Consider the case of a crack problem in an anisotropic monoclinic material under anti-plane deformation as described in Exercise 11.16. Following relation (11.6.6), choose the complex potential form as F(z^∗) = A = √z^∗, where A =–√2 K₃μ/(C₅₅ + μC₄₅) and K₃ is a real constant. Using this form, show that the nonzero displacement and stresses in the vicinity of the crack tip (see Fig. 11.11) are given by:

Note that the parameter K₃ will be related to the stress intensity factor for this case. Verify that shear stress τ _yz vanishes on each side of the crack face, θ =± π These results can be compared to the corresponding solution for the isotropic case given in Exercise 8.44.

Data from exercise 11.16

Consider an anisotropic monoclinic material symmetric about the x,y-plane (see Fig. 11.2)
and subject to an anti-plane deformation specified by u = v = 0, w = w(x,y). Show that in the absence of body forces, the out-of-plane displacement must satisfy the Navier equation:

where z^∗= x + μy and μ are the roots of the equation C₄₄μ² +2C₄₅μ + C₅₅ = 0. Note that for this case, positive definite strain energy implies that C₄₄C₅₅ > C² ₄₅; therefore, the roots will occur in complex conjugate pairs.

Fig 11.2

Equation 11.6.6

Fig 11.11

Fig 11.2

Data from exercise 8.44

Consider the crack problem shown for the anti-plane strain case with u = v = 0, w = w(x,y).
From Section 7.4, the governing equation for the unknown displacement component with
zero body force was given by Laplace’s equation, which in polar coordinates reads.

Transcribed Image Text:

w = K3/2rRe K3 2r Txz = - Tyz - Re K3 Re 2r cose +usine C45 tCH = R { /cose + usine, 1 cose+usine J