22 = ау (sin p(z — X0) + sin p(Z — X0)). where p is a constant and z = x + iy. Rewrite the above equation in the form

ац + а22 = 2ау sin p(x — X0) cosh py. (6)

On x-axis, y = 0 and ац + а22 = 2ау sin p(x — x0). For a Mises solid therefore,

a22 = ау I sin/? (x – xo)H—-= cos/?(x – xo)l,

. 1

ац = ay 1 sin/?(х — xq)——— — cosp(x — xq)

To satisfy ац = 0 at the crack tip x = xt, choose p such that p(xt — x0) = n/6. Note that on the extended crack-line in the plastic zone, ац = а22 except at the point where p(x — x0) = n/2.

To reconcile the chosen plastic field with LEFM field in the elastic domain, choose а11 = а22 = ау at the yield point (x = xY = rY, y = 0). This is possible provided p(xY — x) = n/2, or xY = 3xt — 2×0. Therefore, the plastic zone is lp = xY — xt = 2(xt — x0) = n/3p. To find its value, evaluate P = a22ds and equate it with P = 2aYrY obtained from LEFM. The result is

p = 1 /(хул/3) and the plastic zone size is

2

lp = ~i=rY = —— I — la

^ 2V3W/

with p, lp and xY known, it is easy to evaluate x0 = xY — n/2p and xt = xY — lp.

To find the crack opening in the plastic observe that on the crack surface x = xc, y = yc, ац = а22 = aY and hence the equation of the deformed crack is

sin p(xc — x0) cosh pyc = 0.5.

Since the value of p is already known, the above equation can be used to find the deformed crack surface in the plastic domain. The surface in the elastic domain is obtained from the solution of Singh et al. and for S ^ E, it can be expressed in the form

2S

xe = a cos p, ye = — sin p.

E