A Simple Model with Stiffness and Strength Degradation

It is relatively simple to build a model with mixed properties, by combining the simple rotating crack model given by Eq. (8.5.33) and the foregoing plastic Rankine model. We assume the total strain to be split into the elastic and fracturing parts,

є = єеі + є1 (8.5.48)

and assume further that the fracturing strain єf can be split into a term linear in the stress and a permanent (irreversible) strain tensor ep:

Єf = Cf (7, Pi ® p, + ep (8.5.49)

in which Cf is the inelastic unloading-reloading compliance (which replaces the secant compliance Cm)- The evolution of C? is given in integrated form as a function of the hardening variable є?, while the evolution of Ep is given in incremental form as before:

Подпись: (8.5.50) (8.5.51) dH(jf) – j

del

where rj(e^) and Н(є?) are material functions which will be related in the sequel to the uniaxial stress – strain curve. The evolution of the hardening-softening parameter is deduced from the loading function and the consistency condition which is taken according to the Rankine criterion:

Подпись: (8.5.52)(7, – ф(І*) < 0

Considering the uniaxial tensile test with monotonic stretching, let us call <7, e^ and ep the axial components of the stress, and fracturing and permanent strain tensors, respectively, we obviously have <7, = <7 and, by definition, e^ = єЛ and so Ф(є* ) is again the softening function. On the other hand, p, is a vector coinciding with the axis of the uniaxial tension. And according to (8.5.51), the permanent strain is readily integrated; it has only a nonzero component, namely the axial one,

ep = H{ef) (8.5.53)

Substituting this into (8.5.48) and identifying the axial components (the remaining ones are all zero), we get

Подпись: (8.5.54)єf = гі(є^)ф(є^) + Я(е^)

from which we can solve for r](e^):

ef – H(jf)
ф(є!)

Therefore, given ф(єї) and II(є?), the properties of the material are completely determined. Note that, according to (8.5.53) II{є?) is nothing else than the permanent strain which is obtained when the specimen in a uniaxial test is stretched up to єI and then unloaded. If no further information is available, it may be assumed that this is a fixed proportion a of the maximum inelastic strain, i. e., Н(е?) — а є? (in which a < 1). With this, the flow rules can be rewritten as

cS = {l-a)W) (8-5-56)

. f

kp — a p, <g> p, є (8.5.57)

This constitutes the simplest triaxial generalization of the uniaxial model described in Section 8.4.3, and may also be viewed as a strongly simplified version of Ortiz’s model (1985). However, one useful feature of Ortiz’s model is that it describes softening in compression as well as tension, which is obviously not the case with this simplified version.

Exercises

8.22 Show that є I in (8.5.13) can be written in a general tensorial form as

ef ^ (CnA I CrB)cr (8.5.58)

where A and В are (for constant n) constant fourth-order tensors which are given in cartesian components by

Aijkt ■- runjUkUi, Bijki — ^Siktiirij 4- ~гцп371кщ (В.5.59)

8.23 Consider a fixed-direction crack model with elastic-softening behavior defined by exponential softening in uniaxial tension, a — j[c~t Determine the evolution of the axial and transverse stress components in uniaxial extension, in which єн =- є increases monotonically and all the remaining components of the strain tensor are zero.

8.24 Show that the response for the uniaxial extension in the preceding exercise is identical for fixed and rotating crack models as long as the strain-softening curve is the same.

8.25 Consider a fixed-direction crack model with elastic-softening behavior defined by exponential softening in uniaxial tension, и — }[e~c //£°. Determine the evolution of the axial and transverse strains in a plane stress tension test, in which єі = є increases monotonically and £33 — <722 “ 0, the shear components being zero.

8.26 Consider a fixed-direction crack model that exhibits elastic-softening behavior defined by exponential softening in uniaxial tension, <7 = f’te’£ /£°, and is amenable to the scalar damage model described in Section 8.5.3. Referring to cartesian axes {хм,£2.яз}, consider a process in which the stress components <722 — 033 — <723 — <7i3 “ 0, while the fracturing strain tensor evolves such that ~ cqA, fJ]2 — £q2 in which A increases monotonically starting at A = 0. All the remaining components are zero. Assuming that v — 0.2, determine: (a) the evolution of the stress components; (b) the evolution of the maximum principal stress; and (c) whether a secondary crack forms. [Hint: Note that for A —* 0, the shear component is negligible, and thus the crack band forms normal to xi.}

8.27 Same as the preceding exercise except that є{2 — єоц in which A increases monotonically starting at A — 0 and fi varies in a way to be determined. Determine the upper bound for ц as a function of A so that secondary cracking would not occur. Determine the evolution of the stress for the limiting case. [Answer: !/i| < 2(1 – ix)e2X/r~ Є-Л]

8.28 Generalize the foregoing result to any softening function ф(є!).

8.29 Show that the tangent approach with Sn and St depending only on efN cannot be distinguished from Rots’ incremental approach for proportional microeracking, i. e., for loadings such that = efNm, where m is an arbitrary vector in the crack plane. Find S-J. in terms of St for this particular case.

8.30 Consider a material conforming to Mazars’ isotropic damage model with elastic-softening behavior defined by exponential softening in uniaxial tension, о —■ }’te"£ /£°. Determine the evolution of the axial

and transverse stress components in uniaxial extension, in which en = e increases inonotonically and all the remaining components of the strain tensor are zero.

8.31 Consider a thin layer of a material conforming to Mazars’ isotropic damage model with elastic-softening behavior deli ned by exponential softening in uniaxial tension, a = f[e~£ This layer is sandwiched between two thick plates of an elastic materia] with the same elastic moduli as the adjacent material, and the sandwich is subjected to uniaxial tension normal to the layer. Neglecting end effects, determine the evolution of the stress tensor in the layer as a function of the strain of the layer in the normal direction (assume that the transverse strain in the layer is dictated by the transverse strain in the elastic plates, which is, in turn, dictated by the elastic Poisson effect).