Page 114 - T. Anderson-Fracture Mechanics - Fundamentals and Applns.-CRC (2005)

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94 Fracture Mechanics: Fundamentals and Applications

and the stresses are given by

∗
Γ ij   θ −  1 M
,
θ m (
∗
σ = r  2 + ∑ r m/2 Γ (, ) ) (A2.23)
ij
ij
m=0
where Γ is a function that depends on F and its derivatives. The order of the stress function
polynomial, M, must be sufficient to model the stresses in all regions of the body. When r → 0 ,
the first term in Equation (A2.23) approaches infinity, while the higher-order terms remain finite
(when m = 0) or approach zero (for m > 0). Thus the higher-order terms are negligible close to the
crack-tip, and stress exhibits a 1 r singularity. Note that this result was obtained without assuming
a specific configuration. It can be concluded that the inverse square-root singularity is universal
for cracks in isotropic elastic media.
A further evaluation of Equation (A2.19) and Equation (A2.20) with the appropriate boundary
conditions reveals the precise nature of the function Γ. Recall that Equation (A2.19) contains four,
as yet unspecified, constants; by applying Equation (A2.21), it is possible to eliminate two of these
constants, resulting in

θ
Φ( , ) = r θ r n / +1 c  sin  n  1  θ * − n − 2 sin  n θ − *  + c  cos  n  − 1  θ +1 − cos  n + 1  

2
 3   2   2   4   2   2  
  n + 2   
(A2.24)
for a given value of n. For crack problems, it is more convenient to express the stress function
*
in terms of θ, the angle from the symmetry plane (Figure A2.1). Substituting θ = θ − π into
Equation (A2.24) yields, after some algebra, the following stress function for the first few values
of n:

Φ( , ) = r θ r / 32 s  − cos θ − 1 cos 3 θ + t  − sin θ  − sin 3 θ    s r + 2 [ − 1 cos θ + ] O2 ( r / 5 2 ) + 
 2 3 2 2 2 
 1   1   2
(A2.25)

where s and t are constants to be defined. The stresses are given by
i
i
σ = 1  s θ + 5cos cos 3 θ  + t − 5sin θ  + 3 θ  s − 2 θ O + 4 cos ( r 1 2 / )  (A2.26a)
+
 +3sin
rr
 1
4 r    2 2   1   2 2   2
1
σ θθ = 4 r  s   − θ 2 − 3cos cos 3 2 θ    + 1   − t 3sin θ  2 − 3 2 θ  2 sin 2 θ + 4s O ( r 1 2 / )  (A2.26b)
+
 +3sin



 1
τ = 1  s sin θ  −  1 3 θ + sin t −  θ  + 3 θ  s cos θ O + 2 sin ( r 12 / )  (A2.26c)
 − 3cos
+ 2
θ r
4 r    2 2   1   2 2   2
Note that the constants s in the stress function (Equation (A2.25)) are multiplied by cosine
i
terms while the t are multiplied by sine terms. Thus, the stress function contains symmetric
i
and antisymmetric components, with respect to θ = 0. When the loading is symmetric about
θ = 0, t = 0, while s = 0 for the special case of pure antisymmetric loading. Examples of
i
i
symmetric loading include pure bending and pure tension; in both cases the principal stress is
normal to the crack plane. Therefore, symmetric loading corresponds to Mode I (Figure 2.14);
antisymmetric loading is produced by in-plane shear on the crack faces and corresponds to
Mode II.