# Superposition Methods

One of the advantages of LEFM is that the solutions for different loading cases are additive in stresses, strains, and displacements. Since stress intensity factors arc nothing but parameters of the stress field, the stress intensity factors are also additive. This explains why Irwin’s local approach is more popular than Griffith’s global approach, even though they are generally equivalent. The energy release rates are not additive (although they are square-root additive); however, special care must be taken in problems in which geometrical nonlincarities arise. Such is generally the case when the resultant mode I stress intensity factor at any of the crack tips becomes negative. This implies interpenetration of the faces of the crack which is in reality impossible; instead, there is partial closing of the crack with face-to-face compression, which is a nonlinear phenomenon. Figure 3.1.3 Load decomposition in three-point bent beams.

The literature is replete with problems solvable by superposition. Many cases are quite obvious: the actual loading often is just a superposition of various simpler loadings for which the stress intensity factors are known. This is the case for eccentric tension or compression, of cracked strips which are solved by superposition of a pure tension or compression state and a pure bending state. Sometimes the simple states are not so obvious, as the following example shows.

Example 3.1.4 To obtain the stress intensity factor of a single-edge cracked specimen subjected to three-point bending for arbitrary span-to-depth ratio, Guinea (1990) and Pastor et al. (1995) used the approximate superposition illustrated in Fig. 3.1.3. In this approximation, the solutions for S/D = 4 (Fig. 3.1.3b) and for pure bending (Fig. 3.1.3c) are superposed so that the resulting bending moment distribution over the central part is the same as that for the actual beam (Fig. 3.1.3a). The result may be written as:

4 D

ks/D{a) = kx(a) + — [fc4(a) – k^a)} (3.1.7)

where k^(a) and koa(a) are the solutions for S/D = 4 and oo given in the example 3.1.1. Rearranging, the final expression turns out to be of the form (3.1.1) with ps/o(a) given by

4 D

Ps/D(a) = Poo (a) + – j-[p4(a) “ Poo (a)] (3-1.8)

withp4(a) and Poo(a) given by (3.1.2) and (3.1.3). This solution was checked against existing results in the literature for S/D — 8 (Brown and Srawley 1966) and finite element results using very small singular quarter-node elements for S/D — 8 and 2.5 (Pastor et al. 1995). The results coincided within 1%. 0

A particularly important class of superposition is that in which the effect of remotely applied stresses is first reduced to the effect of a stress distribution over the crack faces and then the stress intensity factor, due to this stress distribution, is obtained by integration of the stress intensity factor due to a point load at an arbitrary location on the crack faces. Let us illustrate these steps by examples; first the reduction to a case with stresses on the crack faces.

Example 3.1.5 Consider a center-cracked panel subjected to remote uniaxial stress a. We may de­compose the whole elastic solution (Fig. 3.1,4a) as the solution for an uncrackcd panel (Fig. 3.1.4b) and the solution for a cracked panel with the faces of the cracks subjected to stresses identical but opposite to those in the uncracked panel (Fig. 3.1.4c). In this way, the remote boundary conditions are satisfied, as well as the boundary conditions on the crack face. Of course, the stress intensity factor for load case (b) is zero (there is no singularity in an uncracked panel) so that one finds that the stress intensity factors of cases (a) and (c) are identical. This particular example proves that the stress intensity I’actor for a center-cracked panel subjected to remote uniaxial stress a is identical to that corresponding to a center-cracked panel with the crack subjected to internal pressure p = a. 0 Figure 3.1.4 Solution for a cracked panel expressed as superposition of the solution for an uncracked panel and the solution for a loaded crack with no remote stresses.

Even if the foregoing examples may seem very simple, the basic procedure is always the same. In particular, this is the most usual method when internal stresses build up due to thermal or moisture gradients. In those cases, the internal stress distribution is first computed for the body without crack (but with non-uniform temperature or moisture distribution), and then equal and opposite stresses are applied on the crack faces (while keeping uniform temperature).

Example 3.1.6 Consider a long center-cracked panel with free ends subjected to heating on both its sides. Assume that at a given instant the temperature profile is parabolic itcross the section: AT = ATo(2x/D)2 (Fig. 3.1.5a). We decompose this state in the state shown in Fig. 3.1.5b, with thermal gradient and no crack, and the state shown in Fig. 3.1,5c, with no thermal gradient and stresses on the crack faces equal and opposite to those in case (b). The stresses in case (b) may be estimated in the classical way by assuming that initially plane sections remain plane. If we call /3 the coefficient of linear thermal expansion, the result for the stress distribution along the crack plane is Except for a change in sign, this is the stress distribution to be applied on the crack faces in state (c). As in the previous example, the stress intensity factor for the original state (a) is equal to that for state (c) because in state (b) there is no crack and thus there is no stress singularity. D

Once the equivalent problem with stresses on the crack faces has been obtained, the problem remains of finding the stress intensity factor for this case. This may be done directly if the so-called Green function for the problem is known. The Green function is no more than the expression for the stress intensity factor engendered by a unit point load applied at any location on the crack face.

To be systematic, we write the stress intensity factor generated by a load-pair Px located at point x on

Figure 3.1.6 Superposition based on Green’s function: (a) Base problem; (b) general problem for a center cracked panel; (c) Superposition corresponding to the general problem.

the crack faces (Fig. 3.1.6a) in the form

p

K, = ^=kc(a, x/D) (3.1.10)

where kc(a, x/D) is the dimensionless Green function (the Green’s function with dimensions includes the factor 1 /b/D). If k. Q is known, the stress intensity factor generated by an arbitrary stress distribution over the crack faces is easily obtained by integration, as shown in the following example (which can easily be generalized to other more complex situations).

Example 3.1.7 Consider the center-cracked panel subjected to a known normal stress distribution over the crack faces, symmetric with respect to the crack plane (Fig. 3.1.6b). Let the stress at relative distance x/D from the crack center be a(x/D). If we subdivide the crack into infinitesimal length elements dx, the clement at x contributes with an elemental concentrated load dPx — balx/D). According to (3.1.10) this produces an infinitesimal stress intensity factor dKi — (cr(x/D)//15)ka(<x, xjD) dx. Now, adding up the contributions from all the elements we get

Ki = J dKj = -jL= J a(x/D)kc(a, x/D) dx (3.1.11)

or, with the change x/D — и

 /

а

a(u) kG(a, u) dn (3.1.12)

■a

To check this approximation, let us consider the previous Examples 3.1.5 and 3.1.6 for the limiting case a = a/D •C 1. In such a limit, the plate may be considered to be infinite, and the function kG{a, u), with и — x/D, simplifies to kc(a, u) =

Substitution of this expression into (3.1.12) gives        (3.1.14)

For the case of uniform tension cr(u) — a = constant, the integration readily delivers the well-known result К і = Os/тха (in the integral, set и = evsini, du — a cost dt). For the case of the parabolic distribution of temperature in Example 3.1.6 we notice that since we are considering a/D 1, and since

x < a, then the term 2(2x/D)2 in (3.1.9) is also negligible. Thus, the stress intensity factor for this case is just Ki = 13ЕАТо/Ш/3. D

The function mo — Ki/Px ~ kc(a, x/D)//D is called the weight function, of which ко is a dimensionless version. Finding the weight function for a particular case is a difficult problem of elasticity theory, which will be briefly outlined in Section 3.5.5. For a systematic approach to the weight function method, see Wu and Carlsson (1991).

Exercises

3.1 A long strip of width D = 400 inm is subjected to variable uniaxial stress with peaks of 14 MPa. In these conditions, edge cracks may be assumed to grow due to fatigue, and the designer wants the strip not to fail before the fatigue cracks are clearly visible. Determine the required toughness (Kic) if the crack length at which the strip fails must be (a) at least 10 mm, (b) at least 50 mm. Give also the values of the fracture energy if the material is a steel (E’ « 200 GPa).1^

3.2 lna long strip of width D — 300 mm. the expected peak stress (uniaxial) is 30 MPa. Ifthere exist welding flaws which resemble a center crack, determine the maximum flaw size allowable if the fracture toughness is 96 MPay’in and (a) the strength safely factor is 1; (b) the strength safety factor is 2. (Hint: make a lirst estimate of a assuming а/О <C 1, and then iterate until 1 % accuracy of the result.)

3.3 Estimate the stress intensity factor for eccentric tension in a single-edge cracked strip. Let P be the load, P and b the strip width and thickness, and e the eccentricity (positive towards the cracked side). Write the results as (a) (P / b/D)k(a) (b) a. v/PA:(o), with an equal to the mean remote tensile stress; (c) same, but with un equal to the maximum remote tensile stress; show that in this latter case, the shape function k(a) for very short cracks tends to the same value as that for a semi-infinite plate with an edge crack.

3.4 A thin slit of length 2c is machined in a large panel of thickness b made of a brittle material. A flat jack of identical length is inserted into the slit and pressure is applied to it until a crack propagates symmetrically. Determine the stress intensity factor at the crack tips for arbitrary crack length 2a and jack pressure p. Show that when o»c, the stress intensity factor approaches that corresponding to a center crack loaded at its center by a pair of forces equal to the jack force.

3.5 In the pressurized panel of the previous exercise, plot the evolution of the pressure in the jack vs. the crack length for quasi-static crack growth. Assume that the initial slit behaves as a crack, and that the fracture toughness Kic is known. (Hint: plot Pfc/Kic vs а/с.)

3.6 A large panel has a center crack of length 2a subjected to a symmetric internal pressure distribution which takes the value po at the crack center and decreases linearly to zero at the crack tips. Determine the stress intensity factor.

3.7 Show that if a center crack in a large panel is subjected to an arbitrary symmetric pressure distribution of the type p = роф(х/а), where ф(х/а) is a dimensionless function of the relative coordinate x/u along the crack, the resulting stress intensity factor is always of the form l<i = кро/Ш. where A: is a constant.