One of the most famous equations in fracture mechanics is the ./-integral, due to J. Rice (1968a). Although in its original derivation, the ./-integral was not directly related to Q, soon after that Rice (1968b) realized that,/ was equal to Q. In the following paragraphs we introduce the./-integral as a particular form of expressing the energy release rate.
In deriving the J integral, the general expression (2.1.3) is used, together with a particular way of expressing the work and the elastic energy and a particular virtual process. To start fixing the main concepts, we first notice that although we have been continuously referring to a given body, any part of a body is another body in mechanical terms. Henceforth, all the equations used so far may be used for any subbody. In a plane case, we can take any contour Г surrounding the crack tip to define a subbody and apply to it the energy balance equations —in particular Eq. (2.1.3)— to find Q. This is what is done in the derivation that follows.
The derivation of the./-integral is done in many books in an Eulerian framework, where the axes and the contour Г move with the crack tip. It is important to realize that ours use the Lagrangian coordinates (coordinates of material points in the initial state), and then the reference subbody defines a closed system. If that were not the case, the flow (transport) terms would have to be included in the energy balance equations.
In the plane case, the elemental external work 6W is (for stress-free crack surfaces) just the work done
where ds is the differential of arc-length along the contour Г, t; the components of the surface traction vectors acting on this boundary, and 14 the components of the displacement vector (summation is implied by repeated indices).
We write the elastic energy U as the integral of the elastic (or strain) energy density, U, throughout the volume of the subbody defined by Г, which, for a plane case reads
U = b ЇЇ dA (2.1.42)
where. Д(Г) stands for the plane area of the subbody.
Substitution of the above equations (2.1.41) and (2.1.42) into Eq. (2.1.3) leads to:
The next step is to evaluate the variations 8 in any virtual elemental process. To obtain the J-integral expression, we select a virtual process in which we translate all the fields (displacement, stress, energy density) a distance 8a parallel to the crack, while extending the crack by this same amount. The variation of the displacement of a given material point, situated at (aq, xf) due to the translation is easily obtained:
8ui(x і, xf) = 14(34 – 8a, xf) – 14(34,34) — —14,1(3:1, xi)8a (2.1.44)
where 14,1 stands for дгч/дх. *
The other variation to be computed is that of the elastic energy — the integral inEq. (2.1.43). It may be evaluated in various ways. Direct analytical treatment using an expression for the elastic energy density similar to the previous equation (2.1.44) is straightforward. However, the solution may be obtained in a much more physical (and graphical) way as follows: Let the cross-hatched area shown in Fig. 2.1.7a be the subbody. Д(Г) defined by the contour Г in its initial situation. When the crack is extended and the fields are translated by 8a, the subbody reaches a state as defined in Fig. 2.1.7b. Because, by construction, the fields in part (b) of the figure are those in part (a) translated, the final energy of the subbody – A(F) coincides with the initial energy of a subbody ■А'(Г’) defined by a contour Г’ obtained by displacing Г a distance 8a towards the left, as shown in Fig. 2.1.7a by a lightly shaded area partially hidden by. А(Г). Therefore, the variation of U between the initial and final states, is equal to the difference of
initial energies between the two bodies Д'(Г’) and. А(Г):
SU – WU'(r’)] – ^И(Г)] (2.1.45)
and, graphically, this energy reduces to the energy contained in the two crescents shown in Fig. 2.1.7c, positive for the lightly shaded part (on the left) and negative for the cross-hatched part (on the right). The result may be expressed as a contour integral using the infinitesimal surface elements and the contour orientation depicted in Fig. 2.1.7d:
Substitutiomof this result and that in Eq. (2.1.44) fon5ui in Eq. (2.1.44), finally leads to the following expression for the energy release rate:
The integral expression in the right hand member is Rice’s ./-integral. This integral can be computed (i. e., it is defined) whenever all points on contour Г are elastic, even in situations where elastic fracture mechanics does not apply. However, the /-integral is equal to the energy release rate, G, only if (1) the nonelastic zone reduces to a point in the interior of Г, (2) the crack faces are traction-free, and (3) the crack is plane and extends in its own plane.
The /-integral as written in 2.1.47 is —because the factor dx2 must take the proper sign— an oriented line integral. It must be performed anti-clockwise, from the lower to the upper face of the crack, to give the correct result. This need may be avoided by realizing that the sign is correctly captured if one writes dx2 = nds where щ is the component along the crack line of the unit outward normal to the contour, and da is the unoriented arc-length differential. The /-integral may then be written as an unoriented contour integral with positive arc-length differential. (This is, in fact, equivalent to considering the line integral as a surface integral per unit thickness, where da = dA/Ь is the lateral area per unit thickness.) The resulting expression is
Although the /-integral can be analytically evaluated in only very few cases (one of which is shown in the forthcoming example), it is a very powerful theoretical tool.
Example 2.1.6 Consider again the pure bent DCB in Fig. 2.1.3a with the same hypotheses as stated in the preceding section. Let us compute the / integral following the path ABCDEF shown in Fig. 2.1.8. Since ni=0 and tj = 0 along BC and DE, the contribution of these two segments to the / integral is zero. So is the contribution from CD if one assumes that this segment is far enough from the crack tip to be stress free. Therefore, only the segments AB and AF contribute to the integral. Moreover, their contribution is identical, by symmetry arguments. If we call a the bending stress, which is the normal stress <тц in the direction of the arm axis and_the only nonzero component of the stress tensor, we have (along FE, for example): (1) тії = — 1; (2 )U — а2/2Е; (3) 11 = —<r (note the sign); and (4) ui, i = £ц — cr/E. Thus
Using о — MzJ2I where z is measured from the center line and da = dz, Eq. (2.1.27) for Q is readily recovered. D
In true LEFM, in which nonlinear behavior and fracture occur at a single mathematical point at the crack tip, the crack must grow statically under constant Q, because the material lias no memory of the previous loading. This implies that the crack growth resistance is a constant: 71 = Gf. So the quasi-static loading path in a Q-a plot is a step function as depicted in Fig. 2.1,9a.
When the LEFM limit is applicable, a loading-cracking-unloading path in the Q-a plot looks as shown in Fig. 2.1.9b. Along the segment OM, Q increases while the crack retains its initial length ao. Along MA, the crack grows under constant Q = Gf. If at point A the specimen is unloaded, the crack will not heal, and so the unloading AO’ will take place at constant crack length a ~ ai down to zero load.
This process may be also plotted in a P-и plot, a much more usual way of plotting experimental results. In such a plot, the constant crack length segments OM and AO’ become constant compliance lines, i. e., straight lines through the origin. The MA segment is an iso-S curve corresponding to Q = Gf, the equation of which is obtained by eliminating the crack length a from equations 2.1.21 and 2.1.22:
G; = ^C'(a) (2.1.50)
The resulting P-и plot for the process shown in Fig. 2.1.9b typically looks as shown in Fig. 2.1.10a, with negative slope for the iso-S curve. However, there exist certain geometries where the iso-S curves display positive slope as depicted in Fig. 2.1.10b.
Example 2.1.7 A DCB specimen similar to that in Fig. 2.1.3b has a thickness b — 10 mm, width 2h = 20 mm, and initial crack length а о = 80 mm, with E = 300 GPa and Gf~ 100 N/m. We want to describe the evolution of the crack by means of the Q(a) and P(u) curves in a quasi-static test in which the displacement is monotonically increased until the crack doubles its initial length and then is decreased until complete unloading.
The 6(a) curve is immediate (Fig. 2.1.11a): the crack length is constant and equal to a о = 80 mm while Q is increasing up to Gf = 100 N/m (segment OM) from this point on, the fracture energy is kept constant at 100 N/m until the crack reaches 160 mm (point /1) where unloading begins at constant crack length down to O’.
To follow the P(u) process we use the approximate results for и and Q from Examples 2.1.2 and 2.1.4.
Initially, a = ao = 80 mm, and the load grows proportionally to the displacement following the first of (2.1.29) with a — a$:
и ~ =F P(N) = 732.4 u(mm) (2.1.51)
where the dimensions in parentheses indicate the units for the load and the displacement.
Point M corresponds to Q = Gf with a — a0; the load Pm at which this condition is met is obtained from the second of (2.1.29):
Gf. – ,J! hr => Г’м = 625 N (2.1.52)
corresponding to a displacement им that, according to (2.1.51), is given by им = 0.8533 mm. From this point on, the crack will grow and the displacement and load will evolve following Eqs. (2.1.29) with G — Gf. The results are:
P(N) — 625 — and «(mm) — 0.8533 % (2.1.53)
which are the parametric equations of the P-и curve during the process M A in Fig. 2.1.11 a. The cartesian equation is obtained by eliminating a from the foregoing equations. The result is P = Pmt/um/u or P(N) = 577.4 u-’^Cmm-‘/2).
Values Pa and Ua at the unloading point A are obtained by setting a = а. д — 2ao in the parametric equations (2.1.53) with the result Рл = 312.5 N and Ua ~ 3.413 mm. The unloading branch then follows as a linear equation down to the origin, і. e., P = Pau/ua or P(N) = 91.55 u(mm). Figure
2.1.1 lb shows the curve followed in the P-и diagram. Q
Figure 2.1.12 Illustrations for exercises 2.1 (left) and 2.8 (right).
2.1 Consider the mechanism in Fig. 2.1.12 consisting of two rigid bars AB and BC joined by a hinge at B, and a spring MN of constant k. The spring is connected to bar AB by a frictional slider D and to the rigid support EC by a rolling support. Let L be the known length of bar AB, a the position of the slider on this bar, P the vertical load applied at point A, and и the vertical displacement of point A. (a) Find the elastic energy of the system as a function of и and a. (b) Find the load as a function of и and a. (c) Find the energy available to displace the slider a unit length, called Q by analogy with crack growth, (d) Show that Q is no more than the component along AB of the force that the spring exerts on the slider, (e) Show that the slider tends to move towards point В whatever the direction of the load P.