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                       46                                    Fracture Mechanics: Fundamentals and Applications


                       configurations. For more complex situations, the stress intensity factor can be estimated by experiment
                       or numerical analysis (see Chapter 12).
                          One configuration for which a closed-form solution exists is a through crack in an infinite plate
                       subjected to a remote tensile stress (Figure 2.3). Since the remote stress σ is perpendicular to the crack
                       plane, the loading is pure Mode I. Linear elastic bodies must undergo proportional stressing, i.e., all
                       the stress components at all locations increase in proportion to the remotely applied forces. Thus the
                       crack-tip stresses must be proportional to the remote stress, and K  ∝ σ. According to Equation (2.37),
                                                                         I
                       stress intensity has units of  stress •  length . Since the only relevant length scale in Figure 2.3 is the
                       crack size, the relationship between K  and the global conditions must have the following form:
                                                    I
                                                         K  I  O = (  a σ  )                     (2.40)


                          The actual solution, which is derived in Appendix 2.3, is given by
                                                          K  I    a = σπ                         (2.41)

                          Thus the amplitude of the crack-tip singularity for this configuration is proportional to the
                       remote stress and the square root of the crack size. The stress intensity factor for Mode II loading
                       of the plate in Figure 2.3 can be obtained by replacing σ in Equation (2.41) by the remotely applied
                       shear stress (see Figure 2.18 and Equation (2.43) below).
                          A related solution is that for a semi-infinite plate with an edge crack (Figure 2.16). Note that
                       this configuration can be obtained by slicing the plate in Figure 2.3 through the middle of the crack.
                       The stress intensity factor for the edge crack is given by

                                                        K  I  .    a = 112σπ                     (2.42)

                       which is similar to Equation (2.41). The 12% increase in K  for the edge crack is caused by different
                                                                     I
                       boundary conditions at the free edge. As Figure 2.17 illustrates, the edge crack opens more because
                       it is less restrained than the through crack, which forms an elliptical shape when loaded.
                          Consider a through crack in an infinite plate where the normal to the crack plane is oriented
                       at an angle β with the stress axis (Figure 2.18(a)). If  β ≠ 0 , the crack experiences combined Mode
                       I and Mode II loading; K  = 0 as long as the stress axis and the crack normal both lie in the plane
                                           III
                       of the plate. If we redefine the coordinate axis to coincide with the crack orientation (Figure 2.18(b)),
                       we see that the applied stress can be resolved into normal and shear components. The stress normal
                       to the crack plane, σ yy  , produces pure Mode I loading, while τ  applies Mode II loading to the
                                                                           x′y′
                                        ′′
                       crack. The stress intensity factors for the plate in Figure 2.18 can be inferred by relating σ yy ′′  and
                       τ  to σ and β through Mohr’s circle:
                       x′y′
                                                       K  I  yy ′′  π  a = σ

                                                         = σ   2  β cos ( )  π  a               (2.43a)

                       and

                                                    K  II  xy ′′  π  a = τ

                                                                   β
                                                       = σ   β sin( )cos( )  π  a               (2.43b)
                       Note that Equation (2.43) reduces to the pure Mode I solution when β = 0. The maximum K II
                       occurs at β = 45°, where the shear stress is also at a maximum. Section 2.11 addresses fracture
                       under mixed mode conditions.
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