Consider a plate with a centrally located crack subject to far field tensile stress S. for the purpose of analysis; choose a Cartesian coordinate system with origin at the crack center, x-axis along the crack and у-axis perpendicular to it. Suppose the crack extends from x = —a0 to x = a0 on у = 0. Let
z = aocoshf, %=a + ij3, г = V—Ї
with a = 0 describing the initial geometry of a crack in elliptical coordinates. It is a special case of general transformation
z = x + iy = z(%) that relates Cartesian coordinates (x, y) to elliptical coordinates (а, в). In the new system, crack surface а = 0 extends from в = 0 to в = 2n. Because of the symmetry, it is sufficient to consider the upper right quadrant (x > 0, у > 0) or (а > 0, 0 < в < n/2) of the plate.
The plate under external tensile stress normal to the crack is expected to develop a non-uniform stress field that can be obtained from solving the equation of equilibrium
daxx daxy = daxy dayy
Эх Эу ’ Эх Эу
These equations are transformed to complex plane then combined and expressed in the form
It has a solution in terms of a stress function Ф such that
The main task now is to find a suitable stress function. The task becomes easier if the material behavior is linear and isotropic. For such a solid,
3ux duy 3и Эм 1 — v — 2kv2 / Э2Ф
Эх dy dz dz E 3z9z/
where ux and uy are the x-andy-components of displacement u (— ux + iuy); k — 0or 1 depending on whether the plate is in plane stress or plane strain. The material is linear if the mechanical parameters E and v are constant. Otherwise, the material is considered non-linear. The above equations show that for a linear solid, the function Ф must be real and bi-harmonic, and hence d2Ф/дzдZ must be real and harmonic. For a linear solid therefore, choose
_ Э2Ф _ Э/ /3/
Gxx Gnu — a a – — "г I A
dzdz dz dz
where f is an analytic function. The above equation can be integrated to yield
9Ф – a df ^-f + ZTz
where g is another analytic function.
Suppose a particular choice of analytic function f and g solves a given boundary value problem for an isotropic linear solid. The same analytic functions can also be used to solve the problem of another isotropic linear solid if it is subject to the same boundary conditions. This feature can be exploited even for non-linear material provided its non-linear stress strain behavior can be replaced by piece-wise linear approximation. In the simplest case, non-linear material response can be replaced by bi-linear stress strain behavior involving mechanical parameters (u, E) and (vn, En). Suppose the first segment I represents elastic behavior and the second segment n involves plastic deformation characterized by the plastic modulus Ep and the ratio vp — 1/2 such that
111 vn v 1
En E + Ер’ En E + 2Ep’
Once the analytic functions f and g are known or have been found, stress field can be obtained from (1). The displacement field in the elastic domain I is obtained from
u=z — Z = C-—-— 2 E
where C is a constant and z is the deformed position of a particle that initially, in the un-deformed configuration, occupied a position Z. The relation
и = z – Z = Cn + ———— 2
yields displacement in the plastic domain. The constants C and Cn, and the functions f and g must be chosen to ensure continuity of displacement and stress across the elastic plastic interface.
In the case of perfectly plastic material, the yield stress is constant; the stress strain curve is linear and horizontal (parallel to the strain axis) with slope En = 0. Since the stress strain curve is linear in the plastic domain, the sum axx + axx can still be considered analytic. However, it is necessary only to choose an analytic function g for determination of the stress field in plastic domain. The yield criterion provides a second condition that can be used for this purpose. It is necessary only to choose an analytic function that determines axx + axx such that the stress field it generates in conjunction with the yield criterion leaves the crack surface free traction. It is also necessary to maintain continuity between the stress fields of the elastic and plastic solutions across the common boundary.