#### Installation — business terrible - 1 part

September 8th, 2015

For many purposes one can assume that the cracks in concrete are parallel and have a fixed direction, which does not change during the loading process. This is a reasonable approximation for those loading processes in which the axes of principal stress and strain directions do not change drastically since first cracking. We can then consider, as in Section 8.4.1, that a cracked zone has formed consisting of an elastic material intersected by an array of densely distributed parallel cracks. Our task is to describe how the equations must be arranged to incoiporate the response of this array of cracks to triaxial loading.

We consider that in an initially isotropic clastic material the maximum principal stress reached the tensile strength, and so a zone of cracked material formed. After that, the cracking process can be idealized as shown in Fig. 8.5.1a, which shows a crack band, formed by layers of elastic material separated by parallel cracks whose normal is defined by the unit vector n. This normal vector is fixed and coincides with the maximum principal stress direction at the onset of cracking. Let s be the mean spacing of the cracks and Aw the mean displacement between the faces of the cracks (Fig. 8.5.1b). We can define the mean cracking strain vector є? as the crack opening per unit crack-band width, i. e.,

Now, take point О in the cracked /.one as a reference (Fig. 8.5.1a). Let xo be its position vector, and let A be any other point within the cracked zone, with position vector x. The distance between О and A measured normal to the cracks is (Fig. 8.5.1a) OA’ = (x – Xo) ■ n, where the dot indicates the scalar (internal) product. Hence, the number of cracks between the two points is (x — xo) • n/s and,

correspondingly, the displacement generated by the cracks is

f (x – x0) • n. _ ,, , , , „

u’ = ————— Aw — [(x — Xo) • n Aw (8.5.2)

Taking the gradient of the displacement function, and then its symmetric part, we get the macroscopic small strain tensor that corresponds to the cracking:

1 S

= – (n ® es + 0 ® n) — {0 ® n) (8.5.3)

where from now on we use T5 to indicate the symmetric part of an arbitrary second-order tensor T. The foregoing result indicates that the cracking strain is not a general symmetric tensor, since it has three of the six possible components identically zero, indeed, taking an orthonormal base {її, s, t] so that the unit vectors s and t are parallel to the cracks (Fig. 8.5.1c), we easily find that

£{t -= e{s – 4 – 0 (8.5.4)

The fracturing strain tensor thus has only three degrees of freedom, corresponding to the three components of the vector 0.

The foregoing equations define the kinematics of the problem. Before getting any further, we must emphasize that a consistent set of rules must be used to properly define the foregoing vectors. It is implied in our sketch in Fig. 8.5.1b that one of the two faces of the crack is taken as reference; then її is taken as the unit normal to that face external to the uncracked material, and 6w is the displacement of the other face of the crack relative to the first. The reader can easily verify that, upon changing the reference to the other face of the crack, n and 0 change sign but ef remains unchanged.

The total strain tensor is obtained by adding up the elastic strain to the fracturing strain:

+ v v, , л s,

є =f —–er ——tr cr 1 + (£■’ ® n) (8.5.5)

E h 4

where E and v are, respectively, the elastic modulus and Poisson’s ratio of the bulk (uncracked) material. This is one of the basic equations of the fixed crack models. Note that it provides six equations, while we need nine equations. Given the strain tensor, we need to compute the stress tensor and the fracturing strain vector, nine components in all. The remaining three equations must relate the crack opening to the stress.

Since the basic internal variable is the vector 0, rather than the fracturing strain tensor eE it is natural to look for a relationship between 0 and the traction vector a on the crack faces, rather than trying to directly connect 0 to the stress tensor a. Therefore, we assume that the cracking behavior of the material is defined by a vectorial relationship of the form:

• <7 = an — Ф{0,п,- ■ ■) (8.5.6)

where Ф must be understood as a functional which, given crack orientation, evolution of 0 and, possibly, some other variables acting as parameters, yields the traction vector on the crack faces. Various definitions of this functional have been proposed and more could be invented. We discuss next some of the possibilities.