Page 366 - Mechanical Behavior of Materials

Section 8.5 Additional Topics on Application of K 367

The contribution from bending may be determined from Fig. 8.13(a):
√ 6M 6Pe
K 2 = F 2 S 2 πa, S 2 = = (8.28)
2
2
b t b t
Hence, the total stress intensity due to the eccentric load is obtained by summing the two solutions
and using substitutions from the previous equations; it is
P 6F 2 e √

K = K 1 + K 2 = F 1 + πa (8.29)
bt b
where the particular a/b that applies is used to separately determine F 1 and F 2 for tension and
bending, respectively.
Ingenious use of superposition sometimes allows handbook solutions to be employed for cases
not obviously included. For example, consider the case of a central crack in a plate with a pair of
prying forces. Values of K for this case, here denoted K 1 , are available from Fig. 8.15. This K 1
can be considered to be the superposition of three loadings, as shown in Fig. 8.23(a). The two cases
denoted K 2 have the same solution, and K 3 is simply the center-cracked plate from Fig. 8.12(a).
Hence, superposition requires that K 1 = K 2 + K 2 − K 3 , where the K 3 loading is subtracted from
2K 2 to obtain K 1 . This allows K 2 to be determined from the known solutions for K 1 and K 3 , with
details being given as the equations for case (a) in Fig. 8.23.
Similar superposition is shown in Fig. 8.23(b) for the related case of an infinite array of collinear
cracks with prying forces. Here, K 1 and K 3 have known closed-form solutions, as given in the case
(b) equations, so that a closed-form expression for K 2 is readily obtained. Figure 8.23 also gives
some approximations for these solutions that are within 10% for the indicated ranges of α = a/b.
Note that K 2 from Fig. 8.23(a) or (b) is of special interest for applications where tension in a sheet
of material is reacted by a concentrated force due to a bolt or a rivet, or a row of bolts or rivets.
In particular, these K 2 solutions provide K B (see Section 8.5.2) for cracks growing from one hole,
or a row of holes, as shown in Fig. 8.24.

8.5.4 Cracks Inclined or Parallel to an Applied Stress

Consider a crack that is inclined to the applied stress, as in Fig. 8.25. Such a situation is difficult
to handle because there is not only an opening (tensile) mode stress intensity K I , but also a sliding
(shear) mode K II , with these varying with θ as shown. A reasonable, but approximate, approach
to such cases is to treat them as opening mode (K I ) situations, with the crack length being the
projection normal to the stress direction—that is, along the x-axis of Fig. 8.25—giving
√
K = S πa cos θ (8.30)
where F = 1 for this particular example of a wide plate.
Stresses parallel to a crack can generally be ignored in calculating the opening mode K.Note
◦
that this is the case in Fig. 8.25, where K I is zero for θ = 90 . As an additional example, consider
a crack in a pressure vessel wall, as in Fig. 8.26. Only the stress σ t that is normal to the plane of the
crack affects K, and the stress parallel to the crack, such as σ x , may be ignored in calculating K.